Instructions Manual to Serway and Jewett's Physics for Scientists and Engineers [6 ed.]

Table of contents :
ANSWERS TO QUESTIONS......Page 3
SOLUTIONS TO PROBLEMS......Page 4
ANSWERS TO EVEN PROBLEMS......Page 21
ANSWERS TO QUESTIONS......Page 22
SOLUTIONS TO PROBLEMS......Page 24
ANSWERS TO EVEN PROBLEMS......Page 54
ANSWERS TO QUESTIONS......Page 56
SOLUTIONS TO PROBLEMS......Page 57
ANSWERS TO EVEN PROBLEMS......Page 78
ANSWERS TO QUESTIONS......Page 80
SOLUTIONS TO PROBLEMS......Page 82
ANSWERS TO EVEN PROBLEMS......Page 115
ANSWERS TO QUESTIONS......Page 117
SOLUTIONS TO PROBLEMS......Page 120
ANSWERS TO EVEN PROBLEMS......Page 156
ANSWERS TO QUESTIONS......Page 157
SOLUTIONS TO PROBLEMS......Page 160
ANSWERS TO EVEN PROBLEMS......Page 189
ANSWERS TO QUESTIONS......Page 191
SOLUTIONS TO PROBLEMS......Page 193
ANSWERS TO EVEN PROBLEMS......Page 214
ANSWERS TO QUESTIONS......Page 215
SOLUTIONS TO PROBLEMS......Page 218
ANSWERS TO EVEN PROBLEMS......Page 250
ANSWERS TO QUESTIONS......Page 251
SOLUTIONS TO PROBLEMS......Page 253
ANSWERS TO EVEN PROBLEMS......Page 283
ANSWERS TO QUESTIONS......Page 285
SOLUTIONS TO PROBLEMS......Page 287
ANSWERS TO EVEN PROBLEMS......Page 323
ANSWERS TO QUESTIONS......Page 325
SOLUTIONS TO PROBLEMS......Page 327
ANSWERS TO EVEN PROBLEMS......Page 348
ANSWERS TO QUESTIONS......Page 349
SOLUTIONS TO PROBLEMS......Page 351
ANSWERS TO EVEN PROBLEMS......Page 380
ANSWERS TO QUESTIONS......Page 381
SOLUTIONS TO PROBLEMS......Page 383
ANSWERS TO EVEN PROBLEMS......Page 410
ANSWERS TO QUESTIONS......Page 411
SOLUTIONS TO PROBLEMS......Page 414
ANSWERS TO EVEN PROBLEMS......Page 438
ANSWERS TO QUESTIONS......Page 439
SOLUTIONS TO PROBLEMS......Page 441
ANSWERS TO EVEN PROBLEMS......Page 471
ANSWERS TO QUESTIONS......Page 473
SOLUTIONS TO PROBLEMS......Page 474
ANSWERS TO EVEN PROBLEMS......Page 495
ANSWERS TO QUESTIONS......Page 497
SOLUTIONS TO PROBLEMS......Page 499
ANSWERS TO EVEN PROBLEMS......Page 522
ANSWERS TO QUESTIONS......Page 523
SOLUTIONS TO PROBLEMS......Page 525
ANSWERS TO EVEN PROBLEMS......Page 548
ANSWERS TO QUESTIONS......Page 549
SOLUTIONS TO PROBLEMS......Page 551
ANSWERS TO EVEN PROBLEMS......Page 573
ANSWERS TO QUESTIONS......Page 575
SOLUTIONS TO PROBLEMS......Page 579
ANSWERS TO EVEN PROBLEMS......Page 599
ANSWERS TO QUESTIONS......Page 601
SOLUTIONS TO PROBLEMS......Page 602
ANSWERS TO EVEN PROBLEMS......Page 630
ANSWERS TO QUESTIONS......Page 631
SOLUTIONS TO PROBLEMS......Page 634
ANSWERS TO EVEN PROBLEMS......Page 661
ANSWERS TO QUESTIONS......Page 663
SOLUTIONS TO PROBLEMS......Page 666
ANSWERS TO EVEN PROBLEMS......Page 690
ANSWERS TO QUESTIONS......Page 691
SOLUTIONS TO PROBLEMS......Page 692
ANSWERS TO EVEN PROBLEMS......Page 711
ANSWERS TO QUESTIONS......Page 713
SOLUTIONS TO PROBLEMS......Page 714
ANSWERS TO EVEN PROBLEMS......Page 737
ANSWERS TO QUESTIONS......Page 739
SOLUTIONS TO PROBLEMS......Page 742
ANSWERS TO EVEN PROBLEMS......Page 765
ANSWERS TO QUESTIONS......Page 767
SOLUTIONS TO PROBLEMS......Page 769
ANSWERS TO EVEN PROBLEMS......Page 789
ANSWERS TO QUESTIONS......Page 791
SOLUTIONS TO PROBLEMS......Page 795
ANSWERS TO EVEN PROBLEMS......Page 822
ANSWERS TO QUESTIONS......Page 823
SOLUTIONS TO PROBLEMS......Page 825
ANSWERS TO EVEN PROBLEMS......Page 845
ANSWERS TO QUESTIONS......Page 847
SOLUTIONS TO PROBLEMS......Page 850
ANSWERS TO EVEN PROBLEMS......Page 873
ANSWERS TO QUESTIONS......Page 875
SOLUTIONS TO PROBLEMS......Page 878
ANSWERS TO EVEN PROBLEMS......Page 900
ANSWERS TO QUESTIONS......Page 901
SOLUTIONS TO PROBLEMS......Page 903
ANSWERS TO EVEN PROBLEMS......Page 924
ANSWERS TO QUESTIONS......Page 926
SOLUTIONS TO PROBLEMS......Page 928
ANSWERS TO EVEN PROBLEMS......Page 950
ANSWERS TO QUESTIONS......Page 952
SOLUTIONS TO PROBLEMS......Page 955
ANSWERS TO EVEN PROBLEMS......Page 974
ANSWERS TO QUESTIONS......Page 976
SOLUTIONS TO PROBLEMS......Page 980
ANSWERS TO EVEN PROBLEMS......Page 1004
ANSWERS TO QUESTIONS......Page 1006
SOLUTIONS TO PROBLEMS......Page 1011
ANSWERS TO EVEN PROBLEMS......Page 1040
ANSWERS TO QUESTIONS......Page 1042
SOLUTIONS TO PROBLEMS......Page 1044
ANSWERS TO EVEN PROBLEMS......Page 1067
ANSWERS TO QUESTIONS......Page 1068
SOLUTIONS TO PROBLEMS......Page 1071
ANSWERS TO EVEN PROBLEMS......Page 1092
ANSWERS TO QUESTIONS......Page 1094
SOLUTIONS TO PROBLEMS......Page 1096
ANSWERS TO EVEN PROBLEMS......Page 1121
ANSWERS TO QUESTIONS......Page 1122
SOLUTIONS TO PROBLEMS......Page 1125
ANSWERS TO EVEN PROBLEMS......Page 1151
ANSWERS TO QUESTIONS......Page 1152
SOLUTIONS TO PROBLEMS......Page 1153
ANSWERS TO EVEN PROBLEMS......Page 1174
ANSWERS TO QUESTIONS......Page 1176
SOLUTIONS TO PROBLEMS......Page 1179
ANSWERS TO EVEN PROBLEMS......Page 1204
ANSWERS TO QUESTIONS......Page 1206
SOLUTIONS TO PROBLEMS......Page 1208
ANSWERS TO EVEN PROBLEMS......Page 1227
ANSWERS TO QUESTIONS......Page 1228
SOLUTIONS TO PROBLEMS......Page 1232
ANSWERS TO EVEN PROBLEMS......Page 1254
ANSWERS TO QUESTIONS......Page 1256
SOLUTIONS TO PROBLEMS......Page 1258
ANSWERS TO EVEN PROBLEMS......Page 1278
ANSWERS TO QUESTIONS......Page 1280
SOLUTIONS TO PROBLEMS......Page 1282
ANSWERS TO EVEN PROBLEMS......Page 1306

Citation preview

INSTRUCTOR'S SOLUTIONS MANUAL FOR

SERWAY AND JEWETT'S

PHYSICS FOR SCIENTISTS AND ENGINEERS SIXTH EDITION

Ralph V. McGrew Broome Community College James A. Currie Weston High School

Australia • Canada • Mexico • Singapore • Spain • United Kingdom • United States

1 Physics and Measurement CHAPTER OUTLINE 1.1 1.2 1.3 1.4 1.5 1.6 1.7

ANSWERS TO QUESTIONS

Standards of Length, Mass, and Time Matter and Model-Building Density and Atomic Mass Dimensional Analysis Conversion of Units Estimates and Order-ofMagnitude Calculations Significant Figures

Q1.1

Atomic clocks are based on electromagnetic waves which atoms emit. Also, pulsars are highly regular astronomical clocks.

Q1.2

Density varies with temperature and pressure. It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard.

Q1.3

People have different size hands. Defining the unit precisely would be cumbersome.

Q1.4

(a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms

Q1.5

(b) and (d). You cannot add or subtract quantities of different dimension.

Q1.6

A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. If an equation is not dimensionally correct, it cannot be correct.

Q1.7

If I were a runner, I might walk or run 10 1 miles per day. Since I am a college professor, I walk about 10 0 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on vacation.

Q1.8

On February 7, 2001, I am 55 years and 39 days old. 55 yr

F 365.25 d I + 39 d = 20 128 dFG 86 400 s IJ = 1.74 × 10 GH 1 yr JK H 1d K

9

s ~ 10 9 s .

Many college students are just approaching 1 Gs. Q1.9

Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10.

Q1.10

The mass of the forty-six chapter textbook is on the order of 10 0 kg .

Q1.11

With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.

1

2

Physics and Measurement

SOLUTIONS TO PROBLEMS Section 1.1

Standards of Length, Mass, and Time

No problems in this section

Section 1.2 P1.1

Matter and Model-Building

From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, Ldiag = L2 + L2 . Thus, since the atoms are separated by a distance

L = 0.200 nm , the diagonal planes are separated by

Section 1.3 *P1.2

1 2 L + L2 = 0.141 nm . 2

Density and Atomic Mass

Modeling the Earth as a sphere, we find its volume as

4 3 4 π r = π 6.37 × 10 6 m 3 3

e

j

3

= 1.08 × 10 21 m 3 . Its

m 5.98 × 10 24 kg = = 5.52 × 10 3 kg m3 . This value is intermediate between the V 1.08 × 10 21 m 3 tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 kg m3 . The average density of the Earth is significantly higher, so higher-density material must be down below the surface. density is then ρ =

P1.3

a

fb

g

e j

With V = base area height V = π r 2 h and ρ =

ρ=

a

fa

ρ = 2.15 × 10 kg m

3

3

9

3

.

m for both. Then ρ iron = 9.35 kg V and V 19.3 × 10 3 kg / m3 = 23.0 kg . = 9.35 kg 7.86 × 10 3 kg / m3

Let V represent the volume of the model, the same in ρ =

ρ gold =

P1.5

F 10 mm I f GH 1 m JK

1 kg m = 2 π r h π 19.5 mm 2 39.0 mm 4

*P1.4

m , we have V

m gold V

V = Vo − Vi =

ρ=

. Next,

ρ gold ρ iron

4 π r23 − r13 3

e

=

m gold 9.35 kg

and m gold

F GH

j

FG IJ e H K

e

4π ρ r23 − r13 m 4 , so m = ρV = ρ π r23 − r13 = V 3 3

j

j

.

I JK

Chapter 1

P1.6

3

4 4 3 π r and the mass is m = ρV = ρ π r 3 . We divide this equation 3 3 for the larger sphere by the same equation for the smaller:

For either sphere the volume is V =

m A ρ 4π rA3 3 rA3 = = = 5. m s ρ 4π rs3 3 rs3

a f

Then rA = rs 3 5 = 4.50 cm 1.71 = 7.69 cm . P1.7

*P1.8

Use 1 u = 1.66 × 10 −24 g .

F 1.66 × 10 GH 1 u F 1.66 × 10 = 55.9 uG H 1u F 1.66 × 10 = 207 uG H 1u

-24

I = 6.64 × 10 JK gI JK = 9.29 × 10 gI JK = 3.44 × 10 g

−24

g .

−23

g .

−22

g .

(a)

For He, m 0 = 4.00 u

(b)

For Fe, m 0

(c)

For Pb, m 0

(a)

The mass of any sample is the number of atoms in the sample times the mass m 0 of one atom: m = Nm 0 . The first assertion is that the mass of one aluminum atom is

-24

−24

m 0 = 27.0 u = 27.0 u × 1.66 × 10 −27 kg 1 u = 4.48 × 10 −26 kg . Then the mass of 6.02 × 10 23 atoms is m = Nm 0 = 6.02 × 10 23 × 4.48 × 10 −26 kg = 0.027 0 kg = 27.0 g . Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be written m = Nm 0 . 0.027 0 kg = 6.02 × 10 23 m 0 , so m 0 =

0.027 kg 6.02 × 10 23

= 4.48 × 10 −26 kg ,

in agreement with the first assertion. (b)

The general equation m = Nm 0 applied to one mole of any substance gives M g = NM u , where M is the numerical value of the atomic mass. It divides out exactly for all substances, giving 1.000 000 0 × 10 −3 kg = N 1.660 540 2 × 10 −27 kg . With eight-digit data, we can be quite sure of the result to seven digits. For one mole the number of atoms is N=

F 1 I 10 GH 1.660 540 2 JK

−3 + 27

= 6.022 137 × 10 23 .

(c)

The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of one molecule of H 2 O is 2 1.008 0 + 15.999 u = 18.0 u. Then the molar mass is 18.0 g .

(d)

For CO 2 we have 12.011 g + 2 15.999 g = 44.0 g as the mass of one mole.

b

g

b

g

4

Physics and Measurement

P1.9

b gFGH 101 kgg IJK = 4.5 × 10 kg I JK = 3.27 × 10 kg .

Mass of gold abraded: ∆m = 3.80 g − 3.35 g = 0.45 g = 0.45 g

F 1.66 × 10 GH 1 u

−27

Each atom has mass m 0 = 197 u = 197 u

3

−4

kg .

−25

Now, ∆m = ∆N m 0 , and the number of atoms missing is ∆N =

∆m

=

m0

4.5 × 10 −4 kg 3.27 × 10 −25 kg

= 1.38 × 10 21 atoms .

The rate of loss is ∆N ∆t ∆N ∆t P1.10

P1.11

=

FG H

1 yr 1.38 × 10 21 atoms 365.25 d 50 yr

IJ FG 1 d IJ FG 1 h IJ FG 1 min IJ K H 24 h K H 60 min K H 60 s K

= 8.72 × 10 11 atoms s .

e

je

j

(a)

m = ρ L3 = 7.86 g cm 3 5.00 × 10 −6 cm

(b)

N=

(a)

The cross-sectional area is

3

= 9.83 × 10 −16 g = 9.83 × 10 −19 kg

9.83 × 10 −19 kg m = = 1.06 × 10 7 atoms m 0 55.9 u 1.66 × 10 −27 kg 1 u

e

j

a

f a

fa

fa

A = 2 0.150 m 0.010 m + 0.340 m 0.010 m = 6.40 × 10

−3

2

m .

f.

The volume of the beam is

ja

e

f

V = AL = 6.40 × 10 −3 m 2 1.50 m = 9.60 × 10 −3 m3 . Thus, its mass is

e

FIG. P1.11

je9.60 × 10 m j = 72.6 kg . F 1.66 × 10 kg I = 9.28 × 10 The mass of one typical atom is m = a55.9 ufG H 1 u JK 3

m = ρV = 7.56 × 10 kg / m

(b)

−3

3

3

−27

0

m = Nm 0 and the number of atoms is N =

−26

kg . Now

72.6 kg m = = 7.82 × 10 26 atoms . −26 m 0 9.28 × 10 kg

Chapter 1

P1.12

(a)

F 1.66 × 10 GH 1 u

−27

The mass of one molecule is m 0 = 18.0 u

kg

I = 2.99 × 10 JK

−26

5

kg . The number of

molecules in the pail is N pail = (b)

1.20 kg m = = 4.02 × 10 25 molecules . m 0 2.99 × 10 −26 kg

Suppose that enough time has elapsed for thorough mixing of the hydrosphere. N both = N pail

F m I = (4.02 × 10 GH M JK pail

25

F 1.20 kg I , GH 1.32 × 10 kg JK

molecules)

total

21

or

N both = 3.65 × 10 4 molecules .

Section 1.4 P1.13

Dimensional Analysis

The term x has dimensions of L, a has dimensions of LT −2 , and t has dimensions of T. Therefore, the equation x = ka m t n has dimensions of

e

L = LT −2

j aTf m

n

or L1 T 0 = Lm T n − 2 m .

The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and m = 1 . Likewise, equating terms in T, we see that n − 2m must equal 0. Thus, n = 2 . The value of k, a dimensionless constant, cannot be obtained by dimensional analysis . *P1.14

(a)

Circumference has dimensions of L.

(b)

Volume has dimensions of L3 .

(c)

Area has dimensions of L2 .

e j

Expression (i) has dimension L L2

1/2

= L2 , so this must be area (c).

Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension L L2 = L3 , so it is (b). Thus, (a) = ii; (b) = iii, (c) = i .

e j

6

Physics and Measurement

P1.15

*P1.16

(a)

This is incorrect since the units of ax are m 2 s 2 , while the units of v are m s .

(b)

This is correct since the units of y are m, and cos kx is dimensionless if k is in m −1 .

(a)

a f

a∝

∑F

or a = k

∑F

represents the proportionality of acceleration to resultant force and m m the inverse proportionality of acceleration to mass. If k has no dimensions, we have a = k

(b) P1.17

In units,

M ⋅L T

2

=

kg ⋅ m s2

F F L M ⋅L , F = , 2 =1 . m T M T2

, so 1 newton = 1 kg ⋅ m s 2 .

Inserting the proper units for everything except G,

LM kg m OP = G kg Ns Q m

2

2

Multiply both sides by m

Section 1.5 *P1.18

.

m3

2

and divide by kg ; the units of G are

kg ⋅ s 2

.

Conversion of Units

a

fa

f

Each of the four walls has area 8.00 ft 12.0 ft = 96.0 ft 2 . Together, they have area

e

4 96.0 ft 2 P1.19

2

2

jFGH 3.128mft IJK

2

= 35.7 m 2 .

Apply the following conversion factors: 1 in = 2.54 cm , 1 d = 86 400 s , 100 cm = 1 m , and 10 9 nm = 1 m

FG 1 H 32

IJ b2.54 cm inge10 m cmje10 K 86 400 s day −2

in day

9

j=

nm m

9.19 nm s .

This means the proteins are assembled at a rate of many layers of atoms each second! *P1.20

8.50 in 3 = 8.50 in 3

FG 0.025 4 m IJ H 1 in K

3

= 1.39 × 10 −4 m 3

Chapter 1

P1.21

Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A ≈ 30 m 50 m = 1 500 m 2 .

a

fa

f

Categorize: We model the lot as a perfect rectangle to use Area = Length × Width. Use the conversion: 1 m = 3.281 ft .

a

Analyze: A = LW = 100 ft

1m I F 1 m IJ = 1 390 m 150 ft fG f FGH 3.281 a J K H 3.281 ft K ft

2

= 1.39 × 10 3 m 2 .

Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m 2 . Unit conversion is a common technique that is applied to many problems. P1.22

(a)

a

fa

fa

f

V = 40.0 m 20.0 m 12.0 m = 9.60 × 10 3 m 3 3

3

b

g

V = 9.60 × 10 m 3.28 ft 1 m (b)

3

= 3.39 × 10 5 ft 3

The mass of the air is

e

je

j

m = ρ air V = 1.20 kg m3 9.60 × 10 3 m3 = 1.15 × 10 4 kg . The student must look up weight in the index to find

e

je

j

Fg = mg = 1.15 × 10 4 kg 9.80 m s 2 = 1.13 × 10 5 N . Converting to pounds,

jb

e

g

Fg = 1.13 × 10 5 N 1 lb 4.45 N = 2.54 × 10 4 lb . P1.23

(a)

Seven minutes is 420 seconds, so the rate is r=

(b)

30.0 gal = 7.14 × 10 −2 gal s . 420 s

Converting gallons first to liters, then to m3 ,

e

r = 7.14 × 10 −2 gal s

jFGH 3.1786galL IJK FGH 10 1 Lm IJK −3

3

r = 2.70 × 10 −4 m3 s . (c)

At that rate, to fill a 1-m3 tank would take t=

F 1m GH 2.70 × 10

3

−4

m

3

IF 1 h I = G J s JK H 3 600 K

1.03 h .

7

8

Physics and Measurement

*P1.24

(a)

(b)

(c)

(d) P1.25

FG 1.609 km IJ = 560 km = 5.60 × 10 m = 5.60 × 10 cm . H 1 mi K F 0.304 8 m IJ = 491 m = 0.491 km = 4.91 × 10 cm . Height of Ribbon Falls = 1 612 ftG H 1 ft K F 0.304 8 m IJ = 6.19 km = 6.19 × 10 m = 6.19 × 10 cm . Height of Denali = 20 320 ftG H 1 ft K F 0.304 8 m IJ = 2.50 km = 2.50 × 10 m = 2.50 × 10 cm . Depth of King’s Canyon = 8 200 ftG H 1 ft K 5

Length of Mammoth Cave = 348 mi

7

4

3

5

3

5

From Table 1.5, the density of lead is 1.13 × 10 4 kg m 3 , so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks. Density is defined as mass per volume, in ρ =

ρ=

23.94 g 2.10 cm 3

F 1 kg I FG 100 cm IJ GH 1 000 g JK H 1 m K

3

m . We must convert to SI units in the calculation. V

= 1.14 × 10 4 kg m3

At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern. One important common-sense check on density values is that objects which sink in water must have a density greater than 1 g cm 3 , and objects that float must be less dense than water. P1.26

It is often useful to remember that the 1 600-m race at track and field events is approximately 1 mile in length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to mI a1 acrefFGH 6401 miacres IJK FGH 1 609 J mi K 2

*P1.27 P1.28

The weight flow rate is 1 200

FG H

ton 2 000 lb h ton

2

= 4.05 × 10 3 m 2 .

IJ FG 1 h IJ FG 1 min IJ = K H 60 min K H 60 s K

667 lb s .

1 mi = 1 609 m = 1.609 km ; thus, to go from mph to km h , multiply by 1.609. (a)

1 mi h = 1.609 km h

(b)

55 mi h = 88.5 km h

(c)

65 mi h = 104.6 km h . Thus, ∆v = 16.1 km h .

P1.29

Chapter 1

(a)

F 6 × 10 $ I F 1 h I FG 1 day IJ F 1 yr I = GH 1 000 $ s JK GH 3 600 s JK H 24 h K GH 365 days JK

(b)

The circumference of the Earth at the equator is 2π 6.378 × 10 3 m = 4.01 × 10 7 m . The length

12

190 years

e

j

of one dollar bill is 0.155 m so that the length of 6 trillion bills is 9.30 × 10 11 m. Thus, the 6 trillion dollars would encircle the Earth 9.30 × 10 11 m = 2.32 × 10 4 times . 4.01 × 0 7 m 1.99 × 10 30 kg mSun = = 1.19 × 10 57 atoms m atom 1.67 × 10 −27 kg

P1.30

N atoms =

P1.31

V = At so t =

V 3.78 × 10 −3 m 3 = = 1.51 × 10 −4 m or 151 µm 2 A 25.0 m

b

a

fe

13.0 acres 43 560 ft 2 acre 1 V = Bh = 3 3 7 3 = 9.08 × 10 ft ,

P1.32

g

j a481 ftf h

or

e

V = 9.08 × 10 7 ft 3

jFGH 2.83 ×110ft

−2

m3

3

I JK

B

FIG. P1.32

= 2.57 × 10 6 m3 P1.33 *P1.34

9

b

ge

jb

g

Fg = 2.50 tons block 2.00 × 10 6 blocks 2 000 lb ton = 1.00 × 10 10 lbs The area covered by water is

a fe

j a fa fe

j

2 = 0.70 4π 6.37 × 10 6 m A w = 0.70 AEarth = 0.70 4π REarth

2

= 3.6 × 10 14 m 2 .

The average depth of the water is

a

fb

g

d = 2.3 miles 1 609 m l mile = 3.7 × 10 3 m . The volume of the water is

e

je

j

V = A w d = 3.6 × 10 14 m 2 3.7 × 10 3 m = 1.3 × 10 18 m 3 and the mass is

e

je

j

m = ρ V = 1 000 kg m3 1.3 × 10 18 m3 = 1.3 × 10 21 kg .

10 P1.35

Physics and Measurement

(a)

d nucleus, scale = d nucleus, real

e

Fd I = 2.40 × 10 m FG 300 ft IJ = 6.79 × 10 jH 1.06 × 10 m K GH d JK e ft jb304.8 mm 1 ft g = 2.07 mm atom, scale

−15

−10

atom, real

d nucleus, scale = 6.79 × 10 −3

(b)

Vatom = Vnucleus

3 4π ratom 3 3 4π rnucleus 3

=

FG r Hr

IJ = FG d K Hd 3

atom

nucleus

IJ = F 1.06 × 10 K GH 2.40 × 10 3

atom

nucleus

−10 −15

m m

I JK

3

= 8.62 × 10 13 times as large *P1.36

scale distance between

P1.37

=

FG real IJ FG scale IJ = e4.0 × 10 H distanceK H factorK

13

jFGH 71..04××1010 mm IJK = −3

km

200 km

9

The scale factor used in the “dinner plate” model is 0.25 m

S=

5

1.0 × 10 lightyears

= 2.5 × 10 −6 m lightyears .

The distance to Andromeda in the scale model will be

e

je

j

Dscale = Dactual S = 2.0 × 10 6 lightyears 2.5 × 10 −6 m lightyears = 5.0 m .

P1.38

P1.39

FG H

F e6.37 × 10 mjb100 cm mg I =G GH 1.74 × 10 cm JJK = 13.4 IJ = FG e6.37 × 10 mjb100 cm mg IJ = 49.1 K GH 1.74 × 10 cm JK

(a)

2 AEarth 4π rEarth r = = Earth 2 A Moon 4π rMoon rMoon

(b)

VEarth = VMoon

3 4π rEarth 3 3 4π rMoon 3

Fr =G Hr

IJ K

8

Moon

3

6

3

Earth

2

6

2

8

To balance, m Fe = m Al or ρ FeVFe = ρ Al VAl

ρ Fe

FG 4IJ π r H 3K

Fe

3

FG 4 IJ π r H 3K FG ρ IJ = a2.00 cmfFG 7.86 IJ H 2.70 K Hρ K

= ρ Al

Al

1/3

rAl = rFe

Fe Al

3

1/3

= 2.86 cm .

−3

ft , or

Chapter 1

P1.40

11

The mass of each sphere is m Al = ρ Al VAl =

4π ρ Al rAl 3 3

m Fe = ρ FeVFe =

4π ρ Fe rFe 3 . 3

and

Setting these masses equal, 4π ρ Al rAl 3 4π ρ Fe rFe 3 ρ = and rAl = rFe 3 Fe . 3 3 ρ Al

Section 1.6 P1.41

Estimates and Order-of-Magnitude Calculations

Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball as a sphere of diameter 0.038 m. The volume of the room is 4 × 4 × 3 = 48 m3 , while the volume of one ball is

FG H

4π 0.038 m 3 2

IJ K

3

= 2.87 × 10 −5 m3 .

48 ~ 10 6 ping-pong balls in the room. 2.87 × 10 −5 As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best 1 packing fraction” is π 2 = 0.74 so that at least 26% of the space will be empty. Therefore, the 6 above estimate reduces to 1.67 × 10 6 × 0.740 ~ 10 6 .

Therefore, one can fit about

P1.42

A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus,

b

gb

gb

g

the tire would make 50 000 mi 5 280 ft mi 1 rev 8 ft = 3 × 10 7 rev ~ 10 7 rev . P1.43

In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least 1 in 2 = 43 × 10 −5 ft 2 . Since 1 acre = 43 560 ft 2 , the number of blades of grass to be expected on a 16 quarter-acre plot of land is about n=

a

fe

j

0.25 acre 43 560 ft 2 acre total area = = 2.5 × 10 7 blades ~ 10 7 blades . area per blade 43 × 10 −5 ft 2 blade

12 P1.44

Physics and Measurement

A typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then approximately 4 × 10 −3 in 3 . Since 1 acre = 43 560 ft 2 , the volume of water required to cover it to a depth of 1 inch is ft a1 acrefa1 inchf = a1 acre ⋅ infFGH 431560 acre

2

I F 144 in I ≈ 6.3 × 10 JK GH 1 ft JK 2

2

6

in 3 .

The number of raindrops required is n= *P1.45

volume of water required 6.3 × 10 6 in 3 = = 1.6 × 10 9 ~ 10 9 . volume of a single drop 4 × 10 −3 in 3

Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then

a fa

fa

fa

f

V = 0.5 1.3 m 0.5 m 0.3 m = 0.10 m3 . The mass of this volume of water is

e

je

j

m water = ρ water V = 1 000 kg m3 0.10 m3 = 100 kg ~ 10 2 kg . Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is

e

je

j

mcopper = ρ copper V = 8 920 kg m3 0.10 m3 = 892 kg ~ 10 3 kg . P1.46

The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminum cans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are ~250 million people, and 365 days in a year, so

e250 × 10

6

jb

g

cans day 365 days year ≅ 10 11 cans

are thrown away or recycled each year. Guessing that each can weighs around 1 10 of an ounce, we estimate this represents

e10 P1.47

11

jb

gb

gb

g

cans 0.1 oz can 1 lb 16 oz 1 ton 2 000 lb ≈ 3.1 × 10 5 tons year . ~ 10 5 tons

Assume: Total population = 10 7 ; one out of every 100 people has a piano; one tuner can serve about 1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore, # tuners ~

F 1 tuner I F 1 piano I (10 GH 1 000 pianos JK GH 100 people JK

7

people) = 100 .

Chapter 1

Section 1.7 *P1.48

13

Significant Figures

METHOD ONE We treat the best value with its uncertainty as a binomial 21.3 ± 0.2 cm 9.8 ± 0.1 cm ,

a f a A = 21.3a9.8f ± 21.3a0.1f ± 0.2a9.8 f ± a0.2 fa0.1f cm

f

2

.

The first term gives the best value of the area. The cross terms add together to give the uncertainty and the fourth term is negligible. A = 209 cm 2 ± 4 cm 2 . METHOD TWO We add the fractional uncertainties in the data.

a

f FGH 210..23 + 90..18 IJK = 209 cm

fa

A = 21.3 cm 9.8 cm ± P1.49

a

f

π r 2 = π 10.5 m ± 0.2 m

(a)

2

± 2% = 209 cm 2 ± 4 cm 2

2

= π (10.5 m) 2 ± 2(10.5 m)(0.2 m) + ( 0.2 m) 2 = 346 m 2 ± 13 m 2

a

f

2π r = 2π 10.5 m ± 0.2 m = 66.0 m ± 1.3 m

(b)

3

P1.50

(a)

P1.51

a f a m = a1.85 ± 0.02f kg

4

(b)

3

(c)

(d)

f

r = 6.50 ± 0. 20 cm = 6.50 ± 0.20 × 10 −2 m

ρ=

m

c hπ r 4 3

3

also,

δ ρ δ m 3δ r = + . ρ m r

In other words, the percentages of uncertainty are cumulative. Therefore,

a f

δ ρ 0.02 3 0.20 = + = 0.103 , 6.50 ρ 1.85 ρ=

and

a

1.85

c hπ e6.5 × 10 4 3

f

−2

j

m

3

= 1.61 × 10 3 kg m 3

a

f

ρ ± δ ρ = 1.61 ± 0.17 × 10 3 kg m3 = 1.6 ± 0.2 × 10 3 kg m3 .

2

14 P1.52

*P1.53

Physics and Measurement

(a)

756.?? 37.2? 0.83 + 2.5? 796.5/ 3/ = 797

(b)

0.003 2 2 s.f. × 356.3 4 s.f. = 1.140 16 = 2 s.f.

(c)

5.620 4 s.f. × π > 4 s.f. = 17.656 = 4 s.f.

a

f

a

a

f a

P1.55

a

f

a

f

f

1.1

17.66

We work to nine significant digits: 1 yr = 1 yr

P1.54

f

F 365.242 199 d I FG 24 h IJ FG 60 min IJ FG 60 s IJ = GH 1 yr JK H 1 d K H 1 h K H 1 min K

31 556 926.0 s .

The distance around is 38.44 m + 19.5 m + 38.44 m + 19.5 m = 115.88 m , but this answer must be rounded to 115.9 m because the distance 19.5 m carries information to only one place past the decimal. 115.9 m

b

V = 2V1 + 2V2 = 2 V1 + V2

g

a fa fa f V = a10.0 mfa1.0 mfa0.090 mf = 0.900 m V = 2e1.70 m + 0.900 m j = 5.2 m U δA 0.12 m = = 0.0063 | 19.0 m A || δ V δw 0.01 m = = 0.010 V = 0.006 + 0.010 + 0.011 = 0.027 = w 1.0 m V | δt 0.1 cm | = = 0.011 | t 9.0 cm W V1 = 17.0 m + 1.0 m + 1.0 m 1.0 m 0.09 m = 1.70 m 3 3

2

3

3

3

FIG. P1.55

1

1

1

3%

1

1

1

Additional Problems P1.56

It is desired to find the distance x such that 1 000 m x = 100 m x (i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen that

a

fb

g

x 2 = 100 m 1 000 m = 1.00 × 10 5 m 2 and therefore x = 1.00 × 10 5 m 2 = 316 m .

Chapter 1

*P1.57

Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has mass

a

m 0 = 197 u

fFGH 1.66 ×110u

−27

kg

I = 3.27 × 10 JK

−25

kg .

So, the number of atoms in the cube is N=

19 300 kg 3.27 × 10 −25 kg

= 5.90 × 10 28 .

The imagined cubical volume of each atom is d3 =

1 m3

= 1.69 × 10 −29 m 3 .

5.90 × 10 28

So d = 2.57 × 10 −10 m .

P1.58

A total P1.59

F I a fe j FG VV IJ e A j = GG V JJ e4π r j H K H K F 3V IJ = 3FG 30.0 × 10 m IJ = 4.50 m =G H r K H 2.00 × 10 m K

Atotal = N A drop =

total

−6

total

2

total 4π r 3 3

drop

drop

3

2

−5

One month is

b

gb

gb

g

1 mo = 30 day 24 h day 3 600 s h = 2.592 × 10 6 s . Applying units to the equation,

e

j e

j

V = 1.50 Mft 3 mo t + 0.008 00 Mft 3 mo 2 t 2 . Since 1 Mft 3 = 10 6 ft 3 ,

e

j e

j

V = 1.50 × 10 6 ft 3 mo t + 0.008 00 × 10 6 ft 3 mo 2 t 2 . Converting months to seconds, V=

e

1.50 × 10 6 ft 3 mo 2.592 × 10 6 s mo

j e

t+

j

0.008 00 × 10 6 ft 3 mo 2

e2.592 × 10

Thus, V [ft 3 ] = 0.579 ft 3 s t + 1.19 × 10 −9 ft 3 s 2 t 2 .

6

s mo

j

2

t2.

15

16 P1.60

P1.61

Physics and Measurement

af

af

α ′(deg)

α (rad)

tan α

sin α

difference

15.0 20.0 25.0 24.0 24.4 24.5 24.6 24.7

0.262 0.349 0.436 0.419 0.426 0.428 0.429 0.431

0.268 0.364 0.466 0.445 0.454 0.456 0.458 0.460

0.259 0.342 0.423 0.407 0.413 0.415 0.416 0.418

3.47% 6.43% 10.2% 9.34% 9.81% 9.87% 9.98% 10.1%

24.6°

2π r = 15.0 m r = 2.39 m h = tan 55.0° r h = 2.39 m tan( 55.0° ) = 3.41 m

a

f

h

55° r

FIG. P1.61 *P1.62

Let d represent the diameter of the coin and h its thickness. The mass of the gold is m = ρV = ρAt = ρ

F 2π d GH 4

2

I JK

+ π dh t

where t is the thickness of the plating.

LM a2.41f MN 4

m = 19.3 2π

2

a fa

+ π 2.41 0.178

= 0.003 64 grams

fOPPe0.18 × 10 j Q −4

cost = 0.003 64 grams × $10 gram = $0.036 4 = 3.64 cents This is negligible compared to $4.98. P1.63

The actual number of seconds in a year is

b86 400 s daygb365.25 day yrg = 31 557 600 s yr . The percent error in the approximation is

eπ × 10

7

j b

s yr − 31 557 600 s yr 31 557 600 s yr

g × 100% =

0.449% .

Chapter 1

P1.64

V = L3 , A = L2 , h = L

(a)

V = A h L3 = L2 L = L3 . Thus, the equation is dimensionally correct. (b)

e

Vrectangular object P1.65

(a)

j = Awh = aAw fh = Ah , where

Vcylinder = π R 2 h = π R 2 h = Ah , where A = π R 2

The speed of rise may be found from v=

(b)

aVol rate of flowf = 16.5 cm (Area:

(a)

a

π D2 4 )

3

π 6 .30 cm 4

f

s 2

= 0.529 cm s .

Likewise, at a 1.35 cm diameter, v=

P1.66

A = Aw

16.5 cm 3 s

a

π 1.35 cm 4

f

2

= 11.5 cm s .

1 cubic meter of water has a mass

e

je

je

j

m = ρV = 1.00 × 10 −3 kg cm3 1.00 m 3 10 2 cm m (b)

3

= 1 000 kg

As a rough calculation, we treat each item as if it were 100% water. cell:

m = ρV = ρ

FG 4 π R IJ = ρFG 1 π D IJ = e1 000 kg m jFG 1 π IJ e1.0 × 10 H3 K H6 K H6 K 3

3

3

−6

j

m

3

= 5.2 × 10 −16 kg kidney: m = ρV = ρ

FG 4 π R IJ = e1.00 × 10 H3 K 3

−3

kg cm3

jFGH 34 π IJK ( 4.0 cm)

3

= 0.27 kg fly:

m=ρ

FG π D hIJ = e1 × 10 H4 K 2

−3

kg cm 3

= 1.3 × 10 −5 kg

P1.67

V20 mpg =

(10 8 cars)(10 4 mi yr ) = 5.0 × 10 10 gal yr 20 mi gal

V25 mpg =

(10 8 cars)(10 4 mi yr ) = 4.0 × 10 10 gal yr 25 mi gal

Fuel saved = V25 mpg − V20 mpg = 1.0 × 10 10 gal yr

jFGH π4 IJK a2.0 mmf a4.0 mmfe10 2

−1

j

cm mm

3

17

18 P1.68

P1.69

Physics and Measurement

F GH

IF JK GH

IF JK GH

IF JK GH

I FG JK H

IJ FG KH

I JK

furlongs 220 yd 0.914 4 m 1 fortnight 1 day 1 hr = 8.32 × 10 −4 m s fortnight 1 furlong 1 yd 14 days 24 hrs 3 600 s This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth. v = 5.00

The volume of the galaxy is

e

j e10

π r 2 t = π 10 21 m

2

19

j

m ~ 10 61 m3 .

If the distance between stars is 4 × 10 16 m , then there is one star in a volume on the order of

e4 × 10 The number of stars is about

P1.70

10

50

3

m star

The density of each material is ρ =

Al:

Cu:

Brass:

Sn:

Fe:

P1.71

10 61 m 3

(a) (b)

ρ=

a

b

4 51.5 g

g

f a3.75 cmf 4b56.3 g g ρ= = π a1.23 cmf a5.06 cmf 4b94.4 g g ρ= = π a1.54 cmf a5.69 cmf 4b69.1 g g ρ= = π a1.75 cmf a3.74 cmf 4b 216.1 g g ρ= = π a1.89 cmf a9.77 cmf π 2.52 cm

j

~ 10 11 stars .

2

9.36

2

8.91

2

7.68

2

7.88

g cm

3

g cm3

FG g IJ is H cm K F g IJ is The tabulated value G 8.92 H cm K The tabulated value 2.70

This would take

e

2% smaller.

3

5% smaller.

cm3 g cm3 g cm

3

FG H

The tabulated value 7.86

3.16 × 10 7 s yr

3 4 3 4 π r = π 5.00 × 10 −7 m = 5.24 × 10 −19 m3 3 3 1 m3 = = 1.91 × 10 18 micrometeorites 5.24 × 10 −19 m3

Vmm =

3

g

b3 600 s hrgb24 hr daygb365.25 days yrg = Vcube Vmm

3

m ~ 10 50 m 3 .

4m m m . = = V π r 2h π D2h

= 2.75

2

16

j

1.91 × 10 18 micrometeorites 3.16 × 10 7 micrometeorites yr

= 6.05 × 10 10 yr .

g cm3

IJ is K

0.3% smaller.

Chapter 1

ANSWERS TO EVEN PROBLEMS 5.52 × 10 3 kg m3 , between the densities of aluminum and iron, and greater than the densities of surface rocks.

P1.34

1.3 × 10 21 kg

P1.36

200 km

P1.4

23.0 kg

P1.38

(a) 13.4; (b) 49.1

P1.6

7.69 cm

P1.8

(a) and (b) see the solution, N A = 6.022 137 × 10 23 ; (c) 18.0 g; (d) 44.0 g

P1.2

P1.40

P1.10

(a) 9.83 × 10 −16 g ; (b) 1.06 × 10 7 atoms

P1.12

(a) 4.02 × 10 25 molecules; (b) 3.65 × 10 4 molecules

P1.14

(a) ii; (b) iii; (c) i

P1.16

(a)

P1.18

35.7 m 2

P1.20

1.39 × 10 −4 m3

P1.22 P1.24

M ⋅L ; (b) 1 newton = 1 kg ⋅ m s 2 T2

5

3

4

(a) 3.39 × 10 ft ; (b) 2.54 × 10 lb 5

rAl = rFe

FG ρ IJ Hρ K

13

Fe Al

P1.42

~ 10 7 rev

P1.44

~ 10 9 raindrops

P1.46

~ 10 11 cans; ~ 10 5 tons

P1.48

a209 ± 4f cm

P1.50

(a) 3; (b) 4; (c) 3; (d) 2

P1.52

(a) 797; (b) 1.1; (c) 17.66

P1.54

115.9 m

P1.56

316 m

P1.58

4.50 m 2

P1.60

see the solution; 24.6°

P1.62

3.64 cents ; no

P1.64

see the solution

P1.66

(a) 1 000 kg; (b) 5.2 × 10 −16 kg ; 0. 27 kg ;

2

7

(a) 560 km = 5.60 × 10 m = 5.60 × 10 cm ; (b) 491 m = 0.491 km = 4.91 × 10 4 cm ; (c) 6.19 km = 6.19 × 10 3 m = 6.19 × 10 5 cm ; (d) 2.50 km = 2.50 × 10 3 m = 2.50 × 10 5 cm

P1.26

4.05 × 10 3 m 2

P1.28

(a) 1 mi h = 1.609 km h ; (b) 88.5 km h ;

1.3 × 10 −5 kg

(c) 16.1 km h

P1.68

8.32 × 10 −4 m s ; a snail

P1.30

1.19 × 10 57 atoms

P1.70

see the solution

P1.32

2.57 × 10 6 m3

19

2 Motion in One Dimension CHAPTER OUTLINE 2.1 2.2 2.3 2.4 2.5 2.6 2.7

Position, Velocity, and Speed Instantaneous Velocity and Speed Acceleration Motion Diagrams One-Dimensional Motion with Constant Acceleration Freely Falling Objects Kinematic Equations Derived from Calculus

ANSWERS TO QUESTIONS Q2.1

If I count 5.0 s between lightning and thunder, the sound has traveled 331 m s 5.0 s = 1.7 km . The transit time for the light is smaller by

b

ga f

3.00 × 10 8 m s = 9.06 × 10 5 times, 331 m s so it is negligible in comparison. Q2.2

Yes. Yes, if the particle winds up in the +x region at the end.

Q2.3

Zero.

Q2.4

Yes. Yes.

Q2.5

No. Consider a sprinter running a straight-line race. His average velocity would simply be the length of the race divided by the time it took for him to complete the race. If he stops along the way to tie his shoe, then his instantaneous velocity at that point would be zero.

Q2.6

We assume the object moves along a straight line. If its average x velocity is zero, then the displacement must be zero over the time interval, according to Equation 2.2. The object might be stationary throughout the interval. If it is moving to the right at first, it must later move to the left to return to its starting point. Its velocity must be zero as it turns around. The graph of the motion shown to the right represents such motion, as the initial and final positions are the same. In an x vs. t graph, the instantaneous velocity at any time t is the slope of the curve at that point. At t 0 in the graph, the slope of the curve is zero, and thus the instantaneous velocity at that time is also zero.

t0

t

FIG. Q2.6 Q2.7

Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the velocity of the particle is unchanging, or is a constant.

21

22

Motion in One Dimension

a

f

Q2.8

Yes. If you drop a doughnut from rest v = 0 , then its acceleration is not zero. A common misconception is that immediately after the doughnut is released, both the velocity and acceleration are zero. If the acceleration were zero, then the velocity would not change, leaving the doughnut floating at rest in mid-air.

Q2.9

No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the recent past.

Q2.10

Yes. Consider throwing a ball straight up. As the ball goes up, its v velocity is upward v > 0 , and its acceleration is directed down v0 a < 0 . A graph of v vs. t for this situation would look like the figure to the right. The acceleration is the slope of a v vs. t graph, and is always negative in this case, even when the velocity is positive.

a

a f

f

t FIG. Q2.10 Q2.11

(a)

Accelerating East

(b)

Braking East

(c)

Cruising East

(d)

Braking West

(e)

Accelerating West

(f)

Cruising West

(g)

Stopped but starting to move East

(h)

Stopped but starting to move West

Q2.12

No. Constant acceleration only. Yes. Zero is a constant.

Q2.13

The position does depend on the origin of the coordinate system. Assume that the cliff is 20 m tall, and that the stone reaches a maximum height of 10 m above the top of the cliff. If the origin is taken as the top of the cliff, then the maximum height reached by the stone would be 10 m. If the origin is taken as the bottom of the cliff, then the maximum height would be 30 m. The velocity is independent of the origin. Since the change in position is used to calculate the instantaneous velocity in Equation 2.5, the choice of origin is arbitrary.

Q2.14

Once the objects leave the hand, both are in free fall, and both experience the same downward acceleration equal to the free-fall acceleration, –g.

Q2.15

They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity equal to vi . This velocity is the same as the velocity of the second ball, so after they fall through equal heights their impact speeds will also be the same.

Q2.16

With h =

1 2 gt , 2

a

f

1 2 g 0.707t . The time is later than 0.5t. 2

(a)

0.5 h =

(b)

The distance fallen is 0.25 h =

a f

1 2 g 0.5t . The elevation is 0.75h, greater than 0.5h. 2

Chapter 2

Q2.17

Above. Your ball has zero initial speed and smaller average speed during the time of flight to the passing point.

SOLUTIONS TO PROBLEMS Section 2.1

Position, Velocity, and Speed

P2.1

v = 2.30 m s

*P2.2

(a) (b)

v=

∆x 57.5 m − 9.20 m = = 16.1 m s ∆t 3.00 s

(c)

v=

∆x 57.5 m − 0 m = = 11.5 m s ∆t 5.00 s

(a)

v=

∆x 20 ft 1m = ∆t 1 yr 3.281 ft

FG H

IJ FG 1 yr IJ = 2 × 10 m s or in particularly windy times K H 3.156 × 10 s K 1 yr ∆x 100 ft F 1 m I F IJ = 1 × 10 m s . = v= G J G H K H ∆t 1 yr 3.281 ft 3.156 × 10 s K −7

7

7

(b)

The time required must have been ∆t =

P2.3

P2.4

−6

FG H

3 000 mi 1 609 m ∆x = v 10 mm yr 1 mi

(a)

v=

∆x 10 m = = 5 ms ∆t 2s

(b)

v=

5m = 1.2 m s 4s

(c)

v=

x 2 − x1 5 m − 10 m = = −2.5 m s 4 s−2 s t 2 − t1

(d)

v=

x 2 − x 1 −5 m − 5 m = = −3.3 m s 7 s−4 s t 2 − t1

(e)

v=

x 2 − x1 0 − 0 = = 0 ms 8−0 t 2 − t1

x = 10t 2 : For

af xamf ts

= 2.0

2.1

3.0

=

44.1

90

40

(a)

v=

∆x 50 m = = 50.0 m s ∆t 1.0 s

(b)

v=

∆x 4.1 m = = 41.0 m s ∆t 0.1 s

IJ FG 10 mm IJ = KH 1 m K 3

5 × 10 8 yr .

23

24 P2.5

Motion in One Dimension

(a)

Let d represent the distance between A and B. Let t1 be the time for which the walker has d the higher speed in 5.00 m s = . Let t 2 represent the longer time for the return trip in t1 d d d −3.00 m s = − . Then the times are t1 = and t 2 = . The average speed t2 5.00 m s 3.00 m s is:

b

v=

v= (b)

Section 2.2 P2.6

(a)

e

2 15.0 m 2 s 2 8.00 m s

j=

b

b

d+d = + 3 .00d m s

d 5.00 m s

g

2d

b8.00 m sgd g e15.0 m s j

g b

2

3.75 m s

Instantaneous Velocity and Speed

e Thus, at t = 3.00 s: x = e3.00 m s ja3.00 sf =

j

At any time, t, the position is given by x = 3.00 m s 2 t 2 . 2

2

i

ja

e

27.0 m .

f

2

At t f = 3.00 s + ∆t : x f = 3.00 m s 2 3.00 s + ∆t , or

b

ja f

g e

x f = 27.0 m + 18.0 m s ∆t + 3.00 m s 2 ∆t (c)

∆t → 0

(a)

F x − x I = lim e18.0 m s + e3.00 m s j∆tj = GH ∆t JK f

i

2

∆t → 0

x f − xi t f − ti

=

a2.0 − 8.0f m = − 6.0 m = a4 − 1.5f s 2.5 s

18.0 m s .

−2.4 m s

The slope of the tangent line is found from points C and D. tC = 1.0 s, x C = 9.5 m and t D = 3.5 s, x D = 0 ,

b

g

b

g

v ≅ −3.8 m s . (c)

.

at ti = 1.5 s , x i = 8.0 m (Point A) at t f = 4.0 s , x f = 2.0 m (Point B) v=

(b)

2

The instantaneous velocity at t = 3.00 s is: v = lim

P2.7

2

She starts and finishes at the same point A. With total displacement = 0, average velocity = 0 .

i

(b)

Total distance = Total time

g

The velocity is zero when x is a minimum. This is at t ≅ 4 s .

FIG. P2.7

Chapter 2

P2.8

(a)

(b)

P2.9

*P2.10

58 m ≅ 2.5 s 54 m At t = 4.0 s, the slope is v ≅ ≅ 3s 49 m At t = 3.0 s, the slope is v ≅ ≅ 3.4 s 36 m At t = 2.0 s , the slope is v ≅ ≅ 4.0 s

At t = 5.0 s, the slope is v ≅

23 m s . 18 m s . 14 m s . 9.0 m s .

∆v 23 m s ≅ ≅ 4.6 m s 2 ∆t 5.0 s

(c)

a=

(d)

Initial velocity of the car was zero .

(a)

v=

(b)

v=

(c)

v=

(d)

v=

(5 − 0 ) m (1 − 0) s

= 5 ms

(5 − 10) m (4 − 2) s

= −2.5 m s

(5 m − 5 m) (5 s − 4 s) 0 − (−5 m) (8 s − 7 s )

= 0

= +5 m s

FIG. P2.9

Once it resumes the race, the hare will run for a time of t=

x f − xi vx

=

1 000 m − 800 m = 25 s . 8 ms

In this time, the tortoise can crawl a distance

a

f

x f − xi = 0.2 m s ( 25 s)= 5.00 m .

25

26

Motion in One Dimension

Section 2.3 P2.11

Acceleration

Choose the positive direction to be the outward direction, perpendicular to the wall. v f = vi + at : a =

P2.12

(a)

a

f

∆v 22.0 m s − −25.0 m s = = 1.34×10 4 m s 2 . ∆t 3.50 ×10−3 s

Acceleration is constant over the first ten seconds, so at the end,

c

h

v f = vi + at = 0 + 2.00 m s 2 (10.0 s)= 20.0 m s . Then a = 0 so v is constant from t = 10.0 s to t = 15.0 s. And over the last five seconds the velocity changes to

c

h

v f = vi + at = 20.0 m s + 3.00 m s 2 (5.00 s)= 5.00 m s . (b)

In the first ten seconds, 1 2 1 2 at = 0 + 0 + 2.00 m s 2 (10.0 s) = 100 m . 2 2

c

x f = x i + vi t +

h

Over the next five seconds the position changes to x f = xi + vi t +

a

f

1 2 at = 100 m + 20.0 m s (5.00 s)+ 0 = 200 m . 2

And at t = 20.0 s , x f = x i + vi t + *P2.13

(a)

a

f

1 2 1 2 at = 200 m + 20.0 m s (5.00 s)+ −3.00 m s 2 (5.00 s) = 262 m . 2 2

c

h

distance traveled . During the first ∆t quarter mile segment, Secretariat’s average speed was

The average speed during a time interval ∆t is v =

v1 =

0.250 mi 1 320 ft = = 52.4 ft s 25.2 s 25.2 s

b35.6 mi hg .

During the second quarter mile segment, v2 =

1 320 ft = 55.0 ft s 24.0 s

b37.4 mi hg .

For the third quarter mile of the race, v3 =

1 320 ft = 55.5 ft s 23.8 s

b37.7 mi hg ,

and during the final quarter mile, v4 = continued on next page

1 320 ft = 57.4 ft s 23.0 s

b39.0 mi hg .

Chapter 2

(b)

27

Assuming that v f = v 4 and recognizing that vi = 0 , the average acceleration during the race was v f − vi

a= P2.14

(a)

=

total elapsed time

57.4 ft s − 0 = 0.598 ft s 2 . ( 25. 2 + 24.0 + 23.8 + 23.0) s

Acceleration is the slope of the graph of v vs t.

a (m/s2) 2.0

For 0 < t < 5.00 s, a = 0 .

1.6

For 15.0 s < t < 20.0 s , a = 0 . For 5.0 s < t < 15.0 s , a =

a=

v f − vi t f − ti

1.0

.

8.00 − (−8.00) 15.0 − 5.00

0.0

= 1.60 m s 2

t (s) 0

5

a=

v f − vi t f − ti

(i)

For 5.00 s < t < 15.0 s , ti = 5.00 s , vi = −8.00 m s , t f = 15.0 s v f = 8.00 m s a=

(ii)

x = 2.00 + 3.00t − t 2 , v = At t = 3.00 s :

v f − vi t f − ti

=

a

f

8.00 − −8.00 = 1.60 m s 2 . 15.0 − 5.00

ti = 0 , vi = −8.00 m s , t f = 20.0 s , v f = 8.00 m s a=

P2.15

15

FIG. P2.14

We can plot a(t ) as shown. (b)

10

v f − vi t f − ti

=

8.00 − (−8.00) 20.0 − 0

dx dv = 3.00 − 2.00t , a = = −2.00 dt dt

(a)

x = ( 2.00 + 9.00 − 9.00) m = 2.00 m

(b)

v = (3.00 − 6.00) m s = −3.00 m s

(c)

a = −2.00 m s 2

= 0.800 m s 2

20

28 P2.16

Motion in One Dimension

(a)

2

At t = 2.00 s , x = 3.00( 2.00) − 2.00( 2.00)+ 3.00 m = 11.0 m.

a f

At t = 3.00 s , x = 3.00 9.00

2

a f

− 2.00 3.00 + 3.00 m = 24.0 m

so v= (b)

∆x 24.0 m − 11.0 m = 13.0 m s . = 3.00 s − 2.00 s ∆t

At all times the instantaneous velocity is v=

d 3.00t 2 − 2.00t + 3.00 = (6.00t − 2.00) m s dt

c

h

At t = 2.00 s , v = 6.00( 2.00)− 2.00 m s = 10.0 m s . At t = 3.00 s , v = 6.00(3.00)− 2.00 m s = 16.0 m s .

P2.17

a=

(d)

At all times a =

(a)

a=

(b)

Maximum positive acceleration is at t = 3 s, and is approximately 2 m s 2 .

(c)

a = 0 , at t = 6 s , and also for t > 10 s .

(d)

Maximum negative acceleration is at t = 8 s, and is approximately −1.5 m s 2 .

Section 2.4 P2.18

∆v 16.0 m s − 10.0 m s = = 6.00 m s 2 ∆t 3.00 s − 2.00 s

(c)

d (6.00 − 2.00)= 6.00 m s 2 . (This includes both t = 2.00 s and t = 3.00 s ). dt

∆v 8.00 m s = = 1.3 m s 2 ∆t 6.00 s

Motion Diagrams

(a) (b) (c) (d) (e) continued on next page

Chapter 2

(f)

Section 2.5 P2.19

29

One way of phrasing the answer: The spacing of the successive positions would change with less regularity. Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the accelerations vectors would vary in magnitude and direction.

One-Dimensional Motion with Constant Acceleration

c

From v 2f = vi2 + 2 ax , we have 10.97 ×10 3 m s

h

2

= 0 + 2 a( 220 m) , so that a = 2.74×10 5 m s 2

which is a = 2.79 ×10 4 times g . P2.20

P2.21

(a)

x f − xi =

(b)

a=

v f − vi t

a

f

1 1 vi + v f t becomes 40 m = vi + 2.80 m s (8.50 s) which yields vi = 6.61 m s . 2 2

c

=

h

2.80 m s − 6.61 m s = −0.448 m s 2 8.50 s

Given vi = 12.0 cm s when x i = 3.00 cm(t = 0) , and at t = 2.00 s , x f = −5.00 cm , 1 1 2 2 at : −5.00 − 3.00 = 12.0( 2.00)+ a( 2.00) 2 2 32.0 a =− = −16.0 cm s 2 . −8.00 = 24.0 + 2 a 2

x f − x i = vi t +

*P2.22

(a)

Let i be the state of moving at 60 mi h and f be at rest

d 0 = b60 mi hg

2 + 2 a x x f − xi v xf2 = v xi

ax

(b)

Similarly,

fFGH 5 1280mift IJK −3 600 mi F 5 280 ft I F 1 h I = G JG J = −21.8 mi h ⋅ s 242 h H 1 mi K H 3 600 s K F 1 609 m IJ FG 1 h IJ = −9.75 m s = −21.8 mi h ⋅ s G H 1 mi K H 3 600 s K 2

a

+ 2 a x 121 ft − 0

2

b

0 = 80 mi h ax = − (c)

i

b

g

2

6 400 5 280

b

a

+ 2 a x 211 ft − 0

422 3 600

g

g

2

.

f

mi h ⋅ s = −22.2 mi h ⋅ s = −9.94 m s 2 .

Let i be moving at 80 mi h and f be moving at 60 mi h .

d i b60 mi hg = b80 mi hg + 2a a211 ft − 121 ftf 2 800b5 280 g a =− mi h ⋅ s = −22.8 mi h ⋅ s = −10.2 m s 2a90fb3 600g 2 + 2 a x x f − xi v xf2 = v xi 2

2

x

x

2

.

30 *P2.23

Motion in One Dimension

(a)

Choose the initial point where the pilot reduces the throttle and the final point where the boat passes the buoy: x i = 0 , x f = 100 m , v xi = 30 m s , v xf = ?, a x = −3.5 m s 2 , t = ? x f = xi + v xi t +

1 axt 2 : 2

a

f

100 m = 0 + 30 m s t +

1 −3.5 m s 2 t 2 2

c

h

c1.75 m s ht − a30 m sft + 100 m = 0 . 2

2

We use the quadratic formula: t=

t=

c

−b ± b 2 − 4ac 2a

h

30 m s ± 900 m 2 s 2 − 4 1.75 m s 2 (100 m)

c

2 1.75 m s

2

h

=

30 m s ± 14.1 m s 3.5 m s 2

= 12.6 s or 4.53 s .

The smaller value is the physical answer. If the boat kept moving with the same acceleration, it would stop and move backward, then gain speed, and pass the buoy again at 12.6 s.

P2.24

e

j

(b)

v xf = v xi + a x t = 30 m s − 3.5 m s 2 4.53 s = 14.1 m s

(a)

Total displacement = area under the v , t curve from t = 0 to 50 s.

a f

b b

ga f b ga f

ga

f

1 50 m s 15 s + 50 m s 40 − 15 s 2 1 + 50 m s 10 s 2 ∆x = 1 875 m ∆x =

(b)

From t = 10 s to t = 40 s , displacement is

(c)

b

ga f b

ga f

1 50 m s + 33 m s 5 s + 50 m s 25 s = 1 457 m . 2 ∆v (50 − 0) m s 0 ≤ t ≤ 15 s : a1 = = = 3.3 m s 2 ∆t 15 s − 0 15 s < t < 40 s : a 2 = 0 ∆x =

40 s ≤ t ≤ 50 s : a 3 = continued on next page

∆v (0 − 50) m s = = −5.0 m s 2 ∆t 50 s − 40 s

FIG. P2.24

Chapter 2

(d)

1 1 a1 t 2 = 3.3 m s 2 t 2 or x1 = 1.67 m s 2 t 2 2 2

c

h

c

h

(i)

x1 = 0 +

(ii)

1 x 2 = (15 s) 50 m s − 0 + 50 m s (t − 15 s) or x 2 = 50 m s t − 375 m 2

(iii)

For 40 s ≤ t ≤ 50 s ,

a

x3 =

f

a

f

FG area under v vs t IJ + 1 a (t − 40 s) + a50 m sf(t − 40 s) H from t = 0 to 40 sK 2 2

3

or x 3 = 375 m + 1 250 m +

ja

1 −5.0 m s 2 t − 40 s 2

e

f + b50 m sgat − 40 sf 2

which reduces to

b

g e

j

x 3 = 250 m s t − 2.5 m s 2 t 2 − 4 375 m .

P2.25

total displacement 1 875 m = = 37.5 m s total elapsed time 50 s

(e)

v=

(a)

Compare the position equation x = 2.00 + 3.00t − 4.00t 2 to the general form x f = xi + vi t +

1 2 at 2

to recognize that x i = 2.00 m, vi = 3.00 m s, and a = −8.00 m s 2 . The velocity equation, v f = vi + at , is then

c

h

v f = 3.00 m s − 8.00 m s 2 t . The particle changes direction when v f = 0 , which occurs at t = time is:

a

x = 2.00 m + 3.00 m s

(b)

fFGH 38 sIJK − c4.00 m s hFGH 38 sIJK 2

3 s . The position at this 8

2

= 2.56 m .

2v 1 2 at , observe that when x f = xi , the time is given by t = − i . Thus, a 2 when the particle returns to its initial position, the time is

From x f = xi + vi t +

t=

c

a

−2 3.00 m s −8.00 m s

and the velocity is v f = 3.00 m s − 8.00 m s 2

2

f=3 s

hFGH 34 sIJK =

4

−3.00 m s .

31

32 *P2.26

Motion in One Dimension

The time for the Ford to slow down we find from 1 v xi + v xf t 2 2 250 m 2 ∆x t= = = 6.99 s . v xi + v xf 71.5 m s + 0

x f = xi +

d

a

i

f

Its time to speed up is similarly t=

2(350 m) 0 + 71.5 m s

= 9.79 s .

The whole time it is moving at less than maximum speed is 6.99 s + 5.00 s + 9.79 s = 21.8 s . The Mercedes travels

a

fb ga

1 1 v xi + v xf t = 0 + 71.5 + 71.5 m s 21.8 s 2 2 = 1 558 m

d

x f = xi +

i

f

while the Ford travels 250 + 350 m = 600 m, to fall behind by 1 558 m − 600 m = 958 m . P2.27

(a)

c

h

vi = 100 m s , a = −5.00 m s 2 , v f = vi + at so 0 = 100 − 5t , v 2f = vi2 + 2 a x f − xi so 2

c

h

0 = (100 ) − 2(5.00) x f − 0 . Thus x f = 1 000 m and t = 20.0 s .

P2.28

(b)

At this acceleration the plane would overshoot the runway: No .

(a)

Take ti = 0 at the bottom of the hill where x i = 0 , vi = 30.0 m s, a = −2.00 m s 2 . Use these values in the general equation x f = xi + vi t + to find

a

1 2 at 2

f 12 c−2.00 m s ht 2

x f = 0 + 30.0t m s + when t is in seconds

c

2

h

x f = 30.0t − t 2 m .

e

j

To find an equation for the velocity, use v f = vi + at = 30.0 m s + −2.00 m s 2 t , v f = (30.0 − 2.00t ) m s . (b)

The distance of travel x f becomes a maximum, x max , when v f = 0 (turning point in the motion). Use the expressions found in part (a) for v f to find the value of t when x f has its maximum value: From v f = (3.00 − 2.00t ) m s , v f = 0 when t = 15.0 s. Then

c

h

2

x max = 30.0t − t 2 m = (30.0)(15.0)−(15.0) = 225 m .

Chapter 2

P2.29

33

In the simultaneous equations:

R| S|x T

v xf = v xi + a x t f

− xi =

R|v = v − c5.60 m s h(4.20 s)U| U| ht V|W we have S|T 62.4 m = 12 cv + v h(4.20 s) V|W . xf

1 v xi + v xf 2

c

So substituting for v xi gives 62.4 m =

2

xi

xi

xf

1 v xf + 56.0 m s 2 ( 4.20 s)+ v xf ( 4.20 s) 2

c

h

14.9 m s = v xf +

1 5.60 m s 2 ( 4.20 s). 2

c

h

Thus v xf = 3.10 m s .

P2.30

Take any two of the standard four equations, such as substitute into the other: v xi = v xf − a x t x f − xi =

R| S|x T

v xf = v xi + a x t f

− xi =

1 v xi + v xf 2

c

U| ht V|W. Solve one for v

1 v xf − a x t + v xf t . 2

c

h

Thus 1 x f − xi = v xf t − a x t 2 . 2 Back in problem 29, 62.4 m = v xf ( 4.20 s)− v xf =

P2.31

v f − vi

(a)

a=

(b)

x f = vi t +

t

=

632

e j= 5 280 3 600

1.40

a

fFGH

1 2 −5.60 m s 2 ( 4. 20 s) 2

c

h

62.4 m − 49.4 m = 3.10 m s . 4.20 s

−662 ft s 2 = −202 m s 2

I a f a fa f JK

5 280 1 2 1 1.40 − 662 1.40 at = 632 2 3 600 2

2

= 649 ft = 198 m

xi ,

and

34 P2.32

Motion in One Dimension

(a)

The time it takes the truck to reach 20.0 m s is found from v f = vi + at . Solving for t yields t=

v f − vi a

=

20.0 m s − 0 m s 2.00 m s 2

= 10.0 s .

The total time is thus 10.0 s + 20.0 s + 5.00 s = 35.0 s . (b)

The average velocity is the total distance traveled divided by the total time taken. The distance traveled during the first 10.0 s is x 1 = vt =

FG 0 + 20.0 IJ(10.0)= 100 m . H 2 K

With a being 0 for this interval, the distance traveled during the next 20.0 s is x 2 = vi t +

1 2 at = ( 20.0)( 20.0)+ 0 = 400 m. 2

The distance traveled in the last 5.00 s is x 3 = vt =

FG 20.0 + 0 IJ(5.00)= 50.0 m. H 2 K

The total distance x = x1 + x 2 + x 3 = 100 + 400 + 50 = 550 m , and the average velocity is x 550 = 15.7 m s . given by v = = t 35.0 P2.33

We have vi = 2.00 ×10 4 m s, v f = 6.00 ×10 6 m s , x f − xi = 1.50 ×10−2 m .

c

c

h

h

2 1.50 ×10−2 m 2 x f − xi 1 = = 4.98 ×10−9 s vi + v f t : t = 4 6 2 vi + v f 2.00 ×10 m s + 6.00 ×10 m s

c

h

(a)

x f − xi =

(b)

v 2f = vi2 + 2 a x x f − xi :

d

ax =

i

v 2f − vi2 2( x f − xi )

e6.00 × 10 =

6

ms

j − e2.00 × 10 2

2(1.50 × 10 −2 m)

4

ms

j

2

= 1.20 × 10 15 m s 2

Chapter 2

*P2.34

(a)

c

h

c

2 v xf2 = v xi + 2 a x x f − x i : 0.01 3 ×10 8 m s

ax (b)

c3×10 =

6

h

2

ms

80 m

= 0 + 2 a x ( 40 m)

h

2

= 1.12 ×10 11 m s 2

We must find separately the time t1 for speeding up and the time t 2 for coasting: x f − xi =

x f − xi =

1 1 v xf + v xi t1 : 40 m = 3 × 10 6 m s + 0 t1 2 2 t1 = 2.67 × 10 −5 s

d

i

e

j

1 1 v xf + v xi t 2 : 60 m = 3 × 10 6 m s + 3 × 10 6 m s t 2 2 2 t 2 = 2.00 × 10 −5 s

d

i

e

j

total time = 4.67 ×10−5 s . *P2.35

(a)

Along the time axis of the graph shown, let i = 0 and f = t m . Then v xf = v xi + a x t gives v c = 0 + am tm am =

(b)

vc . tm

The displacement between 0 and t m is x f − xi = v xi t +

1 vc 2 1 1 axt 2 = 0 + t m = v c tm . 2 tm 2 2

The displacement between t m and t 0 is x f − xi = v xi t +

a

f

1 a x t 2 = v c t0 − tm + 0 . 2

The total displacement is ∆x =

FG H

1 1 v c t m + v c t 0 − v c t m = v c t 0 − tm 2 2

IJ K

.

(c)

For constant v c and t 0 , ∆x is minimized by maximizing t m to t m = t 0 . Then v t 1 ∆x min = v c t 0 − t 0 = c 0 . 2 2

(e)

This is realized by having the servo motor on all the time.

(d)

We maximize ∆x by letting t m approach zero. In the limit ∆x = v c t 0 − 0 = v c t 0 .

(e)

This cannot be attained because the acceleration must be finite.

FG H

IJ K

a

f

35

36 *P2.36

Motion in One Dimension

Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its motion from entry to exit, 1 axt 2 2 1 A = 0 + vi ∆t d + a∆t d2 = v d ∆t d 2 1 v d = vi + a∆t d 2

x f = xi + v xi t +

(a)

The speed halfway through the photogate in space is given by 2 v hs = vi2 + 2 a

FG A IJ = v H 2K

2 i

+ av d ∆t d .

v hs = vi2 + av d ∆t d and this is not equal to v d unless a = 0 . (b)

The speed halfway through the photogate in time is given by v ht = vi + a

FG ∆t IJ and this is H 2K d

equal to v d as determined above. P2.37

(a)

Take initial and final points at top and bottom of the incline. If the ball starts from rest, vi = 0 , a = 0.500 m s 2 , x f − xi = 9.00 m . Then

d

i

ja

e

f

v 2f = vi2 + 2 a x f − xi = 0 2 + 2 0.500 m s 2 9.00 m v f = 3.00 m s . (b)

x f − x i = vi t +

1 2 at 2

1 0.500 m s 2 t 2 2 t = 6.00 s

e

9.00 = 0 +

(c)

Take initial and final points at the bottom of the planes and the top of the second plane, respectively: vi = 3.00 m s, v f = 0 , x f − xi = 15.00 m.

c

h

v 2f = vi2 + 2 a x f − xi gives a= (d)

j

v 2f − vi2

c

2 x f − xi

h

=

a

0 − 3.00 m s 2(15.0 m)

f

2

= −0.300 m s 2 .

Take the initial point at the bottom of the planes and the final point 8.00 m along the second: vi = 3.00 m s, x f − xi = 8.00 m , a = −0.300 m s 2

d

i b

v 2f = vi2 + 2 a x f − xi = 3.00 m s v f = 2.05 m s .

g + 2e−0.300 m s ja8.00 mf = 4.20 m 2

2

2

s2

Chapter 2

P2.38

37

Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sue’s car. For her we have x is = 0 , vis = 30.0 m s , a s = −2.00 m s 2 so her position is given by x s (t )= x is + vis t +

a

f

1 1 a s t 2 = 30.0 m s t + −2.00 m s 2 t 2 . 2 2

c

h

For the van, x iv = 155 m, viv = 5.00 m s , a v = 0 and x v (t )= xiv + viv t +

a

f

1 a v t 2 = 155 + 5.00 m s t + 0 . 2

To test for a collision, we look for an instant t c when both are at the same place: 30.0t c − t c2 = 155 + 5.00t c 0 = t c2 − 25.0t c + 155 . From the quadratic formula 2

tc =

25.0 ± ( 25.0) − 4(155) 2

= 13.6 s or 11.4 s .

The smaller value is the collision time. (The larger value tells when the van would pull ahead again if the vehicles could move through each other). The wreck happens at position

a

f

155 m + 5.00 m s (11.4 s)= 212 m . *P2.39

As in the algebraic solution to Example 2.8, we let t represent the time the trooper has been moving. We graph x car = 45 + 45t

x (km) 1.5 car 1

and 2

x trooper = 1.5t . They intersect at

police officer

0.5

10 t = 31 s .

20

30

FIG. P2.39

40

t (s)

38

Motion in One Dimension

Section 2.6 P2.40

Freely Falling Objects

a

f

Choose the origin y = 0 , t = 0 at the starting point of the ball and take upward as positive. Then yi = 0 , vi = 0 , and a = −g = −9.80 m s 2 . The position and the velocity at time t become: y f − yi = vi t +

1 1 1 2 at : y f = − gt 2 = − 9.80 m s 2 t 2 2 2 2

e

j

and

c

h

v f = vi + at : v f = −gt = − 9.80 m s 2 t . 1 2 9.80 m s 2 (1.00 s) = −4.90 m 2 1 2 at t = 2.00 s : y f = − 9.80 m s 2 ( 2.00 s) = −19.6 m 2 1 2 at t = 3.00 s : y f = − 9.80 m s 2 (3.00 s) = −44.1 m 2

h h h

at t = 1.00 s : y f = −

(b)

at t = 1.00 s : v f = − 9.80 m s 2 (1.00 s)= −9.80 m s at t = 2.00 s : v f at t = 3.00 s : v f

P2.41

c c c

(a)

c h = −c9.80 m s h( 2.00 s)= −19.6 m s = −c9.80 m s h(3.00 s)= −29.4 m s 2 2

Assume that air resistance may be neglected. Then, the acceleration at all times during the flight is that due to gravity, a = −g = −9.80 m s 2 . During the flight, Goff went 1 mile (1 609 m) up and then 1 mile back down. Determine his speed just after launch by considering his upward flight:

d

i

v 2f = vi2 + 2 a y f − yi :

jb

e

0 = vi2 − 2 9.80 m s 2 1 609 m vi = 178 m s .

g

His time in the air may be found by considering his motion from just after launch to just before impact: y f − yi = vi t +

a

f

1 1 2 at : 0 = 178 m s t − −9.80 m s 2 t 2 . 2 2

c

h

The root t = 0 describes launch; the other root, t = 36.2 s , describes his flight time. His rate of pay may then be found from pay rate =

b

gb

g

$1.00 = 0.027 6 $ s 3 600 s h = $99.3 h . 36.2 s

We have assumed that the workman’s flight time, “a mile”, and “a dollar”, were measured to threedigit precision. We have interpreted “up in the sky” as referring to the free fall time, not to the launch and landing times. Both the takeoff and landing times must be several seconds away from the job, in order for Goff to survive to resume work.

Chapter 2

P2.42

39

1 We have y f = − gt 2 + vi t + yi 2

h a

c

f

0 = − 4.90 m s 2 t 2 − 8.00 m s t + 30.0 m . Solving for t, t=

8.00 ± 64.0 + 588 . −9.80

Using only the positive value for t, we find that t = 1.79 s . P2.43

1 2 2 at : 4.00 = (1.50)vi −(4.90)(1.50) and vi = 10.0 m s upward . 2

(a)

y f − yi = vi t +

(b)

v f = vi + at = 10.0 −(9.80)(1.50) = −4.68 m s v f = 4.68 m s downward

P2.44

The bill starts from rest vi = 0 and falls with a downward acceleration of 9.80 m s 2 (due to gravity). Thus, in 0.20 s it will fall a distance of ∆y = vi t −

1 2 2 gt = 0 − 4.90 m s 2 (0. 20 s) = −0.20 m . 2

c

h

a

f

This distance is about twice the distance between the center of the bill and its top edge ≅ 8 cm . Thus, David will be unsuccessful . *P2.45

(a)

From ∆y = vi t +

1 2 at with vi = 0 , we have 2 t=

a f=

2 ∆y

2(−23 m)

a

−9.80 m s 2

c

= 2.17 s .

h

(b)

The final velocity is v f = 0 + −9.80 m s 2 ( 2.17 s)= −21.2 m s .

(c)

The time take for the sound of the impact to reach the spectator is t sound =

∆y v sound

=

23 m = 6.76 ×10−2 s , 340 m s

so the total elapsed time is t total = 2.17 s + 6.76 × 10 −2 s ≈ 2.23 s .

40 P2.46

Motion in One Dimension

At any time t, the position of the ball released from rest is given by y1 = h −

1 2 gt . At time t, the 2

1 2 gt . The time at which the 2 h 1 h first ball has a position of y1 = is found from the first equation as = h − gt 2 , which yields 2 2 2 h h . To require that the second ball have a position of y 2 = at this time, use the second t= g 2 position of the ball thrown vertically upward is described by y 2 = vi t −

equation to obtain

F I GH JK

h h 1 h = vi − g . This gives the required initial upward velocity of the second 2 g 2 g

ball as vi = gh . P2.47

(a)

v f = vi − gt : v f = 0 when t = 3.00 s , g = 9.80 m s 2 . Therefore,

c

h

vi = gt = 9.80 m s 2 (3.00 s)= 29.4 m s . (b)

y f − yi =

1 v f + vi t 2

c

h

y f − yi = *P2.48

(a)

b

f

ga

1 29.4 m s 3.00 s = 44.1 m 2

Consider the upward flight of the arrow.

d i 0 = b100 m sg + 2e −9.8 m s j∆y

2 2 v yf = v yi + 2 a y y f − yi 2

∆y = (b)

2

10 000 m 2 s 2 19.6 m s 2

= 510 m

Consider the whole flight of the arrow. y f = yi + v yi t +

b

1 ayt 2 2

g

0 = 0 + 100 m s t +

1 −9.8 m s 2 t 2 2

e

j

The root t = 0 refers to the starting point. The time of flight is given by t=

P2.49

100 m s 4.9 m s 2

= 20.4 s .

Time to fall 3.00 m is found from Eq. 2.12 with vi = 0 , 3.00 m =

1 9.80 m s 2 t 2 , t = 0.782 s. 2

c

h

(a)

With the horse galloping at 10.0 m s, the horizontal distance is vt = 7.82 m .

(b)

t = 0.782 s

Chapter 2

P2.50

Take downward as the positive y direction. (a)

While the woman was in free fall, ∆y = 144 ft , vi = 0 , and a = g = 32.0 ft s 2 . Thus, ∆y = vi t + before impact is:

1 2 at → 144 ft = 0 + 16.0 ft s 2 t 2 giving t fall = 3.00 s . Her velocity just 2

c

h

c

h

v f = vi + gt = 0 + 32.0 ft s 2 (3.00 s)= 96.0 ft s . (b)

While crushing the box, vi = 96.0 ft s , v f = 0 , and ∆y = 18.0 in. = 1.50 ft . Therefore, a=

(c)

v 2f − vi2

a f

2 ∆y

=

a

0 − 96.0 ft s

f

2

2(1.50 ft )

Time to crush box: ∆t =

= −3.07 ×10 3 ft s 2 , or a = 3.07 ×10 3 ft s 2 upward .

2(1.50 ft) ∆y ∆y = v +v = or ∆t = 3.13 ×10−2 s . f i v 0 + 96.0 ft s 2

P2.51

a f

y = 3.00t 3 : At t = 2.00 s , y = 3.00 2.00

3

= 24.0 m and

vy =

A

dy = 9.00t 2 = 36.0 m s . dt

If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is y b = y bi + vi t −

1 1 2 gt = 24.0 + 36.0t − (9.80)t 2 . 2 2

Setting y b = 0 , 0 = 24.0 + 36.0t − 4.90t 2 . Solving for t, (only positive values of t count), t = 7.96 s . *P2.52

Consider the last 30 m of fall. We find its speed 30 m above the ground: y f = yi + v yi t +

1 ayt 2 2

a f 12 e−9.8 m s ja1.5 sf 2

0 = 30 m + v yi 1.5 s + v yi =

2

−30 m + 11.0 m = −12.6 m s . 1.5 s

Now consider the portion of its fall above the 30 m point. We assume it starts from rest

d

i b−12.6 m sg = 0 + 2e−9.8 m s j∆y 2 2 v yf = v yi + 2 a y y f − yi 2

∆y =

2

160 m 2 s 2 −19.6 m s 2

Its original height was then 30 m + −8.16 m = 38.2 m .

= −8.16 m .

41

42

Motion in One Dimension

Section 2.7 P2.53

(a)

Kinematic Equations Derived from Calculus J=

da = constant dt da = Jdt

z

a = J dt = Jt + c 1 but a = ai when t = 0 so c 1 = ai . Therefore, a = Jt + ai dv dt dv = adt a=

z zb

v = adt =

g

Jt + ai dt =

but v = vi when t = 0, so c 2 = vi and v =

1 2 Jt + ai t + c 2 2

1 2 Jt + ai t + vi 2

dx dt dx = vdt v=

z z FGH

x = vdt =

IJ K

1 2 Jt + ai t + vi dt 2

1 3 1 2 Jt + ai t + vi t + c 3 6 2 x = xi x=

when t = 0, so c 3 = xi . Therefore, x = (b)

a

a 2 = Jt + ai

c

f

2

= J 2 t 2 + ai2 + 2 Jai t

a 2 = ai2 + J 2 t 2 + 2 Jai t

FG H

1 3 1 2 Jt + ai t + vi t + xi . 6 2

h

1 2 a 2 = ai2 + 2 J Jt + ai t 2

IJ K

Recall the expression for v: v =

a

f

1 2 1 Jt + ai t + vi . So v − vi = Jt 2 + ai t . Therefore, 2 2

a

a 2 = ai2 + 2 J v − vi

f

.

Chapter 2

P2.54

(a)

See the graphs at the right. Choose x = 0 at t = 0. At t = 3 s, x =

a

f

1 8 m s (3 s)= 12 m . 2

a

f

At t = 5 s, x = 12 m + 8 m s ( 2 s)= 28 m . At t = 7 s, x = 28 m +

P2.55

a

f

1 8 m s ( 2 s)= 36 m . 2

8 ms = 2.67 m s 2 . 3s For 3 < t < 5 s, a = 0 .

(b)

For 0 < t < 3 s, a =

(c)

For 5 s < t < 9 s , a = −

(d)

At t = 6 s, x = 28 m + 6 m s (1 s)= 34 m .

(e)

At t = 9 s, x = 36 m +

(a)

16 m s = −4 m s 2 . 4s

a

a=

f

a

f

1 −8 m s ( 2 s)= 28 m . 2

FIG. P2.54

dv d = −5.00 ×10 7 t 2 + 3.00 ×10 5 t dt dt

c

h

a = − 10.0 ×10 7 m s 3 t + 3.00 ×10 5 m s 2 Take x i = 0 at t = 0. Then v =

dx dt

z ze t

t

0

0

x − 0 = vdt =

j

−5.00 × 10 7 t 2 + 3.00 × 10 5 t dt

t3 t2 + 3.00 × 10 5 3 2 3 3 7 x = − 1.67 × 10 m s t + 1.50 × 10 5 m s 2 t 2 .

x = −5.00 × 10 7

e

(b)

j e

c

h

The bullet escapes when a = 0 , at − 10.0 ×10 7 m s 3 t + 3.00 ×10 5 m s 2 = 0 t=

(c)

j

c

hc

New v = −5.00 ×10 7 3.00 ×10−3

3.00 ×10 5 s = 3.00 ×10−3 s . 10.0 ×10 7

h + c3.00×10 hc3.00×10 h 2

5

−3

v = −450 m s + 900 m s = 450 m s . (d)

c

hc

x = − 1.67 ×10 7 3.00 ×10−3

h + c1.50×10 hc3.00×10 h 3

x = −0.450 m + 1.35 m = 0.900 m

5

−3 2

43

44 P2.56

Motion in One Dimension

a=

dv = −3.00 v 2 , vi = 1.50 m s dt

Solving for v,

dv = −3.00 v 2 dt

z v

z t

v −2 dv = −3.00 dt

v = vi

− When v =

t =0

1 1 1 1 + = −3.00t or 3.00t = − . v vi v vi

vi 1 , t= = 0.222 s . 2 3.00 vi

Additional Problems *P2.57

a f

The distance the car travels at constant velocity, v 0 , during the reaction time is ∆x 1 = v 0 ∆t r . The time for the car to come to rest, from initial velocity v 0 , after the brakes are applied is t2 =

v f − vi a

=

0 − v0 v =− 0 a a

and the distance traveled during this braking period is

a∆xf

2

= vt 2 =

Fv GH

+ vi

f

2

I t = FG 0 + v IJ FG − v IJ = − v . JK H 2 K H a K 2 a 0

2

0

2 0

Thus, the total distance traveled before coming to a stop is

a f + a ∆x f

sstop = ∆x

*P2.58

(a)

1

2

= v 0 ∆t r −

v 02 . 2a

v 02 (See the solution to Problem 2.57) from the 2a intersection of length s i when the light turns yellow, the distance the car must travel before the light turns red is v2 ∆x = sstop + si = v 0 ∆t r − 0 + si . 2a

If a car is a distance sstop = v 0 ∆t r −

Assume the driver does not accelerate in an attempt to “beat the light” (an extremely dangerous practice!). The time the light should remain yellow is then the time required for the car to travel distance ∆x at constant velocity v 0 . This is v2

∆t light (b)

v s ∆x v 0 ∆t r − 20a + si = = = ∆t r − 0 + i . v0 v0 2 a v0

With si = 16 m, v = 60 km h , a = −2.0 m s 2 , and ∆t r = 1.1 s , ∆t light = 1.1 s −

F 0.278 m s I + 16 m F 1 km h I = G J G J 2e −2.0 m s j H 1 km h K 60 km h H 0.278 m s K 60 km h

2

6. 23 s .

Chapter 2

*P2.59

(a)

(b)

As we see from the graph, from about −50 s to 50 s Acela is cruising at a constant positive velocity in the +x direction. From 50 s to 200 s, Acela accelerates in the +x direction reaching a top speed of about 170 mi/h. Around 200 s, the engineer applies the brakes, and the train, still traveling in the +x direction, slows down and then stops at 350 s. Just after 350 s, Acela reverses direction (v becomes negative) and steadily gains speed in the −x direction.

200 ∆v

100

∆t 100 200 300 400

0 –50 0

t (s)

–100

FIG. P2.59(a)

The peak acceleration between 45 and 170 mi/h is given by the slope of the steepest tangent to the v versus t curve in this interval. From the tangent line shown, we find a = slope =

(c)

45

∆v (155 − 45) mi h = = 2. 2 mi h s = 0.98 m s 2 . (100 − 50) s ∆t

a

Let us use the fact that the area under the v versus t curve equals the displacement. The train’s displacement between 0 and 200 s is equal to the area of the gray shaded region, which we have approximated with a series of triangles and rectangles.

f

200

∆x 0 → 200 s = area 1 + area 2 + area 3 + area 4 + area 5

5

100 0

4 1 2 0

b ga f b ga f + b160 mi hga100 sf 1 + a50 sfb100 mi hg 2 1 + a100 sfb170 mi h − 160 mi hg 2 = 24 000bmi hgasf

3

100 200 300 400

≈ 50 mi h 50 s + 50 mi h 50 s

FIG. P2.59(c)

Now, at the end of our calculation, we can find the displacement in miles by converting hours to seconds. As 1 h = 3 600 s , ∆x 0 → 200 s ≈

F 24 000 mi I asf = GH 3 600 s JK

6.7 mi .

t (s)

46 *P2.60

Motion in One Dimension

Average speed of every point on the train as the first car passes Liz: ∆x 8.60 m = = 5.73 m s. 1.50 s ∆t The train has this as its instantaneous speed halfway through the 1.50 s time. Similarly, halfway 8.60 m through the next 1.10 s, the speed of the train is = 7.82 m s . The time required for the speed 1.10 s to change from 5.73 m/s to 7.82 m/s is 1 1 (1.50 s)+ (1.10 s)= 1.30 s 2 2 so the acceleration is: a x =

P2.61

∆v x 7.82 m s − 5.73 m s = = 1.60 m s 2 . ∆t 1.30 s

The rate of hair growth is a velocity and the rate of its increase is an acceleration. Then mm d v xi = 1.04 mm d and a x = 0.132 . The increase in the length of the hair (i.e., displacement) w during a time of t = 5.00 w = 35.0 d is

FG H

∆x = v xi t +

b

IJ K

1 axt 2 2

f 12 b0.132 mm d ⋅ wga35.0 dfa5.00 wf

ga

∆x = 1.04 mm d 35.0 d + or ∆x = 48.0 mm . P2.62

Let point 0 be at ground level and point 1 be at the end of the engine burn. Let point 2 be the highest point the rocket reaches and point 3 be just before impact. The data in the table are found for each phase of the rocket’s motion.

a f

v 2f − 80.0

(0 to 1)

2

a fb

= 2 4.00 1 000

g

120 = 80.0 +( 4.00)t

c

2

(1 to 2)

0 −(120) = 2(−9.80) x f − xi

h

so

v f = 120 m s

giving

t = 10.0 s

giving

x f − xi = 735 m

0 − 120 = −9.80t giving This is the time of maximum height of the rocket.

a

fb

v 2f − 0 = 2 −9.80 −1 735

(2 to 3)

g

v f = −184 = (−9.80)t (a)

t total = 10 + 12.2 + 18.8 = 41.0 s

(b)

cx

f

− xi

h

total

= 1.73 km

continued on next page

t = 12.2 s

giving

t = 18.8 s

FIG. P2.62

Chapter 2

(c)

v final = −184 m s

0 #1 #2 #3 P2.63

P2.64

t 0.0 10.0 22.2 41.0

Launch End Thrust Rise Upwards Fall to Earth

a

x 0 1 000 1 735 0

f

v 80 120 0 –184

a +4.00 +4.00 –9.80 –9.80

Distance traveled by motorist = 15.0 m s t 1 Distance traveled by policeman = 2.00 m s 2 t 2 2

c

h

(a)

intercept occurs when 15.0t = t 2 , or t = 15.0 s

(b)

v(officer)= 2.00 m s 2 t = 30.0 m s

(c)

x(officer )=

c

h

1 2.00 m s 2 t 2 = 225 m 2

c

h

Area A1 is a rectangle. Thus, A1 = hw = v xi t . 1 1 Area A 2 is triangular. Therefore A 2 = bh = t v x − v xi . 2 2 The total area under the curve is

b

A = A1 + A 2 = v xi t +

bv

x

g

vx vx A2

vxi

g

A1

− v xi t 2

0

and since v x − v xi = a x t

t FIG. P2.64

A = v xi t +

1 axt 2 . 2

The displacement given by the equation is: x = v xi t + same result as above for the total area.

1 a x t 2 , the 2

t

47

48 P2.65

Motion in One Dimension

(a)

Let x be the distance traveled at acceleration a until maximum speed v is reached. If this is achieved in time t1 we can use the following three equations: x=

a

a

f

f

1 v + vi t1 , 100 − x = v 10.2 − t1 and v = vi + at1 . 2

The first two give

FG 1 t IJ v = FG10.2 − 1 t IJ at H 2 K H 2 K 200 a= b20.4 − t gt .

100 = 10.2 −

1

1

For Maggie: a =

1

1

200

a18.4fa2.00f = 200 For Judy: a = a17.4fa3.00f =

(b)

1

5.43 m s 2 3.83 m s 2

v = a1 t

a fa f Judy: v = a3.83fa3.00f =

Maggie: v = 5.43 2.00 = 10.9 m s

(c)

11.5 m s

At the six-second mark x=

f

a fa f + a10.9fa4.00f = 54.3 m a fa f + a11.5fa3.00f = 51.7 m

1 5.43 2.00 2 1 Judy: x = 3.83 3.00 2

Maggie: x =

a

1 2 at1 + v 6.00 − t1 2 2

2

Maggie is ahead by 2.62 m . P2.66

a1 = 0.100 m s 2 1 1 x = 1 000 m = a1 t12 + v1 t 2 + a 2 t 22 2 2 at at 1 1 1 000 = a1 t12 + a1 t1 − 1 1 + a 2 1 1 2 a2 2 a2

FG H

IJ K

FG IJ H K

a 2 = −0.500 m s 2 t = t1 + t 2 and v1 = a1 t1 = −a 2 t 2 2

1 000 = t1 =

t2 =

a1 t1 12.9 = ≈ 26 s −a 2 0.500

FG H

IJ K

a 1 a1 1 − 1 t12 2 a2

20 000 = 129 s 1.20

Total time = t = 155 s

Chapter 2

P2.67

Let the ball fall 1.50 m. It strikes at speed given by

c

h

2 + 2 a x f − xi : v xf2 = v xi

c

h

v xf2 = 0 + 2 −9.80 m s 2 (−1.50 m) v xf = −5.42 m s and its stopping is described by

d

2 v xf2 = v xi + 2 a x x f − xi

b

0 = −5.42 m s ax =

g

2

e

−2.00 × 10

j

+ 2 a x −10 −2 m

−29.4 m 2 s 2 −2

i

m

= +1.47 × 10 3 m s 2 .

Its maximum acceleration will be larger than the average acceleration we estimate by imagining constant acceleration, but will still be of order of magnitude ~ 10 3 m s 2 . *P2.68

(a)

x f = xi + v xi t +

1 a x t 2 . We assume the package starts from rest. 2 −145 m = 0 + 0 +

t=

(b)

x f = xi + v xi t +

1 −9.80 m s 2 t 2 2

c

2(−145 m) −9.80 m s 2

h

= 5. 44 s

1 1 2 a x t 2 = 0 + 0 + −9.80 m s 2 (5.18 s) = −131 m 2 2

c

h

distance fallen = x f = 131 m

e

j

(c)

speed = v xf = v xi + a x t = 0 + −9.8 m s 2 5.18 s = 50.8 m s

(d)

The remaining distance is 145 m − 131.5 m = 13.5 m . During deceleration, v xi = −50.8 m s, v xf = 0, x f − xi = −13.5 m

c

h

2 v xf2 = v xi + 2 a x x f − xi :

a

0 = −50.8 m s ax =

f

2

+ 2 a x (−13.5 m)

−2 580 m 2 s 2 = +95.3 m s 2 = 95.3 m s 2 upward . 2 −13.5 m

a

f

49

50 P2.69

Motion in One Dimension

(a)

1 1 2 at = 50.0 = 2.00t + (9.80)t 2 , 2 2 4.90t 2 + 2.00t − 50.0 = 0 y f = v i1 t +

t=

−2.00 + 2.00 2 − 4( 4.90)(−50.0) 2( 4.90)

Only the positive root is physically meaningful: t = 3.00 s after the first stone is thrown. (b)

1 2 at and t = 3.00 − 1.00 = 2.00 s 2 1 2 substitute 50.0 = vi 2 ( 2.00)+ (9.80)( 2.00) : 2 y f = vi 2 t +

vi2 = 15.3 m s downward (c)

v1 f = vi1 + at = 2.00 +(9.80)(3.00)= 31.4 m s downward v 2 f = vi 2 + at = 15.3 +(9.80)( 2.00)= 34.8 m s downward

P2.70

(a)

1 d = (9.80)t12 2 t1 + t 2 = 2.40 4.90t 22

d = 336 t 2

a

336t 2 = 4.90 2.40 − t 2

− 359.5t 2 + 28.22 = 0

t2 =

359.5 ± 358.75 = 0.076 5 s t2 = 9.80 (b) P2.71

(a)

(d)

359.5 ± 359.5 2 − 4( 4.90)( 28.22) 9.80

d = 336 t 2 = 26.4 m

In walking a distance ∆x , in a time ∆t , the length of rope A is only increased by ∆x sin θ . ∆x sin θ . ∴ The pack lifts at a rate ∆t ∆x x sin θ = v boy = v boy ∆t A

x 2

x + h2

FG IJ HK

d 1 dv v boy dx = + v boy x dt A dt A dt v boy v boy x dA x dA = v = v boy a = v boy − 2 , but A dt A dt A 2 2 2 2 2 2 v boy v boy h h v boy x ∴ a= = 1− 2 = 2 3 2 A A A A x 2 + h2 a=

F GH

(c)

2

1 2 Ignoring the sound travel time, d = (9.80)( 2.40) = 28.2 m , an error of 6.82% . 2

v=

(b)

so

f

2 v boy

h

,0

v boy , 0

I JK

c

h

FIG. P2.71

Chapter 2

P2.72

h = 6.00 m, v boy = 2.00 m s v = However, x = v boy t : ∴ v = (a)

(b)

v boy x ∆x x sin θ = v boy = . 12 A ∆t x 2 + h2

c

2 v boy t

c

2 v boy t2

+h

(a)

=

c 4t

4t 2

+ 36

h

12

.

a f vb m s g

ts 0

0

0.5

0.32

1

0.63

1.5

0.89

2

1.11

2.5

1.28

3

1.41

3.5

1.52

4

1.60

4.5

1.66

5

1.71

FIG. P2.72(a)

From problem 2.71 above, a =

a f aem s j

P2.73

h

2 12

h

2 h 2 v boy

cx

2

+h

h

2 3 2

=

2 h 2 v boy

c

2 v boy t2

+h

h

2 3 2

=

c4t

144 2

+ 36

h

32

.

2

ts 0

0.67

0.5

0.64

1

0.57

1.5

0.48

2

0.38

2.5

0.30

3.

0.24

3.5

0.18

4.

0.14

4.5

0.11

5

0.09

FIG. P2.72(b)

We require x s = x k when t s = t k + 1.00

jb

1 3.50 m s 2 t k + 1.00 2 t k + 1.00 = 1.183t k xs =

e

t k = 5.46 s .

ja

1 4.90 m s 2 5.46 s 2

e

f

2

(b)

xk =

(c)

v k = 4.90 m s 2 5.46 s = 26.7 m s vs

e ja f = e3.50 m s ja6.46 sf = 2

= 73.0 m

22.6 m s

g

2

=

jb g

1 4.90 m s 2 t k 2

e

2

= xk

51

52 P2.74

Motion in One Dimension

Time t (s) 0.00

Height h (m) 5.00

0.25

5.75

0.50

6.40

0.75

6.94

1.00

7.38

1.25

∆h (m)

∆t (s)

v (m/s)

midpt time t (s)

0.75

0.25

3.00

0.13

0.65

0.25

2.60

0.38

0.54

0.25

2.16

0.63

0.44

0.25

1.76

0.88

0.34

0.25

1.36

1.13

0.24

0.25

0.96

1.38

0.14

0.25

0.56

1.63

0.03

0.25

0.12

1.88

–0.06

0.25

–0.24

2.13

–0.17

0.25

–0.68

2.38

–0.28

0.25

–1.12

2.63

–0.37

0.25

–1.48

2.88

–0.48

0.25

–1.92

3.13

–0.57

0.25

–2.28

3.38

–0.68

0.25

–2.72

3.63

–0.79

0.25

–3.16

3.88

–0.88

0.25

–3.52

4.13

–0.99

0.25

–3.96

4.38

–1.09

0.25

–4.36

4.63

–1.19

0.25

–4.76

4.88

7.72

1.50

7.96

1.75

8.10

2.00

8.13

2.25

8.07

2.50

7.90

2.75

7.62

3.00

7.25

3.25

6.77

3.50

6.20

3.75

5.52

4.00

4.73

4.25

3.85

4.50

2.86

4.75

1.77

5.00

0.58 TABLE P2.74

acceleration = slope of line is constant.

a =−1.63 m s 2 = 1.63 m s 2 downward

FIG. P2.74

53

Chapter 2

P2.75

The distance x and y are always related by x 2 + y 2 = L2 . Differentiating this equation with respect to time, we have

y B x

dy dx 2x + 2y =0 dt dt

L

y

dy dx = −v . is v B , the unknown velocity of B; and dt dt From the equation resulting from differentiation, we have

Now

O

FG IJ H K

FG H

A

x

dy x dx x = − (−v). =− dt y dt y But

v

α

FIG. P2.75

IJ K

y v v 3 1 = = 0.577 v . = tanα so v B = v . When α = 60.0° , v B = tan 60.0° 3 x tan α

ANSWERS TO EVEN PROBLEMS (a) 2 × 10 −7 m s ; 1 × 10 −6 m s ;

P2.24

(b) 5 × 10 yr

(a) 1.88 km; (b) 1.46 km; (c) see the solution; (d) (i) x 1 = 1.67 m s 2 t 2 ;

P2.4

(a) 50.0 m s ; (b) 41.0 m s

(ii) x 2 = 50 m s t − 375 m ;

P2.6

(a) 27.0 m ; 2 (b) 27.0 m + 18.0 m s ∆t + 3.00 m s 2 ∆t ;

(iii) x 3

P2.2

8

b

g e

e

ja f

(c) 18.0 m s P2.8

(a), (b), (c) see the solution; 4.6 m s 2 ; (d) 0

P2.10

5.00 m

P2.12

(a) 20.0 m s ; 5.00 m s ; (b) 262 m

P2.14

P2.16

j

b g = b 250 m sgt − e 2.5 m s jt 2

2

− 4 375 m ;

(e) 37.5 m s P2.26

958 m

P2.28

(a) x f = 30.0t − t 2 m; v f = 30.0 − 2t m s ;

e

j

a

f

(b) 225 m 1 a x t 2 ; 3.10 m s 2

P2.30

x f − xi = v xf t −

(a) see the solution; (b) 1.60 m s 2 ; 0.800 m s 2

P2.32

(a) 35.0 s; (b) 15.7 m s

(a) 13.0 m s; (b) 10.0 m s; 16.0 m s;

P2.34

(a) 1.12 × 10 11 m s 2 ; (b) 4.67 × 10 −5 s

P2.36

(a) False unless the acceleration is zero; see the solution; (b) True

(c) 6.00 m s 2 ; (d) 6.00 m s 2 P2.18

see the solution

P2.20

(a) 6.61 m s; (b) −0. 448 m s 2

P2.38

Yes; 212 m; 11.4 s

P2.22

(a) −21.8 mi h ⋅ s = −9.75 m s 2 ;

P2.40

(a) −4.90 m ; −19.6 m; −44.1 m; (b) −9.80 m s; −19.6 m s; −29.4 m s

P2.42

1.79 s

(b) −22.2 mi h ⋅ s = −9.94 m s 2 ; (c) −22.8 mi h ⋅ s = −10.2 m s 2

54

Motion in One Dimension

P2.44

No; see the solution

P2.60

1.60 m s 2

P2.46

The second ball is thrown at speed vi = gh

P2.62

(a) 41.0 s; (b) 1.73 km; (c) −184 m s

P2.48

(a) 510 m; (b) 20.4 s

P2.64

v xi t +

P2.50

(a) 96.0 ft s ;

P2.66

155 s; 129 s

P2.68

(a) 5.44 s; (b) 131 m; (c) 50.8 m s ;

3

2

(b) a = 3.07 × 10 ft s upward ; (c) ∆t = 3.13 × 10 −2 s P2.52

1 a x t 2 ; displacements agree 2

(d) 95.3 m s 2 upward

38.2 m

P2.70

(a) 26.4 m; (b) 6.82%

2

P2.54

(a) and (b) see the solution; (c) −4 m s ; (d) 34 m; (e) 28 m

P2.72

see the solution

P2.56

0.222 s

P2.74

see the solution; a x = −1.63 m s 2

P2.58

(a) see the solution; (b) 6.23 s

3 Vectors CHAPTER OUTLINE 3.1 3.2 3.3 3.4

Coordinate Systems Vector and Scalar Quantities Some Properties of Vectors Components of a Vector and Unit Vectors

ANSWERS TO QUESTIONS Q3.1

No. The sum of two vectors can only be zero if they are in opposite directions and have the same magnitude. If you walk 10 meters north and then 6 meters south, you won’t end up where you started.

Q3.2

No, the magnitude of the displacement is always less than or equal to the distance traveled. If two displacements in the same direction are added, then the magnitude of their sum will be equal to the distance traveled. Two vectors in any other orientation will give a displacement less than the distance traveled. If you first walk 3 meters east, and then 4 meters south, you will have walked a total distance of 7 meters, but you will only be 5 meters from your starting point.

Q3.3

The largest possible magnitude of R = A + B is 7 units, found when A and B point in the same direction. The smallest magnitude of R = A + B is 3 units, found when A and B have opposite directions.

Q3.4

Only force and velocity are vectors. None of the other quantities requires a direction to be described.

Q3.5

If the direction-angle of A is between 180 degrees and 270 degrees, its components are both negative. If a vector is in the second quadrant or the fourth quadrant, its components have opposite signs.

Q3.6

The book’s displacement is zero, as it ends up at the point from which it started. The distance traveled is 6.0 meters.

Q3.7

85 miles. The magnitude of the displacement is the distance from the starting point, the 260-mile mark, to the ending point, the 175-mile mark.

Q3.8

Vectors A and B are perpendicular to each other.

Q3.9

No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude.

55

56

Vectors

Q3.10

Any vector that points along a line at 45° to the x and y axes has components equal in magnitude.

Q3.11

A x = B x and A y = B y .

Q3.12

Addition of a vector to a scalar is not defined. Think of apples and oranges.

Q3.13

One difficulty arises in determining the individual components. The relationships between a vector and its components such as A x = A cos θ , are based on right-triangle trigonometry. Another problem would be in determining the magnitude or the direction of a vector from its components. Again, A = A x2 + A y2 only holds true if the two component vectors, A x and A y , are perpendicular.

Q3.14

If the direction of a vector is specified by giving the angle of the vector measured clockwise from the positive y-axis, then the x-component of the vector is equal to the sine of the angle multiplied by the magnitude of the vector.

SOLUTIONS TO PROBLEMS Section 3.1 P3.1

P3.2

Coordinate Systems

a f a fa f y = r sin θ = a5.50 mf sin 240° = a5.50 mfa −0.866f =

x = r cos θ = 5.50 m cos 240° = 5.50 m −0.5 = −2.75 m

(a)

−4.76 m

x = r cos θ and y = r sin θ , therefore x1 = 2.50 m cos 30.0° , y1 = 2.50 m sin 30.0° , and

a

f

a

f

bx , y g = a2.17 , 1.25f m x = a3.80 mf cos 120° , y = a3.80 mf sin 120° , and bx , y g = a−1.90, 3.29f m . 1

1

2

2

(b) P3.3

2

2

d = ( ∆ x) 2 + ( ∆ y) 2 = 16.6 + 4.16 = 4.55 m

The x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m. (a)

We can use the Pythagorean theorem to find the distance from the origin to the fly. distance = x 2 + y 2 =

(b)

θ = tan −1

FG 1 IJ = 26.6° ; r = H 2K

a2.00 mf + a1.00 mf

2.24 m, 26.6°

2

2

= 5.00 m 2 = 2.24 m

Chapter 3

P3.4

(a)

d=

bx

2

− x1

g + by 2

2

− y1

g

2

=

c2.00 − −3.00 h + a−4.00 − 3.00f 2

2

d = 25.0 + 49.0 = 8.60 m (b)

a2.00f + a−4.00f = 20.0 = F 4.00 IJ = −63.4° = tan G − H 2.00 K 2

r1 =

θ1

r2 =

2

4.47 m

−1

a−3.00f + a3.00f 2

2

= 18.0 = 4.24 m

θ 2 = 135° measured from the +x axis. P3.5

We have 2.00 = r cos 30.0° r=

2.00 = 2.31 cos 30.0°

and y = r sin 30.0° = 2.31 sin 30.0° = 1.15 . P3.6

We have r = x 2 + y 2 and θ = tan −1 (a)

FG y IJ . H xK

The radius for this new point is

a− x f

2

+ y2 = x2 + y2 = r

and its angle is tan −1

(b)

FG y IJ = H −x K

180°− θ .

b g

( −2 x) 2 + ( −2 y) 2 = 2r . This point is in the third quadrant if x , y is in the first quadrant

b g

or in the fourth quadrant if x , y is in the second quadrant. It is at an angle of 180°+ θ . (c)

b g

( 3 x) 2 + ( −3 y) 2 = 3r . This point is in the fourth quadrant if x , y is in the first quadrant

b g

or in the third quadrant if x , y is in the second quadrant. It is at an angle of − θ .

57

58

Vectors

Section 3.2

Vector and Scalar Quantities

Section 3.3

Some Properties of Vectors

P3.7

x 100 m x = 100 m tan 35.0° = 70.0 m

tan 35.0° =

a

f

FIG. P3.7 P3.8

R = 14 km

θ = 65° N of E R

13 km

θ 6 km

1 km

FIG. P3.8 P3.9

− R = 310 km at 57° S of W (Scale: 1 unit = 20 km )

FIG. P3.9 P3.10

(a)

Using graphical methods, place the tail of vector B at the head of vector A. The new vector A + B has a magnitude of 6.1 at 112° from the x-axis.

y

A A+B

(b)

The vector difference A − B is found by placing the negative of vector B at the head of vector A. The resultant vector A − B has magnitude 14.8 units at an angle of 22° from the + x-axis.

—B

B

A—B x O

FIG. P3.10

Chapter 3

P3.11

(a)

d = − 10.0 i = 10.0 m since the displacement is in a

C

straight line from point A to point B. (b)

The actual distance skated is not equal to the straight-line displacement. The distance follows the curved path of the semi-circle (ACB). s=

(c) P3.12

59

5.00 m d

B

A

FIG. P3.11

b g

1 2π r = 5π = 15.7 m 2

If the circle is complete, d begins and ends at point A. Hence, d = 0 .

Find the resultant F1 + F2 graphically by placing the tail of F2 at the head of F1 . The resultant force vector F1 + F2 is of magnitude 9.5 N and at an angle of 57° above the x -axis . y

F1 + F2

F2

F1

x

0 1 2 3 N FIG. P3.12 P3.13

(a)

The large majority of people are standing or sitting at this hour. Their instantaneous foot-tohead vectors have upward vertical components on the order of 1 m and randomly oriented horizontal components. The citywide sum will be ~ 10 5 m upward .

(b)

Most people are lying in bed early Saturday morning. We suppose their beds are oriented north, south, east, west quite at random. Then the horizontal component of their total vector height is very nearly zero. If their compressed pillows give their height vectors vertical components averaging 3 cm, and if one-tenth of one percent of the population are on-duty nurses or police officers, we estimate the total vector height as ~ 10 5 0.03 m + 10 2 1 m

a

3

~ 10 m upward .

f

a f

60 P3.14

Vectors N

Your sketch should be drawn to scale, and should look somewhat like that pictured to the right. The angle from the westward direction, θ, can be measured to be 4° N of W , and the distance R from the

1m W

15.0 meters

θ

R

sketch can be converted according to the scale to be 7.9 m .

3.50 meters

30.0°

8.20 meters

E

S

FIG. P3.14 P3.15

To find these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (Scale: 1 unit = 0.5 m ) (a)

A + B = 5.2 m at 60°

(b)

A – B = 3.0 m at 330°

(c)

B – A = 3.0 m at 150°

(d)

A – 2B = 5.2 m at 300°.

FIG. P3.15 *P3.16

The three diagrams shown below represent the graphical solutions for the three vector sums: R 1 = A + B + C , R 2 = B + C + A , and R 3 = C + B + A . You should observe that R 1 = R 2 = R 3 , illustrating that the sum of a set of vectors is not affected by the order in which the vectors are added. 100 m C

B A

A

B R1

A

R2

C B

FIG. P3.16

R3

C

Chapter 3

P3.17

The scale drawing for the graphical solution should be similar to the figure to the right. The magnitude and direction of the final displacement from the starting point are obtained by measuring d and θ on the drawing and applying the scale factor used in making the drawing. The results should be

(Scale: 1 unit = 20 ft )

d = 420 ft and θ = −3°

FIG. P3.17

Section 3.4 P3.18

Components of a Vector and Unit Vectors

Coordinates of the super-hero are:

a f a f y = a100 mf sina −30.0°f =

x = 100 m cos −30.0° = 86.6 m −50.0 m FIG. P3.18 P3.19

A x = −25.0 A y = 40.0 A = A x2 + A y2 =

a−25.0f + a40.0f 2

2

= 47.2 units

We observe that Ay

tan φ =

.

Ax

FIG. P3.19

So

φ = tan −1

F A I = tan 40.0 = tan a1.60f = 58.0° . GH A JK 25.0 y

−1

x

The diagram shows that the angle from the +x axis can be found by subtracting from 180°:

θ = 180° − 58° = 122° . P3.20

a

f

The person would have to walk 3.10 sin 25.0° = 1.31 km north , and

a

f

3.10 cos 25.0° = 2.81 km east .

61

62 P3.21

P3.22

Vectors

x = r cos θ and y = r sin θ , therefore:

j

b g e

j

b g e

j

x = 12.8 cos 150° , y = 12.8 sin 150° , and x , y = −11.1i + 6.40 j m

(b)

x = 3.30 cos 60.0° , y = 3.30 sin 60.0° , and x , y = 1.65 i + 2.86 j cm

(c)

x = 22.0 cos 215° , y = 22.0 sin 215° , and x , y = −18.0 i − 12.6 j in

a a

f a f f a f a−25.0 mfi + a43.3 mfj

x = d cos θ = 50.0 m cos 120 = −25.0 m y = d sin θ = 50.0 m sin 120 = 43.3 m d=

*P3.23

b g e

(a)

(a)

Her net x (east-west) displacement is −3.00 + 0 + 6.00 = +3.00 blocks, while her net y (northsouth) displacement is 0 + 4.00 + 0 = +4.00 blocks. The magnitude of the resultant displacement is R=

b x g + by g net

2

net

2

=

a3.00f + a4.00f 2

2

= 5.00 blocks

and the angle the resultant makes with the x-axis (eastward direction) is

θ = tan −1

FG 4.00 IJ = tan a1.33f = 53.1° . H 3.00 K −1

The resultant displacement is then 5.00 blocks at 53.1° N of E . (b) *P3.24

The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 blocks .

Let i = east and j = north. The unicyclist’s displacement is, in meters

N

280 j + 220 i + 360 j − 300 i − 120 j + 60 i − 40 j − 90 i + 70 j . R

R = −110 i + 550 j =

a110 mf + a550 mf 2

2

at tan −1

= 561 m at 11.3° west of north .

110 m west of north 550 m

The crow’s velocity is v=

∆ x 561 m at 11.3° W of N = ∆t 40 s

= 14.0 m s at 11.3° west of north .

E FIG. P3.24

Chapter 3

P3.25

63

+x East, +y North

∑ x = 250 +125 cos 30° = 358 m ∑ y = 75 +125 sin 30°−150 = −12.5 m

c∑ xh + c∑ yh = a358f + a−12.5f c∑ yh = − 12.5 = −0.0349 tan θ = c∑ xh 358 2

d=

2

2

2

= 358 m

θ = −2.00° d = 358 m at 2.00° S of E P3.26

The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to Chicago. In equation form: d DC east = d DA east + d AC east = 730 cos 5.00°−560 sin 21.0° = 527 miles. d DC north = d DA north + d AC north = 730 sin 5.00°+560 cos 21.0° = 586 miles. By the Pythagorean theorem, d = ( d DC east ) 2 + ( d DC north ) 2 = 788 mi . Then tan θ =

d DC north = 1.11 and θ = 48.0° . d DC east

Thus, Chicago is 788 miles at 48.0° northeast of Dallas . P3.27

(a)

See figure to the right.

(b)

C = A + B = 2.00 i + 6.00 j + 3.00 i − 2.00 j = 5.00 i + 4.00 j C = 25.0 + 16.0 at tan −1

FG 4 IJ = H 5K

6.40 at 38.7°

D = A − B = 2.00 i + 6.00 j − 3.00 i + 2.00 j = −1.00 i + 8.00 j

a−1.00f + a8.00f at tan FGH −81.00.00 IJK D = 8.06 at b180° − 82.9°g = 8.06 at 97.2° 2

D=

P3.28

2

−1

bx + x + x g + by + y + y g = a3.00 − 5.00 + 6.00f + a 2.00 + 3.00 + 1.00 f F 6.00 IJ = 56.3° θ = tan G H 4.00 K d=

1

2

3

2

2

1

2

−1

2

3

2

= 52.0 = 7.21 m

FIG. P3.27

64 P3.29

Vectors

We have B = R − A : A x = 150 cos120° = −75.0 cm A y = 150 sin 120° = 130 cm R x = 140 cos 35.0° = 115 cm R y = 140 sin 35.0° = 80.3 cm FIG. P3.29

Therefore,

a f

e

j

B = 115 − −75 i + 80.3 − 130 j = 190 i − 49.7 j cm B = 190 2 + 49.7 2 = 196 cm

FG H

θ = tan −1 − P3.30

IJ K

49.7 = −14.7° . 190

A = −8.70 i + 15.0 j and B = 13. 2 i − 6.60 j A − B + 3C = 0 :

3C = B − A = 21.9 i − 21.6 j C = 7.30 i − 7.20 j

or C x = 7.30 cm ; C y = −7.20 cm P3.31

(a)

aA + Bf = e3 i − 2 jj + e− i − 4jj =

2 i − 6 j

(b)

aA − Bf = e3i − 2jj − e− i − 4jj =

4i + 2 j

(c)

A + B = 2 2 + 6 2 = 6.32

(d)

A − B = 4 2 + 2 2 = 4.47

(e)

θ A+B = tan−1 − θ A−B = tan−1

P3.32

(a)

FG 6 IJ = −71.6°= H 2K FG 2 IJ = 26.6° H 4K

288°

D = A + B + C = 2 i + 4j D = 2 2 + 4 2 = 4. 47 m at θ = 63.4°

(b)

E = − A − B + C = −6 i + 6 j E = 6 2 + 6 2 = 8.49 m at θ = 135°

Chapter 3

P3.33

e

j

d1 = −3.50 j m

e

j

d 2 = 8.20 cos 45.0° i + 8.20 sin 45.0° j = 5.80 i + 5.80 j m

e

j

d 3 = −15.0 i m

a

f a

f e−9.20i + 2.30jj m

R = d1 + d 2 + d 3 = −15.0 + 5.80 i + 5.80 − 3.50 j = (or 9.20 m west and 2.30 m north)

The magnitude of the resultant displacement is R = R x2 + R y2 =

FG 2.30 IJ = H −9.20 K

The direction is θ = arctan P3.34

2

2

= 9.48 m .

166° . A = 10.0

Refer to the sketch R = A + B + C = −10.0 i − 15.0 j + 50.0 i = 40.0 i − 15.0 j

a f + a−15.0f

R = 40.0

a−9.20f + a2.30f

2

2 12

R

B = 15.0 C = 50.0

= 42.7 yards

FIG. P3.34 P3.35

F = F1 + F2

(a)

a

a

a f F = 60.0 i + 104j − 20.7 i + 77.3 j = e39.3 i + 181 jj N f

f

a

f

F = 120 cos 60.0° i + 120 sin 60.0° j − 80.0 cos 75.0° i + 80.0 sin 75.0° j

F = 39.3 2 + 181 2 = 185 N

θ = tan −1 (b) P3.36

F3 = − F =

East x 0m 1.41 –0.500 +0.914 R=

FG 181 IJ = H 39.3 K

77.8°

e−39.3 i − 181jj N

West y 4.00 m 1.41 –0.866 4.55 2

2

x + y = 4.64 m at 78.6° N of E

65

66 P3.37

Vectors

A = 3.00 m, θ A = 30.0°

B = 3.00 m , θ B = 90.0°

A x = A cos θ A = 3.00 cos 30.0° = 2.60 m

A y = A sin θ A = 3.00 sin 30.0°= 1.50 m

e

j

A = A x i + A y j = 2.60 i + 1.50 j m Bx = 0 , By = 3.00 m

e

j

A + B = 2.60 i + 1.50 j + 3.00 j = P3.38

B = 3.00 j m

so

e2.60i + 4.50jj m

Let the positive x-direction be eastward, the positive y-direction be vertically upward, and the positive z-direction be southward. The total displacement is then

e

j

e

j

e

j

d = 4.80 i + 4.80 j cm + 3.70 j − 3.70k cm = 4.80 i + 8.50 j − 3.70k cm .

P3.39

2

2

2

(a)

The magnitude is d = ( 4.80) +(8.50) + (−3.70) cm = 10.4 cm .

(b)

Its angle with the y-axis follows from cos θ =

8.50 , giving θ = 35.5° . 10. 4

B = Bx i + By j + Bz k = 4.00 i + 6.00 j + 3.00k B = 4.00 2 + 6.00 2 + 3.00 2 = 7.81

α = cos −1 β = cos −1 γ = cos −1 P3.40

FG 4.00 IJ = H 7.81 K FG 6.00 IJ = H 7.81 K FG 3.00 IJ = H 7.81 K

59.2° 39.8° 67.4°

The y coordinate of the airplane is constant and equal to 7.60 ×10 3 m whereas the x coordinate is given by x = vi t where vi is the constant speed in the horizontal direction. is

At t = 30.0 s we have x = 8.04×10 3 , so vi = 268 m s. The position vector as a function of time

b

g e

j

P = 268 m s t i + 7.60 × 10 3 m j . At t = 45.0 s , P = 1. 21 × 10 4 i + 7.60 × 10 3 j m. The magnitude is

P=

c1.21×10 h + c7.60×10 h 4 2

3 2

m = 1.43 ×10 4 m

and the direction is

θ = arctan

F 7.60×10 I = GH 1.21×10 JK 3

4

32.2° above the horizontal .

Chapter 3

P3.41

P3.42

(a)

A = 8.00 i + 12.0 j − 4.00k

(b)

B=

(c)

C = −3A = −24.0 i − 36.0 j + 12.0k

A = 2.00 i + 3.00 j − 1.00k 4

R = 75.0 cos 240° i + 75.0 sin 240° j + 125 cos 135° i + 125 sin 135° j + 100 cos 160° i + 100 sin 160° j R = −37.5 i − 65.0 j − 88.4i + 88.4j − 94.0 i + 34.2 j R = −220 i + 57.6 j 2

R = (−220 ) + 57.6 2 at arctan

FG 57.6 IJ above the –x-axis H 220 K

R = 227 paces at 165° P3.43

(a)

e5.00 i − 1.00j − 3.00k j m

C=A+B= 2

2

2

C = (5.00) +(1.00) +(3.00) m = 5.92 m (b)

D = 2A − B =

e4.00i − 11.0j + 15.0k j m

2

2

2

D = ( 4.00) +(11.0) +(15.0) m = 19.0 m P3.44

The position vector from radar station to ship is

e

j

e

j

S = 17.3 sin 136° i + 17.3 cos 136° j km = 12.0 i − 12.4 j km. From station to plane, the position vector is

e

j

P = 19.6 sin 153° i + 19.6 cos 153° j + 2.20k km, or

e

j

P = 8.90 i − 17.5 j + 2.20k km. (a)

To fly to the ship, the plane must undergo displacement D = S− P =

(b)

e3.12 i + 5.02 j − 2.20k j km .

The distance the plane must travel is 2

2

2

D = D = (3.12) +(5.02) +( 2.20) km = 6.31 km .

67

68 P3.45

Vectors

The hurricane’s first displacement is is

FG 41.0 km IJ(3.00 h) at 60.0° N of W, and its second displacement H h K

FG 25.0 km IJ(1.50 h) due North. With i H h K

representing east and j representing north, its total

displacement is:

FG 41.0 km cos 60.0°IJ a3.00 hfe− ij + FG 41.0 km sin 60.0°IJ a3.00 hfj + FG 25.0 km IJ a1.50 hfj = 61.5 kme− ij H h K H h K H hK +144 km j 2

2

with magnitude (61.5 km) +(144 km) = 157 km . P3.46

(a)

a

E= (b)

y

e15.1i + 7.72 jj cm a a f e−7.72i + 15.1jj cm

f

a a f e+7.72 i + 15.1jj cm

f

27.0° 27.0° F

FIG. P3.46

A x = −3.00 , A y = 2.00 (a)

A = A x i + A y j = −3.00 i + 2.00 j

(b)

A = A x2 + A y2 = (−3.00) +( 2.00) = 3.61

2

tan θ =

Ay Ax

=

2

2.00 = −0.667 , tan−1 (−0.667)= −33.7° (−3.00)

a

f

θ is in the 2 nd quadrant, so θ = 180°+ −33.7° = 146° . (c)

G E

G = + 17.0 cm sin 27.0° i + 17.0 cm cos 27.0° j G=

P3.47

f

F = − 17.0 cm sin 27.0° i + 17.0 cm cos 27.0° j F=

(c)

a

f

E = 17.0 cm cos 27.0° i + 17.0 cm sin 27.0° j

R x = 0 , R y = −4.00 , R = A + B thus B = R − A and Bx = R x − A x = 0 − (−3.00)= 3.00 , By = R y − A y = −4.00 − 2.00 = −6.00 . Therefore, B = 3.00 i − 6.00 j .

27.0° x

Chapter 3

P3.48

Let +x = East, +y = North, x 300 –175 0 125

P3.49

y 0 303 150 453 y = 74.6° N of E x

(a)

θ = tan−1

(b)

R = x 2 + y 2 = 470 km

(a)

R x = 40.0 cos 45.0°+30.0 cos 45.0° = 49.5

y

R y = 40.0 sin 45.0°−30.0 sin 45.0°+20.0 = 27.1

A

R = 49.5 i + 27.1j (b)

a49.5f + a27.1f = 56.4 F 27.1 IJ = 28.7° θ = tan G H 49.5 K 2

R=

2

O

−1

B

45°

x

45° C

FIG. P3.49 P3.50

Taking components along i and j , we get two equations:

6.00 a − 8.00b + 26.0 = 0 and

−8.00 a + 3.00b + 19.0 = 0 . Solving simultaneously, a = 5.00 , b = 7.00 . Therefore, 5.00A + 7.00B + C = 0 .

69

70

Vectors

Additional Problems P3.51

Let θ represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 180°−θ , magnitude of R is then R = 2 A cos

R

θ /2

θ θ , and . The 2 2

FG θ IJ . [Hint: apply the law of H 2K

cosines to the isosceles triangle and use the fact that B = A .] Again, A, –B, and D = A − B form an isosceles triangle with apex angle θ. Applying the law of cosines and the identity

B θ

A

θ

A D

–B FIG. P3.51

a1 − cosθ f = 2 sin FGH θ2 IJK 2

FG θ IJ . H 2K

gives the magnitude of D as D = 2 A sin The problem requires that R = 100D . Thus, 2 A cos

FG θ IJ = 200 A sinFG θ IJ . This gives tanFG θ IJ = 0.010 and H 2K H 2K H 2K

θ = 1.15° . P3.52

Let θ represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 180°−θ , magnitude of R is then R = 2 A cos

θ θ , and . The 2 2

FG θ IJ . [Hint: apply the law of H 2K

cosines to the isosceles triangle and use the fact that B = A . ] Again, A, –B, and D = A − B form an isosceles triangle with apex angle θ. Applying the law of cosines and the identity

a1 − cosθ f = 2 sin FGH θ2 IJK 2

FG θ IJ . H 2K Fθ I Fθ I The problem requires that R = nD or cosG J = n sinG J giving H 2K H 2K F 1I θ = 2 tan G J . H nK

gives the magnitude of D as D = 2 A sin

−1

FIG. P3.52

Chapter 3

P3.53

(a)

R x = 2.00 , R y = 1.00 , R z = 3.00

(b)

R = R x2 + R y2 + R z2 = 4.00 + 1.00 + 9.00 = 14.0 = 3.74

(c)

cos θ x = cos θ y = cos θ z =

*P3.54

F I GH JK F R I = 74.5° from + y GH R JK F R I = 36.7° from + z GH R JK

Rx R ⇒ θ x = cos−1 x = 57.7° from + x R R Ry R

y

⇒ θ y = cos−1

Rz ⇒ θ z = cos−1 R

z

Take the x-axis along the tail section of the snake. The displacement from tail to head is

a

a

f

f

240 m i + 420 − 240 m cos 180°−105° i − 180 m sin 75° j = 287 m i − 174 mj . 2

2

Its magnitude is (287) +(174) m = 335 m . From v =

distance , the time for each child’s run is ∆t

a fa a fb

fb

g

Inge: ∆t =

distance 335 m h 1 km 3 600 s = = 101 s v 12 km 1 000 m 1 h

Olaf: ∆t =

420 m ⋅ s = 126 s . 3.33 m

ga f

Inge wins by 126 − 101 = 25.4 s . *P3.55

The position vector from the ground under the controller of the first airplane is

a fa f a = e17.4i + 8.11j + 0.8k j km .

fa

f a

f

a fa f a = e16.5 i + 6.02 j + 1.1k j km .

fa

f a

f

r1 = 19.2 km cos 25° i + 19.2 km sin 25° j + 0.8 km k

The second is at r2 = 17.6 km cos 20° i + 17.6 km sin 20° j + 1.1 km k

Now the displacement from the first plane to the second is

e

j

r2 − r1 = −0.863 i − 2.09 j + 0.3k km with magnitude 2

2

2

(0.863 ) +( 2.09) +(0.3) = 2.29 km .

71

72 *P3.56

Vectors

3

Let A represent the distance from island 2 to island 3. The displacement is A = A at 159° . Represent the displacement from 3 to 1 as B = B at 298° . We have 4.76 km at 37° +A + B = 0 .

A 28° B

For x-components

a4.76 kmf cos 37°+ A cos 159°+B cos 298° = 0

37°

1

69° C

2 N E

3.80 km − 0.934 A + 0.469B = 0 FIG. P3.56

B = −8.10 km + 1.99 A For y-components,

a4.76 kmf sin 37°+ A sin 159°+B sin 298° = 0 2.86 km + 0.358 A − 0.883B = 0 (a)

We solve by eliminating B by substitution:

a

f

2.86 km + 0.358 A − 0.883 −8.10 km + 1.99 A = 0 2.86 km + 0.358 A + 7.15 km − 1.76 A = 0 10.0 km = 1.40 A A = 7.17 km

*P3.57

(b)

B = −8.10 km + 1.99(7.17 km)= 6.15 km

(a)

We first express the corner’s position vectors as sets of components

a a

f f

a a

f f

A = 10 m cos 50° i + 10 m sin 50° j = 6.43 m i +7.66 mj B = 12 m cos 30° i + 12 m sin 30° j = 10.4 m i +6.00 mj . The horizontal width of the rectangle is 10.4 m − 6.43 m = 3.96 m . Its vertical height is 7.66 m − 6.00 m = 1.66 m . Its perimeter is 2(3.96 + 1.66) m = 11.2 m . (b)

The position vector of the distant corner is Bx i + A y j = 10.4 mi +7.66 mj = 10.4 2 + 7.66 2 m at 7.66 m = 12.9 m at 36.4° . tan−1 10.4 m

Chapter 3

P3.58

Choose the +x-axis in the direction of the first force. The total force, y in newtons, is then 31 N 12.0 i + 31.0 j − 8.40 i − 24.0 j =

e3.60 ij + e7.00jj N

8.4 N

The magnitude of the total force is 2

x R

.

12 N 35.0° horizontal 24 N

2

(3.60) +(7.00) N = 7.87 N and the angle it makes with our +x-axis is given by tan θ =

(7.00)

(3.60) θ = 62.8° . Thus, its angle counterclockwise from the horizontal is 35.0°+62.8° = 97.8° .

P3.59

FIG. P3.58 ,

d 1 = 100 i d 2 = −300 j

a a

f f

a a

f f

d 3 = −150 cos 30.0° i − 150 sin 30.0° j = −130 i − 75.0 j d = −200 cos 60.0° i + 200 sin 60.0° j = −100 i + 173 j 4

R = d1 + d 2 + d 3 + d 4 =

e−130 i − 202 jj m

a−130f + a−202f = F 202 IJ = 57.2° φ = tan G H 130 K 2

R=

2

240 m FIG. P3.59

−1

θ = 180 + φ = 237°

P3.60

e

j

   dr d 4 i + 3 j − 2 t j = = 0 + 0 − 2 j = − 2.00 m s j dt dt

b

g

The position vector at t = 0 is 4i + 3 j . At t = 1 s , the position is 4i + 1j , and so on. The object is moving straight downward at 2 m/s, so dr represents its velocity vector . dt P3.61

a

f a

f

v = v x i + v y j = 300 + 100 cos 30.0° i + 100 sin 30.0° j

e

j

v = 387 i + 50.0 j mi h v = 390 mi h at 7.37° N of E

73

74 P3.62

Vectors

(a)

e

j

You start at point A: r1 = rA = 30.0 i − 20.0 j m. The displacement to B is rB − rA = 60.0 i + 80.0 j − 30.0 i + 20.0 j = 30.0 i + 100 j .

e

j

You cover half of this, 15.0 i + 50.0 j to move to r2 = 30.0 i − 20.0 j + 15.0 i + 50.0 j = 45.0 i + 30.0 j . Now the displacement from your current position to C is rC − r2 = −10.0 i − 10.0 j − 45.0 i − 30.0 j = −55.0 i − 40.0 j . You cover one-third, moving to 1 r3 = r2 + ∆r23 = 45.0 i + 30.0 j + −55.0 i − 40.0 j = 26.7 i + 16.7 j . 3

e

j

The displacement from where you are to D is rD − r3 = 40.0 i − 30.0 j − 26.7 i − 16.7 j = 13.3 i − 46.7 j . You traverse one-quarter of it, moving to r4 = r3 +

b

g

1 1 rD − r3 = 26.7 i + 16.7 j + 13.3 i − 46.7 j = 30.0 i + 5.00 j . 4 4

e

j

The displacement from your new location to E is rE − r4 = −70.0 i + 60.0 j − 30.0 i − 5.00 j = −100 i + 55.0 j of which you cover one-fifth the distance, −20.0 i + 11.0 j, moving to r4 + ∆r45 = 30.0 i + 5.00 j − 20.0 i + 11.0 j = 10.0 i + 16.0 j . The treasure is at (10.0 m, 16.0 m) . (b)

Following the directions brings you to the average position of the trees. The steps we took numerically in part (a) bring you to rA +

ar then to ar then to

A

+ rB

f+r

2 + rB + rC A

C

−

ar

A +rB

2

3

f+r

−

f =

ar

a

f FGH

r + rB 1 rB − rA = A 2 2

IJ K

rA + rB + rC 3

A + rB + rC

f

r + rB + rC + rD = A 3 4 4 rA + rB +rC + rD f a rA + rB + rC + rD r − r + rB + rC + rD + rE 4 + E = A and last to . 4 5 5

a

D

3

f

This center of mass of the tree distribution is the same location whatever order we take the trees in.

Chapter 3

*P3.63

(a)

75

Let T represent the force exerted by each child. The xcomponent of the resultant force is

af a f a f

T cos 0 + T cos 120°+T cos 240° = T 1 + T −0.5 + T −0.5 = 0 . The y-component is T sin 0 + T sin 120 + T sin 240 = 0 + 0.866T − 0.866T = 0 . FIG. P3.63

Thus,

∑ F = 0.

P3.64

(b)

If the total force is not zero, it must point in some direction. When each child moves one 360° space clockwise, the total must turn clockwise by that angle, . Since each child exerts N the same force, the new situation is identical to the old and the net force on the tire must still point in the original direction. The contradiction indicates that we were wrong in supposing that the total force is not zero. The total force must be zero.

(a)

From the picture, R 1 = a i + bj and R 1 = a 2 + b 2 .

(b)

R 2 = ai + bj + ck ; its magnitude is 2

R1 + c 2 = a 2 + b 2 + c 2 .

FIG. P3.64

76 P3.65

Vectors

Since A + B = 6.00 j , we have

bA

x

g e

j

+ Bx i + A y + B y j = 0 i + 6.00 j FIG. P3.65

giving A x + B x = 0 or A x = −Bx

[1]

A y + B y = 6.00 .

[2]

and

Since both vectors have a magnitude of 5.00, we also have

A x2 + A y2 = Bx2 + By2 = 5.00 2 . From A x = −Bx , it is seen that A x2 = Bx2 . Therefore, A x2 + A y2 = Bx2 + By2 gives

A y2 = By2 . Then, A y = By and Eq. [2] gives A y = By = 3.00 . Defining θ as the angle between either A or B and the y axis, it is seen that cos θ =

Ay A

=

By B

=

3.00 = 0.600 and θ = 53.1° . 5.00

The angle between A and B is then φ = 2θ = 106° .

Chapter 3

*P3.66

Let θ represent the angle the x-axis makes with the horizontal. Since angles are equal if their sides are perpendicular right side to right side and left side to left side, θ is also the angle between the weight and our y axis. The x-components of the forces must add to zero:

x y

0.127 N

Ty

θ θ

−0.150 N sin θ + 0.127 N = 0 .

0.150 N

θ = 57.9°

(b) (a)

FIG. P3.66

The y-components for the forces must add to zero:

a

f

+Ty − 0.150 N cos 57.9° = 0 , Ty = 0.079 8 N . (c) P3.67

The angle between the y axis and the horizontal is 90.0°−57.9°= 32.1° .

The displacement of point P is invariant under rotation of 2 the coordinates. Therefore, r = r ′ and r 2 = r ′ or,

bg

y

b g + by ′g . Also, from the figure, β = θ −α F y′ I F yI ∴tan G J = tan G J − α H xK H x′ K y ′ e j − tan α = x ′ 1 + e j tan α

x 2 + y2 = x′

2

P

y′

2

−1

r

β

−1

α

y x

x′ α

θ

β

t

O

y x

FIG. P3.67

Which we simplify by multiplying top and bottom by x cosα . Then,

x ′ = x cosα + y sinα , y ′ = −x sinα + y cosα . ANSWERS TO EVEN PROBLEMS P3.2

a

fa

f

(a) 2.17 m, 1.25 m ; −1.90 m, 3.29 m ; (b) 4.55 m

P3.16

see the solution

P3.18

86.6 m and –50.0 m

P3.4

(a) 8.60 m; (b) 4.47 m at −63.4° ; 4.24 m at 135°

P3.20

1.31 km north; 2.81 km east

P3.6

(a) r at 180°− θ ; (b) 2r at 180°+ θ ; (c) 3r at –θ

P3.22

−25.0 m i + 43.3 m j

P3.8

14 km at 65° north of east

P3.24

14.0 m s at 11.3° west of north

P3.10

(a) 6.1 at 112°; (b) 14.8 at 22°

P3.26

788 mi at 48.0° north of east

P3.12

9.5 N at 57°

P3.28

7.21 m at 56.3°

P3.14

7.9 m at 4° north of west

P3.30

C = 7.30 cm i − 7.20 cm j

x

77

78

Vectors

P3.50

a = 5.00 , b = 7.00

P3.52

2 tan −1

P3.54

25.4 s

P3.56

(a) 7.17 km; (b) 6.15 km

P3.58

7.87 N at 97.8° counterclockwise from a horizontal line to the right

P3.32

(a) 4.47 m at 63.4°; (b) 8.49 m at 135°

P3.34

42.7 yards

P3.36

4.64 m at 78.6°

P3.38

(a) 10.4 cm; (b) 35.5°

P3.40

1.43 × 10 4 m at 32.2° above the horizontal

P3.42

−220 i + 57.6 j = 227 paces at 165°

P3.44

(a) 3.12 i + 5.02 j − 2.20k km; (b) 6.31 km

P3.60

e j (b) e −7.72 i + 15.1jj cm; (c) e +7.72 i + 15.1jj cm

P3.62

b−2.00 m sgj ; its velocity vector (a) a10.0 m, 16.0 mf ; (b) see the solution

P3.64

(a) R 1 = a i + bj ; R 1 = a 2 + b 2 ;

(a) 74.6° north of east; (b) 470 km

P3.66

P3.46

P3.48

e

j

(a) 15.1i + 7.72 j cm;

FG 1 IJ H nK

(b) R 2 = ai + bj + ck ; R 2 = a 2 + b 2 + c 2 (a) 0.079 8N; (b) 57.9°; (c) 32.1°

4 Motion in Two Dimensions CHAPTER OUTLINE 4.1 4.2 4.3 4.4 4.5 4.6

ANSWERS TO QUESTIONS

The Position, Velocity, and Acceleration Vectors Two-Dimensional Motion with Constant Acceleration Projectile Motion Uniform Circular Motion Tangential and Radial Acceleration Relative Velocity and Relative Acceleration

Q4.1

Yes. An object moving in uniform circular motion moves at a constant speed, but changes its direction of motion. An object cannot accelerate if its velocity is constant.

Q4.2

No, you cannot determine the instantaneous velocity. Yes, you can determine the average velocity. The points could be widely separated. In this case, you can only determine the average velocity, which is v=

Q4.3

(a)

a

v a

(b)

v a

v a

v

∆x . ∆t

a

v

a v

a

v a

v a v

10 i m s

−9.80 j m s 2

Q4.4

(a)

Q4.5

The easiest way to approach this problem is to determine acceleration first, velocity second and finally position. Vertical: In free flight, a y = − g . At the top of a projectile’s trajectory, v y = 0. Using this, the

(b)

d

i

maximum height can be found using v 2fy = viy2 + 2 a y y f − yi . Horizontal: a x = 0 , so v x is always the same. To find the horizontal position at maximum height, one needs the flight time, t. Using the vertical information found previously, the flight time can be found using v fy = viy + a y t . The horizontal position is x f = vix t . If air resistance is taken into account, then the acceleration in both the x and y-directions would have an additional term due to the drag. Q4.6

A parabola.

79

80

Motion in Two Dimensions

Q4.7

The balls will be closest together as the second ball is thrown. Yes, the first ball will always be moving faster, since its flight time is larger, and thus the vertical component of the velocity is larger. The time interval will be one second. No, since the vertical component of the motion determines the flight time.

Q4.8

The ball will have the greater speed. Both the rock and the ball will have the same vertical component of the velocity, but the ball will have the additional horizontal component.

Q4.9

(a)

Q4.10

Straight up. Throwing the ball any other direction than straight up will give a nonzero speed at the top of the trajectory.

Q4.11

No. The projectile with the larger vertical component of the initial velocity will be in the air longer.

Q4.12

The projectile is in free fall. Its vertical component of acceleration is the downward acceleration of gravity. Its horizontal component of acceleration is zero.

Q4.13

(a)

Q4.14

60°. The projection angle appears in the expression for horizontal range in the function sin 2 θ . This function is the same for 30° and 60°.

Q4.15

The optimal angle would be less than 45°. The longer the projectile is in the air, the more that air resistance will change the components of the velocity. Since the vertical component of the motion determines the flight time, an angle less than 45° would increase range.

Q4.16

The projectile on the moon would have both the larger range and the greater altitude. Apollo astronauts performed the experiment with golf balls.

Q4.17

Gravity only changes the vertical component of motion. Since both the coin and the ball are falling from the same height with the same vertical component of the initial velocity, they must hit the floor at the same time.

Q4.18

(a)

yes

no

(b)

(b)

no

no

yes

(c)

(c)

no

yes

(b)

(d)

(d)

yes

(e)

no

no

yes

In the second case, the particle is continuously changing the direction of its velocity vector. Q4.19

The racing car rounds the turn at a constant speed of 90 miles per hour.

Q4.20

The acceleration cannot be zero because the pendulum does not remain at rest at the end of the arc.

Q4.21

(a)

The velocity is not constant because the object is constantly changing the direction of its motion.

(b)

The acceleration is not constant because the acceleration always points towards the center of the circle. The magnitude of the acceleration is constant, but not the direction.

(a)

straight ahead

Q4.22

(b)

in a circle or straight ahead

Chapter 4

Q4.23

v v

v

a a v

a

a

v

Q4.24

a

a v a

81

v

r

a

r

r a aa v v

a

v

r

r v

a

Q4.25

The unit vectors r and θ are in different directions at different points in the xy plane. At a location along the x-axis, for example, r = i and θ = j, but at a point on the y-axis, r = j and θ = − i . The unit vector i is equal everywhere, and j is also uniform.

Q4.26

The wrench will hit at the base of the mast. If air resistance is a factor, it will hit slightly leeward of the base of the mast, displaced in the direction in which air is moving relative to the deck. If the boat is scudding before the wind, for example, the wrench’s impact point can be in front of the mast.

Q4.27

(a)

The ball would move straight up and down as observed by the passenger. The ball would move in a parabolic trajectory as seen by the ground observer.

(b)

Both the passenger and the ground observer would see the ball move in a parabolic trajectory, although the two observed paths would not be the same.

(a)

g downward

Q4.28

(b)

g downward

The horizontal component of the motion does not affect the vertical acceleration.

SOLUTIONS TO PROBLEMS Section 4.1

The Position, Velocity, and Acceleration Vectors

a f

a f

xm 0

P4.1

ym −3 600

−3 000

0

−1 270 −4 270 m (a)

1 270 −2 330 m

Net displacement = x 2 + y 2 = 4.87 km at 28.6° S of W FIG. P4.1

b20.0 m sga180 sf + b25.0 m sga120 sf + b30.0 m sga60.0 sf =

(b)

Average speed =

(c)

Average velocity =

180 s + 120 s + 60.0 s

4.87 × 10 3 m = 13.5 m s along R 360 s

23.3 m s

82 P4.2

*P4.3

Motion in Two Dimensions

e

j

(a)

r = 18.0t i + 4.00t − 4.90t 2 j

(b)

v=

(c)

a=

b18.0 m sgi + 4.00 m s − e9.80 m s jt j 2

(d)

e−9.80 m s j j ra3.00 sf = a54.0 mf i − a32.1 mf j

(e)

v 3.00 s = 18.0 m s i − 25.4 m s j

(f)

a 3.00 s =

2

a

f b

a

f e−9.80 m s j j

g b

g

2

The sun projects onto the ground the x-component of her velocity:

a

f

5.00 m s cos −60.0° = 2.50 m s . P4.4

(a)

From x = −5.00 sin ω t , the x-component of velocity is vx =

FG IJ b H K

g

dx d = −5.00 sin ω t = −5.00ω cos ω t dt dt

and a x =

dv x = +5.00ω 2 sin ω t dt

similarly, v y = and a y =

FG d IJ b4.00 − 5.00 cos ω tg = 0 + 5.00ω sin ω t H dt K

FG d IJ b5.00ω sin ω tg = 5.00ω H dt K

2

cos ω t .

e5.00ω i + 0jj m s e0 i + 5.00ω jj m s .

At t = 0 , v = −5.00ω cos 0 i + 5.00ω sin 0 j = and a = 5.00ω 2 sin 0 i + 5.00ω 2 cos 0 j = (b)

2

a4.00 mfj + a5.00 mfe− sin ω t i − cos ω t jj a5.00 mfω − cos ω t i + sin ω t j a5.00 mfω sin ω t i + cos ω t j

r = x i + yj = v= a=

(c)

2

2

a

f

The object moves in a circle of radius 5.00 m centered at 0 , 4.00 m .

Chapter 4

Section 4.2 P4.5

Two-Dimensional Motion with Constant Acceleration v f = vi + a t

(a)

v f − vi

a=

=

t

r f = ri + v i t +

(b)

e9.00i + 7.00jj − e3.00i − 2.00jj = 3.00

e2.00i + 3.00jj m s

2

1 2 1 a t = 3.00 i − 2.00 j t + 2.00 i + 3.00 j t 2 2 2

e

e

j

j

e

j

e

j

x = 3.00t + t 2 m and y = 1.50t 2 − 2.00t m P4.6

FG IJ e j H K dv F d I a= = G J e −12.0t jj = −12.0 j m s dt H dt K v=

(a)

e

2

j

r = 3.00 i − 6.00 j m; v = −12.0 j m s

(b) P4.7

dr d 3.00 i − 6.00t 2 j = −12.0t j m s = dt dt

e

a f e

j

j

v i = 4.00 i + 1.00 j m s and v 20.0 = 20.0 i − 5.00 j m s ∆ v x 20.0 − 4.00 = m s 2 = 0.800 m s 2 ∆t 20.0 ∆ v y −5.00 − 1.00 ay = = m s 2 = −0.300 m s 2 ∆t 20.0

(a)

ax =

(b)

θ = tan −1

(c)

At t = 25.0 s

FG −0.300 IJ = −20.6° = H 0.800 K

339° from + x axis

a f a fa f a f a fa f

1 1 2 a x t 2 = 10.0 + 4.00 25.0 + 0.800 25.0 = 360 m 2 2 1 1 2 y f = yi + v yi t + a y t 2 = −4.00 + 1.00 25.0 + −0.300 25.0 = −72.7 m 2 2 v xf = v xi + a x t = 4 + 0.8 25 = 24 m s x f = xi + v xi t +

a f v = v + a t = 1 − 0.3a 25f = −6.5 m s Fv I F −6.50 IJ = −15.2° θ = tan G J = tan G H 24.0 K Hv K yf

yi

y

−1

y x

−1

83

84 P4.8

Motion in Two Dimensions

a = 3.00 j m s 2 ; v i = 5.00 i m s ; ri = 0 i + 0 j (a)

r f = ri + v i t +

v f = vi + a t = (b)

LM5.00ti + 1 3.00t jOP m 2 N Q   e5.00i + 3.00tjj m s

1 2 at = 2

2

a f

a fa f e

1 2 t = 2.00 s , r f = 5.00 2.00 i + 3.00 2.00 j = 10.0 i + 6.00 j m 2 so x f = 10.0 m , y f = 6.00 m

j

a f e j + v = a5.00f + a6.00f

v f = 5.00 i + 3.00 2.00 j = 5.00 i + 6.00 j m s 2 v f = v f = v xf

*P4.9

(a)

2

2 yf

2

For the x-component of the motion we have x f = xi + v xi t +

e

j

0.01 m = 0 + 1.80 × 10 7 m s t +

e4 × 10 t= =

14

j e

= 7.81 m s 1 axt 2 . 2

1 8 × 10 14 m s 2 t 2 2

e

j

j

m s 2 t 2 + 1.80 × 10 7 m s t − 10 −2 m = 0

e1.8 × 10 m sj − 4e4 × 10 2e 4 × 10 m s j

−1.80 × 10 7 m s ±

2

7

14

14

je

j

m s 2 −10 −2 m

2

−1.8 × 10 7 ± 1.84 × 10 7 m s 8 × 10 14 m s 2

We choose the + sign to represent the physical situation t= Here y f = yi + v yi t +

4.39 × 10 5 m s 8 × 10

14

ms

2

= 5.49 × 10 −10 s .

1 1 a y t 2 = 0 + 0 + 1.6 × 10 15 m s 2 5.49 × 10 −10 s 2 2

e

e

je

j

2

= 2.41 × 10 −4 m .

j

So, r f = 10.0 i + 0.241 j mm . (b)

e

je

v f = v i + at = 1.80 × 10 7 m s i + 8 × 10 14 m s 2 i + 1.6 × 10 15 m s 2 j 5.49 × 10 −10 s

e

j e j e m sj i + e8.78 × 10 m sj j

j

= 1.80 × 10 7 m s i + 4.39 × 10 5 m s i + 8.78 × 10 5 m s j = (c) (d)

e1.84 × 10

7

5

e1.84 × 10 m sj + e8.78 × 10 m sj = F 8.78 × 10 I = 2.73° Fv I θ = tan G J = tan G v H K H 1.84 × 10 JK 2

7

vf =

−1

y x

−1

5

5

7

2

1.85 × 10 7 m s

j

Chapter 4

Section 4.3 P4.10

Projectile Motion

x = v xi t = vi cos θ i t

b

ga

fa

x = 300 m s cos 55.0° 42.0 s

f

3

x = 7. 23 × 10 m y = v yi t −

b

1 2 1 gt = vi sin θ i t − gt 2 2 2

ga

fa

f 12 e9.80 m s ja42.0 sf 2

y = 300 m s sin 55.0° 42.0 s − P4.11

(a)

2

= 1.68 × 10 3 m

The mug leaves the counter horizontally with a velocity v xi (say). If time t elapses before it hits the ground, then since there is no horizontal acceleration, x f = v xi t , i.e., t=

xf v xi

=

a1.40 mf v xi FIG. P4.11

In the same time it falls a distance of 0.860 m with acceleration downward of 9.80 m s 2 . Then y f = yi + v yi t +

1 1 a y t 2 : 0 = 0.860 m + −9.80 m s 2 2 2

e

jFGH 1.40v m IJK xi

Thus,

e4.90 m s je1.96 m j = 2

v xi = (b)

2

0.860 m

3.34 m s .

The vertical velocity component with which it hits the floor is

e

v yf = v yi + a y t = 0 + −9.80 m s 2

1.40 m I = −4.11 m s . jFGH 3.34 m s JK

Hence, the angle θ at which the mug strikes the floor is given by

θ = tan −1

F v I = tan FG −4.11 IJ = GH v JK H 3.34 K yf xf

−1

−50.9° .

2

.

85

86 P4.12

Motion in Two Dimensions

The mug is a projectile from just after leaving the counter until just before it reaches the floor. Taking the origin at the point where the mug leaves the bar, the coordinates of the mug at any time are x f = v xi t +

1 1 1 a x t 2 = v xi t + 0 and y f = v yi t + a y t 2 = 0 − g t 2 . 2 2 2

When the mug reaches the floor, y f = − h so −h = −

1 2 gt 2

which gives the time of impact as t=

(a)

2h . g

Since x f = d when the mug reaches the floor, x f = v xi t becomes d = v xi

2h giving the g

initial velocity as v xi = d (b)

g . 2h

Just before impact, the x-component of velocity is still v xf = v xi while the y-component is v yf = v yi + a y t = 0 − g

2h . g

Then the direction of motion just before impact is below the horizontal at an angle of

θ = tan −1

Fv GG v H

yf xf

I JJ = tan K

−1

Fg GG Hd

2h g g 2h

I JJ = K

tan −1

FG 2 h IJ HdK

.

Chapter 4

P4.13

(a)

The time of flight of the first snowball is the nonzero root of y f = yi + v yi t1 +

b

gb

g

0 = 0 + 25.0 m s sin 70.0° t1 − t1 =

2( 25.0 m s) sin 70.0° 9.80 m s 2

1 a y t12 2

1 9.80 m s 2 t12 2

e

j

= 4.79 s .

The distance to your target is

b

a

g

f

x f − xi = v xi t1 = 25.0 m s cos 70.0° 4.79 s = 41.0 m . Now the second snowball we describe by y f = yi + v yi t 2 +

b

g

1 ayt2 2 2

e

j

0 = 25.0 m s sin θ 2 t 2 − 4.90 m s 2 t 22

a

f

t 2 = 5.10 s sin θ 2 x f − xi = v xi t 2

b

g

a

f

a

f

41.0 m = 25.0 m s cos θ 2 5.10 s sin θ 2 = 128 m sin θ 2 cos θ 2 0.321 = sin θ 2 cos θ 2 Using sin 2θ = 2 sin θ cos θ we can solve 0.321 =

1 sin 2θ 2 2

2θ 2 = sin −1 0.643 and θ 2 = 20.0° . (b)

a

f

a

t1 − t 2 = 4.79 s − 1.75 s = 3.05 s . P4.14

f

The second snowball is in the air for time t 2 = 5.10 s sin θ 2 = 5.10 s sin 20° = 1.75 s , so you throw it after the first by

From Equation 4.14 with R = 15.0 m , vi = 3.00 m s , θ max = 45.0° ∴g =

vi2 9.00 = = 0.600 m s 2 R 15.0

87

88 P4.15

Motion in Two Dimensions

h= so

b

g

vi2 sin 2θ i vi2 sin 2 θ i ; R= ; 3h = R , g 2g

b

2 3 vi2 sin 2 θ i vi sin 2θ i = g 2g

g

2 sin 2 θ i tan θ i = = 3 sin 2θ i 2 4 = 53.1° . thus θ i = tan −1 3 or

FG IJ HK

*P4.16

(a)

To identify the maximum height we let i be the launch point and f be the highest point:

d

i + 2b− g gb y

2 2 v yf = v yi + 2 a y y f − yi

0= y max =

vi2 vi2

2

sin θ i

max

−0

g

2

sin θ i . 2g

To identify the range we let i be the launch and f be the impact point; where t is not zero: 1 ay t 2 2 1 0 = 0 + vi sin θ i t + − g t 2 2 2 vi sin θ i t= g

y f = yi + v yi t +

b g

1 axt 2 2 2 v sin θ i + 0. d = 0 + vi cos θ i i g

x f = xi + v xi t +

For this rock, d = y max vi2 sin 2 θ i 2 vi2 sin θ i cos θ i = g 2g sin θ i = tan θ i = 4 cos θ i

θ i = 76.0° (b)

Since g divides out, the answer is the same on every planet.

(c)

The maximum range is attained for θ i = 45° : d max vi cos 45° 2 vi sin 45° g = = 2.125 . d gvi cos 76° 2 vi sin 76° So d max =

17d . 8

Chapter 4

P4.17

a f

(a)

x f = v xi t = 8.00 cos 20.0° 3.00 = 22.6 m

(b)

Taking y positive downwards, y f = v yi t +

1 2 gt 2

a f 12 a9.80fa3.00f

y f = 8.00 sin 20.0° 3.00 + (c)

89

a

f

10.0 = 8.00 sin 20.0° t +

2

= 52.3 m .

a f

1 9.80 t 2 2

4.90t 2 + 2.74t − 10.0 = 0 t= *P4.18

a2.74f

−2.74 ±

2

+ 196

= 1.18 s

9.80

We interpret the problem to mean that the displacement from fish to bug is

a

a

f

a

f

f a

f

2.00 m at 30° = 2.00 m cos30° i + 2.00 m sin30° j = 1.73 m i + 1.00 m j. If the water should drop 0.03 m during its flight, then the fish must aim at a point 0.03 m above the bug. The initial velocity of the water then is directed through the point with displacement

a1.73 mfi + a1.03 mfj = 2.015 m at 30.7°. For the time of flight of a water drop we have 1 axt 2 2

x f = xi + v xi t +

b

g

1.73 m = 0 + vi cos 30.7° t + 0 so t=

1.73 m . vi cos 30.7°

The vertical motion is described by y f = yi + v yi t +

1 ayt 2 . 2

The “drop on its path” is 1 −3.00 cm = −9.80 m s 2 2

e

.73 m I jFGH v 1cos J 30.7° K

2

.

i

Thus, vi =

1.73 m 9.80 m s 2 = 2.015 m 12.8 s −1 = 25.8 m s . cos30.7° 2 × 0.03 m

e

j

90 P4.19

Motion in Two Dimensions

(a)

We use the trajectory equation: y f = x f tan θ i −

gx 2f 2 vi2 cos 2 θ i

.

With x f = 36.0 m, vi = 20.0 m s, and θ = 53.0° we find

e9.80 m s ja36.0 mf = 3.94 m. = a36.0 mf tan 53.0°− 2b 20.0 m sg cos a53.0°f 2

2

yf

2

2

The ball clears the bar by

a3.94 − 3.05f m = (b)

0.889 m .

The time the ball takes to reach the maximum height is t1 =

b

f

ga

20.0 m s sin53.0° vi sin θ i = = 1.63 s . g 9.80 m s 2 xf

The time to travel 36.0 m horizontally is t 2 =

t2 =

vix

36.0 m = 2.99 s . ( 20.0 m s) cos 53.0°

a

f

Since t 2 > t1 the ball clears the goal on its way down . P4.20

b

g

The horizontal component of displacement is x f = v xi t = vi cos θ i t . Therefore, the time required to d reach the building a distance d away is t = . At this time, the altitude of the water is vi cos θ i y f = v yi t +

FG H

IJ FG K H

g d d 1 a y t 2 = vi sin θ i − vi cos θ i 2 2 vi cos θ i

IJ K

Therefore the water strikes the building at a height h above ground level of h = y f = d tan θ i −

gd 2 2 vi2 cos 2 θ i

.

2

.

Chapter 4

*P4.21

(a)

For the horizontal motion, we have 1 axt 2 2 24 m = 0 + vi cos 53° 2.2 s + 0 x f = xi + v xi t +

a

fa

f

vi = 18.1 m s . (b)

As it passes over the wall, the ball is above the street by y f = yi + v yi t +

b

fa

ga

f 12 e−9.8 m s ja2.2 sf 2

y f = 0 + 18.1 m s sin 53° 2.2 s +

2

1 ayt 2 2

= 8.13 m .

So it clears the parapet by 8.13 m − 7 m = 1.13 m . (c)

Note that the highest point of the ball’s trajectory is not directly above the wall. For the whole flight, we have from the trajectory equation

g FGH 2v

b

y f = tan θ i x f −

g 2 i

cos

2

Ix θ JK i

2 f

or

F f GG 2 18.19m.8 sm scos g Hb

a

2

6 m = tan 53° x f −

2

2

I Jx . 53° JK

Solving,

e0.041 2 m jx −1

2 f

− 1.33 x f + 6 m = 0

and xf =

a

fa f .

1.33 ± 1.33 2 − 4 0.0412 6

e

2 0.0412 m

−1

j

This yields two results: x f = 26.8 m or 5.44 m The ball passes twice through the level of the roof. It hits the roof at distance from the wall 26.8 m − 24 m = 2.79 m .

2 f

91

92 *P4.22

Motion in Two Dimensions

When the bomb has fallen a vertical distance 2.15 km, it has traveled a horizontal distance x f given by xf =

a3.25 kmf − a2.15 kmf 2

y f = x f tan θ −

2

= 2.437 km

gx 2f 2 vi2 cos 2 θ i

e9.8 m s jb2 437 mg −2 150 m = b 2 437 mg tan θ − 2b 280 m sg cos θ ∴−2 150 m = b 2 437 mg tan θ − a371.19 mfe1 + tan θ j 2

2

i

2

2

i

2

i

i

2

∴ tan θ − 6.565 tan θ i − 4.792 = 0 ∴ tan θ i =

F H

1 6.565 ± 2

a6.565f − 4a1fa−4.792f IK = 3.283 ± 3.945 . 2

Select the negative solution, since θ i is below the horizontal. ∴ tan θ i = −0.662 , θ i = −33.5° P4.23

The horizontal kick gives zero vertical velocity to the rock. Then its time of flight follows from y f = yi + v yi t +

1 ayt 2 2

1 −9.80 m s 2 t 2 2 t = 2.86 s .

−40.0 m = 0 + 0 +

e

j

The extra time 3.00 s − 2.86 s = 0.143 s is the time required for the sound she hears to travel straight back to the player. It covers distance

b343 m sg0.143 s = 49.0 m =

a

where x represents the horizontal distance the rock travels. x = 28.3 m = v xi t + 0t 2 ∴ v xi =

f

x 2 + 40.0 m

28.3 m = 9.91 m s 2.86 s

2

93

Chapter 4

P4.24

From the instant he leaves the floor until just before he lands, the basketball star is a projectile. His 2 2 = v yi + 2 a y y f − yi . vertical velocity and vertical displacement are related by the equation v yf Applying this to the upward part of his flight gives 0 =

2 v yi

d i + 2e −9.80 m s ja1.85 − 1.02f m . From this, 2

v yi = 4.03 m s . [Note that this is the answer to part (c) of this problem.]

ja

e

f

2 For the downward part of the flight, the equation gives v yf = 0 + 2 −9.80 m s 2 0.900 − 1.85 m .

Thus the vertical velocity just before he lands is v yf = −4.32 m s. (a)

His hang time may then be found from v yf = v yi + a y t :

e

j

−4.32 m s = 4.03 m s + −9.80 m s 2 t or t = 0.852 s . (b)

Looking at the total horizontal displacement during the leap, x = v xi t becomes

a

2.80 m = v xi 0.852 s

f

which yields v xi = 3.29 m s . (c)

v yi = 4.03 m s . See above for proof.

(d)

The takeoff angle is: θ = tan −1

F v I = tan F 4.03 m s I = GH v JK GH 3.29 m s JK yi

−1

50.8° .

xi

(e)

Similarly for the deer, the upward part of the flight gives 2 2 v yf = v yi + 2 a y y f − yi :

d

i

ja

e

f

2 + 2 −9.80 m s 2 2.50 − 1.20 m 0 = v yi

so v yi = 5.04 m s .

d

i

ja

e

f

2 2 2 For the downward part, v yf = v yi + 2 a y y f − yi yields v yf = 0 + 2 −9.80 m s 2 0.700 − 2.50 m

and v yf = −5.94 m s.

e

j

The hang time is then found as v yf = v yi + a y t : −5.94 m s = 5.04 m s + −9.80 m s 2 t and t = 1.12 s .

94 *P4.25

Motion in Two Dimensions

The arrow’s flight time to the collision point is t=

x f − xi v xi

=

b

150 m = 5.19 s . 45 m s cos 50°

g

The arrow’s altitude at the collision is y f = yi + v yi t +

b

1 ayt 2 2

f

ga

= 0 + 45 m s sin 50° 5.19 s + (a)

ja

1 −9.8 m s 2 5.19 s 2

e

f

2

= 47.0 m .

The required launch speed for the apple is given by

d

2 2 v yf = v yi + 2 a y y f − yi

e

i

ja

2 0 = v yi + 2 −9.8 m s 2 47 m − 0

f

v yi = 30.3 m s . (b)

The time of flight of the apple is given by v yf = v yi + a y t 0 = 30.3 m s − 9.8 m s 2 t t = 3.10 s . So the apple should be launched after the arrow by 5.19 s − 3.10 s = 2.09 s .

*P4.26

For the smallest impact angle

θ = tan −1

Fv I, GH v JK yf xf

we want to minimize v yf and maximize v xf = v xi . The final y-component 2 2 of velocity is related to v yi by v yf = v yi + 2 gh , so we want to minimize v yi

and maximize v xi . Both are accomplished by making the initial velocity horizontal. Then v xi = v , v yi = 0 , and v yf = 2 gh . At last, the impact angle is

θ = tan −1

Fv I = GH v JK yf xf

tan −1

F GH

2 gh v

I JK

.

FIG. P4.26

Chapter 4

Section 4.4 P4.27

P4.28

Uniform Circular Motion

b

g

v2 , T = 24 h 3 600 s h = 86 400 s R 2π R 2π ( 6.37 × 10 6 m) v= = = 463 m s T 86 400 s

b

a=

g

2

6

6.37 × 10 m

= 0.033 7 m s 2 directed toward the center of Earth .

r = 0.500 m; vt = a=

P4.30

2

20.0 m s v2 ac = = = 377 m s 2 r 1.06 m The mass is unnecessary information.

b463 m sg a= P4.29

95

ac =

a

f

2π r 2π 0.500 m = = 10. 47 m s = 10.5 m s 60 .0 s T 200 rev

a

10. 47 v2 = R 0.5 v2 r

e

f

2

= 219 m s 2 inward

ja

f

v = a c r = 3 9.8 m s 2 9.45 m = 16.7 m s

a

f

Each revolution carries the astronaut over a distance of 2π r = 2π 9.45 m = 59.4 m. Then the rotation rate is 16.7 m s P4.31

(a)

v = rω

a a

fb fb

FG 1 rev IJ = H 59.4 m K gb gb

0.281 rev s .

g g

At 8.00 rev s , v = 0.600 m 8.00 rev s 2π rad rev = 30.2 m s = 9.60π m s . At 6.00 rev s , v = 0.900 m 6.00 rev s 2π rad rev = 33.9 m s = 10.8π m s . 6.00 rev s gives the larger linear speed.

b

(b)

9.60π m s v2 = Acceleration = r 0.600 m

(c)

At 6.00 rev s , acceleration =

g

2

= 1.52 × 10 3 m s 2 .

b10.8π m sg 0.900 m

2

= 1.28 × 10 3 m s 2 .

96 P4.32

Motion in Two Dimensions

The satellite is in free fall. Its acceleration is due to gravity and is by effect a centripetal acceleration. ac = g so v2 = g. r Solving for the velocity, v = rg =

a6 ,400 + 600fe10 mje8.21 m s j = 3

v=

2

7.58 × 10 3 m s

2πr T

and

e

T = 5.80 × 10 3 s

Section 4.5 P4.33

j

3 2π r 2π 7 ,000 × 10 m = = 5.80 × 10 3 s v 7.58 × 10 3 m s

T=

FG 1 min IJ = 96.7 min . H 60 s K

Tangential and Radial Acceleration

We assume the train is still slowing down at the instant in question. ac =

v2 = 1.29 m s 2 r

at =

−40.0 km h 10 3 m km ∆v = 15.0 s ∆t

b

ge

1h 3 600 s

j = −0.741 m s

e1.29 m s j + e−0.741 m s j F a I = tan FG 0.741 IJ GH a JK H 1.29 K 2 2

a = a c2 + a t2 = at an angle of tan −1

je

t

2 2

−1

c

a = 1.48 m s 2 inward and 29.9 o backward P4.34

(a)

at = 0.600 m s 2

(b)

4.00 m s v2 ar = = r 20.0 m

(c)

a = a t2 + a r2 = 1.00 m s 2

b

θ = tan −1

g

2

= 0.800 m s 2

ar = 53.1° inward from path at

2

FIG. P4.33

Chapter 4

P4.35

r = 2.50 m , a = 15.0 m s 2 (a) (b)

ja

e

f

a c = a cos 30.0 o = 15.0 m s 2 cos 30° = 13.0 m s 2 v2 r so v 2 = ra c = 2.50 m 13.0 m s 2 = 32.5 m 2 s 2 ac =

e

j

FIG. P4.35

v = 32.5 m s = 5.70 m s (c)

a 2 = a t2 + a r2 so a t = a 2 − a r2 =

P4.36

97

e15.0 m s j − e13.0 m s j = 2 2

2

7.50 m s 2

(a)

See figure to the right.

(b)

The components of the 20.2 and the 22.5 m s 2 along the rope together constitute the centripetal acceleration:

e

j a

f e

j

a c = 22.5 m s 2 cos 90.0°−36.9° + 20.2 m s 2 cos 36.9° = 29.7 m s 2 (c)

*P4.37

v2 so v = a c r = 29.7 m s 2 1.50 m = 6.67 m s tangent to circle r v = 6.67 m s at 36.9° above the horizontal

a

ac =

f

FIG. P4.36

at

Let i be the starting point and f be one revolution later. The curvilinear motion with constant tangential acceleration is described by ∆ x = v xi t +

1 axt 2 2

1 2 at t 2 4π r

a

θ

ar

2π r = 0 + at =

and v xf = v xi + a x t , v f = 0 + a t t = Then tan θ =

4π r t 2 at 1 = 2 = a r t 16π 2 r 4π

t

FIG. P4.37

2

v 2f 16π 2 r 2 4π r = . The magnitude of the radial acceleration is a r = . r t t 2r

θ = 4.55° .

98

Motion in Two Dimensions

Section 4.6 P4.38

(a)

Relative Velocity and Relative Acceleration

e

a

j

v H = 0 + a H t = 3.00 i − 2.00 j m s 2 5.00 s

e

j

f

v H = 15.0 i − 10.0 j m s

e

a

j

v J = 0 + a j t = 1.00 i + 3.00 j m s 2 5.00 s

e

f

j

v J = 5.00 i +15.0 j m s

e

j

v HJ = v H − v J = 15.0 i − 10.0 j − 5.00 i − 15.0 j m s

e

j

v HJ = 10.0 i − 25.0 j m s v HJ = (10.0) 2 + ( 25.0) 2 m s = 26.9 m s (b)

a

1 1 a H t 2 = 3.00 i − 2.00 j m s 2 5.00 s 2 2   rH = 37.5 i − 25.0 j m

e

rH = 0 + 0 +

e

f

2

j

a

f e

1 2 1.00 i + 3.00 j m s 2 5.00s = 12.5 i + 37.5 j m 2 = r − r = 37.5 i − 25.0 j − 12.5 i − 37.5 j m

e

rJ = rHJ

j

H

j

j

e

J

j

e

j

rHJ = 25.0 i − 62.5 j m rHJ = (c)

2

2

67.3 m

e

j

a HJ = a H − a J = 3.00 i − 2.00 j − 1.00 i − 3.00 j m s 2 a HJ =

*P4.39

a25.0f + a62.5f m = e2.00i − 5.00 jj m s

2

v ce = the velocity of the car relative to the earth. v wc = the velocity of the water relative to the car. v we = the velocity of the water relative to the earth. These velocities are related as shown in the diagram at the right. (a)

(b)

Since v we is vertical, v wc sin 60.0° = v ce = 50.0 km h or v wc = 57.7 km h at 60.0° west of vertical . Since v ce has zero vertical component,

b

vce

vwe 60°

vwc

v we = v ce + v wc FIG. P4.39

g

v we = v wc cos 60.0° = 57.7 km h cos 60.0° = 28.9 km h downward .

Chapter 4

P4.40

The bumpers are initially 100 m = 0.100 km apart. After time t the bumper of the leading car travels 40.0 t, while the bumper of the chasing car travels 60.0t. Since the cars are side by side at time t, we have 0.100 + 40.0t = 60.0t , yielding t = 5.00 × 10 −3 h = 18.0 s .

P4.41

Total time in still water t =

d 2 000 = = 1.67 × 10 3 s . v 1. 20

Total time = time upstream plus time downstream: 1 000 = 1.43 × 10 3 s (1.20 − 0.500 ) 1 000 = = 588 s . 1.20 + 0.500

t up = t down

Therefore, ttotal = 1.43 × 10 3 + 588 = 2.02 × 10 3 s . P4.42

v = 150 2 + 30.0 2 = 153 km h

θ = tan −1 P4.43

99

FG 30.0 IJ = H 150 K

11.3° north of west

For Alan, his speed downstream is c + v, while his speed upstream is c − v . Therefore, the total time for Alan is t1 =

2L L L c + = 2 c+v c−v 1 − v2

.

c

For Beth, her cross-stream speed (both ways) is c2 − v2 . Thus, the total time for Beth is t 2 =

Since 1 −

2L c2 − v2

=

2L c

1−

v2 c2

.

v2 < 1 , t1 > t 2 , or Beth, who swims cross-stream, returns first. c2

100 P4.44

Motion in Two Dimensions

(a)

To an observer at rest in the train car, the bolt accelerates downward and toward the rear of the train. a= tan θ =

b2.50 m sg + b9.80 m sg 2

2.50 m s 2 9.80 m s 2

2

= 10.1 m s 2

= 0.255

θ = 14.3° to the south from the vertical (b) P4.45

a = 9.80 m s 2 vertically downward

Identify the student as the S’ observer and the professor as the S observer. For the initial motion in S’, we have v ′y v ′x

= tan 60.0° = 3 .

Let u represent the speed of S’ relative to S. Then because there is no x-motion in S, we can write v x = v ′x + u = 0 so that v x′ = −u = −10.0 m s . Hence the ball is thrown backwards in S’. Then,

v y = v ′y = 3 v ′x = 10.0 3 m s . Using v y2 = 2 gh we find

e10.0 3 m sj = h= 2e9.80 m s j

FIG. P4.45

2

2

15.3 m .

The motion of the ball as seen by the student in S’ is shown in diagram (b). The view of the professor in S is shown in diagram (c). *P4.46

Choose the x-axis along the 20-km distance. The ycomponents of the displacements of the ship and the speedboat must agree:

b26 km hgt sina40°−15°f = b50 km hgt sin α 11.0 = 12.7° . α = sin −1 50

x

N

25°

40° 15°

α E y

The speedboat should head 15°+12.7° = 27.7° east of north .

FIG. P4.46

Chapter 4

101

Additional Problems *P4.47

b

g

(a)

The speed at the top is v x = vi cos θ i = 143 m s cos 45° = 101 m s .

(b)

In free fall the plane reaches altitude given by

d

i 0 = b143 m s sin 45°g + 2e −9.8 m s jd y − 31 000 ft i F 3.28 ft IJ = 3.27 × 10 ft . y = 31 000 ft + 522 mG H 1m K

2 2 v yf = v yi + 2 a y y f − yi

2

2

3

f

(c)

f

For the whole free fall motion v yf = v yi + a y t

e

j

−101 m s = +101 m s − 9.8 m s 2 t t = 20.6 s (d)

ac =

v2 r

e

j

v = a c r = 0.8 9.8 m s 2 4,130 m = 180 m s P4.48

At any time t, the two drops have identical y-coordinates. The distance between the two drops is then just twice the magnitude of the horizontal displacement either drop has undergone. Therefore,

af b g b

g

d = 2 x t = 2 v xi t = 2 vi cos θ i t = 2 vi t cos θ i . P4.49

After the string breaks the ball is a projectile, and reaches the ground at time t: y f = v yi t + −1.20 m = 0 +

1 −9.80 m s 2 t 2 2

e

j

so t = 0.495 s. Its constant horizontal speed is v x =

x 2.00 m = = 4.04 m s t 0.495 s

b

g

4.04 m s v2 so before the string breaks a c = x = r 0.300 m

2

= 54.4 m s 2 .

1 ayt 2 2

102 P4.50

Motion in Two Dimensions

(a)

b

gd i

y f = tan θ i x f −

g 2 vi2

x 2f

cos 2 θ i

Setting x f = d cos φ , and y f = d sin φ , we have

b

gb

g

d sin φ = tan θ i d cos φ −

Solving for d yields, d =

or d =

(b) P4.51

Setting

b

g 2 vi2

bd cos φ g . 2

2

cos θ i

FIG. P4.50

2 vi2 cos θ i sin θ i cos φ − sin φ cos θ i

2 vi2 cos θ i sin θ i − φ

g cos 2 φ

g

2

g cos φ

.

b

v 2 1 − sin φ dd φ = 0 leads to θ i = 45°+ and d max = i dθ i 2 g cos 2 φ

g

.

Refer to the sketch: (b)

b

g

∆ x = v xi t ; substitution yields 130 = vi cos 35.0° t . 1 ∆ y = v yi t + at 2 ; substitution yields 2

b

a

g

f

1 −9.80 t 2 . 2

20.0 = vi sin 35.0° t +

Solving the above gives t = 3.81 s . (a)

vi = 41.7 m s

(c)

v yf = vi sin θ i − gt , v x = vi cos θ i

FIG. P4.51

a fa f

At t = 3.81 s , v yf = 41.7 sin 35.0°− 9.80 3.81 = −13.4 m s

a

f

v x = 41.7 cos 35.0° = 34.1 m s 2 v f = v x2 + v yf = 36.7 m s .

Chapter 4

P4.52

(a)

The moon’s gravitational acceleration is the probe’s centripetal acceleration: (For the moon’s radius, see end papers of text.) a=

v2 r

v2 1 9.80 m s 2 = 6 1.74 × 10 6 m

e

j

v = 2.84 × 10 6 m 2 s 2 = 1.69 km s (b)

P4.53

(a)

2π r T 2π r 2π (1.74 × 10 6 m) T= = = 6.47 × 10 3 s = 1.80 h v 1.69 × 10 3 m s v=

b

5.00 m s v2 ac = = r 1.00 m

g

2

= 25.0 m s 2

at = g = 9.80 m s 2 (b)

See figure to the right.

(c)

a = a c2 + a t2 =

φ = tan −1

e25.0 m s j + e9.80 m s j

FG a IJ = tan Ha K t

2 2

−1

c

P4.54

9.80 m s 2 25.0 m s 2

2 2

= 26.8 m s 2

= 21.4° FIG. P4.53

x f = vix t = vi t cos 40.0° 10.0 m . vi cos 40.0° At this time, y f should be 3.05 m − 2.00 m = 1.05 m .

Thus, when x f = 10.0 m , t =

Thus, 1.05 m =

bv sin 40.0°g10.0 m + 1 e−9.80 m s jLM 10.0 m OP . v cos 40.0° 2 N v cos 40.0° Q 2

i

2

i

From this, vi = 10.7 m s .

i

103

104 P4.55

Motion in Two Dimensions

The special conditions allowing use of the horizontal range equation applies. For the ball thrown at 45°, D = R 45 =

vi2 sin 90 . g

For the bouncing ball, v 2 sin 2θ D = R1 + R 2 = i + g

vi 2 2

e j

sin 2θ g

where θ is the angle it makes with the ground when thrown and when bouncing. (a)

We require: vi2 vi2 sin 2θ vi2 sin 2θ = + g g 4g 4 5 θ = 26.6°

sin 2θ =

(b)

FIG. P4.55

The time for any symmetric parabolic flight is given by 1 2 gt 2 1 0 = vi sin θ i t − gt 2 . 2

y f = v yi t −

If t = 0 is the time the ball is thrown, then t =

2 vi sin θ i is the time at landing. g

So for the ball thrown at 45.0° t 45 =

2 vi sin 45.0° . g

For the bouncing ball, 2 v sin 26.6° 2 t = t1 + t 2 = i + g

e j sin 26.6° = 3v sin 26.6° . vi 2

i

g

The ratio of this time to that for no bounce is 3 vi sin 26.6 ° g 2 vi sin 45 .0 ° g

=

1.34 = 0.949 . 1.41

g

Chapter 4

P4.56

105

Using the range equation (Equation 4.14) R=

vi2 sin( 2θ i ) g

vi2 . Given R, this yields vi = gR . g If the boy uses the same speed to throw the ball vertically upward, then

the maximum range occurs when θ i = 45° , and has a value R =

v y = gR − gt and y = gR t −

gt 2 2

at any time, t. R , and so the maximum height reached is g

At the maximum height, v y = 0, giving t =

y max P4.57

R g = gR − g 2

F RI GH g JK

2

= R−

R R = . 2 2

Choose upward as the positive y-direction and leftward as the positive x-direction. The vertical height of the stone when released from A or B is

a

vi

f

yi = 1.50 + 1.20 sin 30.0° m = 2.10 m (a)

B

A

The equations of motion after release at A are

a

30° 1.20 m

f

v y = vi sin 60.0°− gt = 1.30 − 9.80t m s

30°

vi

v x = vi cos 60.0° = 0.750 m s

e ∆ x = a0.750t f m −1.30 ± a1.30f When y = 0 , t =

j

y = 2.10 +1.30t − 4.90t 2 m FIG. P4.57

A

2

+ 41.2

−9.80

(b)

a

fa

f

= 0.800 s. Then, ∆ x A = 0.750 0.800 m = 0.600 m .

The equations of motion after release at point B are

a

f

a

f

v y = vi − sin 60.0° − gt = −1.30 − 9.80t m s v x = vi cos 60.0 = 0.750 m s

e

j

yi = 2.10 − 1.30t − 4.90t 2 m . When y = 0 , t =

b

+1.30 ±

1.50 m s v2 = r 1.20 m

a−1.30f −9.80

g

2

+ 41.2

a

fa

f

= 0.536 s. Then, ∆ x B = 0.750 0.536 m = 0.402 m .

2

= 1.87 m s 2 toward the center

(c)

ar =

(d)

After release, a = − g j = 9.80 m s 2 downward

106 P4.58

Motion in Two Dimensions

The football travels a horizontal distance R=

b g = a20.0f sina60.0°f = 35.3 m.

vi2 sin 2θ i g

2

9.80

Time of flight of ball is FIG. P4.58

2 v sin θ i 2( 20.0) sin 30.0° t= i = = 2.04 s . g 9.80

The receiver is ∆ x away from where the ball lands and ∆ x = 35.3 − 20.0 = 15.3 m. To cover this distance in 2.04 s, he travels with a velocity v= P4.59

(a)

15.3 = 7.50 m s in the direction the ball was thrown . 2.04

1 2 g t ; ∆ x = vi t 2 Combine the equations eliminating t: ∆y= −

FG IJ H K

1 ∆x ∆y= − g vi 2

b g = FGH −2g∆ y IJK v

From this, ∆ x

2

2

.

2 i

FIG. P4.59 thus ∆ x = vi

−2 ∆ y −2( −300 ) = 275 = 6.80 × 10 3 = 6.80 km . g 9.80

(b)

The plane has the same velocity as the bomb in the x direction. Therefore, the plane will be 3 000 m directly above the bomb when it hits the ground.

(c)

When φ is measured from the vertical, tan φ =

therefore, φ = tan −1

F ∆ x I = tan F 6 800 I = GH ∆ y JK GH 3 000 JK −1

∆x ∆y

66. 2° .

Chapter 4

*P4.60

(a)

We use the approximation mentioned in the problem. The time to travel 200 m horizontally is ∆x 200 m t= = = 0.200 s . The bullet falls by v x 1,000 m s

(c)

ja f

1 1 a y t 2 = 0 + −9.8 m s 2 0.2 s 2 2

e

∆ y = v yi t + (b)

107

The telescope axis must point below the barrel axis 0.196 m = 0.056 1° . by θ = tan −1 200 m t=

50.0 m = 0.050 0 s . The bullet falls by only 1 000 m s ∆y=

ja

1 −9.8 m s 2 0.05 s 2

e

a

f

2

2

= −0.196 m . barrel axis bullet path scope axis 50

150 200 250 FIG. P4.60(b)

= −0.0122 m .

a

f

f

1 1 200 m , the scope axis points to a location 19.6 cm = 4.90 cm above the 4 4 barrel axis, so the sharpshooter must aim low by 4.90 cm − 1.22 cm = 3.68 cm .

At range 50 m =

(d)

t=

150 m = 0.150 s 1 000 m s

ja a

f

1 2 −9.8 m s 2 0.15 s = 0.110 m 2 150 19.6 cm − 11.0 cm = 3.68 cm . Aim low by 200

∆y=

(e)

t=

e

f

250 m = 0.250 s 1 000 m s

∆y=

ja

1 −9.8 m s 2 0.25 s 2

e

Aim high by 30.6 cm −

f

2

= 0.306 m

a

f

250 19.6 cm = 6.12 cm . 200

(f), (g) Many marksmen have a hard time believing it, but they should aim low in both cases. As in case (a) above, the time of flight is very nearly 0.200 s and the bullet falls below the barrel axis by 19.6 cm on its way. The 0.0561° angle would cut off a 19.6-cm distance on a vertical wall at a horizontal distance of 200 m, but on a vertical wall up at 30° it cuts off distance h as shown, where cos 30° = 19.6 cm h , h = 22.6 cm. The marksman

barrel axis scope 30° 19.6 cm

must aim low by 22.6 cm − 19.6 cm = 3.03 cm . The answer can be obtained by considering limiting cases. Suppose the target is nearly straight above or below you. Then gravity will not cause deviation of the path of the bullet, and one must aim low as in part (c) to cancel out the sighting-in of the telescope.

30°

h

19.6 cm scope axis

bullet hits here

FIG. P4.60(f–g)

108 P4.61

Motion in Two Dimensions

(a)

From Part (c), the raptor dives for 6.34 − 2.00 = 4.34 s undergoing displacement 197 m downward and 10.0 4.34 = 43.4 m forward.

a fa f

∆d v= ∆t

P4.62

(b)

α = tan −1

(c)

197 =

FG −197 IJ = H 43.4 K

a197f + a43.4f 2

2

= 46.5 m s

4.34 −77.6°

1 2 gt , t = 6.34 s 2

FIG. P4.61

Measure heights above the level ground. The elevation y b of the ball follows yb = R + 0 − with x = vi t so y b = R − (a)

gx 2 2 vi2

1 2 gt 2

.

The elevation yr of points on the rock is described by y r2 + x 2 = R 2 . We will have y b = y r at x = 0 , but for all other x we require the ball to be above the rock

surface as in y b > y r . Then y b2 + x 2 > R 2

F R − gx I GH 2 v JK

2

2

+ x2 > R2

2 i

R2 −

gx 2 R vi2

+

g 2x4 4vi4 g 2x4 4vi4

+ x2 > R2 + x2 >

gx 2 R vi2

.

If this inequality is satisfied for x approaching zero, it will be true for all x. If the ball’s parabolic trajectory has large enough radius of curvature at the start, the ball will clear the gR whole rock: 1 > 2 vi

vi > gR . (b)

With vi = gR and y b = 0 , we have 0 = R −

gx 2 2 gR

or x = R 2 . The distance from the rock’s base is x−R=

e

j

2 −1 R .

Chapter 4

P4.63

(a)

While on the incline v 2f − vi2 = 2 a∆x v f − vi = at

a fa f

v 2f − 0 = 2 4.00 50.0 20.0 − 0 = 4.00t v f = 20.0 m s t = 5.00 s (b)

FIG. P4.63

Initial free-flight conditions give us v xi = 20.0 cos 37.0° = 16.0 m s and v yi = −20.0 sin 37.0° = −12.0 m s v xf = v xi since a x = 0

a

fa

f a

v yf = − 2 a y ∆ y + v yi 2 = − 2 −9.80 −30.0 + −12.0

a16.0f + a−27.1f 2

v f = v xf 2 + v yf 2 = (c)

t1 = 5 s ; t 2 =

v yf − v yi ay

=

2

f

2

= −27.1 m s

= 31.5 m s at 59.4° below the horizontal

−27.1 + 12.0 = 1.53 s −9.80

t = t1 + t 2 = 6.53 s (d) P4.64

a f

∆x = v xi t1 = 16.0 1.53 = 24.5 m y 2 = 16 x

Equation of bank: Equations of motion:

x = vi t 1 y = − g t2 2

a1f a 2f a3 f F I GH JK

1 x2 g . Equate y 2 vi2 from the bank equation to y from the equations of motion:

Substitute for t from (2) into (3) y = −

L 1 Fx 16 x = M− g G MN 2 H v

From this, x = 0 or x

3

FIG. P4.64

I OP ⇒ g x − 16x = xF g x − 16I = 0 . JK PQ 4v GH 4v JK F 10 I = 18.8 m . Also, 64v = and x = 4G g H 9.80 JK 1 Fx I 1 a9.80fa18.8f y = − gG J = − 2 Hv K 2 a10.0f = −17.3 m . 2

2

2

2 4

2 i

4 i

2 3

4 i

4 i

1/3

4

2

2

2

2 i

2

109

110 P4.65

Motion in Two Dimensions

(a)

a f

1 2 1 at ; 70.0 = 15.0 t 2 2 2 70.0 = vi t Roadrunner: ∆ x = vi t ;

Coyote:

∆x =

Solving the above, we get vi = 22.9 m s and t = 3.06 s. (b)

At the edge of the cliff,

a fa f

v xi = at = 15.0 3.06 = 45.8 m s . Substituting into ∆ y =

1 a y t 2 , we find 2

a

f

1 −9.80 t 2 2 t = 4.52 s

−100 =

∆ x = v xi t +

a fa

f a fa

f

1 1 2 a x t 2 = 45.8 4.52 s + 15.0 4.52 s . 2 2

Solving, ∆ x = 360 m . (c)

For the Coyote’s motion through the air

a f t = 0 − 9.80a 4.52f =

v xf = v xi + a x t = 45.8 + 15 4.52 = 114 m s v yf = v yi + a y P4.66

−44.3 m s .

Think of shaking down the mercury in an old fever thermometer. Swing your hand through a circular arc, quickly reversing direction at the bottom end. Suppose your hand moves through onequarter of a circle of radius 60 cm in 0.1 s. Its speed is 1 4

a2π fa0.6 mf ≅ 9 m s 0.1 s

and its centripetal acceleration is

v 2 ( 9 m s) 2 ≅ ~ 10 2 m s 2 . r 0.6 m

The tangential acceleration of stopping and reversing the motion will make the total acceleration somewhat larger, but will not affect its order of magnitude.

Chapter 4

P4.67

(a)

∆ x = v xi t , ∆ y = v yi t +

1 2 gt 2

a

111

f

d cos 50.0° = 10.0 cos 15.0° t and

a

f

− d sin 50.0° = 10.0 sin 15.0° t +

a

f

1 −9.80 t 2 . 2

Solving, d = 43.2 m and t = 2.88 s . (b)

Since a x = 0 ,

FIG. P4.67 v xf = v xi = 10.0 cos 15.0° = 9.66 m s

a f

v yf = v yi + a y t = 10.0 sin 15.0°−9.80 2.88 = −25.6 m s . Air resistance would decrease the values of the range and maximum height. As an airfoil, he can get some lift and increase his distance. *P4.68

For one electron, we have y = viy t , D = vix t +

1 1 a x t 2 ≅ a x t 2 , v yf = v yi , and v xf = v xi + a x t ≅ a x t . 2 2

The angle its direction makes with the x-axis is given by

θ = tan −1

v yf v xf

= tan −1

v yi axt

= tan −1

v yi t axt

2

= tan −1

y . 2D

FIG. P4.68

Thus the horizontal distance from the aperture to the virtual source is 2D. The source is at coordinate x = − D . *P4.69

(a)

The ice chest floats downstream 2 km in time t, so that 2 km = v w t . The upstream motion of the boat is described by d = ( v − v w )15 min. The downstream motion is described by 2 km d + 2 km = ( v + v w )( t − 15 min) . We eliminate t = and d by substitution: vw

bv − v g15 min + 2 km = bv + v gFGH 2vkm − 15 minIJK v va15 minf − v a15 minf + 2 km = 2 km + 2 km − va15 minf − v a15 minf v v va30 minf = 2 km v w

w

w

w

w

w

w

2 km vw = = 4.00 km h . 30 min (b)

In the reference frame of the water, the chest is motionless. The boat travels upstream for 15 min at speed v, and then downstream at the same speed, to return to the same point. Thus it travels for 30 min. During this time, the falls approach the chest at speed v w , traveling 2 km. Thus vw =

∆x 2 km = = 4.00 km h . ∆ t 30 min

112 *P4.70

Motion in Two Dimensions

Let the river flow in the x direction. (a)

To minimize time, swim perpendicular to the banks in the y direction. You are in the water for time t in ∆ y = v y t , t =

(b)

80 m = 53.3 s . 1.5 m s

b

g

The water carries you downstream by ∆ x = v x t = 2.50 m s 53.3 s = 133 m . K vw

(c) K vw

K vw K vs

K vs

K K vs + vw

K K vs + vw

K vs

To minimize downstream drift, you should swim so that K K your resultant velocity v s + v w is perpendicular to your K swimming velocity v s relative to the water. This condition is shown in the middle picture. It maximizes the angle between the resultant velocity and the shore. The angle 1.5 m s K between v s and the shore is given by cos θ = , 2.5 m s

K K vs + vw

K vs

K K vs + vw K v w = 2.5 m/s i

θ = 53.1° . (d)

Now v y = v s sin θ = 1.5 m s sin 53.1° = 1.20 m s t=

∆y 80 m = = 66.7 s v y 1.2 m s

b

g

∆ x = v x t = 2.5 m s − 1.5 m s cos 53.1° 66.7 s = 107 m .

θ

Chapter 4

*P4.71

113

Find the highest firing angle θ H for which the projectile will clear the mountain peak; this will yield the range of the closest point of bombardment. Next find the lowest firing angle; this will yield the maximum range under these conditions if both θ H and θ L are > 45° ; x = 2500 m, y = 1800 m , vi = 250 m s .

a f

1 2 1 gt = vi sin θ t − gt 2 2 2 x f = v xi t = vi cos θ t y f = v yi t −

a

f

Thus t=

xf vi cos θ

.

Substitute into the expression for y f x IJ 1 F x − gG a f v cos θ 2 H v cos θ K

y f = vi sin θ

f

f

i

i

2

= x f tan θ −

gx 2f 2 vi2 cos 2 θ

gx 2f 1 2 = tan θ + 1 so y f = x f tan θ − 2 tan 2 θ + 1 and but cos 2 θ 2 vi

e

0=

gx 2f 2 vi2

j

tan 2 θ − x f tan θ +

gx 2f 2 vi2

+ yf .

Substitute values, use the quadratic formula and find tan θ = 3.905 or 1.197 , which gives θ H = 75.6° and θ L = 50.1° .

b

g

Range at θ H =

vi2 sin 2θ H = 3.07 × 10 3 m from enemy ship g 3.07 × 10 3 − 2 500 − 300 = 270 m from shore.

b

g

Range at θ L =

vi2 sin 2θ L = 6.28 × 10 3 m from enemy ship g 6.28 × 10 3 − 2 500 − 300 = 3.48 × 10 3 from shore.

Therefore, safe distance is < 270 m or > 3.48 × 10 3 m from the shore.

FIG. P4.71

114 *P4.72

Motion in Two Dimensions

We follow the steps outlined in Example 4.7, eliminating t =

d cos φ to find vi cos θ

vi sin θ d cos φ gd 2 cos 2 φ − 2 = − d sin φ . vi cos θ 2 vi cos 2 θ Clearing of fractions, 2 vi2 cos θ sin θ cos φ − gd cos 2 φ = −2 vi2 cos 2 θ sin φ . To maximize d as a function of θ, we differentiate through with respect to θ and set

a

f

2 vi2 cos θ cos θ cos φ + 2 vi2 sin θ − sin θ cos φ − g

dd = 0: dθ

a

f

dd cos 2 φ = −2 vi2 2 cos θ − sin θ sin φ . dθ

We use the trigonometric identities from Appendix B4 cos 2θ = cos 2 θ − sin 2 θ and sin φ 1 sin 2θ = 2 sin θ cos θ to find cos φ cos 2θ = sin 2θ sin φ . Next, = tan φ and cot 2θ = give cos φ tan 2θ

a

f

cot 2φ = tan φ = tan 90°−2θ so φ = 90°−2θ and θ = 45°−

φ . 2

ANSWERS TO EVEN PROBLEMS P4.2

e j (b) v = 18.0 i + a 4.00 − 9.80t f j ; (c) a = e −9.80 m s j j ; (d) a54.0 mf i − a32.1 mf j ; (e) b18.0 m sg i − b 25.4 m sg j ; (f) e −9.80 m s j j

(a) r = 18.0t i + 4.00t − 4.90t 2 j ;

P4.8

e

P4.4

e

j

e

e

P4.10

e7.23 × 10

3

j

m, 1.68 × 10 3 m

g horizontally; 2h 2h below the horizontal (b) tan −1 d (a) d

P4.14

0.600 m s 2

P4.16

(a) 76.0°; (b) the same; (c)

(c) a circle of radius 5.00 m centered at 0 , 4.00 m

P4.18

25.8 m s

(a) v = −12.0t j m s ; a = −12.0 j m s 2 ; (b) r = 3.00 i − 6.00 j m; v = −12.0 j m s

P4.20

d tan θ i −

e

j

+5.00 m − sin ω t i − cos ω t j ;

e

j esin ω t i + cos ω t jj ;

v = 5.00 m ω − cos ω t i + sin ω t j ; a = 5.00 m ω 2

a

P4.6

j

FG IJ H K

j

a = 0 i + 5.00ω 2 j m s 2 ; (b) r = 4.00 m j

j

(b) r = 10.0 i + 6.00 j m; 7.81 m s

P4.12

(a) v = −5.00ω i + 0 j m s ;

j

v = 5.00 i + 3.00tj m s ;

2

2

e

(a) r = 5.00t i + 1.50t 2 j m ;

f

e

j

e2 v

gd 2 2 i

cos 2 θ i

j

17d 8

Chapter 4

P4.22

33.5° below the horizontal

P4.48

2vi t cos θ i

P4.24

(a) 0.852 s; (b) 3.29 m s ; (c) 4.03 m s; (d) 50.8°; (e) 1.12 s

P4.50

(a) see the solution;

−1

F GH

2gh

I JK

P4.26

tan

P4.28

0.033 7 m s 2 toward the center of the Earth

v

P4.30

0.281 rev s

P4.32

7.58 × 10 3 m s; 5.80 × 10 3 s

P4.34

(a) 0.600 m s 2 forward; (b) 0.800 m s 2 inward;

φ

(b) θ i = 45°+ ; d max = 2

(a) see the solution; (b) 29.7 m s 2 ; (c) 6.67 m s at 36.9° above the horizontal

P4.38

(a) 26.9 m s ; (b) 67.3 m; (c) 2.00 i − 5.00 j m s 2

e

j

P4.40

18.0 s

P4.42

153 km h at 11.3° north of west

P4.44

(a) 1.69 km s ; (b) 6.47 × 10 3 s

P4.54

10.7 m s

P4.56

R 2

P4.58

7.50 m s in the direction the ball was thrown

P4.60

(a) 19.6 cm; (b) 0.0561°; (c) aim low 3.68 cm; (d) aim low 3.68 cm; (e) aim high 6.12 cm; (f) aim low; (g) aim low

P4.62

(a)

P4.64

a18.8 m; − 17.3 mf

P4.66

see the solution; ~ 10 2 m s 2

P4.68

x = −D

P4.70

(a) at 90° to the bank; (b) 133 m; (c) upstream at 53.1° to the bank; (d) 107 m

P4.72

see the solution

2

(a) 10.1 m s at 14.3° south from the vertical; (b) 9.80 m s 2 vertically downward

P4.46

27.7° east of north

g

2

g cos φ

P4.52

(c) 1.00 m s 2 forward and 53.1° inward P4.36

b

vi2 1 − sin φ

115

gR ; (b)

e

j

2 −1 R

5 The Laws of Motion CHAPTER OUTLINE 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

The Concept of Force Newton’s First Law and Inertial Frames Mass Newton’s Second Law The Gravitational Force and Weight Newton’s Third Law Some Applications of Newton’s Laws Forces of Friction

ANSWERS TO QUESTIONS Q5.1

(a)

The force due to gravity of the earth pulling down on the ball—the reaction force is the force due to gravity of the ball pulling up on the earth. The force of the hand pushing up on the ball—reaction force is ball pushing down on the hand.

(b)

The only force acting on the ball in free-fall is the gravity due to the earth -the reaction force is the gravity due to the ball pulling on the earth.

Q5.2

The resultant force is zero, as the acceleration is zero.

Q5.3

Mistake one: The car might be momentarily at rest, in the process of (suddenly) reversing forward into backward motion. In this case, the forces on it add to a (large) backward resultant.

Mistake two: There are no cars in interstellar space. If the car is remaining at rest, there are some large forces on it, including its weight and some force or forces of support. Mistake three: The statement reverses cause and effect, like a politician who thinks that his getting elected was the reason for people to vote for him. Q5.4

When the bus starts moving, the mass of Claudette is accelerated by the force of the back of the seat on her body. Clark is standing, however, and the only force on him is the friction between his shoes and the floor of the bus. Thus, when the bus starts moving, his feet start accelerating forward, but the rest of his body experiences almost no accelerating force (only that due to his being attached to his accelerating feet!). As a consequence, his body tends to stay almost at rest, according to Newton’s first law, relative to the ground. Relative to Claudette, however, he is moving toward her and falls into her lap. (Both performers won Academy Awards.)

Q5.5

First ask, “Was the bus moving forward or backing up?” If it was moving forward, the passenger is lying. A fast stop would make the suitcase fly toward the front of the bus, not toward the rear. If the bus was backing up at any reasonable speed, a sudden stop could not make a suitcase fly far. Fine her for malicious litigiousness.

Q5.6

It would be smart for the explorer to gently push the rock back into the storage compartment. Newton’s 3rd law states that the rock will apply the same size force on her that she applies on it. The harder she pushes on the rock, the larger her resulting acceleration.

117

118

The Laws of Motion

Q5.7

The molecules of the floor resist the ball on impact and push the ball back, upward. The actual force acting is due to the forces between molecules that allow the floor to keep its integrity and to prevent the ball from passing through. Notice that for a ball passing through a window, the molecular forces weren’t strong enough.

Q5.8

While a football is in flight, the force of gravity and air resistance act on it. When a football is in the process of being kicked, the foot pushes forward on the ball and the ball pushes backward on the foot. At this time and while the ball is in flight, the Earth pulls down on the ball (gravity) and the ball pulls up on the Earth. The moving ball pushes forward on the air and the air backward on the ball.

Q5.9

It is impossible to string a horizontal cable without its sagging a bit. Since the cable has a mass, gravity pulls it downward. A vertical component of the tension must balance the weight for the cable to be in equilibrium. If the cable were completely horizontal, then there would be no vertical component of the tension to balance the weight. Some physics teachers demonstrate this by asking a beefy student to pull on the ends of a cord supporting a can of soup at its center. Some get two burly young men to pull on opposite ends of a strong rope, while the smallest person in class gleefully mashes the center of the rope down to the table. Point out the beauty of sagging suspension-bridge cables. With a laser and an optical lever, demonstrate that the mayor makes the courtroom table sag when he sits on it, and the judge bends the bench. Give them “I make the floor sag” buttons, available to instructors using this manual. Estimate the cost of an infinitely strong cable, and the truth will always win.

Q5.10

As the barbell goes through the bottom of a cycle, the lifter exerts an upward force on it, and the scale reads the larger upward force that the floor exerts on them together. Around the top of the weight’s motion, the scale reads less than average. If the iron is moving upward, the lifter can declare that she has thrown it, just by letting go of it for a moment, so our answer applies also to this case.

Q5.11

As the sand leaks out, the acceleration increases. With the same driving force, a decrease in the mass causes an increase in the acceleration.

Q5.12

As the rocket takes off, it burns fuel, pushing the gases from the combustion out the back of the rocket. Since the gases have mass, the total remaining mass of the rocket, fuel, and oxidizer decreases. With a constant thrust, a decrease in the mass results in an increasing acceleration.

Q5.13

The friction of the road pushing on the tires of a car causes an automobile to move. The push of the air on the propeller moves the airplane. The push of the water on the oars causes the rowboat to move.

Q5.14

As a man takes a step, the action is the force his foot exerts on the Earth; the reaction is the force of the Earth on his foot. In the second case, the action is the force exerted on the girl’s back by the snowball; the reaction is the force exerted on the snowball by the girl’s back. The third action is the force of the glove on the ball; the reaction is the force of the ball on the glove. The fourth action is the force exerted on the window by the air molecules; the reaction is the force on the air molecules exerted by the window. We could in each case interchange the terms ‘action’ and ‘reaction.’

Q5.15

The tension in the rope must be 9 200 N. Since the rope is moving at a constant speed, then the resultant force on it must be zero. The 49ers are pulling with a force of 9 200 N. If the 49ers were winning with the rope steadily moving in their direction or if the contest was even, then the tension would still be 9 200 N. In all of these case, the acceleration is zero, and so must be the resultant force on the rope. To win the tug-of-war, a team must exert a larger force on the ground than their opponents do.

Chapter 5

119

Q5.16

The tension in the rope when pulling the car is twice that in the tug-of-war. One could consider the car as behaving like another team of twenty more people.

Q5.17

This statement contradicts Newton’s 3rd law. The force that the locomotive exerted on the wall is the same as that exerted by the wall on the locomotive. The wall temporarily exerted on the locomotive a force greater than the force that the wall could exert without breaking.

Q5.18

The sack of sand moves up with the athlete, regardless of how quickly the athlete climbs. Since the athlete and the sack of sand have the same weight, the acceleration of the system must be zero.

Q5.19

The resultant force doesn’t always add to zero. If it did, nothing could ever accelerate. If we choose a single object as our system, action and reaction forces can never add to zero, as they act on different objects.

Q5.20

An object cannot exert a force on itself. If it could, then objects would be able to accelerate themselves, without interacting with the environment. You cannot lift yourself by tugging on your bootstraps.

Q5.21

To get the box to slide, you must push harder than the maximum static frictional force. Once the box is moving, you need to push with a force equal to the kinetic frictional force to maintain the box’s motion.

Q5.22

The stopping distance will be the same if the mass of the truck is doubled. The stopping distance will decrease by a factor of four if the initial speed is cut in half.

Q5.23

If you slam on the brakes, your tires will skid on the road. The force of kinetic friction between the tires and the road is less than the maximum static friction force. Anti-lock brakes work by “pumping” the brakes (much more rapidly that you can) to minimize skidding of the tires on the road.

Q5.24

With friction, it takes longer to come down than to go up. On the way up, the frictional force and the component of the weight down the plane are in the same direction, giving a large acceleration. On the way down, the forces are in opposite directions, giving a relatively smaller acceleration. If the incline is frictionless, it takes the same amount of time to go up as it does to come down.

Q5.25

(a)

The force of static friction between the crate and the bed of the truck causes the crate to accelerate. Note that the friction force on the crate is in the direction of its motion relative to the ground (but opposite to the direction of possible sliding motion of the crate relative to the truck bed).

(b)

It is most likely that the crate would slide forward relative to the bed of the truck.

Q5.26

In Question 25, part (a) is an example of such a situation. Any situation in which friction is the force that accelerates an object from rest is an example. As you pull away from a stop light, friction is the force that accelerates forward a box of tissues on the level floor of the car. At the same time, friction of the ground on the tires of the car accelerates the car forward.

120

The Laws of Motion

SOLUTIONS TO PROBLEMS The following problems cover Sections 5.1–5.6. Section 5.1

The Concept of Force

Section 5.2

Newton’s First Law and Inertial Frames

Section 5.3

Mass

Section 5.4

Newton’s Second Law

Section 5.5

The Gravitational Force and Weight

Section 5.6

Newton’s Third Law

P5.1

For the same force F, acting on different masses F = m 1 a1 and F = m2 a2 (a)

m1 a 1 = 2 = 3 m2 a1

(b)

F = m1 + m 2 a = 4m1 a = m1 3.00 m s 2

a

f

c

a = 0.750 m s *P5.2

v f = 880 m s, m = 25.8 kg , x f = 6 m v 2f = 2 ax f = 2 x f F=

P5.3

2

mv 2f 2x f

FG F IJ H mK

= 1.66 ×10 6 N forward

m = 3.00 kg

e

j

a = 2.00 i + 5.00 j m s 2

e6.00i + 15.0jj N a6.00f + a15.0f N =

∑ F = ma = ∑F =

2

2

16.2 N

h

Chapter 5

P5.4

Fg = weight of ball = mg v release = v and time to accelerate = t : a= (a)

∆v v v  = = i ∆t t t

Distance x = vt : x=

(b)

FG v IJ t = H 2K

Fg v  Fp − Fg j = i gt Fg v  i + Fg j gt

Fp =

P5.5

vt 2

e

j

m = 4.00 kg , v i = 3.00 i m s , v 8 = 8.00 i + 10.0 j m s , t = 8.00 s ∆v 5.00 i + 10.0 j a= = m s2 t 8.00 F = ma = 2.50 i + 5.00 j N

e

j

2

2

F = ( 2.50) +(5.00) = 5.59 N P5.6

(a)

Let the x-axis be in the original direction of the molecule’s motion.

e

v f = vi + at: −670 m s = 670 m s + a 3.00 × 10 −13 s

j

a = −4. 47 × 10 15 m s 2 (b)

For the molecule,

∑ F = ma . Its weight is negligible.

e

j

Fwall on molecule = 4.68 × 10 −26 kg −4.47 × 10 15 m s 2 = −2.09 × 10 −10 N G Fmolecule on wall = +2.09 × 10 −10 N

121

122 P5.7

The Laws of Motion

(a)

∑ F = ma and v 2f = vi2 + 2 ax f or a =

v 2f − vi2 2x f

.

Therefore,

∑F = m

ev

2 f

− vi2

j

2x f

∑ F = 9.11 × 10 −31 (b)

LMe7.00 × 10 kg N

5

j − e3.00 × 10 2b0.050 0 mg

m s2

2

5

m s2

j OPQ 2

= 3.64 × 10 −18 N .

The weight of the electron is

c

hc

h

Fg = mg = 9.11×10−31 kg 9.80 m s 2 = 8.93 ×10−30 N The accelerating force is 4.08 ×10 11 times the weight of the electron. P5.8

P5.9

a

f

(a)

Fg = mg = 120 lb = 4.448 N lb (120 lb)= 534 N

(b)

m=

Fg g

=

534 N = 54.5 kg 9.80 m s 2

Fg = mg = 900 N , m =

900 N = 91.8 kg 9.80 m s 2

cF h g

P5.10

on Jupiter

c

h

= 91.8 kg 25.9 m s 2 = 2.38 kN

Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just canceled out by a glass of tomato juice. By subtraction, Fg = mg p and Fg = mg C give

c h

c

p

c h

C

h

∆Fg = m g p − g C . For a person whose mass is 88.7 kg, the change in weight is

b

g

∆Fg = 88.7 kg 9.809 5 − 9.780 8 = 2.55 N . A precise balance scale, as in a doctor’s office, reads the same in different locations because it compares you with the standard masses on its beams. A typical bathroom scale is not precise enough to reveal this difference.

Chapter 5

P5.11

(a)

∑ F = F1 + F2 = e 20.0 i + 15.0 jj N ∑ F = ma:

20.0 i + 15.0 j = 5.00a a = 4.00 i + 3.00 j m s 2

e

j

or

a = 5.00 m s 2 at θ = 36.9° (b)

F2 x = 15.0 cos 60.0° = 7.50 N

FIG. P5.11

F2 y = 15.0 sin 60.0° = 13.0 N

e

j

F2 = 7.50 i + 13.0 j N

∑ F = F1 + F2 = e27.5 i + 13.0 jj N = ma = 5.00a a= P5.12

e5.50i + 2.60jj m s

2

= 6.08 m s 2 at 25.3°

We find acceleration: r f − ri = v i t +

1 2 at 2

a

f j

1 2 4.20 m i − 3.30 mj = 0+ a 1.20 s = 0.720 s 2 a 2 a = 5.83 i − 4.58 j m s 2 .

e

Now ∑ F = ma becomes Fg + F2 = ma

e

b

j

ge

j

F2 = 2.80 kg 5.83 i − 4.58 j m s 2 + 2.80 kg 9.80 m s 2 j F2 = P5.13

(a)

e16.3 i + 14.6 jj N

.

You and the earth exert equal forces on each other: m y g = M e a e . If your mass is 70.0 kg,

a70.0 kg fc9.80 m s h = = 2

ae (b)

5.98 ×10 24 kg

~ 10−22 m s 2 .

You and the planet move for equal times intervals according to x = 50.0 cm high, 2xy ay xe =

=

1 2 at . If the seat is 2

2xe ae

a

f

my 70.0 kg 0.500 m ae ~ 10 −23 m . xy = xy = ay me 5.98 × 10 24 kg

123

124 P5.14

The Laws of Motion

∑ F = ma reads

e−2.00 i + 2.00j + 5.00i − 3.00j − 45.0ij N = me3.75 m s ja 2

where a represents the direction of a

e−42.0 i − 1.00jj N = me3.75 m s ja 2

FG 1.00 IJ below the –x-axis H 42.0 K ∑ F = 42.0 N at 181° = mc3.75 m s ha . ∑F =

2

2

(42.0) +(1.00) N at tan−1 2

For the vectors to be equal, their magnitudes and their directions must be equal. (a)

∴ a is at 181° counterclockwise from the x-axis

(b)

m=

(d)

v f = v i + at = 0 + 3.75 m s 2 at 181° 10.0 s so v f = 37.5 m s at 181°

42.0 N = 11.2 kg 3.75 m s 2

e

j

v f = 37.5 m s cos 181° i + 37.5 m s sin 181° j so v f =

P5.15

(c)

v f = 37.5 2 + 0.893 2 m s = 37.5 m s

(a)

15.0 lb up

(b)

5.00 lb up

(c)

0

Section 5.7 P5.16

e−37.5 i − 0.893 jj m s

Some Applications of Newton’s Laws

dy dx = 10t , v y = = 9t 2 dt dt dv y dv x = 10 , a y = = 18t ax = dt dt vx =

At t = 2.00 s , a x = 10.0 m s 2 , a y = 36.0 m s 2

∑ Fx = ma x : 3.00 kg e10.0 ∑ Fy = ma y : 3.00 kg e36.0

j m s j = 108 N

m s 2 = 30.0 N 2

∑F=

Fx2 + Fy2 = 112 N

Chapter 5

P5.17

m = 1.00 kg

50.0 m

mg = 9.80 N

α

0.200 m tan α = 25.0 m α = 0.458°

0.200 m

T

T

Balance forces,

mg

2T sin α = mg T= P5.18

9.80 N = 613 N 2 sin α

FIG. P5.17

T3 = Fg

(1)

T1 sin θ 1 + T2 sin θ 2 = Fg

(2)

T1 cos θ 1 = T2 cos θ 2

(3)

θ2

θ1

Eliminate T2 and solve for T1 T1 =

bsinθ

Fg cos θ 2 1

cos θ 2 + cos θ 1 sin θ 2

Fg

g

=

Fg cos θ 2

b

sin θ 1 + θ 2

T3 = Fg = 325 N

FG cos 25.0° IJ = 296 N H sin 85.0° K F cos θ IJ = 296 NFG cos 60.0° IJ = =T G H cos 25.0° K H cos θ K

P5.19

1

1

2

See the solution for T1 in Problem 5.18.

T1

θ1

T1 = Fg T2

g

θ2 T3

163 N

FIG. P5.18

T2

125

126 P5.20

The Laws of Motion

(a)

An explanation proceeding from fundamental physical principles will be best for the parents and for you. Consider forces on the bit of string touching the weight hanger as shown in the free-body diagram: Horizontal Forces: Vertical Forces:

∑ Fx = ma x : −Tx + T cos θ = 0

∑ Fy = ma y : −Fg + T sin θ = 0

FIG. P5.20

You need only the equation for the vertical forces to find that the tension in the string is Fg . The force the child feels gets smaller, changing from T to T cos θ , while given by T = sin θ the counterweight hangs on the string. On the other hand, the kite does not notice what you are doing and the tension in the main part of the string stays constant. You do not need a level, since you learned in physics lab to sight to a horizontal line in a building. Share with the parents your estimate of the experimental uncertainty, which you make by thinking critically about the measurement, by repeating trials, practicing in advance and looking for variations and improvements in technique, including using other observers. You will then be glad to have the parents themselves repeat your measurements.

P5.21

Fg

e

(b)

T=

(a)

Isolate either mass

sin θ

=

0.132 kg 9.80 m s 2 sin 46.3°

j=

1.79 N

T + mg = ma = 0 T = mg . The scale reads the tension T, so

FIG. P5.21(a)

e

j

T = mg = 5.00 kg 9.80 m s 2 = 49.0 N . (b)

Isolate the pulley T2 + 2T1 = 0 T2 = 2 T1 = 2mg = 98.0 N .

(c)

∑ F = n + T + mg = 0 FIG. P5.21(b)

Take the component along the incline n x + Tx + mg x = 0 or 0 + T − mg sin 30.0° = 0 T = mg sin 30.0° = = 24.5 N .

a f

mg 5.00 9.80 = 2 2

FIG. P5.21(c)

Chapter 5

P5.22

127

The two forces acting on the block are the normal force, n, and the weight, mg. If the block is considered to be a point mass and the xaxis is chosen to be parallel to the plane, then the free body diagram will be as shown in the figure to the right. The angle θ is the angle of inclination of the plane. Applying Newton’s second law for the accelerating system (and taking the direction up the plane as the positive x direction) we have FIG. P5.22

∑ Fy = n − mg cos θ = 0: n = mg cos θ ∑ Fx = −mg sin θ = ma : a = −g sin θ (a)

When θ = 15.0°

a = −2.54 m s 2 (b)

Starting from rest

d

i

v 2f = vi2 + 2 a x f − xi = 2 ax f

e

ja

f

v f = 2 ax f = 2 −2.54 m s 2 −2.00 m = 3.18 m s P5.23

Choose a coordinate system with i East and j North.

∑ F = ma = 1.00 kg e10.0

j

m s 2 at 30.0°

a5.00 Nfj + F = a10.0 Nf∠30.0° = a5.00 Nfj + a8.66 Nfi 1

∴ F1 = 8.66 N (East ) *P5.24

FIG. P5.23

First, consider the block moving along the horizontal. The only force in the direction of movement is T. Thus, ∑ Fx = ma

a f

T = 5 kg a

n

(1)

Next consider the block that moves vertically. The forces on it are the tension T and its weight, 88.2 N. We have

5 kg

+x T

49 N

T

+y

9 kg Fg = 88.2 N

FIG. P5.24

∑ Fy = ma

a f

88.2 N − T = 9 kg a

(2)

Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be added to give 88.2 N = 14 kg a. Then

b

g

a = 6.30 m s 2 and T = 31.5 N .

128 P5.25

The Laws of Motion

After it leaves your hand, the block’s speed changes only because of one component of its weight:

∑ Fx = ma x Taking v f

− mg sin 20.0° = ma

d

i = 0 , v = 5.00 m s, and a = −g sina 20.0°f gives v 2f = vi2 + 2 a x f − xi .

i

a

2

fc

0 = (5.00) − 2(9.80) sin 20.0° x f − 0

h

or xf = P5.26

FIG. P5.25

25.0 = 3.73 m . 2(9.80) sin 20.0°

a

f

m1 = 2.00 kg , m 2 = 6.00 kg , θ = 55.0° (a)

∑ Fx = m 2 g sin θ − T = m 2 a and T − m1 g = m1 a a=

*P5.27

a

m 2 g sin θ − m1 g = 3.57 m s 2 m1 + m 2 FIG. P5.26

f

(b)

T = m1 a + g = 26.7 N

(c)

Since vi = 0 , v f = at = 3.57 m s 2 ( 2.00 s)= 7.14 m s .

c

h

We assume the vertical bar is in compression, pushing up on the pin with force A, and the tilted bar is in tension, exerting force B on the pin at −50° .

∑ Fx = 0:

−2 500 N cos 30°+ B cos 50° = 0

∑ Fy = 0:

−2 500 N sin 30°+ A − 3.37 × 10 3 N sin 50° = 0

B = 3.37 × 10 3 N A = 3.83 × 10 3 N

Positive answers confirm that B is in tension and A is in compression.

30° 2 500 N

50° A

B

2 500 N cos30°

B cos50° A

2 500 N sin30° FIG. P5.27

B sin50°

Chapter 5

P5.28

First, consider the 3.00 kg rising mass. The forces on it are the tension, T, and its weight, 29.4 N. With the upward direction as positive, the second law becomes

∑ Fy = ma y : T − 29.4 N = a3.00 kg fa

(1)

The forces on the falling 5.00 kg mass are its weight and T, and its acceleration is the same as that of the rising mass. Calling the positive direction down for this mass, we have

∑ Fy = ma y : 49 N − T = a5.00 kg fa

FIG. P5.28

(2)

Equations (1) and (2) can be solved simultaneously by adding them:

a

f a

f

T − 29.4 N + 49.0 N − T = 3.00 kg a + 5.00 kg a (b)

This gives the acceleration as a=

(a)

19.6 N = 2.45 m s 2 . 8.00 kg

Then

a

fc

h

T − 29.4 N = 3.00 kg 2.45 m s 2 = 7.35 N . The tension is T = 36.8 N . (c)

Consider either mass. We have y = vi t +

*P5.29

1 1 2 2 at = 0 + 2.45 m s 2 (1.00 s) = 1.23 m . 2 2

c

h

As the man rises steadily the pulley turns steadily and the tension in the rope is the same on both sides of the pulley. Choose man-pulleyand-platform as the system:

T

∑ Fy = ma y +T − 950 N = 0 T = 950 N . The worker must pull on the rope with force 950 N .

950 N FIG. P5.29

129

130 *P5.30

The Laws of Motion

Both blocks move with acceleration a =

a= (a)

FG m Hm

IJ K

− m1 g: 2 + m1 2

F 7 kg − 2 kg I 9.8 m s GH 7 kg + 2 kg JK

2

= 5.44 m s 2 .

Take the upward direction as positive for m1 .

d

i

2 + 2 a x x f − xi : v xf2 = v xi

b

0 = −2.4 m s xf = −

g + 2e5.44 m s jdx − 0i 2

2

5.76 m 2 s 2

e

2 5.44 m s 2

j

f

= −0.529 m

x f = 0.529 m below its initial level (b)

ja

e

v xf = v xi + a x t: v xf = −2.40 m s + 5.44 m s 2 1.80 s

f

v xf = 7. 40 m s upward P5.31

Forces acting on 2.00 kg block: T − m1 g = m 1 a

(1)

Forces acting on 8.00 kg block: Fx − T = m 2 a (a)

(2)

Eliminate T and solve for a: a=

Fx − m1 g m1 + m 2

a > 0 for Fx > m1 g = 19.6 N . (b)

Eliminate a and solve for T: T=

a

m1 Fx + m 2 g m1 + m 2

f FIG. P5.31

T = 0 for Fx ≤−m 2 g = −78.4 N . (c)

Fx , N ax , m s 2

–100

–78.4

–50.0

0

50.0

100

–12.5

–9.80

–6.96

–1.96

3.04

8.04

Chapter 5

*P5.32

(a)

For force components along the incline, with the upward direction taken as positive,

∑ Fx = ma x :

− mg sin θ = ma x

e

j

a x = − g sin θ = − 9.8 m s 2 sin 35° = −5.62 m s 2 . For the upward motion,

d

2 + 2 a x x f − xi v xf2 = v xi

b

0= 5 m s xf = (b)

i

g + 2e−5.62 m s jdx − 0i 2

2

25 m 2 s 2

e

j

2 5.62 m s 2

f

= 2.22 m .

The time to slide down is given by x f = xi + v xi t +

1 axt 2 2

0 = 2.22 m + 0 + t=

a

1 −5.62 m s 2 t 2 2

e

j

f = 0.890 s .

2 2.22 m 5.62 m s

2

For the second particle, x f = xi + v xi t +

1 axt 2 2

a

f e

ja

0 = 10 m + v xi 0.890 s + −5.62 m s 2 0.890 s v xi =

−10 m + 2.22 m = −8.74 m s 0.890 s

speed = 8.74 m s .

f

2

131

132 P5.33

The Laws of Motion

First, we will compute the needed accelerations:

a1f a 2f a3 f a4f

ay = 0 v yf − v yi 1. 20 m s − 0 = ay = t 0.800 s = 1.50 m s 2

Before it starts to move: During the first 0.800 s:

While moving at constant velocity: a y = 0 v yf − v yi 0 − 1.20 m s = During the last 1.50 s: ay = 1.50 s t = −0.800 m s 2

Newton’s second law is:

FIG. P5.33

∑ Fy = ma y

b

ge

j b

g S = 706 N + b72.0 kg ga

+S − 72.0 kg 9.80 m s 2 = 72.0 kg a y

P5.34

(a)

When a y = 0 , S = 706 N .

(b)

When a y = 1.50 m s 2 , S = 814 N .

(c)

When a y = 0 , S = 706 N .

(d)

When a y =−0.800 m s 2 , S = 648 N .

(a)

Pulley P1 has acceleration a 2 . Since m1 moves twice the distance P1 moves in the same time, m1 has twice the acceleration of P1 , i.e., a1 = 2 a 2 .

(b)

From the figure, and using

∑ F = ma:

m 2 g − T2 = m 2 a 2 T1 = m1 a1 = 2m1 a 2 T2 − 2T1 = 0

y

.

a1f a 2f a3 f

FIG. P5.34

Equation (1) becomes m 2 g − 2T1 = m 2 a 2 . This equation combined with Equation (2) yields

FG H

IJ K

T1 m 2m1 + 2 = m 2 g m1 2 T1 = (c)

m1 m 2 m1 m 2 g and T2 = g . 2m1 + 12 m 2 m1 + 14 m 2

From the values of T1 and T2 we find that a1 =

m2 g T1 = 2m1 + 12 m 2 m1

and a 2 =

m2 g 1 a1 = . 2 4m 1 + m 2

Chapter 5

Section 5.8 *P5.35

Forces of Friction +y

+y

n ground = Fg /2 = 85.0 lb

n tip

22.0° 22.0° F2 F1

f

+x

+x F = 45.8 lb

Fg = 170 lb Free-Body Diagram of Person

22.0°

Free-Body Diagram of Crutch Tip

FIG. P5.35 From the free-body diagram of the person,

∑ Fx = F1 sina22.0°f − F2 sina22.0°f = 0 , which gives F1 = F2 = F . Then, ∑ Fy = 2 F cos 22.0°+85.0 lbs − 170 lbs = 0 yields F = 45.8 lb. (a)

Now consider the free-body diagram of a crutch tip.

∑ Fx = f −( 45.8 lb) sin 22.0°= 0 , or f = 17. 2 lb .

∑ Fy = n tip −( 45.8 lb) cos 22.0°= 0 , which gives n tip = 42.5 lb . For minimum coefficient of friction, the crutch tip will be on the verge of slipping, so f 17.2 lb f = f s max = µ s n tip and µ s = = = 0.404 . n tip 42.5 lb

a f

(b)

As found above, the compression force in each crutch is F1 = F2 = F = 45.8 lb .

133

134 P5.36

The Laws of Motion

For equilibrium: f = F and n = Fg . Also, f = µ n i.e.,

µ= µs =

f F = n Fg 75.0 N = 0.306 25.0 9.80 N

a f

FIG. P5.36

and

µk = P5.37

∑ Fy = ma y :

60.0 N = 0.245 . 25.0(9.80) N

+n − mg = 0 fs ≤ µ sn = µ s mg

This maximum magnitude of static friction acts so long as the tires roll without skidding.

∑ Fx = ma x :

− f s = ma

The maximum acceleration is a = −µ s g . The initial and final conditions are: x i = 0 , vi = 50.0 mi h = 22.4 m s , v f = 0

d

i

v 2f = vi2 + 2 a x f − xi : − vi2 = −2 µ s gx f (a)

xf =

vi2 2 µg

xf =

a22.4 m sf = 2(0.100 )c9.80 m s h

xf =

vi2 2 µg

2

(b)

2

a22.4 m sf = 2(0.600)c9.80 m s h

256 m

2

xf =

2

42.7 m

Chapter 5

P5.38

If all the weight is on the rear wheels, (a)

F = ma: µ s mg = ma But ∆x = so µ s =

2 ∆x : gt 2

µs = (b) *P5.39

(a)

at 2 µ s gt 2 = 2 2

a

fb

2 0.250 mi 1 609 m mi

e9.80 m s ja4.96 sf 2

2

g=

3.34 .

Time would increase, as the wheels would skid and only kinetic friction would act; or perhaps the car would flip over. The person pushes backward on the floor. The floor pushes forward on the person with a force of friction. This is the only horizontal force on the person. If the person’s shoe is on the point of slipping the static friction force has its maximum value.

∑ Fx = ma x : ∑ Fy = ma y :

f = µ sn = ma x n − mg = 0

ma x = µ s mg x f = xi + v xi t +

1 ax t 2 2

e e

j j

a x = µ s g = 0.5 9.8 m s 2 = 4.9 m s 2 1 3 m = 0 + 0 + 4.9 m s 2 t 2 2

FIG. P5.39

t = 1.11 s

(b)

P5.40

xf =

2x f 2(3 m) 1 µ s gt 2 , t = = = 0.875 s 2 µs g (0.8) 9.8 m s 2

c

h

m suitcase = 20.0 kg , F = 35.0 N

∑ Fx = ma x : ∑ Fy = ma y : (a)

−20.0 N + F cos θ = 0 +n + F sin θ − Fg = 0

F cos θ = 20.0 N cos θ =

20.0 N = 0.571 35.0 N

θ = 55.2° (b)

n = Fg − F sin θ = 196 − 35.0(0.821) N n = 167 N

FIG. P5.40

135

136 P5.41

The Laws of Motion

m = 3.00 kg , θ = 30.0° , x = 2.00 m, t = 1.50 s (a)

x=

1 2 at : 2

a

f

1 2 a 1.50 s 2 4.00 a= = 1.78 m s 2 2 1.50

2.00 m =

FIG. P5.41

a f

∑ F = n + f + mg = m a : Along x: 0 − f + mg sin 30.0° = ma

b

f = m g sin 30.0°− a

g

Along y: n + 0 − mg cos 30.0° = 0 n = mg cos 30.0°

a

f

f m g sin 30.0°−a a , µ k = tan 30.0°− = = 0.368 n mg cos 30.0° g cos 30.0°

(b)

µk =

(c)

f = m g sin 30.0°−a , f = 3.00 9.80 sin 30.0°−1.78 = 9.37 N

(d)

v 2f = vi2 + 2 a x f − xi

a

a

f

c

f

h

where x f − xi = 2.00 m

a fa f

v 2f = 0 + 2 1.78 2.00 = 7.11 m 2 s 2 v f = 7.11 m 2 s 2 = 2.67 m s

Chapter 5

*P5.42

First we find the coefficient of friction:

∑ Fy = 0:

n

+n − mg = 0 f = µ sn = µ s mg

∑ Fx = ma x : − µ s mg = −

v 2f

mvi2 2 ∆x

=

vi2

+ 2 a x ∆x = 0

n

f

b

g ja

e

f

mg

mg sin10°

2

88 ft s v2 µs = i = = 0.981 2 g∆x 2 32.1 ft s 2 123 ft

f

mg cos10° FIG. P5.42

Now on the slope

∑ Fy = 0: ∑ Fx = ma x :

+n − mg cos 10° = 0 f s = µ sn = µ s mg cos 10° − µ s mg cos 10°+ mg sin 10° = − ∆x = =

P5.43

mvi2 2 ∆x

vi2 2 g µ s cos 10°− sin 10°

b

g

b88 ft sg = 2e32.1 ft s ja0.981 cos 10°− sin 10°f 2

2

152 ft .

T − f k = 5.00 a (for 5.00 kg mass) 9.00 g − T = 9.00 a (for 9.00 kg mass) Adding these two equations gives:

a f

137

a fa f

9.00 9.80 − 0.200 5.00 9.80 = 14.0 a a = 5.60 m s 2 ∴ T = 5.00 5.60 + 0.200 5.00 9.80

a f

a fa f

= 37.8 N

FIG. P5.43

138 P5.44

The Laws of Motion

Let a represent the positive magnitude of the acceleration −aj of m1 , of the acceleration −a i of m 2 , and of the acceleration +aj of m 3 . Call T12 the tension in the left rope and T23 the tension in the cord on the right. For m1 ,

∑ Fy = ma y

+T12 − m1 g = −m1 a

For m 2 ,

∑ Fx = ma x

−T12 + µ k n + T23 = −m 2 a

and

∑ Fy = ma y

n − m2 g = 0

for m 3 ,

∑ Fy = ma y

T23 − m 3 g = +m 3 a

n

T12

T23 f = µ kn m2 g

T12

T23

m1 g

m3 g

we have three simultaneous equations

b g − 0.350a9.80 N f − T = b1.00 kg ga +T − 19.6 N = b 2.00 kg ga . −T12 + 39.2 N = 4.00 kg a

+T12

23

23

(a)

FIG. P5.44

Add them up:

a

f

+39.2 N − 3. 43 N − 19.6 N = 7.00 kg a

a = 2.31 m s 2 , down for m1 , left for m 2 , and up for m 3 . (b)

a

fc

Now −T12 + 39. 2 N = 4.00 kg 2.31 m s 2

h

T12 = 30.0 N

a

fc

and T23 − 19.6 N = 2.00 kg 2.31 m s 2

h T23 = 24.2 N .

P5.45

(a)

See Figure to the right

(b)

68.0 − T − µm 2 g = m 2 a (Block #2) T − µm1 g = m1 a (Block #1)

T

m1 n1 m1

Adding,

g b g 68.0 a= bm + m g − µg = 1

1.29 m s 2

2

T = m1 a + µm1 g = 27. 2 N

T

m2

f2 = µ k n 2

m1 g = 118 N

68.0 − µ m1 + m 2 g = m1 + m 2 a

F n2

T

f1 = µ k n 1

b

m2

m2 g = 176 N

FIG. P5.45

F

Chapter 5

P5.46

(Case 1, impending upward motion) Setting

∑ Fx = 0:

P cos 50.0°−n = 0

fs , max = µ sn:

fs , max = µ s P cos 50.0°

a

f

= 0.250 0.643 P = 0.161 P Setting

∑ Fy = 0:

a f

P sin 50.0°−0.161P − 3.00 9.80 = 0 Pmax = 48.6 N

(Case 2, impending downward motion) As in Case 1, FIG. P5.46

fs, max = 0.161P Setting

∑ Fy = 0:

a f

P sin 50.0°+0.161P − 3.00 9.80 = 0 Pmin = 31.7 N

*P5.47

y

When the sled is sliding uphill

∑ Fy = ma y :

+n − mg cos θ = 0

∑ Fx = ma x :

+ mg sin θ + µ k mg cos θ = ma up

n

f = µ k n = µ k mg cos θ

mg sin θ

v f = 0 = vi + a up t up

mg cos θ

vi = − a up t up 1 vi + v f t up 2 1 1 2 ∆x = a up t up + 0 t up = a up t up 2 2 ∆x =

d e

x

i

FIG. P5.47

j

When the sled is sliding down, the direction of the friction force is reversed: mg sin θ − µ k mg cos θ = ma down ∆x=

1 2 a down t down . 2

Now t down = 2t up 1 1 2 = a down 2t up a up tup 2 2 a up = 4a down

e j

b

2

g sin θ + µ k g cos θ = 4 g sin θ − µ k g cos θ 5 µ k cos θ = 3 sin θ

µk =

FG 3 IJ tanθ H 5K

g

f

139

140 *P5.48

The Laws of Motion

Since the board is in equilibrium, ∑ Fx = 0 and we see that the normal forces must be the same on both sides of the board. Also, if the minimum normal forces (compression forces) are being applied, the board is on the verge of slipping and the friction force on each side is

a f

f = fs

max

n

f

f

n

= µ sn .

The board is also in equilibrium in the vertical direction, so

∑ Fy = 2 f − Fg = 0 , or

f=

Fg 2

.

Fg = 95.5 N FIG. P5.48

The minimum compression force needed is then n= *P5.49

(a)

f

µs

a

=

Fg 2µ s

=

95.5 N = 72.0 N . 2(0.663)

f

n + F sin 15°− 75 N cos 25° = 0

n

F 15° f s, max

∴ n = 67.97 − 0.259 F fs , max = µ s n = 24.67 − 0.094F 25° For equilibrium: F cos 15°+24.67 − 0.094F − 75 sin 25°= 0 . This gives F = 8.05 N .

75 N FIG. P5.49(a)

(b)

n

F cos 15°−( 24.67 − 0.094F )− 75 sin 25° = 0 .

F

This gives F = 53.2 N .

15° f s, max

25° 75 N

FIG. P5.49(b) (c)

f k = µ k n = 10.6 − 0.040 F . Since the velocity is constant, the net force is zero: F cos 15°−(10.6 − 0.040 F )− 75 sin 25° = 0 . This gives F = 42.0 N .

n

F 15°

fk

25° 75 N

FIG. P5.49(c)

Chapter 5

*P5.50

We must consider separately the disk when it is in contact with the roof and when it has gone over the top into free fall. In the first case, we take x and y as parallel and perpendicular to the surface of the roof:

∑ Fy = ma y :

+n − mg cos θ = 0 n = mg cos θ

then friction is f k = µ k n = µ k mg cos θ

∑ Fx = ma x :

FIG. P5.50

− f k − mg sin θ = ma x

a

f

a x = − µ k g cos θ − g sin θ = −0.4 cos 37°− sin 37° 9.8 m s 2 = −9.03 m s 2

The Frisbee goes ballistic with speed given by

i b

d

2 v xf2 = v xi + 2 a x x f − xi = 15 m s

g + 2e−9.03 m s ja10 m − 0f = 44.4 m 2

2

2

v xf = 6.67 m s For the free fall, we take x and y horizontal and vertical:

d

2 2 = v yi + 2 a y y f − yi v yf

b

i

g + 2e−9.8 m s jdy b4.01 m sg = 6.84 m = 6.02 m +

0 = 6.67 m s sin 37°

2

2

f

− 10 m sin 37°

i

2

yf

19.6 m s 2

Additional Problems P5.51

(a)

see figure to the right

(b)

First consider Pat and the chair as the system. Note that two ropes support the system, and T = 250 N in each rope. Applying ∑ F = ma 2T − 480 = ma , where m =

480 = 49.0 kg . 9.80 FIG. P5.51

Solving for a gives a= (c)

500 − 480 = 0.408 m s 2 . 49.0

∑ F = ma on Pat: 320

∑ F = n + T − 320 = ma , where m = 9.80 = 32.7 kg n = ma + 320 − T = 32.7(0.408)+ 320 − 250 = 83.3 N .

s2

141

142 P5.52

The Laws of Motion

∑ F = ma gives the object’s acceleration ∑ F = e8.00 i − 4.00tjj N 

a=



2.00 kg

m

dv a = 4.00 m s 2 i − 2.00 m s 3 tj = . dt

e

j e

j

Its velocity is

z

z

v

t

dv = v − v i = v − 0 = adt

vi

0

ze t

j e v = e 4.00t m s ji − e1.00t v=

0

2

(a)

j m s jj .

4.00 m s 2 i − 2.00 m s 3 tj dt 2

3

2

We require v = 15.0 m s , v = 225 m 2 s 2 16.0t 2 m 2 s 4 + 1.00t 4 m 2 s 6 = 225 m 2 s 2 1.00t 4 + 16.0 s 2 t 2 − 225 s 4 = 0 t2 =

−16.0 ±

a16.0f − 4a−225f = 9.00 s 2

2.00

t = 3.00 s . Take ri = 0 at t = 0. The position is

z ze t

t

0

0

r = vdt =

e

r = 4.00 m s 2 at t = 3 s we evaluate.

e18.0i − 9.00jj m

(c)

r=

(b)

So r = (18.0) +(9.00) m = 20.1 m

2

2

j e

j

4.00t m s 2 i − 1.00t 2 m s 3 j dt 2

3

j t2 i − e1.00 m s j t3 j 3

2

Chapter 5

*P5.53

(a)

y

Situation A

∑ Fx = ma x : ∑ Fy = ma y :

FA + µ sn − mg sin θ = 0 +n − mg cos θ = 0

a

FA = mg sin θ − µ s cos θ

fs

FA

mg cos θ FIG. P5.53(a)

f.

y

Situation B

∑ Fx = ma x : ∑ Fy = ma y :

FB cos θ + µ s n − mg sin θ = 0

FB

fs

a

a

mg sin θ − µ s cos θ

mg cos θ

mg sin θ

FIG. P5.53(b)

FB cos θ + µ s mg cos θ + µ s FB sin θ − mg sin θ = 0 FB =

f

cos θ + µ s sin θ

f

FA = 2 kg 9.8 m s 2 sin 25°−0.16 cos 25° = 5.44 N FB =

a

f

19.6 N 0.278 = 5.59 N cos 25°+0.16 sin 25°

Student A need exert less force. (d)

FB =

FA F = A cos 25°+0.38 sin 25° 1.07

Student B need exert less force.

b g P − Q = b3 kg ga Q = b 4 kg ga

P5.54

18 N − P = 2 kg a

a f

Adding gives 18 N = 9 kg a so

FIG. P5.54

a = 2.00 m s 2 . (b)

e

j

Q = 4 kg 2 m s 2 = 8.00 N net force on the 4 kg

e

j

P − 8 N = 3 kg 2 m s 2 = 6.00 N net force on the 3 kg and P = 14 N

e

j

18 N − 14 N = 2 kg 2 m s 2 = 4.00 N net force on the 2 kg continued on next page

x

n

− FB sin θ + n − mg cos θ = 0

Substitute n = mg cos θ + FB sin θ to find

(c)

x

n

mg sin θ

Eliminate n = mg cos θ to solve for

(b)

143

144

P5.55

The Laws of Motion

(c)

From above, Q = 8.00 N and P = 14.0 N .

(d)

The 3-kg block models the heavy block of wood. The contact force on your back is represented by Q, which is much less than the force F. The difference between F and Q is the net force causing acceleration of the 5-kg pair of objects. The acceleration is real and nonzero, but lasts for so short a time that it never is associated with a large velocity. The frame of the building and your legs exert forces, small relative to the hammer blow, to bring the partition, block, and you to rest again over a time large relative to the hammer blow. This problem lends itself to interesting lecture demonstrations. One person can hold a lead brick in one hand while another hits the brick with a hammer.

(a)

First, we note that F = T1 . Next, we focus on the mass M and write T5 = Mg . Next, we focus on the bottom pulley and write T5 = T2 + T3 . Finally, we focus on the top pulley and write T4 = T1 + T2 + T3 . Since the pulleys are not starting to rotate and are frictionless, T1 = T3 , and T2 = T3 . From this Mg information, we have T5 = 2T2 , soT2 = . 2 Then T1 = T2 = T3 =

Mg 3 Mg , and T4 = , and 2 2

T5 = Mg . (b)

Since F = T1 , we have F =

Mg . 2 FIG. P5.55

P5.56

We find the diver’s impact speed by analyzing his free-fall motion:

c

h

v 2f = vi2 + 2 ax = 0 + 2 −9.80 m s 2 (−10.0 m) so v f = −14.0 m s. Now for the 2.00 s of stopping, we have v f = vi + at :

a

0 = −14.0 m s + a 2.00 s

f

2

a = +7.00 m s . Call the force exerted by the water on the diver R. Using

e

j

∑ Fy = ma ,

e

+ R − 70.0 kg 9.80 m s 2 = 70.0 kg 7.00 m s 2 R = 1.18 kN .

j

Chapter 5

P5.57

(a)

145

The crate is in equilibrium, just before it starts to move. Let the normal force acting on it be n and the friction force, fs . Resolving vertically: n = Fg + P sin θ FIG. P5.57

Horizontally: P cos θ = fs But, fs ≤ µ sn i.e.,

c

P cos θ ≤ µ s Fg + P sin θ

h

or

a

f

P cos θ − µ s sin θ ≤ µ s Fg . Divide by cos θ :

a

f

P 1− µ s tan θ ≤ µ s Fg sec θ . Then Pminimum =

(b)

P=

µ s Fg sec θ 1 − µ s tan θ

.

0.400(100 N ) sec θ 1 − 0.400 tan θ

b g Pa N f

θ deg

0.00

15.0

30.0

45.0

60.0

40.0

46.4

60.1

94.3

260

If the angle were 68.2° or more, the expression for P would go to infinity and motion would become impossible.

146 P5.58

The Laws of Motion

(a)

Following the in-chapter Example about a block on a frictionless incline, we have

c

h

a = g sin θ = 9.80 m s 2 sin 30.0°

a = 4.90 m s 2 (b)

The block slides distance x on the incline, with sin 30.0° =

c

h

c

0.500 m x

h

x = 1.00 m: v 2f = vi2 + 2 a x f − xi = 0 + 2 4.90 m s 2 (1.00 m) v f = 3.13 m s after time t s =

(c)

Now in free fall y f − yi = v yi t +

2x f vf

=

2(1.00 m) 3.13 m s

= 0.639 s .

1 ayt 2 : 2

b

g

−2.00 = −3.13 m s sin 30.0° t −

e4.90 m s jt + b1.56 m sgt − 2.00 m = 0 2

1 9.80 m s 2 t 2 2

e

j

2

t=

−1.56 m s ±

b1.56 m sg − 4e4.90 m s ja−2.00 mf 2

9.80 m s 2

Only one root is physical t = 0.499 s

b

g

a

f

x f = v x t = 3.13 m s cos 30.0° 0.499 s = 1.35 m (d)

total time = t s + t = 0.639 s + 0.499 s = 1.14 s

(e)

The mass of the block makes no difference.

2

Chapter 5

P5.59

With motion impending, n + T sin θ − mg = 0

b

f = µ s mg − T sin θ

g

and T cos θ − µ s mg + µ sT sin θ = 0

FIG. P5.59

so T=

µ s mg . cos θ + µ s sin θ

To minimize T, we maximize cos θ + µ s sin θ

b

g

d cos θ + µ s sin θ = 0 = − sin θ + µ s cos θ . dθ

*P5.60

(a)

θ = tan−1 µ s = tan−1 0.350 = 19.3°

(b)

T=

a

fc

0.350 1.30 kg 9.80 m s 2 cos 19.3°+0.350 sin 19.3°

(a)

See Figure (a) to the right.

(b)

See Figure (b) to the right.

(c)

For the pin,

∑ Fy = ma y :

h=

4.21 N

a

fc

h

mg = 36.4 kg 9.8 m s 2 = 357 N

C cos θ − 357 N = 0 357 N . C= cos θ

For the foot,

FIG. P5.60(a)

∑ Fy = ma y :

FIG. P5.60(b)

+n B − C cos θ = 0 n B = 357 N .

(d)

For the foot with motion impending,

∑ Fx = ma x :

+ f s − C sin θ s = 0

µ sn B = C sin θ s 357 N cos θ s sin θ s C sin θ s µs = = = tan θ s . nB 357 N

b

(e)

The maximum coefficient is

µ s = tan θ s = tan 50.2° = 1.20 .

g

147

148 P5.61

The Laws of Motion

∑ F = ma For m1 : For m 2 :

T = m1 a T − m2 g = 0

Eliminating T, a=

m2 g m1

For all 3 blocks:

FIG. P5.61

a

f

F = M + m1 + m 2 a =

af

P5.62

e j

1

2

2

1

a f

t2 s2

ts

aM + m + m fFGH mm g IJK

xm

0

0

0

1.02

1.04 0

0.100

1.53

2.34 1

0.200

2.01

4.04 0

0.350

2.64

6.97 0

0.500

3.30

10.89

0.750

3.75

14.06

1.00 FIG. P5.62

From x =

1 2 1 at the slope of a graph of x versus t 2 is a , and 2 2

e

j

a = 2 × slope = 2 0.071 4 m s 2 = 0.143 m s 2 . From a ′ = g sin θ , a ′ = 9.80 m s 2

FG 1.77 4 IJ = 0.137 m s H 127.1 K

2

, different by 4%.

The difference is accounted for by the uncertainty in the data, which we may estimate from the third point as

b

ga f = 18%.

0.350 − 0.071 4 4.04 0.350

Chapter 5

P5.63

a

f

(1)

m1 a − A = T ⇒ a =

T +A m1

(2)

MA = R x = T ⇒ A =

T M

(3)

m2 a = m2 g − T ⇒ T = m2 g − a

(a)

b

g

FIG. P5.63

Substitute the value for a from (1) into (3) and solve for T:

LM F T + AI OP . N GH m JK Q

T = m2 g −

1

Substitute for A from (2):

LM F T + T I OP = N GH m M JK Q

T = m2 g − (b)

1

From (2), A =

1

1

2

a− A =

a

m 2 g m1 + M

a

f

m1 M + m 2 M + m1

f

.

T , Substitute the value of T: M A=

(d)

LM m M N m M + m am

Solve (3) for a and substitute value of T: a=

(c)

m2 g

Mm 2 g m1 M + m 2 m1 + M

a

f

m1 m 2 g m1 M + m 2 m1 + M

a

f

.

1

OP + Mf Q

.

149

150 P5.64

The Laws of Motion

(a), (b) Motion impending n = 49.0 N

n = 49.0 N

f s1 P

5.00 kg

15.0 kg

f s1 f s2

Fg = 49.0 N

147 N

196 N

a

fs1 = µn = 14.7 N

f

fs2 = 0.500 196 N = 98.0 N FIG. P5.64

P = f s1 + f s 2 = 14.7 N + 98.0 N = 113 N (c)

Once motion starts, kinetic friction acts.

a

f

a

f b

g

112.7 N − 0.100 49.0 N − 0.400 196 N = 15.0 kg a 2 a 2 = 1.96 m s 2

a

f b

g

0.100 49.0 N = 5.00 kg a1 a1 = 0.980 m s 2 *P5.65

(a)

Let x represent the position of the glider along the air track. Then z 2 = x 2 + h02 , 12 −1 2 dx 1 2 dz dz is the rate at which string passes = z − h02 2z x = z 2 − h02 , vx = . Now dt 2 dt dt over the pulley, so it is equal to v y of the counterweight.

e

j

e

j a f c

v x = z z 2 − h02 (b) (c)

ax =

h

−1 2

v y = uv y

dv y dv x du d at release from rest, v y = 0 and a x = ua y . = uv y = u + vy dt dt dt dt

80.0 cm , z = 1.60 m , u = z 2 − h02 z For the counterweight sin 30.0° =

e

∑ Fy = ma y :

j

−1 2

e

z = 1.6 2 − 0.8 2

j a1.6f = 1.15 . −1 2

T − 0.5 kg 9.8 m s 2 = −0.5 kga y a y = −2T + 9.8

For the glider

∑ Fx = ma x :

a

f

T cos 30° = 1.00 kg a x = 1.15 a y = 1.15 −2T + 9.8 = −2.31T + 11.3 N 3.18T = 11.3 N T = 3.56 N

Chapter 5

*P5.66

The upward acceleration of the rod is described by y f = yi + v yi t +

1 ayt 2 2

1 a y 8 × 10 −3 s 2 a y = 31.2 m s 2

e

1 × 10 −3 m = 0 + 0 +

j

2

The distance y moved by the rod and the distance x moved by the wedge in the same time are related y y . Then their speeds and by tan 15° = ⇒ x = x tan 15° accelerations are related by

FIG. P5.66

dy dx 1 = dt tan 15° dt and d2x dt

2

=

FG H

IJ K

d2y 1 1 = 31.2 m s 2 = 117 m s 2 . 2 tan 15° dt tan 15°

The free body diagram for the rod is shown. Here H and H ′ are forces exerted by the guide.

∑ Fy = ma y :

n cos 15°− mg = ma y

e

j

e

n cos 15°−0.250 kg 9.8 m s 2 = 0.250 kg 31.2 m s 2 10.3 N = 10.6 N n= cos 15°

j

For the wedge,

∑ Fx = Ma x :

e

−n sin 15°+ F = 0.5 kg 117 m s 2

a

f

j

F = 10.6 N sin 15°+58.3 N = 61.1 N *P5.67

(a)

Consider forces on the midpoint of the rope. It is nearly in equilibrium just before the car begins to move. Take the y-axis in the direction of the force you exert:

∑ Fy = ma y :

−T sin θ + f − T sin θ = 0 T=

(b)

T=

100 N = 410 N 2 sin 7°

f . 2 sin θ

FIG. P5.67

151

152 P5.68

The Laws of Motion

Since it has a larger mass, we expect the 8.00-kg block to move down the plane. The acceleration for both blocks should have the same magnitude since they are joined together by a non-stretching string. Define up the left hand plane as positive for the 3.50-kg object and down the right hand plane as positive for the 8.00-kg object.

∑ F1 = m1 a1 : ∑ F2 = m 2 a 2 :

− m1 g sin 35.0°+T = m1 a

FIG. P5.68

m 2 g sin 35.0°−T = m 2 a

and

a fa f a8.00fa9.80f sin 35.0°−T = 8.00a .

− 3.50 9.80 sin 35.0°+T = 3.50 a

Adding, we obtain

a

f

+45.0 N − 19.7 N = 11.5 kg a . (b)

Thus the acceleration is

a = 2.20 m s 2 . By substitution,

a

fc

h

−19.7 N + T = 3.50 kg 2.20 m s 2 = 7.70 N . (a)

The tension is T = 27.4 N .

P5.69

a = 5.00 m s 2

Choose the x-axis pointing down the slope.

a

v f = vi + at: 30.0 m s = 0 + a 6.00 s 2

f

a = 5.00 m s . Consider forces on the toy.

∑ Fx = ma x :

e

mg sin θ = m 5.00 m s 2

j

θ = 30.7° ∑ Fy = ma y : − mg cos θ + T = 0 T = mg cos θ = 0.100 9.80 cos 30.7° T = 0.843 N

a

fa f

FIG. P5.69

Chapter 5

*P5.70

153

Throughout its up and down motion after release the block has

∑ Fy = ma y :

+n − mg cos θ = 0 n = mg cos θ .

Let R = R x i + R y j represent the force of table on incline. We have

∑ Fx = ma x :

+ R x − n sin θ = 0 R x = mg cos θ sin θ ∑ Fy = ma y : − Mg − n cos θ + Ry = 0

R y = Mg + mg cos 2 θ .

e

j

R = mg cos θ sin θ to the right + M + m cos 2 θ g upward *P5.71

FIG. P5.70

Take +x in the direction of motion of the tablecloth. For the mug:

∑ Fx = ma x

0.1 N = 0.2 kg a x

a x = 0.5 m s 2 .

Relative to the tablecloth, the acceleration of the mug is 0.5 m s 2 − 3 m s 2 = −2.5 m s 2 . The mug reaches the edge of the tablecloth after time given by ∆ x = v xi t +

1 axt 2 2

1 −2.5 m s 2 t 2 2 t = 0.490 s .

e

−0.3 m = 0 +

j

The motion of the mug relative to tabletop is over distance

ja

1 1 a x t 2 = 0.5 m s 2 0.490 s 2 2

e

The tablecloth slides 36 cm over the table in this process.

f

2

= 0.060 0 m .

154 P5.72

The Laws of Motion

∑ Fy = ma y : n − mg cos θ = 0 or

a f n = a82.3 N f cos θ

n = 8.40 9.80 cos θ

∑ Fx = ma x : mg sin θ = ma or a = g sin θ

e

j

a = 9.80 m s 2 sin θ

θ , deg n , N 0.00

82.3

a, m s 2 0.00

5.00

82.0

0.854

10.0

81.1

1.70

15.0

79.5

2.54

20.0

77.4

3.35

25.0

74.6

4.14

30.0

71.3

4.90

35.0

67.4

5.62

40.0

63.1

6.30

45.0

58.2

6.93

50.0

52.9

7.51

55.0

47.2

8.03

60.0

41.2

8.49

65.0

34.8

8.88

70.0

28.2

9.21

75.0

21.3

9.47

80.0

14.3

9.65

85.0

7.17

9.76

90.0

0.00

9.80

FIG. P5.72

At 0°, the normal force is the full weight and the acceleration is zero. At 90°, the mass is in free fall next to the vertical incline.

Chapter 5

P5.73

(a)

Apply Newton’s second law to two points where butterflies are attached on either half of mobile (other half the same, by symmetry) (1) (2) (3) (4)

T2 cos θ 2 − T1 cos θ 1 = 0 T1 sin θ 1 − T2 sin θ 2 − mg = 0 T2 cos θ 2 − T3 = 0 T2 sin θ 2 − mg = 0

Substituting (4) into (2) for T2 sin θ 2 , T1 sin θ 1 − mg − mg = 0 .

FIG. P5.69

Then T1 =

2mg . sin θ 1

Substitute (3) into (1) for T2 cos θ 2 : T3 − T1 cos θ 1 = 0 , T3 = T1 cos θ 1 Substitute value of T1 : T3 = 2mg

2mg cos θ 1 = = T3 . sin θ 1 tan θ 1

From Equation (4), T2 = (b)

mg . sin θ 2

Divide (4) by (3): mg T2 sin θ 2 . = T2 cos θ 2 T3 Substitute value of T3 : tan θ 2 =

FG H

mg tan θ 1 tan θ 1 , θ 2 = tan−1 2 2mg

IJ K

.

Then we can finish answering part (a): T2 = (c)

mg sin

b

tan−1 12

tan θ 1

g

.

D is the horizontal distance between the points at which the two ends of the string are attached to the ceiling. D = 2A cos θ 1 + 2A cos θ 2 + A and L = 5A D=

RS T

LM N

FG H

1 L 2 cos θ 1 + 2 cos tan−1 tan θ 1 2 5

IJ OP + 1UV KQ W

155

156

The Laws of Motion

ANSWERS TO EVEN PROBLEMS P5.2

1.66 × 10 6 N forward

P5.4

P5.42

152 ft

Fg v  vt (a) ; (b) i + Fg j gt 2

P5.44

(a) 2.31 m s 2 down for m1 , left for m 2 and up for m 3 ; (b) 30.0 N and 24.2 N

(a) 4.47 × 10 15 m s 2 away from the wall; (b) 2.09 × 10 −10 N toward the wall

P5.46

Any value between 31.7 N and 48.6 N

P5.48

72.0 N

P5.8

(a) 534 N down; (b) 54.5 kg

P5.50

6.84 m

P5.10

2.55 N for an 88.7 kg person

P5.52

(a) 3.00 s; (b) 20.1 m; (c) 18.0 i − 9.00 j m

P5.12

e16.3i + 14.6jj N

P5.54

(a) 2.00 m s 2 to the right; (b) 8.00 N right on 4 kg; 6.00 N right on 3 kg; 4 N right on 2 kg; (c) 8.00 N between 4 kg and 3 kg; 14.0 N between 2 kg and 3 kg; (d) see the solution

P5.6

F I GH JK

e

j

P5.14

(a) 181°; (b) 11.2 kg; (c) 37.5 m s ; (d) −37.5 i − 0.893 j m s

P5.16

112 N

P5.18

T1 = 296 N ; T2 = 163 N ; T3 = 325 N

P5.56

1.18 kN

P5.20

(a) see the solution; (b) 1.79 N

P5.58

P5.22

(a) 2.54 m s 2 down the incline; (b) 3.18 m s

(a) 4.90 m s 2 ; (b) 3.13 m s at 30.0° below the horizontal; (c) 1.35 m; (d) 1.14 s; (e) No

P5.60

(a) and (b) see the solution; (c) 357 N; (d) see the solution; (e) 1.20

P5.24

see the solution; 6.30 m s 2 ; 31.5 N

P5.62

see the solution; 0.143 m s 2 agrees with

P5.26

(a) 3.57 m s 2 ; (b) 26.7 N; (c) 7.14 m s

P5.28

(a) 36.8 N; (b) 2.45 m s 2 ; (c) 1.23 m

P5.30

(a) 0.529 m; (b) 7.40 m s upward

P5.32

(a) 2.22 m; (b) 8.74 m s

P5.34

(a) a1 = 2 a 2 ; m1 m 2 g m m g ; T2 = 1 2m ; (b) T1 = m2 2m1 + 2 m1 + 42 m2 g m2 g ; a2 = (c) a1 = m2 4 m 2m1 + 2 1 + m2

e

j

P5.36

µ s = 0.306 ; µ k = 0.245

P5.38

(a) 3.34; (b) Time would increase

P5.40

(a) 55.2°; (b) 167 N

0.137 m s 2 P5.64

(a) see the solution; (b) on block one: 49.0 N j − 49.0 N j + 14.7 N i ; on block two: −49.0 N j − 14.7 N i − 147 N j +196 N j − 98.0 N i + 113 N i ; (c) for block one: 0.980 i m s 2 ; for block two: 1.96 m s 2 i

P5.66

61.1 N

P5.68

(a) 2.20 m s 2 ; (b) 27.4 N

P5.70

mg cos θ sin θ to the right

e

j

+ M + m cos 2 θ g upward P5.72

see the solution

6 Circular Motion and Other Applications of Newton’s Laws CHAPTER OUTLINE 6.1

6.2 6.3 6.4 6.5

Q6.4

Newton’s Second Law Applied to Uniform Circular Motion Nonuniform Circular Motion Motion in Accelerated Frames Motion in the Presence of Resistive Forces Numerical Modeling in Particle Dynamics

ANSWERS TO QUESTIONS Q6.1

Mud flies off a rapidly spinning tire because the resultant force is not sufficient to keep it moving in a circular path. In this case, the force that plays a major role is the adhesion between the mud and the tire.

Q6.2

The spring will stretch. In order for the object to move in a circle, the force exerted on the object by the spring must have a mv 2 . Newton’s third law says that the force exerted on size of r the object by the spring has the same size as the force exerted by the object on the spring. It is the force exerted on the spring that causes the spring to stretch.

Q6.3

Driving in a circle at a constant speed requires a centripetal acceleration but no tangential acceleration.

(a)

The object will move in a circle at a constant speed.

(b)

The object will move in a straight line at a changing speed.

Q6.5

The speed changes. The tangential force component causes tangential acceleration.

Q6.6

Consider the force required to keep a rock in the Earth’s crust moving in a circle. The size of the force is proportional to the radius of the circle. If that rock is at the Equator, the radius of the circle through which it moves is about 6400 km. If the rock is at the north pole, the radius of the circle through which it moves is zero!

Q6.7

Consider standing on a bathroom scale. The resultant force on you is your actual weight minus the normal force. The scale reading shows the size of the normal force, and is your ‘apparent weight.’ If you are at the North or South Pole, it can be precisely equal to your actual weight. If you are at the equator, your apparent weight must be less, so that the resultant force on you can be a downward force large enough to cause your centripetal acceleration as the Earth rotates.

Q6.8

A torque is exerted by the thrust force of the water times the distance between the nozzles.

157

158

Circular Motion and Other Applications of Newton’s Laws

Q6.9

I would not accept that statement for two reasons. First, to be “beyond the pull of gravity,” one would have to be infinitely far away from all other matter. Second, astronauts in orbit are moving in a circular path. It is the gravitational pull of Earth on the astronauts that keeps them in orbit. In the space shuttle, just above the atmosphere, gravity is only slightly weaker than at the Earth’s surface. Gravity does its job most clearly on an orbiting spacecraft, because the craft feels no other forces and is in free fall.

Q6.10

This is the same principle as the centrifuge. All the material inside the cylinder tends to move along a straight-line path, but the walls of the cylinder exert an inward force to keep everything moving around in a circular path.

Q6.11

The ball would not behave as it would when dropped on the Earth. As the astronaut holds the ball, she and the ball are moving with the same angular velocity. The ball, however, being closer to the center of rotation, is moving with a slower tangential velocity. Once the ball is released, it acts according to Newton’s first law, and simply drifts with constant velocity in the original direction of its velocity when released—it is no longer “attached” to the rotating space station. Since the ball follows a straight line and the astronaut follows a circular path, it will appear to the astronaut that the ball will “fall to the floor”. But other dramatic effects will occur. Imagine that the ball is held so high that it is just slightly away from the center of rotation. Then, as the ball is released, it will move very slowly along a straight line. Thus, the astronaut may make several full rotations around the circular path before the ball strikes the floor. This will result in three obvious variations with the Earth drop. First, the time to fall will be much larger than that on the Earth, even though the feet of the astronaut are pressed into the floor with a force that suggests the same force of gravity as on Earth. Second, the ball may actually appear to bob up and down if several rotations are made while it “falls”. As the ball moves in a straight line while the astronaut rotates, sometimes she is on the side of the circle on which the ball is moving toward her and other times she is on the other side, where the ball is moving away from her. The third effect is that the ball will not drop straight down to her feet. In the extreme case we have been imagining, it may actually strike the surface while she is on the opposite side, so it looks like it ended up “falling up”. In the least extreme case, in which only a portion of a rotation is made before the ball strikes the surface, the ball will appear to move backward relative to the astronaut as it falls.

Q6.12

The water has inertia. The water tends to move along a straight line, but the bucket pulls it in and around in a circle.

Q6.13

There is no such force. If the passenger slides outward across the slippery car seat, it is because the passenger is moving forward in a straight line while the car is turning under him. If the passenger pushes hard against the outside door, the door is exerting an inward force on him. No object is exerting an outward force on him, but he should still buckle his seatbelt.

Q6.14

Blood pressure cannot supply the force necessary both to balance the gravitational force and to provide the centripetal acceleration, to keep blood flowing up to the pilot’s brain.

Q6.15

The person in the elevator is in an accelerating reference frame. The apparent acceleration due to gravity, “g,” is changed inside the elevator. “g”= g ± a

Q6.16

When you are not accelerating, the normal force and your weight are equal in size. Your body interprets the force of the floor pushing up on you as your weight. When you accelerate in an elevator, this normal force changes so that you accelerate with the elevator. In free fall, you are never weightless since the Earth’s gravity and your mass do not change. It is the normal force—your apparent weight—that is zero.

Chapter 6

Q6.17

159

From the proportionality of the drag force to the speed squared and from Newton’s second law, we derive the equation that describes the motion of the skydiver: m

dv y dt

= mg −

Dρ A 2 vy 2

where D is the coefficient of drag of the parachutist, and A is the projected area of the parachutist’s body. At terminal speed, ay =

dv y dt

F 2mg I = 0 and V G H Dρ A JK T

12

.

When the parachute opens, the coefficient of drag D and the effective area A both increase, thus reducing the speed of the skydiver. Modern parachutes also add a third term, lift, to change the equation to m

dv y dt

= mg −

Dρ A 2 Lρ A 2 vy − vx 2 2

where v y is the vertical velocity, and v x is the horizontal velocity. The effect of lift is clearly seen in the “paraplane,” an ultralight airplane made from a fan, a chair, and a parachute. Q6.18

The larger drop has higher terminal speed. In the case of spheres, the text demonstrates that terminal speed is proportional to the square root of radius. When moving with terminal speed, an object is in equilibrium and has zero acceleration.

Q6.19

Lower air density reduces air resistance, so a tank-truck-load of fuel takes you farther.

Q6.20

Suppose the rock is moving rapidly when it enters the water. The speed of the rock decreases until it reaches terminal velocity. The acceleration, which is upward, decreases to zero as the rock approaches terminal velocity.

Q6.21

The thesis is false. The moment of decay of a radioactive atomic nucleus (for example) cannot be predicted. Quantum mechanics implies that the future is indeterminate. On the other hand, our sense of free will, of being able to make choices for ourselves that can appear to be random, may be an illusion. It may have nothing to do with the subatomic randomness described by quantum mechanics.

160

Circular Motion and Other Applications of Newton’s Laws

SOLUTIONS TO PROBLEMS Section 6.1 P6.1

Newton’s Second Law Applied to Uniform Circular Motion

m = 3.00 kg , r = 0.800 m. The string will break if the tension exceeds the weight corresponding to 25.0 kg, so

a f

Tmax = Mg = 25.0 9.80 = 245 N . When the 3.00 kg mass rotates in a horizontal circle, the tension causes the centripetal acceleration, T=

Then

v2 =

and

0 ≤ v ≤ 65.3

or P6.2

a f

3.00 v 2 mv 2 = . r 0.800

so

a

f a

a f

f

0.800 T 0.800 Tmax 0.800 245 rT = ≤ = = 65.3 m 2 s 2 m 3.00 3.00 3.00

FIG. P6.1

0 ≤ v ≤ 8.08 m s .

v2 , both m and r are unknown but remain constant. Therefore, ∑ F is proportional to v 2 r 2 18.0 and increases by a factor of as v increases from 14.0 m/s to 18.0 m/s. The total force at the 14.0 higher speed is then In

∑F = m

FG IJ H K

F 18.0 I a130 Nf = 215 N .

∑ Ffast = GH 14.0 JK Symbolically, write

Dividing gives

F mI

∑ Fslow = GH r JK b14.0

∑ Ffast ∑ Fslow

=

FG 18.0 IJ H 14.0 K

ms

g

2

2

and

F mI

∑ Ffast = GH r JK b18.0

g

2

ms .

2

, or

F 18.0 I

2

F 18.0 I a130 Nf =

∑ Ffast = GH 14.0 JK ∑ Fslow = GH 14.0 JK

2

215 N .

This force must be horizontally inward to produce the driver’s centripetal acceleration.

Chapter 6

P6.3

P6.4

e

je

2

(a)

(b)

2.20 × 10 6 m s v2 = = 9.13 × 10 22 m s 2 inward a= r 0.530 × 10 −10 m

e

j

Neglecting relativistic effects. F = ma c =

(b)

mv 2 r

e2.998 × 10 m sj kg j a0.480 mf 7

e

(a)

= 8.32 × 10 −8 N inward

2

F = 2 × 1.661 × 10 P6.5

j

9.11 × 0 −31 kg 2.20 × 10 6 m s mv 2 = F= r 0.530 × 10 −10 m

−27

2

= 6.22 × 10 −12 N

static friction

e j

ma i = f i + nj + mg − j

∑ Fy = 0 = n − mg v2 = f = µn = µmg . r 2 50.0 cm s v2 = = 0.085 0 . Then µ = rg 30.0 cm 980 cm s 2 thus n = mg and

∑ Fr = m

b

a

P6.6

(a)

P6.7

fe

∑ Fy = ma y , mg moon down = v = g moon r =

(b)

g

mv 2 down r

e1.52 m s je1.7 × 10 2

e

j

6

j

m + 100 × 10 3 m = 1.65 × 10 3 m s

j

2π 1.8 × 10 6 m 2πr v= ,T= = 6.84 × 10 3 s = 1.90 h T 1.65 × 10 3 m s

n = mg since a y = 0 The force causing the centripetal acceleration is the frictional force f. From Newton’s second law f = ma c =

mv 2 . r

But the friction condition is f ≤ µ sn i.e.,

FIG. P6.7

mv 2 ≤ µ s mg r

a

fe

v ≤ µ s rg = 0.600 35.0 m 9.80 m s 2

j

v ≤ 14.3 m s

161

162

Circular Motion and Other Applications of Newton’s Laws 2

v = r

b86.5 km hge

1h 3 600 s

je

1 000 m 1 km

2

j F 1g I= GH 9.80 m s JK

P6.8

a=

P6.9

T cos 5.00° = mg = 80.0 kg 9.80 m s 2

2

61.0 m

b

ge

0.966 g

j

a68.6 Nfi + a784 Nfj

(a)

T = 787 N : T =

(b)

T sin 5.00° = ma c : a c = 0.857 m s 2 toward the center of the circle. The length of the wire is unnecessary information. We could, on the other hand, use it to find the radius of the circle, the speed of the bob, and the period of the motion.

P6.10

(b)

v=

235 m = 6.53 m s 36.0 s

The radius is given by

1 2πr = 235 m 4 r = 150 m

(a)

F v I toward center GH r JK b6.53 m sg at 35.0° north of west = 2

ar =

2

150 m

e

je

e j

j

= 0.285 m s 2 cos 35.0° − i + sin 35.0° j = −0.233 m s 2 i + 0.163 m s 2 j

(c)

a= =

dv

f

− vi

i

t

e6.53 m s j − 6.53 m s ij

36.0 s = −0.181 m s 2 i + 0.181 m s 2 j

FIG. P6.9

Chapter 6

*P6.11

b ge

j

Fg = mg = 4 kg 9.8 m s 2 = 39.2 N

Ta

1.5 m 2m θ = 48.6° r = 2 m cos 48.6° = 1.32 m

θ

sin θ =

39.2 N

a f

mv 2 ∑ Fx = ma x = r Ta cos 48.6°+Tb cos 48.6° = Ta + Tb =

163

Tb

b gb

4 kg 6 m s

109 N = 165 N cos 48.6°

g

forces

ac

v

2

1.32 m

motion FIG. P6.11

∑ Fy = ma y +Ta sin 48.6°−Tb sin 48.6°−39.2 N = 0 Ta − Tb = (a)

39. 2 N = 52.3 N sin 48.6° To solve simultaneously, we add the equations in Ta and Tb : Ta + Tb + Ta − Tb = 165 N + 52.3 N Ta =

(b) *P6.12

217 N = 108 N 2

Tb = 165 N − Ta = 165 N − 108 N = 56.2 N

v2 . Let f represent the rotation rate. Each revolution carries each bit of metal through distance r 2πr , so v = 2πrf and

ac =

ac =

v2 = 4π 2 rf 2 = 100 g . r

A smaller radius implies smaller acceleration. To meet the criterion for each bit of metal we consider the minimum radius:

F 100 g IJ f =G H 4π r K 2

12

F 100 ⋅ 9.8 m s I =G H 4π a0.021 mf JK 2

2

12

= 34.4

FG H

IJ K

1 60 s = 2.06 × 10 3 rev min . s 1 min

164

Circular Motion and Other Applications of Newton’s Laws

Section 6.2 P6.13

Nonuniform Circular Motion

M = 40.0 kg , R = 3.00 m, T = 350 N (a)

T

Mv 2 R R M

∑ F = 2T − Mg =

gFGH IJK F 3.00 IJ = 23.1 em s j = 700 − a 40.0fa9.80f G H 40.0 K b

v 2 = 2T − Mg v2

2

2

n − Mg = F = n = Mg +

P6.14

(a)

Mv 2 R

child + seat

child alone

FIG. P6.13(a)

FIG. P6.13(b)

IJ K

FG H

Mv 2 23.1 = 40.0 9.80 + = 700 N R 3.00

Consider the forces acting on the system consisting of the child and the seat:

∑ Fy = ma y ⇒ 2T − mg = m v2 = R v= (b)

FG 2T − gIJ Hm K F 2T − gIJ RG Hm K

v2 R

Consider the forces acting on the child alone:

F

∑ Fy = ma y ⇒ n = mGH g + and from above, v 2 = R

v2 R

I JK

FG 2T − gIJ , so Hm K FG H

n=m g+ P6.15

n

Mg

v = 4.81 m s (b)

Mg

T

IJ K

2T − g = 2T . m

Let the tension at the lowest point be T.

∑ F = ma:

T − mg = ma c =

F GH

mv 2 r

I JK L b8.00 m sg T = b85.0 kg gM9.80 m s + 10.0 m MN T=m g+

v2 r

2

2

OP PQ = 1.38 kN > 1 000 N

He doesn’t make it across the river because the vine breaks.

FIG. P6.15

Chapter 6

P6.16

b

(a)

4.00 m s v2 = ac = r 12.0 m

(b)

a = a c2 + a t2 a=

a1.33f + a1.20f 2

g

2

= 1.33 m s 2

2

at an angle θ = tan −1

= 1.79 m s 2

FG a IJ = Ha K c

FIG. P6.16

48.0° inward

t

P6.17

∑ Fy =

mv 2 = mg + n r

But n = 0 at this minimum speed condition, so mv 2 = mg ⇒ v = gr = r P6.18

e9.80 m s ja1.00 mf = 2

3.13 m s .

At the top of the vertical circle, T=m

a

(a)

f a40..00500f − a0.400fa9.80f =

v2 − mg R

2

or T = 0.400 P6.19

FIG. P6.17

8.88 N

v = 20.0 m s,

B

C

n = force of track on roller coaster, and

15 m

10 m

R = 10.0 m . A

Mv 2 ∑ F = R = n − Mg

FIG. P6.19

From this we find n = Mg +

b

ge

500 kg 20.0 m s 2 Mv 2 = 500 kg 9.80 m s 2 + R 10.0 m

b

ge

j

n = 4 900 N + 20 000 N = 2.49 × 10 4 N (b)

At B, n − Mg = −

Mv 2 R

The max speed at B corresponds to n=0 − Mg = −

2 Mv max ⇒ v max = Rg = 15.0 9.80 = 12.1 m s R

a f

j

165

166 P6.20

Circular Motion and Other Applications of Newton’s Laws

b e

g

2

(a)

v2 ac = r

(b)

Let n be the force exerted by the rail.

13.0 m s v2 = = 8.62 m r= a c 2 9.80 m s 2

j

Newton’s law gives Mv 2 Mg + n = r v2 n=M − g = M 2 g − g = Mg , downward r

I JK

F GH

(c)

ac =

v2 r

ac =

b13.0 m sg

b

FIG. P6.20

g

2

20.0 m

= 8.45 m s 2

If the force exerted by the rail is n1 then

Mv 2 = Ma c r n1 = M a c − g which is < 0 since a c = 8.45 m s 2 n1 + Mg =

b

g

Thus, the normal force would have to point away from the center of the curve. Unless they have belts, the riders will fall from the cars. To be safe we must require n1 to be positive. Then a c > g . We need v2 > g or v > rg = r

Section 6.3 P6.21

(a)

a20.0 mfe9.80 m s j , v > 14.0 m s . 2

Motion in Accelerated Frames T

18.0 N

∑ Fx = Ma , a = M = 5.00 kg =

3.60 m s 2

to the right. (b)

If v = const, a = 0, so T = 0 (This is also an equilibrium situation.)

(c)

Someone in the car (noninertial observer) claims that the forces on the mass along x are T and a fictitious force (–Ma). Someone at rest outside the car (inertial observer) claims that T is the only force on M in the x-direction.

5.00 kg

FIG. P6.21

167

Chapter 6

*P6.22

We adopt the view of an inertial observer. If it is on the verge of sliding, the cup is moving on a circle with its centripetal acceleration caused by friction.

∑ Fy = ma y :

+n − mg = 0

∑ Fx = ma x :

f=

n f

mv 2 = µ sn = µ s mg r

ja

e

mg FIG. P6.22

f

v = µ s gr = 0.8 9.8 m s 2 30 m = 15.3 m s

If you go too fast the cup will begin sliding straight across the dashboard to the left. P6.23

The only forces acting on the suspended object are the force of gravity mg and the force of tension T, as shown in the free-body diagram. Applying Newton’s second law in the x and y directions,

or (a)

∑ Fx = T sin θ = ma ∑ Fy = T cos θ − mg = 0

(1)

T cos θ = mg

(2)

T cos θ T sin θ mg FIG. P6.23

Dividing equation (1) by (2) gives tan θ =

a 3.00 m s 2 = = 0.306 . g 9.80 m s 2

Solving for θ, θ = 17.0° (b)

From Equation (1), T=

*P6.24

a

fc a f

h

0.500 kg 3.00 m s 2 ma = = 5.12 N . sin θ sin 17.0°

The water moves at speed v=

a

f

2πr 2π 0.12 m = = 0.104 m s . T 7.25 s

The top layer of water feels a downward force of gravity mg and an outward fictitious force in the turntable frame of reference,

b

mv 2 m 0.104 m s = r 0.12 m

g

2

= m9.01 × 10 −2 m s 2 .

It behaves as if it were stationary in a gravity field pointing downward and outward at tan −1

0.090 1 m s 2 9.8 m s 2

= 0.527° .

Its surface slopes upward toward the outside, making this angle with the horizontal.

168 P6.25

Circular Motion and Other Applications of Newton’s Laws

Fmax = Fg + ma = 591 N Fmin = Fg − ma = 391 N (a)

Adding, 2 Fg = 982 N, Fg = 491 N

(b)

Since Fg = mg , m =

(c)

Subtracting the above equations,

491 N = 50.1 kg 9.80 m s 2

∴ a = 2.00 m s 2

2ma = 200 N P6.26

(a)

∑ Fr = mar mg = g= T=

(b) *P6.27

FG H

mv 2 m 2πR = R R T

IJ K

2

4π 2 R T2 4π 2 R 6.37 × 10 6 m = 2π = 5.07 × 10 3 s = 1.41 h g 9.80 m s 2

speed increase factor =

FG H

IJ K

v new T 2πR Tcurrent 24.0 h = = current = = 17.1 2πR 1.41 h v current Tnew Tnew

The car moves to the right with acceleration a. We find the acceleration of a b of the block relative to the Earth. The block moves to the right also.

∑ Fy = ma y : ∑ Fx = ma x :

+n − mg = 0 , n = mg , f = µ k mg + µ k mg = ma b , a b = µ k g

The acceleration of the block relative to the car is a b − a = µ k g − a . In this frame the block starts from rest and undergoes displacement −A and gains speed according to

d

2 + 2 a x x f − xi v xf2 = v xi

v xf2 (a)

d b

v = 2A a − µ k g

continued on next page

gi

12

b

ga

i

f b

g

= 0 + 2 µ k g − a − A − 0 = 2A a − µ k g .

to the left

Chapter 6

(b)

169

The time for which the box slides is given by 1 v xi + v xf t 2 1 −A = 0 − 2A a − µ k g 2

d i LM d b gi OPQt N F 2A I . t=G H a − µ g JK

∆x =

12

12

k

The car in the Earth frame acquires finals speed v xf = v xi

F 2A I + at = 0 + aG H a − µ g JK

12

. The speed

k

of the box in the Earth frame is then

b

g − a 2 Af b a − µ g g + a 2 A f = ba − µ g g

v be = v bc + v ce = − 2A a − µ k g 12

12

12

k

+a

a

12

*P6.28

b

µ k g 2A

2A a − µ k g

g

12

=

12

k

12

k

2 µ k gA . v

Consider forces on the backpack as it slides in the Earth frame of reference.

∑ Fy = ma y : ∑ Fx = ma x :

We solve for µ k : vt − L =

b

g

b

+n − mg = ma , n = m g + a , f k = µ k m g + a

b

g

− µ k m g + a = ma x

The motion across the floor is described by L = vt +

P6.29

=

12

k

k

=

F 2A I GH a − µ g JK µ g a 2 Af ba − µ g g

a

f

b

g

g

1 1 a x t 2 = vt − µ k g + a t 2 . 2 2

2 vt − L 1 = µk . µk g + a t2 , 2 g + a t2

b

g

b

g

In an inertial reference frame, the girl is accelerating horizontally inward at

b

5.70 m s v2 = r 2.40 m

g

2

= 13.5 m s 2

In her own non-inertial frame, her head feels a horizontally outward fictitious force equal to its mass times this acceleration. Together this force and the weight of her head add to have a magnitude equal to the mass of her head times an acceleration of g2 +

Fv I GH r JK 2

2

=

a9.80f + a13.5f 2

2

m s 2 = 16.7 m s 2

16.7 = 1.71 . 9.80 Thus, the force required to lift her head is larger by this factor, or the required force is

This is larger than g by a factor of

a

f

F = 1.71 55.0 N = 93.8 N .

170 *P6.30

Circular Motion and Other Applications of Newton’s Laws

(a)

The chunk is at radius r =

0.137 m + 0.080 m = 0.054 2 m . Its speed is 4 v=

20 000 2πr = 2π 0.054 2 m = 114 m s 60 s T

b

g

and its acceleration ac =

b

g

2

114 m s v2 = = 2.38 × 10 5 m s 2 horizontally inward r 0.054 2 m

= 2.38 × 10 5 m s 2 (b)

F g I= GH 9.8 m s JK 2

2.43 × 10 4 g .

In the frame of the turning cone, the chunk feels a mv 2 . In this frame its horizontally outward force of r 3.3 cm acceleration is up along the cone, at tan −1 a13 .7 − 8 f cm = 49.2° .

e

+n −

je

mv 2 r

f

49.2°

2

Take the y axis perpendicular to the cone:

∑ Fy = ma y :

a

n

FIG. P6.30(b)

mv 2 sin 49.2° = 0 r

j

n = 2 × 10 −3 kg 2.38 × 10 5 m s 2 sin 49.2° = 360 N (c)

a

f

f = µ k n = 0.6 360 N = 216 N

∑ Fx = ma x :

e2 × 10

−3

je

2

mv cos 49. 2°− f = ma x r

j

e

j

kg 2.38 × 10 5 m s 2 cos 49.2°−216 N = 2 × 10 −3 kg a x

a x = 47.5 × 10 4 m s 2 radially up the wall of the cone P6.31

F 4π R I cos 35.0° = 0.027 6 m s GH T JK We take the y axis along the local vertical. ba g = 9.80 − ba g = 9.78 m s ba g = 0.015 8 m s ar =

2

2

net y net x

θ = arctan

2

e

N

2

r y

2

ax = 0.092 8° ay

35.0° a r

(exaggerated size)

θ g0 a net 35.0°

Equator

FIG. P6.31

Chapter 6

Section 6.4 P6.32

Motion in the Presence of Resistive Forces

m = 80.0 kg , vT = 50.0 m s , mg = (a)

DρAvT2 DρA mg ∴ = 2 = 0.314 kg m 2 2 vT

v = 30.0 m s

At

a=g− (b)

171

DρAv 2 2

m

= 9.80 −

a0.314fa30.0f 80.0

2

= 6. 27 m s 2 downward

At v = 50.0 m s , terminal velocity has been reached. ∑ Fy = 0 = mg − R

b

ge

j

⇒ R = mg = 80.0 kg 9.80 m s 2 = 784 N directed up (c)

v = 30.0 m s DρAv 2 = 0.314 30.0 2

At

a

P6.33

(a)

fa f

2

= 283 N upward

a = g − bv When v = vT , a = 0 and g = bvT

b=

g vT

The Styrofoam falls 1.50 m at constant speed vT in 5.00 s.

P6.34

y 1.50 m = = 0.300 m s 5.00 s t

Thus,

vT =

Then

b=

(b)

At t = 0 , v = 0 and

a = g = 9.80 m s 2 down

(c)

When v = 0.150 m s, a = g − bv = 9.80 m s 2 − 32.7 s −1 0.150 m s = 4.90 m s 2

(a)

ρ=

9.80 m s 2 = 32.7 s −1 0.300 m s

jb

e

g

down

m 1 , A = 0.020 1 m 2 , R = ρ air ADvT2 = mg V 2 m = ρ beadV = 0.830 g cm 3

LM 4 π a8.00 cmf OP = 1.78 kg N3 Q 3

Assuming a drag coefficient of D = 0.500 for this spherical object, and taking the density of air at 20°C from the endpapers, we have

vT =

(b)

v 2f

=

vi2

b ge j = 0.500e1.20 kg m je0.020 1 m j 2 1.78 kg 9.80 m s 2 3

b53.8 m sg = = + 2 gh = 0 + 2 gh : h = 2 g 2e9.80 m s j v 2f

2

2

2

148 m

53.8 m s

172 P6.35

Circular Motion and Other Applications of Newton’s Laws

Since the upward velocity is constant, the resultant force on the ball is zero. Thus, the upward applied force equals the sum of the gravitational and drag forces (both downward): F = mg + bv . The mass of the copper ball is

FG IJ e HK

4πρr 3 4 = π 8.92 × 10 3 kg m3 2.00 × 10 −2 m 3 3

m=

je

j

3

= 0.299 kg .

The applied force is then

a

fa f a

fe

j

F = mg + bv = 0.299 9.80 + 0.950 9.00 × 10 −2 = 3.01 N . P6.36

∑ Fy = ma y +T cos 40.0°− mg = 0

b620 kgge9.80 m s j = 7.93 × 10 2

T=

3

cos 40.0° F ma = ∑ x x

N

− R + T sin 40.0° = 0

e

j

R = 7.93 × 10 3 N sin 40.0° = 5.10 × 10 3 N =

D=

P6.37

(a)

e

2 5.10 × 10 3 N

jFH

kg m s 2 N

IK

1 DρAv 2 2

2R = ρAv 2 1. 20 kg m 2 3.80 m 2 40.0 m s

e

je

jb

e

P6.38

= 1. 40

e

je

j

3.00 × 10 −3 kg 9.80 m s 2 mg = = 1.47 N ⋅ s m vT 2.00 × 10 −2 m s

In the equation describing the time variation of the velocity, we have v = vT 1 − e − bt m

(c)

2

R = vT b = mg

At terminal velocity,

∴b = (b)

g

FIG. P6.36

j

v = 0.632 vT when e − bt m = 0.368

FG m IJ lna0.368f = HbK

2.04 × 10 −3 s

or at time

t=−

At terminal velocity,

R = vT b = mg = 2.94 × 10 −2 N

The resistive force is

a

fe

jb

1 1 DρAv 2 = 0.250 1.20 kg m3 2.20 m 2 27.8 m s 2 2 R = 255 N R=

a=−

255 N R =− = −0.212 m s 2 m 1200 kg

je

g

2

Chapter 6

P6.39

af

a

v t = vi e − ct

(a)

f

v 20.0 s = 5.00 = vi e −20 .0 c , vi = 10.0 m s .

(b)

At t = 40.0 s

FG 1 IJ H 2K v = b10.0 m sge

(c)

v = vi e − ct

s=

So 5.00 = 10.0 e −20 .0 c and −20.0 c = ln

P6.40

∑ F = ma − kmv 2 = m − kdt =

dv v2

c=−

−40 .0 c

c h=

ln

1 2

20.0

b

3.47 × 10 −2 s −1

f

ga

= 10.0 m s 0.250 = 2.50 m s

dv = − cvi e − ct = − cv dt

dv dt

z z t

v

0

v0

− k dt = v −2 dv

a f

−k t − 0 =

v −1 −1

v

=− v0

1 1 + v v0

1 + v 0 kt 1 1 = + kt = v v0 v0 v0 v= 1 + v 0 kt *P6.41

(a)

From Problem 40, v=

v0 dx = dt 1 + v 0 kt

z z x

t

dx = v 0

0

0

z t

dt 1 v 0 kdt = 1 + v 0 kt k 0 1 + v 0 kt

b b

g g

t 1 ln 1 + v 0 kt 0 k 1 x − 0 = ln 1 + v 0 kt − ln 1 k 1 x = ln 1 + v 0 kt k x

x0 =

b

(b)

*P6.42

b

g

g

We have ln 1 + v 0 kt = kx v0 v = kx0 = v 0 e − kx = v 1 + v 0 kt = e kx so v = 1 + v 0 kt e

We write − kmv 2 = −

1 DρAv 2 so 2 k=

e

je

j

−3 3 2 DρA 0.305 1.20 kg m 4.2 × 10 m = = 5.3 × 10 −3 m 2m 2 0.145 kg

b

b

g

v = v 0 e − kx = 40.2 m s e

e

− 5.3 ×10

−3

g

ja

m 18 .3 m

f=

36.5 m s

173

174 P6.43

Circular Motion and Other Applications of Newton’s Laws

a

fa

f

1 DρAv 2 , we estimate that D = 1.00 , ρ = 1.20 kg m3 , A = 0.100 m 0.160 m = 1.60 × 10 −2 m 2 2 and v = 27.0 m s. The resistance force is then

In R =

R=

a fe

jb

1 1.00 1.20 kg m3 1.60 × 10 −2 m 2 27.0 m s 2

je

g

2

= 7.00 N

or

R ~ 10 1 N

Section 6.5

Numerical Modeling in Particle Dynamics

a

f af a

f

Note: In some problems we compute each new position as x t + ∆t = x t + v t + ∆t ∆t , rather than x t + ∆t = x t + v t ∆t as quoted in the text. This method has the same theoretical validity as that presented in the text, and in practice can give quicker convergence.

a

P6.44

f af af (a)

(b)

At v = vT , a = 0, − mg + bvT = 0

af

ts 0 0.005 0.01 0.015

vT =

xm

a f

vms

2 2 1.999 755 1.999 3

0 –0.049 –0.095 55 –0.139 77

e

je

j

3.00 × 10 −3 kg 9.80 m s 2 mg = = 0.980 m s b 3.00 × 10 −2 kg s

a f

b g

F mN –29.4 –27.93 –26.534 –25.2

e

a m s2

j

–9.8 –9.31 –8.844 5 –8.40

. . . we list the result after each tenth iteration 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65

1.990 1.965 1.930 1.889 1.845 1.799 1.752 1.704 1.65 1.61 1.56 1.51 1.46

–0.393 –0.629 –0.770 –0.854 –0.904 –0.935 –0.953 –0.964 –0.970 –0.974 –0.977 –0.978 –0.979

–17.6 –10.5 –6.31 –3.78 –2.26 –1.35 –0.811 –0.486 –0.291 –0.174 –0.110 –0.062 4 –0.037 4

–5.87 –3.51 –2.10 –1.26 –0.754 –0.451 –0.270 –0.162 –0.096 9 –0.058 0 –0.034 7 –0.020 8 –0.012 5

Terminal velocity is never reached. The leaf is at 99.9% of vT after 0.67 s. The fall to the ground takes about 2.14 s. Repeating with ∆t = 0.001 s , we find the fall takes 2.14 s.

Chapter 6

P6.45

(a)

When v = vT , a = 0,

∑ F = − mg + CvT2 = 0

mg =− vT = − C

af

(b)

ts

a f

e4.80 × 10

−4

je

kg 9.80 m s 2

2.50 × 10

−5

kg m

a f

b g

xm 0 0 –0.392 –1.168 –2.30 –3.77 –5.51 –7.48 –9.65 –11.96 –14.4

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2

175

j=

−13.7 m s

e

vms

F mN

a m s2

0 –1.96 –3.88 –5.683 2 –7.306 8 –8.710 7 –9.880 3 –10.823 –11.563 –12.13 –12.56

– 4.704 – 4.608 – 4.327 6 –3.896 5 –3.369 3 –2.807 1 –2.263 5 –1.775 3 –1.361 6 –1.03 –0.762

–9.8 –9.599 9 –9.015 9 –8.117 8 –7.019 3 –5.848 1 –4.715 6 –3.698 6 –2.836 6 –2.14 –1.59

–0.154 –0.029 1 –0.005 42

–0.321 –0.060 6 –0.011 3

j

. . . listing results after each fifth step –13.49 –13.67 –13.71

–27.4 –41.0 –54.7

3 4 5

The hailstone reaches 99% of vT after 3.3 s, 99.95% of vT after 5.0 s, 99.99% of vT after 6.0 s, 99.999% of vT after 7.4 s. P6.46

(a)

At terminal velocity,

∑ F = 0 = − mg + CvT2

C=

(b) (c)

e

mg vT2

b0.142 kgge9.80 m s j = b42.5 m sg 2

=

jb

2

Cv 2 = 7.70 × 10 −4 kg m 36.0 m s

g

2

7.70 × 10 −4 kg m

= 0.998 N

Elapsed Time (s)

Altitude (m)

Speed (m/s)

Resistance Force (N)

Net Force (N)

0.000 00 0.050 00 … 2.950 00 3.000 00 3.050 00 … 6.250 00 6.300 00

0.000 00 1.757 92

Acceleration m s2

36.000 00 35.158 42

–0.998 49 –0.952 35

–2.390 09 –2.343 95

–16.831 58 –16.506 67

48.623 27 48.640 00 48.632 24

0.824 94 0.334 76 –0.155 27

–0.000 52 –0.000 09 0.000 02

–1.392 12 –1.391 69 –1.391 58

–9.803 69 –9.800 61 –9.799 87

1.250 85 –0.106 52

–26.852 97 –27.147 36

0.555 55 0.567 80

–0.836 05 –0.823 80

–5.887 69 –5.801 44

e

j

Maximum height is about 49 m . It returns to the ground after about 6.3 s with a speed of approximately 27 m s .

176 P6.47

Circular Motion and Other Applications of Newton’s Laws

(a)

At constant velocity mg =− vT = − C vT = −

(b)

(a)

b50.0 kg ge9.80 m s j = 2

0.200 kg m

b50.0 kg gb9.80 m sg = 20.0 kg m

−49.5 m s with chute closed and

−4.95 m s with chute open.

We use time increments of 0.1 s for 0 < t < 10 s , then 0.01 s for 10 s < t < 12 s , and then 0.1 s again. time(s) 0 1 2 4 7 10 10.1 10.3 11 12 50 100 145

6.48

∑ F = 0 = − mg + CvT2

height(m) 1000 995 980 929 812 674 671 669 665 659 471 224 0

velocity(m/s) 0 –9.7 –18.6 –32.7 –43.7 –47.7 –16.7 –8.02 –5.09 –4.95 –4.95 –4.95 –4.95

We use a time increment of 0.01 s. time(s) 0 0.100 0.200 0.400 1.00 1.92 2.00 4.00 5.00 6.85

x(m) 0 7.81 14.9 27.1 51.9 70.0 70.9 80.4 81.4 81.8

y(m) 0 5.43 10.2 18.3 32.7 38.5 38.5 26.7 17.7 0

(b)

range = 81.8 m

(c)

So we have maximum range at θ = 15.9°

with θ 30.0° 35.0° 25.0° 20.0° 15.0° 10.0° 17.0° 16.0° 15.5° 15.8° 16.1° 15.9°

we find range 86.410 m 81.8 m 90.181 m 92.874 m 93.812 m 90.965 m 93.732 m 93.839 8 m 93.829 m 93.839 m 93.838 m 93.840 2 m

Chapter 6

P6.49

(a)

∑ F = − mg + Cv 2 = 0 . Thus,

At terminal speed,

C= (b)

(c)

177

mg v2

b0.046 0 kgge9.80 m s j = b44.0 m sg 2

=

2

2.33 × 10 −4 kg m

We set up a spreadsheet to calculate the motion, try different initial speeds, and home in on 53 m s as that required for horizontal range of 155 m, thus:

Time t (s)

x (m)

0.000 0 0.002 7 … 2.501 6 2.504 3 2.506 9 … 3.423 8 3.426 5 3.429 1 … 5.151 6 5.154 3

0.000 0 0.121 1

vx (m/s)

e

ax m s2

j

y (m)

vy

ay

em s j

v=

2

v x2

+

v y2

tan −1

Fv I GH v JK y x

(m/s)

(deg)

45.687 0 –10.565 9 0.000 0 27.451 5 –13.614 6 45.659 0 –10.552 9 0.072 7 27.415 5 –13.604 6

53.300 0 53.257 4

31.000 0 30.982 2

90.194 6 28.937 5 –4.238 8 32.502 4 0.023 5 –9.800 0 90.271 3 28.926 3 –4.235 5 32.502 4 –0.002 4 –9.800 0 90.348 0 28.915 0 –4.232 2 32.502 4 –0.028 4 –9.800 0

28.937 5 28.926 3 28.915 1

0.046 6 –0.004 8 –0.056 3

115.229 8 25.492 6 –3.289 6 28.397 2 –8.890 5 –9.399 9 115.297 4 25.483 9 –3.287 4 28.373 6 –8.915 4 –9.397 7 115.364 9 25.475 1 –3.285 1 28.350 0 –8.940 3 –9.395 4

26.998 4 26.998 4 26.998 4

–19.226 2 –19.282 2 –19.338 2

154.996 8 20.843 8 –2.199 2 0.005 9 –23.308 7 –7.049 8 155.052 0 20.838 0 –2.198 0 –0.055 9 –23.327 4 –7.045 4

31.269 2 31.279 2

–48.195 4 –48.226 2

(m/s)

Similarly, the initial speed is 42 m s . The motion proceeds thus:

Time t (s)

x (m)

0.000 0 0.000 0 0.003 5 0.100 6 … 2.740 5 66.307 8 2.744 0 66.379 7 2.747 5 66.451 6 … 3.146 5 74.480 5 3.150 0 74.549 5 3.153 5 74.618 5 … 5.677 0 118.969 7 5.680 5 119.024 8

vx (m/s)

e

ax m s2

j

y (m)

vy

ay

em s j 2

v=

v x2

+

v y2

tan −1

Fv I GH v JK y x

(m/s)

(deg)

0.000 0 30.826 6 –14.610 3 0.107 9 30.775 4 –14.594 3

42.150 0 42.102 6

47.000 0 46.967 1

20.548 4 –2.137 4 39.485 4 0.026 0 –9.800 0 20.541 0 –2.135 8 39.485 5 –0.008 3 –9.800 0 20.533 5 –2.134 3 39.485 5 –0.042 6 –9.800 0

20.548 5 20.541 0 20.533 5

0.072 5 –0.023 1 –0.118 8

19.715 6 –1.967 6 38.696 3 –3.942 3 –9.721 3 19.708 7 –1.966 2 38.682 5 –3.976 4 –9.720 0 19.701 8 –1.964 9 38.668 6 –4.010 4 –9.718 6

20.105 8 20.105 8 20.105 8

–11.307 7 –11.406 7 –11.505 6

15.739 4 –1.254 0 0.046 5 –25.260 0 –6.570 1 15.735 0 –1.253 3 –0.041 9 –25.283 0 –6.564 2

29.762 3 29.779 5

–58.073 1 –58.103 7

28.746 2 –4.182 9 28.731 6 –4.178 7

(m/s)

The trajectory in (c) reaches maximum height 39 m, as opposed to 33 m in (b). In both, the ball reaches maximum height when it has covered about 57% of its range. Its speed is a minimum somewhat later. The impact speeds are both about 30 m/s.

178

Circular Motion and Other Applications of Newton’s Laws

Additional Problems *P6.50

When the cloth is at a lower angle θ, the radial component of ∑ F = ma reads 68° R

mv 2 n + mg sin θ = . r

mg

At θ = 68.0° , the normal force drops to zero and v2 . g sin 68° = r v = rg sin 68° =

p

p

mg sin68°

mg cos68°

FIG. P6.50

a0.33 mfe9.8 m s j sin 68° = 1.73 m s 2

The rate of revolution is

b

angular speed = 1.73 m s

*P6.51

(a)

b

v = 30 km h

IJ FG 2πr IJ = gFGH 12rev πr K H 2π a0.33 mf K

0.835 rev s = 50.1 rev min .

1 h I F 1 000 m I gFGH 3 600 G J = 8.33 m s s JK H 1 km K

∑ Fy = ma y : +n − mg = −

mv 2 r

L F vI b8.33 m sg n = mG g − J = 1 800 kg M9.8 m s − 20. 4 m MN H rK 2

2

n

2

OP PQ

mg FIG. P6.51

= 1.15 × 10 4 N up (b)

Take n = 0 . Then mg =

mv 2 . r

v = gr =

P6.52

(a)

∑ Fy = ma y = mg − n =

(b)

mv 2 R

e9.8 m s ja20.4 mf = 2

mv 2 R n = mg −

When n = 0 ,

mg =

Then,

v=

mv 2 R

mv 2 R

gR .

14.1 m s = 50.9 km h

Chapter 6

*P6.53

(a)

slope =

(b)

slope =

(c)

(d)

0.160 N − 0 = 0.016 2 kg m 9.9 m 2 s 2 R v

2

=

1 2

DρAv 2 v

1 Dρ A 2

=

2

1 DρA = 0.016 2 kg m 2 2 0.016 2 kg m D= 1.20 kg m3 π 0.105 m

e

179

b

g

ja

f

2

= 0.778

e

je

j

From the table, the eighth point is at force mg = 8 1.64 × 10 −3 kg 9.8 m s 2 = 0.129 N and

b

g

2

horizontal coordinate 2.80 m s . The vertical coordinate of the line is here 0.129 N − 0.127 N 2 = 1.5%. 0.016 2 kg m 2.8 m s = 0.127 N . The scatter percentage is 0.127 N

b

P6.54

gb

g

(e)

The interpretation of the graph can be stated thus: For stacked coffee filters falling at terminal speed, a graph of air resistance force as a function of squared speed demonstrates that the force is proportional to the speed squared within the experimental uncertainty estimated as 2%. This proportionality agrees with that described by the theoretical equation 1 R = DρAv 2 . The value of the constant slope of the graph implies that the drag coefficient 2 for coffee filters is D = 0.78 ± 2% .

(a)

While the car negotiates the curve, the accelerometer is at the angle θ. mv 2 r

Horizontally:

T sin θ =

Vertically:

T cos θ = mg

where r is the radius of the curve, and v is the speed of the car. tan θ =

By division, v2 = g tan θ : Then a c = r

v2 rg

e

a c = 9.80 m s

2

j tan 15.0°

a c = 2.63 m s 2

b23.0 m sg r=

(b)

v2 r= ac

(c)

v 2 = rg tan θ = 201 m 9.80 m s 2 tan 9.00°

2

2.63 m s 2

a

fe

j

v = 17.7 m s

= 201 m

FIG. P6.54

180 P6.55

Circular Motion and Other Applications of Newton’s Laws

Take x-axis up the hill

∑ Fx = ma x :

+T sin θ − mg sin φ = ma T a = sin θ − g sin φ m ∑ Fy = ma y : +T cos θ − mg cos φ = 0 mg cos φ T= cos θ g cos φ sin θ − g sin φ a= cos θ a = g cos φ tan θ − sin φ

b

*P6.56

(a)

a

f

2π 7.46 m = 1.23 m s . The 38 s total force on it must add to

The speed of the bag is

ma c =

b30 kg gb1.23 m sg

fs

y

n ac

2

mg

= 6.12 N

7.46 m

g

FIG. P6.56

∑ Fx = ma x : ∑ Fy = ma y :

fs cos 20 − n sin 20 = 6.12 N

b

ge

j

fs sin 20 + n cos 20 − 30 kg 9.8 m s 2 = 0

f cos 20 − 6.12 N n= s sin 20 Substitute: fs sin 20 + fs

cos 2 20 cos 20 − 6.12 N = 294 N sin 20 sin 20 f s 2.92 = 294 N + 16.8 N

a

f

a f

fs = 106 N (b)

a b

f gb

2π 7.94 m = 1.47 m s 34 s 2 30 kg 1.47 m s = 8.13 N ma c = 7.94 m fs cos 20 − n sin 20 = 8.13 N v=

g

fs sin 20 + n cos 20 = 294 N fs cos 20 − 8.13 N sin 20 cos 2 20 cos 20 − 8.13 N = 294 N fs sin 20 + fs sin 20 sin 20 fs 2.92 = 294 N + 22.4 N

n=

a

a f

fs = 108 N n=

f

a108 Nf cos 20 − 8.13 N = 273 N

sin 20 fs 108 N = 0.396 µs = = n 273 N

x

Chapter 6

P6.57

(a)

Since the centripetal acceleration of a person is downward (toward the axis of the earth), it is equivalent to the effect of a falling elevator. Therefore, Fg′ = Fg −

(b)

181

mv 2 or Fg > Fg′ r

a f

At the poles v = 0 and Fg′ = Fg = mg = 75.0 9.80 = 735 N down.

b

FIG. P6.57

g

At the equator, Fg′ = Fg − ma c = 735 N − 75.0 0.033 7 N = 732 N down. P6.58

(a)

(b)

(c)

P6.59

(a)

Since the object of mass m 2 is in equilibrium,

∑ Fy = T − m 2 g = 0

or

T = m2 g .

The tension in the string provides the required centripetal acceleration of the puck. Thus,

Fc = T = m 2 g .

From

Fc =

we have

v=

b

v = 300 mi h

m1 v 2 R RFc = m1

FG m IJ gR Hm K 2

.

1

gFGH 6088.0.0 mift sh IJK = 440 ft s

At the lowest point, his seat exerts an upward force; therefore, his weight seems to increase. His apparent weight is Fg′ = mg + m (b)

FG H

v2 160 = 160 + r 32.0

IJ a440f K 1 200

2

= 967 lb .

At the highest point, the force of the seat on the pilot is directed down and Fg′ = mg − m

v2 = −647 lb . r

Since the plane is upside down, the seat exerts this downward force. (c)

mv 2 . If we vary the aircraft’s R and v such that the above is true, R then the pilot feels weightless.

When Fg′ = 0 , then mg =

182 P6.60

Circular Motion and Other Applications of Newton’s Laws

For the block to remain stationary,

e

∑ Fy = 0 and ∑ Fx = mar .

j

e

j

n1 = m p + m b g so f ≤ µ s1n1 = µ s1 m p + m b g . mb g

At the point of slipping, the required centripetal force equals the maximum friction force:

e

2 v max

n1

j r = µ em + m j g a0.750fa0.120fa9.80f = 0.939 m s .

∴ mp + mb or v max = µ s1 rg =

s1

p

mp g

fp

b

f

For the penny to remain stationary on the block:

mb g

∑ Fy = 0 ⇒ n 2 − m p g = 0 or n 2 = m p g

∑ Fx = ma r ⇒ f p = m p

and

n2

2

v . r

fp

When the penny is about to slip on the block, f p = f p , max = µ s 2 n 2 or µ s 2 m p g = m p

mp g

2 v max r

mp g

v max = µ s 2 rg =

a0.520fa0.120fa9.80f = 0.782 m s

FIG. P6.60

This is less than the maximum speed for the block, so the penny slips before the block starts to slip. The maximum rotation frequency is Max rpm =

P6.61

P6.62

v=

a

gLM a N

OPFG 60 s IJ = f QH 1 min K

v max 1 rev = 0.782 m s 2πr 2π 0.120 m

b

62.2 rev min .

f

2πr 2π 9.00 m = = 3.77 m s T 15.0 s

a

f

v2 = 1.58 m s 2 r

(a)

ar =

(b)

Flow = m g + a r = 455 N

(c)

Fhigh

(d)

Fmid = m g 2 + a r2 = 397 N upward and at θ = tan −1

b g = mb g − a g = r

328 N ar 1.58 = tan −1 = 9.15° inward . g 9.8

Standing on the inner surface of the rim, and moving with it, each person will feel a normal force exerted by the rim. This inward force causes the 3.00 m s 2 centripetal acceleration: ac =

v2 : r

The period of rotation comes from v = so the frequency of rotation is

e3.00 m s ja60.0 mf = 13.4 m s 2πr 2π a60.0 mf = = 28.1 s T= v 13. 4 m s 1 1 1 F 60 s I = f= = G J = 2.14 rev min T 28.1 s 28.1 s H 1 min K v = ac r =

2πr : T

2

.

Chapter 6

P6.63

(a)

The mass at the end of the chain is in vertical equilibrium. T Thus T cos θ = mg . l = 2.50 m mv 2 Horizontally T sin θ = ma r = θ r

R = 4.00 m

a f r = a 2.50 sin 28.0°+4.00f m = 5.17 m

r

r = 2.50 sin θ + 4.00 m

v2 . Then a r = 5.17 m By division tan θ =

183

mg

FIG. P6.63

ar v2 = g 5.17 g

a fa fa

f

v 2 = 5.17 g tan θ = 5.17 9.80 tan 28.0° m 2 s 2 v = 5.19 m s (b)

T cos θ = mg

b

ge

j

50.0 kg 9.80 m s 2 mg T= = = 555 N cos θ cos 28.0° P6.64

(a)

The putty, when dislodged, rises and returns to the original level in time t. To find t, we use 2v where v is the speed of a point on the rim of the wheel. v f = vi + at : i.e., − v = + v − gt or t = g 2πR 2 v 2πR If R is the radius of the wheel, v = = , so t = . t g v Thus, v 2 = πRg and v = πRg .

(b)

The putty is dislodged when F, the force holding it to the wheel is F=

P6.65

(a)

mv 2 R

n=

f − mg = 0

f = µ sn T=

(b)

v=

f

2πR T

4π 2 Rµ s g

T = 2.54 s #

mv 2 = mπ g . R

n

FG H

IJ K

rev 1 rev 60 s rev = = 23.6 min 2.54 s min min mg

FIG. P6.65

184 P6.66

Circular Motion and Other Applications of Newton’s Laws

Let the x–axis point eastward, the y-axis upward, and the z-axis point southward. (a)

vi2 sin 2θ i g The initial speed of the ball is therefore

The range is Z =

vi =

gZ = sin 2θ i

a9.80fa285f = 53.0 m s sin 96.0°

The time the ball is in the air is found from ∆y = viy t +

b

1 a y t 2 as 2

f e

ga

j

0 = 53.0 m s sin 48.0° t − 4.90 m s 2 t 2 giving t = 8.04 s .

e

j

6 2πR e cos φ i 2π 6.37 × 10 m cos 35.0° = = 379 m s 86 400 s 86 400 s

(b)

vix =

(c)

360° of latitude corresponds to a distance of 2πR e , so 285 m is a change in latitude of ∆φ =

FG S IJ a360°f = FG 285 m IJ a360°f = 2.56 × 10 GH 2π e6.37 × 10 mj JK H 2πR K 6

e

−3

degrees

The final latitude is then φ f = φ i − ∆φ = 35.0°−0.002 56° = 34.997 4° . The cup is moving eastward at a speed v fx =

2πR e cos φ f 86 400 s

, which is larger than the eastward

velocity of the tee by ∆v x = v fx − v fi = =

2πR e 2πR e cos φ f − cos φ i = cos φ i − ∆φ − cos φ i 86 400 s 86 400 s

b

2πR e cos φ i cos ∆φ + sin φ i sin ∆φ − cos φ i 86 400 s

Since ∆φ is such a small angle, cos ∆φ ≈ 1 and ∆v x ≈

∆v x ≈ (d)

b g e

g

e

2πR e sin φ i sin ∆φ . 86 400 s

j sin 35.0° sin 0.002 56° =

2π 6.37 × 10 6 m 86 400 s

ja

f

∆x = ∆v x t = 1.19 × 10 −2 m s 8.04 s = 0.095 5 m = 9.55 cm

1.19 × 10 −2 m s

Chapter 6

P6.67

(a)

If the car is about to slip down the incline, f is directed up the incline.

∑ Fy = n cos θ + f sin θ − mg = 0 where n=

b

g

b

g

θt

2 v min yields R

b

Rg tan θ − µ s

v min =

n

f

mg µ s mg and f = . cos θ 1 + µ s tan θ cos θ 1 + µ s tan θ

∑ Fx = n sin θ − f cos θ = m

Then,

f = µ sn gives

θt

mg

n cos θ

g

f sin θ

.

1 + µ s tan θ

f cos θ

n sin θ

When the car is about to slip up the incline, f is directed down the incline. Then, ∑ Fy = n cos θ − f sin θ − mg = 0 with f = µ sn yields mg

µ s mg mg n= and f = . cos θ 1 − µ s tan θ cos θ 1 − µ s tan θ

b

In this case,

g

∑ Fx = n sin θ + f cos θ = m

If v min =

(c)

v min =

b

Rg tan θ − µ s

b

1 + µ s tan θ

1 − µ s tan θ

g = 0 , then

g

θt

n

2 v max , which gives R

Rg tan θ + µ s

v max =

(b)

b

g

f

.

θ t mg

µ s = tan θ .

a100 mfe9.80 m s jatan 10.0°−0.100f = 1 + a0.100 f tan 10.0° a100 mfe9.80 m s jatan 10.0°+0.100f = 1 − a0.100f tan 10.0°

n cos θ

2

n sin θ

8.57 m s

f cos θ

2

v max =

16.6 m s

f sin θ mg

FIG. P6.67

185

186 P6.68

Circular Motion and Other Applications of Newton’s Laws

(a)

The bead moves in a circle with radius v = R sin θ at a speed of v=

2πr 2πR sin θ = T T

The normal force has an inward radial component of n sin θ and an upward component of n cos θ

∑ Fy = ma y :

n cos θ − mg = 0

or n=

Then

∑ Fx = n sin θ = m

mg cos θ

FIG. P6.68(a)

FG mg IJ sinθ = m FG 2πR sinθ IJ K H cosθ K R sin θ H T

v2 becomes r

which reduces to

g sin θ 4π 2 R sin θ = cos θ T2

This has two solutions:

sin θ = 0 ⇒ θ = 0°

and

cos θ =

2

gT 2 4π 2 R

(1) (2)

If R = 15.0 cm and T = 0.450 s, the second solution yields

e9.80 m s ja0.450 sf cos θ = 4π a0.150 mf 2

2

2

= 0.335 and θ = 70.4°

Thus, in this case, the bead can ride at two positions θ = 70. 4° and θ = 0° . (b)

At this slower rotation, solution (2) above becomes

e9.80 m s ja0.850 sf cos θ = 4π a0.150 mf 2

2

2

= 1.20 , which is impossible.

In this case, the bead can ride only at the bottom of the loop, θ = 0° . The loop’s rotation must be faster than a certain threshold value in order for the bead to move away from the lowest position.

Chapter 6

P6.69

At terminal velocity, the accelerating force of gravity is balanced by frictional drag: mg = arv + br 2 v 2 (a)

e

j e

j

mg = 3.10 × 10 −9 v + 0.870 × 10 −10 v 2

LM 4 π e10 mj OP N3 Q = e3.10 × 10 jv + e0.870 × 10 jv

4.11 × 10 −11

3

−5

m = ρV = 1 000 kg m3

For water,

−9

−10

2

Assuming v is small, ignore the second term on the right hand side: v = 0.013 2 m s . (b)

e

j e

j

mg = 3.10 × 10 −8 v + 0.870 × 10 −8 v 2 Here we cannot ignore the second term because the coefficients are of nearly equal magnitude.

e j e j a3.10f + 4a0.870fa4.11f = 1.03 m s 2a0.870f

4.11 × 10 −8 = 3.10 × 10 −8 v + 0.870 × 10 −8 v 2 v= (c)

e

2

−3.10 ±

j e

j

mg = 3.10 × 10 −7 v + 0.870 × 10 −6 v 2 Assuming v > 1 m s , and ignoring the first term:

e

j

4.11 × 10 −5 = 0.870 × 10 −6 v 2 P6.70

187

v=

FG mg IJ LM1 − expFG −bt IJ OP where expaxf = e H b KN H m KQ

v = 6.87 m s x

is the exponential function. mg b

At t → ∞ ,

v → vT =

At t = 5.54 s

0.500 vT = vT 1 − exp

LM MN

F −ba5.54 sf I OP GH 9.00 kg JK PQ

F −ba5.54 sf I = 0.500 ; GH 9.00 kg JK − b a5.54 sf = ln 0.500 = −0.693 ;

exp

9.00 kg

b=

b9.00 kg ga0.693f = 1.13 m s 5.54 s

b9.00 kg ge9.80 m s j = 2

mg b

(a)

vT =

vT =

(b)

0.750 vT = vT 1 − exp

LM N

FG −1.13t IJ OP H 9.00 s K Q

1.13 kg s

FG −1.13t IJ = 0.250 H 9.00 s K 9.00aln 0.250 f t= s=

exp

−1.13

continued on next page

11.1 s

78.3 m s

188

Circular Motion and Other Applications of Newton’s Laws

(c)

FG IJ LM1 − expFG − bt IJ OP ; H K N H mKQ

mg dx = dt b

z z FGH x

dx =

x0

t

0

mg b

IJ LM1 − expFG −bt IJ OPdt KN H m KQ F GH

I FG IJ = mgt + F m g I LexpFG −bt IJ − 1O JK H K b GH b JK MN H m K PQ F b9.00 kg g e9.80 m s j I 5.54 s J expa−0.693f − 1 x = 9.00 kg e9.80 m s j +G 1.13 kg s G H b1.13 m sg JK x = 434 m + 626 ma −0.500f = 121 m mgt m2 g − bt + x − x0 = exp 2 b m b

t

2

2

0

2

2

At t = 5.54 s ,

P6.71

2

2

∑ Fy = L y − Ty − mg = L cos 20.0°−T sin 20.0°−7.35 N = ma y = 0 ∑ Fx = L x + Tx = L sin 20.0°+T cos 20.0° = m

b

g

v2 r

2

35.0 m s v2 = 0.750 kg = 16.3 N m r 60.0 m cos 20.0°

a

f

∴ L sin 20.0°+T cos 20.0° = 16.3 N L cos 20.0°−T sin 20.0° = 7.35 N cos 20.0° 16.3 N = sin 20.0° sin 20.0° sin 20.0° 7.35 N = L−T cos 20.0° cos20.0° 16.3 N 7.35 N − T cot 20.0°+ tan 20.0° = sin 20.0° cos 20.0° T 3.11 = 39.8 N L+T

a a f

T = 12.8 N

f

FIG. P6.71

Chapter 6

P6.72

(a)

af

a f

2.00

18.9

3.00

42.1

4.00

73.8

(b)

ts dm 4.88 1.00

700 600

6.00 154 7.00 199

500

8.00 246 9.00 296

(c)

347

11.0

399

12.0

452

13.0

505

14.0

558

15.0

611

16.0

664

17.0

717

18.0

770

19.0

823

20.0

876

d (m) 900 800

5.00 112

10.0

400 300 200 100 0

0

2

4

6

8

10 12 14 16 18 20

t (s)

A straight line fits the points from t = 11.0 s to 20.0 s quite precisely. Its slope is the terminal speed. vT = slope =

*P6.73

189

876 m − 399 m = 53.0 m s 20.0 s − 11.0 s dv dx =0−k = − kv dt dt

v = vi − kx implies the acceleration is

a=

Then the total force is

∑ F = ma = ma− kvf ∑ F = −kmv

The resistive force is opposite to the velocity:

.

ANSWERS TO EVEN PROBLEMS P6.2

215 N horizontally inward

P6.4

6.22 × 10 −12 N

P6.6

(a) 1.65 km s ; (b) 6.84 × 10 3 s

P6.8

0.966 g

P6.10

e j (c) e −0.181 i + 0.181 jj m s

(a) −0.233 i + 0.163 j m s 2 ; (b) 6.53 m s ; 2

P6.12

2.06 × 10 3 rev min

P6.14

(a)

P6.16

(a) 1.33 m s 2 ; (b) 1.79 m s 2 forward and 48.0° inward

P6.18

8.88 N

R

FG 2T − gIJ ; (b) 2T upward Hm K

190

Circular Motion and Other Applications of Newton’s Laws

P6.46

(a) 7.70 × 10 −4 kg m; (b) 0.998 N; (c) The ball reaches maximum height 49 m. Its flight lasts 6.3 s and its impact speed is 27 m s .

15.3 m s Straight across the dashboard to the left

P6.48

(a) see the solution; (b) 81.8 m; (c) 15.9°

P6.24

0.527°

P6.50

0.835 rev s

P6.26

(a) 1.41 h; (b) 17.1

P6.52

(a) mg −

P6.28

µk =

P6.54

(a) 2.63 m s 2 ; (b) 201 m; (c) 17.7 m s

P6.30

(a) 2.38 × 10 5 m s 2 horizontally inward = 2.43 × 10 4 g ; (b) 360 N inward perpendicular to the cone; (c) 47.5 × 10 4 m s 2

P6.56

(a) 106 N; (b) 0.396

P6.58

(a) m 2 g ; (b) m 2 g ; (c)

P6.32

(a) 6.27 m s 2 downward ; (b) 784 N up; (c) 283 N up

P6.60

62.2 rev min

P6.62

2.14 rev min

P6.34

(a) 53.8 m s ; (b) 148 m P6.64

P6.36

1.40

(a) v = πRg ; (b) mπ g

P6.38

−0. 212 m s 2

P6.66

(a) 8.04 s; (b) 379 m s; (c) 1.19 cm s ; (d) 9.55 cm

P6.40

see the solution

P6.68

(a) either 70.4° or 0°; (b) 0°

P6.42

36.5 m s

P6.70

(a) 78.3 m s ; (b) 11.1 s; (c) 121 m

P6.44

(a) 0.980 m s ; (b) see the solution

P6.72

(a) and (b) see the solution; (c) 53.0 m s

P6.20

(a) 8.62 m; (b) Mg downward; (c) 8.45 m s 2 , Unless they are belted in, the riders will fall from the cars.

P6.22

a

2 vt − L

b g + agt

f

2

mv 2 ; (b) v = gR R

FG m IJ gR Hm K 2

1

7 Energy and Energy Transfer CHAPTER OUTLINE 7.1 7.2 7.3 7.4 7.5

7.6

7.7 7.8 7.9

Systems and Environments Work Done by a Constant Force The Scalar Product of Two Vectors Work Done by a Varying Force Kinetic Energy and the Work-Kinetic Energy Theorem The Non-Isolated System—Conservation of Energy Situations Involving Kinetic Friction Power Energy and the Automobile

ANSWERS TO QUESTIONS Q7.1

The force is perpendicular to every increment of displacement. Therefore, F ⋅ ∆r = 0 .

Q7.2

(a)

Positive work is done by the chicken on the dirt.

(b)

No work is done, although it may seem like there is.

(c)

Positive work is done on the bucket.

(d)

Negative work is done on the bucket.

(e)

Negative work is done on the person’s torso.

Q7.3

Yes. Force times distance over which the toe is in contact with the ball. No, he is no longer applying a force. Yes, both air friction and gravity do work.

Q7.4

Force of tension on a ball rotating on the end of a string. Normal force and gravitational force on an object at rest or moving across a level floor.

Q7.5

(a)

Tension

(c)

Positive in increasing velocity on the downswing. Negative in decreasing velocity on the upswing.

(b)

Air resistance

Q7.6

No. The vectors might be in the third and fourth quadrants, but if the angle between them is less than 90° their dot product is positive.

Q7.7

The scalar product of two vectors is positive if the angle between them is between 0 and 90°. The scalar product is negative when 90° < θ < 180° .

Q7.8

If the coils of the spring are initially in contact with one another, as the load increases from zero, the graph would be an upwardly curved arc. After the load increases sufficiently, the graph will be linear, described by Hooke’s Law. This linear region will be quite large compared to the first region. The graph will then be a downward curved arc as the coiled spring becomes a completely straight wire. As the load increases with a straight wire, the graph will become a straight line again, with a significantly smaller slope. Eventually, the wire would break.

Q7.9

k ′ = 2 k . To stretch the smaller piece one meter, each coil would have to stretch twice as much as one coil in the original long spring, since there would be half as many coils. Assuming that the spring is ideal, twice the stretch requires twice the force. 191

192

Energy and Energy Transfer

Q7.10

Kinetic energy is always positive. Mass and squared speed are both positive. A moving object can always do positive work in striking another object and causing it to move along the same direction of motion.

Q7.11

Work is only done in accelerating the ball from rest. The work is done over the effective length of the pitcher’s arm—the distance his hand moves through windup and until release.

Q7.12

Kinetic energy is proportional to mass. The first bullet has twice as much kinetic energy.

Q7.13

The longer barrel will have the higher muzzle speed. Since the accelerating force acts over a longer distance, the change in kinetic energy will be larger.

Q7.14

(a)

Kinetic energy is proportional to squared speed. Doubling the speed makes an object's kinetic energy four times larger.

(b)

If the total work on an object is zero in some process, its speed must be the same at the final point as it was at the initial point.

Q7.15

The larger engine is unnecessary. Consider a 30 minute commute. If you travel the same speed in each car, it will take the same amount of time, expending the same amount of energy. The extra power available from the larger engine isn’t used.

Q7.16

If the instantaneous power output by some agent changes continuously, its average power in a process must be equal to its instantaneous power at least one instant. If its power output is constant, its instantaneous power is always equal to its average power.

Q7.17

It decreases, as the force required to lift the car decreases.

Q7.18

As you ride an express subway train, a backpack at your feet has no kinetic energy as measured by you since, according to you, the backpack is not moving. In the frame of reference of someone on the side of the tracks as the train rolls by, the backpack is moving and has mass, and thus has kinetic energy.

Q7.19

The rock increases in speed. The farther it has fallen, the more force it might exert on the sand at the bottom; but it might instead make a deeper crater with an equal-size average force. The farther it falls, the more work it will do in stopping. Its kinetic energy is increasing due to the work that the gravitational force does on it.

Q7.20

The normal force does no work because the angle between the normal force and the direction of motion is usually 90°. Static friction usually does no work because there is no distance through which the force is applied.

Q7.21

An argument for: As a glider moves along an airtrack, the only force that the track applies on the glider is the normal force. Since the angle between the direction of motion and the normal force is 90°, the work done must be zero, even if the track is not level. Against: An airtrack has bumpers. When a glider bounces from the bumper at the end of the airtrack, it loses a bit of energy, as evidenced by a decreased speed. The airtrack does negative work.

Q7.22

Gaspard de Coriolis first stated the work-kinetic energy theorem. Jean Victor Poncelet, an engineer who invaded Russia with Napoleon, is most responsible for demonstrating its wide practical applicability, in his 1829 book Industrial Mechanics. Their work came remarkably late compared to the elucidation of momentum conservation in collisions by Descartes and to Newton’s Mathematical Principles of the Philosophy of Nature, both in the 1600’s.

Chapter 7

SOLUTIONS TO PROBLEMS Section 7.1

Systems and Environments

Section 7.2

Work Done by a Constant Force

P7.1

W = F∆r cos θ = 16.0 N 2.20 m cos 25.0° = 31.9 J

a

(a)

fa

f

(b), (c) The normal force and the weight are both at 90° to the displacement in any time interval. Both do 0 work.

∑ W = 31.9 J + 0 + 0 =

(d) P7.2

31.9 J

The component of force along the direction of motion is

a

f

F cos θ = 35.0 N cos 25.0° = 31.7 N . The work done by this force is

a

f

a

fa

f

W = F cos θ ∆r = 31.7 N 50.0 m = 1.59 × 10 3 J . P7.3

Method One. Let φ represent the instantaneous angle the rope makes with the vertical as it is swinging up from φ i = 0 to φ f = 60° . In an incremental bit of motion from angle φ to φ + dφ , the definition of radian measure implies that

a

f

∆r = 12 m dφ . The angle θ between the incremental displacement and the

b

g

force of gravity is θ = 90°+ φ . Then cos θ = cos 90°+φ = − sin φ . The work done by the gravitational force on Batman is

z

z

f

φ = 60 °

i

φ =0

W = F cos θdr =

b

ga

FIG. P7.3

f

mg − sin φ 12 m dφ

f z sin φ dφ = b−80 kg ge9.8 m s ja12 mfb− cos φ g = a −784 N fa12 mfa− cos 60°+1f = −4.70 × 10 J a

= − mg 12 m

60 °

2

0

60 ° 0

3

Method Two.

b

ge

j

The force of gravity on Batman is mg = 80 kg 9.8 m s 2 = 784 N down. Only his vertical displacement contributes to the work gravity does. His original y-coordinate below the tree limb is –12 m. His final y-coordinate is −12 m cos 60° = −6 m . His change in elevation is

a

f

a

f

−6 m − −12 m = 6 m . The work done by gravity is

a

fa f

W = F∆r cos θ = 784 N 6 m cos 180° = −4.70 kJ .

193

194 P7.4

Energy and Energy Transfer

ja fa f

e

(a)

W = mgh = 3.35 × 10 −5 9.80 100 J = 3.28 × 10 −2 J

(b)

Since R = mg , Wair resistance = −3.28 × 10 −2 J

Section 7.3

The Scalar Product of Two Vectors

P7.5

A = 5.00 ; B = 9.00 ; θ = 50.0° A ⋅ B = AB cos θ = 5.00 9.00 cos 50.0° = 28.9

P7.6

A ⋅ B = A x i + A y j + A z k ⋅ B x i + B y j + Bz k

a fa f

e

je j A ⋅ B = A B e i ⋅ i j + A B e i ⋅ jj + A B e i ⋅ k j + A B e j ⋅ i j + A B e j ⋅ jj + A B e j ⋅ k j + A B ek ⋅ i j + A B ek ⋅ jj + A B ek ⋅ k j x

x

x

y

x z

y x

y

y

y

z x

z y

z

z z

A ⋅ B = A x Bx + A y B y + A z Bz P7.7

P7.8

a fa f

a

fa f

(a)

W = F ⋅ ∆r = Fx x + Fy y = 6.00 3.00 N ⋅ m + −2.00 1.00 N ⋅ m = 16.0 J

(b)

θ = cos −1

16 FG F ⋅ ∆r IJ = cos = H F∆r K ae 6.00f + a−2.00f jea3.00f + a1.00f j −1

2

2

2

2

36.9°

We must first find the angle between the two vectors. It is:

θ = 360°−118°−90.0°−132° = 20.0° Then

a

fb

g

F ⋅ v = Fv cos θ = 32.8 N 0.173 m s cos 20.0° or F ⋅ v = 5.33

P7.9

(a)

N ⋅m J = 5.33 = 5.33 W s s

A = 3.00 i − 2.00 j B = 4.00 i − 4.00 j

(b)

θ = cos −1

B = 3.00 i − 4.00 j + 2.00k A = −2.00 i + 4.00 j

(c)

FIG. P7.8

cos θ =

a fa f

A⋅B −6.00 − 16.0 = AB 20.0 29.0

A = i − 2.00 j + 2.00k B = 3.00 j + 4.00k

A ⋅B 12.0 + 8.00 = cos −1 = 11.3° AB 13.0 32.0

θ = cos −1

a fa f

FG A ⋅ B IJ = cos FG −6.00 + 8.00 IJ = H AB K H 9.00 ⋅ 25.0 K −1

θ = 156°

82.3°

Chapter 7

P7.10

e

j e

A − B = 3.00 i + j − k − − i + 2.00 j + 5.00k

j

A − B = 4.00 i − j − 6.00k

a

f e

je

a

j

f a

f

C ⋅ A − B = 2.00 j − 3.00k ⋅ 4.00 i − j − 6.00k = 0 + −2.00 + +18.0 = 16.0

Section 7.4

Work Done by a Varying Force

z f

P7.11

W = Fdx = area under curve from xi to x f i

x f = 8.00 m

xi = 0

(a)

W = area of triangle ABC = W0 → 8 =

FG 1 IJ × 8.00 m × 6.00 N = H 2K

24.0 J

FIG. P7.11

x f = 10.0 m

xi = 8.00 m

(b)

FG 1 IJ AC × altitude, H 2K

FG 1 IJ CE × altitude, H 2K F 1I = G J × a 2.00 mf × a −3.00 N f = −3.00 J H 2K

W = area of ∆CDE = W8 →10

a

P7.12

f

W0 →10 = W0→ 8 + W8→10 = 24.0 + −3.00 = 21.0 J

(c)

a

f

Fx = 8 x − 16 N (a)

See figure to the right

(b)

Wnet =

a

fa

f a

fa

f

− 2.00 m 16.0 N 1.00 m 8.00 N + = −12.0 J 2 2

FIG. P7.12

195

196 P7.13

Energy and Energy Transfer

z

W = Fx dx and W equals the area under the Force-Displacement curve For the region 0 ≤ x ≤ 5.00 m ,

(a)

W=

a3.00 Nfa5.00 mf = 2

7.50 J

For the region 5.00 ≤ x ≤ 10.0 ,

(b)

a

FIG. P7.13

fa

f

W = 3.00 N 5.00 m = 15.0 J For the region 10.0 ≤ x ≤ 15.0 ,

(c)

W=

a3.00 Nfa5.00 mf = 2

7.50 J

For the region 0 ≤ x ≤ 15.0

(d)

a

f

W = 7.50 + 7.50 + 15.0 J = 30.0 J

z f

P7.14

W = F ⋅ dr = i

zb

5m 0

P7.15

k=

ze

5m

j

4x i + 3 y j N ⋅ dx i

0

x2 4 N m xdx + 0 = 4 N m 2

g

b

g

5m

= 50.0 J 0

a fa f

4.00 9.80 N F Mg = = = 1.57 × 10 3 N m y y 2.50 × 10 −2 m

(a)

For 1.50 kg mass y =

(b)

Work =

a fa f

1.50 9.80 mg = = 0.938 cm k 1.57 × 10 3

1 2 ky 2 1 Work = 1.57 × 10 3 N ⋅ m 4.00 × 10 −2 m 2

e

P7.16

(a)

je

j

= 1.25 J

Spring constant is given by F = kx k=

(b)

2

Work = Favg x =

a

fa

f

a

f

230 N F = = 575 N m x 0.400 m

a

1 230 N 0.400 m = 46.0 J 2

f

Chapter 7

*P7.17

(a)

b

Fapplied = k leaf x + k helper x h = k x + k h x − y 0

197

g

b

g

N N x + 3.60 × 10 5 x − 0.5 m m m 6.8 × 10 5 N x = = 0.768 m 8.85 × 10 5 N m

5 × 10 5 N = 5.25 × 10 5

(b)

W=

FG H

IJ a K

f

N 1 1 1 k x 2 + k h x h2 = 5.25 × 10 5 0.768 m m 2 2 2

2

+

a

f

N 1 3.60 × 10 5 0.268 m m 2

2

= 1.68 × 10 5 J

z f

P7.18

(a)

W = F ⋅ dr W=

i 0.600 m

ze

j

15 000 N + 10 000 x N m − 25 000 x 2 N m 2 dx cos 0°

0

10 000 x 2 25 000 x 3 − W = 15 000 x + 2 3

0.600 m

0

W = 9.00 kJ + 1.80 kJ − 1.80 kJ = 9.00 kJ (b)

Similarly,

b10.0 kN mga1.00 mf − e25.0 kN m ja1.00 mf W = a15.0 kN fa1.00 mf + 2 3 2

2

3

W = 11.7 kJ , larger by 29.6% P7.19

a

f

1 2 k 0.100 m 2 ∴ k = 800 N m and to stretch the spring to 0.200 m requires 4.00 J =

∆W = P7.20

(a)

a fa

1 800 0.200 2

f

2

− 4.00 J = 12.0 J

The radius to the object makes angle θ with the horizontal, so its weight makes angle θ with the negative side of the x-axis, when we take the x–axis in the direction of motion tangent to the cylinder. ∑ Fx = ma x F − mg cos θ = 0 F = mg cos θ

FIG. P7.20

z f

(b)

W = F ⋅ dr i

We use radian measure to express the next bit of displacement as dr = Rdθ in terms of the next bit of angle moved through:

z

π 2

W=

0

mg cos θRdθ = mgR sin θ

a f

W = mgR 1 − 0 = mgR

π 2 0

198 *P7.21

Energy and Energy Transfer

The same force makes both light springs stretch. (a)

The hanging mass moves down by x = x1 + x 2 =

FG H

mg mg 1 1 + = mg + k1 k2 k1 k 2

= 1.5 kg 9.8 m s 2 (b)

IJ K

F 1m + 1m I= GH 1 200 N 1 800 N JK

2.04 × 10 −2 m

We define the effective spring constant as

FG b g H F 1m + 1m I = =G H 1 200 N 1 800 N JK

k=

mg F 1 1 = = + k1 k 2 x mg 1 k1 + 1 k 2

IJ K

−1

−1

*P7.22

720 N m

See the solution to problem 7.21.

FG 1 + 1 IJ Hk k K F1 1I (b) k=G + J Hk k K L F O N kg ⋅ m s = k =M P= = NxQ m m

(a)

x = mg

1

2

−1

1

P7.23

2

2

kg s2

Section 7.5

Kinetic Energy and the Work-Kinetic Energy Theorem

Section 7.6

The Non-Isolated System—Conservation of Energy

P7.24

(a)

KA =

(b)

1 mv B2 = K B : v B = 2

(c)

∑ W = ∆K = K B − K A = 2 mevB2 − v 2A j = 7.50 J − 1.20 J =

(a)

K=

1 1 mv 2 = 0.300 kg 15.0 m s 2 2

(b)

K=

1 0.300 30.0 2

P7.25

b

gb

1 0.600 kg 2.00 m s 2 2K B = m

g

2

= 1.20 J

a2fa7.50f = 0.600

5.00 m s

1

b

a

fa f

gb

2

=

a

g

2

= 33.8 J

fa f a4f = 4a33.8f =

1 0.300 15.0 2

6.30 J

2

135 J

Chapter 7

P7.26

e

199

j

v i = 6.00 i − 2.00 j = m s (a)

vi = vix2 + viy2 = 40.0 m s Ki =

(b)

b

ge

1 1 mvi2 = 3.00 kg 40.0 m 2 s 2 = 60.0 J 2 2

j

v f = 8.00 i + 4.00 j

v 2f = v f ⋅ v f = 64.0 + 16.0 = 80.0 m 2 s 2 1 3.00 ∆K = K f − K i = m v 2f − vi2 = 80.0 − 60.0 = 60.0 J 2 2

e

P7.27

a f

j

Consider the work done on the pile driver from the time it starts from rest until it comes to rest at the end of the fall. Let d = 5.00 m represent the distance over which the driver falls freely, and h = 0.12 m the distance it moves the piling. 1 1 mv 2f − mvi2 2 2 mg h + d cos 0°+ F d cos 180° = 0 − 0 .

∑ W = ∆K :

Wgravity + Wbeam =

b ga f d ia f bmg gah + df = b2 100 kgge9.80 m s ja5.12 mf = F=

so

2

Thus,

driver is upward . P7.28

(a)

a f

a f

(a)

a f

Ki + ∑ W = K f = 0 + ∑W =

(b)

(c)

F=

f

a

f

1 mv 2f − 0 = ∑ W = (area under curve from x = 0 to x = 15.0 m) 2 2 area 2 30.0 J = = 3.87 m s m 4.00 kg

∆K = K f − K i = vf =

P7.29

a

1 mv 2f − 0 = ∑ W = (area under curve from x = 0 to x = 10.0 m) 2 2 area 2 22.5 J = = 3.35 m s m 4.00 kg

∆K = K f − K i = vf =

(c)

a

f

1 mv 2f 2

jb

1 15.0 × 10 −3 kg 780 m s 2

e

g

2

= 4.56 kJ

4.56 × 10 3 J W = = 6.34 kN ∆r cos θ 0.720 m cos 0°

a f v −v b780 m sg − 0 = 422 km s = a= 2x 2a0.720 mf ∑ F = ma = e15 × 10 kg je422 × 10 m s j = 2 f

2

2 i

2

f

(d)

8.78 × 10 5 N . The force on the pile

1 mv 2f − 0 = ∑ W = (area under curve from x = 0 to x = 5.00 m) 2 2 area 2 7.50 J = = 1.94 m s m 4.00 kg

∆K = K f − K i = vf =

(b)

0.120 m

d

−3

3

2

6.34 kN

200 P7.30

Energy and Energy Transfer

(a)

e

Kf = (b)

j

v f = 0.096 3 × 10 8 m s = 2.88 × 10 7 m s 1 1 mv 2f = 9.11 × 10 −31 kg 2.88 × 10 7 m s 2 2

e

Ki + W = K f :

je

j

2

= 3.78 × 10 −16 J

0 + F∆r cos θ = K f

a

f

F 0.028 m cos 0° = 3.78 × 10 −16 J

F = 1.35 × 10 −14 N

∑F =

1.35 × 10 −14 N = 1. 48 × 10 +16 m s 2 9.11 × 10 −31 kg

(c)

∑ F = ma ;

a=

(d)

v xf = v xi + a x t

2.88 × 10 7 m s = 0 + 1.48 × 10 16 m s 2 t

m

e

j

t = 1.94 × 10 −9 s 1 v xi + v xf t 2 1 0.028 m = 0 + 0 + 2.88 × 10 7 m s t 2 x f = xi +

Check:

d

i

e

j

t = 1.94 × 10 −9 s

Section 7.7 P7.31

Situations Involving Kinetic Friction

∑ Fy = ma y :

n − 392 N = 0 n = 392 N

a fa f = F∆r cos θ = a130 fa5.00f cos 0° =

f k = µ k n = 0.300 392 N = 118 N

(a)

WF

(b)

∆Eint = f k ∆x = 118 5.00 = 588 J

650 J

a fa f

a fa f

(c)

Wn = n∆r cos θ = 392 5.00 cos 90° = 0

(d)

W g = mg∆r cos θ = 392 5.00 cos −90° = 0

(e)

∆K = K f − K i = ∑ Wother − ∆Eint 1 mv 2f − 0 = 650 J − 588 J + 0 + 0 = 62.0 J 2

(f)

vf =

a fa f a f

2K f m

=

a

f

2 62.0 J = 1.76 m s 40.0 kg

FIG. P7.31

Chapter 7

P7.32

(a)

vf =

so

(b)

a fe

1 2 1 2 1 kx i − kx f = 500 5.00 × 10 −2 2 2 2 1 1 1 Ws = mv 2f − mvi2 = mv 2f − 0 2 2 2 Ws =

2

j

2

c∑ W h = 2a0.625f m s = 2.00

m

1 1 mvi2 − f k ∆x + Ws = mv 2f 2 2

a

fa fa fb g 1 0.282 J = b 2.00 kg gv 2 2a0.282 f v = m s = 0.531 m s

0 − 0.350 2.00 9.80 0.050 0 J + 0.625 J =

− 0 = 0.625 J

0.791 m s

1 mv 2f 2

2 f

f

P7.33

(a)

2.00

a f = b10.0 kg gd9.80 m s ia5.00 mf cos 110° =

W g = mg cos 90.0°+θ Wg

(b)

FIG. P7.32

2

−168 J

f k = µ k n = µ k mg cos θ ∆Eint = f k = µ k mg cos θ

a

fa

fa fa f

∆Eint = 5.00 m 0.400 10.0 9.80 cos 20.0° = 184 J (c)

a fa f

WF = F = 100 5.00 = 500 J

(d)

∆K = ∑ Wother − ∆Eint = WF + W g − ∆Eint = 148 J

(e)

∆K =

1 1 mv 2f − mvi2 2 2 2 ∆K 2 148 + vi2 = + 1.50 vf = 10.0 m

a f

P7.34

∑ Fy = ma y :

a f a f

a

2

FIG. P7.33

= 5.65 m s

f

n + 70.0 N sin 20.0°−147 N = 0 n = 123 N f k = µ k n = 0.300 123 N = 36.9 N

a

f

a

fa

f

a

fa

f

a

fa

f

(a)

W = F∆r cos θ = 70.0 N 5.00 m cos 20.0° = 329 J

(b)

W = F∆r cos θ = 123 N 5.00 m cos 90.0° = 0 J

(c)

W = F∆r cos θ = 147 N 5.00 m cos 90.0° = 0

(d)

∆Eint = F∆x = 36.9 N 5.00 m = 185 J

(e)

∆K = K f − K i = ∑ W − ∆Eint = 329 J − 185 J = +144 J

a

fa

f

FIG. P7.34

201

202 P7.35

Energy and Energy Transfer

µ k = 0.100 1 K i − f k ∆x + Wother = K f : mvi2 − f k ∆x = 0 2 2 2.00 m s v2 1 ∆x = i = = 2.04 m mvi2 = µ k mg∆x 2 2 µ k g 2 0.100 9.80 vi = 2.00 m s

b a

Section 7.8 *P7.36

Pav =

g

fa f

Power

b

W K f mv 2 0.875 kg 0.620 m s = = = 2 ∆t ∆t ∆t 2 21 × 10 −3 s

e

W t

P=

j

a

g

2

= 8.01 W

fa

f

mgh 700 N 10.0 m = = 875 W t 8.00 s

P7.37

Power =

P7.38

A 1 300-kg car speeds up from rest to 55.0 mi/h = 24.6 m/s in 15.0 s. The output work of the engine is equal to its final kinetic energy,

b

gb

1 1 300 kg 24.6 m s 2 with power P = P7.39

(a)

g

2

= 390 kJ

390 000 J ~ 10 4 W around 30 horsepower. 15.0 s

∑ W = ∆K , but ∆K = 0 because he moves at constant speed. The skier rises a vertical

a

f

distance of 60.0 m sin 30.0° = 30.0 m . Thus,

b

ja

ge

f

Win = −Wg = 70.0 kg 9.8 m s 2 30.0 m = 2.06 × 10 4 J = 20.6 kJ . (b)

The time to travel 60.0 m at a constant speed of 2.00 m/s is 30.0 s. Thus,

Pinput = P7.40

(a)

W 2.06 × 10 4 J = = 686 W = 0.919 hp . ∆t 30.0 s

The distance moved upward in the first 3.00 s is ∆y = vt =

LM 0 + 1.75 m s OPa3.00 sf = 2.63 m . N 2 Q

The motor and the earth’s gravity do work on the elevator car: 1 1 mvi2 + Wmotor + mg∆y cos 180° = mv 2f 2 2 1 2 Wmotor = 650 kg 1.75 m s − 0 + 650 kg g 2.63 m = 1.77 × 10 4 J 2

b

Also, W = P t so P = (b)

gb

g

ga

b

f

W 1.77 × 10 4 J = = 5.91 × 10 3 W = 7.92 hp. t 3.00 s

b

g

When moving upward at constant speed v = 1.75 m s the applied force equals the

b

ge

j = 6.37 × 10 N . Therefore, P = Fv = e6.37 × 10 N jb1.75 m sg = 1.11 × 10

weight = 650 kg 9.80 m s

2

3

3

4

W = 14.9 hp .

Chapter 7

P7.41

203

energy = power × time For the 28.0 W bulb:

a

fe

j

Energy used = 28.0 W 1.00 × 10 4 h = 280 kilowatt ⋅ hrs

a

fb

g

total cost = $17.00 + 280 kWh $0.080 kWh = $39.40 For the 100 W bulb:

a

fe

j

Energy used = 100 W 1.00 × 10 4 h = 1.00 × 10 3 kilowatt ⋅ hrs 1.00 × 10 4 h = 13.3 # bulb used = 750 h bulb

b

g e

jb

g

total cost = 13.3 $0.420 + 1.00 × 10 3 kWh $0.080 kWh = $85.60 Savings with energy-efficient bulb = $85.60 − $39.40 = $46.20

*P7.42

(a)

FG 454 g IJ FG 9 kcal IJ FG 4 186 J IJ = 1.71 × 10 H 1 lb K H 1 g K H 1 kcal K

Burning 1 lb of fat releases energy

1 lb

The mechanical energy output is

e1.71 × 10 Jja0.20f = nF∆r cos θ .

Then

3.42 × 10 6 J = nmg∆y cos 0°

7

J.

7

b ge jb J = ne5.88 × 10 Jj

ga

f

3.42 × 10 6 J = n 50 kg 9.8 m s 2 80 steps 0.150 m 3.42 × 10 6

3

3.42 × 10 6 J = 582 . 5.88 × 10 3 J This method is impractical compared to limiting food intake. where the number of times she must climb the steps is n =

(b)

Her mechanical power output is

P=

*P7.43

(a)

(b)

FG H

IJ K

1 hp W 5.88 × 10 3 J = = 90.5 W = 90.5 W = 0.121 hp . t 65 s 746 W

IJ FG 1 kcal IJ FG 1.30 × 10 J IJ = K H 4 186 J K H 1 gal K 1 h F 10 mi I F 1 kcal I F 1.30 × 10 J I = 776 mi gal . For bicycling G J 400 kcal H h K GH 4 186 J JK GH 1 gal JK

The fuel economy for walking is

FG H

8

1h 3 mi 220 kcal h

8

423 mi gal .

204

Energy and Energy Transfer

Section 7.9 P7.44

Energy and the Automobile

At a speed of 26.8 m/s (60.0 mph), the car described in Table 7.2 delivers a power of P1 = 18.3 kW to the wheels. If an additional load of 350 kg is added to the car, a larger output power of

P2 = P1 + (power input to move 350 kg at speed v) will be required. The additional power output needed to move 350 kg at speed v is:

b g b

g

∆Pout = ∆f v = µ r mg v . Assuming a coefficient of rolling friction of µ r = 0.016 0 , the power output now needed from the engine is

b

gb

ge

jb

g

P2 = P1 + 0.016 0 350 kg 9.80 m s 2 26.8 m s = 18.3 kW + 1.47 kW . With the assumption of constant efficiency of the engine, the input power must increase by the same factor as the output power. Thus, the fuel economy must decrease by this factor:

bfuel economyg = FGH PP IJK bfuel economyg = FGH 18.318+.31.47 IJK b6.40 km Lg or bfuel economy g = 5.92 km L . 1

2

1

2

2

P7.45

(a)

fuel needed = =

(b)

(c)

1 2

mv 2f − 12 mvi2

useful energy per gallon

b900 kggb24.6 m sg = a0.150fe1.34 × 10 J galj

=

b

mv 2f − 0

eff.× energy content of fuel

2

1 2

8

1.35 × 10 −2 gal

73.8 power =

FG 1 gal IJ FG 55.0 mi IJ FG 1.00 h IJ FG 1.34 × 10 H 38.0 mi K H 1.00 h K H 3 600 s K H 1 gal

Additional Problems P7.46

1 2

b g b g At apex, v = b 40.0 m sg cos 30.0° i + 0 j = b34.6 m sgi 1 1 And K = mv = b0.150 kg gb34.6 m sg = 90.0 J 2 2

At start, v = 40.0 m s cos 30.0° i + 40.0 m s sin 30.0° j

2

2

8

J

I a0.150f = JK

8.08 kW

g

Chapter 7

P7.47

b

gb

Concentration of Energy output = 0.600 J kg ⋅ step 60.0 kg

b

gb

1 step I gFGH 1.50 J = 24.0 J m mK

g

F = 24.0 J m 1 N ⋅ m J = 24.0 N

P = Fv

a

f

70.0 W = 24.0 N v v = 2.92 m s P7.48

(a)

a fa f

A ⋅ i = A 1 cos α . But also, A ⋅ i = A x .

a Afa1f cos α = A

Thus, Similarly,

cos β =

and

cos γ =

Ax . A

Ay A Az A

A = A x2 + A y2 + A z2 .

where

P7.49

or cos α =

x

(b)

cos 2 α + cos 2 β + cos 2 γ =

(a)

x = t + 2.00t 3

FG A IJ + FG A IJ + FG A IJ H AK H AK H AK x

2

y

2

z

2

=

A2 =1 A2

Therefore, dx = 1 + 6.00t 2 dt 1 1 K = mv 2 = 4.00 1 + 6.00t 2 2 2 v=

a fe

a12.0tf m s F = ma = 4.00a12.0t f = a 48.0t f N dv = dt

2

(b)

a=

(c)

P = Fv = 48.0t 1 + 6.00t 2 =

(d)

W=

a

z

fe

ze

2.00

2 .00

0

0

Pdt =

j e48.0t + 288t j W 3

j

48.0t + 288 t 3 dt = 1 250 J

j = e2.00 + 24.0t 2

2

j

+ 72.0t 4 J

205

206 *P7.50

Energy and Energy Transfer

(a)

We write F = ax b

a f 5 000 N = aa0.315 mf F 0.315 IJ = 2.44 5=G H 0.129 K

b

1 000 N = a 0.129 m b

b

b

ln 5 = b ln 2.44 ln 5 = 1.80 = b ln 2.44 1 000 N a= = 4.01 × 10 4 N m1.8 = a 1.80 0.129 m

b=

a

(b)

W=

z

z

0. 25 m

0. 25 m

0

0

Fdx =

= 4.01 × 10

4

4.01 × 10 4

N x 2 .8 m1.8 2.8

f

N 1.8 x dx m1.8

0. 25 m

= 4.01 × 10

4

0

a

f

N 0.25 m 2.8 m1.8

2.8

= 294 J *P7.51

The work done by the applied force is

z f

W = Fapplied dx =

z

x max

e

0

i

z

z

x max

j

− − k1 x + k 2 x 2 dx

x max

x2 = k 1 x dx + k 2 x dx = k1 2 0 0 = k1 P7.52

(a)

2 x max

2

+ k2

2

x max 0

x3 + k2 3

xmax 0

3 x max

3

The work done by the traveler is mghs N where N is the number of steps he climbs during the ride. N = (time on escalator)(n) where

h atime on escalatorf = vertical velocity of person

and

vertical velocity of person = v + nhs

Then,

N=

nh v + nhs

and the work done by the person becomes Wperson = continued on next page

mgnhhs v + nhs

Chapter 7

(b)

207

The work done by the escalator is

b

ga f a

fb

ga f

We = power time = force exerted speed time = mgvt h as above. v + nhs

where

t=

Thus,

We =

mgvh . v + nhs

As a check, the total work done on the person’s body must add up to mgh, the work an elevator would do in lifting him.

∑ W = Wperson + We =

It does add up as follows:

P7.53

(a)

(b) *P7.54

∆K =

1 mv 2 − 0 = ∑ W , so 2

v2 =

2W and v = m

b

2W m

W = F ⋅ d = Fx d ⇒ Fx =

W d

During its whole motion from y = 10.0 m to y = −3.20 mm, the force of gravity and the force of the plate do work on the ball. It starts and ends at rest Ki + ∑ W = K f 0 + Fg ∆y cos 0°+ Fp ∆x cos 180° = 0

b

g b g 5 kg e9.8 m s ja10 mf = = 1.53 × 10

mg 10.003 2 m − Fp 0.003 20 m = 0 2

Fp

P7.55

g

mgnhhs mgvh mgh nhs + v + = = mgh v + nhs v + nhs v + nhs

b

F t = m

OPa3.00 sf = PQ

240 W

P = Fv = F vi + at = F 0 +

(b)

P=

2

m

IJ FG F IJ t K HmK

g FGH

(a)

LM a20.0 Nf MN 5.00 kg

3.2 × 10

−3

2

5

N upward

208

Energy and Energy Transfer

z f

*P7.56

(a)

W1 = F1 dx =

z

xi 1 + x a

i

(b)

W2 =

k1 x dx =

xi 1

z

− xi 2 + x a

k 2 x dx =

− xi 2

(c)

1 k 1 x i1 + x a 2

b

1 k 2 − xi 2 + x a 2

b

g

2

g

2

− xi21 =

− xi22 =

e

(a)

j

j

Before the horizontal force is applied, the springs exert equal forces: k 1 xi1 = k 2 xi 2 k1 xi1 k2

1 1 k1 x a2 + k1 x a xi1 + k 2 x a2 − k 2 x a xi 2 2 2 k x 1 1 2 2 = k 1 x a + k 2 x a + k1 x a xi1 − k 2 x a 1 i1 2 2 k2 1 = k1 + k 2 x a2 2

W1 + W2 =

b

*P7.57

e

1 k 2 x a2 − 2 x a xi 2 2

xi 2 =

(d)

1 k1 x a2 + 2 x a xi1 2

g

z ze t

t

0

0

v = a dt =

j

1.16t − 0.21t 2 + 0.24t 3 dt

t2 t3 t4 = 1.16 − 0.21 + 0. 24 2 3 4

t

= 0.58t 2 − 0.07t 3 + 0.06t 4 0

At t = 0 , vi = 0. At t = 2.5 s ,

ja f − e0.07 m s ja2.5 sf + e0.06 m s ja2.5 sf

e

v f = 0.58 m s 3 2.5 s

2

3

4

5

4

= 4.88 m s

Ki + W = K f 0+W = (b)

b

1 1 mv 2f = 1 160 kg 4.88 m s 2 2

g

2

= 1.38 × 10 4 J

At t = 2.5 s ,

e

j

e

ja f + e0.240 m s ja2.5 sf

a = 1.16 m s 3 2.5 s − 0.210 m s 4 2.5 s

2

5

3

Through the axles the wheels exert on the chassis force

∑ F = ma = 1 160 kg 5.34

m s 2 = 6.19 × 10 3 N

and inject power

b

g

P = Fv = 6.19 × 10 3 N 4.88 m s = 3.02 × 10 4 W .

= 5.34 m s 2 .

Chapter 7

P7.58

(a)

The new length of each spring is

x 2 + L2 , so its extension is

x 2 + L2 − L and the force it exerts is k

FH

IK

x 2 + L2 − L toward its

fixed end. The y components of the two spring forces add to zero. Their x components add to F = −2 ik

FH

x 2 + L2 − L

IK

x x 2 + L2

F GH

= −2 kx i 1 −

z

L x 2 + L2

z

f

(b)

0

W = Fx dx

FIG. P7.58

.

F GH

I dx J x +L K ex + L j + kL b1 2g

W = −2 kx 1 −

i

A

z 0

W = −2 k x dx + kL A

ze 0

A

j

2 −1 2

x2 + L

x2 W = −2 k 2

2 x dx

0

A

L

2

2

2

0 2 12

A

W = 2 kL2 + kA 2 − 2 kL A 2 + L2

W = −0 + kA 2 + 2 kL2 − 2 kL A 2 + L2 *P7.59

I JK

For the rocket falling at terminal speed we have

∑ F = ma + R − Mg = 0 Mg = (a)

1 DρAvT2 2

For the rocket with engine exerting thrust T and flying up at the same speed,

∑ F = ma +T − Mg − R = 0 T = 2 Mg The engine power is P = Fv = TvT = 2 MgvT . (b)

For the rocket with engine exerting thrust Tb and flying down steadily at 3vT , 1 2 Rb = DρA 3 vT = 9 Mg 2

b g

∑ F = ma −Tb − Mg + 9 Mg = 0 Tb = 8 Mg The engine power is P = Tv = 8 Mg 3 vT = 24MgvT .

209

210 P7.60

Energy and Energy Transfer

(a)

a fe j e20.5i + 14.3 jj N = a 42.0 N fecos 150° i + sin 150° jj = e−36.4i + 21.0 jj N

F1 = 25.0 N cos 35.0° i + sin 35.0° j = F2

e−15.9 i + 35.3 jj N

(b)

∑ F = F1 + F2 =

(c)

a=

(d)

v f = v i + at = 4.00 i + 2.50 j m s + −3.18 i + 7.07 j m s 2 3.00 s

∑F =

e−3.18 i + 7.07 jj m s e

vf = (e)

m

2

j

e

ja

je

e−5.54i + 23.7 jj m s

r f = ri + v i t +

e

1 2 at 2

f 12 e−3.18i + 7.07 jjem s ja3.00 sf

jb ga e−2.30i + 39.3 jj m

2

r f = 0 + 4.00 i + 2.50 j m s 3.00 s + ∆r = r f = (f)

Kf =

(g)

Kf =

b

1 1 mv 2f = 5.00 kg 2 2

(a)

g a5.54f + a23.7f em s j = 2

1 mvi2 + ∑ F ⋅ ∆r 2 1 2 K f = 5.00 kg 4.00 + 2.50 2 K f = 55.6 J + 1 426 J = 1.48 kJ

b

P7.61

f

2

2

2

1.48 kJ

g a f a f bm sg + a−15.9 Nfa−2.30 mf + a35.3 Nfa39.3 mf

∑ W = ∆K :

2

2

Ws + W g = 0

a

f

j a

f a

1 2 kxi − 0 + mg∆x cos 90°+60° = 0 2 1 2 1.40 × 10 3 N m × 0.100 − 0.200 9.80 sin 60.0° ∆x = 0 2 ∆x = 4.12 m

e

(b)

∑ W = ∆K + ∆Eint :

fa fa

f

Ws + W g − ∆Eint = 0

1 2 kxi + mg∆x cos 150°− µ k mg cos 60° ∆x = 0 2 1 2 1.40 × 10 3 N m × 0.100 − 0.200 9.80 sin 60.0° ∆x − 0.200 9.80 0.400 cos 60.0° ∆x = 0 2 ∆x = 3.35 m

e

j a

f a

fa fa

f

a

fa fa

fa

f

Chapter 7

P7.62

(a)

211

a f Lammf FaNf Lammf

FN 2.00

15.0

14.0

112

4.00

32.0

16.0

126

6.00

49.0

18.0

149

8.00

64.0

20.0

175

10.0

79.0

22.0

190

12.0

98.0 FIG. P7.62

(b)

A straight line fits the first eight points, together with the origin. By least-square fitting, its slope is 0.125 N mm ± 2% = 125 N m ± 2% In F = kx , the spring constant is k =

(c) P7.63

b

F , the same as the slope of the F-versus-x graph. x

f

ga

F = kx = 125 N m 0.105 m = 13.1 N

K i + Ws + W g = K f 1 1 1 1 mvi2 + kx i2 − kx 2f + mg∆x cos θ = mv 2f 2 2 2 2 1 2 1 FIG. P7.63 0 + kxi − 0 + mgxi cos 100° = mv 2f 2 2 1 1 1.20 N cm 5.00 cm 0.050 0 m − 0.100 kg 9.80 m s 2 0.050 0 m sin 10.0° = 0.100 kg v 2 2 2 −3 2 0.150 J − 8.51 × 10 J = 0.050 0 kg v

b

v=

P7.64

ga

fb b

g b g

ge

jb

g

b

0.141 = 1.68 m s 0.050 0

b

gea6.00f − a8.00f jbm sg

b

ge

1 1 m v 2f − vi2 : ∆Eint = − 0.400 kg 2 2

e

j

2

2

(a)

∆Eint = − ∆K = −

(b)

∆Eint = f∆r = µ k mg 2πr :

5.60 J = µ k 0.400 kg 9.80 m s 2 2π 1.50 m

Thus,

µ k = 0.152 .

(c)

g

a f

j a

2

= 5.60 J

f

After N revolutions, the object comes to rest and K f = 0 . 1 mvi2 2

Thus,

∆Eint = − ∆K = −0 + K i =

or

µ k mg N 2πr =

This gives

b8.00 m sg = = N= µ mg a 2πr f a0.152fe9.80 m s j2π a1.50 mf

a f

1 2

k

mvi2

1 mvi2 . 2 2

1 2

2

2.28 rev .

212 P7.65

Energy and Energy Transfer

If positive F represents an outward force, (same as direction as r), then

z

ze

rf

f

W = F ⋅ dr = i

j

2 F0σ 13 r −13 − F0σ 7 r −7 dr

ri

2 F σ 13 r −12 F0σ 7 r −6 − W= 0 −12 −6 W=

− F0σ

13

e

r f−12

− ri−12

6

W = 1.03 × 10

−77

r f−6

rf ri

j + F σ er

− ri−6

0

7

−6 f

− ri−6

6

j= Fσ 0

6

− 1.89 × 10

−134

r f−12

7

r f−6 − ri−6 −

F0 σ 13 −12 r f − ri−12 6

− ri−12

W = 1.03 × 10 −77 1.88 × 10 −6 − 2.44 × 10 −6 10 60 − 1.89 × 10 −134 3.54 × 10 −12 − 5.96 × 10 −8 10 120 W = −2.49 × 10 −21 J + 1.12 × 10 −21 J = −1.37 × 10 −21 J P7.66

P∆t = W = ∆K =

a ∆m f v

2

2

ρ=

The density is

∆m ∆m = . vol A∆x

Substituting this into the first equation and solving for P , since 3

for a constant speed, we get

P=

ρAv 2

Also, since P = Fv,

F=

ρAv 2 . 2

∆x = v, ∆t

FIG. P7.66

.

Our model predicts the same proportionalities as the empirical equation, and gives D = 1 for the drag coefficient. Air actually slips around the moving object, instead of accumulating in front of it. For this reason, the drag coefficient is not necessarily unity. It is typically less than one for a streamlined object and can be greater than one if the airflow around the object is complicated.

z

23 .7

P7.67

We evaluate

375dx by calculating 12.8 x + 3.75 x 3

a

f + 375a0.100f + … 375a0.100f = 0.806 a12.8f + 3.75a12.8f a12.9f + 3.75a12.9f a23.6f + 3.75a23.6f 375 0.100 3

3

3

and

a

f + 375a0.100f + … 375a0.100f = 0.791 . a12.9f + 3.75a12.9f a13.0f + 3.75a13.0f a23.7f + 3.75a23.7f 375 0.100 3

3

3

The answer must be between these two values. We may find it more precisely by using a value for ∆x smaller than 0.100. Thus, we find the integral to be 0.799 N ⋅ m .

Chapter 7

*P7.68

P=

1 Dρπr 2 v 3 2

ja

f b8 m sg

1 1 1.20 kg m3 π 1.5 m 2

e

(a)

Pa =

(b)

24 m s Pb v b3 = 3 = Pa v a 8 ms

e

F GH

I JK

2

3

= 2.17 × 10 3 W

3

= 3 3 = 27

j

Pb = 27 2.17 × 10 3 W = 5.86 × 10 4 W P7.69

(a)

The suggested equation P∆t = bwd implies all of the following cases: (1)

(3)

FG w IJ a2df H 2K F ∆t I F d I P G J = bwG J H 2 K H 2K P∆t = b

(2)

and

(4)

FG ∆t IJ = bFG w IJ d H 2 K H 2K FG P IJ ∆t = bFG w IJ d H 2K H 2K P

v = constant n

d

fk = µ k n

F

w

These are all of the proportionalities Aristotle lists.

FIG. P7.69 (b)

For one example, consider a horizontal force F pushing an object of weight w at constant velocity across a horizontal floor with which the object has coefficient of friction µ k .

∑ F = ma implies that: +n − w = 0 and F − µ k n = 0 so that F = µ k w As the object moves a distance d, the agent exerting the force does work W = Fd cos θ = Fd cos 0° = µ k wd and puts out power P =

W ∆t

This yields the equation P∆t = µ k wd which represents Aristotle’s theory with b = µ k . Our theory is more general than Aristotle’s. Ours can also describe accelerated motion. *P7.70

(a)

So long as the spring force is greater than the friction force, the block will be gaining speed. The block slows down when the friction force becomes the greater. It has maximum speed when kx a − f k = ma = 0.

e1.0 × 10 (b)

3

j

N m x a − 4.0 N = 0

0

x = −4.0 × 10 −3 m 0

By the same logic,

e1.0 × 10

3

j

N m x b − 10.0 N = 0

x = −1.0 × 10 −2 m

FIG. P7.70

213

214

Energy and Energy Transfer

ANSWERS TO EVEN PROBLEMS P7.2

1.59 × 10 3 J

P7.44

5.92 km L

P7.4

(a) 3.28 × 10 −2 J ; (b) −3.28 × 10 −2 J

P7.46

90.0 J

P7.6

see the solution

P7.8

5.33 W

P7.10

16.0

P7.12

(a) see the solution; (b) −12.0 J

P7.48

Ay Ax A ; cos β = ; cos γ = z ; A A A (b) see the solution (a) cos α =

P7.50

(a) a =

P7.52

(a)

40.1 kN ; b = 1.80 ; (b) 294 J m 1.8

mgnhhs mgvh ; (b) v + nhs v + nhs

P7.14

50.0 J

P7.16

(a) 575 N m ; (b) 46.0 J

P7.54

1.53 × 10 5 N upward

P7.18

(a) 9.00 kJ; (b) 11.7 kJ, larger by 29.6%

P7.56

see the solution

P7.20

(a) see the solution; (b) mgR

P7.58

(a) see the solution;

P7.22

mg mg 1 1 + + ; (b) (a) k1 k 2 k1 k2

FG H

IJ K

(b) 2 kL2 + kA 2 − 2 kL A 2 + L2

−1

P7.24

(a) 1.20 J; (b) 5.00 m s ; (c) 6.30 J

P7.26

(a) 60.0 J; (b) 60.0 J

P7.60

e

j

(a) F1 = 20.5 i + 14.3 j N ;

e

j

F2 = −36.4i + 21.0 j N ;

e j (c) e −3.18 i + 7.07 jj m s ; (d) e −5.54i + 23.7 jj m s ; (e) e −2.30 i + 39.3 jj m ; (f) 1.48 kJ; (g) 1.48 kJ (b) −15.9 i + 35.3 j N ;

2

P7.28

(a) 1.94 m s ; (b) 3.35 m s ; (c) 3.87 m s

P7.30

(a) 3.78 × 10 −16 J ; (b) 1.35 × 10 −14 N ; (c) 1.48 × 10 +16 m s 2 ; (d) 1.94 ns

P7.32

(a) 0.791 m s; (b) 0.531 m s

P7.34

(a) 329 J; (b) 0; (c) 0; (d) 185 J; (e) 144 J

P7.36

8.01 W

P7.38

~ 10 4 W

P7.40

(a) 5.91 kW; (b) 11.1 kW

P7.42

No. (a) 582; (b) 90.5 W = 0.121 hp

P7.62

(a) see the solution; (b) 125 N m ± 2% ; (c) 13.1 N

P7.64

(a) 5.60 J; (b) 0.152; (c) 2.28 rev

P7.66

see the solution

P7.68

(a) 2.17 kW; (b) 58.6 kW

P7.70

(a) x = −4.0 mm ; (b) −1.0 cm

8 Potential Energy CHAPTER OUTLINE 8.1 8.2

8.3 8.4

8.5

8.6

Q8.4

Potential Energy of a System The Isolated System—Conservation of Mechanical Energy Conservative and Nonconservative Forces Changes in Mechanical Energy for Nonconservative Forces Relationship Between Conservative Forces and Potential Energy Energy Diagrams and the Equilibrium of a System

ANSWERS TO QUESTIONS Q8.1

The final speed of the children will not depend on the slide length or the presence of bumps if there is no friction. If there is friction, a longer slide will result in a lower final speed. Bumps will have the same effect as they effectively lengthen the distance over which friction can do work, to decrease the total mechanical energy of the children.

Q8.2

Total energy is the sum of kinetic and potential energies. Potential energy can be negative, so the sum of kinetic plus potential can also be negative.

Q8.3

Both agree on the change in potential energy, and the kinetic energy. They may disagree on the value of gravitational potential energy, depending on their choice of a zero point.

(a)

mgh is provided by the muscles.

(b)

No further energy is supplied to the object-Earth system, but some chemical energy must be supplied to the muscles as they keep the weight aloft.

(c)

The object loses energy mgh, giving it back to the muscles, where most of it becomes internal energy.

Q8.5

Lift a book from a low shelf to place it on a high shelf. The net change in its kinetic energy is zero, but the book-Earth system increases in gravitational potential energy. Stretch a rubber band to encompass the ends of a ruler. It increases in elastic energy. Rub your hands together or let a pearl drift down at constant speed in a bottle of shampoo. Each system (two hands; pearl and shampoo) increases in internal energy.

Q8.6

Three potential energy terms will appear in the expression of total mechanical energy, one for each conservative force. If you write an equation with initial energy on one side and final energy on the other, the equation contains six potential-energy terms.

215

216 Q8.7

Potential Energy

(a)

It does if it makes the object’s speed change, but not if it only makes the direction of the velocity change.

(b)

Yes, according to Newton’s second law.

Q8.8

The original kinetic energy of the skidding can be degraded into kinetic energy of random molecular motion in the tires and the road: it is internal energy. If the brakes are used properly, the same energy appears as internal energy in the brake shoes and drums.

Q8.9

All the energy is supplied by foodstuffs that gained their energy from the sun.

Q8.10

Elastic potential energy of plates under stress plus gravitational energy is released when the plates “slip”. It is carried away by mechanical waves.

Q8.11

The total energy of the ball-Earth system is conserved. Since the system initially has gravitational energy mgh and no kinetic energy, the ball will again have zero kinetic energy when it returns to its original position. Air resistance will cause the ball to come back to a point slightly below its initial position. On the other hand, if anyone gives a forward push to the ball anywhere along its path, the demonstrator will have to duck.

Q8.12

Using switchbacks requires no less work, as it does not change the change in potential energy from top to bottom. It does, however, require less force (of static friction on the rolling drive wheels of a car) to propel the car up the gentler slope. Less power is required if the work can be done over a longer period of time.

Q8.13

There is no work done since there is no change in kinetic energy. In this case, air resistance must be negligible since the acceleration is zero.

Q8.14

There is no violation. Choose the book as the system. You did work and the earth did work on the book. The average force you exerted just counterbalanced the weight of the book. The total work on the book is zero, and is equal to its overall change in kinetic energy.

Q8.15

Kinetic energy is greatest at the starting point. Gravitational energy is a maximum at the top of the flight of the ball.

Q8.16

Gravitational energy is proportional to mass, so it doubles.

Q8.17

In stirring cake batter and in weightlifting, your body returns to the same conformation after each stroke. During each stroke chemical energy is irreversibly converted into output work (and internal energy). This observation proves that muscular forces are nonconservative.

Chapter 8

Q8.18

Let the gravitational energy be zero at the lowest point in the motion. If you start the vibration by pushing down on the block (2), its kinetic energy becomes extra elastic potential energy in the spring ( Us ). After the block starts moving up at its lower turning point (3), this energy becomes both kinetic energy (K) and gravitational potential energy ( U g ), and then just gravitational energy when the block is at its greatest height (1). The energy then turns back into kinetic and elastic potential energy, and the cycle repeats.

Q8.19

217

FIG. Q8.18

(a)

Kinetic energy of the running athlete is transformed into elastic potential energy of the bent pole. This potential energy is transformed to a combination of kinetic energy and gravitational potential energy of the athlete and pole as the athlete approaches the bar. The energy is then all gravitational potential of the pole and the athlete as the athlete hopefully clears the bar. This potential energy then turns to kinetic energy as the athlete and pole fall to the ground. It immediately becomes internal energy as their macroscopic motion stops.

(b)

Rotational kinetic energy of the athlete and shot is transformed into translational kinetic energy of the shot. As the shot goes through its trajectory as a projectile, the kinetic energy turns to a mix of kinetic and gravitational potential. The energy becomes internal energy as the shot comes to rest.

(c)

Kinetic energy of the running athlete is transformed to a mix of kinetic and gravitational potential as the athlete becomes projectile going over a bar. This energy turns back into kinetic as the athlete falls down, and becomes internal energy as he stops on the ground.

The ultimate source of energy for all of these sports is the sun. See question 9. Q8.20

Chemical energy in the fuel turns into internal energy as the fuel burns. Most of this leaves the car by heat through the walls of the engine and by matter transfer in the exhaust gases. Some leaves the system of fuel by work done to push down the piston. Of this work, a little results in internal energy in the bearings and gears, but most becomes work done on the air to push it aside. The work on the air immediately turns into internal energy in the air. If you use the windshield wipers, you take energy from the crankshaft and turn it into extra internal energy in the glass and wiper blades and wiper-motor coils. If you turn on the air conditioner, your end effect is to put extra energy out into the surroundings. You must apply the brakes at the end of your trip. As soon as the sound of the engine has died away, all you have to show for it is thermal pollution.

Q8.21

A graph of potential energy versus position is a straight horizontal line for a particle in neutral equilibrium. The graph represents a constant function.

Q8.22

The ball is in neutral equilibrium.

Q8.23

The ball is in stable equilibrium when it is directly below the pivot point. The ball is in unstable equilibrium when it is vertically above the pivot.

218

Potential Energy

SOLUTIONS TO PROBLEMS Section 8.1

Potential Energy of a System

P8.1

With our choice for the zero level for potential energy when the car is at point B,

(a)

UB = 0 . When the car is at point A, the potential energy of the car-Earth system is given by

FIG. P8.1

U A = mgy where y is the vertical height above zero level. With 135 ft = 41.1 m , this height is found as:

a

f

y = 41.1 m sin 40.0° = 26.4 m . Thus,

b

ge

ja

f

U A = 1 000 kg 9.80 m s 2 26.4 m = 2.59 × 10 5 J . The change in potential energy as the car moves from A to B is

U B − U A = 0 − 2.59 × 10 5 J = −2.59 × 10 5 J . (b)

With our choice of the zero level when the car is at point A, we have U A = 0 . The potential energy when the car is at point B is given by U B = mgy where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5 m. Because this distance is now below the zero reference level, it is a negative number. Thus,

b

ge

ja

f

U B = 1 000 kg 9.80 m s 2 −26.5 m = −2.59 × 10 5 J . The change in potential energy when the car moves from A to B is

U B − U A = −2.59 × 10 5 J − 0 = −2.59 × 10 5 J .

Chapter 8

P8.2

(a)

219

We take the zero configuration of system potential energy with the child at the lowest point of the arc. When the string is held horizontal initially, the initial position is 2.00 m above the zero level. Thus,

a

fa

f

U g = mgy = 400 N 2.00 m = 800 J . (b)

From the sketch, we see that at an angle of 30.0° the child is at a vertical height of 2.00 m 1 − cos 30.0° above the lowest point of the arc. Thus,

a

fa

f

a

FIG. P8.2

fa

fa

f

U g = mgy = 400 N 2.00 m 1 − cos 30.0° = 107 J . (c)

The zero level has been selected at the lowest point of the arc. Therefore, U g = 0 at this location.

*P8.3

The volume flow rate is the volume of water going over the falls each second:

a

fb

g

3 m 0.5 m 1.2 m s = 1.8 m3 s The mass flow rate is

m V = ρ = 1 000 kg m3 1.8 m3 s = 1 800 kg s t t

e

je

j

If the stream has uniform width and depth, the speed of the water below the falls is the same as the speed above the falls. Then no kinetic energy, but only gravitational energy is available for conversion into internal and electric energy.

ja f

energy mgy m = = gy = 1 800 kg s 9.8 m s 2 5 m = 8.82 × 10 4 J s t t t The output power is Puseful = efficiency Pin = 0.25 8.82 × 10 4 W = 2.20 × 10 4 W The input power is Pin =

b

g

b e

ge

j

The efficiency of electric generation at Hoover Dam is about 85%, with a head of water (vertical drop) of 174 m. Intensive research is underway to improve the efficiency of low head generators.

Section 8.2 *P8.4

(a)

The Isolated System—Conservation of Mechanical Energy One child in one jump converts chemical energy into mechanical energy in the amount that her body has as gravitational energy at the top of her jump: mgy = 36 kg 9.81 m s 2 0.25 m = 88.3 J . For all of the jumps of the children the energy is

e ja f 12e1.05 × 10 j88.3 J = 1.11 × 10 J . 6

(b)

9

0.01 1.11 × 10 9 J = 1.11 × 10 5 J , making the Richter 100 log E − 4.8 log 1.11 × 10 5 − 4.8 5.05 − 4.8 magnitude = = = 0.2 . 1.5 1.5 1.5

The seismic energy is modeled as E =

220 P8.5

Potential Energy

a f 12 mv 1 g a3.50 Rf = 2 g a Rf + v 2

Ui + K i = U f + K f :

mgh + 0 = mg 2 R +

2

2

v = 3.00 gR

∑F = m

2

v : R

n + mg = m

v2 R

LM v − g OP = m L 3.00 gR − g O = 2.00mg N R Q MN R PQ n = 2.00e5.00 × 10 kg je9.80 m s j 2

n=m

−3

2

= 0.098 0 N downward P8.6

FIG. P8.5

K i + Ui = K f + U f 1 2 m 6.00 m s + 0 = 0 + m 9.80 m s 2 y 2

From leaving ground to the highest point,

b

g

e

b6.00 m sg = ∴y = a2fe9.80 m s j

j

2

The mass makes no difference:

*P8.7

(a)

2

1.84 m

1 1 1 1 mvi2 + kx i2 = mv 2f + kx 2f 2 2 2 2 1 1 2 0 + 10 N m −0.18 m = 0.15 kg v 2f + 0 2 2

f b ga g F 10 N I FG 1 kg ⋅ m IJ = = a0.18 mf G H 0.15 kg ⋅ m JK H 1 N ⋅ s K b

vf (b)

2

K i + U si = K f + U sf 1 0 + 10 N m −0.18 m 2

b

0.162 J = vf =

f

ga

a

b

g

2

b

1 0.15 kg v 2f 2 1 + 10 N m 0. 25 m − 0.18 m 2 =

b

1 0.15 kg v 2f + 0.024 5 J 2

f

g ga

1.47 m s

2 0.138 J = 1.35 m s 0.15 kg

f

2

FIG. P8.7

Chapter 8

*P8.8

The energy of the car is E =

221

1 mv 2 + mgy 2

1 mv 2 + mgd sin θ where d is the distance it has moved along the track. 2 dE dv P= = mv + mgv sin θ dt dt E=

(a)

When speed is constant,

jb

e

g

P = mgv sin θ = 950 kg 9.80 m s 2 2.20 m s sin 30° = 1.02 × 10 4 W (b)

2. 2 m s − 0 dv =a= = 0.183 m s 2 dt 12 s Maximum power is injected just before maximum speed is attained:

b

ge

j

P = mva + mgv sin θ = 950 kg 2.2 m s 0.183 m s 2 + 1.02 × 10 4 W = 1.06 × 10 4 W (c)

*P8.9

(a)

At the top end, 1 1 2.20 m s mv 2 + mgd sin θ = 950 kg 2 2

FG b H

g + e9.80 m s j1 250 m sin 30°IJK = 2

2

5.82 × 10 6 J

Energy of the object-Earth system is conserved as the object moves between the release point and the lowest point. We choose to measure heights from y = 0 at the top end of the string.

eK + U j = eK + U j : g

g

i

0 + mgyi =

f

1 mv 2f + mgy f 2

e9.8 m s ja−2 m cos 30°f = 12 v + e9.8 m s ja−2 mf v = 2e9.8 m s ja 2 mfa1 − cos 30°f = 2.29 m s 2

2 f

2

f

(b)

Choose the initial point at θ = 30° and the final point at θ = 15° :

a

f

f 2 gLacos 15°− cos 30°f = 2e9.8 m s ja 2 mfacos 15°− cos 30°f =

0 + mg − L cos 30° = vf = P8.10

2

a

1 mv 2f + mg − L cos 15° 2

2

1.98 m s

Choose the zero point of gravitational potential energy of the object-spring-Earth system as the configuration in which the object comes to rest. Then because the incline is frictionless, we have EB = E A : K B + U gB + U sB = K A + U gA + U sA

a

f

or

0 + mg d + x sin θ + 0 = 0 + 0 +

Solving for d gives

d=

kx 2 −x . 2mg sin θ

1 2 kx . 2

222 P8.11

Potential Energy

From conservation of energy for the block-spring-Earth system, U gt = U si , or

b0.250 kg ge9.80 m s jh = FGH 12 IJK b5 000 N mga0.100 mf 2

2

This gives a maximum height h = 10.2 m . P8.12

(a)

FIG. P8.11

The force needed to hang on is equal to the force F the trapeze bar exerts on the performer. From the free-body diagram for the performer’s body, as shown, F − mg cos θ = m

v2 A

F = mg cos θ + m

v2 A

or FIG. P8.12

Apply conservation of mechanical energy of the performer-Earth system as the performer moves between the starting point and any later point:

b

g

a

f

mg A − A cos θ i = mg A − A cos θ + Solve for (b)

1 mv 2 2

mv 2 and substitute into the force equation to obtain F = mg 3 cos θ − 2 cos θ i A

b

At the bottom of the swing, θ = 0° so

b

F = mg 3 − 2 cos θ i

b

g

F = 2mg = mg 3 − 2 cos θ i which gives

θ i = 60.0° .

g

g

.

223

Chapter 8

P8.13

Using conservation of energy for the system of the Earth and the two objects (a)

b5.00 kg gga4.00 mf = b3.00 kg gga4.00 mf + 12 a5.00 + 3.00fv

2

v = 19.6 = 4.43 m s (b)

Now we apply conservation of energy for the system of the 3.00 kg object and the Earth during the time interval between the instant when the string goes slack and the instant at which the 3.00 kg object reaches its highest position in its free fall. FIG. P8.13

a f

1 3.00 v 2 = mg ∆y = 3.00 g∆y 2 ∆y = 1.00 m y max = 4.00 m + ∆y = 5.00 m P8.14

m1 > m 2 (a)

m1 gh = v=

(b)

b

g

1 m1 + m 2 v 2 + m 2 gh 2

b bm

g g

2 m1 − m 2 gh 1 + m2

Since m 2 has kinetic energy

1 m 2 v 2 , it will rise an additional height ∆h determined from 2 m 2 g ∆h =

1 m2 v 2 2

or from (a), ∆h =

The total height m 2 reaches is h + ∆h = P8.15

b b

g g

m1 − m 2 h v2 = m1 + m 2 2g

2m1 h . m1 + m 2

The force of tension and subsequent force of compression in the rod do no work on the ball, since they are perpendicular to each step of displacement. Consider energy conservation of the ballEarth system between the instant just after you strike the ball and the instant when it reaches the top. The speed at the top is zero if you hit it just hard enough to get it there. K i + U gi = K f + U gf :

a f a fa f

1 mvi2 + 0 = 0 + mg 2L 2 vi = 4 gL = 4 9.80 0.770 vi = 5.49 m s

initial

final L

vi FIG. P8.15

L

224 *P8.16

Potential Energy

efficiency =

e=

useful output energy useful output power = total input energy total input power

m water gy t

b1 2gm ev tj air

2

=

b

g e Av t j

2 ρ water v water t gy

ρ air πr

2

b

g

2 ρ w v w t gy

=

2

2 3

ρ a πr v

where A is the length of a cylinder of air passing through the mill and v w is the volume of water pumped in time t. We need inject negligible kinetic energy into the water because it starts and ends at rest.

e ja fb g e je j F 1 000 L IJ FG 60 s IJ = 160 L min sG H 1 m K H 1 min K 2

3 v w eρ a πr 2 v 3 0.275 1.20 kg m π 1.15 m 11 m s = = 2 ρ w gy t 2 1 000 kg m 3 9.80 m s 2 35 m

= 2.66 × 10 −3 m3 P8.17

(a)

3

K i + U gi = K f + U gf 1 1 mvi2 + 0 = mv 2f + mgy f 2 2 1 1 1 2 2 2 mv xi + mv yi = mv xf + mgy f 2 2 2 But v xi = v xf , so for the first ball

yf =

2 v yi

2g

=

b1 000 sin 37.0°g 2a9.80f

2

= 1.85 × 10 4 m

and for the second

b1 000g = = 2a9.80f 2

yf (b)

5.10 × 10 4 m

The total energy of each is constant with value

b

gb

1 20.0 kg 1 000 m s 2

g

2

= 1.00 × 10 7 J .

3

Chapter 8

P8.18

In the swing down to the breaking point, energy is conserved: mgr cos θ =

1 mv 2 2

at the breaking point consider radial forces

∑ Fr = mar +Tmax − mg cos θ = m Eliminate

v2 r

v2 = 2 g cos θ r Tmax − mg cos θ = 2mg cos θ Tmax = 3mg cos θ

θ = cos −1

F T I = cos GH 3mg JK max

−1

F I GG 3 2.00 kg44.59.N80 m s JJ ge jK Hb 2

θ = 40.8° *P8.19

(a)

For a 5-m cord the spring constant is described by F = kx , mg = k 1.5 m . For a longer cord of length L the stretch distance is longer so the spring constant is smaller in inverse proportion:

a

f

k=

5 m mg = 3.33 mg L L 1.5 m

eK + U

g

+ Us

j = eK + U i

g

+ Us

j

initial

f

1 0 + mgyi + 0 = 0 + mgy f + kx 2f 2 mg 2 1 2 1 mg yi − y f = kx f = 3.33 xf 2 2 L

d

FIG. P8.19(a)

i

here yi − y f = 55 m = L + x f 1 2 55.0 mL = 3.33 55.0 m − L 2 55.0 mL = 5.04 × 10 3 m 2 − 183 mL + 1.67 L2

a

f

0 = 1.67L2 − 238 L + 5.04 × 10 3 = 0 L=

a fe 2a1.67f

238 ± 238 2 − 4 1.67 5.04 × 10 3

j = 238 ± 152 = 3.33

only the value of L less than 55 m is physical. (b)

mg 25.8 m ∑ F = ma k = 3.33

x max = x f = 55.0 m − 25.8 m = 29.2 m + kx max − mg = ma mg 3.33 29. 2 m − mg = ma 25.8 m a = 2.77 g = 27.1 m s 2

final

25.8 m

225

226 *P8.20

Potential Energy

When block B moves up by 1 cm, block A moves down by 2 cm and the separation becomes 3 cm. v h We then choose the final point to be when B has moved up by and has speed A . Then A has 3 2 2h moved down and has speed v A : 3

eK

A

+ KB + Ug

0+0+0=

j = eK i

A

+ KB + Ug

FG IJ H K

v 1 1 mv A2 + m A 2 2 2

2

+

j

f

mgh mg 2 h − 3 3

mgh 5 = mv A2 3 8 8 gh 15

vA =

Section 8.3 P8.21

Conservative and Nonconservative Forces

b

ge

j

y

Fg = mg = 4.00 kg 9.80 m s 2 = 39.2 N (a)

Work along OAC = work along OA + work along AC = Fg OA cos 90.0°+ Fg AC cos 180°

a f a f = a39.2 N fa5.00 mf + a39.2 N fa5.00 mfa −1f

B

C (5.00, 5.00) m

O

A

= −196 J (b)

W along OBC = W along OB + W along BC = 39.2 N 5.00 m cos 180°+ 39.2 N 5.00 m cos 90.0°

a

fa

f

a

fa

f

FIG. P8.21

= −196 J (c)

a f F 1 IJ = 2 mjG − H 2K

Work along OC = Fg OC cos135°

a

fe

= 39.2 N 5.00 ×

−196 J

The results should all be the same, since gravitational forces are conservative. P8.22

(a)

z

W = F ⋅ dr and if the force is constant, this can be written as

z

d

i

W = F ⋅ dr = F ⋅ r f − ri , which depends only on end points, not path. (b)

z

W = F ⋅ dr =

a

f

ze

je j a f z dx + a4.00 Nf z dy + a 4.00 N fy = 15.0 J + 20.0 J = 35.0 J

3 i + 4j ⋅ dx i + dyj = 3.00 N

5.00 m

W = 3.00 N x 0

5.00 m 0

The same calculation applies for all paths.

5.00 m

5.00 m

0

0

x

Chapter 8

P8.23

z

dx i ⋅ 2 y i + x 2 j =

z

dyj ⋅ 2 y i + x 2 j =

5.00 m

WOA =

(a)

0

z

2 ydx

z

x dy

z

x dy

z

2 ydx

j

5.00 m

j

5 .00 m 2

e

j

5.00 m 2

e

j

5.00 m

e

227

0

WOA = 0

and since along this path, y = 0

5 .00 m

W AC =

0

e

For x = 5.00 m,

W AC = 125 J

and

WOAC = 0 + 125 = 125 J WOB =

(b)

z

dyj ⋅ 2 y i + x 2 j =

z

dx i ⋅ 2 yi + x 2 j =

5 .00 m 0

0

0

WOB = 0

since along this path, x = 0 ,

WBC =

5.00 m 0

0

WBC = 50.0 J

since y = 5.00 m,

WOBC = 0 + 50.0 = 50.0 J WOC =

(c)

WOC =

Since x = y along OC,

ze ze

je

5.00 m

j

(d)

F is nonconservative since the work done is path dependent.

(a)

a ∆K f

a

= ∑ W = W g = mg∆h = mg 5.00 − 3.20 1 1 mv B2 − mv 2A = m 9.80 1.80 2 2 v B = 5.94 m s A→B

f

a

(b)

Wg

A →C

A

a fa f

f

Similarly, vC = v A2 + 2 g 5.00 − 2.00 = 7.67 m s

a

f

= mg 3.00 m = 147 J

2

2 x + x 2 dx = 66.7 J

0

P8.24

j z e2ydx + x dyj

dx i + dyj ⋅ 2 y i + x 2 j =

B 5.00 m

3.20 m

FIG. P8.24

C 2.00 m

228 P8.25

Potential Energy

(a)

e

j

F = 3.00 i + 5.00 j N m = 4.00 kg

e

j

r = 2.00 i − 3.00 j m

a f

a

f

W = 3.00 2.00 + 5.00 −3.00 = −9.00 J The result does not depend on the path since the force is conservative. (b)

W = ∆K

Fa f I GH JK

4.00 4.00 v 2 −9.00 = − 4.00 2 2 so v = (c)

2

32.0 − 9.00 = 3.39 m s 2.00

∆U = −W = 9.00 J

Section 8.4

Changes in Mechanical Energy for Nonconservative Forces

P8.26

U f = K i − K f + Ui

(a)

U f = 30.0 − 18.0 + 10.0 = 22.0 J E = 40.0 J

(b) P8.27

Yes, ∆Emech = ∆K + ∆U is not equal to zero. For conservative forces ∆K + ∆U = 0 .

The distance traveled by the ball from the top of the arc to the bottom is πR . The work done by the non-conservative force, the force exerted by the pitcher, is

a f

∆E = F∆r cos 0° = F πR .

We shall assign the gravitational energy of the ball-Earth system to be zero with the ball at the bottom of the arc. Then becomes or

1 1 mv 2f − mvi2 + mgy f − mgyi 2 2 1 1 mv 2f = mvi2 + mgyi + F πR 2 2

∆Emech =

a f

v f = vi2 + 2 gyi +

a f a15.0f + 2a9.80fa1.20f + 2a30.0fπ a0.600f 0.250

2 F πR = m

2

v f = 26.5 m s *P8.28

The useful output energy is

a

f

d

i

120 Wh 1 − 0.60 = mg y f − yi = Fg ∆y ∆y =

b

g FG J IJ FG N ⋅ m IJ = H W ⋅sKH J K

120 W 3 600 s 0.40 890 N

194 m

Chapter 8

*P8.29

229

As the locomotive moves up the hill at constant speed, its output power goes into internal energy plus gravitational energy of the locomotive-Earth system:

Pt = mgy + f∆r = mg∆r sin θ + f∆r

P = mgv f sin θ + fv f

As the locomotive moves on level track,

F 746 W I = f b27 m sg f = 2.76 × 10 N GH 1 hp JK F 5 m IJ + e2.76 × 10 Njv Then also 746 000 W = b160 000 kg ge9.8 m s jv G H 100 m K P = fvi

4

1 000 hp

2

vf = P8.30

746 000 W 1.06 × 10 5 N

4

f

f

= 7.04 m s

We shall take the zero level of gravitational potential energy to be at the lowest level reached by the diver under the water, and consider the energy change from when the diver started to fall until he came to rest. 1 1 mv 2f − mvi2 + mgy f − mgyi = f k d cos 180° 2 2

∆E =

d

i

0 − 0 − mg yi − y f = − f k d fk = P8.31

d

mg yi − y f

i = b70.0 kg ge9.80 m s ja10.0 m + 5.00 mf = 2

5.00 m

d

m 2 gh − fh =

Ui + K i + ∆Emech = U f + K f :

2.06 kN

1 1 m1 v 2 + m 2 v 2 2 2

f = µn = µm1 g m 2 gh − µm1 gh = v2 =

v=

P8.32

d

i e

∆Emech = K f − K i + U gf − U gi

b

b

g

1 m1 + m 2 v 2 2

gb g

2 m 2 − µm1 hg m1 + m 2

e

FIG. P8.31

ja

f

b

2 9.80 m s 2 1.50 m 5.00 kg − 0.400 3.00 kg 8.00 kg

g

= 3.74 m s

j

But ∆Emech = Wapp − f∆x , where Wapp is the work the boy did pushing forward on the wheels.

d

i e

j 1 = me v − v j + mg a − hf + f∆x 2 1 = a 47.0 f a6. 20f − a1.40f − a 47.0fa9.80fa 2.60 f + a 41.0 fa12.4f 2

Thus,

Wapp = K f − K i + U gf − U gi + f∆x

or

Wapp Wapp

Wapp = 168 J

2 f

2 i

2

2

FIG. P8.32

230 P8.33

Potential Energy

1 1 m v 2f − vi2 = − mvi2 = −160 J 2 2

e

∆K =

(b)

∆U = mg 3.00 m sin 30.0° = 73.5 J

(c)

The mechanical energy converted due to friction is 86.5 J

a

f= (d)

f

FIG. P8.33

86.5 J = 28.8 N 3.00 m

f = µ k n = µ k mg cos 30.0° = 28.8 N 28.8 N µk = = 0.679 5.00 kg 9.80 m s 2 cos 30.0°

b

P8.34

j

(a)

ge

j

Consider the whole motion: K i + U i + ∆Emech = K f + U f (a)

0 + mgyi − f1 ∆x1 − f 2 ∆x 2 =

1 mv 2f + 0 2

b80.0 kg ge9.80 m s j1 000 m − a50.0 Nfa800 mf − b3 600 Nga200 mf = 12 b80.0 kg gv 1 784 000 J − 40 000 J − 720 000 J = b80.0 kg gv 2 2b 24 000 Jg = 24.5 m s v = 2

2 f

2 f

f

(b) (c)

80.0 kg

Yes this is too fast for safety. Now in the same energy equation as in part (a), ∆x 2 is unknown, and ∆x1 = 1 000 m − ∆x 2 :

a

fb

g b

g

784 000 J − 50.0 N 1 000 m − ∆x 2 − 3 600 N ∆x 2 =

b

g

b

gb

1 80.0 kg 5.00 m s 2

g

2

784 000 J − 50 000 J − 3 550 N ∆x 2 = 1 000 J ∆x 2 = (d)

733 000 J = 206 m 3 550 N

Really the air drag will depend on the skydiver’s speed. It will be larger than her 784 N weight only after the chute is opened. It will be nearly equal to 784 N before she opens the chute and again before she touches down, whenever she moves near terminal speed.

P8.35

aK + Uf + ∆E

(a)

a

0+

b

f

= K +U f :

mech

i

1 2 1 kx − f∆x = mv 2 + 0 2 2

ge

1 8.00 N m 5.00 × 10 −2 m 2

(b)

j − e3.20 × 10

e

2 5. 20 × 10 −3 J

v=

231

Chapter 8

5.30 × 10

−3

j=

kg

2

−2

ja

f 12 e5.30 × 10

N 0.150 m =

−3

j

kg v 2

1.40 m s

When the spring force just equals the friction force, the ball will stop speeding up. Here Fs = kx ; the spring is compressed by 3. 20 × 10 −2 N = 0.400 cm 8.00 N m and the ball has moved 5.00 cm − 0.400 cm = 4.60 cm from the start.

(c)

Between start and maximum speed points, 1 2 1 1 kxi − f∆x = mv 2 + kx 2f 2 2 2 1 1 1 −2 2 − 3.20 × 10 −2 4.60 × 10 −2 = 5.30 × 10 −3 v 2 + 8.00 4.00 × 10 −3 8.00 5.00 × 10 2 2 2 v = 1.79 m s

e

P8.36

j e

je

j e

j

e

∑ Fy = n − mg cos 37.0° = 0 ∴ n = mg cos 37.0° = 400 N

a

f

f = µn = 0.250 400 N = 100 N − f∆x = ∆Emech

a−100fa20.0f = ∆U + ∆U + ∆K + ∆K ∆U = m g d h − h i = a50.0fa9.80fa 20.0 sin 37.0°f = 5.90 × 10 ∆U = m g d h − h i = a100 fa9.80fa −20.0f = −1.96 × 10 1 ∆K = m e v − v j 2 m 1 ∆K = m e v − v j = ∆K = 2 ∆K 2 m A

A

A

B

B

f

A

B

B

A

B

i

f

A

B

4

i

2 f

2 f

3

2 i

2 i

B

A

A

A

Adding and solving, ∆K A = 3.92 kJ . FIG. P8.36

j

2

232 P8.37

Potential Energy

(a)

The object moved down distance 1.20 m + x. Choose y = 0 at its lower point. K i + U gi + U si + ∆Emech = K f + U gf + U sf 0 + mgyi + 0 + 0 = 0 + 0 +

1 2 kx 2

b1.50 kg ge9.80 m s ja1.20 m + xf = 12 b320 N mgx 0 = b160 N mgx − a14.7 N fx − 17.6 J 14.7 N ± a −14.7 N f − 4b160 N mga−17.6 N ⋅ mf x= 2b160 N mg 2

2

2

2

x=

14.7 N ± 107 N 320 N m

The negative root tells how high the object will rebound if it is instantly glued to the spring. We want x = 0.381 m (b)

From the same equation,

b1.50 kg ge1.63 m s ja1.20 m + xf = 12 b320 N mgx 2

2

0 = 160 x 2 − 2.44x − 2.93 The positive root is x = 0.143 m . (c)

The equation expressing the energy version of the nonisolated system model has one more term: mgyi − f∆x =

1 2 kx 2

b1.50 kg ge9.80 m s ja1.20 m + xf − 0.700 Na1.20 m + xf = 12 b320 N mgx 2

17.6 J + 14.7 Nx − 0.840 J − 0.700 Nx = 160 N m x 2 160 x 2 − 14.0 x − 16.8 = 0 x=

14.0 ±

a14.0f − 4a160fa−16.8f

x = 0.371 m

2

320

2

Chapter 8

P8.38

233

The total mechanical energy of the skysurfer-Earth system is 1 mv 2 + mgh . 2

Emech = K + U g = Since the skysurfer has constant speed,

a f

dEmech dv dh = mv + mg = 0 + mg − v = − mgv . dt dt dt The rate the system is losing mechanical energy is then dEmech = mgv = 75.0 kg 9.80 m s 2 60.0 m s = 44.1 kW . dt

b

*P8.39

(a)

jb

g

Let m be the mass of the whole board. The portion on the rough surface has mass normal force supporting it is a=

(b)

ge

µ mgx mxg and the frictional force is k = ma . Then L L

µ k gx opposite to the motion. L

In an incremental bit of forward motion dx, the kinetic energy converted into internal µ mgx energy is f k dx = k dx . The whole energy converted is L

z

L µ k mgx µ mg x 2 1 mv 2 = dx = k 2 2 L L 0

L

= 0

µ k mgL 2

v = µ k gL

Section 8.5

Relationship Between Conservative Forces and Potential Energy

P8.40

U = − − Ax + Bx 2 dx =

ze x

(a)

j

0

(b)

∆U = −

z

3.00 m

Fdx =

∆K =

(a)

Ax 2 Bx 3 − 2 3

j a f

e

A 3.00 2 − 2.00

2.00 m

P8.41

mx . The L

FG − 5.00 A + 19.0 BIJ H 2 3 K

z

W = Fx dx =

2

z a2x + 4fdx = FGH 2x2

5 .00 m

2

1

(b)

∆K + ∆U = 0

(c)

∆K = K f −

mv12 2

2

−

+ 4x

a f − a2.00f

B 3.00

3

3

I JK

=

5.00 19.0 A− B 2 3

5 .00 m

= 25.0 + 20.0 − 1.00 − 4.00 = 40.0 J 1

∆U = − ∆K = −W = −40.0 J K f = ∆K +

3

mv12 = 62.5 J 2

234 P8.42

Potential Energy

e e

j e j e

∂ 3x 3 y − 7x ∂U =− = − 9x 2 y − 7 = 7 − 9x 2 y Fx = − ∂x ∂x ∂ 3x 3 y − 7x ∂U =− = − 3 x 3 − 0 = −3 x 3 Fy = − ∂y ∂y

j

j

b g

Thus, the force acting at the point x , y is F = Fx i + Fy j = P8.43

af

e7 − 9x yji − 3x j . 2

3

A r d A A ∂U Fr = − =− = 2 . The positive value indicates a force of repulsion. dr r ∂r r

Ur =

Section 8.6

FG IJ H K

Energy Diagrams and the Equilibrium of a System

P8.44

stable

unstable

neutral

FIG. P8.44 P8.45

(a)

Fx is zero at points A, C and E; Fx is positive at point B and negative at point D.

(b)

A and E are unstable, and C is stable.

(c)

Fx B A

C

E D

FIG. P8.45

x (m)

Chapter 8

P8.46

(a)

235

There is an equilibrium point wherever the graph of potential energy is horizontal: At r = 1.5 mm and 3.2 mm, the equilibrium is stable. At r = 2.3 mm , the equilibrium is unstable. A particle moving out toward r → ∞ approaches neutral equilibrium.

P8.47

(b)

The system energy E cannot be less than –5.6 J. The particle is bound if −5.6 J ≤ E < 1 J .

(c)

If the system energy is –3 J, its potential energy must be less than or equal to –3 J. Thus, the particle’s position is limited to 0.6 mm ≤ r ≤ 3.6 mm .

(d)

K + U = E . Thus, K max = E − U min = −3.0 J − −5.6 J = 2.6 J .

(e)

Kinetic energy is a maximum when the potential energy is a minimum, at r = 1.5 mm .

(f)

−3 J + W = 1 J . Hence, the binding energy is W = 4 J .

(a)

When the mass moves distance x, the length of each spring

a

changes from L to k

FH

IK

f

x 2 + L2 , so each exerts force

x 2 + L2 − L towards its fixed end. The y-components

cancel out and the x components add to: Fx = −2 k

FH

x 2 + L2 − L

IK FG H

x 2

2

x +L

I = −2kx + JK

2 kLx x 2 + L2

FIG. P8.47(a)

Choose U = 0 at x = 0 . Then at any point the potential energy of the system is

z z FGH −2kx + x2kLx+ L IJK dx = 2kz xdx − 2kLz U a x f = kx + 2 kLFH L − x + L IK U a x f = 40.0 x + 96.0FH 1.20 − x + 1.44 IK af

x

U x = − Fx dx = − 0

x

2

0

2

(b)

x

2

2

2

0

x

0

x 2

x + L2

dx

2

2

af

For negative x, U x has the same value as for positive x. The only equilibrium point (i.e., where Fx = 0) is x = 0 . (c)

K i + U i + ∆Emech = K f + U f 1 0 + 0.400 J + 0 = 1.18 kg v 2f + 0 2 v f = 0.823 m s

b

g

FIG. P8.47(b)

236

Potential Energy

Additional Problems P8.48

The potential energy of the block-Earth system is mgh. An amount of energy µ k mgd cos θ is converted into internal energy due to friction on the incline. Therefore the final height y max is found from mgy max = mgh − µ k mgd cos θ where y max sin θ ∴ mgy max = mgh − µ k mgy max cot θ d=

h y max

θ

Solving, y max = P8.49

FIG. P8.48

h 1 + µ k cot θ

.

At a pace I could keep up for a half-hour exercise period, I climb two stories up, traversing forty steps each 18 cm high, in 20 s. My output work becomes the final gravitational energy of the system of the Earth and me,

b

ja

ge

f

mgy = 85 kg 9.80 m s 2 40 × 0.18 m = 6 000 J 6 000 J = ~ 10 2 W . 20 s

making my sustainable power P8.50

v = 100 km h = 27.8 m s The retarding force due to air resistance is R=

a

fe

jb

1 1 DρAv 2 = 0.330 1.20 kg m 3 2.50 m 2 27.8 m s 2 2

je

g

2

= 382 N

Comparing the energy of the car at two points along the hill, K i + U gi + ∆E = K f + U gf or

a f

K i + U gi + ∆We − R ∆s = K f + U gf

where ∆We is the work input from the engine. Thus,

a f d

i e

∆We = R ∆s + K f − K i + U gf − U gi

j

Recognizing that K f = K i and dividing by the travel time ∆t gives the required power input from the engine as

FG ∆W IJ = RFG ∆s IJ + mgFG ∆y IJ = Rv + mgv sinθ H ∆t K H ∆t K H ∆t K P = a382 N fb 27.8 m sg + b1 500 kg ge9.80 m s jb 27.8 m sg sin 3.20° P=

e

2

P = 33.4 kW = 44.8 hp

Chapter 8

P8.51

m = mass of pumpkin R = radius of silo top

∑ Fr = mar ⇒ n − mg cos θ = −m

237

v2 R

When the pumpkin first loses contact with the surface, n = 0 . Thus, at the point where it leaves the surface: v 2 = Rg cos θ .

FIG. P8.51

Choose U g = 0 in the θ = 90.0° plane. Then applying conservation of energy for the pumpkin-Earth system between the starting point and the point where the pumpkin leaves the surface gives K f + U gf = K i + U gi 1 mv 2 + mgR cos θ = 0 + mgR 2 Using the result from the force analysis, this becomes 1 mRg cos θ + mgR cos θ = mgR , which reduces to 2 cos θ =

b g

2 , and gives θ = cos −1 2 3 = 48.2° 3

as the angle at which the pumpkin will lose contact with the surface. P8.52

b

ja

ge

f

(a)

U A = mgR = 0. 200 kg 9.80 m s 2 0.300 m = 0.588 J

(b)

K A + U A = KB + UB K B = K A + U A − U B = mgR = 0.588 J

(c)

(d)

2K B = m

vB =

a

f

2 0.588 J = 2.42 m s 0.200 kg

b

ge

ja

f

FIG. P8.52

UC = mghC = 0.200 kg 9.80 m s 2 0.200 m = 0.392 J

b

K C = K A + U A − U C = mg h A − hC

b

ge

ja

g

f

K C = 0.200 kg 9.80 m s 2 0.300 − 0.200 m = 0.196 J P8.53

b

gb

1 1 mv B2 = 0.200 kg 1.50 m s 2 2

g

2

(a)

KB =

= 0.225 J

(b)

∆Emech = ∆K + ∆U = K B − K A + U B − U A

b g = 0.225 J + b0.200 kg ge9.80 m s ja0 − 0.300 mf = K B + mg hB − hA

2

= 0.225 J − 0.588 J = −0.363 J (c)

It’s possible to find an effective coefficient of friction, but not the actual value of µ since n and f vary with position.

238 P8.54

Potential Energy

The gain in internal energy due to friction represents a loss in mechanical energy that must be equal to the change in the kinetic energy plus the change in the potential energy. Therefore, − µ k mgx cos θ = ∆K +

1 2 kx − mgx sin θ 2

and since vi = v f = 0 , ∆K = 0. Thus,

a fa fa

fa

f a100fa02.200f − a2.00fa9.80fasin 37.0°fa0.200f 2

− µ k 2.00 9.80 cos 37.0° 0.200 =

and we find µ k = 0.115 . Note that in the above we had a gain in elastic potential energy for the spring and a loss in gravitational potential energy. P8.55

(a)

Since no nonconservative work is done, ∆E = 0

k = 100 N/m

Also ∆K = 0

2.00 kg

therefore, Ui = U f

b

g

where Ui = mg sin θ x and U f =

1 2 kx 2

FIG. P8.55

a fa f

a f 2x and solving we find

Substituting values yields 2.00 9.80 sin 37.0° = 100

x = 0.236 m (b)

∑ F = ma . Only gravity and the spring force act on the block, so − kx + mg sin θ = ma For x = 0.236 m ,

a = −5.90 m s 2 . The negative sign indicates a is up the incline. The acceleration depends on position . (c)

U(gravity) decreases monotonically as the height decreases. U(spring) increases monotonically as the spring is stretched. K initially increases, but then goes back to zero.

Chapter 8

P8.56

k = 2.50 × 10 4 N m,

m = 25.0 kg

x A = −0.100 m,

Ug

(a)

Emech = K A + U gA + U sA

x =0

= Us

=0

Emech

1 2 kx A 2 = 25.0 kg 9.80 m s 2 −0.100 m

Emech

1 2.50 × 10 4 2 = −24.5 J + 125 J = 100 J

Emech = 0 + mgx A +

b

f ja N mja −0.100 mf

ge

e

+

(b)

x=0

2

Since only conservative forces are involved, the total energy of the child-pogo-stick-Earth system at point C is the same as that at point A.

b

ge

j

0 + 25.0 kg 9.80 m s 2 xC + 0 = 0 − 24.5 J + 125 J

K C + U gC + U sC = K A + U gA + U sA :

x C = 0.410 m

b

a

g

f

1 25.0 kg v B2 + 0 + 0 = 0 + −24.5 J + 125 J 2 v B = 2.84 m s

(c)

K B + U gB + U sB = K A + U gA + U sA :

(d)

K and v are at a maximum when a = ∑ F m = 0 (i.e., when the magnitude of the upward spring force equals the magnitude of the downward gravitational force). This occurs at x < 0 where

k x = mg

or

x=

Thus,

K = K max at x = −9.80 mm

b25.0 kg ge9.8 m s j = 9.80 × 10 2

(e)

e

2.50 × 10 4 N m

x =−9.80 mm

sA

m

s x =−9.80 mm

2 max

2

4

yielding

−3

j + eU − U j 1 b25.0 kg gv = b25.0 kgge9.80 m s j a−0.100 mf − b−0.009 8 mg 2 1 + e 2.50 × 10 N mj a −0.100 mf − b−0.009 8 mg 2

K max = K A + U gA − U g or

P8.57

239

2

v max = 2.85 m s

∆Emech = − f∆x E f − Ei = − f ⋅ d BC 1 2 kx − mgh = − µmgd BC 2 mgh − 12 kx 2 = 0.328 µ= mgd BC

FIG. P8.57

2

240 P8.58

Potential Energy

(a)

F=−

(b)

F=0

d − x 3 + 2 x 2 + 3 x i = dx

e

j e3 x

2

j

− 4x − 3 i

when x = 1.87 and − 0.535 (c)

The stable point is at

af

x = −0.535 point of minimum U x . The unstable point is at

FIG. P8.58

af

x = 1.87 maximum in U x . P8.59

aK + U f = aK + U f 1 0 + b30.0 kg ge9.80 m s ja0.200 mf + b 250 N mga0.200 mf 2 1 = b50.0 kg gv + b 20.0 kg ge9.80 m s ja0.200 mf sin 40.0° 2 58.8 J + 5.00 J = b 25.0 kg gv + 25.2 J i

f

2

2

2

2

2

v = 1.24 m s P8.60

(a)

FIG. P8.59

Between the second and the third picture, ∆Emech = ∆K + ∆U − µmgd = −

b

1 1 mvi2 + kd 2 2 2

g

b

ge

b

ge

1 1 50.0 N m d 2 + 0.250 1.00 kg 9.80 m s 2 d − 1.00 kg 3.00 m s 2 = 0 2 2 −2.45 ± 21.25 N d= = 0.378 m 50.0 N m (b)

j

j

Between picture two and picture four, ∆Emech = ∆K + ∆U

a f 12 mv − 12 mv 2 v = b3.00 m sg − b1.00 kg g a2.45 Nfa2fa0.378 mf 2

− f 2d =

2 i

2

= 2.30 m s (c)

For the motion from picture two to picture five, ∆Emech = ∆K + ∆U

a

f

b

gb

g

1 2 1.00 kg 3.00 m s 2 9.00 J D= − 2 0.378 m = 1.08 m 2 0.250 1.00 kg 9.80 m s 2 − f D + 2d = −

a

fb

ge

j

a

f

FIG. P8.60

Chapter 8

P8.61

(a)

Initial compression of spring:

ga f

b

1 450 N m ∆x 2 ∴ ∆x = 0. 400 m (b)

2

1 2 1 kx = mv 2 2 2

b

gb

1 0.500 kg 12.0 m s 2

=

g

2

Speed of block at top of track: ∆Emech = − f∆x

FIG. P8.61

FG mgh + 1 mv IJ − FG mgh + 1 mv IJ = − f aπRf H K H K 2 2 b0.500 kgge9.80 m s ja2.00 mf + 12 b0.500 kg gv − 12 b0.500 kggb12.0 m sg = −a7.00 N faπ fa1.00 mf T

2 T

B

2 B

2 T

2

2

0.250 vT2 = 4.21 ∴ vT = 4.10 m s (c)

Does block fall off at or before top of track? Block falls if a c < g ac =

a f

2

4.10 vT2 = = 16.8 m s 2 R 1.00

Therefore a c > g and the block stays on the track . P8.62

Let λ represent the mass of each one meter of the chain and T represent the tension in the chain at the table edge. We imagine the edge to act like a frictionless and massless pulley. (a)

For the five meters on the table with motion impending,

∑ Fy = 0 :

+n − 5 λg = 0

n = 5 λg

b g

fs ≤ µ sn = 0.6 5λg = 3λg

∑ Fx = 0 :

+T − f s = 0

T = fs

T ≤ 3 λg

FIG. P8.62

The maximum value is barely enough to support the hanging segment according to

∑ Fy = 0 :

+T − 3 λg = 0

T = 3 λg

so it is at this point that the chain starts to slide. continued on next page

241

242

Potential Energy

(b)

Let x represent the variable distance the chain has slipped since the start.

a f +n − a5 − x fλg = 0 n = a5 − xfλg f = µ n = 0.4a5 − xfλg = 2λg − 0.4xλg

Then length 5 − x remains on the table, with now

∑ Fy = 0 :

k

k

Consider energies of the chain-Earth system at the initial moment when the chain starts to slip, and a final moment when x = 5 , when the last link goes over the brink. Measure heights above the final position of the leading end of the chain. At the moment the final link slips off, the center of the chain is at y f = 4 meters. Originally, 5 meters of chain is at height 8 m and the middle of the dangling segment is at 3 height 8 − = 6.5 m . 2 K i + U i + ∆Emech = K f + U f :

FG 1 mv + mgyIJ H2 K b5λg g8 + b3λg g6.5 − z b2λg − 0.4xλg gdx = 12 b8λ gv + b8λg g4 b

z f

g

2

0 + m1 gy1 + m 2 gy 2 i − f k dx =

f

i

5 0

2

z

z

5

5

40.0 g + 19.5 g − 2.00 g dx + 0.400 g x dx = 4.00 v 2 + 32.0 g 0

0

5

27.5 g − 2.00 gx 0 + 0.400 g

a f

2 5

x 2

= 4.00 v 2

0

a f

27.5 g − 2.00 g 5.00 + 0.400 g 12.5 = 4.00 v 2 22.5 g = 4.00 v 2

a22.5 mfe9.80 m s j = 2

v= P8.63

4.00

7.42 m s

Launch speed is found from mg

FG 4 hIJ = 1 mv : H5 K 2 2

v = 2g

FG 4 IJ h H 5K

v y = v sin θ

The height y above the water (by conservation of energy for the child-Earth system) is found from mgy =

1 h mv y2 + mg 2 5

FIG. P8.63

1 mv x2 is constant in projectile motion) 2 1 2 h 1 2 h y= vy + = v sin 2 θ + 2g 5 2g 5

(since

y=

LM FG IJ OP sin θ + h = 5 N H KQ

1 4 2g h 2g 5

2

4 h h sin 2 θ + 5 5

Chapter 8

*P8.64

(a)

The length of string between glider and pulley is given by A 2 = x 2 + h02 . Then 2 A Now

(b)

eK

A

a

243

dA dx = 2x + 0. dt dt

f

dA dA x = v y = v x = cos θ v x . is the rate at which string goes over the pulley: A dt dt

+ KB + U g

j = eK i

b

0 + 0 + m B g y 30 − y 45

A

g

+ KB + U g

j

f

1 1 = m A v x2 + m B v y2 2 2

Now y 30 − y 45 is the amount of string that has gone over the pulley, A 30 − A 45 . We have h h h0 h0 sin 30° = 0 and sin 45° = 0 , so A 30 − A 45 = − = 0. 40 m 2 − 2 = 0.234 m . A 30 A 45 sin 30° sin 45° From the energy equation

e

0.5 kg 9.8 m s 2 0.234 m =

1 1 1.00 kg v x2 + 0.500 kg v x2 cos 2 45° 2 2

1.15 J = 1.35 m s 0.625 kg

vx =

P8.65

j

b

g

(c)

v y = v x cos θ = 1.35 m s cos 45° = 0.958 m s

(d)

The acceleration of neither glider is constant, so knowing distance and acceleration at one point is not sufficient to find speed at another point.

b

g

The geometry reveals D = L sin θ + L sin φ , 50.0 m = 40.0 m sin 50°+ sin φ , φ = 28.9° (a)

From takeoff to alighting for the Jane-Earth system

eK + U j + W = eK + U j 1 mv + mg a − L cos θ f + FDa −1f = 0 + mg b− L cos φ g 2 1 50 kg v + 50 kg e9.8 m s ja −40 m cos 50°f − 110 Na50 mf = 50 kg e9.8 m s ja −40 m cos 28.9°f 2 g

wind

i

g

f

2 i

2 i

2

2

1 50 kg vi2 − 1.26 × 10 4 J − 5.5 × 10 3 J = −1.72 × 10 4 J 2

vi = (b)

a

f

2 947 J = 6.15 m s 50 kg

For the swing back

b

g e

a f

a

f f

1 mvi2 + mg − L cos φ + FD +1 = 0 + mg − L cos θ 2 1 130 kg vi2 + 130 kg 9.8 m s 2 −40 m cos 28.9° + 110 N 50 m 2

e

ja

ja

= 130 kg 9.8 m s 2 −40 m cos 50°

f

1 130 kg vi2 − 4.46 × 10 4 J + 5 500 J = −3.28 × 10 4 J 2 vi =

b

g=

2 6 340 J 130 kg

9.87 m s

a

f

244 P8.66

Potential Energy

1 1 mv 2 = kx 2 2 2 2 5.00 kg 1.20 m s mv k= 2 = x 10 −2 m 2

Case I: Surface is frictionless

b

e

je

eK + U j = eK + U j g

g

A

= 7. 20 × 10 2 N m

(b)

11.1 m s v2 = ac = r 6.3 m

(c)

∑ Fy = ma y

g

2

2

B

1 0 + mgy A = mv B2 + 0 2

b

2

j − a0.300fb5.00 kg ge9.80 m s je10

5.00 kg 2 1 v = 7.20 × 10 2 N m 10 −1 m 2 2 v = 0.923 m s (a)

g

µ k = 0.300 1 1 mv 2 = kx 2 − µ k mgx 2 2

Case II: Surface is rough,

*P8.67

gb

e

−1

j

m

j

v B = 2 gy A = 2 9.8 m s 2 6.3 m = 11.1 m s

2

= 19.6 m s 2 up

+n B − mg = ma c

e

j

n B = 76 kg 9.8 m s 2 + 19.6 m s 2 = 2.23 × 10 3 N up

a

f

(d)

W = F∆r cos θ = 2.23 × 10 3 N 0.450 m cos 0° = 1.01 × 10 3 J

(e)

eK + U j g

B

e

+ W = K +Ug

j

D

b

g

1 1 mv B2 + 0 + 1.01 × 10 3 J = mv D2 + mg y D − y B 2 2 1 1 2 76 kg 11.1 m s + 1.01 × 10 3 J = 76 kg v D2 + 76 kg 9.8 m s 2 6.3 m 2 2

b

g

e5.70 × 10

3

e

j

J − 4.69 × 10 3 J 2 76 kg

(f)

eK + U j = eK + U j 1 mv + 0 = 0 + mg b y 2 g

g

D

2 D

(g)

E

j

= v D = 5.14 m s

where E is the apex of his motion

E − yD

g

y E − yD =

b

g

2

5.14 m s v D2 = = 1.35 m 2 g 2 9.8 m s 2

e

j

Consider the motion with constant acceleration between takeoff and touchdown. The time is the positive root of 1 y f = yi + v yi t + a y t 2 2 1 −2.34 m = 0 + 5.14 m s t + −9.8 m s 2 t 2 2 2 4.9t − 5.14t − 2.34 = 0

e

t=

a fa

j

f=

5.14 ± 5.14 2 − 4 4.9 −2.34 9.8

1.39 s

Chapter 8

*P8.68

If the spring is just barely able to lift the lower block from the table, the spring lifts it through no noticeable distance, but exerts on the block a force equal to its weight Mg. The extension of the spring, from Fs = kx , must be Mg k . Between an initial point at release and a final point when the moving block first comes to rest, we have

FG H

IJ K

FG H

IJ K

FG IJ H K

2

0 + mg −

c he−4m j = −m ± M= 2c h Only a positive mass is physical, so we take M = ma3 − 1f = 2m . −m ± m 2 − 4

2

2

1 2

1 2

(a)

FG IJ H K

4mg Mg 1 Mg 1 4mg + k = 0 + mg + k 2 2 k k k k 2 2 2 2 2 2 2 mMg M g 4m g 8m g − + = + k k k 2k 2 M 4m 2 = mM + 2 M2 + mM − 4m 2 = 0 2

K i + U gi + U si = K f + U gf + U sf :

P8.69

245

9m 2

Take the original point where the ball is released and the final point where its upward swing stops at height H and horizontal displacement

a

x = L2 − L − H

f

2

= 2LH − H 2

Since the wind force is purely horizontal, it does work

z

z

Wwind = F ⋅ ds = F dx = F 2LH − H 2

FIG. P8.69

The work-energy theorem can be written: K i + U gi + Wwind = K f + U gf , or 0 + 0 + F 2LH − H 2 = 0 + mgH giving F 2 2LH − F 2 H 2 = m 2 g 2 H 2 Here H = 0 represents the lower turning point of the ball’s oscillation, and the upper limit is at F 2 2L = F 2 + m 2 g 2 H . Solving for H yields

a f e

j

H=

2LF 2 2L = F + m2 g 2 1 + mg F

b

2

g

2

As F → 0 , H → 0 as is reasonable. As F → ∞ , H → 2L , which would be hard to approach experimentally. (b)

H=

a

f

2 2.00 m

b

ge

j

1 + 2.00 kg 9.80 m s 2 14.7 N

continued on next page

2

= 1.44 m

246

Potential Energy

(c)

Call θ the equilibrium angle with the vertical.

∑ Fx = 0 ⇒ T sin θ = F , and ∑ Fy = 0 ⇒ T cos θ = mg Dividing: tan θ =

F 14.7 N = = 0.750 , or θ = 36.9° mg 19.6 N

a

f a

fa

f

Therefore, H eq = L 1 − cos θ = 2.00 m 1 − cos 36.9° = 0.400 m (d)

As F → ∞ , tan θ → ∞ , θ → 90.0° and H eq → L A very strong wind pulls the string out horizontal, parallel to the ground. Thus,

eH j eq

P8.70

max

=L .

Call φ = 180°−θ the angle between the upward vertical and the radius to the release point. Call v r the speed here. By conservation of energy

vi = Rg

The path after string is cut

K i + U i + ∆E = K r + U r 1 1 mvi2 + mgR + 0 = mv r2 + mgR cos φ 2 2 gR + 2 gR = v r2 + 2 gR cos φ

C

R

θ

v r = 3 gR − 2 gR cos φ FIG. P8.70

The components of velocity at release are v x = v r cos φ and v y = v r sin φ so for the projectile motion we have x = vxt

1 y = v y t − gt 2 2

R sin φ = v r cos φt − R cos φ = v r sin φt −

1 2 gt 2

By substitution − R cos φ = v r sin φ

R sin φ g R 2 sin 2 φ − v r cos φ 2 v r2 cos 2 φ

with sin 2 φ + cos 2 φ = 1 ,

b

gR sin 2 φ = 2 v r2 cos φ = 2 cos φ 3 gR − 2 gR cos φ

g

sin 2 φ = 6 cos φ − 4 cos 2 φ = 1 − cos 2 φ 3 cos 2 φ − 6 cos φ + 1 = 0 cos φ =

6 ± 36 − 12 6

Only the – sign gives a value for cos φ that is less than one: cos φ = 0.183 5

φ = 79.43°

so θ = 100.6°

247

Chapter 8

P8.71

Applying Newton’s second law at the bottom (b) and top (t) of the circle gives Tb − mg =

Adding these gives

mv b2

and −Tt − mg = −

R

Tb = Tt + 2mg +

m

e

v b2

−

vt mg

mv t2 R v t2

Tb

j

R mg

Also, energy must be conserved and ∆U + ∆K = 0 So,

e

m v b2 − v t2 2

j + b0 − 2mgRg = 0 and mev

Tt

2 b

− v t2

R

j = 4mg

vb

FIG. P8.71

Substituting into the above equation gives Tb = Tt + 6mg . P8.72

(a)

(b)

Energy is conserved in the swing of the pendulum, and the stationary peg does no work. So the ball’s speed does not change when the string hits or leaves the peg, and the ball swings equally high on both sides. Relative to the point of suspension,

a

Ui = 0, U f = − mg d − L − d

θ L

d Peg

f

From this we find that

a

f

1 − mg 2d − L + mv 2 = 0 2 Also for centripetal motion, mg =

mv 2 where R = L − d . R

Upon solving, we get d =

3L . 5

FIG. P8.72

248 *P8.73

Potential Energy

(a)

At the top of the loop the car and riders are in free fall:

∑ Fy = ma y :

mg down = v = Rg

mv 2 down R

Energy of the car-riders-Earth system is conserved between release and top of loop: K i + U gi = K f + U gf :

0 + mgh =

a f

1 mv 2 + mg 2 R 2

a f

1 Rg + g 2 R 2 h = 2.50 R

gh =

(b)

Let h now represent the height ≥ 2.5 R of the release point. At the bottom of the loop we have mgh =

1 mv b2 2

∑ Fy = ma y :

or

v b2 = 2 gh

n b − mg = n b = mg +

mv b2 up R m 2 gh

b g b g R

a f

1 mv t2 + mg 2 R 2 v t2 = 2 gh − 4 gR

At the top of the loop, mgh =

∑ Fy = ma y :

−n t − mg = −

FIG. P8.73

mv t2

R m 2 gh − 4 gR n t = − mg + R m 2 gh − 5mg nt = R

b

g

b g

Then the normal force at the bottom is larger by n b − n t = mg +

b g − mb2 ghg + 5mg =

m 2 gh R

R

6mg .

Chapter 8

*P8.74

(a)

249

Conservation of energy for the sled-rider-Earth system, between A and C: K i + U gi = K f + U gf

b

1 m 2.5 m s 2 vC = (b)

g

2

ja

e

1 mvC2 + 0 2

b2.5 m sg + 2e9.80 m s ja9.76 mf = 2

2

b

gb

g b

ja

ge

f

1 2 80 kg 2.5 m s + 80 kg 9.80 m s 2 9.76 m − f k ∆x = 0 + 0 2 − f k ∆x = −7.90 × 10 3 J 7.90 × 10 3 J 7.90 × 10 3 N ⋅ m = = 158 N ∆x 50 m

The water exerts a frictional force

fk =

and also a normal force of

n = mg = 80 kg 9.80 m s 2 = 784 N

b

ge

a158 Nf + a784 Nf 2

The magnitude of the water force is (d)

FIG. P8.74(a)

14.1 m s

Incorporating the loss of mechanical energy during the portion of the motion in the water, we have, for the entire motion between A and D (the rider’s stopping point), K i + U gi − f k ∆x = K f + U gf :

(c)

f

+ m 9.80 m s 2 9.76 m =

j

2

= 800 N

The angle of the slide is

θ = sin −1

9.76 m = 10.4° 54.3 m

For forces perpendicular to the track at B,

∑ Fy = ma y :

FIG. P8.74(d)

n B − mg cos θ = 0

b

ge

j

n B = 80.0 kg 9.80 m s 2 cos 10.4° = 771 N (e)

∑ Fy = ma y :

mvC2 r nC = 80.0 kg 9.80 m s 2 +nC − mg =

b ge j b80.0 kggb14.1 m sg + 2

20 m

nC = 1.57 × 10 3 N up

FIG. P8.74(e)

The rider pays for the thrills of a giddy height at A, and a high speed and tremendous splash at C. As a bonus, he gets the quick change in direction and magnitude among the forces we found in parts (d), (e), and (c).

250

Potential Energy

ANSWERS TO EVEN PROBLEMS P8.2

(a) 800 J; (b) 107 J; (c) 0

P8.4

(a) 1.11 × 10 9 J ; (b) 0.2

P8.6

1.84 m

P8.42

e7 − 9x yji − 3x j

P8.44

see the solution

P8.46

(a) r = 1.5 mm and 3.2 mm, stable; 2.3 mm and unstable; r → ∞ neutral; (b) −5.6 J ≤ E < 1 J ; (c) 0.6 mm ≤ r ≤ 3.6 mm ; (d) 2.6 J; (e) 1.5 mm; (f) 4 J

P8.48

see the solution

P8.50

33.4 kW

P8.52

(a) 0.588 J; (b) 0.588 J; (c) 2.42 m s; (d) 0.196 J; 0.392 J

6

P8.8

(a) 10.2 kW; (b) 10.6 kW; (c) 5.82 × 10 J

P8.10

d=

P8.12

(a) see the solution; (b) 60.0°

P8.14

(a)

kx 2 −x 2mg sin θ

b bm

g +m g

2 m1 − m 2 gh 1

2

; (b)

2m1 h m1 + m 2

2

3

P8.16

160 L min

P8.54

0.115

P8.18

40.8°

P8.56

P8.20

FG 8 gh IJ H 15 K

(a) 100 J; (b) 0.410 m; (c) 2.84 m s ; (d) −9.80 mm ; (e) 2.85 m s

P8.58

(a) 3 x 2 − 4x − 3 i ; (b) 1.87; –-0.535;

P8.22

(a) see the solution; (b) 35.0 J

P8.24

(a) v B = 5.94 m s; vC = 7.67 m s ; (b) 147 J

P8.26

(a) U f = 22.0 J ; E = 40.0 J ; (b) Yes. The total

12

mechanical energy changes.

e

j

(c) see the solution P8.60

(a) 0.378 m; (b) 2.30 m s ; (c) 1.08 m

P8.62

(a) see the solution; (b) 7.42 m s

P8.64

(a) see the solution; (b) 1.35 m s ; (c) 0.958 m s ; (d) see the solution

P8.28

194 m

P8.30

2.06 kN up

P8.66

0.923 m s

P8.32

168 J

P8.68

2m

P8.34

(a) 24.5 m s ; (b) yes; (c) 206 m; (d) Air drag depends strongly on speed.

P8.70

100.6°

P8.36

3.92 kJ

P8.72

see the solution

P8.38

44.1 kW

P8.74

(a) 14.1 m s; (b) −7.90 J ; (c) 800 N; (d) 771 N; (e) 1.57 kN up

P8.40

(a)

Ax 2 Bx 3 − ; 2 3 5 A 19B 19B 5 A (b) ∆U = − ; ∆K = − 2 3 3 2

9 Linear Momentum and Collisions CHAPTER OUTLINE 9.1 9.2 9.3 9.4 9.5 9.6 9.7

Linear Momentum and Its Conservation Impulse and Momentum Collisions in One Dimension Two-Dimensional Collisions The Center of Mass Motion of a System of Particles Rocket Propulsion

(c)

ANSWERS TO QUESTIONS Q9.1

No. Impulse, F∆t , depends on the force and the time for which it is applied.

Q9.2

The momentum doubles since it is proportional to the speed. The kinetic energy quadruples, since it is proportional to the speed-squared.

Q9.3

The momenta of two particles will only be the same if the masses of the particles of the same.

Q9.4

(a)

It does not carry force, for if it did, it could accelerate itself.

(b)

It cannot deliver more kinetic energy than it possesses. This would violate the law of energy conservation.

It can deliver more momentum in a collision than it possesses in its flight, by bouncing from the object it strikes.

Q9.5

Provided there is some form of potential energy in the system, the parts of an isolated system can move if the system is initially at rest. Consider two air-track gliders on a horizontal track. If you compress a spring between them and then tie them together with a string, it is possible for the system to start out at rest. If you then burn the string, the potential energy stored in the spring will be converted into kinetic energy of the gliders.

Q9.6

No. Only in a precise head-on collision with momenta with equal magnitudes and opposite directions can both objects wind up at rest. Yes. Assume that ball 2, originally at rest, is struck squarely by an equal-mass ball 1. Then ball 2 will take off with the velocity of ball 1, leaving ball 1 at rest.

Q9.7

Interestingly, mutual gravitation brings the ball and the Earth together. As the ball moves downward, the Earth moves upward, although with an acceleration 10 25 times smaller than that of the ball. The two objects meet, rebound, and separate. Momentum of the ball-Earth system is conserved.

Q9.8

(a)

Linear momentum is conserved since there are no external forces acting on the system.

(b)

Kinetic energy is not conserved because the chemical potential energy initially in the explosive is converted into kinetic energy of the pieces of the bomb. 251

252

Linear Momentum and Collisions

Q9.9

Momentum conservation is not violated if we make our system include the Earth along with the clay. When the clay receives an impulse backwards, the Earth receives the same size impulse forwards. The resulting acceleration of the Earth due to this impulse is significantly smaller than the acceleration of the clay, but the planet absorbs all of the momentum that the clay loses.

Q9.10

Momentum conservation is not violated if we choose as our system the planet along with you. When you receive an impulse forward, the Earth receives the same size impulse backwards. The resulting acceleration of the Earth due to this impulse is significantly smaller than your acceleration forward, but the planet’s backward momentum is equal in magnitude to your forward momentum.

Q9.11

As a ball rolls down an incline, the Earth receives an impulse of the same size and in the opposite direction as that of the ball. If you consider the Earth-ball system, momentum conservation is not violated.

Q9.12

Suppose car and truck move along the same line. If one vehicle overtakes the other, the fastermoving one loses more energy than the slower one gains. In a head-on collision, if the speed of the m + 3m c times the speed of the car, the car will lose more energy. truck is less than T 3mT + m c

Q9.13

The rifle has a much lower speed than the bullet and much less kinetic energy. The butt distributes the recoil force over an area much larger than that of the bullet.

Q9.14

His impact speed is determined by the acceleration of gravity and the distance of fall, in v 2f = vi2 − 2 g 0 − yi . The force exerted by the pad depends also on the unknown stiffness of the pad.

Q9.15

The product of the mass flow rate and velocity of the water determines the force the firefighters must exert.

Q9.16

The sheet stretches and pulls the two students toward each other. These effects are larger for a faster-moving egg. The time over which the egg stops is extended so that the force stopping it is never too large.

Q9.17

(c) In this case, the impulse on the Frisbee is largest. According to Newton’s third law, the impulse on the skater and thus the final speed of the skater will also be largest.

Q9.18

Usually but not necessarily. In a one-dimensional collision between two identical particles with the same initial speed, the kinetic energy of the particles will not change.

Q9.19

g downward.

Q9.20

As one finger slides towards the center, the normal force exerted by the sliding finger on the ruler increases. At some point, this normal force will increase enough so that static friction between the sliding finger and the ruler will stop their relative motion. At this moment the other finger starts sliding along the ruler towards the center. This process repeats until the fingers meet at the center of the ruler.

Q9.21

The planet is in motion around the sun, and thus has momentum and kinetic energy of its own. The spacecraft is directed to cross the planet’s orbit behind it, so that the planet’s gravity has a component pulling forward on the spacecraft. Since this is an elastic collision, and the velocity of the planet remains nearly unchanged, the probe must both increase speed and change direction for both momentum and kinetic energy to be conserved.

b

g

Chapter 9

253

Q9.22

No—an external force of gravity acts on the moon. Yes, because its speed is constant.

Q9.23

The impulse given to the egg is the same regardless of how it stops. If you increase the impact time by dropping the egg onto foam, you will decrease the impact force.

Q9.24

Yes. A boomerang, a kitchen stool.

Q9.25

The center of mass of the balls is in free fall, moving up and then down with the acceleration due to gravity, during the 40% of the time when the juggler’s hands are empty. During the 60% of the time when the juggler is engaged in catching and tossing, the center of mass must accelerate up with a somewhat smaller average acceleration. The center of mass moves around in a little circle, making three revolutions for every one revolution that one ball makes. Letting T represent the time for one cycle and Fg the weight of one ball, we have FJ 0.60T = 3 Fg T and FJ = 5 Fg . The average force exerted by the juggler is five times the weight of one ball.

Q9.26

In empty space, the center of mass of a rocket-plus-fuel system does not accelerate during a burn, because no outside force acts on this system. According to the text’s ‘basic expression for rocket propulsion,’ the change in speed of the rocket body will be larger than the speed of the exhaust relative to the rocket, if the final mass is less than 37% of the original mass.

Q9.27

The gun recoiled.

Q9.28

Inflate a balloon and release it. The air escaping from the balloon gives the balloon an impulse.

Q9.29

There was a time when the English favored position (a), the Germans position (b), and the French position (c). A Frenchman, Jean D’Alembert, is most responsible for showing that each theory is consistent with the others. All are equally correct. Each is useful for giving a mathematically simple solution for some problems.

SOLUTIONS TO PROBLEMS Section 9.1 P9.1

Linear Momentum and Its Conservation

e

(a)

(b)

j

v = 3.00 i − 4.00 j m s

m = 3.00 kg ,

e

j

p = mv = 9.00 i − 12.0 j kg ⋅ m s Thus,

p x = 9.00 kg ⋅ m s

and

p y = −12.0 kg ⋅ m s

a9.00f + a12.0f = 15.0 kg ⋅ m s F p I = tan a−1.33f = 307° GH p JK 2

p = p x2 + p y2 =

θ = tan −1

y x

−1

2

254 P9.2

Linear Momentum and Collisions

(a)

At maximum height v = 0 , so p = 0 .

(b)

Its original kinetic energy is its constant total energy, Ki =

a

f b

1 1 mvi2 = 0.100 kg 15.0 m s 2 2

g

2

= 11.2 J .

At the top all of this energy is gravitational. Halfway up, one-half of it is gravitational and the other half is kinetic:

v=

b

gb

b

g

1 0.100 kg v 2 2 2 × 5.62 J = 10.6 m s 0.100 kg

K = 5.62 J =

g

Then p = mv = 0.100 kg 10.6 m s j p = 1.06 kg ⋅ m s j . P9.3

I have mass 85.0 kg and can jump to raise my center of gravity 25.0 cm. I leave the ground with speed given by

d

i

v 2f − vi2 = 2 a x f − x i :

ja

e

0 − vi2 = 2 −9.80 m s 2 0.250 m

f

vi = 2.20 m s Total momentum of the system of the Earth and me is conserved as I push the earth down and myself up:

j b

e

gb

0 = 5.98 × 10 24 kg v e + 85.0 kg 2.20 m s

g

v e ~ 10 −23 m s P9.4

(a)

For the system of two blocks ∆p = 0 , or

pi = p f

Therefore,

0 = Mv m + 3 M 2.00 m s

Solving gives

v m = −6.00 m s (motion toward the

a fb

g

left). (b)

a f

1 2 1 1 2 + 3 M v 32M = 8. 40 J kx = Mv M 2 2 2

FIG. P9.4

Chapter 9

P9.5

(a)

The momentum is p = mv , so v =

(b)

K=

Section 9.2 *P9.6

1 mv 2 implies v = 2

FG IJ H K

p p 1 1 and the kinetic energy is K = mv 2 = m m m 2 2

2K 2K , so p = mv = m = m m

2

=

Impulse and Momentum

a f

From the impulse-momentum theorem, F ∆t = ∆p = mv f − mvi , the average force required to hold

F=

d

i = b12 kg gb0 − 60 mi hg F 1 m s I = −6.44 × 10 GH 2.237 mi h JK 0.050 s − 0 a ∆t f

m v f − vi

3

N.

Therefore, the magnitude of the needed retarding force is 6.44 × 10 3 N , or 1 400 lb. A person cannot exert a force of this magnitude and a safety device should be used.

*P9.8

p2 . 2m

2mK .

onto the child is

P9.7

255

(a)

z

I = Fdt = area under curve

jb

g

I=

1 1.50 × 10 −3 s 18 000 N = 13.5 N ⋅ s 2

(b)

F=

13.5 N ⋅ s = 9.00 kN 1.50 × 10 −3 s

(c)

From the graph, we see that Fmax = 18.0 kN

e

FIG. P9.7

1 1 mv12 = mgy1 . The rebound speed is given by mgy 2 = mv 22 . The 2 2 impulse of the floor is the change in momentum,

The impact speed is given by

b = me

g

mv 2 up − mv1 down = m v 2 + v1 up

j

2 gh2 + 2 gh1 up

e

= 0.15 kg 2 9.8 m s 2

je

= 1.39 kg ⋅ m s upward

j

0.960 m + 1.25 m up

256 P9.9

Linear Momentum and Collisions

∆p = F∆t

f e j a = ma − v sin 60.0°− v sin 60.0°f = −2mv sin 60.0° = −2b3.00 kg gb10.0 m sga0.866f

∆p y = m v fy − viy = m v cos 60.0° − mv cos 60.0° = 0 ∆p x

= −52.0 kg ⋅ m s Fave = P9.10

P9.11

∆p x −52.0 kg ⋅ m s = = −260 N 0.200 s ∆t

FIG. P9.9

Assume the initial direction of the ball in the –x direction.

ga f b

b

ga fe j

(a)

Impulse, I = ∆p = p f − pi = 0.060 0 40.0 i − 0.060 0 50.0 − i = 5.40 i N ⋅ s

(b)

Work = K f − K i =

b

1 0.060 0 2

g a40.0f − a50.0f 2

2

= −27.0 J

Take x-axis toward the pitcher (a)

b0.200 kg gb15.0 m sga− cos 45.0°f + I = b0.200 kggb40.0 m sg cos 30.0°

pix + I x = p fx :

x

I x = 9.05 N ⋅ s

b0.200 kg gb15.0 m sga− sin 45.0°f + I = b0.200 kg gb40.0 m sg sin 30.0° I = e9.05 i + 6.12 jj N ⋅ s

piy + I y = p fy :

(b)

b

y

f

ga

a

e

Fm = P9.12

f

a

1 1 0 + Fm 4.00 ms + Fm 20.0 ms + Fm 4.00 ms 2 2 Fm × 24.0 × 10 −3 s = 9.05 i + 6.12 j N ⋅ s I=

f

j

e377 i + 255jj N

If the diver starts from rest and drops vertically into the water, the velocity just before impact is found from K f + U gf = K i + U gi 1 2 + 0 = 0 + mgh ⇒ v impact = 2 gh mv impact 2 With the diver at rest after an impact time of ∆t , the average force during impact is given by F=

e

m 0 − v impact ∆t

j = −m

2 gh ∆t

or F =

m 2 gh ∆t

(directed upward).

Assuming a mass of 55 kg and an impact time of ≈ 1.0 s , the magnitude of this average force is

b55 kg g 2e9.8 m s ja10 mf = 770 N , or F = 2

1.0 s

~ 10 3 N .

Chapter 9

P9.13

The force exerted on the water by the hose is F=

b

gb

g

0.600 kg 25.0 m s − 0 ∆p water mv f − mvi = = = 15.0 N . 1.00 s ∆t ∆t

According to Newton's third law, the water exerts a force of equal magnitude back on the hose. Thus, the gardener must apply a 15.0 N force (in the direction of the velocity of the exiting water stream) to hold the hose stationary. *P9.14

(a)

Energy is conserved for the spring-mass system: K i + U si = K f + U sf :

1 2 1 kx = mv 2 + 0 2 2 k v=x m

0+

k larger. m

(b)

From the equation, a smaller value of m makes v = x

(c)

I = p f − p i = mv f = 0 = mx

(d)

From the equation, a larger value of m makes I = x km larger.

(e)

For the glider, W = K f − K i =

k = x km m

1 1 mv 2 − 0 = kx 2 2 2

The mass makes no difference to the work.

Section 9.3 P9.15

Collisions in One Dimension

b200 g gb55.0 m sg = b46.0 g gv + b200 g gb40.0 m sg v = 65.2 m s

*P9.16

bm v

g = bm v + m v g 22.5 g b35 m sg + 300 g b−2.5 m sg = 22.5 gv 1 1

v1 f =

+ m2 v2

i

1 1

2 2 f

1f

+0

37.5 g ⋅ m s = 1.67 m s 22.5 g

FIG. P9.16

257

258 P9.17

Linear Momentum and Collisions

Momentum is conserved 10.0 × 10 −3 kg v = 5.01 kg 0.600 m s

j b

e

gb

g

v = 301 m s P9.18

(a)

mv1i + 3mv 2 i = 4mv f where m = 2.50 × 10 4 kg vf =

P9.19

a f

4.00 + 3 2.00 = 2.50 m s 4

a f

LM N

a f OPQ e

ja

f

1 1 1 4m v 2f − mv12i + 3m v 22i = 2.50 × 10 4 12.5 − 8.00 − 6.00 = −3.75 × 10 4 J 2 2 2

(b)

K f − Ki =

(a)

The internal forces exerted by the actor do not change the total momentum of the system of the four cars and the movie actor

a4mfv = a3mfb2.00 m sg + mb4.00 m sg i

6.00 m s + 4.00 m s = 2.50 m s vi = 4 (b)

(c)

g

b g fb g

a fb

1 1 2 2 3m 2.00 m s + m 4.00 m s − 4 m 2.50 m s 2 2 2.50 × 10 4 kg 2 = 12.0 + 16.0 − 25.0 m s = 37.5 kJ 2

Wactor = K f − K i = Wactor

a fb ja

FIG. P9.19

e

g

2

The event considered here is the time reversal of the perfectly inelastic collision in the previous problem. The same momentum conservation equation describes both processes.

P9.20

v1 , speed of m1 at B before collision. 1 m1 v12 = m1 gh 2

a fa f

v1 = 2 9.80 5.00 = 9.90 m s v1 f , speed of m1 at B just after collision. m − m2 1 v1 f = 1 v1 = − 9.90 m s = −3.30 m s m1 + m 2 3 At the highest point (after collision)

a f

m1 ghmax =

a

1 m1 −3.30 2

f

FIG. P9.20

b−3.30 m sg = 2e9.80 m s j 2

2

hmax =

2

0.556 m

Chapter 9

P9.21

259

(a), (b) Let v g and v p be the velocity of the girl and the plank relative to the ice surface. Then we may say that v g − v p is the velocity of the girl relative to the plank, so that v g − v p = 1.50

(1)

But also we must have m g v g + m p v p = 0 , since total momentum of the girl-plank system is zero relative to the ice surface. Therefore 45.0 v g + 150 v p = 0 , or v g = −3.33 v p Putting this into the equation (1) above gives

FIG. P9.21

−3.33 v p − v p = 1.50 or v p = −0.346 m s

a

f

Then v g = −3.33 −0.346 = 1.15 m s *P9.22

For the car-truck-driver-driver system, momentum is conserved: p 1i + p 2 i = p 1 f + p 2 f :

b4 000 kg gb8 m sgi + b800 kg gb8 m sge− ij = b4 800 kg gv i f

vf =

25 600 kg ⋅ m s = 5.33 m s 4 800 kg

For the driver of the truck, the impulse-momentum theorem is F∆t = p f − pi :

a

f b

gb

g b

gb

g

F 0.120 s = 80 kg 5.33 m s i − 80 kg 8 m s i

e j

F = 1.78 × 10 3 N − i on the truck driver For the driver of the car,

a

f b

gb

g b

gb

ge j

F 0.120 s = 80 kg 5.33 m s i − 80 kg 8 m s − i

F = 8.89 × 10 3 Ni on the car driver , 5 times larger. P9.23

(a)

According to the Example in the chapter text, the fraction of total kinetic energy transferred to the moderator is f2 =

4m1 m 2

bm

1

+ m2

g

2

where m 2 is the moderator nucleus and in this case, m 2 = 12m1 f2 =

b

4m1 12m1

b13m g 1

2

g = 48 = 169

0. 284 or 28.4%

of the neutron energy is transferred to the carbon nucleus. (b)

a fe j Jj = = a0.716 fe1.6 × 10

K C = 0.284 1.6 × 10 −13 J = 4.54 × 10 −14 J Kn

−13

1.15 × 10 −13 J

260 P9.24

Linear Momentum and Collisions

Energy is conserved for the bob-Earth system between bottom and top of swing. At the top the stiff rod is in compression and the bob nearly at rest. 1 Mv b2 + 0 = 0 + Mg 2 A 2 v b2 = g 4A so v b = 2 gA

K i + Ui = K f + U f :

FIG. P9.24

Momentum of the bob-bullet system is conserved in the collision: mv = m P9.25

v + M 2 gA 2

e

j

v=

4M m

gA

At impact, momentum of the clay-block system is conserved, so:

b

g

mv1 = m1 + m 2 v 2 After impact, the change in kinetic energy of the clay-block-surface system is equal to the increase in internal energy:

b b

g g

b

g ge

1 m1 + m 2 v 22 = f f d = µ m1 + m 2 gd 2 1 0.112 kg v 22 = 0.650 0.112 kg 9.80 m s 2 7.50 m 2 v 22 = 95.6 m 2 s 2 v 2 = 9.77 m s

e12.0 × 10 P9.26

−3

b

j b

gb

kg v1 = 0.112 kg 9.77 m s

ja

g

f

FIG. P9.25

v1 = 91.2 m s

We assume equal firing speeds v and equal forces F required for the two bullets to push wood fibers apart. These equal forces act backward on the two bullets. For the first,

K i + ∆Emech = K f

For the second,

pi = p f

1 7.00 × 10 −3 kg v 2 − F 8.00 × 10 −2 m = 0 2 7.00 × 10 −3 kg v = 1.014 kg v f

e

e

j

j b e7.00 × 10 jv =

e

j

g

−3

vf

1.014

b

e

j

K i + ∆Emech = K f :

Substituting for v f ,

1 1 7.00 × 10 −3 v 7.00 × 10 −3 kg v 2 − Fd = 1.014 kg 2 2 1.014

e

b

j

e

−3 1 1 7.00 × 10 −3 2 Fd = 7.00 × 10 v − 2 2 1.014

e

Substituting for v,

e

g

1 1 7.00 × 10 −3 kg v 2 − Fd = 1.014 kg v 2f 2 2

Again,

j

jFGH

Fd = F 8.00 × 10 −2 m 1 −

7.00 × 10 −3 1.014

I JK

j

gFGH

I JK

2

2

v2 d = 7.94 cm

Chapter 9

*P9.27

(a)

261

c∑ ph = c∑ ph , gives a4.0 + 10 + 3.0f kg v = b4.0 kggb5.0 m sg + b10 kg gb3.0 m sg + b3.0 kggb−4.0 m sg .

Using conservation of momentum,

after

before

Therefore, v = +2.24 m s , or 2. 24 m s toward the right . (b)

No . For example, if the 10-kg and 3.0-kg mass were to stick together first, they would move with a speed given by solving

b13 kggv = b10 kg gb3.0 m sg + b3.0 kg gb−4.0 m sg , or v 1

1

= +1.38 m s .

Then when this 13 kg combined mass collides with the 4.0 kg mass, we have

b17 kggv = b13 kggb1.38 m sg + b4.0 kg gb5.0 m sg , and v = +2.24 m s just as in part (a). Coupling order makes no difference.

Section 9.4 P9.28

(a)

Two-Dimensional Collisions First, we conserve momentum for the system of two football players in the x direction (the direction of travel of the fullback).

b90.0 kg gb5.00 m sg + 0 = b185 kg gV cosθ where θ is the angle between the direction of the final velocity V and the x axis. We find V cos θ = 2.43 m s

(1)

Now consider conservation of momentum of the system in the y direction (the direction of travel of the opponent).

b95.0 kg gb3.00 m sg + 0 = b185 kggaV sinθ f which gives,

V sin θ = 1.54 m s

Divide equation (2) by (1)

tan θ =

V = 2.88 m s

Then, either (1) or (2) gives

b b

gb gb

1 90.0 kg 5.00 m s 2 1 K f = 185 kg 2.88 m s 2 Ki =

1.54 = 0.633 2.43

θ = 32.3°

From which

(b)

(2)

g g

b

gb

1 95.0 kg 3.00 m s 2

2

+

2

= 7.67 × 10 2 J

g

2

= 1.55 × 10 3 J

Thus, the kinetic energy lost is 783 J into internal energy.

262 P9.29

Linear Momentum and Collisions

p xf = p xi

b

mvO cos 37.0°+ mv Y cos 53.0° = m 5.00 m s 0.799 vO + 0.602 v Y = 5.00 m s

g

(1)

p yf = p yi mvO sin 37.0°− mv Y sin 53.0° = 0 0.602 vO = 0.799 v Y

(2)

Solving (1) and (2) simultaneously, vO = 3.99 m s and v Y = 3.01 m s . P9.30

p xf = p xi :

a

f

mvO cos θ + mv Y cos 90.0°−θ = mvi vO cos θ + v Y sin θ = vi

p yf = p yi :

FIG. P9.29

a

(1)

f

mvO sin θ − mv Y sin 90.0°−θ = 0 vO sin θ = v Y cos θ

(2)

From equation (2), vO = v Y

FG cosθ IJ H sinθ K

(3) FIG. P9.30

Substituting into equation (1), vY so

F cos θ I + v GH sinθ JK 2

e

Y

sin θ = vi

j

v Y cos 2 θ + sin 2 θ = vi sin θ , and v Y = vi sin θ .

Then, from equation (3), vO = vi cos θ . We did not need to write down an equation expressing conservation of mechanical energy. In the problem situation, the requirement of perpendicular final velocities is equivalent to the condition of elasticity.

Chapter 9

P9.31

The initial momentum of the system is 0. Thus,

a1.20mfv and

Bi

b

= m 10.0 m s

g

v Bi = 8.33 m s

b b g

a

g a

fb

g FG e H

1 1 1 2 2 m 10.0 m s + 1. 20m 8.33 m s = m 183 m 2 s 2 2 2 2 1 1 1 1 2 2 K f = m vG + 1.20m v B = m 183 m 2 s 2 2 2 2 2 Ki =

or

fb g

vG2 + 1.20 v B2 = 91.7 m 2 s 2

e

jIJK

j

(1)

From conservation of momentum,

a

f

mvG = 1.20m v B or

vG = 1.20 v B

(2)

Solving (1) and (2) simultaneously, we find vG = 7.07 m s (speed of green puck after collision) and P9.32

v B = 5.89 m s (speed of blue puck after collision)

We use conservation of momentum for the system of two vehicles for both northward and eastward components. For the eastward direction:

b

g

M 13.0 m s = 2 MV f cos 55.0° For the northward direction: Mv 2i = 2 MV f sin 55.0° Divide the northward equation by the eastward equation to find:

b

g

v 2 i = 13.0 m s tan 55.0° = 18.6 m s = 41.5 mi h Thus, the driver of the north bound car was untruthful.

FIG. P9.32

263

264 P9.33

Linear Momentum and Collisions

By conservation of momentum for the system of the two billiard balls (with all masses equal),

b

g

5.00 m s + 0 = 4.33 m s cos 30.0°+ v 2 fx v 2 fx = 1.25 m s

b

g

0 = 4.33 m s sin 30.0°+ v 2 fy v 2 fy = −2.16 m s v 2 f = 2.50 m s at − 60.0°

FIG. P9.33

Note that we did not need to use the fact that the collision is perfectly elastic. P9.34

(a)

pi = p f

so

p xi = p xf

and

p yi = p yf

From (2),

mvi = mv cos θ + mv cos φ

(1)

0 = mv sin θ + mv sin φ

(2)

sin θ = − sin φ

θ = −φ Furthermore, energy conservation for the system of two protons requires 1 1 1 mvi2 = mv 2 + mv 2 2 2 2 vi v= so 2

so

(b)

Hence, (1) gives vi =

b

2 vi cos θ

g

FIG. P9.34

θ = 45.0°

2

φ = −45.0°

a f v = e3.00 i − 1.20 jj m s

3.00 5.00 i − 6.00 j = 5.00 v

P9.35

m1 v 1i + m 2 v 2i = m1 + m 2 v f :

P9.36

x-component of momentum for the system of the two objects: p1ix + p 2ix = p1 fx + p 2 fx : − mvi + 3mvi = 0 + 3mv 2 x y-component of momentum of the system:

0 + 0 = − mv1 y + 3 mv 2 y

by conservation of energy of the system:

+

also

1 1 1 1 mvi2 + 3mvi2 = mv12y + 3m v 22 x + v 22 y 2 2 2 2 2 vi v2x = 3 v1 y = 3 v 2 y

So the energy equation becomes

4vi2 = 9 v 22 y +

we have

or continued on next page

8 vi2 = 12 v 22 y 3 2 vi v2y = 3

e

4vi2 + 3 v 22 y 3

j

Chapter 9

(a)

The object of mass m has final speed

v1 y = 3 v 2 y =

and the object of mass 3 m moves at

v 22 x + v 22 y =

2 vi

v 22 x + v 22 y =

(b)

θ = tan −1

Fv I GH v JK 2y

θ = tan −1

2x

P9.37

F GH

m 0 = 17.0 × 10 −27 kg

v i = 0 (the parent nucleus)

m1 = 5.00 × 10 −27 kg

v 1 = 6.00 × 10 6 j m s

m 2 = 8.40 × 10 −27 kg

v 2 = 4.00 × 10 6 i m s

(a)

4vi2 2 vi2 + 9 9 2 vi 3

I JK

2 vi 3 = 35.3° 3 2 vi

m1 v 1 + m 2 v 2 + m 3 v 3 = 0

where m 3 = m 0 − m1 − m 2 = 3.60 × 10 −27 kg FIG. P9.37 −27 −27 −27 6 6 × × + × × + × 5.00 10 6.00 10 j 8.40 10 4.00 10 i 3.60 10 v3 = 0

e

v3 = (b)

je

j e

je

j e

j

e−9.33 × 10 i − 8.33 × 10 jj m s 6

1 1 1 m1 v12 + m 2 v 22 + m 3 v 32 2 2 2 1 −27 5.00 × 10 6.00 × 10 6 E= 2

6

E=

LMe N

je

j + e8.40 × 10 je4.00 × 10 j + e3.60 × 10 je12.5 × 10 j OQP 2

−27

6 2

−27

6 2

E = 4.39 × 10 −13 J

Section 9.5 P9.38

The Center of Mass

The x-coordinate of the center of mass is x CM =

∑ m i xi ∑ mi

=

b

0+0+0+0 2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg

g

xCM = 0 and the y-coordinate of the center of mass is yCM =

∑ m i yi ∑ mi

yCM = 1.00 m

=

b2.00 kgga3.00 mf + b3.00 kg ga2.50 mf + b2.50 kg ga0f + b4.00 kg ga−0.500 mf 2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg

265

266 P9.39

Linear Momentum and Collisions

Take x-axis starting from the oxygen nucleus and pointing toward the middle of the V. yCM = 0

Then

x CM =

and

x CM =

∑ mi x i ∑ mi

a

=

f

a

f

0 + 1.008 u 0.100 nm cos 53.0°+1.008 u 0.100 nm cos 53.0° 15.999 u + 1.008 u + 1.008 u

FIG. P9.39

xCM = 0.006 73 nm from the oxygen nucleus *P9.40

Let the x axis start at the Earth’s center and point toward the Moon.

e

j

24 22 8 m1 x1 + m 2 x 2 5.98 × 10 kg 0 + 7.36 × 10 kg 3.84 × 10 m = m1 + m 2 6.05 × 10 24 kg

x CM =

= 4.67 × 10 6 m from the Earth’s center The center of mass is within the Earth, which has radius 6.37 × 10 6 m. P9.41

Let A1 represent the area of the bottom row of squares, A 2 the middle square, and A3 the top pair. A = A1 + A 2 + A 3 M = M1 + M 2 + M 3 M1 M = A1 A A1 = 300 cm 2 , A 2 = 100 cm 2 , A3 = 200 cm 2 , A = 600 cm 2 A 300 cm 2 M M1 = M 1 = M= 2 A 2 600 cm M2 M3 x CM x CM

FG IJ H K F A IJ = 100 cm = MG H A K 600 cm F A IJ = 200 cm = MG H A K 600 cm 2

2 2 2

M=

M 6

FIG. P9.41

M 3 x M + x 2 M 2 + x 3 M 3 15.0 cm = 1 1 = M = 11.7 cm

yCM =

3

1 2

a

f

2

M=

c M h + 5.00 cmc Mh + 10.0 cmc Mh

a

yCM = 13.3 cm

1 6

M

f c Mha25.0 cmf = 13.3 cm

M 5.00 cm + 16 M 15.0 cm + M

1 2

1 3

1 3

Chapter 9

*P9.42

(a)

267

Represent the height of a particle of mass dm within the object as y. Its contribution to the gravitational energy of the object-Earth system is dm gy . The total gravitational energy is 1 Ug = gy dm = g y dm . For the center of mass we have yCM = y dm , so U g = gMyCM . M all mass

z

(b)

a f

z

a

e

fa

f

1 3.6 m 15.7 m 64.8 m = 1.83 × 10 3 m 3 . Its mass is 2 1.83 × 10 3 m 3 = 6.96 × 10 6 kg . Its center of mass is above its base by one-

The volume of the ramp is

ρV = 3 800 kg m3

fa

z

je

j

1 third of its height, yCM = 15.7 m = 5.23 m . Then 3 U g = MgyCM = 6.96 × 10 6 kg 9.8 m s 2 5.23 m = 3.57 × 10 8 J .

e

P9.43

M=

(a)

z

z

0.300 m

0 .300 m

0

0

λdx =

j

50.0 g m + 20.0 x g m 2 dx

M = 50.0 x g m + 10.0 x 2 g m 2

x CM =

(b)

x CM

*P9.44

z

xdm

all mass

M

=

LM MN

1 M

z

0.300 m

λ xdx =

0

0.300 m 0

1 M

z

= 15.9 g

0.300 m

50.0 x g m + 20.0 x 2 g m 2 dx

0

3

20 x g m 2 1 25.0 x 2 g m + = 15.9 g 3

OP PQ

0.300 m

= 0.153 m 0

Take the origin at the center of curvature. We have L = r=

2L

π

1 2πr , 4

. An incremental bit of the rod at angle θ from the x axis has

dm M Mr = , dm = dθ where we have used the rdθ L L definition of radian measure. Now

mass given by

yCM = =

z

z

135 °

1 1 Mr r2 y dm = r sin θ dθ = M all mass M θ = 45 ° L L

FG 2L IJ 1 a− cosθ f HπK L 2

135° 45 °

=

4L

π

2

2L

π

−

4 2L

π

2

=

F1 − 2 2 I L = π GH π JK 2

z

θ x

135 °

sin θ dθ

FIG. P9.44

45 °

FG 1 + 1 IJ = 4 2L H 2 2K π

The top of the bar is above the origin by r = by

y

0.063 5L .

2

2L

π

, so the center of mass is below the middle of the bar

268

Linear Momentum and Collisions

Section 9.6 P9.45

Motion of a System of Particles

(a)

v CM = = v CM =

P9.46

b

m1 v 1 + m 2 v 2 M 2.00 kg 2.00 i m s − 3.00 j m s + 3.00 kg 1.00 i m s + 6.00 j m s M

=

ge

j b

ge

e1.40i + 2.40 jj m s b

ge

j

e7.00i + 12.0jj kg ⋅ m s

p = Mv CM = 5.00 kg 1. 40 i + 2.40 j m s =

(a)

See figure to the right.

(b)

Using the definition of the position vector at the center of mass,

rCM =

2.00 kg + 3.00 kg

e−2.00i − 1.00jj m

FIG. P9.46

The velocity of the center of mass is v CM = v CM =

(d)

m1 r1 + m 2 r2 m1 + m 2

b2.00 kg ga1.00 m, 2.00 mf + b3.00 kg ga−4.00 m, − 3.00 mf

rCM = (c)

j

5.00 kg

(b)

rCM =

P9.47

∑ mi v i

b

gb

g b

gb g

2.00 kg 3.00 m s , 0.50 m s + 3.00 kg 3.00 m s , − 2.00 m s P m1 v 1 + m 2 v 2 = = M m1 + m 2 2.00 kg + 3.00 kg

b

e3.00i − 1.00jj m s

The total linear momentum of the system can be calculated as P = Mv CM or as

P = m1 v 1 + m 2 v 2

Either gives

P=

e15.0 i − 5.00jj kg ⋅ m s

Let x = distance from shore to center of boat A = length of boat x ′ = distance boat moves as Juliet moves toward Romeo The center of mass stays fixed. Before:

h + M cx + h dM + M + M i M ax − x ′f + M cx + − x ′h + M c x + − x ′h = x dM + M + M i F 55.0 + 77.0 IJ = x ′a−80.0 − 55.0 − 77.0f + A a55.0 + 77.0f AG − H 2 2K 2 55.0 A 55.0a 2.70f x′ = = = 0.700 m x CM =

c

Mb x + M J x − B

After:

B

A 2

R

J

R

J

A 2

A 2

R

CM

B

212

212

J

R

A 2

FIG. P9.47

g

Chapter 9

P9.48

(a)

269

Conservation of momentum for the two-ball system gives us:

b

g

b

g

0.200 kg 1.50 m s + 0.300 kg −0.400 m s = 0.200 kg v1 f + 0.300 kg v 2 f Relative velocity equation: v 2 f − v1 f = 1.90 m s

d

0.300 − 0.120 = 0.200 v1 f + 0.300 1.90 + v1 f

Then

v1 f = −0.780 m s

v 2 f = 1.12 m s

v 1 f = −0.780 i m s (b)

v CM =

Before,

i

v 2 f = 1.12 i m s

b0.200 kg gb1.50 m sgi + b0.300 kg gb−0.400 m sgi b

g

0.500 kg

v CM = 0.360 m s i Afterwards, the center of mass must move at the same velocity, as momentum of the system is conserved.

Section 9.7 P9.49

Rocket Propulsion dM dt

e

je

j

Thrust = 2.60 × 10 3 m s 1.50 × 10 4 kg s = 3.90 × 10 7 N

(a)

Thrust = v e

(b)

∑ Fy = Thrust − Mg = Ma :

ja f e

e

j

3.90 × 10 7 − 3.00 × 10 6 9.80 = 3.00 × 10 6 a

a = 3.20 m s 2 *P9.50

(a)

dM 12.7 g = = 6.68 × 10 −3 kg s dt 1.90 s

The fuel burns at a rate Thrust = v e

dM : dt

e

5.26 N = v e 6.68 × 10 −3 kg s

j

v e = 787 m s (b)

F M I: GH M JK

v f − vi = v e ln

g FGH 53.553g .+5 25g .+5 25g .−512g .7 g IJK

b

i

v f − 0 = 797 m s ln

f

v f = 138 m s P9.51

v = v e ln

Mi Mf

(a)

Mi = e v v e M f

Mi = e 5 3.00 × 10 3 kg = 4.45 × 10 5 kg

The mass of fuel and oxidizer is

∆M = Mi − M f = 445 − 3.00 × 10 3 kg = 442 metric tons

(b)

a

e

a

j

f

f

∆M = e 2 3.00 metric tons − 3.00 metric tons = 19.2 metric tons Because of the exponential, a relatively small increase in fuel and/or engine efficiency causes a large change in the amount of fuel and oxidizer required.

F M I = − v lnF M I GH M JK GH M JK F M − kt IJ = − v lnFG 1 − k tIJ = M − kt , so v = − v lnG H M K H M K

From Equation 9.41, v − 0 = v e ln

i

Now, M f

i

i

f

e

f

e

i

e

i

i

M With the definition, Tp ≡ i , this becomes k

F GH

I JK F t IJ = 144 s , v = −b1 500 m sg lnG 1 − H 144 s K af

v t = − v e ln 1 −

With v e = 1 500 m s, and Tp

a f vbm sg

v (m/s)

ts

1220 1780

120

2690

132

3730

FH

dt

p

IK OP F 1 Q = −v G GH 1 − e

t Tp

IF 1 I F v IF 1 JJ GH − T JK = GH T JK GG 1 − K H e

p

p

t Tp

I JJ , or K

ve Tp − t 1 500 m s 144 s − t

a f aem s j 2

0

10.4

20

12.1

40

14.4

60

17.9

80

23.4

100

34.1

120

62.5

132

125

continued on next page

2

a (m/s ) 140 120 100 80 60 40 20 0

t (s)

FIG. P9.52(d)

140

ts

120

With v e = 1 500 m s, and Tp = 144 s , a =

100

(d)

LM N

d − v e ln 1 − Tt

80

af

at =

FIG. P9.52(b)

60

af

dv at = = dt

t (s)

140

80 100

120

808

100

60

2500 2000 1500 1000 500 0

80

488

60

40

40

0 224

40

(c)

0 20

4000 3500 3000

20

(b)

t Tp

20

(a)

0

P9.52

Linear Momentum and Collisions

0

270

Chapter 9

(e)

LM F t I OP LM1 − t OPF − dt I − − = v ln 1 dt v T ln G J z z MN H T K PQ z MN T PQGH T JK LF t I F t I F t I O xat f = v T MG 1 − J lnG 1 − J − G 1 − J P MNH T K H T K H T K PQ F tI xat f = v eT − t j lnG 1 − J + v t H TK t

af

x t = 0 + vdt = 0

t

t

e

e p

p

0

0

p

p

t

e p

e

p

p

p

0

e

p

With v e = 1 500 m s = 1.50 km s , and Tp = 144 s ,

f FGH

IJ K

t + 1.50t 144

a f xakmf

ts

60

22.1

80

42.2

100

71.7

120

115

132

153

100 80 60 40 20 0

t (s)

FIG. P9.52(f) *P9.53

The thrust acting on the spacecraft is

∑ F = b3 500 kg ge2.50 × 10 −6 je9.80

∑ F = ma : thrust =

FG dM IJ v : H dt K e

8.58 × 10 −2 N = ∆M = 4.41 kg

F ∆M I b70 m sg GH 3 600 s JK

j

m s 2 = 8.58 × 10 −2 N

140

9.23

120

40

140 120

100

2.19

80

20

160

60

0

0

0

x (km)

40

a

x = 1.50 144 − t ln 1 −

20

(f)

p

271

272

Linear Momentum and Collisions

Additional Problems P9.54

(a)

When the spring is fully compressed, each cart moves with same velocity v. Apply conservation of momentum for the system of two gliders

b

pi = p f : (b)

g

v=

m1 v 1 + m 2 v 2 = m1 + m 2 v

m1 v 1 + m 2 v 2 m1 + m 2

b

g

1 1 1 1 m1 v12 + m 2 v 22 = m1 + m 2 v 2 + kx m2 2 2 2 2

Only conservative forces act, therefore ∆E = 0 . Substitute for v from (a) and solve for x m . x m2

bm =

xm = (c)

1

g

b

g

b

+ m 2 m1 v12 + m1 + m 2 m 2 v 22 − m1 v1

b

k m1 + m 2

e

m1 m 2 v12 + v 22 − 2 v1 v 2

b

k m1 + m 2

g

j = bv

1

− v2

g

g − bm v g 2

2 2

2

− 2m 1 m 2 v1 v 2

g kbmm m+ m g 1

2

1

2

m1 v 1 + m 2 v 2 = m1 v 1 f + m 2 v 2 f

d

i

d

i

Conservation of momentum:

m1 v 1 − v 1 f = m 2 v 2 f − v 2

Conservation of energy:

1 1 1 1 m1 v12 + m 2 v 22 = m1 v12 f + m 2 v 22 f 2 2 2 2

which simplifies to: Factoring gives

e m dv

j e j i ⋅ dv + v i = m dv

(1)

m1 v12 − v12 f = m 2 v 22 f − v 22 1

1

− v1 f

1

1f

2

2f

id

− v2 ⋅ v2 f + v2

i

and with the use of the momentum equation (equation (1)), this reduces to

dv

or

v1 f = v 2 f + v 2 − v1

1

i d

+ v1 f = v 2 f + v 2

i (2)

Substituting equation (2) into equation (1) and simplifying yields: v2 f =

FG 2m IJ v + FG m Hm +m K Hm 1

1

1

2

IJ K

− m1 v2 + 1 m2 2

Upon substitution of this expression for v 2 f into equation 2, one finds v1 f =

FG m Hm

IJ K

FG H

IJ K

− m2 2m 2 v1 + v2 m1 + m 2 1 + m2 1

Observe that these results are the same as Equations 9.20 and 9.21, which should have been expected since this is a perfectly elastic collision in one dimension.

P9.55

(a)

b60.0 kg g4.00 m s = a120 + 60.0f kgv

Chapter 9

273

f

v f = 1.33 m s i (b)

b

∑ Fy = 0 :

g

n − 60.0 kg 9.80 m s 2 = 0

a

f

f k = µ k n = 0.400 588 N = 235 N f = −235 N i

FIG. P9.55

k

(c)

For the person, pi + I = p f mvi + Ft = mv f

b60.0 kg g4.00 m s − a235 Nft = b60.0 kg g1.33 m s t = 0.680 s (d)

a

f

person:

mv f − mv i = 60.0 kg 1.33 − 4.00 m s = −160 N ⋅ si

cart:

120 kg 1.33 m s − 0 = +160 N ⋅ si

b

g

a

f

(e)

x f − xi =

1 1 4.00 + 1.33 m s 0.680 s = 1.81 m vi + v f t = 2 2

(f)

x f − xi =

1 1 vi + v f t = 0 + 1.33 m s 0.680 s = 0.454 m 2 2

(g)

1 1 1 mv 2f − mvi2 = 60.0 kg 1.33 m s 2 2 2

(h)

1 1 1 mv 2f − mvi2 = 120.0 kg 1.33 m s 2 2 2

(i)

d d

i

b

i

g

b

b

g

2

−

g

2

b

1 60.0 kg 4.00 m s 2

g

2

= −427 J

− 0 = 107 J

The force exerted by the person on the cart must equal in magnitude and opposite in direction to the force exerted by the cart on the person. The changes in momentum of the two objects must be equal in magnitude and must add to zero. Their changes in kinetic energy are different in magnitude and do not add to zero. The following represent two ways of thinking about ’ why. ’ The distance the cart moves is different from the distance moved by the point of application of the friction force to the cart. The total change in mechanical energy for both objects together, − 320 J, becomes +320 J of additional internal energy in this perfectly inelastic collision.

P9.56

The equation for the horizontal range of a projectile is R =

vi2 sin 2θ . Thus, with θ = 45.0° , the initial g

velocity is vi = Rg =

a f

a200 mfe9.80 m s j = 44.3 m s 2

I = F ∆t = ∆p = mvi − 0 Therefore, the magnitude of the average force acting on the ball during the impact is: F=

e

jb

g

46.0 × 10 −3 kg 44.3 m s mvi = = 291 N . ∆t 7.00 × 10 −3 s

274 P9.57

Linear Momentum and Collisions

We hope the momentum of the wrench provides enough recoil so that the astronaut can reach the ship before he loses life support! We might expect the elapsed time to be on the order of several minutes based on the description of the situation. No external force acts on the system (astronaut plus wrench), so the total momentum is constant. Since the final momentum (wrench plus astronaut) must be zero, we have final momentum = initial momentum = 0. m wrench v wrench + m astronaut v astronaut = 0

b

gb

g

0.500 kg 20.0 m s m wrench v wrench =− = −0.125 m s m astronaut 80.0 kg At this speed, the time to travel to the ship is Thus v astronaut = −

t=

30.0 m = 240 s = 4.00 minutes 0.125 m s

The astronaut is fortunate that the wrench gave him sufficient momentum to return to the ship in a reasonable amount of time! In this problem, we were told that the astronaut was not drifting away from the ship when he threw the wrench. However, this is not quite possible since he did not encounter an external force that would reduce his velocity away from the ship (there is no air friction beyond earth’s atmosphere). If this were a real-life situation, the astronaut would have to throw the wrench hard enough to overcome his momentum caused by his original push away from the ship. P9.58

Using conservation of momentum from just before to just after the impact of the bullet with the block:

a

f

m

mvi = M + m v f or

vi =

FG M + m IJ v H m K

f

vi

(1)

M

h

The speed of the block and embedded bullet just after impact may be found using kinematic equations: d = v f t and h = Thus, t =

d

1 2 gt 2

g 2h d = and v f = = d g t 2h

FIG. P9.58

gd 2 2h

Substituting into (1) from above gives vi =

FG M + m IJ H m K

gd 2 . 2h

Chapter 9

*P9.59

(a)

275

Conservation of momentum:

e

j

e

j

0.5 kg 2 i − 3 j + 1k m s + 1.5 kg −1i + 2 j − 3k m s

e

j

= 0.5 kg −1i + 3 j − 8k m s + 1.5 kg v 2 f v2 f =

e−0.5 i + 1.5j − 4k j kg ⋅ m s + e0.5i − 1.5j + 4k j kg ⋅ m s = 1.5 kg

0

The original kinetic energy is 1 1 0.5 kg 2 2 + 3 2 + 1 2 m 2 s 2 + 1.5 kg 1 2 + 2 2 + 3 2 m 2 s 2 = 14.0 J 2 2

e

j

e

j

1 0.5 kg 1 2 + 3 2 + 8 2 m 2 s 2 + 0 = 18.5 J different from the original 2 energy so the collision is inelastic .

e

The final kinetic energy is

(b)

j

We follow the same steps as in part (a):

e−0.5 i + 1.5j − 4k j kg ⋅ m s = 0.5 kge−0.25i + 0.75j − 2k j m s + 1.5 kg v e−0.5i + 1.5j − 4k j kg ⋅ m s + e0.125 i − 0.375 j + 1k j kg ⋅ m s v = 2f

2f

1.5 kg

e−0.250i + 0.750 j − 2.00k j m s

=

We see v 2 f = v 1 f , so the collision is perfectly inelastic . (c)

Conservation of momentum:

e−0.5 i + 1.5j − 4k j kg ⋅ m s = 0.5 kge−1i + 3 j + ak j m s + 1.5 kg v −0.5 i + 1.5 j − 4k j kg ⋅ m s + e0.5 i − 1.5 j − 0.5 ak j kg ⋅ m s e v = 2f

2f

1.5 kg

=

a−2.67 − 0.333 afk m s

Conservation of energy:

a

1 1 0.5 kg 1 2 + 3 2 + a 2 m 2 s 2 + 1.5 kg 2.67 + 0.333 a 2 2 2 = 2.5 J + 0. 25 a + 5.33 J + 1.33 a + 0.083 3 a 2

14.0 J =

e

j

f

2

m2 s 2

0 = 0.333 a 2 + 1.33 a − 6.167 a=

a

fa

−1.33 ± 1.33 2 − 4 0.333 −6.167

f

0.667 a = 2.74 or − 6.74. Either value is possible. ∴ a = 2.74 ,

a fh c = c−2.67 − 0.333a −6.74fhk m s =

v 2 f = −2.67 − 0.333 2.74 k m s = −3.58k m s

∴ a = −6.74 , v 2 f

−0.419 k m s

276 P9.60

Linear Momentum and Collisions

(a)

The initial momentum of the system is zero, which remains constant throughout the motion. Therefore, when m1 leaves the wedge, we must have m 2 v wedge + m1 v block = 0

(b)

or

b3.00 kg gv

so

v wedge = −0.667 m s

wedge

b

gb

g

+ 0.500 kg +4.00 m s = 0

v wedge

+x

Using conservation of energy for the block-wedgeEarth system as the block slides down the smooth (frictionless) wedge, we have

FIG. P9.60

K block + U system + K wedge = K block + U system i

or *P9.61

(a)

0 + m1 gh + 0 =

i

f

LM 1 m a4.00f + 0OP + 1 m a−0.667f N2 Q 2 1

2

2

2

+ K wedge

f

which gives h = 0.952 m .

Conservation of the x component of momentum for the cart-bucket-water system:

b

g

mvi + 0 = m + ρV v (b)

v block = 4.00 m/s

vi =

m + ρV v m

Raindrops with zero x-component of momentum stop in the bucket and slow its horizontal motion. When they drip out, they carry with them horizontal momentum. Thus the cart slows with constant acceleration.

Chapter 9

P9.62

Consider the motion of the firefighter during the three intervals: (1) before, (2) during, and (3) after collision with the platform. (a)

v1

While falling a height of 4.00 m, his speed changes from vi = 0 to v1 as found from

i b

d

v2

g

∆E = K f + U f − K i − U i , or K f = ∆E − U f + K i + U i When the initial position of the platform is taken as the zero level of gravitational potential, we have

a f

1 mv12 = fh cos 180° − 0 + 0 + mgh 2

FIG. P9.62

Solving for v1 gives v1 = (b)

b

2 − fh + mgh m

g = 2c−300a4.00f + 75.0a9.80f4.00h =

6.81 m s

75.0

During the inelastic collision, momentum is conserved; and if v 2 is the speed of the firefighter and platform just after collision, we have mv1 = m + M v 2 or

a

v2 =

f

a f

75.0 6.81 m 1 v1 = = 5.38 m s m + M 75.0 + 20.0

Following the collision and again solving for the work done by non-conservative forces, using the distances as labeled in the figure, we have (with the zero level of gravitational potential at the initial position of the platform): ∆E = K f + U fg + U fs − K i − U ig − U is , or 1 1 − fs = 0 + m + M g − s + ks 2 − m + M v 2 − 0 − 0 2 2

a

fa f

a

f

This results in a quadratic equation in s:

a f

2 000s 2 − 931 s + 300s − 1 375 = 0 or s = 1.00 m

277

278 *P9.63

Linear Momentum and Collisions

(a)

Each object swings down according to mgR =

1 mv12 2

MgR =

1 Mv12 2

a

v1 = 2 gR

f

The collision: −mv1 + Mv1 = + m + M v 2 M−m v2 = v1 M+m Swinging up:

a

f a f a f 2 gRa1 − cos 35°fa M + mf = a M − m f 2 gR f

a

1 M + m v 22 = M + m gR 1 − cos 35° 2 v 2 = 2 gR 1 − cos 35° 0.425 M + 0.425m = M − m 1.425m = 0.575 M m = 0.403 M

P9.64

(b)

No change is required if the force is different. The nature of the forces within the system of colliding objects does not affect the total momentum of the system. With strong magnetic attraction, the heavier object will be moving somewhat faster and the lighter object faster still. Their extra kinetic energy will all be immediately converted into extra internal energy when the objects latch together. Momentum conservation guarantees that none of the extra kinetic energy remains after the objects join to make them swing higher.

(a)

Use conservation of the horizontal component of momentum for the system of the shell, the cannon, and the carriage, from just before to just after the cannon firing. p xf = p xi : or

(b)

m shell v shell cos 45.0°+ m cannon v recoil = 0

a200fa125f cos 45.0°+b5 000gv

recoil

=0

v recoil = −3.54 m s

FIG. P9.64

Use conservation of energy for the system of the cannon, the carriage, and the spring from right after the cannon is fired to the instant when the cannon comes to rest. K f + U gf + U sf = K i + U gi + U si :

0+0+ x max =

1 2 1 2 kx max = mv recoil +0+0 2 2 2 mv recoil = k

e

b5 000ga−3.54f

2

2.00 × 10 4

ja

m = 1.77 m

f

Fs, max = 2.00 × 10 4 N m 1.77 m = 3.54 × 10 4 N

(c)

Fs, max = kx max

(d)

No. The rail exerts a vertical external force (the normal force) on the cannon and prevents it from recoiling vertically. Momentum is not conserved in the vertical direction. The spring does not have time to stretch during the cannon firing. Thus, no external horizontal force is exerted on the system (cannon, carriage, and shell) from just before to just after firing. Momentum of this system is conserved in the horizontal direction during this interval.

Chapter 9

P9.65

(a)

Utilizing conservation of momentum,

b

g

m + m2 = 1 m1

2 gh

279

v1i

m 1 v1 A = m 1 + m 2 v B v1 A

y

v1 A ≅ 6.29 m s (b)

x

Utilizing the two equations, FIG. P9.65

1 2 gt = y and x = v1 A t 2 we combine them to find v1 A =

x 2y g

From the data, v1 A = 6.16 m s Most of the 2% difference between the values for speed is accounted for by the uncertainty 0.01 0.1 1 1 0.1 in the data, estimated as + + + + = 1.1% . 8.68 68.8 263 257 85.3 *P9.66

The ice cubes leave the track with speed determined by mgyi =

e

1 mv 2 ; 2

j

v = 2 9.8 m s 2 1.5 m = 5.42 m s . Its speed at the apex of its trajectory is 5.42 m s cos 40° = 4.15 m s . For its collision with the wall we have mvi + F∆t = mv f

FG H

0.005 kg 4.15 m s + F∆t = 0.005 kg −

IJ K

1 4.15 m s 2

F∆t = −3.12 × 10 −2 kg ⋅ m s The impulse exerted by the cube on the wall is to the right, +3.12 × 10 −2 kg ⋅ m s. Here F could refer to a large force over a short contact time. It can also refer to the average force if we interpret ∆t as 1 s, the time between one cube’s tap and the next’s. 10 Fav =

3.12 × 10 −2 kg ⋅ m s = 0.312 N to the right 0.1 s

280 P9.67

Linear Momentum and Collisions

(a)

Find the speed when the bullet emerges from the block by using momentum conservation:

400 m/s

mvi = MVi + mv The block moves a distance of 5.00 cm. Assume for an approximation that the block quickly reaches its maximum velocity, Vi , and the bullet kept going with a constant velocity, v. The block then compresses the spring and stops.

FIG. P9.67

1 1 MVi2 = kx 2 2 2 Vi = v=

v

5.00 cm

b900 N mge5.00 × 10

j

−2

m

2

= 1.50 m s

1.00 kg

jb

e

g b

gb

5.00 × 10 −3 kg 400 m s − 1.00 kg 1.50 m s mvi − MVi = m 5.00 × 10 −3 kg

g

v = 100 m s (b)

jb mj

1 5.00 × 10 −3 kg 100 m s 2

e 1 + b900 N mge5.00 × 10 2

∆ E = ∆K + ∆ U =

−2

g

2

−

jb

1 5.00 × 10 −3 kg 400 m s 2

e

g

2

2

∆E = −374 J , or there is an energy loss of 374 J . *P9.68

The orbital speed of the Earth is vE =

S

CM

2πr 2π 1.496 × 10 11 m = = 2.98 × 10 4 m s T 3.156 × 10 7 s

E

In six months the Earth reverses its direction, to undergo momentum change

e

je

FIG. P9.68

j

m E ∆v E = 2m E v E = 2 5.98 × 10 24 kg 2.98 × 10 4 m s = 3.56 × 10 25 kg ⋅ m s . Relative to the center of mass, the sun always has momentum of the same magnitude in the opposite direction. Its 6-month momentum change is the same size, mS ∆v S = 3.56 × 10 25 kg ⋅ m s . Then ∆v S =

3.56 × 10 25 kg ⋅ m s 1.991 × 10 30 kg

= 0.179 m s .

Chapter 9

P9.69

(a)

b3.00 kg gb7.00 m sgj + e12.0 Nija5.00 sf = b3.00 kg gv v = e 20.0 i + 7.00 jj m s

p i + Ft = p f :

f

f

(b)

a=

(c)

a=

(d)

∆r = v i t +

(e) (f)

(g) P9.70

v f − vi t

20.0 i + 7.00 j − 7.00 jj m s e a= =

:

5.00 s

∑F :

a=

m

4.00 i m s 2

12.0 N i = 4.00 i m s 2 3.00 kg

ja

f e

ja

1 ∆r = 7.00 m s j 5.00 s + 4.00 m s 2 i 5.00 s 2   ∆r = 50.0 i + 35.0 j m

1 2 at : 2

e

e

f

2

j

e

je

j

W = 12.0 N i ⋅ 50.0 m i + 35.0 mj = 600 J

W = F ⋅ ∆r :

b

ge

1 1 mv 2f = 3.00 kg 20.0 i + 7.00 j ⋅ 20.0 i + 7.00 j m 2 s 2 2 2 1 mv 2f = 1.50 kg 449 m 2 s 2 = 674 J 2

b

je b

gb

1 1 mvi2 + W = 3.00 kg 7.00 m s 2 2

g

2

j

ge

j

+ 600 J = 674 J

b

g

M = 360 kg − 2.50 kg s t .

We find the mass from

b

gb

g

1 500 m s 2.50 kg s 3 750 N Thrust v e dM dt = = = M M M M We find the velocity and position according to Euler, v new = v old + a ∆t from x new = x old + v ∆t and If we take ∆t = 0.132 s , a portion of the output looks like this: We find the acceleration from

a=

a f a f

Time t(s)

Total mass (kg)

Acceleration a m s2

Speed, v (m/s)

Position x(m)

0.000 0.132 0.264 ... 65.868 66.000 66.132 ... 131.736 131.868 132.000

360.00 359.67 359.34

10.4167 10.4262 10.4358

0.0000 1.3750 2.7513

0.0000 0.1815 0.54467

195.330 195.000 194.670

19.1983 19.2308 19.2634

916.54 919.08 921.61

27191 27312 27433

30.660 30.330 30.000

122.3092 123.6400 125.0000

3687.3 3703.5 3719.8

152382 152871 153362

(a)

The final speed is

(b)

The rocket travels

e

j

v f = 3.7 km s 153 km

281

282 P9.71

Linear Momentum and Collisions

The force exerted by the table is equal to the change in momentum of each of the links in the chain. By the calculus chain rule of derivatives, F1 =

a f

dp d mv dm dv = =v +m . dt dt dt dt

We choose to account for the change in momentum of each link by having it pass from our area of interest just before it hits the table, so that v

FIG. P9.71

dm dv ≠ 0 and m = 0. dt dt

Since the mass per unit length is uniform, we can express each link of length dx as having a mass dm: dm =

M dx . L

The magnitude of the force on the falling chain is the force that will be necessary to stop each of the elements dm. F1 = v

FG IJ H K

FG IJ H K

dm M dx M 2 =v = v dt L dt L

After falling a distance x, the square of the velocity of each link v 2 = 2 gx (from kinematics), hence F1 =

2 Mgx . L

The links already on the table have a total length x, and their weight is supported by a force F2 : F2 =

Mgx . L

Hence, the total force on the chain is Ftotal = F1 + F2 =

3 Mgx . L

That is, the total force is three times the weight of the chain on the table at that instant.

Chapter 9

P9.72

A picture one second later differs by showing five extra kilograms of sand moving on the belt.

b

gb

g

(a)

5.00 kg 0.750 m s ∆p x = = 3.75 N ∆t 1.00 s

(b)

The only horizontal force on the sand is belt friction, p xi + f∆t = p xf

so from (c)

∆p x = 3.75 N ∆t

this is

f=

and

Fext = 3.75 N

The belt is in equilibrium:

∑ Fx = ma x :

+ Fext − f = 0

a

f

(d)

W = F∆r cos θ = 3.75 N 0.750 m cos 0° = 2.81 J

(e)

1 1 ∆m v 2 = 5.00 kg 0.750 m s 2 2

a f

(f)

*P9.73

283

x CM =

b

g

2

= 1.41 J

Friction between sand and belt converts half of the input work into extra internal energy.

∑ m i xi ∑ mi

=

c

m1 R +

A 2

h + m a0 f = 2

c

m1 R +

m1 + m 2

A 2

h

y

m1 + m 2

x R

A 2 FIG. P9.73

ANSWERS TO EVEN PROBLEMS P9.2

(a) 0; (b) 1.06 kg ⋅ m s ; upward

P9.20

0.556 m

P9.4

(a) 6.00 m s to the left; (b) 8.40 J

P9.22

1.78 kN on the truck driver; 8.89 kN in the opposite direction on the car driver

P9.6

The force is 6.44 kN P9.24

v=

4M m

gA

P9.8

1.39 kg ⋅ m s upward

P9.10

(a) 5.40 N ⋅ s toward the net; (b) −27.0 J

P9.26

7.94 cm

P9.12

~ 10 3 N upward

P9.28

(a) 2.88 m s at 32.3°; (b) 783 J becomes internal energy

P9.14

(a) and (c) see the solution; (b) small; (d) large; (e) no difference

P9.30

v Y = vi sin θ ; vO = vi cos θ

P9.16

1.67 m s

P9.32

No; his speed was 41.5 mi h

P9.18

(a) 2.50 m s ; (b) 3.75 × 10 4 J

P9.34

(a) v =

vi 2

; (b) 45.0° and –45.0°

284

Linear Momentum and Collisions

2 vi ; (b) 35.3° 3

P9.36

(a)

P9.38

a0, 1.00 mf

P9.40

4.67 × 10 6 m from the Earth’s center

P9.42

(a) see the solution; (b) 3.57 × 10 8 J

2vi ;

v2 f =

P9.44

0.063 5L

P9.46

(a) see the solution; (b) −2.00 m, − 1.00 m ; (c) 3.00 i − 1.00 j m s ;

a f e j (d) e15.0 i − 5.00 jj kg ⋅ m s

P9.48

(a) −0.780 i m s ; 1.12 i m s; (b) 0.360 i m s

P9.50

(a) 787 m s; (b) 138 m s

P9.52

see the solution

P9.54

(a)

m1 v 1 + m 2 v 2 ; m1 + m 2

b

(b) v1 − v 2

(c) v 1 f =

g kbmm m+ m g ; 1

1

2

2

FG m Hm

IJ K

FG H

1

− m1 v2 1 + m2

FG 2m IJ v + FG m Hm +m K Hm 1

1

IJ K

− m2 2m 2 v1 + v2 ; m1 + m 2 1 + m2 1

2

2

IJ K

P9.56

291 N

P9.58

FG M + m IJ H m K

P9.60

(a) −0.667 m s; (b) 0.952 m

P9.62

(a) 6.81 m s; (b) 1.00 m

P9.64

(a) −3.54 m s ; (b) 1.77 m; (c) 35.4 kN; (d) No. The rails exert a vertical force to change the momentum

P9.66

0.312 N to the right

P9.68

0.179 m s

P9.70

(a) 3.7 km s ; (b) 153 km

P9.72

(a) 3.75 N to the right; (b) 3.75 N to the right; (c) 3.75 N; (d) 2.81 J; (e) 1.41 J; (f) Friction between sand and belt converts half of the input work into extra internal energy.

gd 2 2h

10 Rotation of a Rigid Object About a Fixed Axis ANSWERS TO QUESTIONS

CHAPTER OUTLINE 10.1 10.2

10.3 10.4 10.5 10.6 10.7

10.8 10.9

Angular Position, Velocity, and Acceleration Rotational Kinematics: Rotational Motion with Constant Angular Acceleration Angular and Linear Quantities Rotational Energy Calculation of Moments of Inertia Torque Relationship Between Torque and Angular Acceleration Work, Power, and Energy in Rotational Motion Rolling Motion of a Rigid Object

Q10.1

1 rev/min, or

π rad/s. Into the wall (clockwise rotation). α = 0. 30

FIG. Q10.1 Q10.2

+ k , − k

Q10.3

Yes, they are valid provided that ω is measured in degrees per second and α is measured in degrees per second-squared.

Q10.4

The speedometer will be inaccurate. The speedometer measures the number of revolutions per second of the tires. A larger tire will travel more distance in one full revolution as 2πr .

Q10.5

Smallest I is about x axis and largest I is about y axis.

Q10.6

ML2 if the mass was nonuniformly distributed, nor 12 could it be calculated if the mass distribution was not known. The moment of inertia would no longer be

Q10.7

The object will start to rotate if the two forces act along different lines. Then the torques of the forces will not be equal in magnitude and opposite in direction.

Q10.8

No horizontal force acts on the pencil, so its center of mass moves straight down.

Q10.9

You could measure the time that it takes the hanging object, m, to fall a measured distance after being released from rest. Using this information, the linear acceleration of the mass can be calculated, and then the torque on the rotating object and its angular acceleration.

Q10.10

You could use ω = αt and v = at . The equation v = Rω is valid in this situation since a = Rα .

Q10.11

The angular speed ω would decrease. The center of mass is farther from the pivot, but the moment of inertia increases also. 285

286

Rotation of a Rigid Object About a Fixed Axis

Q10.12

The moment of inertia depends on the distribution of mass with respect to a given axis. If the axis is changed, then each bit of mass that makes up the object is a different distance from the axis. In example 10.6 in the text, the moment of inertia of a uniform rigid rod about an axis perpendicular to the rod and passing through the center of mass is derived. If you spin a pencil back and forth about this axis, you will get a feeling for its stubbornness against changing rotation. Now change the axis about which you rotate it by spinning it back and forth about the axis that goes down the middle of the graphite. Easier, isn’t it? The moment of inertia about the graphite is much smaller, as the mass of the pencil is concentrated near this axis.

Q10.13

Compared to an axis through the center of mass, any other parallel axis will have larger average squared distance from the axis to the particles of which the object is composed.

Q10.14

A quick flip will set the hard–boiled egg spinning faster and more smoothly. The raw egg loses mechanical energy to internal fluid friction.

Q10.15

I CM = MR 2 , I CM = MR 2 , I CM =

Q10.16

Yes. If you drop an object, it will gain translational kinetic energy from decreasing gravitational potential energy.

Q10.17

No, just as an object need not be moving to have mass.

Q10.18

No, only if its angular momentum changes.

Q10.19

Yes. Consider a pendulum at its greatest excursion from equilibrium. It is momentarily at rest, but must have an angular acceleration or it would not oscillate.

Q10.20

Since the source reel stops almost instantly when the tape stops playing, the friction on the source reel axle must be fairly large. Since the source reel appears to us to rotate at almost constant angular velocity, the angular acceleration must be very small. Therefore, the torque on the source reel due to the tension in the tape must almost exactly balance the frictional torque. In turn, the frictional torque is nearly constant because kinetic friction forces don’t depend on velocity, and the radius of the axle where the friction is applied is constant. Thus we conclude that the torque exerted by the tape on the source reel is essentially constant in time as the tape plays. v must increase to keep the As the source reel radius R shrinks, the reel’s angular speed ω = R tape speed v constant. But the biggest change is to the reel’s moment of inertia. We model the reel as a roll of tape, ignoring any spool or platter carrying the tape. If we think of the roll of tape as a 1 uniform disk, then its moment of inertia is I = MR 2 . But the roll’s mass is proportional to its base 2 area π R 2 . Thus, on the whole the moment of inertia is proportional to R 4 . The moment of inertia decreases very rapidly as the reel shrinks! The tension in the tape coming into the read-and-write heads is normally dominated by balancing frictional torque on the source reel, according to TR ≈ τ friction . Therefore, as the tape plays the tension is largest when the reel is smallest. However, in the case of a sudden jerk on the tape, the rotational dynamics of the source reel becomes important. If the source reel is full, then the moment of inertia, proportional to R 4 , will be so large that higher tension in the tape will be required to give the source reel its angular acceleration. If the reel is nearly empty, then the same tape acceleration will require a smaller tension. Thus, the tape will be more likely to break when the source reel is nearly full. One sees the same effect in the case of paper towels; it is easier to snap a towel free when the roll is new than when it is nearly empty.

1 1 MR 2 , I CM = MR 2 3 2

Chapter 10

287

Q10.21

The moment of inertia would decrease. This would result in a higher angular speed of the earth, shorter days, and more days in the year!

Q10.22

There is very little resistance to motion that can reduce the kinetic energy of the rolling ball. Even though there is static friction between the ball and the floor (if there were none, then no rotation would occur and the ball would slide), there is no relative motion of the two surfaces—by the definition of “rolling”—and so no force of kinetic friction acts to reduce K. Air resistance and friction associated with deformation of the ball eventually stop the ball.

Q10.23

In the frame of reference of the ground, no. Every point moves perpendicular to the line joining it to the instantaneous contact point. The contact point is not moving at all. The leading and trailing edges of the cylinder have velocities at 45° to the vertical as shown.

v vCM

CM

v P

FIG. Q10.23 Q10.24

The sphere would reach the bottom first; the hoop would reach the bottom last. If each object has the same mass and the same radius, they all have the same torque due to gravity acting on them. The one with the smallest moment of inertia will thus have the largest angular acceleration and reach the bottom of the plane first.

Q10.25

To win the race, you want to decrease the moment of inertia of the wheels as much as possible. Small, light, solid disk-like wheels would be best!

SOLUTIONS TO PROBLEMS Section 10.1 P10.1

(a)

(b)

Angular Position, Velocity, and Acceleration

θ t= 0 = 5.00 rad ω t =0 =

dθ dt

α t=0 =

dω dt

t=0

= 10.0 + 4.00t t = 0 = 10.0 rad s = 4.00 rad s 2

t=0

θ t= 3.00 s = 5.00 + 30.0 + 18.0 = 53.0 rad ω t = 3.00 s =

dθ dt

α t = 3.00 s =

dω dt

t = 3 .00 s

= 10.0 + 4.00t t = 3.00 s = 22.0 rad s = 4.00 rad s 2

t = 3 .00 s

288

Rotation of a Rigid Object About a Fixed Axis

Section 10.2 *P10.2

P10.3

P10.4

Rotational Kinematics: Rotational Motion with Constant Angular Acceleration

ω f = 2.51 × 10 4 rev min = 2.63 × 10 3 rad s

P10.6

2.63 × 10 3 rad s − 0 = 8.22 × 10 2 rad s 2 3.2 s

α=

(b)

1 1 θ f = ω i t + αt 2 = 0 + 8.22 × 10 2 rad s 2 3.2 s 2 2

(a)

α=

(b)

1 1 θ = ω i t + αt 2 = 4.00 rad s 2 3.00 s 2 2

=

t

ja f

e

2

= 4.21 × 10 3 rad

ω − ω i 12.0 rad s = = 4.00 rad s 2 t 3.00 s

ja

e

f

2

= 18.0 rad

ω i = 2 000 rad s , α = −80.0 rad s 2

a fa f

(a)

ω f = ω i + αt = 2 000 − 80.0 10.0 = 1 200 rad s

(b)

0 = ω i + αt t=

P10.5

ω f −ωi

(a)

ωi =

ωi

=

−α

2 000 = 25.0 s 80.0

FG H

IJ FG 2π rad IJ = 10π K H 1.00 rev K 3

100 rev 1 min 1.00 min 60.0 s

ω f −ωi

(a)

t=

(b)

θ f = ωt =

α

=

FG ω H

f

0 − 103π −2.00

+ωi 2

rad s , ω f = 0

s = 5.24 s

IJ t = FG 10π K H6

rad s

IJ FG 10π sIJ = KH 6 K

27.4 rad

ω i = 3 600 rev min = 3.77 × 10 2 rad s θ = 50.0 rev = 3.14 × 10 2 rad and ω f = 0 ω 2f = ω i2 + 2αθ

e

0 = 3.77 × 10 2 rad s

j

2

e

+ 2α 3.14 × 10 2 rad

j

α = −2.26 × 10 2 rad s 2 P10.7

ω = 5.00 rev s = 10.0π rad s . We will break the motion into two stages: (1) a period during which the tub speeds up and (2) a period during which it slows down. 0 + 10.0π rad s 8.00 s = 40.0π rad 2

a f 10.0π rad s + 0 = ωt = a12.0 sf = 60.0π rad 2

While speeding up,

θ 1 = ωt =

While slowing down,

θ2

So,

θ total = θ 1 + θ 2 = 100π rad = 50.0 rev

Chapter 10

P10.8

1 θ f − θ i = ω i t + αt 2 and ω f = ω i + αt are two equations in two unknowns ω i and α 2 1 1 θ f − θ i = ω f − αt t + αt 2 = ω f t − αt 2 2 2 2π rad 1 37.0 rev = 98.0 rad s 3.00 s − α 3.00 s 1 rev 2

d

ω i = ω f − αt :

i

FG H

e

j

232 rad = 294 rad − 4.50 s 2 α :

P10.9

*P10.10

289

(a)

ω=

(b)

∆t =

α=

IJ K

a

f

a

f

2

61.5 rad = 13.7 rad s 2 2 4.50 s

∆θ 1 rev 2π rad = = = 7.27 × 10 −5 rad s ∆t 1 day 86 400 s ∆θ

ω

=

FG H

IJ K = b0.750 rad sgt . For the bone,

107° 2π rad = 2.57 × 10 4 s or 428 min −5 7.27 × 10 rad s 360°

The location of the dog is described by θ d

θb =

1 1 2π rad + 0.015 rad s 2 t 2 . 3 2

We look for a solution to 2π + 0.007 5t 2 3 0 = 0.007 5t 2 − 0.75t + 2.09 = 0

0.75t =

t=

b

g

0.75 ± 0.75 2 − 4 0.007 5 2.09 0.015

= 2.88 s or 97.1 s

2π 2π − 2π + 0.007 5t 2 or if 0.75t = + 2π + 0.007 5t 2 that is, if 3 3 either the dog or the turntable gains a lap on the other. The first equation has

The dog and bone will also pass if 0.75t =

t=

b

ga

0.75 ± 0.75 2 − 4 0.007 5 −4.19 0.015

f = 105 s or − 5.30 s

only one positive root representing a physical answer. The second equation has t=

b

g

0.75 ± 0.75 2 − 4 0.007 5 8.38 0.015

= 12.8 s or 87.2 s .

In order, the dog passes the bone at 2.88 s after the merry-go-round starts to turn, and again at 12.8 s and 26.6 s, after gaining laps on the bone. The bone passes the dog at 73.4 s, 87.2 s, 97.1 s, 105 s, and so on, after the start.

290

Rotation of a Rigid Object About a Fixed Axis

Section 10.3 P10.11

Angular and Linear Quantities

Estimate the tire’s radius at 0.250 m and miles driven as 10 000 per year.

θ=

θ = 6.44 × 10 7

P10.12

P10.13

FG IJ H K F 1 rev IJ = 1.02 × 10 rad yr G H 2π rad K

s 1.00 × 10 4 mi 1 609 m = = 6.44 × 10 7 rad yr 0.250 m 1 mi r 7

rev yr or ~ 10 7 rev yr

v 45.0 m s = = 0.180 rad s r 250 m

(a)

v = rω ; ω =

(b)

45.0 m s v2 = ar = r 250 m

b

g

2

= 8.10 m s 2 toward the center of track

Given r = 1.00 m, α = 4.00 rad s 2 , ω i = 0 and θ i = 57.3° = 1.00 rad (a)

ω f = ω i + αt = 0 + αt

a

f

At t = 2.00 s , ω f = 4.00 rad s 2 2.00 s = 8.00 rad s (b)

b

g

v = rω = 1.00 m 8.00 rad s = 8.00 m s

b

a r = a c = rω 2 = 1.00 m 8.00 rad s

e

g

2

= 64.0 m s 2

j

a t = rα = 1.00 m 4.00 rad s 2 = 4.00 m s 2 The magnitude of the total acceleration is: a = a r2 + a t2 =

e64.0 m s j + e4.00 m s j 2 2

2 2

= 64.1 m s 2

The direction of the total acceleration vector makes an angle φ with respect to the radius to point P:

φ = tan −1

FG a IJ = tan FG 4.00 IJ = H 64.0 K Ha K t

−1

3.58°

c

(c)

a

f e

ja

1 1 θ f = θ i + ω i t + αt 2 = 1.00 rad + 4.00 rad s 2 2.00 s 2 2

f

2

= 9.00 rad

Chapter 10

*P10.14

(a)

Consider a tooth on the front sprocket. It gives this speed, relative to the frame, to the link of the chain it engages: v = rω =

(b)

FG 0.152 m IJ 76 rev min FG 2π rad IJ FG 1 min IJ = H 2 K H 1 rev K H 60 s K v 0.605 m s = 0.07 m = 17.3 rad s r 2

c

h

Consider the wheel tread and the road. A thread could be unwinding from the tire with this speed relative to the frame: v = rω =

(d)

FG 0.673 m IJ 17.3 rad s = H 2 K

ω=

(b)

ω 2f = ω i2 + 2α ∆θ

a f

b25.0 rad sg − 0 = = 2a ∆θ f 2 a1.25 revfb 2π rad rev g 2

ω 2f − ω i2

∆ω

∆t =

(a)

s = vt = 11.0 m s 9.00 s = 99.0 m

α

=

25.0 rad s

(c)

b

θ=

(b)

FG 1 IJ = 1.39 m s H 1 rad K

v 25.0 m s = = 25.0 rad s r 1.00 m

(a)

α=

P10.16

5.82 m s

We did not need to know the length of the pedal cranks, but we could use that information to find the linear speed of the pedals: v = rω = 0.175 m 7.96 rad s

P10.15

0.605 m s

Consider the chain link engaging a tooth on the rear sprocket:

ω= (c)

291

39.8 rad s 2

ga

39.8 rad s 2

= 0.628 s

f

s 99.0 m = = 341 rad = 54.3 rev r 0.290 m

ωf =

vf r

=

22.0 m s = 75.9 rad s = 12.1 rev s 0.290 m

292 P10.17

P10.18

Rotation of a Rigid Object About a Fixed Axis

FG H

IJ K

2π rad 1 200 rev = 126 rad s 1 rev 60.0 s

(a)

ω = 2πf =

(b)

v = ωr = 126 rad s 3.00 × 10 −2 m = 3.77 m s

(c)

a c = ω 2 r = 126

(d)

s = rθ = ωrt = 126 rad s 8.00 × 10 −2 m 2.00 s = 20.1 m

b

ge

j

a f e8.00 × 10 j = 1 260 m s 2

−2

b

2

so a r = 1.26 km s 2 toward the center

ja

ge

f

The force of static friction must act forward and then more and more inward on the tires, to produce both tangential and centripetal acceleration. Its tangential component is m 1.70 m s 2 . Its radially

e

inward component is

mv . This takes the maximum value r

e

j

FG H

mω 2f r = mr ω i2 + 2α∆θ = mr 0 + 2α With skidding impending we have

IJ K

π = mπrα = mπa t = mπ 1.70 m s 2 . 2

e

µs = (a)

e

j

∑ Fy = ma y , + n − mg = 0, n = mg

fs = µ sn = µ s mg = m 2 1.70 m s 2

*P10.19

j

2

1.70 m s 2 g

j

2

e

+ m 2π 2 1.70 m s 2

j

2

1 + π 2 = 0.572

Let RE represent the radius of the Earth. The base of the building moves east at v1 = ω RE where ω is one revolution per day. The top of the building moves east at v 2 = ω RE + h . Its eastward speed relative to the ground is v 2 − v1 = ω h . The object’s time of fall is given by

b

∆y = 0 +

1 2 gt , t = 2

g

2h . During its fall the object’s eastward motion is unimpeded so its g

b

g

deflection distance is ∆x = v 2 − v1 t = ω h

f FGH 92.8sm IJK 2

F I GH JK

2h 2 = ω h3 2 g g

12

.

12

(b)

2π rad 50 m 86 400 s

(c)

The deflection is only 0.02% of the original height, so it is negligible in many practical cases.

a

32

= 1.16 cm

Chapter 10

Section 10.4 P10.20

293

Rotational Energy

m1 = 4.00 kg , r1 = y1 = 3.00 m ; m 2 = 2.00 kg , r2 = y 2 = 2.00 m; m 3 = 3.00 kg , r3 = y 3 = 4.00 m ;

ω = 2.00 rad s about the x-axis (a)

I x = m1 r12 + m 2 r22 + m 3 r32

a f

a f + 3.00a4.00f 1 = a92.0fa 2.00f = 184 J 2

I x = 4.00 3.00 KR =

(b)

1 I xω 2 2

2

+ 2.00 2.00

2

2

= 92.0 kg ⋅ m 2

2

a f = r ω = 2.00a 2.00f = = r ω = 4.00a 2.00f =

FIG. P10.20

v2 v3

K1 =

2

4.00 m s

3

8.00 m s

K = K 1 + K 2 + K 3 = 72.0 + 16.0 + 96.0 = 184 J = P10.21

(a)

a fa f a fa f a fa f

1 1 2 m1 v12 = 4.00 6.00 = 72.0 J 2 2 1 1 2 K 2 = m 2 v 22 = 2.00 4.00 = 16.0 J 2 2 1 1 2 K 3 = m 3 v 32 = 3.00 8.00 = 96.0 J 2 2

v1 = r1ω = 3.00 2.00 = 6.00 m s

1 I xω 2 2

I = ∑ m j r j2

y (m)

j

4

In this case,

3.00 kg

r1 = r2 = r3 = r4 r= I=

2

13.0 m

2.00 kg

2

a3.00 mf + a2.00 mf 2

3

2

= 13.0 m

1

3.00 + 2.00 + 2.00 + 4.00 kg

x (m) 0

1

2

= 143 kg ⋅ m 2 (b)

KR =

jb

1 2 1 Iω = 143 kg ⋅ m 2 6.00 rad s 2 2

e

g

2

2.00 kg

4.00 kg

= 2.57 × 10 3 J FIG. P10.21

3

294 P10.22

Rotation of a Rigid Object About a Fixed Axis

a

I = Mx 2 + m L − x

f

2

a

x

f

dI = 2 Mx − 2m L − x = 0 (for an extremum) dx mL ∴x = M+m d2I = 2m + 2 M ; therefore I is minimum when the axis of dx 2 mL which is also the center rotation passes through x = M+m of mass of the system. The moment of inertia about an axis passing through x is I CM = M

LM mL OP NM + mQ

2

LM N

+m 1−

m M+m

OP L Q 2

2

=

L m

M x

Mm 2 L = µL2 M+m

L−x

FIG. P10.22

Mm . where µ = M+m Section 10.5 P10.23

Calculation of Moments of Inertia

We assume the rods are thin, with radius much less than L. Call the junction of the rods the origin of coordinates, and the axis of rotation the z-axis. For the rod along the y-axis, I =

y

1 mL2 from the table. 3

x

For the rod parallel to the z-axis, the parallel-axis theorem gives I=

FG IJ H K

1 L mr 2 + m 2 2

2

≅

axis of rotation z

1 mL2 4

FIG. P10.23

In the rod along the x-axis, the bit of material between x and x + dx has mass distance r = x 2 +

FG L IJ H 2K

FG m IJ dx and is at H LK

2

from the axis of rotation. The total rotational inertia is:

z FGH

L2

I total =

1 1 L2 mL2 + mL2 + x2 + 3 4 4 −L 2

FG IJ H K

7 m x3 = mL2 + 12 L 3 =

L2

+ −L 2

I FG m IJ dx JK H L K L2

mL x 4 −L 2

2

7 11mL2 mL mL2 + = mL2 + 12 12 4 12

Note: The moment of inertia of the rod along the x axis can also be calculated from the parallel-axis 2 1 L theorem as . mL2 + m 12 2

FG IJ H K

Chapter 10

P10.24

295

Treat the tire as consisting of three parts. The two sidewalls are each treated as a hollow cylinder of inner radius 16.5 cm, outer radius 30.5 cm, and height 0.635 cm. The tread region is treated as a hollow cylinder of inner radius 30.5 cm, outer radius 33.0 cm, and height 20.0 cm. Use I =

1 m R12 + R 22 for the moment of inertia of a hollow cylinder. 2

e

j

Sidewall:

a

f − a0.165 mf e6.35 × 10 mje1.10 × 10 1 = b1.44 kg g a0.165 mf + a0.305 mf = 8.68 × 10 2

m = π 0.305 m I side

2

2

−3

2

2

3

−2

j

kg m 3 = 1.44 kg kg ⋅ m 2

Tread:

a

f − a0.305 mf a0.200 mfe1.10 × 10 kg m j = 11.0 kg 1 = b11.0 kg g a0.330 mf + a0.305 mf = 1.11 kg ⋅ m 2

m = π 0.330 m I tread

2

2

3

3

2

2

2

Entire Tire:

e

j

I total = 2 I side + I tread = 2 8.68 × 10 −2 kg ⋅ m 2 + 1.11 kg ⋅ m 2 = 1.28 kg ⋅ m 2 P10.25

Every particle in the door could be slid straight down into a high-density rod across its bottom, without changing the particle’s distance from the rotation axis of the door. Thus, a rod 0.870 m long with mass 23.0 kg, pivoted about one end, has the same rotational inertia as the door: I=

b

ga

f

1 1 ML2 = 23.0 kg 0.870 m 3 3

2

= 5.80 kg ⋅ m 2 .

The height of the door is unnecessary data. P10.26

Model your body as a cylinder of mass 60.0 kg and circumference 75.0 cm. Then its radius is 0.750 m = 0.120 m 2π and its moment of inertia is

b

ga

f

1 1 MR 2 = 60.0 kg 0.120 m 2 2

2

= 0.432 kg ⋅ m 2 ~ 10 0 kg ⋅ m 2 = 1 kg ⋅ m 2 .

296 P10.27

Rotation of a Rigid Object About a Fixed Axis

For a spherical shell dI =

2 2 4πr 2 dr ρ r 2 dmr 2 = 3 3

e

z z ze

j

e j af 2 rI F 4πr jG 14. 2 − 11.6 J e10 kg m jdr I= H 3 RK R R F 2I F 2I = G J 4π e14. 2 × 10 j − G J 4π e11.6 × 10 j H 3K H K 5 3 6 8π 14 2 11 6 . . I= e10 jR FGH 5 − 6 IJK 3 rI F M = z dm = z 4πr G 14.2 − 11.6 J 10 dr H RK F 14.2 − 11.6 IJ R = 4π × 10 G H3 4K b8π 3ge10 jR b14.2 5 − 11.6 6g = 2 FG .907 IJ = 0.330 I = 4π × 10 R R b14.2 3 − 11.6 4g 3 H 1.83 K MR 2 4πr 2 r 2 ρ r dr 3

I = dI = R

4

3

3

0

3

3

5

3

5

5

R

2

3

0

3

3

3

2

3

5

3

2

∴ I = 0.330 MR 2 *P10.28

(a)

y h hx . The area of the front face = , y= L x L 1 1 is hL. The volume of the plate is hLw . Its density is 2 2 2M M M . The mass of the ribbon is ρ= = 1 = V hLw 2 hLw

By similar triangles,

dm = ρdV = ρywdx =

y

h

x L

2 Mywdx 2 Mhx 2 Mxdx . = dx = hLw hLL L2

FIG. P10.28

The moment of inertia is I=

z

r 2 dm =

x2

x=0

all mass

(b)

z L

z

L

2 Mxdx 2 M 3 2 M L4 ML2 = 2 x dx = 2 = . 2 2 L L 0 L 4

From the parallel axis theorem I = I CM + M I h = I CM + M

FG L IJ H 3K

inertia I CM +

2

FG 2L IJ H3K

2

= I CM +

4ML2 and 9

ML2 . The two triangles constitute a rectangle with moment of 9 1 ML2 1 + = 2 M L2 . Then 2 I CM = ML2 9 3 9

= I CM +

4ML2 + I CM 9

I = I CM +

a f

4ML2 1 8 1 = ML2 + ML2 = ML2 . 9 18 18 2

Chapter 10

*P10.29

297

We consider the cam as the superposition of the original solid disk and a disk of negative mass cut from it. With half the radius, the cut-away part has one-quarter the face area and one-quarter the volume and one-quarter the mass M 0 of the original solid cylinder: M0 −

1 M0 = M 4

M0 =

4 M. 3

By the parallel-axis theorem, the original cylinder had moment of inertia

FG R IJ = 1 M R + M R = 3 M R . H 2K 2 4 4 1F 1 I F R I M R . The whole cam has The negative-mass portion has I = G − M J G J = − H KH 2 K 2 4 32 2

I CM + M 0

2

0

2

0

I=

0

0

0

2

2

M R 2 23 3 23 4 23 1 1 23 23 M0 R 2 − 0 = M0 R 2 = MR 2 = MR 2 and K = Iω 2 = MR 2 ω 2 = MR 2ω 2 . 4 32 32 32 3 24 2 2 24 48

Section 10.6 P10.30

2

Torque

Resolve the 100 N force into components perpendicular to and parallel to the rod, as

a

f

Fpar = 100 N cos 57.0° = 54.5 N and

a

f

Fperp = 100 N sin 57.0° = 83.9 N

The torque of Fpar is zero since its line of action passes through the pivot point.

a

FIG. P10.30

f

The torque of Fperp is τ = 83.9 N 2.00 m = 168 N ⋅ m (clockwise) P10.31

∑ τ = 0.100 ma12.0 N f − 0.250 ma9.00 N f − 0.250 ma10.0 N f =

−3.55 N ⋅ m

The thirty-degree angle is unnecessary information.

FIG. P10.31 P10.32

The normal force exerted by the ground on each wheel is

b

ge

j

1 500 kg 9.80 m s 2 mg n= = = 3 680 N 4 4 The torque of friction can be as large as τ max = f max r = µ sn r = 0.800 3 680 N 0.300 m = 882 N ⋅ m

b g a

fb

ga

f

The torque of the axle on the wheel can be equally as large as the light wheel starts to turn without slipping.

298 P10.33

Rotation of a Rigid Object About a Fixed Axis

In the previous problem we calculated the maximum torque that can be applied without skidding to be 882 N · m. This same torque is to be applied by the frictional force, f, between the brake pad and the rotor for this wheel. Since the wheel is slipping against the brake pad, we use the coefficient of kinetic friction to calculate the normal force.

b g

τ = fr = µ k n r , so n =

Section 10.7 P10.34

(a)

882 N ⋅ m τ = = 8.02 × 10 3 N = 8.02 kN 0.500 0.220 m µ kr

a

fa

f

Relationship Between Torque and Angular Acceleration

b

ge

1 1 MR 2 = 2.00 kg 7.00 × 10 −2 m 2 2 0.600 τ α= = = 122 rad s 2 I 4.90 × 10 −3 ∆ω α= ∆t 2π ∆ω 1 200 60 ∆t = = = 1.03 s 122 α I=

j

2

= 4.90 × 10 −3 kg ⋅ m 2

c h

(b) P10.35

P10.36

∆θ =

b

ga

1 2 1 αt = 122 rad s 1.03 s 2 2

f

2

= 64.7 rad = 10.3 rev

m = 0.750 kg , F = 0.800 N

a

f

(a)

τ = rF = 30.0 m 0.800 N = 24.0 N ⋅ m

(b)

α=

(c)

at

24.0 τ rF = = I mr 2 0.750 30.0

= 0.035 6 rad s 2

a f = αr = 0.035 6a30.0 f = 1.07 m s

ω f = ω i + αt :

2

FIG. P10.35

2

a

f

10.0 rad s = 0 + α 6.00 s 10.00 rad s 2 = 1.67 rad s 2 α= 6.00

∑τ

36.0 N ⋅ m = 21.6 kg ⋅ m 2 2 1.67 rad s

(a)

∑ τ = 36.0 N ⋅ m = Iα :

I=

(b)

ω f = ω i + αt :

0 = 10.0 + α 60.0

α

=

a f

α = −0.167 rad s 2

e

je

j

τ = Iα = 21.6 kg ⋅ m 2 0.167 rad s 2 = 3.60 N ⋅ m (c)

Number of revolutions θ f = θ i + ω i t + During first 6.00 s During next 60.0 s

1 2 αt 2

a fa f a f a fa f FG IJ H K

1 2 1.67 6.00 = 30.1 rad 2 1 2 θ f = 10.0 60.0 − 0.167 60.0 = 299 rad 2 1 rev θ total = 329 rad = 52.4 rev 2π rad

θf =

Chapter 10

P10.37

For m1 , ∑ Fy = ma y :

+n − m 1 g = 0 n1 = m1 g = 19.6 N f k 1 = µ k n1 = 7.06 N

b

∑ Fx = ma x :

g

−7.06 N + T1 = 2.00 kg a

For the pulley,

∑ τ = Iα :

FG IJ H K 1 = b10.0 kg ga 2 = b5.00 kg ga

−T1 R + T2 R = −T1 + T2 −T1 + T2

1 a MR 2 2 R

+n 2 − m 2 g cos θ = 0

For m 2 ,

(1)

(2)

ja

e

f

n 2 = 6.00 kg 9.80 m s 2 cos 30.0° = 50.9 N

FIG. P10.37

fk 2 = µ kn 2 = 18.3 N : −18.3 N − T2 + m 2 sin θ = m 2 a −18.3 N − T2 + 29.4 N = 6.00 kg a (3)

b

(a)

g

Add equations (1), (2), and (3):

b

g

−7.06 N − 18.3 N + 29.4 N = 13.0 kg a a=

e

j

4.01 N = 0.309 m s 2 13.0 kg

T1 = 2.00 kg 0.309 m s 2 + 7.06 N = 7.67 N

(b)

e

j

T2 = 7.67 N + 5.00 kg 0.309 m s 2 = 9.22 N P10.38

b

ga

f

1 1 mR 2 = 100 kg 0.500 m 2 2 ω i = 50.0 rev min = 5.24 rad s I=

α=

ω f −ωi t

=

2

= 12.5 kg ⋅ m 2

0 − 5.24 rad s = −0.873 rad s 2 6.00 s

e

j

τ = Iα = 12.5 kg ⋅ m 2 −0.873 rad s 2 = −10.9 N ⋅ m The magnitude of the torque is given by fR = 10.9 N ⋅ m, where f is the force of friction. 10.9 N ⋅ m 0.500 m

Therefore,

f=

yields

µk =

f 21.8 N = = 0.312 n 70.0 N

and

f = µ kn

FIG. P10.38

299

300 *P10.39

Rotation of a Rigid Object About a Fixed Axis

1

∑ τ = Iα = 2 MR 2α

a

f 12 b80 kggFGH 1.225 mIJK e−1.67 rad s j

f a

2

−135 N 0.230 m + T 0.230 m =

2

T = 21.5 N

Section 10.8 P10.40

Work, Power, and Energy in Rotational Motion

The moment of inertia of a thin rod about an axis through one end is I = kinetic energy is given as KR = with

Ih =

a

f

m h L2h 60.0 kg 2.70 m = 3 3

2

1 ML2 . The total rotational 3

1 1 2 I hω 2h + I mω m 2 2

= 146 kg ⋅ m 2

a f F I GH JK 2π rad F 1 h I = = 1.75 × 10 rad s 1 h GH 3 600 s JK 1 1 = a146 fe1.45 × 10 j + a675 fe1.75 × 10 j 2 2 2

and In addition,

*P10.41

100 kg 4.50 m m L2 = 675 kg ⋅ m 2 Im = m m = 3 3 2π rad 1h ωh = = 1.45 × 10 −4 rad s 12 h 3 600 s

while

ωm

Therefore,

KR

−3

−4 2

−3 2

= 1.04 × 10 −3 J

1 11 E where E = Iω 2 = MR 2ω 2 is the stored energy and 2 22 ∆t 1 P∆x ∆x is the time it can roll. Then MR 2ω 2 = P∆t = and ∆t = 4 v v

The power output of the bus is P =

a

fc 2

h

2

2π MR 2ω 2 v 1 600 kg 0.65 m 4 000 ⋅ 60 s 11.1 m s ∆x = = = 24.5 km . 4P 4 18 ⋅ 746 W

P10.42

a

fa

a

f

f

Work done = F∆r = 5.57 N 0.800 m = 4.46 J 1 1 and Work = ∆K = Iω 2f − Iω i2 2 2 (The last term is zero because the top starts from rest.) Thus, 4.46 J =

1 4.00 × 10 −4 kg ⋅ m 2 ω 2f 2

e

A′

F

j

A

and from this, ω f = 149 rad s . FIG. P10.42

Chapter 10

*P10.43

(a)

g a f a f = 2.28 × 10 kg ⋅ m j b b g e j 1 1 1 F 0.82 m s IJ 0.850 kg gb0.82 m sg + b0.42 kg gb0.82 m sg + e 2.28 × 10 kg ⋅ m jG b H 0.03 m K 2 2 2 +0.42 kg e9.8 m s ja0.7 mf − 0.25b0.85 kg ge9.8 m s ja0.7 mf F v IJ 1 1 1 = b0.85 kg gv + b0.42 kg gv + e 2.28 × 10 kg ⋅ m jG 2 2 2 H 0.03 m K 0.512 J + 2.88 J − 1.46 J = b0.761 kg gv 1 1 2 M R12 + R 22 = 0.35 kg 0.02 m + 0.03 m 2 2 K 1 + K 2 + K rot + U g 2 − f k ∆x = K 1 + K 2 + K rot f

e

I=

2

−4

2

i

2

2

−4

2

2

2

2

2 f

−4

2 f

f

2

2

2 f

vf =

(b) P10.44

ω=

1.94 J = 1.59 m s 0.761 kg

v 1.59 m s = = 53.1 rad s 0.03 m r

We assume the rod is thin. For the compound object

LM N

OP Q

1 2 M rod L2 + m ball R 2 + M ball D 2 3 5 1 2 2 I = 1.20 kg 0.240 m + 2.00 kg 4.00 × 10 −2 m 3 5 I = 0.181 kg ⋅ m 2 I=

a

(a)

f

e

j

2

a

f

+ 2.00 kg 0.280 m

2

K f + U f = K i + U i + ∆E

FG IJ H K

a

f ja

1 2 L + M ball g L + R + 0 Iω + 0 = 0 + M rod g 2 2 1 0.181 kg ⋅ m 2 ω 2 = 1.20 kg 9.80 m s 2 0.120 m + 2.00 kg 9.80 m s 2 0.280 m 2 1 0.181 kg ⋅ m 2 ω 2 = 6.90 J 2

e e

j j

e

(b)

ω = 8.73 rad s

(c)

v = rω = 0.280 m 8.73 rad s = 2.44 m s

(d)

v 2f = vi2 + 2 a y f − yi

a

f

e

f

d

e

i

ja

f

v f = 0 + 2 9.80 m s 2 0.280 m = 2.34 m s The speed it attains in swinging is greater by

2.44 = 1.043 2 times 2.34

ja

f

301

302 P10.45

Rotation of a Rigid Object About a Fixed Axis

(a)

For the counterweight,

∑ Fy = ma y becomes: For the reel

∑ τ = Iα

50.0 − T =

FG 50.0 IJ a H 9.80 K

reads TR = Iα = I I=

where

a R

1 MR 2 = 0.093 8 kg ⋅ m 2 2

We substitute to eliminate the acceleration: 50.0 − T = 5.10

F TR I GH I JK 2

T = 11.4 N a=

d

v f = 2 7.57 6.00 = 9.53 m s

Use conservation of energy for the system of the object, the reel, and the Earth:

aK + U f = aK + U f : i

1 1 mv 2 + Iω 2 2 2 v2 I 2mgh = mv 2 + I 2 = v 2 m + 2 R R

mgh =

f

2mgh m + I2

v=

F I FG IJ GH JK H K 2a50.0 N fa6.00 mf = = 5.10 kg +

R

P10.46

FIG. P10.45

50.0 − 11.4 = 7.57 m s 2 5.10

a f

i

v 2f = vi2 + 2 a x f − xi : (b)

and

0.093 8

a0. 250 f

9.53 m s

2

Choose the zero gravitational potential energy at the level where the masses pass. K f + U gf = K i + U gi + ∆E 1 1 1 m1 v 2 + m 2 v 2 + Iω 2 = 0 + m1 gh1i + m 2 gh2 i + 0 2 2 2

a b

LM a f OPFG IJ N QH K

f

2

a fa f

a fa

1 1 1 v 15.0 + 10.0 v 2 + 3.00 R 2 = 15.0 9.80 1.50 + 10.0 9.80 −1.50 2 2 2 R 1 26.5 kg v 2 = 73.5 J ⇒ v = 2.36 m s 2 P10.47

g

f

From conservation of energy for the object-turntable-cylinder-Earth system,

FG IJ H K

1 v I 2 r

2

+

1 mv 2 = mgh 2

v2 = 2mgh − mv 2 r2 2 gh I = mr 2 −1 v2 I

FG H

IJ K

FIG. P10.47

Chapter 10

P10.48

The moment of inertia of the cylinder is I=

b

f

ga

1 1 mr 2 = 81.6 kg 1.50 m 2 2

2

= 91.8 kg ⋅ m 2

and the angular acceleration of the merry-go-round is found as

α=

a f a

fa

f

50.0 N 1.50 m Fr τ = = = 0.817 rad s 2 . 2 I I 91.8 kg ⋅ m

e

j

At t = 3.00 s, we find the angular velocity

ω = ω i + αt

ja

e

f

ω = 0 + 0.817 rad s 2 3.00 s = 2.45 rad s

P10.49

jb

and K =

1 2 1 Iω = 91.8 kg ⋅ m 2 2. 45 rad s 2 2

(a)

Find the velocity of the CM

e

g

aK + U f = aK + U f i

2

= 276 J .

Pivot

f

R

1 0 + mgR = Iω 2 2 2mgR 2mgR = 3 ω= 2 I 2 mR 4g Rg = 2 3R 3

vCM = R

*P10.50

g

FIG. P10.49

Rg 3

(b)

v L = 2 vCM = 4

(c)

vCM =

(a)

The moment of inertia of the cord on the spool is

2mgR = 2m

Rg

ea

f + a0.09 mf j = 4.16 × 10

1 1 M R12 + R 22 = 0.1 kg 0.015 m 2 2

e

j

e

2

2

−4

kg ⋅ m 2 .

j

The protruding strand has mass 10 −2 kg m 0.16 m = 1.6 × 10 −3 kg and I = I CM + Md 2 =

FG a H

f + a0.09 m + 0.08 mf IJK

1 1 0.16 m ML2 + Md 2 = 1.6 × 10 −3 kg 12 12

2

2

= 4.97 × 10 −5 kg ⋅ m 2 For the whole cord, I = 4.66 × 10 −4 kg ⋅ m 2 . In speeding up, the average power is

P=

(b)

a

fa

E = ∆t

1 2

Iω 2 ∆t

=

a

f

⋅ 2π I fFGH 2 000 J= 60 s K

P = τω = 7.65 N 0.16 m + 0.09 m

FG H

4.66 × 10 −4 kg ⋅ m 2 2 500 ⋅ 2π 2 0.215 s 60 s 401 W

IJ K

2

= 74.3 W

303

304

Rotation of a Rigid Object About a Fixed Axis

Section 10.9 P10.51

P10.52

b

gb

1 1 mv 2 = 10.0 kg 10.0 m s 2 2

(a)

K trans =

(b)

K rot =

(c)

K total = K trans + K rot = 750 J

FG H

1 2 1 1 Iω = mr 2 2 2 2

b

W = K f − K i = K trans + K rot W= or

P10.53

Rolling Motion of a Rigid Object

g

2

IJ FG v IJ = 1 b10.0 kg gb10.0 m sg KH r K 4

g − bK f

2

2

2

trans

+ K rot

FG H

g

i

1 1 1 1 2 Mv 2 + Iω 2 − 0 − 0 = Mv 2 + MR 2 2 2 5 2 2 7 W= Mv 2 10

(a)

= 500 J

FG IJ H K

IJ FG v IJ KH RK

= 250 J

2

τ = Iα mgR sin θ = I CM + mR 2 α

e

a=

j

mgR 2 sin θ I CM + mR

R

2

mgR 2 sin θ

a hoop =

2mR 2

f

1 = g sin θ 2

mgR 2 sin θ

(b)

θ

FIG. P10.53

Rf = Iα f = µn = µmg cos θ Iα f R µ= = = mg cos θ mg cos θ

P10.54

mg

n

2 = a disk = 3 g sin θ 2 3 mR 2 4 The disk moves with the acceleration of the hoop. 3

LM N

OP Q

c

1 1 1 I mv 2 + Iω 2 = m + 2 v2 2 2 2 R Also, Ui = mgh , U f = 0 , K=

2 3

g sin θ

he

1 2

mR 2

2

R mg cos θ where and

j=

1 tan θ 3

v R vi = 0

ω=

since no slipping.

LM N

OP Q

1 I m + 2 v 2 = mgh 2 R 2 gh v2 = 1+ I 2

Therefore, Thus,

e j mR

1 I = mR 2 2

For a disk, So For a ring, Since v disk

v2 =

2 gh 1 + 12

or

v disk =

4 gh 3

2 gh or v ring = gh 2 > v ring , the disk reaches the bottom first. I = mR 2 so v 2 =

Chapter 10

P10.55

305

1 ∆x 3.00 m = = 2.00 m s = 0 + v f 1.50 s 2 ∆t vf 4.00 m s 8.00 rad s v f = 4.00 m s and ω f = = = −2 r 6 . 38 × 10 −2 6.38 × 10 m 2

d

v=

i

e

j

We ignore internal friction and suppose the can rolls without slipping.

e K + K + U j + ∆E b0 + 0 + mgy g + 0 = FGH 12 mv trans

rot

g

mech

i

i

2 f

+

e

= K trans + K rot + U g 1 2 Iω f + 0 2

IJ K

j

f

e j a3.00 mf sin 25.0° = 12 b0.215 kg gb4.00 m sg 2.67 J = 1.72 J + e7 860 s jt

0.215 kg 9.80 m s 2

2

+

FG H

IJ K

1 8.00 rad s I 2 6.38 × 10 −2

2

−2

I= P10.56

(a)

0.951 kg ⋅ m 2 s 2

= 1.21 × 10 −4 kg ⋅ m 2

7 860 s −2

The height of the can is unnecessary data.

Energy conservation for the system of the ball and the Earth between the horizontal section and top of loop: 1 1 1 1 mv 22 + Iω 22 + mgy 2 = mv12 + Iω 12 2 2 2 2

FG H

IJ FG v IJ + mgy KH r K 1F2 IF v I + G mr J G J KH r K 2H3

1 1 2 mv 22 + mr 2 2 2 3 1 mv12 2 5 2 5 v 2 + gy 2 = v12 6 6 =

v 2 = v12 −

2

2

2

2

6 gy 2 = 5

1

b4.03 m sg

2

−

2

FIG. P10.56

f e ja b2.38 m sg = 12.6 m s =

6 9.80 m s 2 0.900 m = 2.38 m s 5 2

v2 2 The centripetal acceleration is 2 >g 0.450 m r Thus, the ball must be in contact with the track, with the track pushing downward on it. (b)

FG H

v 3 = v12 − (c)

IJ FG v IJ + mgy = 1 mv + 1 FG 2 mr IJ FG v IJ KH r K KH r K 2 2H3 6 = b 4.03 m sg − e9.80 m s ja −0.200 mf = 5

1 1 2 mv 32 + mr 2 2 2 3 6 gy 3 5

3

2

3

2 1

2

2

1

2

2

4.31 m s

1 1 mv 22 + mgy 2 = mv12 2 2 v 2 = v12 − 2 gy 2 =

b4.03 m sg − 2e9.80 m s ja0.900 mf = 2

2

−1.40 m 2 s 2

This result is imaginary. In the case where the ball does not roll, the ball starts with less energy than in part (a) and never makes it to the top of the loop.

306

Rotation of a Rigid Object About a Fixed Axis

Additional Problems P10.57

A 1 sin θ = mA 2α 2 3 3 g sin θ α= 2 A 3 g sin θ r at = 2 A 3 g Then r > g sin θ 2 A 2 for r > A 3 1 ∴ About the length of the chimney will have a 3 tangential acceleration greater than g sin θ . mg

FG H

P10.58

FG IJ H K

IJ K

θt

θt

g sin θ

g

FIG. P10.57

The resistive force on each ball is R = DρAv 2 . Here v = rω , where r is the radius of each ball’s path. The resistive torque on each ball is τ = rR , so the total resistive torque on the three ball system is τ total = 3rR . The power required to maintain a constant rotation rate is P = τ totalω = 3rRω . This required power may be written as

a f ω = e3r DAω jρ 2π rad F 10 rev I F 1 min I 1 000π ω= G J = 30.0 rad s 1 rev GH 1 min JK H 60.0 s K F 1 000π IJ ρ P = 3a0.100 mf a0.600 fe 4.00 × 10 m jG H 30.0 s K P = e0.827 m s jρ , where ρ is the density of the resisting medium. P = τ totalω = 3r DρA rω

With

2

3

or (a)

3

3

3

5

−4

2

3

3

In air, ρ = 1.20 kg m3 ,

e

j

and P = 0.827 m 5 s 3 1. 20 kg m 3 = 0.992 N ⋅ m s = 0.992 W

P10.59

at

(b)

In water, ρ = 1 000 kg m3 and P = 827 W .

(a)

W = ∆K =

(b) (c)

1 2 1 2 1 1 Iω f − Iω i = I ω 2f − ω i2 where I = mR 2 2 2 2 2 1 2 2 1.00 kg 0.500 m 8.00 rad s − 0 = 4.00 J 2

e j F 1IF I = G JG Jb f b ga g H 2KH K ω − 0 ωr b8.00 rad sga0.500 mf = = = 1.60 s t= f

α

2.50 m s 2

a

1 θ f = θ i + ω i t + αt 2 ; θ i = 0 ; ω i = 0 2 1 2 1 2.50 m s 2 2 1.60 s = 6.40 rad θ f = αt = 2 2 0.500 m

F GH

Ia JK s = rθ = a0.500 mfa6. 40 radf =

f

3.20 m < 4.00 m Yes

Chapter 10

*P10.60

Start

The quantity of tape is constant. Then the area of the rings you see it fill is constant. This is expressed by

rt

π rt2 − π rs2 = π r 2 − π rs2 + π r22 − π rs2 or r2 = rt2 + rs2 − r 2 is the outer radius of spool 2. (a)

P10.61

(a)

j

Later r2

r

v v and ω 2 = . The rt rs takeup reel must spin at maximum speed. At the end, v v and ω 1 = . The angular r = rs and r2 = rt so ω 2 = rt rs speeds are just reversed.

At the start, r = rt and r2 = rs so ω 1 =

v FIG. P10.60

Since only conservative forces act within the system of the rod and the Earth, ∆E = 0

where I =

K f + U f = K i + Ui

so

FG IJ H K

1 2 L Iω + 0 = 0 + Mg 2 2

1 ML2 3

FIG. P10.61

3g L

ω=

Therefore, (b)

v

v Where the tape comes off spool 1, ω 1 = . Where the r −1 2 v 2 2 . tape joins spool 2, ω 2 = = v rs + rt − r 2 r2

e

(b)

rs

∑ τ = Iα , so that in the horizontal orientation, Mg

FG L IJ = ML H 2K 3

α=

FG L IJ ω H 2K

a x = a r = − rω 2 = −

(d)

Using Newton’s second law, we have R x = Ma x = −

2

3g 2

(c)

= −

2

3g 2L

a y = − a t = − rα = −α

3 Mg 2

R y − Mg = Ma y = −

3 Mg 4

α

Ry =

Mg 4

FG L IJ = H 2K

−

3g 4

307

308 P10.62

Rotation of a Rigid Object About a Fixed Axis

e

j

α = −10.0 rad s 2 − 5.00 rad s 3 t =

z z

ω

dω =

65.0

t

dω dt

−10.0 − 5.00t dt = −10.0t − 2.50t 2 = ω − 65.0 rad s

0

dθ ω= = 65.0 rad s − 10.0 rad s 2 t − 2.50 rad s 3 t 2 dt

e

(a)

j e

j

At t = 3.00 s,

ja

e

f e

je

j

ω = 65.0 rad s − 10.0 rad s 2 3.00 s − 2.50 rad s 3 9.00 s 2 = 12.5 rad s

z z

θ

(b)

0

t

dθ = ω dt =

b

0

z t

e

j e

j

65.0 rad s − 10.0 rad s 2 t − 2.50 rad s 3 t 2 dt

0

g e

j e

j

θ = 65.0 rad s t − 5.00 rad s 2 t 2 − 0.833 rad s 3 t 3 At t = 3.00 s,

b

ga

f e

j

e

j

θ = 65.0 rad s 3.00 s − 5.00 rad s 2 9.00 s 2 − 0.833 rad s 3 27.0 s 3 θ = 128 rad P10.63

The first drop has a velocity leaving the wheel given by

1 mvi2 = mgh1 , so 2

e

ja

f

e

ja

f

v1 = 2 gh1 = 2 9.80 m s 2 0.540 m = 3.25 m s The second drop has a velocity given by v 2 = 2 gh2 = 2 9.80 m s 2 0.510 m = 3.16 m s From ω =

v , we find r

ω1 =

3.16 m s v1 3.25 m s v = = 8.53 rad s and ω 2 = 2 = = 8.29 rad s 0.381 m 0.381 m r r

or

α=

b

g b 2

8.29 rad s − 8.53 rad s ω 22 − ω 12 = 2θ 4π

g

2

= −0.322 rad s 2

Chapter 10

P10.64

At the instant it comes off the wheel, the first drop has a velocity v1 , directed upward. The magnitude of this velocity is found from K i + U gi = K f + U gf 1 mv12 + 0 = 0 + mgh1 or v1 = 2 gh1 2 and the angular velocity of the wheel at the instant the first drop leaves is

ω1 =

2 gh1

v1 = R

R2

Similarly for the second drop: v 2 = 2 gh2 and ω 2 =

v2 = R

. 2 gh2 R2

.

The angular acceleration of the wheel is then 2 gh2

ω 2 − ω 12 = a= 2 2θ P10.65

R2

−

2 gh1 R2

a f

2 2π

=

b

g h2 − h1 2πR

1 1 1 1 Mv 2f + Iω 2f : U f = Mgh f = 0 ; K i = Mvi2 + Iω i2 = 0 2 2 2 2 1 v Ui = Mgh i : f = µN = µMg cos θ ; ω = ; h = d sin θ and I = mr 2 2 r Kf =

b g

(a)

∆E = E f − Ei or − fd = K f + U f − K i − U i 1 1 − fd = Mv 2f + Iω 2f − Mgh 2 2

F GH

I − Mgd sinθ b g JK 2 1L mO M + P v = Mgd sin θ − b µMg cos θ gd or M 2N 2Q bsinθ − µ cosθ g v = 2 Mgd

1 mr 2 − µMg cos θ d = Mv 2 + 2 2

v2 r2

2

2

m 2

vd (b)

+M

L O M = M4 gd sin θ − µ cos θ gP b N am + 2 M f Q

12

v 2f = vi2 + 2 a∆x , v d2 = 2 ad a=

FG H

IJ b K

v d2 M sin θ − µ cos θ = 2g 2d m + 2M

g

2

g

.

309

310 P10.66

Rotation of a Rigid Object About a Fixed Axis

(a)

E=

FG H

IJ e j K

1 2 MR 2 ω 2 2 5

1 2 E = ⋅ 5.98 × 10 24 6.37 × 10 6 2 5

e

(b)

je

LM FG MN H

π I j FGH 862400 JK 2

IJ FG 2π IJ OP K H T K PQ 1 dT = MR a 2π f e −2T j 5 dt 1 F 2π I F −2 I dT = MR G J G J H T K H T K dt 5 F −2 I F 10 × 10 = e 2.57 × 10 JjG H 86 400 s JK GH 3.16 × 10

= 2.57 × 10 29 J

2

dE d 1 2 MR 2 = dt dt 2 5 2

2

2

−3

2

2

−6

29

7

I b86 400 s dayg J sK

s

dE = −1.63 × 10 17 J day dt *P10.67

(a)

ω f = ω i + αt α=

ω f −ωi t

=

2π Tf

− t

e

2π Ti

=

d

2π Ti − T f

i

Ti T f t

j F 1 d I FG 1 yr IJ = ~ 1 d 1 d 100 yr GH 86 400 s JK H 3.156 × 10 s K 2π −10 −3 s

2

7

(b)

−10 −22 s −2

The Earth, assumed uniform, has moment of inertia I=

2 2 MR 2 = 5.98 × 10 24 kg 6.37 × 10 6 m 5 5

e

∑ τ = Iα ~ 9.71 × 10 37

je j = 9.71 × 10 kg ⋅ m e−2.67 × 10 s j = −10 2

−22

2

−2

37

16

kg ⋅ m 2

N⋅m

The negative sign indicates clockwise, to slow the planet’s counterclockwise rotation. (c)

τ = Fd . Suppose the person can exert a 900-N force. d=

τ 2.59 × 10 16 N ⋅ m ~ 10 13 m = 900 N F

This is the order of magnitude of the size of the planetary system.

Chapter 10

P10.68

∆ θ = ωt t=

∆θ

=

∆θ = 31°

c h rev = 0.005 74 s

v

31.0 ° 360° 900 rev 60 s

ω 0.800 m = 139 m s v= 0.005 74 s

ω

d

FIG. P10.68 P10.69

τ f will oppose the torque due to the hanging object:

∑ τ = Iα = TR − τ f :

τ f = TR − Iα

(1)

Now find T, I and α in given or known terms and substitute into equation (1).

b

∑ Fy = T − mg = − ma :

T=m g−a

at 2 2

also ∆y = vi t +

a=

and

g

(2)

2y

(3)

t2

α=

2y a = 2: R Rt

I=

1 R M R2 + 2 2

LM MN

FIG. P10.69

(4)

FG IJ OP = 5 MR H K PQ 8 2

2

(5)

Substituting (2), (3), (4), and (5) into (1),

FG H

τ f =m g−

we find P10.70

(a)

2y t2

IJ R − 5 MR b2yg = RLMmFG g − 2 y IJ − 5 My OP K 8 Rt N H t K 4t Q 2

2

2

2

W = ∆K + ∆ U W = K f − K i + U f − Ui 1 1 1 mv 2 + Iω 2 − mgd sin θ − kd 2 2 2 2 1 2 1 ω I + mR 2 = mgd sin θ + kd 2 2 2

0=

e

ω=

j

2mgd sin θ + kd 2 I + mR 2 FIG. P10.70

(b)

ω= ω=

b

ge

ja

fa

f

a

f

2 0.500 kg 9.80 m s 2 0.200 m sin 37.0° + 50.0 N m 0.200 m

a

f

1.00 kg ⋅ m 2 + 0.500 kg 0.300 m 1.18 + 2.00 = 3.04 = 1.74 rad s 1.05

2

2

311

312 P10.71

Rotation of a Rigid Object About a Fixed Axis

(a)

m 2 g − T2 = m 2 a

b

g

e

j

T2 = m 2 g − a = 20.0 kg 9.80 m s 2 − 2.00 m s 2 = 156 N T1 − m1 g sin 37.0° = m1 a

b

f

ga

T1 = 15.0 kg 9.80 sin 37.0°+2.00 m s 2 = 118 N (b)

FG a IJ H RK bT − T gR = a156 N − 118 Nfa0.250 mf I=

bT

2

g

− T1 R = Iα = I 2

2

1

2.00 m s 2

a

P10.72

2

FIG. P10.71

= 1.17 kg ⋅ m 2

For the board just starting to move,

FG A IJ cosθ = FG 1 mA IJ α H 2K H3 K 3 F gI α = G J cos θ 2 H AK

∑ τ = Iα :

2

mg

3 g cos θ 2 3 The vertical component is a y = a t cos θ = g cos 2 θ 2 If this is greater than g, the board will pull ahead of the ball falling: The tangential acceleration of the end is

a t = Aα =

2 3

FIG. P10.72

(a)

3 2 g cos 2 θ ≥ g gives cos 2 θ ≥ so 2 3

(b)

When θ = 35.3° , the cup will land underneath the release-point of the ball if

cos θ ≥

When A = 1.00 m, and θ = 35.3°

rc = 1.00 m

a

f

θ ≤ 35.3°

and

2 = 0.816 m 3

so the cup should be 1.00 m − 0.816 m = 0.184 m from the moving end P10.73

At t = 0 , ω = 3.50 rad s = ω 0 e 0 . Thus, ω 0 = 3.50 rad s

At t = 9.30 s, ω = 2.00 rad s = ω 0 e −σ a 9.30 s f , yielding σ = 6.02 × 10 −2 s −1

(a)

e

dω d ω 0 e = dt dt At t = 3.00 s,

α=

b

− σt

j = ω a−σ fe 0

ge

−σt

j

α = 3.50 rad s −6.02 × 10 −2 s −1 e

z t

(b)

θ = ω 0 e −σt dt = 0

e

−3.00 6 .02 × 10 −2

j=

−0.176 rad s 2

ω 0 − σt ω e − 1 = 0 1 − e −σt −σ σ

At t = 2.50 s , 3.50 rad s − 6 .02 × 10 −2 ja 2.50 f 1−e e θ= = 8.12 rad = 1.29 rev −2 6.02 × 10 1 s

e

(c)

As t → ∞ , θ →

LM j N

OP Q

3.50 rad s ω0 1 − e −∞ = = 58.2 rad = 9.26 rev σ 6.02 × 10 −2 s −1

e

j

rc = A cos θ

Chapter 10

P10.74

313

Consider the total weight of each hand to act at the center of gravity (mid-point) of that hand. Then the total torque (taking CCW as positive) of these hands about the center of the clock is given by

τ = −m h g

FG L IJ sinθ H 2K h

h

− mm g

FG L IJ sinθ H2K m

m

=−

g m h L h sin θ h + m m Lm sin θ m 2

b

g

If we take t = 0 at 12 o’clock, then the angular positions of the hands at time t are

θ h = ω ht , π rad h 6

where

ωh =

and

θ m = ω mt ,

where

ω m = 2π rad h

Therefore,

τ = −4.90 m s 2 60.0 kg 2.70 m sin

f FGH π6t IJK + 100 kga4.50 mf sin 2πtOPQ L F πt I O τ = −794 N ⋅ mMsinG J + 2.78 sin 2πt P , where t is in hours. H K N 6 Q

or (a)

(i)

At 3:00, t = 3.00 h , so

(ii)

LM N

a

LM FG π IJ + 2.78 sin 6π OP = N H 2K Q

τ = −794 N ⋅ m sin

At 5:15, t = 5 h +

−794 N ⋅ m

15 h = 5.25 h , and substitution gives: 60

τ = −2 510 N ⋅ m

(b)

(iii)

At 6:00,

τ = 0 N ⋅m

(iv)

At 8:20,

τ = −1 160 N ⋅ m

(v)

At 9:45,

τ = −2 940 N ⋅ m

The total torque is zero at those times when

FG πt IJ + 2.78 sin 2πt = 0 H6K

sin

We proceed numerically, to find 0, 0.515 295 5, ..., corresponding to the times 12:00:00 2:33:25 4:58:14 7:27:36 10:02:59

12:30:55 2:56:29 5:30:52 8:03:05 10:27:29

12:58:19 3:33:22 6:00:00 8:26:38 11:01:41

1:32:31 3:56:55 6:29:08 9:03:31 11:29:05

1:57:01 4:32:24 7:01:46 9:26:35

314 *P10.75

Rotation of a Rigid Object About a Fixed Axis

(a)

As the bicycle frame moves forward at speed v, the center of each wheel moves forward at v the same speed and the wheels turn at angular speed ω = . The total kinetic energy of the R bicycle is K = K trans + K rot or K=

b

IJ b K

FG H

g

g FGH

1 1 1 1 m frame + 2m wheel v 2 + 2 I wheel ω 2 = m frame + 2m wheel v 2 + m wheel R 2 2 2 2 2

IJ FG v IJ . KH R K 2

2

This yields K= (b)

b

g

b

1 1 m frame + 3m wheel v 2 = 8.44 kg + 3 0.820 kg 2 2

g b3.35 m sg

2

= 61.2 J .

As the block moves forward with speed v, the top of each trunk moves forward at the same v speed and the center of each trunk moves forward at speed . The angular speed of each 2 v roller is ω = . As in part (a), we have one object undergoing pure translation and two 2R identical objects rolling without slipping. The total kinetic energy of the system of the stone and the trees is K = K trans + K rot or K=

FG IJ H K

1 1 v m stone v 2 + 2 m tree 2 2 2

2

+2

FG 1 I H2

treeω

2

IJ = 1 FG m K 2H

stone

+

IJ K

FG H

1 1 m tree v 2 + m tree R 2 2 2

IJ FG v IJ . K H 4R K

This gives K= P10.76

FG H

IJ K

b

1 3 1 m stone + m tree v 2 = 844 kg + 0.75 82.0 kg 2 4 2

g b0.335 m sg

2

= 50.8 J .

Energy is conserved so ∆U + ∆K rot + ∆K trans = 0

a

f LMN 12 mv

fa

mg R − r cos θ − 1 +

2

OP 1 LM 2 mr OPω Q 2 N5 Q

−0 +

2

2

=0

θ

R

Since rω = v , this gives

ω= or

ω=

a

fa

f

10 R − r 1 − cos θ g 7 r2

a

10 Rg 1 − cos θ 7r

2

f

since R >> r .

FIG. P10.76

2

2

Chapter 10

P10.77

F aI

1

∑ F = T − Mg = − Ma: ∑ τ = TR = Iα = 2 MR 2 GH R JK (a)

Combining the above two equations we find

b

T =M g−a

g

and a=

2T M

FIG. P10.77 T=

thus

Mg 3

FG IJ H K

2T 2 Mg 2 = = g 3 M M 3

(b)

a=

(c)

v 2f = vi2 + 2 a x f − xi

d

i

v 2f = 0 + 2 vf =

FG 2 gIJ ah − 0f H3 K

4 gh 3

For comparison, from conservation of energy for the system of the disk and the Earth we have U gi + K rot i + K trans i = U gf + K rot f + K trans f :

Mgh + 0 + 0 = 0 + vf =

P10.78

(a)

4 gh 3

∑ Fx = F − f = Ma: ∑ τ = fR = Iα Using I =

(b)

FG H

1 1 MR 2 2 2

1 a 2F MR 2 and α = , we find a = 2 3M R

When there is no slipping, f = µ Mg . Substituting this into the torque equation of part (a), we have

µ MgR =

1 F . MRa and µ = 2 3 Mg

IJ FG v IJ KH R K f

2

+

1 Mv f 2 2

315

316 P10.79

Rotation of a Rigid Object About a Fixed Axis

(a)

∆K rot + ∆K trans + ∆U = 0

m

Note that initially the center of mass of the sphere is a distance h + r above the bottom of the loop; and as the mass reaches the top of the loop, this distance above the reference level is 2R − r . The conservation of energy requirement gives

h

a f

a

r

f

1 1 mg h + r = mg 2 R − r + mv 2 + Iω 2 2 2 For the sphere I = gh + 2 gr = 2 gR +

R

P

FIG. P10.79

2 mr 2 and v = rω so that the expression becomes 5

7 2 v 10

(1)

Note that h = hmin when the speed of the sphere at the top of the loop satisfies the condition mv 2

∑ F = mg = aR − r f

a

or v 2 = g R − r

f

Substituting this into Equation (1) gives

a

f

a

f

a

f

hmin = 2 R − r + 0.700 R − r or hmin = 2.70 R − r = 2.70 R (b)

When the sphere is initially at h = 3 R and finally at point P, the conservation of energy equation gives

a

f

1 1 mg 3 R + r = mgR + mv 2 + mv 2 , or 2 5 10 2 2R + r g v = 7

a

f

Turning clockwise as it rolls without slipping past point P, the sphere is slowing down with counterclockwise angular acceleration caused by the torque of an upward force f of static 2 friction. We have ∑ Fy = ma y and ∑ τ = Iα becoming f − mg = − mα r and fr = mr 2 α . 5

FG IJ H K

Eliminating f by substitution yields α = mv 2

∑ Fx = −n = − R − r = −

c h( 2R + r ) mg = 10 7

R−r

5g so that 7r −20mg 7

∑ Fy =

5 − mg 7

(since R >> r )

317

Chapter 10

P10.80

Consider the free-body diagram shown. The sum of torques about the chosen pivot is

F 1 I F a IJ = FG 2 mlIJ a H K H3 K

∑ τ = Iα ⇒ FA = GH 3 ml 2 JK G (a)

CM l 2

b

f

Hx

A

CM

mg

l

g

∑ Fx = maCM ⇒ F + H x = maCM or H x = maCM − F

b

Hy

(1)

CM

A = l = 1.24 m : In this case, Equation (1) becomes 3 14.7 N 3F = = 35.0 m s 2 aCM = 2m 2 0.630 kg

a

pivot

ge

F = 14.7 N

j

Thus, H x = 0.630 kg 35.0 m s 2 − 14.7 N = +7.35 N or FIG. P10.80

H x = 7.35 i N . (b)

A=

1 = 0.620 m : For this situation, Equation (1) yields 2 aCM =

Again,

a

f

3 14.7 N 3F = = 17.5 m s 2 . 4m 4 0.630 kg

b

g

∑ Fx = maCM ⇒ H x = maCM − F , so

b

ge

j

H x = 0.630 kg 17.5 m s 2 − 14.7 N = −3.68 N or H x = −3.68 i N . (c)

If H x = 0, then

F

∑ Fx = maCM ⇒ F = maCM , or aCM = m .

Thus, Equation (1) becomes FA=

P10.81

FG 2 mlIJ FG F IJ so A= 2 l = 2 a1.24 mf = H 3 KH mK 3 3

b

g

0.827 m from the top .

2 Let the ball have mass m and radius r. Then I = mr 2 . If the ball takes four seconds to go down 5 twenty-meter alley, then v = 5 m s . The translational speed of the ball will decrease somewhat as the ball loses energy to sliding friction and some translational kinetic energy is converted to rotational kinetic energy; but its speed will always be on the order of 5.00 m s , including at the starting point. As the ball slides, the kinetic friction force exerts a torque on the ball to increase the angular speed. v When ω = , the ball has achieved pure rolling motion, and kinetic friction ceases. To determine the r elapsed time before pure rolling motion is achieved, consider:

F 2 I L b5.00 m sg r OP which gives MN t PQ

∑ τ = Iα ⇒ b µ k mg gr = GH 5 mr 2 JK M t=

2( 5.00 m s) 2.00 m s = 5µ k g µkg

Note that the mass and radius of the ball have canceled. If µ k = 0.100 for the polished alley, the sliding distance will be given by

b

∆x = vt = 5.00 m s

O L gMM a0.1002f.009.80m ms s PP = 10.2 m or ∆x ~ jQ N e 2

10 1 m .

318 P10.82

Rotation of a Rigid Object About a Fixed Axis

Conservation of energy between apex and the point where the grape leaves the surface: mg∆y =

1 1 mv 2f + Iω 2f 2 2

a

f

FG H

IJ FG v IJ KH R K f

R

2

f

n

F I GH JK

2 7 vf which gives g 1 − cos θ = 10 R

f

∆y = R—R cosθ

θ

1 1 2 mgR 1 − cos θ = mv 2f + mR 2 2 2 5

a

i

(1) mg cosθ

mg sinθ

Consider the radial forces acting on the grape: mg cos θ − n =

mv 2f R

FIG. P10.82 .

At the point where the grape leaves the surface, n → 0 . Thus, mg cos θ =

mv 2f R

or

v 2f R

= g cos θ .

Substituting this into Equation (1) gives g − g cos θ = P10.83

(a)

10 7 and θ = 54.0° . g cos θ or cos θ = 17 10

There are not any horizontal forces acting on the rod, so the center of mass will not move horizontally. Rather, the center of mass drops straight downward (distance h/2) with the rod rotating about the center of mass as it falls. From conservation of energy: K f + U gf = K i + U gi

FG IJ H K I = MgFG h IJ which reduces to JK H 2K

1 1 h 2 or + Iω 2 + 0 = 0 + Mg MvCM 2 2 2

FG H

1 1 1 2 + MvCM Mh 2 2 2 12 vCM = (b)

IJ FG v KH

2

CM h 2

3 gh 4

In this case, the motion is a pure rotation about a fixed pivot point (the lower end of the rod) with the center of mass moving in a circular path of radius h/2. From conservation of energy: K f + U gf = K i + U gi

FG IJ H K

1 2 h or Iω + 0 = 0 + Mg 2 2 1 1 Mh 2 2 3

IJ FG v IJ KH K

vCM =

3 gh 4

FG H

CM h 2

2

= Mg

FG h IJ which reduces to H 2K

Chapter 10

P10.84

(a)

319

Mr 2 where M is the initial mass of R2 mr 2 , + ∆K rot = 0 . Thus, when I = 2

The mass of the roll decreases as it unrolls. We have m = the roll. Since ∆E = 0 , we then have ∆U g + ∆K trans

bmgr − MgRg + mv2 + LM mr2 N 2

Since ω r = v , this becomes v =

e

4g R3 − r 3 3r

2

OP Q

ω2 =0 2

j

2

(b)

Using the given data, we find v = 5.31 × 10 4 m s

(c)

We have assumed that ∆E = 0 . When the roll gets to the end, we will have an inelastic collision with the surface. The energy goes into internal energy . With the assumption we made, there are problems with this question. It would take an infinite time to unwrap the tissue since dr → 0 . Also, as r approaches zero, the velocity of the center of mass approaches infinity, which is physically impossible.

P10.85

(a)

∑ Fx = F + f = MaCM ∑ τ = FR − fR = Iα

b

g

FR − MaCM − F R =

f = MaCM − F = M

(c)

v 2f = vi2 + 2 a x f − xi vf =

8 Fd 3M

Mg IaCM R

FG 4F IJ − F = H 3M K

(b)

d

F

i

aCM = 1 F 3

4F 3M

n

f

FIG. P10.85

320 P10.86

Rotation of a Rigid Object About a Fixed Axis

Call ft the frictional force exerted by each roller backward on the plank. Name as fb the rolling resistance exerted backward by the ground on each roller. Suppose the rollers are equally far from the ends of the plank.

M m

R

m

For the plank,

b

∑ Fx = ma x

F R

FIG. P10.86

g

6.00 N − 2 f t = 6.00 kg a p

The center of each roller moves forward only half as far as the plank. Each roller has acceleration and angular acceleration ap 2

ap 2

ap

=

a5.00 cmf a0.100 mf Then for each,

b g a2 1 f a5.00 cmf + f a5.00 cmf = b 2.00 kg ga5.00 cmf ∑ τ = Iα 2 F1 I f + f = G kg J a H2 K ∑ Fx = ma x

t

So

t

p

+ f t − f b = 2.00 kg

b

b

ap

2

10.0 cm

p

Add to eliminate fb :

b

g

2 f t = 1.50 kg a p (a)

b

g b

g

And 6.00 N − 1.50 kg a p = 6.00 kg a p ap = For each roller, a =

(b)

ap 2

a6.00 Nf = b7.50 kg g

0.800 m s 2

= 0.400 m s 2

b

g

Substituting back, 2 f t = 1.50 kg 0.800 m s 2

Mg 6.00 N

ft = 0.600 N 0.600 N + f b = fb = −0.200 N

1 kg 0.800 m s 2 2

e

j

ft

ft

nt

nt nt

nt

ft

ft

The negative sign means that the horizontal force of ground on each roller is 0.200 N forward

mg

mg

rather than backward as we assumed. fb

nb

fb

nb

FIG. P10.86(b)

Chapter 10

P10.87

Rolling is instantaneous rotation about the contact point P. The weight and normal force produce no torque about this point. Now F1 produces a clockwise torque about P and makes the

321

F3 F2 F4

spool roll forward. Counterclockwise torques result from F3 and F4 , making the

θc

spool roll to the left. The force F2 produces zero torque about point P and does

P

not cause the spool to roll. If F2 were strong enough, it would cause the spool to slide to the right, but not roll. P10.88

FIG. P10.87 F2

The force applied at the critical angle exerts zero torque about the spool’s contact point with the ground and so will not make the spool roll. From the right triangle shown in the sketch, observe that θ c = 90°−φ = 90°− 90°−γ = γ .

b

g

γ R

r Thus, cos θ c = cos γ = . R

r

φ

θc

P FIG. P10.88 P10.89

(a)

Consider motion starting from rest over distance x along the incline:

bK

trans

g

b

+ K rot + U i + ∆E = K trans + K rot + U

0 + 0 + Mgx sin θ + 0 =

a

FG H

1 1 Mv 2 + 2 mR 2 2 2

f

g

f

IJ FG v IJ KH RK

2

+0

2 Mgx sin θ = M + 2m v 2 Since acceleration is constant, v 2 = vi2 + 2 ax = 0 + 2 ax , so

a

f

2 Mgx sin θ = M + 2m 2 ax a=

Mg sin θ M + 2m

a

f

y x

∆x

θ

FIG. P10.88 continued on next page

F1

322

Rotation of a Rigid Object About a Fixed Axis

(c)

Suppose the ball is fired from a cart at rest. It moves with acceleration g sin θ = a x down the incline and a y = − g cos θ perpendicular to the incline. For its range along the ramp, we have y − yi = v yi t − t=

2 v yi g cos θ

x − xi = v xi t + d=0+

d= (b)

1 g cos θt 2 = 0 − 0 2

1 axt 2 2

F GH

I JK

2 4v yi 1 g sin θ 2 2 g cos 2 θ

2 2 v yi sin θ

g cos 2 θ

In the same time the cart moves x − xi = v xi t + 1 2

dc = 0 + dc =

a

1 axt 2 2

F g sinθM I FG 4v IJ GH aM + 2mf JK H g cos θ K 2 yi

2

2

2 2 v yi sin θM

f

g M + 2m cos 2 θ

So the ball overshoots the cart by ∆x = d − d c = ∆x = ∆x =

2 2 v yi

2 sin θ 2 v yi

g cos 2 θ

sin θM +

2 4v yi 2

−

2 sin θM 2 v yi

sin θm −

a

a

g cos 2 θ M + 2m 2 2 v yi

g cos θ M + 2m 2 4mv yi

sin θ

aM + 2mfg cos

2

θ

f

f

sin θM

Chapter 10

P10.90

∑ Fx = ma x reads − f + T = ma . If we take torques around the center of mass, we can use ∑ τ = Iα , which reads + fR 2 − TR1 = Iα . For rolling without slipping, α =

a . By substitution, R2

b

Ia I = T− f R2 R 2 m

fR 22 m − TR1 R 2 m = IT − If

e

j b

f I + mR 22 = T I + mR1 R 2 f=

mg T

f

fR 2 − TR1 =

F I + mR R I T GH I + mR JK 1

2 2

323

g

n FIG. P10.90

g

2

Since the answer is positive, the friction force is confirmed to be to the left.

ANSWERS TO EVEN PROBLEMS P10.28

1 ML2 2

P10.30

168 N ⋅ m clockwise

−226 rad s 2

P10.32

882 N ⋅ m

P10.8

13.7 rad s 2

P10.34

(a) 1.03 s; (b) 10.3 rev

P10.10

(a) 2.88 s; (b) 12.8 s

P10.36

(a) 21.6 kg ⋅ m 2 ; (b) 3.60 N ⋅ m ; (c) 52.4 rev

P10.12

(a) 0.180 rad s; (b) 8.10 m s 2 toward the center of the track

P10.38

0.312

P10.40

1.04 × 10 −3 J

P10.14

(a) 0.605 m s ; (b) 17.3 rad s ; (c) 5.82 m s ; (d) The crank length is unnecessary

P10.42

149 rad s

P10.16

(a) 54.3 rev; (b) 12.1 rev s

P10.44

(a) 6.90 J; (b) 8.73 rad s ; (c) 2.44 m s ; (d) 1.043 2 times larger

P10.18

0.572

P10.46

2.36 m s

P10.20

(a) 92.0 kg ⋅ m 2 ; 184 J ; (b) 6.00 m s ; 4.00 m s ; 8.00 m s ; 184 J

P10.48

276 J

P10.50

(a) 74.3 W; (b) 401 W

P10.22

see the solution P10.52

7 Mv 2 10

P10.54

The disk;

P10.2

(a) 822 rad s 2 ; (b) 4.21 × 10 3 rad

P10.4

(a) 1.20 × 10 2 rad s ; (b) 25.0 s

P10.6

2

P10.24

1.28 kg ⋅ m

P10.26

~ 10 0 kg ⋅ m 2

4 gh versus 3

gh

324 P10.56

Rotation of a Rigid Object About a Fixed Axis

a

f

(a) 2.38 m s ; (b) 4.31 m s; (c) It will not reach the top of the loop.

P10.76

P10.58

(a) 0.992 W; (b) 827 W

P10.78

see the solution

P10.60

see the solution

P10.80

P10.62

(a) 12.5 rad s ; (b) 128 rad

(a) 35.0 m s 2 ; 7.35 i N ; (b) 17.5 m s 2 ; −3.68 i N ;

P10.64 P10.66

b

g h2 − h1 2πR

g

(a) 2.57 × 10 29 J ; (b) −1.63 × 10 17 J day

P10.68

139 m s

P10.70

(a)

P10.72

see the solution

P10.74

(a) −794 N ⋅ m ; −2 510 N ⋅ m; 0; −1 160 N ⋅ m; −2 940 N ⋅ m; (b) see the solution

2mgd sin θ + kd 2 I + mR 2

; (b) 1.74 rad s

7r 2

(c) At 0.827 m from the top. P10.82

2

10 Rg 1 − cos θ

54.0°

e

4g R3 − r 3

j ; (b) 5.31 × 10

4

m s;

P10.84

(a)

P10.86

(a) 0.800 m s 2 ; 0.400 m s 2 ; (b) 0.600 N between each cylinder and the plank; 0.200 N forward on each cylinder by the ground

P10.88

see the solution

P10.90

see the solution; to the left

2

3r (c) It becomes internal energy.

11 Angular Momentum CHAPTER OUTLINE 11.1 11.2 11.3 11.4 11.5 11.6

The Vector Product and Torque Angular Momentum Angular Momentum of a Rotating Rigid Object Conservation of Angular Momentum The Motion of Gyroscopes and Tops Angular Momentum as a Fundamental Quantity

ANSWERS TO QUESTIONS Q11.1

No to both questions. An axis of rotation must be defined to calculate the torque acting on an object. The moment arm of each force is measured from the axis.

Q11.2

A ⋅ B × C is a scalar quantity, since B × C is a vector. Since A ⋅ B is a scalar, and the cross product between a scalar and a vector is not defined, A ⋅ B × C is undefined.

a

f

a

f

a f

Q11.3

e j

(a)

Down–cross–left is away from you: − j × − i = − k

(b)

Left–cross–down is toward you: − i × − j = k

e j

FIG. Q11.3 Q11.4

The torque about the point of application of the force is zero.

Q11.5

You cannot conclude anything about the magnitude of the angular momentum vector without first defining your axis of rotation. Its direction will be perpendicular to its velocity, but you cannot tell its direction in three-dimensional space until an axis is specified.

Q11.6

Yes. If the particles are moving in a straight line, then the angular momentum of the particles about any point on the path is zero.

Q11.7

Its angular momentum about that axis is constant in time. You cannot conclude anything about the magnitude of the angular momentum.

Q11.8

No. The angular momentum about any axis that does not lie along the instantaneous line of motion of the ball is nonzero.

325

326

Angular Momentum

Q11.9

There must be two rotors to balance the torques on the body of the helicopter. If it had only one rotor, the engine would cause the body of the helicopter to swing around rapidly with angular momentum opposite to the rotor.

Q11.10

The angular momentum of the particle about the center of rotation is constant. The angular momentum about any point that does not lie along the axis through the center of rotation and perpendicular to the plane of motion of the particle is not constant in time.

Q11.11

The long pole has a large moment of inertia about an axis along the rope. An unbalanced torque will then produce only a small angular acceleration of the performer-pole system, to extend the time available for getting back in balance. To keep the center of mass above the rope, the performer can shift the pole left or right, instead of having to bend his body around. The pole sags down at the ends to lower the system center of gravity.

Q11.12

The diver leaves the platform with some angular momentum about a horizontal axis through her center of mass. When she draws up her legs, her moment of inertia decreases and her angular speed increases for conservation of angular momentum. Straightening out again slows her rotation.

Q11.13

Suppose we look at the motorcycle moving to the right. Its drive wheel is turning clockwise. The wheel speeds up when it leaves the ground. No outside torque about its center of mass acts on the airborne cycle, so its angular momentum is conserved. As the drive wheel’s clockwise angular momentum increases, the frame of the cycle acquires counterclockwise angular momentum. The cycle’s front end moves up and its back end moves down.

Q11.14

The angular speed must increase. Since gravity does not exert a torque on the system, its angular momentum remains constant as the gas contracts.

Q11.15

Mass moves away from axis of rotation, so moment of inertia increases, angular speed decreases, and period increases.

Q11.16

The turntable will rotate counterclockwise. Since the angular momentum of the mouse-turntable system is initially zero, as both are at rest, the turntable must rotate in the direction opposite to the motion of the mouse, for the angular momentum of the system to remain zero.

Q11.17

Since the cat cannot apply an external torque to itself while falling, its angular momentum cannot change. Twisting in this manner changes the orientation of the cat to feet-down without changing the total angular momentum of the cat. Unfortunately, humans aren’t flexible enough to accomplish this feat.

Q11.18

The angular speed of the ball must increase. Since the angular momentum of the ball is constant, as the radius decreases, the angular speed must increase.

Q11.19

Rotating the book about the axis that runs across the middle pages perpendicular to the binding—most likely where you put the rubber band—is the one that has the intermediate moment of inertia and gives unstable rotation.

Q11.20

The suitcase might contain a spinning gyroscope. If the gyroscope is spinning about an axis that is oriented horizontally passing through the bellhop, the force he applies to turn the corner results in a torque that could make the suitcase swing away. If the bellhop turns quickly enough, anything at all could be in the suitcase and need not be rotating. Since the suitcase is massive, it will want to follow an inertial path. This could be perceived as the suitcase swinging away by the bellhop.

Chapter 11

SOLUTIONS TO PROBLEMS Section 11.1

P11.1

The Vector Product and Torque i

j

M×N= 6

2

k

−1 = −7.00 i + 16.0 j − 10.0k

2 −1 −3 P11.2

a

fa

f

area = A × B = AB sin θ = 42.0 cm 23.0 cm sin 65.0°−15.0° = 740 cm 2

(b)

A + B = 42.0 cm cos 15.0°+ 23.0 cm cos 65.0° i + 42.0 cm sin 15.0°+ 23.0 cm sin 65.0° j A + B = 50.3 cm i + 31.7 cm j

a a

f a f f a f length = A + B = a50.3 cmf + a31.7 cmf 2

P11.3

f a

(a)

(a)

(b)

i

j k

2

3

a

2

f

a

f

= 59.5 cm

A × B = −3 4 0 = −17.0k 0

A × B = A B sin θ 17 = 5 13 sin θ

FG 17 IJ = 70.6° H 5 13 K A ⋅ B = −3.00a6.00f + 7.00a −10.0f + a −4.00fa9.00 f = −124 AB = a−3.00f + a7.00f + a −4.00f ⋅ a6.00f + a −10.0 f + a9.00 f F A ⋅ B IJ = cos a−0.979f = 168° (a) cos G H AB K θ = arcsin

P11.4

2

2

2

−1

j

6.00 −10.0

= 127

k

9.00

a23.0f + a3.00f + a−12.0f = 26.1 FG A × B IJ = sin a0.206f = 11.9° or 168° H AB K

A×B =

(c)

2

7.00 −4.00 = 23.0 i + 3.00 j − 12.0k

A × B = −3.00

sin −1

2

−1

i (b)

2

2

2

2

−1

Only the first method gives the angle between the vectors unambiguously.

327

328 *P11.5

Angular Momentum

a

f a

f

τ = r × F = 0.450 m 0.785 N sin 90°−14° up × east = 0.343 N ⋅ m north

FIG. P11.5 P11.6

The cross-product vector must be perpendicular to both of the factors, so its dot product with either factor must be zero:

e

je

j

Does 2 i − 3 j + 4k ⋅ 4i + 3 j − k = 0 ? 8 − 9 − 4 = −5 ≠ 0

No . The cross product could not work out that way. P11.7

A × B = A ⋅ B ⇒ AB sin θ = AB cos θ ⇒ tan θ = 1 or

θ = 45.0° i

P11.8

j k

a f a f a f a−7.00 N ⋅ mfk

(a)

τ = r × F = 1 3 0 = i 0 − 0 − j 0 − 0 + k 2 − 9 = 3 2 0

(b)

The particle’s position vector relative to the new axis is 1 i + 3 j − 6 j = 1 i − 3 j . i j k

a

f

τ = 1 −3 0 = 11.0 N ⋅ m k 3 2 0

P11.9

B

F3 = F1 + F2 The torque produced by F3 depends on the perpendicular distance OD, therefore translating the point of application of F3 to any other point along

F3 D O

BC will not change the net torque . A

C

F1

FIG. P11.9

F2

Chapter 11

*P11.10

i × i = 1 ⋅ 1 ⋅ sin 0° = 0

j i

j × j and k × k are zero similarly since the vectors being multiplied are parallel.

k

329

i × j = k

j × i = − k

j × k = i

k × j = − i

k × i = j

i × k = − j

i × j = 1 ⋅ 1 ⋅ sin 90° = 1 FIG. P11.10

Section 11.2 P11.11

Angular Momentum

L = ∑ mi vi ri

b

y

gb

f b

ga

gb

ga

f

= 4.00 kg 5.00 m s 0.500 m + 3.00 kg 5.00 m s 0.500 m

3.00 kg

2

L = 17.5 kg ⋅ m s , and

e

j

L = 17.5 kg ⋅ m 2 s k

x 1.00 m 4.00 kg

FIG. P11.11 P11.12

L=r×p

ge e j b j L = e −8.10k − 13.9k j kg ⋅ m s = e −22.0 kg ⋅ m sjk L = 1.50 i + 2.20 j m × 1.50 kg 4.20 i − 3.60 j m s 2

P11.13

e

j

r = 6.00 i + 5.00tj m so

v=

e

dr = 5.00 j m s dt

j

p = mv = 2.00 kg 5.00 j m s = 10.0 j kg ⋅ m s i

and

2

L = r × p = 6.00 0

j 5.00t 10.0

k 0 = 0

e60.0 kg ⋅ m sjk 2

330 P11.14

Angular Momentum

mv 2 r

∑ Fx = ma x

T sin θ =

∑ Fy = ma y

T cos θ = mg

So

2

sin θ v = cos θ rg

v = rg

θ

l

sin θ cos θ

L = rmv sin 90.0° L = rm rg

m

sin θ cos θ

FIG. P11.14

sin θ cos θ r = A sin θ , so L = m 2 gr 3

L=

P11.15

m 2 gA 3

sin 4 θ cos θ

The angular displacement of the particle around the circle is θ = ωt =

vt . R

y v

The vector from the center of the circle to the mass is then R cos θ i + R sin θ j .

R

m

θ Q

P

The vector from point P to the mass is r = R i + R cos θ i + R sin θ j

LMFG NH

r = R 1 + cos

FG vt IJ IJ i + sinFG vt IJ jOP H R KK H R K Q

FIG. P11.15

The velocity is v= So

FG IJ H K

FG IJ H K

dr vt  vt  = − v sin i + v cos j dt R R

L = r × mv

a

f L F vt I O mvRk McosG J + 1P N H RK Q

L = mvR 1 + cos ωt i + sin ωtj × − sin ωt i + cos ωtj L= P11.16

(a)

The net torque on the counterweight-cord-spool system is:

b

ge

j

τ = r × F = 8.00 × 10 −2 m 4.00 kg 9.80 m s 2 = 3.14 N ⋅ m . (b)

L = r × mv + Iω

(c)

τ=

b

L = Rmv +

g

dL = 0.400 kg ⋅ m a dt

a=

FG IJ FG H K H

IJ K

1 v M MR 2 =R m+ v= 2 R 2

3.14 N ⋅ m = 7.85 m s 2 0.400 kg ⋅ m

b0.400 kg ⋅ mgv

x

Chapter 11

P11.17

(a) (b)

vi = vxi i

zero At the highest point of the trajectory, vi

v 2 sin 2θ 1 x= R= i and 2 2g y = hmax =

b

vi sin θ

Lv =M MN

=

2 i

g

b

v2 R

2

FIG. P11.17

b

sin 2θ  vi sin θ i+ 2g 2g

− m vi sin θ

θ

O

2g

L 1 = r1 × mv 1

(c)

331

g jOP × mv PQ 2



xi i

g v cos θ k 2

i

2g

v 2 sin 2θ L 2 = R i × mv 2 , where R = i g

e

j

= mR i × vi cos θ i − vi sin θ j

− mvi3 sin 2θ sin θ  = − mRvi sin θ k = k g (d) P11.18

The downward force of gravity exerts a torque in the –z direction.

Whether we think of the Earth’s surface as curved or flat, we interpret the problem to mean that the plane’s line of flight extended is precisely tangent to the mountain at its peak, and nearly parallel to the wheat field. Let the positive x direction be eastward, positive y be northward, and positive z be vertically upward. (a)

a

f e

j p = mv = 12 000 kg e −175 i m sj = −2.10 × 10 i kg ⋅ m s L = r × p = e 4.30 × 10 k mj × e −2.10 × 10 i kg ⋅ m sj = e −9.03 × 10 r = 4.30 km k = 4.30 × 10 3 m k

6

3

(b)

6

a

9

j

kg ⋅ m 2 s j

f

No . L = r p sin θ = mv r sin θ , and r sin θ is the altitude of the plane. Therefore, L = constant as the plane moves in level flight with constant velocity.

(c)

Zero . The position vector from Pike’s Peak to the plane is anti-parallel to the velocity of the plane. That is, it is directed along the same line and opposite in direction. Thus, L = mvr sin180° = 0 .

332 P11.19

Angular Momentum

The vector from P to the falling ball is 1 r = ri + v i t + at 2 2

FG H

m

IJ K

1 2  r = A cos θ i + A sin θ j + 0 − gt j 2

e

j

l

θ

The velocity of the ball is P

v = v i + at = 0 − gtj

L = r × mv

So

LMe N

IJ OP e j KQ

FG H

1 2  L = m A cos θ i + A sin θ j + 0 − gt j × − gtj 2

j

FIG. P11.19

L = − mAgt cos θ k P11.20

In the vertical section of the hose, the water has zero angular momentum about our origin (point O between the fireman’s feet). As it leaves the nozzle, a parcel of mass m has angular momentum:

a

fb

L = r × mv = mrv sin 90.0° = m 1.30 m 12.5 m s

e

2

j

vf

g

L = 16.3 m s m

1.30 m

The torque on the hose is the rate of change in angular momentum. Thus,

τ=

jb

O

g

dL dm = 16.3 m 2 s = 16.3 m 2 s 6.31 kg s = 103 N ⋅ m dt dt

e

j

e

vi

FIG. P11.20

Section 11.3 *P11.21 P11.22

K=

Angular Momentum of a Rotating Rigid Object 1 2 1 I 2ω 2 L2 Iω = = 2 2 I 2I

The moment of inertia of the sphere about an axis through its center is I=

b

ga

f

2 2 MR 2 = 15.0 kg 0.500 m 5 5

2

= 1.50 kg ⋅ m 2

Therefore, the magnitude of the angular momentum is

e

jb

g

L = Iω = 1.50 kg ⋅ m 2 3.00 rad s = 4.50 kg ⋅ m 2 s Since the sphere rotates counterclockwise about the vertical axis, the angular momentum vector is directed upward in the +z direction.

e

j

Thus, L = 4.50 kg ⋅ m 2 s k .

Chapter 11

P11.23

(a)

(b)

FG 1 MR IJω = 1 b3.00 kg ga0.200 mf b6.00 rad sg = H2 K 2 L1 F RI O L = Iω = M MR + M G J Pω H 2 K PQ MN 2 3 = b3.00 kg ga0.200 mf b6.00 rad sg = 0.540 kg ⋅ m s 4 2

2

L = Iω =

2

2

2

P11.24

2

The total angular momentum about the center point is given by L = I hω h + I mω m Ih =

and

I m3 =

In addition,

ωh =

f

2

a

= 146 kg ⋅ m 2

f

m m L2m 100 kg 4.50 m = 3 3

2

= 675 kg ⋅ m 2

F I GH JK 2π rad F 1 h I ω = = 1.75 × 10 rad s 1 h GH 3 600 s JK L = 146 kg ⋅ m e1.45 × 10 rad sj + 675 kg ⋅ m e1.75 × 10 2π rad 1h = 1.45 × 10 −4 rad s 12 h 3 600 s −3

m

−4

2

Thus,

2

−3

rad s

j

L = 1.20 kg ⋅ m 2 s

or (a)

a

m h L2h 60.0 kg 2.70 m = 3 3

with

while

P11.25

0.360 kg ⋅ m 2 s

I=

a

1 m1 L2 + m 2 0.500 12

f

a f

2

=

a

fa f

1 0.100 1.00 12

2

a

+ 0.400 0.500

f

2

= 0.108 3 kg ⋅ m 2

L = Iω = 0.108 3 4.00 = 0.433 kg ⋅ m 2 s (b)

I=

a

fa f

1 1 m1 L2 + m 2 R 2 = 0.100 1.00 3 3

2

a f

+ 0.400 1.00

2

= 0.433

a f

L = Iω = 0.433 4.00 = 1.73 kg ⋅ m 2 s *P11.26

∑ Fx = ma x :

+ fs = ma x

We must use the center of mass as the axis in

af a

f a

n

f

∑ τ = Iα :

Fg 0 − n 77.5 cm + fs 88 cm = 0

∑ Fy = ma y :

+n − Fg = 0

Fg

88 cm

fs

155 cm 2 FIG. P11.26

We combine the equations by substitution:

a f a f e9.80 m s j77.5 cm = 8.63 m s =

− mg 77.5 cm + ma x 88 cm = 0 2

ax

88 cm

2

333

334 *P11.27

Angular Momentum

v2 = ω 2r r

We require a c = g =

e9.80 m s j = 0.313 rad s 2

g = r

ω=

100 m

a

2

f

I = Mr = 5 × 10 4 kg 100 m

2

= 5 × 10 8 kg ⋅ m 2

(a)

L = Iω = 5 × 10 8 kg ⋅ m 2 0.313 s = 1.57 × 10 8 kg ⋅ m 2 s

(c)

∑ τ = Iα =

d

I ω f −ωi

i

∆t

∑ τ∆t = Iω f − Iω i = L f − Li This is the angular impulse-angular momentum theorem. (b)

Section 11.4 P11.28

(a)

∆t =

1.57 × 10 8 kg ⋅ m 2 s = 6.26 × 10 3 s = 1.74 h 2 125 N 100 m

a

fa

f

From conservation of angular momentum for the system of two cylinders:

1

g

+ I 2 ω f = I 1ω i

Kf =

so

P11.29

∑τ

=

Conservation of Angular Momentum

bI (b)

Lf −0

Iiω i = I f ω f :

b

g

1 I 1 + I 2 ω 2f 2 Kf Ki

=

1 2

bI

1

+ I2

2 1 2 I 1ω i

or

ωf =

and

Ki =

g FG I ω IJ HI +I K 1

1

2

i

2

=

ω 2 = 7.14 rev min

1 I 1ω i2 2

I1 which is less than 1 . I1 + I 2

e250 kg ⋅ m jb10.0 rev ming = 250 kg ⋅ m 2

I1 ωi I1 + I 2

2

a

f

+ 25.0 kg 2.00 m

2

ω2

Chapter 11

P11.30

(a)

335

The total angular momentum of the system of the student, the stool, and the weights about the axis of rotation is given by

e j

I total = I weights + I student = 2 mr 2 + 3.00 kg ⋅ m 2 Before:

r = 1.00 m .

Thus,

I i = 2 3.00 kg 1.00 m

After:

r = 0.300 m

Thus,

I f = 2 3.00 kg 0.300 m

b

f

ga

b

2

+ 3.00 kg ⋅ m 2 = 9.00 kg ⋅ m 2

f

ga

2

+ 3.00 kg ⋅ m 2 = 3.54 kg ⋅ m 2

We now use conservation of angular momentum.

I f ω f = I iω i

ωf =

or

(b)

P11.31

F I I ω = FG 9.00 IJ b0.750 rad sg = GH I JK H 3.54 K i

f

i

jb

g

jb

g

Ki =

1 1 I iω i2 = 9.00 kg ⋅ m 2 0.750 rad s 2 2

Kf =

1 1 I f ω 2f = 3.54 kg ⋅ m 2 1.91 rad s 2 2

e

e

1.91 rad s

2

= 2.53 J

2

= 6.44 J

Let M = mass of rod and m = mass of each bead. From Iiω i = I f ω f , we have

(a)

LM 1 MA N 12

2

OP Q

+ 2mr12 ω i =

LM 1 MA N 12

2

OP Q

+ 2mr22 ω f

When A = 0.500 m , r1 = 0.100 m , r2 = 0.250 m , and with other values as stated in the problem, we find

ω f = 9.20 rad s . (b)

Since there is no external torque on the rod,

L = constant and ω is unchanged . *P11.32

Let M represent the mass of all the ribs together and L the length of each. The original moment of 1 inertia is ML2 . The final effective length of each rib is L sin 22.5° and the final moment of inertia is 3 1 2 M L sin 22.5° angular momentum of the umbrella is conserved: 3

a

f

1 1 ML2ω i = ML2 sin 2 22.5° ω f 3 3 1.25 rad s ωf = = 8.54 rad s sin 2 22.5°

336 P11.33

Angular Momentum

(a)

The table turns opposite to the way the woman walks, so its angular momentum cancels that of the woman. From conservation of angular momentum for the system of the woman and the turntable, we have L f = Li = 0 so,

L f = I womanω woman + I tableω table = 0

and

ω table = − ω table

woman

woman rv woman

I table

500 kg ⋅ m 2

a

f

1 1 2 2 m woman v woman + Iω table 2 2

work done = ∆K = K f − 0 = W=

P11.34

I woman m r2 ω woman = − woman I table I table

ω table = 0.360 rad s counterclockwise

or (b)

F I FG v IJ = − m I JK GH JK H r K 60.0 kg a 2.00 mfb1.50 m sg =− = −0.360 rad s F GH

b

gb

1 60 kg 1.50 m s 2

g

2

+

jb

1 500 kg ⋅ m 2 0.360 rad s 2

e

g

2

= 99.9 J

When they touch, the center of mass is distant from the center of the larger puck by yCM =

a

f

0 + 80.0 g 4.00 cm + 6.00 cm = 4.00 cm 120 g + 80.0 g

e

jb

je

g

(a)

L = r1 m1 v1 + r2 m 2 v 2 = 0 + 6.00 × 10 −2 m 80.0 × 10 −3 kg 1.50 m s = 7.20 × 10 −3 kg ⋅ m 2 s

(b)

The moment of inertia about the CM is

FG 1 m r + m d IJ + FG 1 m r + m d IJ H2 K H2 K 1 I = b0.120 kg ge6.00 × 10 mj + b0.120 kg ge 4.00 × 10 j 2 1 + e80.0 × 10 kg je 4.00 × 10 mj + e80.0 × 10 kg je6.00 × 10 2 I=

2 1 1

2 1 1

2 2 2

−2

−3

2 2 2

2

−2

−2 2

2

−3

I = 7.60 × 10 −4 kg ⋅ m 2 Angular momentum of the two-puck system is conserved: L = Iω

ω=

L 7.20 × 10 −3 kg ⋅ m 2 s = = 9. 47 rad s I 7.60 × 10 −4 kg ⋅ m 2

−2

m

j

2

Chapter 11

P11.35

(a)

Li = mvA

∑ τ ext = 0 , so L f

337

= Li = mvA

a f F m IJ v =G H m + MK

Lf = m + M vfA vf

(b)

f

FG H

IJ K

Fraction of K lost =

1 2

m2 M +m 2

mv 2 − 12 1 2

v2

mv

=

FIG. P11.35 M M+m

For one of the crew, mv 2 = mω i2 r r

∑ Fr = mar :

n=

We require

n = mg , so ω i =

Now,

Iiω i = I f ω f

g r

a

f

5.00 × 10 8 kg ⋅ m 2 + 150 × 65.0 kg × 100 m

F 5.98 × 10 I GH 5.32 × 10 JK 8 8

(a)

(b)

2

a

f

g = 5.00 × 10 8 kg ⋅ m 2 + 50 × 65.0 kg 100 m r

2

ωf

g g = ω f = 1.12 r r a r = ω 2f r = 1.26 g = 12.3 m s 2

Now, P11.37

v

1 mv 2 2 1 K f = M + m v 2f 2 m vf = v ⇒ velocity of the bullet M+m and block Ki =

a

P11.36

l M

Consider the system to consist of the wad of clay and the cylinder. No external forces acting on this system have a torque about the center of the cylinder. Thus, angular momentum of the system is conserved about the axis of the cylinder.

L f = Li :

Iω = mvi d

or

LM 1 MR N2

Thus,

ω=

2

OP Q

+ mR 2 ω = mvi d 2mvi d

a M + 2 m fR

2

FIG. P11.37

.

No . Some mechanical energy changes to internal energy in this perfectly inelastic collision.

338 *P11.38

Angular Momentum

(a)

Let ω be the angular speed of the signboard when it is vertical. 1 2 Iω = Mgh 2 1 1 1 ML2 ω 2 = Mg L 1 − cos θ ∴ 2 3 2

FG H

a

IJ K

a

3 g 1 − cos θ L

∴ω =

f

f

Mg

m

ja

e

θ

v

f

3 9.80 m s 2 1 − cos 25.0°

=

FIG. P11.38

0.50 m

= 2.35 rad s (b)

I f ω f = Iiω i − mvL represents angular momentum conservation ∴

FG 1 ML H3

2

∴ω f = =

(c)

1 3

1 3

IJ K

+ mL2 ω f =

1 ML2ω i − mvL 3

MLω i − mv

c

1 3

h

M+m L

b2.40 kg ga0.5 mfb2.347 rad sg − b0.4 kg gb1.6 m sg = b2.40 kgg + 0.4 kg a0.5 mf 1 3

0. 498 rad s

Let hCM = distance of center of mass from the axis of rotation. hCM =

b2.40 kg ga0.25 mf + b0.4 kgga0.50 mf = 0.285 7 m . 2.40 kg + 0.4 kg

Apply conservation of mechanical energy:

aM + mfgh a1 − cos θ f = 12 FGH 13 ML + mL IJKω L c M + mhL ω OP ∴θ = cos M1 − MN 2aM + mfgh PQ R| b2.40 kg g + 0.4 kg a0.50 mf b0.498 rad sg = cos S1 − |T 2b2.40 kg + 0.4 kg ge9.80 m s jb0.285 7 mg 2

CM

1 3

−1

2

2

2

2

CM

−1

1 3

2

2

= 5.58°

1 L 2

2

U| V| W

Chapter 11

P11.39

339

The meteor will slow the rotation of the Earth by the largest amount if its line of motion passes farthest from the Earth’s axis. The meteor should be headed west and strike a point on the equator tangentially. Let the z axis coincide with the axis of the Earth with +z pointing northward. Then, conserving angular momentum about this axis, ∑ L f = ∑ Li ⇒ Iω f = Iω i + mv × r or Thus,

2 2 MR 2ω f k = MR 2ω i k − mvRk 5 5 mvR 5mv or = ωi −ω f = 2 2 2 MR 5 MR

ωi −ω f =

e

je j = 5.91 × 10 kg je6.37 × 10 mj

5 3.00 × 10 13 kg 30.0 × 10 3 m s

e

2 5.98 × 10

24

−14

6

rad s

∆ω max ~ 10 −13 rad s

Section 11.5 *P11.40

The Motion of Gyroscopes and Tops

Angular momentum of the system of the spacecraft and the gyroscope is conserved. The gyroscope and spacecraft turn in opposite directions. 0 = I 1ω 1 + I 2 ω 2 :

θ t

− I 1ω 1 = I 2

b

g

−20 kg ⋅ m 2 −100 rad s = 5 × 10 5 kg ⋅ m 2 t=

*P11.41

I=

2.62 × 10 5 s = 131 s 2 000

2 2 MR 2 = 5.98 × 10 24 kg 6.37 × 10 6 m 5 5

e

je

j

2

= 9.71 × 10 37 kg ⋅ m 2

F 2π rad I = 7.06 × 10 kg ⋅ m s GH 86 400 s JK F 2π rad I FG 1 yr IJ F 1 d I = kg ⋅ m sjG H 2.58 × 10 yr JK H 365.25 d K GH 86 400 s JK

L = Iω = 9.71 × 10 37 kg ⋅ m 2

e

τ = Lω p = 7.06 × 10 33

Section 11.6 P11.42

FG 30° IJ FG π rad IJ H t K H 180° K

33

2

2

2

4

5.45 × 10 22 N ⋅ m

Angular Momentum as a Fundamental Quantity 6.626 1 × 10 −34 J ⋅ s

(a)

L=

h h = mvr so v = 2πmr 2π

(b)

K=

1 1 mv 2 = 9.11 × 10 −31 kg 2.19 × 10 6 m s 2 2

(c)

ω=

= L 1.055 × 10 −34 J ⋅ s = = I mr 2 9.11 × 10 −31 kg 0.529 × 10 −10 m

e

e

v=

je

je

e

je

j

2π 9.11 × 10 -31 kg 0.529 × 10 −10 m

j

2

j

= 2.18 × 10 −18 J

2

= 4.13 × 10 16 rad s

= 2.19 × 10 6 m s

340

Angular Momentum

Additional Problems *P11.43

First, we define the following symbols: I P = moment of inertia due to mass of people on the equator I E = moment of inertia of the Earth alone (without people) ω = angular velocity of the Earth (due to rotation on its axis) 2π T= = rotational period of the Earth (length of the day) ω R = radius of the Earth The initial angular momentum of the system (before people start running) is

b

g

Li = I P ω i + I E ω i = I P + I E ω i When the Earth has angular speed ω, the tangential speed of a point on the equator is v t = Rω . Thus, when the people run eastward along the equator at speed v relative to the surface of the Earth, vp v =ω + . their tangential speed is v p = v t + v = Rω + v and their angular speed is ω P = R R The angular momentum of the system after the people begin to run is

FG H

L f = I Pω p + I Eω = I P ω +

IJ K

b

g

I v v + I Eω = I P + I E ω + P . R R

d

i

Since no external torques have acted on the system, angular momentum is conserved L f = Li ,

b

g

b

g

IPv = I P + I E ω i . Thus, the final angular velocity of the Earth is R IPv IPv ω = ωi − = ω i 1 − x = , where x ≡ . I P + I E Rω i IP + IE R

giving I P + I E ω +

b

a f

g

b

The new length of the day is T = day is ∆T = T − Ti ≈ Ti x = Ti

LM MN bI

P

2π

ω

=

g

T 2π = i ≈ Ti 1 + x , so the increase in the length of the ωi 1− x 1− x

OP g PQ

a f

a f

IPv Ti2 I P v 2π , this may be written as ∆T ≈ . Since ω i = . Ti + I E Rω i 2π I P + I E R

b

To obtain a numeric answer, we compute

jb

e

I P = m p R 2 = 5.5 × 10 9 70 kg

g e6.37 × 10 mj 6

2

= 1.56 × 10 25 kg ⋅ m 2

2

= 9.71 × 10 37 kg ⋅ m 2 .

and IE =

2 2 m E R 2 = 5.98 × 10 24 kg 6.37 × 10 6 m 5 5

e

je

j

e8.64 × 10 sj e1.56 × 10 kg ⋅ m jb2.5 m sg = Thus, ∆T ≈ 2π e1.56 × 10 + 9.71 × 10 j kg ⋅ m e6.37 × 10 mj 4

25

2

25

37

2

2

6

7.50 × 10 −11 s .

g

Chapter 11

*P11.44

(a)

341

bK + U g = bK + U g s A

s B

1 0 + mgy A = mv B2 + 0 2

e

j

v B = 2 gy A = 2 9.8 m s 2 6.30 m = 11.1 m s (b)

L = mvr = 76 kg 11.1 m s 6.3 m = 5.32 × 10 3 kg ⋅ m 2 s toward you along the axis of the channel.

(c)

The wheels on his skateboard prevent any tangential force from acting on him. Then no torque about the axis of the channel acts on him and his angular momentum is constant. His legs convert chemical into mechanical energy. They do work to increase his kinetic energy. The normal force acts forward on his body on its rising trajectory, to increase his linear momentum.

(d)

L = mvr

(e)

v=

5.32 × 10 3 kg ⋅ m 2 s = 12.0 m s 76 kg 5.85 m

eK + U j + W = e K + U j 1 1 76 kg b11.1 m sg + 0 + W = 76 kg b12.0 m sg 2 2 g

g

B

C

2

2

+ 76 kg 9.8 m s 2 0.45 m

W = 5.44 kJ − 4.69 kJ + 335 J = 1.08 kJ (f)

eK + U j = e K + U j 1 1 76 kg b12.0 m sg + 0 = 76 kgv 2 2 g

C

g

D

2

2 D

+ 76 kg 9.8 m s 2 5.85 m

v D = 5.34 m s (g)

Let point E be the apex of his flight:

eK + U j = e K + U j 1 76 kg b5.34 m sg + 0 = 0 + 76 kg e9.8 m s jb y 2 by − y g = 1.46 m g

D

g

E

2

E

(h)

E

− yD

g

D

For the motion between takeoff and touchdown 1 y f = yi + v yi t + a y t 2 2 −2.34 m = 0 + 5.34 m s t − 4.9 m s 2 t 2 t=

(i)

2

a fa f =

−5.34 ± 5.34 2 + 4 4.9 2.34 −9.8

1. 43 s

This solution is more accurate. In chapter 8 we modeled the normal force as constant while the skateboarder stands up. Really it increases as the process goes on.

342 P11.45

Angular Momentum

(a)

I = ∑ mi ri2 =m

FG 4d IJ H3K

2

+m

FG d IJ H 3K

2

+m

FG 2d IJ H3K

2

m

m

1

d2 = 7m 3

2d 3

2 d

Think of the whole weight, 3mg, acting at the center of gravity.

τ =r×F=

FG d IJ e− ij × 3mge− jj = bmgdgk H 3K

3g τ 3mgd = = counterclockwise 2 I 7md 7d

(c)

α=

(d)

a = αr =

FG 3 g IJ FG 2d IJ = H 7d K H 3 K

2g up 7

The angular acceleration is not constant, but energy is.

a K + U f + ∆E = a K + U f F dI 1 0 + a3m f g G J + 0 = Iω H 3K 2 i

(e)

maximum kinetic energy = mgd

(f)

ωf =

(g)

L f = Iω f =

(h)

vf =ω fr =

6g 7d 7md 2 3

6g = 7d

6g d = 7d 3

FG 14 g IJ H3K

2 gd 21

12

md 3 2

f

2 f

3 d

FIG. P11.45 (b)

m

P

+0

Chapter 11

P11.46

(a)

af b

343

g

The radial coordinate of the sliding mass is r t = 0.012 5 m s t . Its angular momentum is

b

gb

gb

gb

g

2

L = mr 2ω = 1.20 kg 2.50 rev s 2π rad rev 0.012 5 m s t 2

e

j

L = 2.95 × 10 −3 kg ⋅ m 2 s 3 t 2

or

The drive motor must supply torque equal to the rate of change of this angular momentum:

τ=

b

ja f b0.005 89 Wgt

dL = 2.95 × 10 −3 kg ⋅ m 2 s 3 2t = dt

e

f

ga

(b)

τ f = 0.005 89 W 440 s = 2.59 N ⋅ m

(c)

P = τω = 0.005 89 W t 5π rad s =

(d)

P f = 0.092 5 W s 440 s = 40.7 W

(e)

T =m

(f)

W=

(g)

b

gb

b

g b0.092 5 W sgt

f

ga

v2 = mrω 2 = 1.20 kg 0.012 5 m s t 5π rad s r

b

z

zb

440 s

440 s

0

0

Pdt =

gb

gb

g

0.092 5 W s tdt =

ja

g

2

1 0.092 5 J s 2 440 s 2

e

b3.70 N sgt

=

f

2

= 8.96 kJ

The power the brake injects into the sliding block through the string is

b

gb

g b

g

Pb = F ⋅ v = Tv cos 180° = − 3.70 N s t 0.012 5 m s = − 0.046 3 W s t = Wb =

z

440 s

0

0

Pb dt = −

g

0.046 3 W s tdt

b

ga

1 0.046 3 W s 440 s 2

=− (h)

zb

440 s

dWb dt

∑ W = W + Wb = 8.96 kJ − 4.48 kJ =

f

2

= −4.48 kJ

4.48 kJ

Just half of the work required to increase the angular momentum goes into rotational kinetic energy. The other half becomes internal energy in the brake. P11.47

Using conservation of angular momentum, we have

e j

e j

Laphelion = Lperihelion or mra2 ω a = mrp2 ω p . vp

e j vr = emr j r

Thus, mra2

a

a

ra v a = rp v p or v a =

2 p

rp ra

giving

p

vp =

b

g

0.590 AU 54.0 km s = 0.910 km s . 35.0 AU

344 P11.48

Angular Momentum

(a)

∑ τ = MgR − MgR =

(b)

∑τ =

0

dL , and since dt

∑τ = 0 , L =

constant.

Since the total angular momentum of the system is zero, the monkey and bananas move upward with the same speed at any instant, and he will not reach the bananas (until they get tangled in the pulley). Also, since the tension in the rope is the same on both sides, Newton’s second law applied to the monkey and bananas give the same acceleration upwards.

FIG. P11.48 P11.49

(a)

τ = r × F = r F sin180° = 0 Angular momentum is conserved. L f = Li mrv = mri vi v=

b g

m ri vi mv 2 = r r3

ri vi r

2

(b)

T=

(c)

The work is done by the centripetal force in the negative direction. Method 1:

b g W = z F ⋅ dA = − z Tdr ′ = − z dr ′ = 2a r ′ f Fr I mbr v g F 1 1I 1 mv G − 1J = − J= G 2 Hr r K 2 Hr K r

ri

i i

b g ar ′ f

FIG. P11.49

m ri vi

2 i

Method 2:

(d)

m ri vi

3

2

2

W = ∆K =

2

2 i

2 r

2

ri

2 i 2

F GH

I JK

r2 1 1 1 mv 2 − mvi2 = mvi2 i2 − 1 2 2 2 r

Using the data given, we find

v = 4.50 m s

T = 10.1 N

W = 0.450 J

Chapter 11

P11.50

(a)

Angular momentum is conserved:

F GH

FG H

mvi d 1 d = Md 2 + m 2 12 2

ω=

(b)

m

IJ IJω KK 2

ω vi

6mvi Md + 3md

d

O

O

(a)

1 The original energy is mvi2 . 2

(b)

FIG. P11.50

The final energy is

F GH

I 36m v JK aMd + 3mdf 2 2 i

1 2 1 1 md 2 Iω = Md 2 + 2 2 12 4

2

=

3m 2 vi2 d . 2 Md + 3md

a

f

The loss of energy is 3m 2 vi2 d mMvi2 d 1 = mvi2 − 2 2 Md + 3md 2 Md + 3md

a

f a

f

and the fractional loss of energy is

a

mMvi2 d 2

2 Md + 3md P11.51

(a)

f

mvi2

=

M . M + 3m

FG d IJ H 2K L = 2b75.0 kg gb5.00 m sga5.00 mf Li = m1 v1i r1i + m 2 v 2i r2i = 2mv i

Li = 3 750 kg ⋅ m 2 s (b)

345

1 1 m1 v12i + m 2 v 22i 2 2 1 75.0 kg 5.00 m s Ki = 2 2 Ki =

FG IJ b H K

gb

g

2

= 1.88 kJ

(c)

Angular momentum is conserved: L f = Li = 3 750 kg ⋅ m 2 s

(d)

vf =

(e)

Kf = 2

(f)

W = K f − K i = 5.62 kJ

Lf

=

3 750 kg ⋅ m 2 s = 10.0 m s 2 75.0 kg 2.50 m

d i b

2 mr f

f

ga

FG 1 IJ b75.0 kggb10.0 m sg H 2K

2

= 7.50 kJ

FIG. P11.51

346 P11.52

Angular Momentum

(a)

(b) (c)

*P11.53

LM FG d IJ OP = Mvd N H 2K Q F1 I K = 2G Mv J = Mv H2 K Li = 2 Mv

2

2

L f = Li = Mvd Lf

(d)

Mvd = = 2v vf = 2 Mr f 2 M d4

(e)

Kf = 2

(f)

W = K f − K i = 3 Mv 2

FIG. P11.52

ch

FG 1 Mv IJ = Ma2 vf H2 K 2 f

2

= 4Mv 2

The moment of inertia of the rest of the Earth is I=

2 2 MR 2 = 5.98 × 10 24 kg 6.37 × 10 6 m 5 5

e

j

2

= 9.71 × 10 37 kg ⋅ m 2 .

For the original ice disks, I=

1 1 Mr 2 = 2.30 × 10 19 kg 6 × 10 5 m 2 2

e

j

2

= 4.14 × 10 30 kg ⋅ m 2 .

For the final thin shell of water, I=

2 2 Mr 2 = 2.30 × 10 19 kg 6.37 × 10 6 m 3 3

e

j

2

= 6.22 × 10 32 kg ⋅ m 2 .

Conservation of angular momentum for the spinning planet is expressed by Iiω i = I f ω f

π 2π = e6. 22 × 10 + 9.71 × 10 j e4.14 × 10 + 9.71 × 10 j 86 2400 s b86 400 s + δ g F1 + δ I F 1 + 4.14 × 10 I = F 1 + 6.22 × 10 I GH 86 400 s JK GH 9.71 × 10 JK GH 9.71 × 10 JK 30

37

32

30

32

37

37

δ 6.22 × 10 32 4.14 × 10 30 = − 86 400 s 9.71 × 10 37 9.71 × 10 37 δ = 0.550 s

37

Chapter 11

P11.54

For the cube to tip over, the center of mass (CM) must rise so that it is over the axis of rotation AB. To do this, the CM must be raised a

e

distance of a

j

CM

2 −1 .

e

1 I cubeω 2 2

j

∴ Mga

2 −1 =

D

A

From conservation of angular momentum,

F GH

4a 8 Ma mv = 3 3 mv ω= 2 Ma

F GH

2

P11.55

C

Iω JK

D B

4a/3 A

I JK

1 8 Ma 2 m 2 v 2 = Mga 2 3 4M 2 a 2 M 3 ga m

v=

347

e

2 −1

e

2 −1

j

FIG. P11.54

j

Angular momentum is conserved during the inelastic collision. Mva = Iω

ω=

Mva 3 v = I 8a

The condition, that the box falls off the table, is that the center of mass must reach its maximum height as the box rotates, hmax = a 2 . Using conservation of energy: 1 2 Iω = Mg a 2 − a 2

e

F GH

j

I FG 3v IJ = Mgea JK H 8 a K 16 v = gae 2 − 1j 3 O L ga v = 4M e 2 − 1jP 3 Q N 1 8 Ma 2 2 3

2

2 −a

j

FIG. P11.55

2

12

P11.56

(a)

The net torque is zero at the point of contact, so the angular momentum before and after the collision must be equal.

FG 1 MR IJ ω = FG 1 MR IJω + eMR jω H2 K H2 K 2

(b)

∆E = E

e

1 1 2 2

MR 2

ωi 2 3

je j

Rω i 2 3

e j e MR jω

+ 12 M 1 1 2 2

2

i

2

2 i

−

e

1 1 2 2

2

j

MR 2 ω i2

= −

ω=

2 3

ωi 3

348 P11.57

Angular Momentum

∆p Rω i Mv MRω = = = µMg µMg 3 µg f

(a)

∆t =

(b)

W = ∆K =

µMgx =

1 2 1 Iω = MR 2ω i2 2 18

1 MR 2ω i2 18

x=

(See Problem 11.56)

R 2ω i2 18 µg

ANSWERS TO EVEN PROBLEMS P11.2

(a) 740 cm 2 ; (b) 59.5 cm

P11.32

8.54 rad s

P11.4

(a) 168°; (b) 11.9° principal value; (c) Only the first is unambiguous.

P11.34

(a) 7.20 × 10 −3 kg ⋅ m 2 s ; (b) 9.47 rad s

P11.36

12.3 m s 2

P11.38

(a) 2.35 rad s; (b) 0.498 rad s ; (c) 5.58°

P11.40

131 s

P11.42

(a) 2.19 × 10 6 m s ; (b) 2.18 × 10 −18 J ;

P11.6

No; see the solution

P11.8

(a) −7.00 N ⋅ m k ; (b) 11.0 N ⋅ m k

P11.10

see the solution

P11.12

e−22.0 kg ⋅ m sjk

P11.14

see the solution

P11.16

(a) 3.14 N ⋅ m ; (b) 0. 400 kg ⋅ m v ;

a

f

(c) 4.13 × 10 16 rad s

2

P11.44

b

(c) 7.85 m s P11.18

a

f

g

2

e

j

(a) +9.03 × 10 9 kg ⋅ m 2 s south; (b) No;

P11.46

b g b g (e) b3.70 N sgt ; (f) 8.96 kJ; (g) −4.48 kJ (a) 0.005 89 W t ; (b) 2.59 N ⋅ m ; (c) 0.092 5 W s t ; (d) 40.7 W ;

(c) 0

(h) +4.48 kJ

P11.20

103 N ⋅ m

P11.22

e4.50 kg ⋅ m sj up

P11.24

(a) 11.1 m s; (b) 5.32 × 10 3 kg ⋅ m 2 s ; (c) see the solution; (d) 12.0 m s; (e) 1.08 kJ ; (f) 5.34 m s; (g) 1.46 m; (h) 1.43 s; (i) see the solution

2

2

1.20 kg ⋅ m s perpendicularly into the clock face

P11.26

8.63 m s 2

P11.28

Kf I1 Iω = (a) 1 i ; (b) Ki I1 + I 2 I1 + I 2

P11.30

(a) 1.91 rad s ; (b) 2.53 J; 6.44 J

P11.48

(a) 0; (b) 0; no

P11.50

(a)

P11.52

(a) Mvd ; (b) Mv 2 ; (c) Mvd ; (d) 2v; (e) 4Mv 2 ; (f) 3 Mv 2

6mvi M ; (b) M + 3m Md + 3md

P11.54

M 3 ga m

P11.56

(a)

e

2 −1

j

ωi 2 ∆E =− ; (b) E 3 3

12 Static Equilibrium and Elasticity CHAPTER OUTLINE 12.1 12.2 12.3 12.4

The Conditions for Equilibrium More on the Center of Gravity Examples of Rigid Objects in Static Equilibrium Elastic Properties of Solids

ANSWERS TO QUESTIONS Q12.1

When you bend over, your center of gravity shifts forward. Once your CG is no longer over your feet, gravity contributes to a nonzero net torque on your body and you begin to rotate.

Q12.2

Yes, it can. Consider an object on a spring oscillating back and forth. In the center of the motion both the sum of the torques and the sum of the forces acting on the object are (separately) zero. Again, a meteoroid flying freely through interstellar space feels essentially no forces and keeps moving with constant velocity.

Q12.3

No—one condition for equilibrium is that

∑ F = 0 . For this to

be true with only a single force acting on an object, that force would have to be of zero magnitude; so really no forces act on that object. Q12.4

(a)

Consider pushing up with one hand on one side of a steering wheel and pulling down equally hard with the other hand on the other side. A pair of equal-magnitude oppositelydirected forces applied at different points is called a couple.

(b)

An object in free fall has a non-zero net force acting on it, but a net torque of zero about its center of mass.

Q12.5

No. If the torques are all in the same direction, then the net torque cannot be zero.

Q12.6

(a)

Yes, provided that its angular momentum is constant.

(b)

Yes, provided that its linear momentum is constant.

Q12.7

A V-shaped boomerang, a barstool, an empty coffee cup, a satellite dish, and a curving plastic slide at the edge of a swimming pool each have a center of mass that is not within the bulk of the object.

Q12.8

Suspend the plywood from the nail, and hang the plumb bob from the nail. Trace on the plywood along the string of the plumb bob. Now suspend the plywood with the nail through a different point on the plywood, not along the first line you drew. Again hang the plumb bob from the nail and trace along the string. The center of gravity is located halfway through the thickness of the plywood under the intersection of the two lines you drew.

349

350

Static Equilibrium and Elasticity

Q12.9

The center of gravity must be directly over the point where the chair leg contacts the floor. That way, no torque is applied to the chair by gravity. The equilibrium is unstable.

Q12.10

She can be correct. If the dog stands on a relatively thick scale, the dog’s legs on the ground might support more of its weight than its legs on the scale. She can check for and if necessary correct for this error by having the dog stand like a bridge with two legs on the scale and two on a book of equal thickness—a physics textbook is a good choice.

Q12.11

If their base areas are equal, the tall crate will topple first. Its center of gravity is higher off the incline than that of the shorter crate. The taller crate can be rotated only through a smaller angle before its center of gravity is no longer over its base.

Q12.12

The free body diagram demonstrates that it is necessary to have friction on the ground to counterbalance the normal force of the wall and to keep the base of the ladder from sliding. Interestingly enough, if there is friction on the floor and on the wall, it is not possible to determine whether the ladder will slip from the equilibrium conditions alone.

FIG. Q12.12 Q12.13

When you lift a load with your back, your back muscles must supply the torque not only to rotate your upper body to a vertical position, but also to lift the load. Since the distance from the pivot—your hips—to the load—essentially your shoulders—is great, the force required to supply the lifting torque is very large. When lifting from your knees, your back muscles need only keep your back straight. The force required to do that is much smaller than when lifting with your back, as the torque required is small, because the moment arm of the load is small—the line of action of the load passes close to your hips. When you lift from your knees, your much stronger leg and hip muscles do the work.

Q12.14

Shear deformation.

Q12.15

The vertical columns experience simple compression due to gravity acting upon their mass. The horizontal slabs, however, suffer significant shear stress due to gravity. The bottom surface of a sagging lintel is under tension. Stone is much stronger under compression than under tension, so horizontal slabs are more likely to fail.

351

Chapter 12

SOLUTIONS TO PROBLEMS Section 12.1 P12.1

The Conditions for Equilibrium

To hold the bat in equilibrium, the player must exert both a force and a torque on the bat to make

∑ Fx = ∑ Fy = 0

and

F 0.600 m

∑τ = 0

∑ Fy = 0 ⇒ F − 10.0 N = 0 , or the player must exert a net

O

upward force of F = 10.0 N 10.0 N

To satisfy the second condition of equilibrium, the player must exert an applied torque τ a to make

FIG. P12.1

∑ τ = τ a − a0.600 mfa10.0 N f = 0 . Thus, the required torque is τ a = +6.00 N ⋅ m or 6.00 N ⋅ m counterclockwise

P12.2

Use distances, angles, and forces as shown. The conditions of equilibrium are:

∑ Fy = 0 ⇒ ∑ Fx = 0 ⇒ ∑τ = 0 ⇒

Fy

Fx − R x = 0 Fy A cos θ − Fg

Fx

l

Fy + R y − Fg = 0

FG A IJ cos θ − F A sinθ = 0 H 2K x

Ry Fg

θ

Rx O

FIG. P12.2 P12.3

Take torques about P.

L

∑ τ p = −n0 MN

OP Q

LM N

OP Q

A A + d + m1 g + d + m b gd − m 2 gx = 0 2 2

m1

We want to find x for which n 0 = 0 . x=

bm g + m ggd + m g 1

b

m2 g

1

A 2

=

bm

1

g

+ mb d m2

mb g

m1 g

O

m2 g d

A 2

m2

P CG

+ m1 2A

x nO

nP A FIG. P12.3

352

Static Equilibrium and Elasticity

Section 12.2 P12.4

More on the Center of Gravity

The hole we can count as negative mass xCG =

m 1 x1 − m 2 x 2 m1 − m 2

Call σ the mass of each unit of pizza area.

xCG = xCG = P12.5

c h c− h − σπ c h

σπR 2 0 − σπ σπR 2 R 8 3 4

=

R 2 2

R 2

R 2 2

R 6

The coordinates of the center of gravity of piece 1 are

4.00 cm

x1 = 2.00 cm and y1 = 9.00 cm . The coordinates for piece 2 are

18.0 cm

1

x 2 = 8.00 cm and y 2 = 2.00 cm .

2

The area of each piece is

12.0 cm

A1 = 72.0 cm 2 and A 2 = 32.0 cm 2 .

FIG. P12.5

And the mass of each piece is proportional to the area. Thus,

∑ m i xi xCG = ∑ mi

e72.0 cm ja2.00 cmf + e32.0 cm ja8.00 cmf = = 2

2

72.0 cm 2 + 32.0 cm 2

3.85 cm

and yCG =

∑ m i yi ∑ mi

e72.0 cm ja9.00 cmf + e32.0 cm ja2.00 cmf = 2

=

2

104 cm 2

6.85 cm .

4.00 cm

Chapter 12

P12.6

Let σ represent the mass-per-face area. A vertical strip at position x, with width dx and height

ax − 3.00f

y

2

1.00 m

has mass

9

a

f

y = (x — 3.00)2/9

2

σ x − 3.00 dx dm = . 9 The total mass is

x

z

a f

z

3.00

2

σ x − 3 dx M = dm = 9 x =0

FG σ IJ z ex − 6 x + 9jdx H 9K F σ I L x 6x + 9xOP M = G JM − H 9 KN 3 2 Q 3.00

M=

353

x 0

2

FIG. P12.6

0

3

3.00 m

dx

3.00

2

=σ

0

The x-coordinate of the center of gravity is xCG =

P12.7

z

xdm M

=

1 9σ

z

3.00

a f

2

σx x − 3 dx =

0

σ 9σ

ze

3.00

j

x 3 − 6 x 2 + 9 x dx =

0

LM N

1 x 4 6x3 9x 2 − + 9 4 3 2

OP Q

3.00

= 0

6.75 m = 0.750 m 9.00

Let the fourth mass (8.00 kg) be placed at (x, y), then xCG = 0 = x=− Similarly,

yCG = 0 =

a3.00fa4.00f + m axf 4

12.0 + m 4

12.0 = −1.50 m 8.00

a3.00fa4.00f + 8.00byg 12.0 + 8.00

y = −1.50 m P12.8

In a uniform gravitational field, the center of mass and center of gravity of an object coincide. Thus, the center of gravity of the triangle is located at x = 6.67 m , y = 2.33 m (see the Example on the center of mass of a triangle in Chapter 9). The coordinates of the center of gravity of the three-object system are then:

b6.00 kg ga5.50 mf + b3.00 kgga6.67 mf + b5.00 kg ga−3.50 mf a6.00 + 3.00 + 5.00f kg

xCG =

∑ m i xi ∑ mi

xCG =

35.5 kg ⋅ m = 2.54 m and 14.0 kg

yCG =

∑ m i yi ∑ mi

yCG =

66.5 kg ⋅ m = 4.75 m 14.0 kg

=

=

b6.00 kg ga7.00 mf + b3.00 kgga2.33 mf + b5.00 kgga+3.50 mf 14.0 kg

354

Static Equilibrium and Elasticity

Section 12.3 P12.9

Examples of Rigid Objects in Static Equilibrium

∑ τ = 0 = mga3r f − Tr 2T − Mg sin 45.0° = 0

3r

bg

Mg sin 45.0° 1 500 kg g sin 45.0° = 2 2 = 530 9.80 N

T=

a fa f

m=

T 530 g = = 177 kg 3g 3g

m

1 500 kg

θ = 45° FIG. P12.9 *P12.10

(a)

For rotational equilibrium of the lowest rod about its point of support,

m1 = 9.00 g

+12.0 g g 3 cm − m1 g 4 cm (b)

For the middle rod, + m 2 2 cm − 12.0 g + 9.0 g 5 cm = 0

b

(c)

P12.11

g

m 2 = 52.5 g

For the top rod, 52.5 g + 12.0 g + 9.0 g 4 cm − m3 6 cm = 0

b

∑τ = 0 .

g

m3 = 49.0 g

Fg → standard weight

24.0 cm

26.0 cm

Fg′ → weight of goods sold

a

f a f F 13 I F = F′ G J H 12 K F F − F ′ I 100 = F 13 − 1I × 100 = GH F ′ JK GH 12 JK Fg 0.240 = Fg′ 0.260 g

Fg

F′g

g

g

g

FIG. P12.11 8.33%

g

*P12.12

(a)

Consider the torques about an axis perpendicular to the page and through the left end of the horizontal beam.

T

∑ τ = +aT sin 30.0°fd − a196 N fd = 0 ,

V

giving T = 392 N . (b)

30.0°

H

196 N

d

FIG. P12.12

a

f

From

∑ Fx = 0 ,

From

∑ Fy = 0 , V + T sin 30.0°−200 N = 0 , or V = 196 N − a392 N f sin 30.0° =

H − T cos 30.0° = 0 , or H = 392 N cos 30.0° = 339 N to the right . 0 .

355

Chapter 12

P12.13

(a)

∑ Fx = f − n w = 0 ∑ Fy = n g − 800 N − 500 N = 0

nw

Taking torques about an axis at the foot of the ladder,

a800 Nfa4.00 mf sin 30.0°+a500 Nfa7.50 mf sin 30.0° −n a15.0 cmf cos 30.0° = 0

500 N ng

w

Solving the torque equation, nw

a4.00 mfa800 Nf + a7.50 mfa500 Nf tan 30.0° = 268 N . =

800 N

f A

15.0 m

Next substitute this value into the Fx equation to find

f = n w = 268 N Solving the equation

FIG. P12.13

in the positive x direction.

∑ Fy = 0 , n g = 1 300 N in the positive y direction.

(b)

In this case, the torque equation

∑τ A = 0

gives:

a9.00 mfa800 Nf sin 30.0°+a7.50 mfa500 Nf sin 30.0°−a15.0 mfbn g sin 60.0° = 0 w

or

n w = 421 N .

Since f = n w = 421 N and f = fmax = µn g , we find fmax 421 N = = 0.324 . 1 300 N ng

µ=

P12.14

(a)

∑ Fx = f − n w = 0 ∑ Fy = n g − m1 g − m 2 g = 0

nw

(1)

m2 g

(2)

F LI

∑ τ A = −m1 g GH 2 JK cos θ − m 2 gx cos θ + n w L sin θ = 0 From the torque equation, nw =

LM 1 m g + FG x IJ m g OP cot θ N2 H LK Q 1

Then, from equation (1): and from equation (2): (b)

m1 g

2

L1 F xI O f = n = M m g + G J m g P cot θ N2 H LK Q n = bm + m g g w

g

1

1

2

2

If the ladder is on the verge of slipping when x = d , then

µ=

f

x=d

ng

=

e

m1 2

+

m2d L

j cot θ

m1 + m 2

.

θ

f A

ng

FIG. P12.14

356 P12.15

Static Equilibrium and Elasticity

(a)

Taking moments about P,

aR sin 30.0°f0 + aR cos 30.0°fa5.00 cmf − a150 Nfa30.0 cmf = 0 R = 1 039.2 N = 1.04 kN The force exerted by the hammer on the nail is equal in magnitude and opposite in direction:

1.04 kN at 60° upward and to the right. (b)

FIG. P12.15

f = R sin 30.0°−150 N = 370 N n = R cos 30.0° = 900 N

a

f a

f

Fsurface = 370 N i + 900 N j P12.16

See the free-body diagram at the right. When the plank is on the verge of tipping about point P, the normal force n1 goes to zero. Then, summing torques about point P gives

∑ τ p = −mgd + Mgx = 0

or

F mI x = G Jd . H MK

From the dimensions given on the free-body diagram, observe that d = 1.50 m Thus, when the plank is about to tip, x= P12.17

F 30.0 kg I a1.50 mf = GH 70.0 kg JK

Mg 3.00 m

x P

n1

mg

d

6.00 m

FIG. P12.16

0.643 m .

Torque about the front wheel is zero.

a

fb g a

fb g

0 = 1.20 m mg − 3.00 m 2 Fr Thus, the force at each rear wheel is

Fr = 0.200mg = 2.94 kN . The force at each front wheel is then Ff =

mg − 2 Fr = 4. 41 kN . 2

FIG. P12.17

n2

1.50 m

Chapter 12

P12.18

∑ Fx = Fb − Ft + 5.50 N = 0

(1)

357

5.50 N

∑ Fy = n − mg = 0 Summing torques about point O, 10.0 m

∑ τ O = Ft a1.50 mf − a5.50 mfa10.0 mf = 0

mg

Ft

which yields Ft = 36.7 N to the left

1.50 m

Then, from Equation (1), Fb

O

Fb = 36.7 N − 5.50 N = 31.2 N to the right

n

FIG. P12.18 P12.19

P12.20

(a)

Te sin 42.0° = 20.0 N

Te = 29.9 N

(b)

Te cos 42.0° = Tm

Tm = 22.2 N

Relative to the hinge end of the bridge, the cable is attached horizontally out a distance x = 5.00 m cos 20.0° = 4.70 m and

a f vertically down a distance y = a5.00 mf sin 20.0° = 1.71 m . The cable then makes the following angle with the horizontal:

θ = tan −1 (a)

LM a12.0 + 1.71f m OP = 71.1° . N 4.70 m Q

af

a

f

f

f

a

which yields T = 35.5 kN

∑ Fx = 0 ⇒ Rx − T cos 71.1° = 0 or

(c)

a

f

b

R x = 35.5 kN cos 71.1° = 11.5 kN right

g

∑ Fy = 0 ⇒ R y − 19.6 kN + T sin 71.1°−9.80 kN = 0 Thus,

a

f

R y = 29. 4 kN − 35.5 kN sin 71.1° = −4.19 kN = 4.19 kN down

y

19.6 kN 9.80 kN

f

−9.80 kN 7.00 m cos 20.0° = 0

(b)

20.0°

Rx

7.00 m

−T cos 71.1° 1.71 m + T sin 71.1° 4.70 m

a

x

5.00 m

R x 0 + R y 0 − 19.6 kN 4.00 m cos 20.0°

a

T

4.00 m

Take torques about the hinge end of the bridge:

af

Ry

FIG. P12.20

358 *P12.21

Static Equilibrium and Elasticity

(a)

We model the horse as a particle. The drawbridge will fall out from under the horse.

α = mg =

(b)

e

1 2

A cos θ 0 1 3

mA

j

3 9.80 m s 2 cos 20.0°

a

2 8.00 m

f

1 2 Iω = mgh 2 1 1 1 ∴ ⋅ mA 2ω 2 = mg ⋅ A 1 − sin θ 0 2 3 2

b

∴ω = (c)

Rx

θ0 A

3g cos θ 0 2A

=

2

Ry

= 1.73 rad s

mg

2

FIG. P12.21(a)

g

e

ja

3 9.80 m s 2 3g 1 − sin θ 0 = 1 − sin 20° = 1.56 rad s A 8.00 m

b

g

f

The linear acceleration of the bridge is:

a

fe

Ry

θ0

1 1 a = Aα = 8.0 m 1.73 rad s 2 = 6.907 m s 2 2 2

j

The force at the hinge + the force of gravity produce the acceleration a = 6.907 m s 2 at right angles to the bridge.

b

ge

a

Rx

mg

FIG. P12.21(c)

j

R x = ma x = 2 000 kg 6.907 m s 2 cos 250° = −4.72 kN

Ry − mg = ma y

j b

e

g

e

j

∴ R y = m g + a y = 2 000 kg 9.80 m s 2 + 6.907 m s 2 sin 250° = 6.62 kN

e

j

Thus: R = −4.72 i + 6.62 j kN . (d)

Rx = 0 a=ω

2

FG 1 AIJ = b1.56 rad sg a4.0 mf = 9.67 m s H2 K 2

R y − mg = ma

b

ge

j

Ry 2

∴ R y = 2 000 kg 9.8 m s 2 + 9.67 m s 2 = 38.9 kN Thus: R y = 38.9 j kN

a mg FIG. P12.21(d)

Rx

Chapter 12

P12.22

Call the required force F, with components Fx = F cos 15.0° and Fx

center of the wheel by the handles.

R

F cos 15.0°−n x

b

b nx

8.00 cm

Just as the wheel leaves the ground, the ground exerts no force on it.

∑ Fx = 0 : ∑ Fy = 0 :

400 N

Fy

Fy = − F sin 15.0° , transmitted to the

ny

a

a

distances

(1)

forces

− F sin 15.0°−400 N + n y = 0 (2)

FIG. P12.22

Take torques about its contact point with the brick. The needed distances are seen to be:

a

f

b = R − 8.00 cm = 20.0 − 8.00 cm = 12.0 cm 2

2

a = R − b = 16.0 cm (a)

a

∑τ = 0 :

f

− Fx b + Fy a + 400 N a = 0 , or

a

f

a

f

a

fa

f

F − 12.0 cm cos 15.0°+ 16.0 cm sin 15.0° + 400 N 16.0 cm = 0 F=

so (b)

6 400 N ⋅ cm = 859 N 7.45 cm

Then, using Equations (1) and (2),

a

f

n x = 859 N cos 15.0° = 830 N and

a

f

n y = 400 N + 859 N sin 15.0° = 622 N n = n x2 + n y2 = 1.04 kN

θ = tan −1

F n I = tan a0.749f = GH n JK y

−1

36.9° to the left and upward

x

*P12.23

When x = x min , the rod is on the verge of slipping, so

b g

f = fs From

max

= µ sn = 0.50n .

∑ Fx = 0 , n − T cos 37° = 0 , or n = 0.799T .

a

f

∑ Fy = 0 ,

Using

∑τ = 0

a

2.0 m

37°

n f

x Fg

Fg

2.0 m

FIG. P12.23

Thus, f = 0.50 0.799T = 0.399T From

f + T sin 37°−2 Fg = 0 , or 0.399T − 0.602T − 2 Fg = 0 , giving T = 2.00 Fg .

for an axis perpendicular to the page and through the left end of the beam gives

f e j

359

a

f

− Fg ⋅ x min − Fg 2.0 m + 2 Fg sin 37° 4.0 m = 0 , which reduces to x min = 2.82 m .

360 P12.24

Static Equilibrium and Elasticity

x=

3L 4

L

If the CM of the two bricks does not lie over the edge, then the bricks balance. If the lower brick is placed

L over the edge, then the 4

second brick may be placed so that its end protrudes over the edge. P12.25

x

3L 4

FIG. P12.24

To find U, measure distances and forces from point A. Then, balancing torques,

a0.750fU = 29.4a2.25f

U = 88.2 N

To find D, measure distances and forces from point B. Then, balancing torques,

a0.750fD = a1.50fa29.4f Also, notice that U = D + Fg , so *P12.26

D = 58.8 N

∑ Fy = 0 .

Consider forces and torques on the beam.

∑ Fx = 0 : ∑ Fy = 0 : ∑τ = 0 :

R cos θ − T cos 53° = 0 R sin θ + T sin 53°−800 N = 0

aT sin 53°f8 m − a600 Nfx − a200 Nf4 m = 0

600 Nx + 800 N ⋅ m = 93.9 N m x + 125 N . As x increases from 2 m, this expression 8 m sin 53° grows larger.

b

(a)

Then T =

(b)

From substituting back,

g

R cos θ = 93.9 x + 125 cos 53° R sin θ = 800 N − 93.9 x + 125 sin 53° Dividing, tan θ =

800 N R sin θ = − tan 53°+ 93.9 x +125 cos 53° R cos θ

a

f F 32 − 1IJ tan θ = tan 53° G H 3x + 4 K

As x increases the fraction decreases and θ decreases . continued on next page

Chapter 12

(c)

To find R we can work out R 2 cos 2 θ + R 2 sin 2 θ = R 2 . From the expressions above for R cos θ and R sin θ ,

a

R 2 = T 2 cos 2 53°+T 2 sin 2 53°−1 600 NT sin 53°+ 800 N R 2 = T 2 − 1 600T sin 53°+640 000

a

R 2 = 93.9 x + 125

f

2

a

f

2

f

− 1 278 93.9 x + 125 + 640 000

e

R = 8 819 x 2 − 96 482 x + 495 678

j

12

At x = 0 this gives R = 704 N . At x = 2 m , R = 581 N . At x = 8 m , R = 537 N . Over the range of possible values for x, the negative term −96 482x dominates the positive term

8 819 x 2 , and R decreases as x increases.

Section 12.4 P12.27

∆L F =Y A Li ∆L =

P12.28

Elastic Properties of Solids

(a)

a fa fa f e je j

200 9.80 4.00 FLi = = 4.90 mm AY 0.200 × 10 −4 8.00 × 10 10 stress =

F F = 2 A πr

a

f FGH d2 IJK

e

8

F = stress π

F = 1.50 × 10

2

F 2.50 × 10 m I N m jπ G H 2 JK -2

2

2

F = 73.6 kN (b)

a

∆L =

*P12.29

f

stress = Y strain =

Y∆L Li

astressfL = e1.50 × 10 i

Y

The definition of Y =

8

ja

f=

N m 2 0.250 m

1.50 × 10

10

N m

2

2.50 mm

stress means that Y is the slope of the graph: strain Y=

300 × 10 6 N m 2 = 1.0 × 10 11 N m 2 . 0.003

361

362 P12.30

Static Equilibrium and Elasticity

Count the wires. If they are wrapped together so that all support nearly equal stress, the number should be 20.0 kN = 100 . 0.200 kN Since cross-sectional area is proportional to diameter squared, the diameter of the cable will be

a1 mmf P12.31

100 ~ 1 cm .

From the defining equation for the shear modulus, we find ∆x as

e

ja

f

5.00 × 10 −3 m 20.0 N hf ∆x = = = 2.38 × 10 −5 m −4 6 2 2 SA 3.0 × 10 N m 14.0 × 10 m

e

je

j

or ∆x = 2.38 × 10 −2 mm . P12.32

The force acting on the hammer changes its momentum according to

a f

mvi + F ∆t = mv f so F = Hence, F =

30.0 kg −10.0 m s − 20.0 m s 0.110 s

m v f − vi ∆t

.

= 8.18 × 10 3 N .

By Newton’s third law, this is also the magnitude of the average force exerted on the spike by the hammer during the blow. Thus, the stress in the spike is: stress =

and the strain is: strain = P12.33

(a)

a fa f = π e5.00 × 10

F 8.18 × 10 3 N = = 1.97 × 10 7 N m 2 b0.023 0 mg2 A

π

4

stress 1.97 × 10 7 N m 2 = = 9.85 × 10 −5 . Y 20.0 × 10 10 N m 2

F = A stress

F −3

m

j e4.00 × 10 2

8

N m2

j

3.0 ft

= 3.14 × 10 4 N (b)

t A

The area over which the shear occurs is equal to the circumference of the hole times its thickness. Thus,

a f

e

je

j

A = 2πr t = 2π 5.00 × 10 −3 m 5.00 × 10 −3 m = 1.57 × 10

af

−4

e

m

FIG. P12.33

2

je

j

So, F = A Stress = 1.57 × 10 −4 m 2 4.00 × 10 8 N m 2 = 6.28 × 10 4 N .

Chapter 12

P12.34

363

Let the 3.00 kg mass be mass #1, with the 5.00 kg mass, mass # 2. Applying Newton’s second law to each mass gives: m1 a = T − m1 g

(1)

m2 a = m2 g − T

and

(2)

where T is the tension in the wire. Solving equation (1) for the acceleration gives: a = and substituting this into equation (2) yields:

T − g, m1

m2 T − m2 g = m2 g − T . m1

Solving for the tension T gives

b

gb

ge

j

2 2m1 m 2 g 2 3.00 kg 5.00 kg 9.80 m s = = 36.8 N . 8.00 kg m 2 + m1

T=

From the definition of Young’s modulus, Y =

FLi , the elongation of the wire is: A ∆L

a f a36.8 Nfa2.00 mf TL ∆L = = YA e2.00 × 10 N m jπ e2.00 × 10 i

11

P12.35

2

−3

j

m

2

= 0.029 3 mm .

Consider recompressing the ice, which has a volume 1.09V0 .

F ∆V IJ = −e2.00 × 10 N m ja−0.090f = ∆P = − BG 1.09 HV K 9

2

1.65 × 10 8 N m 2

i

*P12.36

B=−

∆P ∆V Vi

=−

∆PVi ∆V

e

j

1.13 × 10 8 N m 2 1 m3 ∆PVi =− = −0.053 8 m 3 B 0.21 × 10 10 N m 2

(a)

∆V = −

(b)

The quantity of water with mass 1.03 × 10 3 kg occupies volume at the bottom 1 m 3 − 0.053 8 m 3 = 0.946 m 3 . So its density is

(c) *P12.37

1.03 × 10 3 kg 0.946 m

3

= 1.09 × 10 3 kg m3 .

With only a 5% volume change in this extreme case, liquid water is indeed nearly incompressible.

Part of the load force extends the cable and part compresses the column by the same distance ∆A : F= ∆A =

YA A A ∆A Ys As ∆A + AA As F YA A A AA

+

Ys As As

= 8.60 × 10

−4

=

e

8 500 N

7 ×10 10 π 0.162 4 2 − 0 .161 4 2

a f

4 3. 25

m

j + 20 ×10 π b0.012 7 g 4a 5.75 f 10

2

364

Static Equilibrium and Elasticity

Additional Problems *P12.38

(a)

The beam is perpendicular to the wall, since 3 2 + 4 2 = 5 2 . Then sin θ =

(b)

∑ τ hinge = 0 :

a f

f

+T sin θ 3 m − 250 N 10 m = 0 T=

(c)

a

4m ; θ = 53.1° . 5m

2 500 Nm = 1.04 × 10 3 N 3 m sin 53.1°

1.04 × 10 3 N T = = 0.126 m k 8.25 × 10 3 N m The cable is 5.126 m long. From the law of cosines, x=

a fa

5.126 m

4m

α

f

4 2 = 5.126 2 + 3 2 − 2 3 5.126 cos θ

θ = cos −1 (d)

2

2

3m

2

3 + 5.126 − 4 = 51.2° 2 3 5.126

a fa

f

θ

FIG. P12.38

From the law of sines, the angle the hinge makes with the wall satisfies

sin α sin 51.2° = 5.126 m 4m

sin α = 0.998 58

∑ τ hinge = 0

a f

a

fa

f

+T 3 m sin 51.2°−250 N 10 m 0.998 58 = 0 T = 1.07 × 10 3 N x=

(e)

1.07 × 10 3 N = 0.129 m 8.25 × 10 3 N m

θ = cos −1 (f)

3 2 + 5.129 2 − 4 2 = 51.1° 2 3 5.129

a fa

f

Now the answers are self-consistent: sin 51.1° = 0.998 51 4m T 3 m sin 51.1°−250 N 10 m 0.998 51 = 0 sin α = 5.129 m

a f

a

fa

f

3

T = 1.07 × 10 N x = 0.129 5 m

θ = 51.1° P12.39

Let n A and n B be the normal forces at the points of support. Choosing the origin at point A with

∑ Fy = 0 and ∑ τ = 0,

A

B

we find:

e

j e j −e3.00 × 10 jb g g15.0 − e8.00 × 10 jb g g25.0 + n a50.0 f = 0

n A + n B − 8.00 × 10 4 g − 3.00 × 10 4 g = 0 and 4

4

B

15.0 m 50.0 m

FIG. P12.39

The equations combine to give n A = 5.98 × 10 5 N and b B = 4.80 × 10 5 N .

365

Chapter 12

P12.40

When the concrete has cured and the pre-stressing tension has been released, the rod presses in on the concrete and with equal force, T2 , the concrete produces tension in the rod. (a)

Thus, ∆L = or (b)

f FGH ∆LL IJK

a

In the concrete: stress = 8.00 × 10 6 N m 2 = Y ⋅ strain = Y

astressfL = e i

ja

8.00 × 10 6 N m 2 1.50 m 9

Y

30.0 × 10 N m

i

f

2

∆L = 4.00 × 10 −4 m = 0.400 mm .

In the concrete: stress =

T2 = 8.00 × 10 6 N m 2 , so Ac

e

je

j

T2 = 8.00 × 10 6 N m 2 50.0 × 10 −4 m 2 = 40.0 kN (c)

FG IJ H K

T2 Li T2 ∆L = Ysteel so ∆L = AR Li A R Ysteel

For the rod:

e4.00 × 10 Nja1.50 mf ∆L = = 2.00 × 10 e1.50 × 10 m je20.0 × 10 N m j 4

−4

10

2

m = 2.00 mm

(d)

The rod in the finished concrete is 2.00 mm longer than its unstretched length. To remove stress from the concrete, one must stretch the rod 0.400 mm farther, by a total of 2. 40 mm .

(e)

For the stretched rod around which the concrete is poured:

FG H

IJ K

FG ∆L IJ A Y H L K F 2.40 × 10 m I e1.50 × 10 m je20.0 × 10 T =G H 1.50 m JK ∆Ltotal T1 = Ysteel AR Li

or T1 =

−3

−4

1

*P12.41

−3

2

total i

2

With A as large as possible, n1 and n 2 will both be large. The equality sign in f 2 ≤ µ sn 2 will be true, but the less-than sign in f1 < µ sn1 . Take torques about the lower end of the pole. n 2 A cos θ + Fg

R steel

FG 1 AIJ cos θ − f A sinθ = 0 H2 K f

1 Fg = 0 2

Since n 2 > 0 , it is necessary that 1 − 0.576 tan θ < 0 1 = 1.736 0.576 ∴θ > 60.1° ∴ tan θ >

∴A =

f2 A

n2

Setting f 2 = 0.576n 2 , the torque equation becomes

a

j

N m 2 = 48.0 kN

θ

2

n 2 1 − 0.576 tan θ +

10

7.80 ft d < = 9.00 ft sin θ sin 60.1°

n1

Fg

θ f1 FIG. P12.41

d

366 P12.42

Static Equilibrium and Elasticity

Call the normal forces A and B. They make angles α and β with the vertical.

∑ Fx = 0: ∑ Fy = 0:

Mg

A sin α − B sin β = 0

A

A cos α − Mg + B cos β = 0 β

α

A sin α sin β

Substitute B =

A cos α + A cos β

b

sin α = Mg sin β

g

Mg

A cos α sin β + sin α cos β = Mg sin β sin β A = Mg sin α + β

b

sin α B = Mg sin α + β

b

P12.43

B

(a)

See the diagram.

(b)

If x = 1.00 m , then

B sin α

A sin α

g

B cos α

A cos α

g

FIG. P12.42 T

Ry 60.0°

Rx O

∑ τ O = a−700 N fa1.00 mf − a200 N fa3.00 mf

a fa f +aT sin 60.0°fa6.00 mf = 0

x 700 N

− 80.0 N 6.00 m

3.00 m

200 N

80.0 N

3.00 m

FIG. P12.43

Solving for the tension gives: T = 343 N .

(c)

From

∑ Fx = 0 , R x = T cos 60.0° =

From

∑ Fy = 0 , R y = 980 N − T sin 60.0° =

171 N . 683 N .

If T = 900 N :

∑ τ O = a−700 N fx − a 200 N fa3.00 mf − a80.0 N fa6.00 mf + Solving for x gives: x = 5.13 m .

a900 Nf sin 60.0° a6.00 mf = 0 .

367

Chapter 12

P12.44

(a)

Sum the torques about top hinge:

∑ τ = 0:

T cos 30.0° C

af af

af

C 0 + D 0 + 200 N cos 30.0° 0

a

f

+200 N sin 30.0° 3.00 m

a

f a

1.80 m

f

−392 N 1.50 m + A 1.80 m

af 160 N bright g

Giving A =

392 N

A

+B 0 = 0

1.50 m

1.50 m

B

.

FIG. P12.44

∑ Fx = 0 :

(b)

T sin 30.0°

D

−C − 200 N cos 30.0°+ A = 0 C = 160 N − 173 N = −13.2 N In our diagram, this means 13.2 N to the right .

∑ Fy = 0 : +B + D − 392 N + 200 N sin 30.0° = 0

(c)

b g

B + D = 392 N − 100 N = 292 N up

Given C = 0: Take torques about bottom hinge to obtain

(d)

af af a

f af

a

f

a

f

a

f

A 0 + B 0 + 0 1.80 m + D 0 − 392 N 1.50 m + T sin 30.0° 3.00 m + T cos 30.0° 1.80 m = 0 so T = P12.45

Using

588 N ⋅ m = 192 N . 1.50 m + 1.56 m

a

f

∑ Fx = ∑ Fy = ∑ τ = 0, choosing the origin at the left end

of the beam, we have (neglecting the weight of the beam)

∑ Fx = Rx − T cos θ = 0 , ∑ Fy = Ry + T sin θ − Fg = 0 , and

∑ τ = − Fg aL + d f + T sin θ a2L + d f = 0.

Solving these equations, we find: (a)

T=

(b)

Rx

a f sin θ a 2L + d f F aL + d f cot θ = Fg L + d

g

2L + d

Ry =

Fg L 2L + d

FIG. P12.45

368 P12.46

Static Equilibrium and Elasticity

∑ τ point 0 = 0 gives T sin 25.0°

aT cos 25.0°fFGH 34A sin 65.0°IJK + aT sin 25.0°fFGH 34A cos 65.0°IJK FA I = b 2 000 N gaA cos 65.0°f + b1 200 N gG cos 65.0°J H2 K

l

3l 4 1 200 N

From which, T = 1 465 N = 1.46 kN From

65.0°

H

∑ Fx = 0 ,

b

g

H = T cos 25.0° = 1 328 N toward right = 1.33 kN From

2 000 N

T cos 25.0°

V FIG. P12.46

∑ Fy = 0 ,

b

g

V = 3 200 N − T sin 25.0° = 2 581 N upward = 2.58 kN P12.47

We interpret the problem to mean that the support at point B is frictionless. Then the support exerts a force in the x direction and FBy = 0

∑ Fx = FBx − FAx = 0

b

g and ∑ τ = −b3 000 g ga 2.00f − b10 000 g ga6.00f + F a1.00 f = 0 . FAy − 3 000 + 10 000 g = 0

Bx

These equations combine to give

FIG. P12.47 5

FAx = FBx = 6.47 × 10 N FAy = 1.27 × 10 5 N P12.48

a

f

n= M+m g H = f

a

H

f

H max = f max = µ s m + M g mgL cos 60.0°+ Mgx cos 60.0°− HL sin 60.0° 2 µ m + M tan 60.0° m x H tan 60.0° m = − = s − 2M 2M L Mg M

∑τ A = 0 =

a

3 1 = µ s tan 60.0°− = 0.789 2 4

x

f

Mg mg

n

60.0° A

f

FIG. P12.48

Chapter 12

P12.49

From the free-body diagram, the angle T makes with the rod is

T 20°

θ = 60.0°+20.0° = 80.0° and the perpendicular component of T is T sin 80.0°. Summing torques around the base of the rod,

a

∑ τ = 0:

fb

a

g

f

10 000 N

− 4.00 m 10 000 N cos 60°+T 4.00 m sin 80° = 0 T=

∑ Fx = 0 :

b10 000 Ng cos 60.0° = sin 80.0°

FV

5.08 × 10 3 N

60°

FH − T cos 20.0° = 0

FH

FH = T cos 20.0° = 4.77 × 10 3 N

∑ Fy = 0 :

FIG. P12.49

FV + T sin 20.0°−10 000 N = 0

b

g

and FV = 10 000 N − T sin 20.0° = 8.26 × 10 3 N P12.50

Choosing the origin at R, (1) (2) (3)

R

∑ Fx = + R sin 15.0°−T sin θ = 0 ∑ Fy = 700 − R cos 15.0°+T cos θ = 0 ∑ τ = −700 cos θ a0.180f + T b0.070 0g = 0

Solve the equations for θ from (3), T = 1 800 cos θ from (1), R =

T 90°

15.0°

1 800 sin θ cos θ sin 15.0°

1 800 sin θ cos θ cos 15.0° Then (2) gives 700 − + 1 800 cos 2 θ = 0 sin 15.0° or

θ

cos 2 θ + 0.388 9 − 3.732 sin θ cos θ = 0

θ

18.0 cm 25.0 cm

n FIG. P12.50

Squaring, cos 4 θ − 0.880 9 cos 2 θ + 0.010 13 = 0 Let

u = cos 2 θ then using the quadratic equation, u = 0.011 65 or 0.869 3

Only the second root is physically possible, ∴θ = cos −1 0.869 3 = 21.2° ∴ T = 1.68 × 10 3 N P12.51

and

Choosing torques about R, with −

a

f a

R = 2.34 × 10 3 N

∑τ = 0

fFGH IJK a

f

2L L 350 N + T sin 12.0° − 200 N L = 0 . 2 3

From which, T = 2.71 kN . Let R x = compression force along spine, and from R x = Tx = T cos 12.0° = 2.65 kN .

∑ Fx = 0

FIG. P12.51

369

370 P12.52

Static Equilibrium and Elasticity

(a)

(b)

Just three forces act on the rod: forces perpendicular to the sides of the trough at A and B, and its weight. The lines of action of A and B will intersect at a point above the rod. They will have no torque about this point. The rod’s weight will cause a torque about the point of intersection as in Figure 12.52(a), and the rod will not be in equilibrium unless the center of the rod lies vertically below the intersection point, as in Figure 12.52(b). All three forces must be concurrent. Then the line of action of the weight is a diagonal of the rectangle formed by the trough and the normal forces, and the rod’s center of gravity is vertically above the bottom of the trough.

2

AO =

So cos θ = (a)

2

L 1+

2

cos 2 30.0 ° cos 2 60.0 °

2

F cos GH cos

2

30.0° 2 60.0°

Fg

O

FIG. P12.52(a)

I JK

B Fg

θ A 30.0°

L = 2

60.0° O

FIG. P12.52(b)

AO 1 = and θ = 60.0° . 2 L

Locate the origin at the bottom left corner of the cabinet and let x = distance between the resultant normal force and the front of the cabinet. Then we have

∑ Fx = 200 cos 37.0°− µn = 0 ∑ Fy = 200 sin 37.0°+n − 400 = 0 ∑ τ = na0.600 − xf − 400a0.300f + 200 sin 37.0° a0.600 f

(2)

−200 cos 37.0° 0.400 = 0

(3)

a

From (2), From (3),

From (1), (b)

A

In Figure (b), AO cos 30.0° = BO cos 60.0° and L2 = AO + BO = AO + AO

P12.53

B

(1)

f

n = 400 − 200 sin 37.0° = 280 N

a

f

72.2 − 120 + 280 0.600 − 64.0 280 x = 20.1 cm to the left of the front edge x=

µk =

200 cos 37.0° = 0.571 280

In this case, locate the origin x = 0 at the bottom right corner of the cabinet. Since the cabinet is about to tip, we can use ∑ τ = 0 to find h:

∑ τ = 400a0.300f − a300 cos 37.0°fh = 0

h=

FIG. P12.53

120 = 0.501 m 300 cos 37.0°

Chapter 12

P12.54

(a), (b) Use the first diagram and sum the torques about the lower front corner of the cabinet. ∑ τ = 0 ⇒ − F 1.00 m + 400 N 0.300 m = 0

a

f a fa a400 Nfa0.300 mf = 120 N yielding F = 1.00 m = 0 ⇒ − F f + 120 N = 0 , ∑ x

∑ Fy = 0 ⇒ −400 N + n = 0 ,

0.300 m F

f

400 N 1.00 m

f = 120 N n = 400 N

or so

f

f 120 N Thus, µ s = = = 0.300 . n 400 N (c)

n

Apply F ′ at the upper rear corner and directed so θ + φ = 90.0° to obtain the largest possible lever arm.

θ = tan −1

F’

θ

FG 1.00 m IJ = 59.0° H 0.600 m K

400 N

θ f

a1.00 mf + a0.600 mf + a400 Nfa0.300 mf = 0 2

φ

1.00 m

Thus, φ = 90.0°−59.0° = 31.0° . Sum the torques about the lower front corner of the cabinet: −F′

371

2

n 0.600 m

120 N ⋅ m = 103 N . so F′ = 1.17 m Therefore, the minimum force required to tip the cabinet is

FIG. P12.54

103 N applied at 31.0° above the horizontal at the upper left corner . P12.55

(a)

We can use

∑ Fx = ∑ Fy = 0 and ∑ τ = 0 with pivot point at

the contact on the floor. Then

P T

∑ Fx = T − µ sn = 0 ,

L/2

∑ Fy = n − Mg − mg = 0, and

FL

Mg

I

∑ τ = MgaL cos θ f + mg GH 2 cos θ JK − T aL sin θ f = 0

L/2 mg

Solving the above equations gives M=

FG H

m 2 µ s sin θ − cos θ 2 cos θ − µ s sin θ

n

θ

IJ K

f

FIG. P12.55

This answer is the maximum vaue for M if µ s < cot θ . If µ s ≥ cot θ , the mass M can increase without limit. It has no maximum value, and part (b) cannot be answered as stated either. In the case µ s < cot θ , we proceed. (b)

At the floor, we have the normal force in the y-direction and frictional force in the xdirection. The reaction force then is

b g

R = n 2 + µ sn

2

=

a M + mf g

1 + µ s2 .

At point P, the force of the beam on the rope is

b g

F = T 2 + Mg

2

a

= g M 2 + µ s2 M + m

f

2

.

372 P12.56

Static Equilibrium and Elasticity

(a)

The height of pin B is

1000 N

a10.0 mf sin 30.0° = 5.00 m .

B 10.0 m

The length of bar BC is then

nA

nC 45.0°

30.0°

BC =

5.00 m = 7.07 m. sin 45.0°

C

A

FIG. P12.56(a)

Consider the entire truss:

∑ Fy = n A − 1 000 N + nC = 0 ∑ τ A = −b1 000 N g10.0 cos 30.0°+nC 10.0 cos 30.0°+7.07 cos 45.0°

=0

Which gives nC = 634 N . Then, n A = 1 000 N − nC = 366 N . (b)

(c)

Suppose that a bar exerts on a pin a force not along the length of the bar. Then, the pin exerts on the bar a force with a component perpendicular to the bar. The only other force on the bar is the pin force on the other end. For ∑ F = 0 , this force must also have a component perpendicular to the bar. Then, the total torque on the bar is not zero. The contradiction proves that the bar can only exert forces along its length.

FIG. P12.56(b)

Joint A:

CAB

∑ Fy = 0 : −C AB sin 30.0°+366 N = 0 , so

C AB = 732 N

A TAC nA = 366 N

∑ Fx = 0 : −C AB cos 30.0°+TAC = 0

a

1000 N

f

TAC = 732 N cos 30.0° = 634 N 30.0°

Joint B:

a

f

∑ Fx = 0 : 732 N cos 30.0°−CBC cos 45.0° = 0 C BC =

a732 Nf cos 30.0° = cos 45.0°

897 N

CAB = 732 N

B 45.0° CBC

FIG. P12.56(c)

Chapter 12

P12.57

From geometry, observe that cos θ =

1 4

θ = 75.5°

and

For the left half of the ladder, we have

∑ Fx = T − R x = 0 ∑ Fy = R y + n A − 686 N = 0 ∑ τ top = 686 Na1.00 cos 75.5°f + T a2.00 sin 75.5°f

(1)

−n A 4.00 cos 75.5° = 0

(3)

a

f

(2) FIG. P12.57

For the right half of the ladder we have

∑ Fx = Rx − T = 0 ∑ Fy = nB − Ry = 0 ∑ τ top = nB a 4.00 cos 75.5°f − T a 2.00 sin 75.5°f = 0

(4) (5)

Solving equations 1 through 5 simultaneously yields: (a)

T = 133 N

(b)

n A = 429 N

and

n B = 257 N

(c)

R x = 133 N

and

R y = 257 N

The force exerted by the left half of the ladder on the right half is to the right and downward. P12.58 (a)

x CG = = yCG =

∑ m i xi ∑ mi

b1 000 kg g10.0 m + b125 kgg0 + b125 kg g0 + b125 kg g20.0 m = 1 375 kg

9.09 m

b1 000 kg g10.0 m + b125 kgg20.0 m + b125 kgg20.0 m + b125 kg g0 1 375 kg

= 10.9 m (b)

By symmetry, x CG = 10.0 m There is no change in yCG = 10.9 m

(c) P12.59

vCG =

FG 10.0 m − 9.09 m IJ = H 8.00 s K

0.114 m s

Considering the torques about the point at the bottom of the bracket yields:

b0.050 0 mga80.0 Nf − Fb0.060 0 mg = 0 so

F = 66.7 N .

373

374 P12.60

Static Equilibrium and Elasticity

When it is on the verge of slipping, the cylinder is in equilibrium. and f1 = n 2 = µ sn1 f 2 = µ sn 2 ∑ Fx = 0 :

∑ Fy = 0 : ∑ τ = 0:

P + n1 + f 2 = Fg P = f1 + f 2

As P grows so do f1 and f 2 n 1 Therefore, since µ s = , f1 = 1 2 2 n then (1) P + n1 + 1 = Fg 4 5 So P + n1 = Fg 4 3 Therefore, P = Fg 8 P12.61

(a)

(b) P12.62

(a)

and

a f

F A ∆L Li

=

FIG. P12.60 (2)

FG IJ H K

becomes

F = k ∆L , Young’s modulus is Y = Thus, Y =

n 2 n1 = 2 4 n n 3 P = 1 + 1 = n1 2 4 4 5 4 P+ P = Fg 4 3 f2 =

and

or

8 P = Fg 3

FLi A ∆L

a f

kLi YA and k = A Li

z

za

∆L

∆L

0

0

W = − Fdx = −

f

− kx dx =

YA Li

z

∆L

xdx = YA

a ∆L f

0

2

2 Li

Take both balls together. Their weight is 3.33 N and their CG is at their contact point.

P1

∑ Fx = 0 : + P3 − P1 = 0 P2 = 3.33 N ∑ Fy = 0 : + P2 − 3.33 N = 0 ∑ τ A = 0: − P3 R + P2 R − 3.33 NaR + R cos 45.0°f

a

f

+ P1 R + 2 R cos 45.0° = 0

Substituting,

a

f a

fa + P Ra1 + 2 cos 45.0°f = 0

− P1 R + 3.33 N R − 3.33 N R 1 + cos 45.0°

3.33 N P3

Fg

f

1

P2

a3.33 Nf cos 45.0° = 2 P cos 45.0° 1

P1 = 1.67 N so P3 = 1.67 N (b)

FIG. P12.62(a)

Take the upper ball. The lines of action of its weight, of P1 , and of the normal force n exerted by the lower ball all go through its center, so for rotational equilibrium there can be no frictional force.

∑ Fx = 0 : n cos 45.0°− P1 = 0

1.67 N = 2.36 N cos 45.0° ∑ Fy = 0 : n sin 45.0°−1.67 N = 0 gives the same result

n=

1.67 N

n cos 45.0°

n sin 45.0°

FIG. P12.62(b)

P1

Chapter 12

P12.63

∑ Fy = 0 :

375

+380 N − Fg + 320 N = 0 Fg = 700 N

Take torques about her feet:

a

∑ τ = 0:

f a

f a

f

−380 N 2.00 m + 700 N x + 320 N 0 = 0 x = 1.09 m

P12.64

FIG. P12.63

The tension in this cable is not uniform, so this becomes a fairly difficult problem. dL F = L YA At any point in the cable, F is the weight of cable below that point. Thus, F = µgy where µ is the mass per unit length of the cable. Then, ∆y =

z FGH

Li 0

IJ K

z

L

2 µg i 1 µgLi dL dy = ydy = 2 YA L YA 0

a fa fa f e je j a10.0 − 1.00f m s = F ∆v I F = mG J = b1.00 kg g H ∆t K 0.002 s 2

∆y =

P12.65

(a)

(b) (c)

2. 40 9.80 500 1 = 0.049 0 m = 4.90 cm 2 2.00 × 10 11 3.00 × 10 −4

stress =

4 500 N

4 500 N F = = 4.50 × 10 6 N m 2 0.010 m 0.100 m A

a

fa

f

Yes . This is more than sufficient to break the board.

376 P12.66

Static Equilibrium and Elasticity

The CG lies above the center of the bottom. Consider a disk of water at height y above the bottom. Its radius is

fFGH 30.0y cm IJK = 25.0 cm + 3y yI yI yI F F F Its area is π G 25.0 cm + J . Its volume is π G 25.0 cm + J dy and its mass is πρ G 25.0 cm + J H K3 H K3 H 3K a

25.0 cm + 35.0 − 25.0 cm

2

2

whole mass of the water is

M=

z

z

30 .0 cm

30 .0 cm

y =0

0

dm =

F GH

I JK

50.0 y y 2 + dy 3 9

πρ 625 +

L 50.0 y y O + P M = πρ M625 y + 6 27 PQ MN L 50.0a30.0f a30.0f OP + M = πρ M625a30.0f + 6 27 PQ MN M = π e10 kg cm je 27 250 cm j = 85.6 kg 2

3

30.0

0

2

−3

3

3

3

The height of the center of gravity is yCG =

z

30 .0 cm y =0

= πρ

ydm M

z

30 .0 cm 0

F 625y + 50.0 y GH 3

2

+

I JK

y 3 dy 9 M

LM OP MN PQ 50.0a30.0 f a30.0f πρ L 625a30.0f = + + M 2 9 36 M MN π e10 kg cm j 453 750 cm = πρ 625 y 2 50.0 y 3 y 4 = + + 2 9 36 M 2

−3

30 .0 cm

0

3

3

4

yCG

M 1.43 × 10 3 kg ⋅ cm = = 16.7 cm 85.6 kg

4

OP PQ

2

dy . The

Chapter 12

P12.67

Let θ represent the angle of the wire with the vertical. The radius of the circle of motion is r = 0.850 m sin θ . For the mass:

a

f

v2 = mrω 2 r T sin θ = m 0.850 m sin θ ω 2

a

a f f a f a f π e3.90 × 10 mj e7.00 × 10 N m je1.00 × 10 j AY ⋅ astrainf ω= = ma0.850 mf b1.20 kg ga0.850 mf a

2

P12.68

10

mg

FIG. P12.67

T Further, = Y ⋅ strain or T = AY ⋅ strain A Thus, AY ⋅ strain = m 0.850 m ω 2 , giving

or

θ r

f

−4

T

θ

∑ Fr = mar = m

377

−3

2

ω = 5.73 rad s .

For the bridge as a whole:

D

B

∑ τ A = n A a0f − a13.3 kNfa100 mf + nE a 200 mf = 0

so

nE =

a13.3 kNfa100 mf = 200 m

A

6.66 kN

E C

nA

∑ Fy = n A − 13.3 kN + n E = 0 gives

100 m

100 m 13.3 kN

n A = 13.3 kN − n E = 6.66 kN At Pin A:

∑ Fy = − FAB sin 40.0°+6.66 kN = 0 or

b

6.66 kN = 10.4 kN compression sin 40.0° ∑ Fx = FAC − 10.4 kN cos 40.0° = 0 so

FAB =

a

f

a

f

FAB

a

FAC = 10.4 kN cos 40.0° = 7.94 kN tension At Pin B:

40.0°

g

FAC

f

∑ Fy = a10.4 kN f sin 40.0°− FBC sin 40.0° = 0

a

Thus, FBC = 10. 4 kN tension

a

40.0°

f

b

b g = 10.4 kN atensionf = 7.94 kN atensionf

By symmetry: FDE = FAB = 10.4 kN compression

FEC = FAC

We can check by analyzing Pin C:

∑ Fx = +7.94 kN − 7.94 kN = 0 or 0 = 0 ∑ Fy = 2a10.4 kNf sin 40.0°−13.3 kN = 0 which yields 0 = 0 .

40.0° FBC

FBD = 2 10.4 kN cos 40.0° = 15.9 kN compression

FDC = FBC

FBD

f

∑ Fx = FAB cos 40.0°+ FBC cos 40.0°− FBD = 0

and

nA = 6.66 kN

FAB = 10.4 kN

g 10.4 kN

10.4 kN 40.0°

40.0°

7.94 kN

7.94 kN

13.3 kN

FIG. P12.68

nE

378 P12.69

Static Equilibrium and Elasticity

Member AC is not in pure compression or tension. It also has shear forces present. It exerts a downward force S AC and a tension force FAC on Pin A and on Pin C. Still, this member is in equilibrium.

SAC

25.0 m

FAC

FAC C

A

′ = 0 ⇒ FAC = FAC ′ ∑ Fx = FAC − FAC ∑ τ A = 0: −a14.7 kNfa25.0 mf + S ′AC a50.0 mf = 0 or

14.7 kN

S ′AC = 7.35 kN ∑ Fy = S AC − 14.7 kN + 7.35 kN = 0 ⇒ S AC = 7.35 kN

D

B

Then S AC = S ′AC and we have proved that the loading by the car A is equivalent to one-half the weight of the car pulling down on n A each of pins A and C, so far as the rest of the truss is concerned. For the Bridge as a whole:

SAC

25.0 m

E C 75.0 m

nE

25.0 m 14.7 kN

∑ τ A = 0:

a

fa

f a

f

7.35 kN

− 14.7 kN 25.0 m + n E 100 m = 0 n E = 3.67 kN

FAB 30.0°

∑ Fy = n A − 14.7 kN + 3.67 kN = 0

FAC

n A = 11.0 kN At Pin A:

nA = 11.0 kN

∑ Fy = −7.35 kN + 11.0 kN − FAB sin 30.0° = 0

b

FAB = 7.35 kN compression

FBD

g

30.0°

∑ Fx = FAC − a7.35 kN f cos 30.0° = 0

a

FAC = 6.37 kN tension At Pin B:

7.35 kN

60.0° FBC

f

4.24 kN

∑ Fy = −a7.35 kN f sin 30.0°− FBC sin 60.0° = 0

60.0°

a f ∑ F = a7.35 kNf cos 30.0°+a 4.24 kN f cos 60.0°− F F = 8.49 kN bcompressiong FBC = 4.24 kN tension x

BD

FCD 60.0°

6.37 kN

FCE

=0 7.35 kN

BD

At Pin C:

∑ Fy = a 4.24 kNf sin 60.0°+ FCD sin 60.0°−7.35 kN = 0

a

FCD = 4.24 kN tension

f

30.0°

∑ Fx = −6.37 kN − a4.24 kNf cos 60.0°+a 4.24 kNf cos 60.0°+ FCE = 0

a

f

FCE = 6.37 kN tension

At Pin E:

6.37 kN 3.67 kN

FIG. P12.69

∑ Fy = − FDE sin 30.0°+3.67 kN = 0

b

FDE = 7.35 kN compression or ∑ Fx = −6.37 kN − FDE cos 30.0° = 0 which gives FDE = 7.35 kN as before.

FDE

g

379

Chapter 12

P12.70

(1) (2)

ph = Iω

ω p

p = MvCM

h

vCM

If the ball rolls without slipping, Rω = vCM So, h = P12.71

(a)

2 Iω Iω I = = = R 5 p MvCM MR

FIG. P12.70

If the acceleration is a, we have Px = ma and Py + n − Fg = 0 . Taking the origin at the center of

H L d

gravity, the torque equation gives

a

f

CG

Py L − d + Px h − nd = 0 .

P h

Solving these equations, we find

(b)

F d − ah I . L GH g JK ah e 2.00 m s ja1.50 mf = = If P = 0 , then d =

(c)

Using the given data, Px = −306 N and Py = 553 N .

Py =

n

Fg

Fgy

FIG. P12.71

2

y

9.80 m s 2

g

e

0.306 m .

j

Thus, P = −306 i + 553 j N . *P12.72

When the cyclist is on the point of tipping over forward, the normal force on the rear wheel is zero. Parallel to the plane we have f1 − mg sin θ = ma . Perpendicular to the plane, n1 − mg cos θ = 0 . Torque about the center of mass:

af a

f a

mg

f

mg 0 − f1 1.05 m + n1 0.65 m = 0 .

f1

Combining by substitution,

FIG. P12.72

ma = f1 − mg sin θ =

FG H

n1 0.65 m 0.65 m − mg sin θ = mg cos θ − mg sin θ 1.05 m 1.05 m

IJ K

0.65 − sin 20° = 2.35 m s 2 1.05 When the car is on the point of rolling over, the normal force on its inside wheels is zero. a = g cos 20°

*P12.73

∑ Fy = ma y :

n − mg = 0

∑ Fx = ma x :

f=

n1

mg h

mv 2 R

Take torque about the center of mass: fh − n Then by substitution

2 mv max

R

h−

mgd =0 2

f

d = 0. 2 v max =

mg

d gdR 2h

FIG. P12.73

A wider wheelbase (larger d) and a lower center of mass (smaller h) will reduce the risk of rollover.

380

Static Equilibrium and Elasticity

ANSWERS TO EVEN PROBLEMS P12.2

Fy + R y − Fg = 0 ; Fx − R x = 0 ; Fy A cos θ − Fg

FG A IJ cos θ − F A sinθ = 0 H 2K

P12.40

(a) 0.400 mm; (b) 40.0 kN; (c) 2.00 mm; (d) 2.40 mm; (e) 48.0 kN

P12.42

at A: Mg

P12.44

(a) 160 N to the right; (b) 13.2 N to the right; (c) 292 N up; (d) 192 N

x

sin β sin α ; at B: Mg sin α + β sin α + β

b

g

b

g

P12.4

see the solution

P12.6

0.750 m

P12.8

a2.54 m, 4.75 mf

P12.10

(a) 9.00 g; (b) 52.5 g; (c) 49.0 g

P12.46

1.46 kN ; 1.33 i + 2.58 j kN

P12.12

(a) 392 N; (b) 339 i + 0 j N

e

P12.48

0.789

P12.14

(a) f =

P12.50

T = 1.68 kN ; R = 2.34 kN; θ = 21.2°

P12.52

(a) see the solution; (b) 60.0°

P12.54

(a) 120 N; (b) 0.300; (c) 103 N at 31.0° above the horizontal to the right

P12.56

(a), (b) see the solution; (c) C AB = 732 N ; TAC = 634 N ; C BC = 897 N

P12.58

(a) 9.09 m, 10.9 m ; (b) 10.0 m, 10.9 m ; (c) 0.114 m s to the right

ng

j

LM m g + m gx OP cot θ ; N2 L Q e + j cot θ = bm + m g g ; (b) µ = m +m 1

2

m1 2

1

2

m2d L

1

2

P12.16

see the solution; 0.643 m

P12.18

36.7 N to the left ; 31.2 N to the right

P12.20

(a) 35.5 kN; (b) 11.5 kN to the right; (c) 4.19 kN down

P12.22

(a) 859 N; (b) 104 kN at 36.9° above the horizontal to the left

P12.24

3L 4

P12.60

e

a

j

f

a

f

3 Fg 8

P12.62

(a) P1 = 1.67 N ; P2 = 3.33 N ; P3 = 1.67 N ; (b) 2.36 N

P12.26

(a) see the solution; (b) θ decreases ; (c) R decreases

P12.64

4.90 cm

P12.28

(a) 73.6 kN; (b) 2.50 mm

P12.66

16.7 cm above the center of the bottom

P12.30

~ 1 cm

P12.68

P12.32

9.85 × 10 −5

P12.34

0.029 3 mm

C AB = 10.4 kN ; TAC = 7.94 kN ; TBC = 10.4 kN ; C BD = 15.9 kN ; C DE = 10.4 kN ; TDC = 10. 4 kN ; TEC = 7.94 kN

P12.36

(a) −0.053 8 m3 ; (b) 1.09 × 10 3 kg m3 ; (c) Yes, in most practical circumstances

P12.38

(a) 53.1°; (b) 1.04 kN; (c) 0.126 m, 51.2°; (d) 1.07 kN; (e) 0.129 m, 51.1°; (f) 51.1°

P12.70

2 R 5

P12.72

2.35 m s 2

13 Universal Gravitation CHAPTER OUTLINE 13.1 13.2 13.3 13.4 13.5 13.6 13.7

Newton’s Law of Universal Gravitation Measuring the Gravitational Constant Free-Fall Acceleration and the Gravitational Force Kepler’s Laws and the Motion of Planets The Gravitational Field Gravitational Potential Energy Energy Considerations in Planetary and Satellite Motion

ANSWERS TO QUESTIONS Q13.1

Because g is the same for all objects near the Earth’s surface. The larger mass needs a larger force to give it just the same acceleration.

Q13.2

To a good first approximation, your bathroom scale reading is unaffected because you, the Earth, and the scale are all in free fall in the Sun’s gravitational field, in orbit around the Sun. To a precise second approximation, you weigh slightly less at noon and at midnight than you do at sunrise or sunset. The Sun’s gravitational field is a little weaker at the center of the Earth than at the surface subsolar point, and a little weaker still on the far side of the planet. When the Sun is high in your sky, its gravity pulls up on you a little more strongly than on the Earth as a whole. At midnight the Sun pulls down on you a little less strongly than it does on the Earth below you. So you can have another doughnut with lunch, and your bedsprings will still last a little longer.

Q13.3

Kepler’s second law states that the angular momentum of the Earth is constant as the Earth orbits the sun. Since L = mωr , as the orbital radius decreases from June to December, then the orbital speed must increase accordingly.

Q13.4

Because both the Earth and Moon are moving in orbit about the Sun. As described by Fgravitational = ma centripetal , the gravitational force of the Sun merely keeps the Moon (and Earth) in a nearly circular orbit of radius 150 million kilometers. Because of its velocity, the Moon is kept in its orbit about the Earth by the gravitational force of the Earth. There is no imbalance of these forces, at new moon or full moon.

Q13.5

Air resistance causes a decrease in the energy of the satellite-Earth system. This reduces the diameter of the orbit, bringing the satellite closer to the surface of the Earth. A satellite in a smaller orbit, however, must travel faster. Thus, the effect of air resistance is to speed up the satellite!

Q13.6

Kepler’s third law, which applies to all planets, tells us that the period of a planet is proportional to r 3 2 . Because Saturn and Jupiter are farther from the Sun than Earth, they have longer periods. The Sun’s gravitational field is much weaker at a distant Jovian planet. Thus, an outer planet experiences much smaller centripetal acceleration than Earth and has a correspondingly longer period.

381

382 Q13.7

Universal Gravitation

Ten terms are needed in the potential energy: U = U 12 + U 13 + U 14 + U 15 + U 23 + U 24 + U 25 + U 34 + U 35 + U 45 . With N particles, you need

N

∑ ai − 1f = i =1

Q13.8

N2 − N terms. 2

No, the escape speed does not depend on the mass of the rocket. If a rocket is launched at escape speed, then the total energy of the rocket-Earth system will be zero. When the separation distance GM E m 1 = 0 , the mass becomes infinite U = 0 the rocket will stop K = 0 . In the expression mv 2 − 2 r m of the rocket divides out.

a

f

a

f

Q13.9

It takes 100 times more energy for the 10 5 kg spacecraft to reach the moon than the 10 3 kg spacecraft. Ideally, each spacecraft can reach the moon with zero velocity, so the only term that need be analyzed is the change in gravitational potential energy. U is proportional to the mass of the spacecraft.

Q13.10

The escape speed from the Earth is 11.2 km/s and that from the Moon is 2.3 km/s, smaller by a factor of 5. The energy required—and fuel—would be proportional to v 2 , or 25 times more fuel is required to leave the Earth versus leaving the Moon.

Q13.11

The satellites used for TV broadcast are in geosynchronous orbits. The centers of their orbits are the center of the Earth, and their orbital planes are the Earth’s equatorial plane extended. This is the plane of the celestial equator. The communication satellites are so far away that they appear quite close to the celestial equator, from any location on the Earth’s surface.

Q13.12

For a satellite in orbit, one focus of an elliptical orbit, or the center of a circular orbit, must be located at the center of the Earth. If the satellite is over the northern hemisphere for half of its orbit, it must be over the southern hemisphere for the other half. We could share with Easter Island a satellite that would look straight down on Arizona each morning and vertically down on Easter Island each evening.

Q13.13

The absolute value of the gravitational potential energy of the Earth-Moon system is twice the kinetic energy of the moon relative to the Earth.

Q13.14

In a circular orbit each increment of displacement is perpendicular to the force applied. The dot product of force and displacement is zero. The work done by the gravitational force on a planet in an elliptical orbit speeds up the planet at closest approach, but negative work is done by gravity and the planet slows as it sweeps out to its farthest distance from the Sun. Therefore, net work in one complete orbit is zero.

Q13.15

Every point q on the sphere that does not lie along the axis connecting the center of the sphere and the particle will have companion point q’ for which the components of the gravitational force perpendicular to the axis will cancel. Point q’ can be found by rotating the sphere through 180° about the axis. The forces will not necessarily cancel if the mass is not uniformly distributed, unless the center of mass of the non-uniform sphere still lies along the axis.

q

Fpq Fpq q’ (behind the sphere)

FIG. Q13.15

p

Chapter 13

Q13.16

Speed is maximum at closest approach. Speed is minimum at farthest distance.

Q13.17

Set the universal description of the gravitational force, Fg = Fg = ma gravitational , where M X and R X

383

GM X m

, equal to the local description, R X2 are the mass and radius of planet X, respectively, and m is the

mass of a “test particle.” Divide both sides by m. Q13.18

The gravitational force of the Earth on an extra particle at its center must be zero, not infinite as one interpretation of Equation 13.1 would suggest. All the bits of matter that make up the Earth will pull in different outward directions on the extra particle.

Q13.19

Cavendish determined G. Then from g =

Q13.20

The gravitational force is conservative. An encounter with a stationary mass cannot permanently speed up a spacecraft. Jupiter is moving. A spacecraft flying across its orbit just behind the planet will gain kinetic energy as the planet’s gravity does net positive work on it.

Q13.21

Method one: Take measurements from an old kinescope of Apollo astronauts on the moon. From the motion of a freely falling object or from the period of a swinging pendulum you can find the acceleration of gravity on the moon’s surface and calculate its mass. Method two: One could determine the approximate mass of the moon using an object hanging from an extremely sensitive balance, with knowledge of the position and distance of the moon and the radius of the Earth. First weigh the object when the moon is directly overhead. Then weigh of the object when the moon is just rising or setting. The slight difference between the measured weights reveals the cause of tides in the Earth’s oceans, which is a difference in the strength of the moon’s gravity between different points on the Earth. Method three: Much more precisely, from the motion of a spacecraft in orbit around the moon, its mass can be determined from Kepler’s third law.

Q13.22

The spacecraft did not have enough fuel to stop dead in its high-speed course for the Moon.

GM , one may determine the mass of the Earth. R2

SOLUTIONS TO PROBLEMS Section 13.1 P13.1

Newton’s Law of Universal Gravitation

For two 70-kg persons, modeled as spheres, Fg =

P13.2

F = m1 g =

g=

Gm 2 r

2

Gm1 m 2 r

2

e6.67 × 10 =

−11

jb

gb

N ⋅ m 2 kg 2 70 kg 70 kg

a2 mf

2

g

~ 10 −7 N .

Gm1 m 2

=

r2

e6.67 × 10

−11

je a100 mf

N ⋅ m 2 kg 2 4.00 × 10 4 × 10 3 kg 2

j=

2.67 × 10 −7 m s 2

384 P13.3

Universal Gravitation

(a)

At the midpoint between the two objects, the forces exerted by the 200-kg and 500-kg objects are oppositely directed,

(b)

Gm1 m 2

and from

Fg =

we have

∑F =

r2

b

gb

G 50.0 kg 500 kg − 200 kg

a0.200 mf

2

g=

2.50 × 10 −5 N toward the 500-kg object.

At a point between the two objects at a distance d from the 500-kg objects, the net force on the 50.0-kg object will be zero when

b gb g = Gb50.0 kggb500 kg g d a0.400 m − df

G 50.0 kg 200 kg 2

d = 0.245 m

or P13.4

2

m1 + m 2 = 5.00 kg F =G

m 2 = 5.00 kg − m1 m1 m 2 r

2

b5.00 kg gm g bm − 3.00 kg gbm

e

⇒ 1.00 × 10 −8 N = 6.67 × 10 −11 N ⋅ m 2 kg 2

2 1 − m1 =

b

e1.00 × 10

−8

6.67 × 10

je

N 0.040 0 m 2

−11

2

N ⋅ m kg

2

j m ba50..00200kgm−f m g 1

1

2

j = 6.00 kg

2

Thus, m12 − 5.00 kg m1 + 6.00 kg = 0 or

1

1

g

− 2.00 kg = 0

giving m1 = 3.00 kg, so m 2 = 2.00 kg . The answer m1 = 2.00 kg and m 2 = 3.00 kg is physically equivalent. P13.5

The force exerted on the 4.00-kg mass by the 2.00-kg mass is directed upward and given by F24 = G

m4m2  j = 6.67 × 10 −11 N ⋅ m 2 kg 2 2 r24

e

j b4.00a3kg.00gbm2.00f kg g j 2

= 5.93 × 10 −11 j N The force exerted on the 4.00-kg mass by the 6.00-kg mass is directed to the left F64 = G

m 4 m6 2 r64

e− ij = e−6.67 × 10

−11

N ⋅ m 2 kg 2

j b4.00a4kg.00gbm6.00f kg g i 2

FIG. P13.5

= −10.0 × 10 −11 iN Therefore, the resultant force on the 4.00-kg mass is F4 = F24 + F64 =

e−10.0i + 5.93 jj × 10

−11

N .

Chapter 13

P13.6

(a)

385

The Sun-Earth distance is 1.496 × 10 11 m and the Earth-Moon distance is 3.84 × 10 8 m , so the distance from the Sun to the Moon during a solar eclipse is 1.496 × 10 11 m − 3.84 × 10 8 m = 1.492 × 10 11 m M S = 1.99 × 10 30 kg

The mass of the Sun, Earth, and Moon are

M E = 5.98 × 10 24 kg M M = 7.36 × 10 22 kg

and

We have FSM =

(b)

(c)

FEM =

FSE

e6.67 × 10

e6.67 × 10 =

Gm1 m 2 r2

−11

e6.67 × 10 je1.99 × 10 je7.36 × 10 j = = e1.492 × 10 j −11

30

22

11 2

je

je

j=

je

j=

N ⋅ m 2 kg 2 5.98 × 10 24 7.36 × 10 22

e3.84 × 10 j

8 2

−11

je

N ⋅ m 2 kg 2 1.99 × 10 30 5.98 × 10 24

e1.496 × 10 j

11 2

4.39 × 10 20 N

1.99 × 10 20 N

3.55 × 10 22 N

Note that the force exerted by the Sun on the Moon is much stronger than the force of the Earth on the Moon. In a sense, the Moon orbits the Sun more than it orbits the Earth. The Moon’s path is everywhere concave toward the Sun. Only by subtracting out the solar orbital motion of the Earth-Moon system do we see the Moon orbiting the center of mass of this system.

Section 13.2

P13.7

Measuring the Gravitational Constant

b

ge

j

1.50 kg 15.0 × 10 −3 kg GMm −11 2 2 F= = 6.67 × 10 N ⋅ m kg = 7.41 × 10 −10 N 2 −2 r2 4.50 × 10 m

e

j

e

j

386 P13.8

Universal Gravitation

Let θ represent the angle each cable makes with the vertical, L the cable length, x the distance each ball scrunches in, and d = 1 m the original distance between them. Then r = d − 2 x is the separation of the balls. We have

∑ Fy = 0 :

T cos θ − mg = 0

∑ Fx = 0 :

T sin θ −

Then

tan θ =

Gmm =0 r2

FIG. P13.8

Gmm r 2 mg

x 2

L −x

2

=

a

Gm

g d − 2x

f

a

x d − 2x

2

f

2

=

Gm 2 L − x2 . g

Gm is numerically small. There are two possibilities: either x is small or else d − 2 x is g

The factor small.

Possibility one: We can ignore x in comparison to d and L, obtaining

e6.67 × 10 xa1 mf = 2

−11

jb

N ⋅ m 2 kg 2 100 kg

e9.8 m s j 2

e

g 45 m

x = 3.06 × 10 −8 m.

j

The separation distance is r = 1 m − 2 3.06 × 10 −8 m = 1.000 m − 61.3 nm . Possibility two: If d − 2 x is small, x ≈ 0.5 m and the equation becomes

a0.5 mfr = e6.67 × 10 b9N.8 ⋅Nm kgkgg jb100 kgg a45 mf − a0.5 mf −11

2

2

2

2

2

r = 2.74 × 10 −4 m .

For this answer to apply, the spheres would have to be compressed to a density like that of the nucleus of atom.

Section 13.3 P13.9

a=

Free-Fall Acceleration and the Gravitational Force MG

b 4R g E

2

=

9.80 m s 2 = 0.613 m s 2 16.0

toward the Earth.

e j = 4 πGρR 4 πR 3

GM Gρ 3 g= 2 = R R2

P13.10 If

gM 1 = = gE 6

then

g ρM = M gE ρE

3

4πGρ M R M 3 4πGρ E RE 3

F I F R I = FG 1 IJ a4f = GH JK GH R JK H 6 K E

M

2 . 3

Chapter 13

P13.11

(a)

At the zero-total field point, so

(b)

GmM E rE2

387

GmM M

=

rM2

MM r 7.36 × 10 22 = rE = E 24 ME 9 .01 5.98 × 10 r rE + rM = 3.84 × 10 8 m = rE + E 9.01 3.84 × 10 8 m rE = = 3.46 × 10 8 m 1.11 rM = rE

At this distance the acceleration due to the Earth’s gravity is gE =

GM E rE2

e6.67 × 10 =

−11

je mj

N ⋅ m 2 kg 2 5.98 × 10 24 kg

e3.46 × 10

8

j

2

g E = 3.34 × 10 −3 m s 2 directed toward the Earth

Section 13.4 P13.12

Kepler’s Laws and the Motion of Planets

b

g

3 2πr 2π 384 400 × 10 m = = 1.02 × 10 3 m s . T 27.3 × 86 400 s

(a)

v=

(b)

In one second, the Moon falls a distance

b

g

e e

j a f j 2

3 1 1 v 2 2 1 1.02 × 10 x = at 2 = t = × 1.00 2 2 r 2 3.844 × 10 8

2

= 1.35 × 10 −3 m = 1.35 mm .

The Moon only moves inward 1.35 mm for every 1020 meters it moves along a straight-line path. P13.13

Applying Newton’s 2nd Law, GMM

a 2r f

2

=

∑ F = ma yields Fg = ma c for each star:

Mv 2 r

M=

or

4v 2 r . G

We can write r in terms of the period, T, by considering the time and distance of one complete cycle. The distance traveled in one orbit is the circumference of the stars’ common orbit, so 2πr = vT . Therefore M=

e

FG IJ H K

4v 2 r 4v 2 vT = G G 2π

ja 3

fb

FIG. P13.13

g

3 2 v 3 T 2 220 × 10 m s 14.4 d 86 400 s d = = 1.26 × 10 32 kg = 63.3 solar masses so, M = −11 2 2 πG π 6.67 × 10 N ⋅ m kg

e

j

388 P13.14

Universal Gravitation

Since speed is constant, the distance traveled between t1 and t 2 is equal to the distance traveled between t 3 and t 4 . The area of a triangle is equal to one-half its (base) width across one side times its (height) dimension perpendicular to that side. So

b

g

b

1 1 bv t 2 − t1 = bv t 4 − t 3 2 2

g

states that the particle’s radius vector sweeps out equal areas in equal times. P13.15

T2 =

M=

4π 2 a 3 GM 4π 2 a 3 GT 2

(Kepler’s third law with m r ,

∆a =

Across the planet,

∆g 2 ∆a 2. 22 × 10 −6 m s 2 = = = 2. 26 × 10 −7 g g 9.80 m s 2

d

3

Energy conservation for the two-sphere system from release to contact: −

Gmm Gmm 1 1 =− + mv 2 + mv 2 R 2r 2 2

Gm (a)

(b)

FG 1 − 1 IJ = v H 2r R K

FG L 1 − 1 OIJ H MN 2r R PQK

2

12

v = Gm

The injected impulse is the final momentum of each sphere,

FG L 1 − 1 OIJ H MN 2r R PQK

mv = m 2 2 Gm

12

LM N

= Gm 3

FG 1 − 1 IJ OP H 2r R K Q

12

.

If they now collide elastically each sphere reverses its velocity to receive impulse

LM N

a f

mv − − mv = 2mv = 2 Gm 3 P13.50

2GM M r

LM 1 MN ad − r f

FG 1 − 1 IJ OP H 2r R K Q

12

Momentum is conserved: m1 v 1i + m 2 v 2i = m1 v 1 f + m 2 v 2 f 0 = Mv 1 f + 2 Mv 2 f v2 f = −

1 v1 f 2

Energy is conserved:

aK + U f + ∆E = aK + U f i

0− −

f

Gm 1 m 2 Gm1 m 2 1 1 + 0 = m1 v12 f + m 2 v 22 f − rf ri 2 2

a f

a fFGH

GM 2 M 1 1 1 v1 f = Mv12 f + 2 M 12 R 2 2 2

v1 f =

2 GM R 3

v2 f =

IJ K

2

−

a f

GM 2 M 4R

1 1 GM v1 f = R 2 3

Chapter 13

P13.51

e1.25 × 10 =

(a)

v2 ac = r

(b)

diff = 10.2 − 9.90 = 0.312 m s 2 =

ac

6.67 × 10

P13.52

(a)

11

−11

2

ms

1.53 × 10

N ⋅ m kg

11

j

2

= 10.2 m s 2

m

GM r2

e0.312 m s je1.53 × 10 mj M= 2

6

401

2

= 1.10 × 10 32 kg

2

FIG. P13.51 GM E

The free-fall acceleration produced by the Earth is g =

r2

= GM E r −2 (directed downward)

a f

dg = GM E −2 r −3 = −2GM E r −3 . dr

Its rate of change is

The minus sign indicates that g decreases with increasing height. dg 2GM E =− . dr RE3

At the Earth’s surface, (b)

For small differences, ∆g ∆r

*P13.53

=

∆g h

=

2GM E

∆g =

Thus,

RE3

e

ja

je

2GM E h

f=

2 6.67 × 10 −11 N ⋅ m 2 kg 2 5.98 × 10 24 kg 6.00 m

RE3

1.85 × 10 −5 m s 2

(c)

∆g =

(a)

Each bit of mass dm in the ring is at the same distance from the object at A. The separate GmM ring Gmdm to the system energy add up to − . When the object is at A, contributions − r r this is

e6.37 × 10 mj 6

3

−6.67 × 10 −11 N ⋅ m 2 1 000 kg 2.36 × 10 20 kg kg 2 (b)

8

8

2

= −7.04 × 10 4 J .

When the object is at the center of the ring, the potential energy is −

(c)

e1 × 10 mj + e2 × 10 mj 2

6.67 × 10 −11 N ⋅ m 2 1 000 kg 2.36 × 10 20 kg kg 2 1 × 10 8 m

= −1.57 × 10 5 J .

Total energy of the object-ring system is conserved:

e K + U j = eK + U j g

g

A

B

1 0 − 7.04 × 10 J = 1 000 kgv B2 − 1.57 × 10 5 J 2 4

vB =

F 2 × 8.70 × 10 J I GH 1 000 kg JK 4

12

= 13.2 m s

402 P13.54

Universal Gravitation

To approximate the height of the sulfur, set mv 2 = mg Io h 2

h = 70 000 m

GM = 1.79 m s 2 r2

g Io =

a fb

g

b

v = 2 1.79 70 000 ≈ 500 m s over 1 000 mi h

v = 2 g Io h

g

A more precise answer is given by 1 GMm GMm =− mv 2 − 2 r1 r2 1 2 v = 6.67 × 10 −11 8.90 × 10 22 2

e

P13.55

*P13.56

je

jFGH 1.82 1× 10

−

6

1 1.89 × 10

6

IJ K

From the walk, 2πr = 25 000 m. Thus, the radius of the planet is r = From the drop:

∆y =

so,

g=

a

1 2 1 gt = g 29. 2 s 2 2

a

f

f = 3.28 × 10

2 1.40 m

a

29.2 s

f

2

2

v = 492 m s 25 000 m = 3.98 × 10 3 m 2π

= 1.40 m −3

m s2 =

MG r2

∴ M = 7.79 × 10 14 kg

The distance between the orbiting stars is d = 2r cos 30° = 3 r since 3 . The net inward force on one orbiting star is cos 30° = 2 Gmm GMm Gmm mv 2 30 °+ + 30 ° = cos cos r d2 r2 d2 2 2 Gm 2 cos 30° GM 4π r + 2 = r rT 2 3r 2 m 4π 2 r 3 +M = G T2 3

FG H

T2 =

r

d

60°

F F

4π 2 r 3

G M+

30°

r

IJ K

e j F r T = 2π G GH GeM + m 3 3

m 3

P13.57

30°

F

I JJ jK

12

FIG. P13.56

For a 6.00 km diameter cylinder, r = 3 000 m and to simulate 1 g = 9.80 m s 2 v2 = ω 2r r g = 0.057 2 rad s ω= r g=

The required rotation rate of the cylinder is

1 rev 110 s

(For a description of proposed cities in space, see Gerard K. O’Neill in Physics Today, Sept. 1974.)

Chapter 13

P13.58

(a)

G has units

N ⋅ m 2 kg ⋅ m ⋅ m 2 m3 = = kg 2 s 2 ⋅ kg 2 s 2 ⋅ kg

and dimensions G =

L3 . T2 ⋅ M

The speed of light has dimensions of c = as angular momentum or h =

M ⋅ L2 . T

L , and Planck’s constant has the same dimensions T

We require G p c q h r = L , or L3 p T −2 p M − p Lq T − q M r L2 r T − r = L1 M 0 T 0 . Thus, 3 p + q + 2r = 1 −2 p − q − r = 0 −p + r = 0 which reduces (using r = p ) to

3p + q + 2p = 1 −2 p − q − p = 0

These equations simplify to Then, 5 p − 3 p = 1 , yielding p =

5 p + q = 1 and q = −3 p . 1 3 1 , q = − , and r = . 2 2 2

Therefore, Planck length = G 1 2 c − 3 2 h1 2 . (b) P13.59

e6.67 × 10 j e3 × 10 j e6.63 × 10 j −11 1 2

8 −3 2

−34 1 2

e

= 1.64 × 10 −69

Gm p m 0 1 2 = m 0 v esc 2 R v esc =

2Gm p R

4 With m p = ρ πR 3 , we have 3 v esc = = So, v esc ∝ R .

403

2Gρ 34 πR 3 R 8πGρ R 3

j

12

= 4.05 × 10 −35 m ~ 10 −34 m

404 *P13.60

Universal Gravitation

For both circular orbits, GM E m

∑ F = ma :

=

r2

v=

mv 2 r

GM E r

FIG. P13.60

e6.67 × 10 N ⋅ m kg je5.98 × 10 e6.37 × 10 m + 2 × 10 mj

kg

j=

7.79 × 10 3 m s .

N ⋅ m 2 kg 2 5.98 × 10 24 kg

j=

7.85 × 10 3 m s .

−11

(a)

(b)

The original speed is vi =

The final speed is

vi =

2

2

6

e6.67 × 10

−11

e

24

5

je

6

j

6.47 × 10 m

The energy of the satellite-Earth system is K +Ug =

GM E m GM E m 1 GM E GM E 1 = m − =− mv 2 − 2 2 2r r r r

g=

−3.04 × 10 9 J .

jb

g=

−3.08 × 10 9 J .

Originally

(d)

Finally

Ef = −

(e)

Thus the object speeds up as it spirals down to the planet. The loss of gravitational energy is so large that the total energy decreases by

−11

N ⋅ m 2 kg 2 5.98 × 10 24 kg 100 kg

e

i

e6.67 × 10

je

jb

(c)

e6.67 × 10 E =−

6

j

2 6.57 × 10 m −11

je

N ⋅ m 2 kg 2 5.98 × 10 24 kg 100 kg

e

6

j

2 6. 47 × 10 m

e

j

Ei − E f = −3.04 × 10 9 J − −3.08 × 10 9 J = 4.69 × 10 7 J . (f)

The only forces on the object are the backward force of air resistance R, comparatively very small in magnitude, and the force of gravity. Because the spiral path of the satellite is not perpendicular to the gravitational force, one component of the gravitational force pulls forward on the satellite to do positive work and make its speed increase.

405

Chapter 13

P13.61

(a)

At infinite separation U = 0 and at rest K = 0 . Since energy of the two-planet system is conserved we have, 0=

Gm1 m 2 1 1 m1 v12 + m 2 v 22 − d 2 2

(1)

The initial momentum of the system is zero and momentum is conserved. 0 = m1 v1 − m 2 v 2

Therefore,

b

Therefore, (a)

K1 =

1 m1 v12 = 1.07 × 10 32 J 2

b

g

g

d

and

v 2 = 2.58 × 10 3 m s

and

K2 =

1 m 2 v 22 = 2.67 × 10 31 J 2

The net torque exerted on the Earth is zero. Therefore, the angular momentum of the Earth is conserved; mra v a = mrp v p and v a = v p

(b)

b

2G d m1 + m 2

Substitute given numerical values into the equation found for v1 and v 2 in part (a) to find v1 = 1.03 × 10 4 m s

P13.62

v 2 = m1

and

g

2G m 1 + m 2

b g

v r = v1 − − v 2 =

Relative velocity (b)

2G d m1 + m 2

v1 = m 2

Combine equations (1) and (2):

(2)

Kp =

F r I = e3.027 × 10 GH r JK p

je

e

je

ms

a

1 1 mv p2 = 5.98 × 10 24 3.027 × 10 4 2 2

e

4

j

2

471 I JK = jFGH 11..521

2.93 × 10 4 m s

= 2.74 × 10 33 J

je

j

6.673 × 10 −11 5.98 × 10 24 1.99 × 10 30 GmM =− = −5.40 × 10 33 J Up = − 11 rp 1.471 × 10 (c)

Using the same form as in part (b), K a = 2.57 × 10 33 J and U a = −5.22 × 10 33 J . Compare to find that K p + U p = −2.66 × 10 33 J and K a + U a = −2.65 × 10 33 J . They agree.

406 P13.63

Universal Gravitation

(a)

The work must provide the increase in gravitational energy W = ∆U g = U gf − U gi GM E M p

=−

rf GM E M p

=−

RE + y

= GM E M p =

+ +

F1 GH R

F 6.67 × 10 GH kg

GM E M p ri GM E M p RE 1 RE + y

−

E

−11

N ⋅ m2

2

I JK

I 5.98 × 10 JK e

24

jb

kg 100 kg

gFGH 6.37 ×110

6

m

−

W = 850 MJ (b)

In a circular orbit, gravity supplies the centripetal force: GM E M p

bR Then,

E

+y

Mpv2

=

g bR 2

E

+y

g

1 1 GM E M p Mp v2 = 2 2 RE + y

b

g

So, additional work = kinetic energy required =

e

je

jb

−11 N ⋅ m 2 5.98 × 10 24 kg 100 kg 1 6.67 × 10 2 kg 2 7.37 × 10 6 m

e je

j

∆W = 2.71 × 10 9 J P13.64

Centripetal acceleration comes from gravitational acceleration. v 2 M c G 4π 2 r 2 = 2 = r r T 2r GM c T 2 = 4π 2 r 3

e6.67 × 10 ja20fe1.99 × 10 je5.00 × 10 j −11

−3 2

30

= 4π 2 r 3

rorbit = 119 km P13.65

e

j

15 2πr 2π 30 000 × 9.46 × 10 m = = 7 × 10 15 s = 2 × 10 8 yr v 2.50 × 10 5 m s

(a)

T=

(b)

4π 2 30 000 × 9.46 × 10 15 m 4π 2 a 3 M= = GT 2 6.67 × 10 −11 N ⋅ m 2 kg 2 7.13 × 10 15 s

e

e

j

3

je

M = 1.34 × 10 11 solar masses ~ 10 11 solar masses The number of stars is on the order of 10 11 .

j

2

= 2.66 × 10 41 kg

g

1 7.37 × 10 6 m

IJ K

Chapter 13

P13.66

(a)

From the data about perigee, the energy of the satellite-Earth system is E= or

(b)

407

6.67 × 10 je5.98 × 10 ja1.60f j e 7.02 × 10

a fe

GM E m 1 1 = 1.60 8.23 × 10 3 mv p2 − rp 2 2

2

−11

24

−

6

E = −3.67 × 10 7 J

b

ge

je

j

L = mvr sin θ = mv p rp sin 90.0° = 1.60 kg 8.23 × 10 3 m s 7.02 × 10 6 m = 9.24 × 10 10 kg ⋅ m 2 s

(c)

Since both the energy of the satellite-Earth system and the angular momentum of the Earth are conserved, at apogee we must have

1 GMm =E mv a2 − 2 ra

and

mv a ra sin 90.0° = L .

Thus,

6.67 × 10 −11 5.98 × 10 24 1.60 1 2 = −3.67 × 10 7 J 1.60 v a − 2 rs

and

b1.60 kg gv r

e

a f

a a

ja f

je

= 9.24 × 10 10 kg ⋅ m 2 s .

e

ja fa f

je

6.67 × 10 −11 5.98 × 10 24 1.60 1.60 v a 1 2 Solving simultaneously, = −3.67 × 10 7 1.60 v a − 2 9.24 × 10 10

a f

0.800 v a2 − 11 046 v a + 3.672 3 × 10 7 = 0

which reduces to

va =

so

11 046 ±

b11 046g − 4a0.800fe3.672 3 × 10 j . 2a0.800 f 2

7

This gives v a = 8 230 m s or 5 580 m s . The smaller answer refers to the velocity at the apogee while the larger refers to perigee. ra =

Thus,

(d)

b

ge

j

The major axis is 2a = rp + ra , so the semi-major axis is a=

(e)

9.24 × 10 10 kg ⋅ m 2 s L = = 1.04 × 10 7 m . mv a 1.60 kg 5.58 × 10 3 m s

T=

4π 2 a 3 = GM E

1 7.02 × 10 6 m + 1.04 × 10 7 m = 8.69 × 10 6 m 2

e

j

e

j

4π 2 8.69 × 10 6 m

e6.67 × 10

T = 8 060 s = 134 min

−11

je

3

N ⋅ m 2 kg 2 5.98 × 10 24 kg

j

408 *P13.67

Universal Gravitation

Let m represent the mass of the meteoroid and vi its speed when far away. No torque acts on the meteoroid, so its angular momentum is conserved as it moves between the distant point and the point where it grazes the Earth, moving perpendicular to the radius: Li = L f :

FIG. P13.67

mri × v i = mr f × v f

b

g

m 3 RE vi = mRE v f v f = 3 vi Now energy of the meteoroid-Earth system is also conserved: GM E m 1 1 mvi2 + 0 = mv 2f − 2 2 RE

eK + U j = eK + U j : g

g

i

f

GM E 1 2 1 vi = 9 vi2 − 2 2 RE

e j

GM E = 4vi2 : RE *P13.68

vi =

GM E 4 RE

From Kepler’s third law, minimum period means minimum orbit size. The “treetop satellite” in Figure P13.35 has minimum period. The radius of the satellite’s circular orbit is essentially equal to the radius R of the planet.

FG H R R e 4π R j GρV = GMm

∑ F = ma :

2

=

mv 2 m 2πR = R R T 2

2

IJ K

2

2

RT 2 4 4π 2 R 3 G ρ πR 3 = 3 T2

FG H

IJ K

The radius divides out: T 2 Gρ = 3π P13.69

T=

3π Gρ

If we choose the coordinate of the center of mass at the origin, then 0=

bMr

2

− mr1

g

M+m

and

Mr2 = mr1

(Note: this is equivalent to saying that the net torque must be zero and the two experience no angular acceleration.) For each mass F = ma so mr1ω 12 =

MGm d2

and

Mr2ω 22 =

MGm d2

FIG. P13.69

b

g

Combining these two equations and using d = r1 + r2 gives r1 + r2 ω 2 = with ω 1 = ω 2 = ω 2π and T =

ω

we find T 2 =

4π 2 d 3 G M+m

a

f

.

aM + mfG d2

409

Chapter 13

P13.70

The gravitational force exerted on m 2 by the Earth (mass m1 ) accelerates m 2 according to: Gm1 m 2 . The equal magnitude force exerted on the Earth by m 2 produces negligible m2 g 2 = r2 acceleration of the Earth. The acceleration of relative approach is then

(a)

g2 =

Gm1 r

2

=

e6.67 × 10

−11

je mj

N ⋅ m 2 kg 2 5.98 × 10 24 kg

e1.20 × 10

7

2

j=

2.77 m s 2 .

Again, m 2 accelerates toward the center of mass with g 2 = 2.77 m s 2 . Now the Earth accelerates toward m 2 with an acceleration given as

(b)

m1 g 1 = g1 =

Gm1 m 2 r2

Gm 2 r

2

e6.67 × 10 =

−11

je mj

N ⋅ m 2 kg 2 2.00 × 10 24 kg

e1.20 × 10

7

2

j = 0.926 m s

2

The distance between the masses closes with relative acceleration of

g rel = g 1 + g 2 = 0.926 m s 2 + 2.77 m s 2 = 3.70 m s 2 . P13.71

Initial Conditions and Constants: Mass of planet:

5.98 × 10 24 kg

Radius of planet: Initial x: Initial y: Initial v x :

6.37 × 10 6 m 0.0 planet radii 2.0 planet radii +5 000 m/s

Time interval:

10.9 s

Initial

vy :

0.0 m/s FIG. P13.71

t (s) 0.0 10.9 21.7 32.6 … 5 431.6 5 442.4 5 453.3 5 464.1 … 10 841.3 10 852.2 10 863.1

x (m)

y (m)

r (m)

vx (m/s)

vy

ax

(m/s)

em s j 2

ay

em s j 2

0.0 54 315.3 108 629.4 162 941.1

12 740 000.0 12 740 000.0 12 739 710.0 12 739 130.0

12 740 000.0 12 740 115.8 12 740 173.1 12 740 172.1

5 000.0 4 999.9 4 999.7 4 999.3

0.0 –26.7 –53.4 –80.1

0.000 0 –0.010 0 –0.021 0 –0.031 0

–2.457 5 –2.457 4 –2.457 3 –2.457 2

112 843.8 31 121.4 –50 603.4 –132 324.3

–8 466 816.0 –8 467 249.7 –8 467 026.9 –8 466 147.7

8 467 567.9 8 467 306.9 8 467 178.2 8 467 181.7

–7 523.0 –7 523.2 –7 522.8 –7 521.9

–39.9 20.5 80.9 141.4

–0.074 0 –0.020 0 0.033 0 0.087 0

5.562 5 5.563 3 5.563 4 5.562 8

–108 629.0 –54 314.9 0.4

12 739 134.4 12 739 713.4 12 740 002.4

12 739 597.5 12 739 829.2 12 740 002.4

4 999.9 5 000.0 5 000.0

53.3 26.6 –0.1

0.021 0 0.010 0 0.000 0

–2.457 5 –2.457 5 –2.457 5

The object does not hit the Earth ; its minimum radius is 1.33 RE . Its period is 1.09 × 10 4 s . A circular orbit would require a speed of 5.60 km s .

410

Universal Gravitation

ANSWERS TO EVEN PROBLEMS P13.2

2.67 × 10 −7 m s 2

P13.40

(a) 10.0 m s 2 ; (b) 21.8 km s

P13.4

3.00 kg and 2.00 kg

P13.42

11.8 km s

P13.6

(a) 4.39 × 10 20 N toward the Sun; (b) 1.99 × 10 20 N toward the Earth; (c) 3.55 × 10 22 N toward the Sun

P13.44

GM E m 12 RE

P13.8

see the solution; either 1 m − 61.3 nm or 2.74 × 10 −4 m

P13.46

(a) v 0

F GM IJ =G H r K

(c) r f =

P13.10

2 3

P13.12

(a) 1.02 km s ; (b) 1.35 mm

P13.14

see the solution

P13.16

1.27

P13.18

Planet Y has turned through 1.30 revolutions

P13.20

1.63 × 10 4 rad s

P13.22

18.2 ms

P13.24

(a) 1.31 × 10 17 N toward the center; (b) 2.62 × 10 12 N kg

E

12

; (b) vi =

5

GM E 1 2 r

e j

25r 7

P13.48

2.26 × 10 −7

P13.50

2 GM 1 GM ; 3 R 3 R

4

;

P13.52

(a), (b) see the solution; (c) 1.85 × 10 −5 m s 2

P13.54

492 m s

P13.56

see the solution

P13.58

(a) G 1 2 c − 3 2 h1 2 ; (b) ~ 10 −34 m

P13.60

(a) 7.79 km s; (b) 7.85 km s;(c) −3.04 GJ ; (d) −3.08 GJ ; (e) loss = 46.9 MJ ; (f) A component of the Earth’s gravity pulls forward on the satellite in its downward banking trajectory.

P13.62

(a) 29.3 km s ; (b) K p = 2.74 × 10 33 J ;

P13.26

(a) −4.77 × 10 9 J ; (b) 569 N down; (c) 569 N up

P13.28

2.52 × 10 7 m

P13.30

2.82 × 10 9 J

P13.32

(a) see the solution; (b) 340 s

P13.34

(a) 42.1 km s ; (b) 2.20 × 10 11 m

P13.64

119 km

P13.36

1.58 × 10 10 J

P13.66

(a) −36.7 MJ ; (b) 9.24 × 10 10 kg ⋅ m 2 s ; (c) 5.58 km s; 10.4 Mm; (d) 8.69 Mm; (e) 134 min

P13.68

see the solution

P13.70

(a) 2.77 m s 2 ; (b) 3.70 m s 2

P13.38

b

U p = −5.40 × 10 33 J ;(c) K a = 2.57 × 10 33 J; U a = −5.22 × 10 33 J; yes

g bGM g ; (b) bGM g b R + hg ; L R + 2 h OP − 2π R m (c) GM m M MN 2 R bR + hg PQ b86 400 sg (a) 2π RE + h E

E

−1 2 E −1 2

3 2

12

E

2

E

E

E

2 E

2

The satellite should be launched from the Earth’s equator toward the east.

14 Fluid Mechanics CHAPTER OUTLINE 14.1 14.2 14.3 14.4 14.5 14.6 14.7

Pressure Variation of Pressure with Depth Pressure Measurements Buoyant Forces and Archimede’s Principle Fluid Dynamics Bernoulli’s Equation Other Applications of Fluid Dynamics

ANSWERS TO QUESTIONS Q14.1

The weight depends upon the total volume of glass. The pressure depends only on the depth.

Q14.2

Both must be built the same. The force on the back of each dam is the average pressure of the water times the area of the dam. If both reservoirs are equally deep, the force is the same.

FIG. Q14.2 Q14.3

If the tube were to fill up to the height of several stories of the building, the pressure at the bottom of the depth of the tube of fluid would be very large according to Equation 14.4. This pressure is much larger than that originally exerted by inward elastic forces of the rubber on the water. As a result, water is pushed into the bottle from the tube. As more water is added to the tube, more water continues to enter the bottle, stretching it thin. For a typical bottle, the pressure at the bottom of the tube can become greater than the pressure at which the rubber material will rupture, so the bottle will simply fill with water and expand until it bursts. Blaise Pascal splintered strong barrels by this method.

Q14.4

About 1 000 N: that’s about 250 pounds.

Q14.5

The submarine would stop if the density of the surrounding water became the same as the average density of the submarine. Unfortunately, because the water is almost incompressible, this will be much deeper than the crush depth of the submarine.

Q14.6

Yes. The propulsive force of the fish on the water causes the scale reading to fluctuate. Its average value will still be equal to the total weight of bucket, water, and fish.

Q14.7

The boat floats higher in the ocean than in the inland lake. According to Archimedes’s principle, the magnitude of buoyant force on the ship is equal to the weight of the water displaced by the ship. Because the density of salty ocean water is greater than fresh lake water, less ocean water needs to be displaced to enable the ship to float. 411

412

Fluid Mechanics

Q14.8

In the ocean, the ship floats due to the buoyant force from salt water. Salt water is denser than fresh water. As the ship is pulled up the river, the buoyant force from the fresh water in the river is not sufficient to support the weight of the ship, and it sinks.

Q14.9

Exactly the same. Buoyancy equals density of water times volume displaced.

Q14.10

At lower elevation the water pressure is greater because pressure increases with increasing depth below the water surface in the reservoir (or water tower). The penthouse apartment is not so far below the water surface. The pressure behind a closed faucet is weaker there and the flow weaker from an open faucet. Your fire department likely has a record of the precise elevation of every fire hydrant.

Q14.11

As the wind blows over the chimney, it creates a lower pressure at the top of the chimney. The smoke flows from the relatively higher pressure in front of the fireplace to the low pressure outside. Science doesn’t suck; the smoke is pushed from below.

Q14.12

The rapidly moving air above the ball exerts less pressure than the atmospheric pressure below the ball. This can give substantial lift to balance the weight of the ball.

Q14.13

The ski–jumper gives her body the shape of an airfoil. She deflects downward the air stream as it rushes past and it deflects her upward by Newton’s third law. The air exerts on her a lift force, giving her a higher and longer trajectory. To say it in different words, the pressure on her back is less than the pressure on her front. FIG. Q14.13

Q14.14

The horizontal force exerted by the outside fluid, on an area element of the object’s side wall, has equal magnitude and opposite direction to the horizontal force the fluid exerts on another element diametrically opposite the first.

Q14.15

The glass may have higher density than the liquid, but the air inside has lower density. The total weight of the bottle can be less than the weight of an equal volume of the liquid.

Q14.16

Breathing in makes your volume greater and increases the buoyant force on you. You instinctively take a deep breath if you fall into the lake.

Q14.17

No. The somewhat lighter barge will float higher in the water.

Q14.18

The level of the pond falls. This is because the anchor displaces more water while in the boat. A floating object displaces a volume of water whose weight is equal to the weight of the object. A submerged object displaces a volume of water equal to the volume of the object. Because the density of the anchor is greater than that of water, a volume of water that weighs the same as the anchor will be greater than the volume of the anchor.

Q14.19

The metal is more dense than water. If the metal is sufficiently thin, it can float like a ship, with the lip of the dish above the water line. Most of the volume below the water line is filled with air. The mass of the dish divided by the volume of the part below the water line is just equal to the density of water. Placing a bar of soap into this space to replace the air raises the average density of the compound object and the density can become greater than that of water. The dish sinks with its cargo.

Chapter 14

413

Q14.20

The excess pressure is transmitted undiminished throughout the container. It will compress air inside the wood. The water driven into the wood raises its average density and makes if float lower in the water. Add some thumbtacks to reach neutral buoyancy and you can make the wood sink or rise at will by subtly squeezing a large clear–plastic soft–drink bottle. Bored with graph paper and proving his own existence, René Descartes invented this toy or trick.

Q14.21

The plate must be horizontal. Since the pressure of a fluid increases with increasing depth, other orientations of the plate will give a non-uniform pressure on the flat faces.

Q14.22

The air in your lungs, the blood in your arteries and veins, and the protoplasm in each cell exert nearly the same pressure, so that the wall of your chest can be in equilibrium.

Q14.23

Use a balance to determine its mass. Then partially fill a graduated cylinder with water. Immerse the rock in the water and determine the volume of water displaced. Divide the mass by the volume and you have the density.

Q14.24

When taking off into the wind, the increased airspeed over the wings gives a larger lifting force, enabling the pilot to take off in a shorter length of runway.

Q14.25

Like the ball, the balloon will remain in front of you. It will not bob up to the ceiling. Air pressure will be no higher at the floor of the sealed car than at the ceiling. The balloon will experience no buoyant force. You might equally well switch off gravity.

Q14.26

Styrofoam is a little more dense than air, so the first ship floats lower in the water.

Q14.27

We suppose the compound object floats. In both orientations it displaces its own weight of water, so it displaces equal volumes of water. The water level in the tub will be unchanged when the object is turned over. Now the steel is underwater and the water exerts on the steel a buoyant force that was not present when the steel was on top surrounded by air. Thus, slightly less wood will be below the water line on the block. It will appear to float higher.

Q14.28

A breeze from any direction speeds up to go over the mound and the air pressure drops. Air then flows through the burrow from the lower entrance to the upper entrance.

Q14.29

Regular cola contains a considerable mass of dissolved sugar. Its density is higher than that of water. Diet cola contains a very small mass of artificial sweetener and has nearly the same density as water. The low–density air in the can has a bigger effect than the thin aluminum shell, so the can of diet cola floats.

Q14.30

(a)

Lowest density: oil; highest density: mercury

(b)

The density must increase from top to bottom.

(a)

Since the velocity of the air in the right-hand section of the pipe is lower than that in the middle, the pressure is higher.

(b)

The equation that predicts the same pressure in the far right and left-hand sections of the tube assumes laminar flow without viscosity. Internal friction will cause some loss of mechanical energy and turbulence will also progressively reduce the pressure. If the pressure at the left were not higher than at the right, the flow would stop.

Q14.31

414 Q14.32

Fluid Mechanics

Clap your shoe or wallet over the hole, or a seat cushion, or your hand. Anything that can sustain a force on the order of 100 N is strong enough to cover the hole and greatly slow down the escape of the cabin air. You need not worry about the air rushing out instantly, or about your body being “sucked” through the hole, or about your blood boiling or your body exploding. If the cabin pressure drops a lot, your ears will pop and the saliva in your mouth may boil—at body temperature—but you will still have a couple of minutes to plug the hole and put on your emergency oxygen mask. Passengers who have been drinking carbonated beverages may find that the carbon dioxide suddenly comes out of solution in their stomachs, distending their vests, making them belch, and all but frothing from their ears; so you might warn them of this effect.

SOLUTIONS TO PROBLEMS Section 14.1 P14.1

Pressure

e

M = ρ ironV = 7 860 kg m3 M = 0.111 kg

P14.2

jLMN 34 π b0.015 0 mg OPQ 3

The density of the nucleus is of the same order of magnitude as that of one proton, according to the assumption of close packing:

ρ=

m 1.67 × 10 −27 kg ~ ~ 10 18 kg m3 . 3 −15 V 4π 10 m 3

e

j

With vastly smaller average density, a macroscopic chunk of matter or an atom must be mostly empty space.

a f

50.0 9.80 F = A π 0.500 × 10 −2

= 6.24 × 10 6 N m 2

P14.3

P=

P14.4

Let Fg be its weight. Then each tire supports so yielding

P14.5

e

j

2

Fg 4

,

F Fg = A 4A Fg = 4 AP = 4 0.024 0 m 2 200 × 10 3 N m 2 = 1.92 × 10 4 N P=

e

je

j

The Earth’s surface area is 4πR 2 . The force pushing inward over this area amounts to

e

j

e

j

F = P0 A = P0 4πR 2 . This force is the weight of the air: Fg = mg = P0 4πR 2 so the mass of the air is

m=

e

P0 4πR 2 g

j = e1.013 × 10

5

jLMN e

j OPQ

N m 2 4π 6.37 × 10 6 m 9.80 m s 2

2

= 5.27 × 10 18 kg .

Chapter 14

Section 14.2 P14.6

415

Variation of Pressure with Depth

e

jb

je

g

P = P0 + ρgh = 1.013 × 10 5 Pa + 1 024 kg m3 9.80 m s 2 1 000 m

(a)

P = 1.01 × 10 7 Pa (b)

The gauge pressure is the difference in pressure between the water outside and the air inside the submarine, which we suppose is at 1.00 atmosphere.

Pgauge = P − P0 = ρgh = 1.00 × 10 7 Pa The resultant inward force on the porthole is then

a

F = Pgauge A = 1.00 × 10 7 Pa π 0.150 m P14.7

Fel = Ffluid and

h=

= 7.09 × 10 5 N .

kx ρgA −3

2

3

3

2

−2

= O j PQ

m

2

15 000 F = 2 200 3.00

e

1.62 m FIG. P14.7

F1 F = 2 A1 A 2

Since the pressure is the same on both sides, In this case,

P14.9

2

kx = ρghA

or

e1 000 N m je5.00 × 10 mj h= e10 kg m je9.80 m s jLMNπ e1.00 × 10 P14.8

f

F2 = 225 N

or

j

Fg = 80.0 kg 9.80 m s 2 = 784 N When the cup barely supports the student, the normal force of the ceiling is zero and the cup is in equilibrium.

e

j

Fg = F = PA = 1.013 × 10 5 Pa A A=

Fg P

=

784 = 7.74 × 10 −3 m 2 5 1.013 × 10

FIG. P14.9 P14.10

(a)

Suppose the “vacuum cleaner” functions as a high–vacuum pump. The air below the brick will exert on it a lifting force

LM e N

j OPQ =

F = PA = 1.013 × 10 5 Pa π 1.43 × 10 −2 m (b)

2

65.1 N .

The octopus can pull the bottom away from the top shell with a force that could be no larger than

b

g

e

je

ja

f LNMπ e1.43 × 10

F = PA = P0 + ρgh A = 1.013 × 10 5 Pa + 1 030 kg m3 9.80 m s 2 32.3 m F = 275 N

−2

j OQP

m

2

416 P14.11

Fluid Mechanics

The excess water pressure (over air pressure) halfway down is

e

ja

je

f

Pgauge = ρgh = 1 000 kg m3 9.80 m s 2 1.20 m = 1.18 × 10 4 Pa . The force on the wall due to the water is

ja

e

fa

f

F = Pgauge A = 1.18 × 10 4 Pa 2.40 m 9.60 m = 2.71 × 10 5 N horizontally toward the back of the hole. P14.12

P14.13

The pressure on the bottom due to the water is Pb = ρgz = 1.96 × 10 4 Pa So,

Fb = Pb A = 5.88 × 10 6 N

On each end,

F = PA = 9.80 × 10 3 Pa 20.0 m 2 = 196 kN

On the side,

F = PA = 9.80 × 10 3

2

588 kN

In the reference frame of the fluid, the cart’s acceleration causes a fictitious force to act backward, as if a the acceleration of gravity were g 2 + a 2 directed downward and backward at θ = tan −1 from the g d vertical. The center of the spherical shell is at depth below the air bubble and the pressure there is 2

F I GH JK

P = P0 + ρg eff h = P0 + P14.14

e j Pae60.0 m j =

1 ρd g 2 + a 2 . 2

The air outside and water inside both exert atmospheric pressure, so only the excess water pressure ρgh counts for the net force. Take a strip of hatch between depth h and h + dh . It feels force

a

f

dF = PdA = ρgh 2.00 m dh . (a)

1.00 m

The total force is

z

F = dF =

z

2.00 m

2.00 m

a

f

ρgh 2.00 m dh

h = 1.00 m

FIG. P14.14

a2.00 mf a2.00 mf − a1.00 mf = e1 000 kg m je9.80 m s j f 2 F = 29.4 kN bto the right g The lever arm of dF is the distance a h − 1.00 mf from hinge to strip: τ = z dτ = z ρgha 2.00 mfa h − 1.00 mfdh Lh h O τ = ρg a 2.00 mfM − a1.00 mf P 2 Q N3 F 7.00 m − 3.00 m I τ = e1 000 kg m je9.80 m s ja 2.00 mfG JK 2 H 3 h2 F = ρg 2.00 m 2

a

(b)

2.00 m

2.00 m

3

2

2

1.00 m

2.00 m

h = 1.00 m

3

2

2.00 m

1.00 m

3

2

τ = 16.3 kN ⋅ m counterclockwise

3

3

2

Chapter 14

P14.15

417

The bell is uniformly compressed, so we can model it with any shape. We choose a sphere of diameter 3.00 m. The pressure on the ball is given by: P = Patm + ρ w gh so the change in pressure on the ball from when it is on the surface of the ocean to when it is at the bottom of the ocean is ∆P = ρ w gh . In addition: ∆V =

ρ ghV 4πρ w ghr 3 −V∆P =− w =− , where B is the Bulk Modulus. 3B B B

∆V = −

je jb a3fe14.0 × 10 Paj

e

ga

f

4π 1 030 kg m3 9.80 m s 2 10 000 m 1.50 m 10

3

= −0.010 2 m3

Therefore, the volume of the ball at the bottom of the ocean is V − ∆V =

a

f

4 π 1.50 m 3

3

− 0.010 2 m 3 = 14.137 m3 − 0.010 2 m 3 = 14.127 m 3 .

This gives a radius of 1.499 64 m and a new diameter of 2.999 3 m. Therefore the diameter decreases by 0.722 mm .

Section 14.3 P14.16

(a)

Pressure Measurements We imagine the superhero to produce a perfect vacuum in the straw. Take point 1 at the water surface in the basin and point 2 at the water surface in the straw: P1 + ρgy1 = P2 + ρgy 2

e

je

j

1.013 × 10 5 N m 2 + 0 = 0 + 1 000 kg m 3 9.80 m s 2 y 2 (b) P14.17

y 2 = 10.3 m

No atmosphere can lift the water in the straw through zero height difference.

P0 = ρgh h=

P0 10.13 × 10 5 Pa = = 10.5 m ρg 0.984 × 10 3 kg m3 9.80 m s 2

e

je

j

No. Some alcohol and water will evaporate. The equilibrium vapor pressures of alcohol and water are higher than the vapor pressure of mercury.

FIG. P14.17

418 P14.18

Fluid Mechanics

(a)

Using the definition of density, we have hw =

(b)

100 g m water = = 20.0 cm 2 A 2 ρ water 5.00 cm 1.00 g cm 3

e

j

Sketch (b) at the right represents the situation after the water is added. A volume A 2 h2 of mercury has been displaced by water in the right tube. The additional volume of mercury now in the left tube is A1 h . Since the total volume of mercury has not changed,

b

A 2 h2 = A1 h

g

FIG. P14.18

h2 =

or

A1 h A2

(1)

At the level of the mercury–water interface in the right tube, we may write the absolute pressure as: P = P0 + ρ water ghw The pressure at this same level in the left tube is given by

b

g

P = P0 + ρ Hg g h + h2 = P0 + ρ water ghw which, using equation (1) above, reduces to

LM N

ρ Hg h 1 + or h =

ρ water hw

e

ρ Hg 1 +

A1 A2

j

OP Q

A1 = ρ water hw A2

.

e1.00 g cm ja20.0 cmf = Thus, the level of mercury has risen a distance of h = e13.6 g cm jc1 + h 3

3

10 .0 5.00

0.490 cm

above the original level.

b

g

P14.19

∆P0 = ρg∆h = −2.66 × 10 3 Pa :

P = P0 + ∆P0 = 1.013 − 0.026 6 × 10 5 Pa = 0.986 × 10 5 Pa

P14.20

Let h be the height of the water column added to the right side of the U–tube. Then when equilibrium is reached, the situation is as shown in the sketch at right. Now consider two points, A and B shown in the sketch, at the level of the water–mercury interface. By Pascal’s Principle, the absolute pressure at B is the same as that at A. But, PA = P0 + ρ w gh + ρ Hg gh2 and

b

h1 water

h

B

h2

Mercury A

g

PB = P0 + ρ w g h1 + h + h2 . Thus, from PA = PB , ρ w h1 + ρ w h + ρ w h2 = ρ w h + ρ Hg h2 , or h1 =

LM ρ Nρ

Hg w

OP Q

a

fa

f

− 1 h2 = 13.6 − 1 1.00 cm = 12.6 cm .

FIG. P14.20

Chapter 14

*P14.21

P = P0 + ρgh The gauge pressure is

(a)

ja

e

f

P − P0 = ρgh = 1 000 kg 9.8 m s 2 0.160 m = 1.57 kPa = 1.57 × 10 3 Pa

419

FG 1 atm IJ H 1.013 × 10 Pa K 5

= 0.015 5 atm . It would lift a mercury column to height 1 568 Pa P − P0 h= = = 11.8 mm . ρg 13 600 kg m 3 9.8 m s 2

e

(b)

je

j

Increased pressure of the cerebrospinal fluid will raise the level of the fluid in the spinal tap.

(c)

Blockage of the fluid within the spinal column or between the skull and the spinal column would prevent the fluid level from rising.

Section 14.4 P14.22

Buoyant Forces and Archimede’s Principle

(a)

The balloon is nearly in equilibrium: − Fg ∑ Fy = ma y ⇒ B − Fg

e j

helium

e j

payload

=0

ρ air gV − ρ helium gV − m payload g = 0 This reduces to m payload = ρ air − ρ helium V = 1.29 kg m3 − 0.179 kg m 3 400 m 3

or

b

g e

je

j

m payload = 444 kg (b)

Similarly, m payload = ρ air − ρ hydrogen V = 1.29 kg m3 − 0.089 9 kg m3 400 m 3

e

j e

je

j

m payload = 480 kg The air does the lifting, nearly the same for the two balloons. P14.23

At equilibrium

∑ F = 0 or

where The applied force,

B is the buoyant force. Fapp = B − mg

b

g

B = Vol ρ water g

where

a f So, F = aVolf g b ρ g b 4 F = π e1.90 × 10 mj e9.80 m s je10 kg m 3 F = bm + ρ V g g must be equal to F = ρ Vg

and

app

water

−2

app

P14.24

Fapp + mg = B

g

3

m = Vol ρ ball . 4 − ρ ball = πr 3 g ρ water − ρ ball 3

s

2

3

b

w

3

g

FIG. P14.23

j

− 84.0 kg m 3 = 0.258 N

Since V = Ah , m + ρ s Ah = ρ w Ah and A =

b

m ρw − ρs h

g

FIG. P14.24

420

Fluid Mechanics

P14.25

(a)

Before the metal is immersed:

∑ Fy = T1 − Mg = 0 or

b

ge

T1 = Mg = 1.00 kg 9.80 m s 2

scale

j

= 9.80 N

B

T1

(b)

After the metal is immersed:

∑ Fy = T2 + B − Mg = 0

or

b

Mg

g

Mg

T2 = Mg − B = Mg − ρ w V g V=

M

ρ

=

1.00 kg 2 700 kg m3

a

e

T2 = Mg − B = 9.80 N − 1 000 kg m3 (a)

b

FIG. P14.25

Thus,

*P14.26

T2

Fg

∑ Fy = 0 :

(b)

F jGH 2 7001.00kgkgm

3

I 9.80 m s = j JK e 2

6.17 N .

−15 N − 10 N + B = 0

B = 25.0 N T

B

FIG. P14.26(a)

(c)

The oil pushes horizontally inward on each side of the block.

(d)

String tension increases . The oil causes the water below to be under greater pressure, and the water pushes up more strongly on the bottom of the block.

(e)

Consider the equilibrium just before the string breaks:

15 N

−15 N − 60 N + 25 N+ Boil = 0 Boil = 50 N 60 N

For the buoyant force of the water we have B = ρVg

jb

e

g

25 N = 1 000 kg m3 0.25Vblock 9.8 m s 2

Vblock = 1.02 × 10 −2 m3

25 N

Boil

FIG. P14.26(e)

For the buoyant force of the oil

e

j e

j

50 N = 800 kg m3 f e 1.02 × 10 −2 m 3 9.8 m s 2 f e = 0.625 = 62.5% (f)

e

j e

j

−15 N + 800 kg m3 f f 1.02 × 10 −2 m 3 9.8 m s 2 = 0

15 N

f f = 0.187 = 18.7% Boil FIG. P14.26(f)

Chapter 14

P14.27

421

P = P0 + ρgh Taking P0 = 1.013 × 10 5 N m 2 and h = 5.00 cm

(a)

we find

Ptop = 1.017 9 × 10 5 N m 2

For h = 17.0 cm, we get

Pbot = 1.029 7 × 10 5 N m 2

Since the areas of the top and bottom are

A = 0.100 m

we find

Ftop = Ptop A = 1.017 9 × 10 3 N

and

Fbot = 1.029 7 × 10 3 N

T + B − Mg = 0

(b)

e

a

f

je

2

= 10 −2 m 2

je

B = ρ w Vg = 10 3 kg m3 1.20 × 10 −3 m3 9.80 m s 2 = 11.8 N

and

Mg = 10.0 9.80 = 98.0 N

Therefore,

T = Mg − B = 98.0 − 11.8 = 86.2 N

a f

b

FIG. P14.27

j

where

g

Fbot − Ftop = 1.029 7 − 1.017 9 × 10 3 N = 11.8 N

(c)

which is equal to B found in part (b). P14.28

Consider spherical balloons of radius 12.5 cm containing helium at STP and immersed in air at 0°C and 1 atm. If the rubber envelope has mass 5.00 g, the upward force on each is B − Fg ,He − Fg , env = ρ air Vg − ρ HeVg − m env g

b gFGH 34 πr IJK g − m g L4 O = a1. 29 − 0.179 f kg m M π a0.125 mf Pe9.80 m s j − 5.00 × 10 3 N Q 3

Fup = ρ air − ρ He Fup

env

3

3

2

−3

e

j

kg 9.80 m s 2 = 0.040 1 N

If your weight (including harness, strings, and submarine sandwich) is

e

j

70.0 kg 9.80 m s 2 = 686 N 686 N = 17 000 ~ 10 4 . 0.040 1 N

you need this many balloons: P14.29

(a)

e

a

j

e

ja

f

3

But B = Weight of block = mg = ρ woodVwood g = 0.650 g cm 3 20.0 cm g

a f a fa fa f 20.0 − h = 20.0a0.650 f so h = 20.0a1 − 0.650f = 7.00 cm 3

0.650 20.0 g = 1.00 20.0 20.0 20.0 − h g

(b)

f

According to Archimedes, B = ρ water Vwater g = 1.00 g cm 3 20.0 × 20.0 × 20.0 − h g

B = Fg + Mg where M = mass of lead

a f a f M = a1.00 − 0.650fa 20.0f = 0.350a 20.0 f 3

3

1.00 20.0 g = 0.650 20.0 g + Mg 3

3

= 2 800 g = 2.80 kg

422 *P14.30

Fluid Mechanics

(a)

The weight of the ball must be equal to the buoyant force of the water: 4 3 πrouter g 3

1.26 kgg = ρ water router (b)

F 3 × 1.26 kg I =G H 4π 1 000 kg m JK

13

3

= 6.70 cm

The mass of the ball is determined by the density of aluminum:

FG 4 πr H3

IJ K FG IJ ea H K

4 − πri3 3 4 1.26 kg = 2 700 kg m3 π 0.067 m 3 m = ρ Al V = ρ Al

3 0

f

3

− ri3

j

1.11 × 10 −4 m 3 = 3.01 × 10 −4 m 3 − ri3

e

ri = 1.89 × 10 −4 m 3 *P14.31

j

13

= 5.74 cm

Let A represent the horizontal cross-sectional area of the rod, which we presume to be constant. The rod is in equilibrium:

∑ Fy = 0 :

− mg + B = 0 = − ρ 0 Vwhole rod g + ρ fluidVimmersed g

a

f

ρ 0 ALg = ρA L − h g The density of the liquid is *P14.32

ρ=

ρ 0L . L−h

We use the result of Problem 14.31. For the rod floating in a liquid of density 0.98 g cm 3 ,

ρ = ρ0

L L−h

0.98 g cm3 =

a

3

ρ 0L L − 0.2 cm

f

e

j

0.98 g cm L − 0.98 g cm 3 0.2 cm = ρ 0 L For floating in the dense liquid,

ρ 0L

1.14 g cm 3 = 1.14 g cm 3 (a)

By substitution,

aL − 1.8 cmf − e1.14 g cm j1.8 cm = ρ L 3

a

0

f

a f

1.14L − 1.14 1.8 cm = 0.98L − 0.2 0.98 0.16L = 1.856 cm L = 11.6 cm

(b)

Substituting back,

a

ρ 0 = 0.963 g cm (c)

f

0.98 g cm3 11.6 cm − 0.2 cm = ρ 0 11.6 cm 3

ρ 0L is not of the form ρ = a + bh , equal-size L−h steps of ρ do not correspond to equal-size steps of h.

The marks are not equally spaced. Because ρ =

Chapter 14

P14.33

P14.34

The balloon stops rising when

bρ

Therefore,

V=

air

g

− ρ He gV = Mg

and

400 M = ρ air − ρ He 1.25 e −1 − 0.180

bρ

air

423

g

− ρ He V = M ,

V = 1 430 m3

Since the frog floats, the buoyant force = the weight of the frog. Also, the weight of the displaced water = weight of the frog, so

ρ oozeVg = m frog g or

m frog = ρ oozeV = ρ ooze

FG H

IJ e K

1 4 3 2π πr = 1.35 × 10 3 kg m3 6.00 × 10 −2 m 2 3 3

j e

j

3

Hence, m frog = 0.611 kg . P14.35

B = Fg V = ρ sphere gV 2 1 ρ sphere = ρ H 2O = 500 kg m3 2 4 ρ glycerin g V − ρ sphere gV = 0 10 10 ρ glycerin = 500 kg m3 = 1 250 kg m3 4

ρ H 2O g

FG H

IJ K

e

P14.36

FIG. P14.35

j

Constant velocity implies zero acceleration, which means that the submersible is in equilibrium under the gravitational force, the upward buoyant force, and the upward resistance force:

∑ Fy = ma y = 0

e

j

− 1. 20 × 10 4 kg + m g + ρ w gV + 1 100 N = 0

where m is the mass of the added water and V is the sphere’s volume. 1.20 × 10 4 kg + m = 1.03 × 10 3 so P14.37

LM 4 π a1.50f OP + 1 100 N N3 Q 9.8 m s 3

2

m = 2.67 × 10 3 kg

By Archimedes’s principle, the weight of the fifty planes is equal to the weight of a horizontal slice of water 11.0 cm thick and circumscribed by the water line:

a f 50e 2.90 × 10 kg j g = e1 030 kg m j g a0.110 mf A ∆B = ρ water g ∆V 4

3

giving A = 1.28 × 10 4 m 2 . The acceleration of gravity does not affect the answer.

424

Fluid Mechanics

Section 14.5

Fluid Dynamics

Section 14.6

Bernoulli’s Equation

P14.38

By Bernoulli’s equation,

b b

g g

dm = ρAv = 1 000π 5.00 × 10 −2 dt

e

P14.39

b

g

1 1 1 000 v 2 = 6.00 × 10 4 N m 2 + 1 000 16 v 2 2 2 1 2.00 × 10 4 N m 2 = 1 000 15 v 2 2 v = 1.63 m s

8.00 × 10 4 N m 2 +

jb 2

FIG. P14.38

g

1.63 m s = 12.8 kg s

Assuming the top is open to the atmosphere, then P1 = P0 . Note P2 = P0 . Flow rate = 2.50 × 10 −3 m3 min = 4.17 × 10 −5 m3 s . (a)

so A1 >> A 2 Assuming v1 = 0 ,

v1 4mk ), the system will be overdamped and will not oscillate.

Period decreases.

(b)

Period increases.

(c)

No change.

Chapter 15

441

Q15.16

Yes. An oscillator with damping can vibrate at resonance with amplitude that remains constant in time. Without damping, the amplitude would increase without limit at resonance.

Q15.17

The phase constant must be π rad .

Q15.18

Higher frequency. When it supports your weight, the center of the diving board flexes down less than the end does when it supports your weight. Thus the stiffness constant describing the center of 1 k is greater the board is greater than the stiffness constant describing the end. And then f = 2π m for you bouncing on the center of the board.

FG IJ H K

Q15.19

The release of air from one side of the parachute can make the parachute turn in the opposite direction, causing it to release air from the opposite side. This behavior will result in a periodic driving force that can set the parachute into side-to-side oscillation. If the amplitude becomes large enough, the parachute will not supply the needed air resistance to slow the fall of the unfortunate skydiver.

Q15.20

An imperceptibly slight breeze may be blowing past the leaves in tiny puffs. As a leaf twists in the wind, the fibers in its stem provide a restoring torque. If the frequency of the breeze matches the natural frequency of vibration of one particular leaf as a torsional pendulum, that leaf can be driven into a large-amplitude resonance vibration. Note that it is not the size of the driving force that sets the leaf into resonance, but the frequency of the driving force. If the frequency changes, another leaf will be set into resonant oscillation.

Q15.21

We assume the diameter of the bob is not very small compared to the length of the cord supporting it. As the water leaks out, the center of mass of the bob moves down, increasing the effective length of the pendulum and slightly lowering its frequency. As the last drops of water dribble out, the center of mass of the bob hops back up to the center of the sphere, and the pendulum frequency quickly increases to its original value.

SOLUTIONS TO PROBLEMS Section 15.1 P15.1

Motion of an Object Attached to a Spring

(a)

Since the collision is perfectly elastic, the ball will rebound to the height of 4.00 m and then repeat the motion over and over again. Thus, the motion is periodic .

(b)

To determine the period, we use: x =

1 2 gt . 2

The time for the ball to hit the ground is t =

a

a

f

2 4.00 m 2x = = 0.909 s g 9.80 m s 2

f

This equals one-half the period, so T = 2 0.909 s = 1.82 s . (c)

No . The net force acting on the ball is a constant given by F = − mg (except when it is in contact with the ground), which is not in the form of Hooke’s law.

442

Oscillatory Motion

Section 15.2 P15.2

P15.3

Mathematical Representation of Simple Harmonic Motion

π 6

IJ K

x = 5.00 cm cos 2t +

(b)

v=

dx π = − 10.0 cm s sin 2t + dt 6

(c)

a=

π dv = − 20.0 cm s 2 cos 2t + dt 6

(d)

A = 5.00 cm

a

g FGH

b

IJ K

j FGH

e

f a

IJ K

f FGH π6 IJK =

a

At t = 0 ,

x = 5.00 cm cos

At t = 0 ,

v = −5.00 cm s

At t = 0 ,

a = −17.3 cm s 2

and

T=

f

b

2π

ω

=

4.33 cm

2π = 3.14 s 2

g

x = 4.00 m cos 3.00πt + π Compare this with x = A cos ωt + φ to find (a)

ω = 2π f = 3.00π or

*P15.4

f FGH

a

(a)

T=

f = 1.50 Hz

1 = 0.667 s f

(b)

A = 4.00 m

(c)

φ = π rad

(d)

x t = 0.250 s = 4.00 m cos 1.75π = 2.83 m

(a)

The spring constant of this spring is

a

f a

f a

f

k=

F 0.45 kg 9.8 m s 2 = = 12.6 N m x 0.35 m

we take the x-axis pointing downward, so φ = 0 x = A cos ωt = 18.0 cm cos (d)

12.6 kg 0.45 kg ⋅ s 2

84. 4 s = 18.0 cm cos 446.6 rad = 15.8 cm

a f

Now 446.6 rad = 71 × 2π + 0.497 rad . In each cycle the object moves 4 18 = 72 cm , so it has

a

f a

f

moved 71 72 cm + 18 − 15.8 cm = 51.1 m . (b)

By the same steps, k = x = A cos

(e)

a f

0. 44 kg 9.8 m s 2 = 12.1 N m 0.355 m

k 12.1 t = 18.0 cm cos 84.4 = 18.0 cm cos 443.5 rad = −15.9 cm m 0.44

443.5 rad = 70 2π + 3.62 rad

a

f

Distance moved = 70 72 cm + 18 + 15.9 cm = 50.7 m (c)

The answers to (d) and (e) are not very different given the difference in the data about the two vibrating systems. But when we ask about details of the future, the imprecision in our knowledge about the present makes it impossible to make precise predictions. The two oscillations start out in phase but get completely out of phase.

Chapter 15

P15.5

(a)

At t = 0 , x = 0 and v is positive (to the right). Therefore, this situation corresponds to x = A sin ωt and

v = vi cos ωt

Since f = 1.50 Hz ,

ω = 2π f = 3.00π

a

a

f

v max = vi = Aω = 2.00 3.00π = 6.00π cm s = 18.8 cm s The particle has this speed at t = 0 and next at

(c)

a

a max = Aω 2 = 2.00 3.00π

f

2

P15.6

t=

T 1 = s 2 3

t=

3 T = 0.500 s 4

= 18.0π 2 cm s 2 = 178 cm s 2

This positive value of acceleration first occurs at (d)

f

x = 2.00 cm sin 3.00π t

Also, A = 2.00 cm, so that (b)

443

2 s and A = 2.00 cm, the particle will travel 8.00 cm in this time. 3 3 Hence, in 1.00 s = T , the particle will travel 8.00 cm + 4.00 cm = 12.0 cm . 2

Since T =

FG H

IJ K

af

FG v IJ sin ωt HωK i

The proposed solution

x t = xi cos ωt +

implies velocity

v=

dx = − x iω sin ωt + vi cos ωt dt

and acceleration

a=

v dv = − x iω 2 cos ωt − viω sin ωt = −ω 2 x i cos ωt + i sin ωt = −ω 2 x ω dt

FG H

FG IJ H K

IJ K

(a)

The acceleration being a negative constant times position means we do have SHM, and its angular frequency is ω. At t = 0 the equations reduce to x = xi and v = vi so they satisfy all the requirements.

(b)

v 2 − ax = − x iω sin ωt + vi cos ωt

b

g − e− x ω 2

i

2

jFGH

cos ωt − vi sin ωt xi cos ωt +

FG v IJ sin ωtIJ HωK K i

v 2 − ax = xi2ω 2 sin 2 ωt − 2 xi viω sin ωt cos ωt + vi2 cos 2 ωt + x i2ω 2 cos 2 ωt + x i viω cos ωt sin ωt + xi viω sin ωt cos ωt + vi2 sin 2 ωt = x i2ω 2 + vi2 So this expression is constant in time. On one hand, it must keep its original value vi2 − ai xi . On the other hand, if we evaluate it at a turning point where v = 0 and x = A , it is A 2ω 2 + 0 2 = A 2ω 2 . Thus it is proved. P15.7

(a)

T=

12.0 s = 2.40 s 5

(b)

f=

1 1 = = 0.417 Hz T 2. 40

(c)

ω = 2π f = 2π 0.417 = 2.62 rad s

a

f

444 *P15.8

Oscillatory Motion

The mass of the cube is

ja

e

f

m = ρV = 2.7 × 10 3 kg m3 0.015 m

3

= 9.11 × 10 −3 kg

The spring constant of the strip of steel is k= f=

P15.9

f=

ω 1 = π 2 2π

k m

14.3 N F = = 52.0 N m x 0.027 5 m

ω 2π

k=

x = A cos ωt

52 kg

k 1 = m 2π

1 2π

T=

or

Solving for k,

*P15.10

=

s 2 9.11 × 10 −3 kg

= 12.0 Hz

1 m = 2π f k 4π 2 m

A = 0.05 m

T

2

=

b

4π 2 7.00 kg

a2.60 sf

2

g=

40.9 N m . a = − Aω 2 cos ωt

v = − Aω sin ωt

If f = 3 600 rev min = 60 Hz , then ω = 120π s −1

a

f

v max = 0.05 120π m s = 18.8 m s P15.11

(a)

ω=

k = m

8.00 N m = 4.00 s −1 0.500 kg

From this we find that

(b)

t=

FG 1 IJ sin FG x IJ and when H 4.00 K H 10.0 K −1

Using t =

f

2

m s 2 = 7.11 km s 2

a

a f a = −160 sina 4.00t f cm s v = 40.0 cos 4.00t cm s

v max = 40.0 cm s 2

amax = 160 cm s 2 .

x = 6.00 cm, t = 0.161 s.

a f a = −160 sin 4.00a0.161f = FG 1 IJ sin FG x IJ H 4.00 K H 10.0 K −1

when x = 0 , t = 0 and when

x = 8.00 cm, t = 0.232 s.

Therefore,

∆t = 0.232 s .

f

x = 10.0 sin 4.00t cm .

so position is given by

v = 40.0 cos 4.00 0.161 = 32.0 cm s

We find

(c)

a

amax = 0.05 120π

−96.0 cm s 2 .

Chapter 15

P15.12

445

m = 1.00 kg , k = 25.0 N m, and A = 3.00 cm. At t = 0 , x = −3.00 cm (a)

ω=

k = m

25.0 = 5.00 rad s 1.00 2π 2π T= = = 1.26 s ω 5.00

so that, (b)

b

g

v max = Aω = 3.00 × 10 −2 m 5.00 rad s = 0.150 m s

b

amax = Aω 2 = 3.00 × 10 −2 m 5.00 rad s (c)

g

2

= 0.750 m s 2

Because x = −3.00 cm and v = 0 at t = 0 , the required solution is x = − A cos ωt

a

f

x = −3.00 cos 5.00t cm

or

a a

f f

dx = 15.0 sin 5.00t cm s dt dv a= = 75.0 cos 5.00t cm s 2 dt v=

P15.13

The 0.500 s must elapse between one turning point and the other. Thus the period is 1.00 s.

ω=

b

2π = 6. 28 s T

f

ga

and v max = ωA = 6.28 s 0.100 m = 0.628 m s . P15.14

(a)

v max = ωA A=

(b)

Section 15.3 P15.15

(a)

v max

ω

=

v

ω

x = − A sin ωt = −

FG v IJ sin ωt HωK

Energy of the Simple Harmonic Oscillator Energy is conserved for the block-spring system between the maximum-displacement and the half-maximum points:

aK + U f = aK + U f 1 b6.50 N mga0.100 mf 2 i

32.5 mJ = k = m

0+

f

b

2

1 m 0.300 m s 2

b

=

1 m 0.300 m s 2

g

+ 8.12 mJ

2

6.50 N m = 3.46 rad s 0.542 kg

(b)

ω=

(c)

amax = Aω 2 = 0.100 m 3.46 rad s

b

g

2

g

2

+

1 2 1 1 kA = mv 2 + kx 2 2 2 2

b

ge

1 6.50 N m 5.00 × 10 −2 m 2

m=

a

2 24.4 mJ 9.0 × 10

∴T =

= 1.20 m s 2

j

2π

ω

=

−2

f

2

m s2

2

= 0.542 kg

2π rad = 1.81 s 3.46 rad s

446 P15.16

P15.17

Oscillatory Motion

b

P15.19

k = mω 2 = 0.200 kg 25.1 rad s

(b)

E=

kA 2 ⇒A= 2

2E = k

2

126

Choose the car with its shock-absorbing bumper as the system; by conservation of energy, k = 3.16 × 10 −2 m m

e

v=x

j

e

j

(a) (b)

v max = Aω

(c)

a max = Aω 2 = 3.50 × 10 −2 m 22.4 s −1

(a)

E=

(b)

v = ω A2 − x2 =

e

b

j

2

ge

250 = 22.4 s −1 0.500

v max = 0.784 m s

= 17.5 m s 2

1 2 1 kA = 35.0 N m 4.00 × 10 −2 m 2 2

j

2

= 28.0 mJ

k A2 − x2 m

35.0

e4.00 × 10 j − e1.00 × 10 j = 1.02 m s 50.0 × 10 1 1 1 1 mv = kA − kx = a35.0 fLe 4.00 × 10 j − e3.00 × 10 j O = MN PQ 2 2 2 2 v=

(c)

= 0.153 J k = m

ω=

where

5.00 × 10 6 = 2.23 m s 10 3

2

−2 kA 2 250 N m 3.50 × 10 m E= = 2 2

.

P15.20

g = 126 N m 2a 2.00f = 0.178 m

(a)

1 1 mv 2 = kx 2 : 2 2

P15.18

2π 2π = = 25.1 rad s T 0. 250

m = 200 g , T = 0.250 s, E = 2.00 J ; ω =

−2 2

−3

2

2

−2 2

−2 2

2

(d)

1 2 1 kx = E − mv 2 = 15.8 mJ 2 2

(a)

k=

(b)

ω=

(c)

v max = ωA = 50.0 0. 200 = 1.41 m s at x = 0

(d)

amax = ω 2

(e)

E=

(f)

v = ω A 2 − x 2 = 50.0

(g)

a = ω 2 x = 50.0

−2 2

F 20.0 N = = 100 N m x 0.200 m k = 50.0 rad s m

1 2 kA 2

f=

so

a f A = 50.0a0.200 f = 10.0 m s 1 = a100 fa0.200 f = 2.00 J 2

2

at x = ± A

2

FG 0.200 IJ = H 3 K

a

8 0.200 9

f

2

3.33 m s 2

= 1.33 m s

ω 2π

= 1.13 Hz

12.2 mJ

Chapter 15

P15.21

(a)

E=

a f

1 2 1 kA , so if A ′ = 2 A , E ′ = k A ′ 2 2

2

=

a f

1 k 2A 2

2

447

= 4E

Therefore E increases by factor of 4 .

*P15.22

(b)

v max =

k A , so if A is doubled, v max is doubled . m

(c)

a max =

k A , so if A is doubled, a max also doubles . m

(d)

T = 2π

(a)

y f = yi + v yi t +

m is independent of A, so the period is unchanged . k

1 ayt 2 2 1 −11 m = 0 + 0 + −9.8 m s 2 t 2 2

e

t= (b)

j

22 m ⋅ s 2 = 1.50 s 9.8 m

Take the initial point where she steps off the bridge and the final point at the bottom of her motion.

eK + U

g

+ Us

j = eK + U i

g

+ Us

j

f

1 0 + mgy + 0 = 0 + 0 + kx 2 2 1 65 kg 9.8 m s 2 36 m = k 25 m 2 k = 73. 4 N m

a

(c)

The spring extension at equilibrium is x =

f

2

F 65 kg 9.8 m s 2 = = 8.68 m , so this point is k 73.4 N m

11 + 8.68 m = 19.7 m below the bridge and the amplitude of her oscillation is 36 − 19.7 = 16.3 m . k = m

73.4 N m = 1.06 rad s 65 kg

(d)

ω=

(e)

Take the phase as zero at maximum downward extension. We find what the phase was 25 m higher when x = −8.68 m: In x = A cos ωt ,

FG H

t −8.68 m = 16.3 m cos 1.06 s t = −2.01 s

IJ K

16.3 m = 16.3 m cos 0 t 1.06 = −122° = −2.13 rad s

Then +2.01 s is the time over which the spring stretches. (f)

total time = 1.50 s + 2.01 s = 3.50 s

448 P15.23

Oscillatory Motion

Model the oscillator as a block-spring system. v2 + ω 2x2 = ω 2 A2

From energy considerations, v max = ωA and v =

ωA 2

From this we find x 2 = P15.24

3 2 A 4

so

FG ωA IJ H2K

and

x=

2

+ ω 2x2 = ω 2 A2

3 A = ±2.60 cm where A = 3.00 cm 2

The potential energy is

a f

1 2 1 2 kx = kA cos 2 ωt . 2 2

Us = The rate of change of potential energy is

a f

a f

dU s 1 2 1 = kA 2 cos ωt −ω sin ωt = − kA 2ω sin 2ωt . 2 2 dt (a)

This rate of change is maximal and negative at 2ωt =

Then, t =

π π π , 2ωt = 2π + , or in general, 2ωt = 2nπ + for integer n. 2 2 2

a

f

π 4n + 1 π 4n + 1 = 4ω 4 3.60 s −1

a

f

e

j

For n = 0 , this gives t = 0.218 s while n = 1 gives t = 1.09 s . All other values of n yield times outside the specified range. (b)

Section 15.4 P15.25

dU s dt

= max

1 2 1 kA ω = 3.24 N m 5.00 × 10 −2 m 2 2

b

ge

j e3.60 s j = 2

−1

14.6 mW

Comparing Simple Harmonic Motion with Uniform Circular Motion

(a)

The motion is simple harmonic because the tire is rotating with constant velocity and you are looking at the motion of the bump projected in a plane perpendicular to the tire.

(b)

Since the car is moving with speed v = 3.00 m s , and its radius is 0.300 m, we have:

ω=

3.00 m s = 10.0 rad s . 0.300 m

Therefore, the period of the motion is: T=

2π

ω

=

2π

b10.0 rad sg =

0.628 s .

Chapter 15

P15.26

The angle of the crank pin is θ = ωt . Its x-coordinate is

ω Piston

x = A cos θ = A cos ωt

A

where A is the distance from the center of the wheel to the crank pin. This is of the form x = A cos ωt + φ , so the yoke and piston rod move with simple harmonic motion.

b

Section 15.5 P15.27

(a)

P15.28

P15.29

g

FIG. P15.26

The Pendulum T = 2π L=

(b)

x = –A

gT 2 4π 2

L g

e9.80 m s ja12.0 sf 2

=

Tmoon = 2π

4π 2 L

= 2π

g moon

2

= 35.7 m

35.7 m 1.67 m s 2

= 29.1 s

The period in Tokyo is

TT = 2π

LT gT

and the period in Cambridge is

TC = 2π

LC gC

We know

TT = TC = 2.00 s

For which, we see

LT LC = gT gC

or

g C LC 0.994 2 = = = 1.001 5 g T LT 0.992 7

The swinging box is a physical pendulum with period T = 2π

I . mgd

The moment of inertia is given approximately by I=

1 mL2 (treating the box as a rod suspended from one end). 3

Then, with L ≈ 1.0 m and d ≈

T ≈ 2π

1 3

L , 2

mL2

mg

ch L 2

= 2π

a

f

2 1.0 m 2L = 2π = 1.6 s or T ~ 10 0 s . 2 3g 3 9.8 m s

e

j

x ( t)

449

450 P15.30

P15.31

Oscillatory Motion

ω=

2π : T

T=

ω=

g : L

L=

2π

=

ω

2π = 1.42 s 4. 43

g 9.80 = = 0.499 m 2 ω2 4.43

a f

Using the simple harmonic motion model: A = rθ = 1 m 15°

π = 0. 262 m 180°

g 9.8 m s 2 = = 3.13 rad s L 1m

ω= (a)

v max = Aω = 0. 262 m 3.13 s = 0.820 m s

(b)

a max = Aω 2 = 0.262 m 3.13 s

b

a tan = rα (c)

g

2

= 2.57 m s 2

a tan 2.57 m s 2 = = 2.57 rad s 2 r 1m

α=

FIG. P15.31

F = ma = 0.25 kg 2.57 m s 2 = 0.641 N

More precisely, (a)

1 mv 2 and 2 ∴ v max = 2 gL 1 − cos θ = 0.817 m s mgh =

a

(b)

f

f

Iα = mgL sin θ

α max =

P15.32

a

h = L 1 − cos θ

mgL sin θ 2

mL

=

g sin θ i = 2.54 rad s 2 L

a fa

f

(c)

Fmax = mg sin θ i = 0.250 9.80 sin 15.0° = 0.634 N

(a)

The string tension must support the weight of the bob, accelerate it upward, and also provide the restoring force, just as if the elevator were at rest in a gravity field 9.80 + 5.00 m s 2

a

T = 2π

L 5.00 m = 2π g 14.8 m s 2

T = 3.65 s (b)

T = 2π

(c)

g eff =

5.00 m

e9.80 m s

2

− 5.00 m s 2

j

= 6.41 s

e9.80 m s j + e5.00 m s j

T = 2π =

2 2

5.00 m 11.0 m s 2

2 2

= 4.24 s

= 11.0 m s 2

f

Chapter 15

P15.33

Referring to the sketch we have

x R For small displacements, tan θ ≈ sin θ mg F=− and x = − kx R Since the restoring force is proportional to the displacement from equilibrium, the motion is simple harmonic motion. F = − mg sin θ

tan θ =

and

Comparing toF = − mω 2 x shows ω =

P15.34

T=

(a)

g . R

k = m

total measured time 50

a f Period, T asf Length, L m

4

1.000 0.750 0.500 3

1.996 1.732 1.422

2

L 4π 2 L T = 2π so g= g T2 The calculated values for g are:

(b)

af

0

1.996 1.732 1.422

g m s2

9.91

j

9.87

From T 2 = Thus, g =

0.25

0.5

0.75

9.76

this agrees with the accepted value of g = 9.80 m s 2 within 0.5%.

F 4π I L , the slope of T GH g JK 2

2

versus L graph =

4π 2 = 4.01 s 2 m . g

4π 2 = 9.85 m s 2 . This is the same as the value in (b). slope

f = 0. 450 Hz , d = 0.350 m, and m = 2.20 kg T=

1 ; f

T = 2π I =T2

I ; mgd mgd 4π

2

=

T2 =

FG 1 IJ H fK

2

1.0 L, m

FIG. P15.34

Thus, g ave = 9.85 m s 2 (c)

1

Period, T s

e

FIG. P15.33

T2, s2

The measured periods are:

P15.35

451

4π 2 I mgd

mgd 4π

2

=

a fa f = e0.450 s j

2.20 9.80 0.350 4π

2

−1 2

0.944 kg ⋅ m 2 FIG. P15.35

452 P15.36

Oscillatory Motion

(a)

The parallel-axis theorem: I = I CM + Md 2 = =M

FG 13 m IJ H 12 K

a

f

1 1 ML2 + Md 2 = M 1.00 m 12 12

2

a

f

+ M 1.00 m

2

2

e

j

M 13 m 2 13 m I = 2π = 2π = 2.09 s T = 2π 12 Mg 1.00 m Mgd 12 9.80 m s 2 (b)

(a)

e

j

FIG. P15.36

1.00 m 9.80 m s

2

difference =

= 2.01 s

2.09 s − 2.01 s = 4.08% 2.01 s

The parallel axis theorem says directly I = I CM + md 2 so

(b)

f

For the simple pendulum T = 2π

P15.37

a

T = 2π

eI

I = 2π mgd

CM

+ md 2

j

mgd

When d is very large T → 2π

d gets large. g

When d is very small T → 2π

I CM gets large. mgd

So there must be a minimum, found by

j bmgdg F 1I FG 1 IJ eI mg + 2π bmgd g = 2π e I + md j G − J bmgd g H 2K H 2K −π e I + md jmg 2π md mgd = + =0 + + I md mgd I md mgd b g b g e j e j

dT d =0= 2π I CM + md 2 dd dd

e

12

−1 2

2 12

CM

−1 2

2

CM

2 12

CM

−3 2

3 2

CM

2 12

3 2

This requires − I CM − md 2 + 2md 2 = 0 or P15.38

ICM = md 2 .

We suppose the stick moves in a horizontal plane. Then,

b

f

ga

1 1 2.00 kg 1.00 m mL2 = 12 12 I T = 2π I=

κ

κ=

4π 2 I T

2

=

e

4π 2 0.167 kg ⋅ m 2

a180 sf

2

j=

2

= 0.167 kg ⋅ m 2

203 µN ⋅ m

CM

+ md 2

j

−1 2

2md

Chapter 15

P15.39

e

je

(a)

I = 5.00 × 10 −7 kg ⋅ m 2

(b)

I

d 2θ = −κθ ; dt 2

e

Section 15.6

θ

π I JK jFGH 0.2250

FIG. P15.39 2

= 3.16 × 10 −4

N⋅m rad

Damped Oscillations 1 1 mv 2 + kx 2 2 2 2 dE d x = mv 2 + kxv dt dt 2 md x = − kx − bv dt 2 dE = v − kx − bv + kvx dt dE = − bv 2 < 0 dt

E=

The total energy is Taking the time-derivative, Use Equation 15.31:

a

Thus, P15.41

2

κ 2π =ω = I T

κ = Iω 2 = 5.00 × 10 −7

P15.40

j

T = 0.250 s, I = mr 2 = 20.0 × 10 −3 kg 5.00 × 10 −3 m

f

b

θ i = 15.0°

g

θ t = 1 000 = 5.50° x1 000 Ae − bt 2 m 5.50 − b 1 000 g = = =e b 15.0 xi A

x = Ae − bt 2 m

2m

FG 5.50 IJ = −1.00 = −bb1 000g H 15.0 K 2m

ln ∴ P15.42

b = 1.00 × 10 −3 s −1 2m

b

x = Ae − bt 2 m cos ωt + φ

Show that is a solution of

− kx − b

where

ω=

b

x = Ae − bt 2 m cos ωt + φ

FG H

g

g

2

dx d x =m 2 dt dt

FG IJ H K

k b − m 2m

(1)

2

.

(2)

IJ b g b g K d x b L FG − b IJ cosbωt + φ g − Ae ω sinbωt + φ gOP =− Ae H 2m K 2m MN dt Q L FG − b IJω sinbωt + φ g + Ae ω cosbωt + φ gOP − M Ae H 2m K N Q dx b = Ae − bt 2 m − cos ωt + φ − Ae − bt 2 mω sin ωt + φ dt 2m 2

− bt 2 m

2

− bt 2 m

continued on next page

(3) (4)

− bt 2 m

− bt 2 m

2

(5)

453

454

Oscillatory Motion

Substitute (3), (4) into the left side of (1) and (5) into the right side of (1); 2

b g 2bm Ae cosbωt + φ g + bωAe sinbωt + φ g bL FG − b IJ cosbωt + φ g − Ae ω sinbωt + φ gOP = − M Ae H 2m K 2N Q b cosbωt + φ g + Ae ω sinbωt + φ g − mω Ae 2 Compare the coefficients of Ae cosbωt + φ g and Ae sinbωt + φ g : F k b I = −k + b b bF b I b = − G− − mω = − mG − cosine-term: − k + J K H 2m 2 2m 4m 2m H m 4m JK − kAe − bt 2 m cos ωt + φ +

− bt 2 m

− bt 2 m

− bt 2 m

− bt 2 m

− bt 2 m

− bt 2 m

2

− bt 2 m

− bt 2 m

2

sine-term:

bω = +

2

2

2

2

2

af af

b b ω + ω = bω 2 2

b

g

Since the coefficients are equal, x = Ae − bt 2 m cos ωt + φ is a solution of the equation. *P15.43

The frequency if undamped would be ω 0 = (a)

k = m

With damping

ω = ω 02 −

FG b IJ H 2m K

2

=

2.05 × 10 4 N m = 44.0 s. 10.6 kg

FG 44 1 IJ − FG 3 kg IJ H s K H s 2 10.6 kg K

= 1 933.96 − 0.02 = 44.0 f= (b)

b

ω 44.0 = = 7.00 Hz 2π 2π s

2

1 s

g

In x = A 0 e − bt 2 m cos ωt + φ over one cycle, a time T = A0 e − b 2π

2 mω

2

2π

ω

, the amplitude changes from A0 to

for a fractional decrease of

A 0 − A 0 e − πb mω = 1 − e −π 3 a10.6⋅44.0 f = 1 − e −0 .020 2 = 1 − 0.979 98 = 0.020 0 = 2.00% . A0 (c)

The energy is proportional to the square of the amplitude, so its fractional rate of decrease is twice as fast: E= We specify

1 2 1 2 − 2 bt 2 m = E0 e − bt m . kA = kA 0 e 2 2 0.05E0 = E0 e − 3 t 10.6 0.05 = e − 3 t 10.6 e + 3 t 10 .6 = 20 3t = ln 20 = 3.00 10.6 t = 10.6 s

Chapter 15

Section 15.7 P15.44

(a)

Forced Oscillations For resonance, her frequency must match f0 =

(b)

ω0 1 = 2π 2π

4.30 × 10 3 N m = 2.95 Hz . 12.5 kg

k 1 = m 2π

dx dv = − Aω sin ωt , and a = = − Aω 2 cos ωt , the maximum acceleration dt dt is Aω 2 . When this becomes equal to the acceleration due to gravity, the normal force exerted on her by the mattress will drop to zero at one point in the cycle:

From x = A cos ωt , v =

2

Aω = g P15.45

or

b g

F = 3.00 cos 2π t N 2π = 2π rad s T

(a)

ω=

(b)

In this case,

A=

g

ω2

=

g k m

e9.80 m s jb12.5 kgg = A= 2

gm = k

4.30 × 10 3 N m

and

k = 20.0 N m

so

T = 1.00 s

ω0 =

k = m

2.85 cm

20.0 = 3.16 rad s 2.00

The equation for the amplitude of a driven oscillator,

P15.46

455

FG F IJ eω H mK 0

2

− ω 02

j

−1

a f

3 4π 2 − 3.16 2

with b = 0, gives

A=

Thus

A = 0.050 9 m = 5.09 cm .

F0 cos ωt − kx = m

b

x = A cos ωt + φ

d2x dt 2

ω0 =

=

2 −1

k m

(1)

g

(2)

b

dx = − Aω sin ωt + φ dt

g

d2x = − Aω 2 cos ωt + φ dt 2

b

(3)

g

(4)

b g e j b j cosbωt + φ g = F cos ωt

Substitute (2) and (4) into (1):

F0 cos ωt − kA cos ωt + φ = m − Aω 2 cos ωt + φ

Solve for the amplitude:

ekA − mAω

2

0

These will be equal, provided only that φ must be zero and kA − mAω 2 = F0 Thus, A =

F0 m

c h−ω k m

2

g

456 P15.47

Oscillatory Motion

From the equation for the amplitude of a driven oscillator with no damping, F0 m

A=

eω

2

− ω 02

j

2

e

ω = 2π f = 20.0π s −1

e

F0 = mA ω F0 =

P15.48

j

FG 40.0 IJ e2.00 × 10 jb3 950 − 49.0g = H 9.80 K −2

eω

2

− ω 02

j + b bω m g 2

k 200 = = 49.0 s −2 40 .0 m 9.80

c h

318 N

A=

2

Fext m

eω

2

− ω 02

j

2

=

e

Fext m

±ω

2

− ω 02

j

=±

ω 2 = ω 02 ±

This yields

ω = 8.23 rad s or ω = 4.03 rad s

Then,

f=

ω 2π

Fext m

ω 2 − ω 02

Fext m k Fext 6.30 N m 1.70 N = ± = ± 0.150 kg A m mA 0.150 kg 0.440 m

Thus,

b

gives either f = 1.31 Hz

or

f

ga

f = 0.641 Hz

The beeper must resonate at the frequency of a simple pendulum of length 8.21 cm: f=

*P15.50

− ω 02

ω 02 =

Fext m

A=

With b = 0,

P15.49

2

j

1 2π

g 1 = L 2π

9.80 m s 2 = 1.74 Hz . 0.082 1 m

For the resonance vibration with the occupants in the car, we have for the spring constant of the suspension f=

1 2π

k m

e j d1 130 kg + 4b72.4 kg gi = 1.82 × 10 F 4b72. 4 kg ge9.8 m s j x= = = 1.56 × 10 m k = 4π 2 f 2 m = 4π 2 1.8 s −1

2

2

Now as the occupants exit

k

1.82 × 10

5

kg s

2

−2

5

kg s 2

Chapter 15

457

Additional Problems P15.51

Let F represent the tension in the rod. pivot (a)

At the pivot, F = Mg + Mg = 2 Mg A fraction of the rod’s weight Mg

FG y IJ as well as the H LK

P L

weight of the ball pulls down on point P. Thus, the tension in the rod at point P is F = Mg

FG y IJ + Mg = H LK

FG H

Mg 1 +

y L

IJ K

y

.

M FIG. P15.51

(b)

Relative to the pivot, I = I rod + I ball =

1 4 ML2 + ML2 = ML2 3 3

I where m = 2 M and d is the distance from the mgd pivot to the center of mass of the rod and ball combination. Therefore, For the physical pendulum, T = 2π

d=

For L = 2.00 m, T =

P15.52

(a)

Total energy =

M

4π 3

c h + ML = 3L and T = 2π L 2

M+M

a

4

f=

2 2.00 m 9.80 m s

2

b

4 3

ML2

a 2 M f gc h 3L 4

4π 3

=

2L . g

2.68 s .

f

ga

1 2 1 kA = 100 N m 0.200 m 2 2

2

= 2.00 J

At equilibrium, the total energy is:

b

g

b

g

b

g

1 1 m1 + m 2 v 2 = 16.0 kg v 2 = 8.00 kg v 2 . 2 2 Therefore,

b8.00 kg gv

2

= 2.00 J , and v = 0.500 m s .

This is the speed of m1 and m 2 at the equilibrium point. Beyond this point, the mass m 2 moves with the constant speed of 0.500 m/s while mass m1 starts to slow down due to the restoring force of the spring. continued on next page

458

Oscillatory Motion

(b)

The energy of the m1 -spring system at equilibrium is:

b

gb

1 1 m1 v 2 = 9.00 kg 0.500 m s 2 2 This is also equal to Therefore,

g

2

= 1.125 J .

a f

1 2 k A′ , where A ′ is the amplitude of the m1 -spring system. 2

a fa f

1 100 A ′ 2

2

= 1.125 or A ′ = 0.150 m. m1 = 1.885 s k

The period of the m1 -spring system is T = 2π

1 T = 0.471 s after it passes the equilibrium point for the spring to become fully 4 stretched the first time. The distance separating m1 and m 2 at this time is: and it takes

D=v

P15.53

F d xI GH dt JK 2

FG T IJ − A ′ = 0.500 m s a0.471 sf − 0.150 m = 0.085 6 = H 4K µs

= Aω 2

2

B

max

P

fmax = µ sn = µ s mg = mAω 2 A=

8.56 cm .

µsg = 6.62 cm ω2

n f

B mg

FIG. P15.53 P15.54

The maximum acceleration of the oscillating system is a max = Aω 2 = 4π 2 Af 2 . The friction force exerted between the two blocks must be capable of accelerating block B at this rate. Thus, if Block B is about to slip,

e

f = fmax = µ sn = µ s mg = m 4π 2 Af 2 P15.55

j

or

µs g . 4π 2 f 2

A=

Deuterium is the isotope of the element hydrogen with atoms having nuclei consisting of one proton and one neutron. For brevity we refer to the molecule formed by two deuterium atoms as D and to the diatomic molecule of hydrogen-1 as H.

MD = 2MH

ωD = ωH

k MD k MH

=

MH = MD

1 2

fD =

fH 2

= 0.919 × 10 14 Hz

Chapter 15

P15.56

1 1 mv 2 + IΩ 2 , 2 2 where Ω is the rotation rate of the ball about its center of mass. Since the center of the ball moves along a circle of radius 4R, its displacement from equilibrium is s = 4R θ and its speed is ds dθ = 4R v= . Also, since the ball rolls without dt dt slipping,

459

The kinetic energy of the ball is K =

5R

a f

FG IJ H K

v=

θ

ds = RΩ dt

Ω=

so

R h

FG IJ H K

v dθ =4 R dt

s

FIG. P15.56

The kinetic energy is then K= =

FG H

1 dθ m 4R 2 dt

IJ K

2

FG IJ H K

112mR 2 dθ 10 dt

+

FG H

1 2 mR 2 2 5

IJ FG 4 dθ IJ K H dt K

2

2

a

f

When the ball has an angular displacement θ, its center is distance h = 4R 1 − cos θ higher than when at the equilibrium position. Thus, the potential energy is U g = mgh = 4mgR 1 − cos θ . For small

a

f

angles, 1 − cos θ ≈

a

f

2

θ (see Appendix B). Hence, U g ≈ 2mgRθ 2 , and the total energy is 2 E = K +Ug =

FG IJ H K

112mR 2 dθ 10 dt

2

+ 2mgRθ 2 .

FG IJ H K 28 R d θ d θ F 5 g IJθ . + gθ = 0 , or This reduces to = −G H 28R K 5 dt dt Since E = constant in time, 2

FG IJ H K

112 mR 2 dθ d 2θ dE dθ . + 4mgRθ =0= 5 dt dt dt 2 dt 2

2

2

With the angular acceleration equal to a negative constant times the angular position, this is in the 5g . defining form of a simple harmonic motion equation with ω = 28 R The period of the simple harmonic motion is then T =

2π

ω

= 2π

28 R . 5g

460 P15.57

Oscillatory Motion

(a) Li

L a

h

a

FIG. P15.57(a) (b)

T = 2π

π 1 dL dT = dt g L dt

L g

af

We need to find L t and

dL . From the diagram in (a), dt

FG IJ H K

1 dh a h dL =− − ; . 2 dt 2 2 dt

L = Li + But

(1)

dM dV dh =ρ = − ρA . Therefore, dt dt dt

IJ K

(2)

FG 1 IJ FG dM IJ t = L − L H 2 ρA K H dt K

(3)

FG H

1 dM dL dh 1 dM =− = ; ρA dt dt dt 2 ρA dt

z

L

Also,

dL =

Li

i

Substituting Equation (2) and Equation (3) into Equation (1):

F GH

π 1 g 2 ρa 2

dT = dt (c)

I FG dM IJ JK H dt K

1 Li +

1 2 ρa 2

c ht dM dt

.

Substitute Equation (3) into the equation for the period. 2π

T=

g

Li +

FG IJ H K

1 dM t 2 dt 2 ρa

Or one can obtain T by integrating (b):

F GH

I FG dM IJ z JK H dt K z L + dt c ht L 2 OPL O 1 F dM I π F 1 I F dM I M T −T = L + t− L P M G J G J G J g H 2 ρa K H dt K M PQ N c h PQMN 2 ρa H dt K L 2π 1 F dM I , so T = L + G Jt . g 2 ρa H dt K g Ti

dT =

T

i

But Ti = 2π

i

t

π 1 g 2 ρa 2

0

2

i

2

i

1 2 ρa 2

1 2 ρa 2

dM dt

dM dt

i

2

i

Chapter 15

P15.58

ω=

(a) P15.59

k 2π = m T 2

k =ω m =

4π 2 m

m′ =

(b)

T2

a f

k T′

= m

4π 2

FG T ′ IJ HTK

2

Hy

We draw a free-body diagram of the pendulum. The force H exerted by the hinge causes no torque about the axis of rotation.

τ = Iα

Hx h

d 2θ = −α dt 2

and

τ = MgL sin θ + kxh cos θ = − I

kx

Lθ

x

2

d θ dt 2

k m

mg

For small amplitude vibrations, use the approximations: sin θ ≈ θ , cos θ ≈ 1, and x ≈ s = hθ . Therefore,

F GH

(a)

I JK

MgL + kh 2 d 2θ θ = −ω 2θ = − I dt 2

b

g

ω=

ML2 1 2π

we have at t = 0

v = −ωA sin φ = − v max

This requires φ = 90° , so

x = A cos ωt + 90°

And this is equivalent to

x = − A sin ωt

Numerically we have

ω=

and v max = ωA

20 m s = 10 s −1 A

In

a

MgL + kh 2 ML2

f

50 N m = 10 s −1 0.5 kg

e

a

= 2π f

g

v = −ωA sin ωt + φ

j

f e

A=2m

j

x = −2 m sin 10 s −1 t

So

(b)

b

MgL + kh 2

In x = A cos ωt + φ ,

k = m

L sinθ

FIG. P15.59

f= *P15.60

2

1 1 1 mv 2 + kx 2 = kA 2 , 2 2 2

implies

FG H

1 2 1 kx = 3 mv 2 2 2

11 2 1 2 1 2 kx + kx = kA 32 2 2 x=±

continued on next page

IJ K

3 A = ±0.866 A = ±1.73 m 4

4 2 x = A2 3

hcosθ

461

462

Oscillatory Motion

(c)

ω=

(d)

In

g L

g

L=

a

ω2

=

9.8 m s 2

e10 s j

−1 2

f e

= 0.098 0 m

j

x = −2 m sin 10 s −1 t

the particle is at x = 0 at t = 0 , at 10t = π s , and so on. The particle is at

x=1 m

when

−

with solutions

e10 s jt = − π6

1 = sin 10 s −1 t 2

e

j

−1

e10 s jt = π + π6 , and so on. FπI The minimum time for the motion is ∆t in 10 ∆t = G J s H 6K FπI ∆t = G J s = 0.052 4 s H 60 K −1

P15.61

(a)

FIG. P15.60(d)

At equilibrium, we have

F LI

∑ τ = 0 − mg GH 2 JK + kx0 L where x 0 is the equilibrium compression. After displacement by a small angle,

FIG. P15.61

F LI

F LI

∑ τ = − mg GH 2 JK + kxL = −mg GH 2 JK + kbx0 − Lθ gL = − kθL2 But,

1

∑ τ = Iα = 3 mL2

d 2θ . dt 2

So

d 2θ 3k = − θ. 2 m dt

The angular acceleration is opposite in direction and proportional to the displacement, so 3k we have simple harmonic motion with ω 2 = . m (b)

f=

ω 1 = 2π 2π

3k 1 = m 2π

b

g=

3 100 N m 5.00 kg

1.23 Hz

Chapter 15

*P15.62

463

As it passes through equilibrium, the 4-kg object has speed v max = ωA =

100 N m k A= 2 m = 10.0 m s. m 4 kg

In the completely inelastic collision momentum of the two-object system is conserved. So the new 10-kg object starts its oscillation with speed given by

b

g b g b

g

4 kg 10 m s + 6 kg 0 = 10 kg v max v max = 4.00 m s (a)

The new amplitude is given by

1 1 2 = kA 2 mv max 2 2

b

10 kg 4 m s

g = b100 N mgA 2

2

A = 1.26 m

(b)

(c)

Thus the amplitude has decreased by

2.00 m − 1.26 m = 0.735 m

The old period was

T = 2π

4 kg m = 2π = 1.26 s k 100 N m

The new period is

T = 2π

10 2 s = 1.99 s 100

The period has increased by

1.99 m − 1.26 m = 0.730 s

The old energy was

1 1 2 = 4 kg 10 m s mv max 2 2

The new mechanical energy is

1 10 kg 4 m s 2

b gb

b

gb

g

2

g

2

= 200 J

= 80 J

The energy has decreased by 120 J .

P15.63

(d)

The missing mechanical energy has turned into internal energy in the completely inelastic collision.

(a)

T=

(b)

E=

(c)

At maximum angular displacement

2π

ω

= 2π

L = 3.00 s g

a fa f

1 1 mv 2 = 6.74 2.06 2 2

a

h = L − L cos θ = L 1 − cos θ

2

f

= 14.3 J

mgh =

1 mv 2 2

cos θ = 1 −

h L

h=

v2 = 0.217 m 2g

θ = 25.5°

464 P15.64

Oscillatory Motion

One can write the following equations of motion: T − kx = 0

(describes the spring)

mg − T ′ = ma = m

a

f

R T′ − T = I

2

d x dt 2

(for the hanging object)

d 2θ I d 2 x = dt 2 R dt 2

(for the pulley) with I =

FIG. P15.64

1 MR 2 2

Combining these equations gives the equation of motion

FG m + 1 MIJ d x + kx = mg . H 2 K dt 2

2

af

mg mg (where arises because of the extension of the spring due to k k the weight of the hanging object), with frequency The solution is x t = A sin ωt +

f=

P15.65

ω 2π

=

k 1 = m + 12 M 2π

1 2π

(a)

For M = 0

f = 3.56 Hz

(b)

For M = 0.250 kg

f = 2.79 Hz

(c)

For M = 0.750 kg

f = 2.10 Hz

100 N m . 0.200 kg + 12 M

Suppose a 100-kg biker compresses the suspension 2.00 cm. Then,

k=

F 980 N = = 4.90 × 10 4 N m x 2.00 × 10 −2 m

If total mass of motorcycle and biker is 500 kg, the frequency of free vibration is f=

1 2π

1 k = m 2π

4.90 × 10 4 N m = 1.58 Hz 500 kg

If he encounters washboard bumps at the same frequency, resonance will make the motorcycle bounce a lot. Assuming a speed of 20.0 m/s, we find these ridges are separated by 20.0 m s 1.58 s −1

= 12.7 m ~ 10 1 m .

In addition to this vibration mode of bouncing up and down as one unit, the motorcycle can also vibrate at higher frequencies by rocking back and forth between front and rear wheels, by having just the front wheel bounce inside its fork, or by doing other things. Other spacing of bumps will excite all of these other resonances.

Chapter 15

P15.66

(a)

For each segment of the spring dK = Also,

vx =

x v A

a f

1 dm v x2 . 2 dm =

and

m dx . A

FIG. P15.66

Therefore, the total kinetic energy of the block-spring system is K=

(b)

ω=

k m eff

Therefore,

P15.67

(a)

∑ F = −2T sin θ j

z FGH A

0

I JK

FG H

IJ K

x2v2 m m 2 1 dx = M+ v . 2 2 3 A A

FG H

IJ K

1 1 m 2 m eff v 2 = M+ v 2 2 3

and

T=

1 1 Mv 2 + 2 2

2π

ω

= 2π

M + m3 k

.

where θ = tan −1

FG y IJ H LK

Therefore, for a small displacement y sin θ ≈ tan θ = L (b)

and

FIG. P15.67

−2Ty ∑ F = L j

The total force exerted on the ball is opposite in direction and proportional to its displacement from equilibrium, so the ball moves with simple harmonic motion. For a spring system,

∑ F = −kx

becomes here

Therefore, the effective spring constant is

2T and L

∑F = − ω=

2T y. L k = m

2T . mL

465

466 P15.68

Oscillatory Motion

(a)

Assuming a Hooke’s Law type spring, F = Mg = kx and empirically Mg = 1.74x − 0.113 k = 1.74 N m ± 6% .

so

(b)

M , kg 0.020 0

x, m 0.17

Mg , N 0.196

0.040 0

0.293

0.392

0.050 0

0.353

0.49

0.060 0

0.413

0.588

0.070 0

0.471

0.686

0.080 0

0.493

0.784

We may write the equation as theoretically T2 =

4π 2 4π 2 M+ ms 3k k

FIG. P15.68

and empirically T 2 = 21.7 M + 0.058 9 k=

so

4π 2 = 1.82 N m ± 3% 21.7

Time, s T , s 7.03 0.703

M , kg 0.020 0

T 2 , s2 0.494

9.62

0.962

0.040 0

0.925

10.67

1.067

0.050 0

1.138

11.67

1.167

0.060 0

1.362

12.52

1.252

0.070 0

1.568

13.41

1.341

0.080 0

1.798

The k values 1.74 N m ± 6% and so (c)

1.82 N m ± 3% differ by 4% they agree.

Utilizing the axis-crossing point,

ms = 3

FG 0.058 9 IJ kg = H 21.7 K

8 grams ± 12%

in agreement with 7.4 grams.

Chapter 15

P15.69

(a)

∆K + ∆U = 0 Thus, K top + U top = K bot + U bot

M R

where K top = U bot = 0 1 2 Iω , but 2 h = R − R cos θ = R 1 − cos θ

Therefore, mgh =

a

θ

f

v R MR 2 mr 2 and I = + + mR 2 2 2 Substituting we find

ω=

2

2

2

(b)

M m

M m

2

+

r2 R2

r2 R2

+2

1 2

dCM =

af

mR + M 0 m+M

MR 2 + 12 mr 2 + mR 2 mgR

We require Ae − bt 2 m =

A 2

bt = ln 2 or 2m The spring constant is irrelevant.

or

(b)

2

2

I mT gd CM

T = 2π

T = 2π

(a)

2

2

mT = m + M

P15.70

e + bt 2 m = 2 0.100 kg s t = 0.693 2 0.375 kg

b

∴ t = 5.20 s

g

We can evaluate the energy at successive turning points, where 1 1 1 1 1 2 cos ωt + φ = ±1 and the energy is kx 2 = kA 2 e − bt 2 m . We require kA 2 e − bt 2 m = kA 2 2 2 2 2 m ln 2 0.375 kg 0.693 ∴t = = = 2.60 s . or e + bt m = 2 b 0.100 kg s

b

g

a

(c)

m FIG. P15.69

2

so

θ

v

f 12 FGH MR2 + mr2 + mR IJK Rv L M mr + m OPv mgRa1 − cos θ f = M + N 4 4R 2 Q a1 − cos θ f v = 4 gR e + + 2j Rg a1 − cos θ f v=2 a

mgR 1 − cos θ =

and

467

f

1 2 kA , the fractional rate of change of energy over time is 2 2 d 1 dE dA dA 1 dt 2 kA dt dt 2 k 2 A dt = 1 = = 2 2 1 kA 2 E A kA 2 2

From E =

e

j

a f

two times faster than the fractional rate of change in amplitude.

FG H

IJ K

468 P15.71

Oscillatory Motion

(a)

When the mass is displaced a distance x from equilibrium, spring 1 is stretched a distance x1 and spring 2 is stretched a distance x 2 . By Newton’s third law, we expect k1 x1 = k 2 x 2 . When this is combined with the requirement that x = x1 + x 2 ,

FIG. P15.71

LM k OPx Nk + k Q L k k OP x = ma F =M Nk + k Q x1 =

we find The force on either spring is given by

(b)

2

1

2

This is in the form

F = k eff x = ma

and

T = 2π

b

m k1 + k 2 m = 2π k eff k1 k 2

g

In this case each spring is distorted by the distance x which the mass is displaced. Therefore, the restoring force is

b

g

F = − k1 + k 2 x so that P15.72

1

1 2

1

where a is the acceleration of the mass m.

2

T = 2π

b

k eff = k1 + k 2

and

m k1 + k 2

g

.

Let A represent the length below water at equilibrium and M the tube’s mass:

∑ Fy = 0 ⇒ − Mg + ρπ r 2 Ag = 0 . Now with any excursion x from equilibrium

a f

− Mg + ρπ r 2 A − x g = Ma . Subtracting the equilibrium equation gives − ρπ r 2 gx = Ma a=−

F ρπ r g I x = −ω x GH M JK 2

2

The opposite direction and direct proportionality of a to x imply SHM with angular frequency

ω= T=

ρπ r 2 g M 2π

ω

=

FG 2 IJ HrK

πM ρg

Chapter 15

P15.73

For θ max = 5.00° , the motion calculated by the Euler method agrees quite precisely with the prediction of θ max cos ωt . The period is T = 2.20 s . Time, t (s) 0.000 0.004 0.008 … 0.544 0.548 0.552 … 1.092 1.096 1.100 1.104 … 1.644 1.648 1.652 … 2.192 2.196 2.200 2.204

Angle, θ (°) 5.000 0 4.999 3 4.998 0

Ang. speed (°/s) 0.000 0 –0.163 1 –0.326 2

Ang. Accel. ° s2

e j

θ max cos ωt

–40.781 5 –40.776 2 –40.765 6

5.000 0 4.999 7 4.998 7

0.056 0 –0.001 1 –0.058 2

–14.282 3 –14.284 2 –14.284 1

–0.457 6 0.009 0 0.475 6

0.081 0 0.023 9 –0.033 3

–4.999 4 –5.000 0 –5.000 0 –4.999 3

–0.319 9 –0.156 8 0.006 3 0.169 4

40.776 5 40.781 6 40.781 4 40.775 9

–4.998 9 –4.999 8 –5.000 0 –4.999 6

–0.063 8 0.003 3 0.060 4

14.282 4 14.284 2 14.284 1

0.439 7 –0.027 0 –0.493 6

–0.071 6 –0.014 5 0.042 7

4.999 4 5.000 0 5.000 0 4.999 3

0.313 7 0.150 6 –0.012 6 –0.175 7

–40.776 8 –40.781 7 –40.781 3 –40.775 6

4.999 1 4.999 9 5.000 0 4.999 4

For θ max = 100° , the simple harmonic motion approximation θ max cos ωt diverges greatly from the Euler calculation. The period is T = 2.71 s , larger than the small-angle period by 23%. Time, Angle, t (s) θ (°) 0.000 100.000 0 0.004 99.992 6 0.008 99.977 6 … 1.096 –84.744 9 1.100 –85.218 2 1.104 –85.684 0 … 1.348 –99.996 0 1.352 –100.000 8 1.356 –99.998 3 … 2.196 40.150 9 2.200 41.045 5 2.204 41.935 3 … 2.704 99.998 5 2.708 100.000 8 2.712 99.995 7

Ang. speed (°/s) 0.000 0 –1.843 2 –3.686 5

Ang. Accel. ° s2

e j

θ max cos ωt

–460.606 6 –460.817 3 –460.838 2

100.000 0 99.993 5 99.973 9

–120.191 0 –118.327 2 –116.462 0

465.948 8 466.286 9 466.588 6

–99.995 4 –99.999 8 –99.991 1

–3.053 3 –1.210 0 0.633 2

460.812 5 460.805 7 460.809 3

–75.797 9 –75.047 4 –74.287 0

224.867 7 223.660 9 222.431 8

–301.713 2 –307.260 7 –312.703 5

99.997 1 99.999 3 99.988 5

2.420 0 0.576 8 –1.266 4

–460.809 0 –460.805 7 –460.812 9

12.642 2 11.507 5 10.371 2

FIG. P15.73

469

470 *P15.74

Oscillatory Motion

(a)

The block moves with the board in what we take as the positive x direction, stretching the spring until the spring force −kx is equal in magnitude to the maximum force of static µ mg . friction µ sn = µ s mg . This occurs at x = s k

(b)

Since v is small, the block is nearly at the rest at this break point. It starts almost immediately to move back to the left, the forces on it being −kx and + µ k mg . While it is sliding the net force exerted on it can be written as − kx + µ k mg = − kx +

FG H

IJ K

kµ k mg µ mg = −k x − k = − kx rel k k

where x rel is the excursion of the block away from the point

µ k mg . k

Conclusion: the block goes into simple harmonic motion centered about the equilibrium µ mg position where the spring is stretched by k . k (d)

The amplitude of its motion is its original displacement, A =

b

g

µ s mg µ k mg − . It first comes to k k

2 µ k − µ s mg µ k mg . Almost immediately at this point it −A= k k latches onto the slowly-moving board to move with the board. The board exerts a force of static friction on the block, and the cycle continues. rest at spring extension

(c)

The graph of the motion looks like this:

FIG. P15.74(c) (e)

b

g

2 A 2 µ s − µ k mg = . v kv The time for which the block is springing back is one half a cycle of simple harmonic motion,

The time during each cycle when the block is moving with the board is

F GH

I JK

1 m m =π . We ignore the times at the end points of the motion when the speed of 2π 2 k k 2A the block changes from v to 0 and from 0 to v. Since v is small compared to , these

π

times are negligible. Then the period is T= continued on next page

b

g

2 µ s − µ k mg kv

+π

m . k

m k

Chapter 15

(f)

T=

a

fb

b0.024 m sgb12 N mg f=

Then

*P15.75

ge

2 0.4 − 0.25 0.3 kg 9.8 m s 2

b

g

2 µ s − µ k mg

j +π

0.3 kg = 3.06 s + 0.497 s = 3.56 s 12 N m

1 = 0.281 Hz . T +π

m increases as m increases, so the frequency decreases . k

(g)

T=

(h)

As k increases, T decreases and f increases .

(i)

As v increases, T decreases and f increases .

(j)

As µ s − µ k increases, T increases and f decreases .

(a)

Newton’s law of universal gravitation is

F=−

Thus,

F=−

Which is of Hooke’s law form with

k=

(b)

kv

b

471

g

GMm r

=−

2

FG H

IJ K

Gm 4 3 πr ρ r2 3

FG 4 πρGmIJ r H3 K

4 πρGm 3

FG 4 IJ πρGmr = ma H 3K F 4I a = − G J πρGr = −ω r H 3K

The sack of mail moves without friction according to

−

2

Since acceleration is a negative constant times excursion from equilibrium, it executes SHM with

ω=

4πρG 3

and period

T=

The time for a one-way trip through the earth is

T = 2

We have also

g=

so

g 4ρG = 3 πR e

b g

and

2π

ω

3π ρG

=

3π 4 ρG GM e R e2

=

G 4πR e3 ρ 3 R e2

(a) 4.33 cm; (b) −5.00 cm s ;

P15.6

see the solution

P15.8

12.0 Hz

P15.10

18.8 m s; 7.11 km s 2

2

(c) −17.3 cm s ; (d) 3.14 s; 5.00 cm P15.4

4 πρGR e 3

Re 6.37 × 10 6 m T =π =π = 2.53 × 10 3 s = 42.2 min . 2 2 g 9.8 m s

ANSWERS TO EVEN PROBLEMS P15.2

=

(a) 15.8 cm; (b) −15.9 cm; (c) see the solution; (d) 51.1 m; (e) 50.7 m

472

Oscillatory Motion

(a) 1.26 s; (b) 0.150 m s; 0.750 m s 2 ; 15 cm sin 5t ; (c) x = −3 cmcos 5t ; v = s 75 cm a= cos 5t s2

P15.42

see the solution

P15.44

(a) 2.95 Hz; (b) 2.85 cm

P15.46

see the solution

P15.48

either 1.31 Hz or 0.641 Hz

P15.14

F vI (a) ; (b) x = − G J sin ωt HωK ω

P15.50

1.56 cm

P15.16

(a) 126 N m; (b) 0.178 m

P15.52

(a) 0.500 m s ; (b) 8.56 cm

P15.18

(a) 0.153 J; (b) 0.784 m s; (c) 17.5 m s 2

P15.54

A=

P15.20

(a) 100 N m; (b) 1.13 Hz; (c) 1.41 m s at x = 0 ;

P15.56

see the solution

(f) 1.33 m s ; (g) 3.33 m s 2

P15.58

(a) k =

P15.22

(a) 1.50 s; (b) 73.4 N m; (c) 19.7 m below the bridge; (d) 1.06 rad s; (e) 2.01 s; (f) 3.50 s

P15.60

(a) x = −2 m sin 10t ; (b) at x ± 1.73 m; (c) 98.0 mm; (d) 52.4 ms

P15.24

(a) 0.218 s and 1.09 s; (b) 14.6 mW

P15.62

P15.26

The position of the piston is given by x = A cos ωt .

(a) decreased by 0.735 m; (b) increased by 0.730 s; (c) decreased by 120 J; (d) see the solution

P15.64

(a) 3.56 Hz ; (b) 2.79 Hz; (c) 2.10 Hz

P15.66

(a)

P15.68

see the solution; (a) k = 1.74 N m ± 6% ; (b) 1.82 N m ± 3%; they agree; (c) 8 g ± 12%; it agrees

P15.70

(a) 5.20 s; (b) 2.60 s; (c) see the solution

P15.72

see the solution; T =

P15.74

see the solution; (f) 0.281 Hz ; (g) decreases; (h) increases; (i) increases; (j) decreases

P15.12

FG H

FG H

IJ K

IJ K

v

µsg 4π 2 f 2

2

(d) 10.0 m s at x = ± A ; (e) 2.00 J;

P15.28

gC = 1.001 5 gT

P15.30

1.42 s; 0.499 m

P15.32

(a) 3.65 s; (b) 6.41 s; (c) 4.24 s

P15.34

(a) see the solution; (b), (c) 9.85 m s 2 ; agreeing with the accepted value within 0.5%

P15.36

(a) 2.09 s; (b) 4.08%

P15.38

203 µN ⋅ m

P15.40

see the solution

FG IJ H K

4π 2 m T′ ; (b) m ′ = m 2 T T

a

FG H

2

f a f

IJ K

M + m3 1 m 2 M+ v ; (b) T = 2π 2 3 k

FG 2 IJ HrK

πM ρg

16 Wave Motion CHAPTER OUTLINE 16.1 16.2 16.3 16.4 16.5 16.6

Propagation of a Disturbance Sinusoidal Waves The Speed of Waves on Strings Reflection and Transmission Rate of Energy Transfer by Sinusoidal Waves on Strings The Linear Wave Equation

ANSWERS TO QUESTIONS Q16.1

As the pulse moves down the string, the particles of the string itself move side to side. Since the medium—here, the string—moves perpendicular to the direction of wave propagation, the wave is transverse by definition.

Q16.2

To use a slinky to create a longitudinal wave, pull a few coils back and release. For a transverse wave, jostle the end coil side to side.

Q16.3

From v =

Q16.4

It depends on from what the wave reflects. If reflecting from a less dense string, the reflected part of the wave will be right side up.

T

µ

, we must increase the tension by a factor of 4.

2π vA

Q16.5

Yes, among other things it depends on. v max = ωA = 2π fA =

Q16.6

Since the frequency is 3 cycles per second, the period is

Q16.7

Amplitude is increased by a factor of

Q16.8

The section of rope moves up and down in SHM. Its speed is always changing. The wave continues on with constant speed in one direction, setting further sections of the rope into up-and-down motion.

Q16.9

Each element of the rope must support the weight of the rope below it. The tension increases with

λ

. Here v is the speed of the wave.

1 second = 333 ms. 3

2 . The wave speed does not change.

height. (It increases linearly, if the rope does not stretch.) Then the wave speed v =

T

µ

increases

with height. Q16.10

The difference is in the direction of motion of the elements of the medium. In longitudinal waves, the medium moves back and forth parallel to the direction of wave motion. In transverse waves, the medium moves perpendicular to the direction of wave motion.

473

474

Wave Motion

Q16.11

Slower. Wave speed is inversely proportional to the square root of linear density.

Q16.12

As the wave passes from the massive string to the less massive string, the wave speed will increase according to v =

T

. The frequency will remain unchanged. Since v = fλ , the wavelength must

µ

increase. Q16.13

Higher tension makes wave speed higher. Greater linear density makes the wave move more slowly.

Q16.14

The wave speed is independent of the maximum particle speed. The source determines the maximum particle speed, through its frequency and amplitude. The wave speed depends instead on properties of the medium.

Q16.15

Longitudinal waves depend on the compressibility of the fluid for their propagation. Transverse waves require a restoring force in response to sheer strain. Fluids do not have the underlying structure to supply such a force. A fluid cannot support static sheer. A viscous fluid can temporarily be put under sheer, but the higher its viscosity the more quickly it converts input work into internal energy. A local vibration imposed on it is strongly damped, and not a source of wave propagation.

Q16.16

Let ∆t = ts − t p represent the difference in arrival times of the two waves at a station at distance

d = v s ts = v p t p

F1 1I from the hypocenter. Then d = ∆tG − J Hv v K s

−1

. Knowing the distance from the first

p

station places the hypocenter on a sphere around it. A measurement from a second station limits it to another sphere, which intersects with the first in a circle. Data from a third non-collinear station will generally limit the possibilities to a point. Q16.17

The speed of a wave on a “massless” string would be infinite!

SOLUTIONS TO PROBLEMS Section 16.1 P16.1

Propagation of a Disturbance

Replace x by x − vt = x − 4.5t 6 to get y= 2 x − 4.5t + 3

a

f

Chapter 16

475

P16.2

FIG. P16.2 P16.3

a

5.00 e −a x + 5 t f is of the form f x + vt 2

f

so it describes a wave moving to the left at v = 5.00 m s . P16.4

(a)

The longitudinal wave travels a shorter distance and is moving faster, so it will arrive at point B first.

(b)

The wave that travels through the Earth must travel

e

j

a distance of

2 R sin 30.0° = 2 6.37 × 10 6 m sin 30.0° = 6.37 × 10 6 m

at a speed of

7 800 m/s

Therefore, it takes

6.37 × 10 6 m = 817 s 7 800 m s

The wave that travels along the Earth’s surface must travel

FG π radIJ = 6.67 × 10 H3 K

a distance of

s = Rθ = R

at a speed of

4 500 m/s

Therefore, it takes

6.67 × 10 6 = 1 482 s 4 500

The time difference is

665 s = 11.1 min

6

m

476 P16.5

Wave Motion

b

g b

where t is the travel time for the faster wave.

a fb g b ga f b4.50 km sga17.3 sf = 23.6 s or t = a7.80 − 4.50f km s and the distance is d = b7.80 km sga 23.6 sf = 184 km Then, 7.80 − 4.50 km s t = 4.50 km s 17.3 s

Section 16.2 P16.6

.

Sinusoidal Waves

Using data from the observations, we have λ = 1.20 m and f =

8.00 12.0 s

fFGH 128..000 s IJK =

a

Therefore, v = λf = 1.20 m

P16.7

0.800 m s

f=

40.0 vibrations 4 = Hz 30.0 s 3

λ=

v 42.5 cm s = 4 = 31.9 cm = 0.319 m f 3 Hz

a

fa

v=

425 cm = 42.5 cm s 10.0 s

f

P16.8

v = fλ = 4.00 Hz 60.0 cm = 240 cm s = 2.40 m s

P16.9

y = 0.020 0 m sin 2.11x − 3.62t in SI units

A = 2.00 cm

k = 2.11 rad m

λ=

2π = 2.98 m k

ω = 3.62 rad s

f=

ω = 0.576 Hz 2π

b

v = fλ = P16.10

ga

The distance the waves have traveled is d = 7.80 km s t = 4.50 km s t + 17.3 s

b

g a

f

ω 2π 3.62 = = 1.72 m s 2π k 2.11

g a

f

y = 0.005 1 m sin 310 x − 9.30t SI units v=

ω 9.30 = = 0.030 0 m s k 310

s = vt = 0.300 m in positive x - direction

f

Chapter 16

*P16.11

a

f db

g b

(a)

The transverse velocity is

∂y = − Aω cos kx − ωt ∂t

Its maximum magnitude is

Aω = 12 cm 31.4 rad s = 3.77 m s

ay =

(b)

∂v y ∂t

=

a

b

a

c

fh

f g

a

∂ − Aω cos kx − ωt = − Aω 2 sin kx − ωt ∂t

a

f

fe

Aω 2 = 0.12 m 31.4 s −1

The maximum value is P16.12

gi

From y = 12.0 cm sin 1.57 rad m x − 31.4 rad s t

a

f a

j

2

= 118 m s 2

f

At time t, the phase of y = 15.0 cm cos 0.157 x − 50.3t at coordinate x is

b

g b

g

φ = 0.157 rad cm x − 50.3 rad s t . Since 60.0° = φB = φ A ±

π rad , or (since x A = 0 ), 3

π rad , the requirement for point B is that 3

b0.157 rad cmgx − b50.3 rad sgt = 0 − b50.3 rad sgt ± π3 rad . B

This reduces to x B = P16.13

±π rad = ±6.67 cm . 3 0.157 rad cm

b

a

g

f

y = 0.250 sin 0.300 x − 40.0t m

a

Compare this with the general expression y = A sin kx − ωt (a)

A = 0.250 m

(b)

ω = 40.0 rad s

(c)

k = 0.300 rad m

(d)

λ=

(e)

v = fλ =

(f)

The wave moves to the right, in + x direction .

2π 2π = = 20.9 m k 0.300 rad m

FG ω IJ λ = FG 40.0 rad s IJ a20.9 mf = H 2π K H 2π K

133 m s

f

477

478 P16.14

Wave Motion

(a)

See figure at right.

(b)

T=

2π

=

ω

y (cm)

2π = 0.125 s 50.3

10 0

This agrees with the period found in the example in the text.

t (s) 0.1

—10

0.2

FIG. P16.14 P16.15

(a)

k=

φ = −0.785

In general, Assuming

b

(a)

y (mm) 0.2 0.1 0.0 –0.1 –0.2

t=0 0.2 0.4

x (mm)

FIG. P16.16(a) (b)

2π = 18.0 rad m λ 0.350 m 1 1 T= = = 0.083 3 s f 12.0 s k=

2π

=

ω = 2π f = 2π 12.0 s = 75.4 rad s

f b ga y = A sinb kx + ωt + φ g specializes to y = 0.200 m sinb18.0 x m + 75.4t s + φ g

v = fλ = 12.0 s 0.350 m = 4.20 m s

(c)

at x = 0 , t = 0 we require

b g

−3.00 × 10 −2 m = 0. 200 m sin +φ

φ = −8.63° = −0.151 rad so

b g a0.200 mf sinb18.0 x m + 75.4t s − 0.151 radg

y x, t =

g

y = 0.080 0 sin 7.85 x + 6π t − 0.785 m

Therefore, P16.16

−1

or

b g

Or (where y 0 , t = 0 at t = 0 )

λ

=

2π

then we require that

Therefore,

(b)

2π

a0.800 mf = 7.85 m ω = 2π f = 2π a3.00f = 6.00π rad s y = A sina kx + ωt f y = b0.080 0g sinb7.85 x + 6π t g m y = 0.080 0 sinb7.85 x + 6π t + φ g yb x, 0g = 0 at x = 0.100 m 0 = 0.080 0 sinb0.785 + φ g

A = y max = 8.00 cm = 0.080 0 m

Chapter 16

P16.17

f FGH π8 x + 4π tIJK

a

y = 0.120 m sin

(a)

(b)

v=

dy : dt

a=

dv : dt

fa f FGH π8 x + 4π tIJK va0.200 s, 1.60 mf = −1.51 m s Fπ I a = a −0.120 mfa 4π f sinG x + 4π tJ H8 K aa0.200 s, 1.60 mf = 0 a

x = 0.120 4π cos

2

π 2π = : 8 λ 2π ω = 4π = : T k=

λ = 16.0 m T = 0.500 s v=

P16.18

479

(a)

λ 16.0 m = = 32.0 m s T 0.500 s

b g b g yb0 , 0g = A sin φ = 0.020 0 m y x , t = A sin kx + ωt + φ

Let us write the wave function as

dy dt

= Aω cos φ = −2.00 m s 0, 0

ω=

Also,

2π 2π = = 80.0 π s T 0.025 0 s

A 2 = xi2 +

FG v IJ = b0.020 0 mg + FG 2.00 m s IJ HωK H 80.0 π s K i

2

2

2

A = 0.021 5 m (b)

A sin φ 0.020 0 = −2 = −2.51 = tan φ A cos φ 80 .0π

a

f

Your calculator’s answer tan −1 −2.51 = −1.19 rad has a negative sine and positive cosine, just the reverse of what is required. You must look beyond your calculator to find

φ = π − 1.19 rad = 1.95 rad

b

g

(c)

v y, max = Aω = 0.021 5 m 80.0π s = 5.41 m s

(d)

λ = v x T = 30.0 m s 0.025 0 s = 0.750 m

b

k=

2π

λ

=

ga

2π = 8.38 m 0.750 m

b g b

g b

f

ω = 80.0π s

y x , t = 0.021 5 m sin 8.38 x rad m + 80.0π t rad s + 1.95 rad

g

480 P16.19

Wave Motion

(a)

f=

v

=

λ

b1.00 m sg = 2.00 m

b

0.500 Hz

g

ω = 2π f = 2π 0.500 s = 3.14 rad s 2π

2π = 3.14 rad m 2.00 m

(b)

k=

(c)

y = A sin kx − ωt + φ becomes

λ

b

g

a0.100 mf sinb3.14x m − 3.14t s + 0g

y= (d)

=

For x = 0 the wave function requires

a

f b

a

f b

y = 0.100 m sin −3.14t s (e)

(f)

g

y = 0.100 m sin 4.71 rad − 3.14 t s vy =

g

∂y = 0.100 m − 3.14 s cos 3.14x m − 3.14t s ∂t

b

g b

g

The cosine varies between +1 and –1, so

b

v y ≤ 0.314 m s P16.20

g

a0.100 mf sina1.00 rad − 20.0tf

(a)

at x = 2.00 m , y =

(b)

y = 0.100 m sin 0.500 x − 20.0t = A sin kx − ωt

a

f a

f

so ω = 20.0 rad s and f =

Section 16.3 P16.21

P16.22

ω 2π

a

f

= 3.18 Hz

The Speed of Waves on Strings

The down and back distance is 4.00 m + 4.00 m = 8.00 m .

a

f

The speed is then

v=

d total 4 8.00 m T = = 40.0 m s = t 0.800 s µ

Now,

µ=

0.200 kg = 5.00 × 10 −2 kg m 4.00 m

So

T = µv 2 = 5.00 × 10 −2 kg m 40.0 m s

The mass per unit length is: µ =

jb

e

g

2

= 80.0 N

0.060 0 kg = 1.20 × 10 −2 kg m . 5.00 m

b

gb

The required tension is: T = µv 2 = 0.012 0 kg m 50.0 m s

g

2

= 30.0 N .

Chapter 16

P16.23

v=

P16.24

(a)

T

µ

1 350 kg ⋅ m s 2

=

5.00 × 10 −3 kg m

= 520 m s

a f

ω = 2π f = 2π 500 = 3 140 rad s , k =

j b

e

y = 2.00 × 10 −4 m sin 16.0 x − 3 140t v = 196 m s =

(b)

481

ω 3 140 = = 16.0 rad m v 196

g

T 4.10 × 10 −3 kg m

T = 158 N P16.25

P16.26

T

Mg

T = Mg is the tension;

v=

Then,

MgL L2 = 2 m t

and

g=

=

µ

m L

MgL L = is the wave speed. m t

=

e e

j

1.60 m 4.00 × 10 −3 kg Lm = = 1.64 m s 2 Mt 2 3.00 kg 3.61 × 10 −3 s 2

j

T

v=

µ

T = µv 2 = ρAv 2 = ρπr 2 v 2

ja fe

e

j b200 m sg

T = 8 920 kg m3 π 7.50 × 10 −4 m

2

2

T = 631 N P16.27

Since µ is constant, µ =

T2

v 22

=

T2

P16.28

T1

v12

and

Fv I =G J Hv K 2

1

2

F 30.0 m s I a6.00 Nf = T =G H 20.0 m s JK

The period of the pendulum is T = 2π

2

1

13.5 N .

L g

Let F represent the tension in the string (to avoid confusion with the period) when the pendulum is vertical and stationary. The speed of waves in the string is then: v=

Mg MgL F = m = µ m L

Since it might be difficult to measure L precisely, we eliminate so v =

Mg T g Tg = m 2π 2π

M . m

L=

T g 2π

482 P16.29

Wave Motion

If the tension in the wire is T, the tensile stress is Stress =

T A

a

f

T = A stress .

so

The speed of transverse waves in the wire is v=

T

µ

=

a

A Stress m L

f=

Stress m AL

=

Stress m Volume

=

Stress

ρ

where ρ is the density. The maximum velocity occurs when the stress is a maximum: v max = P16.30

2.70 × 10 8 Pa = 185 m s . 7 860 kg m 3

mg = 2T sin θ

From the free-body diagram

T= The angle θ is found from

mg 2 sin θ

cos θ =

3L 8 L 2

=

3 4 FIG. P16.30

∴θ = 41.4° (a)

v=

T

v=

µ

v=

or (b) P16.31

mg = 2 µ sin 41. 4°

F 30.4 GH

I J kg K

ms

F I 9.80 m s GG JJ × ° 2 8 . 00 10 kg m sin 41 . 4 e j H K 2

−3

m

m

m = 3.89 kg

v = 60.0 = 30.4 m and

The total time is the sum of the two times.

µ L =L v T Let A represent the cross-sectional area of one wire. The mass of one wire can be written both as m = ρV = ρAL and also as m = µL .

In each wire

t=

Then we have

µ = ρA =

Thus,

t=L

For copper,

For steel, The total time is

πρd 2 4

F πρd I GH 4T JK L aπ fb8 920ge1.00 × 10 t = a 20.0 fM MM a4fa150f N L aπ fb7 860ge1.00 × 10 t = a30.0fM MM a4fa150f N 2

12

0.137 + 0.192 = 0.329 s

j OP PP Q j OP PP Q

12 −3 2

= 0.137 s

12 −3 2

= 0.192 s

Chapter 16

P16.32

Refer to the diagrams. From the free-body diagram of point A:

∑ Fy = 0 ⇒ T1 sin θ = Mg

∑ Fx = 0 ⇒ T1 cos θ = T

and

Combining these equations to eliminate T1 gives the tension in the Mg . string connecting points A and B as: T = tan θ

v=

T

=

µ

Mg tan θ m L

=

D L/4

L/4

θ

A

d

B

MgL m tan θ

θ d

L/2 M

The speed of transverse waves in this segment of string is then

483

M

T1

θ

A

T

and the time for a pulse to travel from A to B is t=

*P16.33

L 2

v

=

f has units Hz = 1 s , so T =

(a)

Mg

mL tan θ . 4Mg

FIG. P16.32

1 has units of seconds, s . For the other T we have T = µv 2 , f

kg m 2 kg ⋅ m = = N . m s2 s2

with units

The first T is period of time; the second is force of tension.

(b)

Section 16.4

Reflection and Transmission

Problem 7 in Chapter 18 can be assigned with this section.

Section 16.5 P16.34

f=

P= P16.35

Rate of Energy Transfer by Sinusoidal Waves on Strings v

λ

=

30.0 = 60.0 Hz 0.500

FG H

ω = 2π f = 120π rad s

IJ a K

1 1 0.180 µω 2 A 2 v = 120π 2 2 3.60

f a0.100f a30.0f = 2

2

1.07 kW

Suppose that no energy is absorbed or carried down into the water. Then a fixed amount of power is spread thinner farther away from the source, spread over the circumference 2π r of an expanding circle. The power-per-width across the wave front

P 2π r is proportional to amplitude squared so amplitude is proportional to

P . 2π r

484 P16.36

Wave Motion

T = constant; v =

T

µ

; P=

1 µω 2 A 2 v 2

(a)

If L is doubled, v remains constant and P is constant .

(b)

If A is doubled and ω is halved, P ∝ ω 2 A 2 remains constant .

(c)

If λ and A are doubled, the product ω 2 A 2 ∝

A2

remains constant, so

λ2

P remains constant . (d)

If L and λ are halved, then ω 2 ∝

1

is quadrupled, so P is quadrupled .

λ2

(Changing L doesn’t affect P ). P16.37

A = 5.00 × 10 −2 m

T

v=

Therefore,

P=

µ = 4.00 × 10 −2 kg m

1 µω 2 A 2 v : 2

µ

ω2 =

P = 300 W

T = 100 N

= 50.0 m s

a f

2 300 2P = µA 2 v 4.00 × 10 −2 5.00 × 10 −2

e

je

j a50.0f 2

ω = 346 rad s ω = 55.1 Hz f= 2π P16.38

µ = 30.0 g m = 30.0 × 10 −3 kg m λ = 1.50 m f = 50.0 Hz:

ω = 2π f = 314 s −1

2 A = 0.150 m:

A = 7.50 × 10 −2 m

(a)

FG 2π x − ωtIJ Hλ K y = e7.50 × 10 j sina 4.19 x − 314t f

y = A sin

−2

P16.39

IJ W ja f e7.50 × 10 j FGH 4314 .19 K

1 1 µω 2 A 2 v = 30.0 × 10 −3 314 2 2

e

FIG. P16.38

(b)

P=

(a)

v = fλ =

(b)

λ=

2π 2π = m = 7.85 m k 0.800

(c)

f=

50.0 = 7.96 Hz 2π

(d)

P=

1 1 µω 2 A 2 v = 12.0 × 10 −3 50.0 2 2

−2 2

2

ω 2π ω 50.0 m s = 62.5 m s = = k 0.800 2π k

e

ja f a0.150f a62.5f W = 2

2

21.1 W

P = 625 W

Chapter 16

*P16.40

FG H

Comparing y = 0.35 sin 10πt − 3πx + k= (a)

The rate of energy transport is

jb

1 1 µω 2 A 2 v = 75 × 10 −3 kg m 10π s 2 2

e

g a0.35 mf 3.33 m s = 2

2

15.1 W .

The energy per cycle is Eλ = P T =

P16.41

IJ with y = A sinbkx − ωt + φ g = A sinbωt − kx − φ + π g we have K

3π λ ω 10π s = = = 3.33 m s . , ω = 10π s , A = 0.35 m . Then v = fλ = 2π f m 2π k 3π m

P= (b)

π 4

485

jb

1 1 µω 2 A 2 λ = 75 × 10 −3 kg m 10π s 2 2

e

g a0.35 mf 2

2

2π m = 3.02 J . 3π

Originally, 1 µω 2 A 2 v 2 1 T P0 = µω 2 A 2 µ 2

P0 =

1 P0 = ω 2 A 2 Tµ 2 The doubled string will have doubled mass-per-length. Presuming that we hold tension constant, it can carry power larger by 2 times. 2 P0 = *P16.42

1 2 2 ω A T 2µ 2

As for a strong wave, the rate of energy transfer is proportional to the square of the amplitude and to the speed. We write P = FvA 2 where F is some constant. With no absorption of energy, 2 2 = Fv mudfill A mudfill Fv bedrock A bedrock

v bedrock A = mudfill = v mudfill A bedrock The amplitude increases by 5.00 times.

25 v mudfill =5 v mudfill

486

Wave Motion

Section 16.6 P16.43

The Linear Wave Equation

a

f

(a)

A = 7.00 + 3.00 4.00 yields A = 40.0

(b)

In order for two vectors to be equal, they must have the same magnitude and the same direction in three-dimensional space. All of their components must be equal. Thus, 7.00 i + 0 j + 3.00k = A i + Bj + Ck requires A = 7.00 , B = 0 , and C = 3.00 .

(c)

In order for two functions to be identically equal, they must be equal for every value of every variable. They must have the same graphs. In

a

f

a

f

A + B cos Cx + Dt + E = 0 + 7.00 mm cos 3.00 x + 4.00t + 2.00 , the equality of average values requires that A = 0 . The equality of maximum values requires B = 7.00 mm . The equality for the wavelength or periodicity as a function of x requires C = 3.00 rad m . The equality of period requires D = 4.00 rad s , and the equality of zero-crossings requires E = 2.00 rad .

*P16.44

∂2y

The linear wave equation is

∂x

1 ∂2 y v 2 ∂t 2

If

y = e b a x − vt f

then

∂y ∂y = − bve b a x − vt f and = be b a x − vt f ∂t ∂x ∂2 y ∂t 2

= b 2 v 2 e b a x − vt f and

∂2y

Therefore,

P16.45

=

2

∂t 1 ∂2y

The linear wave equation is

a

To show that y = ln b x − vt

2

∂2y ∂x

= b 2 e b a x − vt f

∂x 2

, demonstrating that e b a x − vt f is a solution

2

∂2y

=

v 2 ∂t 2

f

= v2

∂2y

∂x 2

is a solution, we find its first and second derivatives with respect to x

and t and substitute into the equation. ∂2y

∂y 1 = − bv ∂t b x − vt

fa f

a ∂y = ba x − vt f ∂x

−1

=

∂t 2 ∂2y

b

∂x

e j a f 2

−v 1 ∂2y 1 Then 2 2 = 2 v ∂t v x − vt

2

=−

2

a f

−1 − bv

a

f

2

b x − vt b

f

b 2 x − vt

=−

1

ax − vtf

2

2

a

=

∂2y ∂x 2

=− 2

v2

ax − vtf

=−

2

1

ax − vtf

2

so the given wave function is a solution.

Chapter 16

P16.46

(a)

From y = x 2 + v 2 t 2 , ∂2y

∂y = 2x ∂x

evaluate

∂x 2 ∂2y

∂y = v 2 2t ∂t Does

∂2y ∂t 2

equation. Note

So (c)

a

1 x + vt 2

a

∂t 2

=2 = 2v 2

1 ∂2y ? v 2 ∂t 2

=

By substitution: 2 =

(b)

487

f

2

+

1 2 v 2 and this is true, so the wave function does satisfy the wave v2

a

1 x − vt 2

f 12 ax + vtf

f x + vt =

f

2

2

1 2 1 1 1 x + xvt + v 2 t 2 + x 2 − xvt + v 2 t 2 2 2 2 2 2 2 2 = x + v t as required. =

a

f 12 ax − vtf

and g x − vt =

2

.

y = sin x cos vt makes ∂2y

∂y = cos x cos vt ∂x

∂x 2 ∂2y

∂y = − v sin x sin vt ∂t Then

∂2y ∂x

2

=

∂t 2

= − sin x cos vt = − v 2 sin x cos vt

1 ∂2 y v 2 ∂t 2

−1 2 v sin x cos vt which is true as required. v2 Note sin x + vt = sin x cos vt + cos x sin vt

becomes − sin x cos vt =

a

f

a

f

sin x − vt = sin x cos vt − cos x sin vt .

a f a f 1 f a x + vtf = sina x + vt f 2

So sin x cos vt = f x + vt + g x − vt with and

a

f

g x − vt =

a

1 sin x − vt 2

f

.

Additional Problems P16.47

Assume a typical distance between adjacent people ~ 1 m . Then the wave speed is

v=

∆x 1 m ~ ~ 10 m s ∆t 0.1 s

Model the stadium as a circle with a radius of order 100 m. Then, the time for one circuit around the stadium is T=

e j

2 2π r 2π 10 ~ = 63 s ~ 1 min . v 10 m s

488 P16.48

Wave Motion

a

f

Compare the given wave function y = 4.00 sin 2.00 x − 3.00t cm to the general form

a

f

y = A sin kx − ωt to find

P16.49

(a)

amplitude A = 4.00 cm = 0.040 0 m

(b)

k=

(c)

ω = 2π f = 3.00 s −1 and f = 0.477 Hz

(d)

T=

(e)

The minus sign indicates that the wave is traveling in the positive x -direction .

(a)

Let u = 10π t − 3π x +

2π

λ

= 2.00 cm −1 and λ = π cm = 0.031 4 m

1 = 2.09 s f

π 4

du dx = 10π − 3π = 0 at a point of constant phase dt dt dx 10 = = 3.33 m s 3 dt The velocity is in the positive x -direction .

f FGH

y 0.100 , 0 = 0.350 m sin −0.300π +

(c)

k=

(d) *P16.50

g a

b

(b)

(a)

2π

λ

IJ K

π = −0.054 8 m = −5.48 cm 4

= 3π : λ = 0.667 m

ω = 2π f = 10π : f = 5.00 Hz

fa f FGH 0.175 m = a0.350 mf sin b99.6 rad sgt ∴ sin b99.6 rad sgt = 0.5 vy =

∂y π = 0.350 10π cos 10π t − 3π x + ∂t 4

a

IJ K

a fa

f

v y, max = 10π 0.350 = 11.0 m s

The smallest two angles for which the sine function is 0.5 are 30° and 150°, i.e., 0.523 6 rad and 2.618 rad. 99.6 rad s t1 = 0.523 6 rad , thus t1 = 5.26 ms

b g b99.6 rad sgt

2

= 2.618 rad , thus t 2 = 26.3 ms

∆t ≡ t 2 − t1 = 26.3 ms − 5.26 ms = 21.0 ms (b) P16.51

Distance traveled by the wave =

FG ω IJ ∆t = FG 99.6 rad s IJ e21.0 × 10 sj = H k K H 1.25 rad m K −3

The equation v = λf is a special case of speed = (cycle length)(repetition rate).

e

jb

g

Thus, v = 19.0 × 10 −3 m frame 24.0 frames s = 0.456 m s .

1.68 m .

Chapter 16

P16.52

Assuming the incline to be frictionless and taking the positive x-direction to be up the incline:

∑ Fx = T − Mg sinθ = 0

T = Mg sin θ

or the tension in the string is

v=

The speed of transverse waves in the string is then

The time interval for a pulse to travel the string’s length is ∆t =

P16.53

µ

=

g

+ Us

j

top

0 + Mgx + 0 + 0 = 0 + 0 + x=

b

2 Mg k

ge

j

e

+ ∆E = K + U g + U s 1 2 kx 2

(a)

T = kx = 2 Mg = 2 2.00 kg 9.80 m s 2 = 39.2 N

(b)

L = L0 + x = L0 +

(c)

v=

2 Mg k 39.2 N L = 0.500 m + = 0.892 m 100 N m

v=

T

µ

=

TL m

39.2 N × 0.892 m 5.0 × 10 −3 kg

v = 83.6 m s Mgx =

1 2 kx 2

(a)

T = kx = 2 Mg

(b)

L = L0 + x = L0 +

(c)

v=

T

µ

=

TL = m

2 Mg k

FG H

2 Mg 2 Mg L0 + m k

IJ K

Mg sin θ m L

=

L m =L = v MgL sin θ

Energy is conserved as the block moves down distance x:

eK + U

P16.54

T

j

bottom

MgL sin θ m mL Mg sin θ

489

490 P16.55

Wave Motion

T

v=

(b)

From Equation 16.21, P =

µ

=

80.0 N

(a)

e5.00 × 10

−3

= 179 m s

j

kg 2.00 m

FG IJ H K

1 v µvω 2 A 2 and ω = 2π 2 λ

FG IJ H K

1 2πv µvA 2 λ 2

P=

2π 2

P=

FH

2

=

2π 2 µA 2 v 3

λ2

IK b0.040 0 mg b179 m sg a0.160 mf

5.00 ×10 −3 kg 2.00 m

2

3

2

P = 1.77 × 10 4 W = 17.7 kW P16.56

v=

T

µ

Now v = fλ implies v =

ω so that k

FG IJ H K

µ ω m= g k *P16.57

µv 2 . g

and in this case T = mg ; therefore, m =

2

0.250 kg m

=

9.80 m s 2

LM 18π s OP N 0.750π m Q −1

−1

Let M = mass of block, m = mass of string. For the block,

2

= 14.7 kg .

∑ F = ma implies T =

speed of a wave on the string is then T

v= t=

µ

(a)

µ= v=

M m

m M m = M

0.003 2 kg = 0.084 3 rad 0.450 kg

dm dx = ρA = ρA dL dx T

µ

=

T = ρA

T ρ ax + b

a

With all SI units, v =

(b)

= rω

m r

r 1 = v ω

θ = ωt = P16.58

Mω 2 r

=

v x= 0 =

v x=10 .0 =

e

f

=

T

j

ρ 10 x + 10 −2 10 −4

b2 700ge0 + 10 je10 j −2

24.0 −2

−4

j

ρ 10 x + 10 −2 cm 2

−3

24.0

b2 700ge10

e

T −3

ms

= 94.3 m s

je j

+ 10 −2 10 −4

= 66.7 m s

mv b2 = mω 2 r . The r

Chapter 16

P16.59

v=

T

where µ Therefore, dx , so that But v = dt

T = µxg , the weight of a length x, of rope. v = gx dx dt = gx t=

and

z

L 0

P16.60

dx gx

=

1

x 1 2

g

L

= 2 0

L g

FG mxg IJ + Mg , so the wave speed is: H LK T TL F MgL IJ = dx . = = xg + G v= H m K dt µ m L F MgL IJ OP dx 1 xg + b MgL m g t= Then t = z dt = z M xg + G g N H m KQ MgL I 2 LF LF m+M − MI F MgL IJ OP −G t = MG Lg + t=2 J JK H m K PQ g MNH m K g GH m L F m − 0I When M = 0 , as in the previous problem, t=2 J = 2 Lg g GH m K F 1 m − 1 m + …I F mI As m → 0 we expand m + M = M G 1 + J = M G 1 + H MK H 2 M 8 M JK F M + em M j − em M j + … − M I L G JJ to obtain t=2 gG m H K LF1 m I mL = t≈2 J G gH2 MK Mg

At distance x from the bottom, the tension is T =

(a)

t

L

0

0

(b)

1 2 x=L

−1 2

1 2

12

2

2

1 2

(a)

x=0

12

12

(c)

P16.61

491

2

1 8

32

The speed in the lower half of a rope of length L is the same function of distance (from the L bottom end) as the speed along the entire length of a rope of length . 2 L L′ with L ′ = Thus, the time required = 2 g 2

FG IJ H K

and the time required = 2

F GH

L L = 0.707 2 2g g

I JK

.

It takes the pulse more that 70% of the total time to cover 50% of the distance. (b)

By the same reasoning applied in part (a), the distance climbed in τ is given by d = L L , we find the distance climbed = . g 4 1 In half the total trip time, the pulse has climbed of the total length. 4 For τ =

t = 2

gτ 2 . 4

492 P16.62

P16.63

Wave Motion

(a)

v=

ω 15.0 = = 5.00 m s in positive x -direction k 3.00

(b)

v=

15.0 = 5.00 m s in negative x -direction 3.00

(c)

v=

15.0 = 7.50 m s in negative x -direction 2.00

(d)

v=

12.0 1 2

= 24.0 m s in positive x -direction T A ∆L L

Young’s modulus for the wire may be written as Y =

, where T is the tension maintained in the

wire and ∆L is the elongation produced by this tension. Also, the mass density of the wire may be

µ . A The speed of transverse waves in the wire is then

expressed as ρ =

v=

T

µ

=

T A

µ

=

Y

A

c h ∆L L

ρ

∆L ρv 2 = . L Y If the wire is aluminum and v = 100 m s, the strain is

and the strain in the wire is

jb

e

2.70 × 10 3 kg m3 100 m s ∆L = L 7.00 × 10 10 N m 2 *P16.64

(a)

g

2

= 3.86 × 10 −4 .

Consider a short section of chain at the top of the loop. A freebody diagram is shown. Its length is s = R 2θ and its mass is µR2θ . In the frame of reference of the center of the loop, Newton’s second law is mv 02 µR 2θv 02 2T sin θ down = down = ∑ Fy = ma y R R

θ

θ

a f

2θ

T

R

FIG. P16.64(a)

For a very short section, sin θ = θ and T = µ v 02 . T

(b)

The wave speed is v =

(c)

In the frame of reference of the center of the loop, each pulse moves with equal speed clockwise and counterclockwise.

µ

= v0 .

v v0

v0 FIG. P16.64(c-1)

continued on next page

v0

v

T

Chapter 16

493

In the frame of reference of the ground, once pulse moves backward at speed v 0 + v = 2 v 0 and the other forward at v 0 − v = 0 . The one pulse makes two revolutions while the loop makes one revolution and the other pulse does not move around the loop. If it is generated at the six-o’clock position, it will stay at the six-o’clock position.

v0

v0

v0

FIG. P16.64(c-2) P16.65

(a)

Assume the spring is originally stationary throughout, extended to have a length L much greater than its equilibrium length. We start moving one end forward with the speed v at which a wave propagates on the spring. In this way we create a single pulse of compression that moves down the length of the spring. For an increment of spring with length dx and mass dm, just as the pulse swallows it up, ∑ F = ma becomes kdx = adm or But

k dm dx

= a.

dm k = µ so a = . µ dx dv v v2 = when vi = 0. But L = vt , so a = . dt t L

Also, a =

Equating the two expressions for a, we have

(b)

Using the expression from part (a) v =

F T I F 2T IJ v=G J =G H µK H µ K F T I F 2T IJ v′ = G J = G H µ ′ K H 3µ K

(a)

0

0

12

= v0

= v0

0

∆t left =

L 2

v

∆t right =

= L 2

v′

L 2 v0 2 =

=

L 2 v0

2 3

2 where v 0 ≡

12

0

(b)

µ

12

12

P16.66

kL

∆t 0 2 2 =

∆t left + ∆t right = 0.966 ∆t 0

=

µ

2 3

v2 or v = L

kL2 = m

FG T IJ Hµ K

12

0

0

2 3

= 0.354∆t 0 where ∆t 0 ≡

∆t 0 2

=

k

= 0.612 ∆t 0

L v0

kL

µ

.

b100 N mga2.00 mf 0.400 kg

2

= 31.6 m s .

494 P16.67

P16.68

Wave Motion

af

FG IJ H K

1 1 ω µω 3 2 −2 bx A0 e µω 2 A 2 v = µω 2 A02 e −2 bx = 2 2 2k k

(a)

P x =

(b)

P 0 =

(c)

P x = e −2 bx P 0

v=

af

µω 3 2 A0 2k

af af

4 450 km = 468 km h = 130 m s 9.50 h

b g e j µ a x f is a linear function, so it is of the form To have µ a0f = µ we require b = µ . Then 2

130 m s v2 d= = = 1 730 m g 9.80 m s 2 *P16.69

(a)

0

af µ aL f = µ

µ x = mx + b

0

a f bµ

µ x =

Then (b)

= mL + µ 0

µL − µ0 L

m=

so

L

L

g

− µ0 x L

+ µ0

dx dx , the time required to move from x to x + dx is . The time required to move v dt from 0 to L is

From v =

∆t =

z z z FGH b L

dx L dx 1 = = T v T 0 0 µ

z

L

af

µ x dx

0

I FG µ − µ IJ dxF ∆t = JK H L K GH µ T I 1 L I F b µ − µ gx 1 F ∆t = +µ J G J G L T H µ − µ KH K 2L ∆t = eµ − µ j 3 T bµ − µ g 2Le µ − µ je µ + µ µ + µ j ∆t = 3 T e µ − µ je µ + µ j 2L F µ + µ µ + µ I ∆t = G µ + µ JK 3 TH 1

L 0

g

12

µL − µ0 x + µ0 L

0

L

32

L

L

0

0

0

L

0

L

0

L

32 L

32 0

L

L

L

0

L

L

L

0

0

0

0

3 2 0

0

0

L

L L − µ0

IJ K

Chapter 16

495

ANSWERS TO EVEN PROBLEMS P16.2

see the solution

P16.4

(a) the P wave; (b) 665 s

P16.6

0.800 m s

P16.8

2.40 m s

P16.10

0.300 m in the positive x-direction

P16.12

±6.67 cm

P16.14

(a) see the solution; (b) 0.125 s; in agreement with the example

P16.16

(a) see the solution; (b) 18.0 m ; 83.3 ms ; 75.4 rad s ; 4.20 m s ; (c) 0.2 m sin 18 x + 75.4t − 0.151

a

P16.18

f a

b g

b0.021 5 mg sinb8.38x + 80.0π t + 1.95g

P16.20

(a) see the solution; (b) 3.18 Hz

P16.22

30.0 N

P16.24

(a) y = 0.2 mm sin 16 x − 3 140t ; (b) 158 N

P16.26

631 N

P16.28

v=

P16.30

F m I (a) v = G 30.4 H s ⋅ kg JK

P16.32

a

Tg 2π

f b

P16.40

(a) 15.1 W ; (b) 3.02 J

P16.42

The amplitude increases by 5.00 times

P16.44

see the solution

P16.46

(a) see the solution; 1 1 2 2 x + vt + x − vt ; (b) 2 2 1 1 (c) sin x + vt + sin x − vt 2 2

a

a

f

a

f

a

f

P16.48

(a) 0.040 0 m; (b) 0.031 4 m ; (c) 0.477 Hz; (d) 2.09 s; (e) positive x -direction

P16.50

(a) 21.0 ms ; (b) 1.68 m

P16.52

∆t =

P16.54

(a) 2Mg ; (b) L0 +

g

M m

f

(a) y = 0.075 0 sin 4.19 x − 314t ; (b) 625 W

f

(a) 0.021 5 m; (b) 1.95 rad; (c) 5.41 m s ; (d) y x , t =

g a

b

P16.38

(c)

f

mL Mg sin θ 2 Mg ; k 2 Mg 2 Mg L0 + m k

P16.56

14.7 kg

P16.58

(a) v =

FG H

e

IJ K

T −7

ρ 10 x + 10 −6

j

in SI units;

(b) 94.3 m s; 66.7 m s m ; (b) 3.89 kg

mL tan θ 4Mg

P16.34

1.07 kW

P16.36

(a), (b), (c) P is constant ; (d) P is quadrupled

P16.60

see the solution

P16.62

(a) 5.00 i m s ; (b) −5.00 i m s ; (c) −7.50 i m s ; (d) 24.0 i m s

P16.64

(a) µ v 02 ; (b) v 0 ; (c) One travels 2 rev and the other does not move around the loop.

496

Wave Motion

F 2T IJ = v 2 ; (a) v = G Hµ K F 2T IJ = v 2 ; (b) 0.966∆t v′ = G 3 H 3µ K 12

P16.66

0

P16.68

0

0

12

0

0

0

0

130 m s ; 1.73 km

17 Sound Waves CHAPTER OUTLINE 17.1 17.2 17.3 17.4 17.5 17.6

Speed of Sound Waves Periodic Sound Waves Intensity of Periodic Sound Waves The Doppler Effect Digital Sound Recording Motion Picture Sound

ANSWERS TO QUESTIONS Q17.1

Sound waves are longitudinal because elements of the medium—parcels of air—move parallel and antiparallel to the direction of wave motion.

Q17.2

We assume that a perfect vacuum surrounds the clock. The sound waves require a medium for them to travel to your ear. The hammer on the alarm will strike the bell, and the vibration will spread as sound waves through the body of the clock. If a bone of your skull were in contact with the clock, you would hear the bell. However, in the absence of a surrounding medium like air or water, no sound can be radiated away. A larger-scale example of the same effect: Colossal storms raging on the Sun are deathly still for us. What happens to the sound energy within the clock? Here is the answer: As the sound wave travels through the steel and plastic, traversing joints and going around corners, its energy is converted into additional internal energy, raising the temperature of the materials. After the sound has died away, the clock will glow very slightly brighter in the infrared portion of the electromagnetic spectrum.

1 meter from the sonic ranger, then the sensor would have to measure how long it 2 would take for a sound pulse to travel one meter. Since sound of any frequency moves at about 343 m s, then the sonic ranger would have to be able to measure a time difference of under 0.003 seconds. This small time measurement is possible with modern electronics. But it would be more expensive to outfit sonic rangers with the more sensitive equipment than it is to print “do not 1 use to measure distances less than meter” in the users’ manual. 2

Q17.3

If an object is

Q17.4

The speed of sound to two significant figures is 340 m s. Let’s assume that you can measure time to 1 second by using a stopwatch. To get a speed to two significant figures, you need to measure a 10 time of at least 1.0 seconds. Since d = vt , the minimum distance is 340 meters.

Q17.5

The frequency increases by a factor of 2 because the wave speed, which is dependent only on the medium through which the wave travels, remains constant.

497

498

Sound Waves

Q17.6

When listening, you are approximately the same distance from all of the members of the group. If different frequencies traveled at different speeds, then you might hear the higher pitched frequencies before you heard the lower ones produced at the same time. Although it might be interesting to think that each listener heard his or her own personal performance depending on where they were seated, a time lag like this could make a Beethoven sonata sound as if it were written by Charles Ives.

Q17.7

Since air is a viscous fluid, some of the energy of sound vibration is turned into internal energy. At such great distances, the amplitude of the signal is so decreased by this effect you re unable to hear it.

Q17.8

We suppose that a point source has no structure, and radiates sound equally in all directions (isotropically). The sound wavefronts are expanding spheres, so the area over which the sound energy spreads increases according to A = 4π r 2 . Thus, if the distance is tripled, the area increases by a factor of nine, and the new intensity will be one-ninth of the old intensity. This answer according to the inverse-square law applies if the medium is uniform and unbounded. For contrast, suppose that the sound is confined to move in a horizontal layer. (Thermal stratification in an ocean can have this effect on sonar “pings.”) Then the area over which the sound energy is dispersed will only increase according to the circumference of an expanding circle: A = 2π rh , and so three times the distance will result in one third the intensity. In the case of an entirely enclosed speaking tube (such as a ship’s telephone), the area perpendicular to the energy flow stays the same, and increasing the distance will not change the intensity appreciably.

Q17.9

He saw the first wave he encountered, light traveling at 3.00 × 10 8 m s . At the same moment, infrared as well as visible light began warming his skin, but some time was required to raise the temperature of the outer skin layers before he noticed it. The meteor produced compressional waves in the air and in the ground. The wave in the ground, which can be called either sound or a seismic wave, traveled much faster than the wave in air, since the ground is much stiffer against compression. Our witness received it next and noticed it as a little earthquake. He was no doubt unable to distinguish the P and S waves. The first air-compression wave he received was a shock wave with an amplitude on the order of meters. It transported him off his doorstep. Then he could hear some additional direct sound, reflected sound, and perhaps the sound of the falling trees.

Q17.10

A microwave pulse is reflected from a moving object. The waves that are reflected back are Doppler shifted in frequency according to the speed of the target. The receiver in the radar gun detects the reflected wave and compares its frequency to that of the emitted pulse. Using the frequency shift, the speed can be calculated to high precision. Be forewarned: this technique works if you are either traveling toward or away from your local law enforcement agent!

Q17.11

As you move towards the canyon wall, the echo of your car horn would be shifted up in frequency; as you move away, the echo would be shifted down in frequency.

Q17.12

Normal conversation has an intensity level of about 60 dB.

Q17.13

A rock concert has an intensity level of about 120 dB. A cheering crowd has an intensity level of about 90 dB. Normal conversation has an intensity level of about 50–60 dB. Turning a page in the textbook has an intensity level of about 10–20 dB.

Chapter 17

499

Q17.14

One would expect the spectra of the light to be Doppler shifted up in frequency (blue shift) as the star approaches us. As the star recedes in its orbit, the frequency spectrum would be shifted down (red shift). While the star is moving perpendicular to our line of sight, there will be no frequency shift at all. Overall, the spectra would oscillate with a period equal to that of the orbiting stars.

Q17.15

For the sound from a source not to shift in frequency, the radial velocity of the source relative to the observer must be zero; that is, the source must not be moving toward or away from the observer. The source can be moving in a plane perpendicular to the line between it and the observer. Other possibilities: The source and observer might both have zero velocity. They might have equal velocities relative to the medium. The source might be moving around the observer on a sphere of constant radius. Even if the source speeds up on the sphere, slows down, or stops, the frequency heard will be equal to the frequency emitted by the source.

Q17.16

Wind can change a Doppler shift but cannot cause one. Both v o and v s in our equations must be interpreted as speeds of observer and source relative to the air. If source and observer are moving relative to each other, the observer will hear one shifted frequency in still air and a different shifted frequency if wind is blowing. If the distance between source and observer is constant, there will never be a Doppler shift.

Q17.17

If the object being tracked is moving away from the observer, then the sonic pulse would never reach the object, as the object is moving away faster than the wave speed. If the object being tracked is moving towards the observer, then the object itself would reach the detector before reflected pulse.

Q17.18

New-fallen snow is a wonderful acoustic absorber as it reflects very little of the sound that reaches it. It is full of tiny intricate air channels and does not spring back when it is distorted. It acts very much like acoustic tile in buildings. So where does the absorbed energy go? It turns into internal energy—albeit a very small amount.

Q17.19

As a sound wave moves away from the source, its intensity decreases. With an echo, the sound must move from the source to the reflector and then back to the observer, covering a significant distance.

Q17.20

The observer would most likely hear the sonic boom of the plane itself and then beep, baap, boop. Since the plane is supersonic, the loudspeaker would pull ahead of the leading “boop” wavefront before emitting the “baap”, and so forth. “How are you?” would be heard as “?uoy era woH”

Q17.21

This system would be seen as a star moving in an elliptical path. Just like the light from a star in a binary star system, described in the answer to question 14, the spectrum of light from the star would undergo a series of Doppler shifts depending on the star’s speed and direction of motion relative to the observer. The repetition rate of the Doppler shift pattern is the period of the orbit. Information about the orbit size can be calculated from the size of the Doppler shifts.

SOLUTIONS TO PROBLEMS Section 17.1

Speed of Sound Waves

b

ga

f

P17.1

Since v light >> v sound : d ≈ 343 m s 16. 2 s = 5.56 km

P17.2

v=

B

ρ

=

2.80 × 10 10 = 1.43 km s 13.6 × 10 3

500 P17.3

Sound Waves

a20.0 m − 1.75 mf = 5.32 × 10

Sound takes this time to reach the man:

−2

343 m s

so the warning should be shouted no later than before the pot strikes. Since the whole time of fall is given by y =

1 2 gt : 2

s

0.300 s + 5.32 × 10 −2 s = 0.353 s 18.25 m =

1 9.80 m s 2 t 2 2

e

j

t = 1.93 s

P17.4

the warning needs to come

1.93 s − 0.353 s = 1.58 s

into the fall, when the pot has fallen

1 9.80 m s 2 1.58 s 2

to be above the ground by

20.0 m − 12.2 m = 7.82 m

(a)

At 9 000 m, ∆T =

e

ja

f

2

= 12.2 m

FG 9 000 IJ a−1.00° Cf = −60.0° C so T = −30.0° C . H 150 K

Using the chain rule:

a

fFGH IJK

a

f

dv dv dT dx dv dT v dv 1 , so dt = 247 s = =v = v 0.607 = dt dT dx dt dT dx v 150 247

f z dvv z L 331.5 + 0.607a30.0f OP Fv I t = a 247 sf lnG J = a 247 sf ln M Hv K N 331.5 + 0.607a−30.0f Q t

a

vf

dt = 247 s

0

vi

f

i

t = 27.2 s for sound to reach ground. (b)

t=

9 000 h = = 25.7 s v 331.5 + 0.607 30.0

a f

It takes longer when the air cools off than if it were at a uniform temperature. *P17.5

Let x 1 represent the cowboy’s distance from the nearer canyon wall and x 2 his distance from the farther cliff. The sound for the first echo travels distance 2 x1 . For the second, 2 x 2 . For the third, 2 x 2 − 2 x1 2 x1 + 2 x 2 − 2 x 2 2 x1 + 2 x 2 . For the fourth echo, 2 x1 + 2 x 2 + 2 x1 . Then = 1.92 s and = 1.47 s . 340 m s 340 m s 2x2 1 = 1.92 s + 1.47 s ; x 2 = 576 m. Thus x 1 = 340 m s 1.47 s = 250 m and 2 340 m s (a)

(b)

So x 1 + x 2 = 826 m

b

2 x 1 + 2 x 2 + 2 x 1 − 2 x1 + 2 x 2 340 m s

g=

1.47 s

Chapter 17

P17.6

It is easiest to solve part (b) first: (b)

The distance the sound travels to the plane is d s = h 2 +

FG h IJ H 2K

2

=

h 5 . 2

The sound travels this distance in 2.00 s, so

b ga f 2a686 mf = 614 m . giving the altitude of the plane as h = ds =

h 5 = 343 m s 2.00 s = 686 m 2

5

(a)

a

Thus, the speed of the plane is: v =

Section 17.2 P17.7

*P17.8

λ=

*P17.9

P17.10

26° C = 346 m s 273° C

Let t represent the time for the echo to return. Then 1 1 vt = 346 m s 24 × 10 −3 s = 4.16 m . 2 2

Let ∆t represent the duration of the pulse: ∆t =

(c)

307 m = 153 m s . 2.00 s

340 m s v = = 5.67 mm f 60.0 × 10 3 s −1

d= (b)

h = 307 m . 2

Periodic Sound Waves

The sound speed is v = 331 m s 1 + (a)

f

The distance the plane has traveled in 2.00 s is v 2.00 s =

L = 10 λ =

b

10λ 10 λ 10 10 = = = = 0.455 µs . v fλ f 22 × 10 6 1 s

g

10 v 10 346 m s = = 0.157 mm f 22 × 10 6 1 s

v 1 500 m s = = 1.50 mm f 10 6 s 1 500 m s = 75.0 µm If f = 20 MHz , λ = 2 × 10 7 s

If f = 1 MHz , λ =

∆Pmax = ρvω smax smax

e

j

4.00 × 10 −3 N m 2 ∆Pmax = = = 1.55 × 10 −10 m ρvω 1.20 kg m3 343 m s 2π 10.0 × 10 3 s −1

e

jb

ga fe

j

501

502 P17.11

Sound Waves

A = 2.00 µm

(a)

2π = 0.400 m = 40.0 cm 15.7 ω 858 = 54.6 m s v= = k 15.7

λ=

P17.12

a fb

g a fe

j

(b)

s = 2.00 cos 15.7 0.050 0 − 858 3.00 × 10 −3 = −0.433 µm

(c)

v max = Aω = 2.00 µm 858 s −1 = 1.72 mm s

b

ge

j F π x − 340π t IJ (SI units) ∆P = a1.27 Paf sinG Hm s K

(a)

The pressure amplitude is: ∆Pmax = 1.27 Pa .

P17.13

(b)

ω = 2π f = 340π s, so f = 170 Hz

(c)

k=

(d)

v = λf = 2.00 m 170 Hz = 340 m s

k=

ω=

2π

=

λ

2π v

λ

a

smax = 2π

fa

f

2π = 62.8 m −1 0.100 m

=

ω = 2π f =

k=

= π m , giving λ = 2.00 m

λ

a

Therefore,

P17.14

2π

f

b

2π 343 m s

g = 2.16 × 10

4

−1

s a0.100 mf ∆P = a0.200 Paf sin 62.8 x m − 2.16 × 10

2π v

λ

=

b

2π 343 m s

a0.100 mf

g = 2.16 × 10

a jb

4

f ge

0.200 Pa ∆Pmax = = 2. 25 × 10 −8 m 3 ρvω 1.20 kg m 343 m s 2.16 × 10 4 s −1

e

=

2π

j

−1

max

P17.15

ts .

rad s

a0.100 mf = 62.8 m Therefore, s = s cosb kx − ω t g = e 2.25 × 10 λ

4

FG 2π v IJ s H λK 2π a1.20fa343 f e5.50 × 10 j = =

∆Pmax = ρvω smax = ρv 2πρv 2 smax λ= ∆Pmax

−8

j e

max 2

0.840

j

m cos 62.8 x m − 2.16 × 10 4 t s .

−6

5.81 m

503

Chapter 17

P17.16

(a)

The sound “pressure” is extra tensile stress for one-half of each cycle. When it becomes 0.500% 13.0 × 10 10 Pa = 6.50 × 10 8 Pa , the rod will break. Then, ∆Pmax = ρvω smax

a

fe

j

smax = (b)

6.50 × 10 8 N m 2 ∆Pmax = = 4.63 mm . ρvω 8.92 × 10 3 kg m3 5 010 m s 2π 500 s

a

jb

e

From s = smax cos kx − ωt

f

a

f

b

g

f

ga

jb

gb

1 1 1 2 2 ρv ωsmax = ρvvmax = 8.92 × 10 3 kg m 3 5 010 m s 14.5 m s 2 2 2 = 4.73 × 10 9 W m 2

I=

(c)

*P17.17

g

∂s = −ωsmax sin kx − ωt ∂t v max = ωsmax = 2π 500 s 4.63 mm = 14.5 m s v=

b

gb

af

e

g

2

Let P x represent absolute pressure as a function of x. The net force to the right on the chunk of air is + P x A − P x + ∆x A . Atmospheric ∂∆P ∆xA . pressure subtracts out, leaving − ∆P x + ∆x + ∆P x A = − P x + ∆x A Px A ∂x ∂2s The mass of the air is ∆m = ρ∆V = ρA∆x and its acceleration is 2 . So ∂t FIG. P17.17 Newton’s second law becomes 2 ∂∆P ∂ s − ∆xA = ρA∆x 2 ∂x ∂t ∂ ∂s ∂2s −B − =ρ 2 ∂x ∂x ∂t

af a f a f af

FG H

af

a

f

IJ K

B ∂2s ∂2s = ρ ∂x 2 ∂t 2 Into this wave equation as a trial solution we substitute the wave function s x , t = smax cos kx − ωt we find ∂s = − ksmax sin kx − ωt ∂x ∂2s = − k 2 smax cos kx − ωt 2 ∂x ∂s = +ωsmax sin kx − ωt ∂t ∂2s = −ω 2 smax cos kx − ωt ∂t 2 B B ∂2s ∂2s = becomes − k 2 smax cos kx − ωt = −ω 2 smax cos kx − ωt ρ ∂x 2 ∂t 2 ρ

b g

a

f

a

f

a

This is true provided

B 4π 2

ρ λ2

f

f

a

B

ρ

.

f

= 4π 2 f 2 .

The sound wave can propagate provided it has λ2 f 2 = v 2 = speed v =

f

f

a

a

a

B

ρ

; that is, provided it propagates with

504

Sound Waves

Section 17.3 *P17.18

Intensity of Periodic Sound Waves

The sound power incident on the eardrum is ℘ = IA where I is the intensity of the sound and A = 5.0 × 10 −5 m 2 is the area of the eardrum. (a)

At the threshold of hearing, I = 1.0 × 10 −12 W m 2 , and

e

je

j

℘ = 1.0 × 10 −12 W m 2 5.0 × 10 −5 m 2 = 5.00 × 10 −17 W . (b)

At the threshold of pain, I = 1.0 W m 2 , and

e I= JK

je

j

℘ = 1.0 W m 2 5.0 × 10 −5 m 2 = 5.00 × 10 −5 W . P17.19

FG I IJ = 10 logF 4.00 × 10 66.0 dB GH 1.00 × 10 HI K F I I 70.0 dB = 10 log G H 1.00 × 10 W m JK W m j10 b Therefore, I = e1.00 × 10 −6

β = 10 log

−12

0

P17.20

(a)

−12

−12

(b)

I=

2

70 .0 10

2

g=

1.00 × 10 −5 W m 2 .

2 ∆Pmax , so 2ρ v

jb

e

ge

∆Pmax = 2 ρvI = 2 1.20 kg m3 343 m s 1.00 × 10 −5 W m 2

j

∆Pmax = 90.7 mPa P17.21

I= (a)

1 2 ρω 2 smax v 2 At f = 2 500 Hz , the frequency is increased by a factor of 2.50, so the intensity (at constant

a f = 6.25 . Therefore, 6.25a0.600f = 3.75 W m smax ) increases by 2.50

(b) P17.22

2

.

0.600 W m 2

The original intensity is I 1 = (a)

2

1 2 2 ρω 2 smax v = 2π 2 ρvf 2 smax 2

If the frequency is increased to f ′ while a constant displacement amplitude is maintained, the new intensity is

b g

2

2 I 2 = 2π 2 ρv f ′ smax so

continued on next page

b g

FG IJ H K

2 2 I 2 2π ρv f ′ smax f′ = = 2 2 2 I1 f 2π ρvf smax

2

or I 2 =

FG f ′ IJ H fK

2

I1 .

Chapter 17

(b)

If the frequency is reduced to f ′ = intensity is

505

f while the displacement amplitude is doubled, the new 2

I 2 = 2π 2 ρv

FG f IJ b2s g H 2K 2

max

2

2 = 2π 2 ρvf 2 smax = I1

or the intensity is unchanged . *P17.23

(a)

For the low note the wavelength is λ = For the high note λ =

v 343 m s = = 2.34 m . f 146.8 s

343 m s = 0.390 m . 880 s

880 Hz = 5.99 nearly 146.8 Hz equal to a small integer. This fact is associated with the consonance of the notes D and A. We observe that the ratio of the frequencies of these two notes is

(b)

β = 10 dB log I=

F GH 10

I −12

W m

2

I = 75 dB gives I = 3.16 × 10 JK

−5

W m2

2 ∆Pmax 2 ρv

jb

e

g

∆Pmax = 3.16 × 10 −5 W m 2 2 1.20 kg m3 343 m s = 0.161 Pa for both low and high notes. (c)

b

g

1 1 2 2 ρv ωsmax = ρv 4π 2 f 2 smax 2 2 I smax = 2 2π ρvf 2 for the low note, I=

smax =

3.16 × 10 −5 W m 2

1 2π 1.20 kg m 343 m s 146.8 s 2

3

6.24 × 10 −5 m = 4.25 × 10 −7 m 146.8 for the high note, 6.24 × 10 −5 smax = m = 7.09 × 10 −8 m 880 =

(d)

*P17.24

146.8 880 = = 1.093 , the 134.3 804.9 wavelengths and displacement amplitudes are made 1.093 times larger, and the pressure amplitudes are unchanged.

With both frequencies lower (numerically smaller) by the factor

The power necessarily supplied to the speaker is the power carried away by the sound wave: P=

b

1 ρAv ωsmax 2

e

g

2

2 = 2π 2 ρAvf 2 smax

j FGH 0.082 m IJK b343 m sgb600 1 sg e0.12 × 10

= 2π 2 1.20 kg m3 π

2

2

−2

j

m

2

= 21.2 W

506 P17.25

Sound Waves

(a)

I 1 = 1.00 × 10 −12 W m 2 10 b

e

or

j

I 1 = 1.00 × 10

e

−4

I 2 = 1.00 × 10

W m

−12

β 1 10

g = e1.00 × 10 −12

j

W m 2 10 80.0 10

2

W m 2 10 b

j

−4.5

β 2 10

g = e1.00 × 10 −12 −5

2

j

W m 2 10 75.0 10

2

or I 2 = 1.00 × 10 W m = 3.16 × 10 W m When both sounds are present, the total intensity is

I = I 1 + I 2 = 1.00 × 10 −4 W m 2 + 3.16 × 10 −5 W m 2 = 1.32 × 10 −4 W m 2 . (b)

The decibel level for the combined sounds is

β = 10 log *P17.26

(a)

(b)

F 1.32 × 10 GH 1.00 × 10

−4

W m2

−12

2

W m

I = 10 log 1.32 × 10 = e j JK 8

81.2 dB .

v and f is the same for all three waves. Since the speed is smallest in air, λ is f 1 493 m s smallest in air. It is larger by = 4.51 times in water and by 331 m s 5 950 = 18.0 times in iron . 331 We have λ =

From I =

1 2 ρvω 2 smax ; smax = 2

2I 0

ρvω 02

, smax is smallest in iron, larger in water by

7 860 ⋅ 5 950 7 860 ⋅ 5 950 ρ iron v iron = = 5.60 times , and larger in air by = 331 times . 1 000 ⋅ 1 493 1.29 ⋅ 331 ρ water v water (c)

From I =

2 ∆Pmax ; ∆Pmax = 2 Iρv , ∆Pmax is smallest in air, larger in water by 2 ρv

1 000 ⋅ 1 493 = 59.1 times , and larger in iron by 1.29 ⋅ 331 (d)

b

7 860 ⋅ 5 950 = 331 times . 1.29 ⋅ 331

g

331 m s 2π v v2π = = = 0.331 m in air f ω 2 000 π s 1 493 m s 5 950 m s λ= = 1.49 m in water λ= = 5.95 m in iron 1 00 0 s 1 000 s

λ=

smax =

2I0

ρvω 02

=

2 × 10 −6 W m 2

e1.29 kg m jb331 m sgb6 283 1 sg 3

2

smax =

2 × 10 −6 1 = 1.84 × 10 −10 m in water 1 000 1 493 6 283

smax =

2 × 10 −6 1 = 3.29 × 10 −11 m in iron 7 860 5 950 6 283

b

b

∆Pmax = 2 Iρv =

g

g 2e10

−6

je

j

W m 2 1.29 kg m3 331 m s = 0.029 2 Pa in air

b g b7 860gb5 950g =

∆Pmax = 2 × 10 −6 1 000 1 493 = 1.73 Pa in water ∆Pmax = 2 × 10 −6

= 1.09 × 10 −8 m in air

9.67 Pa in iron

Chapter 17

P17.27

(a)

120 dB = 10 dB log I = 1.00 W m 2 = r=

℘ = 4π I

LM MN 10

I −12

W m

2

℘

OP PQ

4π r 2 6.00 W

e

4π 1.00 W m 2

= 0.691 m

j

We have assumed the speaker is an isotropic point source. (b)

0 dB = 10 dB log

F GH 10

−12

I JK

I W m2

I = 1.00 × 10 −12 W m 2 r=

℘ = 4π I

6.00 W

e

4π 1.00 × 10 -12 W m 2

j

= 691 km

We have assumed a uniform medium that absorbs no energy. P17.28

We begin with β 2 = 10 log

FG I IJ , and β HI K 2 0

1

I ℘ ℘ , and I 1 = , giving 2 I1 4π r22 4π r12

Then, β 2 − β 1 = 10 log P17.29

FG r IJ Hr K 1

2

2

= 20 log

1 0

β2 Also, I 2 =

FG I IJ , so HI K FI I − β = 10 log G J . HI K Fr I =G J . Hr K

= 10 log

2

1

1

2

1 2

FG r IJ Hr K 1

.

2

Since intensity is inversely proportional to the square of the distance,

a f a fa f 2

I4 =

10.0 ∆P 2 1 I 0. 4 and I 0. 4 = max = = 0.121 W m 2 . 100 2 ρv 2 1.20 343

The difference in sound intensity level is ∆β = 10 log

FG I HI

4 km

0.4 km

At 0.400 km,

β 0. 4 = 10 log At 4.00 km,

IJ = 10a−2.00f = −20.0 dB . K

F 0.121 W m I = 110.8 dB . GH 10 W m JK 2

−12

2

a

f

β 4 = β 0.4 + ∆β = 110.8 − 20.0 dB = 90.8 dB .

Allowing for absorption of the wave over the distance traveled,

b

ga

f

β ′4 = β 4 − 7.00 dB km 3.60 km = 65.6 dB . This is equivalent to the sound intensity level of heavy traffic.

507

508 P17.30

Sound Waves

Let r1 and r2 be the distance from the speaker to the observer that hears 60.0 dB and 80.0 dB, respectively. Use the result of problem 28,

β 2 − β 1 = 20 log Thus, log

FG r IJ , to obtain 80.0 − 60.0 = 20 logFG r IJ . Hr K Hr K 1

1

2

2

FG r IJ = 1 , so r Hr K 1

= 10.0r2 . Also: r1 + r2 = 110 m , so

1

2

10.0r2 + r2 = 110 m giving r2 = 10.0 m , and r1 = 100 m . P17.31

We presume the speakers broadcast equally in all directions. rAC = 3.00 2 + 4.00 2 m = 5.00 m 1.00 × 10 −3 W ℘ I= = = 3.18 × 10 −6 W m 2 2 4π r 2 4π 5.00 m

(a)

a

β = 10 dB log

f

F 3.18 × 10 W m I GH 10 W m JK −6

2

−12

2

β = 10 dB 6.50 = 65.0 dB rBC = 4. 47 m

(b)

I=

1.50 × 10 −3 W

a

f

4π 4.47 m

β = 10 dB log

2

= 5.97 × 10 −6 W m 2

F 5.97 × 10 I GH 10 JK −6

−12

β = 67.8 dB I = 3.18 µW m 2 + 5.97 µW m 2

(c)

β = 10 dB log

P17.32

In I =

F 9.15 × 10 I = GH 10 JK −6

69.6 dB

−12

℘ 1 , intensity I is proportional to 2 , 2 r 4π r

so between locations 1 and 2:

I 2 r12 = . I 1 r22 2 2 2 1

b g , intensity is proportional to s , so II = ss . F s I F r I F 1I F r I Then, G J = G J or G J = G J , giving r = 2r = 2a50.0 mf = 100 m . H s K H r K H 2K H r K But, r = a50.0 mf + d yields d = 86.6 m .

In I =

1 ρv ωsmax 2

2

2 max

2

2

2

2

1

1

2

2

2

1 2

2

2

2

1

2

1

Chapter 17

P17.33

β = 10 log

FG I IJ H 10 K

I = 10 b

−12

β 10

g e10 −12 j

509

W m2

I a120 dB f = 1.00 W m 2 ; I a100 dB f = 1.00 × 10 −2 W m 2 ; I a10 dB f = 1.00 × 10 −11 W m 2 (a)

℘ = 4π r 2 I so that r12 I 1 = r22 I 2 r2

(b) P17.34

P17.35

r2

FI =r G HI FI =r G HI 1

1

1 2

1 2

IJ K IJ K

12

a

f

1.00 = 30.0 m 1.00 × 10 −2

a

f

1.00 = 9.49 × 10 5 m 1.00 × 10 −11

= 3.00 m 12

= 3.00 m

a

f e7.00 × 10 2

ja

−2

(a)

E =℘t = 4π r 2 It = 4π 100 m

(b)

β = 10 log

(a)

The sound intensity inside the church is given by

F 7.00 × 10 I = GH 1.00 × 10 JK −2

−12

f

W m 2 0.200 s = 1.76 kJ

108 dB

FG I IJ HI K F I I 101 dB = a10 dBf lnG H 10 W m JK I = 10 e10 W m j = 10 W m β = 10 ln

0

−12

10.1

−12

2

−1.90

2

2

= 0.012 6 W m 2

We suppose that sound comes perpendicularly out through the windows and doors. Then, the radiated power is

e

je

j

℘ = IA = 0.012 6 W m 2 22.0 m 2 = 0.277 W . Are you surprised by how small this is? The energy radiated in 20.0 minutes is

b

ga

E =℘t = 0.277 J s 20.0 min (b)

60.0 s I fFGH 1.00 J= min K

332 J .

If the ground reflects all sound energy headed downward, the sound power, ℘ = 0.277 W , covers the area of a hemisphere. One kilometer away, this area is

b

g

2

A = 2π r 2 = 2π 1 000 m = 2π × 10 6 m 2 . The intensity at this distance is I=

0. 277 W ℘ = = 4.41 × 10 −8 W m 2 A 2π × 10 6 m 2

and the sound intensity level is

a

f FGH 14..0041××1010

β = 10 dB ln

−8

W m2

−12

2

W m

I= JK

46.4 dB .

510 *P17.36

Sound Waves

Assume you are 1 m away from your lawnmower and receiving 100 dB sound from it. The intensity I of this sound is given by 100 dB = 10 dB log −12 ; I = 10 −2 W m 2 . If the lawnmower 10 W m2 ℘ radiates as a point source, its sound power is given by I = . 4π r 2

a f

2

℘ = 4π 1 m 10 −2 W m 2 = 0.126 W Now let your neighbor have an identical lawnmower 20 m away. You receive from it sound with 0.126 W intensity I = = 2.5 × 10 −5 W m 2 . The total sound intensity impinging on you is 2 4π 20 m

a

10

−2

f

W m + 2.5 × 10 −5 W m 2 = 1.002 5 × 10 −2 W m 2 . So its level is 2

10 dB log

1.002 5 × 10 −2 10 −12

= 100.01 dB .

If the smallest noticeable difference is between 100 dB and 101 dB, this cannot be heard as a change from 100 dB.

Section 17.4 P17.37

f′= f

The Doppler Effect

bv ± v g bv ± v g O S

(a)

(b)

P17.38

(a)

a343 + 40.0f = 338 Hz a343 + 20.0f a343 + 20.0f = 483 Hz f ′ = 510 a343 + 40.0f F 115 min I = 12.0 rad s ω = 2π f = 2π G H 60.0 s min JK v = ωA = b12.0 rad sge1.80 × 10 mj = f ′ = 320

−3

max

(b)

The heart wall is a moving observer. f′= f

(c)

0.021 7 m s

FG v + v IJ = b2 000 000 HzgFG 1 500 + 0.021 7 IJ = H v K H 1 500 K O

2 000 028.9 Hz

Now the heart wall is a moving source. f ′′ = f ′

FG v IJ = b2 000 029 HzgF 1 500 I = GH 1 500 − 0.021 7 JK Hv−v K s

2 000 057.8 Hz

Chapter 17

P17.39

f′=

Approaching ambulance:

511

f

b1 − v vg S

f

f ′′ =

Departing ambulance:

d1 − b− v vgi F vI F vI 560G 1 − J = 480G 1 + J H vK H vK S

S

Since f ′ = 560 Hz and f ′′ = 480 Hz

S

vS = 80.0 v 80.0 343 m s = 26.4 m s vS = 1 040

1 040

a f

P17.40

(a)

The maximum speed of the speaker is described by 1 1 2 = kA 2 mv max 2 2 v max =

a

f

20.0 N m 0.500 m = 1.00 m s 5.00 kg

k A= m

The frequencies heard by the stationary observer range from ′ =f fmin

FG v IJ to f ′ Hv+v K

max

max

=f

FG v IJ Hv−v K max

where v is the speed of sound.

(b)

β = 10 dB log

F 343 m s I = GH 343 m s + 1.00 m s JK F 343 m s I = = 440 HzG H 343 m s − 1.00 m s JK

fmin ′ = 440 Hz

439 Hz

fmax ′

441 Hz

FG I IJ = 10 dB logF ℘ 4π r I GH I JK HI K 2

0

0

The maximum intensity level (of 60.0 dB) occurs at r = rmin = 1.00 m . The minimum intensity level occurs when the speaker is farthest from the listener (i.e., when r = rmax = rmin + 2 A = 2.00 m).

F ℘ I − 10 dB logF ℘ I GH 4π I r JK GH 4π I r JK F ℘ 4π I r I = 10 dB logF r I . β or −β = 10 dB log G GH r JK JK ℘ H 4π I r This gives: 60.0 dB − β = 10 dB log a 4.00f = 6.02 dB , and β = 54.0 dB Thus, β max − β min = 10 dB log

max

min

2 0 min

2 0 min

min

2 0 max

2 0 max

2 max 2 min

min

.

512 P17.41

Sound Waves

f′= f

F 340 I GH 340 − b−9.80t g JK 485a340f + a 485fd9.80t i = a512 fa340 f F 512 − 485 IJ 340 = 1.93 s t =G H 485 K 9.80

FG v IJ Hv−v K

485 = 512

s

fall

f

f

d1 =

1 2 gt f = 18.3 m : 2

t return =

18.3 = 0.053 8 s 340

The fork continues to fall while the sound returns. t total fall = t f + treturn = 1.93 s + 0.053 8 s = 1.985 s

P17.42

b

g

d total =

1 2 gt total fall = 19.3 m 2

a

f

m −10° C = 325 m s s⋅° C

(a)

v = 331 m s + 0.6

(b)

Approaching the bell, the athlete hears a frequency of After passing the bell, she hears a lower frequency of

*P17.43

(a)

FG v + v IJ H v K F v + b− v g I f ′′ = f G H v JK f′= f

O

O

The ratio is

f ′′ v − vO 5 = = f ′ v + vO 6

which gives 6 v − 6 v o = 5 v + 5 v o or

vO =

a

v 325 m s = = 29.5 m s 11 11

f

Sound moves upwind with speed 343 − 15 m s . Crests pass a stationary upwind point at frequency 900 Hz. Then

λ=

v 328 m s = = 0.364 m f 900 s

(b)

By similar logic,

λ=

343 + 15 m s v = = 0.398 m f 900 s

(c)

The source is moving through the air at 15 m/s toward the observer. The observer is stationary relative to the air. f′= f

a

f

FG v + v IJ = 900 HzFG 343 + 0 IJ = H 343 − 15 K H v−v K o

941 Hz

s

(d)

The source is moving through the air at 15 m/s away from the downwind firefighter. Her speed relative to the air is 30 m/s toward the source. f′= f

FG v + v IJ = 900 HzF 343 + 30 I = 900 HzFG 373 IJ = GH 343 − a−15f JK H 358 K H v−v K o s

938 Hz

Chapter 17

*P17.44

The half-angle of the cone of the shock wave is θ where

θ = sin −1

FG v Hv

sound source

IJ = sin FG 1 IJ = 41.8° . H 1.5 K K −1

φ

As shown in the sketch, the angle between the direction of propagation of the shock wave and the direction of the plane’s velocity is

The half angle of the shock wave cone is given by sin θ =

vS =

P17.46

θ = sin −1

P17.47

(b)

sin θ

=

v light vS

.

2.25 × 10 8 m s = 2.82 × 10 8 m s sin 53.0°

a

f

v 1 = sin −1 = 46.4° vS 1.38

sin θ =

v 1 = ; θ = 19.5° vS 3.00

tan θ =

h h ; x= x tan θ

x= (a)

v light

v shock FIG. P17.44

φ = 90°−θ = 90°−41.8° = 48.2° . P17.45

20 000 m = 5.66 × 10 4 m = 56.6 km tan 19.5°

It takes the plane t =

5.66 × 10 4 m x = = 56.3 s to travel this distance. vS 3.00 335 m s

b

g

x

θ t=0

θ h

h

Observer a.

v plane

θ

Observer hears the boom b. FIG. P17.47(a)

513

514

Sound Waves

Section 17.5

Digital Sound Recording

Section 17.6

Motion Picture Sound

*P17.48

For a 40-dB sound, 40 dB = 10 dB log

LM MN 10

I = 10 −8 W m 2 =

2 ∆Pmax 2 ρv

−12

OP PQ

I W m2

jb

e

g

∆Pmax = 2 ρvI = 2 1.20 kg m 2 343 m s 10 −8 W m 2 = 2.87 × 10 −3 N m 2

*P17.49

code =

(b)

For sounds of 40 dB or softer, too few digital words are available to represent the wave form with good fidelity.

(c)

In a sound wave ∆P is negative half of the time but this coding scheme has no words available for negative pressure variations.

65 536 = 7

28.7 N m 2

If the source is to the left at angle θ from the direction you are facing, the sound must travel an extra distance d sin θ to reach your right ear as shown, where d is the distance between your ears. The d sin θ . Then delay time is ∆t in v = ∆t

θ = sin −1

*P17.50

2.87 × 10 −3 N m 2

(a)

103 dB = 10 dB log (a)

b

g

343 m s 210 × 10 v∆ t = sin −1 d 0.19 m

LM MN 10

I −12

W m

2

I = 2.00 × 10 −2 W m 2 =

−6

s

= 22.3° left of center .

θ

θ ear

ear

FIG. P17.49

OP PQ ℘ 4π r 2

=

a

℘

f

4π 1.6 m

2

℘ = 0.642 W (b)

efficiency =

sound output power 0.642 W = = 0.004 28 total input power 150 W

Additional Problems P17.51

Model your loud, sharp sound impulse as a single narrow peak in a graph of air pressure versus time. It is a noise with no pitch, no frequency, wavelength, or period. It radiates away from you in all directions and some of it is incident on each one of the solid vertical risers of the bleachers. Suppose that, at the ambient temperature, sound moves at 340 m/s; and suppose that the horizontal width of each row of seats is 60 cm. Then there is a time delay of 0.6 m

b340 m sg = 0.002 s continued on next page

Chapter 17

515

between your sound impulse reaching each riser and the next. Whatever its material, each will reflect much of the sound that reaches it. The reflected wave sounds very different from the sharp pop you made. If there are twenty rows of seats, you hear from the bleachers a tone with twenty crests, each separated from the next in time by

a

f = 0.004 s . b340 m sg 2 0.6 m

This is the extra time for it to cross the width of one seat twice, once as an incident pulse and once again after its reflection. Thus, you hear a sound of definite pitch, with period about 0.004 s, frequency 1 ~ 300 Hz 0.003 5 s wavelength

b

g

340 m s v = = 1.2 m ~ 10 0 m f 300 s

λ=

b

and duration

g

a

f

20 0.004 s ~ 10 −1 s . P17.52

v 343 m s = = 0.232 m f 1 480 s −1

(a)

λ=

(b)

β = 81.0 dB = 10 dB log

e

j

LM MN 10

I −12

W m

2

OP PQ

I = 10 −12 W m 2 10 8.10 = 10 −3.90 W m 2 = 1.26 × 10 −4 W m 2 = smax =

(c)

P17.53

λ′ =

2I = ρ vω 2

e

2 1.26 × 10 −4 W m 2

e1.20 kg m jb343 m sg4π e1 480 s j 3

v 343 m s = = 0.246 m f ′ 1 397 s −1

Since cos 2 θ + sin 2 θ = 1 ,

j

−1 2

2

1 2 ρvω 2 smax 2

= 8.41 × 10 −8 m

∆λ = λ ′ − λ = 13.8 mm

sin θ = ± 1 − cos 2 θ (each sign applying half the time)

a

f

a

∆P = ∆Pmax sin kx − ωt = ± ρvω smax 1 − cos 2 kx − ωt

f

2 2 2 ∆P = ± ρvω smax − smax − s2 cos 2 kx − ωt = ± ρvω smax

Therefore P17.54

a

f

The trucks form a train analogous to a wave train of crests with speed v = 19.7 m s 2 and unshifted frequency f = = 0.667 min −1 . 3.00 min (a)

(b)

The cyclist as observer measures a lower Doppler-shifted frequency: 19.7 + −4. 47 v + vo = 0.667 min −1 = 0.515 min f′= f v 19.7

FG IJ e H K F v + v ′ IJ = e0.667 min f ′′ = f G H v K o

jFGH a f IJK jFGH 19.7 +19a.−71.56f IJK =

−1

0.614 min

The cyclist’s speed has decreased very significantly, but there is only a modest increase in the frequency of trucks passing him.

516

Sound Waves

ja

f

P17.55

v=

2d vt 1 :d= = 6.50 × 10 3 m s 1.85 s = 6.01 km t 2 2

P17.56

(a)

The speed of a compression wave in a bar is

e

Y

v= (b)

ρ

20.0 × 10 10 N m 2 7 860 kg m

3

= 5.04 × 10 3 m s .

The signal to stop passes between layers of atoms as a sound wave, reaching the back end of the bar in time t=

(c)

=

0.800 m L = = 1.59 × 10 −4 s . v 5.04 × 10 3 m s

As described by Newton’s first law, the rearmost layer of steel has continued to move forward with its original speed vi for this time, compressing the bar by

b

ge

j

∆L = vi t = 12.0 m s 1.59 × 10 −4 s = 1.90 × 10 −3 m = 1.90 mm . (d)

The strain in the rod is:

(e)

The stress in the rod is:

σ =Y

∆L 1.90 × 10 −3 m = = 2.38 × 10 −3 . L 0.800 m

FG ∆L IJ = e20.0 × 10 HLK

10

je

j

N m 2 2.38 × 10 −3 = 476 MPa .

Since σ > 400 MPa , the rod will be permanently distorted. (f)

We go through the same steps as in parts (a) through (e), but use algebraic expressions rather than numbers: The speed of sound in the rod is v =

Y

ρ

.

The back end of the rod continues to move forward at speed vi for a time of t = traveling distance ∆L = vi t after the front end hits the wall. The strain in the rod is:

∆L vi t ρ = = vi . L L Y

The stress is then: σ = Y

FG ∆L IJ = Yv HLK

ρ i

Y

ρ L =L , v Y

= vi ρY .

For this to be less than the yield stress, σ y , it is necessary that

vi ρY < σ y or vi
20.0 cm, the next two modes will be observed, corresponding to f = or L 2 = P18.47

3v 5v = 0.502 m and L3 = = 0.837 m . 4f 4f

We suppose these are the lowest resonances of the enclosed air columns.

λ = 0.670 m 2

For one,

λ=

v 343 m s = = 1.34 m f 256 s −1

length = d AA =

For the other,

λ=

v 343 m s = = 0.780 m f 440 s −1

length = 0.390 m

So, (b)

original length = 1.06 m

λ = 2d AA = 2.12 m

P18.48

3v 5v and f = . 4L 2 4L3

343 m s = 162 Hz 2.12 m

(a)

f=

(a)

For the fundamental mode of an open tube, L=

(b)

v = 331 m s 1 +

343 m s λ v = = = 0.195 m . 2 2 f 2 880 s −1

e

j

a−5.00f = 328 m s 273

We ignore the thermal expansion of the metal. f= The flute is flat by a semitone.

v

λ

=

328 m s v = = 841 Hz 2L 2 0.195 m

a

f

Chapter 18

Section 18.6 P18.49

541

Standing Waves in Rod and Plates 5 100 v = = 1.59 kHz 2L 2 1.60

(a)

f=

(b)

Since it is held in the center, there must be a node in the center as well as antinodes at the ends. The even harmonics have an antinode at the center so only the odd harmonics are

a fa f

present. (c) P18.50

f=

3 560 v′ = = 1.11 kHz 2L 2 1.60

When the rod is clamped at one-quarter of its length, the vibration pattern reads ANANA and the rod length is L = 2d AA = λ . Therefore, L =

Section 18.7 P18.51

a fa f

v 5 100 m s = = 1.16 m f 4 400 Hz

Beats: Interference in Time

f ∝v∝ T

f new = 110

540 = 104.4 Hz 600

∆f = 5.64 beats s P18.52

(a)

The string could be tuned to either 521 Hz or 525 Hz from this evidence.

(b)

Tightening the string raises the wave speed and frequency. If the frequency were originally 521 Hz, the beats would slow down. Instead, the frequency must have started at 525 Hz to become 526 Hz .

(c)

From f =

v

λ

=

T µ 2L

=

1 T 2L µ

FG IJ H K

f2 T2 f = and T2 = 2 f1 T1 f1

2

T1 =

FG 523 Hz IJ H 526 Hz K

2

T1 = 0.989T1 .

The fractional change that should be made in the tension is then fractional change =

T1 − T2 = 1 − 0.989 = 0.011 4 = 1.14% lower. T1

The tension should be reduced by 1.14% .

542 P18.53

Superposition and Standing Waves

For an echo f ′ = f

bv + v g the beat frequency is f bv − v g s s

b

= f′− f .

Solving for fb . gives fb = f

b2 v g s

bv − v g when approaching wall. 2a1.33f = a 256f a343 − 1.33f = 1.99 Hz beat frequency s

(a)

fb

(b)

When he is moving away from the wall, v s changes sign. Solving for v s gives vs =

*P18.54

2 - foot pipes produces actual frequencies of 131 Hz and 196 Hz and a 3 combination tone at 196 − 131 Hz = 65.4 Hz , so this pair supplies the so-called missing fundamental. The 4 and 2-foot pipes produce a combination tone 262 − 131 Hz = 131 Hz , so this does not work. 2 The 2 and 2 - foot pipes produce a combination tone at 262 − 196 Hz = 65.4 Hz , so this works. 3 2 Also, 4, 2 , and 2 - foot pipes all playing together produce the 65.4-Hz combination tone. 3

Using the 4 and 2

a

Section 18.8 P18.55

a fa f a fa f

5 343 fb v = = 3.38 m s . 2 f − fb 2 256 − 5

f

a

a

f

f

Non-Sinusoidal Wave Patterns

We list the frequencies of the harmonics of each note in Hz: Note A C# E

1 440.00 554.37 659.26

2 880.00 1 108.7 1 318.5

Harmonic 3 1 320.0 1 663.1 1 977.8

4 1 760.0 2 217.5 2 637.0

The second harmonic of E is close the the third harmonic of A, and the fourth harmonic of C# is close to the fifth harmonic of A. P18.56

We evaluate s = 100 sin θ + 157 sin 2θ + 62.9 sin 3θ + 105 sin 4θ +51.9 sin 5θ + 29.5 sin 6θ + 25.3 sin 7θ where s represents particle displacement in nanometers and θ represents the phase of the wave in radians. As θ advances by 2π , time advances by (1/523) s. Here is the result: FIG. P18.56

5 2 200.0 2 771.9 3 296.3

Chapter 18

543

Additional Problems P18.57

f = 87.0 Hz speed of sound in air: v a = 340 m s (a)

ja

e

f

v = fλ b = 87.0 s −1 0.400 m

λb =

v = 34.8 m s

(b)

λ a = 4L va = λ a f

UV W

L=

340 m s va = = 0.977 m 4 f 4 87.0 s −1

e

j

FIG. P18.57 *P18.58

(a)

Use the Doppler formula f′= f

bv ± v g . bv ∓ v g 0 s

With f1′ = frequency of the speaker in front of student and f 2′ = frequency of the speaker behind the student.

m s + 1.50 m sg = 458 Hz f b343b343 m s − 0g b343 m s − 1.50 m sg = 454 Hz f ′ = a 456 Hzf b343 m s + 0g

a

f1′ = 456 Hz

2

Therefore, fb = f1′ − f 2′ = 3.99 Hz . (b)

The waves broadcast by both speakers have λ =

v 343 m s = = 0.752 m . The standing wave f 456 s

λ = 0.376 m . The student walks from one maximum to the next in 2 0.376 m 1 = 0.251 s , so the frequency at which she hears maxima is f = = 3.99 Hz . time ∆t = T 1.50 m s

between them has d AA =

P18.59

Moving away from station, frequency is depressed: 343 f ′ = 180 − 2.00 = 178 Hz : 178 = 180 343 − −v Solving for v gives Therefore,

v=

a2.00fa343f

a f

178 v = 3.85 m s away from station

Moving toward the station, the frequency is enhanced: 343 f ′ = 180 + 2.00 = 182 Hz : 182 = 180 343 − v 2.00 343 Solving for v gives 4= 182 Therefore, v = 3.77 m s toward the station

a fa f

544 P18.60

Superposition and Standing Waves

v=

a48.0fa2.00f = 141 m s 4.80 × 10 −3

d NN = 1.00 m ; λ = 2.00 m ; f =

λa = P18.61

v

λ

= 70.7 Hz

v a 343 m s = = 4.85 m f 70.7 Hz

Call L the depth of the well and v the speed of sound.

f λ4 = a2n − 1f 4vf = a2n4− 511fb.5343s m sg e j a2n + 1fb343 m sg λ v = a 2n + 1f = L = 2an + 1f − 1 4 4f 4e60.0 s j a2n − 1fb343 m sg = a2n + 1fb343 m sg 4e51.5 s j 4e60.0 s j a

L = 2n − 1

Then for some integer n

1

−1

1

2

and for the next resonance

2

Thus,

−1

and we require an integer solution to The equation gives n =

−1

−1

2n + 1 2n − 1 = 60.0 51.5

111.5 = 6.56 , so the best fitting integer is n = 7 . 17

Then

L=

and

L=

af b g = 21.6 m 4e51.5 s j 2a7 f + 1 b343 m sg = 21.4 m 4e60.0 s j 2 7 − 1 343 m s −1

−1

suggest the best value for the depth of the well is 21.5 m . P18.62

The second standing wave mode of the air in the pipe reads ANAN, with d NA = so

λ = 2.33 m

and

f=

v

λ

=

343 m s = 147 Hz 2.33 m

For the string, λ and v are different but f is the same.

λ 0.400 m = d NN = 2 2 so

λ = 0.400 m

a

fa

f

T

v = λf = 0. 400 m 147 Hz = 58.8 m s =

e

jb

T = µv 2 = 9.00 × 10 −3 kg m 58.8 m s

g

µ 2

= 31.1 N

λ 1.75 m = 3 4

Chapter 18

P18.63

(a)

Since the first node is at the weld, the wavelength in the thin wire is 2L or 80.0 cm. The frequency and tension are the same in both sections, so f=

(b)

1 2L

T

µ

=

1 2f thin wire.

so L ′ =

(a)

1 2 0.400

a

4.60 = 59.9 Hz . 2.00 × 10 −3

f

As the thick wire is twice the diameter, the linear density is 4 times that of the thin wire.

µ ′ = 8.00 g m

P18.64

545

T µ′

L′ =

LM 1 OP N a2fa59.9f Q

4.60 = 20.0 cm half the length of the 8.00 × 10 −3

For the block:

∑ Fx = T − Mg sin 30.0° = 0 so T = Mg sin 30.0° = (b)

1 Mg . 2

The length of the section of string parallel to the incline is h = 2 h . The total length of the string is then 3h . sin 30.0°

FIG. P18.64

m 3h

(c)

The mass per unit length of the string is

µ=

(d)

The speed of waves in the string is

v=

(e)

In the fundamental mode, the segment of length h vibrates as one loop. The distance between adjacent nodes is then d NN =

f=

3 Mgh 2m

v

λ

=

1 3 Mgh = 2h 2m

3 Mg 8mh

λ= h .

The period of the standing wave of 3 nodes (or two loops) is T=

(h)

FG Mg IJ FG 3 h IJ = H 2 KH m K

When the vertical segment of string vibrates with 2 loops (i.e., 3 nodes), then h = 2 the wavelength is

(f)

µ

=

λ = h , so the wavelength is λ = 2h . 2

The frequency is

(g)

T

e

j

fb = 1.02 f − f = 2.00 × 10 −2 f =

e2.00 × 10 j −2

3 Mg 8mh

1 λ 2m = =h = 3 Mgh f v

2mh 3 Mg

FG λ IJ and H 2K

546 P18.65

Superposition and Standing Waves

(a)

f=

n 2L

T

so

f′ L L 1 = = = f L ′ 2L 2

µ

The frequency should be halved to get the same number of antinodes for twice the length. (b)

n′ T = n T′

FG IJ = LM n OP H K Nn + 1 Q L n OP T T′ = M Nn + 1 Q T ′ F nf ′L ′ I =G J T H n ′fL K T′ n = T n′

so

2

2

2

The tension must be

(c)

FG IJ H K

T′ 3 = T 2⋅2

P18.66

2

f ′ n ′L T ′ = f nL ′ T

so

2

T′ 9 = T 16

to get twice as many antinodes.

0.010 0 kg = 5.00 × 10 −3 kg m : For the wire, µ = 2.00 m

T

v=

µ

e200 kg ⋅ m s j 2

=

5.00 × 10 −3 kg m

v = 200 m s If it vibrates in its simplest state, d NN = 2.00 m =

f=

v

λ

b200 m sg = 50.0 Hz

=

4.00 m

(a)

The tuning fork can have frequencies 45.0 Hz or 55.0 Hz .

(b)

If f = 45.0 Hz , v = fλ = 45.0 s 4.00 m = 180 m s .

b

b

g

Then, T = v 2 µ = 180 m s

g e5.00 × 10 2

or if f = 55.0 Hz , T = v 2 µ = f 2 P18.67

λ : 2

j λ µ = b55.0 sg a 4.00 mf e5.00 × 10 −3

kg m = 162 N 2

2

2

−3

j

kg m = 242 N .

We look for a solution of the form

a

f

a

f

b g = A sina 2.00 x − 10.0t f cos φ + A cosa 2.00 x − 10.0t f sin φ

5.00 sin 2.00 x − 10.0t + 10.0 cos 2.00 x − 10.0 t = A sin 2.00 x − 10.0 t + φ

This will be true if both

5.00 = A cos φ and 10.0 = A sin φ ,

requiring

a5.00f + a10.0f 2

2

= A2

A = 11.2 and φ = 63.4°

a

The resultant wave 11.2 sin 2.00 x − 10.0t + 63.4°

f

is sinusoidal.

Chapter 18

P18.68

P18.69

2π

and ω = 2π f =

2π v

FG 2π x IJ cosFG 2π vt IJ H λK H λ K

b g

y x , t = 2 A sin kx cos ωt = 2 A sin

(a)

With k =

(b)

For the fundamental vibration,

λ 1 = 2L

so

y1 x , t = 2 A sin

(c)

For the second harmonic λ 2 = L and

y 2 x , t = 2 A sin

(d)

In general, λ n =

(a)

Let θ represent the angle each slanted rope makes with the vertical.

λ

λ

:

2L and n

b g

FG π x IJ cosFG π vt IJ HLK H L K

b g

FG 2π x IJ cosFG 2π vt IJ H LK H L K

b g

FG nπ x IJ cosFG nπ vt IJ H LK H L K

yn x , t = 2 A sin

In the diagram, observe that: sin θ =

1.00 m 2 = 1.50 m 3

or θ = 41.8° . Considering the mass,

∑ Fy = 0 : 2T cos θ = mg

b12.0 kg ge9.80 m s j = or T =

FIG. P18.69

2

2 cos 41.8°

(b)

*P18.70

d AA =

78.9 N T

v=

For the standing wave pattern shown (3 loops),

d=

3 λ 2

or

λ=

2 2.00 m = 1.33 m 3

Thus, the required frequency is

f=

µ

a

v

λ

=

=

78.9 N = 281 m s 0.001 00 kg m

The speed of transverse waves in the string is

f

281 m s = 211 Hz 1.33 m

λ = 7.05 × 10 −3 m is the distance between antinodes. 2

Then λ = 14.1 × 10 −3 m and f =

v

λ

=

3.70 × 10 3 m s 14.1 × 10

−3

m

547

= 2.62 × 10 5 Hz .

The crystal can be tuned to vibrate at 2 18 Hz , so that binary counters can derive from it a signal at precisely 1 Hz.

FIG. P18.70

548

Superposition and Standing Waves

ANSWERS TO EVEN PROBLEMS P18.2

see the solution

P18.38

0.656 m; 1.64 m

P18.4

5.66 cm

P18.40

3 kHz; see the solution

P18.6

0.500 s

P18.42

∆t =

P18.8

(a) 3.33 rad; (b) 283 Hz

P18.10

(a) The number is the greatest f 1 + ; integer ≤ d v 2

P18.44

L = 0.252 m, 0.504 m, 0.757 m, … , n 0.252 m for n = 1, 2 , 3 , …

P18.46

0.502 m; 0.837 m

P18.48

(a) 0.195 m; (b) 841 m

P18.50

1.16 m

P18.52

(a) 521 Hz or 525 Hz; (b) 526 Hz; (c) reduce by 1.14%

FG IJ H K d − bn − 1 2g b v f g = 2bn − 1 2 gb v f g 2

(b) Ln

2

2

where

n = 1, 2 , … , n max P18.12

λ ; 2 (b) along the hyperbola 9 x 2 − 16 y 2 = 144 (a) ∆x =

a

f

π r 2v 2 Rf

a

f

2 2 4-foot and 2 -foot ; 2 and 2 - foot; and 3 3 all three together

P18.14

(a) 2n + 1 π m for n = 0 , 1, 2 , 3 , …; (b) 0.029 4 m

P18.54

P18.16

see the solution

P18.56

see the solution

P18.18

see the solution

P18.58

(a) and (b) 3.99 beats s

P18.20

15.7 Hz

P18.60

4.85 m

P18.22

(a) 257 Hz; (b) 6

P18.62

31.1 N

P18.24

(a) 495 Hz; (b) 990 Hz

P18.64

(a)

P18.26

19.976 kHz

P18.28

3.84%

3 Mgh 1 m Mg ; (b) 3h; (c) ; (d) ; 2m 2 3h 3 Mg 2mh (e) ; (g) h; ; (f) 8mh 3 Mg

e

(h) 2.00 × 10 −2

j

3 Mg 8mh

P18.30

291 Hz

P18.32

0.352 Hz

P18.66

(a) 45.0 Hz or 55.0 Hz; (b) 162 N or 242 N

P18.34

see the solution

P18.68

see the solution

P18.36

(a) 531 Hz; (b) 42.5 mm

P18.70

262 kHz

19 Temperature CHAPTER OUTLINE 19.1 19.2 19.3

19.4 19.5

Temperature and the Zeroth Law of Thermodynamics Thermometers and the Celsius Temperature Scale The Constant-Volume Gas Thermometer and the Absolute Temperature Scale Thermal Expansion of Solids and Liquids Macroscopic Description of an Ideal Gas

ANSWERS TO QUESTIONS Q19.1

Two objects in thermal equilibrium need not be in contact. Consider the two objects that are in thermal equilibrium in Figure 19.1(c). The act of separating them by a small distance does not affect how the molecules are moving inside either object, so they will still be in thermal equilibrium.

Q19.2

The copper’s temperature drops and the water temperature rises until both temperatures are the same. Then the metal and the water are in thermal equilibrium.

Q19.3

The astronaut is referring to the temperature of the lunar surface, specifically a 400°F difference. A thermometer would register the temperature of the thermometer liquid. Since there is no atmosphere in the moon, the thermometer will not read a realistic temperature unless it is placed into the lunar soil.

Q19.4

Rubber contracts when it is warmed.

Q19.5

Thermal expansion of the glass bulb occurs first, since the wall of the bulb is in direct contact with the hot water. Then the mercury heats up, and it expands.

Q19.6

If the amalgam had a larger coefficient of expansion than your tooth, it would expand more than the cavity in your tooth when you take a sip of your ever-beloved coffee, resulting in a broken or cracked tooth! As you ice down your now excruciatingly painful broken tooth, the amalgam would contract more than the cavity in your tooth and fall out, leaving the nerve roots exposed. Isn’t it nice that your dentist knows thermodynamics?

Q19.7

The measurements made with the heated steel tape will be too short—but only by a factor of 5 × 10 −5 of the measured length.

Q19.8

(a)

One mole of H 2 has a mass of 2.016 0 g.

(b)

One mole of He has a mass of 4.002 6 g.

(c)

One mole of CO has a mass of 28.010 g.

Q19.9

The ideal gas law, PV = nRT predicts zero volume at absolute zero. This is incorrect because the ideal gas law cannot work all the way down to or below the temperature at which gas turns to liquid, or in the case of CO 2 , a solid. 549

550

Temperature

Q19.10

Call the process isobaric cooling or isobaric contraction. The rubber wall is easy to stretch. The air inside is nearly at atmospheric pressure originally and stays at atmospheric pressure as the wall moves in, just maintaining equality of pressure outside and inside. The air is nearly an ideal gas to start with, but PV = nRT soon fails. Volume will drop by a larger factor than temperature as the water vapor liquefies and then freezes, as the carbon dioxide turns to snow, as the argon turns to slush, and as the oxygen liquefies. From the outside, you see contraction to a small fraction of the original volume.

Q19.11

Cylinder A must be at lower pressure. If the gas is thin, it will be at one-third the absolute pressure of B.

Q19.12

At high temperature and pressure, the steam inside exerts large forces on the pot and cover. Strong latches hold them together, but they would explode apart if you tried to open the hot cooker.

Q19.13

(a)

The water level in the cave rises by a smaller distance than the water outside, as the trapped air is compressed. Air can escape from the cave if the rock is not completely airtight, and also by dissolving in the water.

(b)

The ideal cave stays completely full of water at low tide. The water in the cave is supported by atmospheric pressure on the free water surface outside.

(a)

(b) FIG. Q19.13

Q19.14

Absolute zero is a natural choice for the zero of a temperature scale. If an alien race had bodies that were mostly liquid water—or if they just liked its taste or its cleaning properties—it is conceivable that they might place one hundred degrees between its freezing and boiling points. It is very unlikely, on the other hand, that these would be our familiar “normal” ice and steam points, because atmospheric pressure would surely be different where the aliens come from.

Q19.15

As the temperature increases, the brass expands. This would effectively increase the distance, d, from the pivot point to the center of mass of the pendulum, and also increase the moment of inertia of the pendulum. Since the moment of inertia is proportional to d 2 , and the period of a physical I pendulum is T = 2π , the period would increase, and the clock would run slow. mgd

Q19.16

As the water rises in temperature, it expands. The excess volume would spill out of the cooling system. Modern cooling systems have an overflow reservoir to take up excess volume when the coolant heats up and expands.

Q19.17

The coefficient of expansion of metal is larger than that of glass. When hot water is run over the jar, both the glass and the lid expand, but at different rates. Since all dimensions expand, there will be a certain temperature at which the inner diameter of the lid has expanded more than the top of the jar, and the lid will be easier to remove.

Chapter 19

Q19.18

The sphere expands when heated, so that it no longer fits through the ring. With the sphere still hot, you can separate the sphere and ring by heating the ring. This more surprising result occurs because the thermal expansion of the ring is not like the inflation of a blood-pressure cuff. Rather, it is like a photographic enlargement; every linear dimension, including the hole diameter, increases by the same factor. The reason for this is that the atoms everywhere, including those around the inner circumference, push away from each other. The only way that the atoms can accommodate the greater distances is for the circumference—and corresponding diameter—to grow. This property was once used to fit metal rims to wooden wagon and horse-buggy wheels. If the ring is heated and the sphere left at room temperature, the sphere would pass through the ring with more space to spare.

551

FIG. Q19.18

SOLUTIONS TO PROBLEMS Section 19.1

Temperature and the Zeroth Law of Thermodynamics

No problems in this section

Section 19.2

Thermometers and the Celsius Temperature Scale

Section 19.3

The Constant-Volume Gas Thermometer and the Absolute Temperature Scale

P19.1

Since we have a linear graph, the pressure is related to the temperature as P = A + BT , where A and B are constants. To find A and B, we use the data

a f 1.635 atm = A + a78.0° CfB

0.900 atm = A + −80.0° C B

(1) (2)

Solving (1) and (2) simultaneously, we find

A = 1.272 atm

and

B = 4.652 × 10 −3 atm ° C

Therefore,

P = 1.272 atm + 4.652 × 10 −3 atm ° C T

(a)

P = 0 = 1.272 atm + 4.652 × 10 −3 atm ° C T

At absolute zero which gives

e

j

e

j

T = −274° C .

(b)

At the freezing point of water P = 1.272 atm + 0 = 1.27 atm .

(c)

And at the boiling point P = 1.272 atm + 4.652 × 10 −3 atm ° C 100° C = 1.74 atm .

e

ja

f

552 P19.2

Temperature

P1V = nRT1 and P2 V = nRT2 imply that

P19.3

P19.4

P2 T2 = P1 T1

a

fa

f

(a)

P2 =

0.980 atm 273 K + 45.0 K P1T2 = = 1.06 atm 273 + 20.0 K T1

(b)

T3 =

T1 P3 P1

a f a293 K fa0.500 atmf = 149 K = = 0.980 atm

−124° C FIG. P19.2

a

f

9 9 TC + 32.0° F = −195.81 + 32.0 = −320° F 5 5

(a)

TF =

(b)

T = TC + 273.15 = −195.81 + 273.15 = 77.3 K

(a)

To convert from Fahrenheit to Celsius, we use

TC =

and the Kelvin temperature is found as

T = TC + 273 = 310 K

(b)

In a fashion identical to that used in (a), we find TC = −20.6° C T = 253 K

and P19.5

P19.6

b

FG 212° F − 32.0° F IJ = H 100° C − 0.00° C K

(a)

∆T = 450° C = 450° C

(b)

∆T = 450° C = 450 K

810° F

a f 100° C = aa60.0° Sf + b Subtracting, 100° C = aa75.0° Sf Require

0.00° C = a −15.0° S + b

a

a = 1.33 C° S° .

f

Then 0.00° C = 1.33 −15.0° S C°+ b b = 20.0° C .

b

g

So the conversion is TC = 1.33 C° S° TS + 20.0° C . P19.7

(a)

T = 1 064 + 273 = 1 337 K melting point T = 2 660 + 273 = 2 933 K boiling point

(b)

g a

f

5 5 TF − 32.0 = 98.6 − 32.0 = 37.0° C 9 9

∆T = 1 596° C = 1 596 K . The differences are the same.

Chapter 19

Section 19.4 P19.8

553

Thermal Expansion of Solids and Liquids

α = 1.10 × 10 −5 ° C −1 for steel

e

a

j

f

∆L = 518 m 1.10 × 10 −5 ° C −1 35.0° C − −20.0° C = 0.313 m P19.9

The wire is 35.0 m long when TC = −20.0° C .

b

∆L = Liα T − Ti

g

a f a f for Cu. ∆L = a35.0 mfe1.70 × 10 aC°f jc35.0° C − a −20.0° C fh = −1

α = α 20.0° C = 1.70 × 10 −5 C°

−1

−5

a

fe

ja

+3.27 cm

f

P19.10

∆L = Liα ∆T = 25.0 m 12.0 × 10 −6 C° 40.0° C = 1.20 cm

P19.11

For the dimensions to increase, ∆L = αLi ∆T

a

fa

1.00 × 10 −2 cm = 1.30 × 10 −4 ° C −1 2.20 cm T − 20.0° C

f

T = 55.0° C *P19.12 P19.13

*P19.14

ja

e

fa

f

∆L = αLi ∆T = 22 × 10 −6 C° 2.40 cm 30° C = 1.58 × 10 −3 cm

a

fa

f

a

fa

f

(a)

∆L = αLi ∆T = 9.00 × 10 −6 ° C −1 30.0 cm 65.0° C = 0.176 mm

(b)

∆L = αLi ∆T = 9.00 × 10 −6 ° C −1 1.50 cm 65.0° C = 8.78 × 10 −4 cm

(c)

∆V = 3αVi ∆T = 3 9.00 × 10 −6 ° C −1

e

F jGH 30.0aπ fa4 1.50f

Ia JK

2

f

cm3 65.0° C = 0.093 0 cm3

The horizontal section expands according to ∆L = αLi ∆T .

ja

e

fa

f

∆x = 17 × 10 −6 ° C −1 28.0 cm 46.5° C − 18.0° C = 1.36 × 10 −2 cm The vertical section expands similarly by

ja

e

FIG. P19.14

fa

f

∆y = 17 × 10 −6 ° C −1 134 cm 28.5° C = 6.49 × 10 −2 cm . The vector displacement of the pipe elbow has magnitude ∆r = ∆x 2 + ∆ y 2 =

a0.136 mmf + a0.649 mmf 2

2

= 0.663 mm

and is directed to the right below the horizontal at angle

θ = tan −1

FG ∆y IJ = tan FG 0.649 mm IJ = 78.2° H 0.136 mm K H ∆x K −1

∆r = 0.663 mm to the right at 78.2° below the horizontal

554 P19.15

Temperature

b

g

b

L Al 1 + α Al ∆T = LBrass 1 + α Brass ∆T L Al − LBrass ∆T = LBrassα Brass − L Alα Al

(a)

∆T =

g

a10.01 − 10.00f a10.00fe19.0 × 10 j − a10.01fe24.0 × 10 j −6

−6

∆T = −199° C so T = −179° C. This is attainable. ∆T =

(b)

a10.02 − 10.00f a10.00fe19.0 × 10 j − a10.02fe24.0 × 10 j −6

−6

∆T = −396° C so T = −376° C which is below 0 K so it cannot be reached. P19.16

g a50.0° Cf

jb

e

∆A = 2 17.0 × 10 −6 ° C −1 0.080 0 m

∆A = 2αAi ∆T :

(a)

2

∆A = 1.09 × 10 −5 m 2 = 0.109 cm 2 (b)

The length of each side of the hole has increased. Thus, this represents an increase in the area of the hole.

b

g

e

e

P19.17

∆V = β − 3α Vi ∆T = 5.81 × 10 −4 − 3 11.0 × 10 −6

P19.18

(a)

a

f

jjb50.0 galga20.0f =

0.548 gal

a

5.050 cm = 5.000 cm 1 + 24.0 × 10 −6 ° C −1 T − 20.0° C

L = Li 1 + α∆T :

f

T = 437° C (b)

L Al = LBrass for some ∆T , or

We must get

b

g

b

Li , Al 1 + α Al ∆T = Li , Brass 1 + α Brass ∆T

e

g

j

e

j

5.000 cm 1 + 24.0 × 10 −6 ° C −1 ∆T = 5.050 cm 1 + 19.0 × 10 −6 ° C −1 ∆T Solving for ∆T , ∆T = 2 080° C , T = 3 000° C

so

This will not work because aluminum melts at 660° C . P19.19

b

a

g

f

(a)

V f = Vi 1 + β∆T = 100 1 + 1.50 × 10 −4 −15.0 = 99.8 mL

(b)

∆Vacetone = βVi ∆T

b

b

∆Vflask = βVi ∆T

g

g

acetone

Pyrex

b

= 3αVi ∆T

g

Pyrex

for same Vi , ∆T , ∆Vacetone β acetone 1.50 × 10 −4 1 = = = −6 ∆Vflask β flask × 6 . 40 10 −2 3 3.20 × 10

e

j

The volume change of flask is about 6% of the change in the acetone’s volume .

Chapter 19

P19.20

(a),(b)

555

The material would expand by ∆L = αLi ∆T , ∆L = α∆T , but instead feels stress Li F Y∆L = = Yα∆T = 7.00 × 10 9 N m 2 12.0 × 10 −6 C° A Li

e

a f a30.0° Cf

j

−1

= 2.52 × 10 6 N m 2 . This will not break concrete. P19.21

(a)

b

g

∆V = Vt β t ∆T − VAl β Al ∆T = β t − 3α Al Vi ∆T

e

j

e

ja

= 9.00 × 10 −4 − 0.720 × 10 −4 ° C −1 2 000 cm3 60.0° C

∆V = 99.4 cm3 (b)

f

overflows.

The whole new volume of turpentine is

ja

e

f

2 000 cm3 + 9.00 × 10 −4 ° C −1 2 000 cm3 60.0° C = 2 108 cm3

so the fraction lost is

99.4 cm3 = 4.71 × 10 −2 3 2 108 cm

and this fraction of the cylinder’s depth will be empty upon cooling:

a

f

4.71 × 10 −2 20.0 cm = 0.943 cm . *P19.22

The volume of the sphere is VPb =

a

f

4 3 4 π r = π 2 cm 3 3

3

= 33.5 cm3 .

The amount of mercury overflowing is

e

j

overflow = ∆VHg + ∆VPb − ∆Vglass = β Hg VHg + β PbVPb − β glassVglass ∆T where Vglass = VHg + VPb is the initial volume. Then

e j e j e j e 1 1 L O = Ma182 − 27f10 118 cm + a87 − 27f10 33.5 cm P 40° C = 0.812 cm C° C° N Q

j

overflow = β Hg − β glass VHg + β Pb − β glass VPb ∆T = β Hg − 3α glass VHg + 3α Pb − 3α glass VPb ∆T −6

P19.23

In

−6

3

3

F Y∆L = require ∆L = αLi ∆T A Li

F = Yα∆T A F 500 N = ∆T = −4 2 AYα 2.00 × 10 m 20.0 × 10 10 N m 2 11.0 × 10 −6 C°

e

∆T = 1.14° C

je

je

j

3

556 *P19.24

Temperature

Model the wire as contracting according to ∆L = αLi ∆T and then stretching according to ∆L Y F stress = = Y = αLi ∆T = Yα∆T . A Li Li

e

1 45° C = 396 N C°

j

(a)

F = YAα∆T = 20 × 10 10 N m 2 4 × 10 −6 m 2 11 × 10 −6

(b)

∆T =

3 × 10 8 N m 2 stress = = 136° C Yα 20 × 10 10 N m 2 11 × 10 −6 C°

e

j

To increase the stress the temperature must decrease to 35° C − 136° C = −101° C . (c) *P19.25

The original length divides out, so the answers would not change.

The area of the chip decreases according to

∆A = γA1 ∆T = A f − Ai

b

a

g

A f = Ai 1 + γ∆T = Ai 1 + 2α∆T

f

The star images are scattered uniformly, so the number N of stars that fit is proportional to the area.

a

f

ja

e

f

Then N f = N i 1 + 2α∆T = 5 342 1 + 2 4.68 × 10 −6 ° C −1 −100° C − 20° C = 5 336 star images .

Section 19.5 P19.26

P19.27

Macroscopic Description of an Ideal Gas

a

fe

n=

(b)

N = nN A

(a)

Initially, PV i i = n i RTi

a = a 2.99 molfe6.02 × 10

j

fa

f

j a1.00 atmfV = n Ra10.0 + 273.15f K P b0. 280V g = n Ra 40.0 + 273.15f K

23

molecules mol = 1.80 × 10 24 molecules i

Finally, Pf V f = n f RT f

f

i

i

i

0.280 Pf

giving

313.15 K = 1.00 atm 283.15 K Pf = 3.95 atm

or

Pf = 4.00 × 10 5 Pa abs. .

Dividing these equations,

(b)

je

9.00 atm 1.013 × 10 5 Pa atm 8.00 × 10 −3 m 3 PV = = 2.99 mol 8.314 N ⋅ mol K 293 K RT

(a)

a f P a1.02fb0.280V g = n Ra85.0 + 273.15f K

After being driven

d

i

i

Pd = 1.121Pf = 4.49 × 10 5 Pa P19.28

a fa f a fa f

3 150 0.100 3 PV = = 884 balloons 3 4π r P ′ 4π 0.150 3 1.20 If we have no special means for squeezing the last 100 L of helium out of the tank, the tank will be full of helium at 1.20 atm when the last balloon is inflated. The number of balloons is then reduced 0.100 m3 3 to to 884 − = 877 . 3 4π 0.15 m PV = NP ′V ′ =

e

a

4 3 π r NP ′ : 3

j f

N=

Chapter 19

P19.29

The equation of state of an ideal gas is PV = nRT so we need to solve for the number of moles to find N.

e

ja

fa ga

fa

f

1.01 × 10 5 N m 2 10.0 m 20.0 m 30.0 m PV = = 2. 49 × 10 5 mol n= RT 8.314 J mol ⋅ K 293 K

b

e

5

N = nN A = 2.49 × 10 mol 6.022 × 10 *P19.30

(a)

PV i i = n i RTi =

23

f

j

molecules mol = 1.50 × 10 29 molecules

mi RTi M

e

j

3

4.00 × 10 −3 kg 1.013 × 10 5 N 4π 6.37 × 10 6 m mole ⋅ K MPV i i = mi = RTi 8.314 Nm 50 K mole m2 3 = 1.06 × 10 21 kg (b)

Pf V f PV i i

=

n f RT f ni RTi

F 1.06 × 10 kg + 8.00 × 10 kg I T GH JK 50 K 1.06 × 10 kg F 1 IJ = 56.9 K T = 100 K G H 1.76 K nRT F 9.00 g I F 8.314 J I F 773 K I =G P= G J G J V H 18.0 g mol K H mol K K H 2.00 × 10 m JK = 21

2 ⋅1 =

20

f

21

f

P19.31

P19.32

P19.33

557

−3

a

1.61 MPa = 15.9 atm

fa f

(a)

T2 = T1

P2 = 300 K 3 = 900 K P1

(b)

T2 = T1

P2 V2 = 300 2 2 = 1 200 K P1V1

∑ Fy = 0 :

3

a fa f

b

g

ρ out gV − ρ in gV − 200 kg g = 0

bρ

out

ge

− ρ in 400 m

3

j = 200 kg

The density of the air outside is 1.25 kg m3 . n P = From PV = nRT , V RT The density is inversely proportional to the temperature, and the density of the hot air is

jFGH 283T K IJK e1.25 kg m jFGH 1 − 283T K IJK e400 m j = 200 kg e

ρ in = 1.25 kg m3

Then

in

3

3

in

283 K 1− = 0.400 Tin 283 K 0.600 = Tin = 472 K Tin

FIG. P19.33

558 *P19.34

P19.35

Temperature

Consider the air in the tank during one discharge process. We suppose that the process is slow enough that the temperature remains constant. Then as the pressure drops from 2.40 atm to 1.20 atm, the volume of the air doubles. During the first discharge, the air volume changes from 1 L to 2 L. Just 1 L of water is expelled and 3 L remains. In the second discharge, the air volume changes from 2 L to 4 L and 2 L of water is sprayed out. In the third discharge, only the last 1 L of water comes out. Were it not for male pattern dumbness, each person could more efficiently use his device by starting with the tank half full of water. (a)

PV = nRT

e je j b ga f m = nM = a 41.6 molfb 28.9 g molg =

1.013 × 10 5 Pa 1.00 m3 PV n= = = 41.6 mol RT 8.314 J mol ⋅ K 293 K (b)

1.20 kg , in agreement with the tabulated density of

3

1.20 kg m at 20.0°C. *P19.36

e

j

2

The void volume is 0.765Vtotal = 0.765π r 2 = 0.765π 1.27 × 10 −2 m 0.2 m = 7.75 × 10 −5 m3 . Now for the gas remaining PV = nRT n=

P19.37

(a)

b

n=

PV = nRT

e

j ga

−5 5 2 3 PV 12.5 1.013 × 10 N m 7.75 × 10 m = = 3.96 × 10 −2 mol RT 8.314 Nm mole K 273 + 25 K

f

PV RT

a

fe 3

−3 5 PVM 1.013 × 10 Pa 0.100 m 28.9 × 10 kg mol = m = nM = RT 8.314 J mol ⋅ K 300 K

b

f

ga

j

m = 1.17 × 10 −3 kg

P19.38

e

j

(b)

Fg = mg = 1.17 × 10 −3 kg 9.80 m s 2 = 11.5 mN

(c)

F = PA = 1.013 × 10 5 N m 2 0.100 m

(d)

The molecules must be moving very fast to hit the walls hard.

ja

e

At depth,

P = P0 + ρgh

At the surface,

P0 V f = nRT f :

Therefore

and

2

= 1.01 kN

PVi = nRTi P0 V f

bP + ρghgV

F T I FG P + ρgh IJ GH T JK H P K F 293 K IJ FG 1.013 × 10 = 1.00 cm G H 278 K K GH

V f = Vi Vf

f

f

0

i

=

Tf Ti

0

i

0

3

V f = 3.67 cm3

5

e

je

ja

f IJ JK

Pa + 1 025 kg m 3 9.80 m s 2 25.0 m 1.013 × 10 5 Pa

Chapter 19

P19.39

mf

PV = nRT :

mi

=

nf

m f = mi

so

Pf V f RTi Pf = RT f PV Pi i i

=

ni

FP I GH P JK f

i

∆m = mi − m f = mi P19.40

559

F P − P I = 12.0 kgFG 41.0 atm − 26.0 atm IJ = GH P JK H 41.0 atm K i

f

4.39 kg

i

My bedroom is 4 m long, 4 m wide, and 2.4 m high, enclosing air at 100 kPa and 20° C = 293 K . Think of the air as 80.0% N 2 and 20.0% O 2 . Avogadro’s number of molecules has mass

a0.800fb28.0 g molg + a0.200fb32.0 g molg = 0.028 8 kg mol F mI PV = nRT = G J RT H MK PVM e1.00 × 10 N m je38.4 m jb0.028 8 kg molg = = 45.4 kg m= RT b8.314 J mol ⋅ K ga293 K f

Then

5

gives *P19.41

2

3

~ 10 2 kg

The CO 2 is far from liquefaction, so after it comes out of solution it behaves as an ideal gas. Its molar mass is M = 12.0 g mol + 2 16.0 g mol = 44.0 g mol . The quantity of gas in the cylinder is m sample 6.50 g = = 0.148 mol n= M 44.0 g mol

b

Then

PV = nRT

gives

V=

g

b

ga

nRT 0.148 mol 8.314 J mol ⋅ K 273 K + 20 K = P 1.013 × 10 5 N m 2

e

je

je

j

f FG 1 N ⋅ m IJ F 10 L I = H 1 J K GH 1 m JK 3

3

3.55 L

P19.42

10 −9 Pa 1.00 m 3 6.02 × 10 23 molecules mol PVN A = = 2.41 × 10 11 molecules N= RT 8.314 J K ⋅ mol 300 K

P19.43

P0 V = n1 RT1 =

b

FG m IJ RT H MK F m IJ RT P V = n RT = G HMK P VM F 1 1I − J m −m = G R HT T K 0

2

1

2

1

1

2

2

2

0

1

2

ga

f

560 P19.44

Temperature

(a)

Initially the air in the bell satisfies P0 Vbell = nRTi or

a

f

P0 2.50 m A = nRTi

(1)

When the bell is lowered, the air in the bell satisfies

a

f

Pbell 2.50 m − x A = nRT f

(2)

where x is the height the water rises in the bell. Also, the pressure in the bell, once it is lowered, is equal to the sea water pressure at the depth of the water level in the bell.

a

a

f

f

Pbell = P0 + ρg 82.3 m − x ≈ P0 + ρg 82.3 m

(3)

The approximation is good, as x < 2.50 m. Substituting (3) into (2) and substituting nR from (1) into (2),

a

fa

f

P0 + ρg 82.3 m 2.50 m − x A = P0 Vbell

Tf Ti

.

Using P0 = 1 atm = 1.013 × 10 5 Pa and ρ = 1.025 × 10 3 kg m3

L O fMM TT FGH1 + ρga82P.3 mf IJK PP N Q LM 277.15 K F e1.025 × 10 kg m je9.80 m s ja82.3 mf I G1 + JJ = a 2.50 mfM1 − 293.15 K G 1.013 × 10 N m K H MN a

x = 2.50 m 1 −

−1

f

0

0

3

3

−1

2

2

5

OP PP Q

x = 2.24 m (b)

If the water in the bell is to be expelled, the air pressure in the bell must be raised to the water pressure at the bottom of the bell. That is,

a

f

Pbell = P0 + ρg 82.3 m 5

e

ja

je

f

= 1.013 × 10 Pa + 1.025 × 10 3 kg m 3 9.80 m s 2 82.3 m 5

Pbell = 9.28 × 10 Pa = 9.16 atm

Additional Problems P19.45

The excess expansion of the brass is

b

a f a f ∆a ∆Lf = 2.66 × 10 m

∆ ∆L = 19.0 − 11.0 × 10 −6 −4

(a)

The rod contracts more than tape to a length reading 0.950 0 m − 0.000 266 m = 0.949 7 m

(b)

g a° Cf a0.950 mfa35.0° Cf

∆Lrod − ∆Ltape = α brass − α steel Li ∆T

0.950 0 m + 0.000 266 m = 0.950 3 m

−1

Chapter 19

P19.46

At 0°C, 10.0 gallons of gasoline has mass,

ρ=

from

m V

jb

e

m = ρV = 730 kg m3 10.0 gal

80 m I = 27.7 kg gFGH 0.003 1.00 gal JK 3

The gasoline will expand in volume by

b

ga

f

∆V = βVi ∆T = 9.60 × 10 −4 ° C −1 10.0 gal 20.0° C − 0.0° C = 0.192 gal At 20.0°C,

10.192 gal = 27.7 kg 10.0 gal = 27.7 kg

F 10.0 gal I = 27.2 kg GH 10.192 gal JK

The extra mass contained in 10.0 gallons at 0.0°C is 27.7 kg − 27.2 kg = 0.523 kg . P19.47

Neglecting the expansion of the glass, ∆h = ∆h =

V β∆T A

b0.250 cm 2g e1.82 × 10 π e 2.00 × 10 cmj 3

4 3π

−3

2

−4

ja

f

° C −1 30.0° C = 3.55 cm

FIG. P19.47 P19.48

(a)

The volume of the liquid increases as ∆V = Vi β∆T . The volume of the flask increases as ∆Vg = 3αVi ∆T . Therefore, the overflow in the capillary is Vc = Vi ∆T β − 3α ; and in the

b

capillary Vc = A∆h . Therefore, ∆h = (b)

Vi β − 3α ∆T . A

b

g

b g

For a mercury thermometer

β Hg = 1.82 × 10 −4 ° C −1

and for glass,

3α = 3 × 3.20 × 10 −6 ° C −1

Thus

β − 3α ≈ β

or

α 0 , so Q < 0g + – + bW < 0 , ∆E > 0 since ∆E < 0 for B → C → A ; so Q > 0g = − P bV − V g = −3.00 atmb0.400 − 0.090 0 g m int

BC

int

int

B

C

3

B

P(atm)

= −94.2 kJ ∆Eint = Q + W

int

3.0

a

f

Eint, C − Eint, B = 100 − 94.2 kJ Eint, C − Eint, B = 5.79 kJ

1.0

Since T is constant,

B

C

A

0.090 0.20

D

0.40

Eint, D − Eint, C = 0

b

a

g

FIG. P20.32

f

WDA = − PD VA − VD = −1.00 atm 0.200 − 1.20 m3 = +101 kJ

a

f

Eint, A − Eint, D = −150 kJ + +101 kJ = −48.7 kJ

d

i d

i d

Now, Eint, B − Eint, A = − Eint, C − Eint, B + Eint, D − Eint, C + Eint, A − Eint, D Eint, B − Eint, A = − 5.79 kJ + 0 − 48.7 kJ = 42.9 kJ

i

1.2

V(m 3)

587

Chapter 20

*P20.33

1 2 πr . The arrow in Figure P20.33 2 looks like a semicircle when the scale makes 1.2 L fill the same space as 100 kPa. Its area is

The area of a true semicircle is

a

fa

f

je

500 300

1 1 π 2.4 L 200 kPa = π 2.4 × 10 −3 m3 2 × 10 5 N m 2 . 2 2

e

P(kPa)

j

The work on the gas is

0

z

A

1.2

B

3.6

B

W = − PdV = − area under the arch shown in the graph A

FG 1 π 2.4a200f J + 3 × 10 N m H2 = −b754 J + 1 440 Jg = −2 190 J 5

=−

2

4.8 × 10 −3 m3

FIG. P20.33

IJ K

∆Eint = Q + W = 5 790 J − 2 190 J = 3.60 kJ

Section 20.6 P20.34

(a)

Some Applications of the First Law of Thermodynamics

FG V IJ = − P V lnFG V IJ HV K HV K F W I = b0.025 0g expLM −3 000 OP = so V = V expG + MN 0.025 0e1.013 × 10 j PQ H P V JK f

W = −nRT ln

f

i

i

P20.35

f

Pf V f

f

5

f

e

j= g

1.013 × 10 5 Pa 0.025 0 m 3

Tf =

(a)

∆Eint = Q − P∆V = 12.5 kJ − 2.50 kPa 3.00 − 1.00 m 3 = 7.50 kJ

(b)

b

1.00 mol 8.314 J K ⋅ mol

(a)

0.007 65 m 3

305 K

a

f

V1 V2 = T1 T2 V 3.00 300 K = 900 K T2 = 2 T1 = 1.00 V1

a

P20.36

i

(b)

nR

=

f

f

W = − P∆V = − P 3αV∆T

e

f

L jMM e N

= − 1.013 × 10 5 N m 2 3 24.0 × 10 −6 ° C −1

O F I jGH 2.70 ×110.00 kgkg m JK a18.0° CfPP Q

W = −48.6 mJ

b

gb

ga

f

(b)

Q = cm∆T = 900 J kg⋅° C 1.00 kg 18.0° C = 16.2 kJ

(c)

∆Eint = Q + W = 16. 2 kJ − 48.6 mJ = 16.2 kJ

3

3

6.0

V (L)

588 P20.37

Heat and the First Law of Thermodynamics

OP a f LM MN e je j PQ F 18.0 g I = W = −a1.00 molfb8.314 J K ⋅ molga373 K f + e1.013 × 10 N m jG H 10 g m JK Q = mL = 0.018 0 kg e 2.26 × 10 J kg j = 40.7 kJ b

g

W = − P∆V = − P Vs − Vw = −

P nRT 18.0 g +P P 1.00 g cm 3 10 6 cm 3 m3 5

2

6

3

−3.10 kJ

6

v

∆Eint = Q + W = 37.6 kJ P20.38

(a)

The work done during each step of the cycle equals the negative of the area under that segment of the PV curve. W = WDA + W AB + WBC + WCD

b

g

b

g

W = − Pi Vi − 3Vi + 0 − 3 Pi 3Vi − Vi + 0 = −4PV i i

P20.39

(b)

The initial and final values of T for the system are equal. Therefore, ∆Eint = 0 and Q = −W = 4PV i i .

(c)

W = −4PV i i = −4nRTi = −4 1.00 8.314 273 = −9.08 kJ

(a)

3 PV i i = Pf V f = nRT = 2.00 mol 8.314 J K ⋅ mol 300 K = 4.99 × 10 J

(b) (c)

a fa

fa f

b

ga

FIG. P20.38

f

3

Vi =

nRT 4.99 × 10 J = 0.400 atm Pi

Vf =

nRT 4.99 × 10 3 J 1 = = Vi = 0.041 0 m3 1.20 atm Pf 3

FG V IJ = −e4.99 × 10 j lnFG 1 IJ = H 3K HV K

z

W = − PdV = −nRT ln

f

3

+5.48 kJ

i

∆Eint = 0 = Q + W Q = −5.48 kJ

P20.40

∆Eint, ABC = ∆Eint, AC (a)

(conservation of energy)

∆Eint, ABC = Q ABC + WABC

(First Law)

Q ABC = 800 J + 500 J = 1 300 J (b)

WCD = − PC ∆VCD , ∆VAB = − ∆VCD , and PA = 5 PC 1 1 Then, WCD = PA ∆VAB = − W AB = 100 J 5 5 (+ means that work is done on the system)

(c)

WCDA = WCD so that QCA = ∆Eint, CA − WCDA = −800 J − 100 J = −900 J (– means that energy must be removed from the system by heat)

(d)

∆Eint, CD = ∆Eint, CDA − ∆Eint, DA = −800 J − 500 J = −1 300 J and QCD = ∆Eint, CD − WCD = −1 300 J − 100 J = −1 400 J

FIG. P20.40

Chapter 20

Section 20.7 P20.41

Energy Transfer Mechanisms

∆T L P L 10.0 W 0.040 0 m k= = = 2.22 × 10 −2 W m⋅° C A∆T 1.20 m 2 15.0° C

P = kA

b

g

a

P20.42

f kA∆T b0.800 W m⋅° C ge3.00 m ja 25.0° C f P= = = 1.00 × 10

P20.43

In the steady state condition,

PAu = PAg

so that

k Au A Au

In this case

A Au = A Ag

2

6.00 × 10 −3 m

L

FG ∆T IJ H ∆x K

Au

4

= k Ag A Ag

W = 10.0 kW

FG ∆T IJ H ∆x K

Ag

∆x Au = ∆x Ag

a f = aT − 30.0f

∆TAu = 80.0 − T ∆TAg

and

FIG. P20.43

where T is the temperature of the junction. Therefore, k Au 80.0 − T = k Ag T − 30.0

a

P=

A∆T

∑k

Li

i

*P20.45

f

e6.00 m ja50.0° Cf 2

=

i

e

j

2 4.00 × 10 −3 m

0.800 W m⋅° C + 5.00 × 10 −3 m

0.023 4 W m⋅° C

= 1.34 kW

We suppose that the area of the transistor is so small that energy flow by heat from the transistor directly to the air is negligible compared to energy conduction through the mica.

P = kA

bT − T g h

c

L

Th = Tc + P20.46

a

T = 51.2° C

And

P20.44

f

e

j

1.50 W 0.085 2 × 10 −3 m PL = 35.0° C + = 67.9° C kA 0.075 3 W m⋅° C 8.25 × 6.25 10 −6 m 2

b

ga

f

From Table 20.4, (a)

R = 0.890 ft 2 ⋅° F ⋅ h Btu

(b)

The insulating glass in the table must have sheets of glass less than estimate the R-value of a 0.250-inch air space as Then for the double glazing

LM N

Rb = 0.890 + (c)

b

g

1 inch thick. So we 8

0.250 times that of the thicker air space. 3.50

FG 0.250 IJ 1.01 + 0.890OP ft ⋅° F ⋅ h = H 3.50 K Q Btu 2

1.85

ft 2 ⋅° F ⋅ h . Btu

Since A and T2 − T1 are constants, heat flow is reduced by a factor of

1.85 = 2.08 . 0.890

589

590 P20.47

Heat and the First Law of Thermodynamics

jLNM e

e

j OQPa0.965fb5 800 K g 2

P = σAeT 4 = 5.669 6 × 10 −8 W m 2 ⋅ K 4 4π 6.96 × 10 8 m

4

P = 3.77 × 10 26 W P20.48

Suppose the pizza is 70 cm in diameter and A = 2.0 cm thick, sizzling at 100°C. It cannot lose heat by conduction or convection. It radiates according to P = σAeT 4 . Here, A is its surface area,

a

f

A = 2π r 2 + 2π rA = 2π 0.35 m

2

a

fa

f

+ 2π 0.35 m 0.02 m = 0.81 m 2 .

Suppose it is dark in the infrared, with emissivity about 0.8. Then

e

ja fa

je

P = 5.67 × 10 −8 W m 2 ⋅ K 4 0.81 m 2 0.80 373 K

f

4

= 710 W ~ 10 3 W .

If the density of the pizza is half that of water, its mass is

ja

e

f a0.02 mf = 4 kg .

m = ρV = ρπ r 2 A = 500 kg m3 π 0.35 m

2

Suppose its specific heat is c = 0.6 cal g⋅° C . The drop in temperature of the pizza is described by:

d

Q = mc T f − Ti

P= dT f dt P20.49

dT f dQ = mc −0 dt dt =

710 J s P = = 0.07 ° C s ~ 10 −1 K s mc 4 kg 0.6 ⋅ 4 186 J kg ⋅° C

b gb

g

P = σAeT 4 2.00 W = 5.67 × 10 −8 W m 2 ⋅ K 4 0.250 × 10 −6 m 2 0.950 T 4

e

ja

je

e

T = 1.49 × 10 14 K 4 P20.50

i

j

14

f

= 3.49 × 10 3 K

We suppose the earth below is an insulator. The square meter must radiate in the infrared as much energy as it absorbs, P = σAeT 4 . Assuming that e = 1.00 for blackbody blacktop:

e

ja f

je

1 000 W = 5.67 × 10 −8 W m 2 ⋅ K 4 1.00 m 2 1.00 T 4

e

T = 1.76 × 10 10 K P20.51

j

4 14

= 364 K (You can cook an egg on it.)

The sphere of radius R absorbs sunlight over the area of its day hemisphere, projected as a flat circle perpendicular to the light: π R 2 . It radiates in all directions, over area 4π R 2 . Then, in steady state,

Pin = Pout

e

j

e

j

e 1 340 W m 2 π R 2 = eσ 4π R 2 T 4 The emissivity e, the radius R, and π all cancel.

L 1 340 W m Therefore, T = M MN 4e5.67 × 10 W m 2

−8

2

OP ⋅K jP Q 4

14

= 277 K = 4° C .

Chapter 20

591

Additional Problems P20.52

77.3 K = –195.8°C is the boiling point of nitrogen. It gains no heat to warm as a liquid, but gains heat to vaporize:

b

ge

j

Q = mL v = 0.100 kg 2.01 × 10 5 J kg = 2.01 × 10 4 J . The water first loses heat by cooling. Before it starts to freeze, it can lose

b

gb

ga

f

Q = mc∆T = 0.200 kg 4 186 J kg⋅° C 5.00° C = 4.19 × 10 3 J .

e

j

The remaining 2.01 × 10 4 − 4.19 × 10 3 J = 1.59 × 10 4 J that is removed from the water can freeze a mass x of water: Q = mL f

e

1.59 × 10 4 J = x 3.33 × 10 5 J kg

j

x = 0.047 7 kg = 47.7 g of water can be frozen P20.53

The increase in internal energy required to melt 1.00 kg of snow is

b

ge

j

∆Eint = 1.00 kg 3.33 × 10 5 J kg = 3.33 × 10 5 J The force of friction is

b

ge

j

f = µn = µmg = 0. 200 75.0 kg 9.80 m s 2 = 147 N

According to the problem statement, the loss of mechanical energy of the skier is assumed to be equal to the increase in internal energy of the snow. This increase in internal energy is

a

f

∆Eint = f∆r = 147 N ∆r = 3.33 × 10 5 J

∆r = 2.27 × 10 3 m .

and P20.54

(a)

The energy thus far gained by the copper equals the energy loss by the silver. Your down parka is an excellent insulator. Qcold = −Q hot or

d i = −m c dT − T i b9.00 g gb387 J kg⋅° Cga16.0° Cf = −b14.0 g gb234 J kg⋅° CgdT dT − 30.0° Ci = −17.0° C mCu c Cu T f − Ti

f

so (b)

Ag Ag

Cu

f

i Ag

f

− 30.0° C

i

FG dT IJ H dt K

= − mCu c Cu

Ag

T f , Ag = 13.0° C .

Differentiating the energy gain-and-loss equation gives: m Ag c Ag

FG dT IJ H dt K FG dT IJ H dt K

Ag

FG IJ = − 9.00 gb387 J kg⋅° Cg b+0.500 ° C sg H K 14.0 gb234 J kg⋅° Cg −0.532 ° C s b negative sign ⇒ decreasing temperatureg

=− Ag

= Ag

mCu c Cu dT m Ag c Ag dt

Cu

Ag

FG dT IJ H dt K

Cu

592 P20.55

Heat and the First Law of Thermodynamics

(a)

Before conduction has time to become important, the energy lost by the rod equals the energy gained by the helium. Therefore, mL v He = mc ∆T Al or so

b g c h bρVL g = cρVc ∆T h cρVc ∆T h V = bρL g e2.70 g cm je62.5 cm jb0.210 cal g⋅° Cga295.8° Cf V = e0.125 g cm je2.09 × 10 J kg jb1.00 cal 4.186 Jgb1.00 kg 1 000 g g v He

Al

Al

He

v He

3

He

3

3

4

VHe = 1.68 × 10 4 cm3 = 16.8 liters (b)

The rate at which energy is supplied to the rod in order to maintain constant temperatures is given by dT 295.8 K P = kA = 31.0 J s ⋅ cm ⋅ K 2.50 cm 2 = 917 W dx 25.0 cm This power supplied to the helium will produce a “boil-off” rate of

FG IJ b H K

jFGH

ge

a

fe je

j

IJ K

917 W 10 3 g kg P = = 351 cm 3 s = 0.351 L s ρL v 0.125 g cm3 2.09 × 10 4 J kg

e

*P20.56

j

At the equilibrium temperature Teq the diameters of the sphere and ring are equal:

e

j 5.01 cm + 5.01 cme 2.40 × 10

e

j 1 ° C jeT − T j = 5.00 cm + 5.00 cme1.70 × 10

d s + d sα Al Teq − Ti = d r + d r α Cu Teq − 15° C −5

eq

i

−5

je

1 ° C Teq − 15° C

0.01° C + 1.202 4 × 10 −4 Teq − 1.202 4 × 10 −4 Ti = 8.5 × 10 −5 Teq − 1.275 × 10 −3 ° C 1.127 5 × 10 −2 ° C + 3.524 × 10 −5 Teq = 1.202 4 × 10 −4 Ti 319.95° C + Teq = 3.412 0Ti At the equilibrium temperature, the energy lost is equal to the energy gained:

e

j

e

m s c Al Teq − Ti = − m r cCu Teq − 15° C

e

j

j

e

10.9 g 0.215 cal g ⋅° C Teq − Ti = −25 g 0.092 4 cal g⋅° C Teq − 15° C 2.343 5Teq − 2.343 5Ti = 34.65° C − 2.31Teq 4.653 5Teq = 34.65° C + 2.343 5Ti Solving by substitution,

b

g

4.653 5 3.412 0Ti − 319.95° C = 34.65° C + 2.343 5Ti 15.877 7Ti − 1 488.89° C = 34.65° C + 2.343 5Ti 1 523.54° C = 113° C 13.534

(b)

Ti =

(a)

Teq = −319.95 + 3.412 0 112.57 = 64.1° C

a

f

j

j

Chapter 20

P20.57

b g

Q = mc∆T = ρV c∆T so that when a constant temperature difference ∆T is maintained, the rate of adding energy to the liquid is P =

P . ρR∆T

and the specific heat of the liquid is c = P20.58

(a)

Work done by the gas is the negative of the area under the PV curve W = − Pi

(b)

FG IJ H K

dQ dV =ρ c∆T = ρRc∆T dt dt

FG V − V IJ = H2 K i

i

+

PV i i . 2

z

In this case the area under the curve is W = − PdV . Since the process is isothermal,

FG V IJ = nRT H 4K F dV IJ bPV g = − PV lnFG V 4 IJ = PV ln 4 W =− z G HVK HV K PV = PV i i = 4 Pi

i

i

Vi 4

and

i i

i i

i

Vi

i

FIG. P20.58

i i

= +1.39 PV i i (c) P20.59

The area under the curve is 0 and W = 0 .

Call the initial pressure P1 . In the constant volume process 1 → 2 the work is zero. P1V1 = nRT1 P2 V2 = nRT2 so

FG H

IJ a f K

P2 V2 T2 1 = ; T2 = 300 K 1 = 75.0 K 4 P1V1 T1

Now in 2 → 3

z 3

b

g

W = − PdV = − P2 V3 − V2 = − P3V3 + P2V2 2

a

fb

ga

W = −nRT3 + nRT2 = − 1.00 mol 8.314 J mol ⋅ K 300 K − 75.0 K W = −1.87 kJ

f

593

594 *P20.60

Heat and the First Law of Thermodynamics

The initial moment of inertia of the disk is

ja

f

1 1 1 1 4 MR 2 = ρVR 2 = ρπR 2 tR 2 = 8 920 kg m3 π 28 m 1.2 m = 1.033 × 10 10 kg ⋅ m 2 2 2 2 2

e

The rotation speeds up as the disk cools off, according to I iω i = I f ω f 2 1 1 1 MRi2ω i = MR 2f ω f = MRi2 1 − α ∆T ω f 2 2 2 1 1 ω f = ωi = 25 rad s 2 1 − α ∆T 1 − 17 × 10 −6 1 ° C 830° C

c

c

(a)

h

h

e

j

2

= 25.720 7 rad s

The kinetic energy increases by 1 1 1 1 1 I f ω 2f − I iω i2 = I iω iω f − I iω i2 = I iω i ω f − ω i 2 2 2 2 2 1 10 2 = 1.033 × 10 kg ⋅ m 25 rad s 0.720 7 rad s = 9.31 × 10 10 J 2

d

i

b

b

g

f

ga

(b)

∆Eint = mc∆T = 2.64 × 10 7 kg 387 J kg⋅° C 20° C − 850° C = −8.47 × 10 12 J

(c)

As 8.47 × 10 12 J leaves the fund of internal energy, 9.31 × 10 10 J changes into extra kinetic energy, and the rest, 8.38 × 10 12 J is radiated.

*P20.61

The loss of mechanical energy is GM E m 1 1 = 670 kg 1.4 × 10 4 m s mvi2 + RE 2 2

e

j

2

+

6.67 × 10 −11 Nm 2 5.98 × 10 24 kg 670 kg kg 2 6.37 × 10 6 m

= 6.57 × 10 10 J + 4.20 × 10 10 J = 1.08 × 10 11 J One half becomes extra internal energy in the aluminum: ∆Eint = 5.38 × 10 10 J. To raise its temperature to the melting point requires energy mc∆T = 670 kg 900

a

fh

J 660 − −15° C = 4.07 × 10 8 J . kg ° C

c

To melt it, mL = 670 kg 3.97 × 10 5 J kg = 2.66 × 10 8 J . To raise it to the boiling point,

b

gb

g

mc∆T = 670 1 170 2 450 − 600 J = 1.40 × 10 9 J . To boil it, mL = 670 kg 1.14 × 10 7 J kg = 7.64 × 10 9 J . Then

b

gd

i

5.38 × 10 10 J = 9.71 × 10 9 J + 670 1 170 T f − 2 450° C J ° C T f = 5.87 × 10 4 ° C

Chapter 20

P20.62

a

fb

g

(a)

Fv = 50.0 N 40.0 m s = 2 000 W

(b)

Energy received by each object is 1 000 W 10 s = 10 4 J = 2 389 cal . The specific heat of iron

ga f

b

is 0.107 cal g ⋅° C , so the heat capacity of each object is 5.00 × 10 3 × 0.107 = 535.0 cal ° C. ∆T = P20.63

2 389 cal = 4.47° C 535.0 cal ° C

The power incident on the solar collector is

j a

e

f

Pi = IA = 600 W m 2 π 0.300 m

2

= 170 W .

For a 40.0% reflector, the collected power is Pc = 67.9 W. The total energy required to increase the temperature of the water to the boiling point and to evaporate it is Q = cm∆T + mLV :

b

ga

f

Q = 0.500 kg 4 186 J kg ⋅° C 80.0° C + 2.26 × 10 6 J kg = 1.30 × 10 6 J .

The time interval required is ∆t = P20.64

595

Q

Pc

=

1.30 × 10 6 J = 5.31 h . 67.9 W

FIG. P20.63

From Q = mLV the rate of boiling is described by

P=

Q LV m = ∆t ∆t

∴

P m = ∆t LV

Model the water vapor as an ideal gas

FG m IJ RT H MK P V m F RT I = G J ∆t ∆t H M K P F RT I P Av = G J L HMK

P0 V0 = nRT = 0

0

v=

V

b

ga

f

1 000 W 8.314 J mol ⋅ K 373 K P RT = MLV P0 A 0.018 0 kg mol 2.26 × 10 6 J kg 1.013 × 10 5 N m 2 2.00 × 10 −4 m 2

v = 3.76 m s

b

ge

je

je

j

596 P20.65

Heat and the First Law of Thermodynamics

Energy goes in at a constant rate P . For the period from 50.0 min to 60.0 min, Q = mc∆T

a a

f b f

gb

T°( C)

f

ga

P 10.0 min = 10 kg + mi 4 186 J kg⋅° C 2.00° C − 0° C (1) P 10.0 min = 83.7 kJ + 8.37 kJ kg mi

b

g

a

f

e

Substitute P =

e

mi 3.33 × 10 5 J kg 50.0 min

e

mi 3.33 × 10 5 J kg

j

3.00

1.00

For the period from 0 to 50.0 min, Q = mi L f

P 50.0 min = mi 3.33 × 10 5 J kg

2.00

0.00

j

20.0

j into Equation (1) to find b

40.0

60.0 t (min)

FIG. P20.65

g

= 83.7 kJ + 8.37 kJ kg mi 5.00 83.7 kJ = 1.44 kg mi = 66.6 − 8.37 kJ kg

a

P20.66

(a)

f

The block starts with K i =

b

gb

1 1 mvi2 = 1.60 kg 2.50 m s 2 2

g

2

= 5.00 J

All this becomes extra internal energy in ice, melting some according to “Q ” = m ice L f . Thus, the mass of ice that melts is m ice = For the block:

“Q ” K i 5.00 J = = = 1.50 × 10 −5 kg = 15.0 mg . Lf L f 3.33 × 10 5 J kg

Q = 0 (no energy flows by heat since there is no temperature difference)

W = −5.00 J ∆Eint = 0 (no temperature change) and

∆K = −5.00 J

For the ice,

Q=0

W = +5.00 J ∆Eint = +5.00 J and (b)

∆K = 0

Again, K i = 5.00 J and m ice = 15.0 mg For the block of ice: Q = 0; ∆Eint = +5.00 J ; ∆K = −5.00 J so W = 0 . For the copper, nothing happens: Q = ∆Eint = ∆K = W = 0 .

continued on next page

Chapter 20

Again, K i = 5.00 J. Both blocks must rise equally in temperature.

(c)

∆T =

“Q ” = mc∆T :

“Q ” 5.00 J = = 4.04 × 10 −3 ° C mc 2 1.60 kg 387 J kg ⋅° C

b

gb

g

At any instant, the two blocks are at the same temperature, so for both Q = 0. For the moving block:

∆K = −5.00 J

and

∆Eint = +2.50 J

so

W = −2.50 J

For the stationary block:

∆K = 0

and

∆Eint = +2.50 J

so

W = +2.50 J

For each object in each situation, the general continuity equation for energy, in the form ∆K + ∆Eint = W + Q , correctly describes the relationship between energy transfers and changes in the object’s energy content. P20.67

A = Aend walls + A ends of attic + A side walls + Aroof

a f LMN 12 × 4.00 m × a4.00 mf tan 37.0°OPQ F 4.00 m IJ +2a10.0 m × 5.00 mf + 2a10.0 mfG H cos37.0° K

A = 2 8.00 m × 5.00 m + 2 2 ×

A = 304 m 2

P=

e

ja

je

f

4.80 × 10 −4 kW m⋅° C 304 m 2 25.0° C kA∆T = = 17.4 kW = 4.15 kcal s L 0.210 m

b

gb

g

Thus, the energy lost per day by heat is 4.15 kcal s 86 400 s = 3.59 × 10 5 kcal day . The gas needed to replace this loss is

P20.68

3.59 × 10 5 kcal day 9 300 kcal m3

= 38.6 m3 day .

FG IJ H K

∆T LρAdx = kA dt x Lρ Lρ

e

z

z

8.00

∆t

4.00

0

xdx = k∆T dt

2 8.00

x 2

= k∆T∆t

4.00

je

3.33 × 10 5 J kg 917 kg m3

∆t = 3.66 × 10 4 s = 10.2 h

F b0.080 0 mg − b0.040 0 mg jGG 2 H 2

2

I JJ = b2.00 W m⋅° Cga10.0° Cf∆t K

597

598 P20.69

Heat and the First Law of Thermodynamics

W = W AB + WBC + WCD + WDA

z

z

z

P

z

B

C

D

A

A

B

C

D

P2

W = − PdV − PdV − PdV − PdV W = −nRT1

z

z

B

z

C

z

D

FG V IJ − P bV HV K B

2

1

C

FG V IJ − P bV HV K

g

− VB − nRT2 ln

2

1

C

P1 A

− VD

g

VB P1 V P = and 2 = 2 V1 P2 VC P1

FP I

FP I

FP I

D

A V1

Now P1VA = P2 VB and P2 VC = P1VD , so only the logarithmic terms do not cancel out. Also,

C

A

dV dV − P2 dV − nRT2 − P1 dV V V A B C D

W = −nRT1 ln

B

V2

FIG. P20.69

FP I

FP I

∑ W = −nRT1 lnGH P1 JK − nRT2 lnGH P2 JK = +nRT1 lnGH P2 JK − nRT2 lnGH P2 JK = −nRbT2 − T1 g lnGH P2 JK 1

1

1

2

1

Moreover P1V2 = nRT2 and P1V1 = nRT1

∑W = P20.70

b

g FGH PP IJK

− P1 V2 − V1 ln

2

1

For a cylindrical shell of radius r, height L, and thickness dr, the equation for thermal conduction, dQ dT = − kA dt dx

becomes

Under equilibrium conditions, dT = −

F GH

1 dQ dt 2π kL

I FG dr IJ JK H r K

g

dQ is constant; therefore, dt Tb − Ta = −

and

F GH

I FG IJ JK H K

1 dQ b ln dt 2π kL a

b g b g

dQ 2π kL Ta − Tb = dt ln b a

But Ta > Tb , so P20.71

b

dQ dT = − k 2π rL dt dr

From problem 70, the rate of energy flow through the wall is

b g b g

dQ 2π kL Ta − Tb = dt ln b a

e

jb

ga

−5 dQ 2π 4.00 × 10 cal s ⋅ cm⋅° C 3 500 cm 60.0° C = dt ln 256 cm 250 cm

b

g

f

dQ = 2. 23 × 10 3 cal s = 9.32 kW dt This is the rate of energy loss from the plane by heat, and consequently is the rate at which energy must be supplied in order to maintain a constant temperature.

FIG. P20.71

V

Chapter 20

P20.72

599

Qcold = −Q hot

b dT

Q Al = − Q water + Qcalo

or

m Al c Al

f

− Ti

i

Al

g

b

gd

= − m w c w + m c c c T f − Ti

i

w

b0.200 kg gc a+39.3° Cf = − 0.400 kgb4 186 J kg⋅° Cg + 0.040 0 kgb630 J kg⋅° Cg a−3.70° Cf Al

c Al = *P20.73

6.29 × 10 3 J = 800 J kg⋅° C 7.86 kg⋅° C

e

a

j

fb

(a)

P = σAeT 4 = 5.67 × 10 −8 W m 2 K 4 5.1 × 10 14 m 2 0.965 5 800 K

(b)

Tavg = 0.1 4 800 K + 0.9 5 890 K = 5.78 × 10 3 K

b

g

b

e

= 3.16 × 10 22 W

5 800 − 5 781 = 0.327% . 5 800

g j e j b W 0.9e5.1 × 10 j0.965b5 890g = 3.17 × 10 W

P = 5.67 × 10 −8 W m 2 K 4 0.1 5.1 × 10 14 m 2 0.965 4 800 K +5.67 × 10 −8

4

g

This is cooler than 5 800 K by

(c)

g

14

This is larger than 3.158 × 10 22 W by

4

4

22

1.29 × 10 20 W = 0.408%. 3.16 × 10 22 W

ANSWERS TO EVEN PROBLEMS P20.2

0.105°C

P20.22

liquid lead at 805°C

P20.4

87.0°C

P20.24

(a) −12.0 MJ ; (b) +12.0 MJ

P20.6

The energy input to the water is 6.70 times larger than the laser output of 40.0 kJ.

P20.26

−nR T2 − T1

P20.8

88.2 W

P20.28

(a) 567 J ; (b) 167 J

P20.10

(a) 25.8°C ; (b) no

P20.30

(a) 12.0 kJ; (b) −12.0 kJ

P20.32

42.9 kJ

P20.34

(a) 7.65 L; (b) 305 K

bm =

Al c Al

g

+ m c c w Tc + m h c w Th

b

g

P20.12

Tf

P20.14

(a) 380 K ; (b) 206 kPa

P20.36

(a) −48.6 mJ ; (b) 16.2 kJ; (c) 16.2 kJ

P20.16

12.9 g

P20.38

(a) −4PV i i ; (b) +4PV i i ; (c) −9.08 kJ

P20.18

(a) all the ice melts; 40.4°C ; (b) 8.04 g melts; 0°C

P20.40

(a) 1 300 J ; (b) 100 J ; (c) −900 J ; (d) −1 400 J

P20.20

34.0 km

P20.42

10.0 kW

m Al c Al + m c c w + m h c w

600

Heat and the First Law of Thermodynamics

P20.44

1.34 kW

P20.46

(a) 0.890 ft 2 ⋅° F ⋅ h Btu ; (b) 1.85 (c) 2.08

P20.48

(a) ~ 10 3 W ; (b) ~ −10 −1 K s

P20.50

364 K

P20.52

47.7 g

P20.54

(a) 13.0°C ; (b) −0.532 ° C s

P20.56

(a) 64.1°C ; (b) 113°C

P20.58

see the solution (a)

P20.60

10

ft 2 ⋅° F ⋅ h ; Btu

1 PV i i ; (c) 0 i i ; (b) 1.39 PV 2

(a) 9.31 × 10 J ; (b) −8.47 × 10 (c) 8.38 × 10 12 J

12

J;

P20.62

(a) 2 000 W ; (b) 4.47°C

P20.64

3.76 m s

P20.66

(a) 15.0 mg ; block: Q = 0; W = −5.00 J ; ∆Eint = 0 ; ∆K = −5.00 J ; ice: Q = 0; W = 5.00 J ; ∆Eint = 5.00 J ; ∆K = 0 (b) 15.0 mg ; block: Q = 0; W = 0 ; ∆Eint = 5.00 J ; ∆K = −5.00 J ; metal: Q = 0; W = 0 ; ∆Eint = 0 ; ∆K = 0 (c) 0.004 04°C ; moving block: Q = 0; W = −2.50 J ; ∆Eint = 2.50 J ; ∆K = −5.00 J ; stationary block: Q = 0; W = 2.50 J ; ∆Eint = 2.50 J ; ∆K = 0

P20.68

10.2 h

P20.70

see the solution

P20.72

800 J kg ⋅° C

21 The Kinetic Theory of Gases ANSWERS TO QUESTIONS

CHAPTER OUTLINE 21.1 21.2 21.3 21.4 21.5 21.6 21.7

Molecular Model of an Ideal Gas Molar Specific Heat of an Ideal Gas Adiabatic Processes for an Ideal Gas The Equipartition of Energy The Boltzmann Distribution Law Distribution of Molecular Speeds Mean Free Path

Q21.1

The molecules of all different kinds collide with the walls of the container, so molecules of all different kinds exert partial pressures that contribute to the total pressure. The molecules can be so small that they collide with one another relatively rarely and each kind exerts partial pressure as if the other kinds of molecules were absent. If the molecules collide with one another often, the collisions exactly conserve momentum and so do not affect the net force on the walls.

Q21.2

The helium must have the higher rms speed. According to Equation 21.4, the gas with the smaller mass per atom must have the higher average speed-squared and thus the higher rms speed.

Q21.3

Yes. As soon as the gases are mixed, they come to thermal equilibrium. Equation 21.4 predicts that the lighter helium atoms will on average have a greater speed than the heavier nitrogen molecules. Collisions between the different kinds of molecules gives each kind the same average kinetic energy of translation.

Q21.4

If the average velocity were non-zero, then the bulk sample of gas would be moving in the direction of the average velocity. In a closed tank, this motion would result in a pressure difference within the tank that could not be sustained.

Q21.5

The alcohol evaporates, absorbing energy from the skin to lower the skin temperature.

Q21.6

Partially evacuating the container is equivalent to letting the remaining gas expand. This means that the gas does work, making its internal energy and hence its temperature decrease. The liquid in the container will eventually reach thermal equilibrium with the low pressure gas. This effect of an expanding gas decreasing in temperature is a key process in your refrigerator or air conditioner.

Q21.7

Since the volume is fixed, the density of the cooled gas cannot change, so the mean free path does not change. The collision frequency decreases since each molecule of the gas has a lower average speed.

Q21.8

The mean free path decreases as the density of the gas increases.

Q21.9

The volume of the balloon will decrease. The pressure inside the balloon is nearly equal to the constant exterior atmospheric pressure. Then from PV = nRT , volume must decrease in proportion to the absolute temperature. Call the process isobaric contraction. 601

602

The Kinetic Theory of Gases

Q21.10

The dry air is more dense. Since the air and the water vapor are at the same temperature, they have the same kinetic energy per molecule. For a controlled experiment, the humid and dry air are at the same pressure, so the number of molecules per unit volume must be the same for both. The water molecule has a smaller molecular mass (18.0 u) than any of the gases that make up the air, so the humid air must have the smaller mass per unit volume.

Q21.11

Suppose the balloon rises into air uniform in temperature. The air cannot be uniform in pressure because the lower layers support the weight of all the air above them. The rubber in a typical balloon is easy to stretch and stretches or contracts until interior and exterior pressures are nearly equal. So as the balloon rises it expands. This is an isothermal expansion, with P decreasing as V increases by the same factor in PV = nRT . If the rubber wall is very strong it will eventually contain the helium at higher pressure than the air outside but at the same density, so that the balloon will stop rising. More likely, the rubber will stretch and break, releasing the helium to keep rising and “boil out” of the Earth’s atmosphere.

Q21.12

A diatomic gas has more degrees of freedom—those of vibration and rotation—than a monatomic gas. The energy content per mole is proportional to the number of degrees of freedom.

Q21.13

(a)

Average molecular kinetic energy increases by a factor of 3.

(b)

The rms speed increases by a factor of

(c)

Average momentum change increases by

(d)

Rate of collisions increases by a factor of

(e)

Pressure increases by a factor of 3.

3. 3. 3 since the mean free path remains unchanged.

Q21.14

They can, as this possibility is not contradicted by any of our descriptions of the motion of gases. If the vessel contains more than a few molecules, it is highly improbable that all will have the same speed. Collisions will make their speeds scatter according to the Boltzmann distribution law.

Q21.15

Collisions between molecules are mediated by electrical interactions among their electrons. On an atomic level, collisions of billiard balls work the same way. Collisions between gas molecules are perfectly elastic. Collisions between macroscopic spheres can be very nearly elastic. So the hardsphere model is very good. On the other hand, an atom is not ‘solid,’ but has small-mass electrons moving through empty space as they orbit the nucleus.

Q21.16

As a parcel of air is pushed upward, it moves into a region of lower pressure, so it expands and does work on its surroundings. Its fund of internal energy drops, and so does its temperature. As mentioned in the question, the low thermal conductivity of air means that very little heat will be conducted into the now-cool parcel from the denser but warmer air below it.

Q21.17

A more massive diatomic or polyatomic molecule will generally have a lower frequency of vibration. At room temperature, vibration has a higher probability of being excited than in a less massive molecule. The absorption of energy into vibration shows up in higher specific heats.

SOLUTIONS TO PROBLEMS Section 21.1 P21.1

Molecular Model of an Ideal Gas

F = Nm P=

a

8.00 sin 45.0°− −8.00 sin 45.0° ∆v = 500 5.00 × 10 −3 kg ∆t 30.0 s

e

F = 1.57 N m 2 = 1.57 Pa A

j

f

ms

= 0.943 N

Chapter 21

e5.00 × 10 j 2e4.68 × 10

−26

23

P21.2

P21.3

F=

jb

kg 300 m s

1.00 s 14.0 N F and P = = = 17.6 kPa . A 8.00 × 10 −4 m 2

603

g = 14.0 N

We first find the pressure exerted by the gas on the wall of the container. NkT 3 N A k BT 3 RT 3 8.314 N ⋅ m mol ⋅ K 293 K = = = = 9.13 × 10 5 Pa P= 3 −3 V V V 8.00 × 10 m Thus, the force on one of the walls of the cubical container is

b

e

f

ga

je

j

F = PA = 9.13 × 10 5 Pa 4.00 × 10 −2 m 2 = 3.65 × 10 4 N . P21.4

F GH

I JK

2 N mv 2 , so that 3V 2

Use Equation 21.2, P =

K av =

mv 2 3 PV = where N = nN A = 2 N A 2 2N

K av =

3 8.00 atm 1.013 × 10 5 Pa atm 5.00 × 10 −3 m 3 3 PV = 2 2N A 2 2 mol 6.02 × 10 23 molecules mol

b

a

g

a

fe

je

fe

j

j

K av = 5.05 × 10 −21 J molecule P21.5

P=

2N KE 3V

d i

Equation 21.2

e

je

j

−3 5 3 PV 3 1.20 × 10 4.00 × 10 = = 2.00 × 10 24 molecules N= 2 KE 2 3.60 × 10 −22

d i

n= P21.6

e

j

24

2.00 × 10 molecules N = = 3.32 mol N A 6.02 × 10 23 molecules mol

One mole of helium contains Avogadro’s number of molecules and has a mass of 4.00 g. Let us call m the mass of one atom, and we have N A m = 4.00 g mol 4.00 g mol m= = 6.64 × 10 −24 g molecule or 6.02 × 10 23 molecules mol

m = 6.64 × 10 −27 kg

P21.7

a

f f

5 4 PV 1.013 × 10 Pa 3 π 0.150 m = N= k BT 1.38 × 10 −23 J K 293 K

(a)

PV = Nk BT :

(b)

K=

(c)

For helium, the atomic mass is

e

ja

3

= 3.54 × 10 23 atoms

ja f

3 3 k BT = 1.38 × 10 −23 293 J = 6.07 × 10 −21 J 2 2

e

m=

4.00 g mol molecules mol

−27

kg molecule

6.02 × 10

m = 6.64 × 10 1 3 mv 2 = k BT : 2 2

23

∴ v rms =

3 k BT = 1.35 km s m

= 6.64 × 10 −24 g molecule

604 P21.8

The Kinetic Theory of Gases

v= vO v He

3 k BT m M He 4.00 1 = = = 32.0 8.00 MO

vO = P21.9

1 350 m s 8.00

= 477 m s

ja

f

(a)

K=

3 3 k BT = 1.38 × 10 −23 J K 423 K = 8.76 × 10 −21 J 2 2

(b)

K=

1 2 = 8.76 × 10 −21 J mv rms 2

e

so

v rms =

For helium,

m=

1.75 × 10 −20 J m

(1)

4.00 g mol 6.02 × 10 23 molecules mol

= 6.64 × 10 −24 g molecule

m = 6.64 × 10 −27 kg molecule 39.9 g mol

m=

Similarly for argon,

6.02 × 10 23 molecules mol

= 6.63 × 10 −23 g molecule

m = 6.63 × 10 −26 kg molecule Substituting in (1) above,

P21.10

(a)

we find for helium,

v rms = 1.62 km s

and for argon,

v rms = 514 m s

PV = nRT =

Nmv 2 3

The total translational kinetic energy is

3 3 PV = 3.00 × 1.013 × 10 5 5.00 × 10 −3 = 2.28 kJ 2 2

je j 3 k T 3 RT 3a8.314fa300f mv = = = = 6. 21 × 10 2 2 2N 2e6.02 × 10 j F 1 N m I FG 1 J IJ = 1 J m 1 Pa = a1 PafG H 1 Pa JK H 1 N ⋅ m K Etrans =

(b)

2

e

B

23

A

P21.11

(a)

(b)

Nmv 2 = Etrans : 2

2

−21

J

3

For a monatomic ideal gas, Eint =

3 nRT 2

For any ideal gas, the energy of molecular translation is the same, Etrans = Thus, the energy per volume is

3 3 nRT = PV . 2 2

Etrans 3 = P . 2 V

Chapter 21

Section 21.2 P21.12

Eint = ∆Eint

P21.13

P21.14

Molar Specific Heat of an Ideal Gas 3 nRT 2 3 3 = nR∆T = 3.00 mol 8.314 J mol ⋅ K 2.00 K = 74.8 J 2 2

a

fb

f

ga

We us the tabulated values for C P and C V

b

f

ga

(a)

Q = nC P ∆T = 1.00 mol 28.8 J mol ⋅ K 420 − 300 K = 3.46 kJ

(b)

∆Eint = nCV ∆T = 1.00 mol 20.4 J mol ⋅ K 120 K = 2.45 kJ

(c)

W = −Q + ∆Eint = −3.46 kJ + 2.45 kJ = −1.01 kJ

b

f

ga

The piston moves to keep pressure constant. Since V =

nRT , then P

nR∆T for a constant pressure process. P 2Q Q Q = = Q = nC P ∆T = n CV + R ∆T so ∆T = 7nR n CV + R n 5 R 2 + R ∆V =

b g b g b nR F 2Q I 2Q 2 QV ∆V = G J= = P H 7nR K 7 P 7 nRT e4.40 × 10 Jja5.00 Lf 2 ∆V = = 2.52 L 7 a1.00 molfb8.314 J mol ⋅ K ga300 K f

and

g

3

Thus, P21.15

V f = Vi + ∆V = 5.00 L + 2.52 L = 7.52 L

n = 1.00 mol, Ti = 300 K (b)

Since V = constant, W = 0

(a)

∆Eint = Q + W = 209 J + 0 = 209 J

(c)

∆Eint = nCV ∆T = n so

FG 3 RIJ ∆T H2 K

∆T =

a fb

f

2 209 J 2 ∆Eint = = 16.8 K 3nR 3 1.00 mol 8.314 J mol ⋅ K

a

T = Ti + ∆T = 300 K + 16.8 K = 317 K

g

605

606 P21.16

The Kinetic Theory of Gases

(a)

Consider heating it at constant pressure. Oxygen and nitrogen are diatomic, so C P = Q = nC P ∆T =

e

FG IJ H K N m je100 m j a1.00 K f =

7 7 PV ∆T nR∆T = 2 2 T

5 2 7 1.013 × 10 Q= 2 300 K

(b)

(a)

Ug gy

=

e

1.18 × 10 5 J

j

9.80 m s 2 2.00 m

We assume that the bulb does not expand. Then this is a constant-volume heating process. PV The quantity of the gas is n = i . The energy input is Q = P ∆t = nCV ∆T so RTi

P∆t P∆tRTi = . nCV PVC i V

FG H

The final temperature is T f = Ti + ∆T = Ti 1 + The final pressure is Pf = Pi

P21.18

(a)

(b)

F GH

FG H

Tf

= Pi 1 +

Ti

IJ K

P∆tR . PVC i V

IJ K

P∆tR . PVC i V

I = 1.18 atm J s ⋅ mol ⋅ K 1.013 × 10 N 4π a0.05 mf 12.5 J K F 1.00 mol I = 719 J kg ⋅ K = 0.719 kJ kg ⋅ K 5 5 C = R = b8.314 J mol ⋅ K gG 2 2 H 0.028 9 kg JK F PV IJ m = Mn = M G H RT K F 200 × 10 Pae0.350 m j I m = b0.028 9 kg molgG GH b8.314 J mol ⋅ K ga300 K f JJK = 0.811 kg Pf = 1 atm 1 +

3.60 J 4 s 8.314 J ⋅ m 2 3 mol ⋅ K

3

5

V

3

(c)

118 kJ

= 6.03 × 10 3 kg

∆T =

(b)

3

U g = mgy m=

*P21.17

3

We consider a constant volume process where no work is done.

b

f

ga

Q = mCV ∆T = 0.811 kg 0.719 kJ kg ⋅ K 700 K − 300 K = 233 kJ (d)

7R 2

We now consider a constant pressure process where the internal energy of the gas is increased and work is done.

FG 7R IJ ∆T = mFG 7C IJ ∆T b g H2K H 5K L7 O Q = 0.811 kg M b0.719 kJ kg ⋅ K gPa 400 K f = 327 kJ N5 Q

Q = mC P ∆T = m CV + R ∆T = m

V

Chapter 21

P21.19

Consider 800 cm3 of (flavored) water at 90.0 °C mixing with 200 cm3 of diatomic ideal gas at 20.0°C: Qcold = −Q hot

d

i dT

a f i = −bρV g

m air c P , air T f − Ti , air = − m w c w ∆T

or

a∆T f

w

− m air c P , air

=

− Ti , air

f

mwcw

w

a90.0° C − 20.0° Cf b ρ V gc

c air P , air

w w

w

where we have anticipated that the final temperature of the mixture will be close to 90.0°C. 7 The molar specific heat of air is C P, air = R 2 7 R 7 1.00 mol So the specific heat per gram is = 8.314 J mol ⋅ K = 1.01 J g⋅° C c P, air = 2 M 2 28.9 g

F I FG IJ b g G JK H K H 1.20 × 10 g cm je 200 cm j b1.01 J g⋅° C ga70.0° C f a∆T f = − e e1.00 g cm je800 cm j b4.186 J kg⋅° Cg a∆T f ≈ −5.05 × 10 ° C −3

3

w

or

3

3

3

−3

w

The change of temperature for the water is between 10 −3 ° C and 10 −2 ° C . P21.20

b

Q = nC P ∆T

g

isobaric

b

+ nCV ∆T

g

isovolumetric

In the isobaric process, V doubles so T must double, to 2Ti . In the isovolumetric process, P triples so T changes from 2Ti to 6Ti . Q =n P21.21

607

FG 7 RIJ b2T − T g + nFG 5 RIJ b6T − 2T g = 13.5nRT = H2 K H2 K i

i

i

i

i

13.5 PV

In the isovolumetric process A → B , W = 0 and Q = nCV ∆T = 500 J 500 J = n

FG 3R IJ bT H2K

B

TB = 300 K +

g

− TA or TB = TA +

a

2 500 J 3nR

f

a f = 340 K 3a1.00 molfb8.314 J mol ⋅ K g 2 500 J

In the isobaric process B → C , Q = nC P ∆T = Thus,

b

g

5nR TC − TB = −500 J . 2

a

f

2 500 J 1 000 J = 340 K − = 316 K 5nR 5 1.00 mol 8.314 J mol ⋅ K

(a)

TC = TB −

(b)

The work done on the gas during the isobaric process is

a

b

fb

g a

g

fb

ga

WBC = − PB ∆V = −nR TC − TB = − 1.00 mol 8.314 J mol ⋅ K 316 K − 340 J or

WBC = +200 J

The work done on the gas in the isovolumetric process is zero, so in total Won gas = +200 J .

f

608 *P21.22

The Kinetic Theory of Gases

(a)

At any point in the heating process, Pi = kVi and P = kV = nRTi

Pf =

Vi

2

2Vi = 2 Pi and T f =

Pf V f nR

z f

(b)

The work input is W = − PdV = − i

=

2 Pi 2Vi = 4Ti . nR

z

nRTi

2Vi

Vi

Vi

2

nRT V 2 VdV = − 2 i 2 Vi

b

P21.23

(a)

a

2Vi

=− Vi

nRTi 2Vi

2

e4V

i

2

3 − Vi2 = − nRTi . 2

j

g

5 15 R 4Ti − Ti = + nRTi . The heat input 2 2

The change in internal energy, is ∆Eint = nCV ∆T = n is Q = ∆Eint − W =

Pi nRTi V . At the end, V= Vi Vi2

f

18 nRTi = 9 1 mol RTi . 2

The heat required to produce a temperature change is Q = n1C 1 ∆T + n 2 C 2 ∆T The number of molecules is N 1 + N 2 , so the number of “moles of the mixture” is n1 + n 2 and Q = n1 + n 2 C∆T ,

b

g

n C + n 2C 2 C= 1 1 . n1 + n 2

so

(b)

m

Q = ∑ n i C i ∆T = i =1

F ∑ n I C∆T GH JK m

i =1

i

m

∑ n i Ci

i =1 m

C=

∑ ni i =1

Section 21.3

P21.24

(a)

(b) (c)

Adiabatic Processes for an Ideal Gas γ PV i i

=

Tf

Pf V f

Ti

=

Pf V fγ

PV i i

=

so

f

f

Tf

i

i

Ti

Since the process is adiabatic, C P R + CV = , CV CV

b

∆Eint = nCV ∆T = 0.016 0 mol and

i

Vi

FG P IJ FG V IJ = a20.0fa0.118f H P KH V K

Since γ = 1.40 =

FPI =G J HP K

Vf

f

1γ

=

FG 1.00 IJ H 20.0 K

57

= 0.118

= 2.35

Q=0 CV =

5 R and ∆T = 2.35Ti − Ti = 1.35Ti 2

gFGH 52 IJK b8.314 J mol ⋅ K g 1.35a300 K f =

W = −Q + ∆Eint = 0 + 135 J = +135 J .

135 J

Chapter 21

P21.25

(a)

γ γ PV i i = Pf V f

Pf

(b)

FV I =PG J HV K i

Ti =

FG 12.0 IJ H 30.0 K

= 5.00 atm

f

e

1.40

= 1.39 atm

je

j

5.00 1.013 × 10 5 Pa 12.0 × 10 −3 m 3 PV i i = = 365 K nR 2.00 mol 8.314 J mol ⋅ K

Tf = (c)

γ

i

Pf V f

=

nR

b

e

g

je

1.39 1.013 × 10 Pa 30.0 × 10 −3 m 3 5

b

2.00 mol 8.314 J mol ⋅ K

g

j=

253 K

The process is adiabatic: Q = 0 C P R + CV 5 = , CV = R 2 CV CV 5 = nCV ∆T = 2.00 mol 8.314 J mol ⋅ K 2

γ = 1.40 = ∆Eint

FG b H

gIJK a253 K − 365 K f =

−4.66 kJ

W = ∆Eint − Q = −4.66 kJ − 0 = −4.66 kJ P21.26

Vi = π

F 2.50 × 10 GH 2

−2

m

I JK

2

0.500 m = 2.45 × 10 −4 m 3

The quantity of air we find from PV i i = nRTi n=

e

je

1.013 × 10 5 Pa 2.45 × 10 −4 m 3 PV i i = RTi 8.314 J mol ⋅ K 300 K

b

ga

f

j

n = 9.97 × 10 −3 mol Adiabatic compression: Pf = 101.3 kPa + 800 kPa = 901.3 kPa (a)

γ γ PV i i = Pf V f

V f = Vi

FPI GH P JK

1γ

i

= 2. 45 × 10 −4 m 3

f

FG 101.3 IJ H 901.3 K

57

V f = 5.15 × 10 −5 m 3 (b)

Pf V f = nRT f Tf

FPI =T =T G J PV P HP K F 101.3 IJ b g = = 300 K G H 901.3 K i

Pf V f i i

i

1γ

Pf

i

i

f

F P Ib =T G J HP K i

1 γ −1

g

i

f

5 7 −1

Tf (c)

560 K

The work put into the gas in compressing it is ∆Eint = nCV ∆T

e

W = 9.97 × 10 −3 mol W = 53.9 J continued on next page

j 52 b8.314 J mol ⋅ K ga560 − 300f K

609

610

The Kinetic Theory of Gases

Now imagine this energy being shared with the inner wall as the gas is held at constant volume. The pump wall has outer diameter 25.0 mm + 2.00 mm + 2.00 mm = 29.0 mm , and volume

LMπ e14.5 × 10 mj − π e12.5 × 10 mj OP4.00 × 10 N Q and mass ρV = e7.86 × 10 kg m je6.79 × 10 m j = 53.3 g 2

−3

2

−3

3

−6

3

−2

m = 6.79 × 10 −6 m 3

3

The overall warming process is described by 53.9 J = nC V ∆T + mc∆T

e j 52 b8.314 J mol ⋅ K gdT − 300 K i +e53.3 × 10 kg jb 448 J kg ⋅ K gdT − 300 K i 53.9 J = b0.207 J K + 23.9 J K gdT − 300 K i 53.9 J = 9.97 × 10 −3 mol

ff

−3

ff

ff

T ff − 300 K = 2.24 K

P21.27

Tf Ti

FV I =G J HV K

γ −1

i

=

f

FG 1 IJ H 2K

0. 400

If Ti = 300 K , then T f = 227 K .

*P21.28

(a)

In

γ PV i i

=

Pf V fγ

we have Pf

FV I =PG J HV K i

f

Pf Then

Pf V f PV i i = Ti Tf

γ

i

F 0.720 m I =PG H 0.240 m JK 3

i

T f = Ti

1.40

3

Pf V f PV i i

= 4.66 Pi

a f 13 = 1.55

= Ti 4.66

The factor of increase in temperature is the same as the factor of increase in internal energy, Eint, f = 1.55 . according to Eint = nCV T . Then Eint, i (b)

FV I = =G J In T PV HV K F 0.720 m I 2=G H V JK

γ

Tf

Pf V f

i

Vf

i

i i

f

Vi

3

FV I =G J HV K

0. 40

f

3

0.720 m = 2 1 0 . 4 = 2 2.5 = 5.66 Vf Vf =

0.720 m 3 = 0.127 m3 5.66

i

f

γ −1

we have

611

Chapter 21

P21.29

(a)

See the diagram at the right.

(b)

PBVBγ = PC VCγ

P B

3 Pi

γ γ 3 PV i i = PV i C

Adiabatic

e j e j = 2.19a 4.00 L f = 8.77 L

VC = 3 1 γ Vi = 3 5 7 Vi = 2.19Vi VC (c)

Pi

PBVB = nRTB = 3 PV i i = 3nRTi

a

f

C

A

VC

Vi = 4 L

TB = 3Ti = 3 300 K = 900 K (d)

After one whole cycle, TA = Ti = 300 K .

(e)

In AB, Q AB = nCV ∆V = n

V(L)

FIG. P21.29

FG 5 RIJ b3T − T g = a5.00fnRT H2 K i

i

i

QBC = 0 as this process is adiabatic

g a f

b

PC VC = nRTC = Pi 2.19Vi = 2.19 nRTi so

TC = 2.19Ti QCA = nC P ∆T = n

FG 7 RIJ bT − 2.19T g = a−4.17fnRT H2 K i

i

For the whole cycle,

i

a

f

a

f

Q ABCA = Q AB + QBC + QCA = 5.00 − 4.17 nRTi = 0.829 nRTi

b ∆E g

int ABCA

= 0 = Q ABCA + W ABCA

a

f

a

f

W ABCA = −Q ABCA = − 0.829 nRTi = − 0.829 PV i i

a

fe

je

j

W ABCA = − 0.829 1.013 × 10 5 Pa 4.00 × 10 −3 m 3 = −336 J P21.30

(a)

See the diagram at the right.

(b)

PBVBγ = PC VCγ

P B

3Pi

Adiabatic

γ γ 3 PV i i = PV i C

VC = 3 1 γ Vi = 3 5 7 Vi = 2.19Vi (c)

PBVB = nRTB = 3 PV i i = 3nRTi TB = 3Ti

(d)

After one whole cycle, TA = Ti

Pi

A

C VC

Vi

FIG. P21.30 continued on next page

af

V L

612

The Kinetic Theory of Gases

(e)

In AB, Q AB = nCV ∆T = n

FG 5 RIJ b3T − T g = a5.00fnRT H2 K i

i

i

QBC = 0 as this process is abiabatic

b g F7 I = nC ∆T = nG RJ bT − 2.19T g = −4.17nRT H2 K

PC VC = nRTC = Pi 2.19Vi = 2.19nRTi so TC = 2.19Ti QCA

P

i

i

i

For the whole cycle,

a

f

Q ABCA = Q AB + QBC + QCA = 5.00 − 4.17 nRTi = 0.830nRTi

b ∆E g

int ABCA

= 0 = Q ABCA + W ABCA

W ABCA = −Q ABCA = −0.830nRTi = −0.830 PV i i P21.31

(a)

The work done on the gas is

z

Vb

Wab = − PdV . Va

For the isothermal process, Wab ′ = −nRTa

z FGH

Vb′ Va

IJ K

1 dV V

FG V IJ = nRT lnFG V IJ . HV K HV K = 5.00 molb8.314 J mol ⋅ K ga 293 K f lna10.0 f

Wab ′ = −nRTa ln Thus, Wab′

b′

a

a

b′

FIG. P21.31

Wab′ = 28.0 kJ . (b)

For the adiabatic process, we must first find the final temperature, Tb . Since air consists primarily of diatomic molecules, we shall use

γ air = 1.40 and C V , air =

a

f

5 R 5 8.314 = = 20.8 J mol ⋅ K . 2 2

Then, for the adiabatic preocess Tb

FV I =T G J HV K

γ −1

a

a

a f

= 293 K 10.0

b

0. 400

= 736 K .

Thus, the work done on the gas during the adiabatic process is

b

g = b−0 + nC ∆T g = nC bT − T g = 5.00 molb 20.8 J mol ⋅ K ga736 − 293f K = 46.0 kJ

Wab −Q + ∆Eint or

Wab

continued on next page

ab

V

ab

V

b

a

.

613

Chapter 21

(c)

For the isothermal process, we have Pb ′ Vb ′ = PaVa . Thus, Pb ′ = Pa

FG V IJ = 1.00 atma10.0f = HV K a

10.0 atm .

b′

For the adiabatic process, we have Pb Vbγ = Pa Vaγ . Thus, Pb = Pa P21.32

FG V IJ HV K

γ

a

a f

= 1.00 atm 10.0

b

1.40

= 25.1 atm .

We suppose the air plus burnt gasoline behaves like a diatomic ideal gas. We find its final absolute pressure:

e

21.0 atm 50.0 cm3

FG 1 IJ H 8K

Pf = 21.0 atm

j

75

75

e

= Pf 400 cm3

j

75

= 1.14 atm

Now Q = 0

d

and W = ∆Eint = nCV T f − Ti ∴W =

i

5 5 5 nRT f − nRTi = Pf V f − PV i i 2 2 2

d

i

FIG. P21.32

F 1.013 × 10 N m I 10 5 W = 1.14 atme 400 cm j − 21.0 atme50.0 cm j G 2 H 1 atm JK e 3

3

W = −150 J The output work is −W = +150 J The time for this stroke is

P=

F GH

1 1 min 4 2 500

−W 150 J = = 25.0 kW ∆t 6.00 × 10 −3 s

I FG 60 s IJ = 6.00 × 10 JK H 1 min K

−3

s

5

2

−6

m3 cm 3

j

614

The Kinetic Theory of Gases

Section 21.4 P21.33

The Equipartition of Energy

The heat capacity at constant volume is nC V . An ideal gas of diatomic molecules has three degrees of freedom for translation in the x, y, and z directions. If we take the y axis along the axis of a molecule, then outside forces cannot excite rotation about this axis, since they have no lever arms. Collisions will set the molecule spinning only about the x and z axes. (a)

If the molecules do not vibrate, they have five degrees of freedom. Random collisions put 1 equal amounts of energy k BT into all five kinds of motion. The average energy of one 2 5 molecule is k BT . The internal energy of the two-mole sample is 2 N

FG 5 k TIJ = nN FG 5 k TIJ = nFG 5 RIJ T = nC T . H2 K H2 K H2 K B

The molar heat capacity is C V =

A

B

V

5 R and the sample’s heat capacity is 2

nCV = n

FG 5 RIJ = 2 molFG 5 b8.314 J mol ⋅ K gIJ H2 K H2 K

nCV = 41.6 J K For the heat capacity at constant pressure we have

g FGH 52 R + RIJK = 72 nR = 2 molFGH 72 b8.314 J mol ⋅ K gIJK

b

nC P = n C V + R = n nC P = 58.2 J K (b)

In vibration with the center of mass fixed, both atoms are always moving in opposite directions with equal speeds. Vibration adds two more degrees of freedom for two more terms in the molecular energy, for kinetic and for elastic potential energy. We have

and

P21.34

FG 7 RIJ = H2 K F9 I = nG RJ = H2 K

nCV = n

58.2 J K

nC P

74.8 J K

FG k T IJ = f FG nRT IJ H2K H 2K 1 F dE I 1 = G J = fR n H dT K 2

(1)

Eint = Nf

B

(2)

CV

int

(3)

C P = CV + R =

(4)

γ =

CP f + 2 = CV f

b

g

1 f +2 R 2

Chapter 21

P21.35

Rotational Kinetic Energy =

1 2 Iω 2

615

Cl

I = 2mr 2 , m = 35.0 × 1.67 × 10 −27 kg , r = 10 −10 m I = 1.17 × 10 −45 kg ⋅ m 2 ∴ K rot =

FIG. P21.35

1 2 Iω = 2.33 × 10 −21 J 2

Section 21.5

The Boltzmann Distribution Law

Section 21.6

Distribution of Molecular Speeds

P21.36

(a)

Cl

ω = 2.00 × 10 12 s −1

The ratio of the number at higher energy to the number at lower energy is e −∆E kBT where ∆E is the energy difference. Here,

a

∆E = 10.2 eV

10 JI fFGH 1.601×eV JK = 1.63 × 10 −19

−18

J

and at 0°C,

ja

e

f

k BT = 1.38 × 10 −23 J K 273 K = 3.77 × 10 −21 J . Since this is much less than the excitation energy, nearly all the atoms will be in the ground state and the number excited is

e

JI 1.63 × 10 j FGH −3.77 J = e2.70 × 10 je JK × 10 −18

2.70 × 10 25 exp

25

−21

−433

.

This number is much less than one, so almost all of the time no atom is excited . (b)

At 10 000°C,

e

j

k BT = 1.38 × 10 −23 J K 10 273 K = 1.42 × 10 −19 J . The number excited is JI 1.63 × 10 e2.70 × 10 j expFGH −1.42 J = e2.70 × 10 je JK × 10 25

−18

−19

25

−11.5

= 2.70 × 10 20 .

616 P21.37

The Kinetic Theory of Gases

(a)

v av =

(b)

ev j

∑ ni v i N

2

av

=

∑ ni vi2

P21.38

(a)

(b) P21.39

= 54.9 m 2 s 2

N

ev j 2

so v rms = (c)

af af af af af a f

1 1 2 + 2 3 + 3 5 + 4 7 + 3 9 + 2 12 = 6.80 m s 15

=

av

= 54.9 = 7.41 m s

v mp = 7.00 m s Vrms, 35 Vrms, 37

3 RT M 35

=

3 RT M 37

=

F 37.0 g mol I GH 35.0 g mol JK 35

The lighter atom,

12

= 1.03

Cl , moves faster. dN v = 0 to find dv

In the Maxwell Boltzmann speed distribution function take

F m I 4π N G H 2π k T JK

F GH

32

exp −

B

mv 2 2 k BT

I F 2v − 2mv I = 0 JK GH 2k T JK 3

B

and solve for v to find the most probable speed. Reject as solutions

v = 0 and v = ∞

Retain only

2−

Then

v mp =

mv 2 =0 k BT 2 k BT m

P21.40

The most probable speed is v mp =

P21.41

(a)

From v av =

P21.42

6.64 × 10

−27

kg

f=

132 m s .

8 k BT πm

we find the temperature as T =

e

ja

e

2 1.38 × 10 −23 J K 4.20 K

2 k BT = m

e

e

8 1.38 × 10

je

π 6.64 × 10 −27 kg 2.37 × 10 3 m s

(b)

T=

At 0°C,

1 3 2 = k BT0 mv rms0 2 2

e

8 1.38 × 10 −23 J mol ⋅ K

At the higher temperature,

b

1 m 2 v rms0 2

g

je

π 6.64 × 10 −27 kg 1.12 × 10 4 m s

2

j =

a

j

−23

J mol ⋅ K

j

2

= 1.06 × 10 3 K

3 k BT 2

f

T = 4T0 = 4 273 K = 1 092 K = 819° C .

j

2

= 2.37 × 10 4 K

Chapter 21

*P21.43

(a)

(b)

From the Boltzmann distribution law, the number density of molecules with gravitational energy mgy is n 0 e − mgy k BT . These are the molecules with height y, so this is the number per volume at height y as a function of y.

b g=e

n y n0

− mgy k BT

=e

= e − Mgy

e

N A k BT

= e − Mgy

je

RT

je

− 28 .9 × 10 −3 kg mol 9.8 m s 2 11 × 10 3 m

j b8.314 J mol⋅K ga 293 K f

= e −1.279 = 0.278 *P21.44

(a)

We calculate

z

∞

e − mgy

k BT

z

∞

dy =

e − mgy

k BT

=−

FG − mgdy IJ F − k T I H k T K GH mg JK k T k T =− 0 − 1f = a mg mg B

B

y=0

0

k BT − mgy e mg

∞

k BT

B

B

0

Using Table B.6 in the appendix

z

∞

ye − mgy

k BT

dy =

0

z z

∞

Then y =

ye − mgy

0 ∞

k BT

dy =

e − mgy

k BT

dy

F k TI bmg k T g GH mg JK 1!

2

=

2

B

.

B

bk T mg g B

k BT mg

2

=

k BT . mg

0

(b)

Section 21.7 P21.45

(a)

y=

b

k BT 8.314 J 283 K s 2 RT = = = 8.31 × 10 3 m M N A g Mg mol ⋅ K 28.9 × 10 −3 kg 9.8 m

g

Mean Free Path

FG N IJ RT and N = PVN so that RT HN K e1.00 × 10 ja133fa1.00fe6.02 × 10 j = N= a8.314fa300f A

PV =

A

−10

(b)

A=

f=

23

3.21 × 10 12 molecules

1 1.00 m3 V = = 2 12 2 12 nV π d 2 Nπ d 2 3.21 × 10 12 molecules π 3.00 × 10 −10 m

e

A = 779 km (c)

617

v = 6.42 × 10 −4 s −1 A

je

j a 2f 2

12

618 P21.46

The Kinetic Theory of Gases

The average molecular speed is v=

8 k BT 8 k B N AT = πm π N Am

v=

8 RT πM

b π e 2.016 × 10

g

8 8.314 J mol ⋅ K 3.00 K

v=

−3

kg mol

j

v = 178 m s (a)

The mean free path is 1

A=

2

2π d n V

=

1

e

j

2π 0. 200 × 10 −9 m

2

1 m3

A = 5.63 × 10 18 m The mean free time is A 5.63 × 10 18 m = = 3.17 × 10 16 s = 1.00 × 10 9 yr . 178 m s v (b)

Now nV is 10 6 times larger, to make A smaller by 10 6 times:

A = 5.63 × 10 12 m . Thus, P21.47

A = 3.17 × 10 10 s = 1.00 × 10 3 yr . v

From Equation 21.30, A = For an ideal gas, nV = k BT

Therefore, A =

P21.48

A=

2π d 2 nV

d = 3.60 × 10 −10 m

2π d 2 nV

N P = V k BT

2π d 2 P −1

1

, as required.

nV = nV =

P k BT 1.013 × 10 5

e1.38 × 10 ja293f −23

= 2.51 × 10 25 m3

∴ A = 6.93 × 10 −8 m, or about 193 molecular diameters .

Chapter 21

P21.49

k BT

Using P = nV k BT , Equation 21.30 becomes A =

e1.38 × 10

−23

ja

J K 293 K

(1)

2π Pd 2

f

= 9.36 × 10 −8 m

(a)

A=

(b)

Equation (1) shows that P1 A1 = P2 A 2 . Taking P1 A1 from (a) and with A 2 = 1.00 m, we find

e

je

5

2π 1.013 × 10 Pa 3.10 × 10

P2 = (c)

−10

j

m

2

a1.00 atmfe9.36 × 10

−8

j=

m

1.00 m

9.36 × 10 −8 atm .

For A 3 = 3.10 × 10 −10 m , we have

a1.00 atmfe9.36 × 10 P = 3

3.10 × 10

−10

−8

j=

m

m

302 atm .

Additional Problems P21.50

(a)

n=

PV (1.013 × 10 5 Pa)( 4.20 m × 3.00 m × 2.50 m) = = 1.31 × 10 3 mol ( 8.314 J mol ⋅ K )( 293 K ) RT

e

je

N = nN A = 1.31 × 10 3 mol 6.02 × 10 23 molecules mol

j

N = 7.89 × 10 26 molecules

jb

e

g

(b)

m = nM = 1.31 × 10 3 mol 0.028 9 kg mol = 37.9 kg

(c)

1 3 3 m 0 v 2 = k BT = 1.38 × 10 −23 J k 293 K = 6.07 × 10 −21 J molecule 2 2 2

(d)

For one molecule, m0 =

f

0.028 9 kg mol M = = 4.80 × 10 −26 kg molecule N A 6.02 × 10 23 molecules mol

v rms =

(e),(f)

ja

e

e

j=

2 6.07 × 10 −21 J molecule 4.80 × 10

−26

kg molecule

FG 5 RIJ T = 5 PV H2 K 2 5 = e1.013 × 10 Paje31.5 m j = 2

503 m s

Eint = nCV T = n Eint

5

3

7.98 MJ

619

620 P21.51

The Kinetic Theory of Gases

(a)

Pf = 100 kPa Vf =

nRT f

=

Pf

T f = 400 K

b

ga

2.00 mol 8.314 J mol ⋅ K 400 K 3

100 × 10 Pa

a f

f = 0.066 5 m

3

= 66.5 L

a

fb ga f W = − P∆V = −nR∆T = −a 2.00 molfb8.314 J mol ⋅ K ga100 K f =

∆Eint = 3.50 nR∆T = 3.50 2.00 mol 8.314 J mol ⋅ K 100 K = 5.82 kJ −1.66 kJ

Q = ∆Eint − W = 5.82 kJ + 1.66 kJ = 7.48 kJ (b)

T f = 400 K Pf = Pi

(c)

V f = Vi =

b

f

ga

nRTi 2.00 mol 8.314 J mol ⋅ K 300 K = = 0.049 9 m 3 = 49.9 L Pi 100 × 10 3 Pa

F T I = 100 kPaFG 400 K IJ = GH T JK H 300 K K f

z

W = − PdV = 0 since V = constant

133 kPa

i

∆Eint = 5.82 kJ as in part (a)

Q = ∆Eint − W = 5.82 kJ − 0 = 5.82 kJ

Pf = 120 kPa

T f = 300 K

V f = Vi

F P I = 49.9 LFG 100 kPa IJ = GH P JK H 120 kPa K i

a f

∆Eint = 3.50 nR∆T = 0 since T = constant

41.6 L

f

F I = −nRT lnFG P IJ GH JK HP K F 100 kPa IJ = +909 J W = −a 2.00 molfb8.314 J mol ⋅ K ga300 K f lnG H 120 kPa K

z

z

Vf

W = − PdV = −nRTi

Vi

Vf dV = −nRTi ln V Vi

i

i

f

Q = ∆Eint − W = 0 − 910 J = −909 J (d)

Pf = 120 kPa

γ =

C P CV + R 3.50 R + R 4.50 9 = = = = 3.50 R 3.50 7 CV CV

FPI F 100 kPa IJ = = : so V = V G J = 49.9 L G H 120 kPa K HP K F P V IJ = 300 K FG 120 kPa IJ FG 43.3 L IJ = 312 K T =T G H 100 kPa K H 49.9 L K H PV K ∆E = a3.50fnR∆T = 3.50a 2.00 molfb8.314 J mol ⋅ K ga12.4 K f = 722 J Q = 0 badiabatic process g 1γ

Pf V fγ

f

γ PV i i

i

f

f

f

i i

int

W = −Q + ∆Eint = 0 + 722 J = +722 J

i

i

f

79

43.3 L

Chapter 21

P21.52

(a)

The average speed v av is just the weighted average of all the speeds.

af a f a f a f a f a f a f= a 2 + 3 + 5 + 4 + 3 + 2 + 1f

2 v + 3 2 v + 5 3 v + 4 4v + 3 5 v + 2 6 v + 1 7 v

v av = (b)

3.65 v

First find the average of the square of the speeds, 2 v av

=

a f + 3a2vf + 5a3vf + 4a4vf + 3a5vf + 2a6 vf + 1a7 vf

2v

2

2

2

2

2

2

2+3+5+4+3+2+1

2

= 15.95 v 2 .

2 = 3.99 v . The root-mean square speed is then v rms = v av

(c)

The most probable speed is the one that most of the particles have; i.e., five particles have speed 3.00 v .

(d)

PV =

1 2 Nmv av 3

a

f

F GH

2 20 m 15.95 v mv 2 = 106 Therefore, P = 3 V V

(e)

1 1 2 = m 15.95 v 2 = 7.98mv 2 . mv av 2 2

e

j

z f

(a)

PV γ = k . So, W = − PdV = − k i

(b)

.

The average kinetic energy for each particle is K=

P21.53

I JK

z f

i

dV Pf V f − PV i i = γ γ −1 V

dEint = dQ + dW and dQ = 0 for an adiabatic process.

d

i

Therefore, W = + ∆Eint = nCV T f − Ti . To show consistency between these 2 equations, consider that γ = Therefore,

C 1 = V. γ −1 R

Using this, the result found in part (a) becomes

d

W = Pf V f − PV i i Also, for an ideal gas

i CR . V

PV = nT so that W = nCV T f − Ti . R

d

i

CP and C P − CV = R . CV

621

622 *P21.54

The Kinetic Theory of Gases

d

W = nCV T f − Ti

(a)

i

3 −2 500 J = 1 mol 8.314 J mol ⋅ K T f − 500 K 2

d

i

T f = 300 K γ γ PV i i = Pf V f

(b)

F nRT IJ PG H P K

γ

i

i

b g

Pi Pf = Pi *P21.55

b g

γ γ −1

γ γ −1

=

Tf

I b5 3gb3 2

FT GH T JK f

Pf

i

γ

f

f

i

Ti

F nRT I =P G H P JK

Tiγ Pi1 −γ = T fγ Pf1−γ

f

T I b g F P =PG J HT K g F 300 IJ = 1.00 atm = 3.60 atmG H 500 K f

f

i

γ γ −1

i

5 2

Let the subscripts ‘1’ and ‘2’ refer to the hot and cold compartments, respectively. The pressure is higher in the hot compartment, therefore the hot compartment expands and the cold compartment contracts. The work done by the adiabatically expanding gas is equal and opposite to the work done by the adiabatically compressed gas. nR nR T1i − T1 f = − T2i − T2 f γ −1 γ −1

d

i

d

∴ T1 f + T2 f = T1i + T2 i = 800 K Consider the adiabatic changes of the gases.

P1i V1γi = P1 f V1γf and P2i V2γi = P2 f V2γ f ∴

∴

P1i V1γi

P2 iV2γi

=

P1 f V1γf P2 f V2γ f

F I GH JK

V1 f P1i = P2 i V2 f

γ

, since V1i = V2i and P1 f = P2 f

F GH

nRT1 f P1 f nRT1i V1i ∴ = nRT2 i V2 i nRT2 f P2 f ∴

∴

F GH FT =G HT

T1 f T1i = T2i T2 f T1 f T2 f

1i 2i

I JK IJ K

I JK

γ

, using the ideal gas law

γ

, since V1i = V2i and P1 f = P2 f 1γ

=

FG 550 K IJ H 250 K K

1 1.4

= 1.756 (2)

Solving equations (1) and (2) simultaneously gives T1 f = 510 K, T2 f = 290 K .

i (1)

Chapter 21

*P21.56

623

The work done by the gas on the bullet becomes its kinetic energy:

b

1 1 mv 2 = 1.1 × 10 −3 kg 120 m s 2 2 The work on the gas is

2

= 7.92 J .

1 Pf V f − PV i i = −7.92 J. γ −1

d

FV I . GH V JK L FV I O 1 So −7.92 J = P MV G J − V P . PQ 0. 40 M H V K N

g

i

γ

γ Also Pf V fγ = PV i i

i

Pf = Pi

f

γ

i

i

f

i

f

And V f = 12 cm3 + 50 cm 0.03 cm 2 = 13.5 cm3 . Then Pi =

P21.57

a f LM13.5 cm c h N

−7.92 J 0.40 10 6 cm 3 m 3 3

12 1.40 13 .5

= − 12 cm O QP 3

5.74 × 10 6 Pa = 56.6 atm .

The pressure of the gas in the lungs of the diver must be the same as the absolute pressure of the water at this depth of 50.0 meters. This is:

je ja F 1.00 atm IJ = 5.98 atm PaG H 1.013 × 10 Pa K e

f

P = P0 + ρgh = 1.00 atm + 1.03 × 10 3 kg m 3 9.80 m s 2 50.0 m or

P = 1.00 atm + 5.05 × 10 5

5

If the partial pressure due to the oxygen in the gas mixture is to be 1.00 atmosphere (or the fraction 1 1 of the total pressure) oxygen molecules should make up only of the total number of 5.98 5.98 molecules. This will be true if 1.00 mole of oxygen is used for every 4.98 mole of helium. The ratio by weight is then

a4.98 mol Hefb4.003 g mol Hegg = b1.00 mol O gb2 × 15.999 g mol O gg 2

P21.58

(a)

0.623 .

2

Maxwell’s speed distribution function is Nv

F m I = 4π N G H 2π k T JK

3 2

v 2 e − mv

2

2 k BT

B

With

N = 1.00 × 10 4 , 0.032 kg M = = 5.32 × 10 −26 kg m= N A 6.02 × 10 23 T = 500 K

and

k B = 1.38 × 10 −23 J molecule ⋅ K

e

j

this becomes N v = 1.71 × 10 −4 v 2 e

e

j

− 3 .85 × 10 −6 v 2

To the right is a plot of this function for the range 0 ≤ v ≤ 1 500 m s. FIG. P21.58(a) continued on next page

624

The Kinetic Theory of Gases

(b)

The most probable speed occurs where N v is a maximum. From the graph, v mp ≈ 510 m s

(c)

v av =

8 k BT = πm

e

ja f = π e5.32 × 10 j

8 1.38 × 10 −23 500 −26

575 m s

Also, v rms = (d)

3 k BT = m

e

ja f =

3 1.38 × 10 −23 500 5.32 × 10

−26

624 m s 300 m s ≤ v ≤ 600 m s

The fraction of particles in the range

z

600

N v dv

is

300

where

N = 10 4

N

and the integral of N v is read from the graph as the area under the curve. This is approximately 4 400 and the fraction is 0.44 or 44% . P21.59

(a)

Since pressure increases as volume decreases (and vice versa),

LM OP N Q

1 dV dV < 0 and − > 0. V dP dP (b)

For an ideal gas, V =

FG H

IJ K

1 d nRT nRT and κ 1 = − . P V dP P

If the compression is isothermal, T is constant and

κ1 = − (c)

IJ K

For an adiabatic compression, PV γ = C (where C is a constant) and

κ2 =−

(d)

FG H

1 1 nRT − 2 = . V P P

κ1 = γ =

FG IJ H K

1 d C V dP P

1γ

=

FG IJ H K

1 1 C1 γ 1 P1 γ = 1 γ +1 = . γP V γ P b1 γ g+1 γP

1 1 = = 0.500 atm −1 2.00 atm P

a

f

CP 5 and for a monatomic ideal gas, γ = , so that 3 CV

κ2 =

1 = γP

5 3

a

1 = 0.300 atm −1 2.00 atm

f

Chapter 21

P21.60

(a)

B

The speed of sound is v =

ρ

where B = −V

dP . dV

According to Problem 59, in an adiabatic process, this is B =

a f a f

1

κ2

= γP .

nRT M PM m s nM = = = where m s is the sample mass. Then, the speed of sound V V V RT RT

Also, ρ =

B

in the ideal gas is v =

ρ

b

= γP

ga

FG RT IJ = H PM K

1.40 8.314 J mol ⋅ K 293 K

v=

γRT . M

f=

(b)

344 m s 0.028 9 kg mol This nearly agrees with the 343 m/s listed in Table 17.1.

(c)

We use k B =

γ kB N AT γ k BT γRT R and M = mN A : v = = = . M mN A m NA

The most probable molecular speed is

2k BT , m

8k BT , and the rms speed is πm

the average speed is

3k BT . m

All are somewhat larger than the speed of sound. P21.61

n=

(a)

(b)

(c)

(d)

1.20 kg m = = 41.5 mol M 0.028 9 kg mol Vi = Pf Pi

a

fb

Vf Vi

so V f

FP I F 400 IJ = V G J = e0.514 m jG H 200 K HPK f

i

2

3

2

= 2.06 m 3

i

e400 × 10 Paje2.06 m j = 2.38 × 10 K nR a41.5 molfb8.314 J mol ⋅ K g F P I 2V = − 2 F P I V − V W = − z PdV = −C z V dV = − G j G Je 3 HV K H V JK 3 2 F 200 × 10 Pa I L W =− G e2.06 m j − a0.514 mf OQP = −4.80 × 10 J 3 H 0.514 m JK NM Tf =

Pf V f

3

3

3

=

Vf

Vf

Vi

Vi

12

3

(e)

f

ga

41.5 mol 8.314 J mol ⋅ K 298 K nRTi = = 0.514 m 3 Pi 200 × 10 3 Pa

=

625

a

∆Eint = nCV ∆T = 41.5 mol

i 12 i

3 3 2

3 2 Vf

i 12 i

Vi

32 f

3 2 i

3 2

fLMN 52 b8.314 J mol ⋅ K gOPQe2.38 × 10

5

3

j

− 298 K

∆Eint = 1.80 × 10 6 J Q = ∆Eint − W = 1.80 × 10 6 J + 4.80 × 10 5 J = 2.28 × 10 6 J = 2.28 MJ

626 P21.62

The Kinetic Theory of Gases

b

1 1 1 mvi2 − mv 2f = 0.142 kg 2 2 2

The ball loses energy

g a47.2f − a42.5f 2

2

m 2 s 2 = 29.9 J

g a19.4 mf = 0.083 4 m PV 1.013 × 10 Pae0.083 4 m j = n= RT b8.314 J mol ⋅ K ga293 K f = 3.47 mol b

V = π 0.037 0 m

The air volume is

2

3

5

and its quantity is

3

The air absorbs energy according to Q = nC P ∆T ∆T =

So

P21.63

af

N v v = 4π N Note that

Thus,

Q = nC P 3. 47 mol

F −mv I GH 2k T JK F 2k T IJ v =G H m K F m I N a vf = 4π N G H 2π k T JK F v I ee N avf =G J N ev j H v K

F m I GH 2π k T JK

32

B

For

B

12

v

v2e

e− v

2

2 v mp

j

2 1 − v 2 v mp

j

mp

mp

v mp 50

a f = FG 1 IJ N e v j H 50 K Nv v v

3 2

B

v

v=

0. 296° C

B

mp

v

c hb8.314 J mol ⋅ K g =

2

v 2 exp

2

And

29.9 J

7 2

2

e

b g

1 − 1 50

2

= 1.09 × 10 −3

mp

The other values are computed similarly, with the following results:

v v mp 1 50 1 10 1 2 1 2 10 50

af

Nv v

e j

N v v mp

1.09 × 10 −3 2.69 × 10 −2 0.529 1.00 0.199 1.01 × 10 −41 1.25 × 10 −1 082

To find the last value, note:

a50f e

2 1 − 2 500

10 log 2 500 e

= 2 500 e −2 499

aln10 fb−2 499 ln10 g = 10 log 2 50010 −2 499 ln10 = 10 log 2 500 − 2 499 ln 10 = 10 −1 081.904

Chapter 21

P21.64

(a)

627

The effect of high angular speed is like the effect of a very high gravitational field on an atmosphere. The result is:

The larger-mass molecules settle to the outside while the region at smaller r has a higher concentration of low-mass molecules. (b)

Consider a single kind of molecules, all of mass m. To cause the centripetal acceleration of the molecules between r and r + dr , the pressure must increase outward according to ∑ Fr = mar . Thus,

a

f

b

ge j

PA − P + dP A = − nmA dr rω 2

where n is the number of molecules per unit volume and A is the area of any cylindrical surface. This reduces to dP = nmω 2 rdr . But also P = nk BT , so dP = k BTdn . Therefore, the equation becomes

z

z

n

r

dn mω 2 dn mω 2 = = rdr giving rdr or n k BT n k BT 0 n 0

af

ln n

n n0

F I GH JK

mω 2 r 2 = k BT 2

FG n IJ = mω r H n K 2k T 2

ln

0

P21.65

Then,

=

0

and solving for n: n = n 0 e mr

B

2 2 = as v av First find v av

2 v av

2

r

2

ω 2 2 k BT

.

z

∞

1 m . v 2 N v dv . Let a = N0 2 k BT

4 Nπ − 1 2 a 3 2

∞

N

0

z

v 4 e − av

2

dv

= 4 a 3 2 π −1 2

π 3 k BT = a m

3 8a2

2 = The root-mean square speed is then v rms = v av

3 k BT . m

To find the average speed, we have v av

*P21.66

e

∞ 4Na 3 2 π −1 2 1 = vN v dv = N0 N

z

j

z

∞

2

v 3 e − av dv =

0

4 a 3 2 π −1 2 = 2a 2

8 k BT . πm

dP for the function implied by PV = nRT = constant , and also for the different dV function implied by PV γ = constant . We can use implicit differentiation: We want to evaluate

From PV = constant From PV γ = constant Therefore, The theorem is proved.

P

FG dP IJ H dV K FG dP IJ H dV K

dV dP +V =0 dV dV

PγV γ −1 + V γ

FG dP IJ H dV K

dP =0 dV

=γ adiabat

FG dP IJ H dV K

isotherm

=− isotherm

=− adiabat

P V

γP V

628 P21.67

The Kinetic Theory of Gases

e

je

j

1.013 × 10 5 Pa 5.00 × 10 −3 m 3 PV = = 0.203 mol RT 8.314 J mol ⋅ K 300 K

(a)

n=

(b)

TB = TA

b

f

ga

FG P IJ = 300 K FG 3.00 IJ = H 1.00 K HP K B

900 K

A

TC = TB = 900 K VC = VA

(c)

C

15.0 L

A

a

FIG. P21.67

fb

f ga

ga

3 3 nRTA = 0.203 mol 8.314 J mol ⋅ K 300 K = 760 J 2 2 3 3 = Eint, C = nRTB = 0.203 mol 8.314 J mol ⋅ K 900 K = 2.28 kJ 2 2

Eint, A = Eint, B

FG T IJ = 5.00 LFG 900 IJ = H 300 K HT K a

(d) A B C

P (atm) 1.00 3.00 1.00

fb

V(L) 5.00 5.00 15.00

f

Eint (kJ) 0.760 2.28 2.28

T(K) 300 900 900

(e)

For the process AB, lock the piston in place and put the cylinder into an oven at 900 K. For BC, keep the sample in the oven while gradually letting the gas expand to lift a load on the piston as far as it can. For CA, carry the cylinder back into the room at 300 K and let the gas cool without touching the piston.

(f)

For AB:

a

f

∆Eint = Eint, B − Eint, A = 2.28 − 0.760 kJ = 1.52 kJ

W= 0 Q = ∆Eint − W = 1.52 kJ

For BC:

FG V IJ HV K

∆Eint = 0 , W = −nRTB ln

a

fb

C

B

ga

f a f

W = − 0. 203 mol 8.314 J mol ⋅ K 900 K ln 3.00 = −1.67 kJ Q = ∆Eint − W = 1.67 kJ For CA:

a f W = − P∆V = −nR∆T = −a0.203 molfb8.314 J mol ⋅ K ga −600 K f = ∆Eint = Eint, A − Eint, C = 0.760 − 2.28 kJ = −1.52 kJ

1.01 kJ

Q = ∆Eint − W = −1.52 kJ − 1.01 kJ = −2.53 kJ (g)

We add the amounts of energy for each process to find them for the whole cycle. Q ABCA = +1.52 kJ + 1.67 kJ − 2.53 kJ = 0.656 kJ W ABCA = 0 − 1.67 kJ + 1.01 kJ = −0.656 kJ

b ∆E g

int ABCA

= +1.52 kJ + 0 − 1.52 kJ = 0

P21.68

(a)

00 mol I F 6.02 × 10 molecules I b10 000 g gFGH 1.18.0 JK = g JK GH 1.00 mol 23

Chapter 21

629

3.34 × 10 26 molecules

(b)

After one day, 10 −1 of the original molecules would remain. After two days, the fraction would be 10 −2 , and so on. After 26 days, only 3 of the original molecules would likely remain, and after 27 days , likely none.

(c)

The soup is this fraction of the hydrosphere:

F 10.0 kg I . GH 1.32 × 10 kg JK 21

Therefore, today’s soup likely contains this fraction of the original molecules. The number of original molecules likely in the pot again today is:

F 10.0 kg I 3.34 × 10 GH 1.32 × 10 kg JK e 21

P21.69

26

j

molecules = 2.53 × 10 6 molecules .

1 GmM GM . Since the free-fall acceleration at the surface is g = 2 , this can mv 2 = 2 RE RE 1 GmM 2 also be written as: mv = = mgRE . 2 RE

(a)

For escape,

(b)

For O 2 , the mass of one molecule is m=

0.032 0 kg mol 6.02 × 10 23 molecules mol

= 5.32 × 10 −26 kg molecule .

FG 3 k T IJ , the temperature is H 2 K e5.32 × 10 kgje9.80 m s je6.37 × 10 mj = mgR = T= 15 k 15e1.38 × 10 J mol ⋅ K j B

Then, if mgRE = 10

−26

6

2

E

−23

B

P21.70

(a)

1.60 × 10 4 K .

For sodium atoms (with a molar mass M = 32.0 g mol ) 1 3 mv 2 = k BT 2 2

FG IJ H K

1 M 3 v 2 = k BT 2 NA 2 v rms =

(b)

t=

d v rms

=

3 RT = M

0.010 m = 20 ms 0.510 m s

b

ge

3 8.314 J mol ⋅ K 2.40 × 10 −4 K 23.0 × 10

−3

kg

j=

0.510 m s

630

The Kinetic Theory of Gases

ANSWERS TO EVEN PROBLEMS P21.2

17.6 kPa

P21.42

819°C

P21.4

5.05 × 10 −21 J molecule

P21.44

(a) see the solution; (b) 8.31 km

P21.6

6.64 × 10 −27 kg

P21.46

(a) 5.63 × 10 18 m; 1.00 × 10 9 yr ; (b) 5.63 × 10 12 m; 1.00 × 10 3 yr

P21.8

477 m s P21.48

193 molecular diameters

P21.10

(a) 2.28 kJ; (b) 6.21 × 10 −21 J

P21.50

(a) 7.89 × 10 26 molecules; (b) 37.9 kg ;

P21.12

74.8 J

P21.14

7.52 L

P21.16

(a) 118 kJ ; (b) 6.03 × 10 3 kg

P21.18

(a) 719 J kg ⋅ K ; (b) 0.811 kg ; (c) 233 kJ; (d) 327 kJ

(c) 6.07 × 10 −21 J molecule ; (d) 503 m s; (e) 7.98 MJ ; (f) 7.98 MJ

P21.20

13.5 PV

P21.22

(a) 4Ti ; (b) 9 1 mol RTi

P21.24

(a) 0.118 ; (b) 2.35 ; (c) 0; 135 J ; 135 J

P21.26

(a) 5.15 × 10 −5 m 3 ; (b) 560 K ; (c) 2.24 K

P21.28

(a) 1.55 ; (b) 0.127 m3

P21.30

(a) see the solution; (b) 2.19Vi ; (c) 3Ti ; (d) Ti ; (e) −0.830 PV i i

P21.32

25.0 kW

P21.34

see the solution

P21.36

(a) No atom, almost all the time; (b) 2.70 × 10 20

P21.38

(a) 1.03 ; (b)

P21.40

132 m s

a

35

f

Cl

P21.52

(a) 3.65 v; (b) 3.99 v; (c) 3.00 v; mv 2 ; (e) 7.98mv 2 (d) 106 V

P21.54

(a) 300 K ; (b) 1.00 atm

P21.56

5.74 × 10 6 Pa

P21.58

(a) see the solution; (b) 5.1 × 10 2 m s; (c) v av = 575 m s ; v rms = 624 m s ; (d) 44%

P21.60

(a) see the solution; (b) 344 m s nearly agreeing with the tabulated value; (c) see the solution; somewhat smaller than each

P21.62

0.296°C

P21.64

see the solution

P21.66

see the solution

P21.68

(a) 3.34 × 10 26 molecules ; (b) during the 27th day; (c) 2.53 × 10 6 molecules

P21.70

(a) 0.510 m s ; (b) 20 ms

F GH

I JK

22 Heat Engines, Entropy, and the Second Law of Thermodynamics CHAPTER OUTLINE 22.1

22.2 22.3 22.4 22.5 22.6 22.7 22.8

Heat Engines and the Second Law of Thermodynamics Heat Pumps and Refrigerators Reversible and Irreversible Processes The Carnot Engine Gasoline and Diesel Engines Entropy Entropy Changes in Irreversible Processes Entropy on a Microscopic Scale

ANSWERS TO QUESTIONS Q22.1

First, the efficiency of the automobile engine cannot exceed the Carnot efficiency: it is limited by the temperature of burning fuel and the temperature of the environment into which the exhaust is dumped. Second, the engine block cannot be allowed to go over a certain temperature. Third, any practical engine has friction, incomplete burning of fuel, and limits set by timing and energy transfer by heat.

Q22.2

It is easier to control the temperature of a hot reservoir. If it cools down, then heat can be added through some external means, like an exothermic reaction. If it gets too hot, then heat can be allowed to “escape” into the atmosphere. To maintain the temperature of a cold reservoir, one must remove heat if the reservoir gets too hot. Doing this requires either an “even colder” reservoir, which you also must maintain, or an endothermic process.

Q22.3

A higher steam temperature means that more energy can be extracted from the steam. For a constant temperature heat sink at Tc , and steam at Th , the efficiency of the power plant goes as Th − Tc T = 1 − c and is maximized for a high Th . Th Th

Q22.4

No. Any heat engine takes in energy by heat and must also put out energy by heat. The energy that is dumped as exhaust into the low-temperature sink will always be thermal pollution in the outside environment. So-called ‘steady growth’ in human energy use cannot continue.

Q22.5

No. The first law of thermodynamics is a statement about energy conservation, while the second is a statement about stable thermal equilibrium. They are by no means mutually exclusive. For the particular case of a cycling heat engine, the first law implies Q h = Weng + Q c , and the second law implies Q c > 0.

Q22.6

Take an automobile as an example. According to the first law or the idea of energy conservation, it must take in all the energy it puts out. Its energy source is chemical energy in gasoline. During the combustion process, some of that energy goes into moving the pistons and eventually into the mechanical motion of the car. Clearly much of the energy goes into heat, which, through the cooling system, is dissipated into the atmosphere. Moreover, there are numerous places where friction, both mechanical and fluid, turns mechanical energy into heat. In even the most efficient internal combustion engine cars, less than 30% of the energy from the fuel actually goes into moving the car. The rest ends up as useless heat in the atmosphere. 631

632

Heat Engines, Entropy, and the Second Law of Thermodynamics

Q22.7

Suppose the ambient temperature is 20°C. A gas can be heated to the temperature of the bottom of the pond, and allowed to cool as it blows through a turbine. The Carnot efficiency of such an engine ∆T 80 = = 22%. is about e c = Th 373

Q22.8

No, because the work done to run the heat pump represents energy transferred into the house by heat.

Q22.9

A slice of hot pizza cools off. Road friction brings a skidding car to a stop. A cup falls to the floor and shatters. Your cat dies. Any process is irreversible if it looks funny or frightening when shown in a videotape running backwards. The free flight of a projectile is nearly reversible.

Q22.10

Below the frost line, the winter temperature is much higher than the air or surface temperature. The earth is a huge reservoir of internal energy, but digging a lot of deep trenches is much more expensive than setting a heat-exchanger out on a concrete pad. A heat pump can have a much higher coefficient of performance when it is transferring energy by heat between reservoirs at close to the same temperature.

Q22.11

(a)

When the two sides of the semiconductor are at different temperatures, an electric potential (voltage) is generated across the material, which can drive electric current through an external circuit. The two cups at 50°C contain the same amount of internal energy as the pair of hot and cold cups. But no energy flows by heat through the converter bridging between them and no voltage is generated across the semiconductors.

(b)

A heat engine must put out exhaust energy by heat. The cold cup provides a sink to absorb output or wasted energy by heat, which has nowhere to go between two cups of equally warm water.

Q22.12

Energy flows by heat from a hot bowl of chili into the cooler surrounding air. Heat lost by the hot stuff is equal to heat gained by the cold stuff, but the entropy decrease of the hot stuff is less than the entropy increase of the cold stuff. As you inflate a soft car tire at a service station, air from a tank at high pressure expands to fill a larger volume. That air increases in entropy and the surrounding atmosphere undergoes no significant entropy change. The brakes of your car get warm as you come to a stop. The shoes and drums increase in entropy and nothing loses energy by heat, so nothing decreases in entropy.

Q22.13

(a)

For an expanding ideal gas at constant temperature, ∆S =

(b)

For a reversible adiabatic expansion ∆Q = 0 , and ∆S = 0 . An ideal gas undergoing an irreversible adiabatic expansion can have any positive value for ∆S up to the value given in part (a).

FG IJ H K

V ∆Q = nR ln 2 . V1 T

Q22.14

The rest of the Universe must have an entropy change of +8.0 J/K, or more.

Q22.15

Even at essentially constant temperature, energy must flow by heat out of the solidifying sugar into the surroundings, to raise the entropy of the environment. The water molecules become less ordered as they leave the liquid in the container to mix into the whole atmosphere and hydrosphere. Thus the entropy of the surroundings increases, and the second law describes the situation correctly.

Chapter 22

633

Q22.16

To increase its entropy, raise its temperature. To decrease its entropy, lower its temperature. “Remove energy from it by heat” is not such a good answer, for if you hammer on it or rub it with a blunt file and at the same time remove energy from it by heat into a constant temperature bath, its entropy can stay constant.

Q22.17

An analogy used by Carnot is instructive: A waterfall continuously converts mechanical energy into internal energy. It continuously creates entropy as the organized motion of the falling water turns into disorganized molecular motion. We humans put turbines into the waterfall, diverting some of the energy stream to our use. Water flows spontaneously from high to low elevation and energy spontaneously flows by heat from high to low temperature. Into the great flow of solar radiation from Sun to Earth, living things put themselves. They live on energy flow, more than just on energy. A basking snake diverts energy from a high-temperature source (the Sun) through itself temporarily, before the energy inevitably is radiated from the body of the snake to a low-temperature sink (outer space). A tree builds organized cellulose molecules and we build libraries and babies who look like their grandmothers, all out of a thin diverted stream in the universal flow of energy crashing down to disorder. We do not violate the second law, for we build local reductions in the entropy of one thing within the inexorable increase in the total entropy of the Universe. Your roommate’s exercise puts energy into the room by heat.

Q22.18

(a)

Entropy increases as the yeast dies and as energy is transferred from the hot oven into the originally cooler dough and then from the hot bread into the surrounding air.

(b)

Entropy increases some more as you metabolize the starches, converting chemical energy into internal energy.

Q22.19

Either statement can be considered an instructive analogy. We choose to take the first view. All processes require energy, either as energy content or as energy input. The kinetic energy which it possessed at its formation continues to make the Earth go around. Energy released by nuclear reactions in the core of the Sun drives weather on the Earth and essentially all processes in the biosphere. The energy intensity of sunlight controls how lush a forest or jungle can be and how warm a planet is. Continuous energy input is not required for the motion of the planet. Continuous energy input is required for life because energy tends to be continuously degraded, as heat flows into lower-temperature sinks. The continuously increasing entropy of the Universe is the index to energy-transfers completed.

Q22.20

The statement is not true. Although the probability is not exactly zero that this will happen, the probability of the concentration of air in one corner of the room is very nearly zero. If some billions of molecules are heading toward that corner just now, other billions are heading away from the corner in their random motion. Spontaneous compression of the air would violate the second law of thermodynamics. It would be a spontaneous departure from thermal and mechanical equilibrium.

Q22.21

Shaking opens up spaces between jellybeans. The smaller ones more often can fall down into spaces below them. The accumulation of larger candies on top and smaller ones on the bottom implies a small increase in order, a small decrease in one contribution to the total entropy, but the second law is not violated. The total entropy increases as the system warms up, its increase in internal energy coming from the work put into shaking the box and also from a bit of gravitational energy loss as the beans settle compactly together.

634

Heat Engines, Entropy, and the Second Law of Thermodynamics

SOLUTIONS TO PROBLEMS Section 22.1 P22.1

P22.2

Heat Engines and the Second Law of Thermodynamics Weng

e=

(b)

Q c = Q h − Weng = 360 J − 25.0 J = 335 J

Qh

=

25.0 J = 0.069 4 or 6.94% 360 J

(a)

Weng = Q h − Q c = 200 J e=

Weng

Qc

=1−

Qh

Qh

(1)

= 0.300

(2)

From (2), Q c = 0.700 Q h

(3)

Solving (3) and (1) simultaneously, we have

P22.3

(a)

Q h = 667 J and

(b)

Q c = 467 J .

(a)

We have e =

Weng Qh

=

Qh − Qc Qh

=1−

Qc Qh

= 0.250

with Q c = 8 000 J, we have Q h = 10.7 kJ (b)

Weng = Q h − Q c = 2 667 J and from P =

*P22.4

Weng ∆t

, we have ∆t =

Weng

P

=

2 667 J = 0.533 s . 5 000 J s

We have Q hx = 4Q hy , Weng x = 2Weng y and Q cx = 7Q cy . As well as Q hx = Weng x + Q cx and Q hy = Weng y + Q cy . Substituting, 4Q hy = 2Weng y + 7Q cy 4Q hy = 2Weng y + 7Q hy − 7Weng y 5Weng y = 3Q hy (b)

ey =

(a)

ex =

Weng y Q hy Weng x Q hx

=

=

3 = 60.0% 5 2Weng y 4Q hy

=

a

f

2 0.600 = 0.300 = 30.0% 4

Chapter 22

*P22.5

(a)

The input energy each hour is

e7.89 × 10

3

jb

= 1.18 × 10 J h g 601min h F 1 L IJ = 29.4 L h J hjG H 4.03 × 10 J K 9

J revolution 2 500 rev min

e

implying fuel input 1.18 × 10 9

7

Q h = Weng + Q c . For a continuous-transfer process we may divide by time to have

(b)

Q h Weng Q c = + ∆t ∆t ∆t Weng Q h Q c Useful power output = = − ∆t ∆t ∆t

F 7.89 × 10 J − 4.58 × 10 J I 2 500 rev 1 min = 1.38 × 10 GH revolution revolution JK 1 min 60 s F 1 hp IJ = 185 hp P = 1.38 × 10 WG H 746 W K P 1.38 × 10 J s F 1 rev I P = τω ⇒ τ = = ω b2 500 rev 60 sg GH 2π rad JK = 527 N ⋅ m Q 4.58 × 10 J F 2 500 rev I = G J = 1.91 × 10 W ∆t revolution H 60 s K 3

=

5

eng

3

c

5

e

je

j

Q c = mL f = 15 × 10 −3 kg 1.18 × 10 4 J kg = 177 J

The heat to melt 15.0 g of Hg is

The energy absorbed to freeze 1.00 g of aluminum is

e

je

j

Q h = mL f = 10 −3 kg 3.97 × 10 5 J / kg = 397 J Weng = Q h − Q c = 220 J

and the work output is

e=

Section 22.2

Weng Qh

=

220 J = 0.554 , or 55.4% 397 J

Th − Tc 933 K − 243.1 K = = 0.749 = 74.9% 933 K Th

The theoretical (Carnot) efficiency is

P22.7

W

eng

(d) P22.6

5

5

eng

(c)

3

Heat Pumps and Refrigerators

b

g

COP refrigerator =

Qc W

(a)

If Q c = 120 J and COP = 5.00 , then W = 24.0 J

(b)

Heat expelled = Heat removed + Work done. Q h = Q c + W = 120 J + 24 J = 144 J

635

636 P22.8

Heat Engines, Entropy, and the Second Law of Thermodynamics

Qc Q . Therefore, W = c . W 3.00 The heat removed each minute is COP = 3.00 =

b b

gb gb

f b f

ga ga

ge

QC = 0.030 0 kg 4 186 J kg ° C 22.0° C + 0.030 0 kg 3.33 × 10 5 J kg t + 0.030 0 kg 2 090 J kg ° C 20.0° C = 1.40 × 10 4 J min or,

Qc = 233 J s. t

Thus, the work done per sec = P =

P22.9

j

233 J s = 77.8 W . 3.00

(a)

FG 10.0 Btu IJ FG 1055 J IJ FG 1 h IJ FG 1 W IJ = H h ⋅ W K H 1 Btu K H 3 600 s K H 1 J s K

(b)

Coefficient of performance for a refrigerator:

(c)

With EER 5, 5

2.93

Btu 10 000 Btu h = : P h⋅ W

aCOPf P=

refrigerator

10 000 Btu h 5 Btu h⋅W

= 2 000 W = 2.00 kW

a fb g Cost = e3.00 × 10 kWhjb0.100 $ kWhg = $300 P ∆t = 2.00 kW 1 500 h = 3.00 × 10 3 kWh

Energy purchased is

3

With EER 10, 10

Btu 10 000 Btu h = : P h⋅ W

P=

10 000 Btu h 10 Btu h⋅W

= 1 000 W = 1.00 kW

a fb g Cost = e1.50 × 10 kWhjb0.100 $ kWhg = $150 P ∆t = 1.00 kW 1 500 h = 1.50 × 10 3 kWh

Energy purchased is

3

Thus, the cost for air conditioning is

Section 22.3

half as much with EER 10

Reversible and Irreversible Processes

No problems in this section

Section 22.4 P22.10

The Carnot Engine

Weng T and When e = e c , 1 − c = Th Qh

(a)

Qh

Weng ∆t Qh ∆t

=1−

Tc Th

FH IK ∆t e1.50 × 10 Wjb3 600 sg = = Weng ∆t

5

T

1−

1 − Tc

h

293 773

Q h = 8.69 × 10 8 J = 869 MJ (b)

Qc = Qh −

F W I ∆t = 8.69 × 10 − e1.50 × 10 jb3 600g = 3.30 × 10 GH ∆t JK eng

8

5

8

J = 330 MJ

Chapter 22

P22.11

Tc = 703 K

P22.13

∆T 1 440 = = 67.2% Th 2 143

(a)

ec =

(b)

Q h = 1.40 × 10 5 J , Weng = 0.420 Q h

P= P22.12

Th = 2 143 K

Weng ∆t

=

5.88 × 10 4 J = 58.8 kW 1s ∆T 120 K = = 0.253 Th 473 K

The Carnot efficiency of the engine is

ec =

At 20.0% of this maximum efficiency,

e = 0.200 0.253 = 0.050 6

From the definition of efficiency

Weng = Q h e

and

Qh =

a

Isothermal expansion at

Th = 523 K

Isothermal compression at

Tc = 323 K

f

Weng e

=

10.0 kJ = 197 kJ 0.050 6

Gas absorbs 1 200 J during expansion. (a) (b) P22.14

*P22.15

FG T IJ = 1 200 JFG 323 IJ = 741 J H 523 K HT K = Q − Q = b1 200 − 741g J = 459 J c

Qc = Qh Weng

h

h

c

Tc Th

We use

ec = 1 −

as

0.300 = 1 −

From which,

Th = 819 K = 546° C

573 K Th

The efficiency is

ec = 1 −

Then

Tc = Th Qh ∆t

(a)

Qc ∆t Qh ∆t

=

a a

f f

Q c Th 273 + 100 K = 15. 4 W = 19.6 W ∆t Tc 273 + 20 K

Q h = Weng + Q c The useful power output is

(b)

Qc Tc =1− Th Qh

Qh =

F Q I ∆t = mL GH ∆t JK h

V

Weng ∆t m=

=

Qh ∆t

−

Qc ∆t

= 19.6 W − 15.4 W = 4.20 W

gFGH

I JK

Q h ∆t 3 600 s = 19.6 J s = 3.12 × 10 −2 kg ∆t LV 2.26 × 10 6 J kg

b

637

638 P22.16

Heat Engines, Entropy, and the Second Law of Thermodynamics

a a

f f

The Carnot summer efficiency is

e c ,s = 1 −

273 + 20 K Tc =1− = 0.530 Th 273 + 350 K

And in winter,

ec ,w = 1 −

283 = 0.546 623

FG 0.546 IJ = 0.330 or 33.0% H 0.530 K F P V I = F PV I . In an adiabatic process, P V = PV . Also, G H T JK GH T JK F P Ib g Dividing the second equation by the first yields T = T G J HPK

Then the actual winter efficiency is

0.320

γ

P22.17

(a)

f

γ f

f

γ i i

γ

f

i i

f

i

f

Since γ =

γ −1 γ

.

i

γ −1 2 5 = = 0.400 and we have for Argon, γ 5 3

b

T f = 1 073 K (b)

i

f

× 10 Pa I gFGH 1300 J .50 × 10 Pa K 3

0. 400

6

= 564 K .

∆Eint = nCV ∆T = Q − Weng = 0 − Weng , so Weng = −nCV ∆T , and the power output is

P= =

Weng t

=

−nCV ∆T or t

b−80.0 kg ge

1.00 mol 0 .039 9 kg

jc hb8.314 J mol ⋅ K gb564 − 1 073gK 3 2

60.0 s 5

P = 2.12 × 10 W = 212 kW

P22.18

Tc 564 K =1− = 0.475 or 47.5% 1 073 K Th

(c)

eC = 1 −

(a)

emax = 1 −

(b)

P=

Weng ∆t

Tc 278 =1− = 5.12 × 10 −2 = 5.12% Th 293 = 75.0 × 10 6 J s

From e = (c)

jb

e

g

Weng = 75.0 × 10 6 J s 3 600 s h = 2.70 × 10 11 J h

Therefore, Weng Qh

we find

Qh =

Weng e

=

2.70 × 10 11 J h 5.12 × 10 −2

= 5.27 × 10 12 J h = 5.27 TJ h

As fossil-fuel prices rise, this way to use solar energy will become a good buy.

Chapter 22

*P22.19

(a)

e=

Weng1 + Weng2

e1 Q1 h + e 2 Q 2 h Q h1

=

Q h1

Now Q 2 h = Q1 c = Q1 h − Weng 1 = Q h1 − e1Q1 h . So e =

(b)

b

e 1 Q 1 h + e 2 Q 1 h − e1 Q 1 h Q1 h

e = e 1 + e 2 − e1 e 2 = 1 −

g=

e1 + e 2 − e 1 e 2 .

FG H

Ti T T +1− c − 1− i Th Ti Th

IJ FG 1 − T IJ = 2 − T KH T K T c

i

i

h

−

Tc T T T T −1+ i + c − c = 1− c Ti Th Ti Th Th

The combination of reversible engines is itself a reversible engine so it has the Carnot efficiency. (c)

With Weng2 = Weng1 , e = 1−

FG H

Tc T = 2 1− i Th Th

Weng1 + Weng2 Q1 h

IJ K

2T Tc =1− i Th Th 2Ti = Th + Tc

0−

Ti =

(d)

b

1 Th + Tc 2

e1 = e 2 = 1 −

g

Ti T =1− c Th Ti

Ti2 = Tc Th

b g

Ti = ThTc P22.20

12

The work output is Weng = We are told e = 0.200 = and

Weng Qh

b

b

5.00 m s 1 mt Qh 2

g

b

b

5.00 m s 1 Substitute Q h = m t 2 0.200 Then,

2

6.50 m s 300 K 1 = mt eC = 1 − 2 Th Qh

300 K = 0.200 1− Th

F GG H

1 2 1 2

g

g

2

.

b g m b5.00 m sg t

.

2

m t 6.50 m s

300 K = 0.338 Th 300 K Th = = 453 K 0.662

1−

g

1 2 m train 5.00 m s . 2

2 2

I JJ K

=

2Weng1 Q1 h

= 2 e1

639

640 P22.21

Heat Engines, Entropy, and the Second Law of Thermodynamics

For the Carnot engine, e c = 1 −

Tc 300 K =1− = 0.600 . 750 K Th

Weng

Also,

ec =

so

Qh =

and

Q c = Q h − Weng = 250 J − 150 J = 100 J .

Qh

.

Weng ec

=

150 J = 250 J . 0.600

FIG. P22.21 (a)

Qh =

Weng eS

=

150 J = 214 J 0.700

Q c = Q h − Weng = 214 J − 150 J = 64.3 J (b)

Q h, net = 214 J − 250 J = −35.7 J Q c, net = 64.3 J − 100 J = −35.7 J The net flow of energy by heat from the cold to the hot reservoir without work input, is impossible.

(c)

(d)

Weng

For engine S:

Q c = Q h − Weng =

so

Weng =

and

Q h = Q c + Weng = 233 J + 100 J = 333 J .

Qc 1 eS

−1

=

eS

FIG. P22.21(b)

− Weng .

100 J = 233 J . 1 0 .700 − 1

Q h, net = 333 J − 250 J = 83.3 J Wnet = 233 J − 150 J = 83.3 J Q c,net = 0 The output of 83.3 J of energy from the heat engine by work in a cyclic process without any exhaust by heat is impossible. FIG. P22.21(d)

(e)

Both engines operate in cycles, so

∆SS = ∆SCarnot = 0 .

For the reservoirs,

∆S h = −

Qh Th

Thus, ∆Stotal = ∆SS + ∆SCarnot + ∆S h + ∆S c = 0 + 0 − A decrease in total entropy is impossible.

and ∆S c = +

Qc Tc

.

83.3 J 0 + = −0.111 J K . 750 K 300 K

Chapter 22

P22.22

(a)

641

First, consider the adiabatic process D → A :

FG V IJ = 1 400 kPaFG 10.0 L IJ H 15.0 L K HV K FG nRT IJ V = FG nRT IJ V HV K HV K F V IJ = 720 K FG 10.0 IJ = T =T G H 15.0 K HV K γ

PD VDγ = PA VAγ so

PD = PA D

Also

D

53

A

γ

γ

A

D

A

A

γ −1

or

D

A

= 712 kPa .

D

23

A

549 K .

D

Now, consider the isothermal process C → D : TC = TD = 549 K .

FG V IJ = LMP FG V IJ OPFG V IJ = P V H V K MN H V K PQH V K V V 1 400 kPaa10.0 L f = = 445 kPa 24.0 La15.0 L f γ

PC = PD

D C

A

A

D

D

C

A

C

γ

A γ −1 D

53

PC

23

Next, consider the adiabatic process B → C : PBVBγ = PC VCγ . But, PC =

PA VAγ

VC VDγ −1

FG V IJ V = F P V I V which reduces to V = V V V H V K GH V V JK F V IJ = 1 400 kPaFG 10.0 L IJ = 875 kPa . Finally, P = P G H 16.0 L K HV K A

Hence, PA

B

B

A

γ B

A

C

γ

A γ −1 D

γ

B

C

A C

15.0 L

D

A B

State A B C D (b)

FG V IJ . HV K 10.0 La 24.0 L f = = 16.0 L

from above. Also considering the isothermal process, PB = PA

P(kPa) 1 400 875 445 712

For the isothermal process A → B : so Q = −W = nRT ln

V(L) 10.0 16.0 24.0 15.0

T(K) 720 720 549 549 ∆Eint = nCV ∆T = 0

FG V IJ = 2.34 molb8.314 J mol ⋅ K ga720 K f lnFG 16.0 IJ = H 10.0 K HV K B

+6.58 kJ .

A

For the adiabatic process B → C :

b

LM 3 b8.314 J mol ⋅ K gOPa549 − 720f K = N2 Q = 0 + a −4.98 kJf = −4.98 kJ .

g

∆Eint = nCV TC − TB = 2.34 mol and W = −Q + ∆Eint continued on next page

Q= 0 −4.98 kJ

A B

.

642

Heat Engines, Entropy, and the Second Law of Thermodynamics

∆Eint = nCV ∆T = 0

For the isothermal process C → D : and Q = −W = nRT ln

FG V IJ = 2.34 molb8.314 J mol ⋅ K ga549 K f lnFG 15.0 IJ = H 24.0 K HV K D

−5.02 kJ .

C

Finally, for the adiabatic process D → A :

b

g

∆Eint = nCV TA − TD = 2.34 mol

Q= 0

LM 3 b8.314 J mol ⋅ K gOPa720 − 549f K = N2 Q

+4.98 kJ

and W = −Q + ∆Eint = 0 + 4.98 kJ = +4.98 kJ . Process A→B B→C C→D D→ A ABCDA

Q(kJ) +6.58 0 –5.02 0 +1.56

W(kJ) –6.58 –4.98 +5.02 +4.98 –1.56

∆Eint (kJ) 0 –4.98 0 +4.98 0

The work done by the engine is the negative of the work input. The output work Weng is given by the work column in the table with all signs reversed. (c)

e=

Weng Qh

ec = 1 −

P22.23

aCOPf

refrig

P22.24

aCOPf

heat pump

P22.25

(a)

=

=

−W ABCD 1.56 kJ = = 0.237 or 23.7% 6.58 kJ Q A→B

Tc 549 =1− = 0.237 or 23.7% 720 Th

Tc 270 = = 9.00 ∆T 30.0 =

Qc + W W

=

Th 295 = = 11.8 ∆T 25

For a complete cycle, ∆Eint = 0 and

W = Qh − Q c = Q c

LM bQ g OP MN Q − 1PQ . h c

We have already shown that for a Carnot cycle (and only for a Carnot cycle)

Therefore,

(b)

W = Qc

LM T − T OP N T Q h

c

COP =

Qc

=

Th . Tc

.

c

We have the definition of the coefficient of performance for a refrigerator, Using the result from part (a), this becomes

Qh

Tc . Th − Tc

COP =

Qc W

.

Chapter 22

P22.26

643

COP = 0.100COPCarnot cycle Qh

or

W Qh W

FQ I F I 1 = 0.100G GH W JK H Carnot efficiency JK F T IJ = 0.100FG 293 K IJ = 1.17 = 0.100G H 293 K − 268 K K HT −T K h

= 0.100

Carnot cycle

h

h

c

FIG. P22.26 Thus, 1.17 joules of energy enter the room by heat for each joule of work done. P22.27

P22.28

Qc Tc 4.00 = = 0.013 8 = ∆T 289 W ∴ W = 72.2 J per 1 J energy removed by heat.

aCOPf

Carnot refrig

=

A Carnot refrigerator runs on minimum power. Q t Q t Q Q For it: h = c so h = c . Th Tc Th Tc Solving part (b) first:

P22.29

FG IJ b H K

IJ e K

gFGH

jFGH

IJ K

(b)

Q h Q c Th 298 K 1h = = 8.00 MJ h = 8.73 × 10 6 J h = 2. 43 kW 273 K 3 600 s t t Tc

(a)

8.00 × 10 6 J h W Qh Qc = − = 2.43 kW − = 204 W t t t 3 600 s h

e=

W = 0.350 Qh

W = 0.350Q h

Qh = W + Qc

Q c = 0.650Q h Q c 0.650Q h = = 1.86 COP refrigerator = W 0.350Q h

b

*P22.30

g

To have the same efficiencies as engines, 1 − between reservoirs with the same ratio becomes

Thp Thp − Tcp

=

Tcp Thp

Tcp Thp

=

=1−

Tcr the pump and refrigerator must operate Thr

Tcr , which we define as r. Now COPp = 1.50COPr Thr

Thp 3 Tcr 2 3r 2 3 rThr or = ,r= . , = 2 Thr − Tcr 3 Thp − rThp 2 Thr − rThr 1 − r 1 − r

(a)

COPr =

2 r = 3 1−r 1−

(b)

COPp =

1 1 = = 3.00 1 − r 1 − 32

(c)

e=1−r =1−

2 3

= 2.00

2 = 33.3% 3

644

Heat Engines, Entropy, and the Second Law of Thermodynamics

Section 22.5 P22.31

(a)

Gasoline and Diesel Engines γ γ PV i i = Pf V f

FV I F 50.0 cm I = P G J = e3.00 × 10 PajG H 300 cm JK HV K γ

Pf

(b)

i

3

6

i

z

Vi

= 244 kPa

3

f

W = PdV

1.40

P = Pi

Vi

FG V IJ HVK i

γ

Integrating,

F 1 IJ PV LM1 − FG V IJ W =G H γ − 1 K MN H V K

γ −1

i

i i

f

OP PQ = a2.50fe3.00 × 10 Paje5.00 × 10 6

−5

= 192 J P22.32

Compression ratio = 6.00 , γ = 1.40

FG V IJ HV K F 1 IJ e=1−G H 6.00 K γ −1

(a)

2

Efficiency of an Otto-engine e = 1 −

1

(b) P22.33

0 . 400

= 51. 2% .

If actual efficiency e ′ = 15.0% losses in system are e − e ′ = 36.2% .

eOtto = 1 −

1

bV V g 1

γ −1

=1−

2

1

a6.20fb7 5−1g

=1−

1

a6.20f

0. 400

eOtto = 0.518 We have assumed the fuel-air mixture to behave like a diatomic gas. Now e =

Weng Qh

=

Weng t Qh t

746 W 1 hp Q h Weng t = = 102 hp t e 0.518 Qh = 146 kW t Q h = Weng + Q c Qc t Qc t

=

Q h Weng − t t

F 746 W I = GH 1 hp JK

= 146 × 10 3 W − 102 hp

70.8 kW

L F 50.0 cm I OP m jM1 − G MN H 300 cm JK PQ 3

3

3

0. 400

Chapter 22

P22.34

(a), (b) The quantity of gas is n=

e

je

j

100 × 10 3 Pa 500 × 10 −6 m 3 PA VA = 0.020 5 mol = RTA 8.314 J mol ⋅ K 293 K

Eint, A =

b

f

ga

5 5 5 nRTA = PA VA = 100 × 10 3 Pa 500 × 10 −6 m 3 = 125 J 2 2 2

e

je

FG V IJ = e100 × 10 Paja8.00f HV K γ

In process AB, PB = PA

e

3

A

j

1.40

= 1.84 × 10 6 Pa

B

je

j

1.84 × 10 6 Pa 500 × 10 −6 m3 8.00 PBVB = 673 K = TB = nR 0.020 5 mol 8.314 J mol ⋅ K Eint, B = so

b

b

gb

g

gb

f

ga

5 5 nRTB = 0.020 5 mol 8.314 J mol ⋅ K 673 K = 287 J 2 2

∆Eint, AB = 287 J − 125 J = 162 J = Q − Wout = 0 − Wout

W AB = −162 J

Process BC takes us to:

b

gb

gb

g

0.020 5 mol 8.314 J mol ⋅ K 1 023 K nRTC = = 2.79 × 10 6 Pa VC 62.5 × 10 −6 m 3 5 5 Eint, C = nRTC = 0.020 5 mol 8.314 J mol ⋅ K 1 023 K = 436 J 2 2 Eint, BC = 436 J − 287 J = 149 J = Q − Wout = Q − 0 PC =

b

gb

gb

g

QBC = 149 J

In process CD:

FG V IJ = e2.79 × 10 PajFG 1 IJ = 1.52 × 10 Pa H 8.00 K HV K e1.52 × 10 Paje500 × 10 m j = 445 K P V = = nR b0.020 5 molgb8.314 J mol ⋅ K g 5 5 = nRT = b0.020 5 molgb8.314 J mol ⋅ K ga 445 K f = 2 2 γ

PD = PC

6

C

5

D

5

TD

1.40

−6

3

D D

Eint, D

D

190 J

∆Eint, CD = 190 J − 436 J = −246 J = Q − Wout = 0 − Wout WCD = 246 J and

∆Eint, DA = Eint, A − Eint, D = 125 J − 190 J = −65.0 J = Q − Wout = Q − 0 QDA = −65.0 J

continued on next page

645

646

Heat Engines, Entropy, and the Second Law of Thermodynamics

For the entire cycle, ∆Eint, net = 162 J + 149 − 246 − 65.0 = 0 . The net work is Weng = −162 J + 0 + 246 J + 0 = 84.3 J Q net = 0 + 149 J + 0 − 65.0 J = 84.3 J The tables look like: State A B C D A

T(K) 293 673 1 023 445 293

P(kPa) 100 1 840 2 790 152 100

V(cm3 ) 500 62.5 62.5 500 500

Process AB BC CD DA ABCDA

Q(J) 0 149 0 –65.0 84.3

output W(J) –162 0 246 0 84.3

∆Eint (J) 162 149 –246 –65.0 0

(c)

The input energy is Q h = 149 J , the waste is Q c = 65.0 J , and Weng = 84.3 J .

(d)

The efficiency is: e =

(e)

Let f represent the angular speed of the crankshaft. Then

Weng Qh

=

84.3 J = 0.565 . 149 J

obtain work in the amount of 84.3 J/cycle: 1 000 J s = f=

Section 22.6 P22.35

Eint (J) 125 287 436 190 125

FG f IJ b84.3 J cycleg H 2K

2 000 J s = 23.7 rev s = 1.42 × 10 3 rev min 84.3 J cycle

Entropy

For a freezing process, ∆S =

b

ge

f is the frequency at which we 2

j

5 ∆Q − 0.500 kg 3.33 × 10 J kg = = −610 J K . 273 K T

Chapter 22

P22.36

647

At a constant temperature of 4.20 K, 20.5 kJ kg Lv ∆Q = = T 4.20 K 4.20 K ∆S = 4.88 kJ kg ⋅ K ∆S =

FG IJ H K F 353 IJ = 46.6 cal K = ∆S = 250 g b1.00 cal g⋅° C g lnG H 293 K

z z f

P22.37

∆S =

i

*P22.38

(a)

T

f Tf dQ mcdT = = mc ln T T Ti T i

195 J K

The process is isobaric because it takes place under constant atmospheric pressure. As described by Newton’s third law, the stewing syrup must exert the same force on the air as the air exerts on it. The heating process is not adiabatic (energy goes in by heat), isothermal (T goes up), isovolumetic (it likely expands a bit), cyclic (it is different at the end), or isentropic (entropy increases). It could be made as nearly reversible as you wish, by not using a kitchen stove but a heater kept always just incrementally higher in temperature than the syrup. The process would then also be eternal, and impractical for food production.

(b)

The final temperature is 220° F = 212° F + 8° F = 100° C + 8° F

FG 100 − 0° C IJ = 104° C . H 212 − 32° F K

For the mixture,

b

ga

Q = m1 c 1 ∆T + m 2 c 2 ∆T = 900 g 1 cal g⋅° C + 930 g 0.299 cal g⋅° C 104.4° C − 23° C = 9.59 × 10 4 cal = 4.02 × 10 5 J (c)

Consider the reversible heating process described in part (a):

z zb f

f

g

T b g T T F 4.186 J IJ FG 1° C IJ lnFG 273 + 104 IJ = 900a1f + 930a0.299f bcal ° C gG H 1 cal K H 1 K K H 273 + 23 K = b 4 930 J K g0.243 = 1.20 × 10 J K

∆S =

i

dQ = T

m1 c 1 + m 2 c 2 dT

= m1 c 1 + m 2 c 2 ln

i

3

f

i

f

648 *P22.39

Heat Engines, Entropy, and the Second Law of Thermodynamics

We take data from the description of Figure 20.2 in section 20.3, and we assume a constant specific heat for each phase. As the ice is warmed from –12°C to 0°C, its entropy increases by

z f

∆S =

i

z

z

273 K

273 K

mc ice dT dQ 273 K = = mc ice T −1 dT = mc ice ln T 261 K T T 261 K 261 K

b

f

ga

b

IJ IJ gFGH FGH 273 261 K K

∆S = 0.027 0 kg 2 090 J kg ⋅° C ln 273 K − ln 261 K = 0.027 0 kg 2 090 J kg⋅° C ln ∆S = 2.54 J K As the ice melts its entropy change is

e

j

5 Q mL f 0.027 0 kg 3.33 × 10 J kg ∆S = = = = 32.9 J K T T 273 K

As liquid water warms from 273 K to 373 K,

z f

∆S =

i

FG T IJ = 0.027 0 kgb4 186 J kg⋅° Cg lnFG 373 IJ = 35.3 J K H 273 K HT K

mc liquid dT

f

= mc liquid ln

T

i

As the water boils and the steam warms, ∆S = ∆S =

FG IJ H K

Tf mL v + mc steam ln T Ti

e

0.027 0 kg 2.26 × 10 6 J kg 373 K

j + 0.027 0 kgb2 010 J kg⋅° Cg lnFG 388 IJ = 164 J K + 2.14 J K H 373 K

The total entropy change is

a2.54 + 32.9 + 35.3 + 164 + 2.14f J K =

236 J K .

We could equally well have taken the values for specific heats and latent heats from Tables 20.1 and 20.2. For steam at constant pressure, the molar specific heat in Table 21.2 implies a specific heat of 1 mol = 1 970 J kg ⋅ K , nearly agreeing with 2 010 J kg ⋅ K . 35.4 J mol ⋅ K 0.018 kg

b

Section 22.7

gFGH

I JK

Entropy Changes in Irreversible Processes

F GH

1 000 1 000 Q 2 Q1 − = − T2 T1 290 5 700

I JK

P22.40

∆S =

J K = 3.27 J K

P22.41

The car ends up in the same thermodynamic state as it started, so it undergoes zero changes in entropy. The original kinetic energy of the car is transferred by heat to the surrounding air, adding to the internal energy of the air. Its change in entropy is ∆S =

1 2

mv 2 T

=

a f

750 20.0 293

2

J K = 1.02 kJ K .

Chapter 22

P22.42

649

c iron = 448 J kg⋅° C ; c water = 4 186 J kg⋅° C

b

gd

i b

gb

gd

Qcold = −Q hot :

4.00 kg 4 186 J kg⋅° C T f − 10.0° C = − 1.00 kg 448 J kg⋅° C T f − 900° C

which yields

T f = 33.2° C = 306.2 K ∆S =

z

306.2 K 283 K

i

z

c water m water dT 306.2 K c iron m iron dT + T T 1 173 K

FG 306.2 IJ + c m lnFG 306.2 IJ H 283 K H 1 173 K ∆S = b 4 186 J kg ⋅ K gb 4.00 kg gb0.078 8g + b 448 J kg ⋅ K gb1.00 kg ga −1.34f ∆S = c water m water ln

iron

iron

∆S = 718 J K P22.43

Sitting here writing, I convert chemical energy, in ordered molecules in food, into internal energy that leaves my body by heat into the room-temperature surroundings. My rate of energy output is equal to my metabolic rate, 2 500 kcal d =

FG H

IJ K

2 500 × 10 3 cal 4.186 J = 120 W . 86 400 s 1 cal

My body is in steady state, changing little in entropy, as the environment increases in entropy at the rate ∆S Q T Q ∆t 120 W = = = = 0. 4 W K ~ 1 W K . 293 K ∆t ∆t T When using powerful appliances or an automobile, my personal contribution to entropy production is much greater than the above estimate, based only on metabolism. P22.44

b

V=

(b)

∆Eint = nCV ∆T =

(c)

W =0

(d)

∆Sargon =

b

so

ga

f

ge

j

F 40.0 gm I L 3 b8.314 J mol ⋅ K gOa−200° Cf = GH 39.9 g mol JK MN 2 PQ

i

∆S bath =

−2.50 kJ

Q = ∆Eint = −2.50 kJ

FG IJ H K F 40.0 g I L 3 b8.314 J mol ⋅ K gO lnFG 273 IJ = =G PQ H 473 K H 39.9 g mol JK MN 2

z f

(e)

gb

40.0 g 8.314 J mol ⋅ K 473 K nRTi = 39.4 × 10 −3 m3 = 39.4 L = 3 Pi 39.9 g mol 100 × 10 Pa

(a)

Tf dQ = nCV ln T Ti

−6.87 J K

2.50 kJ = +9.16 J K 273 K

The total change in entropy is ∆Stotal = ∆Sargon + ∆S bath = −6.87 J K + 9.16 J K = +2.29 J K ∆Stotal > 0 for this irreversible process.

650 P22.45

Heat Engines, Entropy, and the Second Law of Thermodynamics

∆S = nR ln

F V I = R ln 2 = GH V JK f

5.76 J K

i

There is no change in temperature . FIG. P22.45 P22.46

F V I = b0.044 0ga2fR ln 2 GH V JK ∆S = 0.088 0a8.314f ln 2 = 0.507 J K ∆S = nR ln

f

i

FIG. P22.46 P22.47

For any infinitesimal step in a process on an ideal gas, dEint = dQ + dW :

dQ = dEint − dW = nCV dT + PdV = nCV dT +

and

dQ dT dV = nCV + nR T T V

If the whole process is reversible,

∆S =

z f

i

Also, from the ideal gas law,

a

∆S = 1.00 mol

Tf Ti

=

dQr = T

z FGH f

i

nCV

IJ K

nRTdV V

FG IJ H K

Pf V f PV i i

040 0g I F 0.040 0 I + a1.00 molfb8.314 J mol ⋅ K g lnG fLMN 32 b8.314 J mol ⋅ K gOPQ lnFGH aa12..0000fbfb00..025 J 0g K H 0.025 0 JK

= 18.4 J K P22.48

F T I + nR lnF V I GH T JK GH V JK L5 O F 2P ⋅ 2V IJ + a1.00 molfb8.314 J mol ⋅ K g lnFG 2V IJ = a1.00 molfM b8.314 J mol ⋅ K gP lnG HVK N2 Q H PV K

∆S = nC V ln

FG IJ H K

Tf Vf dT dV + nR = nCV ln + nR ln T V Ti Vi

f

f

i

i

∆S = 34.6 J K

Chapter 22

Section 22.8 P22.49

P22.50

P22.51

651

Entropy on a Microscopic Scale

(a)

A 12 can only be obtained one way 6 + 6

(b)

A 7 can be obtained six ways: 6 + 1 , 5 + 2 , 4 + 3 , 3 + 4 , 2 + 5 , 1 + 6

(a)

The table is shown below. On the basis of the table, the most probable result of a toss is 2 heads and 2 tails .

(b)

The most ordered state is the least likely state. Thus, on the basis of the table this is either all heads or all tails .

(c)

The most disordered is the most likely state. Thus, this is 2 heads and 2 tails . Result All heads 3H, 1T 2H, 2T 1H, 3T All tails

Possible Combinations HHHH THHH, HTHH, HHTH, HHHT TTHH, THTH, THHT, HTTH, HTHT, HHTT HTTT, THTT, TTHT, TTTH TTTT

Total 1 4 6 4 1

(a)

Result All red 2R, 1G 1R, 2G All green

Possible Combinations RRR RRG, RGR, GRR RGG, GRG, GGR GGG

Total 1 3 3 1

(b)

Result All red 4R, 1G 3R, 2G

Possible Combinations RRRRR RRRRG, RRRGR, RRGRR, RGRRR, GRRRR RRRGG, RRGRG, RGRRG, GRRRG, RRGGR, RGRGR, GRRGR, RGGRR, GRGRR, GGRRR GGGRR, GGRGR, GRGGR, RGGGR, GGRRG, GRGRG, RGGRG, GRRGG, RGRGG, RRGGG RGGGG, GRGGG, GGRGG, GGGRG, GGGGR GGGGG

Total 1 5

2R, 3G 1R, 4G All green

10 10 5 1

Additional Problems P22.52

The conversion of gravitational potential energy into kinetic energy as the water falls is reversible. But the subsequent conversion into internal energy is not. We imagine arriving at the same final state by adding energy by heat, in amount mgy, to the water from a stove at a temperature infinitesimally above 20.0°C. Then, ∆S =

z

e

je

ja

f

3 3 2 dQ Q mgy 5 000 m 1 000 kg m 9.80 m s 50.0 m = = = = 8.36 × 10 6 J K . T T T 293 K

652 P22.53

Heat Engines, Entropy, and the Second Law of Thermodynamics

H ET so if all the electric energy is converted into internal energy, the steady-state ∆t condition of the house is described by H ET = Q .

Pelectric =

(a)

(b)

Therefore,

Pelectric =

For a heat pump,

aCOPf

Q = 5 000 W ∆t =

Carnot

Th 295 K = = 10.92 ∆T 27 K

a

f

Actual COP = 0.6 10.92 = 6.55 =

Qh W

=

Q h ∆t W ∆t

Therefore, to bring 5 000 W of energy into the house only requires input power

P22.54

Pheat pump =

W Q h ∆t 5 000 W = = = 763 W ∆t COP 6.56

f

e

Q c = mc∆T + mL + mc∆T =

b

ga

j

b

ga

Q c = 0.500 kg 4 186 J kg ⋅° C 10° C + 0.500 kg 3.33 × 10 5 J kg + 0.500 kg 2 090 J kg⋅° C 20° C Q c = 2.08 × 10 5 J Qc W W=

P22.55

b

g

= COPc refrigerator =

b

Q c Th − Tc Tc

Tc Th − Tc

g = e2.08 × 10 Jj 20.0° C − a−20.0° Cf = a273 − 20.0f K 5

∆S hot =

−1 000 J 600 K

∆Scold =

+750 J 350 K

(a)

∆SU = ∆S hot + ∆Scold = 0.476 J K

(b)

ec = 1 −

T1 = 0.417 T2

b

g

Weng = e c Q h = 0.417 1 000 J = 417 J (c)

Wnet = 417 J − 250 J = 167 J

b

g

T1 ∆SU = 350 K 0.476 J K = 167 J

32.9 kJ

f

Chapter 22

*P22.56

(a)

653

The energy put into the engine by the hot reservoir is dQ h = mcdTh . The energy put into the

LM F MN GH

a f

cold reservoir by the engine is dQ c = − mcdTc = 1 − e dQ h = 1 − 1 − −

dTc dTh = Tc Th

z

dT dT − = T T T

Tf Tc

IJ OPmcdT . Then K PQ h

z

Tf

h

− ln T ln

Tc Th

Tf Tc

T

= ln T T f

h

Tf

Tc = ln Tf Th

T f2 = Tc Th

b g

T f = ThTc (b)

d

12

i

d

Then Q h = Weng + Q c .

d i d i = mceT − T T − T T + T j = mceT − 2 T T + T j = mce T − T j FV I For an isothermal process, Q = nRT lnG J HV K Therefore, Q = nRb3T g ln 2 F 1I Q = nRbT g lnG J and H 2K 3 = nRbT − 3T g For the constant volume processes, Q = ∆E 2 3 = nRb3T − T g and Q = ∆E 2 Weng = mc Th − T f − mc T f − Tc

P22.57

(a)

h

h c

h

h

h c

c

c

c

h

2

c

2

1

1

i

3

i

2

int, 2

4

int, 4

i

i

i

i

The net energy by heat transferred is then Q = Q1 + Q 2 + Q 3 + Q 4 or (b)

i

The hot reservoir loses energy Q h = mc Th − T f . The cold reservoir gains Q c = mc T f − Tc .

FIG. P22.57

Q = 2nRTi ln 2 .

A positive value for heat represents energy transferred into the system. Therefore,

a

Q h = Q1 + Q 4 = 3nRTi 1 + ln 2

f

Since the change in temperature for the complete cycle is zero, ∆Eint = 0 and Weng = Q Therefore, the efficiency is

ec =

Weng Qh

=

2 ln 2 Q = = 0.273 Q h 3 1 + ln 2

a

f

654 P22.58

Heat Engines, Entropy, and the Second Law of Thermodynamics

(a)

Weng t

8

= 1.50 × 10 Waelectrical

LM OP = = , Q mL f MN 0.150 PQ∆t , Weng t

and L = 33.0 kJ g = 33.0 × 10 6 J kg

LM W t OP ∆t N 0.150 Q L e1.50 × 10 Wjb86 400 s dayg m= = 2 620 metric tons day 0.150e33.0 × 10 J kg je10 kg metric tonj Cost = b$8.00 metric tongb 2 618 metric tons day gb365 days yr g m=

eng

8

6

(b)

3

Cost = $7.65 million year (c)

First find the rate at which heat energy is discharged into the water. If the plant is 15.0% efficient in producing electrical energy then the rate of heat production is Qc t Then,

Qc t

=

=

FW GH t

eng

I FG 1 − 1IJ = e1.50 × 10 WjFG 1 − 1IJ = 8.50 × 10 JK H e K H 0.150 K 8

8

W.

mc∆T and t Q

c 8.50 × 10 8 J s m = t = = 4.06 × 10 4 kg s . t c∆T 4 186 J kg ⋅° C 5.00° C

b

P22.59

Weng T = ec = 1 − c = Th Qh Q h = Weng + Q c :

Weng ∆t Qh ∆t

:

Qh ∆t Qc ∆t Qc ∆t

Q c = mc∆T :

Qc ∆t

=

=

f

ga

P Th P = Th − Tc 1 − Tc Th

b

Qh ∆t

g

−

Weng ∆t

=

P Th P Tc −P = Th − Tc Th − Tc

=

FG ∆m IJ c∆T = P T H ∆t K T − T c

h

c

P Tc ∆m = Th − Tc c∆T ∆t

b

g

e

ja

f

1.00 × 10 9 W 300 K ∆m = = 5.97 × 10 4 kg s ∆t 200 K 4 186 J kg ⋅° C 6.00° C

b

ga

f

Chapter 22

P22.60

Weng T = ec = 1 − c = Th Qh

Weng ∆t Qh ∆t

Qh

P

=

=

P Th Th − Tc

e1 − j F Q I −P = P T Q =G ∆t H ∆t JK T −T ∆t

Tc Th

c

h

c

h

c

Q c = mc∆T , where c is the specific heat of water. Qc

Therefore,

∆t

∆m = ∆t

and

P22.61

(a)

=

FG ∆m IJ c∆T = P T H ∆t K T − T c

h

c

P Tc Th − Tc c∆T

b

g

a f a f 5 98.6° F = a98.6 − 32.0f° C = a37.0 + 273.15f K = 310.15 K 9 dQ dT F 310.15 IJ = 54.86 cal K ∆S =z = b 453.6 g gb1.00 cal g ⋅ K g × z = 453.6 lnG H 274.82 K T T Q a310.15 − 274.82f = −51.67 cal K ∆S =− = −a 453.6 fa1.00f 35.0° F =

5 35.0 − 32.0 ° C = 1.67 + 273.15 K = 274.82 K 9

310.15

ice water

274.82

body

Tbody

310.15

∆Ssystem = 54.86 − 51.67 = 3.19 cal K (b)

a453.6fa1fbT

F

g e

ja fb

− 274.82 = 70.0 × 10 3 1 310.15 − TF

g

Thus,

b70.0 + 0.453 6g × 10 T = a70.0fa310.15f + b0.453 6ga274.82f × 10 3

F

3

and TF = 309.92 K = 36.77° C = 98.19° F

FG 309.92 IJ = 54.52 cal K H 274.82 K F 310.15 IJ = −51.93 cal K = −e70.0 × 10 j lnG H 309.92 K

∆Sice ′ water = 453.6 ln ∆S body ′

3

∆Ssys ′ = 54.52 − 51.93 = 2.59 cal K which is less than the estimate in part (a).

655

656 P22.62

Heat Engines, Entropy, and the Second Law of Thermodynamics

(a)

For the isothermal process AB, the work on the gas is W AB = − PA VA ln

FG V IJ HV K B

A

e

j FGH 1050..00 IJK

je

W AB = −5 1.013 × 10 5 Pa 10.0 × 10 −3 m3 ln W AB = −8.15 × 10 3 J

where we have used 1.00 atm = 1.013 × 10 5 Pa 1.00 L = 1.00 × 10 −3 m 3

and

ja

e

FIG. P22.62

f

WBC = − PB ∆V = − 1.013 × 10 5 Pa 10.0 − 50.0 × 10 −3 m3 = +4.05 × 10 3 J WCA = 0 and Weng = −W AB − WBC = 4.11 × 10 3 J = 4.11 kJ (b)

Since AB is an isothermal process, ∆Eint, AB = 0 and

Q AB = −W AB = 8.15 × 10 3 J

For an ideal monatomic gas,

CV =

3R 5R and C P = 2 2

TB = TA =

e

je

j

1.013 × 10 5 50.0 × 10 −3 PBVB 5.05 × 10 3 = = nR R R

e

je

j

1.013 × 10 5 10.0 × 10 −3 PC VC 1.01 × 10 3 = = TC = nR R R

Also,

QCA = nC V ∆T = 1.00

FG 3 RIJ FG 5.05 × 10 H 2 KH

3

so the total energy absorbed by heat is Q AB + QCA = 8.15 kJ + 6.08 kJ = 14.2 kJ . (c)

QBC = nC P ∆T = QBC =

(d)

e=

a

5 1.013 × 10 5 2

Weng Qh

e

=

f

5 5 nR∆T = PB ∆VBC 2 2

j a10.0 − 50.0f × 10

Weng Q AB + QCA

=

−3

= −1.01 × 10 4 J = −10.1 kJ

4.11 × 10 3 J = 0.289 or 28.9% 1.42 × 10 4 J

I JK

− 1.01 × 10 3 = 6.08 kJ R

657

Chapter 22

*P22.63

Like a refrigerator, an air conditioner has as its purpose the removal of energy by heat from the cold reservoir. Tc 280 K = = 14.0 20 K Th − Tc

Its ideal COP is

COPCarnot =

(a)

0.400 14.0 = 5.60 =

a f

Its actual COP is

5.60

Qc Qh − Qc

=

Q c ∆t Q h ∆t − Q c ∆t

Qh Q Q − 5.60 c = c ∆t ∆t ∆t

a

f

5.60 10.0 kW = 6.60 Weng

Q Qc and c = 8.48 kW ∆t ∆t

Qh Qc − = 10.0 kW − 8.48 kW = 1.52 kW ∆t ∆t

(b)

Q h = Weng + Q c :

(c)

The air conditioner operates in a cycle, so the entropy of the working fluid does not change. The hot reservoir increases in entropy by

=

∆t

Qh Th

=

e10.0 × 10

3

jb

J s 3 600 s

300 K

g = 1.20 × 10

5

J K

The cold room decreases in entropy by ∆S = −

Qc Tc

=−

e8.48 × 10

3

jb

J s 3 600 s

280 K

g = −1.09 × 10

The net entropy change is positive, as it must be:

+1.20 × 10 5 J K − 1.09 × 10 5 J K = 1.09 × 10 4 J K (d)

COPCarnot =

We suppose the actual COP is

0.400 11.2 = 4.48

a f

As a fraction of the original 5.60, this is drop by 20.0% .

z

Vf

P22.64

(a)

W=

Vi

(b)

Tc 280 K = = 11.2 25 K Th − Tc

The new ideal COP is

PdV = nRT

z

2 Vi Vi

a f

4. 48 = 0.800 , so the fractional change is to 5.60

FG IJ H K

2Vi dV = 1.00 RT ln = RT ln 2 V Vi

The second law refers to cycles.

5

J K

658 P22.65

Heat Engines, Entropy, and the Second Law of Thermodynamics

At point A, PV i i = nRTi

and

n = 1.00 mol

At point B, 3PV i i = nRTB

so

TB = 3Ti

and

TC = 6Ti

so

TD = 2Ti

b3 P gb2V g = nRT At point D, P b 2V g = nRT

At point C,

i

i

i

C

i

D

The heat for each step in the cycle is found using C V = CP =

5R : 2

3R and 2

b g = nC b6T − 3T g = 7.50nRT = nC b 2T − 6T g = −6nRT = nC bT − 2T g = −2.50nRT

Q AB = nCV 3Ti − Ti = 3nRTi QBC QCD QDA

i

V

i

i

P

i

Qleaving = Q c = QCD + QDA = 8.50nRTi

(c)

Actual efficiency,

e=

(d)

Carnot efficiency,

ec = 1 −

i

z z

Qh − Qc Qh

= 0.190

Tc T = 1 − i = 0.833 6Ti Th

z

FG IJ H K

f f Tf nC P dT dQ T = = nC P T −1 dT = nC P ln T T f = nC P ln T f − ln Ti = nC P ln i T T Ti i i

∆S = nC P ln (a)

i

Qentering = Q h = Q AB + QBC = 10.5nRTi

(b)

∆S =

FIG. P22.65

i

i

Therefore,

i

*P22.67

i

(a)

f

*P22.66

P

F PV GH nR

f

d

i

I JK

nR = nC P ln 3 PVi

The ideal gas at constant temperature keeps constant internal energy. As it puts out energy by work in expanding it must take in an equal amount of energy by heat. Thus its entropy increases. Let Pi , Vi , Ti represent the state of the gas before the isothermal expansion. Let PC , VC , Ti represent the state after this process, so that PV i i = PC VC . Let Pi , 3Vi , T f represent the state after the adiabatic compression.

b g

γ

Then

PC VCγ = Pi 3Vi

Substituting

PC =

gives

γ −1 = Pi 3 γ Viγ PV i i VC

Then

VCγ −1 = 3 γ Viγ −1 and

continued on next page

PV i i VC

e

j VC = 3γ Vi

bγ −1g

Chapter 22

The work output in the isothermal expansion is

z

z

C

C

W = PdV = nRTi V −1 dV = nRTi ln i

i

FG V IJ = nRT lne3 HV K C

i

i

659

γ I H γ − 1 JK ln 3

b g j = nRTi FG

γ γ −1

This is also the input heat, so the entropy change is ∆S =

P22.68

IJ K

Since

C P = γ C V = CV + R

we have

bγ − 1gC

and

CP =

V

= R , CV =

R γ −1

γR γ −1

∆S = nC P ln 3

Then the result is (b)

FG H

γ Q = nR ln 3 γ −1 T

The pair of processes considered here carry the gas from the initial state in Problem 66 to the final state there. Entropy is a function of state. Entropy change does not depend on path. Therefore the entropy change in Problem 66 equals ∆Sisothermal + ∆S adiabatic in this problem. Since ∆Sadiabatic = 0, the answers to Problems 66 and 67 (a) must be the same.

Simply evaluate the maximum (Carnot) efficiency. eC =

∆T 4.00 K = = 0.014 4 277 K Th

The proposal does not merit serious consideration. P22.69

The heat transfer over the paths CD and BA is zero since they are adiabatic.

b

g

Over path BC: QBC = nC P TC − TB > 0

b

g

Over path DA: QDA = nCV TA − TD < 0

P

Adiabatic Processes B

C

Therefore, Q c = QDA and Q h = Q BC

D

The efficiency is then

bT Q bT 1 LT − T O e=1− M P γ NT −T Q e=1−

Qc

=1−

h

D

A

C

B

g − T gC

A

D

− TA C V

C

B

P

Vi

3Vi FIG. P22.69

V

660 P22.70

Heat Engines, Entropy, and the Second Law of Thermodynamics

(a)

Use the equation of state for an ideal gas nRT P 1.00 8.314 600 VA = = 1.97 × 10 −3 m 3 5 25.0 1.013 × 10

V=

VC

a fa f e j 1.00a8.314fa 400f = = 1.013 × 10 5

32.8 × 10 −3 m3 FIG. P22.70

Since AB is isothermal,

PA VA = PBVB

and since BC is adiabatic,

PBVBγ = PC VCγ

Combining these expressions, VB

LF P I V OP = MG J MNH P K V PQ C

γ C

A

A

b g

1 γ −1

LF 1.00 I e32.8 × 10 m j = MG MMH 25.0 JK 1.97 × 10 m N

OPb PP Q

LF 25.0 I e1.97 × 10 m j = MG MMH 1.00 JK 32.8 × 10 m N

OPb PP Q

3 1.40

−3

−3

3

1 0. 400

g

VB = 11.9 × 10 −3 m3

Similarly,

VD

LF P I V OP = MG J MNH P K V PQ

b g

A

γ 1 γ −1 A

C

C

or

VD = 5.44 × 10 −3 m3

Since AB is isothermal,

PA VA = PBVB

and

PB = PA

Also, CD is an isothermal and PD

−3

FG V IJ = 25.0 atmF 1.97 × 10 GH 11.9 × 10 HV K F V IJ = 1.00 atmF 32.8 × 10 =P G GH 5.44 × 10 HV K A B

C

C

3 1.40

−3

D

−3 −3 −3 −3

I JK m I J= m K

3

1 0 . 400

g

m3 = 4.14 atm m3 3

3

6.03 atm

Solving part (c) before part (b): ec = 1 −

Tc 400 K =1− = 0.333 600 K Th

(c)

For this Carnot cycle,

(b)

Energy is added by heat to the gas during the process AB. For the isothermal process, ∆Eint = 0 . and the first law gives

Q AB = −W AB = nRTh ln

FG V IJ HV K B

A

b

ga

f FGH 111.97.9 IJK = 8.97 kJ

or

Q h = Q AB = 1.00 mol 8.314 J mol ⋅ K 600 K ln

Then, from

e=

Weng Qh

a

f

the net work done per cycle is Weng = e c Q h = 0.333 8.97 kJ = 2.99 kJ .

Chapter 22

P22.71

(a)

20.0°C

(b)

∆S = mc ln

(c) (d)

Tf T1

+ mc ln

Tf T2

gLM N

b

= 1.00 kg 4.19 kJ kg ⋅ K ln

Tf T1

+ ln

661

OP = b4.19 kJ K g lnFG 293 ⋅ 293 IJ H 283 303 K T Q

Tf

2

∆S = +4.88 J K Yes . Entropy has increased.

ANSWERS TO EVEN PROBLEMS P22.2

(a) 667 J ; (b) 467 J

P22.4

(a) 30.0% ; (b) 60.0%

P22.6

55.4%

P22.8

77.8 W

P22.10

(a) 869 MJ ; (b) 330 MJ

P22.12

197 kJ

P22.14

546°C

P22.16

33.0%

P22.18

(a) 5.12%; (b) 5.27 TJ h; (c) see the solution

P22.34

(a), (b) see the solution; (c) Q h = 149 J ; Q c = 65.0 J ; Weng = 84.3 J ; (d) 56.5%; (e) 1.42 × 10 3 rev min

P22.36

4.88 kJ kg ⋅ K

P22.38

(a) isobaric; (b) 402 kJ; (c) 1.20 kJ K

P22.40

3.27 J K

P22.42

718 J K

P22.44

(a) 39.4 L ; (b) −2.50 kJ; (c) −2.50 kJ; (d) −6.87 J K ; (e) +9.16 J K

P22.46

0.507 J K

P22.48

34.6 J K

P22.50

(a) 2 heads and 2 tails ; (b) All heads or all tails; (c) 2 heads and 2 tails

P22.20

453 K

P22.22

(a), (b) see the solution; (c) 23.7%; see the solution

P22.24

11.8

P22.52

8.36 MJ K

P22.26

1.17 J

P22.54

32.9 kJ

P22.28

(a) 204 W ; (b) 2.43 kW

P22.56

see the solution

P22.30

(a) 2.00 ; (b) 3.00 ; (c) 33.3%

P22.58

(a) 2.62 × 10 3 tons d ; (b) $7.65 million yr ;

P22.32

(a) 51.2% ; (b) 36.2%

(c) 4.06 × 10 4 kg s

662 P22.60

Heat Engines, Entropy, and the Second Law of Thermodynamics

P Tc Th − Tc c∆T

b

g

P22.62

(a) 4.11 kJ ; (b) 14.2 kJ; (c) 10.1 kJ; (d) 28.9%

P22.64

see the solution

P22.66

nC P ln 3

P22.68

no; see the solution

P22.70

(a) A B C D

P, atm 25.0 4.14 1.00 6.03

(b) 2.99 kJ ; (c) 33.3%

V, L 1.97 11.9 32.8 5.44

23 Electric Fields CHAPTER OUTLINE 23.1 23.2 23.3 23.4 23.5

23.6 23.7

Properties of Electric Charges Charging Objects by Induction Coulomb’s Law The Electric Field Electric Field of a Continuous Charge Distribution Electric Field Lines Motion of Charged Particles in a Uniform Electric Field

ANSWERS TO QUESTIONS Q23.1

A neutral atom is one that has no net charge. This means that it has the same number of electrons orbiting the nucleus as it has protons in the nucleus. A negatively charged atom has one or more excess electrons.

Q23.2

When the comb is nearby, molecules in the paper are polarized, similar to the molecules in the wall in Figure 23.5a, and the paper is attracted. During contact, charge from the comb is transferred to the paper by conduction. Then the paper has the same charge as the comb, and is repelled.

Q23.3

The clothes dryer rubs dissimilar materials together as it tumbles the clothes. Electrons are transferred from one kind of molecule to another. The charges on pieces of cloth, or on nearby objects charged by induction, can produce strong electric fields that promote the ionization process in the surrounding air that is necessary for a spark to occur. Then you hear or see the sparks.

Q23.4

To avoid making a spark. Rubber-soled shoes acquire a charge by friction with the floor and could discharge with a spark, possibly causing an explosion of any flammable material in the oxygenenriched atmosphere.

Q23.5

Electrons are less massive and more mobile than protons. Also, they are more easily detached from atoms than protons.

Q23.6

The electric field due to the charged rod induces charges on near and far sides of the sphere. The attractive Coulomb force of the rod on the dissimilar charge on the close side of the sphere is larger than the repulsive Coulomb force of the rod on the like charge on the far side of the sphere. The result is a net attraction of the sphere to the rod. When the sphere touches the rod, charge is conducted between the rod and the sphere, leaving both the rod and the sphere like-charged. This results in a repulsive Coulomb force.

Q23.7

All of the constituents of air are nonpolar except for water. The polar water molecules in the air quite readily “steal” charge from a charged object, as any physics teacher trying to perform electrostatics demonstrations in the summer well knows. As a result—it is difficult to accumulate large amounts of excess charge on an object in a humid climate. During a North American winter, the cold, dry air allows accumulation of significant excess charge, giving the potential (pun intended) for a shocking (pun also intended) introduction to static electricity sparks. 1

2

Electric Fields

Q23.8

Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses) of two particles, and inversely proportional to the square of the separation distance. An electrical force exhibits the same proportionalities, with charge as the intrinsic property. Differences: The electrical force can either attract or repel, while the gravitational force as described by Newton’s law can only attract. The electrical force between elementary particles is vastly stronger than the gravitational force.

Q23.9

No. The balloon induces polarization of the molecules in the wall, so that a layer of positive charge exists near the balloon. This is just like the situation in Figure 23.5a, except that the signs of the charges are reversed. The attraction between these charges and the negative charges on the balloon is stronger than the repulsion between the negative charges on the balloon and the negative charges in the polarized molecules (because they are farther from the balloon), so that there is a net attractive force toward the wall. Ionization processes in the air surrounding the balloon provide ions to which excess electrons in the balloon can transfer, reducing the charge on the balloon and eventually causing the attractive force to be insufficient to support the weight of the balloon.

Q23.10

The electric field due to the charged rod induces a charge in the aluminum foil. If the rod is brought towards the aluminum from above, the top of the aluminum will have a negative charge induced on it, while the parts draping over the pencil can have a positive charge induced on them. These positive induced charges on the two parts give rise to a repulsive Coulomb force. If the pencil is a good insulator, the net charge on the aluminum can be zero.

Q23.11

So the electric field created by the test charge does not distort the electric field you are trying to measure, by moving the charges that create it.

Q23.12

With a very high budget, you could send first a proton and then an electron into an evacuated region in which the field exists. If the field is gravitational, both particles will experience a force in the same direction, while they will experience forces in opposite directions if the field is electric. On a more practical scale, stick identical pith balls on each end of a toothpick. Charge one pith ball + and the other –, creating a large-scale dipole. Carefully suspend this dipole about its center of mass so that it can rotate freely. When suspended in the field in question, the dipole will rotate to align itself with an electric field, while it will not for a gravitational field. If the test device does not rotate, be sure to insert it into the field in more than one orientation in case it was aligned with the electric field when you inserted it on the first trial.

Q23.13

The student standing on the insulating platform is held at the same electrical potential as the generator sphere. Charge will only flow when there is a difference in potential. The student who unwisely touches the charged sphere is near zero electrical potential when compared to the charged sphere. When the student comes in contact with the sphere, charge will flow from the sphere to him or her until they are at the same electrical potential.

Q23.14

An electric field once established by a positive or negative charge extends in all directions from the charge. Thus, it can exist in empty space if that is what surrounds the charge. There is no material at point A in Figure 23.23(a), so there is no charge, nor is there a force. There would be a force if a charge were present at point A, however. A field does exist at point A.

Q23.15

If a charge distribution is small compared to the distance of a field point from it, the charge distribution can be modeled as a single particle with charge equal to the net charge of the distribution. Further, if a charge distribution is spherically symmetric, it will create a field at exterior points just as if all of its charge were a point charge at its center.

Chapter 23

3

Q23.16

The direction of the electric field is the direction in which a positive test charge would feel a force when placed in the field. A charge will not experience two electrical forces at the same time, but the vector sum of the two. If electric field lines crossed, then a test charge placed at the point at which they cross would feel a force in two directions. Furthermore, the path that the test charge would follow if released at the point where the field lines cross would be indeterminate.

Q23.17

Both figures are drawn correctly. E1 and E 2 are the electric fields separately created by the point charges q1 and q 2 in Figure 23.14 or q and –q in Figure 23.15, respectively. The net electric field is the vector sum of E1 and E 2 , shown as E. Figure 23.21 shows only one electric field line at each point away from the charge. At the point location of an object modeled as a point charge, the direction of the field is undefined, and so is its magnitude.

Q23.18

The electric forces on the particles have the same magnitude, but are in opposite directions. The electron will have a much larger acceleration (by a factor of about 2 000) than the proton, due to its much smaller mass.

Q23.19

The electric field around a point charge approaches infinity as r approaches zero.

Q23.20

Vertically downward.

Q23.21

Four times as many electric field lines start at the surface of the larger charge as end at the smaller charge. The extra lines extend away from the pair of charges. They may never end, or they may terminate on more distant negative charges. Figure 23.24 shows the situation for charges +2q and –q.

Q23.22

At a point exactly midway between the two changes.

Q23.23

Linear charge density, λ, is charge per unit length. It is used when trying to determine the electric field created by a charged rod. Surface charge density, σ, is charge per unit area. It is used when determining the electric field above a charged sheet or disk. Volume charge density, ρ, is charge per unit volume. It is used when determining the electric field due to a uniformly charged sphere made of insulating material.

Q23.24

Yes, the path would still be parabolic. The electrical force on the electron is in the downward direction. This is similar to throwing a ball from the roof of a building horizontally or at some angle with the vertical. In both cases, the acceleration due to gravity is downward, giving a parabolic trajectory.

Q23.25

No. Life would be no different if electrons were + charged and protons were – charged. Opposite charges would still attract, and like charges would repel. The naming of + and – charge is merely a convention.

Q23.26

If the antenna were not grounded, electric charges in the atmosphere during a storm could place the antenna at a high positive or negative potential. The antenna would then place the television set inside the house at the high voltage, to make it a shock hazard. The wire to the ground keeps the antenna, the television set, and even the air around the antenna at close to zero potential.

Q23.27

People are all attracted to the Earth. If the force were electrostatic, people would all carry charge with the same sign and would repel each other. This repulsion is not observed. When we changed the charge on a person, as in the chapter-opener photograph, the person’s weight would change greatly in magnitude or direction. We could levitate an airplane simply by draining away its electric charge. The failure of such experiments gives evidence that the attraction to the Earth is not due to electrical forces.

4

Electric Fields

Q23.28

In special orientations the force between two dipoles can be zero or a force of repulsion. In general each dipole will exert a torque on the other, tending to align its axis with the field created by the first dipole. After this alignment, each dipole exerts a force of attraction on the other.

SOLUTIONS TO PROBLEMS Section 23.1 *P23.1

(a)

Properties of Electric Charges The mass of an average neutral hydrogen atom is 1.007 9u. Losing one electron reduces its mass by a negligible amount, to

e

j

1.007 9 1.660 × 10 −27 kg − 9.11 × 10 −31 kg = 1.67 × 10 −27 kg . Its charge, due to loss of one electron, is

e

j

0 − 1 −1.60 × 10 −19 C = +1.60 × 10 −19 C . (b)

By similar logic, charge = +1.60 × 10 −19 C

e

j

mass = 22.99 1.66 × 10 −27 kg − 9.11 × 10 −31 kg = 3.82 × 10 −26 kg (c)

charge of Cl − = −1.60 × 10 −19 C

e

j

mass = 35.453 1.66 × 10 −27 kg + 9.11 × 10 −31 kg = 5.89 × 10 −26 kg (d)

e

j

charge of Ca ++ = −2 −1.60 × 10 −19 C = +3.20 × 10 −19 C

e

j e

j

mass = 40.078 1.66 × 10 −27 kg − 2 9.11 × 10 −31 kg = 6.65 × 10 −26 kg (e)

e

e

mass = 14.007 1.66 × 10 −27 (f)

j kg j + 3e9.11 × 10

charge of N 3 − = 3 −1.60 × 10 −19 C = −4.80 × 10 −19 C

e

−31

j

j

kg = 2.33 × 10 −26 kg

charge of N 4 + = 4 1.60 × 10 −19 C = +6.40 × 10 −19 C

e

j e

j

mass = 14.007 1.66 × 10 −27 kg − 4 9.11 × 10 −31 kg = 2.32 × 10 −26 kg (g)

We think of a nitrogen nucleus as a seven-times ionized nitrogen atom.

e

j

charge = 7 1.60 × 10 −19 C = 1.12 × 10 −18 C

e

mass = 14.007 1.66 × 10 (h)

−27

j e

j

kg − 7 9.11 × 10 −31 kg = 2.32 × 10 −26 kg

charge = −1.60 × 10 −19 C

b

g

mass = 2 1.007 9 + 15.999 1.66 × 10 −27 kg + 9.11 × 10 −31 kg = 2.99 × 10 −26 kg

Chapter 23

P23.2

F 10.0 grams I FG 6.02 × 10 GH 107.87 grams mol JK H

(a)

N=

(b)

# electrons added =

IJ FG 47 electrons IJ = K H atom K

2.62 × 10 24

Q 1.00 × 10 −3 C = = 6.25 × 10 15 e 1.60 × 10 −19 C electron

2.38 electrons for every 10 9 already present .

or

Section 23.2

Charging Objects by Induction

Section 23.3

Coulomb’s Law

P23.3

atoms mol

23

If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person contains N≅

F 70 000 grams I FG 6.02 × 10 GH 18 grams mol JK H

molecules mol

23

IJ FG 10 protons IJ ≅ 2.3 × 10 K H molecule K

28

protons .

With an excess of 1% electrons over protons, each person has a charge

e

je j e3.7 × 10 j = e9 × 10 j

q = 0.01 1.6 × 10 −19 C 2.3 × 10 28 = 3.7 × 10 7 C .

So

F = ke

q1 q 2 r2

7 2

9

0.6 2

N = 4 × 10 25 N ~ 10 26 N .

This force is almost enough to lift a weight equal to that of the Earth:

e

j

Mg = 6 × 10 24 kg 9.8 m s 2 = 6 × 10 25 N ~ 10 26 N . *P23.4

The force on one proton is F =

e8.99 × 10 P23.5

(a)

(b)

9

F 1.6 × 10 N ⋅ m C jG H 2 × 10

Fe =

Fg =

2

k e q1 q 2 r2

r2

r2

−19

−15

e8.99 × 10 =

Gm1 m 2

k e q1 q 2 C m

9

e6.67 × 10 =

I JK

away from the other proton. Its magnitude is

2

= 57.5 N .

je

N ⋅ m 2 C 2 1.60 × 10 −19 C

e

3.80 × 10

−11

−10

j

m

2

je

j

2

= 1.59 × 10 −9 N

N ⋅ m 2 C 2 1.67 × 10 −27 kg

e3.80 × 10

−10

j

m

2

j

2

= 1.29 × 10 −45 N

The electric force is larger by 1.24 × 10 36 times . (c)

If k e q = m

q1 q 2 r

2

=G

m1 m 2 r2

brepulsiong

with q1 = q 2 = q and m1 = m 2 = m , then

6.67 × 10 −11 N ⋅ m 2 kg 2 G = = 8.61 × 10 −11 C kg . ke 8.99 × 10 9 N ⋅ m 2 C 2

5

6

Electric Fields

P23.6

We find the equal-magnitude charges on both spheres: q1 q 2

F = ke

= ke

r2

q2 r2

a

1.00 × 10 4 N = 1.05 × 10 −3 C . 8.99 × 10 9 N ⋅ m 2 C 2

f

F = 1.00 m ke

q=r

so

The number of electron transferred is then N xfer =

1.05 × 10 −3 C

= 6.59 × 10 15 electrons .

1.60 × 10 −19 C e −

The whole number of electrons in each sphere is

F 10.0 g I e6.02 × 10 GH 107.87 g mol JK

N tot =

23

je

j

atoms mol 47 e − atom = 2.62 × 10 24 e − .

The fraction transferred is then f=

P23.7

F GH

I JK

N xfer 6.59 × 10 15 = = 2.51 × 10 −9 = 2.51 charges in every billion. 24 N tot 2.62 × 10

F1 = k e

F2 = k e

e8.99 × 10 =

q1 q 2 r

2

q1 q 2 r2

=

e8.99 × 10

je

je

N ⋅ m 2 C 2 7.00 × 10 −6 C 2.00 × 10 −6 C

9

9

N ⋅ m2

j = 0.503 N

a0.500 mf C je7.00 × 10 C je 4.00 × 10 C j = 1.01 N a0.500 mf 2

−6

2

−6

2

Fx = 0.503 cos 60.0°+1.01 cos 60.0° = 0.755 N Fy = 0.503 sin 60.0°−1.01 sin 60.0° = −0.436 N

a

f a

f

F = 0.755 N i − 0.436 N j = 0.872 N at an angle of 330°

FIG. P23.7

P23.8

F = ke

P23.9

(a)

q1 q 2 r2

e8.99 × 10 =

9

je

N ⋅ m 2 C 2 1.60 × 10 −19 C

e

j

2 6.37 × 10 6 m

j e6.02 × 10 j 2

23 2

2

= 514 kN

The force is one of attraction . The distance r in Coulomb’s law is the distance between centers. The magnitude of the force is F=

(b)

k e q1 q 2 r2

e

= 8.99 × 10 9 N ⋅ m 2 C 2

12.0 × 10 C je18.0 × 10 C j = je a0.300 mf −9

−9

2

2.16 × 10 −5 N .

The net charge of −6.00 × 10 −9 C will be equally split between the two spheres, or −3.00 × 10 −9 C on each. The force is one of repulsion , and its magnitude is F=

k e q1 q 2 r

2

e

= 8.99 × 10 9 N ⋅ m 2 C 2

3.00 × 10 C je3.00 × 10 C j = je a0.300 mf −9

−9

2

8.99 × 10 −7 N .

Chapter 23

P23.10

x from the left end of the rod. This bead

Let the third bead have charge Q and be located distance will experience a net force given by F=

b g

k e 3q Q x

2

i+

7

b g e− i j . ad − x f ke q Q

2

The net force will be zero if

3 1 = 2 x d−x

a

f

2

, or d − x =

x 3

.

This gives an equilibrium position of the third bead of x = 0.634d . The equilibrium is stable if the third bead has positive charge .

P23.11

kee2

e

(a)

F=

(b)

We have F =

r2

= 8.99 × 10 N ⋅ m

2

2

−10

C

j

m

2

j

2

= 8.22 × 10 −8 N

mv 2 from which r

v=

P23.12

e1.60 × 10 C j e0.529 × 10

−19

9

e

j=

8.22 × 10 −8 N 0.529 × 10 −10 m

Fr = m

9.11 × 10

−31

The top charge exerts a force on the negative charge

kg k e qQ d 2 + 2 x 2

ch

2.19 × 10 6 m s .

which is directed upward and to the

FG d IJ to the x-axis. The two positive charges together exert force H 2x K F 2 k qQ I FG a− xfi IJ −2 k qQ d GG JJ G = ma or for x 1 , the components perpendicular to the x-axis add to zero. The total field is

(b)

P23.24

E=∑

b g cosθ =

nk e Q n i 2

R +x

2

e

k eQx i 2

R + x2

j

3 2

.

FIG. P23.23

A circle of charge corresponds to letting n grow beyond all bounds, but the result does not depend on n. Smearing the charge around the circle does not change its amount or its distance from the field point, so it does not change the field . keq r2

r=

keq a2

e− ij + a2kaqf e− ij + a3kaqf e− ij + … = − ka qi FGH1 + 21 + 31 + …IJK = e

2

e

2

e 2

2

2

−

π 2ke q i 6a2

Chapter 23

Section 23.5 P23.25

Electric Field of a Continuous Charge Distribution

e b g a f a f a f a

je

j

8.99 × 10 9 22.0 × 10 −6 ke Q keλ k eQ = = = E= 0.290 0.140 + 0.290 d +d d +d d +d

fa

f

E = 1.59 × 10 6 N C , directed toward the rod.

P23.26

E=

z

k e dq x2

E = keλ 0

z

∞ x0

P23.27

P23.28

P23.29

E=

ex

, where dq = λ 0 dx

FG 1 IJ H xK

dx x

= x0

keλ 0 x0

The direction is − i or left for λ 0 > 0

e8.99 × 10 je75.0 × 10 jx = 6.74 × 10 x ex + 0.100 j ex + 0.010 0j −6

9

j

+ a2

∞

= keλ 0 −

2

k e xQ 2

FIG. P23.25

3 2

=

5

2 32

2

32

2

(a)

At x = 0.010 0 m ,

E = 6.64 × 10 6 i N C = 6.64i MN C

(b)

At x = 0.050 0 m ,

E = 2.41 × 10 7 i N C = 24.1i MN C

(c)

At x = 0.300 m ,

E = 6.40 × 10 6 i N C = 6. 40 i MN C

(d)

At x = 1.00 m ,

E = 6.64 × 10 5 i N C = 0.664i MN C

z

E = dE =

E=

LM k λ x dxe− ij OP z MMN x PPQ = −k λ x i z x

∞

e

ex

k e Qx + a2

e

3

x0

2

∞

0 0

j

0 0

−3

F GH

dx = − k e λ 0 x 0 i −

x0

3 2

For a maximum,

dE = Qk e dx

LM MM ex N

a

x 2 + a 2 − 3 x 2 = 0 or x =

2

1 2

+ a2

j

3 2

−

OP =0 P + x a e j PQ 3x 2

2 52

2

.

Substituting into the expression for E gives E=

k eQa

2

e aj 3 2

2 32

=

k eQ 3

3 2

a

2

=

2 k eQ 3 3a

2

=

Q 6 3π ∈0 a 2

.

∞

1 2x

2

x0

I= JK

keλ 0 −i 2 x0

e j

11

12

Electric Fields

P23.30

F GH

e

E = 2π 8.99 × 10

P23.31

x

E = 2π k eσ 1 −

2

x +R 9

2

I JK

je7.90 × 10

−3

F jGG 1 − H

I JJ = 4.46 × 10 FG 1 − H + a0.350f K x

x2

2

(a)

At x = 0.050 0 m ,

E = 3.83 × 10 8 N C = 383 MN C

(b)

At x = 0.100 m ,

E = 3.24 × 10 8 N C = 324 MN C

(c)

At x = 0.500 m ,

E = 8.07 × 10 7 N C = 80.7 MN C

(d)

At x = 2.00 m ,

E = 6.68 × 10 8 N C = 6.68 MN C

(a)

From Example 23.9: E = 2π k eσ 1 −

F GH

σ=

I J + 0.123 K x

8

x2

I JK

x x2 + R2

Q = 1.84 × 10 −3 C m 2 πR 2

ja

e

f

E = 1.04 × 10 8 N C 0.900 = 9.36 × 10 7 N C = 93.6 MN C

b

appx: E = 2π k eσ = 104 MN C about 11% high (b)

F jGH

e

E = 1.04 × 10 8 N C 1 −

I = 1.04 × 10 J e cm K

30.0 cm

jb

g

8 N C 0.004 96 = 0.516 MN C 30.0 + 3.00 Q 5.20 × 10 −6 appx: E = k e 2 = 8.99 × 10 9 = 0.519 MN C about 0.6% high 2 r 0.30

e

P23.32

g

2

2

b

j a f

LM MN L = 2π k σ M1 − MN

The electric field at a distance x is

Ex = 2π k eσ 1 −

This is equivalent to

Ex

For large x,

R2 x2

x2 + R2

OP PQ

1

e

1+

> R ,

1 x2 + R2 2

Ex

≈

1 x2

, so

Ex ≈

e

2

2

e

2

k eQ x2

2

e

2

e

2

2

for a disk at large distances

2

2

2

Chapter 23

P23.33

z

z

Due to symmetry

Ey = dEy = 0 , and Ex = dE sin θ = k e

where

dq = λds = λrdθ ,

so that,

Ex =

z

keλ π k λ sin θdθ = e − cos θ r 0 r

a

f

π

=

0

z

dq sin θ r2

2k e λ r

q L and r = . π L 2 8.99 × 10 9 N ⋅ m 2 C 2 7.50 × 10 −6 C π 2 k e qπ = Ex = . 2 L2 0.140 m

FIG. P23.33

λ=

where

e

Thus,

je f

a

13

j

7

Ex = 2.16 × 10 N C .

Solving,

e

j

Since the rod has a negative charge, E = −2.16 × 10 7 i N C = −21.6 i MN C . P23.34

(a)

We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has Qdx and produces, at the chosen point, a field charge h kex Qdx dE = i. 3 2 h x2 + R2

e

j

The total field is

z z

E=

dE =

d+h

e k Qi ex + R j E= 2h b− 1 2g d

all charge

k eQxdx

h x2 + R

j

2 3 2

d+h 2 −1 2

2

e

(b)

i=

ze

k eQi d + h 2 x + R2 2h x=d

k Qi = e h

x=d

LM MM d Ne

j

−3 2

2 xdx

OP − P +R j ead + hf + R j PQ 1

2

1

2 12

2

2

12

Think of the cylinder as a stack of disks, each with thickness dx, charge per-area σ =

dE =

So,

Qdx . One disk produces a field π R2h

2π k eQdx

π R2h

F GG 1 − H ex

I Ji . + R j JK F 2 k Qdx G 1− R h G H ex x

2

2 12

I JJ i E= z dE = z +R j K L O 2 k Qi M 2 k Qi L 1 E= M z dx − 2 z ex + R j 2xdxPP = R h Mx R h MN MN Q 2 k Qi L O d + h − d − ead + hf + R j + ed + R j P E= M R h N Q 2 k Qi L O h + ed + R j − ead + hf + R j P E= R h MN Q d+h

all charge

e 2

x =d

e

2

d+h

d+h

d

x=d

2

2

2 12

x

2 12

2

2 −1 2

2

e 2

e 2

Qdx , and chargeh

2

e 2

12

2

2 12

2

2

12

e

2 2 1 x +R d+h − d 2 12

j

1 2 d+h

d

OP PP Q

14

Electric Fields

P23.35

(a)

The electric field at point P due to each element of length dx, is k dq dE = 2 e 2 and is directed along the line joining the element to x +y point P. By symmetry, Ex = dEx = 0

and since

dq = λdx ,

E = Ey = dEy = dE cos θ

where

cos θ =

z

z

z

ze 2

Therefore,

E = 2 k e λy

0

P23.36

dx 2

x +y

j

2 32

y 2

x + y2

. FIG. P23.35

2 k e λ sin θ 0 . y

=

θ 0 = 90°

Ey =

2k e λ . y

(b)

For a bar of infinite length,

(a)

The whole surface area of the cylinder is A = 2π r 2 + 2π rL = 2π r r + L .

j b

e

and

a f

g

Q = σA = 15.0 × 10 −9 C m 2 2π 0.025 0 m 0.025 0 m + 0.060 0 m = 2.00 × 10 −10 C (b)

For the curved lateral surface only, A = 2π rL .

j b

e

f

ga

Q = σA = 15.0 × 10 −9 C m 2 2π 0.025 0 m 0.060 0 m = 1.41 × 10 −10 C

P23.37

j b

e

g b0.060 0 mg = 2

(c)

Q = ρV = ρπ r 2 L = 500 × 10 −9 C m 3 π 0.025 0 m

(a)

Every object has the same volume, V = 8 0.030 0 m

e

je

a

f

3

5.89 × 10 −11 C

= 2.16 × 10 −4 m3 .

j

For each, Q = ρV = 400 × 10 −9 C m 3 2.16 × 10 −4 m3 = 8.64 × 10 −11 C (b)

We must count the 9.00 cm 2 squares painted with charge: (i)

6 × 4 = 24 squares

e

j e

j

j e

j

j e

j

j e

j

Q = σA = 15.0 × 10 −9 C m 2 24.0 9.00 × 10 −4 m 2 = 3.24 × 10 −10 C (ii)

34 squares exposed

e

Q = σA = 15.0 × 10 −9 C m 2 34.0 9.00 × 10 −4 m 2 = 4.59 × 10 −10 C (iii)

34 squares

e

Q = σA = 15.0 × 10 −9 C m 2 34.0 9.00 × 10 −4 m 2 = 4.59 × 10 −10 C (iv)

32 squares

e

Q = σA = 15.0 × 10 −9 C m 2 32.0 9.00 × 10 −4 m 2 = 4.32 × 10 −10 C (c)

(i)

total edge length:

e = e80.0 × 10

Q = λ = 80.0 × 10 −12 (ii)

Q=λ

continued on next page

b g C mj24 × b0.030 0 mg = C mj44 × b0.030 0 mg =

= 24 × 0.030 0 m

−12

5.76 × 10 −11 C 1.06 × 10 −10 C

Chapter 23

Section 23.6

e

j b

g

e

j b

g

(iii)

Q = λ = 80.0 × 10 −12 C m 64 × 0.030 0 m = 1.54 × 10 −10 C

(iv)

Q = λ = 80.0 × 10 −12 C m 40 × 0.030 0 m = 0.960 × 10 −10 C

Electric Field Lines P23.39

P23.38

FIG. P23.38

P23.40

(a)

(b) P23.41

(a)

FIG. P23.39

q1 −6 1 = = − 3 q 2 18

q1 is negative, q 2 is positive The electric field has the general appearance shown. It is zero at the center , where (by symmetry) one can see that the three charges individually produce fields that cancel out. In addition to the center of the triangle, the electric field lines in the second figure to the right indicate three other points near the middle of each leg of the triangle where E = 0 , but they are more difficult to find mathematically.

(b)

You may need to review vector addition in Chapter Three. The electric field at point P can be found by adding the electric field vectors due to each of the two lower point charges: E = E 1 + E 2 . The electric field from a point charge is E = k e

q r2

r.

As shown in the solution figure at right, E1 = k e E2 = ke

q a2 q a2

to the right and upward at 60° to the left and upward at 60°

E = E1 + E 2 = k e = 1.73 k e

q a2

j

q a

2

ecos 60° i + sin 60° jj + e− cos 60° i + sin 60° jj = k

FIG. P23.41 e

q a2

e

j

2 sin 60° j

15

16

Electric Fields

Section 23.7 P23.42

Motion of Charged Particles in a Uniform Electric Field qE m

F = qE = ma

a=

v f = vi + at

vf =

qEt m

e1.602 × 10 ja520fe48.0 × 10 j = =

4.39 × 10 6 m s

e1.602 × 10 ja520fe48.0 × 10 j =

2.39 × 10 3 m s

−19

ve

electron:

−9

−31

9.11 × 10 in a direction opposite to the field −19

vp =

proton:

P23.43

P23.44

−9

1.67 × 10 −27 in the same direction as the field

a f

qE 1.602 × 10 −19 640 = = 6.14 × 10 10 m s 2 m 1.67 × 10 −27

(a)

a=

(b)

v f = vi + at

(c)

x f − xi =

(d)

K=

(a)

−19 6.00 × 10 5 qE 1.602 × 10 a= = = 5.76 × 10 13 m s so a = −5.76 × 10 13 i m s 2 m 1.67 × 10 −27

(b)

v f = vi + 2 a x f − xi

e

1 vi + v f t 2

d

i

xf =

e

e

je

je

e

d

e

je

j

2

= 1.20 × 10 −15 J

j

j

jb

g

v i = 2.84 × 10 6 i m s

v f = vi + at

e

j

0 = 2.84 × 10 6 + −5.76 × 10 13 t

t = 4.93 × 10 −8 s

The required electric field will be in the direction of motion . Work done = ∆K so,

− Fd = −

which becomes

eEd = K

and

E=

j

i

e

P23.45

t = 1.95 × 10 −5 s

1 1.20 × 10 6 1.95 × 10 −5 = 11.7 m 2

1 1 mv 2 = 1.67 × 10 −27 kg 1.20 × 10 6 m s 2 2

0 = vi2 + 2 −5.76 × 10 13 0.070 0 (c)

j

1.20 × 10 6 = 6.14 × 10 10 t

1 mvi2 (since the final velocity = 0 ) 2

K . ed

Chapter 23

P23.46

17

The acceleration is given by

d

a f

i

v 2f = vi2 + 2 a x f − x i or

v 2f = 0 + 2 a − h .

Solving

a=−

Now

∑ F = ma :

Therefore

qE = −

(a)

v 2f 2h

F GH

.

mv 2f 2h

− mg j + qE = −

mv 2f j

I JK

2h

.

+ mg j .

Gravity alone would give the bead downward impact velocity

ja

e

f

2 9.80 m s 2 5.00 m = 9.90 m s . To change this to 21.0 m/s down, a downward electric field must exert a downward electric force.

P23.47

F GH

I JK

(b)

(a)

t=

(b)

ay =

(c)

I LM b21.0 m sg JK M 2a5.00 mf N

OP PQ

2

− 9.80 m s 2 = 3.43 µC

0.050 0 x = = 1.11 × 10 −7 s = 111 ns v x 4.50 × 10 5

e

je

j

−19 9.60 × 10 3 qE 1.602 × 10 = = 9.21 × 10 11 m s 2 −27 m 1.67 × 10

e

y f − yi = v yi t +

*P23.48

F GH

2 1.00 × 10 −3 kg N ⋅ s 2 m vf −g = q= E 2h 1.00 × 10 4 N C kg ⋅ m

j

1 ayt 2 : 2

yf =

1 9.21 × 10 11 1.11 × 10 −7 2

e

je

e

j

2

= 5.68 × 10 −3 m = 5.68 mm

je

j

v yf = v yi + a y t = 9.21 × 10 11 1.11 × 10 −7 = 1.02 × 10 5 m s

v x = 4.50 × 10 5 m s

jb

ge j e j ∑ F = e 2 × 10 kg ⋅ m s je− jj = 1 × 10 m s − j . a= and moves with acceleration: e je j m 2 × 10 kg Its x-component of velocity is constant at e1.00 × 10 m sj cos 37° = 7.99 × 10 m s . Thus it moves in a e

The particle feels a constant force: F = qE = 1 × 10 −6 C 2 000 N C − j = 2 × 10 −3 N − j −3

2

13

−16

5

2

4

parabola opening downward. The maximum height it attains above the bottom plate is described by

d

i

2 2 = v yi + 2 a y y f − yi : v yf

e

0 = 6.02 × 10 4 m s

y f = 1.81 × 10 −4 m . continued on next page

j − e2 × 10 2

13

jd

m s2 y f − 0

i

18

Electric Fields

Since this is less than 10 mm, the particle does not strike the top plate, but moves in a symmetric parabola and strikes the bottom plate after a time given by y f = yi + v yi t +

1 ayt 2 2

e

j

0 = 0 + 6.02 × 10 4 m s t +

1 −1 × 10 13 m s 2 t 2 2

e

j

since t > 0 ,

t = 1.20 × 10 −8 s .

The particle’s range is

x f = xi + v x t = 0 + 7.99 × 10 4 m s 1.20 × 10 −8 s = 9.61 × 10 −4 m .

e

je

j

In sum, The particle strikes the negative plate after moving in a parabola with a height of 0.181 mm and a width of 0.961 mm. P23.49

vi = 9.55 × 10 3 m s ay =

(a)

e

^

ja f j

−19 720 eE 1.60 × 10 = = 6.90 × 10 10 m s 2 − 27 m 1.67 × 10

e

vi2 sin 2θ = 1. 27 × 10 −3 m so that ay

R=

e9.55 × 10 j

3 2

6.90 × 10

sin 2θ 10

sin 2θ = 0.961 t=

(b)

FIG. P23.49

= 1.27 × 10 −3

θ = 36.9°

R R = vix vi cos θ

90.0°−θ = 53.1°

If θ = 36.9° , t = 167 ns .

If θ = 53.1° , t = 221 ns .

Additional Problems *P23.50

The two given charges exert equal-size forces of attraction on each other. If a third charge, positive or negative, were placed between them they could not be in equilibrium. If the third charge were at a point x > 15 cm , it would exert a stronger force on the 45 µ C than on the −12 µ C , and could not produce equilibrium for both. Thus the third charge must be at x = − d < 0 . Its equilibrium requires

b

k e q 12 µ C d

2

g = k qb45 µ Cg a15 cm + df e

2

15 cm + d = 1.94d

FG 15 cm + d IJ H d K

2

=

d q

x=0 – –12 µC

15 cm x + 45 µ C

FIG. P23.50

45 = 3.75 12

d = 16.0 cm .

The third charge is at x = −16.0 cm . The equilibrium of the −12 µ C requires

b

g = k b45 µ Cg12 µ C a16.0 cmf a15 cmf

k e q 12 µ C

2

e

2

q = 51.3 µ C .

All six individual forces are now equal in magnitude, so we have equilibrium as required, and this is the only solution.

Chapter 23

P23.51

jb

e

while the e − has acceleration

ae =

(a)

e1.60 × 10

jb

C 640 N C kg

FG H

IJ K

1 1 1 a p t 2 + a e t 2 = 1 837 a p t 2 . 2 2 2 1 4.00 cm 2 = 21.8 µm . d = apt = 2 1 837

g = 1.12 × 10

14

m s 2 = 1 836 a p .

1 a p t 2 ), knowing: 2

The distance from the positive plate to where the meeting occurs equals the distance the 1 sodium ion travels (i.e., d Na = a Na t 2 ). This is found from: 2 eE eE 1 1 1 1 2 2 t2 + t2 . 4.00 cm = a Na t + aCl t : 4.00 cm = 2 2 2 22.99 u 2 35. 45 u 1 1 1 This may be written as 4.00 cm = a Na t 2 + 0.649 a Na t 2 = 1.65 a Na t 2 2 2 2 1 4.00 cm 2 = 2.43 cm . so d Na = a Na t = 2 1.65

FG H

(a)

−31

We want to find the distance traveled by the proton (i.e., d =

Thus, (b)

−19

9.110 × 10

4.00 cm =

P23.52

g

The proton moves with acceleration

−19 C 640 N C qE 1.60 × 10 = = 6.13 × 10 10 m s 2 ap = m 1.673 × 10 −27 kg

IJ K b

FG H

IJ K

g

FG H

IJ K

The field, E1 , due to the 4.00 × 10 −9 C charge is in the –x direction. 8.99 × 10 9 N ⋅ m 2 C 2 −4.00 × 10 −9 C keq E1 = 2 r = i 2 r 2.50 m

e

a

je f

j

FIG. P23.52(a)

= −5.75 i N C

Likewise, E 2 and E3 , due to the 5.00 × 10 −9 C charge and the 3.00 × 10 −9 C charge are E2 = E3

ke q r

2

e8.99 × 10 r=

e8.99 × 10 =

9

je j i = 11.2 N C i a2.00 mf N ⋅ m C je3.00 × 10 C j i = 18.7 N C i a1.20 mf 9

N ⋅ m 2 C 2 5.00 × 10 −9 C 2

2

−9

2

2

E R = E1 + E 2 + E 3 = 24.2 N C in +x direction. (b)

E2 E3

ke q

b ge j = r = b11. 2 N C ge + jj r k q = r = b5.81 N C ge −0.371i +0.928 jj r

E1 =

r2 ke q

r = −8.46 N C 0.243 i + 0.970 j

2

e 2

Ex = E1 x + E3 x = −4.21i N C

ER = 9.42 N C

Ey = E1 y + E2 y + E3 y = 8.43 j N C

θ = 63.4° above − x axis

FIG. P23.52(b)

19

20

Electric Fields

*P23.53

(a)

Each ion moves in a quarter circle. The electric force causes the centripetal acceleration.

∑ F = ma (b)

qE =

mv 2 R

E=

d

For the x-motion,

2 v xf2 = v xi + 2 a x x f − xi

0 = v 2 + 2ax R

ax = −

Ex = −

mv 2 qR

i

v 2 Fx qE x = = m 2R m

mv 2 . Similarly for the y-motion, 2 qR

v 2 = 0 + 2ay R

ay = +

v 2 qE y = m 2R

Ey =

mv 2 2 qR

The magnitude of the field is E x2 + E y2 = P23.54

at 135° counterclockwise from the x -axis .

2 qR

From the free-body diagram shown,

∑ Fy = 0 :

T cos 15.0° = 1.96 × 10 −2 N .

So

T = 2.03 × 10 −2 N .

From or P23.55

mv 2

(a)

∑ Fx = 0 , we have q=

qE = T sin 15.0°

e

j

2.03 × 10 −2 N sin 15.0° T sin 15.0° = = 5.25 × 10 −6 C = 5.25 µC . E 1.00 × 10 3 N C

FIG. P23.54

Let us sum force components to find

∑ Fx = qEx − T sin θ = 0 , and ∑ Fy = qEy + T cos θ − mg = 0 . Combining these two equations, we get q=

e1.00 × 10 ja9.80f eE cot θ + E j a3.00 cot 37.0°+5.00f × 10 −3

mg

x

=

y

5

= 1.09 × 10 −8 C

= 10.9 nC (b)

From the two equations for T=

Free Body Diagram FIG. P23.55

∑ Fx

and

∑ Fy

qEx = 5.44 × 10 −3 N = 5.44 mN . sin 37.0°

we also find

Chapter 23

P23.56

21

This is the general version of the preceding problem. The known quantities are A, B, m, g, and θ. The unknowns are q and T. The approach to this problem should be the same as for the last problem, but without numbers to substitute for the variables. Likewise, we can use the free body diagram given in the solution to problem 55. Again, Newton’s second law: and qA , into Eq. (2), sin θ

Substituting T =

(a)

(b)

∑ Fx = −T sin θ + qA = 0

(1)

∑ Fy = +T cos θ + qB − mg = 0

(2)

qA cos θ + qB = mg . sin θ mg A cot θ + B

Isolating q on the left,

q=

a

Substituting this value into Eq. (1),

T=

a A cosθ + B sinθ f

mgA

f

.

.

If we had solved this general problem first, we would only need to substitute the appropriate values in the equations for q and T to find the numerical results needed for problem 55. If you find this problem more difficult than problem 55, the little list at the first step is useful. It shows what symbols to think of as known data, and what to consider unknown. The list is a guide for deciding what to solve for in the analysis step, and for recognizing when we have an answer. P23.57

F=

k e q1 q 2 r

2

tan θ =

:

15.0 60.0

θ = 14.0°

e8.99 × 10 je10.0 × 10 j F = a0.150f e8.99 × 10 je10.0 × 10 j F = a0.600f e8.99 × 10 je10.0 × 10 j F = a0.619f

−6 2

9

1

2

−6 2

9

3

2

= 2.50 N

−6 2

9

2

= 40.0 N

2

= 2.35 N

Fx = − F3 − F2 cos 14.0° = −2.50 − 2.35 cos 14.0° = −4.78 N Fy = − F1 − F2 sin 14.0° = −40.0 − 2.35 sin 14.0° = −40.6 N Fnet = Fx2 + Fy2 = tan φ =

Fy Fx

φ = 263°

=

−40.6 −4.78

a4.78f + a40.6f 2

2

= 40.9 N

FIG. P23.57

22

Electric Fields

P23.58

d cos 30.0° = 15.0 cm, 15.0 cm d= cos 30.0°

From Figure A: or

θ = sin −1

From Figure B:

θ = sin −1

FG d IJ H 50.0 cm K F 15.0 cm I = 20.3° GH 50.0 cmacos 30.0°f JK

Figure A

Fq

or

= tan θ mg Fq = mg tan 20.3°

From Figure C:

Fq = 2 F cos 30.0°

(1)

LM k q OP cos 30.0° MN a0.300 mf PQ Combining equations (1) and (2), L k q OP cos 30.0° = mg tan 20.3° 2M MN a0.300 mf PQ mg a0.300 mf tan 20.3° q = 2

e

Fq = 2

e

2

(2) Figure B

2

2

2

2

2 k e cos 30.0°

e2.00 × 10 kg je9.80 m s ja0.300 mf tan 20.3° = 2e8.99 × 10 N ⋅ m C j cos 30.0° −3

q

2

2

2

9

2

2

Figure C

q = 4.20 × 10 −14 C 2 = 2.05 × 10 −7 C = 0.205 µC P23.59

Charge

FIG. P23.58

Q resides on each block, which repel as point charges: 2

F= Q=

Solving for Q, *P23.60

P23.61

b gb g = kbL − L g . L k bL − L g . 2L

ke Q 2 Q 2

i

2

i

ke

If we place one more charge q at the 29th vertex, the total force on the central charge will add up to k qQ k e qQ toward vertex 29 . F28 charges = zero: F28 charges + e 2 away from vertex 29 = 0 a a2 According to the result of Example 23.7, the left-hand rod creates this field at a distance d from its right-hand end: k eQ E= d 2a + d

a

dF = F= F=

f

k eQQ dx 2a d d + 2a

a

k eQ 2a

2

+ k eQ 4a

2

FIG. P23.61

f

IJ z a f FGH K F k Q I lnF FG − ln 2a + b + ln b IJ = k Q ln b = G H b K b − 2a ab − 2 afab + 2af H 4a JK GH b 4a b

k Q2 dx 1 2a + x = e − ln x x + 2a x 2a 2a x=b− 2a 2

e

2

2

b

b−2a

2

e

2

2

2

b2 − 4a 2

I JK

23

Chapter 23

P23.62

b

g

At equilibrium, the distance between the charges is r = 2 0.100 m sin 10.0° = 3.47 × 10 −2 m Now consider the forces on the sphere with charge +q , and use

∑ Fy = 0 : ∑ Fx = 0 :

∑ Fy = 0 :

mg cos 10.0° = F2 − F1 = T sin 10.0°

T cos 10.0° = mg , or T =

(1)

Fnet

(2)

θθ L –q

+q

r

Fnet is the net electrical force on the charged sphere. Eliminate T from (2) by use of (1). mg sin 10.0° Fnet = = mg tan 10.0° = 2.00 × 10 −3 kg 9.80 m s 2 tan 10.0° = 3.46 × 10 −3 N cos 10.0°

e

je

j

Fnet is the resultant of two forces, F1 and F2 . F1 is the attractive force on +q exerted by −q , and F2 is the force exerted on +q by the external electric field.

FIG. P23.62

Fnet = F2 − F1 or F2 = Fnet + F1

e5.00 × 10 Cje5.00 × 10 Cj = 1.87 × 10 C j e3.47 × 10 mj −8

e

9

F1 = 8.99 × 10 N ⋅ m

2

−8

2

−3

2

−2

N

Thus, F2 = Fnet + F1 yields F2 = 3.46 × 10 −3 N + 1.87 × 10 −2 N = 2.21 × 10 −2 N and F2 = qE , or E =

P23.63

z

Q = λd =

z

F2 2. 21 × 10 −2 N = = 4. 43 × 10 5 N C = 443 kN C . q 5.00 × 10 −8 C

90 .0 °

λ 0 cos θRdθ = λ 0 R sin θ

−90 .0 °

90 .0 ° −90.0 °

a f

= λ 0 R 1 − −1 = 2 λ 0 R

λ = 10.0 µC m so b ga f F b3.00 µCgeλ cos θRdθ j I 1 1 F b3.00 µC gbλd g I cos θ = dF = G JJ G J 4π ∈ G 4π ∈ H R R K H K e3.00 × 10 Cje10.0 × 10 C mj cos θdθ F = z e8.99 × 10 N ⋅ m C j a0.600 mf 8.99a30.0f F = e10 Nj z FGH 12 + 12 cos 2θ IJK dθ 0.600 F1 1 I = 0.707 N Downward. F = a0.450 N fG θ + sin 2θ JK H2 4

Q = 12.0 µC = 2 λ 0 0.600 m = 12.0 µC

0

2

0

y

2

0

90.0 °

y

2

0

−6

−6

9

2

2

2

−90 .0 °

y

−3

−π 2

−π 2

cosθ

0 –1

π 2

π 2

y

1

1 0

0°

360°

cos2θ 0°

360°

FIG. P23.63

Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0 . P23.64

At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can be located by determining the angle θ corresponding to equilibrium. In terms of lengths s, attractive force

1 a 3 , and r, shown in Figure P23.64, the charge at the origin exerts an 2

es + continued on next page

k eQq 1 2

a 3

j

2

24

Electric Fields

The other two charges exert equal repulsive forces of magnitude of the two repulsive forces add, balancing the attractive force, Fnet = k eQq 1 2

k eQq r2

. The horizontal components

LM 2 cosθ MM r − es + N 2

a

1 2

r=

The equilibrium condition, in terms of θ, is

Fnet =

Thus the equilibrium value of θ satisfies

2 cos θ sin 2 θ

sin θ

2

a

s=

From Figure P23.64

OP P=0 3j P Q

1

1 a cot θ 2

FG 4 IJ k QqFG 2 cos θ sin θ − H a K GH e 2

2

e

e

3 + cot θ

j

2

I JJ = 0 . 3 + cot θ j K 1

2

=1.

One method for solving for θ is to tabulate the left side. To three significant figures a value of θ corresponding to equilibrium is 81.7°. The distance from the vertical side of the triangle to the equilibrium position is 1 s = a cot 81.7° = 0.072 9 a . 2

θ

2 cos θ sin 2 θ

60°

FIG. P23.64

e

3 + cot θ

j

2

4

70°

2.654

80°

1.226

90°

0

81°

1.091

81.5°

1.024

81.7°

0.997

A second zero-field point is on the negative side of the x-axis, where θ = −9.16° and s = −3.10 a . P23.65

(a)

(b)

From the 2Q charge we have

Fe − T2 sin θ 2 = 0 and mg − T2 cos θ 2 = 0 .

Combining these we find

Fe T sin θ 2 = 2 = tan θ 2 . mg T2 cos θ 2

From the Q charge we have

Fe = T1 sin θ 1 = 0 and mg − T1 cos θ 1 = 0 .

Combining these we find

Fe T sin θ 1 = 1 = tan θ 1 or θ 2 = θ 1 . mg T1 cos θ 1

Fe =

k e 2QQ r

2

=

2 k eQ

FIG. P23.65

2

r2

If we assume θ is small then

tan θ ≈

r 2

.

Substitute expressions for Fe and tan θ into either equation found in part (a) and solve for r.

F I GH JK

F GH

Fe 2k Q 2 1 4k eQ 2 r = tan θ then e 2 ≈ and solving for r we find r ≈ mg 2 mg mg r

I JK

13

.

25

Chapter 23

P23.66

(a)

The distance from each corner to the center of the square is

FG L IJ + FG L IJ H 2K H 2K 2

2

=

L 2

z

+q

.

+q

L/2

L/2 +q

The distance from each positive charge to −Q is then z2 +

x –Q

+q

L2 . Each positive charge exerts a force directed 2

FIG. P23.66 k eQq

along the line joining q and −Q , of magnitude

z 2 + L2 2

.

z

The line of force makes an angle with the z-axis whose cosine is

2

z + L2 2

The four charges together exert forces whose x and y components add to zero, while the 4k e Qqz F= − k z-components add to 3 2 2 z + L2 2

e

(b)

For z >> L , the magnitude of this force is Fz = −

4k eQqz

eL 2 j 2

32

F 4a2f k Qq I z = ma GH L JK 3 2

=−

3

e

j

z

Therefore, the object’s vertical acceleration is of the form a z = −ω 2 z with ω 2 =

af

42

32

k eQq

3

mL

=

k eQq 128 mL3

.

Since the acceleration of the object is always oppositely directed to its excursion from equilibrium and in magnitude proportional to it, the object will execute simple harmonic motion with a period given by T=

P23.67

(a)

2π

ω

=

2π

a128f

14

mL3 = k e Qq

mL3 . k eQq

π

a8 f

14

FG H

IJ K

qE , m which is constant and directed downward. Therefore, it behaves like a simple pendulum in the presence of a modified uniform gravitational field with a period given by:

The total non-contact force on the cork ball is: F = qE + mg = m g +

T = 2π

L 0.500 m = 2π g + qE m 9.80 m s 2 + 2.00 × 10 −6 C 1.00 × 10 5 N C 1.00 × 10 −3 kg

e

je

j

= 0.307 s

(b)

Yes . Without gravity in part (a), we get T = 2π T = 2π

0.500 m

e2.00 × 10 Cje1.00 × 10 −6

5

j

L qE m

N C 1.00 × 10 −3 kg

= 0.314 s (a 2.28% difference).

26

Electric Fields

P23.68

The bowl exerts a normal force on each bead, directed along the radius line or at 60.0° above the horizontal. Consider the free-body diagram of the bead on the left:

∑ Fy = n sin 60.0°− mg = 0 , mg . sin 60.0°

or

n=

Also,

∑ Fx = − Fe + n cos 60.0° = 0 , keq 2

or

R

2

F mg I RG H k 3 JK

q=

Thus,

= n cos 60.0° =

n

mg mg = . tan 60.0° 3

Fe mg

12

.

FIG. P23.68

e

P23.69

(a)

60.0°

There are 7 terms which contribute: 3 are s away (along sides) 3 are 1 is

1

2s away (face diagonals) and sin θ = 3s away (body diagonal) and sin φ =

= cos θ

2 1 3

. FIG. P23.69

The component in each direction is the same by symmetry. F=

P23.70

ke q2 s2

LM1 + 2 + 1 OPei + j + kj = N 2 2 3 3Q ke q2

keq 2 s2

a1.90fei + j + kj

(b)

F = Fx2 + Fy2 + Fz2 = 3. 29

(a)

Zero contribution from the same face due to symmetry, opposite face contributes

FG k q sin φIJ where Hr K

4

e 2

sin φ =

s r

r=

s2

away from the origin

FG s IJ + FG s IJ H 2K H 2K

E=4

2

k e qs r3

=

2

+ s 2 = 1.5 s = 1.22s

4

a1.22f

3

keq s2

= 2.18

keq s2

FIG. P23.70 (b)

The direction is the k direction.

27

Chapter 23

P23.71

k e xQ

E = Ex =

The field on the axis of the ring is calculated in Example 23.8,

The force experienced by a charge −q placed along the axis of the ring is

ex + a j LM x F = − k Qq MM ex + a j N F k Qq IJ x F = −G Ha K e

2 3 2

2

e

and when x d , the length of line falling within the sphere is 2 R 2 − d 2 so

ΦE =

2λ R 2 − d 2 ∈0

.

Q−6q ∈0

, of which one-sixth goes

34

Gauss’s Law

P24.20

Φ E , hole = E ⋅ A hole =

FG k Q IJ eπ r HR K e

2

F e8.99 × 10 N ⋅ m C je10.0 × 10 Cj I JJ π e1.00 × 10 j GG a0.100 mf H K 2

9

2

=

−6

2

2

−3

j

m

2

Φ E, hole = 28.2 N ⋅ m 2 C P24.21

ΦE =

qin 170 × 10 −6 C = = 1.92 × 10 7 N ⋅ m 2 C ∈0 8.85 × 10 −12 C 2 N ⋅ m 2 1.92 × 10 7 N ⋅ m 2 C 1 ΦE = 6 6

(a)

bΦ g

(b)

Φ E = 19.2 MN ⋅ m 2 C

(c)

E one face

=

bΦ g

E one face

= 3.20 MN ⋅ m 2 C

The answer to (a) would change because the flux through each face of the cube would not be equal with an asymmetric charge distribution. The sides of the cube nearer the charge would have more flux and the ones further away would have less. The answer to (b) would remain the same, since the overall flux would remain the same.

P24.22

No charge is inside the cube. The net flux through the cube is zero. Positive flux comes out through the three faces meeting at g. These three faces together fill solid angle equal to one-eighth of a sphere as seen from q, and together pass 1 q . Each face containing a intercepts equal flux going into the cube: flux 8 ∈0

FG IJ H K

0 = Φ E , net Φ E , abcd =

Section 24.3 P24.23

FIG. P24.22

q = 3 Φ E , abcd + 8 ∈0 −q 24 ∈0

Application of Gauss’s Law to Various Charge Distributions

The charge distributed through the nucleus creates a field at the surface equal to that of a point k q charge at its center: E = e2 r

e8.99 × 10 Nm C je82 × 1.60 × 10 E= a208f 1.20 × 10 m 9

2

13

E = 2.33 × 10 21 N C

2

−15

−19

C

j

2

away from the nucleus

Chapter 24

P24.24

(a)

E=

(b)

E=

k eQr a3

= 0

e8.99 × 10 je26.0 × 10 ja0.100f = 365 kN C a a0.400f k Q e8.99 × 10 je 26.0 × 10 j = = 1.46 MN C E= r a0.400f k eQr 3

−6

9

=

3

−6

9

e

(c)

E=

(d)

2

2

k eQ r2

e8.99 × 10 je26.0 × 10 j = = a0.600f −6

9

649 kN C

2

The direction for each electric field is radially outward .

*P24.25

mg = qE = q

0

P24.26

−12 0.01 9.8 Q 2 ∈0 mg 2 8.85 × 10 = = = −2.48 µC m 2 A q −0.7 × 10 −6

0

jb

e

2 8.99 × 10 9 Q 2.40 2k e λ 4 E= 3.60 × 10 = 0.190 r Q = +9.13 × 10 −7 C = +913 nC

(a)

g

E= 0

(b) *P24.27

ja fa f

e

FG σ IJ = qFG Q A IJ H2∈ K H 2∈ K

The volume of the spherical shell is

a

f − a0.20 mf

4 π 0.25 m 3

3

3

= 3.19 × 10 −2 m3 .

Its charge is

e

je

j

ρV = −1.33 × 10 −6 C m 3 3.19 × 10 −2 m3 = −4.25 × 10 −8 C . The net charge inside a sphere containing the proton’s path as its equator is −60 × 10 −9 C − 4. 25 × 10 −8 C = −1.02 × 10 −7 C . The electric field is radially inward with magnitude ke q r

2

=

q ∈0 4π r

2

=

e

8.99 × 10 9 Nm 2 1.02 × 10 −7 C 2

a

f

C 0.25 m

2

j = 1.47 × 10

4

N C.

For the proton

∑ F = ma

F eEr IJ v=G HmK

eE = 12

mv 2 r

F 1.60 × 10 =G GH

−19

e

j

C 1.47 × 10 4 N C 0.25 m 1.67 × 10 −27 kg

I JJ K

12

= 5.94 × 10 5 m s .

35

36

Gauss’s Law

P24.28

e

σ = 8.60 × 10 −6 C cm 2 E=

jFGH 100mcm IJK

2

= 8.60 × 10 −2 C m 2

σ 8.60 × 10 −2 = = 4.86 × 10 9 N C away from the wall 2 ∈0 2 8.85 × 10 −12

e

j

The field is essentially uniform as long as the distance from the center of the wall to the field point is much less than the dimensions of the wall. P24.29

If ρ is positive, the field must be radially outward. Choose as the gaussian surface a cylinder of length L and radius r, contained inside the charged rod. Its volume is π r 2 L and it encloses charge ρπ r 2 L . Because the charge distribution is long, no electric flux passes through the circular end caps; E ⋅ dA = EdA cos 90.0° = 0 . The curved surface has E ⋅ dA = EdA cos 0° , and E must be the same strength everywhere over the curved surface. q ρπ r 2 L Gauss’s law, E ⋅ dA = , becomes E dA = . ∈0 ∈0 Curved

z

FIG. P24.29

z

Surface

Now the lateral surface area of the cylinder is 2π rL :

b g

E 2π r L = *P24.30

ρπ r 2 L . ∈0

ρr radially away from the cylinder axis . 2 ∈0

E=

Thus,

Let ρ represent the charge density. For the field inside the sphere at r1 = 5 cm we have E1 4π r12 =

q inside 4π r13 ρ = ∈0 3 ∈0

E1 =

e

je

r1 ρ 3 ∈0

j

−12 C 2 −86 × 10 3 N 3 ∈0 E1 3 8.85 × 10 ρ= = = −4.57 × 10 −5 C m 3 . r1 0.05 m Nm 2 C Now for the field outside at r3 = 15 cm 4π r23 ρ E3 4π r32 = 3 ∈0

a

f e−4.57 × 10 Cj = 8.99 × 10

k 4 π 0.10 m E3 = 2e r3 3

3

−5

m

3

9

e

Nm 2 −1.91 × 10 −7 C

a0.15 mf C 2

2

j = −7.64 × 10

4

NC

E 3 = 76. 4 kN C radially inward P24.31

E= 0

(a)

E=

(b) P24.32

k eQ r2

e8.99 × 10 je32.0 × 10 j = 7.19 MN C = a0.200f −6

9

E = 7.19 MN C radially outward

2

The distance between centers is 2 × 5.90 × 10 −15 m . Each produces a field as if it were a point charge at its center, and each feels a force as if all its charge were a point at its center. F=

k e q1 q 2 r2

e

a46f e1.60 × 10 Cj C j e2 × 5.90 × 10 mj 2

9

= 8.99 × 10 N ⋅ m

2

−19

2

−15

2

2

= 3.50 × 10 3 N = 3.50 kN

Chapter 24

P24.33

Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart with a 0.05 m separation on the ends of strings making angles of 10° with the vertical. (a)

mg

∑ Fy = T cos 10°−mg = 0 ⇒ T = cos 10° ∑ Fx = T sin 10°− Fe = 0 ⇒ Fe = T sin 10° , so Fe =

FG mg IJ sin 10° = mg tan 10° = b0.001 kgge9.8 m s j tan 10° H cos 10° K 2

Fe ≈ 2 × 10 −3 N ~ 10 −3 N or 1 mN

(b)

Fe =

keq 2 r2

e8.99 × 10 N ⋅ m N≈ a0.25 mf 9

2 × 10

−3

2

j

C 2 q2

2

q ≈ 1. 2 × 10 −7 C ~ 10 −7 C or 100 nC

*P24.34

37

keq

(c)

E=

(d)

ΦE =

r

≈

2

e8.99 × 10

9

je

N ⋅ m 2 C 2 1.2 × 10 −7 C

a0.25 mf

2

N C ~ 10 kN C

4 3 πa ρ 3

ρ=

3Q 4π a 3

The flux is that created by the enclosed charge within radius r: ΦE =

(b)

4

q 1.2 × 10 −7 C ≈ = 1. 4 × 10 4 N ⋅ m 2 C ~ 10 kN ⋅ m 2 C ∈0 8.85 × 10 −12 C 2 N ⋅ m 2

The charge density is determined by Q = (a)

j ≈ 1.7 × 10

ΦE =

q in 4π r 3 ρ 4π r 3 3Q Qr 3 = = = ∈0 3 ∈0 3 ∈0 4π a 3 ∈0 a 3 Q . Note that the answers to parts (a) and (b) agree at r = a . ∈0

(c)

ΦE Q ∈0

0

0

a FIG. P24.34(c)

r

FIG. P24.33

38

Gauss’s Law

P24.35

(a)

e

je

j

9 2 2 2.00 × 10 −6 C 7.00 m 2 k e λ 2 8.99 × 10 N ⋅ m C = E= 0.100 m r

E = 51.4 kN C , radially outward (b)

b

g

Φ E = EA cos θ = E 2π rA cos 0°

j a

e

fb

ga f

Φ E = 5.14 × 10 4 N C 2π 0.100 m 0.020 0 m 1.00 = 646 N ⋅ m 2 C P24.36

(a)

(b)

ρ=

Q 5.70 × 10 −6 = = 2.13 × 10 −2 C m 3 3 4π 3 4π a 0.040 0 3 3

b

g

FG 4 π r IJ = e2.13 × 10 H3 K F4 I = ρ G π r J = e 2.13 × 10 H3 K

qin = ρ

3

−2

qin

3

−2

jFGH 34 π IJK b0.020 0g jFGH 34 π IJK b0.040 0g

3

= 7.13 × 10 −7 C = 713 nC

3

= 5.70 µC

9.00 × 10 −6 C m 2 σ = = 508 kN C , upward 2 ∈0 2 8.85 × 10 −12 C 2 N ⋅ m 2

P24.37

E=

P24.38

Note that the electric field in each case is directed radially inward, toward the filament.

j

e

je

j

e

je

j

e

je

j

(a)

E=

−6 9 2 2 2 k e λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C m = = 16.2 MN C 0.100 m r

(b)

E=

−6 9 2 2 2 k e λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C m = = 8.09 MN C 0.200 m r

(c)

−6 9 2 2 2 k e λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C m = = 1.62 MN C E= 1.00 m r

Section 24.4 P24.39

e

z

Conductors in Electrostatic Equilibrium

b g

EdA = E 2π rl =

qin ∈0

E=

q in l λ = 2π ∈0 r 2π ∈0 r

(a)

r = 3.00 cm

E= 0

(b)

r = 10.0 cm

E=

(c)

r = 100 cm

E=

e

30.0 × 10 −9

ja

2π 8.85 × 10 −12 0.100

e

30.0 × 10 −9

f=

= ja f

2π 8.85 × 10 −12 1.00

5 400 N C , outward

540 N C , outward

Chapter 24

P24.40

σ= P24.41

EA =

From Gauss’s Law,

ja

Q ∈0

39

f

Q =∈0 E = 8.85 × 10 −12 −130 = −1.15 × 10 −9 C m 2 = −1.15 nC m 2 A

e

σ conductor for the field outside the aluminum looks ∈0

The fields are equal. The Equation 24.9 E =

σ insulator for the field around glass. But its charge will spread out to 2 ∈0 Q cover both sides of the aluminum plate, so the density is σ conductor = . The glass carries charge 2A Q Q only on area A, with σ insulator = . The two fields are the same in magnitude, and both are A 2 A ∈0 perpendicular to the plates, vertically upward if Q is positive. different from Equation 24.8 E =

*P24.42

(a)

All of the charge sits on the surface of the copper sphere at radius 15 cm. The field inside is zero .

(b)

The charged sphere creates field at exterior points as if it were a point charge at the center: E=

P24.43

ke q r

2

away =

e8.99 × 10

9

je

Nm 2 40 × 10 −9 C 2

a

f

C 0.17 m

(c)

e8.99 × 10 E=

(d)

All three answers would be the same.

(a)

E=

9

je

2

Nm 2 40 × 10 −9 C 2

a

f

C 0.75 m

σ ∈0

2

e

j outward =

j outward =

je

1.24 × 10 4 N C outward

639 N C outward

j

σ = 8.00 × 10 4 8.85 × 10 −12 = 7.08 × 10 −7 C m 2

σ = 708 nC m 2 , positive on one face and negative on the other.

P24.44

e

σ=

(a)

E= 0 k eQ

(b)

E=

(c)

E= 0

(d)

ja

f

Q 2 Q = σA = 7.08 × 10 −7 0.500 C A Q = 1.77 × 10 −7 C = 177 nC , positive on one face and negative on the other.

(b)

E=

r2

k eQ r2

e8.99 × 10 je8.00 × 10 j = 7.99 × 10 = b0.030 0g

7

NC

E = 79.9 MN C radially outward

e8.99 × 10 je4.00 × 10 j = 7.34 × 10 = b0.070 0g

6

NC

E = 7.34 MN C radially outward

−6

9

2

−6

9

2

40

Gauss’s Law

P24.45

The charge divides equally between the identical spheres, with charge like point charges at their centers: F=

P24.46

b gb g = k Q aL + R + Rf 4aL + 2Rf

ke Q 2 Q 2

2

e

2

2

=

e

8.99 × 10 9 N ⋅ m 2 60.0 × 10 −6 C

a

f

4 C 2 2.01 m

j

Q on each. Then they repel 2

2

2

= 2.00 N .

The electric field on the surface of a conductor varies inversely with the radius of curvature of the surface. Thus, the field is most intense where the radius of curvature is smallest and vice-versa. The local charge density and the electric field intensity are related by E= (a)

σ

σ =∈0 E .

or

∈0

Where the radius of curvature is the greatest,

e

je

j

σ =∈0 Emin = 8.85 × 10 −12 C 2 N ⋅ m 2 2.80 × 10 4 N C = 248 nC m 2 . (b)

Where the radius of curvature is the smallest,

e

je

j

σ =∈0 Emax = 8.85 × 10 −12 C 2 N ⋅ m 2 5.60 × 10 4 N C = 496 nC m 2 . P24.47

(a)

(b)

P24.48

(a)

(b) P24.49

(a)

Inside surface: consider a cylindrical surface within the metal. Since E inside the conducting shell is zero, the total charge inside the gaussian surface must be zero, so the inside charge/length = − λ . 0 = λA + qin

so

qin = −λ A

Outside surface:

The total charge on the metal cylinder is

2λA = qin + qout

qout = 2λA + λA

so the outside charge/length is

E=

E=

e

r k eQ r2

r

e8.99 × 10 je6.40 × 10 j = = a0.150f −6

9

2

2.56 MN C , radially inward

The charge density on each of the surfaces (upper and lower) of the plate is:

E=

FG IJ H K

e

E=

j

−8 1 q 1 4.00 × 10 C = = 8.00 × 10 −8 C m 2 = 80.0 nC m 2 . 2 A 2 0.500 m 2

a

f

FG σ IJ k = F 8.00 × 10 C m I k = b9.04 kN Cgk H ∈ K GH 8.85 × 10 C N ⋅ m JK −8

0

(c)

3λ radially outward 2π ∈0 r

E=0

σ=

(b)

b g = 6k λ =

2 k e 3λ

3λ .

b−9.04 kN Cgk

−12

2

2

2

Chapter 24

P24.50

(a)

The charge +q at the center induces charge −q on the inner surface of the conductor, where its surface density is: −q σa = . 4π a 2

(b)

The outer surface carries charge Q + q with density

σb = P24.51

Q+q

41

.

4π b 2

Use Gauss’s Law to evaluate the electric field in each region, recalling that the electric field is zero everywhere within conducting materials. The results are: E = 0 inside the sphere and within the material of the shell

P24.52

E = ke

Q between the sphere and shell, directed radially inward r2

E = ke

2Q outside the shell, directed radially outward . r2

Charge

−Q is on the outer surface of the sphere .

Charge

+Q is on the inner surface of the shell ,

and

+2Q is on the outer surface of the shell.

An approximate sketch is given at the right. Note that the electric field lines should be perpendicular to the conductor both inside and outside.

FIG. P24.52

Section 24.5 P24.53

(a)

Formal Derivation of Gauss‘s Law Uniform E, pointing radially outward, so Φ E = EA . The arc length is ds = Rdθ , and the circumference is 2π r = 2π R sin θ

z

A = 2π rds =

zb

θ

z

θ

g

a

2π R sin θ Rdθ = 2π R 2 sin θdθ = 2π R 2 − cos θ

0

f

θ 0

0

a

f

a

1 Q Q ΦE = ⋅ 2π R 2 1 − cos θ = 1 − cos θ 4π ∈0 R 2 2 ∈0

a

f

[independent of R!]

f

(b)

For θ = 90.0° (hemisphere): Φ E =

Q Q 1 − cos 90° = . 2 ∈0 2 ∈0

(c)

For θ = 180° (entire sphere): Φ E =

Q Q 1 − cos 180° = 2 ∈0 ∈0

a

b

= 2π R 2 1 − cos θ

f

[Gauss’s Law].

g

FIG. P24.53

42

Gauss’s Law

Additional Problems P24.54

E = ay i + bzj + cxk E = ay i + cxk

In general, In the xy plane, z = 0 and

z

ze

Φ E = E ⋅ dA =

w

z

(b)

x=0

y =h

x=0

j

x2 Φ E = ch xdx = ch 2 x=0

(a)

y=0

ay i + cxk ⋅ k dA

w

P24.55

z

y

x=w

chw 2 = 2

dA = hdx

x

FIG. P24.54

qin = +3Q − Q = +2Q The charge distribution is spherically symmetric and qin > 0 . Thus, the field is directed radially outward . k e qin

for r ≥ c .

E=

(d)

Since all points within this region are located inside conducting material, E = 0 for

r

2

=

2 k eQ

(c)

r2

b < r < c.

z

(e)

Φ E = E ⋅ dA = 0 ⇒ qin =∈0 Φ E = 0

(f)

qin = +3Q

(g)

E=

(h)

qin = ρV =

(i)

E=

(j)

From part (d), E = 0 for b < r < c . Thus, for a spherical gaussian surface with b < r < c , qin = +3Q + qinner = 0 where qinner is the charge on the inner surface of the conducting shell. This yields qinner = −3Q .

(k)

Since the total charge on the conducting shell is q net = qouter + qinner = −Q , we have

k e qin r

2

k e qin r

2

=

3 k eQ r2

(radially outward) for a ≤ r < b .

F +3Q I FG 4 π r IJ = +3Q r GH π a JK H 3 K a k F r I r = +3Q J = 3 k Q (radially outward) for 0 ≤ r ≤ a . G r H a K a 4 3

e 2

3

3

3

3

3

3

e

3

b g

qouter = −Q − qinner = −Q − −3Q = +2Q . (l)

This is shown in the figure to the right.

E

a

b

c

FIG. P24.55(l)

r

Chapter 24

43

P24.56

The sphere with large charge creates a strong field to polarize the other sphere. That means it pushes the excess charge over to the far side, leaving charge of the opposite sign on the near side. This patch of opposite charge is smaller in amount but located in a stronger external field, so it can feel a force of attraction that is larger than the repelling force felt by the larger charge in the weaker field on the other side.

P24.57

(a)

(b)

z

e

j

E ⋅ dA = E 4π r 2 =

qin ∈0

FG 4 π r IJ H3 K 3

For r < a ,

qin = ρ

so

E=

For a < r < b and c < r ,

qin = Q .

So

Q E= . 4π r 2 ∈0

For b ≤ r ≤ c ,

E = 0 , since E = 0 inside a conductor.

ρr . 3 ∈0 FIG. P24.57

Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the conductor, the total charge enclosed by a spherical surface of radius b ≤ r ≤ c must be zero. q1 + Q = 0

Therefore,

and

σ1 =

q1 4π b

2

=

−Q . 4π b 2

Let q 2 = induced charge on the outside surface of the hollow sphere. Since the hollow sphere is uncharged, we require q Q σ2 = 1 2 = . q1 + q 2 = 0 and 4π c 4π c 2 P24.58

z

e

j

E ⋅ dA = E 4π r 2 =

(a)

e−3.60 × 10

3

qin ∈0

j a

f

N C 4π 0.100 m

2

=

Q 8.85 × 10

−12

C 2 N ⋅ m2

aa < r < bf

Q = −4.00 × 10 −9 C = −4.00 nC (b)

We take Q ′ to be the net charge on the hollow sphere. Outside c, Q + Q′ 2 +2.00 × 10 2 N C 4π 0.500 m = r>c 8.85 × 10 −12 C 2 N ⋅ m 2

e

j a

f

a f

Q + Q ′ = +5.56 × 10 −9 C , so Q ′ = +9.56 × 10 −9 C = +9.56 nC (c)

For b < r < c : E = 0 and qin = Q + Q1 = 0 where Q1 is the total charge on the inner surface of the hollow sphere. Thus, Q1 = −Q = +4.00 nC . Then, if Q 2 is the total charge on the outer surface of the hollow sphere,

Q 2 = Q ′ − Q1 = 9.56 nC − 4.0 nC = +5.56 nC .

44

Gauss’s Law

*P24.59

y

The vertical velocity component of the moving charge increases according to dv y

m

dv y dx = qE y . m dx dt

= Fy

dt

v

0

q

Now

dx = v x has the nearly constant value v. So dt

dv y =

q Ey dx mv

z

vy x

Q FIG. P24.59

q ∞ Ey dx . mv −∞

0

vx

d

z

vy

v y = dv y =

θ

The radially outward compnent of the electric field varies along the x axis, but is described by

z

∞

z

∞

Ey dA =

−∞

−∞

z

∞

So

Ey dx =

−∞

tan θ = P24.60

vy v

b g

Ey 2π d dx =

=

Q . ∈0

qQ Q and v y = . The angle of deflection is described by 2π d ∈0 mv 2π d ∈0 qQ 2π ∈0 dmv

θ = tan −1

2

qQ 2π ∈0 dmv 2

.

First, consider the field at distance r < R from the center of a uniform sphere of positive charge Q = + e with radius R.

b

e

g

j

4π r 2 E =

(a)

qin ρV = = ∈0 ∈0

F GH

+e 4π 3 3 R

I JK

4 3π

r3

∈0

so E =

F e I r directed outward GH 4π ∈ R JK 0

3

The force exerted on a point charge q = − e located at distance r from the center is then F = qE = − e

F e I r = −F e I r = GH 4π ∈ R JK GH 4π ∈ R JK 2

0

3

3

0

− Kr .

k e2 e2 = e3 3 4π ∈0 R R

(b)

K=

(c)

Fr = m e a r = −

F k e I r , so a GH R JK e

2

3

r

=−

F k e I r = −ω r GH m R JK e

e

2

3

2

Thus, the motion is simple harmonic with frequency

(d)

f = 2.47 × 10

15

1 Hz = 2π

e8.99 × 10

9

je

f=

ω 1 = 2π 2π

N ⋅ m 2 C 2 1.60 × 10 −19 C

e9.11 × 10

−31

j

j

2

kg R 3

which yields R 3 = 1.05 × 10 −30 m3 , or R = 1.02 × 10 −10 m = 102 pm .

kee2

me R3

.

Chapter 24

P24.61

The field direction is radially outward perpendicular to the axis. The field strength depends on r but not on the other cylindrical coordinates θ or z. Choose a Gaussian cylinder of radius r and length L. If r < a , ΦE =

E=

qin ∈0

and

λ 2π r ∈0

or

b

g

E 2π rL =

E=

b

E=

λ

g

b

E=

r

ar < a f

.

e

j

λL + ρπ r 2 − a 2 L ∈0

e

λ + ρπ r 2 − a 2 2π r ∈0

g

E 2π rL =

If r > b ,

λL ∈0

2π r ∈0

E 2π rL =

If a < r < b ,

P24.62

45

j r

e

aa < r < bf .

j

λL + ρπ b 2 − a 2 L ∈0

e

λ + ρπ b 2 − a 2 2π r ∈0

j r

ar > b f

.

Consider the field due to a single sheet and let E+ and E− represent the fields due to the positive and negative sheets. The field at any distance from each sheet has a magnitude given by Equation 24.8: E+ = E− = (a)

σ . 2 ∈0

To the left of the positive sheet, E+ is directed toward the left and E− toward the right and the net field over this region is E = 0 .

(b)

In the region between the sheets, E+ and E− are both directed toward the right and the net field is E=

(c)

σ ∈0

to the right .

FIG. P24.62

To the right of the negative sheet, E+ and E− are again oppositely directed and E = 0 .

46

Gauss’s Law

P24.63

The magnitude of the field due to the each sheet given by Equation 24.8 is E= (a)

σ directed perpendicular to the sheet. 2 ∈0

In the region to the left of the pair of sheets, both fields are directed toward the left and the net field is

σ

E=

∈0

to the left .

(b)

In the region between the sheets, the fields due to the individual sheets are oppositely directed and the net field is E= 0 .

(c)

In the region to the right of the pair of sheets, both are fields are directed toward the right and the net field is

σ to the right . ∈0

E= P24.64

FIG. P24.63

The resultant field within the cavity is the superposition of two fields, one E + due to a uniform sphere of positive charge of radius 2a, and the other E − due to a sphere of negative charge of radius a centered within the cavity.

F GH

I JK

4 π r 3ρ = 4π r 2 E+ 3 ∈0 –−

F GH

I JK

4 π r13 ρ = 4π r12 E− 3 ∈0

Since r = a + r1 ,

so

E+ =

ρr ρr r = 3 ∈0 3 ∈0

so

E− =

ρ r1 −ρ − r1 = r1 . 3 ∈0 3 ∈0

a f

−ρ r − a E− = 3 ∈0 E = E+ + E− =

*P24.65

b g

Thus,

Ex = 0

and

Ey =

ρa 3 ∈0

FIG. P24.64

ρr ρr ρa ρa ρa  − + = = 0 i + j. 3 ∈0 3 ∈0 3 ∈0 3 ∈0 3 ∈0

at all points within the cavity.

Consider the charge distribution to be an unbroken charged spherical shell with uniform charge density σ and a circular disk with charge per area −σ . The total field is that due to the whole sphere, 4π R 2σ σ σ σ Q = = outward plus the field of the disk − = radially inward. The total 2 2 ∈ ∈ ∈0 2 2 4π ∈0 R 4πε 0 R 0 0 field is

σ σ σ outward . − = ∈0 2 ∈0 2 ∈0

Chapter 24

P24.66

The electric field throughout the region is directed along x; therefore, E will be perpendicular to dA over the four faces of the surface which are perpendicular to the yz plane, and E will be parallel to dA over the two faces which are parallel to the yz plane. Therefore,

e

Φ E = − Ex

x=a

j A + eE

x x=a+c

jA = −e3 + 2 a jab + e3 + 2aa + cf jab = 2abca2 a + cf . 2

2

Substituting the given values for a, b, and c, we find Φ E = 0.269 N ⋅ m 2 C .

FIG. P24.66

Q =∈0 Φ E = 2.38 × 10 −12 C = 2.38 pC P24.67

z

e

j

E ⋅ dA = E 4π r 2 =

qin ∈0

z

R

e

0

(b)

AR 5 5

j

4π Ar 5 5

AR 5 . 5 ∈0 r 2

and

E=

For r < R ,

qin = Ar 2 4π r 2 dr =

z r

e

0

Ar 3 . 5 ∈0

E=

and

P24.68

j

qin = Ar 2 4π r 2 dr = 4π

For r > R ,

(a)

The total flux through a surface enclosing the charge Q is disk is

Q . The flux through the ∈0

z

Φ disk = E ⋅ dA where the integration covers the area of the disk. We must evaluate this integral 1Q to find how b and R are related. In the figure, take dA to be and set it equal to 4 ∈0 the area of an annular ring of radius s and width ds. The flux through dA is

b

FIG. P24.68

g

E ⋅ dA = EdA cos θ = E 2π sds cos θ . The magnitude of the electric field has the same value at all points within the annular ring, E=

1 Q 1 Q = 4π ∈0 r 2 4π ∈0 s 2 + b 2

and

cos θ =

b b = 2 r s + b2

e

j

12

.

Integrate from s = 0 to s = R to get the flux through the entire disk. Φ E , disk =

Qb 2 ∈0

ze

R 0

sds s2 + b 2

j

= 32

The flux through the disk equals This is satisfied if R = 3 b .

LM e N

Qb − s2 + b 2 2 ∈0

j OPQ 12

R

= 0

LM MM e N

Q b 1− 2 ∈0 R2 + b2

Q b provided that 4 ∈0 R2 + b2

e

j

12

=

OP j PPQ 12

1 . 2

47

48

Gauss’s Law

P24.69

z

E ⋅ dA =

qin 1 = ∈0 ∈0

z

z r

0

a 4π r 2 dr r

4π a r 4π a r 2 E 4π r 2 = rdr = ∈0 0 ∈0 2 E=

a = constant magnitude 2 ∈0

(The direction is radially outward from center for positive a; radially inward for negative a.) P24.70

z

z

1 ρdV . We ∈0 use a gaussian surface which is a cylinder of radius r, length A , and is coaxial with the charge distribution.

In this case the charge density is not uniform, and Gauss’s law is written as

(a)

b

g

When r < R , this becomes E 2π rA =

ρ0 ∈0

z FGH r

a−

0

E ⋅ dA =

IJ K

r dV . The element of volume is a cylindrical b

shell of radius r, length A , and thickness dr so that dV = 2π rAdr .

b

g FGH 2π r∈ Aρ 2

E 2π rA = (b)

0

b

(a)

I FG a − r IJ so inside the cylinder, E = JK H 2 3b K

FG H

ρ 0r 2r a− 2 ∈0 3b

When r > R , Gauss’s law becomes

g

E 2π rA = P24.71

0

ρ0 ∈0

z FGH a − br IJK b2π rAdrg or outside the cylinder, E =

FG H

R

IJ K

.

ρ0R2 2R a− 2 ∈0 r 3b

0

IJ K

Consider a cylindrical shaped gaussian surface perpendicular to the yz plane with one end in the yz plane and the other end containing the point x: Use Gauss’s law:

z

E ⋅ dA =

. y

qin ∈0

gaussian surface

By symmetry, the electric field is zero in the yz plane and is perpendicular to dA over the wall of the gaussian cylinder. Therefore, the only contribution to the integral is over the end cap containing the point x :

z

a f

(b)

z

ρ Ax q E ⋅ dA = in or EA = ∈0 ∈0

so that at distance x from the mid-line of the slab, E =

a=

a f

FG H

x x

ρx . ∈0

IJ K

−e E ρe F = =− x me me m e ∈0

FIG. P24.71

The acceleration of the electron is of the form

a = −ω 2 x with ω =

Thus, the motion is simple harmonic with frequency

f=

ω 1 = 2π 2π

ρe . m e ∈0

ρe . m e ∈0

Chapter 24

P24.72

Consider the gaussian surface described in the solution to problem 71. (a)

d , 2 1 E ⋅ dA = dq ∈0

z

CA ∈0

EA = E=

(b)

3

Cd 24 ∈0

For − E=

(a)

dq = ρ dV = ρAdx = CAx 2 dx

For x >

z

P24.73

49

z

d 2

x 2 dx =

0

FG IJ F d I H K GH 8 JK 3

1 CA 3 ∈0

E=

or

d d ; 24 ∈0 2

E ⋅ dA =

Cx 3  i for x > 0 ; 3 ∈0

E=−

z

E=−

Cd 3  d i for x < − 24 ∈0 2

z

1 CA x 2 CAx 3 dq = x dx = 3 ∈0 ∈0 ∈0 0 Cx 3  i for x < 0 3 ∈0

A point mass m creates a gravitational acceleration

g=−

z

Gm r2

r at a distance r. Gm

e

j

4π r 2 = −4π Gm . r2 Since the r has divided out, we can visualize the field as unbroken field lines. The same flux would go through any other closed surface around the mass. If there are several or no masses inside a closed surface, each creates field to make its own contribution to the net flux according to

The flux of this field through a sphere is

z

(b)

g ⋅ dA = −

g ⋅ dA = −4π Gm in .

Take a spherical gaussian surface of radius r. The field is inward so g ⋅ dA = g 4π r 2 cos 180° = − g 4π r 2

z

and Then, Or, since

4 −4π Gm in = −4π G π r 3 ρ . 3 4 4 3 2 − g 4π r = −4π G π r ρ and g = π rρG . 3 3 M EGr ME M EGr , g= or g = inward . ρ= 4 3 RE3 RE3 3 π RE

ANSWERS TO EVEN PROBLEMS P24.2

355 kN ⋅ m 2 C

P24.10

(a) −55.6 nC ; (b) The negative charge has a spherically symmetric distribution.

P24.4

(a) −2.34 kN ⋅ m 2 C ; (b) +2.34 kN ⋅ m 2 C ; (c) 0

P24.12

(a)

P24.14

(a) 1.36 MN ⋅ m 2 C ; (b) 678 kN ⋅ m 2 C ; (c) No; see the solution.

P24.6

q ∈0

P24.8

ERh

q q ; (b) ; (c) Plane and square 2 ∈0 2 ∈0 both subtend a solid angle of a hemisphere at the charge.

50

Gauss’s Law

P24.16 P24.18

1.77 pC m3 positive

P24.46

(a) 248 nC m 2 ; (b) 496 nC m 2

Q−6q

P24.48

(a) 2.56 MN C radially inward; (b) 0

P24.50

(a)

P24.52

see the solution

P24.54

chw 2 2

P24.56

see the solution

P24.58

(a) −4.00 nC; (b) +9.56 nC ; (c) +4.00 nC and +5.56 nC

P24.60

(a, b) see the solution; (c)

6 ∈0

P24.20

28. 2 N ⋅ m 2 C

P24.22

−q 24 ∈0

P24.24

(a) 0; (b) 365 kN C ; (c) 1.46 MN C; (d) 649 kN C

P24.26

(a) 913 nC ; (b) 0

P24.28

4.86 GN C away from the wall. It is constant close to the wall

P24.30

76.4 kN C radially inward

P24.32

3.50 kN

−q 4π a

2

; (b)

Q+q 4π b 2

1 2π

ke e2

me R3

;

(d) 102 pm

P24.34 P24.36 P24.38

(a)

3

Qr Q ; (c) see the solution ; (b) ∈0 ∈0 a 3

713 nC ; (b) 5.70 µC (a) 16.2 MN C toward the filament; (b) 8.09 MN C toward the filament; (c) 1.62 MN C toward the filament 2

P24.40

−1.15 nC m

P24.42

(a) 0; (b) 12.4 kN C radially outward; (c) 639 N C radially outward; (d) Nothing would change.

P24.44

(a) 0; (b) 79.9 MN C radially outward; (c) 0; (d) 7.34 MN C radially outward

σ to the right; (c) 0 ∈0

P24.62

(a) 0; (b)

P24.64

see the solution

P24.66

0.269 N ⋅ m 2 C ; 2.38 pC

P24.68

see the solution

P24.70

(a)

P24.72

(a) E =

FG H

IJ K

FG H

ρ 0r ρ R2 2r 2R ; (b) 0 a− a− 2 ∈0 3b 2 ∈0 r 3b

IJ K

Cd 3  d i for x > ; 24 ∈0 2 3 Cd  d E=− i for x < − ; 24 ∈0 2 3 Cx  Cx 3  i for x > 0 ; E = − i for x < 0 (b) E = 3 ∈0 3 ∈0

25 Electric Potential CHAPTER OUTLINE 25.1 25.2 25.3

25.4

25.5

25.6 25.7 25.8

Potential Difference and Electric Potential Potential Difference in a Uniform Electric Field Electric Potential and Potential Energy Due to Point Charges Obtaining the Value of the Electric Field from the Electric Potential Electric Potential Due to Continuous Charge Distributions Electric Potential Due to a Charged Conductor The Milliken Oil Drop Experiment Application of Electrostatistics

ANSWERS TO QUESTIONS Q25.1

When one object B with electric charge is immersed in the electric field of another charge or charges A, the system possesses electric potential energy. The energy can be measured by seeing how much work the field does on the charge B as it moves to a reference location. We choose not to visualize A’s effect on B as an action-at-a-distance, but as the result of a twostep process: Charge A creates electric potential throughout the surrounding space. Then the potential acts on B to inject the system with energy.

Q25.2

The potential energy increases. When an outside agent makes it move in the direction of the field, the charge moves to a region of lower electric potential. Then the product of its negative charge with a lower number of volts gives a higher number of joules. Keep in mind that a negative charge feels an electric force in the opposite direction to the field, while the potential is the work done on the charge to move it in a field per unit charge.

Q25.3

To move like charges together from an infinite separation, at which the potential energy of the system of two charges is zero, requires work to be done on the system by an outside agent. Hence energy is stored, and potential energy is positive. As charges with opposite signs move together from an infinite separation, energy is released, and the potential energy of the set of charges becomes negative.

Q25.4

The charge can be moved along any path parallel to the y-z plane, namely perpendicular to the field.

Q25.5

The electric field always points in the direction of the greatest change in electric potential. This is ∂V ∂V ∂V , Ey = − and Ez = − . implied by the relationships Ex = − ∂x ∂y ∂z

Q25.6

(a)

The equipotential surfaces are nesting coaxial cylinders around an infinite line of charge.

(b)

The equipotential surfaces are nesting concentric spheres around a uniformly charged sphere.

Q25.7

If there were a potential difference between two points on the conductor, the free electrons in the conductor would move until the potential difference disappears.

51

52

Electric Potential

Q25.8

No. The uniformly charged sphere, whether hollow or solid metal, is an equipotential volume. Since there is no electric field, this means that there is no change in electrical potential. The potential at every point inside is the same as the value of the potential at the surface.

Q25.9

Infinitely far away from a line of charge, the line will not look like a point. In fact, without any distinguishing features, it is not possible to tell the distance from an infinitely long line of charge. Another way of stating the answer: The potential would diverge to infinity at any finite distance, if it were zero infinitely far away.

Q25.10

The smaller sphere will. In the solution to the example referred to, equation 1 states that each will q have the same ratio of charge to radius, . In this case, the charge density is a surface charge r q , so the smaller-radius sphere will have the greater charge density. density, 4π r 2

Q25.11

The main factor is the radius of the dome. One often overlooked aspect is also the humidity of the air—drier air has a larger dielectric breakdown strength, resulting in a higher attainable electric potential. If other grounded objects are nearby, the maximum potential might be reduced.

Q25.12

The intense—often oscillating—electric fields around high voltage lines is large enough to ionize the air surrounding the cables. When the molecules recapture their electrons, they release that energy in the form of light.

Q25.13

A sharp point in a charged conductor would imply a large electric field in that region. An electric discharge could most easily take place at that sharp point.

Q25.14

Use a conductive box to shield the equipment. Any stray electric field will cause charges on the outer surface of the conductor to rearrange and cancel the stray field inside the volume it encloses.

Q25.15

No charge stays on the inner sphere in equilibrium. If there were any, it would create an electric field in the wire to push more charge to the outer sphere. All of the charge is on the outer sphere. Therefore, zero charge is on the inner sphere and 10.0 µC is on the outer sphere.

Q25.16

The grounding wire can be touched equally well to any point on the sphere. Electrons will drain away into the ground and the sphere will be left positively charged. The ground, wire, and sphere are all conducting. They together form an equipotential volume at zero volts during the contact. However close the grounding wire is to the negative charge, electrons have no difficulty in moving within the metal through the grounding wire to ground. The ground can act as an infinite source or sink of electrons. In this case, it is an electron sink.

SOLUTIONS TO PROBLEMS Section 25.1 P25.1

Potential Difference and Electric Potential

∆V = −14.0 V ∆V =

W , Q

e

je

j

jb

g

and

Q = − N A e = − 6.02 × 10 23 1.60 × 10 −19 = −9.63 × 10 4 C

so

W = Q∆V = −9.63 × 10 4 C −14.0 J C = 1.35 MJ

e

Chapter 25

P25.2

53

a f

7.37 × 10 −17 = q 115

∆K = q ∆V q = 6. 41 × 10 −19 C

P25.3

(a)

Energy of the proton-field system is conserved as the proton moves from high to low potential, which can be defined for this problem as moving from 120 V down to 0 V.

K i + Ui + ∆Emech = K f + U f

0 + qV + 0 =

1 mv p2 + 0 2

e1.60 × 10

C 120 V

−19

ja

fFGH 1 V1 J⋅ C IJK = 12 e1.67 × 10

−27

j

kg v p2

v p = 1.52 × 10 5 m s (b)

The electron will gain speed in moving the other way, from Vi = 0 to V f = 120 V :

K i + Ui + ∆Emech = K f + U f 0+0+0= 0=

1 mv e2 + qV 2

jb

1 9.11 × 10 −31 kg v e2 + −1.60 × 10 −19 C 120 J C 2

e

j e

g

v e = 6.49 × 10 6 m s P25.4

W = ∆K = − q∆V 0−

1 9.11 × 10 −31 kg 4.20 × 10 5 m s 2

e

je

j

2

e

j

= − −1.60 × 10 −19 C ∆V

From which, ∆V = −0.502 V .

Section 25.2 P25.5

(a)

Potential Difference in a Uniform Electric Field We follow the path from (0, 0) to (20.0 cm, 0) to (20.0 cm, 50.0 cm). ∆U = − (work done) ∆U = − (work from origin to (20.0 cm, 0)) – (work from (20.0 cm, 0) to (20.0 cm, 50.0 cm)) Note that the last term is equal to 0 because the force is perpendicular to the displacement.

b g

e

jb

ga

f

∆U = − qEx ∆x = − 12.0 × 10 −6 C 250 V m 0.200 m = −6.00 × 10 −4 J (b)

P25.6

E=

∆V =

∆U 6.00 × 10 −4 J =− = −50.0 J C = −50.0 V q 12.0 × 10 −6 C

∆V 25.0 × 10 3 J C = = 1.67 × 10 6 N C = 1.67 MN C d 1.50 × 10 −2 m

54

Electric Potential

P25.7

∆U = −

jLNMe1.40 × 10 m sj − e3.70 × 10 = e−1.60 × 10 j∆V

1 1 m v 2f − vi2 = − 9.11 × 10 −31 kg 2 2

e

j

e

+6.23 × 10 −18

∆U = q∆V :

5

2

6

ms

j OQP = 6.23 × 10 2

−18

J

−19

∆V = −38.9 V. The origin is at highest potential. P25.8

jb

e

g

(a)

∆V = Ed = 5.90 × 10 3 V m 0.010 0 m = 59.0 V

(b)

1 mv 2f = q∆V : 2

ja f

1 9.11 × 10 −31 v 2f = 1.60 × 10 −19 59.0 2

e

j

e

v f = 4.55 × 10 6 m s

P25.9

z

z

z

B

C

B

A

A

C

VB − VA = − E ⋅ ds = − E ⋅ ds − E ⋅ ds

a f z dy − aE cos 90.0°f = a325fa0.800 f = +260 V

VB − VA = − E cos 180° VB − VA

z

0 .500

0. 400

−0.300

−0 . 200

dx

FIG. P25.9 *P25.10

Assume the opposite. Then at some point A on some equipotential surface the electric field has a nonzero component Ep in the plane of the surface. Let a test charge start from point A and move

z

B

some distance on the surface in the direction of the field component. Then ∆V = − E ⋅ ds is nonzero. A

The electric potential charges across the surface and it is not an equipotential surface. The contradiction shows that our assumption is false, that Ep = 0 , and that the field is perpendicular to the equipotential surface. P25.11

(a)

Arbitrarily choose V = 0 at 0. Then at other points

V = − Ex

and

U e = QV = −QEx .

Between the endpoints of the motion,

bK + U

s

+ Ue

g = bK + U i

s

+ Ue

g

f

1 2 2QE 0 + 0 + 0 = 0 + kx max − QEx max so x max = . 2 k (b)

At equilibrium,

∑ Fx = − Fs + Fe = 0 or

kx = QE .

So the equilibrium position is at x = continued on next page

QE . k

FIG. P25.11

Chapter 25

(c)

d2x

∑ Fx = − kx + QE = m dt 2

The block’s equation of motion is

.

QE QE , , or x = x ′ + k k so the equation of motion becomes: d 2 x + QE k QE d 2 x′ k =− −k x′ + + QE = m x′ . , or 2 k m dt dt 2 x′ = x −

Let

FG H

b

IJ K

g

FG IJ H K

This is the equation for simple harmonic motion a x′ = −ω 2 x ′

(d)

with

ω=

The period of the motion is then

T=

bK + U

g

b

+ U e i + ∆Emech = K + U s + U e

s

0 + 0 + 0 − µ k mgx max = 0 +

b

2 QE − µ k mg

x max = P25.12

g

g

k . m 2π

ω

= 2π

m . k

f

1 2 kx max − QEx max 2

k 1 ayt 2 2

y f − yi = v yi t +

For the entire motion,

1 ayt 2 2 2mvi − mg − qE = − t m 2 vi −g E= q t 0 − 0 = vi t +

∑ Fy = ma y :

FG H

d

IJ K

2 2 = v yi + 2 a y y f − yi v yf

For the upward flight:

FG H

0 = vi2 + 2 −

2 vi t

IJ by K

so

ay = −

and

E=−

and

y max =

i

max

−0

g

FG IJ FG IJ FG 1 v tIJ z H K H KH 4 K I L1 2.00 kg F 2b 20.1 m sg O ∆V = − 9.80 m s J M b 20.1 m sga 4.10 sfP = G 4.10 s 5.00 × 10 C H Q KN4 ∆V = −

FG H

IJ K

m 2 vi − g j. q t 1 vi t 4

y

ymax

E ⋅ dy = +

0

max m 2 vi m 2 vi −g y = −g q t q t 0

i

2

−6

P25.13

2 vi t

40.2 kV

Arbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rod length, V = − Ed and U e = − λLEd . (a)

aK + U f = a K + U f i

0+0= v= (b)

f

1 µLv 2 − λLEd 2

2 λEd

µ

The same.

=

e

jb

ga

f=

2 40.0 × 10 −6 C m 100 N C 2.00 m

b0.100 kg mg

0.400 m s

55

56

Electric Potential

P25.14

Arbitrarily take V = 0 at point P. Then (from Equation 25.8) the potential at the original position of the charge is − E ⋅ s = − EL cos θ . At the final point a, V = − EL . Suppose the table is frictionless: K +U i = K +U f

a

f a

f

0 − qEL cos θ = v=

Section 25.3 P25.15

1 mv 2 − qEL 2

a

f

2 qEL 1 − cos θ = m

jb

e

fa

ga

2 2.00 × 10 −6 C 300 N C 1.50 m 1 − cos 60.0° 0.010 0 kg

f=

0.300 m s

Electric Potential and Potential Energy Due to Point Charges

e

je

j

e

je

j

8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C q = = 1.44 × 10 −7 V . r 1.00 × 10 −2 m

(a)

The potential at 1.00 cm is V1 = k e

(b)

8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C q The potential at 2.00 cm is V2 = k e = = 0.719 × 10 −7 V . r 2.00 × 10 −2 m Thus, the difference in potential between the two points is ∆V = V2 − V1 = −7.19 × 10 −8 V .

(c)

The approach is the same as above except the charge is −1.60 × 10 −19 C . This changes the sign of each answer, with its magnitude remaining the same. That is, the potential at 1.00 cm is −1.44 × 10 −7 V . The potential at 2.00 cm is −0.719 × 10 −7 V , so ∆V = V2 − V1 = 7.19 × 10 −8 V .

P25.16

(a)

Since the charges are equal and placed symmetrically, F = 0 .

(b)

Since F = qE = 0 , E = 0 .

(c)

q V = 2 k e = 2 8.99 × 10 9 N ⋅ m 2 C 2 r

e

× 10 C I jFGH 2.000.800 m JK −6

V = 4.50 × 10 4 V = 45.0 kV P25.17

(a)

(b)

E=

Q 4π ∈0 r 2

V=

Q 4π ∈0 r

r=

V 3 000 V = = 6.00 m E 500 V m

V = −3 000 V = Q=

Q 4π ∈0 6.00 m

a

−3 000 V

e8.99 × 10

9

V ⋅m C

f

j

a6.00 mf =

−2.00 µC

FIG. P25.16

Chapter 25

P25.18

Ex =

(a)

k e q1 x2

+

ke q2

ax − 2.00f

2

=0

Ex = k e

becomes

a

Dividing by k e ,

2 qx 2 = q x − 2.00

Therefore E = 0

when

f

2

F + q + −2 q I = 0 . GH x ax − 2.00f JK 2

2

x 2 + 4.00 x − 4.00 = 0 . x=

−4.00 ± 16.0 + 16.0 = −4.83 m . 2

(Note that the positive root does not correspond to a physically valid situation.) V=

(b)

k e q1 k q + e 2 =0 x 2.00 − x

FG + q − 2 q IJ = 0 . H x 2.00 − x K 2 qx = qa 2.00 − xf .

or

V = ke

when

x = 0.667 m

For x < 0

x = −2.00 m .

Again solving for x, For 0 ≤ x ≤ 2.00 V = 0 and

P25.19

V = ∑k i

−2 q q = . x 2−x

qi ri

e

je

V = 8.99 × 10 9 7.00 × 10 −6

1 1 O −1 − + jLMN 0.010 P 0 0.010 0 0.038 7 Q

V = −1.10 × 10 7 V = −11.0 MV FIG. P25.19 P25.20

(a)

e

je

je f

j

5.00 × 10 −9 C −3.00 × 10 −9 C 8.99 × 10 9 V ⋅ m C qQ U= = = −3.86 × 10 −7 J 4π ∈0 r 0.350 m

a

The minus sign means it takes 3.86 × 10 −7 J to pull the two charges apart from 35 cm to a much larger separation. (b)

V=

Q1 Q2 + 4π ∈0 r1 4π ∈0 r2

e5.00 × 10 Cje8.99 × 10 = −9

0.175 m

V = 103 V

9

V ⋅m C

57

j + e−3.00 × 10 Cje8.99 × 10 −9

0.175 m

9

V ⋅m C

j

58

Electric Potential

P25.21

U e = q 4V1 + q 4V2 + q 4V3 = q 4

e

U e = 10.0 × 10 −6 C

je 2

FG 1 IJ FG q H 4π ∈ K H r 0

1

+

1

8.99 × 10 9 N ⋅ m 2 C 2

q 2 q3 + r2 r3

IJ K

F 1 jGG 0.600 m + 0.1501 m + 0.600 m 1+ 0.150 m a f a f H 2

2

I JJ K

U e = 8.95 J P25.22

V=

(a)

FG IJ H K

k e q1 k e q 2 k q + =2 e r1 r2 r

F e8.99 × 10 N ⋅ m C je2.00 × 10 Cj I GG JJ 1 00 + 0 500 . m . m a a f f H K 9

V=2

2

−6

2

2

2

V = 3.22 × 10 4 V = 32.2 kV

e

je

j

U = qV = −3.00 × 10 −6 C 3.22 × 10 4 J C = −9.65 × 10 −2 J

(b) P25.23

FIG. P25.22

U = U1 + U 2 + U 3 + U 4

b

g b

U = 0 + U12 + U 13 + U 23 + U 14 + U 24 + U 34 U =0+ U=

k eQ 2 k e Q 2 + s s

k eQ s

2

FG 4 + 2 IJ = H 2K

FG 1 + 1IJ + k Q FG 1 + H 2 K s H e

5.41

k eQ s

2

g 1 2

IJ K

+1

2

FIG. P25.23

FG H

An alternate way to get the term 4 +

IJ is to recognize that there are 4 side pairs and 2 face 2K

2

diagonal pairs. P25.24

Each charge creates equal potential at the center. The total potential is: V =5

P25.25

(a)

LM k b− qg OP = MN R PQ e

−

5keq . R

Each charge separately creates positive potential everywhere. The total potential produced by the three charges together is then the sum of three positive terms. There is no point located at a finite distance from the charges, at which this total potential is zero.

(b)

V=

2k e q ke q keq + = a a a

Chapter 25

P25.26

59

Consider the two spheres as a system. (a)

e j

m1 v1 m2

Conservation of momentum:

0 = m1 v1 i + m 2 v 2 − i or v 2 =

By conservation of energy,

0=

and

k e q1 q 2 k e q1 q 2 1 1 m12 v12 − = m1 v12 + 2 2 m2 r1 + r2 d

v1 =

d

=

b g

k e − q1 q 2 1 1 m1 v12 + m 2 v 22 + r1 + r2 2 2

FG 1 − 1 IJ b g H r + r dK 2b0.700 kg ge8.99 × 10 N ⋅ m C je 2 × 10 C je3 × 10 C j F 1 G H 8 × 10 b0.100 kg gb0.800 kg g

2 m 2 k e q1 q 2 m1 m1 + m 2

1

2

9

v1 =

b g

k e − q1 q 2

2

−6

2

−6

−3

m

−

1 1.00 m

IJ K

= 10.8 m s v2 = (b)

b

g

m1 v1 0.100 kg 10.8 m s = = 1.55 m s 0.700 kg m2

If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) .

P25.27

Consider the two spheres as a system. (a)

e j

Conservation of momentum:

0 = m 1 v1 i + m 2 v 2 − i

or

v2 =

By conservation of energy,

0=

and

k e q1 q 2 k e q1 q 2 1 1 m12 v12 − = m1 v12 + . 2 2 m2 r1 + r2 d

b g

k e − q1 q 2

v1 =

v2 = (b)

m1 v1 . m2 d

=

b g

k e − q1 q 2 1 1 m1 v12 + m 2 v 22 + r1 + r2 2 2

IJ K 2m k q q F 1 G m bm + m g H r + r

2 m 2 k e q1 q 2 m 1 m1 + m 2

b

FG m IJ v Hm K 1

2

1

=

FG 1 gHr +r 1

−

2

1 d

1 e 1 2

2

1

2

1

2

−

1 d

IJ K

If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) .

60

Electric Potential

*P25.28

(a)

In an empty universe, the 20-nC charge can be placed at its location with no energy investment. At a distance of 4 cm, it creates a potential V1 =

e

je

j

8.99 × 10 9 N ⋅ m 2 C 2 20 × 10 −9 C k e q1 = = 4.50 kV . 0.04 m r

To place the 10-nC charge there we must put in energy

e

je

j

U12 = q 2 V1 = 10 × 10 −9 C 4.5 × 10 3 V = 4.50 × 10 −5 J . Next, to bring up the –20-nC charge requires energy

b

U 23 + U 13 = q3 V2 + q3 V1 = q 3 V2 + V1

g

e

= −20 × 10 −9 C 8.99 × 10 9 N ⋅ m 2 C 2

× 10 C 20 × 10 C I + jFGH 100.04 m 0.08 m JK −9

−9

= −4.50 × 10 −5 J − 4.50 × 10 −5 J The total energy of the three charges is

U12 + U 23 + U13 = −4.50 × 10 −5 J . (b)

The three fixed charges create this potential at the location where the fourth is released:

e

V = V1 + V2 + V3 = 8.99 × 10 9 N ⋅ m 2 C 2

F jGH

20 × 10 −9 0.04 2 + 0.03 2

+

10 × 10 −9 20 × 10 −9 − 0.03 0.05

I Cm JK

V = 3.00 × 10 3 V Energy of the system of four charged objects is conserved as the fourth charge flies away:

FG 1 mv + qV IJ = FG 1 mv + qV IJ H2 K H2 K 1 0 + e 40 × 10 C je3.00 × 10 V j = e 2.00 × 10 2 2e1.20 × 10 Jj = 3.46 × 10 m s v= 2

2

i

f

−9

3

−13

j

kg v 2 + 0

−4

4

2 × 10 −13 kg

*P25.29

The original electrical potential energy is U e = qV = q

ke q . d

In the final configuration we have mechanical equilibrium. The spring and electrostatic forces on k q2 k q each charge are − k 2d + q e 2 = 0 . Then k = e 3 . In the final configuration the total potential 18d 3d

a f a f 1 k q + qV = a 2d f 2 18d

keq 4 keq2 = . The missing energy must have become internal 3 3d 9 d k q 2 4k q 2 energy, as the system is isolated: e = e + ∆Eint d 9d

energy is

∆Eint =

1 2 kx 2

5 keq2 . 9 d

e

2

2

+q

Chapter 25

P25.30

af

V x =

(a)

b g

k +Q k e Q1 k e Q 2 + = e + r1 r2 x2 + a2

af

2 k eQ

kQ = e 2 2 a x +a

V x =

af 2 b k Q ag = b x ag

F 2 GG H b x ag

2

b g + a − af

k e +Q x2

I JJ +1K

2

V x e

2

+1 FIG. P25.30(a)

bg

V y =

(b)

b g

b g

k e Q 1 k e Q 2 k e +Q k e −Q + = + r1 r2 y−a y+a

b g k aQ FGH y a1− 1 − y a1+ 1 IJK F 1 − 1 I V b yg = G bk Q ag H y a − 1 y a + 1 JK e

V y =

e

FIG. P25.30(b) P25.31

V=

e

je

j

8.99 × 10 9 N ⋅ m 2 C 2 8.00 × 10 −9 C kQ k eQ 72.0 V ⋅ m = so r = e = . V V V r

For V = 100 V , 50.0 V, and 25.0 V, r = 0.720 m, 1.44 m, and 2.88 m . The radii are inversely proportional to the potential. P25.32

Using conservation of energy for the alpha particle-nucleus system, we have

K f + U f = K i + Ui .

But

Ui =

and

ri ≈ ∞.

Thus,

Ui = 0 .

Also

K f = 0 ( v f = 0 at turning point),

so

U f = Ki

ri

k e qα qgold

or

rmin =

k e qα qgold

rmin 2 k e qα qgold mα vα2

=

=

1 mα vα2 2

e

ja fa fe kg je 2.00 × 10

2 8.99 × 10 9 N ⋅ m 2 C 2 2 79 1.60 × 10 −19 C

e6.64 × 10

−27

7

ms

j

2

j

2

= 2.74 × 10 −14 m = 27.4 fm .

61

62

Electric Potential

P25.33

P25.34

Using conservation of energy k e eQ k e qQ 1 = + mv 2 we have: 2 r1 r2 2 k e eQ 1 1 − m r1 r2

FG H

IJ K

a2fe8.99 × 10

N ⋅ m 2 C 2 −1.60 × 10 −19 C 10 −9 C

which gives:

v=

or

v=

Thus,

v = 7.26 × 10 6 m s . k e qi q j

9

je

9.11 × 10

e

, summed over all pairs of i , j where i ≠ j .

rij

2

2

2

FIG. P25.34

−6 2

Each charge moves off on its diagonal line. All charges have equal speeds. ∑ K +U i = ∑ K +U f

a

0+

f

a

2

2

4k e q 2k q + e L 2L

f F 1 I 4k q = 4G mv J + H 2 K 2L 2

FG 2 + 1 IJ k q = 2mv H 2K L F 1 IJ k q v = G1 + H 8 K mL e

2

e

e

2

+

2k e q 2 2 2L

2

2

A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 × 6 = 12 face diagonal pairs separated by 2s and 4 interior diagonal pairs separated 3s . U=

Section 25.4 P25.37

2

2

9

P25.36

kg

jF 1 − 1 I . GH 0.030 0 m 0.020 0 m JK

b g L qb−2 qg + b−2 qgb3qg + b2 qgb3qg + qb2 qg + qb3qg + 2 qb−2 qg OP U=k M a b a MN b a +b a + b PQ L −2 − 6 + 6 + 2 + 3 − 4 OP U=k q M N 0.400 0.200 0.400 0.200 0.447 0.447 Q L 4 − 4 − 1 OP = −3.96 J U = e8.99 × 10 je6.00 × 10 j M N 0.400 0.200 0.447 Q U=∑

e

P25.35

je

−31

LM N

OP Q

keq2 k q2 12 4 + = 22.8 e 12 + s s 2 3 Obtaining the Value of the Electric Field from the Electric Potential

b

g

V = a + bx = 10.0 V + −7.00 V m x (a)

(b)

At x = 0 ,

V = 10.0 V

At x = 3.00 m ,

V = −11.0 V

At x = 6.00 m ,

V = −32.0 V

E=−

b

g

dV = − b = − −7.00 V m = 7.00 N C in the + x direction dx

Chapter 25

P25.38

k eQ R dV = 0 Er = − dr

(a)

For r < R

V=

(b)

For r ≥ R

V=

k eQ r kQ kQ dV = − − e2 = e2 Er = − dr r r

FG H

P25.39

IJ K

V = 5 x − 3 x 2 y + 2 yz 2

b

Evaluate E at 1, 0 , − 2

g

a fa f af a f −4yz = −4a0fa −2f = 0 + E = a −5 f + a −5 f + 0 =

∂V = −5 + 6 xy = −5 + 6 1 0 = −5 ∂x ∂V 2 2 = +3 x 2 − 2 z 2 = 3 1 − 2 −2 = −5 Ey = − ∂y Ex = −

Ez = −

∂V = ∂z

E = E x2 + E y2 P25.40

2

2 z

2

7.07 N C

∆V ∆s

E A > EB since E =

(a)

2

a f

6−2 V ∆V =− = 200 N C down 2 cm ∆s

(b)

EB = −

(c)

The figure is shown to the right, with sample field lines sketched in. FIG. P25.40

P25.41

Ey = −

Ey =

Section 25.5 P25.42

LM MMN

∂V ∂ k eQ =− ln ∂y ∂y

LM MN

k eQ 1− y

F GG H

2

+

y

y2 2

+ y2 +

+ y2

2

+ y2

OP PQ =

I OP JJ P K PQ k eQ y

2

+ y2

Electric Potential Due to Continuous Charge Distributions

∆V = V2 R − V0 =

k eQ

a f

R2 + 2R

2

−

k eQ k eQ = R R

FG 1 − 1IJ = H 5 K

−0.553

k eQ R

63

64

Electric Potential

P25.43

LM λ OP = C ⋅ FG 1 IJ = N x Q m H mK

(a)

α =

(b)

V = ke

z

C m2

z

LM N

z

FG H

L dq λdx xdx L = ke = k eα = k eα L − d ln 1 + r r d+x d 0

IJ OP KQ FIG. P25.43

P25.44

V=

z

k e dq = ke r

z

αxdx

b

b2 + L 2 − x

g

2

L −x. 2

Let z =

Then x = V = k eα

L − z , and dx = − dz 2

zb

ga f = − k αL

L 2 − z − dz 2

b +z

e

2

2

z

dz 2

b +z

2

+ k eα

z

zdz 2

b +z

2

=−

FH

LMF L I F L I OP GMH 2 − xJK + GH 2 − xJK + b P + k α FGH L2 − xIJK + b N Q L O L FL I k αL M L 2 − L + b L 2 g + b P F LI V=− ln M + k α M G − LJ + b − G J P H K H 2K 2 2 M MN L 2 + bL 2g + b PQ N L O k αL M b + eL 4j − L 2 P V= − ln M 2 MN b + eL 4j + L 2 PPQ L

2

k αL V = − e ln 2

2

L

2

2

e

0

0

2

2

e

2

2

2

2

2

2

2

2

e

IK

k eαL ln z + z 2 + b 2 + k eα z 2 + b 2 2

2

+ b2

OP PQ

e

P25.45

z

1 4π ∈0

V = dV =

z

dq r

All bits of charge are at the same distance from O. So V =

P25.46

dV =

FG IJ e H K

1 Q = 8.99 × 10 9 N ⋅ m 2 C 2 4π ∈0 R k e dq

z b

a

−6

−1.51 MV .

where dq = σdA = σ 2π rdr

r 2 + x2

V = 2πσk e

.50 × 10 C I = jFGH −70.140 m π JK

rdr 2

r + x2

= 2π k eσ

LM N

x2 + b2 − x2 + a2

OP Q FIG. P25.46

Chapter 25

P25.47

V = ke

z

z

a f

V = − k e λ ln − x V = k e ln

Section 25.6 P25.48

z

z

3R −R dq λdx λds λdx = ke + ke + ke −x r R x R all charge semicircle −3 R −R −3 R

+

keλ 3R π R + k e λ ln x R R

a

3R + k e λ π + k e ln 3 = k e λ π + 2 ln 3 R

f

Electric Potential Due to a Charged Conductor

Substituting given values into V =

ke q r

7.50 × 10 3 V =

e8.99 × 10

9

j

N ⋅ m2 C 2 q

0.300 m

.

Substituting q = 2.50 × 10 −7 C , N=

P25.49

(a)

2.50 × 10 −7 C = 1.56 × 10 12 electrons . 1.60 × 10 −19 C e −

E= 0 ;

e

je

j

8.99 × 10 9 26.0 × 10 −6 ke q = = 1.67 MV V= 0.140 R (b)

e8.99 × 10 je26.0 × 10 j = 5.84 MN C r a0.200f k q e8.99 × 10 je 26.0 × 10 j = = 1.17 MV V= E=

keq 2

−6

9

=

2

away

−6

9

e

(c)

E= V=

R

0.200

keq

e8.99 × 10 je26.0 × 10 j = = a0.140f

R2

−6

9

ke q = 1.67 MV R

2

11.9 MN C away

65

66

Electric Potential

*P25.50

(a)

Both spheres must be at the same potential according to where also

q1 + q 2 = 1.20 × 10 −6 C .

Then

q1 =

k e q1 k e q 2 = r1 r2

q 2 r1 r2

q 2 r1 + q 2 = 1.20 × 10 −6 C r2 1.20 × 10 −6 C = 0.300 × 10 −6 C on the smaller sphere 1 + 6 cm 2 cm

q2 =

q1 = 1.20 × 10 −6 C − 0.300 × 10 −6 C = 0.900 × 10 −6 C V= (b)

e

je

j

8.99 × 10 9 N ⋅ m 2 C 2 0.900 × 10 −6 C k e q1 = = 1.35 × 10 5 V r1 6 × 10 −2 m

Outside the larger sphere, E1 =

k e q1 r12

r=

V1 1.35 × 10 5 V r= r = 2. 25 × 10 6 V m away . 0.06 m r1

Outside the smaller sphere, 1.35 × 10 5 V r = 6.74 × 10 6 V m away . 0.02 m

E2 =

The smaller sphere carries less charge but creates a much stronger electric field than the larger sphere.

Section 25.7

The Milliken Oil Drop Experiment

Section 25.8

Application of Electrostatistics

P25.51

(a)

Emax = 3.00 × 10 6 V m = 6

k eQ

a

r

2

=

f

FG IJ HK

FG IJ HK

k eQ 1 1 = Vmax r r r

Vmax = Emax r = 3.00 × 10 0.150 = 450 kV (b)

P25.52

V=

k eQmax r2

= Emax

RSor k Q T r e

max

= Vmax

UV W

ke q k q V and E = e2 . Since E = , r r r

(b)

r=

6.00 × 10 5 V V = = 0.200 m and E 3.00 × 10 6 V m

(a)

q=

Vr = 13.3 µC ke

Q max =

a

6 Emax r 2 3.00 × 10 0.150 = ke 8.99 × 10 9

f

2

= 7.51 µC

Chapter 25

67

Additional Problems

P25.53

q q U = qV = k e 1 2 = 8.99 × 10 9 r12

P25.54

(a)

e

a38fa54f 1.60 × 10 j a5.50 +e6.20f × 10 j

−19 2 −15

= 4.04 × 10 −11 J = 253 MeV

To make a spark 5 mm long in dry air between flat metal plates requires potential difference

e

je

j

∆V = Ed = 3 × 10 6 V m 5 × 10 −3 m = 1.5 × 10 4 V ~ 10 4 V . The area of your skin is perhaps 1.5 m 2 , so model your body as a sphere with this surface area. Its radius is given by 1.5 m 2 = 4π r 2 , r = 0.35 m . We require that you are at the potential found in part (a):

(b)

ke q r

V=

q=

a

f FG H

1.5 × 10 4 V 0.35 m Vr J = k e 8.99 × 10 9 N ⋅ m 2 C 2 V ⋅ C

IJ FG N ⋅ m IJ KH J K

q = 5.8 × 10 −7 C ~ 10 −6 C .

P25.55

P25.56

e

je

j

2

e

je

j

2

(a)

−19 9 k e q1 q 2 − 8.99 × 10 1.60 × 10 = U= r 0.052 9 × 10 −9

(b)

−19 9 k e q1 q 2 − 8.99 × 10 1.60 × 10 U= = r 2 2 0.052 9 × 10 −9

(c)

U=

e

j

= −4.35 × 10 −18 J = −27.2 eV

= −6.80 eV

k e q1 q 2 − k e e 2 = = 0 r ∞

From Example 25.5, the potential created by the ring at the electron’s starting point is Vi =

k eQ x i2

+a

2

=

b

g

xi2

2

k e 2πλa +a

while at the center, it is V f = 2π k e λ . From conservation of energy,

b g

v 2f

1 m e v 2f + − eV f 2

i I 4π ek λ F 2e a 1− = V −V i= G d GH x + a JJK m m 4π e1.60 × 10 je8.99 × 10 je1.00 × 10 j F 0.200 GG1 − = 9.11 × 10 H a0.100f + a0.200f

0 + − eVi =

e

f

d

e

i

2 i

e

−19

v 2f

v f = 1.45 × 10 7 m s

9

−31

2

−7

2

2

I JJ K

68

Electric Potential

*P25.57

b g = 2V

V0 − −V0

0 . d d Assume the ball swings a small distance x to the right. It moves to a place where the voltage created 2V by the plates is lower by − Ex = − 0 x . Its ground connection maintains it at V = 0 by allowing d 2V x k q 2V xR . Then the ball charge q to flow from ground onto the ball, where − 0 + e = 0 q= 0 d R ked

The plates create uniform electric field to the right in the picture, with magnitude

feels electric force F = qE =

4V02 xR ked 2

to the right. For equilibrium this must be balanced by the

horizontal component of string tension according to T cos θ = mg tan θ =

4V02 xR

=

2

k e d mg

F GH

k d 2 mg x for small x. Then V0 = e L 4RL

I JK

T sin θ =

12

4V02 xR ked 2

.

If V0 is less than this value, the only equilibrium position of the ball is hanging straight down. If V0 exceeds this value the ball will swing over to one plate or the other. P25.58

(a)

Take the origin at the point where we will find the potential. One ring, of width dx, has Qdx charge and, according to Example 25.5, creates potential h k eQdx dV = . h x2 + R2 The whole stack of rings creates potential

z

V=

dV =

all charge

(b)

z

d+h d

k eQdx h x2 + R2

=

FH

k eQ ln x + x 2 + R 2 h

A disk of thickness dx has charge

IK

d+h

=

d

F GG H

a f

2

d + h + d + h + R2 k eQ ln h d + d2 + R2

Qdx Qdx and charge-per-area . According to h π R2h

Example 25.6, it creates potential Qdx π R2h Integrating, dV = 2π k e

V=

2 k eQ 2

z

W = Vdq 0

where V =

ke q . R

Therefore, W =

2

2

k eQ 2 . 2R

2

e 2

2

e 2

Q

P25.59

IK

x2 + R2 − x .

z R h FH x + R dx − xdxIK = 2Rk Qh LMN 12 x x + R L kQM ad + hf ad + hf + R − d d + R − 2dh − h R hM NM

d+h d

V=

FH

2

2

2

2

+

FH IK OP Q F d + h + ad + hf + R I OP lnG GH d + d + R JJK PPQ

R2 x2 ln x + x 2 + R 2 − 2 2 2

2

+R

2

2

d+h

d

2

2

I JJ K

.

Chapter 25

P25.60

The positive plate by itself creates a field E =

69

36.0 × 10 −9 C m 2 σ = 2.03 kN C away = 2 ∈0 2 8.85 × 10 −12 C 2 N ⋅ m 2

e

j

from the + plate. The negative plate by itself creates the same size field and between the plates it is in the same direction. Together the plates create a uniform field 4.07 kN C in the space between. (a)

Take V = 0 at the negative plate. The potential at the positive plate is then V −0=−

zb

12.0 cm

g

−4.07 kN C dx .

0

e

ja

f

The potential difference between the plates is V = 4.07 × 10 3 N C 0.120 m = 488 V . (b)

FG 1 mv + qV IJ = FG 1 mv + qV IJ H2 K H2 K 1 qV = e1.60 × 10 C ja 488 V f = mv 2 2

2

i

f

−19

(c)

v f = 306 km s

(d)

v 2f = vi2 + 2 a x f − xi

d

e3.06 × 10

5

ms

j

2

2 f

= 7.81 × 10 −17 J

i

a

f

= 0 + 2 a 0.120 m

a = 3.90 × 10 11 m s 2

P25.61

(e)

∑ F = ma = e1.67 × 10 −27

je

j

(f)

E=

(a)

VB − VA = − E ⋅ ds and the field at distance r from a uniformly

kg 3.90 × 10 11 m s 2 = 6.51 × 10 −16 N

F 6.51 × 10 −16 N = = 4.07 kN C q 1.60 × 10 −19 C

z

B

A

charged rod (where r > radius of charged rod) is E=

λ 2π ∈0 r

2keλ . r

=

In this case, the field between the central wire and the coaxial cylinder is directed perpendicular to the line of charge so that VB − VA = −

z

rb ra

or

FG IJ H K

r 2keλ dr = 2 k e λ ln a , r rb

FG r IJ Hr K

∆V = 2 k e λ ln

continued on next page

a

b

.

FIG. P25.61

70

Electric Potential

(b)

From part (a), when the outer cylinder is considered to be at zero potential, the potential at a distance r from the axis is V = 2 k e λ ln

FG r IJ . HrK a

The field at r is given by E=−

FG H

r ∂V = −2 k e λ ra ∂r

b ∆V F 1 I E= GJ lnbr r g H r K a

P25.62

(a)

a 2

e

∆V . ln ra rb

But, from part (a), 2k e λ = Therefore,

IJ FG − r IJ = 2k λ . KH r K r g

.

b

From Problem 61, E=

∆V 1 . ln ra rb r

b

g

We require just outside the central wire 5.50 × 10 6 V m =

50.0 × 10 3 V ln 0.850 m rb

b

FG 1 IJ gHr K b

e110 m jr lnFGH 0.850r m IJK = 1 . −1

or

b

b

We solve by homing in on the required value

a f jr lnFGH 0.850r m IJK rb m

e110 m

−1

b

0.0100

0.00100

0.00150

0.00145

0.00143

0.00142

4.89

0.740

1.05

1.017

1.005

0.999

b

Thus, to three significant figures,

rb = 1.42 mm . (b)

At ra , E=

P25.63

z

r2

b

V2 − V1 = − E ⋅ dr = − r1

V2 − V1 =

FG 1 IJ = g H 0.850 m K

50.0 kV ln 0.850 m 0.001 42 m

z

r2 r1

λ 2π ∈0 r

FG IJ H K

r −λ ln 2 r1 2π ∈0

dr

9.20 kV m .

71

Chapter 25

*P25.64

Take the illustration presented with the problem as an initial picture. No external horizontal forces act on the set of four balls, so its center of mass stays fixed at the location of the center of the square. As the charged balls 1 and 2 swing out and away from each other, balls 3 and 4 move up with equal y-components of velocity. The maximum-kineticenergy point is illustrated. System energy is conserved:

v 1 +

P25.65

CM

3

v

2 + 4

v

FIG. P25.64

keq2 keq 2 1 1 1 1 = + mv 2 + mv 2 + mv 2 + mv 2 3a 2 2 2 2 a 2k e q 2 = 2mv 2 3a

v

keq2 3 am

v=

g k rbqg + k br−2 qg

b

e

V x , y, z =

For the given charge distribution,

e

1

2

ax + Rf + y + z and r = x + y + z . V b x , y , zg = 0 F1 2I k qG − J = 0 , or 2r = r . Hr r K 4a x + R f + 4y + 4z = x + y + z F8 I F4 I x + y + z + G RJ x + a0fy + a0fz + G R J = 0 . H3 K H3 K 2

r1 =

where The surface on which is given by

e

1

2

which may be written in the form:

2

2

2

2

2

2

1

2

2

This gives:

2

2

2

2

2

2

2

2

2

b

[1]

g

The general equation for a sphere of radius a centered at x 0 , y 0 , z0 is:

bx − x g + by − y g + bz − z g − a = 0 x + y + z + b −2 x gx + b −2 y gy + b −2 z gz + e x

or

2

2

2

0

2

0

2

0

0

2

2

0

0

2 0

j

+ y 02 + z 02 − a 2 = 0 .

[2]

Comparing equations [1] and [2], it is seen that the equipotential surface for which V = 0 is indeed a sphere and that: −2 x 0 = Thus, x 0 = −

4 8 R ; −2 y 0 = 0 ; −2 z 0 = 0 ; x 02 + y 02 + z 02 − a 2 = R 2 . 3 3

FG H

IJ K

4 16 4 2 4 2 R , y 0 = z 0 = 0 , and a 2 = − R = R . 3 9 3 9

The equipotential surface is therefore a sphere centered at

FG − 4 R, 0 , 0IJ H 3 K

, having a radius

2 R . 3

72

Electric Potential

P25.66

(a)

E A = 0 (no charge within)

From Gauss’s law,

EB = k e

e1.00 × 10 j = FG 89.9 IJ V m e j Hr K r r bq + q g = e8.99 × 10 j e−5.00 × 10 j = FG − 45.0 IJ V m H r K r r −8

qA

= 8.99 × 10 9

2

2

2

−9

EC = k e

VC = k e

(b)

A

bq

A

B

+ qB r

∴ At r2 , V = −

9

2

g = e8.99 × 10 j e−5.00 × 10 j = FG − 45.0 IJ V H r K r −9

9

45.0 = −150 V 0.300

Inside r2 , VB = −150 V +

∴ At r1 , V = −450 + P25.67

2

FG H

z r

IJ FG −450 + 89.9 IJ V K H r K

89.9 1 1 = dr = −150 + 89.9 − 2 r 0 . 300 r2 r

89.9 = +150 V so VA = +150 V . 0.150

From Example 25.5, the potential at the center of the ring is kQ Vi = e and the potential at an infinite distance from the ring is R V f = 0 . Thus, the initial and final potential energies of the point charge-ring system are: U i = QVi =

k eQ 2 R

FIG. P25.67

U f = QV f = 0 .

and

From conservation of energy,

K f + U f = K i + Ui

P25.68

or

k Q2 1 Mv 2f + 0 = 0 + e R 2

giving

vf =

V = ke

z

a+L a

λdx x2 + b2

2 k eQ 2 MR

.

= k λ ln LM x + e x N e

2

+b

2

j OPQ

= a

L a + L + aa + Lf k λ ln M MMN a + a + b

2

a +L

e

2

2

+ b2

OP PPQ

Chapter 25

*P25.69

(a)

(b)

V=

keq ke q keq − = r2 − r1 r1 r2 r1 r2

b

g

From the figure, for r >> a ,

r2 − r1 ≅ 2 a cos θ .

Then

V≅

Er = −

73

ke q k p cos θ 2 a cos θ ≅ e 2 . r1 r2 r

2 k e p cos θ ∂V = ∂r r3

FG IJ H K

1 ∂ In spherical coordinates, the θ component of the gradient is . r ∂θ Eθ = −

Therefore,

FIG. P25.69

FG IJ H K

k p sin θ 1 ∂V . = e 3 r ∂θ r

a f 2rk p E a90°f = 0 , E a0°f = 0 k p E a90°f = . r

For r >> a

E r 0° =

and

e 3

r

θ

and

θ

e 3

These results are reasonable for r >> a . Their directions are as shown in Figure 25.13 (c).

af

However, for r → 0 , E 0 → ∞. This is unreasonable, since r is not much greater than a if it is 0. (c)

V=

and

ex

k e py 2

+ y2

j

Ex = −

3 2

3 k e pxy ∂V = 52 2 ∂x x + y2

e

j

e

2 2 ∂V k e p 2 y − x = Ey = − 5 2 ∂y x2 + y2

e

j

j

74

Electric Potential

P25.70

Inside the sphere, Ex = Ey = Ez = 0 .

j IK L O F 3I So E = − M0 + 0 + E a z G − J e x + y + z j 2 x fP = 3E a xze x a K H 2 N Q ∂V ∂ I = − F V − E z + E a ze x + y + z j E =− K ∂y ∂y H F 3I E = − E a zG − J e x + y + z j 2 y = 3E a yze x + y + z j H 2K ∂V F 3I E =− = E − E a zG − J e x + y + z j a 2 zf − E a e x + y + z j H 2K ∂z E = E + E a e 2 z − x − y je x + y + z j Ex = −

Outside,

F H

∂V ∂ =− V0 − E0 z + E0 a 3 z x 2 + y 2 + z 2 ∂x ∂x

x

0

y

0

y

3

0

z

0

z

P25.71

0

2

2

0

3

0

0

e

3

0

2

3

2

3

2

2

2

2

0

3

3

2

+ y2 + z2

j

−5 2

2 −5 2

2 −5 2

2

0

3

2

2

2 −3 2

2 −5 2

2

k e dq

For an element of area which is a ring of radius r and width dr, dV =

b

2

2 −3 2

0

2

2

2 −5 2

2

2

2 −5 2

−3 2

.

r 2 + x2

g

dq = σdA = Cr 2π rdr and

b

V = C 2π k e

gz

R 0

P25.72

L b gMM N

r 2 dr

= C π k e R R 2 + x 2 + x 2 ln

r 2 + x2

dU = Vdq where the potential V =

F GH R +

x R2 + x2

I OP JK P Q

.

ke q . r

e

j

The element of charge in a shell is dq = ρ (volume element) or dq = ρ 4π r 2 dr and the charge q in a sphere of radius r is

z r

q = 4πρ r 2 dr = ρ 0

F 4π r I . GH 3 JK 3

Substituting this into the expression for dU, we have

FG k q IJ dq = k ρFG 4π r IJ FG 1 IJ ρe4π r dr j = k FG 16π IJ ρ r dr HrK H 3 K H 3 KH r K F 16π I ρ r dr = k F 16π I ρ R U = z dU = k G GH 15 JK H 3 JK z dU =

3

e

e

2

e

2

R

2

2

4

e

2

e

2

2 4

5

0

3 k eQ 2 4 But the total charge, Q = ρ π R 3 . Therefore, U = . 5 R 3

Chapter 25

*P25.73

(a)

The whole charge on the cube is 3 q = 100 × 10 −6 C m3 0.1 m = 10 −7 C . Divide up the cube into

ja

e

f

64 or more elements. The little cube labeled a creates at P ke q . The others in the potential 2 64 6.25 + 1.25 2 + 1.25 2 10 −2 m horizontal row behind it contribute

e

keq

64 10 −2

F G mj H

1 8.75 2 + 3.125

+

1 11.25 2 + 3.125

1

+

13.75 2

I. J + 3.125 K

d c

b a P

The little cubes in the rows containing b and c add

e

2k e q

64 10

−2

Le6.25 + 1.25 + 3.75 j + e8.75 + 15.625j mj MN OP + e11.25 + 15.625j + e13.75 + 15.625 j Q 2

2 −1 2

2

−1 2

2

2

e

keq

64 10 −2

LMe6.25 mj N

2

+ 28.125

The whole potential at P is

j

−1 2

−1 2

1.25 cm

−1 2

2

and the bits in row d make potential at P

e

+ … + 13.75 2 + 28.125

j OQP . −1 2

8.987 6 × 10 9 Nm 2 × 10 −7 C

e

FIG. P25.73

j

C 2 64 10 −2 m

b1.580 190g4 =

8 876 V . If we use

more subdivisions of the large cube, we get the same answer to four digits. (b)

75

A sphere centered at the same point would create potential k e q 8.987 6 × 10 9 Nm 2 × 10 −7 C = = 8 988 V , larger by 112 V . r C2 10 −1 m

ANSWERS TO EVEN PROBLEMS P25.2

6.41 × 10 −19 C

P25.4

−0.502 V

P25.6

1.67 MN C

P25.8

(a) 59.0 V ; (b) 4.55 Mm s

P25.10

see the solution

P25.12

40.2 kV

P25.14

0.300 m s

P25.16

(a) 0; (b) 0; (c) 45.0 kV

P25.18

(a) −4.83 m ; (b) 0.667 m and −2.00 m

P25.20

(a) −386 nJ ; (b) 103 V

P25.22

(a) 32.2 kV ; (b) −96.5 mJ

P25.24

−

P25.26

(a) 10.8 m s and 1.55 m s ; (b) greater

P25.28

(a) −45.0 µ J ; (b) 34.6 km s

P25.30

see the solution

P25.32

27.4 fm

P25.34

−3.96 J

P25.36

22.8

5k e q R

keq 2 s

76

Electric Potential

k eQ

P25.60

P25.38

(a) 0; (b)

P25.40

(a) larger at A; (b) 200 N C down; (c) see the solution

P25.42

−0.553

P25.44

L k αL M − ln M 2 MN

r

radially outward

2

(d) 390 Gm s 2 toward the negative plate; (e) 6.51 × 10 −16 N toward the negative plate; (f) 4.07 kN C toward the negative plate

k eQ R

e

LM N

b

2

b2

+ eL 4j − L 2 OP P + e L 4j + L 2 P Q 2

P25.62

(a) 1.42 mm ; (b) 9. 20 kV m

P25.64

Fk q I GH 3am JK

2

x2 + b 2 − x2 + a2

P25.66

OP Q

P25.46

2π k eσ

P25.48

1.56 × 10 12 electrons

P25.50

(a) 135 kV ; (b) 2.25 MV m away from the

P25.54

(a) ~ 10 V ; (b) ~ 10

P25.56

14.5 Mm s

−6

C P25.70

e 2

2

2

B

2

2

2

2

2

OP I PP JJ P K PQ

2

e j ; E = 3E a yze x + y + z j ; E a e2z − x − y j E =E + outside and ex + y + z j z

0

3

2

0

E = 0 inside P25.72

2

Ex = 3E0 a 3 xz x 2 + y 2 + z 2

0

2

2

2

2

2

e

y

d + h + d + h + R2 k eQ ln h d + d 2 + R2 2

FG 89.9 IJ V m radially Hr K F 45.0 IJ V m radially outward; E = G − H r K outward; 89.9 I F (b) V = 150 V ; V = G −450 + J V; H r K F 45.0 IJ V V = G− H r K L a + L + aa + Lf + b OP k λ ln M MMN a + a + b PPQ (a) E A = 0 ; E B =

C

2

(a)

12

A

P25.68

F a f IJ ; GG JK H LMad + hf ad + hf + R − d d + R kQ M F d + h + a d + hf + R (b) R h MM−2dh − h + R lnG GH d + d + R MN

P25.58

2

2

(a) 13.3 µC ; (b) 0.200 m 4

e

C

large sphere and 6.74 MV m away from the small sphere P25.52

(a) 488 V ; (b) 7.81 × 10 −17 J ; (c) 306 km s ;

3 k eQ 2 5 R

−5 2

2 −5 2

2

3

2

2

2

2

2 5 2

2

26 Capacitance and Dielectrics CHAPTER OUTLINE 26.1 26.2 26.3 26.4 26.5 26.6 26.7

Q26.4

ANSWERS TO QUESTIONS

Definition of Capacitance Calculating Capacitance Combinations of Capacitors Energy Stored in a Charged Capacitor Capacitors with Dielectrics Electric Dipole in an Electric Field An Atomic Description of Dielectrics

Q26.1

Nothing happens to the charge if the wires are disconnected. If the wires are connected to each other, charges in the single conductor which now exists move between the wires and the plates until the entire conductor is at a single potential and the capacitor is discharged.

Q26.2

336 km. The plate area would need to be

Q26.3

The parallel-connected capacitors store more energy, since they have higher equivalent capacitance.

1 m2 . ∈0

Seventeen combinations: Individual

C1 , C 2 , C 3

Parallel

C 1 + C 2 + C 3 , C 1 + C 2 , C1 + C 3 , C 2 + C 3

FG 1 + 1 IJ + C , FG 1 + 1 IJ + C , FG 1 + 1 IJ + C HC C K HC C K HC C K FG 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJ HC +C C K HC +C C K HC +C C K FG 1 + 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJ HC C C K HC C K HC C K HC C K −1

Series-Parallel

1

−1

3

2

1

−1

2

3

−1

1

2

3

1

2

3

1

3

−1

1

3

−1

2

−1

Series

2

2

3

−1

1

2

1

−1

2

3

1

−1

3

Q26.5

This arrangement would decrease the potential difference between the plates of any individual capacitor by a factor of 2, thus decreasing the possibility of dielectric breakdown. Depending on the application, this could be the difference between the life or death of some other (most likely more expensive) electrical component connected to the capacitors.

Q26.6

No—not just using rules about capacitors in series or in parallel. See Problem 72 for an example. If connections can be made to a combination of capacitors at more than two points, the combination may be irreducible. 77

78

Capacitance and Dielectrics

Q26.7

A capacitor stores energy in the electric field between the plates. This is most easily seen when using a “dissectable” capacitor. If the capacitor is charged, carefully pull it apart into its component pieces. One will find that very little residual charge remains on each plate. When reassembled, the capacitor is suddenly “recharged”—by induction—due to the electric field set up and “stored” in the dielectric. This proves to be an instructive classroom demonstration, especially when you ask a student to reconstruct the capacitor without supplying him/her with any rubber gloves or other insulating material. (Of course, this is after they sign a liability waiver).

Q26.8

The work you do to pull the plates apart becomes additional electric potential energy stored in the capacitor. The charge is constant and the capacitance decreases but the potential difference increases 1 to drive up the potential energy Q∆V . The electric field between the plates is constant in strength 2 but fills more volume as you pull the plates apart.

Q26.9

A capacitor stores energy in the electric field inside the dielectric. Once the external voltage source is removed—provided that there is no external resistance through which the capacitor can discharge—the capacitor can hold onto this energy for a very long time. To make the capacitor safe to handle, you can discharge the capacitor through a conductor, such as a screwdriver, provided that you only touch the insulating handle. If the capacitor is a large one, it is best to use an external resistor to discharge the capacitor more slowly to prevent damage to the dielectric, or welding of the screwdriver to the terminals of the capacitor.

Q26.10

The work done, W = Q∆V , is the work done by an external agent, like a battery, to move a charge through a potential difference, ∆V . To determine the energy in a charged capacitor, we must add the work done to move bits of charge from one plate to the other. Initially, there is no potential difference between the plates of an uncharged capacitor. As more charge is transferred from one plate to the other, the potential difference increases as shown in Figure 26.12, meaning that more work is needed to transfer each additional bit of charge. The total work is the area under the curve of 1 Figure 26.12, and thus W = Q∆V . 2

Q26.11

Energy is proportional to voltage squared. It gets four times larger.

Q26.12

Let C = the capacitance of an individual capacitor, and C s represent the equivalent capacitance of the group in series. While being charged in parallel, each capacitor receives charge

e

ja

f

Q = C∆Vcharge = 500 × 10 −4 F 800 V = 0.400 C . While being discharged in series, (or 10 times the original voltage).

∆Vdischarge =

Q Q 0. 400 C = = = 8.00 kV C s C 10 5.00 × 10 −5 F

Q26.13

Put a material with higher dielectric strength between the plates, or evacuate the space between the plates. At very high voltages, you may want to cool off the plates or choose to make them of a different chemically stable material, because atoms in the plates themselves can ionize, showing thermionic emission under high electric fields.

Q26.14

The potential difference must decrease. Since there is no external power supply, the charge on the capacitor, Q, will remain constant—that is assuming that the resistance of the meter is sufficiently large. Adding a dielectric increases the capacitance, which must therefore decrease the potential difference between the plates.

Q26.15

Each polar molecule acts like an electric “compass” needle, aligning itself with the external electric field set up by the charged plates. The contribution of these electric dipoles pointing in the same direction reduces the net electric field. As each dipole falls into a configuration of lower potential energy it can contribute to increasing the internal energy of the material.

Chapter 26

79

Q26.16

The material of the dielectric may be able to support a larger electric field than air, without breaking down to pass a spark between the capacitor plates.

Q26.17

The dielectric strength is a measure of the potential difference per unit length that a dielectric can withstand without having individual molecules ionized, leaving in its wake a conducting path from plate to plate. For example, dry air has a dielectric strength of about 3 MV/m. The dielectric constant in effect describes the contribution of the electric dipoles of the polar molecules in the dielectric to the electric field once aligned.

Q26.18

In water, the oxygen atom and one hydrogen atom considered alone have an electric dipole moment that points from the hydrogen to the oxygen. The other O-H pair has its own dipole moment that points again toward the oxygen. Due to the geometry of the molecule, these dipole moments add to have a non-zero component along the axis of symmetry and pointing toward the oxygen. A non-polarized molecule could either have no intrinsic dipole moments, or have dipole moments that add to zero. An example of the latter case is CO 2 . The molecule is structured so that each CO pair has a dipole moment, but since both dipole moments have the same magnitude and opposite direction—due to the linear geometry of the molecule—the entire molecule has no dipole moment.

Q26.19

Heating a dielectric will decrease its dielectric constant, decreasing the capacitance of a capacitor. When you heat a material, the average kinetic energy per molecule increases. If you refer back to the answer to Question 26.15, each polar molecule will no longer be nicely aligned with the applied electric field, but will begin to “dither”—rock back and forth—effectively decreasing its contribution to the overall field.

Q26.20

The primary choice would be the dielectric. You would want to chose a dielectric that has a large dielectric constant and dielectric strength, such as strontium titanate, where κ ≈ 233 (Table 26.1). A convenient choice could be thick plastic or mylar. Secondly, geometry would be a factor. To maximize capacitance, one would want the individual plates as close as possible, since the capacitance is proportional to the inverse of the plate separation—hence the need for a dielectric with a high dielectric strength. Also, one would want to build, instead of a single parallel plate capacitor, several capacitors in parallel. This could be achieved through “stacking” the plates of the capacitor. For example, you can alternately lay down sheets of a conducting material, such as aluminum foil, sandwiched between your sheets of insulating dielectric. Making sure that none of the conducting sheets are in contact with their next neighbors, connect every other plate together. Figure Q26.20 illustrates this idea.

Dielectric

Conductor FIG. Q26.20 This technique is often used when “home-brewing” signal capacitors for radio applications, as they can withstand huge potential differences without flashover (without either discharge between plates around the dielectric or dielectric breakdown). One variation on this technique is to sandwich together flexible materials such as aluminum roof flashing and thick plastic, so the whole product can be rolled up into a “capacitor burrito” and placed in an insulating tube, such as a PVC pipe, and then filled with motor oil (again to prevent flashover).

80

Capacitance and Dielectrics

SOLUTIONS TO PROBLEMS Section 26.1 P26.1

P26.2

P26.4

e

ja

f

e

ja

f

(a)

Q = C∆V = 4.00 × 10 −6 F 12.0 V = 4.80 × 10 −5 C = 48.0 µC

(b)

Q = C∆V = 4.00 × 10 −6 F 1.50 V = 6.00 × 10 −6 C = 6.00 µC

(a)

C=

(b)

∆V =

Section 26.2 P26.3

Definition of Capacitance

E=

Q 10.0 × 10 −6 C = = 1.00 × 10 −6 F = 1.00 µF ∆V 10.0 V Q 100 × 10 −6 C = = 100 V C 1.00 × 10 −6 F

Calculating Capacitance keq r2

e4.90 × 10 N Cja0.210 mf e8.99 × 10 N ⋅ m C j 4

q=

:

2

9

2

= 0.240 µC

q 0.240 × 10 −6 = = 1.33 µC m 2 A 4π 0.120 2

(a)

σ=

(b)

C = 4π ∈0 r = 4π 8.85 × 10 −12 0.120 = 13.3 pF

(a)

C = 4π ∈0 R C R= = k eC = 8.99 × 10 9 N ⋅ m 2 C 2 1.00 × 10 −12 F = 8.99 mm 4π ∈0

a

f

ja

e

f

e

P26.5

2

je

e

je

j

j=

4π 8.85 × 10 −12 C 2 2.00 × 10 −3 m

(b)

C = 4π ∈0 R =

(c)

Q = CV = 2.22 × 10 −13 F 100 V = 2.22 × 10 −11 C

(a)

Q1 R1 = Q 2 R2

ja

e

FG H

Q1 + Q 2 = 1 +

Q 2 = 2.00 µC (b)

N ⋅m

V1 = V2 =

2

0. 222 pF

f

IJ K

R1 Q 2 = 3.50Q 2 = 7.00 µC R2

Q1 = 5.00 µC

Q1 Q 2 5.00 µC = = = 8.99 × 10 4 V = 89.9 kV −1 9 C1 C 2 8.99 × 10 m F 0.500 m

e

j a

f

Chapter 26

P26.6

a fe

je

j

−12 3 2 κ ∈0 A 1.00 8.85 × 10 C 1.00 × 10 m C= = d N ⋅ m 2 800 m

a

f

2

= 11.1 nF

The potential between ground and cloud is

ja

e

f

∆V = Ed = 3.00 × 10 6 N C 800 m = 2.40 × 10 9 V

a f e

je

j

Q = C ∆V = 11.1 × 10 −9 C V 2.40 × 10 9 V = 26.6 C P26.7

∆V = Ed

(a)

E=

E=

(b)

20.0 V 1.80 × 10 −3 m

= 11.1 kV m

σ ∈0

e

je

j

σ = 1.11 × 10 4 N C 8.85 × 10 −12 C 2 N ⋅ m 2 = 98.3 nC m 2

e

jb

je

g

(c)

8.85 × 10 −12 C 2 N ⋅ m 2 7.60 cm 2 1.00 m 100 cm ∈0 A = C= d 1.80 × 10 −3 m

(d)

∆V =

Q C

a

fe

j

Q = 20.0 V 3.74 × 10 −12 F = 74.7 pC P26.8

C=

κ ∈0 A = 60.0 × 10 −15 F d

a fe

je

−12 21.0 × 10 −12 κ ∈0 A 1 8.85 × 10 = C 60.0 × 10 −15 d = 3.10 × 10 −9 m = 3.10 nm

d=

P26.9

Q=

d=

a f

∈0 A ∆V d

a f=

∈0 ∆V

σ

j

a f

∈ ∆V Q =σ = 0 A d

e8.85 × 10

e30.0 × 10

−9

−12

ja

C 2 N ⋅ m 2 150 V

je

f

C cm 2 1.00 × 10 4 cm 2 m 2

j

= 4.42 µm

2

= 3.74 pF

81

82

Capacitance and Dielectrics

P26.10

With θ = π , the plates are out of mesh and the overlap area is zero. With

π R2 . By proportion, the 2 π −θ R2 effective area of a single sheet of charge is 2 When there are two plates in each comb, the number of adjoining sheets of positive and negative charge is 3, as shown in the sketch. When there are N plates on each comb, the number of parallel capacitors is 2 N − 1 and the total capacitance is θ = 0 , the overlap area is that of a semi-circle,

a

a2 N − 1f ∈ aπ − θ f R A = f ∈distance d 2

a

C = 2N − 1

P26.11

(a)

(b)

C=

0

0

effective

A 2 k e ln

f

2

50.0

c h 2e8.99 × 10 j lnc h = F bI ∆V = 2 k λ lnG J Method 1: H aK b a

=

9

7. 27 2.58

2

a2 N − 1f ∈ aπ − θ fR 0

=

FIG. P26.10

2

.

d

2.68 nF

e

q 8.10 × 10 −6 C = = 1.62 × 10 −7 C m A 50.0 m 7.27 = 3.02 kV ∆V = 2 8.99 × 10 9 1.62 × 10 −7 ln 2.58

λ=

e

Method 2: P26.12

∆V =

j FGH

je

IJ K

Q 8.10 × 10 −6 = = 3.02 kV C 2.68 × 10 −9

Let the radii be b and a with b = 2 a . Put charge Q on the inner conductor and – Q on the outer. Electric field exists only in the volume between them. The potential of the inner sphere is Va = that of the outer is Vb =

k eQ . Then b

FG H

Va − Vb =

k e Q k eQ Q b−a − = a b 4π ∈0 ab

Here C =

4π ∈0 2 a 2 = 8π ∈0 a a

IJ and C = Q K V −V a

a=

4π ∈0 ab . b−a

C . 8π ∈0

Volume =

The intervening volume is

b

=

Volume =

FG H

b

ga

f

e

7 20.0 × 10 −6 384π

2

e8.85 × 10

(a)

C=

0.070 0 0.140 ab = = 15.6 pF ke b − a 8.99 × 10 9 0.140 − 0.070 0

(b)

C=

Q ∆V

a f e

∆V =

jb

IJ FG IJ K H K C N ⋅ mj = 2.13 × 10 C N ⋅m j

C3 4 3 4 3 4 4 7C 3 π b − π a = 7 π a3 = 7 π 3 3 3 = 3 3 3 3 8 π ∈0 384π 2 ∈30

The outer sphere is 360 km in diameter. P26.13

k eQ ; a

g

Q 4.00 × 10 −6 C = = 256 kV C 15.6 × 10 −12 F

−12

3

2

2

2 3

16

m3 .

Chapter 26

P26.14

P26.15

∑ Fy = 0 :

T cos θ − mg = 0

∑ Fx = 0 :

T sin θ − Eq = 0

Dividing,

tan θ =

so

E=

and

∆V = Ed =

mg tan θ q mgd tan θ . q

e

je

j

C = 4π ∈0 R = 4π 8.85 × 10 −12 C N ⋅ m 2 6.37 × 10 6 m = 7.08 × 10 −4 F

Section 26.3 P26.16

Eq mg

(a)

Combinations of Capacitors Capacitors in parallel add. Thus, the equivalent capacitor has a value of

C eq = C1 + C 2 = 5.00 µF + 12.0 µF = 17.0 µF . (b)

The potential difference across each branch is the same and equal to the voltage of the battery.

∆V = 9.00 V (c)

b

f

ga

Q5 = C∆V = 5.00 µF 9.00 V = 45.0 µC

b

ga

f

and Q12 = C∆V = 12.0 µF 9.00 V = 108 µC P26.17

(a)

In series capacitors add as 1 1 1 1 1 = + = + C eq C 1 C 2 5.00 µF 12.0 µF

(c)

and

C eq = 3.53 µF .

The charge on the equivalent capacitor is

Q eq = C eq ∆V = 3.53 µF 9.00 V = 31.8 µC .

b

ga

f

Each of the series capacitors has this same charge on it.

(b)

So

Q1 = Q 2 = 31.8 µC .

The potential difference across each is

∆V1 =

Q1 31.8 µC = = 6.35 V C1 5.00 µF

and

∆V2 =

Q 2 31.8 µC = = 2.65 V . C 2 12.0 µF

83

84

Capacitance and Dielectrics

P26.18

The circuit reduces first according to the rule for capacitors in series, as shown in the figure, then according to the rule for capacitors in parallel, shown below.

FG H

C eq = C 1 +

P26.19

IJ K

1 1 11 + = C = 1.83C 2 3 6

FIG. P26.18

C p = C1 + C 2

1 1 1 = + C s C1 C 2

Substitute C 2 = C p − C1

C p − C1 + C1 1 1 1 = + = . C s C1 C p − C1 C1 C p − C1

Simplifying,

C 12 − C 1C p + C pC s = 0 .

C1 =

⇒

C p ± C p2 − 4C p C s 2

e

=

j

1 1 2 Cp ± C p − C pCs 2 4

We choose arbitrarily the + sign. (This choice can be arbitrary, since with the case of the minus sign, we would get the same two answers with their names interchanged.) C1 =

C 2 = C p − C1 = P26.20

b

g

b

1 1 2 1 1 Cp + C p − C p C s = 9.00 pF + 9.00 pF 2 4 2 4

b

g − b9.00 pFgb2.00 pFg = 2

6.00 pF

g

1 1 2 1 Cp − C p − C p C s = 9.00 pF − 1.50 pF = 3.00 pF 2 4 2

C p = C1 + C 2 and

1 1 1 = + . C s C1 C 2

Substitute

C 2 = C p − C1 :

Simplifying, and

C p − C1 + C1 1 1 1 = + = . C s C1 C p − C1 C1 C p − C1

e

j

C 12 − C 1C p + C pC s = 0 C1 =

C p ± C p2 − 4C pC s 2

=

1 1 2 Cp + C p − C pCs 2 4

where the positive sign was arbitrarily chosen (choosing the negative sign gives the same values for the capacitances, with the names reversed). Then, from

C 2 = C p − C1 C2 =

1 1 2 Cp − C p − C pC s . 2 4

Chapter 26

P26.21

(a)

85

1 1 1 = + C s 15.0 3.00 C s = 2.50 µF C p = 2.50 + 6.00 = 8.50 µF C eq =

(b)

FG 1 + 1 IJ H 8.50 µF 20.0 µF K b

−1

ga

= 5.96 µF

f

Q = C∆V = 5.96 µF 15.0 V = 89.5 µC on 20.0 µF Q 89.5 µC = = 4.47 V C 20.0 µF 15.0 − 4.47 = 10.53 V ∆V =

b

f

ga

Q = C∆V = 6.00 µF 10.53 V = 63.2 µC on 6.00 µF

89.5 − 63.2 = 26.3 µC on 15.0 µF and 3.00 µF *P26.22

FIG. P26.21

(a)

Capacitors 2 and 3 are in parallel and present equivalent capacitance 6C. This is in series −1 1 1 = 2C . with capacitor 1, so the battery sees capacitance + 3C 6C

(b)

If they were initially unchanged, C 1 stores the same charge as C 2 and C 3 together. With

LM N

OP Q

greater capacitance, C 3 stores more charge than C 2 . Then Q1 > Q3 > Q 2 .

b

g

(c)

The C 2||C 3 equivalent capacitor stores the same charge as C 1 . Since it has greater Q implies that it has smaller potential difference across it than C 1 . In capacitance, ∆V = C parallel with each other, C 2 and C 3 have equal voltages: ∆V1 > ∆V2 = ∆V3 .

(d)

If C 3 is increased, the overall equivalent capacitance increases. More charge moves through the battery and Q increases. As ∆V1 increases, ∆V2 must decrease so Q 2 decreases. Then Q 3 must increase even more: Q3 and Q1 increase; Q 2 decreases .

P26.23

Q so ∆V Q = 120 µC and

6.00 × 10 −6 =

C=

and or

Q1 = 120 µC − Q 2 Q ∆V = : C 120 − Q 2 Q = 2 6.00 3.00

a3.00fb120 − Q g = a6.00fQ 2

360 = 40.0 µC Q2 = 9.00

Q 20.0

120 − Q 2 Q 2 = C1 C2

2

Q1 = 120 µC − 40.0 µC = 80.0 µC

FIG. P26.23

86

Capacitance and Dielectrics

P26.24

In series , to reduce the effective capacitance:

(a)

1 1 1 = + µ µ 32.0 F 34.8 F C s 1 = 398 µF Cs = 2.51 × 10 −3 µF In parallel , to increase the total capacitance:

(b)

29.8 µF + C p = 32.0 µF C p = 2.20 µF P26.25

100

nC =

1 1 + C1 + " C + C   

=

100 nC

n capacitors

nC =

*P26.26

100C so n 2 = 100 and n = 10 n

For C 1 connected by itself, C 1 ∆V = 30.8 µC where ∆V is the battery voltage: ∆V = For C 1 and C 2 in series:

F GH 1 C

30.8 µC . C1

I JK

1 ∆V = 23.1 µC 1 + 1 C2

substituting,

30.8 µC 23.1 µC 23.1 µC = + C1 C1 C2

C 1 = 0.333C 2 .

For C 1 and C 3 in series:

F GH 1 C

I JK

1 ∆V = 25.2 µC + 1 1 C3

30.8 µC 25. 2 µC 25.2 µC = + C1 C1 C3 For all three: Q=

F GH 1 C

C 1 = 0.222C 3 .

I JK

C 1 ∆V 1 30.8 µC ∆V = = = 19.8 µC . 1 1 1 1 0 C C C C C C .333 + 0.222 + + + + + 1 2 3 1 2 1 3

This is the charge on each one of the three. P26.27

FG 1 + 1 IJ = 3.33 µF H 5.00 10.0 K = 2a3.33f + 2.00 = 8.66 µF = 2a10.0f = 20.0 µF F 1 + 1 IJ = 6.04 µF =G H 8.66 20.0 K

Cs = C p1 C p2

−1

−1

C eq

FIG. P26.27

Chapter 26

P26.28

a f e

ja

f

Q eq = C eq ∆V = 6.04 × 10 −6 F 60.0 V = 3.62 × 10 −4 C Q eq

Q p1 = Q eq , so ∆Vp1 =

e

C p1

=

3.62 × 10 −4 C = 41.8 V 8.66 × 10 −6 F

ja

j e

f

Q 3 = C 3 ∆Vp1 = 2.00 × 10 −6 F 41.8 V = 83.6 µC

P26.29

Cs =

FG 1 + 1 IJ H 5.00 7.00 K

−1

= 2.92 µF

C p = 2.92 + 4.00 + 6.00 = 12.9 µF

FIG. P26.29 *P26.30

According to the suggestion, the combination of capacitors shown is equivalent to

Then

1 1 1 1 C + C0 + C0 + C + C0 = + + = C C0 C + C0 C0 C0 C + C0

b

+ C 02

C 0C

g

2

= 2C + 3C 0 C

2

2C + 2C 0 C − C 02 = 0 C=

FIG. P26.30

e j

−2C 0 ± 4C 02 + 4 2C 02 4

Only the positive root is physical C=

Section 26.4 P26.31

P26.32

C0 2

e

j

3 −1

Energy Stored in a Charged Capacitor

a f

2

=

1 3.00 µF 12.0 V 2

a f

2

=

1 3.00 µF 6.00 V 2

(a)

U=

1 C ∆V 2

(b)

U=

1 C ∆V 2

U=

b

ga

f

2

= 216 µJ

b

ga

f

2

= 54.0 µJ

1 C∆V 2 2

∆V =

2U = C

a

2 300 J

f

30 × 10 −6 C V

= 4. 47 × 10 3 V

87

88

Capacitance and Dielectrics

P26.33

U=

a f

1 C ∆V 2

2

The circuit diagram is shown at the right.

C p = C1 + C 2 = 25.0 µF + 5.00 µF = 30.0 µF

(a)

U=

ja f

1 30.0 × 10 −6 100 2

e

2

= 0.150 J

F 1 + 1 IJ = FG 1 + 1 IJ C =G H C C K H 25.0 µF 5.00 µF K 1 U = C a ∆V f 2 2a0.150f 2U = = 268 V ∆V = −1

(b)

s

1

−1

= 4.17 µF

2

2

4.17 × 10 −6

C

P26.34

Use U =

∈ A 1 Q2 and C = 0 . 2 C d

If d 2 = 2d1 , C 2 = *P26.35

e

je

j

(a)

Q = C∆V = 150 × 10 −12 F 10 × 10 3 V = 1.50 × 10 −6 C

(b)

U=

a f

1 C ∆V 2

u=

2

2U = C

∆V =

P26.36

1 C 1 . Therefore, the stored energy doubles . 2

e

2 250 × 10 −6 J 150 × 10

−12

F

j=

U 1 = ∈0 E 2 V 2

1.00 × 10 −7 1 = 8.85 × 10 −12 3 000 V 2

jb

e

g

2

e

V = 2.51 × 10 −3 m 3 = 2.51 × 10 −3 m 3 P26.37

1.83 × 10 3 V

z

LI J= jFGH 1 000 m K 3

W = U = Fdx so F =

F I GH JK

F GH

I JK

dU d Q 2 d Q2x Q2 = = = 2 ∈0 A dx dx 2C dx 2 ∈0 A

2.51 L

FIG. P26.33

Chapter 26

P26.38

With switch closed, distance d ′ = 0.500d and capacitance C ′ =

a f

a f e

ja

f

(a)

Q = C ′ ∆V = 2C ∆V = 2 2.00 × 10 −6 F 100 V = 400 µC

(b)

The force stretching out one spring is

a f

4C 2 ∆V Q2 F= = 2 ∈0 A 2 ∈0 A

a f = 2Ca∆V f = ∈ A b dgd d

2

2C 2 ∆ V

2

k=

a f FG 4 IJ = 8Ca∆V f H dK d

F 2C ∆V = x d

2

2

=

2

H ET =

The energy transferred is

2

.

0

One spring stretches by distance x =

P26.39

∈0 A 2 ∈0 A = = 2C . d′ d

d , so 4

ja

e

8 2.00 × 10 −6 F 100 V

e

8.00 × 10 −3 m

a

j

2

fe

f

2

= 2.50 kN m .

1 1 Q∆V = 50.0 C 1.00 × 10 8 V = 2.50 × 10 9 J 2 2

j

and 1% of this (or ∆Eint = 2.50 × 10 7 J ) is absorbed by the tree. If m is the amount of water boiled away,

*P26.40

b

f e

ga

∆Eint = m 4 186 J kg⋅° C 100° C − 30.0° C + m 2.26 × 10 6 J kg = 2.50 × 10 7 J

giving

m = 9.79 kg .

a f

1 C ∆V 2

2

+

a f

1 C ∆V 2

2

a f

= C ∆V

2

(a)

U=

(b)

The altered capacitor has capacitance C ′ =

a f a f a f C2 a∆V ′f

C ∆V + C ∆V = C ∆V ′ +

FG H

1 4 ∆V U′ = C 2 3

(c) (d)

P26.41

j

then

IJ K

2

FG H

11 4∆V + C 22 3

IJ K

2

=

C . The total charge is the same as before: 2

∆V ′ =

a ∆V f 4C

4 ∆V . 3

2

3

The extra energy comes from work put into the system by the agent pulling the capacitor plates apart.

a f

U=

1 C ∆V 2

U=

1 R 2 ke

FG H

2

where C = 4π ∈0 R =

IJ FG k Q IJ KH R K e

2

=

k eQ 2 2R

kQ kQ R and ∆V = e − 0 = e R R ke

89

90

Capacitance and Dielectrics

*P26.42

(a)

b

g

2

q12 1 q12 1 q 22 1 1 Q − q1 The total energy is U = U 1 + U 2 = + = + . 2 C 1 2 C 2 2 4π ∈0 R1 2 4π ∈0 R 2 dU = 0: dq1

For a minimum we set

b

ga f

1 2 q1 1 2 Q − q1 + −1 = 0 2 4π ∈0 R1 2 4π ∈0 R 2 R 2 q1 = R1Q − R1 q1 Then q 2 = Q − q1 =

(b)

q1 =

R1Q R1 + R 2

R 2Q = q2 . R1 + R 2

V1 =

k e q1 k e R1Q k eQ = = R1 R1 R1 + R 2 R1 + R 2

V2 =

ke q2 k e R2 Q k eQ = = R2 R 2 R1 + R 2 R1 + R 2

b

g

b

g

and V1 − V2 = 0 .

Section 26.5

P26.43

P26.44

Capacitors with Dielectrics

e

je

j

(a)

−12 F m 1.75 × 10 −4 m 2 κ ∈0 A 2.10 8.85 × 10 C= = = 8.13 × 10 −11 F = 81.3 pF d 4.00 × 10 −5 m

(b)

∆Vmax = Emax d = 60.0 × 10 6 V m 4.00 × 10 −5 m = 2.40 kV

e

je

j

Q max = C∆Vmax , ∆Vmax = Emax d .

but

κ ∈0 A . d

Also,

C=

Thus,

Q max =

(a)

κ ∈0 A Emax d = κ ∈0 AEmax . d

b

g

With air between the plates, κ = 1.00 and

Emax = 3.00 × 10 6 V m .

Therefore,

e

je

je

j

Q max = κ ∈0 AEmax = 8.85 × 10 −12 F m 5.00 × 10 −4 m 2 3.00 × 10 6 V m = 13.3 nC . (b)

With polystyrene between the plates, κ = 2.56 and Emax = 24.0 × 10 6 V m .

e

je

je

j

Q max = κ ∈0 AEmax = 2.56 8.85 × 10 −12 F m 5.00 × 10 −4 m 2 24.0 × 10 6 V m = 272 nC

Chapter 26

P26.45

C= or

κ ∈0 A d 95.0 × 10

−9

=

jb

e

g

3.70 8.85 × 10 −12 0.070 0 A 0.025 0 × 10 −3

A = 1.04 m P26.46

Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet of plastic between 2.54 cm . Then, them. Suppose the plastic has κ ≅ 3 , Emax ~ 10 7 V m and thickness 1 mil = 1 000

e

je

j

−12 C 2 N ⋅ m 2 0. 4 m 2 κ ∈0 A 3 8.85 × 10 ~ 10 −6 F ~ C= −5 d 2.54 × 10 m

e

je

j

∆Vmax = Emax d ~ 10 7 V m 2.54 × 10 −5 m ~ 10 2 V P26.47

C=

Originally,

(a)

a f

e8.85 × 10

−12

je

a f

∈0 A ∆V

i

ja

C 2 N ⋅ m 2 25.0 × 10 −4 m 2 250 V

e1.50 × 10

−2

.

d

j

m

f=

369 pC

Finally, Cf =

κ ∈0 A Q = d ∆V

a f

Cf =

e

−2

f

=

Qd κ ∈0 A

a f

Originally,

Ui =

1 C ∆V 2

Finally,

Uf =

1 C f ∆V 2

e8.85 × 10 ∆U = −

2 i

−12

C

2

0

a f

2 f

i

κ ∈0 Ad

=

∆U = U f − U i =

So,

je e1.50 × 10 mj ∈ Aa ∆V f d a ∆V f = =

80.0 8.85 × 10 −12 C 2 N ⋅ m 2 25.0 × 10 −4 m 2

f

a ∆V f (c)

. i

The charge is the same before and after immersion, with value Q = Q=

(b)

∈0 A Q = ∆V d

a f

∈0 A ∆V

2 i

2d =

κ

−2

=

a f

κ ∈0 A ∆V 2dκ 2

2 i

=

a f

∈0 A ∆V 2 dκ

2 i

.

a f aκ − 1f

− ∈0 A ∆V

2 dκ 2

118 pF

250 V = 3.12 V . 80.0

.

2 i

je25.0 × 10 m ja250 V f a79.0f = 2e1.50 × 10 mja80.0f N⋅m

−4

2

i

j=

2

−45.5 nJ .

91

92

Capacitance and Dielectrics

P26.48

a fe

je

j

−12 1.00 × 10 −4 m 2 κ ∈0 A 173 8.85 × 10 = = 1.53 nF d 0.100 × 10 −3 m

(a)

C = κC 0 =

(b)

The battery delivers the free charge

a f e

ja

f

Q = C ∆V = 1.53 × 10 −9 F 12.0 V = 18.4 nC . (c)

The surface density of free charge is Q 18.4 × 10 −9 C σ= = = 1.84 × 10 −4 C m 2 . A 1.00 × 10 −4 m 2 The surface density of polarization charge is

FG H

σp =σ 1− (d)

We have E = E=

P26.49

IJ FG K H

IJ K

1 1 =σ 1− = 1.83 × 10 −4 C m 2 . κ 173 E0

κ

and E0 =

∆V ; hence, d

∆V 12.0 V = = 694 V m . κd 173 1.00 × 10 −4 m

a fe

j

The given combination of capacitors is equivalent to the circuit diagram shown to the right. Put charge Q on point A. Then,

b

g

b

g

b

g

Q = 40.0 µF ∆VAB = 10.0 µF ∆VBC = 40.0 µF ∆VCD .

FIG. P26.49

So, ∆VBC = 4∆VAB = 4∆VCD , and the center capacitor will break down first, at ∆VBC = 15.0 V . When this occurs, ∆VAB = ∆VCD =

b

g

1 ∆VBC = 3.75 V 4

and VAD = VAB + VBC + VCD = 3.75 V + 15.0 V + 3.75 V = 22.5 V .

Section 26.6 P26.50

(a)

Electric Dipole in an Electric Field The displacement from negative to positive charge is

e

j

e

j

e

j

2 a = −1.20 i + 1.10 j mm − 1.40 i − 1.30 j mm = −2.60 i + 2.40 j × 10 −3 m. The electric dipole moment is

e

je

j

p = 2aq = 3.50 × 10 −9 C −2.60 i + 2.40 j × 10 −3 m = (b)

e

j

e−9.10 i + 8.40jj × 10

e

j

τ = p × E = −9.10 i + 8.40 j × 10 −12 C ⋅ m × 7.80 i − 4.90 j × 10 3 N C

e

j

τ = +44.6k − 65.5k × 10 −9 N ⋅ m = −2.09 × 10 −8 N ⋅ mk continued on next page

−12

C⋅m .

Chapter 26

(c)

e

j

e

93

j

U = − p ⋅ E = − −9.10 i + 8.40 j × 10 −12 C ⋅ m ⋅ 7.80 i − 4.90 j × 10 3 N C

a

f

U = 71.0 + 41.2 × 10 −9 J = 112 nJ

a9.10f + a8.40f a7.80f + a4.90f

p=

(d)

E=

2

2

× 10 −12 C ⋅ m = 12.4 × 10 −12 C ⋅ m

2

2

× 10 3 N C = 9.21 × 10 3 N C

U max = p E = 114 nJ,

U min = −114 nJ

U max − U min = 228 nJ P26.51

(a)

Let x represent the coordinate of the negative charge. Then x + 2 a cos θ is the coordinate of the positive charge. The force on the negative charge is F− = −qE x i . The force on the positive charge is F+

(b)

F-

af dE = + qEa x + 2 a cos θ fi ≈ qEa xfi + q a2 a cosθ fi . dx

The force on the dipole is altogether

F = F− + F+ = q

The balloon creates field along the x-axis of Thus,

a f

−2 k e q dE = . dx x3

a fe

F+

p

θ

E

FIG. P26.51(a)

a

f

dE dE 2 a cos θ i = p cos θ i . dx dx

ke q  i. x2

je a f

j

−2 8.99 × 10 9 2.00 × 10 −6 dE = = −8.78 MN C ⋅ m 3 dx 0.160

At x = 16.0 cm ,

e

je

j

F = 6.30 × 10 −9 C ⋅ m −8.78 × 10 6 N C ⋅ m cos 0° i = −55.3 i mN

Section 26.7 P26.52

An Atomic Description of Dielectrics

2π rAE = so

qin ∈0

E=

λ 2π r ∈0

z

λ max

z

r2

r2

r1

r1

∆V = − E ⋅ d r =

FG IJ H K

r λ λ dr = ln 1 r2 2π r ∈0 2π ∈0

= Emax rinner

2π ∈0

e

je

j FGH 025.200.0 IJK

∆V = 1. 20 × 10 6 V m 0.100 × 10 −3 m ln ∆Vmax = 579 V

FIG. P26.52

94

Capacitance and Dielectrics

P26.53

(a)

Consider a gaussian surface in the form of a cylindrical pillbox with ends of area A ′ 11.0 lna1.10f = lna1.10f H aK −0.10 ln

11.0

where we have reversed the direction of the inequality because we multiplied the whole expression by –1 to remove the negative signs. Comparing the arguments of the logarithms on both sides of the inequality, we see that,

a f

b > 1.10 a

11.0

= 2.85 .

Thus, if b > 2.85 a , the increase in capacitance is less than 10% and it is more effective to increase A .

Chapter 26

103

ANSWERS TO EVEN PROBLEMS P26.2

(a) 1.00 µF ; (b) 100 V

P26.4

(a) 8.99 mm ; (b) 0.222 pF ; (c) 22.2 pC

P26.6

11.1 nF ; 26.6 C

P26.8

3.10 nm

P26.10

a2 N − 1f ∈ aπ − θ fR 0

2.13 × 10 16 m3

P26.14

mgd tan θ q

P26.18

P26.20 P26.22

2

R1Q R 2Q and q 2 = ; R1 + R 2 R1 + R 2 (b) see the solution

P26.44

(a) 13.3 nC ; (b) 272 nC

P26.46

~ 10 −6 F and ~ 10 2 V for two 40 cm by 100 cm sheets of aluminum foil sandwiching a thin sheet of plastic.

P26.48

(a) 1.53 nF ; (b) 18.4 nC ; (c) 184 µC m 2

P26.50

e

j

(a) −9.10 i + 8. 40 j pC ⋅ m ; (b) −20.9 nN ⋅ mk ; (c) 112 nJ ; (d) 228 nJ

C p2

− C p C s and

4

Cp 2

−

C p2 4

− C pC s

P26.52

579 V

P26.54

(a) 3.33 µF ; (b) ∆V3 = 60.0 V ; ∆V6 = 30.0 V ; ∆V2 = 60.0 V ; ∆V4 = 30.0 V ; (c) Q 3 = Q6 = 180 µC ; Q 2 = Q 4 = 120 µC ; (d) 13.4 mJ

(d) Q 3 and Q1 increase and Q 2 decreases P26.24

(a) 398 µF in series; (b) 2.20 µF in parallel

P26.56

189 kV

P26.26

19.8 µC

P26.58

(a) 40.0 µJ ; (b) 500 V

P26.28

83.6 µC

P26.60

yes; 1.00 Mm s

P26.30

e

P26.62

23.3 V ; 26.7 V

P26.64

(a)

P26.32 P26.34

;

(a) q1 =

(a) 2C ; (b) Q1 > Q3 > Q 2 ; (c) ∆V1 > ∆V2 = ∆V3 ;

3 −1

2

P26.42

1.83C +

2

free; 183 µC m 2 induced; (d) 694 V m

(a) 17.0 µF ; (b) 9.00 V ; (c) 45.0 µC and 108 µC

Cp

a ∆V f ; (c) 4C a f ; (b) 4∆V 3 3

(a) C ∆V

(d) Positive work is done on the system by the agent pulling the plates apart.

2

d

P26.12

P26.16

P26.40

j C2

0

4.47 kV energy doubles

P26.36

2.51 × 10 −3 m3 = 2.51 L

P26.38

(a) 400 µC ; (b) 2.50 kN m

(b)

a f

∈0 A 2 + Ax κ − 1

a f

∈0 ∆V

d 2

a f;

A 2 + Ax κ − 1 2d

a f a f 2

;

∈0 ∆V A κ − 1 to the left ; 2d (d) 1.55 mN left (c)

104

Capacitance and Dielectrics

Gasoline has 194 times the specific energy content of the battery, and 727 000 times that of the capacitor.

P26.72

3.00 µF

P26.74

see the solution

P26.68

see the solution; 45 V

P26.76

see the solution

P26.70

2 3

P26.66

27 Current and Resistance CHAPTER OUTLINE 27.1 27.2 27.3 27.4 27.5 27.6

Electric Current Resistance A Model for Electrical Conduction Resistance and Temperature Superconductors Electric Power

ANSWERS TO QUESTIONS Q27.1

Individual vehicles—cars, trucks and motorcycles—would correspond to charge. The number of vehicles that pass a certain point in a given time would correspond to the current.

Q27.2

Voltage is a measure of potential difference, not of current. “Surge” implies a flow—and only charge, in coulombs, can flow through a system. It would also be correct to say that the victim carried a certain current, in amperes.

Q27.3

Geometry and resistivity. In turn, the resistivity of the material depends on the temperature.

Q27.4

Resistance is a physical property of the conductor based on the material of which it is made and its size and shape, including the locations where current is put in and taken out. Resistivity is a physical property only of the material of which the resistor is made.

Q27.5

The radius of wire B is 3 times the radius of wire A, to make its cross–sectional area 3 times larger.

Q27.6

Not all conductors obey Ohm’s law at all times. For example, consider an experiment in which a variable potential difference is applied across an incandescent light bulb, and the current is measured. At very low voltages, the filament follows Ohm’s law nicely. But then long before the ∆V becomes non-linear, because the resistivity is temperaturefilament begins to glow, the plot of I dependent.

Q27.7

A conductor is not in electrostatic equilibrium when it is carrying a current, duh! If charges are placed on an isolated conductor, the electric fields established in the conductor by the charges will cause the charges to move until they are in positions such that there is zero electric field throughout the conductor. A conductor carrying a steady current is not an isolated conductor—its ends must be connected to a source of emf, such as a battery. The battery maintains a potential difference across the conductor and, therefore, an electric field in the conductor. The steady current is due to the response of the electrons in the conductor due to this constant electric field.

105

106

Current and Resistance

Q27.8

The bottom of the rods on the Jacob’s Ladder are close enough so that the supplied voltage is sufficient to produce dielectric breakdown of the air. The initial spark at the bottom includes a tube of ionized air molecules. Since this tube containing ions is warmer than the air around it, it is buoyed up by the surrounding air and begins to rise. The ions themselves significantly decrease the resistivity of the air. They significantly lower the dielectric strength of the air, marking longer sparks possible. Internal resistance in the power supply will typically make its terminal voltage drop, so that it cannot produce a spark across the bottom ends of the rods. A single “continuous” spark, therefore will rise up, becoming longer and longer, until the potential difference is not large enough to sustain dielectric breakdown of the air. Once the initial spark stops, another one will form at the bottom, where again, the supplied potential difference is sufficient to break down the air.

Q27.9

The conductor does not follow Ohm’s law, and must have a resistivity that is current-dependent, or more likely temperature-dependent.

Q27.10

A power supply would correspond to a water pump; a resistor corresponds to a pipe of a certain diameter, and thus resistance to flow; charge corresponds to the water itself; potential difference corresponds to difference in height between the ends of a pipe or the ports of a water pump.

Q27.11

The amplitude of atomic vibrations increases with temperature. Atoms can then scatter electrons more efficiently.

Q27.12

In a metal, the conduction electrons are not strongly bound to individual ion cores. They can move in response to an applied electric field to constitute an electric current. Each metal ion in the lattice of a microcrystal exerts Coulomb forces on its neighbors. When one ion is vibrating rapidly, it can set its neighbors into vibration. This process represents energy moving though the material by heat.

Q27.13

The resistance of copper increases with temperature, while the resistance of silicon decreases with increasing temperature. The conduction electrons are scattered more by vibrating atoms when copper heats up. Silicon’s charge carrier density increases as temperature increases and more atomic electrons are promoted to become conduction electrons.

Q27.14

A current will continue to exist in a superconductor without voltage because there is no resistance loss.

Q27.15

Superconductors have no resistance when they are below a certain critical temperature. For most superconducting materials, this critical temperature is close to absolute zero. It requires expensive refrigeration, often using liquid helium. Liquid nitrogen at 77 K is much less expensive. Recent discoveries of materials that have higher critical temperatures suggest the possibility of developing superconductors that do not require expensive cooling systems.

Q27.16

In a normal metal, suppose that we could proceed to a limit of zero resistance by lengthening the average time between collisions. The classical model of conduction then suggests that a constant applied voltage would cause constant acceleration of the free electrons, and a current steadily increasing in time. On the other hand, we can actually switch to zero resistance by substituting a superconducting wire for the normal metal. In this case, the drift velocity of electrons is established by vibrations of atoms in the crystal lattice; the maximum current is limited; and it becomes impossible to establish a potential difference across the superconductor.

Q27.17

Because there are so many electrons in a conductor (approximately 10 28 electrons m 3 ) the average velocity of charges is very slow. When you connect a wire to a potential difference, you establish an electric field everywhere in the wire nearly instantaneously, to make electrons start drifting everywhere all at once.

Chapter 27

107

Q27.18

Current moving through a wire is analogous to a longitudinal wave moving through the electrons of the atoms. The wave speed depends on the speed at which the disturbance in the electric field can be communicated between neighboring atoms, not on the drift velocities of the electrons themselves. If you leave a direct-current light bulb on for a reasonably short time, it is likely that no single electron will enter one end of the filament and leave at the other end.

Q27.19

More power is delivered to the resistor with the smaller resistance, since P =

Q27.20

The 25 W bulb has a higher resistance. The 100 W bulb carries more current.

Q27.21

One ampere–hour is 3 600 coulombs. The ampere–hour rating is the quantity of charge that the battery can lift though its nominal potential difference.

Q27.22

Choose the voltage of the power supply you will use to drive the heater. Next calculate the required ∆V 2 resistance R as . Knowing the resistivity ρ of the material, choose a combination of wire length

P

and cross–sectional area to make

∆V 2 . R

FG A IJ = FG R IJ . You will have to pay for less material if you make both H AK H ρ K

A and A smaller, but if you go too far the wire will have too little surface area to radiate away the energy; then the resistor will melt.

SOLUTIONS TO PROBLEMS Section 27.1 P27.1

I=

Electric Current ∆Q ∆t

N= P27.2

ja

e

f

∆Q = I∆t = 30.0 × 10 −6 A 40.0 s = 1.20 × 10 −3 C

1.20 × 10 −3 C Q = = 7.50 × 10 15 electrons e 1.60 × 10 −19 C electron

The molar mass of silver = 107.9 g mole and the volume V is

a fa

f e

je

j

V = area thickness = 700 × 10 −4 m 2 0.133 × 10 −3 m = 9.31 × 10 −6 m 3 .

e

je

j

The mass of silver deposited is m Ag = ρV = 10.5 × 10 3 kg m3 9.31 × 10 −6 m 3 = 9.78 × 10 −2 kg . And the number of silver atoms deposited is

e

N = 9.78 × 10 −2 kg I=

10 atoms I F 1 000 g I jFGH 6.02 ×107.9 JK GH 1 kg JK = 5.45 × 10 g 23

∆V 12.0 V = = 6.67 A = 6.67 C s 1.80 Ω R

∆t =

e

je

j

23

atoms

5.45 × 10 23 1.60 × 10 −19 C ∆Q Ne = = = 1.31 × 10 4 s = 3.64 h I I 6.67 C s

108 P27.3

Current and Resistance

af

z t

e

Q t = Idt = I 0τ 1 − e − t τ 0

P27.4

af

j j a0.632fI τ

e

(a)

Q τ = I 0 τ 1 − e −1 =

(b)

Q 10τ = I 0τ 1 − e −10 =

(c)

Q ∞ = I 0τ 1 − e −∞ = I 0τ

af

e

(a)

Using

kee2

(b)

The time for the electron to revolve around the proton once is:

a f

j b0.999 95gI τ

e

r2

0

0

j

kee2 = 2.19 × 10 6 m s . mr

mv 2 , we get: v = r

=

e

j

−11 m 2π r 2π 5.29 × 10 t= = = 1.52 × 10 −16 s . v 2.19 × 10 6 m s

e

j

The total charge flow in this time is 1.60 × 10 −19 C , so the current is I=

P27.5

1.60 × 10 −19 C 1.52 × 10 −16 s

The period of revolution for the sphere is T = revolving charge is I =

P27.6

2π

ω

, and the average current represented by this

q qω = . 2π T

q = 4t 3 + 5t + 6

e

A = 2.00 cm 2

a

J=

(b)

I=

1.00 m I J jFGH 100 cm K

f

I 1.00 s =

(a)

P27.7

= 1.05 × 10 −3 A = 1.05 mA .

dq dt

2

= 2.00 × 10 −4 m 2

e

= 12t 2 + 5 t = 1.00 s

j

t =1.00 s

= 17.0 A

I 17.0 A = = 85.0 kA m 2 A 2.00 × 10 −4 m 2

dq dt

a100 Af sinFGH 120sπ t IJK dt −100 C L F π I O +100 C = 0.265 C q= cosG J − cos 0 P = M H K 120π N 2 Q 120π

z z

q = dq = Idt =

z

1 240 s 0

Chapter 27

P27.8

5.00 A I = A π 4.00 × 10 −3 m

= 99.5 kA m 2

(a)

J=

(b)

J2 =

1 I 1 I = J1 ; 4 A 2 4 A1

A1 =

1 A 2 so π 4.00 × 10 −3 4

e

j

2

e

e

j

2

=

1 π r22 4

j

r2 = 2 4.00 × 10 −3 = 8.00 × 10 −3 m = 8.00 mm P27.9

8.00 × 10 −6 A I = A π 1.00 × 10 −3 m

= 2.55 A m 2

(a)

J=

(b)

From J = nev d , we have

n=

(c)

∆Q , we have From I = ∆t

6.02 × 10 23 1.60 × 10 −19 C ∆Q N A e = = = 1. 20 × 10 10 s . ∆t = I I 8.00 × 10 −6 A

e

j

2

2.55 A m 2 J = = 5.31 × 10 10 m −3 . ev d 1.60 × 10 −19 C 3.00 × 10 8 m s

e

je

e

j

je

j

(This is about 382 years!) P27.10

(a)

K=

The speed of each deuteron is given by

e2.00 × 10 je1.60 × 10 Jj = 12 e2 × 1.67 × 10 6

−19

−27

j

kg v 2 and

1 mv 2 2

v = 1.38 × 10 7 m s . q t

The time between deuterons passing a stationary point is t in

I=

10.0 × 10 −6 C s = 1.60 × 10 −19 C t or

t = 1.60 × 10 −14 s .

e

je

j

So the distance between them is vt = 1.38 × 10 7 m s 1.60 × 10 −14 s = 2.21 × 10 −7 m . (b)

One nucleus will put its nearest neighbor at potential

e

je

j

8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C keq = = 6.49 × 10 −3 V . V= r 2.21 × 10 −7 m This is very small compared to the 2 MV accelerating potential, so repulsion within the beam is a small effect.

109

110 P27.11

Current and Resistance

We use I = nqAv d n is the number of charge carriers per unit volume, and is identical to the number of atoms per unit volume. We assume a contribution of 1 free electron per atom in the relationship above. For aluminum, which has a molar mass of 27, we know that Avogadro’s number of atoms, N A , has a mass of 27.0 g. Thus, the mass per atom is 27.0 g 27.0 g = = 4.49 × 10 −23 g atom . NA 6.02 × 10 23 n=

Thus,

density of aluminum 2.70 g cm 3 = mass per atom 4.49 × 10 −23 g atom

n = 6.02 × 10 22 atoms cm3 = 6.02 × 10 28 atoms m3 . vd =

or,

v d = 0.130 mm s .

Section 27.2 *P27.12

I=

P27.14

(a)

e

je

je

j

Resistance E

J = σE =

P27.13

ρ

=

0.740 V m 2.44 × 10 −8 Ω ⋅ m

FG 1 Ω ⋅ A IJ = H 1V K

3.03 × 10 7 A m 2

∆V 120 V = = 0.500 A = 500 mA R 240 Ω Applying its definition, we find the resistance of the rod, R=

(b)

∆V 15.0 V = = 3 750 Ω = 3.75 kΩ . I 4.00 × 10 −3 A

ρA . Solving for A A and substituting numerical values for R, A, and the value of ρ given for carbon in Table 27.1, we obtain

The length of the rod is determined from the definition of resistivity: R =

A=

P27.15

5.00 A I = = 1.30 × 10 −4 m s 28 3 − nqA 6.02 × 10 m 1.60 × 10 −19 C 4.00 × 10 −6 m 2

Therefore,

RA

ρ

e3.75 × 10 Ωje5.00 × 10 = e3.50 × 10 Ω ⋅ mj

−6

3

−5

m2

j=

536 m .

∆V = IR and

ρA R= : A IρA ∆V = : A

F 1.00 m I = 6.00 × 10 A = a0.600 mmf G H 1 000 mm JK ∆VA a0.900 V fe6.00 × 10 m j I= = ρA e5.60 × 10 Ω ⋅ mja1.50 mf 2

2

−7

−8

I = 6.43 A

2

−7

m2

Chapter 27

P27.16

J=

I

πr

2

= σE =

a

f

σ = 55.3 Ω ⋅ m P27.17

(a)

b

3.00 A

−1

ρ=

2

1

σ

b

= σ 120 N C

M = ρ d V = ρ d AA

we obtain:

A=

V=

M . ρdA MR

A=

g

= 0.018 1 Ω ⋅ m

Given

Thus,

(b)

g

π 0.012 0 m

ρr ρd

where

ρ d ≡ mass density,

Taking ρ r ≡ resistivity,

R=

e1.00 × 10 ja0.500f e1.70 × 10 je8.92 × 10 j

=

−8

3

M

π ρdA

=

e

1.00 × 10 −3

π 8.92 × 10

3

ja1.82f

diameter = 280 µm .

The volume of the gram of gold is given by ρ = V=

m

ρ

=

10 −3 kg

m V

e

j

= 5.18 × 10 −8 m3 = A 2.40 × 10 3 m

19.3 × 10 3 kg m3

A = 2.16 × 10 −11 m 2 R= P27.19

(a)

e

j

−8 3 ρ A 2.44 × 10 Ω ⋅ m 2.4 × 10 m = = 2.71 × 10 6 Ω A 2.16 × 10 −11 m 2

Suppose the rubber is 10 cm long and 1 mm in diameter.

e

je

j

−1 13 ρA 4ρA 4 10 Ω ⋅ m 10 m R= ~ = = ~ 10 18 Ω 2 2 − 3 A πd π 10 m

e

e

j

je

j

(b)

4 1.7 × 10 −8 Ω ⋅ m 10 −3 m 4ρA R= ~ ~ 10 −7 Ω 2 −2 π d2 π 2 × 10 m

(c)

I=

e

I~

∆V 10 2 V ~ 18 ~ 10 −16 A R 10 Ω 10 2 V 10 −7 Ω

~ 10 9 A

j

M . ρd

r = 1.40 × 10 −4 m .

The diameter is twice this distance: *P27.18

A = 1.82 m .

π r 2A =

or

r=

ρrA ρrA ρ ρ A2 = = r d . A M ρdA M

−3

M , ρd

Thus,

111

112 P27.20

Current and Resistance

F 90.0 g I The distance between opposite faces of the cube is A = G H 10.5 g cm JK R=

ρA ρA ρ 1.59 × 10 −8 Ω ⋅ m = = = = 7.77 × 10 −7 Ω = 777 nΩ −2 A A2 A 2.05 × 10 m

∆V 1.00 × 10 −5 V = = 12.9 A R 7.77 × 10 −7 Ω 10.5 g cm 3 6.02 × 10 23 electrons mol n= 107.87 g mol I=

(b)

e

e

n = 5.86 × 10 22 electrons cm 3

I = nqvA and v =

P27.22

Originally, R =

ρ Al A

b g

π rAl

2

=

b g

π rCu

P27.23

J = σE

P27.24

R=

ρ=

3

6

3

e

28

m3

jb

je

Finally, R f =

b g = ρA =

ρA3 3A

9A

g

2

= 3.28 µm s

R . 9

2

σ=

so

−13 A m2 J 6.00 × 10 = = 6.00 × 10 −15 Ω ⋅ m E 100 V m

a

ρ 1 A1 ρ 2 A 2 ρ 1 A1 + ρ 2 A 2 + = A1 A2 d2

e4.00 × 10 R= Section 27.3

jFGH 1.001×.0010m cm IJK = 5.86 × 10

12.9 C s I = nqA 5.86 × 10 28 m3 1.60 × 10 −19 C 0.020 5 m

ρA . A

ρ Cu A

j

ρ Al 2.82 × 10 −8 = = 1.29 ρ Cu 1.70 × 10 −8

rAl = rCu

P27.25

= 2.05 cm .

3

(a)

P27.21

13

−3

ja

f e

ja

f=

j je

j

Ω ⋅ m 0.250 m + 6.00 × 10 −3 Ω ⋅ m 0.400 m

e3.00 × 10

−3

j

m

2

f

−1

378 Ω

A Model for Electrical Conduction m nq 2 τ

so

e je

9.11 × 10 −31 m τ= = = 2.47 × 10 −14 s ρnq 2 1.70 × 10 −8 8. 49 × 10 28 1.60 × 10 −19

e

vd =

qE τ m

e1.60 × 10 jEe2.47 × 10 j −19

so

7.84 × 10 −4 =

Therefore,

E = 0.181 V m .

9.11 × 10 −31

−14

Chapter 27

P27.26

113

n is unaffected

(a)

J =

(b)

I ∝I A

so it doubles . J = nev d

(c)

so v d

τ=

(d)

doubles .

mσ is unchanged as long as σ does not change due to a temperature change in the nq 2

conductor. P27.27

From Equation 27.17,

τ=

me

9.11 × 10 −31

=

e8.49 × 10 je1.60 × 10 j e1.70 × 10 j A = vτ = e8.60 × 10 m sje 2.47 × 10 sj = 2.12 × 10 2

nq ρ

28

P27.28

−8

−14

5

Section 27.4

−19 2

= 2. 47 × 10 −14 s −8

m = 21.2 nm

Resistance and Temperature

At the low temperature TC we write

RC =

∆V = R0 1 + α TC − T0 IC

At the high temperature Th ,

Rh =

∆V ∆V = = R0 1 + α Th − T0 . Ih 1A

b

g

where T0 = 20.0° C .

b

a∆V f a1.00 Af = 1 + e3.90 × 10 a∆V f I 1 + e3.90 × 10 F 1.15 IJ = 1.98 A I = a1.00 A fG H 0.579 K

g ja38.0f ja−108f

−3

Then

−3

C

and

P27.29

C

a f

R = R 0 1 + α ∆T

gives

Solving, And, the final temperature is

a

f e

j

.

140 Ω = 19.0 Ω 1 + 4.50 × 10 −3 ° C ∆T .

∆T = 1.42 × 10 3 ° C = T − 20.0° C . T = 1.44 × 10 3 ° C .

114 P27.30

Current and Resistance

b

g

b

R = R c + Rn = R c 1 + α c T − T0 + Rn 1 + α n T − T0

b

g

b

g

0 = Rc α c T − T0 + Rnα n T − T0 so R c = − Rn R = − Rn

αn + Rn αc

F α IJ F α IJ = RG 1 − R = RG 1 − H αK H αK L e0.400 × 10 ° Cj OP = 10.0 kΩ M1 − MN e−0.500 × 10 ° Cj PQ −1

Rn

n

n

−3

Rn = 5.56 kΩ P27.31

−1

−1

−3

Rn

αn αc

c

c

c

g

R c = 4.44 kΩ

and

b

g e

a

j

f

(a)

ρ = ρ 0 1 + α T − T0 = 2.82 × 10 −8 Ω ⋅ m 1 + 3.90 × 10 −3 30.0° = 3.15 × 10 −8 Ω ⋅ m

(b)

J=

(c)

F π d I = 6.35 × 10 I = JA = J G H 4 JK e

(d)

n=

E

ρ

=

0.200 V m 3.15 × 10 −8 Ω ⋅ m 2

= 6.35 × 10 6 A m 2

LM π e1.00 × 10 A m j MM 4 N 2

6

6.02 × 10 23 electrons

e

6

26.98 g 2.70 × 10 g m

3

j

e

−4

j OP = PP Q

m

2

49.9 mA

= 6.02 × 10 28 electrons m 3

j

6.35 × 10 6 A m 2 J = = 659 µm s vd = ne 6.02 × 10 28 electrons m3 1.60 × 10 −19 C

e

(e) P27.32

b

je

j

f

ga

∆V = EA = 0.200 V m 2.00 m = 0.400 V

For aluminum,

R=

b

ga

α E = 3.90 × 10 −3 ° C −1

(Table 27.1)

α = 24.0 × 10 −6 ° C −1

(Table 19.1)

f

b a

g a f

fFGH

I JK

1 + α E ∆T ρA ρ 0 1 + α E ∆T A 1 + α∆T 1.39 = = R0 = 1.71 Ω = 1.234 Ω 2 A T . α + 1 ∆ 1 002 4 A 1 + α ∆T

a

f

Chapter 27

P27.33

R = R 0 1 + αT R − R0 = R0α∆T R − R0 = α∆T = 5.00 × 10 −3 25.0 = 0.125 R0

e

P27.34

j

a

Assuming linear change of resistance with temperature, R = R0 1 + α∆T

a

ja

f e

f

f

R77 K = 1.00 Ω 1 + 3.92 × 10 −3 −216° C = 0.153 Ω .

P27.35

a

f

1

FG ρ Hρ

W

IJ K

ρ = ρ 0 1 + α∆T or

∆TW =

Require that ρ W = 4ρ 0 Cu so that

∆TW =

Therefore,

TW = 47.6° C + T0 = 67.6° C .

Section 27.5

αW

−1

0W

F 1 GH 4.50 × 10

I FG 4e1.70 × 10 j − 1IJ = 47.6° C . JK ° C JK GH 5.60 × 10 −8

−3

Superconductors

Problem 48 in Chapter 43 can be assigned with this section. Section 27.6 P27.36

I=

Electric Power

P ∆V

=

and R = *P27.37 P27.38

600 W = 5.00 A 120 V

∆V 120 V = = 24.0 Ω . I 5.00 A

e

b

gb

g

P = 0.800 1 500 hp 746 W hp = 8.95 × 10 5 W

b

8.95 × 10 5 = I 2 000

P = I∆V P27.39

j

P = I∆V = 500 × 10 −6 A 15 × 10 3 V = 7.50 W

g

I = 448 A

The heat that must be added to the water is

b

gb

f

ga

Q = mc∆T = 1.50 kg 4 186 J kg ° C 40.0° C = 2.51 × 10 5 J . Thus, the power supplied by the heater is

P=

W Q 2.51 × 10 5 J = = = 419 W ∆t ∆t 600 s

a∆V f = a110 V f 2

and the resistance is R =

P

2

419 W

= 28.9 Ω .

−8

115

116 *P27.40

Current and Resistance

The battery takes in energy by electric transmission

a fa f

e

FG 3 600 s IJ = 469 J . H 1h K

j

P∆t = ∆V I ∆t = 2.3 J C 13.5 × 10 −3 C s 4.2 h It puts out energy by electric transmission

a∆V fIa∆tf = 1.6 J C e18 × 10

−3

FG 3 600 s IJ = 249 J. H 1h K

j

C s 2. 4 h

useful output 249 J = = 0.530 total input 469 J

(a)

efficiency =

(b)

The only place for the missing energy to go is into internal energy: 469 J = 249 J + ∆Eint ∆Eint = 221 J

(c)

We imagine toasting the battery over a fire with 221 J of heat input: Q = mc∆T ∆T =

P27.41

221 J Q = mc 0.015 kg

kg ° C = 15.1° C 975 J

a f R = FG ∆V IJ = FG 140 IJ = 1.361 b g R H ∆V K H 120 K F P − P IJ a100%f = FG P − 1IJ a100%f = a1.361 − 1f100% = ∆% = G H P K HP K ∆V P = P0 ∆V0

2

2

2

2

0

0

0

36.1%

0

a f a∆VR f = 500 W a110 V f = 24.2 Ω R= a500 Wf 2

P27.42

P = I ∆V =

2

a24.2 Ωfπ e2.50 × 10 =

−4

m

j

2

(a)

ρ R= A A

(b)

R = R0 1 + α∆T = 24.2 Ω 1 + 0.400 × 10 −3 1 180 = 35.6 Ω

A=

so

RA

ρ

e

a∆V f = a110f 2

P=

R

35.6

2

= 340 W

1.50 × 10 −6 Ω ⋅ m

jb

g

= 3.17 m

Chapter 27

P27.43

R=

e

117

j

−6 ρA 1.50 × 10 Ω ⋅ m 25.0 m = = 298 Ω 2 A π 0.200 × 10 −3 m

e

j

a

fa

f

∆V = IR = 0.500 A 298 Ω = 149 V ∆V 149 V = = 5.97 V m 25.0 m A

(a)

E=

(b)

P = ∆V I = 149 V 0.500 A = 74.6 W

(c)

R = R0 1 + α T − T0 = 298 Ω 1 + 0.400 × 10 −3 ° C 320° C = 337 Ω I=

a f a

fa

b

g

a a

f f

f

e

149 V ∆V = = 0.443 A R 337 Ω

a f a

fa

j

f

P = ∆V I = 149 V 0.443 A = 66.1 W P27.44

P27.45

a f a f a

∆U = q ∆V = It ∆V = 55.0 A ⋅ h 12.0 V

(b)

Cost = 0.660 kWh

FG $0.060 0 IJ = H 1 kWh K

fFGH 11AC⋅ s IJK FGH 1 V1 J⋅ C IJK FGH 1 W1 J⋅ s IJK = 660 W ⋅ h =

0.660 kWh

3.96¢

a f ∆V = IR a∆V f = a10.0f = 0.833 W P= P = I ∆V

2

R

*P27.46

fa

(a)

(a)

2

120

The resistance of 1 m of 12-gauge copper wire is

ρA ρA = R= A π d 2

b g

2

=

4ρ A

πd

2

=

e

j

4 1.7 × 10 −8 Ω ⋅ m 1 m

e

π 0.205 3 × 10

−2

j

m

2

= 5.14 × 10 −3 Ω .

a

f

2

The rate of internal energy production is P = I∆V = I 2 R = 20 A 5.14 × 10 −3 Ω = 2.05 W . (b)

PAl = I 2 R = PAl ρ Al = PCu ρ Cu

I 2 4ρ Al A

π d2

PAl =

2.82 × 10 −8 Ω ⋅ m 2.05 W = 3.41 W 1.7 × 10 −8 Ω ⋅ m

Aluminum of the same diameter will get hotter than copper.

118 *P27.47

Current and Resistance

The energy taken in by electric transmission for the fluorescent lamp is

a

sI fFGH 3 600 J = 3.96 × 10 J 1h K F $0.08 IJ FG k IJ FG W ⋅ s IJ FG h IJ = $0.088 JG H kWh K H 1 000 K H J K H 3 600 s K 6

P∆t = 11 J s 100 h cost = 3.96 × 10 6

For the incandescent bulb,

a

sI fFGH 3 600 J = 1.44 × 10 1h K F $0.08 IJ = $0.32 JG H 3.6 × 10 J K

P∆t = 40 W 100 h cost = 1.44 × 10 7

7

J

6

saving = $0.32 − $0.088 = $0.232 P27.48

The total clock power is

e270 × 10

jFGH

6

clocks 2.50

Js clock

IJ FG 3 600 s IJ = 2.43 × 10 KH 1 h K

12

J h.

Wout , the power input to the generating plants must be: Qin

From e =

Q in Wout ∆ t 2.43 × 10 12 J h = = = 9.72 × 10 12 J h ∆t e 0.250 and the rate of coal consumption is

e

Rate = 9.72 × 10 12 J h P27.49

a f a

fa

.00 kg coal I jFGH 133.0 J = 2.95 × 10 × 10 J K 6

5

kg coal h = 295 metric ton h .

f

P = I ∆V = 1.70 A 110 V = 187 W

a

fa

f

Energy used in a 24-hour day = 0.187 kW 24.0 h = 4.49 kWh

FG $0.060 0 IJ = $0.269 = 26.9¢ H kWh K P = I∆V = a 2.00 A fa120 V f = 240 W ∆E = b0.500 kg gb 4 186 J kg⋅° C ga77.0° C f = 161 kJ

∴ cost = 4.49 kWh P27.50

int

∆t =

∆Eint

P

=

1.61 × 10 5 J = 672 s 240 W

Chapter 27

P27.51

At operating temperature,

a

fa

f

(a)

P = I∆V = 1.53 A 120 V = 184 W

(b)

Use the change in resistance to find the final operating temperature of the toaster.

a

R = R0 1 + α∆T

f

120 120 1 + 0.400 × 10 −3 ∆T = 1.53 1.80

e

∆T = 441° C *P27.52

j

T = 20.0° C + 441° C = 461° C

You pay the electric company for energy transferred in the amount E = P ∆t (a)

(b)

(c) P27.53

119

7 d I F 86 400 s I F 1 J I fFGH 1 week JK GH 1 d JK GH 1 W ⋅ s JK = 48.4 MJ F 7 d IJ FG 24 h IJ FG k IJ = 13.4 kWh P ∆t = 40 Wa 2 weeksfG H 1 week K H 1 d K H 1 000 K F 7 d IJ FG 24 h IJ FG k IJ FG 0.12 $ IJ = $1.61 P ∆t = 40 Wa 2 weeksfG H 1 week K H 1 d K H 1 000 K H kWh K

a

P ∆t = 40 W 2 weeks

h I F k I F 0.12 $ I fFGH 601min JK GH 1 000 JK GH kWh JK = $0.005 82 F 1 h IJ FG k IJ FG 0.12 $ IJ = $0.416 P ∆t = 5 200 Wa 40 minfG H 60 min K H 1 000 K H kWh K

a

P ∆t = 970 W 3 min

= 0.582¢

Consider a 400-W blow dryer used for ten minutes daily for a year. The energy transferred to the dryer is

b

gb

FG 1 kWh IJ ≈ 20 kWh . H 3.6 × 10 J K

f

ga

P ∆t = 400 J s 600 s d 365 d ≈ 9 × 10 7 J

6

We suppose that electrically transmitted energy costs on the order of ten cents per kilowatt-hour. Then the cost of using the dryer for a year is on the order of

a

fb

g

Cost ≅ 20 kWh $0.10 kWh = $2 ~ $1 .

120

Current and Resistance

Additional Problems P27.54

(a)

I= R=

(b)

(c)

(d)

∆V R ∆V

a ∆V f R ∆V f a a120 V f R= = 2

P = I∆V =

so

a f = a120 V f 2

2

25.0 W

P

2

= 576 Ω

and

100 W

P

2

= 144 Ω

P 25.0 W Q 1.00 C = = 0.208 A = = ∆V 120 V ∆t ∆t 1.00 C ∆t = = 4.80 s 0.208 A The bulb takes in charge at high potential and puts out the same amount of charge at low potential. I=

∆U 1.00 J 1.00 J = ∆t = = 0.040 0 s ∆t ∆t 25.0 W The bulb takes in energy by electrical transmission and puts out the same amount of energy by heat and light.

P = 25.0 W =

b

gb

f

ga

∆U = P∆t = 25.0 J s 86 400 s d 30.0 d = 64.8 × 10 8 J The electric company sells energy .

FG $0.070 0 IJ FG k IJ FG W ⋅ s IJ FG h IJ = $1.26 H kWh K H 1 000 K H J K H 3 600 s K $0.070 0 F kWh I Cost per joule = G J = $1.94 × 10 J kWh H 3.60 × 10 J K

Cost = 64.8 × 10 6 J

−8

6

*P27.55

The original stored energy is Ui =

1 1 Q2 . Q∆Vi = 2 2 C

(a)

When the switch is closed, charge Q distributes itself over the plates of C and 3C in parallel, Q presenting equivalent capacitance 4C. Then the final potential difference is ∆V f = for 4C both.

(b)

The smaller capacitor then carries charge C∆V f = charge 3C

(c)

Q 3Q . = 4C 4

The smaller capacitor stores final energy capacitor possesses energy

(d)

Q Q C= . The larger capacitor carries 4C 4

FG IJ H K

1 Q 3C 2 4C

2

=

1 C ∆V f 2

d i

2

=

FG IJ H K

1 Q C 2 4C

2

=

Q2 . The larger 32C

3Q 2 . 32C

Q 2 3Q 2 Q 2 + = . The loss of potential energy is the energy 32C 32C 8C 3Q 2 Q2 Q2 appearing as internal energy in the resistor: = + ∆Eint ∆Eint = . 8C 2C 8C

The total final energy is

Chapter 27

P27.56

vd = v=

P27.57

I = nqv d A = nqv d π r 2

We find the drift velocity from I nqπ r

x t

2

t=

=

1 000 A 8.49 × 10

28

m

−3

e1.60 × 10

−19

je

C π 10

m

= 2.34 × 10 −4 m s

1 dρ . ρ dT

We begin with the differential equation

α=

(a)

z z

Separating variables, ln

2

200 × 10 3 m x = = 8.54 × 10 8 s = 27.0 yr v 2.34 × 10 −4 m s

ρ

dρ

ρ

ρ0

FG ρ IJ = α bT − T g and Hρ K

T

= αdT T0

ρ = ρ 0 eα bT −T0 g .

0

0

a

f

From the series expansion e x ≅ 1 + x , x ∆V3 > ∆V1 > ∆V2 . 3

(b)

Based on the reasoning above the potential differences are ε 2ε 4ε 2ε . ∆V1 = , ∆V2 = , ∆V3 = , ∆V4 = 3 9 9 3

(c)

All the current goes through resistor 1, so it gets the most. The current then splits at the parallel combination. Resistor 4 gets more than half, because the resistance in that branch is less than in the other branch. Resistors 2 and 3 have equal currents because they are in series. The ranking by current is I 1 > I 4 > I 2 = I 3 .

(d)

Resistor 1 has a current of I. Because the resistance of 2 and 3 in series is twice that of resistor 4, twice as much current goes through 4 as through 2 and 3. The current through I 2I the resistors are I 1 = I , I 2 = I 3 = , I 4 = . 3 3

continued on next page

Chapter 28

(e)

Increasing resistor 3 increases the equivalent resistance of the entire circuit. The current in the circuit, which is the current through resistor 1, decreases. This decreases the potential difference across resistor 1, increasing the potential difference across the parallel combination. With a larger potential difference the current through resistor 4 is increased. With more current through 4, and less in the circuit to start with, the current through resistors 2 and 3 must decrease. To summarize, I 4 increases and I 1 , I 2 , and I 3 decrease .

(f)

If resistor 3 has an infinite resistance it blocks any current from passing through that branch, and the circuit effectively is just resistor 1 and resistor 4 in series with the battery. The circuit 3 now has an equivalent resistance of 4R. The current in the circuit drops to of the original 4 4 current because the resistance has increased by . All this current passes through resistors 1 3 3I 3I , I2 = I3 = 0, I4 = . and 4, and none passes through 2 or 3. Therefore I 1 = 4 4

Section 28.3 P28.20

Kirchhoff’s Rules

a f a fa f

+15.0 − 7.00 I1 − 2.00 5.00 = 0 5.00 = 7.00 I 1

so

I 1 = 0.714 A

so

I 2 = 1.29 A

I 3 = I 1 + I 2 = 2.00 A 0.714 + I 2 = 2.00

a f

a f

+ε − 2.00 1.29 − 5.00 2.00 = 0 P28.21

139

FIG. P28.20

ε = 12.6 V

We name currents I 1 , I 2 , and I 3 as shown. From Kirchhoff’s current rule, I 3 = I 1 + I 2 . Applying Kirchhoff’s voltage rule to the loop containing I 2 and I 3 ,

a f a f 8.00 = a 4.00fI + a6.00fI

12.0 V − 4.00 I 3 − 6.00 I 2 − 4.00 V = 0 3

2

Applying Kirchhoff’s voltage rule to the loop containing I 1 and I 2 ,

a f

a f

− 6.00 I 2 − 4.00 V + 8.00 I1 = 0

a8.00fI

1

a f

= 4.00 + 6.00 I 2 .

FIG. P28.21

Solving the above linear system, we proceed to the pair of simultaneous equations:

RS8 = 4I + 4I T8I = 4 + 6 I 1

1

2

+ 6I2

or

2

RS8 = 4I + 10I TI = 1.33I − 0.667 1

2

2

1

and to the single equation 8 = 4I 1 + 13.3 I 1 − 6.67 I1 = and

14.7 V = 0.846 A . 17.3 Ω

I 3 = I1 + I 2

Then give

a

f

I 2 = 1.33 0.846 A − 0.667 I 1 = 846 mA, I 2 = 462 mA, I 3 = 1.31 A .

All currents are in the directions indicated by the arrows in the circuit diagram.

140 P28.22

Direct Current Circuits

The solution figure is shown to the right.

FIG. P28.22 P28.23

We use the results of Problem 28.21. (a)

a f a fa f a12.0 Vfa1.31 Af120 s = 1.88 kJ .

∆U = ∆V I∆t = 4.00 V −0.462 A 120 s = −222 J .

By the 4.00-V battery: By the 12.0-V battery:

(b)

a f a8.00 Ωf120 s = 687 J a0.462 Af a5.00 Ωf120 s = 128 J . a0.462 Af a1.00 Ωf120 s = 25.6 J . a1.31 Af a3.00 Ωf120 s = 616 J . a1.31 Af a1.00 Ωf120 s = 205 J . I 2 R∆t = 0.846 A

By the 8.00-Ω resistor:

.

2

By the 5.00-Ω resistor:

2

By the 1.00-Ω resistor:

2

By the 3.00-Ω resistor:

2

By the 1.00-Ω resistor: (c)

2

−222 J + 1.88 kJ = 1.66 kJ from chemical to electrical. 687 J + 128 J + 25.6 J + 616 J + 205 J = 1.66 kJ from electrical to internal.

P28.24

We name the currents I 1 , I 2 , and I 3 as shown.

[2]

f a f 80.0 − I a 4.00 kΩf − 60.0 − I a3.00 kΩf = 0

[3]

I 2 = I1 + I 3

[1]

(a)

a

70.0 − 60.0 − I 2 3.00 kΩ − I 1 2.00 kΩ = 0 3

2

Substituting for I 2 and solving the resulting simultaneous equations yields

bthrough R g I = 2.69 mA bthrough R g I = 3.08 mA bthrough R g ∆V = −60.0 V − a3.08 mA fa3.00 kΩf = I 1 = 0.385 mA

(b)

1

3

3

2

2

cf

Point c is at higher potential.

−69.2 V

FIG. P28.24

Chapter 28

P28.25

Label the currents in the branches as shown in the first figure. Reduce the circuit by combining the two parallel resistors as shown in the second figure. Apply Kirchhoff’s loop rule to both loops in Figure (b) to obtain:

and

a2.71RfI + a1.71RfI a1.71RfI + a3.71RfI 1

2

= 250

1

2

= 500 .

(a)

With R = 1 000 Ω, simultaneous solution of these equations yields: I 1 = 10.0 mA and

I 2 = 130.0 mA .

From Figure (b),

Vc − Va = I 1 + I 2 1.71R = 240 V .

Thus, from Figure (a),

I4 =

b

ga

f

Vc − Va 240 V = = 60.0 mA . 4R 4 000 Ω

(b)

Finally, applying Kirchhoff’s point rule at point a in Figure (a) gives:

FIG. P28.25

I = I 4 − I 1 = 60.0 mA − 10.0 mA = +50.0 mA , or P28.26

I = 50.0 mA from point a to point e .

Name the currents as shown in the figure to the right. Then w + x + z = y . Loop equations are −200 w − 40.0 + 80.0 x = 0

FIG. P28.26

−80.0 x + 40.0 + 360 − 20.0 y = 0 +360 − 20.0 y − 70.0 z + 80.0 = 0 Eliminate y by substitution.

Eliminate x. Eliminate z = 17.5 − 13.5 w to obtain

R|x = 2.50w + 0.500 S|400 − 100 x − 20.0 w − 20.0 z = 0 T440 − 20.0 w − 20.0 x − 90.0 z = 0 RS350 − 270 w − 20.0 z = 0 T430 − 70.0w − 90.0z = 0 430 − 70.0 w − 1 575 + 1 215 w = 0 w=

Now

70.0 = 1.00 A upward in 200 Ω . 70.0

z = 4.00 A upward in 70.0 Ω x = 3.00 A upward in 80.0 Ω y = 8.00 A downward in 20.0 Ω

and for the 200 Ω,

a

fa

f

∆V = IR = 1.00 A 200 Ω = 200 V .

141

142 P28.27

Direct Current Circuits

Using Kirchhoff’s rules,

b g b g 10.0 + a1.00fI − a0.060 0fI = 0

12.0 − 0.010 0 I 1 − 0.060 0 I 3 = 0 2

and

3

I1 = I 2 + I 3

a f a f 10.0 + a1.00fI − b0.060 0 gI = 0

12.0 − 0.010 0 I 2 − 0.070 0 I 3 = 0 2

FIG. P28.27

3

Solving simultaneously, I 2 = 0.283 A downward in the dead battery and

I 3 = 171 A downward in the starter.

The currents are forward in the live battery and in the starter, relative to normal starting operation. The current is backward in the dead battery, tending to charge it up. P28.28

a f a fb g = a1.00fI + a1.00fI + a5.00fb I − I + I g = a3.00fb I − I g + a5.00fb I − I + I g

∆Vab = 1.00 I 1 + 1.00 I 1 − I 2 ∆Vab ∆Vab

1

2

1

1

1

2

2

Let I = 1.00 A , I 1 = x , and I 2 = y . Then, the three equations become:

FIG. P28.28

∆Vab = 2.00 x − y , or y = 2.00 x − ∆Vab ∆Vab = −4.00 x + 6.00 y + 5.00 and ∆Vab = 8.00 − 8.00 x + 5.00 y . Substituting the first into the last two gives: 7.00 ∆Vab = 8.00 x + 5.00 and 6.00 ∆Vab = 2.00 x + 8.00 . Solving these simultaneously yields ∆Vab = Then, R ab = P28.29

27 V ∆Vab = 17 1.00 A I

27 V. 17

R ab =

or

27 Ω . 17

We name the currents I 1 , I 2 , and I 3 as shown. (a)

I1 = I 2 + I 3 Counterclockwise around the top loop,

a

f a

f

12.0 V − 2.00 Ω I 3 − 4.00 Ω I 1 = 0 . Traversing the bottom loop,

a

f a

f

8.00 V − 6.00 Ω I 2 + 2.00 Ω I 3 = 0 4 1 1 I 1 = 3.00 − I 3 , I 2 = + I 3 , and I 3 = 909 mA . 3 3 2 (b)

a

fa

f

Va − 0.909 A 2.00 Ω = Vb Vb − Va = −1.82 V

FIG. P28.29

Chapter 28

P28.30

143

We apply Kirchhoff’s rules to the second diagram. 50.0 − 2.00 I 1 − 2.00 I 2 = 0

(1)

20.0 − 2.00 I 3 + 2.00 I 2 = 0

(2)

I1 = I 2 + I 3

(3)

Substitute (3) into (1), and solve for I 1 , I 2 , and I 3 I 1 = 20.0 A ; I 2 = 5.00 A ; I 3 = 15.0 A . Then apply P = I 2 R to each resistor:

a

f a f a2.00 Ωf = F 5.00 AIJ a4.00 Ωf = 25.0 W P =G H 2 K

a2.00 Ωf : a4.00 Ωf :

P = I 12 2.00 Ω = 20.0 A

1

P28.32

FIG. P28.30

(Half of I 2 goes through each)

a

f a

P = I 32 2.00 Ω = 15.0 A

3

P28.31

800 W

2

a2.00 Ωf : Section 28.4

2

f a2.00 Ωf = 2

450 W .

RC Circuits

e

je

j

(a)

RC = 1.00 × 10 6 Ω 5.00 × 10 −6 F = 5.00 s

(b)

Q = Cε = 5.00 × 10 −6 C 30.0 V = 150 µC

(c)

It =

(a)

I t = − I 0 e − t RC 5.10 × 10 −6 C Q = = 1.96 A I0 = RC 1 300 Ω 2.00 × 10 −9 F

ja

e

af

FG H

f

IJ K

LM MN e

ε − t RC −10.0 30.0 = exp e 6 R 1.00 × 10 1.00 × 10 6 5.00 × 10 −6

je

OP = j PQ

4.06 µA

FIG. P28.31

af

b

ge j L −9.00 × 10 s OP I at f = −a1.96 A f exp M MN b1 300 Ωge2.00 × 10 Fj PQ = −61.6 mA L −8.00 × 10 s OP = b5.10 µC g exp M qat f = Qe MN a1 300 Ωfe2.00 × 10 Fj PQ = 0.235 µC −6

−9

(b)

(c) P28.33

U=

−6

− t RC

−9

The magnitude of the maximum current is I 0 = 1.96 A .

a f

1 C ∆V 2

2

and ∆V =

Q . C

Q2 and when the charge decreases to half its original value, the stored energy is one2C 1 quarter its original value: U f = U 0 . 4 Therefore, U =

144 P28.34

Direct Current Circuits

so

qt = 1 − e − t RC Q

0.600 = 1 − e −0 .900 RC

or

e −0.900 RC = 1 − 0.600 = 0.400

thus

RC =

af

a

−0.900 = ln 0.400 RC *P28.35

f

−0.900 = 0.982 s . ln 0.400

a

f

We are to calculate

z

∞

e −2 t RC dt = −

0

P28.36

af

q t = Q 1 − e − t RC

z

∞

IJ K

FG H

RC −2 t RC 2dt RC −2 t RC − =− e e RC 2 0 2

e

je

∞ 0

=−

j

(a)

τ = RC = 1.50 × 10 5 Ω 10.0 × 10 −6 F = 1.50 s

(b)

τ = 1.00 × 10 5 Ω 10.0 × 10 −6 F = 1.00 s

(c)

The battery carries current

10.0 V = 200 µA . 50.0 × 10 3 Ω

The 100 kΩ carries current of magnitude

I = I 0 e − t RC =

e

je

j

FG 10.0 V IJ e H 100 × 10 Ω K 200 µA + b100 µA ge .

(a)

− t 1.00 s

3

.

− t 1.00 s

So the switch carries downward current P28.37

RC −∞ RC RC e − e0 = − 0−1 = + . 2 2 2

Call the potential at the left junction VL and at the right VR . After a “long” time, the capacitor is fully charged. VL = 8.00 V because of voltage divider: 10.0 V = 2.00 A 5.00 Ω VL = 10.0 V − 2.00 A 1.00 Ω = 8.00 V IL =

a

fa f F 2.00 Ω IJ a10.0 V f = 2.00 V =G H 2.00 Ω + 8.00 Ω K

Likewise,

VR

or

IR =

10.0 V = 1.00 A 10.0 Ω

a

f a

fa

FIG. P28.37(a)

f

VR = 10.0 V − 8.00 Ω 1.00 A = 2.00 V .

(b)

Therefore,

∆V = VL − VR = 8.00 − 2.00 = 6.00 V .

Redraw the circuit

R=

and so

1 = 3.60 Ω 1 9.00 Ω + 1 6.00 Ω

b

g b

g

RC = 3.60 × 10 −6 s 1 e − t RC = 10 t = RC ln 10 = 8. 29 µs .

FIG. P28.37(b)

Chapter 28

*P28.38

(a)

af

af

t = RC ln

FG H

∆V0 t = ln RC ∆V

∆V0 = e + t RC ∆V

∆V = e − t RC ∆V0

P28.39

3 000 V

We model the person’s body and street shoes as shown. For the discharge to reach 100 V, q t = Qe − t RC = C∆V t = C∆V0 e − t RC

FG ∆V IJ = 5 000 × 10 Ωe230 × 10 H ∆V K 6

0

e

−12

150 pF

IJ K

000 I JK = j FGH 3100

F ln

t = 1 × 10 6 V A 230 × 10 −12 C V ln 30 = 782 µs

(a)

τ = RC = 4.00 × 10 6 Ω 3.00 × 10 −6 F = 12.0 s

(b)

I=

je

3.91 s

ε − t RC 12.0 = e e − t 12.0 s 6 R 4.00 × 10

a f

q = 36.0 µC 1 − e − t 12 .0 ∆V0 =

FIG. P28.38(a)

j

q = Cε 1 − e − t RC = 3.00 × 10 −6 12.0 1 − e − t 12.0

P28.40

5 000 M Ω

80 pF

j

(b)

e

I = 3.00 µAe − t 12 .0

Q C

FIG. P28.39

b g

Then, if q(t ) = Q e − t RC

∆V ( t ) = ∆V0 e − t RC

and

b∆V g = e

∆V ( t )

− t RC .

0

b g

When ∆V ( t ) = 1 ∆V0 , then 2

e − t RC = −

Section 28.5 P28.41

FG IJ H K

t 1 = ln = − ln 2 . RC 2

R=

Thus,

1 2

t C ln 2

a f

.

Electrical Meters

e

j

∆V = I g rg = I − I g R p , or R p =

I g rg

eI − I

g

a f j eI − I j =

I g 60.0 Ω g

Therefore, to have I = 0.100 A = 100 mA when I g = 0.500 mA : Rp =

a0.500 mAfa60.0 Ωf = 99.5 mA

145

0.302 Ω . FIG. P28.41

146 P28.42

Direct Current Circuits

a

f e

j

Applying Kirchhoff’s loop rule, − I g 75.0 Ω + I − I g R p = 0 . Therefore, if I = 1.00 A when I g = 1.50 mA , Rp =

a f = e1.50 × 10 Aja75.0 Ωf = eI − I j 1.00 A − 1.50 × 10 A −3

I g 75.0 Ω

−3

g

P28.43

Series Resistor → Voltmeter

0.113 Ω .

b

25.0 = 1.50 × 10 −3 Rs + 75.0

∆V = IR :

FIG. P28.42

g

Rs = 16.6 kΩ .

Solving,

FIG. P28.43 P28.44

(a)

In Figure (a), the emf sees an equivalent resistance of 200.00 Ω . 6.000 0 V 200.00 Ω = 0.030 000 A

I=

6.0000 V

A

V

A

V

180.00 Ω

180.00 Ω

180.00 Ω

(a)

(b)

(c)

FIG. P28.44

(b)

20.000 Ω

20.000 Ω

20.000 Ω

b

f

ga

The terminal potential difference is

∆V = IR = 0.030 000 A 180.00 Ω = 5.400 0 V .

In Figure (b),

R eq =

F 1 + 1 I GH 180.00 Ω 20 000 Ω JK

−1

= 178.39 Ω.

The equivalent resistance across the emf is 178.39 Ω + 0.500 00 Ω + 20.000 Ω = 198.89 Ω . The ammeter reads and the voltmeter reads

I=

ε R

=

6.000 0 V = 0.030 167 A 198.89 Ω

b

ga F 1 + 1 I GH 180.50 Ω 20 000 Ω JK

f

∆V = IR = 0.030 167 A 178.39 Ω = 5.381 6 V . −1

(c)

In Figure (c),

= 178.89 Ω .

Therefore, the emf sends current through

Rtot = 178.89 Ω + 20.000 Ω = 198.89 Ω.

The current through the battery is

I=

but not all of this goes through the ammeter.

6.000 0 V = 0.030 168 A 198.89 Ω

b

ga

f

The voltmeter reads

∆V = IR = 0.030 168 A 178.89 Ω = 5.396 6 V .

The ammeter measures current

I=

∆V 5.396 6 V = = 0.029 898 A . R 180.50 Ω

The connection shown in Figure (c) is better than that shown in Figure (b) for accurate readings.

Chapter 28

P28.45

Consider the circuit diagram shown, realizing that I g = 1.00 mA . For the 25.0 mA scale:

a24.0 mAfbR

1

g a

fa

+ R 2 + R3 = 1.00 mA 25.0 Ω

f

FG H

FIG. P28.45

IJ K

25.0 R1 + R 2 + R3 = Ω. 24.0

or

(1)

(2)

For the 100 mA scale:

a49.0 mAfbR + R g = a1.00 mAfb25.0 Ω + R g 49.0b R + R g = 25.0 Ω + R . a99.0 mAfR = a1.00 mAfb25.0 Ω + R + R g

or

99.0 R1 = 25.0 Ω + R 2 + R3 .

(3)

For the 50.0 mA scale:

1

or

1

2

3

2

3

1

2

3

Solving (1), (2), and (3) simultaneously yields R1 = 0.260 Ω , R 2 = 0.261 Ω , R3 = 0.521 Ω . P28.46

∆V = IR (a)

jb

e

20.0 V = 1.00 × 10 −3 A R1 + 60.0 Ω

g

R1 = 1.994 × 10 4 Ω = 19.94 kΩ

P28.47

FIG. P28.46

e

jb

g

R 2 = 30.0 kΩ

e

jb

g

R3 = 50.0 kΩ

(b)

50.0 V = 1.00 × 10 −3 A R 2 + R1 + 60.0 Ω

(c)

100 V = 1.00 × 10 −3 A R3 + R1 + 60.0 Ω

e

ja

Ammeter:

I g r = 0.500 A − I g 0.220 Ω

or

I g r + 0.220 Ω = 0.110 V

Voltmeter:

2.00 V = I g r + 2 500 Ω

a

f

b

f (1)

g

(2)

Solve (1) and (2) simultaneously to find: I g = 0.756 mA and r = 145 Ω . FIG. P28.47

Section 28.6

Household Wiring and Electrical Safety

FG ρ IJ = a1.00 Af e1.70 × 10 Ω ⋅ mja16.0 ftfb0.304 8 m ftg = H AK π e0.512 × 10 mj 2

P28.48

147

2

2

(a)

P =I R=I

(b)

P = I 2 R = 100 0.101 Ω = 10.1 W

−8

−3

a

f

2

0.101 W

148 P28.49

Direct Current Circuits

(a)

P = I∆V :

1 500 W P = = 12.5 A . ∆V 120 V 750 W I= = 6.25 A . 120 V I=

So for the Heater, For the Toaster,

I=

And for the Grill, (b)

1 000 W = 8.33 A . 120 V

12.5 + 6.25 + 8.33 = 27.1 A The current draw is greater than 25.0 amps, so this circuit breaker would not be sufficient.

P28.50

2 2 I Al R Al = I Cu RCu

P28.51

(a)

RCu I Cu = R Al

I Al =

so

ρ Cu 1.70 20.0 = 0.776 20.0 = 15.5 A I Cu = 2.82 ρ Al

a f

a f

Suppose that the insulation between either of your fingers and the conductor adjacent is a chunk of rubber with contact area 4 mm 2 and thickness 1 mm. Its resistance is

e

je

j

10 13 Ω ⋅ m 10 −3 m ρ ≈ ≈ 2 × 10 15 Ω . A 4 × 10 −6 m 2

R=

The current will be driven by 120 V through total resistance (series) 2 × 10 15 Ω + 10 4 Ω + 2 × 10 15 Ω ≈ 5 × 10 15 Ω . It is: I =

(b)

∆V 120 V ~ ~ 10 −14 A . 15 R 5 × 10 Ω

Vh , where Vh 2 is the potential of the “hot” wire. The potential difference between your finger and thumb is ∆V = IR ~ 10 −14 A 10 4 Ω ~ 10 −10 V . So the points where the rubber meets your fingers are

The resistors form a voltage divider, with the center of your hand at potential

e

je

j

at potentials of ~

Vh + 10 −10 V 2

and

~

Vh − 10 −10 V . 2

Additional Problems P28.52

The set of four batteries boosts the electric potential of each bit of charge that goes through them by 4 × 1.50 V = 6.00 V . The chemical energy they store is

a

fb

g

∆U = q∆V = 240 C 6.00 J C = 1 440 J . ∆V 6.00 V = = 0.030 0 A . 200 Ω R

The radio draws current

I=

So, its power is

P = ∆V I = 6.00 V 0.030 0 A = 0.180 W = 0.180 J s.

a f a

Then for the time the energy lasts, we have P = We could also compute this from I =

Q : ∆t

fb

E : ∆t

∆t = ∆t =

g

E

P

=

1 440 J = 8.00 × 10 3 s . 0.180 J s

240 C Q = = 8.00 × 10 3 s = 2. 22 h . I 0.030 0 A

149

Chapter 28

P28.53

aR + r f = FGH εP IJK R . R + a 2 r − x fR − r = 0 . R + a 2.40 − xfR − 1.44 = 0 , −a 2.40 − x f ± a 2.40 − x f R=

ε ε 2R , so P = I 2 R = or 2 R+r R+r ε2 2 , then R + r = xR or Let x ≡ P With r = 1. 20 Ω, this becomes I=

a f

a

2

f

2

With

(b)

For

2

− 5.76

2

ε = 9.20 V and R=

2

2

which has solutions of (a)

2

+4.21 ±

P = 12.8 W , x = 6.61 :

a4.21f

2

− 5.76

2

= 3.84 Ω or

ε = 9.20 V and +1.59 ±

.

0.375 Ω .

P = 21.2 W , x ≡

a1.59f

2

ε2 = 3.99 P

− 5.76

1.59 ± −3.22 = . 2 2 The equation for the load resistance yields a complex number, so there is no resistance R=

that will extract 21.2 W from this battery. The maximum power output occurs when R = r = 1.20 Ω, and that maximum is: Pmax = P28.54

ε2 = 17.6 W . 4r

Using Kirchhoff’s loop rule for the closed loop, +12.0 − 2.00 I − 4.00 I = 0 , so I = 2.00 A

a

fa

f a fa

f

Vb − Va = +4.00 V − 2.00 A 4.00 Ω − 0 10.0 Ω = −4.00 V . Thus, ∆Vab = 4.00 V and point a is at the higher potential . P28.55

*P28.56

I=

ε 3R

Pseries = ε I =

I=

3ε R

Pparallel = ε I =

(a)

Req = 3 R

(b)

Req =

(c)

Nine times more power is converted in the parallel connection.

(a)

We model the generator as a constant-voltage power supply. Connect two light bulbs across it in series. Each bulb is designed to P 100 W = = 0.833 A . Each has resistance carry current I = ∆V 120 V ∆V 120 V = = 144 Ω . In the 240-V circuit the equivalent R= 0.833 A I resistance is 144 Ω + 144 Ω = 288 Ω . The current is ∆V 240 V = = 0.833 A and the generator delivers power I= 288 Ω R P = I∆V = 0.833 A 240 V = 200 W .

1 R = 3 1R + 1R + 1R

b g b g b g

a

continued on next page

f

ε2 3R 3ε 2 R

FIG. P28.56(a)

150

Direct Current Circuits

(b)

The hot pot is designed to carry current I=

P ∆V

=

28.8 Ω

500 W = 4.17 A . 120 V

144 Ω

240 V

It has resistance R=

FIG. P28.56(b)

∆V 120 V = = 28.8 Ω . 4.17 A I

4.17 A = 5 , we can place five light bulbs in parallel and the hot 0.833 A pot in series with their combination. The current in the generator is then 4.17 A and it In terms of current, since

a

f

delivers power P = I∆V = 4.17 A 240 V = 1 000 W . P28.57

The current in the simple loop circuit will be I = (a)

∆Vter = ε − Ir =

(b)

I=

(c)

P = I2R = ε 2

εR R+r

ε R+r

Then 2R = R + r P28.58

.

and

∆Vter → ε as R → ∞ .

and

I→

ε as R → 0 . r

ε2 − 2ε 2 R dP = + 3 dR R+r R+r

R

aR + r f

ε R+r

a

2

and

f a

f

2

FIG. P28.57

=0

R=r .

af

e

j

The potential difference across the capacitor

∆V t = ∆Vmax 1 − e − t RC .

Using 1 Farad = 1 s Ω ,

4.00 V = 10.0 V 1 − e

Therefore,

0.400 = 1.00 − e

Or

e

Taking the natural logarithm of both sides,

−

and

R=−

fLMN

a

e

j

− 3.00 ×10 5 Ω R

a

− 3.00 s

e

j

− 3 .00 × 10 5 Ω R

f Re10.0 ×10

−6

sΩ

j O.

PQ

.

= 0.600 .

3.00 × 10 5 Ω = ln 0.600 R

a

f

3.00 × 10 5 Ω = +5.87 × 10 5 Ω = 587 kΩ . ln 0.600

a

f

Chapter 28

P28.59

Let the two resistances be x and y.

Ps 225 W = = 9.00 Ω 2 2 I 5.00 A

Then,

Rs = x + y =

and

Rp =

so

x 9.00 Ω − x = 2.00 Ω x + 9.00 Ω − x

a

a

y = 9.00 Ω − x

f

x

Pp xy 50.0 W = 2 = = 2.00 Ω 2 x+y I 5.00 A

f

a

a

y

x

f

y

x 2 − 9.00 x + 18.0 = 0 .

f

Factoring the second equation,

ax − 6.00fax − 3.00f = 0

so

x = 6.00 Ω or x = 3.00 Ω .

Then, y = 9.00 Ω − x gives

y = 3.00 Ω or y = 6.00 Ω .

FIG. P28.59

The two resistances are found to be 6.00 Ω and 3.00 Ω . P28.60

Let the two resistances be x and y. Then, Rs = x + y =

Pp xy Ps = = and R . p x + y I2 I2

From the first equation, y = becomes

e

x Ps I 2 − x

e

2

j

x + Ps I − x

j

=

Pp I

2

Using the quadratic formula, x =

Then, y =

(a)

(b)

FG P IJ x + P P HI K I

s p 4

s 2

Ps ± Ps 2 − 4Ps Pp 2I 2

Ps + Ps 2 − 4Ps Pp 2I 2

and

= 0.

FIG. P28.60

.

Ps − Ps 2 − 4Ps Pp 2I 2

.

c ∑ R h − bε + ε g = 0 40.0 V − a 4.00 A f a 2.00 + 0.300 + 0.300 + RfΩ − a6.00 + 6.00 f V = 0 ;

ε−I

1

2

g a

fa

2

f

2 2

2

For the limiting resistor,

b

so

R = 4.40 Ω

a f a2.00 Ωf = 32.0 W . P = I R = a 4.00 A f a0.600 Ωf = 9.60 W . P = a 4.00 A f a 4.40 Ωf = 70.4 W . P = I 2 R = 4.00 A

Inside the supply, Inside both batteries together,

(c)

y

Ps ∓ Ps 2 − 4Ps Pp Ps . − = x gives y 2I 2 I2

The two resistances are P28.61

x

Ps − x , and the second I2 or x 2 −

y

x

P = I ε 1 + ε 2 = 4.00 A 6.00 + 6.00 V = 48.0 W

151

152 *P28.62

Direct Current Circuits

(a)

∆V1 = ∆V2

I 1 R1 = I 2 R 2

I R R + R1 I = I 1 + I 2 = I1 + 1 1 = I1 2 R2 R2 I1 = I2 = (b)

R2

I2

FIG. P28.62(a)

I 1 R1 IR1 = = I2 R2 R1 + R 2

b

The power delivered to the pair is P = I12 R1 + I 22 R2 = I 12 R1 + I − I1 dP we want to find I 1 such that = 0. dI 1

ga f

dP = 2 I 1 R1 + 2 I − I 1 −1 R 2 = 0 dI 1 I1 =

I1

I

IR 2 R1 + R 2

b

R1

g R . For minimum power 2

2

I 1 R1 − IR 2 + I 1 R 2 = 0

IR 2 R1 + R 2

This is the same condition as that found in part (a). P28.63

Let Rm = measured value, R = actual value, (a)

I R = current through the resistor R I = current measured by the ammeter. (a)

b

g

When using circuit (a), I R R = ∆V = 20 000 I − I R or R = 20 000 But since I =

∆V ∆V and I R = , we have Rm R

I R = I R Rm

FIG. P28.63

bR − R g . m

R = 20 000

When R > Rm , we require

bR − R g ≤ 0.050 0 .

b

When using circuit (b), But since I R =

∆V , Rm

When Rm > R , we require From (2) we find

(b)

R

and

Rm

(1)

m

R

g

Therefore, Rm ≥ R 1 − 0.050 0 and from (1) we find (b)

LM I − 1OP . NI Q

R ≤ 1 050 Ω .

a

f

I R R = ∆V − I R 0.5 Ω .

a

f

Rm = 0.500 + R .

bR

m

−R

R

g ≤ 0.050 0 .

R ≥ 10.0 Ω .

(2)

Chapter 28

P28.64

FG H

dE ε −1 RC e . = P = εI = ε dt R

Then the total energy put out by the battery is

z z

z

dE =

FG H

fz

∞ ε2 t − RC exp − R RC 0

a

dE =

IJ FG − dt IJ = −ε C expFG − t IJ K H RC K H RC K

IJ K

FG H

∞

ε2 t exp − dt R RC t=0

∞

2

= −ε 2 C 0 − 1 = ε 2C . 0

FG H

IJ K

The power delivered to the resistor is

ε2 dE 2t . = P = ∆VR I = I 2 R = R 2 exp − dt RC R

So the total internal energy appearing in the resistor is

z z

IJ z expFG − 2t IJ FG − 2dt IJ = − ε C expFG − 2t IJ z K H RC K H RC K 2 H RC K 1 The energy finally stored in the capacitor is U = C a ∆V f 2 dE =

FG H

∞

ε2 RC − R 2

battery. (a)

2

0

conserved ε 2 C =

P28.65

IJ K

The battery supplies energy at a changing rate

153

e

I=

0 2

=

1 2 Cε . Thus, energy of the circuit is 2

j L F ja10.0 V fM1 − e N

−10.0

e 2.00 ×10 je1.00 ×10 j O = −6

6

PQ

9.93 µC

FG IJ H K

FG 10.0 V IJ e = 3.37 × 10 A = 33.7 nA H 2.00 × 10 Ω K dU d F 1 q I F q I dq F q I =G J = G JI = dt dt GH 2 C JK H C K dt H C K dU F 9.93 × 10 C I = e3.37 × 10 Aj = 3.34 × 10 W = 334 nW dt GH 1.00 × 10 C V JK P = Iε = e3.37 × 10 A ja10.0 V f = 3.37 × 10 W = 337 nW −5.00

6

−8

2

−6

−8

−6

battery

IJ K

ε 2C ε 2C 0−1 = . 2 2

dq ∆V − t RC = e dt R

I=

(d)

=−

FG H

ε2 2t exp − dt R RC 0

q = C∆V 1 − e − t RC

e

(c)

∞

∞

1 2 1 ε C + ε 2C and resistor and capacitor share equally in the energy from the 2 2

q = 1.00 × 10 −6

(b)

dE =

−8

−7

−7

154 P28.66

Direct Current Circuits

Start at the point when the voltage has just reached

2 ∆V 3

2 ∆V and is 3 decaying towards 0 V with a time constant R 2 C

and the switch has just closed. The voltage is

∆V

Voltage controlled switch

a f LMN 32 ∆V OPQe . 1 We want to know when ∆V atf will reach ∆V . 3 1 L2 O Therefore, ∆V = M ∆V P e 3 N3 Q

R2

− t R 2C

∆VC t =

+

R1

C

V ∆Vc

C

− t R2C

1 2

or

e − t R 2C =

or

t1 = R 2 C ln 2 .

FIG. P28.66 1 After the switch opens, the voltage is ∆V , increasing toward ∆V with time constant R1 + R 2 C : 3

b

LM 2 ∆V OPe N3 Q

af

∆VC t = ∆V −

.

af

2 ∆V 3 2 2 ∆V = ∆V − ∆Ve − t b R1 + R2 gC 3 3

b

g

t 2 = R1 + R 2 C ln 2

So

(a)

g

∆VC t =

When

P28.67

b

− t R1 + R 2 C

g

or

e − t b R1 + R2 gC =

and

T = t1 + t 2 =

First determine the resistance of each light bulb: P =

a∆V f = a120 Vf 2

R=

bR

1

g

+ 2 R 2 C ln 2 .

2

R

2

60.0 W

P

a∆V f

1 . 2

= 240 Ω .

We obtain the equivalent resistance Req of the network of light FIG. P28.67

bulbs by identifying series and parallel equivalent resistances: Req = R1 +

1 = 240 Ω + 120 Ω = 360 Ω. + 1 R3

b1 R g b g 2

a∆V f = a120 Vf 2

The total power dissipated in the 360 Ω is

(b)

P=

The current through the network is given by P = I 2 Req : I = The potential difference across R1 is

Req

360 Ω

2

= 40.0 W .

P 40.0 W 1 = = A. 360 Ω 3 Req

∆V1 = IR1 =

FG 1 AIJ a240 Ωf = H3 K

The potential difference ∆V23 across the parallel combination of R 2 and R3 is ∆V23 = IR 23 =

I= 1 FG 1 AIJ FG H 3 K H b1 240 Ωg + b1 240 Ωg JK

40.0 V .

80.0 V .

Chapter 28

*P28.68

(a)

155

With the switch closed, current exists in a simple series circuit as shown. The capacitors carry no current. For R 2 we have P 2.40 V ⋅ A = = 18.5 mA . P = I 2 R2 I= R2 7 000 V A The potential difference across R1 and C 1 is

jb

e

g

∆V = IR1 = 1.85 × 10 −2 A 4 000 V A = 74.1 V . The charge on C 1

ja

e

FIG. P28.68(a)

f

Q = C1 ∆V = 3.00 × 10 −6 C V 74.1 V = 222 µC . The potential difference across R 2 and C 2 is

jb

e

g

∆V = IR 2 = 1.85 × 10 −2 A 7 000 Ω = 130 V . The charge on C 2

ja

e

f

Q = C 2 ∆V = 6.00 × 10 −6 C V 130 V = 778 µC . The battery emf is

b

g

b

g

IR eq = I R1 + R 2 = 1.85 × 10 −2 A 4 000 + 7 000 V A = 204 V . (b)

In equilibrium after the switch has been opened, no current exists. The potential difference across each resistor is zero. The full 204 V appears across both capacitors. The new charge C 2

e

ja

f

Q = C 2 ∆V = 6.00 × 10 −6 C V 204 V = 1 222 µC for a change of 1 222 µC − 778 µC = 444 µC . *P28.69

FIG. P28.68(b)

The battery current is

a150 + 45 + 14 + 4f mA = 213 mA .

(a)

The resistor with highest resistance is that carrying 4 mA. Doubling its resistance will reduce the current it carries to 2 mA. Then the total current is

FIG. P28.69

211 = a150 + 45 + 14 + 2f mA = 211 mA , nearly the same as before. The ratio is 213

(b)

The resistor with least resistance carries 150 mA. Doubling its resistance changes this current to 75 mA and changes the total to 138 75 + 45 + 14 + 4 mA = 138 mA . The ratio is = 0.648 , representing a much larger 213 reduction (35.2% instead of 0.9%).

a

(c)

0.991 .

f

This problem is precisely analogous. As a battery maintained a potential difference in parts (a) and (b), a furnace maintains a temperature difference here. Energy flow by heat is analogous to current and takes place through thermal resistances in parallel. Each resistance can have its “R-value” increased by adding insulation. Doubling the thermal resistance of the attic door will produce only a negligible (0.9%) saving in fuel. Doubling the thermal resistance of the ceiling will produce a much larger saving. The ceiling originally has the smallest thermal resistance.

156

*P28.70

Direct Current Circuits

From the hint, the equivalent resistance of

RT +

That is,

RT +

.

1 = R eq 1 RL + 1 R eq RL R eq

= R eq

RL + R eq

2 RT RL + RT R eq + RL R eq = RL R eq + R eq 2 − RT R eq − RT RL = 0 R eq

R eq =

a fb 2a1f

RT ± RT2 − 4 1 − RT RL

g

Only the + sign is physical: R eq =

P28.71

1 2

FH

IK

4RT RL + RT2 + RT .

For example, if

RT = 1 Ω.

And

RL = 20 Ω , R eq = 5 Ω.

(a)

After steady-state conditions have been reached, there is no DC current through the capacitor.

b

g

I R3 = 0 steady-state .

Thus, for R3 :

For the other two resistors, the steady-state current is simply determined by the 9.00-V emf across the 12-kΩ and 15-kΩ resistors in series: For R1 and R 2 : (b)

ε

IbR

1 + R2

g = R1 + R 2

=

9.00 V

a12.0 kΩ + 15.0 kΩf =

b

After the transient currents have ceased, the potential difference across C is the same as the potential difference across R 2 = IR 2 because there is no voltage drop across R3 . Therefore, the charge Q on C is

b

a f

Q = C ∆V

R2

= 50.0 µC .

continued on next page

b g b

g

gb

ga

= C IR 2 = 10.0 µF 333 µA 15.0 kΩ

g

333 µA steady-state .

f FIG. P28.71(b)

Chapter 28

(c)

When the switch is opened, the branch containing R1 is no longer part of the circuit. The capacitor discharges through R 2 + R3 with a time constant of

b g g a

b

fb

g

R 2 + R3 C = 15.0 kΩ + 3.00 kΩ 10.0 µF = 0.180 s . The initial current I i in this discharge circuit is determined by the initial potential difference across the capacitor applied to R 2 + R3 in series: Ii =

a f ∆V

bR

2

b

C

+ R3

g

=

g b

b a g

f f

ga

FIG. P28.71(c)

333 µA 15.0 kΩ IR 2 = = 278 µA . R 2 + R3 15.0 kΩ + 3.00 kΩ

Thus, when the switch is opened, the current through R 2 changes instantaneously from 333 µA (downward) to 278 µA (downward) as shown in the graph. Thereafter, it decays according to

I R 2 = I i e − t b R2 + R3 gC = (d)

b278 µAge

a

− t 0 .180 s

f afor t > 0f

The charge q on the capacitor decays from Qi to q = Qi e −

b

g

.

Qi according to 5

t R 2 + R3 C

Qi − t 0.180 s g = Qi e b 5 5 = e t 0.180 s t 180 ms t = 0.180 s ln 5 = 290 ms ln 5 =

a

*P28.72

(a)

First let us flatten the circuit on a 2-D plane as shown; then reorganize it to a format easier to read. Notice that the five resistors on the top are in the same connection as those in Example 28.5; the same argument tells us that the middle resistor can be removed without affecting the circuit. The remaining resistors over the three parallel branches have equivalent resistance R eq =

(b)

fa f

FG 1 + 1 + 1 IJ H 20 20 10 K

−1

= 5.00 Ω .

So the current through the battery is ∆V 12.0 V = = 2.40 A . R eq 5.00 Ω

FIG. P28.72(a)

157

158 P28.73

Direct Current Circuits

∆V = ε e − t RC

FG ε IJ = FG 1 IJ t . H ∆V K H RC K F ε IJ versus t should A plot of lnG H ∆V K

so

ln

be a straight line with slope equal 1 to . RC Using the given data values: FIG. P28.73 (a)

A least-square fit to this data yields the graph above.

∑ xi = 282 , ∑ xi yi = 244,

∑ = 1.86 × 10 ∑ yi = 4.03 , x i2

4

af

,

4.87

N=8

11.1

19.4 c∑ x y h − c∑ x hc∑ y h = 0.011 8 30.8 N e∑ x j − c∑ x h 46.6 67.3 ∑ x jc∑ y h − c∑ x hc∑ x y h e = 0.088 2 Intercept = 102.2 N e∑ x j − c∑ x h F ε IJ = b0.011 8gt + 0.088 2 The equation of the best fit line is: lnG H ∆V K Slope =

N

i i

i

i

i

i

2 i

(b)

i

2

2 i

2 i

i i

2

i

Thus, the time constant is and the capacitance is

P28.74

(a)

1 R1 . 2

(b)

FG H

5.55

0.109

4.93

0.228

4.34

0.355

3.72

0.509

3.09

0.695

2.47

0.919

1.83

1.219

.

R1

a

IJ K

If R1 = 13.0 Ω and R 2 = 6.00 Ω, then R x = 2.75 Ω .

Ry

Rx

(1)

1 1 1 R1 + R x , or R x = R 2 − R1 . 2 2 4

b

c

Ry

For the second measurement, the equivalent circuit is shown in Figure 2. 1 (2) R ac = R 2 = R y + R x . Thus, 2 Substitute (1) into (2) to obtain: R2 =

0

1 1 = = 84.7 s slope 0.011 8 τ 84.7 s = 8.47 µF . C= = R 10.0 × 10 6 Ω

R ab = R1 = R y + R y = 2 R y Ry =

b

ln ε ∆V

∆V V 6.19

τ = RC =

For the first measurement, the equivalent circuit is as shown in Figure 1.

so

af

ts 0

Figure 1 a Ry

R2

Ry

Figure 2

FIG. P28.74

The antenna is inadequately grounded since this exceeds the limit of 2.00 Ω .

c Rx

g

Chapter 28

P28.75

The total resistance between points b and c is: 2.00 kΩ 3.00 kΩ R= = 1. 20 kΩ . 2.00 kΩ + 3.00 kΩ The total capacitance between points d and e is: C = 2.00 µF + 3.00 µF = 5.00 µF .

a

fa

2.00 kΩ

f

b

3.00 kΩ

The potential difference between point d and e in this series RC circuit at any time is:

a

f

∆V = ε 1 − e − t RC = 120.0 V 1 − e −1 000 t

6

S

a

and q 2

f b ga = C a ∆V f = b3.00 µFga120.0 V f 1 − e

*P28.76

(a)

2

C1 = 2.00 µF d

e

C2 = 3.00 µF 120 V

+

-

f

. FIG. P28.75

Therefore, the charge on each capacitor between points d and e is: q1 = C1 ∆V = 2.00 µF 120.0 V 1 − e −1 000 t

c

159

6

=

b240 µCg 1 − e = b360 µC g 1 − e

−1 000 t 6

−1 000 t 6

−1 000 t 6

.

Let i represent the current in the battery and i c the current charging the capacitor. Then i − i c is q the current in the voltmeter. The loop rule applied to the inner loop is +ε − iR − = 0 . The loop C dq dq rule for the outer perimeter is ε − iR − i − i c r = 0 . With i c = , this becomes ε − iR − ir + r = 0 . dt dt q ε by substitution to obtain Between the two loop equations we eliminate i = − R RC

b g

a

ε − R+r

fFGH Rε − RCq IJK + dqdt r = 0

dq R+r R+r ε+ q+ r =0 R RC dt q r Rr dq − ε+ + =0 R+r C R + r dt

ε−

This is the differential equation required. (b)

To solve we follow the same steps as on page 875.

FG H

ε rC dq ε R + r R+r = − q=− q− dt R RrC RrC R+r

z q

0

z

t dq R+r =− dt q − ε rC R + r RrC 0

a

f

IJ K

F ε rc IJ lnG q − H R+rK

q

=− 0

F q − ε rc aR + r f I = − R + r t q − ε rc = − ε rc e lnG R+r R+r H − ε rc aR + r f JK RrC r Rr where R = q= Cε e1 − e j r+R R+r − t Req C

a f

− R + r RrC t

eq

The voltage across the capacitor is VC = (c)

t

R+r t RrC 0

q r −t R C = ε 1 − e eq . C r+R

e

j

r rε ε 1−0 = . If the switch is then opened, r+R r+R the capacitor discharges through the voltmeter. Its voltage decays exponentially according rε − t rC . to e r+R

As t → ∞ the capacitor voltage approaches

a f

160

Direct Current Circuits

ANSWERS TO EVEN PROBLEMS P28.2

(a) 1.79 A ; (b) 10.4 V

P28.42

0.113 Ω

P28.4

(a) 12.4 V ; (b) 9.65 V

P28.44

P28.6

(a) 17.1 Ω ; (b) 1.99 A in 4 Ω and 9 Ω; 1.17 A in 7 Ω; 0.818 A in 10 Ω

(a) 30.000 mA , 5.400 0 V ; (b) 30.167 mA , 5.381 6 V ; (c) 29.898 mA ; 5.396 6 V

P28.8

29.5 V

P28.46

see the solution

P28.10

(a) see the solution; (b) no

P28.48

(a) 0.101 W; (b) 10.1 W

P28.12

see the solution

P28.50

15.5 A

P28.14

R1 = 1.00 kΩ; R 2 = 2.00 kΩ ; R3 = 3.00 kΩ

P28.52

2.22 h

P28.16

470 Ω and 220 Ω

P28.54

a is 4.00 V higher

P28.18

(a) 11.0 Ω ; (b) and (d) see the solution; (c) 220 Ω; (e) Parallel

P28.56

(a) see the solution; 833 mA; 200 W; (b) see the solution; 4.17 A; 1.00 kW

P28.20

I 1 = 714 mA ; I 2 = 1.29 A ; ε = 12.6 V

P28.58

587 kΩ

P28.22

see the solution

P28.60

P28.24

(a) 0.385 mA in R1 ; 2.69 mA in R3 ; 3.08 mA in R 2 ; (b) c higher by 69.2 V

P28.62

(a) I 1 =

1.00 A up in 200 Ω ; 4.00 A up in 70 Ω ; 3.00 A up in 80 Ω ; 8.00 A down in 20 Ω; 200 V

P28.64

see the solution

P28.28

see the solution

P28.66

bR

P28.30

800 W to the left-hand resistor; 25.0 W to each 4 Ω; 450 W to the right-hand resistor

P28.68

(a) 222 µC ; (b) increase by 444 µC

P28.32

(a) −61.6 mA ; (b) 0.235 µC ; (c) 1.96 A

P28.70

see the solution

P28.34

0.982 s

P28.72

(a) 5.00 Ω; (b) 2.40 A

P28.36

(a) 1.50 s; (b) 1.00 s; (c) 200 µA + 100 µA e − t 1.00 s

P28.74

(a) R x = R 2 −

P28.76

(a) and (b) see the solution; (c)

P28.26

b

g

P28.38

(a) 3.91 s; (b) 0.782 ms

P28.40

t C ln 2

Ps + Ps 2 − 4Ps Pp

and

2I 2

Ps − Ps 2 − 4Ps Pp 2I 2

IR 2 IR1 ; ; I2 = R + R 1 2 1 + R2 (b) see the solution

1

bR

g

g

+ 2 R 2 C ln 2

R1 ; (b) no; R x = 2.75 Ω 4 rε − t rC e r+R

29 Magnetic Fields CHAPTER OUTLINE 29.1 29.2 29.3 29.4 29.5

29.6

Magnetic Fields and Forces Magnetic Force Acting on a Current-Carrying Conductor Torque on a Current Loop in a Uniform Magnetic Field Motion of a Charged Particle in a Uniform Magnetic Field Applications Involving Charged Particles Moving in a Magnetic Field The Hall Effect

ANSWERS TO QUESTIONS Q29.1

The force is in the +y direction. No, the proton will not continue with constant velocity, but will move in a circular path in the x-y plane. The magnetic force will always be perpendicular to the magnetic field and also to the velocity of the proton. As the velocity changes direction, the magnetic force on the proton does too.

Q29.2

If they are projected in the same direction into the same magnetic field, the charges are of opposite sign.

Q29.3

Not necessarily. If the magnetic field is parallel or antiparallel to the velocity of the charged particle, then the particle will experience no magnetic force.

Q29.4

One particle veers in a circular path clockwise in the page, while the other veers in a counterclockwise circular path. If the magnetic field is into the page, the electron goes clockwise and the proton counterclockwise.

Q29.5

Send the particle through the uniform field and look at its path. If the path of the particle is parabolic, then the field must be electric, as the electric field exerts a constant force on a charged particle. If you shoot a proton through an electric field, it will feel a constant force in the same direction as the electric field—it’s similar to throwing a ball through a gravitational field. If the path of the particle is helical or circular, then the field is magnetic—see Question 29.1. If the path of the particle is straight, then observe the speed of the particle. If the particle accelerates, then the field is electric, as a constant force on a proton with or against its motion will make its speed change. If the speed remains constant, then the field is magnetic—see Question 29.3.

Q29.6

Similarities: Both can alter the velocity of a charged particle moving through the field. Both exert forces directly proportional to the charge of the particle feeling the force. Positive and negative charges feel forces in opposite directions. Differences: The direction of the electric force is parallel or antiparallel to the direction of the electric field, but the direction of the magnetic force is perpendicular to the magnetic field and to the velocity of the charged particle. Electric forces can accelerate a charged particle from rest or stop a moving particle, but magnetic forces cannot.

161

162 Q29.7

Magnetic Fields

a

f

Since FB = q v × B , then the acceleration produced by a magnetic field on a particle of mass m is q aB = v × B . For the acceleration to change the speed, a component of the acceleration must be in m the direction of the velocity. The cross product tells us that the acceleration must be perpendicular to the velocity, and thus can only change the direction of the velocity.

a

f

Q29.8

The magnetic field in a cyclotron essentially keeps the charged particle in the electric field for a longer period of time, and thus experiencing a larger change in speed from the electric field, by forcing it in a spiral path. Without the magnetic field, the particle would have to move in a straight line through an electric field over a distance that is very large compared to the size of the cyclotron.

Q29.9

(a)

The qv × B force on each electron is down. Since electrons are negative, v × B must be up. With v to the right, B must be into the page, away from you.

(b)

Reversing the current in the coils would reverse the direction of B, making it toward you. Then v × B is in the direction right × toward you = down, and qv × B will make the electron beam curve up.

Q29.10

If the current is in a direction parallel or antiparallel to the magnetic field, then there is no force.

Q29.11

Yes. If the magnetic field is perpendicular to the plane of the loop, then it exerts no torque on the loop.

Q29.12

If you can hook a spring balance to the particle and measure the force on it in a known electric field, F then q = will tell you its charge. You cannot hook a spring balance to an electron. Measuring the E acceleration of small particles by observing their deflection in known electric and magnetic fields can tell you the charge-to-mass ratio, but not separately the charge or mass. Both an acceleration produced by an electric field and an acceleration caused by a magnetic field depend on the q properties of the particle only by being proportional to the ratio . m

Q29.13

If the current loop feels a torque, it must be caused by a magnetic field. If the current loop feels no torque, try a different orientation—the torque is zero if the field is along the axis of the loop.

Q29.14

The Earth’s magnetic field exerts force on a charged incoming cosmic ray, tending to make it spiral around a magnetic field line. If the particle energy is low enough, the spiral will be tight enough that the particle will first hit some matter as it follows a field line down into the atmosphere or to the surface at a high geographic latitude.

FIG. Q29.14 Q29.15

The net force is zero, but not the net torque.

Q29.16

Only a non-uniform field can exert a non-zero force on a magnetic dipole. If the dipole is aligned with the field, the direction of the resultant force is in the direction of increasing field strength.

Chapter 29

163

Q29.17

The proton will veer upward when it enters the field and move in a counter-clockwise semicircular arc. An electron would turn downward and move in a clockwise semicircular arc of smaller radius than that of the proton, due to its smaller mass.

Q29.18

Particles of higher speeds will travel in semicircular paths of proportionately larger radius. They will take just the same time to travel farther with their higher speeds. As shown in Equation 29.15, the time it takes to follow the path is independent of particle’s speed.

Q29.19

The spiral tracks are left by charged particles gradually losing kinetic energy. A straight path might be left by an uncharged particle that managed to leave a trail of bubbles, or it might be the imperceptibly curving track of a very fast charged particle.

Q29.20

No. Changing the velocity of a particle requires an accelerating force. The magnetic force is proportional to the speed of the particle. If the particle is not moving, there can be no magnetic force on it.

Q29.21

Increase the current in the probe. If the material is a semiconductor, raising its temperature may increase the density of mobile charge carriers in it.

SOLUTIONS TO PROBLEMS Section 29.1 P29.1

Magnetic Fields and Forces

(a)

up

(b)

out of the page, since the charge is negative.

(c)

no deflection

(d)

into the page

FIG. P29.1 P29.2

At the equator, the Earth’s magnetic field is horizontally north. Because an electron has negative charge, F = qv × B is opposite in direction to v × B . Figures are drawn looking down. (a)

Down × North = East, so the force is directed West .

(b)

North × North = sin 0° = 0 : Zero deflection .

(c)

West × North = Down, so the force is directed Up .

(d)

Southeast × North = Up, so the force is Down .

(a)

(c) FIG. P29.2

(d)

164 P29.3

Magnetic Fields

e j Therefore, B = B e− k j which indicates the FB = qv × B ; FB − j = − e v i × B

negative z direction . FIG. P29.3

P29.4

e

je

je

j

FB = qvB sin θ = 1.60 × 10 −19 C 3.00 × 10 6 m s 3.00 × 10 −1 T sin 37.0°

(a)

FB = 8.67 × 10 −14 N a=

(b)

P29.5

F 8.67 × 10 −14 N = = 5.19 × 10 13 m s 2 m 1.67 × 10 −27 kg

e

je

j

F = ma = 1.67 × 10 −27 kg 2.00 × 10 13 m s 2 = 3.34 × 10 −14 N = qvB sin 90° B=

F 3.34 × 10 −14 N = = 2.09 × 10 −2 T −19 7 qv 1.60 × 10 C 1.00 × 10 m s

e

je

j

The right-hand rule shows that B must be in the −y direction to yield a force in the +x direction when v is in the z direction. P29.6

First find the speed of the electron. ∆K =

P29.7

P29.8

1 mv 2 = e∆V = ∆U : 2

v=

e

2 e∆ V = m

e

jb

2 1.60 × 10 −19 C 2 400 J C

e9.11 × 10 ja

je

−31

kg

j

FIG. P29.5

g = 2.90 × 10

7

ms

f

(a)

FB, max = qvB = 1.60 × 10 −19 C 2.90 × 10 7 m s 1.70 T = 7.90 × 10 −12 N

(b)

FB, min = 0 occurs when v is either parallel to or anti-parallel to B.

e

je

ja

f

FB = qvB sin θ

so

8.20 × 10 −13 N = 1.60 × 10 −19 C 4.00 × 10 6 m s 1.70 T sin θ

sin θ = 0.754

and

θ = sin −1 0.754 = 48.9° or 131° .

Gravitational force: Electric force: Magnetic force:

a

f

e

je j F = qE = e −1.60 × 10 C jb100 N C downg = 1.60 × 10 N up .  . F = qv × B = e −1.60 × 10 C je6.00 × 10 m s E j × e50.0 × 10 N ⋅ s C ⋅ m N j Fg = mg = 9.11 × 10 −31 kg 9.80 m s 2 = 8.93 × 10 −30 N down . e

B

−19

−17

−19

FB = −4.80 × 10 −17 N up = 4.80 × 10 −17 N down .

6

−6

Chapter 29

P29.9

FB = qv × B i

j

k

a

f a f a f

v × B = +2 −4 +1 = 12 − 2 i + 1 + 6 j + 4 + 4 k = 10 i + 7 j + 8k +1 +2 −3 v × B = 10 2 + 7 2 + 8 2 = 14.6 T ⋅ m s

jb

e

g

FB = q v × B = 1.60 × 10 −19 C 14.6 T ⋅ m s = 2.34 × 10 −18 N P29.10

jb

e

g e

j

qE = −1.60 × 10 −19 C 20.0 N C k = −3.20 × 10 −18 N k

∑ F = qE + qv × B = ma

e−3.20 × 10 −e3.20 × 10 e1.92 × 10

j e j e N jk − e1.92 × 10 C ⋅ m sji × B = e1.82 × 10 N jk C ⋅ m sji × B = −e5.02 × 10 N jk

−18

−18

−15

je

j

N k − 1.60 × 10 −19 C 1.20 × 10 4 m s i × B = 9.11 × 10 −31 2.00 × 10 12 m s 2 k −15

−18

−18

The magnetic field may have any x -component . Bz = 0 and By = −2.62 mT .

Section 29.2 P29.11

Magnetic Force Acting on a Current-Carrying Conductor

FB = ILB sin θ

with FB = Fg = mg

mg = ILB sin θ so

m g = IB sin θ L

I = 2.00 A

100 cm m m = 0.500 g cm = 5.00 × 10 −2 kg m . L 1 000 g kg

b

and

e5.00 × 10

Thus

−2

I gFGH JK ja9.80f = a2.00fB sin 90.0°

B = 0.245 Tesla with the direction given by right-hand rule: eastward .

a

fa

f a

f

fa

fa

e−2.88jj N

P29.12

FB = IA × B = 2.40 A 0.750 m i × 1.60 T k =

P29.13

(a)

FB = ILB sin θ = 5.00 A 2.80 m 0.390 T sin 60.0° = 4.73 N

(b)

FB = 5.00 A 2.80 m 0.390 T sin 90.0° = 5.46 N

(c)

FB = 5.00 A 2.80 m 0.390 T sin 120° = 4.73 N

a

a

fa

fa

f

a

fa

fa

f

f

FIG. P29.11

165

166 P29.14

Magnetic Fields

FB A I=

=

mg I A × B = A A

b

F

ge

j

0.040 0 kg m 9.80 m s 2 mg = = 0.109 A BA 3.60 T Bin

The direction of I in the bar is to the right .

FIG. P29.14 P29.15

a

f e j e j ej + K g + ∆E = b K The work-energy theorem is bK trasn

0 + 0 + Fs cos θ = IdBL cos 0° = v=

P29.16

trans

+ K rot

g

d f

I

1 1 mv 2 + Iω 2 2 2

FG H

IJ FG v IJ and IdBL = 3 mv KH RK 4 4a 48.0 A fa0.120 mfa0.240 Tfa0.450 mf = 3b0.720 kg g 2

1 1 1 mv 2 + mR 2 2 2 2

4IdBL = 3m

rot i

L y

2

x 1.07 m s .

f e j e j ej + K g + ∆E = bK The work-energy theorem is bK

z FIG. P29.15

a

The rod feels force FB = I d × B = Id k × B − j = IdB i . trans

0 + 0 + Fs cos θ = IdBL cos 0° = P29.17

B

The rod feels force FB = I d × B = Id k × B − j = IdB i .

rot i

trans

+ K rot

g

f

1 1 mv 2 + Iω 2 2 2

FG H

1 1 1 mv 2 + mR 2 2 2 2

IJ FG v IJ KH RK

2

and v =

4IdBL . 3m

The magnetic force on each bit of ring is Ids × B = IdsB radially inward and upward, at angle θ above the radial line. The radially inward components tend to squeeze the ring but all cancel out as forces. The upward components IdsB sin θ all add to I 2π rB sin θ up .

FIG. P29.17

Chapter 29

P29.18

For each segment, I = 5.00 A and B = 0.020 0 N A ⋅ m j . Segment

P29.19

a f

FB = I A × B

A

ab

−0.400 m j

bc

0.400 m k

cd

−0.400 m i + 0.400 m j

da

0.400 m i − 0.400 m k

0

a40.0 mNfe− ij a40.0 mNfe−k j a40.0 mNfek + ij

FIG. P29.18

Take the x-axis east, the y-axis up, and the z-axis south. The field is

b

g

e j b

g

e j

B = 52.0 µT cos 60.0° − k + 52.0 µT sin 60.0° − j .

e j

ej

The current then has equivalent length: L ′ = 1.40 m − k + 0.850 m j

b

ge

j e

j

FB = IL ′ × B = 0.035 0 A 0.850 j − 1.40k m × −45.0 j − 26.0 k 10 −6 T

e

j

e j

FB = 3.50 × 10 −8 N −22.1i − 63.0 i = 2.98 × 10 −6 N − i = 2.98 µN west

Section 29.3 P29.20

(a)

Torque on a Current Loop in a Uniform Magnetic Field 2π r = 2.00 m so

r = 0.318 m

j a

f

ja

f

e

µ = IA = 17.0 × 10 −3 A π 0.318 (b)

m 2 = 5.41 mA ⋅ m 2

τ = µ ×B so

P29.21

2

e

τ = 5.41 × 10 −3 A ⋅ m 2 0.800 T = 4.33 mN ⋅ m

a

f

τ = µB sin θ so 4.60 × 10 −3 N ⋅ m = µ 0.250 sin 90.0°

µ = 1.84 × 10 −2 A ⋅ m 2 = 18.4 mA ⋅ m 2

FIG. P29.19 .

167

168 P29.22

Magnetic Fields

Let θ represent the unknown angle; L, the total length of the wire; and d, the length of one side of the square coil. Then, using the definition of magnetic moment and the right-hand rule in Figure 29.15, we find

(a)

FG L IJ d I at angle θ with the horizontal. H 4d K ∑ τ = bµ × Bg − br × mgg = 0 FG ILBd IJ sina90.0°−θ f − FG mgd IJ sinθ = 0 H 4 K H 2K FG mgd IJ sinθ = FG ILBd IJ cos θ H 4 K H 2K F ILB I = tan FG a3.40 Afa4.00 mfb0.010 0 Tg IJ = θ = tan G GH 2b0.100 kg ge9.80 m s j JK H 2mg JK

µ = NAI :

2

µ=

At equilibrium,

and

−1

−1

2

τm =

(b) P29.23

FG ILBd IJ cosθ = 1 a3.40 Afa4.00 mfb0.010 0 Tga0.100 mf cos 3.97° = H 4 K 4

τ = NBAI sin φ

a

ja

fe

3.39 mN ⋅ m

f

τ = 100 0.800 T 0.400 × 0.300 m 2 1.20 A sin 60° τ = 9.98 N ⋅ m Note that φ is the angle between the magnetic moment and the B field. The loop will rotate so as to align the magnetic moment with the B field. Looking down along the y-axis, the loop will rotate in a clockwise direction. P29.24

FIG. P29.23

From τ = µ × B = IA × B , the magnitude of the torque is IAB sin 90.0°. (a)

Each side of the triangle is

40.0 cm . 3

Its altitude is 13.3 2 − 6.67 2 cm = 11.5 cm and its area is 1 A = 11.5 cm 13.3 cm = 7.70 × 10 −3 m 2 . 2 Then τ = 20.0 A 7.70 × 10 −3 m 2 0.520 N ⋅ s C ⋅ m = 80.1 mN ⋅ m .

a

fa

a

(b)

f

fe

jb

Each side of the square is 10.0 cm and its area is 100 cm 2 = 10 −2 m 2 .

a

fe

ja

f

τ = 20.0 A 10 −2 m 2 0.520 T = 0.104 N ⋅ m (c)

0.400 m = 0.063 7 m 2π A = π r 2 = 1.27 × 10 −2 m 2

r=

a

fe

ja

f

τ = 20.0 A 1.27 × 10 −2 m 2 0.520 = 0.132 N ⋅ m (d)

g

The circular loop experiences the largest torque.

3.97° .

Chapter 29

P29.25

Choose U = 0 when the dipole moment is at θ = 90.0° to the field. The field exerts torque of magnitude µB sin θ on the dipole, tending to turn the dipole moment in the direction of decreasing θ. According to Equations 8.16 and 10.22, the potential energy of the dipole-field system is given by U −0 =

z

θ

a

µB sin θ dθ = µB − cos θ

f

90 .0 °

P29.26

169

(a)

θ 90 .0 °

= − µB cos θ + 0

U = −µ ⋅ B .

or

The field exerts torque on the needle tending to align it with the field, so the minimum energy orientation of the needle is: pointing north at 48.0° below the horizontal

e

je

j

where its energy is U min = − µB cos 0° = − 9.70 × 10 −3 A ⋅ m 2 55.0 × 10 −6 T = −5.34 × 10 −7 J . It has maximum energy when pointing in the opposite direction, south at 48.0° above the horizontal

e

je

j

where its energy is U max = − µB cos 180° = + 9.70 × 10 −3 A ⋅ m 2 55.0 × 10 −6 T = +5.34 × 10 −7 J .

P29.27

e

j

(b)

U min + W = U max : W = U max − U min = +5.34 × 10 −7 J − −5.34 × 10 −7 J = 1.07 µJ

(a)

τ = µ × B,

τ = µ × B = µB sin θ = NIAB sin θ

so

a

f b

τ max = NIAB sin 90.0° = 1 5.00 A π 0.050 0 m (b)

g e3.00 × 10 Tj = 2

−3

118 µN ⋅ m

U = − µ ⋅ B, so − µB ≤ U ≤ + µB

a f

a

f b

Since µB = NIA B = 1 5.00 A π 0.050 0 m

g e3.00 × 10 Tj = 118 µJ , 2

−3

the range of the potential energy is: −118 µJ ≤ U ≤ +118 µJ . *P29.28

(a)

τ = µ × B = NIAB sin θ

e

ja

fb

g F 2π rad IJ FG 1 min IJ = N ⋅ mb3 600 rev mingG H 1 rev K H 60 s K

τ max = 80 10 −2 A 0.025 m ⋅ 0.04 m 0.8 N A ⋅ m sin 90° = 6.40 × 10 −4 N ⋅ m (b)

Pmax = τ maxω = 6.40 × 10 −4

(c)

In one half revolution the work is

b g = 2 NIAB = 2e6.40 × 10 N ⋅ mj = 1.28 × 10 J In one full revolution, W = 2e1.28 × 10 Jj = 2.56 × 10 W = U max − U min = − µB cos 180°− − µB cos 0° = 2 µB −4

−3

−3

(d)

Pavg =

W 2.56 × 10 −3 J = = 0.154 W ∆t 1 60 s

b g

The peak power in (b) is greater by the factor

π . 2

−3

J .

0.241 W

170

Magnetic Fields

Section 29.4 P29.29

Motion of a Charged Particle in a Uniform Magnetic Field B = 50.0 × 10 −6 T; v = 6.20 × 10 6 m s

(a)

Direction is given by the right-hand-rule: southward FB = qvB sin θ

e

je

je

j

FB = 1.60 × 10 −19 C 6.20 × 10 6 m s 50.0 × 10 −6 T sin 90.0°

FIG. P29.29

= 4.96 × 10 −17 N

e

(b)

P29.30

je

1.67 × 10 −27 kg 6.20 × 10 6 m s mv 2 mv 2 = F= so r = r F 4.96 × 10 −17 N

a f

1 mv 2 = q ∆V 2

P29.31

2

= 1.29 km

ja

1 3.20 × 10 −26 kg v 2 = 1.60 × 10 −19 C 833 V 2

e

j

e

The magnetic force provides the centripetal force: qvB sin θ = r=

j

e e

je jb

f

v = 91.3 km s

mv 2 r

j g

3.20 × 10 −26 kg 9.13 × 10 4 m s mv = = 1.98 cm . qB sin 90.0° 1.60 × 10 −19 C 0.920 N ⋅ s C ⋅ m

For each electron, q vB sin 90.0° =

mv 2 eBr . and v = r m

The electrons have no internal structure to absorb energy, so the collision must be perfectly elastic: K=

1 1 1 mv12i + 0 = mv12 f + mv 22 f 2 2 2

K=

e 2 B 2 R 22 e 2 B 2 R12 e2B2 2 1 1 R1 + R 22 m + = m 2 2 2m m2 m2

K=

P29.32

F GH

e

I JK

F GH

jb

I JK

e

g b0.010 0 mg + b0.024 0 mg

e 1.60 × 10 −19 C 0.044 0 N ⋅ s C ⋅ m

e

2 9.11 × 10

We begin with qvB =

−31

kg

j

2

2

2

qRB mv 2 , so v = . R m

The time to complete one revolution is T = Solving for B, B =

j

2π m = 6.56 × 10 −2 T . qT

2π R 2π R 2π m = = . v qRB m qB

= 115 keV

Chapter 29

P29.33

a f

q ∆V =

1 mv 2 2

Also, qvB =

mv 2 r

or

v=

so

r=

a f

2 q ∆V . m

rp2 =

Therefore,

rα2

(a)

(b)

a f

2 m p ∆V eB 2 2 m d ∆V

p

2

d

p

α

2

α

or

qRB = mv .

But

L = mvR = qR 2 B .

Therefore,

R=

Thus,

v=

ja

e

e1.60 × 10

2 p

−19

je

C 1.00 × 10 −3 T

j

= 0.050 0 m = 5.00 cm .

L 4.00 × 10 −25 J ⋅ s = = 8.78 × 10 6 m s . −31 mR 9.11 × 10 kg 0.050 0 m

jb

e

g

f

P29.36

1 mv 2 = q ∆V 2

so

v=

so

r=

a f

2 q ∆V m

a f

m 2 q ∆V m qB

a f

r2 =

m 2 ∆V ⋅ q B2

and

ar ′ f

m=

qB 2 r 2 2 ∆V

and

am ′f = bq′2gBa∆Varf′f

a f

2

4.00 × 10 −25 J ⋅ s

L = qB

P29.35

a f

2 p

mv 2 R

qvB =

mv qB

2

p

2

−19 C 5.20 T qB 1.60 × 10 ω= = = 4.98 × 10 8 rad s m 1.67 × 10 −27 kg

r=

p

2

We begin with

a f

qB 2

rα = rd = 2 rp .

The conclusion is:

P29.34

a f.

2 m ∆V

a f = 2e2m ja∆V f = 2FG 2m a∆V f IJ = 2r q B eB H eB K F 2 m a ∆V f I = 2 r . 2m a ∆V f 2e 4m ja ∆V f = = = 2G q B a2efB H eB JK

rd2 =

and

a f

2 q ∆V = m

mv m = qB qB

2

=

m ′ 2 ∆V ⋅ q′ B2 2

2

a f FG IJ FG 2R IJ H KH R K 2

so

m ′ q′ r ′ 2e = ⋅ 2 = m q r e

2

= 8

171

172 P29.37

Magnetic Fields

E= and

1 mv 2 = e∆V 2 evB sin 90° = B=

mv m = eR eR

mv 2 R 2 e∆ V 1 = m R

1 B= 5.80 × 10 10 m *P29.38

(a)

e

2m∆V e

je

2 1.67 × 10 −27 kg 10.0 × 10 6 V 1.60 × 10

−19

C

j=

7.88 × 10 −12 T v

At the moment shown in Figure 29.21, the particle must be moving upward in order for the magnetic force on it to be

+

into the page, toward the center of this turn of its spiral path. Throughout its motion it circulates clockwise. (b)

B

FIG. P29.38(a)

After the particle has passed the middle of the bottle and moves into the region of increasing magnetic field, the magnetic force on it has a component to the left (as well as a radially inward component) as shown. This force in the –x direction slows and reverses the particle’s motion along the axis.

v B

F

FIG. P29.38(b) (c)

The magnetic force is perpendicular to the velocity and does no work on the particle. The particle keeps constant kinetic energy. As its axial velocity component decreases, its tangential velocity component increases.

(d)

The orbiting particle constitutes a loop of current in the yz q plane and therefore a magnetic dipole moment I A = A T in the –x direction. It is like a little bar magnet with its N pole on the left.

(e)

Problem 17 showed that a nonuniform magnetic field exerts a net force on a magnetic dipole. When the dipole is aligned opposite to the external field, the force pushes it out of the region of stronger field. Here it is to the left, a force of repulsion of one magnetic south pole on another south pole.

+

N

FIG. P29.38(d) B N

S

S

FIG. P29.38(e)

S

Chapter 29

P29.39

e

ja

je

7.94 × 10 −3 m 1.60 × 10 −19 C 1.80 T rqB = m= v 4.60 × 10 5 m s

mv so r= qB

m = 4.97 × 10 −27 kg

F 1u GH 1.66 × 10

−27

I= J kg K

The particle is singly ionized: either a tritium ion,

Section 29.5 P29.40

2.99 u

+ 3 1H

, or a helium ion,

+ 3 2 He

.

Applications Involving Charged Particles Moving in a Magnetic Field

FB = Fe so

qvB = qE

where

v=

2K and K is kinetic energy of the electron. m

E = vB =

P29.41

f

K=

2K B= m

a f

so

mv 2 r

r=

1 mv 2 = q ∆V 2

FB = qv × B =

9.11 × 10

j

(b)

r235 = 8. 23 cm

1.60 × 10

m 238 = m 235

−19

j b0.015 0g =

244 kV m

a f

a f

2 q ∆V m 1 = B B

mv m = qB q

e

r238 =

−31

2 q ∆V m

v=

2 238 × 1.66 × 10 −27 2 000

(a)

r238 = r235

a fe

2 750 1.60 × 10 −19

a f

2m ∆V q

FG 1 IJ = 8.28 × 10 H 1.20 K

−2

m = 8.28 cm

238.05 = 1.006 4 235.04

The ratios of the orbit radius for different ions are independent of ∆V and B. P29.42

E 2 500 V m = = 7.14 × 10 4 m s . B 0.035 0 T

In the velocity selector:

v=

In the deflection chamber:

2.18 × 10 −26 kg 7.14 × 10 4 m s mv r= = = 0.278 m . qB 1.60 × 10 −19 C 0.035 0 T

e

e

je

jb

g

j

173

174 P29.43

Magnetic Fields

(a)

FB = qvB =

mv 2 R

ja

e

f

−19 C 0.450 T v qBR qB 1.60 × 10 ω= = = = = 4.31 × 10 7 rad s −27 R mR m kg 1.67 × 10

(b)

P29.44

K=

v=

ja

e

fa

f

−19 C 0.450 T 1.20 m qBR 1.60 × 10 = = 5.17 × 10 7 m s m 1.67 × 10 −27 kg

1 mv 2 : 2

j 12 e1.67 × 10 kg jv kg je8.07 × 10 m sj mv e1.67 × 10 r= = = 0.162 m qB e1.60 × 10 Cja5.20 Tf

e34.0 × 10

je

eV 1.60 × 10 −19 J eV =

6

−27

7

v = 8.07 × 10 m s

*P29.45

−27

2

7

−19

Note that the “cyclotron frequency” is an angular speed. The motion of the proton is described by

∑ F = ma : q vB sin 90° = qB=m

(a)

(b)

(c) (d)

mv 2 r

v = mω r

e1.60 × 10 Cjb0.8 N ⋅ s C ⋅ mg FG kg ⋅ m IJ = 7.66 × 10 rad s ω= = H N ⋅s K m e1.67 × 10 kgj F 1 IJ = 2.68 × 10 m s v = ω r = e7.66 × 10 rad sja0.350 mfG H 1 rad K F 1 eV IJ = 3.76 × 10 1 1 K = mv = e1.67 × 10 kg je 2.68 × 10 m sj G 2 2 H 1.6 × 10 J K −19

qB

7

7

−27

2

7

2

−27

7

2

−19

a

P29.46

θ =ωt

FB = qvB = B=

eV

The proton gains 600 eV twice during each revolution, so the number of revolutions is 3.76 × 10 6 eV = 3.13 × 10 3 revolutions . 2 600 eV

(e)

6

t=

FG H

mv 2 r

jb

IJ K

θ 3.13 × 10 3 rev 2π rad = 2.57 × 10 −4 s = ω 7.66 × 10 7 rad s 1 rev

4.80 × 10 −16 kg ⋅ m s mv = = 3.00 T qr 1.60 × 10 −19 C 1 000 m

e

f

g

Chapter 29

P29.47

θ = tan −1

FG 25.0 IJ = 68.2° H 10.0 K

R=

and

1.00 cm = 1.08 cm . sin 68.2°

Ignoring relativistic correction, the kinetic energy of the electrons is 1 mv 2 = q∆V 2

From Newton’s second law

e

2 q ∆V = 1.33 × 10 8 m s . m

v=

so

mv 2 = qvB , we find the magnetic field R

je je

FIG. P29.47

j

9.11 × 10 −31 kg 1.33 × 10 8 m s mv B= = = 70.1 mT . qR 1.60 × 10 −19 C 1.08 × 10 −2 m

Section 29.6 P29.48

e

The Hall Effect 1 nq

(a)

RH ≡

(b)

∆VH = B=

P29.49

j

so

n=

1 1 = = 7.44 × 10 28 m −3 − 19 qRH C 0.840 × 10 −10 m3 C 1.60 × 10

e

je

j

IB nqt

b

nqt ∆VH I

g = e7.44 × 10

28

je

je

je

m −3 1.60 × 10 −19 C 0.200 × 10 −3 m 15.0 × 10 −6 V 20.0 A

j=

1.79 T

IB , and given that I = 50.0 A , B = 1.30 T , and t = 0.330 mm, the number of charge nqt carriers per unit volume is

Since ∆VH =

n=

IB = 1.28 × 10 29 m −3 e ∆VH t

b

g

The number density of atoms we compute from the density: n0 =

F GH

8.92 g 1 mole cm 3 63.5 g

I F 6.02 × 10 atoms I F 10 cm I = 8.46 × 10 JK GH mole JK GH 1 m JK 23

3

6

3

So the number of conduction electrons per atom is n 1.28 × 10 29 = = 1.52 n 0 8.46 × 10 28

28

atom m3

175

176 P29.50

Magnetic Fields

∆VH =

(a)

IB nqt

nqt 0.080 0 T B = = = 1.14 × 10 5 T V . ∆VH 0.700 × 10 −6 V I

so

B=

Then, the unknown field is

e

je

FG nqt IJ b∆V g HIK H

j

B = 1.14 × 10 5 T V 0.330 × 10 −6 V = 0.037 7 T = 37.7 mT . nqt = 1.14 × 10 5 T V I

(b)

e

n = 1.14 × 10 5 T V

P29.51

B=

b

nqt ∆VH I

g = e8.49 × 10

e

n = 1.14 × 10 5 T V

so

j e1.60 × 10

0.120 A −19

je

C 2.00 × 10

je

−3

j

m

j qtI

= 4.29 × 10 25 m −3 .

je

je

m −3 1.60 × 10 −19 C 5.00 × 10 −3 m 5.10 × 10 −12 V

28

j

8.00 A

B = 4.33 × 10 −5 T = 43.3 µT

Additional Problems P29.52

(a)

The boundary between a region of strong magnetic field and a region of zero field cannot be perfectly sharp, but we ignore the thickness of the transition zone. In the field the electron moves on an arc of a circle:

∑ F = ma : q vB sin 90° =

mv 2 r

e

je

j

−19 C 10 −3 N ⋅ s C ⋅ m q B 1.60 × 10 v =ω = = = 1.76 × 10 8 rad s r m 9.11 × 10 −31 kg

e

j

FIG. P29.52(a)

The time for one half revolution is, from

∆ θ = ω∆ t ∆t =

(b)

∆θ

ω

=

π rad = 1.79 × 10 −8 s . 1.76 × 10 8 rad s

The maximum depth of penetration is the radius of the path. Then

ja

e

f

v = ω r = 1.76 × 10 8 s −1 0.02 m = 3.51 × 10 6 m s

and K=

1 1 mv 2 = 9.11 × 10 −31 kg 3.51 × 10 6 m s 2 2

e

= 35.1 eV .

je

j

2

= 5.62 × 10 −18 J =

5.62 × 10 −18 J ⋅ e 1.60 × 10 −19 C

Chapter 29

P29.53

(a)

Define vector h to have the downward direction of the current, and vector L to be along the pipe into the page as shown. The electric current experiences a magnetic force .

a

f

I h × B in the direction of L. (b)

The sodium, consisting of ions and electrons, flows along the pipe transporting no net charge. But inside the section of length L, electrons drift upward to constitute downward electric current J × area = J Lw .

FIG. P29.53

a f

The current then feels a magnetic force I h × B = JLwhB sin 90° . This force along the pipe axis will make the fluid move, exerting pressure F JLwhB = = JLB . area hw P29.54

∑ Fy = 0 :

+n − mg = 0

∑ Fx = 0 :

− µ k n + IB sin 90.0° = 0 B=

P29.55

b a

ge

j

2 µ k mg 0.100 0.200 kg 9.80 m s = = 39.2 mT 10.0 A 0.500 m Id

fa

f

The magnetic force on each proton, FB = qv × B = qvB sin 90° downward perpendicular to velocity, causes centripetal acceleration, guiding it into a circular path of radius r, with mv 2 r mv . r= qB qvB =

and

We compute this radius by first finding the proton’s speed:

Now,

(b)

K=

1 mv 2 2

v=

2K = m

FIG. P29.55

e

je

2 5.00 × 10 6 eV 1.60 × 10 −19 J eV 1.67 × 10

e

−27

je

kg

j

j = 3.10 × 10

7

m s.

1.67 × 10 −27 kg 3.10 × 10 7 m s mv = = 6.46 m . r= qB 1.60 × 10 −19 C 0.050 0 N ⋅ s C ⋅ m

e

jb

g

From the figure, observe that 1.00 m 1m = r 6.46 m α = 8.90°

sin α =

(a)

The magnitude of the proton momentum stays constant, and its final y component is

e

je

j

− 1.67 × 10 −27 kg 3.10 × 10 7 m s sin 8.90° = −8.00 × 10 −21 kg ⋅ m s .

177

178 P29.56

Magnetic Fields

e j e

j

If B = Bx i + By j + Bz k , FB = qv × B = e vi i × Bx i + By j + Bz k = 0 + evi B y k − evi Bz j .

(a)

Since the force actually experienced is FB = Fi j , observe that B x could have any value , B y = 0 , and B z = −

Fi . evi

IJ e j FGH K F F I = qv × B = − ee v i j × G B i + 0 j − kJ = ev K H

(b)

If v = −vi i , then

F FB = qv × B = e − vi i × Bx i + 0 j − i k = − Fi j . evi

(c)

If q = − e and v = vi i , then

FB

i

x

i

i

− Fi j .

Reversing either the velocity or the sign of the charge reverses the force. P29.57

(a)

The net force is the Lorentz force given by

a f j e4i − 1j − 2k j + e2 i + 3 j − 1k j × e2 i + 4j + 1k j N

F = qE + qv × B = q E + v × B

e

F = 3.20 × 10 −19

Carrying out the indicated operations, we find:

e3.52i − 1.60jj × 10 N . 3.52 FG F IJ = cos FG H FK GH a3.52f + a1.60f −18

F=

θ = cos

(b)

P29.58

−1

−1

x

2

2

I JJ = K

24.4°

A key to solving this problem is that reducing the normal force will reduce F the friction force: FB = BIL or B = B . IL When the wire is just able to move,

∑ Fy = n + FB cos θ − mg = 0

so

n = mg − FB cos θ

and

f = µ mg − FB cos θ .

Also,

∑ Fx = FB sin θ − f = 0

so FB sin θ = f :

FB sin θ = µ mg − FB cos θ and FB =

We minimize B by minimizing FB :

dFB cos θ − µ sin θ = µmg = 0 ⇒ µ sin θ = cos θ . 2 dθ sin θ + µ cos θ

b

b

b gb

g

FIG. P29.58

g

g

FG 1 IJ = tan a5.00f = 78.7° for the smallest field, and H µK F F µg I bm Lg =G J B= IL H I K sin θ + µ cos θ L a0.200fe9.80 m s j OP 0.100 kg m =M B MN 1.50 A PQ sin 78.7°+a0.200f cos 78.7° = 0.128 T

Thus, θ = tan −1

−1

B

2

min

Bmin = 0.128 T pointing north at an angle of 78.7° below the horizontal

µmg . sin θ + µ cos θ

Chapter 29

*P29.59

179

The electrons are all fired from the electron gun with the same speed v in Ui = K f

qV =

a− efa−∆V f = 12 m v

1 mv 2 2

e

2

v=

2 e∆V me

For φ small, cos φ is nearly equal to 1. The time T of passage of each electron in the chamber is given by d = vT

T=d

FG m IJ H 2 e∆ V K e

12

Each electron moves in a different helix, around a different axis. If each completes just one revolution within the chamber, it will be in the right place to pass through the exit port. Its transverse velocity component v ⊥ = v sin φ swings around according to F⊥ = ma ⊥ qv ⊥ B sin 90° = Then *P29.60

mv ⊥2 r

FG IJ H K

2π m e B e

eB =

12

=

me v⊥ 2π = m eω = m e r T

d

a2 ∆V f

12

B=

FG H

2π 2m e ∆V d e

T=

IJ K

FG H

m e 2π me =d 2 e∆ V eB

IJ K

12

12

.

Let vi represent the original speed of the alpha particle. Let vα and v p represent the particles’ speeds after the collision. We have conservation of momentum 4m p vi = 4m p vα + m p v p and the relative velocity equation vi − 0 = v p − vα . Eliminating vi , 4 v p − 4 v α = 4 vα + v p

3 v p = 8 vα

vα =

3 vp . 8

For the proton’s motion in the magnetic field,

∑ F = ma

ev p B sin 90° =

m p v p2

eBR = vp . mp

R

For the alpha particle, 2 evα B sin 90° = P29.61

4m p vα2

rα =

rα

2 m p vα

rα =

eB

2m p 3 2m p 3 eBR 3 vp = R . = eB 8 4 eB 8 m p

Let ∆x 1 be the elongation due to the weight of the wire and let ∆x 2 be the additional elongation of the springs when the magnetic field is turned on. Then Fmagnetic = 2 k∆x 2 where k is the force constant of mg . (The factor 2 is 2 ∆x 1 included in the two previous equations since there are 2 springs in parallel.) Combining these two equations, we find

the spring and can be determined from k =

Fmagnetic = 2

FG mg IJ ∆x H 2 ∆x K 1

2

=

mg∆x 2 ; but FB = I L × B = ILB . ∆x 1

a

fa fe a fb ge

FIG. P29.61

j

0.100 9.80 3.00 × 10 −3 mg∆x 2 24.0 V Therefore, where I = = 2.00 A , B = = = 0.588 T . IL∆x1 12.0 Ω 2.00 0.050 0 5.00 × 10 −3

j

180 P29.62

Magnetic Fields

Suppose the input power is

a

f

I ~ 1 A = 10 0 A .

120 W = 120 V I :

FG 1 min IJ FG 2π rad IJ ~ 200 rad s H 60 s K H 1 rev K 20 W = τω = τ b 200 rad sg τ ~ 10 N ⋅ m . A ~ 10 m . a3 cmf × a4 cmf, or

ω = 2 000 rev min

Suppose

−1

and the output power is

−3

Suppose the area is about

2

B ~ 10 −1 T .

Suppose that the field is

Then, the number of turns in the coil may be found from τ ≅ NIAB :

b

ge

je

j

0.1 N ⋅ m ~ N 1 C s 10 −3 m 2 10 −1 N ⋅ s C ⋅ m

N ~ 10

giving *P29.63

3

.

The sphere is in translational equilibrium, thus fs − Mg sin θ = 0 .

G µ

(1)

θ

G B

The sphere is in rotational equilibrium. If torques are taken about the center of the sphere, the magnetic field produces a clockwise torque of magnitude µB sin θ , and the frictional force a counterclockwise torque of magnitude fs R , where R is the radius of the sphere. Thus: fs R − µB sin θ = 0 .

(2)

From (1): fs = Mg sin θ . Substituting this in (2) and canceling out sin θ , one obtains

µB = MgR .

b

ge

j f

(3)

fs I

θ

Mg FIG. P29.63

0.08 kg 9.80 m s 2 Mg = = 0.713 A . The current must be π NBR π 5 0.350 T 0.2 m counterclockwise as seen from above.

Now µ = NIπ R 2 . Thus (3) gives I =

P29.64

fa

Call the length of the rod L and the tension in each wire alone

∑ Fx = T sin θ − ILB sin 90.0° = 0 ∑ Fy = T cos θ − mg = 0 , tan θ =

P29.65

a fa

ILB IB = mg mL g

b g

∑ F = ma or qvB sin 90.0° =

or

T sin θ = ILB

or

T cos θ = mg

or

B=

mv 2 r

∴ the angular frequency for each ion is ∆f = f12 − f14 =

FG H

IJ e K e

bm Lgg tanθ = I

qB v =ω = = 2π f and r m

ja

λg tan θ I

fF 1 − 1 I G J j H 12.0 u 14.0 u K

1.60 × 10 −19 C 2.40 T qB 1 1 = − 2π m12 m14 2π 1.66 × 10 −27 kg u

∆f = f12 − f14 = 4.38 × 10 5 s −1 = 438 kHz

T . Then, at equilibrium: 2

Chapter 29

P29.66

Let v x and v ⊥ be the components of the velocity of the positron parallel to and perpendicular to the direction of the magnetic field. (a)

The pitch of trajectory is the distance moved along x by the positron during each period, T (see Equation 29.15)

fFGH 2πBqm IJK e5.00 × 10 jacos 85.0°fa2π fe9.11 × 10 j = p= 0.150e1.60 × 10 j a

p = v x T = v cos 85.0°

−19

(b)

FIG. P29.66

−31

6

From Equation 29.13,

1.04 × 10 −4 m

r=

mv ⊥ mv sin 85.0° = Bq Bq

r=

e9.11 × 10 je5.00 × 10 jasin 85.0°f = a0.150fe1.60 × 10 j −31

P29.67

6

−19

τ = IAB where the effective current due to the orbiting electrons is

I=

∆q q = ∆t T

and the period of the motion is

T=

2π R . v

The electron’s speed in its orbit is found by requiring

keq2 R

2

=

mv 2 or R

Substituting this expression for v into the equation for T, we find

e9.11 × 10 je5.29 × 10 j = 1.52 × 10 s . T = 2π e1.60 × 10 j e8.99 × 10 j F q I 1.60 × 10 π e5.29 × 10 j a0.400f = Therefore, τ = G J AB = H T K 1.52 × 10

mR 3 q2 ke

9

−19

−11 2

−16

Use the equation for cyclotron frequency ω =

e1.60 × 10 Cje5.00 × 10 Tj = a2π fe5.00 rev 1.50 × 10 sj −19

T = 2π

ke . mR

−16

−19 2

m=

v=q

−11 3

−31

P29.68

1.89 × 10 −4 m

qB qB qB = or m = ω 2π f m

−2

−3

3.70 × 10 −24 N ⋅ m .

3.82 × 10 −25 kg .

181

182 P29.69

Magnetic Fields

(a)

K=

1 mv 2 = 6.00 MeV = 6.00 × 10 6 eV 1.60 × 10 −19 J eV 2

e

je

j

K = 9.60 × 10 −13 J

e

2 9.60 × 10

v=

1.67 × 10

FB = qvB = R=

mv = qB

−13

−27

J

j = 3.39 × 10

kg

θ' x x x x x B in = 1.00 T x x x x x 45° x x x x x 45° x x x x x R x x x x x x x x x x 45.0° x x x x x

x 7

ms v

2

mv so R 1.67 × 10 −27 kg 3.39 × 10 7 m s

e

e1.60 × 10

je

−19

ja

C 1.00 T

f

FIG. P29.69

j = 0.354 m

a

f

Then, from the diagram, x = 2 R sin 45.0° = 2 0.354 m sin 45.0° = 0.501 m

P29.70

(b)

From the diagram, observe that θ ′ = 45.0° .

(a)

See graph to the right. The Hall voltage is directly proportional to the magnetic field. A least-square fit to the data gives the equation of the best fitting line as:

120 100 80

∆V H (µV)

60 40 20

e

∆VH = 1.00 × 10

−4

j

V TB .

0 0

0.2

0.4

0.6

0.8

1.0

1.2

B (T)

(b)

Comparing the equation of the line which fits the data best to ∆VH =

FIG. P29.70

F 1 IB GH nqt JK

observe that:

I I = 1.00 × 10 −4 V T, or t = . nqt nq 1.00 × 10 −4 V T

e

j

Then, if I = 0.200 A , q = 1.60 × 10 −19 C , and n = 1.00 × 10 26 m −3 , the thickness of the sample is t=

e1.00 × 10

0.200 A

26

m

−3

je1.60 × 10

−19

je

C 1.00 × 10 −4 V T

j

= 1.25 × 10 −4 m = 0.125 mm .

Chapter 29

P29.71

(a)

183

The magnetic force acting on ions in the blood stream will deflect positive charges toward point A and negative charges toward point B. This separation of charges produces an electric field directed from A toward B. At equilibrium, the electric force caused by this field must balance the magnetic force, so

FG ∆V IJ HdK e160 × 10 Vj ∆V = v= Bd b0.040 0 Tge3.00 × 10 qvB = qE = q

FIG. P29.71

−6

or

−3

j

m

= 1.33 m s .

No . Negative ions moving in the direction of v would be deflected toward point B, giving

(b)

A a higher potential than B. Positive ions moving in the direction of v would be deflected toward A, again giving A a higher potential than B. Therefore, the sign of the potential difference does not depend on whether the ions in the blood are positively or negatively charged. P29.72

When in the field, the particles follow a circular path mv 2 mv , so the radius of the path is: r = according to qvB = r qB (a)

qBh mv , that is, when v = , the qB m particle will cross the band of field. It will move in a full semicircle of radius h, leaving the field at 2 h, 0 , 0 with velocity v = − vj .

When r = h =

b

(b)

g

f

FIG. P29.72

qBh mv , the particle will move in a smaller semicircle of radius r = < h . It will m qB leave the field at 2r, 0 , 0 with velocity v f = − vj .

When v
h , centered at qB m h r, 0 , 0 . The arc subtends an angle given by θ = sin −1 . It will leave the field at the point r with coordinates r 1 − cos θ , h, 0 with velocity v = v sin θ i + v cos θ j .

When v >

b

FG IJ HK

g

a

f

f

ANSWERS TO EVEN PROBLEMS P29.2

(a) west; (b) no deflection; (c) up; (d) down

P29.4

(a) 86.7 fN ; (b) 51.9 Tm s 2

P29.6

(a) 7.90 pN ; (b) 0

P29.8

Gravitational force: 8.93 × 10 −30 N down; Electric force: 16.0 aN up ; Magnetic force: 48.0 aN down

P29.10

By = −2.62 mT ; Bz = 0; Bx may have any value

184

Magnetic Fields

P29.12

e−2.88jj N

P29.14

109 mA to the right

P29.16

FG 4IdBL IJ H 3m K

P29.18

12

e j = 40.0 mN e − k j ; F = a 40.0 mN fe i + k j

P29.50

(a) 37.7 mT ; (b) 4.29 × 10 25 m3

P29.52

(a) 17.9 ns; (b) 35.1 eV

P29.54

39.2 mT

P29.56

(a) Bx is indeterminate. By = 0 ; Bz =

Fab = 0; Fbc = 40.0 mN − i ; Fcd

da

(b) −Fi j ; (c) −Fi j P29.58

128 mT north at an angle of 78.7° below the horizontal

P29.20

(a) 5.41 mA ⋅ m 2 ; (b) 4.33 mN ⋅ m

P29.22

(a)3.97° ; (b) 3.39 mN ⋅ m

P29.24

(a) 80.1 mN ⋅ m ; (b) 104 mN ⋅ m ; (c) 132 mN ⋅ m ; (d) The torque on the circle.

P29.62

P29.26

(a) minimum: pointing north at 48.0° below the horizontal; maximum: pointing south at 48.0° above the horizontal; (b) 1.07 µJ

P29.64

λ g tan θ I

P29.66

(a) 0.104 mm; (b) 0.189 mm

P29.68

3.82 × 10 −25 kg

P29.70

(a) see the solution; empirically, ∆VH = 100 µ V T B ; (b) 0.125 mm

P29.28

(a) 640 µN ⋅ m ; (b) 241 mW; (c) 2.56 mJ; (d) 154 mW

P29.30

1.98 cm

P29.32

65.6 mT

P29.34

(a) 5.00 cm ; (b) 8.78 Mm s

P29.36

m′ =8 m

P29.38

see the solution

P29.40

244 kV m

P29.42

278 mm

P29.44

162 mm

P29.46

3.00 T

P29.48

(a) 7.44 × 10 28 m3 ; (b) 1.79 T

P29.60

P29.72

− Fi ; evi

3R 4 B ~ 10 −1 T; τ ~ 10 −1 N ⋅ m; I ~ 1 A ; A ~ 10 −3 m 2 ; N ~ 10 3

b

g

qBh ; The particle moves in a m semicircle of radius h and leaves the field with velocity −vj; (b) The particle moves in a smaller mv semicircle of radius , attaining final qB velocity −vj; (a) v =

(c) The particle moves in a circular arc of mv , leaving the field with radius r = qB velocity v sin θ i + v cos θ j where

θ = sin −1

FG h IJ HrK

30 Sources of the Magnetic Field CHAPTER OUTLINE 30.1 30.2

30.3 30.4 30.5 30.6 30.7

30.8 30.9

The Biot-Savart Law The Magnetic Force Between Two Parallel Conductors Ampère’s Law The Magnetic Field of a Solenoid Magnetic Flux Gauss’s Law in Magnetism Displacement Current and the General Form of Ampère’s Law Magnetism in Matter The Magnetic Field of the Earth

ANSWERS TO QUESTIONS Q30.1

It is not. The magnetic field created by a single loop of current resembles that of a bar magnet—strongest inside the loop, and decreasing in strength as you move away from the loop. Neither is it in a uniform direction—the magnetic field lines loop though the loop!

Q30.2

No magnetic field is created by a stationary charge, as the rate of flow is zero. A moving charge creates a magnetic field.

Q30.3

The magnetic field created by wire 1 at the position of wire 2 is into the paper. Hence, the magnetic force on wire 2 is in direction down × into the paper = to the right, away from wire 1. Now wire 2 creates a magnetic field into the page at the location of wire 1, so wire 1 feels force up ×into the paper = left, away from wire 2.

FIG. Q30.3

185

186 Q30.4

Sources of the Magnetic Field

No total force, but a torque. Let wire one carry current in the y direction, toward the top of the page. Let wire two be a millimeter above the plane of the paper and carry current to the right, in the x direction. On the left-hand side of wire one, wire one creates magnetic field in the z direction, which exerts force in the i × k = − j direction on wire two. On the right-hand side, wire one produces magnetic field in

2

e j

the − k direction and makes a i × − k = + j force of equal magnitude act 1

on wire two. If wire two is free to move, its center section will twist counterclockwise and then be attracted to wire one.

FIG. Q30.4 Q30.5

Ampère’s law is valid for all closed paths surrounding a conductor, but not always convenient. There are many paths along which the integral is cumbersome to calculate, although not impossible. Consider a circular path around but not coaxial with a long, straight current-carrying wire.

Q30.6

The Biot-Savart law considers the contribution of each element of current in a conductor to determine the magnetic field, while for Ampère’s law, one need only know the current passing through a given surface. Given situations of high degrees of symmetry, Ampère’s law is more convenient to use, even though both laws are equally valid in all situations.

Q30.7

If the radius of the toroid is very large compared to its cross-sectional area, then the field is nearly uniform. If not, as in many transformers, it is not.

Q30.8

Both laws use the concept of flux—the “flow” of field lines through a surface to determine the field strength. They also both relate the integral of the field over a closed geometrical figure to a fundamental constant multiplied by the source of the appropriate field. The geometrical figure is a surface for Gauss’s law and a line for Ampère’s.

Q30.9

Apply Ampère’s law to the circular path labeled 1 in the picture. Since there is no current inside this path, the magnetic field inside the tube must be zero. On the other hand, the current through path 2 is the current carried by the conductor. Therefore the magnetic field outside the tube is nonzero.

FIG. Q30.9 Q30.10

Q30.11

The magnetic field inside a long solenoid is given by B = (a)

If the length

(b)

If N is doubled, the magnetic field is doubled.

µ 0 NI

.

is doubled, the field is cut in half.

The magnetic flux is Φ B = BA cos θ . Therefore the flux is maximum when B is perpendicular to the loop of wire. The flux is zero when there is no component of magnetic field perpendicular to the loop—that is, when the plane of the loop contains the x axis.

Chapter 30

187

Q30.12

Maxwell included a term in Ampère’s law to account for the contributions to the magnetic field by changing electric fields, by treating those changing electric fields as “displacement currents.”

Q30.13

M measures the intrinsic magnetic field in the nail. Unless the nail was previously magnetized, then M starts out from zero. H is due to the current in the coil of wire around the nail. B is related to the sum of M and H. If the nail is aluminum or copper, H makes the dominant contribution to B, but M can add a little in the same or in the opposite direction. If the nail is iron, as it becomes magnetized M can become the dominant contributor to B.

Q30.14

Magnetic domain alignment creates a stronger external magnetic field. The field of one piece of iron in turn can align domains in another iron sample. A nonuniform magnetic field exerts a net force of attraction on magnetic dipoles aligned with the field.

Q30.15

The shock misaligns the domains. Heating will also decrease magnetism.

Q30.16

Magnetic levitation is illustrated in Figure Q30.31. The Earth’s magnetic field is so weak that the floor of his tomb should be magnetized as well as his coffin. Alternatively, the floor of his tomb could be made of superconducting material, which exerts a force of repulsion on any magnet.

Q30.17

There is no magnetic material in a vacuum, so M must be zero. Therefore B = µ 0 H in a vacuum.

Q30.18

Atoms that do not have a permanent magnetic dipole moment have electrons with spin and orbital magnetic moments that add to zero as vectors. Atoms with a permanent dipole moment have electrons with orbital and spin magnetic moments that show some net alignment.

Q30.19

The magnetic dipole moment of an atom is the sum of the dipole moments due to the electrons’ orbital motions and the dipole moments due to the spin of the electrons.

Q30.20

M and H are in opposite directions. Section 30.8 argues that all atoms should be thought of as weakly diamagnetic due to the effect of an external magnetic field on the motions of atomic electrons. Paramagnetic and ferromagnetic effects dominate whenever they exist.

Q30.21

The effects of diamagnetism are significantly smaller than those of paramagnetism.

Q30.22

When the substance is above the Curie temperature, random thermal motion of the molecules prevents the formation of domains. It cannot be ferromagnetic, but only paramagnetic.

Q30.23

A ferromagnetic substance is one in which the magnetic moments of the atoms are aligned within domains, and can be aligned macroscopically. A paramagnetic substance is one in which the magnetic moments are not naturally aligned, but when placed in an external magnetic field, the molecules line their magnetic moments up with the external field. A diamagnetic material is one in which the magnetic moments are also not naturally aligned, but when placed in an external magnetic field, the molecules line up to oppose the external magnetic field.

Q30.24

(a)

B increases slightly

(b)

B decreases slightly

(c)

B increases significantly

Equations 30.33 and 30.34 indicate that, when each metal is in the solenoid, the total field is B = µ 0 1 + χ H . Table 30.2 indicates that B is slightly greater than µ 0 H for aluminum and slightly less for copper. For iron, the field can be made thousands of times stronger, as seen in Example 30.10.

b

g

188

Sources of the Magnetic Field

Q30.25

A “hard” ferromagnetic material requires much more energy per molecule than a “soft” ferromagnetic material to change the orientation of the magnetic dipole moments. This way, a hard ferromagnetic material is more likely to retain its magnetization than a soft ferromagnetic material.

Q30.26

The medium for any magnetic recording should be a hard ferromagnetic substance, so that thermal vibrations and stray magnetic fields will not rapidly erase the information.

Q30.27

If a soft ferromagnetic substance were used, then the magnet would not be “permanent.” Any significant shock, a heating/cooling cycle, or just rotating the magnet in the Earth’s magnetic field would decrease the overall magnetization by randomly aligning some of the magnetic dipole moments.

Q30.28

You can expect a magnetic tape to be weakly attracted to a magnet. Before you erase the information on the tape, the net magnetization of a macroscopic section of the tape would be nearly zero, as the different domains on the tape would have opposite magnetization, and be more or less equal in number and size. Once your external magnet aligns the magnetic moments on the tape, there would be a weak attraction, but not like that of picking up a paper clip with a magnet. A majority of the mass of the tape is non-magnetic, and so the gravitational force acting on the tape will likely be larger than the magnetic attraction.

Q30.29

To magnetize the screwdriver, stroke one pole of the magnet along the blade of the screwdriver several or many times. To demagnetize the screwdriver, drop it on a hard surface a few times, or heat it to some high temperature.

Q30.30

The north magnetic pole is near the south geographic pole. Straight up.

Q30.31

(a)

The magnets repel each other with a force equal to the weight of one of them.

(b)

The pencil prevents motion to the side and prevents the magnets from rotating under their mutual torques. Its constraint changes unstable equilibrium into stable.

(c)

Most likely, the disks are magnetized perpendicular to their flat faces, making one face a north pole and the other a south pole. One disk has its north pole on the top side and the other has its north pole on the bottom side.

(d)

Then if either were inverted they would attract each other and stick firmly together.

SOLUTIONS TO PROBLEMS Section 30.1

The Biot-Savart Law

b

g

µ 0 I µ 0 q v 2π R = = 12.5 T 2R 2R

P30.1

B=

P30.2

4π × 10 −7 T ⋅ m A 1.00 × 10 4 A µ0I = = 2.00 × 10 −5 T = 20.0 µT B= 2π R 2π 100 m

e

a

je

f

j

Chapter 30

P30.3

B=

(a)

FG H

4µ 0 I π 3π cos − cos 4π a 4 4

189

IJ where a = K 2

is the distance from any side to the center. B=

4.00 × 10 −6 0.200

F GH

I JK

2 2 = 2 2 × 10 −5 T = 28.3 µT into the paper + 2 2 FIG. P30.3

For a single circular turn with 4 = 2π R ,

(b)

ja f a f

e

4π 2 × 10 −7 10.0 µ 0 I µ 0π I = = = 24.7 µT into the paper B= 2R 4 4 0. 400

e

ja

f

4π × 10 −7 1.00 A µ0I = = 2.00 × 10 −7 T 2π r 2π 1.00 m

P30.4

B=

P30.5

For leg 1, ds × r = 0 , so there is no contribution to the field from this segment. For leg 2, the wire is only semi-infinite; thus,

a

B=

F GH

f

I JK

µ0I 1 µ0I = into the paper . 2 2π x 4π x

FIG. P30.5 P30.6

We can think of the total magnetic field as the superposition of the field due to the long straight wire µ I (having magnitude 0 and directed into the page) and the field due to the circular loop (having 2π R µ0I magnitude and directed into the page). The resultant magnetic field is: 2R 1 µ0I B= 1+ directed into the page . π 2R

FG H

P30.7

IJ K

b

g

For the straight sections ds × r = 0 . The quarter circle makes one-fourth the field of a full loop: B=

1 µ0I µ0I = into the paper 4 2R 8R

B=

ja 8b0.030 0 mg

e4π × 10

−7

T ⋅ m A 5.00 A

f=

26.2 µT into the paper

190 P30.8

Sources of the Magnetic Field

Along the axis of a circular loop of radius R, B=

µ 0 IR 2

e B L 1 =M B MN bx Rg

or

2 x2 + R2

0

2

B0 ≡

where

j

32

OP + 1 PQ

3 2

µ0I . 2R

FIG. P30.8 x R 0.00 1.00 2.00 3.00 4.00 5.00

*P30.9

B B0 1.00 0.354 0.0894 0.0316 0.0143 0.00754

Wire 1 creates at the origin magnetic field B1 =

(a)

µ0I µ I right hand rule = 0 1 2π r 2π a

=

If the total field at the origin is according to B 2 =

µ 0 I1 j 2π a

µ I 2µ 0 I1 j = 0 1 j + B 2 then the second wire must create field 2π a 2π a

µ 0 I1 µ I j= 0 2 2π a 2π 2 a

.

a f

Then I 2 = 2 I 1 out of the paper = 2 I 1 k . (b)

The other possibility is B1 + B 2 = B2 =

*P30.10

µ I 2 µ 0 I1 − j = 0 1 j + B 2 . Then 2π a 2π a

e j

µ I 3µ 0 I1 −j = 0 2 2π a 2π 2 a

e j

a f

e j

I 2 = 6 I 1 into the paper = 6 I 1 − k

Every element of current creates magnetic field in the same direction, into the page, at the center of the arc. The upper straight portion creates one-half of the field that an infinitely long straight wire would create. The curved portion creates one quarter of the field that a circular loop produces at its 1 µ0I center. The lower straight segment also creates field . 2 2π r The total field is

F 1 µ I + 1 µ I + 1 µ I I into the page = GH 2 2π r 4 2r 2 2π r JK F 0.284 15µ I IJ into the page. =G H r K

B=

0

0

0

0

FG H

IJ K

µ0I 1 1 + into the plane of the paper 2r π 4

Chapter 30

*P30.11

(a)

191

Above the pair of wires, the field out of the page of the 50 A

e j

current will be stronger than the − k field of the 30 A current, so they cannot add to zero. Between the wires, both produce fields into the page. They can only add to zero below the wires, at coordinate y = − y . Here the total field is B=

µ0I 2π r

+

µ0I 2π r

:

FIG. P30.11

OP LM 50 A 30 A − + k k MN d y + 0.28 mi e j y e jPQ 50 y = 30d y + 0.28 mi 50b− y g = 30b0.28 m − y g −20 y = 30a0. 28 mf at y = −0.420 m

0=

µ0 2π

At y = 0.1 m the total field is B =

(b)

4π × 10 −7 T ⋅ m A 2π

B=

µ0I 2π r

+

µ0I 2π r

:

F 50 A e−kj + 30 A e−kjI = 1.16 × 10 Te−kj . JK GH a0.28 − 0.10f m 0.10 m −4

The force on the particle is

e

je

je j e

je j

e j

F = qv × B = −2 × 10 −6 C 150 × 10 6 m s i × 1.16 × 10 −4 N ⋅ s C ⋅ m − k = 3.47 × 10 −2 N − j . (c)

P30.12

dB = B= B=

e j

j

Fe = 3.47 × 10 −2 N + j = qE = −2 × 10 −6 C E .

So

E = −1.73 × 10 4 j N C .

µ0I d × r 4π r 2

µ0I 4π

e

We require

F GH

1 6

FG H

2π a a

2

−

IJ K

1 6

2π b b

2

I JK

µ0I 1 1 directed out of the paper − 12 a b

192 *P30.13

Sources of the Magnetic Field

(a)

We use equation 30.4. For the distance a from the wire to a , a = 0. 288 7L . One the field point we have tan 30° = L 2 wire contributes to the field at P

µ0I µ0I cos θ 1 − cos θ 2 = cos 30°− cos 150° 4π a π 4 0. 288 7L

ga

b g b µ I a1.732 f 1.50 µ I = = . πL π L 4 b0.288 7 g

B=

0

f

a P I

F GH

a

a f b g a f b g

f IJ K

a

FIG. P30.13(a)

I JK

As we showed in part (a), one whole side of the triangle µ I 1.732 creates field at the center 0 . Now one-half of one 4π a nearby side of the triangle will be half as far away from point Pb and have a geometrically similar situation. Then it µ I 1.732 2 µ 0 I 1.732 = . The two halfcreates at Pb field 0 4π a 4π a 2 sides shown crosshatched in the picture create at Pb field 2 µ 0 I 1.732 4µ 0 I 1.732 6µ 0 I = = . The rest of the 2 4π a πL 4π 0.288 7L triangle will contribute somewhat more field in the same direction, so we already have a proof that the field at Pb is

F GH

L 2 θ1

0

Each side contributes the same amount of field in the same direction, which is perpendicularly into the paper in the 1.50 µ 0 I 4.50 µ 0 I = . picture. So the total field is 3 πL πL (b)

θ2

f

a

a

f

FIG. P30.13(b)

stronger . P30.14

Apply Equation 30.4 three times:

F I toward you + µ I F GH JK G 4π d H I µ I F −d + − cos 180°J toward you G 4π a H d + a K µ I FH a + d − d a + d IK

B=

µ0I d cos 0 − 2 4π a d + a2

0

0

2

B=

0

2

2

2

2

2π ad a 2 + d 2

2

away from you

Pb

a d2 + a2

+

a d 2 + a2

I away from you JK

193

Chapter 30

P30.15

Take the x-direction to the right and the y-direction up in the plane of the paper. Current 1 creates at P a field

e

−7

ja

2.00 × 10 T ⋅ m 3.00 A µ I B1 = 0 = 2π a A 0.050 0 m

b

g

5.00 cm

I1

P

f

B1

B2

13.0 cm 12.0 cm

B1 = 12.0 µT downward and leftward, at angle 67.4° below the –x axis. Current 2 contributes

I2

e2.00 × 10 T ⋅ mja3.00 Af clockwise perpendicular to 12.0 cm Aa0.120 mf −7

B2 =

FIG. P30.15

B 2 = 5.00 µT to the right and down, at angle –22.6° Then,

Section 30.2 P30.16

b ge j b B = b−11.1 µTg j − b1.92 µTg j = b −13.0 µTg j

ge

B = B1 + B 2 = 12.0 µT − i cos 67.4°− j sin 67.4° + 5.00 µT i cos 22.6°− j sin 22.6°

The Magnetic Force Between Two Parallel Conductors

Let both wires carry current in the x direction, the first at y = 0 and the second at y = 10.0 cm .

(a)

e

ja f

I2 = 8.00 A

f

4π × 10 −7 T ⋅ m A 5.00 A µ0I B= k= k 2π r 2π 0.100 m

a

y

I1 = 5.00 A

a

fa

f e

j e

je j

FB = I 2 × B = 8.00 A 1.00 m i × 1.00 × 10 −5 T k = 8.00 × 10 −5 N − j

B=

e j e

ja f

f

4π × 10 −7 T ⋅ m A 8.00 A µ0I −k = − k = 1.60 × 10 −5 T − k 2π r 2π 0.100 m

a

e j e

je j

B = 1.60 × 10 −5 T into the page (d)

a

fa

x

FIG. P30.16(a)

FB = 8.00 × 10 −5 N toward the first wire

(c)

y = 10.0 cm

z

B = 1.00 × 10 −5 T out of the page (b)

j

f e

je j e

je j

FB = I 1 × B = 5.00 A 1.00 m i × 1.60 × 10 −5 T − k = 8.00 × 10 −5 N + j

FB = 8.00 × 10 −5 N towards the second wire

194 P30.17

Sources of the Magnetic Field

By symmetry, we note that the magnetic forces on the top and bottom segments of the rectangle cancel. The net force on the vertical segments of the rectangle is (using Equation 30.11)

FG 1 − 1 IJ i = µ I I FG − a IJ i H c + a c K 2π H cac + af K e4π × 10 N A ja5.00 Afa10.0 Afa0.450 mf F −0.150 m I i F= GH a0.100 mfa0.250 mf JK 2π F = e −2.70 × 10 i j N µ 0 I1 I 2 2π

F = F1 + F2 =

−7

0 1 2

2

−5

or *P30.18

F = 2.70 × 10 −5 N toward the left .

FIG. P30.17

To attract, both currents must be to the right. The attraction is described by F = I 2 B sin 90° = I 2 So

µ0I 2π r

F I GjG 4π × 10 2πNa0⋅.s5 Cm⋅fm a20 Af JJ = 40.0 A j K He

F 2π r = 320 × 10 −6 N m µ 0 I1

e

I2 =

FIG. P30.18

−7

Let y represent the distance of the zero-field point below the upper wire. Then

B=

µ0I 2π r

+

b

µ0I 2π r

g

20 0.5 m − y = 40 y

0=

F GH

µ 0 20 A 40 A away + toward y 2π 0.5 m − y

a

b

ga

g b

fIJK

f

20 0.5 m = 60 y

y = 0.167 m below the upper wire *P30.19

Carrying oppositely directed currents, wires 1 and 2 repel each other. If wire 3 were between them, it would have to repel either 1 or 2, so the force on that wire could not be zero. If wire 3 were to the right of wire 2, it would feel a larger force exerted by 2 than that exerted by 1, so the total force on 3 could not be zero. Therefore wire 3 must be to the left of both other wires as shown. It must carry downward current so that it can attract wire 2. (a)

For the equilibrium of wire 3 we have F1 on 3 = F2 on 3

a

f

1.5 20 cm + d = 4d (b)

a

f

Wire 3

Wire 1

Wire 2

1.50 A

I3

4.00 A

20 cm

d

FIG. P30.19

a f

µ 0 1.50 A I 3 µ 0 4 A I3 = 2π d 2π 20 cm + d 30 cm d= = 12.0 cm to the left of wire 1 2.5

a

f

For the equilibrium of wire 1,

a

f f

a fa f a f

µ 0 I 3 1.5 A µ 4 A 1.5 A = 0 2π 12 cm 2π 20 cm

a

I3 =

12 4 A = 2.40 A down 20

We know that wire 2 must be in equilibrium because the forces on it are equal in magnitude to the forces that it exerts on wires 1 and 3, which are equal because they both balance the equal-magnitude forces that 1 exerts on 3 and that 3 exerts on 1.

Chapter 30

P30.20

The separation between the wires is

a

f

a = 2 6.00 cm sin 8.00° = 1.67 cm. (a)

Because the wires repel, the currents are in opposite directions .

(b)

Because the magnetic force acts horizontally,

µ I2 FB = 0 = tan 8.00° Fg 2π amg mg 2π a

I2 =

Section 30.3 P30.21

µ0

FIG. P30.20

tan 8.00° so I = 67.8 A .

Ampère’s Law

Each wire is distant from P by

a0.200 mf cos 45.0° = 0.141 m. Each wire produces a field at P of equal magnitude:

e

ja

f

2.00 × 10 −7 T ⋅ m A 5.00 A µ0I = = 7.07 µT . BA = 2π a 0.141 m

a

f

Carrying currents into the page, A produces at P a field of 7.07 µT to the left and down at –135°, while B creates a field to the right and down at – 45°. Carrying currents toward you, C produces a field downward and to the right at – 45°, while D’s contribution is downward and to the left. The total field is then

b

FIG. P30.21

g

4 7.07 µT sin 45.0° = 20.0 µT toward the bottom of the page P30.22

µ0I at the proton’s location. And we have a 2π d balance between the weight of the proton and the magnetic force

Let the current I be to the right. It creates a field B =

e j e j

mg − j + qv − i ×

e

µ0I k = 0 at a distance d from the wire 2π d

ej

je

je

je

j

1.60 × 10 −19 C 2.30 × 10 4 m s 4π × 10 −7 T ⋅ m A 1.20 × 10 −6 A qvµ 0 I d= = = 5.40 cm 2π mg 2π 1.67 × 10 −27 kg 9.80 m s 2

e

je

j

195

196 P30.23

Sources of the Magnetic Field

µ 0Ia , where I a is the net current 2π ra through the area of the circle of radius ra . In this case, I a = 1.00 A out of the page (the current in the inner conductor), so 4π × 10 −7 T ⋅ m A 1.00 A = 200 µT toward top of page . Ba = 2π 1.00 × 10 −3 m µ I Similarly at point b : Bb = 0 b , where I b is the net current through the area of the circle having 2π rb radius rb . From Ampere’s law, the magnetic field at point a is given by Ba =

e

ja

e

f

j

Taking out of the page as positive, I b = 1.00 A − 3.00 A = −2.00 A , or I b = 2.00 A into the page. Therefore, 4π × 10 −7 T ⋅ m A 2.00 A = 133 µT toward bottom of page . Bb = 2π 3.00 × 10 −3 m

e

P30.24

ja

e

j

f

µ0I , the field will be one-tenth as large at a ten-times larger distance: 400 cm 2π r

(a)

In B =

(b)

B=

(c)

Call r the distance from cord center to field point and 2d = 3.00 mm the distance between conductors. µ I 1 µ I 2d 1 B= 0 − = 0 2 2π r − d r + d 2π r − d 2 3.00 × 10 −3 m −10 −7 so r = 1.26 m 7.50 × 10 T = 2.00 × 10 T ⋅ m A 2.00 A 2 r − 2.25 × 10 −6 m 2 The field of the two-conductor cord is weak to start with and falls off rapidly with distance.

a

4π × 10 −7 T ⋅ m 2.00 A µ0I µ I k + 0 − k so B = 2π r1 2π r2 2π A

e j

FG H

IJ K

ja

e

(d)

f FG 1 − 1 IJ = H 0.398 5 m 0.401 5 m K

f e

7.50 nT

j

The cable creates zero field at exterior points, since a loop in Ampère’s law encloses zero total current. Shall we sell coaxial-cable power cords to people who worry about biological damage from weak magnetic fields?

P30.25

(a)

One wire feels force due to the field of the other ninety-nine. B=

ja fb

e

ge j

j

−7 −2 µ 0 I 0 r 4π × 10 T ⋅ m A 99 2.00 A 0.200 × 10 m = = 3.17 × 10 −3 T 2 −2 2π R 2 2π 0.500 × 10 m

e

This field points tangent to a circle of radius 0.200 cm and exerts force F = I × B toward the center of the bundle, on the single hundredth wire: F FB (b)

a

fe

j

= IB sin θ = 2.00 A 3.17 × 10 −3 T sin 90° = 6.34 mN m = 6.34 × 10 −3 N m inward

B ∝ r , so B is greatest at the outside of the bundle. Since each wire carries the same current, F is greatest at the outer surface .

FIG. P30.25

Chapter 30

P30.26

(a)

(b) *P30.27

e

ja fe j a f T ⋅ m A ja900fe14.0 × 10 A j =

Binner

4π × 10 −7 T ⋅ m A 900 14.0 × 10 3 A µ 0 NI = = = 3.60 T 2π r 2π 0.700 m

Bouter

2 × 10 −7 µ 0 NI = = 2π r

e

197

3

1.30 m

1.94 T

We assume the current is vertically upward. (a)

Consider a circle of radius r slightly less than R. It encloses no current so from

z

b g

B ⋅ ds = µ 0 I inside

B 2π r = 0

we conclude that the magnetic field is zero .

b

g

Now let the r be barely larger than R. Ampere’s law becomes B 2π R = µ 0 I ,

(b)

B=

so

µ0I . 2π R

FIG. P30.27(a) tangent to the wall of the cylinder in a counterclockwise sense .

The field’s direction is (c)

Consider a strip of the wall of width dx and length . Its width is so small compared to 2π R that the field at its location would be essentially unchanged if the current in the strip were turned off. Idx The current it carries is I s = up. 2π R The force on it is µ0I µ I 2 dx Idx F = Is × B = up × into page = 0 2 2 radially inward . 2π R 2π R 4π R

F GH

I JK

FIG. P30.27(c)

The pressure on the strip and everywhere on the cylinder is P=

µ I 2 dx F = 02 2 = A 4π R dx

µ0I 2

b2π Rg

2

inward .

The pinch effect makes an effective demonstration when an aluminum can crushes itself as it carries a large current along its length.

z

2π rB

ja

e

2π 1.00 × 10 −3 0.100

f=

P30.28

From B⋅ d = µ 0 I ,

P30.29

Use Ampère’s law, B ⋅ ds = µ 0 I . For current density J, this becomes

z

z

B ⋅ ds = µ 0 J ⋅ dA .

(a)

I=

µ0

=

4π × 10

−7

500 A .

z

For r1 < R , this gives

B 2π r1 = µ 0

za

r1

fb

g

br 2π rdr and

0

FIG. P30.29

µ 0 br12 B= for r1 < R or inside the cylinder . 3

b

(b)

g

When r2 > R , Ampère’s law yields

b2π r gB = µ z abr fb2π rdr g = 2πµ3bR R

2

0

0

or B =

µ 0 bR 3 for r2 > R or outside the cylinder . 3r2

b

g

0

3

,

198 P30.30

Sources of the Magnetic Field

(a)

See Figure (a) to the right.

(b)

At a point on the z axis, the contribution from each wire has µ0I magnitude B = and is perpendicular to the line 2π a 2 + z 2 from this point to the wire as shown in Figure (b). Combining (Currents are into the paper) fields, the vertical components cancel while the horizontal Figure (a) components add, yielding By = 2

F GH 2π

µ0I 2

2

a +z

I JK

sin θ =

µ0I 2

π a +z

2

F GH

z

I = µ Iz JK π ea + z j 0

2

a +z

2

2

2

The condition for a maximum is: dBy dz

=

a f + µ I = 0 , or µ I ea π ea π ea + z j π ea + z j − µ 0 Iz 2 z

2 2

2

2

2

j =0 +z j

2

− z2

2

2

0

0

Figure (b)

Thus, along the z axis, the field is a maximum at d = a .

Section 30.4 P30.31 *P30.32

FIG. P30.30

The Magnetic Field of a Solenoid

B = µ0

N

I so I =

e

j

1.00 × 10 −4 T 0.400 m B = = 31.8 mA µ 0 n 4π × 10 −7 T ⋅ m A 1 000

e

j

Let the axis of the solenoid lie along the y–axis from y = 0 to y = . We will determine the field at y = a . This point will be inside the solenoid if 0 < a < and outside if a < 0 or a > . We think of solenoid as formed of rings, each of thickness dy. Now I is the symbol for the current in each turn of N N wire and the number of turns per length is dy and . So the number of turns in the ring is

FG IJ H K

the current in the ring is I ring = I

FG IJ H K

FG N IJ dy . Now we use the result of Example 30.3 for the field created H K

by one ring: Bring =

µ 0 I ring R 2

e

2 x2 + R2

j

32

where x is the name of the distance from the center of the ring, at location y, to the field point x = a − y . Each ring creates field in the same direction, along our y–axis, so the whole field of the solenoid is µ 0 I ring R 2 µ 0 I N dyR 2 dy µ 0 INR 2 = = B = ∑ Bring = ∑ . 32 3 2 3 2 2 2 2 2 2 all rings 0 2 a−y 0 2 a−y 2 x +R + R2 + R2

e

j

z eb

b g g j

To perform the integral we change variables to u = a − y . B=

µ 0 INR 2 2

ze

a− a

− du

u2 + R2

j

32

and then use the table of integrals in the appendix: continued on next page

z eb

g

j

Chapter 30 a−

µ INR 2 −u B= 0 2 2 R u2 + R2

(a)

(b)

If

µ 0 IN 2

= a

LM MMN

a a2 + R2

a−

−

aa − f

2

+ R2

199

OP PPQ

is much larger than R and a = 0,

we have B ≅

LM MN

OP PQ

µ 0 IN µ IN − = 0 . 0− 2 2 2

This is just half the magnitude of the field deep within the solenoid. We would get the same result by substituting a = to describe the other end. P30.33

The field produced by the solenoid in its interior is given by

e j e

B = µ 0 nI − i = 4π × 10 −7 T ⋅ m A

e

jFGH 1030.0m IJK a15.0 Afe− ij −2

j

B = − 5.65 × 10 −2 T i The force exerted on side AB of the square current loop is

a

bF g

= IL × B = 0. 200 A

bF g

= 2.26 × 10 −4 N k

B AB

B AB

e

f e2.00 × 10

−2

j e

je j

m j × 5.65 × 10 −2 T − i

j

Similarly, each side of the square loop experiences a force, lying in the plane of the loop, of 226 µN directed away from the center . From the above result, it is seen that the net torque exerted on the square loop by the field of the solenoid should be zero. More formally, the magnetic dipole moment of the square loop is given by

a

fe

FIG. P30.33

j e− ij = −80.0 µA ⋅ m i The torque exerted on the loop is then τ = µ × B = e −80.0 µA ⋅ m i j × e −5.65 × 10 µ = IA = 0.200 A 2.00 × 10 −2 m

2

2

2

Section 30.5 P30.34

a

f

(a)

bΦ g

(b)

The net flux out of the closed surface is zero: Φ B

B flat

= B ⋅ A = Bπ R 2 cos 180 − θ = − Bπ R 2 cos θ

b g

B curved

(a)

flat

b g

+ ΦB

curved

= 0.

= Bπ R 2 cos θ

z

e

j e

j

2

Φ B = B ⋅ dA = B ⋅ A = 5 i + 4 j + 3k T ⋅ 2.50 × 10 −2 m i Φ B = 3.12 × 10 −3 T ⋅ m 2 = 3.12 × 10 −3 Wb = 3.12 mWb

(b)

bΦ g

B total

z

j

Ti = 0

Magnetic Flux

bΦ g P30.35

−2

= B ⋅ dA = 0 for any closed surface (Gauss’s law for magnetism)

200 P30.36

Sources of the Magnetic Field

(a)

Φ B = B ⋅ A = BA where A is the cross-sectional area of the solenoid.

FG µ NI IJ eπ r j = 7.40 µWb H K F µ NI IJ π er − r j = B ⋅ A = BA = G H K L e4π × 10 T ⋅ m Aja300fa12.0 Af OP =M PQπ a8.00f − a4.00f e10 MN a0.300 mf

(b)

ΦB

2

0

ΦB =

2 2

0

2 1

−7

ΦB

Section 30.6

2

2

−3

m

j

2

= 2.27 µWb

Gauss’s Law in Magnetism

No problems in this section

Section 30.7 P30.37

P30.38

Displacement Current and the General Form of Ampère’s Law

a

0.100 A dΦ E dQ dt I = = = = 11.3 × 10 9 V ⋅ m s ∈0 ∈0 8.85 × 10 −12 C 2 N ⋅ m 2 dt

(b)

I d =∈0

dΦ E = I = 0.100 A dt

dQ dt dΦ E d I = = EA = ∈0 ∈0 dt dt

a f

(a)

dE I = = 7.19 × 10 11 V m ⋅ s dt ∈0 A

(b)

z

B ⋅ ds =∈0 µ 0

B=

Section 30.8 P30.39

f

(a)

(a)

LM N

dΦ E d Q ⋅π r 2 so 2π rB =∈0 µ 0 dt dt ∈0 A

a

fe a f

j

OP Q

−2 µ 0 Ir µ 0 0.200 5.00 × 10 = = 2.00 × 10 −7 T 2 2A 2π 0.100

Magnetism in Matter I=

ev 2π r

µ = IA =

F ev I π r GH 2π r JK

2

= 9.27 × 10 −24 A ⋅ m 2

The Bohr model predicts the correct magnetic moment. However, the “planetary model” is seriously deficient in other regards. (b)

Because the electron is (–), its [conventional] current is clockwise, as seen from above, and µ points downward .

FIG. P30.39

Chapter 30

F N I I so I = b2π r gB = 2π a0.100 mfa1.30 Tf = GH 2π r JK µN 5 000e 4π × 10 Wb A ⋅ mja 470f

P30.40

B = µnI = µ

P30.41

Assuming a uniform B inside the toroid is equivalent to assuming NI r

Q32.17

The condition for critical damping must be investigated to design a circuit for a particular purpose. For example, in building a radio receiver, one would want to construct the receiving circuit so that it is underdamped. Then it can oscillate in resonance and detect the desired signal. Conversely, when designing a probe to measure a changing signal, such free oscillations are undesirable. An electrical vibration in the probe would constitute “ringing” of the system, where the probe would measure an additional signal—that of the probe itself! In this case, one would want to design a probe that is critically damped or overdamped, so that the only signal measured is the one under study. Critical damping represents the threshold between underdamping and overdamping. One must know the condition for it to meet the design criteria for a project.

Q32.18

An object cannot exert a net force on itself. An object cannot create momentum out of nothing. A coil can induce an emf in itself. When it does so, the actual forces acting on charges in different parts of the loop add as vectors to zero. The term electromotive force does not refer to a force, but to a voltage.

4L 4L , then the oscillator is overdamped—it will not oscillate. If R < , then the oscillator is C C underdamped and can go through several cycles of oscillation before the radiated signal falls below background noise.

SOLUTIONS TO PROBLEMS Section 32.1 P32.1 P32.2

Self-Inductance

ε =L

jFGH

IJ K

∆I 1.50 A − 0.200 A = 3.00 × 10 −3 H = 1.95 × 10 −2 V = 19.5 mV ∆t 0.200 s

e

Treating the telephone cord as a solenoid, we have:

ja f e

e

j

2

−7 −3 µ N 2 A 4π × 10 T ⋅ m A 70.0 π 6.50 × 10 m = L= 0 A 0.600 m

fFGH

a

0 − 0.500 A ∆I = −2.00 H ∆t 0.010 0 s

P32.3

ε = −L

P32.4

L=

P32.5

ε back = −ε = L

I FG 1 V ⋅ s IJ = JK H 1 H ⋅ A K

NΦ B LI → ΦB = = 240 nT ⋅ m 2 I N

a

b

2

= 1.36 µH .

100 V

through each turn

g

ja

fa f

dI d =L I max sin ω t = LωI max cos ω t = 10.0 × 10 −3 120π 5.00 cos ω t dt dt

f b

g a

e

f a f

ε back = 6.00π cos 120π t = 18.8 V cos 377 t

242 P32.6

Inductance

From ε = L From L =

FG ∆I IJ , we have H ∆t K

L=

NΦ B , we have I

ΦB

a fe

P32.8

N

ε =L

e

j e

500

f=

19.2 µT ⋅ m 2 .

j

(a)

At t = 1.00 s ,

ε = 360 mV

(b)

At t = 4.00 s ,

ε = 180 mV

(c)

ε = 90.0 × 10 −3 2t − 6 = 0

ja

e

f

t = 3.00 s .

FG 450 IJ b0.040 0 Ag = H 0.120 K

(a)

B = µ 0 nI = µ 0

(b)

Φ B = BA = 3.33 × 10 −8 T ⋅ m 2

(c)

L=

188 µT

NΦ B = 0.375 mH I

B and Φ B are proportional to current; L is independent of current

a f e 2

*P32.11

ja

H 4.00 A

dI d 2 = 90.0 × 10 −3 t − 6t V dt dt

(d)

P32.10

−3

−4 µ N 2 A µ 0 420 3.00 × 10 L= 0 = = 4.16 × 10 −4 H 0.160 A dI dI −ε −175 × 10 −6 V ε = −L → = = = −0.421 A s dt dt L 4.16 × 10 −4 H

when P32.9

g LI e 2. 40 × 10 = = j

2

P32.7

b

ε 24.0 × 10 −3 V = = 2.40 × 10 −3 H . 10.0 A s ∆I ∆t

(a)

−3 µ N 2 A µ 0 120 π 5.00 × 10 = L= 0 0.090 0 A

(b)

Φ ′B =

j

2

= 15.8 µH

µm µ N2A ΦB → L = m = 800 1.58 × 10 −5 H = 12.6 mH µ0 A

e

j

We can directly find the self inductance of the solenoid:

ε = −L

dI dt

+0.08 V = − L

0 − 1.8 A 0.12 s

L = 5.33 × 10 −3 Vs A =

e j µ µ N π F 200 m I = = G J A H 2π N K

µ0N2A . A

Here A = π r 2 , 200 m = N 2π r , and A = N 10 −3 m . Eliminating extra unknowns step by step, we have 5.33 × 10 A=

−3

4 × 10

−3

µ N 2π r 2 Vs A = 0 A WbmA

5.33 × 10 −3 AVs

= 0.750 m

0

2

2

0 40 000

4π A

m2

=

e

j

10 −7 40 000 m 2 Tm A

A

Chapter 32

P32.12

L=

243

NΦ B NBA NA µ 0 NI µ0N 2 A = ≈ ⋅ = I I I 2π R 2π R

FIG. P32.12 P32.13

ε = ε 0 e − kt = − L dI = −

ε 0 − kt e dt L

dI dt

ε 0 − kt dq e = kL dt ε Q = 20 . k L

I=

If we require I → 0 as t → ∞ , the solution is

z z

Q = Idt =

Section 32.2 P32.14

I=

∞

ε 0 − kt ε e dt = − 20 kL k L 0

RL Circuits

ε R

e1 − e j : − Rt L

0.900

ε R

=

ε R

1 − e − R a3.00 s f

2.50 H

FG Ra3.00 sf IJ = 0.100 H 2.50 H K

exp − R=

2.50 H ln 10.0 = 1.92 Ω 3.00 s

a f ε e1 −Re j

I (A)

−t τ

P32.15

(a)

At time t,

It =

where

L τ = = 0.200 s . R

After a long time,

I max =

e

ε 1 − e −∞ R

1

0

R

a0.500f Rε = ε e1 − e R

so

0.500 = 1 − e − t 0. 200 s .

− t 0. 200 s

j a

e

Isolating the constants on the right, ln e − t 0. 200 s = ln 0.500

(b)

0.5

j=ε .

At I t = 0.500 I max

af

j

t (s) 0

f

t = −0.693 0.200 s

and solving for t,

−

or

t = 0.139 s .

Similarly, to reach 90% of I max ,

0.900 = 1 − e − t τ

and

t = −τ ln 1 − 0.900 .

Thus,

t = − 0.200 s ln 0.100 = 0.461 s .

a

a

f a

I max

f

f

0.2

0.4

FIG. P32.15

0.6

244 P32.16

Inductance

Taking τ = IR + L

L , R

dI = 0 will be true if dt

Because τ = P32.17

P32.18

0

−t τ

L = 2.00 × 10 −3 s = 2.00 ms R

τ=

(b)

I = I max 1 − e − t τ =

(c)

I max =

I=

I 0 Re − t τ

L , we have agreement with 0 = 0 . R

(a)

.00 V I J e1 − e j FGH 64.00 ΩK

e

−0. 250 2.00

ε 6.00 V = = 1.50 A R 4.00 Ω

0.800 = 1 − e

(d)

FG IJ H K F 1I + Le I e jG − J = 0 . H τK

dI 1 = I0 e−t τ − τ dt

I = I 0 e −t τ :

− t 2.00 ms

a

f a

j=

0.176 A

FIG. P32.17

f

→ t = − 2.00 ms ln 0.200 = 3.22 ms

120 ε 1 − e −t τ = 1 − e −1.80 7.00 = 3.02 A R 9.00

e

∆VR

j

e j = IR = a3.02fa9.00f = 27.2 V

∆VL = ε − ∆VR = 120 − 27.2 = 92.8 V P32.19

Note: It may not be correct to call the voltage or emf across a coil a “potential difference.” Electric potential can only be defined for a conservative electric field, and not for the electric field around an inductor. (a) ∆VR = IR = 8.00 Ω 2.00 A = 16.0 V

a

fa

f

and

∆VL = ε − ∆VR = 36.0 V − 16.0 V = 20.0 V .

Therefore,

∆VR 16.0 V = = 0.800 . ∆VL 20.0 V

a

fa

FIG. P32.19

f

∆VR = IR = 4.50 A 8.00 Ω = 36.0 V

(b)

∆VL = ε − ∆VR = 0 P32.20

a

f

After a long time, 12.0 V = 0. 200 A R . Thus, R = 60.0 Ω . Now, τ =

jb

e

g

L gives R

L = τ R = 5.00 × 10 −4 s 60.0 V A = 30.0 mH . P32.21

e

e jFGH − τ1 IJK

dI = − I max e − t τ dt

j

I = I max 1 − e − t τ :

τ=

L 15.0 H = = 0.500 s : R 30.0 Ω

ε dI R = I max e − t τ and I max = dt L R

dI R ε 100 V = I max e 0 = = = 6.67 A s dt L L 15.0 H

(a)

t=0:

(b)

t = 1.50 s :

dI ε − t τ = e = 6.67 A s e −1.50 a0.500 f = 6.67 A s e −3.00 = 0.332 A s dt L

b

g

b

g

Chapter 32

P32.22

e

j

I = I max 1 − e − t τ :

0.980 = 1 − e −3.00 ×10

−3

−3

τ

0.020 0 = e −3.00 ×10

τ =− L , so R

τ= P32.23

τ

3.00 × 10 −3 = 7.67 × 10 −4 s ln 0.020 0

b

g

FIG. P32.22

ja f

e

L = τ R = 7.67 × 10 −4 10.0 = 7.67 mH

Name the currents as shown. By Kirchhoff’s laws: I1 = I 2 + I 3

(1)

+10.0 V − 4.00 I 1 − 4.00 I 2 = 0

(2)

a f dIdt = 0 3

+10.0 V − 4.00 I 1 − 8.00 I 3 − 1.00

(3)

From (1) and (2),

+10.0 − 4.00 I 1 − 4.00 I 1 + 4.00 I 3 = 0

and

I 1 = 0.500 I 3 + 1.25 A .

Then (3) becomes

10.0 V − 4.00 0.500 I 3 + 1.25 A − 8.00 I 3 − 1.00

FIG. P32.23

a f dIdt = 0 b g a1.00 HfFGH dIdt IJK + a10.0 ΩfI = 5.00 V . 3

3

3

We solve the differential equation using Equations 32.6 and 32.7: .00 V I a f FGH 510.0 J 1− e a ΩK

I 1 = 1.25 + 0.500 I 3 = P32.24

P32.25

a0.500 Af 1 − e 1.50 A − a0.250 A fe f

− 10.0 Ω t 1.00 H

I3 t =

=

−10 t s

L L = , we get R = R C

3.00 H = 1.00 × 10 3 Ω = 1.00 kΩ . 3.00 × 10 −6 F

(a)

Using τ = RC =

(b)

τ = RC = 1.00 × 10 3 Ω 3.00 × 10 −6 F = 3.00 × 10 −3 s = 3.00 ms

e

je

−10 t s

j

For t ≤ 0 , the current in the inductor is zero . At t = 0 , it starts to grow from zero toward 10.0 A with time constant

τ=

a

f

10.0 mH L = = 1.00 × 10 −4 s . R 100 Ω

a

f

e j a10.0 Afe1 − e At t = 200 µs , I = a10.00 A fe1 − e j = 8.65 A .

For 0 ≤ t ≤ 200 µs , I = I max 1 − e − t τ =

−10 000 t s

j

.

−2.00

Thereafter, it decays exponentially as I = I 0 e

a

f

a

f

I = 8.65 A e −10 000b t − 200 µsg s = 8.65 A e −10 000 t

− t′ τ

FIG. P32.25

, so for t ≥ 200 µs ,

s + 2.00

e

j

= 8.65 e 2.00 A e −10 000 t s =

a63.9 Afe

−10 000 t s

.

245

246 P32.26

Inductance

ε

I=

(b)

Initial current is 1.00 A: ∆V12 = 1.00 A 12.00 Ω = 12.0 V

R

=

12.0 V = 1.00 A 12.0 Ω

(a)

a

fa

a

f

fb

g

∆V1 200 = 1.00 A 1 200 Ω = 1.20 kV ∆VL = 1.21 kV . (c)

dI R = − I max e − Rt L dt L dI − L = ∆VL = I max R e − Rt L . dt

I = I max e − Rt L : and

b

12.0 V = 1 212 V e −1 212 t

so

9.90 × 10 −3 = e −606 t .

2.00

t = 7.62 ms .

L 0.140 = = 28.6 ms R 4.90 ε 6.00 V = 1.22 A I max = = R 4.90 Ω

τ=

(a)

e

I = I max 1 − e − t τ

e

P32.28

g

Solving Thus, P32.27

FIG. P32.26

−t τ

j

so

a

f

j

t = −τ ln 0.820 = 5.66 ms

= 0.820 :

e

e

0. 220 = 1.22 1 − e − t τ

j a

fe

j

(b)

I = I max 1 − e −10.0 0.028 6 = 1.22 A 1 − e −350 = 1.22 A

(c)

I = I max e − t τ

(a)

and

0.160 = 1.22 e − t τ

so

t = −τ ln 0.131 = 58.1 ms .

a

(c)

f

For a series connection, both inductors carry equal currents at every instant, so same for both. The voltage across the pair is dI dI dI = L1 + L 2 Leq so dt dt dt

(b)

FIG. P32.27

dI dI dI = L1 1 = L 2 2 = ∆VL dt dt dt ∆VL ∆VL ∆VL = + Thus, Leq L1 L2 Leq

where and

dI is the dt

Leq = L1 + L 2 . I = I 1 + I 2 and

dI dI 1 dI 2 = + . dt dt dt

1 1 1 = + . Leq L1 L 2

dI dI dI + Req I = L1 + IR1 + L 2 + IR 2 dt dt dt dI are separate quantities under our control, so functional equality requires Now I and dt both Leq = L1 + L 2 and Req = R1 + R 2 . Leq

continued on next page

Chapter 32

∆V = Leq

(d)

dI dI dI dI dI 1 dI 2 = + + Req I = L1 1 + R1 I 1 = L 2 2 + R 2 I 2 where I = I 1 + I 2 and . dt dt dt dt dt dt

We may choose to keep the currents constant in time. Then,

1 1 1 = + . Req R1 R 2

We may choose to make the current swing through 0. Then,

1 1 1 = + . Leq L1 L 2

This equivalent coil with resistance will be equivalent to the pair of real inductors for all other currents as well.

Section 32.3 P32.29

L=

P32.30

(a)

Energy in a Magnetic Field

e

j

−4 NΦ B 200 3.70 × 10 1 1 = = 42.3 mH so U = LI 2 = 0.423 H 1.75 A I 1.75 2 2

a

f

e

The magnetic energy stored in the field equals u times the volume of the solenoid (the volume in which B is non-zero).

j a0.260 mfπ b0.031 0 mg

a f LMN e

2

= 6.32 kJ

j OPQ 2

68.0 π 0.600 × 10 −2 N2A L = µ0 = µ0 = 8.21 µH A 0.080 0 1 1 2 U = LI 2 = 8. 21 × 10 −6 H 0.770 A = 2.44 µJ 2 2

ja

e

(a)

U=

(b)

I=

FG IJ H K

ε 1 2 1 LI = L 2 2 2R

FG ε IJ 1 − e b H RK

g

− RL t

R t = ln 2 L u =∈0

z

∞

*P32.34

= 0.064 8 J .

j

2

P32.33

2

2

e

P32.32

f

4.50 T B2 = = 8.06 × 10 6 J m3 . 2 µ 0 2 1.26 × 10 −6 T ⋅ m A

U = uV = 8.06 × 10 6 J m3

P32.31

fa

The magnetic energy density is given by

µ=

(b)

a

0

2

f

=

Lε 2 8R2 so so

E2 = 44.2 nJ m3 2

e −2 Rt L dt = −

z

a0.800fa500f = 27.8 J 8a30.0f ε FεI = G J 1− e b g 2R H R K 2

=

2

− RL t

t=

u=

FG H

IJ K

→e

b g

− RL t

=

1 2

L 0.800 ln 2 = ln 2 = 18.5 ms 30.0 R

B2 = 995 µJ m3 2µ 0

L ∞ −2 Rt L −2 Rdt L −2 Rt L e e =− L 2R 0 2R

∞ 0

=−

a f

L −∞ L L e − e0 = 0−1 = 2R 2R 2R

e

j

247

248 P32.35

Inductance

a

fa

1 2 1 LI = 4.00 H 0.500 A 2 2

(a)

U=

(b)

When the current is 1.00 A, Kirchhoff’s loop rule reads

f

2

U = 0.500 J

a

fa

f

+22.0 V − 1.00 A 5.00 Ω − ∆VL = 0 .

Then ∆VL = 17.0 V . The power being stored in the inductor is I∆VL = 1.00 A 17.0 V = 17.0 W .

a

a

P32.36

fa

P = I∆V = 0.500 A 22.0 V

(c)

f

fa

P = 11.0 W I=

From Equation 32.7,

ε

e1 − e j . − Rt L

R

ε = 2.00 A . R At that time, the inductor is fully energized and P = I ∆V = 2.00 A 10.0 V = 20.0 W .

(a)

The maximum current, after a long time t , is

I=

a f a

P32.37

a

f a5.00 Ωf = 2

(b)

Plost = I 2 R = 2.00 A

(c)

Pinductor = I ∆Vdrop = 0

(d)

U=

e j a10.0 Hfa2.00 Af =

LI 2 2

u =∈0

Therefore

∈0

E2 2

E2 B2 = 2 2µ 0

6.80 × 10 5 V m 3.00 × 10 8 m s

fa

f

20.0 W

2

2

We have

B = E ∈0 µ 0 =

P32.38

FIG. P32.35

f

= 20.0 J B2 . 2µ 0

and

u=

so

B 2 =∈0 µ 0 E 2

= 2.27 × 10 −3 T .

The total magnetic energy is the volume integral of the energy density, u = Because B changes with position, u is not constant. For B = B0

FG R IJ HrK

2

,

u=

B2 . 2µ 0

FB GH 2 µ

2 0 0

I FG R IJ JK H r K

4

.

Next, we set up an expression for the magnetic energy in a spherical shell of radius r and thickness dr. Such a shell has a volume 4π r 2 dr , so the energy stored in it is

j FGH 2π µB R IJK rdr .

e

2 0

dU = u 4π r 2 dr =

0

4

2

We integrate this expression for r = R to r = ∞ to obtain the total magnetic energy outside the sphere. This gives U=

2π B02 R 3

µ0

=

e

j e6.00 × 10 mj e1.26 × 10 T ⋅ m Aj

2π 5.00 × 10 −5 T

2

−6

6

3

= 2.70 × 10 18 J .

Chapter 32

Section 32.4 P32.39

249

Mutual Inductance

af

I1 t = I max e −αt sin ω t with I max = 5.00 A , α = 0.025 0 s −1 , and ω = 377 rad s dI 1 = I max e −α t −α sin ω t + ω cos ω t . dt

b

g

a fh

a fh

dI 1 = 5.00 A s e −0.020 0 − 0.025 0 sin 0.800 377 + 377 cos 0.800 377 dt

b

At t = 0.800 s ,

g

b

g c

c

dI 1 = 1.85 × 10 3 A s . dt Thus, ε 2 = −M

P32.40

ε 2 = −M

bε g

2 max

dI 1 : dt

M=

−ε 2 +3.20 V = = 1.73 mH . dI 1 dt 1.85 × 10 3 A s

dI 1 = − 1.00 × 10 −4 H 1.00 × 10 4 A s cos 1 000t dt

e

j b

je

g

= 1.00 V

ε2

M=

P32.42

Assume the long wire carries current I. Then the magnitude of the magnetic field it generates at µ I distance x from the wire is B = 0 , and this field passes perpendicularly through the plane of the 2π x loop. The flux through the loop is

dI 1 dt

z

=

96.0 mV = 80.0 mH 1.20 A s

P32.41

z

za

f

Φ B = B ⋅ dA = BdA = B Adx =

FG H

z

IJ K

µ 0 IA 1.70 mm dx µ 0 IA 1.70 = ln . 2π 0. 400 mm x 2π 0.400

The mutual inductance between the wire and the loop is then M=

FG H

a f e

IJ K

M = 7.81 × 10 −10 H = 781 pH

P32.43

je

ja f

1 4π × 10 −7 T ⋅ m A 2.70 × 10 −3 m N 2 Φ 12 N 2 µ 0 IA N µ A 1.70 = = 2 0 1.45 = ln 1.45 I1 2π I 0. 400 2π 2π

e

j

(a)

−6 N B Φ BA 700 90.0 × 10 = = 18.0 mH M= 3.50 IA

(b)

−6 Φ A 400 300 × 10 = = 34.3 mH LA = IA 3.50

(c)

ε B = −M

e

j

a

fb

dI A = − 18.0 mH 0.500 A s = −9.00 mV dt

g

250 *P32.44

Inductance

The large coil produces this field at the center of the small coil:

N 1 µ 0 I 1 R12

e

2 x 2 + R12

j

32

. The field is normal to

the area of the small coil and nearly uniform over this area, so it produces flux N 1 µ 0 I 1 R12

Φ 12 =

e

2

2x +

j

3 2 R12

π R 22 through the face area of the small coil. When current I 1 varies, this is the

emf induced in the small coil:

ε 2 = −N2

P32.45

2 2 N 1 N 2 πµ 0 R12 R 22 dI 1 dI 1 N 1 N 2 πµ 0 R12 R 22 d N 1 µ 0 R1 π R 2 I = − = − M = so M . 1 3 2 32 dt 2 x 2 + R 2 3 2 dt dt 2 x 2 + R12 2 x 2 + R12 1

e

j

e

e

j

With I = I 1 + I 2 , the voltage across the pair is: ∆V = − L1

dI 1 dI dI dI dI − M 2 = − L 2 2 − M 1 = − Leq . dt dt dt dt dt dI 1 ∆V M dI 2 = + dt L1 L1 dt

−

So,

−L 2

and

a f

(a)

1 2

+ M2

j dIdt

2

b

FIG. P32.45

g

[1]

g

[2]

= ∆V L1 − M .

dI 2 ∆V M dI 1 = + dt L 2 L 2 dt

By substitution,

−

leads to

e− L L

+ M2

j dIdt = ∆V bL

2

Adding [1] to [2],

e− L L

+ M2

j dIdt = ∆V bL

+ L2 − 2 M .

So,

Leq = −

L1 L 2 − M 2 ∆V = . dI dt L1 + L 2 − 2 M

Section 32.5

1 2

1 2

1

1

−M .

g

Oscillations in an LC Circuit

b g

At different times, U C

I max =

a f

C ∆V L

max

(b)

2

dI 2 M ∆V M dI 2 + + = ∆V dt L1 L1 dt

e− L L

P32.46

j

=

max

b g

= UL

max

so

LM 1 Ca∆V f OP N2 Q 2

= max

1.00 × 10 −6 F 40.0 V = 0.400 A . 10.0 × 10 −3 H

a

f

FG 1 LI IJ H2 K 2

max

Chapter 32

P32.47 P32.48

LM 1 Ca∆V f OP N2 Q 2

= max

FG 1 LI IJ H2 K 2

max

b g

so ∆VC

max

=

251

L 20.0 × 10 −3 H I max = 0.100 A = 20.0 V C 0.500 × 10 −6 F

a

f

When the switch has been closed for a long time, battery, resistor, and coil carry constant current I max =

ε

. When the switch is opened, R current in battery and resistor drops to zero, but the coil carries this same current for a moment as oscillations begin in the LC loop.

We interpret the problem to mean that the voltage amplitude of these 1 1 2 2 . oscillations is ∆V , in C ∆V = LI max 2 2

a f e0.500 × 10 Fja150 Vf a250 Ωf C a ∆V f C a ∆V f R = = Then, L = ε I a50.0 Vf 2

2

2

2 max

P32.49

2

P32.51

P32.52

2

= 0.281 H .

This radio is a radiotelephone on a ship, according to frequency assignments made by international treaties, laws, and decisions of the National Telecommunications and Information Administration. 1 The resonance frequency is f0 = . 2π LC 1 1 Thus, C= = = 608 pF . 2 2 −6 6 2π f 0 L 2π 6.30 × 10 Hz 1.05 × 10 H

b

P32.50

2

−6

2

FIG. P32.50

f=

1 2π LC

L=

:

1

(a)

f=

(b)

Q = Qmax

(c)

I=

(a)

f=

(b) (c)

1

b 2π f g C 2

=

e

j e

1

a f e8.00 × 10 j

2π 120

2

1

−6

−6

2π

j

= 0.220 H

= 135 Hz

b0.082 0 Hge17.0 × 10 Fj cos ω t = b180 µCg cosb847 × 0.001 00g =

2π LC

a fa f a

119 µC

f

dQ = −ω Qmax sin ω t = − 847 180 sin 0.847 = −114 mA dt 1

=

1

a0.100 Hfe1.00 × 10 Fj Q = Cε = e1.00 × 10 Fja12.0 V f = 12.0 µC 2π LC

−6

2π

= 503 Hz

−6

1 2 1 2 Cε = LI max 2 2 I max = ε

(d)

=

g

FIG. P32.52

C 1.00 × 10 −6 F = 12 V = 37.9 mA L 0.100 H

At all times

U=

ja

1 2 1 Cε = 1.00 × 10 −6 F 12.0 V 2 2

e

f

2

= 72.0 µJ .

252 P32.53

Inductance

ω=

1 LC

1

=

a3.30 Hfe840 × 10

Q = Q max cos ω t , I = 2

e 105 × 10 =

e

−6

je

cos 1.899 × 10 4 rad s 2.00 × 10 −3 s

UC =

Q 2C

(b)

UL =

2 Qmax sin 2 ω t 1 2 1 2 LI = Lω 2 Qmax sin 2 ω t = 2 2 2C

(c)

Section 32.6

(a)

e105 × 10 Cj =

j

2

e 2e840 × 10

jj

2

= 6.03 J

b g

je

sin 2 1.899 × 10 4 rad s 2.00 × 10 −3 s −12

F

j

j=

0.529 J

U total = UC + U L = 6.56 J

The RLC Circuit

ωd =

(b)

Rc =

(a)

ω0 =

(b)

ωd =

(c)

e

2 840 × 10 −12

b g

FG IJ H K

1 R − 2L LC

∆ω

ω0

2

fd =

Therefore,

P32.56

j

(a)

UL

P32.55

F

= 1.899 × 10 4 rad s

dQ = −ω Qmax sin ω t dt

−6

P32.54

−12

=

e

1

je

2.20 × 10 −3 1.80 × 10 −6

F 7.60 −G j GH 2e2.20 × 10

−3

I J j JK

2

= 1.58 × 10 4 rad s

ωd = 2.51 kHz . 2π

4L = 69.9 Ω C 1 LC

=

1

a0.500fe0.100 × 10 j −6

FG IJ H K

1 R − 2L LC

2

= 4.47 krad s

= 4.36 krad s

= 2.53% lower

Choose to call positive current clockwise in Figure 32.21. It drains charge from the capacitor dQ . A clockwise trip around the circuit then gives according to I = − dt dI Q + − IR − L = 0 C dt Q dQ d dQ R+L + + = 0 , identical with Equation 32.28. C dt dt dt

Chapter 32

*P32.57

The period of damped oscillation is T = capacitor is Q = Qmax e − RT

2L

= Qmax e −2π R

after one oscillation it is U = U 0 e e 2π R

Lω d

=

1 0.99

b

Lω d =

2 Lω d

. The energy is proportional to the charge squared, so

= 0.99U 0 . Then

g

F GH

R2 2π 2 Ω 1 = 1 250 Ω = L − 2 LC 4L 0.001 005

1.563 × 10 6 Ω 2 =

2π . After one oscillation the charge returning to the ωd

− 2π R Lω d

2π 2 Ω = ln 1.010 1 = 0.001 005 Lω d

a f

2Ω L − C 4

I JK

12

2

L = 1.563 × 10 6 Ω 2 C We are also given

1

ω = 2π × 10 3 s = LC =

LC

1

e2π × 10 sj 3

2

= 2.533 × 10 −8 s 2

Solving simultaneously, C = 2.533 × 10 −8 s 2 L L2 = 1.563 × 10 6 Ω 2 L = 0.199 H 2.533 × 10 −8 s 2 2.533 × 10 −8 s 2 C= = 127 nF = C 0.199 H P32.58

(a)

Q = Qmax e − Rt 2 L cos ω d t

so

I max ∝ e − Rt 2 L

0.500 = e − Rt 2 L

and

Rt = − ln 0.500 2L

so

Q = 0.500Qmax = 0.707Qmax

t=−

(b)

a

f

t=−

a

f

FG IJ H K

2L 2L ln 0.500 = 0.693 R R

2 U 0 ∝ Qmax and U = 0.500U 0

a

f

FG IJ H K

2L 2L ln 0.707 = 0.347 R R

253

(half as long)

254

Inductance

Additional Problems *P32.59

(a)

Let Q represent the magnitude of the opposite charges on the plates of a parallel plate capacitor, the two plates having area A and separation d. The negative plate creates electric Q Q2 toward itself. It exerts on the positive plate force F = field E = toward the 2 ∈0 A 2 ∈0 A Q . The energy density is negative plate. The total field between the plates is ∈0 A uE =

Q2 Q2 1 1 ∈0 E 2 = ∈0 2 2 = . Modeling this as a negative or inward pressure, we 2 2 ∈0 A 2 ∈0 A 2

have for the force on one plate F = PA = (b)

Q2 , in agreement with our first analysis. 2 ∈0 A 2

The lower of the two current sheets shown creates µ J above it magnetic field B = 0 s − k . Let A and w 2 represent the length and width of each sheet. The upper sheet carries current J s w and feels force

e j

Js Js

y x

µ J µ wAJ s2  F = IA × B = J s wA 0 s i × − k = 0 j. 2 2

e j

The force per area is P = (c)

z

µ 0 J s2 F = . Aw 2

FIG. P32.59(b)

µ 0 Js  µ 0 Js  −k + − k = µ 0 J s k , with 2 2 . Outside the space they enclose, the fields of the separate sheets are

Between the two sheets the total magnetic field is magnitude B = µ 0 J s

e j

e j

in opposite directions and add to zero .

P32.60

1

B2 =

µ 02 J s2 µ 0 J s2 = 2µ 0 2

(d)

uB =

(e)

This energy density agrees with the magnetic pressure found in part (b).

2µ 0

With Q = Qmax at t = 0 , the charge on the capacitor at any time is Q = Q max cos ω t where ω = The energy stored in the capacitor at time t is then 2 Q 2 Qmax cos 2 ω t = U 0 cos 2 ω t . = 2C 2C 1 1 cos ω t = and When U = U 0 , 4 2

U=

Therefore, The inductance is then:

t LC

=

π 3

or

1 ω t = π rad . 3 t2 π 2 = . LC 9 L=

9t 2 . π 2C

1 LC

.

Chapter 32

P32.61

a

ε L = −L

(b)

Q = Idt =

(c)

z za t

t

0

0

f

20.0t dt = 10.0t 2

−Q −10.0t 2 = = − 10.0 MV s 2 t 2 C 1.00 × 10 −6 F

e

e e jt

j

j

2

−10.0t 2 Q2 1 2 1 2 ≥ LI , or When ≥ 1.00 × 10 −3 20.0t , −6 2C 2 2 2 1.00 × 10

e

then 100 t 4 ≥ 400 × 10 −9 P32.62

f

d 20.0t dI = − 1.00 mH = −20.0 mV dt dt

(a)

∆VC =

fa

(a)

ε L = −L

(b)

I=

2

a f

dI d = −L Kt = − LK dt dt

dQ , dt

z z t

t

0

0

Q = Idt = Ktdt =

so

∆VC =

When

f

. The earliest time this is true is at t = 4.00 × 10 −9 s = 63.2 µs .

b g

1 C ∆VC 2

2

=

1 2 Kt 2

−Q Kt 2 = − C 2C

F GH

I JK

1 K 2t 4 1 = L K 2t2 C 2 2 2 4C

1 2 LI , 2

e

j

t = 2 LC

Thus

P32.63

ja

j e

and

(c)

FG IJ H K

1 Q2 1 Q = 2 C 2C 2

2

+

1 2 LI 2

so

The flux through each turn of the coil is

I=

3Q 2 . 4CL

ΦB =

LI Q = N 2N

3L C

where N is the number of turns. P32.64

B=

(a)

255

µ 0 NI 2π r

z

Φ B = BdA = L=

z b

a

FG IJ HK

z

µ 0 NI µ NIh b dr µ 0 NIh b = hdr = 0 ln a 2π r 2π a r 2π

FG IJ HK

NΦ B µ0N2h b ln = I a 2π

a f b0.010 0g lnFG 12.0 IJ = 91.2 µH H 10.0 K 2π µ N F A I µ a500 f F 2.00 × 10 = G J = 2π GH 0.110 2π H R K

µ 0 500

(b)

L=

(c)

Lappx

FIG. P32.64

0

2

2

0

2

−4

m2

I= JK

90.9 µH , only 0.3% different.

256 P32.65

Inductance

(a)

B=

So the coil creates flux through itself

Φ B = BA cos θ =

When the current it carries changes,

ε L = −N

a

f

2π r = 3 0.3 m

e

2

2 R +0

L≈

so (b)

Nµ 0 IR 2

At the center,

j

2 3 2

=

Nµ 0 I . 2R

Nµ 0 I π π R 2 cos 0° = Nµ 0 IR . 2R 2

FG IJ H K

dΦ B π dI dI ≈ −N Nµ 0 R = − L 2 dt dt dt

π 2 N µ0R . 2

r ≈ 0.14 m π L ≈ 1 2 4π × 10 −7 T ⋅ m A 0.14 m = 2.8 × 10 −7 H 2 L ~ 100 nH

so

ja

e je

(c) P32.66

(a)

−7 L 2.8 × 10 V ⋅ s A = = 1.0 × 10 −9 s R 270 V A

L ~ 1 ns R

If unrolled, the wire forms the diagonal of a 0.100 m (10.0 cm) rectangle as shown. The length of this rectangle is L′ =

a9.80 mf − a0.100 mf 2

2

f

9.80 m

0.100 m

L′

.

FIG. P32.66(a) The mean circumference of each turn is C = 2π r ′ , where r ′ = radius of each turn. The number of turns is then:

a a

f a f

f

2

2

N=

9.80 m − 0.100 m L′ = = 127 . C 2π 24.0 + 0.644 2 × 10 −3 m

(b)

R=

−8 ρA 1.70 × 10 Ω ⋅ m 10.0 m = = 0.522 Ω 2 A π 0.322 × 10 −3 m

(c)

L=

µN 2 A 800 µ 0 L ′ = A′ A′ C

L=

ja

e

e

e

800 4π × 10 −7 0.100 m

24.0 + 0.644 mm is the mean 2

f

j

FG IJ π ar ′f H K j FG a9.80 mf − a0.100 mf IJ GH π a24.0 + 0.644f × 10 m JK 2

L = 7.68 × 10 −2 H = 76.8 mH

2

2

2

−3

2

π

LMFG 24.0 + 0.644 IJ × 10 NH 2 K

−3

m

OP Q

2

Chapter 32

P32.67

257

From Ampere’s law, the magnetic field at distance r ≤ R is found as:

b g

e j

B 2π r = µ 0 J π r 2 = µ 0

F I GH π R

2

I eπ r j , or B = µ Ir JK 2π R 2

0

2

.

The magnetic energy per unit length within the wire is then

z

F I GH JK

z

R µ I2 R µ I 2 R4 µ0I2 U B2 = = . 2π rdr = 0 4 r 3 dr = 0 4 A 0 2µ 0 4 16π 4π R 0 4π R

b

g

This is independent of the radius of the wire. P32.68

The primary circuit (containing the battery and solenoid) is an RL circuit with R = 14.0 Ω , and

jb

e

ge

j

2

−7 12 500 1.00 × 10 −4 µ N 2 A 4π × 10 L= 0 = = 0.280 H . A 0.070 0

(a)

The time for the current to reach 63.2% of the maximum value is the time constant of the circuit:

τ=

(b)

L 0.280 H = = 0.020 0 s = 20.0 ms . R 14.0 Ω

The solenoid’s average back emf is where

f

Thus, (c)

FIG. P32.68 FG ∆I IJ = LFG I − 0 IJ H ∆t K H ∆t K F ∆V IJ = 0.632FG 60.0 V IJ = 2.71 A . = 0.632G I = 0.632 I HRK H 14.0 Ω K F 2.71 A I = 37.9 V . ε = a0.280 H fG H 0.020 0 s JK f

εL = L

max

L

The average rate of change of flux through each turn of the overwrapped concentric coil is the same as that through a turn on the solenoid:

jb

a f e

ga

fe

4π × 10 −7 T ⋅ m A 12 500 0.070 0 m 2.71 A 1.00 × 10 −4 m 2 ∆Φ B µ 0 n ∆I A = = 0.020 0 s ∆t ∆t

j

= 3.04 mV (d)

The magnitude of the average induced emf in the coil is ε L = N

FG ∆Φ IJ and magnitude of H ∆t K

the average induced current is I=

FG H

IJ K

ε L N ∆Φ B 820 = = 3.04 × 10 −3 V = 0.104 A = 104 mA . R R ∆t 24.0 Ω

e

j

B

258 P32.69

Inductance

b g ε − b I + I gR

ε − I + I 2 R1 − I 2 R2 = 0 .

Left-hand loop: Outside loop:

2

1

−L

dI =0. dt

dI =0. dt This is of the same form as Equation 32.6, so its solution is of the same form as Equation 32.7: ε′ It = 1 − e − R ′t L . R′ Eliminating I 2 gives

ε ′ − IR ′ − L

af

But R ′ =

b b

af

It =

g g j.

ε 1 − e − R ′t L R1

e

When switch is closed, steady current I 0 = 1.20 A . When the switch is opened after being closed a long time, the current in the right loop is

I = I 0 e − R2 t so

e Rt L =

Therefore, P32.71

j

ε ′ ε R 2 R1 + R 2 ε = . = R ′ R1 R 2 R1 + R 2 R1

R1 R 2 R2ε and ε ′ = , so R1 + R 2 R1 + R 2

Thus P32.70

e

FIG. P32.69

(a)

I0 I

L

a

FG IJ H K

I Rt = ln 0 . L I

and

fa

f

1.00 Ω 0.150 s R2 t = = 0.095 6 H = 95.6 mH . L= ln I 0 I ln 1.20 A 0.250 A

b g

b

FIG. P32.70

g

While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of the circuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outer loop of the circuit. Applying Kirchhoff’s loop rule to this loop gives:

a

f

e

j

+ ε 0 − 2.00 + 6.00 × 10 3 Ω 9.00 × 10 −3 A = 0 + ε 0 = 72.0 V with end b at the higher potential (b)

FIG. P32.71(b) (c)

After the switch is opened, the current around the outer loop decays as

I = I max e − Rt L with I max = 9.00 mA , R = 8.00 kΩ , and L = 0.400 H . Thus, when the current has reached a value I = 2.00 mA , the elapsed time is: t=

FG L IJ lnFG I IJ = FG 0.400 H IJ lnFG 9.00 IJ = 7.52 × 10 H R K H I K H 8.00 × 10 Ω K H 2.00 K max

3

−5

s = 75.2 µs .

Chapter 32

P32.72

(a)

IL = 0 ∆V = ε 0 + – L

The instant after the switch is closed, the situation is as shown in the circuit diagram of Figure (a). The requested quantities are: I L = 0 , IC =

ε0 ε , IR = 0 R R

∆VR = ε 0

∆VC= 0

∆VL = ε 0 , ∆VC = 0 , ∆VR = ε 0 (b)

I R = ε 0 /R

Q=0

+ε – 0

I C = ε 0 /R

After the switch has been closed a long time, the steady-state conditions shown in Figure (b) will exist. The currents and voltages are:

Figure (a) IL = 0 ∆V = 0 + – L

I L = 0 , IC = 0 , I R = 0

IR = 0

Q = Cε 0

∆VL = 0 , ∆VC = ε 0 , ∆VR = 0

∆VR = 0

∆VC= ε 0

+ε – 0

Figure (b) FIG. P32.72 P32.73

When the switch is closed, as shown in figure (a), the current in the inductor is I: 12.0 − 7.50 I − 10.0 = 0 → I = 0.267 A . When the switch is opened, the initial current in the inductor remains at 0.267 A.

a0.267 AfR ≤ 80.0 V

IR = ∆V :

R ≤ 300 Ω

(a)

(b) FIG. P32.73

P32.74

jb

e

ge

j

2

(a)

−7 −4 2 µ N 2 A 4π × 10 T ⋅ m A 1 000 1.00 × 10 m L1 = 0 1 = = 2.51 × 10 −4 H = 251 µH A1 0.500 m

(b)

M= M=

(c)

b

e4π × 10

−7

jb

ga fe

T ⋅ m A 1 000 100 1.00 × 10 −4 m 2 0.500 m

ε 1 = −M

j = 2.51 × 10

dI 2 dI dQ1 M dI 2 , or I 1 R1 = − M 2 and I 1 = =− dt dt dt R1 dt

M Q1 = − R1 Q1 =

g

N 2 Φ 2 N 2 Φ 1 N 2 BA N 2 µ 0 N 1 A 1 I 1 A µ 0 N 1 N 2 A = = = = I1 I1 I1 I1 A1

z

tf

dI 2 = −

0

e2.51 × 10

−5

MI 2 i M M I 2 f − I 2i = − 0 − I 2i = R1 R1 R1

d

ja

H 1.00 A

1 000Ω

i

f = 2.51 × 10

b

−8

g

C = 25.1 nC

−5

H = 25.1 µH

259

260 P32.75

Inductance

It has a magnetic field, and it stores energy, so L =

(b)

Every field line goes through the rectangle between the conductors.

(c)

Φ = LI so

L=

Φ 1 = I I

z

I2

is non-zero.

w− a

BdA

y=a

F µ I + µ I I = 2 µ Ix dy = 2µ x ln y GH 2π y 2π bw − yg JK I z 2π y 2π µ x F w − aI L= lnG H a JK . π L=

1 I

z

w−a

xdy

0

0

0

0

a

w−a

. a

0

Thus P32.76

2U

(a)

For an RL circuit,

af

I t = I max e

b g

− RL t

af

It R − RL t = 1 − 10 −9 = e b g ≅ 1 − t I max L

:

e3.14 × 10 je10 j = = b2.50 yrge3.16 × 10 s yrj −8

R t = 10 −9 L

so

Rmax

−9

7

3.97 × 10 −25 Ω .

(If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm 2 , its resistance would be at least 10 −6 Ω ). P32.77

a

fe

1 2 1 LI = 50.0 H 50.0 × 10 3 A 2 2

j

2

= 6.25 × 10 10 J

(a)

UB =

(b)

Two adjacent turns are parallel wires carrying current in the same direction. Since the loops have such large radius, a one-meter section can be regarded as straight. B=

Then one wire creates a field of

µ0I . 2π r

This causes a force on the next wire of F = IAB sin θ F = IA

giving

Evaluating the force,

e

F = 4π × 10

µ0I µ AI 2 sin 90° = 0 . 2π r 2π r

a1.00 mf 50.0 × 10 A j 2π ea0.250 mf j 3

−7

N A

2

2

= 2 000 N .

Chapter 32

P32.78

P 1.00 × 10 9 W = = 5.00 × 10 3 A ∆V 200 × 10 3 V

P = I∆V

I=

From Ampere’s law,

B 2π r = µ 0 I enclosed or B =

(a)

261

b g

µ 0 I enclosed . 2π r

I enclosed = 5.00 × 10 3 A

At r = a = 0.020 0 m ,

FIG. P32.73

(b)

(c)

−7

je

T ⋅ m A 5.00 × 10 3 A

B=

At r = b = 0.050 0 m ,

I enclosed = I = 5.00 × 10 3 A

and

e4π × 10 B=

z

U = udV =

z b

a

e4π × 10 U= (d)

e4π × 10

and

−7

b

−7

2

0

2µ 0

b

2π 0.050 0 m 2

4π

je

T ⋅ m A 5.00 × 10 3 A 4π

je

T ⋅ m A 5.00 × 10 3 A

a f b2π rAdr g = µ I A

Br

g

2π 0.020 0 m

g

j = 0.050 0 T =

50.0 mT .

j = 0.020 0 T =

20.0 mT .

FG IJ HK

z b

b dr µ 0 I 2 A = ln a r 4π a

j e1 000 × 10 mj lnFG 5.00 cmIJ = 2.29 × 10 H 2.00 cm K 2

3

6

J = 2.29 MJ

The magnetic field created by the inner conductor exerts a force of repulsion on the current in the outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a small rectangular section of the outer cylinder of length A and width w.

F w I e5.00 × 10 AjGH 2π b0.050 0 mg JK 3

It carries a current of and experiences an outward force

e5.00 × 10 Ajw Ae20.0 × 10 Tj sin 90.0° . 2π b0.050 0 mg F F e5.00 × 10 A je 20.0 × 10 Tj = = The pressure on it is P = = A wA 2π b0.050 0 mg 3

F = IAB sin θ =

−3

3

−3

318 Pa .

262 P32.79

Inductance

jb

e

f

ga

4π × 10 −7 T ⋅ m A 1 400 2.00 A µ 0 NI = = 2.93 × 10 −3 T upward 1.20 m A

(a)

B=

(b)

2.93 × 10 −3 T B2 u= = = 3.42 J m3 2 µ 0 2 4π × 10 −7 T ⋅ m A

e

e

j

2

j e

b

jFGH 1 N1 J⋅ m IJK = 3.42 N m

2

g

= 3.42 Pa

(c)

To produce a downward magnetic field, the surface of the superconductor must carry a clockwise current.

(d)

The vertical component of the field of the solenoid exerts an inward force on the superconductor. The total horizontal force is zero. Over the top end of the solenoid, its field diverges and has a radially outward horizontal component. This component exerts upward force on the clockwise superconductor current. The total force on the core is upward . You can think of it as a force of repulsion between the solenoid with its north end pointing up, and the core, with its north end pointing down.

(e)

a

fLMN e

j OPQ =

F = PA = 3.42 Pa π 1.10 × 10 −2 m

2

1.30 × 10 −3 N

Note that we have not proven that energy density is pressure. In fact, it is not in some cases; Equation 21.2 shows that the pressure is two-thirds of the translational energy density in an ideal gas.

ANSWERS TO EVEN PROBLEMS P32.2

1.36 µH

P32.4

240 nWb

P32.6

19.2 µWb

P32.8

(a) 360 mV ; (b) 180 mV ; (c) t = 3.00 s

P32.10

(a) 15.8 µH ; (b) 12.6 mH

P32.12

P32.26

(a) 1.00 A ; (b) ∆V12 = 12.0 V , ∆V1 200 = 1.20 kV , ∆VL = 1.21 kV ; (c) 7.62 ms

P32.28

(a), (b), and (c) see the solution; (d) yes; see the solution

P32.30

(a) 8.06 MJ m3 ; (b) 6.32 kJ

P32.32

(a) 27.8 J ; (b) 18.5 ms

see the solution

P32.34

see the solution

P32.14

1.92 Ω

P32.36

(a) 20.0 W ; (b) 20.0 W ; (c) 0;(d) 20.0 J

P32.16

see the solution

P32.18

92.8 V

P32.20

30.0 mH

P32.22

7.67 mH

P32.24

(a) 1.00 kΩ; (b) 3.00 ms

P32.38

2π B02 R 3

µ0

P32.40

1.00 V

P32.42

781 pH

P32.44

M=

= 2.70 × 10 18 J

N1 N 2 πµ 0 R12 R 22

e

2 x 2 + R12

j

32

Chapter 32

P32.46

400 mA

P32.48

281 mH

P32.50

220 mH

P32.52

(a) 503 Hz ; (b) 12.0 µC ; (c) 37.9 mA ; (d) 72.0 µJ

P32.54

(a) 2.51 kHz; (b) 69.9 Ω

P32.56

see the solution

P32.58

(a) 0.693

P32.60

P32.62

P32.64

(a) see the solution; (b) 91.2 µH ; (c) 90.9 µH , 0.3% smaller

P32.66

(a) 127; (b) 0.522 Ω; (c) 76.8 mH

P32.68

(a) 20.0 ms; (b) 37.9 V; (c) 3.04 mV; (d) 104 mA

P32.70

95.6 mH

P32.72

(a) I L = 0 , I C =

P32.74

(a) 251 µH ; (b) 25.1 µH ; (c) 25.1 nC

P32.76

3.97 × 10 −25 Ω

P32.78

(a) 50.0 mT; (b) 20.0 mT; (c) 2.29 MJ; (d) 318 Pa

FG 2L IJ ; (b) 0.347FG 2L IJ H RK H RK

9t 2

ε0 ε , IR = 0 , R R ∆VL = ε 0 , ∆VC = 0 , ∆VR = ε 0 ; (b) I L = 0 , I C = 0 , I R = 0 , ∆VL = 0 , ∆VC = ε 0 , ∆VR = 0

2

π C (a) ε L = − LK ; (b) ∆Vc = (c) t = 2 LC

− Kt 2 ; 2C

263

33 Alternating Current Circuits CHAPTER OUTLINE 33.1 33.2 33.3 33.4 33.5 33.6 33.7 33.8 33.9

AC Sources Resistors in an AC Circuit Inductors in an AC Circuit Capacitors in an AC Circuit The RLC Series Circuit Power in an AC Circuit Resonance in a Series RLC Circuit The Transformer and Power Transmission Rectifiers and Filters

ANSWERS TO QUESTIONS Q33.1

If the current is positive half the time and negative half the time, the average current can be zero. The rms current is not zero. By squaring all of the values of the current, they all become positive. The average (mean) of these positive values is also positive, as is the square root of the average.

Q33.2

∆Vavg =

Q33.3

AC ammeters and voltmeters read rms values. With an oscilloscope you can read a maximum voltage, or test whether the average is zero.

Q33.4

Suppose the voltage across an inductor varies sinusoidally. Then the current in the inductor will have its instantaneous 1 peak positive value cycle after the voltage peaks. The voltage 4 1 is zero and going positive cycle (90°) before the current is 4 zero and going positive.

∆Vmax ∆Vmax , ∆Vrms = 2 2

Q33.5

If it is run directly from the electric line, a fluorescent light tube can dim considerably twice in every cycle of the AC current that drives it. Looking at one sinusoidal cycle, the voltage passes through zero twice. We don’t notice the flickering due to a phenomenon called retinal imaging. We do not notice that the lights turn on and off since our retinas continue to send information to our brains after the light has turned off. For example, most TV screens refresh at between 60 to 75 times per second, yet we do not see the evening news flickering. Home video cameras record information at frequencies as low as 30 frames per second, yet we still see them as continuous action. A vivid display of retinal imaging is that persistent purple spot you see after someone has taken a picture of you with a flash camera.

Q33.6

The capacitive reactance is proportional to the inverse of the frequency. At higher and higher frequencies, the capacitive reactance approaches zero, making a capacitor behave like a wire. As the frequency goes to zero, the capacitive reactance approaches infinity—the resistance of an open circuit.

Q33.7

The second letter in each word stands for the circuit element. For an inductor L, the emf ε leads the current I—thus ELI. For a capacitor C, the current leads the voltage across the device. In a circuit in which the capacitive reactance is larger than the inductive reactance, the current leads the source emf—thus ICE. 265

266

Alternating Current Circuits

Q33.8

The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagrams throughout the chapter. Kirchhoff’s loop rule is true at any instant, but the voltages across different circuit elements are not simultaneously at their maximum values. Do not forget that an inductor can induce an emf in itself and that the voltage across it is 90° ahead of the current in the circuit in phase.

Q33.9

In an RLC series circuit, the phase angle depends on the source frequency. At very low frequency the capacitor dominates the impedance and the phase angle is near –90°. The phase angle is zero at the resonance frequency, where the inductive and capacitive reactances are equal. At very high frequencies φ approaches +90° .

Q33.10

−90° ≤ φ ≤ 90° . The extremes are reached when there is no significant resistance in the circuit.

Q33.11

The resistance remains unchanged, the inductive resistance doubles, and the capacitive reactance is reduced by one half.

Q33.12

The power factor, as seen in equation 33.29, is the cosine of the phase angle between the current and applied voltage. Maximum power will be delivered if ∆V and I are in phase. If ∆V and I are 90° out of phase, the source voltage drives a net current of zero in each cycle and the average power is zero.

Q33.13

The person is doing work at a rate of P = Fv cos θ . One can consider the emf as the “force” that moves the charges through the circuit, and the current as the “speed” of the moving charges. The cos θ factor measures the effectiveness of the cause in producing the effect. Theta is an angle in real space for the vacuum cleaner and phi is the analogous angle of phase difference between the emf and the current in the circuit.

Q33.14

As mentioned in Question 33.5, lights that are powered by alternating current flicker or get slightly brighter and dimmer at twice the frequency of the AC power source. Even if you tried using two banks of lights, one driven by AC 180° of phase from the other, you would not have a stable light source, but one that exhibits a “ripple” in intensity.

Q33.15

In 1881, an assassin shot President James Garfield. The bullet was lost in his body. Alexander Graham Bell invented the metal detector in an effort to save the President’s life. The coil is preserved in the Smithsonian Institution. The detector was thrown off by metal springs in Garfield’s mattress, a new invention itself. Surgeons went hunting for the bullet in the wrong place and Garfield died.

Q33.16

As seen in Example 33.8, it is far more economical to transmit at high voltage than at low voltage, as the I 2 R loss on the transmission line is significantly lower. Transmitting power at high voltage permits the use of step-down transformers to make “low” voltages and high currents available to the end user.

Q33.17

Insulation and safety limit the voltage of a transmission line. For an underground cable, the thickness and dielectric strength of the insulation between the conductors determines the maximum voltage that can be applied, just as with a capacitor. For an overhead line on towers, the designer must consider electrical breakdown of the surrounding air, possible accidents, sparking across the insulating supports, ozone production, and inducing voltages in cars, fences, and the roof gutters of nearby houses. Nuisance effects include noise, electrical noise, and a prankster lighting a hand-held fluorescent tube under the line.

Q33.18

No. A voltage is only induced in the secondary coil if the flux through the core changes in time.

Chapter 33

Q33.19

267

This person needs to consider the difference between the power delivered by a power plant and I 2 R losses in transmission lines. At lower voltages, transmission lines must carry higher currents to transmit the same power, as seen in Example 33.8. The high transmitted current at low voltage actually results in more internal energy production than a lower current at high voltage. In his 2 ∆V , the ∆V does not represent the line voltage but the potential difference between the formula R ends of one conductor. This is very small when the current is small.

a f

Q33.20

The Q factor determines the selectivity of the radio receiver. For example, a receiver with a very low Q factor will respond to a wide range of frequencies and might pick up several adjacent radio stations at the same time. To discriminate between 102.5 MHz and 102.7 MHz requires a high-Q circuit. Typically, lowering the resistance in the circuit is the way to get a higher quality resonance.

Q33.21

Both coils are wrapped around the same core so that nearly all of the magnetic flux created by the primary passes through the secondary coil, and thus induces current in the secondary when the current in the primary changes.

Q33.22

The frequency of a DC signal is zero, making the capacitive reactance at DC infinite. The capacitor then acts as an open switch. An AC signal has a non-zero frequency, and thus the capacitive reactance is finite, allowing a signal to pass from Circuit A to Circuit B.

SOLUTIONS TO PROBLEMS Section 33.1

AC Sources

Section 33.2

Resistors in an AC Circuit

af

b g

b g

a f a283 V f sina628tf

P33.1

∆v t = ∆Vmax sin ω t = 2 ∆Vrms sin ω t = 200 2 sin 2π 100 t =

P33.2

∆Vrms =

P33.3

170 V 2

(a)

P=

(b)

R=

= 120 V

b∆V g rms

2

R

a120 Vf 100 W

→R=

a120 Vf

2

75.0 W

= 193 Ω

2

= 144 Ω

Each meter reads the rms value. ∆Vrms = I rms =

100 V 2

= 70.7 V

∆Vrms 70.7 V = = 2.95 A R 24.0 Ω FIG. P33.3

268 P33.4

Alternating Current Circuits

∆v R = ∆Vmax sin ω t

(a)

b

a

g

f

∆v R = 0.250 ∆Vmax , so sin ω t = 0. 250 , or ω t = sin −1 0.250 . The smallest angle for which this is true is ω t = 0.253 rad Thus, if t = 0.010 0 s ,

ω=

0.253 rad = 25.3 rad s . 0.010 0 s

b

a

t= P33.5

2.89 rad = 0.114 s . 25.3 rad s

i R = I max sin ω t

b

b0.007 00gω = sin a0.600f = 0.644 −1

b

f = 14.6 Hz .

so

g

P = I rms ∆Vrms and ∆Vrms = 120 V for each bulb (parallel circuit), so: I1 = I 2 = I3 =

P1 ∆Vrms 120 V 150 W = = 1.25 A , and R1 = = = 96.0 Ω = R 2 ∆Vrms 120 V 1.25 A I1

P3 ∆Vrms 100 W 120 V = = 0.833 A , and R3 = = = 144 Ω . ∆Vrms 120 V 0.833 A I3

∆Vmax = 15.0 V and Rtotal = 8. 20 Ω + 10.4 Ω = 18.6 Ω I max =

∆Vmax 15.0 V = = 0.806 A = 2 I rms Rtotal 18.6 Ω

2 Pspeaker = I rms Rspeaker =

Section 33.3 P33.8

g

0.600 = sin ω 0.007 00 .

becomes

and ω = 91.9 rad s = 2π f

P33.7

f

f

Thus,

P33.6

a

g

The second time when ∆v R = 0.250 ∆Vmax , ω t = sin −1 0.250 again. For this occurrence, ω t = π − 0.253 rad = 2.89 rad (to understand why this is true, recall the identity sin π − θ = sin θ from trigonometry). Thus,

(b)

FG 0.806 A IJ a10.4 Ωf = H 2 K 2

3.38 W

Inductors in an AC Circuit

For I max = 80.0 mA , I rms =

bX g

L min

=

80.0 mA 2

= 56.6 mA

Vrms 50.0 V = = 884 Ω I rms 0.056 6 A

X L = 2π fL → L =

XL 884 Ω ≥ ≥ 7.03 H 2π f 2π 20.0

a f

Chapter 33

P33.9

XL =

(a)

L=

ω=

P33.10

ω

=

13.3 = 0.042 4 H = 42.4 mH 2π 50.0

a f

∆Vmax 100 = = 40.0 Ω 2.50 I max

XL 40.0 = = 942 rad s L 42.4 × 10 −3

a

f

a

At 50.0 Hz, X L = 2π 50.0 Hz L = 2π 50.0 Hz

I max =

P33.11

XL

XL =

(b)

∆Vmax 100 = = 13.3 Ω I max 7.50

b

2 ∆Vrms

∆Vmax = XL

XL

g = 2 a100 V f = 45.0 Ω

fFGH 2πXa60.0 Hzf IJK = 5060..00 a54.0 Ωf = 45.0 Ω

3.14 A .

IJ a f a fb K b ge i atf = a5.60 A f sina1.59 radf = 5.60 A af

iL t =

FG H

L 60.0 Hz

g

80.0 V sin 65.0π 0.015 5 − π 2 ∆Vmax π = sin ω t − ωL 2 65.0π rad s 70.0 × 10 −3 H

j

L

P33.12

b

g

ω = 2π f = 2π 60.0 s = 377 rad s

b

gb

g

X L = ω L = 377 s 0.020 0 V ⋅ s A = 7.54 Ω I rms =

∆Vrms 120 V = = 15.9 A XL 7.54 Ω

a f F 2π a60.0f ⋅ 1 s IJ = a22.5 Af sin 120° = 19.5 A sin ω t = a 22.5 A f sinG H s 180 K 1 = b0.020 0 V ⋅ s A ga19.5 A f = 3.80 J 2

I max = 2 I rms = 2 15.9 A = 22.5 A

af

i t = I max

P33.13

U=

1 2 Li 2

L=

NΦ B X ∆VL , max where Φ B is the flux through each turn. NΦ B , max = LI max = L ω I XL

NΦ B , max =

2

d

d

2 ∆VL , rms 2π f

i=

FG T ⋅ C ⋅ m IJ FG N ⋅ m IJ FG J IJ = 2π a60.0f H N ⋅ s K H J K H V ⋅ C K

120 V ⋅ s

0.450 T ⋅ m 2 .

i

269

270

Alternating Current Circuits

Section 33.4 P33.14

Capacitors in an AC Circuit 1 1 : < 175 Ω 2π f C 2π f 22.0 × 10 −6

XC =

(a)

e

1

e

j

f > 41.3 Hz

X C , φ is positive; so voltage leads the current .

a

f

For the source-capacitor circuit, the rms source voltage is ∆Vs = 25.1 mA X C . For the circuit with

a

resistor, ∆Vs = 15.7 mA

a

f

2

f

R +

X C2

inductor, ∆Vs = 68.2 mA X L − X C

b

∆Vs = I R 2 + X L − X C

a25.1 mAfX

C

I = 19.3 mA

=I

g

a f = a 25.1 mA f X

= 25.1 mA X C . This gives R = 1.247 X C . For the circuit with ideal

2

b1.247 X g + b0.368 X g C

2

C

2

C.

So X L − X C = 0.368 0 X C . Now for the full circuit

272 P33.25

Alternating Current Circuits

XC =

1 1 = = 1.33 × 10 8 Ω 2π f C 2π 60.0 Hz 20.0 × 10 −12 F

Z=

e50.0 × 10 Ωj + e1.33 × 10 Ωj

a

8

rms body

e

je

j

1 1 = = 49.0 Ω ω C 2π 50.0 65.0 × 10 −6

a fe j X = ω L = 2π a50.0fe185 × 10 j = 58.1 Ω Z = R + b X − X g = a 40.0f + a58.1 − 49.0f −3

2

I max =

L

C

2

2

2

= 41.0 Ω

∆Vmax 150 = = 3.66 A 41.0 Z

FIG. P33.26

a fa f

(a)

∆VR = I max R = 3.66 40 = 146 V

(b)

∆VL = I max X L = 3.66 58.1 = 212.5 = 212 V

(c)

∆VC = I max X C = 3.66 49.0 = 179.1 V = 179 V

(d)

∆VL − ∆VC = 212.5 − 179.1 = 33.4 V

a fa f

a fa f

R = 300 Ω

FG 500 s IJ a0.200 Hf = 200 Ω Hπ K L F 500 s IJ e11.0 × 10 FjOP 1 = M 2π G X = K ωC N H π Q Z = R + b X − X g = 319 Ω and F X − X IJ = 20.0° φ = tan G H R K −1

C

2

−1

XL = 200 Ω

−1

X L = ω L = 2π

*P33.28

≈ 1.33 × 10 8 Ω

= I rms R body = 3.77 × 10 −5 A 50.0 × 10 3 Ω = 1.88 V

L

P33.27

2

5 000 V ∆Vrms = = 3.77 × 10 −5 A 8 Z 1.33 × 10 Ω

b ∆V g XC =

j

2

3

I rms =

P33.26

fe

L

L

C

−6

−1

= 90.9 Ω

XL - XC = 109 Ω

2

{

XC = 90.9 Ω

Z

φ R = 300 Ω

C

FIG. P33.27

Let X c represent the initial capacitive reactance. Moving the plates to half their original separation 1 doubles the capacitance and cuts X C = in half. For the current to double, the total impedance ωC must be cut in half: Zi = 2Z f ,

b

R 2 + R − XC

g

2

F GH

FG H

= 4 R2 + R −

b

R 2 + X L − XC XC 2

IJ IJ KK 2

2 R 2 − 2 RX C + X C2 = 8 R 2 − 4RX C + X C2 XC = 3 R

g

2

FG H

= 2 R2 + XL −

XC 2

IJ K

2

,

Chapter 33

P33.29

a

fa

f

X L = 2π 100 Hz 20.5 H = 1.29 × 10 4 Ω ∆Vrms 200 V = = 50.0 Ω Z= I rms 4.00 A

(a)

bX

− XC

L

g

2

b

= Z 2 − R 2 = 50.0 Ω

X L − X C = 1.29 × 10 4 Ω − ∆VL , rms = I rms X L

(b)

g − b35.0 Ωg 2

2

1 = ±35.7 Ω 2π 100 Hz C

b g = a 4.00 A fe1.29 × 10 Ωj = 4

C = 123 nF or 124 nF

FIG. P33.29

51.5 kV

Notice that this is a very large voltage!

Section 33.6 P33.30

Power in an AC Circuit

b

gb g 1 = b1 000 sge50.0 × 10 F j X = ωC Z = R + bX − X g Z = a 40.0f + a50.0 − 20.0f = 50.0 Ω b∆V g = 100 V I = (a)

X L = ω L = 1 000 s 0.050 0 H = 50.0 Ω −6

C

2

L

C

−1

= 20.0 Ω

2

2

FIG. P33.30

2

rms

rms

50.0 Ω

Z I rms = 2.00 A

φ = arctan

FG X H

b

g

IJ K

− XC R 30.0 Ω φ = arctan = 36.9° 40.0 Ω

P33.31

L

a

f

(b)

P = ∆Vrms I rms cos φ = 100 V 2.00 A cos 36.9° = 160 W

(c)

2 PR = I rms R = 2.00 A 40.0 Ω = 160 W

a

f

2

ω = 1 000 rad s ,

R = 400 Ω ,

∆Vmax = 100 V ,

ω L = 500 Ω ,

Z= R I max =

2

F 1 I + Gω L − H ω C JK

C = 5.00 × 10 −6 F ,

F 1 I = 200 Ω GH ω C JK

2

= 400 2 + 300 2 = 500 Ω

∆Vmax 100 = = 0.200 A Z 500

2 The average power dissipated in the circuit is P = I rms R=

a0.200 Af a400 Ωf = 2 2

P=

L = 0.500 H

8.00 W

F I IR. GH 2 JK 2 max

273

274 P33.32

Alternating Current Circuits

b

Z = R 2 + X L − XC

bX

L

g

g

or X L − X C = Z 2 − R 2

g a75.0 Ωf − a45.0 Ωf = 60.0 Ω FG X − X IJ = tan FG 60.0 Ω IJ = 53.1° H 45.0 Ω K H R K 2

− XC =

φ = tan −1

b

2

L

2

−1

C

∆Vrms 210 V = = 2.80 A Z 75.0 Ω P = ∆Vrms I rms cos φ = 210 V 2.80 A cos 53.1° = 353 W I rms =

b

P33.33

g

b

g

a f

2 I rms R

tan φ =

(b)

X L − XC R

f a

f

a

f

b

a f

3

2

so

1.29 × 10 = 9.00 R

becomes

tan −37.0° =

a

gb g g = a20.0f + a9.42f

f

X L − XC : 16

and

R = 16.0 Ω .

so

X L − X C = −12.0 Ω .

X L = ω L = 2π 60.0 s 0.025 0 H = 9.42 Ω

b

Z = R 2 + X L − XC

2

2

2

Ω = 22.1 Ω

∆Vrms 120 V = = 5.43 A 22.1 Ω Z

(a)

I rms =

(b)

φ = tan −1

(c)

We require φ = 0 . Thus, X L = X C :

9. 42 Ω =

and

C = 281 µF .

FG 9.42 IJ = 25.2° H 20.0 K

b

(d)

b

b ∆V g

rms b

=

power factor = cos φ = 0.905 .

so

rms b

b

e

1

rms b

2 rms d

R

=

b

a20.0 Ωfa120 Vfa5.43 Afa0.905f =

Consider a two-wire transmission line: I rms =

P

FG P IJ b2R g = P 100 H ∆V K ρd b ∆V g = =

P 100

2

Thus,

1

rms

R1

rms

A

or

R1 =

or

A=

2

200P

and the diameter is

109 V R1

2 Rline = and power loss = I rms

∆Vrms

j

2π 60.0 s −1 C

b ∆V g =

g bI g cos φ Rb ∆V g b I g cos φ

Pb = Pd or ∆Vrms rms d

P33.35

fa

P = I rms ∆Vrms cos φ = 9.00 180 cos −37.0° = 1.29 × 10 3 W

(a)

P=

P33.34

a

2r =

.

∆Vrms

b∆V g rms

2

R1

200P

a f

π 2r 4

RL

2

=

200 ρP d

b ∆V g rms

800 ρP d

a f

π ∆V

2

.

2

FIG. P33.35

Chapter 33

P33.36

One-half the time, the left side of the generator is positive, the top diode conducts, and the bottom diode switches off. The power supply sees resistance

LM 1 + 1 OP N 2R 2R Q

−1

= R and the power is

b∆V g rms

R

LM 1 + 1 OP N 3R R Q

−1

=

7R 4

.

R

P=

and

~

Section 33.7 P33.37

P33.38

rms

2

Req rms

=

b

4 ∆Vrms

∆V

R + 4 ∆Vrms

FIG. P33.36

2

.

7R

b

2

g

g

2

7R

2

=

Resonance in a Series RLC Circuit

e

1

j

ω 0 = 2π 99.7 × 10 6 = 6.26 × 10 8 rad s = C=

b ∆V g

b ∆V g

The overall time average power is:

1

ω 02 L

=

1

At resonance,

LC

= 1.82 pF

e6.26 × 10 j e1.40 × 10 j 8 2

−6

1 1 = 2π f L and 2π f C 2π f

b gL 2

=C .

The range of values for C is 46.5 pF to 419 pF . *P33.39

(a)

(b)

1

f=

2π LC

1 1 A C = 6.33 × 10 −13 F = 2 2 − 2 10 12 As 4π f L 4π 10 s 400 × 10 Vs

C=

κ ∈0 A κ ∈0 A 2 = d d

A=

e

FG Cd IJ Hκ ∈ K 0

(c)

FG IJ H K

C=

2

12

=

j

F 6.33 × 10 F × 10 GH 1 × 8.85 × 10 −13

−3

−12

R

2

The other half of the time the right side of the generator is positive, the upper diode is an open circuit, and the lower diode has zero resistance. The equivalent resistance is then Req = R +

2R

R

F

mm

I JK

12

= 8.46 × 10 −3 m

X L = 2π f L = 2π × 10 10 s × 400 × 10 −12 Vs A = 25.1 Ω

b

11 ∆Vrms 14R

g

2

.

275

276 P33.40

P33.41

Alternating Current Circuits

L = 20.0 mH , C = 1.00 × 10 −7 , R = 20.0 Ω , ∆Vmax = 100 V (a)

The resonant frequency for a series –RLC circuit is f =

(b)

At resonance,

I max =

(c)

From Equation 33.38,

Q=

(d)

∆VL , max = X L I max = ω 0 LI max = 2.24 kV 1

The resonance frequency is ω 0 = XL = ω L =

FG H

IJ L = 2 LC K 2

b

Z = R 2 + X L − XC

g

2

LC

1 = 3.56 kHz . LC

∆Vmax = 5.00 A . R

ω 0L = 22.4 . R

. Thus, if ω = 2ω 0 ,

L C

= R 2 + 2.25

1 2π

FG L IJ H CK

1 LC 1 L = = 2C 2 C ωC

and

XC =

so

I rms =

∆Vrms = Z

∆Vrms

b g

2

R + 2.25 L C

and the energy delivered in one period is E = P ∆t :

b∆V g R FG 2π IJ = b∆V g RC eπ E= R + 2.25bL C g H ω K R C + 2.25L rms

2

rms

2

2

2

b

4π ∆Vrms

j

LC =

g RC 2

LC

.

2

4R C + 9.00L

With the values specified for this circuit, this gives: E=

P33.42

a

f a10.0 Ωfe100 × 10 Fj e10.0 × 10 Hj 4a10.0 Ωf e100 × 10 F j + 9.00e10.0 × 10 H j

4π 50.0 V

2

32

−6

2

−6

FG H

IJ L = 2 LC K 2

1

b

LC

.

L C

Then Z = R 2 + X L − X C

g

2

12

−3

The resonance frequency is ω 0 = XL = ω L =

−3

= R 2 + 2.25

FG L IJ H CK

= 242 mJ .

Thus, if

ω = 2ω 0 ,

and

XC =

so

I rms =

LC 1 L 1 = = . 2C 2 C ωC ∆Vrms = Z

and the energy delivered in one period is

b∆V g R FG 2π IJ = b∆V g RC eπ E = P∆t = R + 2.25bL C g H ω K R C + 2.25L rms

2

2

rms

2

2

j

LC =

b

4π ∆Vrms 2

g RC 2

4R C + 9.00L

LC

.

∆Vrms 2

b g

R + 2.25 L C

Chapter 33

P33.43

1

For the circuit of Problem 22, ω 0 =

1

=

= 251 rad s

e160 × 10 Hje99.0 × 10 Fj ω L b 251 rad sge160 × 10 H j = = 0.591 Q= LC

−3

−6

−3

0

68.0 Ω

R

For the circuit of Problem 23, Q =

.

ω 0L 1 L 1 460 × 10 −3 H L = = = = 0.987 . R R LC R C 150 Ω 21.0 × 10 −6 F

The circuit of Problem 23 has a sharper resonance.

Section 33.8 P33.44

The Transformer and Power Transmission

a

f

1 120 V = 9.23 V 13

(a)

∆V2 , rms =

(b)

∆V1 , rms I 1 , rms = ∆V2 , rms I 2 , rms

a120 Vfa0.350 Af = a9.23 V fI

42.0 W = 4.55 A for a transformer with no energy loss. 9.23 V

I 2 , rms =

P = 42.0 W from part (b).

(c)

P33.45

b ∆V g

=

b∆V g

=

out max

out rms

P33.46

(a)

2 , rms

b

N2 ∆Vin N1

g

max

a971 Vf = 2 , rms

FG 2 000 IJ a170 Vf = 971 V H 350 K

687 V

2

d ∆V

=

i = NN d∆V i 2

1

d

N2 =

1 , rms

i

d

i

(b)

I 1 , rms ∆V1 , rms = I 2 , rms ∆V2, rms

(c)

0.950 I 1 , rms ∆V1, rms = I 2 , rms ∆V2, rms

d

i

d

b2 200ga80f =

I 1, rms =

i

I 1 , rms =

110

1 600 windings

a1.50fb2 200g = 110

a1.20fb2 200g = 110a0.950f

30.0 A

25.3 A

277

278 P33.47

Alternating Current Circuits

The rms voltage across the transformer primary is N1 ∆V2, rms N2

d

i

so the source voltage is ∆Vs , rms = I 1 , rms Rs +

d∆V The secondary current is

2, rms

d

RL

i

N1 ∆V2 , rms . N2

d

i

FIG. P33.47

i , so the primary current is

N 2 ∆V2 , rms = I 1 , rms . N1 RL Then ∆Vs , rms =

and Rs =

P33.48

i

+

N 1 RL

N 1 RL

d

N 2 ∆V2 , rms

F ∆V i GGH

s , rms

N2 ∆V1 , rms N1

d

−

d

N 1 ∆V2 , rms

i

N2

d

N 1 ∆V2 , rms N2

i

i IJ = 5a50.0 Ωf FG 80.0 V − 5a25.0 V f IJ = JK 2a25.0 Vf H 2 K

∆V2 , rms =

(b)

I 2 , rms ∆V2 , rms = 0.900 I 1, rms ∆V1, rms

d i d i 120 V I J a120 Vf e10.0 × 10 Vj = 0.900FGH 24.0 ΩK 3

∆V2 , rms

I 2, rms = 54.0 mA

10.0 × 10 3 V = 185 kΩ 0.054 A

(c)

Z2 =

(a)

R = 4.50 × 10 −4 Ω M 6.44 × 10 5 m = 290 Ω and I rms =

I 2 , rms

e

2 Ploss = I rms

(b) (c)

=

87.5 Ω .

N 2 ∆V2 , rms 10.0 × 10 3 V = = = 83.3 120 V N 1 ∆V1 , rms

(a)

I 2, rms

P33.49

d

N 2 ∆V2 , rms Rs

je j R = a10.0 A f a 290 Ωf = 29.0 kW

P 5.00 × 10 6 W = = 10.0 A ∆Vrms 5.00 × 10 5 V

2

Ploss 2.90 × 10 4 = = 5.80 × 10 −3 P 5.00 × 10 6 It is impossible to transmit so much power at such low voltage. Maximum power transfer occurs when load resistance equals the line resistance of 290 Ω , and is

e4.50 × 10 Vj 2 ⋅ 2a 290 Ωf 3

2

= 17.5 kW far below the required 5 000 kW.

Chapter 33

Section 33.9 *P33.50

(a)

279

Rectifiers and Filters Input power = 8 W

a f

Useful output power = I∆V = 0.3 A 9 V = 2.7 W useful output 2.7 W = = 0.34 = 34% efficiency = total input 8W (b)

Total input power = Total output power 8 W = 2.7 W + wasted power wasted power = 5.3 W

(c)

*P33.51

(a)

I= a fa fFGH 861400d s IJK FGH 11WsJ IJK = 1.29 × 10 JFGH 3.6$0.135 J × 10 J K 8

E = P ∆t = 8 W 6 31 d

2

The input voltage is ∆Vin = IZ = I R + ∆Vout = IR . The gain ratio is

(b)

(c)

(a)

b

∆Vout 1 → ∞ and → 0 ωC ∆Vin

As ω → ∞ ,

∆Vout 1 R → 0 and → = 1 ωC ∆Vin R

1 = 2

=I R

2

∆Vout IR = ∆Vin I R2 + 1 ω C

As ω → 0 ,

R2 +

P33.52

X C2

F1I +G H ω C JK g

$4.8

6

=

2

2

. The output voltage is R

b

R2 + 1 ω C

g

.

2

R

b

R2 + 1 ω C

g

2

1 = 4R 2 ω 2C 2

ω 2C 2 =

1 3R2

ω = 2π f =

1

f=

3 RC

b

The input voltage is ∆Vin = IZ = I R 2 + X C2 = I R 2 + 1 ω C ∆Vout = IX C =

I ωC ∆Vout I = . The gain ratio is ∆Vin ωC I R2 + 1 ω C

b

g

g

2

2

1 2π 3 RC

. The output voltage is =

1 ωC

b

R2 + 1 ω C

g

2

.

1 ωC ∆Vout 1 → ∞ and R becomes negligible in comparison. Then → = 1 . As ωC ∆Vin 1 ωC ∆Vout 1 → 0 and ω → ∞, → 0 . ωC ∆Vin

(b)

As ω → 0 ,

(c)

1 = 2 f=

1 ωC 2

b

R + 1 ωC 3 2π RC

g

2

R2 +

F1I GH ω C JK

2

=

4 ω C2 2

R 2ω 2 C 2 = 3

ω = 2π f =

3 RC

280 P33.53

Alternating Current Circuits

∆Vout = ∆Vin

For this RC high-pass filter,

(a)

When then

R 2

R + X C2

.

∆Vout = 0.500 , ∆Vin 0.500 Ω

a0.500 Ωf

2

+ X C2

= 0.500 or X C = 0.866 Ω.

If this occurs at f = 300 Hz, the capacitance is C= (b)

1 1 = = 6.13 × 10 −4 F = 613 µF . 2π f X C 2π 300 Hz 0.866 Ω

a

fa

f

With this capacitance and a frequency of 600 Hz, XC =

1

a

fe

2π 600 Hz 6.13 × 10 −4 F

∆Vout = ∆Vin

R 2

R +

X C2

=

j

= 0.433 Ω

0.500 Ω

a0.500 Ωf + a0.433 Ωf 2

2

= 0.756 . FIG. P33.53

P33.54

For the filter circuit,

(a)

(b)

∆Vout = ∆Vin

XC 2

R + X C2

.

1 1 = = 3.32 × 10 4 Ω 2π f C 2π 600 Hz 8.00 × 10 −9 F

At f = 600 Hz,

XC =

and

∆Vout = ∆Vin

a

fe

j

3.32 × 10 4 Ω

a90.0 Ωf + e3.32 × 10 Ωj 2

4

2

≈ 1.00 .

1 1 = = 33.2 Ω 3 2π f C 2π 600 × 10 Hz 8.00 × 10 −9 F

At f = 600 kHz ,

XC =

and

∆Vout = ∆Vin

e

je

33.2 Ω

a90.0 Ωf + a33.2 Ωf 2

2

= 0.346 .

j

281

Chapter 33

P33.55

∆Vout = ∆Vin

(a)

2

R

b

R + XL − XC

g

2

1 = 4 8.00 Ω

At 200 Hz:

a

a8.00 Ωf

2

.

f + 400π L − 1 400π C a8.00 Ωf + LM8 000π L − 8 0001 π C OP = 4a8.00 Ωf . N Q 2

2

2

At 4 000 Hz:

2

1 = −13.9 Ω . 400π C 1 = +13.9 Ω. 8 000π L − 8 000π C 400π L −

At the low frequency, X L − X C < 0 . This reduces to For the high frequency half-voltage point,

When X L = X C ,

(c)

X L = X C requires

(d)

At 200 Hz,

FG H

∆Vout ∆Vout = ∆Vin ∆Vin f0 =

1 2π LC

IJ K

= 1.00 .

2π

e

5.80 × 10

∆Vout R 1 = = and X C > X L , ∆Vin Z 2

−4

je

H 5.46 × 10 −5 F

j

= 894 Hz .

R

φ

XL - XC

so the phasor diagram is as shown:

φ = − cos −1

FG R IJ = − cos FG 1 IJ so H ZK H 2K

or

Z

−1

φ ∆Vin

or

φ

φ

∆Vout

R

At f0 , X L = X C so

FIG. P33.55(d)

∆Vout and ∆Vin have a phase difference of 0° . ∆Vout R 1 = = and X L − X C > 0 . ∆Vin Z 2

Thus, φ = cos −1

(e)

FG 1 IJ = 60.0° H 2K

∆Vout lags ∆Vin by 60.0° .

or

At 200 Hz and at 4 kHz,

d ∆V P=

i = db1 2g∆V

R

in, rms

d ∆V At f , P =

R

(f)

i = d ∆V

R

a10.0 Vf = 1.56 W 8a8.00 Ωf a10.0 V f = 6.25 W . = 2a8.00 Ωf Hj = 0.408 . 2

in, max

R

2

out, rms

0

i = b1 2g b1 2g∆V 2

2

out, rms

i = b1 2g ∆V 2

in, rms

R

in, max

R

a

fe

−4 ω 0 L 2π f0 L 2π 894 Hz 5.80 × 10 = = We take: Q = 8.00 Ω R R

2

∆Vout

∆Vin

Z

XL - XC

∆Vout leads ∆Vin by 60.0° .

At 4 000 Hz,

[2]

max

1

=

[1]

C = 54.6 µF and L = 580 µH .

Solving Equations (1) and (2) simultaneously gives (b)

FIG. P33.55(a)

2

2

=

2

.

282

Alternating Current Circuits

Additional Problems P33.56

af

The equation for ∆v t during the first period (using y = mx + b ) is:

a f 2b∆VT gt − ∆V max

∆v t =

a ∆v f a ∆v f

z ∆vatf dt = b∆VT g z LMNT2 t − 1OPQ dt b∆V g FG T IJ 2t T − 1 = b∆V g a+1f − a−1f = b∆V g = H 2K 3 6 3 T a∆vf = b∆V g = ∆V

1 = T

2 ave

2 ave

2

max

2 T

0

2

FIG. P33.56

0

3 t =T

2

max

2

3

3

max

2

t=0

2

max

=

LC

2

max

3

ave

1

ω0 =

T

max

∆Vrms =

P33.57

max

3

1

b0.050 0 Hge5.00 × 10 Fj −6

= 2 000 s −1

so the operating angular frequency of the circuit is

ω=

ω0 = 1 000 s −1 . 2

b∆V g Rω R ω + L eω − ω j a400f a8.00fb1 000g P= a8.00f b1 000g + b0.050 0g a1.00 − 4.00f × 10

Using Equation 33.37, P =

rms

2

2

2

*P33.58

2

2

bQ ≈ 12.5g

2 2 0

FIG. P33.57

2

2

2

2

2

2

6 2

= 56.7 W .

The angular frequency is ω = 2π 60 s = 377 s . When S is open, R, L, and C are in series with the source:

b

R 2 + X L − XC

20 V I g = FGH ∆VI IJK = FGH 0.183 J AK 2

s

2

2

= 1.194 × 10 4 Ω 2 .

When S is in position 1, a parallel combination of two R’s presents equivalent resistance with L and C:

FG R IJ + b X H 2K 2

L

− XC

20 V I g = FGH 0.298 J AK 2

2

(1) R , in series 2

= 4.504 × 10 3 Ω 2 .

(2)

When S is in position 2, the current by passes the inductor. R and C are in series with the source: R 2 + X C2 =

FG 20 V IJ H 0.137 A K

2

= 2.131 × 10 4 Ω 2 .

Take equation (1) minus equation (2): 3 2 R = 7. 440 × 10 3 Ω 2 4 continued on next page

R = 99.6 Ω

(3)

Chapter 33

283

(only the positive root is physical.) Now equation (3) gives

a f

X C = 2.131 × 10 4 − 99.6

b

C = ωX C

g

−1

b

2 12

Ω = 106.7 Ω =

g

−1

= 377 s 106.7 Ω

1 (only the positive root is physical.) ωC

= 2.49 × 10 −5 F = C .

Now equation (1) gives

a f

X L − X C = ± 1.194 × 10 4 − 99.6

2 12

Ω = ±44.99 Ω

X L = 106.7 Ω + 44.99 Ω = 61.74 Ω or 151.7 Ω = ω L L=

P33.59

XL

ω

= 0.164 H or 0.402 H = L

The resistance of the circuit is R =

∆V 12.0 V = = 19.0 Ω . I 0.630 A

The impedance of the circuit is Z =

∆Vrms 24.0 V = = 42.1 Ω . 0.570 A I rms

Z 2 = R 2 + ω 2 L2 L= *P33.60

1

ω

Z2 − R2 =

1 377

a42.1f − a19.0f 2

2

= 99.6 mH

The lowest-frequency standing-wave state is NAN. The distance between the clamps we represent λ T as L = d NN = . The speed of transverse waves on the string is v = fλ = = f 2L . The magnetic µ 2 force on the wire oscillates at 60 Hz, so the wire will oscillate in resonance at 60 Hz.

b g

T 2 = 60 s 4L2 0.019 kg m

e

j

T = 274 kg ms 2 L2

Any values of T and L related according to this expression will work, including if L = 0.200 m T = 10.9 N . We did not need to use the value of the current and magnetic field. If we assume the subsection of wire in the field is 2 cm wide, we can find the rms value of the magnetic force:

a fa

fb

g

FB = IAB sin θ = 9 A 0.02 m 0.015 3T sin 90° = 2.75 mN . So a small force can produce an oscillation of noticeable amplitude if internal friction is small. P33.61

(a)

When ω L is very large, the bottom branch carries negligible current. Also, negligible compared to 200 Ω and top branch.

(b)

Now

1 will be ωC

45.0 V = 225 mA flows in the power supply and the 200 Ω

1 → ∞ and ω L → 0 so the generator and bottom branch carry 450 mA . ωC

284 P33.62

Alternating Current Circuits

(a)

With both switches closed, the current goes only through generator and resistor.

af

it =

(b)

(c)

(d)

P=

∆Vmax cos ω t R

b

1 ∆Vmax 2 R

af

g

2

∆Vmax

it =

2

2 2

R +ω L

1 , so ω 0C

C=

(e)

At this resonance frequency,

(f)

U=

b g

1 C ∆VC 2

U max =

2

=

1 2 2 CI X C 2

b

b

1 2 1 ∆Vmax = LI max = L 2 2 R2

U max

(h)

Now ω = 2ω 0 = So φ = arctan

(i)

Now ω L =

(a)

I R, rms =

2 LC

0

0

1 . ω 02 L

g

g

2

b ∆V g L

1 = ω 02 C 2

max 2

2

2R

2

.

F ω L − b1 ω Cg I = arctanF 2 GH GH R JK

1 1 2 ωC

F ω L − b1 ω Cg I . GH R JK

Z= R .

1 2 1 ∆Vmax CI max X C2 = C 2 2 R2

(g)

(b)

FG ω L IJ OP H R KQ

0 = φ = arctan

For We require ω 0 L =

P33.63

LM N

cos ω t + arctan

FIG. P33.62

1

ω=

2LC

=

b g

LC− 12 R

LC

I= JK

F 3 LI GH 2 R C JK

arctan

.

ω0 . 2

∆Vrms 100 V = = 1.25 A 80.0 Ω R

IR

φ

The total current will lag the applied voltage as seen in the phasor IL diagram at the right. ∆Vrms 100 V I L , rms = = = 1.33 A XL 2π 60.0 s −1 0.200 H

e

ja

Thus, the phase angle is: φ = tan −1

∆V

FI GH I

f

L , rms R , rms

I

FIG. P33.63

I = tan FG 1.33 A IJ = JK H 1.25 A K −1

46.7° .

Chapter 33

P33.64

285

Suppose each of the 20 000 people uses an average power of 500 W. (This means 12 kWh per day, or $36 per 30 days at 10¢ per kWh). Suppose the transmission line is at 20 kV. Then 20 000 500 W P I rms = ~ 10 3 A . = 20 000 V ∆Vrms

b

f

ga

If the transmission line had been at 200 kV, the current would be only ~ 10 2 A . P33.65

R = 200 Ω , L = 663 mH , C = 26.5 µF , ω = 377 s −1 , ∆Vmax = 50.0 V

ω L = 250 Ω , (a)

F 1 I = 100 Ω , Z = GH ω C JK

g

2

= 250 Ω

∆Vmax 50.0 V = = 0. 200 A 250 Ω Z X − XC φ = tan −1 L = 36.8° ( ∆V leads I) R I max =

FG H

P33.66

b

R 2 + X L − XC

IJ K

(b)

∆VR, max = I max R = 40.0 V at φ = 0°

(c)

∆VC , max =

(d)

∆VL , max = I maxω L = 50.0 V at φ = +90.0° ( ∆V leads I)

I max = 20.0 V ωC

at φ = −90.0° (I leads ∆V )

af b

L = 2.00 H , C = 10.0 × 10 −6 F , R = 10.0 Ω , ∆v t = 100 sin ω t (a)

g

The resonant frequency ω 0 produces the maximum current and thus the maximum power delivery to the resistor. 1 1 ω0 = = = 224 rad s LC 2.00 10.0 × 10 −6

a fe

j

(b)

b∆V g = a100f = P= 2R 2a10.0f

(c)

I rms =

max

2

2

∆Vrms = Z

2 I rms R=

1 2 I rms 2

e j

500 W

∆Vrms

b

d

R2 + ω L − 1 ω C

max

R

gi

2

and

or

bI g

rms max

b ∆V g rms 2

Z

=

∆Vrms R

2

R=

b

1 ∆Vrms 2 R2

g

2

R.

F 1 I This occurs where Z = 2 R : R + Gω L − H ω C JK = 2R ω L C − 2 Lω C − R ω C + 1 = 0 or L C ω − e 2LC + R C jω + 1 = 0 LMa2.00f e10.0 × 10 j OPω − LM2a2.00fe10.0 × 10 j + a10.0f e10.0 × 10 j OPω + 1 = 0 . N Q N Q 2

2

4 2

2

2

2

2

−6 2

2

2

2

2

4

2

−6

2

2

2

4

2

−6 2

2

2

2

Solving this quadratic equation, we find that

ω 2 = 51 130 , or 48 894

ω 1 = 48 894 = 221 rad s

ω 2 = 51 130 = 226 rad s .

and

286 P33.67

Alternating Current Circuits

(a)

From Equation 33.41,

N 1 ∆V1 = . N 2 ∆V2

Let input impedance

Z1 =

so that

N 1 Z1 I 1 = . N 2 Z2 I 2

∆V1 I1

Z2 =

But from Eq. 33.42,

I 1 ∆V2 N 2 = = . I 2 ∆V1 N 1

N1 = N2

So, combining with the previous result we have

P33.68

2 P = I rms R=

FG ∆V IJ H Z K rms

Z1 Z2

a120 V f a40.0 Ωf : Z = Z 2

2

R , so 250 W =

2

a120f a40.0f a40.0f + 2π f a0.185f − 1 2π f e65.0 × 10 j

F GH

R2 + ω L −

2

250 =

1=

2

2

−6

2 304 f 2 1 600 f 2 + 1.351 1 f 4 − 5 692.3 f 2 + 5 995 300

2

f =

6 396.3 ±

.

8 000 = 31.6 8.00

N1 Z1 = = N2 Z2

(b)

∆V2 I2

and the output impedance

1 ωC

I JK

2

576 000 f 2

and

250 =

so

1.351 1 f 4 − 6 396.3 f 2 + 5 995 300 = 0

e

1 600 f 2 + 1.162 4 f 2 − 2 448.5

b6 396.3g − 4b1.351 1gb5 995 300g = 3 446.5 or 1 287.4 2b1.351 1g 2

f = 58.7 Hz or 35.9 Hz P33.69

IR =

∆Vrms ; R

IL =

∆Vrms ; ωL

b

g

2

OPF 1 I QGH ∆V R JK

(b)

tan φ =

IC − IL 1 1 = ∆Vrms − IR XC X L

C

OP Q

−1

LM N

I rms =

1 − XL

bω Cg

= ∆Vrms

(a)

L1 tan φ = R M NX

∆Vrms

FG 1 IJ + FGω C − 1 IJ H R K H ω LK

I R2

+ IC − IL

IC =

2

2

rms

FIG. P33.69

j

2

Chapter 33

P33.70

(a)

F GH

1 1 + ωC− 2 ω L R

I rms = ∆Vrms

b

∆Vrms → ∆Vrms 1

f=

(b)

2π LC

=

g

max

I JK

2

when ω C =

1 ωL

1 2π 200 × 10

−3

287

e

H 0.150 × 10 −6 F

j

= 919 Hz

∆Vrms 120 V = = 1.50 A 80.0 Ω R ∆Vrms 120 V = = 1.60 A IL = − ωL 374 s 1 0.200 H IR =

ja

e

b g a

FIG. P33.70

f

fe je j I = I + b I − I g = a1.50f + b0.006 73 − 1.60g = L I − I OP = tan L 0.006 73 − 1.60 O = −46.7° φ = tan M MN 1.50 PQ N I Q

I C = ∆Vrms ω C = 120 V 374 s −1 0.150 × 10 −6 F = 6.73 mA

(c) (d)

2 R

rms

−1

C

C

2

L

2

2

2.19 A

−1

L

R

The current is lagging the voltage . P33.71

(a)

X L = X C = 1 884 Ω when f = 2 000 Hz 1 884 Ω XL = = 0.150 H and 2π f 4 000π rad s 1 1 = = 42.2 nF C= 2π f X C 4 000π rad s 1 884 Ω L=

(b)

XL

b g b = 2π f a0.150 H f

Z=

a40.0 Ωf + b X 2

gb

XC =

L

− XC

Impedence, Ω

FIG. P33.71(b)

g

2

g

1

b2π f ge4.22 × 10 Fj −8

f (Hz) X L ( Ω) 283 300

X C ( Ω) Z ( Ω) 12 600 1 2300

600

565

6 280

5 720

800

754

4 710

3 960

1 000

942

3 770

2 830

1 500

1 410

2 510

1 100

2 000

1 880

1 880

40

3 000

2 830

1 260

1 570

4 000

3 770

942

2 830

6 000

5 650

628

5 020

10 000

9 420

377

9 040

288 P33.72

Alternating Current Circuits

1

ω0 =

ω ωL Ω ω0 999.0 0.9990 999.1 0.9991 999.3 0.9993 999.5 0.9995 999.7 0.9997 999.9 0.9999 1.0000 1000 1.0001 1000.1 1.0003 1000.3 1.0005 1000.5 1.0007 1000.7 1.0009 1000.9 1.0010 1001

a f

= 1.00 × 10 6 rad s

LC

For each angular frequency, we find

b

Z = R2 + ω L − 1 ω C then I =

1.00 V Z 2

a

g

2

f

P = I 1.00 Ω .

and

The full width at half maximum is:

b

g

1.000 5 − 0.999 5 ω 0 ∆ω ∆f = = 2π 2π 3 −1 1.00 × 10 s ∆f = = 159 Hz 2π while R 2π L

=

e

1.00 Ω

2π 1.00 × 10 −3 H

j

a f Z aΩf

a f

1 Ω ωC 1001.0

2.24

0.19984

1000.9

2.06

0.23569

1000.7

1.72

0.33768

1000.5

1.41

0.49987

1000.3

1.17

0.73524

1000.1

1.02

0.96153

1000.0

1.00

1.00000

999.9

1.02

0.96154

999.7

1.17

0.73535

999.5

1.41

0.50012

999.3

1.72

0.33799

999.1

2.06

0.23601

999.0

2.24

0.20016

P = I 2R W

= 159 Hz . 1.0 0.8 I 2R (W)

0.6 0.4 0.2 0.0 0.996

0.998

1 ω/ω 0

1.002

1.004

FIG. P33.72 P33.73

∆Vout = ∆Vin

(a)

R 2

b

R + 1 ωC

g

2

=

R 2

b

R + 1 2π f C

g

C 2

∆Vout 1 1 = when =R 3 . 2 V ωC ∆ in Hence, f =

ω 1 = = 1.84 kHz . 2π 2π RC 3

continued on next page

∆Vin

R

FIG. P33.73

∆Vout

Chapter 33

(b)

289

Log Gain versus Log Frequency 0 –1 Log ∆ V out/∆ V in

–2 –3 –4 0

1

2

3

4

5

6

Log f

FIG. P33.73(b)

ANSWERS TO EVEN PROBLEMS P33.2

(a) 193 Ω ; (b) 144 Ω

P33.32

353 W

P33.4

(a) 25.3 rad/s; (b) 0.114 s

P33.34

(a) 5.43 A ; (b) 0.905 ; (c) 281 µF ; (d) 109 V

P33.6

1.25 A and 96.0 Ω for bulbs 1 and 2; 0.833 A and 144 Ω for bulb 3

P33.36

P33.8

7.03 H or more

P33.38

46.5 pF to 419 pF

P33.10

3.14 A

P33.40

P33.12

3.80 J

(a) 3.56 kHz; (b) 5.00 A ; (c) 22.4 ; (d) 2.24 kV

P33.14

P33.16

(a) greater than 41.3 Hz ; (b) less than 87.5 Ω

b

2C ∆Vrms

g

P33.18

–32.0 A

P33.20

2.79 kHz

P33.22

(a) 109 Ω ; (b) 0.367 A ; (c) I max = 0.367 A , ω = 100 rad s, φ = −0.896 rad

P33.24

19.3 mA

P33.26

(a) 146 V ; (b) 212 V ; (c) 179 V ; (d) 33.4 V

P33.28

XC = 3 R

P33.30

(a) 2.00 A ; (b) 160 W ; (c) see the solution

P33.42

b

11 ∆Vrms

g

2

14R

b

4π ∆Vrms

g RC 2

LC

2

4R C + 9L

P33.44

(a) 9.23 V ; (b) 4.55 A ; (c) 42.0 W

P33.46

(a) 1 600 turns ; (b) 30.0 A ; (c) 25.3 A

P33.48

(a) 83.3 ; (b) 54.0 mA ; (c) 185 kΩ

P33.50

(a) 0.34; (b) 5.3 W; (c) $4.8

P33.52

(a) see the solution; (b) 1; 0; (c)

P33.54

(a) 1.00; (b) 0.346

P33.56

see the solution

P33.58

R = 99.6 Ω, C = 24.9 µF , L = 164 mH or 402 mH

3 2π RC

290 P33.60

P33.62

Alternating Current Circuits

L = 0.200 m and T = 10.9 N , or any values related by T = 274 kg ms 2 L2

e

j

a f ∆VR cos ω t ; (b) P = b∆V2R g ; L ∆V F ω L IJ OP ; cos Mω t + tan G (c) iat f = H R KQ N R +ω L b∆V g L ; 1 (d) C = ; (e) Z = R ; (f) max

max

(a) i t =

(g) (i)

b

ω 02 L 2 ∆Vmax L 2

g

2R

1 2LC

~ 10 3 A

P33.66

(a) 224 rad s ; (b) 500 W ; (c) 221 rad s and 226 rad s

P33.68

either 58.7 Hz or 35.9 Hz

P33.70

(a) 919 Hz ; (b) I R = 1.50 A , I L = 1.60 A , I C = 6.73 mA ; (c) 2.19 A ; (d) −46.7° ; current lagging

P33.72

see the solution

2

−1

max

2

P33.64

2 2

max 2

2

2R

; (h) tan −1

F 3 LI; GH 2R C JK

34 Electromagnetic Waves CHAPTER OUTLINE 34.1 34.2 34.3 34.4 34.5

34.6

Maxwell’s Equations and Hertz’s Discoveries Plane Electromagnetic Waves Energy Carried by Electromagnetic Waves Momentum and Radiation Pressure Production of Electromagnetic Waves by an Antenna The Spectrum of Electromagnetic Waves

ANSWERS TO QUESTIONS Q34.1

Radio waves move at the speed of light. They can travel around the curved surface of the Earth, bouncing between the ground and the ionosphere, which has an altitude that is small when compared to the radius of the Earth. The distance across the lower forty-eight states is approximately 5 000 km, requiring a 5 × 10 6 m ~ 10 −2 s . To go halfway around the transit time of 3 × 10 8 m s Earth takes only 0.07 s. In other words, a speech can be heard on the other side of the world before it is heard at the back of a large room.

Q34.2

The Sun’s angular speed in our sky is our rate of rotation, 360° = 15° h . In 8.3 minutes it moves west by 24 h 1h θ = ω t = 15° h 8.3 min = 2.1° . This is about four 60 min times the angular diameter of the Sun.

b

gFGH

IJ a K

f

Q34.3

Energy moves. No matter moves. You could say that electric and magnetic fields move, but it is nicer to say that the fields at one point stay at that point and oscillate. The fields vary in time, like sports fans in the grandstand when the crowd does the wave. The fields constitute the medium for the wave, and energy moves.

Q34.4

No. If a single wire carries DC current, it does not emit electromagnetic waves. In this case, there is a constant magnetic field around the wire. Alternately, if the cable is a coaxial cable, it ideally does not emit electromagnetic waves even while carrying AC current.

Q34.5

Acceleration of electric charge.

Q34.6

The changing magnetic field of the solenoid induces eddy currents in the conducting core. This is accompanied by I 2 R conversion of electrically-transmitted energy into internal energy in the conductor.

Q34.7

A wire connected to the terminals of a battery does not radiate electromagnetic waves. The battery establishes an electric field, which produces current in the wire. The current in the wire creates a magnetic field. Both fields are constant in time, so no electromagnetic induction or “magneto-electric induction” happens. Neither field creates a new cycle of the other field. No wave propagation occurs. 291

292 Q34.8 Q34.9

Electromagnetic Waves

No. Static electricity is just that: static. Without acceleration of the charge, there can be no electromagnetic wave. Sound

Light

The world of sound extends to the top of the atmosphere and stops there; sound requires a material medium. Sound propagates by a chain reaction of density and pressure disturbances recreating each other. Sound in air moves at hundreds of meters per second. Audible sound has frequencies over a range of three decades (ten octaves) from 20 Hz to 20 kHz. Audible sound has wavelengths of ordinary size (1.7 cm to 17 m). Sound waves are longitudinal.

The universe of light fills the whole universe. Light moves through materials, but faster in a vacuum. Light propagates by a chain reaction of electric and magnetic fields recreating each other. Light in air moves at hundreds of millions of meters per second. Visible light has frequencies over a range of less than one octave, from 430 to 750 Terahertz. Visible light has wavelengths of very small size (400 nm to 700 nm). Light waves are transverse.

Sound and light can both be reflected, refracted, or absorbed to produce internal energy. Both have amplitude and frequency set by the source, speed set by the medium, and wavelength set by both source and medium. Sound and light both exhibit the Doppler effect, standing waves, beats, interference, diffraction, and resonance. Both can be focused to make images. Both are described by wave functions satisfying wave equations. Both carry energy. If the source is small, their intensities both follow an inverse-square law. Both are waves. Q34.10

The Poynting vector S describes the energy flow associated with an electromagnetic wave. The direction of S is along the direction of propagation and the magnitude of S is the rate at which electromagnetic energy crosses a unit surface area perpendicular to the direction of S.

Q34.11

Photons carry momentum. Recalling what we learned in Chapter 9, the impulse imparted to a particle that bounces elastically is twice that imparted to an object that sticks to a massive wall. Similarly, the impulse, and hence the pressure exerted by a photon reflecting from a surface must be twice that exerted by a photon that is absorbed.

Q34.12

Different stations have transmitting antennas at different locations. For best reception align your rabbit ears perpendicular to the straight-line path from your TV to the transmitting antenna. The transmitted signals are also polarized. The polarization direction of the wave can be changed by reflection from surfaces—including the atmosphere—and through Kerr rotation—a change in polarization axis when passing through an organic substance. In your home, the plane of polarization is determined by your surroundings, so antennas need to be adjusted to align with the polarization of the wave.

Q34.13

You become part of the receiving antenna! You are a big sack of salt water. Your contribution usually increases the gain of the antenna by a few tenths of a dB, enough to noticeably improve reception.

Q34.14

On the TV set, each side of the dipole antenna is a 1/4 of the wavelength of the VHF radio wave. The electric field of the wave moves free charges in the antenna in electrical resonance, giving maximum current in the center of the antenna, where the cable connects it to the receiver.

Q34.15

The loop antenna is essentially a solenoid. As the UHF radio wave varies the magnetic field inside the loop, an AC emf is induced in the loop as described by Faraday’s and Lenz’s laws. This signal is then carried down a cable to the UHF receiving circuit in the TV. An excellent reference for antennas and all things radio is the ARRL Handbook.

Chapter 34

293

Q34.16

The voltage induced in the loop antenna is proportional to the rate of change of the magnetic field in the wave. A wave of higher frequency induces a larger emf in direct proportion. The instantaneous voltage between the ends of a dipole antenna is the distance between the ends multiplied by the electric field of the wave. It does not depend on the frequency of the wave.

Q34.17

The radiation resistance of a broadcast antenna is the equivalent resistance that would take the same power that the antenna radiates, and convert it into internal energy.

Q34.18

Consider a typical metal rod antenna for a car radio. The rod detects the electric field portion of the carrier wave. Variations in the amplitude of the incoming radio wave cause the electrons in the rod to vibrate with amplitudes emulating those of the carrier wave. Likewise, for frequency modulation, the variations of the frequency of the carrier wave cause constant-amplitude vibrations of the electrons in the rod but at frequencies that imitate those of the carrier.

Q34.19

The frequency of EM waves in a microwave oven, typically 2.45 GHz, is chosen to be in a band of frequencies absorbed by water molecules. The plastic and the glass contain no water molecules. Plastic and glass have very different absorption frequencies from water, so they may not absorb any significant microwave energy and remain cool to the touch.

Q34.20

People of all the world’s races have skin the same color in the infrared. When you blush or exercise or get excited, you stand out like a beacon in an infrared group picture. The brightest portions of your face show where you radiate the most. Your nostrils and the openings of your ear canals are bright; brighter still are just the pupils of your eyes.

Q34.21

Light bulbs and the toaster shine brightly in the infrared. Somewhat fainter are the back of the refrigerator and the back of the television set, while the TV screen is dark. The pipes under the sink show the same weak sheen as the walls until you turn on the faucets. Then the pipe on the right turns very black while that on the left develops a rich glow that quickly runs up along its length. The food on your plate shines; so does human skin, the same color for all races. Clothing is dark as a rule, but your bottom glows like a monkey’s rump when you get up from a chair, and you leave behind a patch of the same blush on the chair seat. Your face shows you are lit from within, like a jack-olantern: your nostrils and the openings of your ear canals are bright; brighter still are just the pupils of your eyes.

Q34.22

Welding produces ultraviolet light, along with high intensity visible and infrared.

Q34.23

12.2 cm waves have a frequency of 2.46 GHz. If the Q value of the phone is low (namely if it is cheap), and your microwave oven is not well shielded (namely, if it is also cheap), the phone can likely pick up interference from the oven. If the phone is well constructed and has a high Q value, then there should be no interference at all.

294

Electromagnetic Waves

SOLUTIONS TO PROBLEMS Section 34.1 *P34.1

(a)

Maxwell’s Equations and Hertz’s Discoveries The rod creates the same electric field that it would if stationary. We apply Gauss’s law to a cylinder of radius r = 20 cm and length : q E ⋅ dA = inside ∈0 λl E 2π rl cos 0° = ∈0

z

b g

E=

(b)

e

FIG. P34.1 2

35 × 10 C m N ⋅ m λ j = 3.15 × 10 3 j N C . radially outward = 2π ∈0 r 2π 8.85 × 10 −12 C 2 0.2 m

e

ja

f

e

je

j

The charge in motion constitutes a current of 35 × 10 −9 C m 15 × 10 6 m s = 0.525 A . This current creates a magnetic field. B=

(c)

j

−9

µ0I 2π r

=

e4π × 10

ja 2π a0. 2 mf

−7

T ⋅ m A 0.525 A

fk=

5.25 × 10 −7 k T

The Lorentz force on the electron is F = qE + qv × B

e

je j e e− jj N + 2.02 × 10 e+ jj N =

j FGH

je

F = −1.6 × 10 −19 C 3.15 × 10 3 j N C + −1.6 × 10 −19 C 240 × 10 6 i m s × 5.25 × 10 −7 k F = 5.04 × 10 −16

Section 34.2 P34.2

(a)

−17

e j

4.83 × 10 −16 − j N

Plane Electromagnetic Waves Since the light from this star travels at 3.00 × 10 8 m s the last bit of light will hit the Earth in

6.44 × 10 18 m 8

3.00 × 10 m s

= 2.15 × 10 10 s = 680 years .

Therefore, it will disappear from the sky in the year 2 004 + 680 = 2.68 × 10 3 C.E. . The star is 680 light-years away. ∆x 1. 496 × 10 11 m = = 499 s = 8.31 min v 3 × 10 8 m s

(b)

∆t =

(c)

8 ∆x 2 3.84 × 10 m = = 2.56 s ∆t = v 3 × 10 8 m s

(d)

6 ∆x 2π 6.37 × 10 m ∆t = = = 0.133 s v 3 × 10 8 m s

(e)

∆t =

e

j

e

j

∆x 10 × 10 3 m = = 3.33 × 10 −5 s v 3 × 10 8 m s

N⋅s C ⋅m

IJ K

Chapter 34

1

P34.3

v=

P34.4

E =c B or

1

=

κµ 0 ∈0

1.78

c = 0.750 c = 2.25 × 10 8 m s

220 = 3.00 × 10 8 B

so B = 7.33 × 10 −7 T = 733 nT . P34.5

fλ = c

(a)

a

f

f 50.0 m = 3.00 × 10 8 m s

or

f = 6.00 × 10 6 Hz = 6.00 MHz .

so E =c B

(b)

or

22.0 = 3.00 × 10 8 Bmax

so

Bmax = −73.3k nT .

k=

(c)

2π

λ

=

2π = 0.126 m −1 50.0

e

j

ω = 2π f = 2π 6.00 × 10 6 s −1 = 3.77 × 10 7 rad s

and

b

g

e

j

B = Bmax cos kx − ω t = −73.3 cos 0.126 x − 3.77 × 10 7 t k nT . P34.6

ω = 2π f = 6.00π × 10 9 s −1 = 1.88 × 10 10 s −1 k=

2π

λ

=

ω 6.00π × 10 9 s −1 = = 20.0π = 62.8 m −1 c 3.00 × 10 8 m s

b

g e

E = 300 V m cos 62.8 x − 1.88 × 10 10 t

P34.7

j

Bmax =

B=

100 V m E = = 3.33 × 10 −7 T = 0.333 µT c 3.00 × 10 8 m s

(b)

λ=

2π 2π = = 0.628 µm k 1.00 × 10 7 m −1

(c)

f=

λ

=

3.00 × 10 8 m s 6. 28 × 10

−7

m

= 4.77 × 10 14 Hz

g e

B = 1.00 µT cos 62.8 x − 1.88 × 10 10 t

(a)

c

b

300 V m E = = 1.00 µT c 3.00 × 10 8 m s

j

295

296 P34.8

Electromagnetic Waves

b

E = Emax cos kx − ω t

g ga f ga f ge j ga f

∂E = − Emax sin kx − ω t k ∂x ∂E = − Emax sin kx − ω t −ω ∂t ∂2E = − Emax cos kx − ω t k 2 ∂x 2 ∂2E = − Emax cos kx − ω t −ω ∂t 2

b b

b b

∂E

We must show:

∂x 2

= µ 0 ∈0

∂2E ∂t 2

.

b

e j

2

a f

g

2

b

g

− k 2 Emax cos kx − ω t = − µ 0 ∈0 −ω Emax cos kx − ω t .

That is,

k2

But this is true, because

ω

2

F1I =G J H fλ K

2

=

1 c2

= µ 0 ∈0 .

The proof for the wave of magnetic field follows precisely the same steps. P34.9

In the fundamental mode, there is a single loop in the standing wave between the plates. Therefore, the distance between the plates is equal to half a wavelength.

a

f

λ = 2L = 2 2.00 m = 4.00 m f=

Thus,

P34.10

c

λ

d A to A = 6 cm ± 5% =

λ = 12 cm ± 5%

a

=

3.00 × 10 8 m s = 7.50 × 10 7 Hz = 75.0 MHz . 4.00 m

λ 2

fe

j

v = λ f = 0.12 m ± 5% 2.45 × 10 9 s −1 = 2.9 × 10 8 m s ± 5%

Section 34.3

Energy Carried by Electromagnetic Waves

P34.11

S=I=

P34.12

Sav =

1 000 W m 2 Energy I =u= = = 3.33 µJ m3 c 3.00 × 10 8 m s Unit Volume

U Uc = = uc At V

P 4.00 × 10 3 W = 2 4π r 4π 4.00 × 1 609 m

b

g

2

= 7.68 µW m 2

Emax = 2 µ 0 cSav = 0.076 1 V m

b

ga

f

b

g

∆Vmax = Emax L = 76.1 mV m 0.650 m = 49.5 mV amplitude

or 35.0 mV (rms)

Chapter 34

P34.13

a

3

P 250 × 10 W = 2 4π r 4π 8.04 × 10 3 W

I=

e

100 W

a

P34.16

f

4π 1.00 m

2

j

2

= 307 µW m 2

= 7.96 W m 2

I = 2.65 × 10 −8 J m3 = 26.5 nJ m 3 c

u=

P34.15

g

r = 5.00 mi 1 609 m mi = 8.04 × 10 3 m S=

P34.14

fb

297

(a)

uE =

1 u = 13.3 nJ m3 2

(b)

uB =

1 u = 13.3 nJ m3 2

(c)

I = 7.96 W m 2

Power output = (power input)(efficiency). Thus,

Power input =

and

A=

I=

Power out 1.00 × 10 6 W = = 3.33 × 10 6 W eff 0.300

P 3.33 × 10 6 W = = 3.33 × 10 3 m 2 . I 1.00 × 10 3 W m 2

2 Bmax c P = 2µ 0 4π r 2

Bmax =

F P GH 4π r

I FG 2 µ IJ = e10.0 × 10 ja2fe4π × 10 j = JK H c K 4π 5.00 × 10 3.00 × 10 e je j −7

3

2

0

3 2

8

5.16 × 10 −10 T

Since the magnetic field of the Earth is approximately 5 × 10 −5 T , the Earth’s field is some 100 000 times stronger. P34.17

(a)

P = I 2 R = 150 W

jb

e

g

A = 2π rL = 2π 0.900 × 10 −3 m 0.080 0 m = 4.52 × 10 −4 m 2

(b)

S=

P = 332 kW m 2 A

B=

µ 0 1.00 µ0I = = 222 µT 2π r 2π 0.900 × 10 −3

E=

∆V IR 150 V = = = 1.88 kV m ∆x 0.080 0m L

e

Note:

(points radially inward)

a f

S=

EB

µ0

j

= 332 kW m 2

298 *P34.18

Electromagnetic Waves

e

j

2

2 3 × 10 6 V m Emax I= = 2 µ 0 c 2 4π × 10 −7 T ⋅ m A 3 × 10 8 m s

(a)

e

je

FG J IJ FG C IJ FG T ⋅ C ⋅ m IJ FG N ⋅ m IJ j H V ⋅CK H A ⋅sKH N ⋅s KH J K 2

I = 1.19 × 10 10 W m 2

P34.19

e

10

F 5 × 10 W m jπ G H 2

−3

2

m

I JK

2

= 2.34 × 10 5 W

(b)

P = IA = 1.19 × 10

(a)

E ⋅ B = 80.0 i + 32.0 j − 64.0k N C ⋅ 0.200 i + 0.080 0 j + 0.290 k µT

e jb g e E ⋅ B = a16.0 + 2.56 − 18.56f N ⋅ s C ⋅ m = 2

S=

(b)

1

µ0

E×B=

2

j

0

e80.0 i + 32.0 j − 64.0kj N C × e0.200 i + 0.080 0 j + 0.290kj µT 4π × 10 −7 T ⋅ m A

e6.40k − 23.2 j − 6.40k + 9.28i − 12.8 j + 5.12 ij × 10 S=

−6

W m2

4π × 10 −7

e11.5i − 28.6 jj W m

S= *P34.20

2

= 30.9 W m 2 at −68.2° from the +x axis.

The energy put into the water in each container by electromagnetic radiation can be written as eP ∆t = eIA∆t where e is the percentage absorption efficiency. This energy has the same effect as heat in raising the temperature of the water: eIA∆t = mc∆T = ρVc∆T ∆T =

eI 2 ∆t eI∆t = ρ c ρ 3c

where is the edge dimension of the container and c the specific heat of water. For the small container, 0.7 25 × 10 3 W m 2 480 s ∆T = = 33.4° C . 10 3 kg m3 0.06 m 4 186 J kg ⋅° C

e

e

ja

For the larger, ∆T =

P34.21

j

f

e

j

0.91 25 J s ⋅ m 2 480 s

e0.12 m j4 186 J ° C 2

= 21.7° C .

We call the current I rms and the intensity I. The power radiated at this frequency is

b

gb

g

P = 0.010 0 ∆Vrms I rms =

b

0.010 0 ∆Vrms

g

2

R

= 1.31 W .

If it is isotropic, the intensity one meter away is I=

P 1.31 W = A 4π 1.00 m

Bmax =

a

2µ 0 I = c

f

2

= 0.104 W m 2 = Sav =

e

je

c 2 Bmax 2µ 0

2 4π × 10 −7 T ⋅ m A 0.104 W m 2 8

3.00 × 10 m s

j=

29.5 nT

Chapter 34

P34.22

(a)

efficiency =

(b)

Sav =

FG H

299

IJ K

useful power output 700 W × 100% = × 100% = 50.0% total power input 1 400 W

P 700 W = = 2.69 × 10 5 W m 2 A 0.068 3 m 0.038 1 m

b

gb

g

Sav = 269 kW m 2 toward the oven chamber

(c)

Sav =

2 Emax 2µ 0 c

e

je

je

j

Emax = 2 4π × 10 −7 T ⋅ m A 3.00 × 10 8 m s 2.69 × 10 5 W m 2 = 1.42 × 10 4 V m = 14.2 kV m

P34.23

Emax : c

7.00 × 10 5 N C

(a)

Bmax =

(b)

E2 I = max : 2µ 0 c

(c)

I=

(a)

e10.0 × 10 j W I= π e0.800 × 10 mj

(b)

uav =

(a)

E = cB = 3.00 × 10 8 m s 1.80 × 10 −6 T = 540 V m

Bmax =

3.00 × 10 8 m s

= 2.33 mT

e7.00 × 10 j I= = 650 MW m 2e 4π × 10 je3.00 × 10 j Lπ O P = IA = e6.50 × 10 W m jM e1.00 × 10 mj P = N4 Q 5 2

2

−7

P : A

8

8

2

−3

2

510 W

−3

P34.24

P34.25

−3

2

= 4.97 kW m 2

I 4.97 × 10 3 J m 2 ⋅ s = = 16.6 µJ m3 c 3.00 × 10 8 m s

e

B2

je

e1.80 × 10 j =

j

−6 2

= 2.58 µJ m3

(b)

uav =

(c)

Sav = cuav = 3.00 × 10 8 2.58 × 10 −6 = 773 W m 2

(d)

This is 77.3% of the intensity in Example 34.5 . It may be cloudy, or the Sun may be setting.

µ0

4π × 10

e

−7

je

j

300

Electromagnetic Waves

Section 34.4 P34.26

P34.27

P34.28

Momentum and Radiation Pressure 2Sav c

The pressure P upon the mirror is

P=

where A is the cross-sectional area of the beam and

Sav =

The force on the mirror is then

F = PA =

Therefore,

F=

For complete absorption, P =

(a)

P . A

e

FG IJ H K

2 P 2P A= . c A c

2 100 × 10 −3

e

3 × 10

8

j

j=

6.67 × 10 −10 N .

25.0 S = = 83.3 nPa . c 3.00 × 10 8

e

2 1 340 W m 2

The radiation pressure is

8

3.00 × 10 m s

j = 8.93 × 10

2

−6

N m2 .

Multiplying by the total area, A = 6.00 × 10 5 m 2 gives: F = 5.36 N . (b)

The acceleration is:

a=

5.36 N F = = 8.93 × 10 −4 m s 2 . m 6 000 kg

(c)

It will arrive at time t where

d=

1 2 at 2

t=

or

P34.29

I=

(a)

(b)

(c)

2d = a

2 Emax P = π r 2 2µ 0 c

Emax =

b

P 2µ 0 c πr

2

15 × 10 −3 J s 3.00 × 10 8 m s p=

g=

1.90 kN C

a1.00 mf =

50.0 pJ

U 5 × 10 −11 = = 1.67 × 10 −19 kg ⋅ m s c 3.00 × 10 8

e

j

2 3.84 × 10 8 m

e

8.93 × 10

−4

m s2

j

= 9.27 × 10 5 s = 10.7 days .

Chapter 34

P34.30

(a)

If PS is the total power radiated by the Sun, and rE and rM are the radii of the orbits of the planets Earth and Mars, then the intensities of the solar radiation at these planets are: IE = IM =

and

PS 4π rE2 PS

.

2 4π rM

FG r IJ = e1 340 W m jF 1.496 × 10 GH 2.28 × 10 Hr K 2

IM = IE

Thus, (b)

2

E

11

11

M

I J mK m

Mars intercepts the power falling on its circular face:

e

jLNM e

j e

If Mars behaves as a perfect absorber, it feels pressure P = and force

(d)

F = PA =

2

= 577 W m 2 .

j OQP =

2 PM = I M π R M = 577 W m 2 π 3.37 × 10 6 m

(c)

301

2

2.06 × 10 16 W .

SM I M = c c

P IM 2.06 × 10 16 W 2 π RM = M = = 6.87 × 10 7 N . 8 c c 3.00 × 10 m s

e

j

The attractive gravitational force exerted on Mars by the Sun is Fg =

GM S M M rM2

e6.67 × 10 =

−11

je

je

N ⋅ m 2 kg 2 1.991 × 10 30 kg 6.42 × 10 23 kg

e2.28 × 10 mj 11

2

j = 1.64 × 10

21

N

which is ~ 10 13 times stronger than the repulsive force of part (c). P34.31

(a)

The total energy absorbed by the surface is U=

(b)

FG 1 IIJ At = LM 1 e750 W m jOPe0.500 × 1.00 m ja60.0 sf = H 2 K N2 Q 2

2

11.3 kJ .

The total energy incident on the surface in this time is 2U = 22.5 kJ , with U = 11.3 kJ being absorbed and U = 11.3 kJ being reflected. The total momentum transferred to the surface is

b g a F U I F 2U IJ = 3U = 3e11.3 × 10 Jj = p=G J +G H c K H c K c 3.00 × 10 m s

p = momentum from absorption + momentum from reflection 3

8

1.13 × 10 −4 kg ⋅ m s

f

302 * P34.32

Electromagnetic Waves

The radiation pressure on the disk is P = Then F =

af

af

H x 0 + H y 0 − mgr sin θ +

∑τ = 0

mg

π r 2 Ir =0 c 2 π 0.4 m 10 7 W s 2 s

a

f g a

b

Hy

PA

fe

r

F 1 kg m I j GH 1 W s JK

π r 2I = sin −1 mgc 0.024 kg m 2 9.8 m 3 × 10 8 m

−1

Hx

π r 2I . c

Take torques about the hinge:

θ = sin

S I F F = = = . c c A π r2

2

3

θ

= sin −1 0.071 2 = 4.09°

FIG. P34.32

Section 34.5 P34.33

P34.34

Production of Electromagnetic Waves by an Antenna

λ=

c = 536 m f

so

h=

λ = 134 m 4

λ=

c = 188 m f

so

h=

λ = 46.9 m 4

P=

a∆V f R

2

a f

or P ∝ ∆V

2

af

∆y

∆V = − Ey ⋅ ∆y = E y ⋅ cos θ

θ

receiving antenna

∆V ∝ cos θ so P ∝ cos 2 θ

a

f

a

f

a

f

(a)

θ = 15.0° : P = Pmax cos 2 15.0° = 0.933 Pmax = 93.3%

(b)

θ = 45.0° : P = Pmax cos 2 45.0° = 0.500Pmax = 50.0%

(c)

θ = 90.0° : P = Pmax cos 2 90.0° = 0

FIG. P34.34

Chapter 34

P34.35

(a)

303

Constructive interference occurs when d cos θ = nλ for some integer n.

F I GH JK

λ λ =n = 2n d λ 2 n = 0 , ± 1, ± 2 , …

cos θ = n

∴ strong signal @ θ = cos −1 0 = 90° , 270° (b)

Destructive interference occurs when d cos θ =

FG 2n + 1 IJ λ : H 2 K

cos θ = 2n + 1

a f

∴ weak signal @ θ = cos −1 ±1 = 0° , 180°

FIG. P34.35 P34.36

For the proton,

∑ F = ma

mv 2 . R 2πR 2π m = T= . v qB qvB sin 90.0° =

yields

The period of the proton’s circular motion is therefore:

*P34.37

The frequency of the proton’s motion is

f=

1 . T

The charge will radiate electromagnetic waves at this frequency, with

λ=

2π mc c = cT = . f qB

(a)

b

g

1 µ 0 J max cos kx − ω t k applies for x > 0 , since it describes a wave 2 1 moving in the i direction. The electric field direction must satisfy S = E × B as i = j × k so

The magnetic field B =

µ0

the direction of the electric field is j when the cosine is positive. For its magnitude we have 1 E = cB , so altogether we have E = µ 0 cJ max cos kx − ω t j . 2

b

(b)

S=

1

E×B=

b

g

g

1 1 2 2 µ 0 cJ max cos 2 kx − ω t i µ0 4

µ0 1 2 S = µ 0 cJ max cos 2 kx − ω t i 4

b

g

(c)

The intensity is the magnitude of the Poynting vector averaged over one or more cycles. The 1 1 2 average of the cosine-squared function is , so I = µ 0 cJ max . 2 8

(d)

J max =

8I = µ0c

e

8 570 W m 2 4π × 10

−7

j

bTm Ag3 × 10

8

ms

= 3.48 A m

304

Electromagnetic Waves

Section 34.6 P34.38

P34.39

P34.40

The Spectrum of Electromagnetic Waves

From the electromagnetic spectrum chart and accompanying text discussion, the following identifications are made: Frequency, f

Wavelength, λ =

2 Hz = 2 × 10 0 Hz 2 kHz = 2 × 10 3 Hz 2 MHz = 2 × 10 6 Hz 2 GHz = 2 × 10 9 Hz 2 THz = 2 × 10 12 Hz 2 PHz = 2 × 10 15 Hz 2 EHz = 2 × 10 18 Hz 2 ZHz = 2 × 10 21 Hz 2 YHz = 2 × 10 24 Hz

150 Mm 150 km 150 m 15 cm 150 µm 150 nm 150 pm 150 fm 150 am

Wavelength, λ

Frequency, f =

2 km = 2 × 10 3 m 2 m = 2 × 10 0 m 2 mm = 2 × 10 −3 m 2 µm = 2 × 10 −6 m 2 nm = 2 × 10 −9 m 2 pm = 2 × 10 −12 m 2 fm = 2 × 10 −15 m 2 am = 2 × 10 −18 m

1.5 × 10 5 Hz 1.5 × 10 8 Hz 1.5 × 10 11 Hz 1.5 × 10 14 Hz 1.5 × 10 17 Hz 1.5 × 10 20 Hz 1.5 × 10 23 Hz 1.5 × 10 26 Hz

f=

(a) (b)

c

λ

=

3.00 × 10 8 m s 5.50 × 10 −7 m f=

λ

=

Classification Radio Radio Radio Microwave Infrared Ultraviolet X-ray Gamma ray Gamma ray

c

λ

Classification Radio Radio Microwave Infrared Ultraviolet/X-ray X-ray/Gamma ray Gamma ray Gamma ray

= 5.45 × 10 14 Hz

3 × 10 8 m s ~ 10 8 Hz 1.7 m

radio wave

1 000 pages, 500 sheets, is about 3 cm thick so one sheet is about 6 × 10 −5 m thick. f=

P34.41

c

c f

3.00 × 10 8 m s 6 × 10 −5 m

~ 10 13 Hz

infrared

(a)

fλ = c

gives

e5.00 × 10

19

(b)

fλ = c

gives

e4.00 × 10

9

j

λ = 6.00 × 10 −12 m = 6.00 pm

j

λ = 0.075 m = 7.50 cm

Hz λ = 3.00 × 10 8 m s : Hz λ = 3.00 × 10 8 m s :

Chapter 34

P34.42

P34.43

(a)

λ=

c 3.00 × 10 8 m s = = 261 m f 1 150 × 10 3 s −1

so

180 m = 0.690 wavelengths 261 m

(b)

λ=

8 c 3.00 × 10 m s = = 3.06 m f 98.1 × 10 6 s −1

so

180 m = 58.9 wavelengths 3.06 m

Time to reach object =

b

g e

305

1 1 total time of flight = 4.00 × 10 −4 s = 2.00 × 10 −4 s . 2 2

e

je

j

j

Thus, d = vt = 3.00 × 10 8 m s 2.00 × 10 −4 s = 6.00 × 10 4 m = 60.0 km .

P34.44

The time for the radio signal to travel 100 km is:

∆t r =

The sound wave travels 3.00 m across the room in:

∆t s =

100 × 10 3 m 8

3.00 × 10 m s

= 3.33 × 10 −4 s .

3.00 m = 8.75 × 10 −3 s . 343 m s

Therefore, listeners 100 km away will receive the news before the people in the news room by a total time difference of

∆t = 8.75 × 10 −3 s − 3.33 × 10 −4 s = 8.41 × 10 −3 s . P34.45

8 c 3.00 × 10 m s = = 4.00 × 10 6 m . f 75.0 Hz

The wavelength of an ELF wave of frequency 75.0 Hz is

λ=

The length of a quarter-wavelength antenna would be

L = 1.00 × 10 6 m = 1.00 × 10 3 km

or

L = 1 000 km

b

.621 mi I gFGH 01.00 J= km K

Thus, while the project may be theoretically possible, it is not very practical. P34.46

(a)

For the AM band,

λ max = λ min =

(b)

For the FM band,

λ max = λ min =

c f min c fmax c f min c fmax

= =

= =

3.00 × 10 8 m s 540 × 10 3 Hz 3.00 × 10 8 m s 1 600 × 10 3 Hz 3.00 × 10 8 m s 88.0 × 10 6 Hz 3.00 × 10 8 m s 108 × 10 6 Hz

= 556 m = 187 m .

= 3.41 m = 2.78 m .

621 mi .

306

Electromagnetic Waves

Additional Problems P34.47

(a)

P = SA :

(b)

cB 2 S = max 2µ 0 S=

P34.48

jLMN e

e

P = 1 340 W m 2 4π 1.496 × 10 11 m

2 Emax 2µ 0 c

2 µ 0S = c

e

j OPQ = 2

3.77 × 10 26 W

je

2 4π × 10 −7 N A 2 1 340 W m 2

j=

3.35 µT

so

Bmax =

so

Emax = 2 µ 0 cS = 2 4π × 10 −7 3.00 × 10 8 1 340 = 1.01 kV m

8

3.00 × 10 m s

e

jb

je

g

Suppose you cover a 1.7 m-by-0.3 m section of beach blanket. Suppose the elevation angle of the Sun is 60°. Then the target area you fill in the Sun’s field of view is

a1.7 mfa0.3 mf cos 30° = 0.4 m . U = IAt = e1 340 W m j a0.6 fa0.5fe0.4 m j b3 600 sg ~ 10 2

P34.49

Now I =

P U = A At

(a)

ε=−

af

2

a

dΦ B d =− BA cos θ dt dt

f

2

ε = −A

af

b

6

J .

g

b

d Bmax cos ω t cos θ = ABmaxω sin ω t cos θ dt

g

ε t = 2π 2 r 2 fBmax cos θ sin 2π f t

ε t = 2π fBmax A sin 2π f t cos θ

ε max = 2π 2 r 2 f Bmax cos θ

Thus,

where θ is the angle between the magnetic field and the normal to the loop. (b)

If E is vertical, B is horizontal, so the plane of the loop should be vertical and the plane should contain the line of sight of the transmitter .

P34.50

(a)

Fgrav =

GM S m 2

=

FG GM HR 2

S

3

R where M S = mass of Sun, r = radius of particle and R = distance from Sun to particle. Since

Frad =

Sπ r 2 , c

FG H

Frad 1 = Fgrav r (b)

IJ LMρFG 4 π r IJ OP KN H 3 KQ

IJ FG 3SR IJ ∝ 1 . K H 4cGM ρ K r 2

S

From the result found in part (a), when Fgrav = Frad , we have r =

r=

3SR 2 4cGM S ρ

e

je

j

3 214 W m 2 3.75 × 10 11 m

e

je

je

2

je

4 6.67 × 10 −11 N ⋅ m 2 kg 2 1.991 × 10 30 kg 1 500 kg m3 3.00 × 10 8 m s

= 3.78 × 10 −7 m

j

Chapter 34

P34.51

Emax = 6.67 × 10 −16 T c

(a)

Bmax =

(b)

Sav =

(c)

P = Sav A = 1.67 × 10 −14 W

(d)

F = PA =

2 Emax = 5.31 × 10 −17 W m 2 2µ 0 c

FG S IJ A = HcK av

5.56 × 10 −23 N (≅ the weight of FIG. P34.51

3 000 H atoms!) P34.52

(a)

307

j a

e

f

The power incident on the mirror is: PI = IA = 1 340 W m 2 π 100 m

2

e

= 4.21 × 10 7 W .

j

The power reflected through the atmosphere is PR = 0.746 4.21 × 10 7 W = 3.14 × 10 7 W .

PR 3.14 × 10 7 W = A π 4.00 × 10 3 m

= 0.625 W m 2

(b)

S=

(c)

Noon sunshine in Saint Petersburg produces this power-per-area on a horizontal surface:

e

j

2

PN = 0.746 1 340 W m 2 sin 7.00° = 122 W m 2 . A

e

j

The radiation intensity received from the mirror is

F 0.625 W m I 100% = GH 122 W m JK 2

2

0.513% of that from the noon Sun in January.

1 2 ∈0 Emax 2

2u = 95.1 mV m ∈0

P34.53

u=

P34.54

The area over which we model the antenna as radiating is the lateral surface of a cylinder,

Emax =

ja

e

f

A = 2π r = 2π 4.00 × 10 −2 m 0.100 m = 2.51 × 10 −2 m 2 . (a)

The intensity is then: S =

(b)

The standard is:

e

0.600 W P = = 23.9 W m 2 . A 2.51 × 10 −2 m 2

0.570 mW cm 2 = 0.570 mW cm 2

jFGH 1.001.00× 10mWW IJK FGH 1.001×.0010m cm IJK = 5.70 W m −3

While it is on, the telephone is over the standard by

4

2

2

23.9 W m 2 5.70 W m 2

2

= 4.19 times .

.

308 P34.55

Electromagnetic Waves

(a)

Bmax = k=

2π

λ

175 V m Emax = = 5.83 × 10 −7 T 8 c 3.00 × 10 m s =

2π = 419 rad m 0.015 0 m

ω = kc = 1.26 × 10 11 rad s

Since S is along x, and E is along y, B must be in the z direction . (That is S ∝ E × B .)

P34.56

Emax Bmax = 40.6 W m 2 2µ 0

(b)

S av =

(c)

Pr =

(d)

∑ F = PA = e2.71 × 10 a=

S av =

e40.6 W m ji 2

2S = 2.71 × 10 −7 N m 2 c

m

−7

m

je

N m 2 0.750 m 2

0.500 kg

j = 4.06 × 10

−7

m s2

Of the intensity

S = 1 340 W m 2

the 38.0% that is reflected exerts a pressure

P1 =

2Sr 2 0.380 S = . c c

The absorbed light exerts pressure

P2 =

S a 0.620S = . c c

a

a=

e406 nm s ji 2

f

Altogether the pressure at the subsolar point on Earth is

P34.57

e

j

2 1.38S 1.38 1 340 W m = P1 + P2 = = = 6.16 × 10 −6 Pa c 3.00 × 10 8 m s

(a)

Ptotal

(b)

1.01 × 10 5 N m 2 Pa = = 1.64 × 10 10 times smaller than atmospheric pressure Ptotal 6.16 × 10 −6 N m 2

(a)

P=

F I = A c

a = 3.03 × 10 −9 m s 2 and

100 J s IA P = = = 3.33 × 10 −7 N = 110 kg a c c 3.00 × 10 8 m s

x=

1 2 at 2

b

2x = 8.12 × 10 4 s = 22.6 h a

t= (b)

F=

b

g b

gb

g b

g

b

g

0 = 107 kg v − 3.00 kg 12.0 m s − v = 107 kg v − 36.0 kg ⋅ m s + 3.00 kg v v=

36.0 = 0.327 m s 110

t = 30.6 s

g

Chapter 34

P34.58

The mirror intercepts power

j a

e

f

P = I 1 A1 = 1.00 × 10 3 W m 2 π 0.500 m

2

309

= 785 W .

In the image, 785 W

(a)

I2 =

P : A2

I2 =

(b)

I2 =

2 Emax so 2µ 0 c

Emax = 2 µ 0 cI 2 = 2 4π × 10 −7 3.00 × 10 8 6.25 × 10 5 = 21.7 kN C

b

0.400P∆t = mc∆T

a

= 625 kW m 2

2

e

Bmax = (c)

g

π 0.020 0 m

f b

je

je

Emax = 72.4 µT c

gb

ga

0. 400 785 W ∆t = 1.00 kg 4 186 J kg⋅° C 100° C − 20.0° C ∆t = P34.59

f

5

3.35 × 10 J = 1.07 × 10 3 s = 17.8 min 314 W

Think of light going up and being absorbed by the bead which presents a face area π rb2 . The light pressure is P =

P34.60

j

S I = . c c

Iπ rb2 4 = mg = ρ π rb3 g 3 c

(a)

F =

(b)

P = IA = 8.32 × 10 7 W m 2 π 2.00 × 10 −3 m

e

and

je

I=

F GH

4ρ gc 3m 3 4π ρ

j

2

I JK

13

= 8.32 × 10 7 W m 2

= 1.05 kW

Think of light going up and being absorbed by the bead, which presents face area π rb2 . S I F . If we take the bead to be perfectly absorbing, the light pressure is P = av = = c c A (a)

F = Fg so

I=

F c Fg c mgc = = . A A π rb2

From the definition of density,

ρ=

m m = V 4 3 π rb3

so

b g 1 F b 4 3gπ ρ I =G r H m JK mgc F 4π ρ I I= G J π H 3m K

13

.

b

Substituting for rb ,

(b)

P = IA

23

F 4ρ I F m I = gc G J G J H 3 K HπK

FG IJ H K

4π r 2 ρ gc 3m P= 3 4πρ

23

13

13

F GH

4ρgc 3m = 3 4π ρ

I JK

13

.

310 P34.61

Electromagnetic Waves

c 3.00 × 10 8 m s = = 1.50 cm f 20.0 × 10 9 s −1

(a)

λ=

(b)

U = P ∆t = 25.0 × 10 3 J s 1.00 × 10 −9 s = 25.0 × 10 −6 J

a f e

je

j

= 25.0 µJ

(c)

uav =

FIG. P34.61

U U = V π r2

U

=

25.0 × 10 −6 J

=

e j eπ r jca∆tf π b0.060 0 mg e3.00 × 10 2

2

je

m s 1.00 × 10 −9 s

8

j

uav = 7.37 × 10 −3 J m3 = 7.37 mJ m 3

(d)

Bmax =

(e)

P34.62

(a)

(b)

2uav = ∈0

Emax =

e

2 7.37 × 10 −3 J m3 8.85 × 10

−12

C N ⋅ m2

= 4.08 × 10 4 V m = 40.8 kV m

Emax 4.08 × 10 4 V m = = 1.36 × 10 −4 T = 136 µT c 3.00 × 10 8 m s

F = PA =

FG S IJ A = u H cK

av A

jb

e

= 7.37 × 10 −3 J m3 π 0.060 0 m

On the right side of the equation,

eC

2

N ⋅m

2

j

2

2

jbm sg

3

= 8.33 × 10 −5 N = 83.3 µN

N ⋅ m2 ⋅ C 2 ⋅ m2 ⋅ s3

=

2

4

C ⋅s ⋅m

jb

e

F = ma = qE or

F = ma c = m

e

C2 m s2

g

3

=

N ⋅m J = = W. s s

g

−19 C 100 N C qE 1.60 × 10 = = 1.76 × 10 13 m s 2 . a= −31 m 9.11 × 10 kg

The radiated power is then:

(c)

j

2

F v I = qvB so GH r JK 2

P=

v=

q2 a2 6π ∈0 c 3

e1.60 × 10 j e1.76 × 10 j = 6π e8.85 × 10 je3.00 × 10 j −19 2

13 2

−12

8 3

= 1.75 × 10 −27 W .

qBr . m

ja

e

2

fa j 2

f

1.60 × 10 −19 0.350 0.500 v2 q 2B2r = = = 5.62 × 10 14 m s 2 . The proton accelerates at a = 2 2 − 27 r m 1.67 × 10

e

The proton then radiates P =

P34.63

P=

q2 a2 6π ∈0 c 3

e1.60 × 10 j e5.62 × 10 j = 6π e8.85 × 10 je3.00 × 10 j −19 2

14 2

−12

8 3

= 1.80 × 10 −24 W .

P S Power 60.0 W = = = = 6.37 × 10 −7 Pa c Ac 2π r c 2π 0.050 0 m 1.00 m 3.00 × 10 8 m s

b

ga

fe

j

Chapter 34

P34.64

b g

FG IJ H K e3.00 × 10 jb0.060 0g = P Therefore, θ = = 2 cκ 2e3.00 × 10 je1.00 × 10 j F = PA =

P AA P P SA = = ,τ =F = , and τ = κθ . c c c 2 2c −3

−11

8

P34.65

P34.66

The light intensity is

I = Sav =

The light pressure is

P=

For the asteroid,

PA = ma

3.00 × 10 −2 deg .

E2 . 2µ 0 c

1 S E2 = = ∈0 E 2 . c 2µ 0 c 2 2 and

a=

∈0 E 2 A . 2m

f = 90.0 MHz , Emax = 2.00 × 10 −3 V m = 200 mV m (a)

λ=

c = 3.33 m f

1 = 1.11 × 10 −8 s = 11.1 ns f E Bmax = max = 6.67 × 10 −12 T = 6.67 pT c

T=

(b)

b

g

E = 2.00 mV m cos 2π

e

FG x − t IJ j H 3.33 m 11.1 ns K j

2

b

(c)

2.00 × 10 −3 E2 = 5.31 × 10 −9 W m 2 I = max = 2 µ 0 c 2 4π × 10 −7 3.00 × 10 8

(d)

I = cuav

(e)

2 5.31 × 10 −9 2I P= = = 3.54 × 10 −17 Pa c 3.00 × 10 8

e

so

a fe

je

j

uav = 1.77 × 10 −17 J m 3

j

g

B = 6.67 pT k cos 2π

FG x − t IJ H 3.33 m 11.1 ns K

311

312 *P34.67

Electromagnetic Waves

(a)

m = ρV = ρ

F 6m IJ r =G H ρ 4π K

14 3 πr 23

F 6b8.7 kg g I =G GH e990 kg m j4π JJK = = 2π a0.161 mf = 0.163 m 13

13

0.161 m

3

1 4π r 2 2

2

2

(b)

A=

(c)

I = eσ T 4 = 0.970 5.67 × 10 −8 W m 2 ⋅ K 4 304 K

(d)

P = IA = 470 W m 2 0.163 m 2 = 76.8 W

(e)

I=

e

= 470 W m 2

g

12

e

je

je

= 8π × 10 −7 Tm A 3 × 10 8 m s 470 W m 2

j

12

= 595 N C

Emax = cBmax Bmax =

(h)

4

2 Emax 2µ 0 c

b

(g)

f

j

Emax = 2 µ 0 cI (f)

ja

e

595 N C 3 × 10 8 m s

= 1.98 µT

The sleeping cats are uncharged and nonmagnetic. They carry no macroscopic current. They are a source of infrared radiation. They glow not by visible-light emission but by infrared emission. Each kitten has radius rk

b

g

2π 0.072 8 m

2

F 6a0.8f IJ =G H 990 × 4π K

13

= 0.072 8 m and radiating area

= 0.033 3 m 2 . Eliza has area 2π

e

j

FG 6a5.5f IJ H 990 × 4π K

23

= 0.120 m 2 . The total glowing

area is 0.120 m 2 + 4 0.033 3 m 2 = 0.254 m 2 and has power output

e

2

j

2

P = IA = 470 W m 0.254 m = 119 W . P34.68

(a)

(b)

At steady state, Pin = Pout and the power radiated out is Pout = eσAT 4 . .

e

j

e

j

Thus,

0.900 1 000 W m 2 A = 0.700 5.67 × 10 −8 W m 2 ⋅ K 4 AT 4

or

L 900 W m T=M MN 0.700e5.67 × 10 W m 2

−8

2

⋅K4

OP j PQ

14

= 388 K = 115° C .

The box of horizontal area A, presents projected area A sin 50.0° perpendicular to the sunlight. Then by the same reasoning,

e

j

e

j

0.900 1 000 W m 2 A sin 50.0° = 0.700 5.67 × 10 −8 W m 2 ⋅ K 4 AT 4

L e900 W m j sin 50.0° OP T=M MN 0.700e5.67 × 10 W m ⋅ K j PQ 2

or

−8

2

4

14

= 363 K = 90.0° C .

Chapter 34

P34.69

313

We take R to be the planet’s distance from its star. The planet, of radius r, presents a

projected area π r 2 perpendicular to the starlight. It radiates over area 4π r 2 .

e j

e

F 6.00 × 10 GH 4π R

R=

23

W

2

e

j

eI in π r 2 = eσ 4π r 2 T 4

At steady-state, Pin = Pout :

I π r = eσ 4π r T JK e j e j 2

2

4

so that 6.00 × 10 23 W = 16πσR 2 T 4

6.00 × 10 23 W 6.00 × 10 23 W = 4 16πσ T 16π 5.67 × 10 −8 W m 2 ⋅ K 4 310 K

ja

e

f

4

= 4.77 × 10 9 m = 4.77 Gm .

ANSWERS TO EVEN PROBLEMS P34.2

(a) 2.68 × 10 3 AD ; (b) 8.31 min ; (c) 2.56 s; (d) 0.133 s; (e) 33.3 µs

P34.4

733 nT

P34.6

E = 300 V m cos 62.8 x − 1.88 × 10 10 t ;

b g e j B = b1.00 µTg cose62.8 x − 1.88 × 10 t j 10

P34.8

see the solution

P34.10

2.9 × 10 8 m s ± 5%

P34.12

49.5 mV

P34.14

3

P34.30

(a) 577 W m 2 ; (b) 2.06 × 10 16 W ; (c) 68.7 MN ; (d) The gravitational force is ~ 10 13 times stronger and in the opposite direction.

P34.32

4.09°

P34.34

(a) 93.3% ; (b) 50.0% ; (c) 0

P34.36

3

(a) 13.3 nJ m ; (b) 13.3 nJ m ;

516 pT, ~ 10 5 times stronger than the Earth’s field

P34.18

(a) 11.9 GW m 2 ; (b) 234 kW

P34.20

33.4°C for the smaller container and 21.7°C for the larger

P34.22

(a) 50.0% ; (b) 269 kW m 2 toward the oven chamber ; (c) 14. 2 kV m

P34.24

(a) 4.97 kW m 2 ; (b) 16.6 µJ m 3

P34.26

667 pN

P34.28

(a) 5.36 N ; (b) 893 µm s 2 ; (c) 10.7 days

eB

P34.38

see the solution

P34.40

(a) ~ 10 8 Hz radio wave; (b) ~ 10 13 Hz infrared light

P34.42

(a) 0.690 wavelengths ; (b) 58.9 wavelengths

P34.44

The radio audience gets the news 8.41 ms sooner.

P34.46

(a) 187 m to 556 m; (b) 2.78 m to 3.41 m

P34.48

~ 10 6 J

P34.50

(a) see the solution; (b) 378 nm

P34.52

(a) 31.4 MW; (b) 0.625 W m 2 ; (c) 0.513%

P34.54

(a) 23.9 W m 2 ; (b) 4.19 times the standard

P34.56

(a) 6.16 µPa ; (b) 1.64 × 10 10 times less than atmospheric pressure

(c) 7.96 W m 2 P34.16

2π m p c

314 P34.58

P34.60 P34.62

Electromagnetic Waves

(a) 625 kW m 2 ; (b) 21.7 kN C and 72.4 µT ; (c) 17.8 min

F 16mρ I (a) G H 9π JK 2

13

F 16π mρ I gc ; (b) G H 9 JK

(a) see the solution; (b) 17.6 Tm s 2 , 1.75 × 10 −27 W ; (c) 1.80 × 10 −24 W

P34.64

2

3.00 × 10 −2 deg

2

P34.66

(a) 3.33 m, 11.1 ns , 6.67 pT ;

b g FGH 3.33x m − 11.1t ns IJK j ; F x − t IJ ; B = b6.67 pTgk cos 2π G H 3.33 m 11.1 ns K

(b) E = 2.00 mV m cos 2π

13

r 2 gc

(c) 5.31 nW m 2 ; (d) 1.77 × 10 −17 J m3 ; (e) 3.54 × 10 −17 Pa P34.68

(a) 388 K; (b) 363 K

35 The Nature of Light and the Laws of Geometric Optics CHAPTER OUTLINE 35.1 35.2 35.3 35.4 35.5 35.6 35.7 35.8 35.9

The Nature of Light Measurements of the Speed of Light The Ray Approximation in Geometric Optics Reflection Refraction Huygen’s Principle Dispersion and Prisms Total Internal Reflection Fermat‘s Principle

Q35.3

ANSWERS TO QUESTIONS Q35.1

The ray approximation, predicting sharp shadows, is valid for λ 0 , γ is negative and multiple reflections from each mirror will occur before the incident and reflected rays intersect.

327

Chapter 35

Section 35.6 *P35.28

Huygen’s Principle

(a)

For the diagrams of contour lines and wave fronts and rays, see Figures (a) and (b) below. As the waves move to shallower water, the wave fronts bend to become more nearly parallel to the contour lines.

(b)

For the diagrams of contour lines and wave fronts and rays, see Figures (c) and (d) below. We suppose that the headlands are steep underwater, as they are above water. The rays are everywhere perpendicular to the wave fronts of the incoming refracting waves. As shown, the rays bend toward the headlands and deliver more energy per length at the headlands.

(a) Contour lines

(b) Wave fronts and rays

(c) Contour lines

(d) Wave fronts and rays

FIG. P35.28

Section 35.7 P35.29

Dispersion and Prisms

From Fig 35.21

n v = 1.470 at 400 nm

and

n r = 1.458 at 700 nm.

Then

1.00 sin θ = 1. 470 sin θ v

and

1.00 sin θ = 1. 458 sin θ r

FG sinθ IJ − sin FG sinθ IJ H 1.458 K H 1.470 K FG sin 30.0° IJ − sin FG sin 30.0° IJ = 0.171° H 1.458 K H 1.470 K

δ r − δ v = θ r − θ v = sin −1 ∆δ = sin −1 P35.30

a

−1

−1

f

n 700 nm = 1.458 (a) (b)

a1.00f sin 75.0° = 1.458 sinθ

2

; θ 2 = 41.5°

θ1

Let

θ 3 + β = 90.0° , θ 2 + α = 90.0° then α + β + 60.0° = 180° .

So

60.0°−θ 2 − θ 3 = 0 ⇒ 60.0°−41.5° = θ 3 = 18.5° .

(c)

1.458 sin 18.5° = 1.00 sin θ 4

(d)

γ = θ 1 − θ 2 + β − 90.0°−θ 4

b

g

θ 4 = 27.6°

b

g γ = 75.0°−41.5°+a90.0°−18.5°f − a90.0°−27.6°f =

42.6°

α

60.0°

β θ2

θ3

FIG. P35.30

θ4

γ

328 P35.31

The Nature of Light and the Laws of Geometric Optics

Taking Φ to be the apex angle and δ min to be the angle of minimum deviation, from Equation 35.9, the index of refraction of the prism material is n= Solving for δ min ,

P35.32

b

g

sin Φ + δ min 2

δ min

b g F ΦI = 2 sin G n sin J − Φ = 2 sin a 2.20f sina 25.0°f − 50.0° = H 2K sin Φ 2 −1

−1

b

g b

g

Note for use in every part:

Φ + 90.0°−θ 2 + 90.0°−θ 3 = 180°

so

θ3 = Φ −θ 2 . α = θ1 −θ 2 .

At the first surface the deviation is At exit, the deviation is

β = θ 4 −θ3 .

The total deviation is therefore

δ = α + β = θ1 +θ 4 −θ 2 −θ3 = θ1 +θ 4 − Φ .

(a)

86.8° .

At entry:

n1 sin θ 1 = n 2 sin θ 2

Thus,

θ 3 = 60.0°−30.0° = 30.0° .

At exit:

1.50 sin 30.0° = 1.00 sin θ 4

FIG. P35.32

FG sin 48.6° IJ = 30.0° . H 1.50 K

or

θ 2 = sin −1

or

θ 4 = sin −1 1.50 sin 30.0° = 48.6°

a

f

so the path through the prism is symmetric when θ 1 = 48.6° . (b)

δ = 48.6°+48.6°−60.0° = 37.2°

(c)

At entry:

(d)

At exit:

sin 45.6° ⇒ θ 2 = 28. 4° 1.50 sin θ 4 = 1.50 sin 31.6° ⇒ θ 4 = 51.7°

At entry:

sin θ 2 =

At exit: P35.33

sin θ 2 =

a

f

sin 51.6° ⇒ θ 2 = 31.5° 1.50 sin θ 4 = 1.50 sin 28.5° ⇒ θ 4 = 45.7°

At the first refraction,

a

f

θ 3 = 60.0°−28.4° = 31.6° .

δ = 45.6°+51.7°−60.0° = 37.3° . θ 3 = 60.0°−31.5° = 28.5° .

δ = 51.6°+45.7°−60.0° = 37.3° .

1.00 sin θ 1 = n sin θ 2 .

The critical angle at the second surface is given by n sin θ 3 = 1.00 :

FG 1.00 IJ = 41.8° . H 1.50 K

or

θ 3 = sin −1

But,

θ 2 = 60.0°−θ 3 .

it is necessary that

θ 2 > 18.2° .

Since sin θ 1 = n sin θ 2 , this becomes

sin θ 1 > 1.50 sin 18.2° = 0.468

or

θ 1 > 27.9° .

FIG. P35.33 Thus, to avoid total internal reflection at the second surface (i.e., have θ 3 < 41.8° )

Chapter 35

P35.34

At the first refraction, 1.00 sin θ 1 = n sin θ 2 . The critical angle at the second surface is given by

FG IJ H K g + b90.0°−θ g + Φ = 180°

n sin θ 3 = 1.00 , or

θ 3 = sin −1

But

b90.0°−θ

which gives

θ 2 = Φ −θ3 .

2

θ1

1.00 . n

Φ

θ2

θ3

3

FIG. P35.34

FG 1.00 IJ and avoid total internal reflection at the second surface, HnK F 1.00 IJ . θ > Φ − sin G it is necessary that HnK L F 1.00 IJ OP Since sin θ = n sin θ , this requirement becomes sin θ > n sin MΦ − sin G H n KQ N F L F 1.00 IJ OPIJ . θ > sin G n sin MΦ − sin G or H n K QK H N θ > sin FH n − 1 sin Φ − cos ΦIK . Through the application of trigonometric identities,

Thus, to have θ 3 < sin −1

−1

2

1

−1

1

2

−1

1

−1

1

P35.35

For the incoming ray,

sin θ 2 =

Using the figure to the right,

bθ g

2 violet

bθ g

2 red

For the outgoing ray, and sin θ 4 = n sin θ 3 :

sin θ 1 . n

= sin −1

bθ g bθ g

4 violet

n sin θ = 1 . From Table 35.1, (a)

θ = sin −1

(b)

θ = sin −1

(c)

θ = sin −1

FG 1 IJ = 24.4° H 2.419 K FG 1 IJ = 37.0° H 1.66 K FG 1 IJ = 49.8° H 1.309 K

FIG. P35.35

= sin −1 1.66 sin 32.52° = 63.17°

= sin −1 1.62 sin 31.78° = 58.56° .

b g

P35.36

FG sin 50.0° IJ = 27.48° H 1.66 K FG sin 50.0° IJ = 28.22° . H 1.62 K

= sin −1

The angular dispersion is the difference ∆θ 4 = θ 4

Total Internal Reflection

2

θ 3 = 60.0°−θ 2

4 red

Section 35.8

−1

violet

b g

− θ4

red

= 63.17°−58.56° = 4.61° .

329

330 P35.37

P35.38

The Nature of Light and the Laws of Geometric Optics

FG n IJ Hn K FG 1.333 IJ = H 2.419 K FG 1.333 IJ = H 1.66 K

sin θ c =

n2 : n1

θ c = sin −1

(a)

Diamond:

θ c = sin −1

(b)

Flint glass:

θ c = sin −1

(c)

Ice:

Since n 2 > n1 , there is no critical angle .

sin θ c =

n air 1.00 = = 0.735 n pipe 1.36

2

1

33.4°

53.4°

θ c = 47.3°

Geometry shows that the angle of refraction at the end is φ = 90.0°−θ c = 90.0°−47.3° = 42.7° . 1.00 sin θ = 1.36 sin 42.7°

Then, Snell’s law at the end,

FIG. P35.38

θ = 67.2° .

gives

The 2-µm diameter is unnecessary information. P35.39

sin θ c =

n2 n1

b

gb

g

n 2 = n1 sin 88.8° = 1.000 3 0.999 8 = 1.000 08 *P35.40

(a)

A ray along the inner edge will escape if any ray escapes. Its angle of R−d and by n sin θ > 1 sin 90° . Then incidence is described by sin θ = R n R−d nd . >1 nR − nd > R nR − R > nd R> n−1 R

a

(b)

(c) P35.41

FIG. P35.39

f

As d → 0 , Rmin → 0 .

This is reasonable.

As n increases, Rmin decreases.

This is reasonable.

As n decreases toward 1, Rmin increases.

This is reasonable.

Rmin =

e

j=

1.40 100 × 10 −6 m 0.40

350 × 10 −6 m

From Snell’s law, n1 sin θ 1 = n 2 sin θ 2 . At the extreme angle of viewing, θ 2 = 90.0°

a1.59fbsinθ g = a1.00f sin 90.0° . 1

So

θ 1 = 39.0° .

Therefore, the depth of the air bubble is rp rd R 2 , then sin θ 2 cannot be equal to nR1 . The ray considered in part (a) R2 undergoes total internal reflection. In this case a ray escaping the atmosphere as shown here is responsible for the apparent radius of the glowing sphere and R3 = R 2 .

sin θ c =

341

4t n2 − 1

2

or n = 1 +

P

d /4 d

2

.

FG 4t IJ HdK

FIG. P35.69 2

.

.

a f= a1.52f − 1

4 0.600 cm 2

B'

t

2.10 cm .

Since violet light has a larger index of refraction, it will lead to a smaller critical angle and the inner edge of the white halo will be tinged with violet light.

342 P35.70

The Nature of Light and the Laws of Geometric Optics

From the sketch, observe that the angle of incidence at point A is the same as the prism angle θ at point O. Given that θ = 60.0° , application of Snell’s law at point A gives 1.50 sin β = 1.00 sin 60.0° or β = 35.3° . From triangle AOB, we calculate the angle of incidence (and reflection) at point B.

b

g b

FIG. P35.70

g

θ = 90.0°− β + 90.0°−γ = 180° so

γ = θ − β = 60.0°−35.3° = 24.7° .

Thus the angle of incidence at point C is

b90.0°−γ g + b90.0°−δ g + a90.0°−θ f = 180° . δ = a90.0°−θ f − γ = 30.0°−24.7° = 5.30° .

Finally, Snell’s law applied at point C gives

1.00 sin φ = 1.50 sin 5.30°

or

φ = sin −1 1.50 sin 5.30° = 7.96° .

Now, using triangle BCQ:

P35.71

(a)

a

f

Given that θ 1 = 45.0° and θ 2 = 76.0° . Snell’s law at the first surface gives n sin α = 1.00 sin 45.0°

(1)

Observe that the angle of incidence at the second surface is

β = 90.0°−α . Thus, Snell’s law at the second surface yields

a

f

n sin β = n sin 90.0°−α = 1.00 sin 76.0° or

(b)

n cos α = sin 76.0° .

FIG. P35.71

(2) sin 45.0° = 0.729 sin 76.0°

Dividing Equation (1) by Equation (2),

tan α =

or

α = 36.1° .

Then, from Equation (1),

n=

sin 45.0° sin 45.0° = = 1.20 . sin α sin 36.1°

From the sketch, observe that the distance the light travels in the plastic is d = the speed of light in the plastic is v = ∆t =

a

f

c , so the time required to travel through the plastic is n

1.20 0.500 m d nL = = = 3.40 × 10 −9 s = 3.40 ns . v c sin α 3.00 × 10 8 m s sin 36.1°

e

j

L . Also, sin α

Chapter 35

P35.72

sin θ 1

sin θ 2

0.174

0.131

343

sin θ 1 sin θ 2 1.330 4

0.342

0.261

1.312 9

0.500

0.379

1.317 7

0.643

0.480

1.338 5

0.766

0.576

1.328 9

0.866

0.647

1.339 0

0.940

0.711

1.322 0

0.985

0.740

1.331 5

The straightness of the graph line demonstrates Snell’s proportionality. The slope of the line is

n = 1.327 6 ± 0.01

and

n = 1.328 ± 0.8% .

FIG. P35.72

ANSWERS TO EVEN PROBLEMS P35.2

227 Mm s

P35.30

(a) 41.5° ; (b) 18.5°; (c) 27.6°; (d) 42.6°

P35.4

(a) see the solution; (b) 300 Mm s

P35.32

(a) see the solution; (b) 37.2°; (c) 37.3° ; (d) 37.3°

P35.6

(a) 1.94 m; (b) 50.0° above the horizontal : antiparallel to the incident ray

P35.34

sin −1

P35.36

(a) 24.4° ; (b) 37.0° ; (c) 49.8°

P35.38

67.2

FH

n 2 − 1 sin Φ − cos Φ

IK

P35.8

five times by the right-hand mirror and six times by the left-hand mirror

P35.10

25.5°; 442 nm

P35.12

(a) 474 THz ; (b) 422 nm; (c) 200 Mm s

P35.40

(a)

P35.14

22.5°

P35.42

P35.16

(a) 181 Mm s ; (b) 225 Mm s ; (c) 136 Mm s

(a) 10.7°; (b) air; (c) Sound falling on the wall from most directions is 100% reflected.

P35.44

54.8° east of north

P35.46

(a)

P35.48

see the solution

P35.50

see the solution

P35.52

(a) 45.0°; (b) yes; see the solution

P35.54

3.79 m

P35.18

3.39 m −1

P35.20

θ 1 = tan n

P35.22

106 ps

P35.24

23.1°

P35.26

(a) 58.9° ; (b) Only if θ 1 = θ 2 = 0

P35.28

see the solution

nd ; (b) yes; (c) 350 µm n −1

FG H

IJ K

h n +1 n +1 ; (b) larger by times 2 2 c

344 P35.56 P35.58

The Nature of Light and the Laws of Geometric Optics

(a) 0.042 6 ; (b) no difference

θ = sin −1

P35.68

(a) nR1 ; (b) R 2

P35.70

7.96°

P35.72

see the solution; n = 1.328 ± 0.8%

0.706

P35.60

(a) 2ω m R ; (b) 2ω m

P35.62

164 s

P35.64

36.5°

x2 + d2 d

LM L F NR H

P35.66

2

n 2 R 2 − L2 − R 2 − L2

IK OP Q

36 Image Formation CHAPTER OUTLINE Images Formed by Flat Mirrors 36.2 Images Formed by Spherical Mirrors 36.3 Images Formed by Refraction 36.4 Thin Lenses 36.5 Lens Aberrations 36.6 The Camera 36.7 The Eye 36.8 The Simple Magnifier 36.9 The Compound Microscope 36.10 The Telescope

ANSWERS TO QUESTIONS

36.1

Q36.1

The mirror shown in the textbook picture produces an inverted image. It actually reverses top and bottom. It is not true in the same sense that “Most mirrors reverse left and right.” Mirrors don’t actually flip images side to side—we just assign the labels “left” and “right” to images as if they were real people mimicking us. If you stand face to face with a real person and raise your left hand, then he or she would have to raise his or her right hand to “mirror” your movement. Try this while facing a mirror. For sake of argument, let’s assume you are facing north and wear a watch on your left hand, which is on the western side. If you raise your left hand, you might say that your image raises its right hand, based on the labels we assign to other people. But your image raises its western-side hand, which is the hand with the watch.

Q36.2

With a concave spherical mirror, for objects beyond the focal length the image will be real and inverted. For objects inside the focal length, the image will be virtual, upright, and magnified. Try a shaving or makeup mirror as an example.

Q36.3

With a convex spherical mirror, all images of real objects are upright, virtual and smaller than the object. As seen in Question 36.2, you only get a change of orientation when you pass the focal point—but the focal point of a convex mirror is on the non-reflecting side!

Q36.4

The mirror equation and the magnification equation apply to plane mirrors. A curved mirror is made flat by increasing its radius of curvature without bound, so that its focal length goes to infinity. 1 1 1 1 1 From + = = 0 we have = − ; therefore, p = − q . The virtual image is as far behind the mirror p q f p q q p as the object is in front. The magnification is M = − = = 1 . The image is right side up and actual p p size.

Q36.5

Stones at the bottom of a clear stream always appears closer to the surface because light is refracted away from the normal at the surface. Example 36.8 in the textbook shows that its apparent depth is three quarters of its actual depth.

345

346 Q36.6

Image Formation

For definiteness, we consider real objects (p > 0 ). q 1 1 1 to be negative, q must be positive. This will happen in = − if p > f , if the q f p p object is farther than the focal point.

(a)

For M = −

(b)

For M = −

(c)

For a real image, q must be positive. As in part (a), it is sufficient for p to be larger than f.

(d)

For q < 0 we need p < f .

(e)

For M > 1 , we consider separately q If M = − < −1, we need p

q to be positive, q must be negative. p 1 1 1 From = − we need p < f . q f p

From

q >1 p

1 1 < q p

1 1 1 + > p p f

or

2 1 > p f

or

p 0 f

or

−p > q.

we may require q < 0 , since then gives

q>p

or 1 1 1 + = , p q f

Now if −

M < −1 and M > 1 . q >1 or p

1 1 > − as required p q

1 1 1 = − we need p< f. q f p Thus the overall condition for an enlarged image is simply p < 2 f .

For q < 0 in

(f) Q36.7

For M < 1 , we have the reverse of part (e), requiring p > 2 f .

Using the same analysis as in Question 36.6 except f < 0 . (a)

Never.

(b)

Always.

(c)

Never, for light rays passing through the lens will always diverge.

(d)

Always.

(e)

Never.

(f)

Always.

Chapter 36

Q36.8

347

We assume the lens has a refractive index higher than its surroundings. For the biconvex lens in 1 1 − are positive and f > 0 . For the Figure 36.27(a), R1 > 0 and R 2 < 0 . Then all terms in n − 1 R1 R 2 other two lenses in part (a) of the figure, R1 and R 2 are both positive but R1 is less than R 2 . Then 1 1 > and the focal length is again positive. R1 R 2

a fFGH

IJ K

For the biconcave lens and the plano-concave lens in Figure 36.27(b), R1 < 0 and R 2 > 0 . Then 1 1 − and the focal length is negative. For the middle lens in part (b) both terms are negative in R1 R 2 1 1 of the figure, R1 and R 2 are both positive but R1 is greater than R 2 . Then < and the focal R1 R 2 length is again negative. Q36.9

Both words are inverted. However OXIDE has up-down symmetry whereas LEAD does not.

Q36.10

An infinite number. In general, an infinite number of rays leave each point of any object and travel in all directions. Note that the three principal rays that we use for imaging are just a subset of the infinite number of rays. All three principal rays can be drawn in a ray diagram, provided that we extend the plane of the lens as shown in Figure Q36.10.

O

F

F

I

FIG. Q36.10 Q36.11

In this case, the index of refraction of the lens material is less than that of the surrounding medium. Under these conditions, a biconvex lens will be diverging.

Q36.12

Chromatic aberration arises because a material medium’s refractive index can be frequency dependent. A mirror changes the direction of light by reflection, not refraction. Light of all wavelengths follows the same path according to the law of reflection, so no chromatic aberration happens.

Q36.13

This is a convex mirror. The mirror gives the driver a wide field of view and an upright image with the possible disadvantage of having objects appear diminished. Your brain can then interpret them as farther away than the objects really are.

Q36.14

As pointed out in Question 36.11, if the converging lens is immersed in a liquid with an index of refraction significantly greater than that of the lens itself, it will make light from a distant source diverge. This is not the case with a converging (concave) mirror, as the law of reflection has nothing to do with the indices of refraction.

348 Q36.15

Image Formation

As in the diagram, let the center of curvature C of the fishbowl and the bottom of the fish define the optical axis, intersecting the fishbowl at vertex V. A ray from the top of the fish that reaches the bowl surface along a radial line through C has angle of incidence zero and angle of refraction zero. This ray exits from the bowl unchanged in direction. A ray from the top of the fish to V is refracted to bend away from the normal. Its extension back inside the fishbowl determines the location of the image and the characteristics of the image. The image is upright, virtual, and enlarged.

C

V

O I

FIG. Q36.15 Q36.16

Because when you look at the in your rear view mirror, the apparent left-right inversion clearly displays the name of the AMBULANCE behind you. Do not jam on your brakes when a MIAMI city bus is right behind you.

Q36.17

The entire image is visible, but only at half the intensity. Each point on the object is a source of rays that travel in all directions. Thus, light from all parts of the object goes through all unblocked parts of the lens and forms an image. If you block part of the lens, you are blocking some of the rays, but the remaining ones still come from all parts of the object.

Q36.18

With the meniscus design, when you direct your gaze near the outer circumference of the lens you receive a ray that has passed through glass with more nearly parallel surfaces of entry and exit. Thus, the lens minimally distorts the direction to the object you are looking at. If you wear glasses, turn them around and look through them the wrong way to maximize this distortion.

Q36.19

The eyeglasses on the left are diverging lenses that correct for nearsightedness. If you look carefully at the edge of the person’s face through the lens, you will see that everything viewed through these glasses is reduced in size. The eyeglasses on the right are converging lenses, which correct for farsightedness. These lenses make everything that is viewed through them look larger.

Q36.20

The eyeglass wearer’s eye is at an object distance from the lens that is quite small—the eye is on the order of 10 −2 meter from the lens. The focal length of an eyeglass lens is several decimeters, positive or negative. Therefore the image distance will be similar in magnitude to the object distance. The onlooker sees a sharp image of the eye behind the lens. Look closely at the left side of Figure Q36.19 and notice that the wearer’s eyes seem not only to be smaller, but also positioned a bit behind the plane of his face—namely where they would be if he was not wearing glasses. Similarly, in the right half of Figure Q36.19, his eyes seem to be in front of the plane of his face and magnified. We as observers take this light information coming from the object through the lens and perceive or photograph the image as if it were an object.

Q36.21

In the diagram, only two of the three principal rays have been used to locate images to reduce the amount of visual clutter. The upright shaded arrows are the objects, and the correspondingly numbered inverted arrows are the images. As you can see, object 2 is closer to the focal point than object 1, and image 2 is farther to the left than image 1.

O1

O2 F

C

V I2 I1

FIG. Q36.21

Chapter 36

349

Q36.22

Absolutely. Only absorbed light, not transmitted light, contributes internal energy to a transparent object. A clear lens can stay ice-cold and solid as megajoules of light energy pass through it.

Q36.23

One can change the f number either by changing the focal length (if using a “zoom” lens) or by changing the aperture of the camera lens. As the f number increases, the exposure time required increases also, as both increasing the focal length or decreasing the aperture decreases the light intensity reaching the film.

Q36.24

Make the mirror an efficient reflector (shiny). Make it reflect to the image even rays far from the axis, by giving it a parabolic shape. Most important, make it large in diameter to intercept a lot of solar power. And you get higher temperature if the image is smaller, as you get with shorter focal length; and if the furnace enclosure is an efficient absorber (black).

Q36.25

For the explanation, we ignore the lens and consider two objects. Hold your two thumbs parallel and extended upward in front of you, at different distances from your nose. Alternately close your left eye and your right eye. You see both thumbs jump back and forth against the background of more distant objects. Parallax by definition is this apparent motion of a stationary object (one thumb) caused by motion of the observer (jumping from right eye to left eye). Your nearer thumb jumps by a larger angle against the background than your farther thumb does. They will jump by the same amount only if they are equally distant from your face. The method of parallax for adjusting one object so that it is the same distance away from you as another object will work even if one ’object’ is an image.

Q36.26

The artist’s statements are accurate, perceptive, and eloquent. The image you see is “almost one’s whole surroundings,” including things behind you and things farther in front of you than the globe is, but nothing eclipsed by the opaque globe or by your head. For example, we cannot see Escher’s index and middle fingers or their reflections in the globe. The point halfway between your eyes is indeed the focus in a figurative sense, but it is not an optical focus. The principal axis will always lie in a line that runs through the center of the sphere and the bridge of your nose. Outside the globe, you are at the center of your observable universe. If you wink at the ball, the center of the looking-glass world hops over to the location of the image of your open eye.

Q36.27

The three mirrors, two of which are shown as M and N in the figure to the right, reflect any incident ray back parallel to its original direction. When you look into the corner you see image I 3 of yourself.

FIG. Q36.21

350 Q36.28

Image Formation

You have likely seen a Fresnel mirror for sound. The diagram represents first a side view of a band shell. It is a concave mirror for sound, designed to channel sound into a beam toward the audience in front of the band shell. Sections of its surface can be kept at the right orientations as they are pushed around inside a rectangular box to form an auditorium with good diffusion of sound from stage to audience, with a floor plan suggested by the second part of the diagram.

FIG. Q36.28

SOLUTIONS TO PROBLEMS Section 36.1 P36.1

Images Formed by Flat Mirrors

I stand 40 cm from my bathroom mirror. I scatter light, which travels to the mirror and back to me in time 0.8 m ~ 10 −9 s 8 3 × 10 m s showing me a view of myself as I was at that look-back time. I’m no Dorian Gray!

P36.2

The virtual image is as far behind the mirror as the choir is in front of the mirror. Thus, the image is 5.30 m behind the mirror. The image of the choir is 0.800 m + 5.30 m = 6.10 m from the organist. Using similar triangles:

View Looking Down South image of choir mirror

h′ 6.10 m = 0.600 m 0.800 m or

a

6.10 m I fFGH 0.800 J= mK

h′ = 0.600 m

0.600 m

4.58 m .

Organist 0.800 m

5.30 m

FIG. P36.2

h'

Chapter 36

P36.3

351

The flatness of the mirror is described

R=∞, f =∞ 1 =0. f

by and

By our general mirror equation, 1 1 1 + = =0 p q f q = −p .

or

FIG. P36.3

Thus, the image is as far behind the mirror as the person is in front. The magnification is then M=

−q h′ =1= p h

h′ = h = 70.0 inches .

so

The required height of the mirror is defined by the triangle from the person’s eyes to the top and bottom of his image, as shown. From the geometry of the triangle, we see that the mirror height must be: h′

F p I = h ′F p I = h ′ . GH p − q JK GH 2p JK 2

Thus, the mirror must be at least 35.0 inches high . P36.4

A graphical construction produces 5 images, with images I 1 and I 2 directly into the mirrors from the object O, and and

bO, I , I g bI , I , I g 2

3

4

1

5

forming the vertices of equilateral triangles.

FIG. P36.4 P36.5

(1)

The first image in the left mirror is 5.00 ft behind the mirror, or 10.0 ft from the position of the person.

(2)

The first image in the right mirror is located 10.0 ft behind the right mirror, but this location is 25.0 ft from the left mirror. Thus, the second image in the left mirror is 25.0 ft behind the mirror, or 30.0 ft from the person.

(3)

The first image in the left mirror forms an image in the right mirror. This first image is 20.0 ft from the right mirror, and, thus, an image 20.0 ft behind the right mirror is formed. This image in the right mirror also forms an image in the left mirror. The distance from this image in the right mirror to the left mirror is 35.0 ft. The third image in the left mirror is, thus, 35.0 ft behind the mirror, or 40.0 ft from the person.

352 *P36.6

Image Formation

(a)

The flat mirrors have R→∞ and

f → ∞.

The upper mirror M 1 produces a virtual, actual sized image I 1 according to 1 1 1 1 + = = =0 p1 q 1 f ∞ q1 = − p 1 with M 1 = −

q1 = +1 . p1

As shown, this image is above the upper mirror. It is the object for mirror M 2 , at object distance p 2 = p1 + h . The lower mirror produces a virtual, actualsize, right-side-up image according to 1 1 + =0 p 2 q2

b

q 2 = − p 2 = − p1 + h with M 2 = −

FIG. P36.6

g

q2 = +1 and M overall = M 1 M 2 = 1. p2

Thus the final image is at distance p1 + h behind the lower mirror. (b) (c)

It is virtual . Upright

(d)

With magnification +1 .

(e)

It does not appear to be reversed left and right. In a top view of the periscope, parallel rays from the right and left sides of the object stay parallel and on the right and left.

Chapter 36

Section 36.2 P36.7

Images Formed by Spherical Mirrors

For a concave mirror, both R and f are positive. f=

We also know that

(a)

R = 10.0 cm . 2

1 1 1 1 1 3 = − = − = q f p 10.0 cm 40.0 cm 40.0 cm

q = 13.3 cm

and

M=

q 13.3 cm =− = −0.333 . 40.0 cm p

The image is 13.3 cm in front of the mirror, real, and inverted . (b)

1 1 1 1 1 1 = − = − = q f p 10.0 cm 20.0 cm 20.0 cm

q = 20.0 cm

and

M=

q 20.0 cm =− = −1.00 . 20.0 cm p

The image is 20.0 cm in front of the mirror, real, and inverted . (c)

1 1 1 1 1 = − = − =0 q f p 10.0 cm 10.0 cm Thus,

q = infinity.

No image is formed . The rays are reflected parallel to each other. P36.8

1 1 1 1 1 = − =− − q f p 0.275 m 10.0 m

gives

q = −0.267 m .

and

diminished

Thus, the image is virtual . M=

−q −0.267 =− = 0.026 7 10.0 m p

a f

Thus, the image is upright +M

c M < 1h .

353

354 P36.9

Image Formation

(a)

1 1 2 + = p q R

gives

1 1 2 + = 30.0 cm q −40.0 cm

1 2 1 =− − = −0.083 3 cm −1 40.0 cm 30.0 cm q

so

q = −12.0 cm

a

M=

(b)

P36.10

f

−12.0 cm −q =− = 0.400 . p 30.0 cm

1 1 2 + = p q R

gives

1 1 2 + = −40.0 cm 60.0 cm q

1 2 1 =− − = −0.066 6 cm −1 40.0 cm 60.0 cm q

so

q = −15.0 cm

a

M=

(c)

a

f

a

f

f

−15.0 cm −q =− = 0.250 . p 60.0 cm

Since M > 0 , the images are upright .

With radius 2.50 m, the cylindrical wall is a highly efficient mirror for sound, with focal length f=

R = 1.25 m . 2

In a vertical plane the sound disperses as usual, but that radiated in a horizontal plane is concentrated in a sound image at distance q from the back of the niche, where 1 1 1 + = p q f

so

1 1 1 + = 2.00 m q 1.25 m

q = 3.33 m . P36.11

(a)

(b)

(c)

1 1 2 + = p q R

becomes

1 2 1 = − q 60.0 cm 90.0 cm

q = 45.0 cm

and

M=

1 1 2 + = p q R

becomes

1 2 1 = − q 60.0 cm 20.0 cm

q = −60.0 cm

and

M=

−q 45.0 cm =− = −0.500 . 90.0 cm p

a a

f f

−60.0 cm −q =− = 3.00 . p 20.0 cm

The image (a) is real, inverted and diminished. That of (b) is virtual, upright, and enlarged. The ray diagrams are similar to Figure 36.15(a) and 36.15(b) in the text, respectively.

FIG. P36.11

Chapter 36

P36.12

355

For a concave mirror, R and f are positive. Also, for an erect image, M is positive. Therefore, q M = − = 4 and q = −4p . p 1 1 1 1 1 1 3 = + becomes = − = ; from which, p = 30.0 cm . f p q 40.0 cm p 4p 4p

*P36.13

The ball is a convex mirror with R = −4. 25 cm and R f = = −2.125 cm. We have 2 M=

q 3 =− 4 p

O

3 p 4 1 1 1 + = p q f

I F

q=−

FIG. P36.13

1 1 1 + = p − 3 4 p −2.125 cm

b g

3 4 1 − = 3 p 3 p −2.125 cm 3 p = 2.125 cm p = 0.708 cm in front of the sphere. The image is upright, virtual, and diminished. *P36.14

(a)

(b)

M = −4 = −

q p

q = 4p

q − p = 0.60 m = 4p − p

p = 0.2 m

1 1 1 1 1 = + = + f p q 0.2 m 0.8 m

f = 160 mm

M=+

q 1 =− 2 p

p = −2 q

q + p = 0.20 m = − q + p = − q − 2 q q = −66.7 mm

p = 133 mm

1 1 2 1 1 + = = + p q R 0.133 m −0.066 7 m

R = −267 mm

q = 0.8 m

C

356 * P36.15

Image Formation

M=−

q p

a

f

q = − Mp = −0.013 30 cm = −0.39 cm 1 1 1 2 + = = p q f R 1 1 2 + = 30 cm −0.39 cm R 2 R= = −0.790 cm −2.53 m −1

FIG. P36.15

The cornea is convex, with radius of curvature 0.790 cm . *P36.16

With q h ′ +4.00 cm = = +0.400 = − 10.0 cm h p q = −0.400 p

M=

the image must be virtual. (a)

It is a convex mirror that produces a diminished upright virtual image.

(b)

We must have p + q = 42.0 cm = p − q p = 42.0 cm + q p = 42.0 cm − 0. 400 p p=

42.0 cm = 30.0 cm 1.40

The mirror is at the 30.0 cm mark . (c)

1 1 1 1 1 1 + = = + = = − 0.050 0 cm p q f 30 cm −0.4 30 cm f

a

f = −20.0 cm

f

The ray diagram looks like Figure 36.15(c) in the text. P36.17

(a)

b

g

q = p + 5.00 m and, since the image must be real, M=−

q = −5 p

or

p + 5.00 m = 5 p

Therefore, or

p = 1.25 m

From

1 1 2 + = , p q R

q = 5p .

and

q = 6.25 m . R=

a fa f

2 pq 2 1.25 6.25 = 1. 25 + 6.25 p+q

a

= 2.08 m concave (b)

FIG. P36.17

f

From part (a), p = 1.25 m ; the mirror should be 1.25 m in front of the object.

Chapter 36

P36.18

Assume that the object distance is the same in both cases (i.e., her face is the same distance from the hubcap regardless of which way it is turned). Also realize that the near image ( q = −10.0 cm ) occurs when using the convex side of the hubcap. Applying the mirror equation to both cases gives: (concave side: R = R ,

a

(convex side: R = − R ,

(1)

f

q = −10.0 cm ) 1 1 2 − =− p 10.0 R 2 p − 10.0 cm = . R 10.0 cm p

or (a)

q = −30.0 cm ) 1 1 2 − = p 30.0 R 2 30.0 cm − p = R 30.0 cm p

or

a

(2)

f

Equating Equations (1) and (2) gives:

or

30.0 cm − p = p − 10.0 cm 3.00 p = 15.0 cm .

Thus, her face is 15.0 cm from the hubcap. (b)

Using the above result ( p = 15.0 cm ) in Equation (1) gives: 2 30.0 cm − 15.0 cm = R 30.0 cm 15.0 cm 2 1 = R 30.0 cm

a

or and

fa

f

R = 60.0 cm .

The radius of the hubcap is 60.0 cm . *P36.19

357

(a)

The flat mirror produces an image according to 1 1 1 2 + = = p q f R

1 1 1 + = =0 24 cm q ∞

q = −24.0 m.

The image is 24.0 m behind the mirror, distant from your eyes by 1.55 m + 24.0 m = 25.6 m . (b)

The image is the same size as the object, so

θ=

h 1.50 m = = 0.058 7 rad . d 25.6 m

(c)

1 1 2 + = p q R

q=

1 = −0.960 m − 1 1 m − 1 24 m

1 1 2 + = −2 m 24 m q

a

This image is distant from your eyes by continued on next page

f

b

g b

g

1.55 m + 0.960 m = 2.51 m .

358

Image Formation

(d)

The image size is given by M =

q h′ =− h p

h′ = − h

θ′ =

So its angular size at your eye is (e)

(a)

h ′ 0.06 m = = 0.023 9 rad . d 2.51 m

Your brain assumes that the car is 1.50 m high and calculate its distance as d′ =

P36.20

IJ K

FG H

q −0.960 m = 0.060 0 m. = −1.50 m p 24 m

h 1.50 m = = 62.8 m . θ ′ 0.023 9

The image starts from a point whose height above the mirror vertex is given by 1 1 1 2 + = = p q f R

1 1 1 + = . 3.00 m q 0.500 m

Therefore,

q = 0.600 m .

As the ball falls, p decreases and q increases. Ball and image pass when q1 = p1 . When this is true, 1 1 1 2 + = = p1 p1 0.500 m p1

p1 = 1.00 m.

or

As the ball passes the focal point, the image switches from infinitely far above the mirror to infinitely far below the mirror. As the ball approaches the mirror from above, the virtual image approaches the mirror from below, reaching it together when p 2 = q 2 = 0 . (b)

The falling ball passes its real image when it has fallen 3.00 m − 1.00 m = 2.00 m =

a

f

2 2.00 m 1 2 gt , or when t = = 0.639 s . 2 9.80 m s 2

The ball reaches its virtual image when it has traversed 3.00 m − 0 = 3.00 m =

Section 36.3 P36.21

a

f

2 3.00 m 1 2 gt , or at t = = 0.782 s . 2 9.80 m s 2

Images Formed by Refraction

n1 n 2 n 2 − n 1 + = = 0 and R → ∞ p q R q=−

a

f

n2 1 50.0 cm = −38.2 cm p=− 1.309 n1

Thus, the virtual image of the dust speck is 38.2 cm below the top surface of the ice.

Chapter 36

P36.22

359

When R → ∞ , the equation describing image formation at a single refracting surface becomes

FG n IJ . We use this to locate the final images of the two surfaces of the glass plate. First, find Hn K the image the glass forms of the bottom of the plate. F 1.33 IJ a8.00 cmf = −6.41 cm q = −G H 1.66 K q = −p

2

1

B1

This virtual image is 6.41 cm below the top surface of the glass of 18.41 cm below the water surface. Next, use this image as an object and locate the image the water forms of the bottom of the plate. 1.00 18.41 cm = −13.84 cm qB2 = − or 13.84 cm below the water surface. 1.33 Now find image the water forms of the top surface of the glass. 1 q3 = − or 9.02 cm below the water surface. 12.0 cm = −9.02 cm 1.33

FG IJ a H K FG IJ a H K

f

f

Therefore, the apparent thickness of the glass is ∆t = 13.84 cm − 9.02 cm = 4.82 cm . P36.23

From Equation 36.8

n1 n 2 n 2 − n 1 + = . p q R

Solve for q to find

q=

In this case,

n1 = 1.50 , n 2 = 1.00 , R = −15.0 cm

and

p = 10.0 cm .

So

q=

n1 n 2 n 2 − n1 + = p q R

b

a1.00fa−15.0 cmfa10.0 cmf a10.0 cmfa1.00 − 1.50f − a1.50fa−15.0 cmf = −8.57 cm .

so

1.00 1.40 1.40 − 1.00 + = ∞ 21.0 mm 6.00 mm 0.066 7 = 0.066 7 .

and They agree. P36.25

g

apparent depth is 8.57 cm .

Therefore, the P36.24

n 2 Rp . p n 2 − n 1 − n1 R

The image is inverted, real and diminished.

n1 n 2 n 2 − n 1 + = p q R

becomes

1.00 1.50 1.50 − 1.00 1 + = = p q 6.00 cm 12.0 cm 1.50

(a)

1.00 1.50 1 + = 20.0 cm q 12.0 cm

or

q=

b1.00 12.0 cmg − b1.00 20.0 cmg =

(b)

1.00 1.50 1 + = 10.0 cm q 12.0 cm

or

q=

b

(c)

1.00 1.50 1 + = 3.0 cm q 12.0 cm

or

q=

1.50

g b

g=

1.00 12.0 cm − 1.00 10.0 cm 1.50

b1.00 12.0 cmg − b1.00 3.0 cmg =

45.0 cm

−90.0 cm

−6.00 cm

360 P36.26

Image Formation

p = ∞ and q = +2 R 1.00 n 2 n 2 − 1.00 + = p q R 0+

n 2 n 2 − 1.00 = R 2R

n 2 = 2.00

so

FIG. P36.26 P36.27

n p n1 n 2 n 2 − n 1 + = becomes q = − 2 . p q R n1

For a plane surface,

Thus, the magnitudes of the rate of change in the image and object positions are related by dq n 2 dp . = dt n1 dt If the fish swims toward the wall with a speed of 2.00 cm s , the speed of the image is given by v image =

Section 36.4 P36.28

dq 1.00 = 2.00 cm s = 1.50 cm s . dt 1.33

b

g

Thin Lenses

Let R1 = outer radius and R 2 = inner radius

OP a Q

a fLM N

fLMN

OP Q

1 1 1 1 1 = n−1 − = 1.50 − 1 − = 0.050 0 cm −1 2.00 m 2.50 cm f R1 R 2 so

P36.29

(a)

f = 20.0 cm .

a fLM N

OP a Q

fLM N

OP fQ

1 1 1 1 1 = n−1 − = 0.440 − 12.0 cm −18.0 cm f R1 R 2

a

f = 16.4 cm

(b)

a

fLM N

OP fQ

1 1 1 = 0.440 − f 18.0 cm −12.0 cm

f = 16.4 cm

a

FIG. P36.29

Chapter 36

P36.30

For a converging lens, f is positive. We use

(a)

1 1 1 + = . p q f

1 1 1 1 1 1 = − = − = q f p 20.0 cm 40.0 cm 40.0 cm M=−

q = 40.0 cm

q 40.0 =− = −1.00 40.0 p

The image is real, inverted , and located 40.0 cm past the lens. (b)

1 1 1 1 1 = − = − =0 q f p 20.0 cm 20.0 cm

q = infinity

No image is formed. The rays emerging from the lens are parallel to each other. (c)

1 1 1 1 1 1 = − = − =− q f p 20.0 cm 10.0 cm 20.0 cm M=−

a

q = −20.0 cm

f

−20.0 q =− = 2.00 p 10.0

The image is upright, virtual and 20.0 cm in front of the lens. P36.31

(a)

1 1 1 1 1 = − = − q f p 25.0 cm 26.0 cm

q = 650 cm

The image is real, inverted, and enlarged . (b)

1 1 1 1 1 = − = − q f p 25.0 cm 24.0 cm

q = −600 cm

The image is virtual, upright, and enlarged . P36.32

(a)

1 1 1 + = : p q f so

1 1 1 + = 32.0 cm 8.00 cm f

f = 6.40 cm q 8.00 cm =− = −0.250 32.0 cm p

(b)

M=−

(c)

Since f > 0 , the lens is converging .

361

362 P36.33

Image Formation

We are looking at an enlarged, upright, virtual image: M=

q h′ =2=− h p

1 1 1 + = p q f

a

f

−2.84 cm q =− = +1. 42 cm 2 2

so

p=−

gives

1 1 1 + = 1.42 cm −2.84 cm f

a

f

f = 2.84 cm .

FIG. P36.33 P36.34

(a)

(b)

1 1 1 + = : p q f

1 1 1 + = p −30.0 cm 12.5 cm

p = 8.82 cm

M=−

a

f

−30.0 q =− = 3.40 , upright p 8.82

See the figure to the right. FIG. P36.34(b)

P36.35

1 1 1 + = : p q f

p −1 + q −1 = constant

We may differentiate through with respect to p:

−1p −2 − 1q −2

dq =0 dp

dq q2 = − 2 = −M 2 . dp p P36.36

P36.37

M=

The image is inverted:

−q −1.80 m h′ = = −75.0 = h 0.024 0 m p

(b)

q + p = 3.00 m = 75.0 p + p

p = 39.5 mm

(a)

q = 2.96 m

1 1 1 1 1 = + = + f p q 0.039 5 m 2.96 m

(a)

1 1 1 + = p q f

1 1 1 + = −32.0 cm 20.0 cm q

so

q=−

a

FG 1 + 1 IJ H 20.0 32.0 K

q = 75.0 p .

f = 39.0 mm

f

−1

= −12.3 cm

The image is 12.3 cm to the left of the lens.

a

f

−12.3 cm q =− = 0.615 p 20.0 cm

(b)

M=−

(c)

See the ray diagram to the right.

FIG. P36.37

Chapter 36

*P36.38

In 1 1 1 + = p q f p −1 + q −1 = constant, we differentiate with respect to time dp dq −1 p −2 − 1 q −2 =0 dt dt dq − q 2 dp . = 2 dt p dt

e j

e j

We must find the momentary image location q: 1 1 1 + = 20 m q 0.3 m q = 0.305 m . Now

*P36.39

a

f

2

0.305 m dq =− 5 m s = −0.001 16 m s = 1.16 mm s toward the lens . 2 dt 20 m

a

(a)

1 1 1 + = p q f

(b)

M=

f

q h′ =− h p

1 1 1 + = 480 cm q 7.00 cm h′ =

a

q = 7.10 cm

fa

f

− hq − 5.00 mm 7.10 cm = = −0.074 0 mm p 480 cm

diameter of illuminated spot = 74.0 µm

P36.40

af

0.100 W 4 P P4 = = 2 A πd π 74.0 × 10 −6 m

= 2.33 × 10 7 W m 2

(c)

I=

(a)

1 1 1 1 1 = n−1 − = 1.50 − 1 − 15.0 cm −12.0 cm f R1 R 2

e

a fLM N

OP a Q

j

2

fLM N

a

(b)

The square is imaged as a trapezoid. FIG. P36.40(b) continued on next page

OP fQ

or

f = 13.3 cm

363

364

Image Formation

(c)

To find the area, first find q R and qL , along with the heights hR′ and hL′ , using the thin lens equation. 1 1 1 + = pR qR f

becomes

1 1 1 + = 20.0 cm q R 13.3 cm

or

q R = 40.0 cm

F − q I = a10.0 cmfa−2.00f = −20.0 cm GH p JK

hR′ = hM R = h

R

R

1 1 1 + = 30.0 cm qL 13.3 cm

a

or

fa

qL = 24.0 cm

f

hL′ = hM L = 10.0 cm −0.800 = −8.00 cm Thus, the area of the image is: P36.41

(a)

Area = q R − qL hL′ +

q=d−p.

The image distance is: 1 1 1 + = p q f

Thus,

1 q R − q L hR′ − hL′ = 224 cm 2 . 2

becomes

This reduces to a quadratic equation:

1 1 1 + = . p d−p f

a f

p 2 + −d p + fd = 0 p=

which yields:

d ± d 2 − 4 fd 2

=

d d2 ± − fd . 2 4

d , both solutions are meaningful and the two solutions are not equal to each 4 other. Thus, there are two distinct lens positions that form an image on the screen.

Since f
> f , f − p ≈ − p . Then,

(c)

Suppose the telescope observes the space station at the zenith:

(b)

=

hf . f −p

(b)

h′ = − P36.55

fp

q f h′ =− =− h p p− f

h′ =

gives

g

a

fa

f

108.6 m 4.00 m hf =− = −1.07 mm . p 407 × 10 3 m

Call the focal length of the objective f o and that of the eyepiece − f e . The distance between the lenses is f o − f e . The objective forms a real diminished inverted image of a very distant object at q1 = f o . This image is a virtual object for the eyepiece at p 2 = − f e . For it

becomes

1 1 1 1 =0 + = , q − fe q − fe 2

q2 = ∞ .

and (a)

1 1 1 + = p q f

The user views the image as virtual . Letting h′ represent the height of the first image, h′ h′ and θ = θo = . The angular fo fe

θ0

F0

θ0

F0

magnification is m=

(c)

I L1

f θ h′ fe = = 0 . θ o h′ f o fe

f Here, f o − f e = 10.0 cm and o = 3.00 . fe Thus,

fe =

fo 3.00

and

Fe

f e = −5.00 cm

O L2

FIG. P36.55 and

Fe

θ

2 f o = 10.0 cm . 3

f o = 15.0 cm f e = 5.00 cm

h’

369

Chapter 36

P36.56

Let I 0 represent the intensity of the light from the nebula and θ 0 its angular diameter. With the h′ as h′ = −θ o 2 000 mm . first telescope, the image diameter h′ on the film is given by θ o = − fo

b g L π a200 mmf OP , and the light The light power captured by the telescope aperture is P = I A = I M MN 4 PQ L π a200 mmf OPa1.50 minf . energy focused on the film during the exposure is E = P ∆t = I M MN 4 PQ 2

1

0

1

0

2

1

1

1

0

Likewise, the light power captured by the aperture of the second telescope is

LM π a60.0 mmf OP and the light energy is E MN 4 PQ

2

= I0

the same light energy per unit area, it is necessary that

a

f

I 0 π 60.0 mm

a

2

f

π θ o 900 mm

4 ∆t 2 2

4

=

a

2

f 4 a1.50 minf . b2 000 mmg 4

I 0 π 200 mm

π θo

LM π a60.0 mmf OP∆t . Therefore, to have MN 4 PQ 2

2

P2 = I 0 A 2 = I 0

2

2

The required exposure time with the second telescope is

a200 mmf a900 mmf a1.50 minf = a60.0 mmf b2 000 mmg 2

∆t 2 =

2

2

2

3.38 min .

Additional Problems P36.57

Only a diverging lens gives an upright diminished image. The image is virtual and

d = p− q = p+q: p= f=

P36.58

M=−

a

f

−M + 1 1 − M 1 1 1 1 1 + = = + = = − Mp p q f p − Mp − Md

d : 1−M − Md

a1 − Mf

q so q = − Mp and d = p − Mp p

2

=

a

fa

f=

− 0.500 20.0 cm

a1 − 0.500f

2

−40.0 cm .

If M < 1 , the lens is diverging and the image is virtual. M=− p=

q p

d : 1−M

so

2

q = − Mp

d =p− q = p+q d = p − Mp

and

a

f a

f

1−M −M + 1 1 1 1 1 1 + = = + = = p q f p − Mp − Mp − Md

b

g

2

f=

− Md

a1 − M f

2

.

2

.

If M > 1 , the lens is converging and the image is still virtual. Now

d = −q − p .

We obtain in this case

f=

Md

a M − 1f

370 P36.59

Image Formation

(a)

a fFGH

IJ K F 1 − 1 IJ 1 = a1.66 − 1fG −65.0 cm H 50.0 cm R K

A

1 1 1 = n−1 − f R1 R 2

(b)

B

C

R2

R2 = 23.1 cm

so

FIG. P36.59

The distance along the axis from B to A is

a

f

R1 − R12 − 2.00 cm

2

2.00 cm

R1

2

1 1 1 = + R 2 50.0 cm 42.9 cm

D

a50.0 cmf − a2.00 cmf 2

= 50.0 cm −

2

= 0.040 0 cm .

Similarly, the axial distance from C to D is 23.1 cm −

a23.1 cmf − a2.00 cmf 2

2

= 0.086 8 cm .

Then, AD = 0.100 cm − 0.040 0 cm + 0.086 8 cm = 0.147 cm . *P36.60

We consider light entering the rod. The surface of entry is convex to the object rays, so R1 = +4.50 cm n1 n 2 n 2 − n 1 + = p 1 q1 R1

1.33 1.50 1.50 − 1.33 + = 100 cm q1 4.50 cm

1.50 = 0.037 8 cm − 0.013 3 cm = 0.024 5 cm q1

q1 = 61.3 cm

O1

V

I2

O2

The first image is real, inverted and diminished. To find its magnification we can use two similar triangles in the ray diagram with their vertices meeting at the center of curvature: h1′ h1 = 100 cm + 4.5 cm 61.3 cm − 4.5 cm

I1

C

C

V

FIG. P36.60

h1′ = −0.543 . h1

Now the first image is a real object for the second surface at object distance from its vertex 75.0 cm + 4.50 cm + 4.50 cm − 61.3 cm = 22.7 cm 1.50 1.33 1.33 − 1.50 + = 22.7 cm q 2 −4.50 cm 1.33 = 0.037 8 cm − 0.066 0 cm = −0.028 2 cm q2 q 2 = −47.1 cm (a)

The final image is inside the rod, 47.1 cm from the second surface .

(b)

It is virtual, inverted, and enlarged . Again by similar triangles meeting at C we have h2 h2′ = 22.7 cm − 4.5 cm 47.1 cm − 4.5 cm

h2′ = 2.34 . h2

Since h2 = h1′ , the overall magnification is M 1 M 2 =

a

fa f

h1′ h2′ h2′ = = −0.543 2.34 = −1.27 . h1 h2 h1

Chapter 36

*P36.61

(a)

1 1 1 1 1 = − = − q1 f1 p1 5 cm 7.5 cm q 15 cm = −2 M1 = − 1 = − 7.5 cm p1

∴ q1 = 15 cm

a f

M = M1 M 2 ∴M2 = −

371

∴ 1 = −2 M 2

q 1 =− 2 2 p2

1 1 1 + = p 2 q2 f2

∴

∴ p 2 = 2 q2 1 1 1 + = 2 q 2 q 2 10 cm

∴ q 2 = 15 cm, p 2 = 30 cm

p1 + q1 + p 2 + q 2 = 7.5 cm + 15 cm + 30 cm + 15 cm = 67.5 cm (b)

1 1 1 1 + = = p1′ q1′ f1 5 cm Solve for q1′ in terms of p1′ : q1′ =

5 p1′ p1′ − 5

(1)

q1′ 5 , using (1). =− p1′ p1′ − 5 q′ 3 M′ ∴ M 2′ = = − p1′ − 5 = − 2 M ′ = M 1′ M 2′ 5 M 1′ p 2′ 3 ∴ q ′2 = p 2′ p1′ − 5 (2) 5 1 1 1 1 + = = Substitute (2) into the lens equation and obtain p ′2 in terms of p1′ : p ′2 q 2′ f 2 10 cm 10 3 p1′ − 10 . (3) p 2′ = 3 p1′ − 5 Substituting (3) in (2), obtain q 2′ in terms of p1′ : q ′2 = 2 3 p1′ − 10 . (4) Now, p1′ + q1′ + p ′2 + q 2′ = a constant. Using (1), (3) and (4), and the value obtained in (a): 10 3 p1′ − 10 5 p1′ + p1′ + + 2 3 p1′ − 10 = 67.5 . p1′ − 5 3 p′ − 5 This reduces to the quadratic equation 21p1′ 2 − 322.5 p1′ + 1 212.5 = 0 , which has solutions p1′ = 8.784 cm and 6.573 cm. Case 1: p1′ = 8.784 cm ∴ p1′ − p1 = 8.784 cm − 7.5 cm = 1.28 cm. From (4): q ′2 = 32.7 cm ∴ q ′2 − q 2 = 32.7 cm − 15 cm = 17.7 cm. Case 2: p1′ = 6.573 cm ∴ p1′ − p1 = 6.573 cm − 7.5 cm = −0.927 cm . From (4): q 2′ = 19. 44 cm ∴ q ′2 = q 2 = 19.44 cm − 15 cm = 4.44 cm. From these results it is concluded that: The lenses can be displaced in two ways. The first lens can be moved 1.28 cm farther from M 1′ = −

b

b

b b

b

g

g

g g

b

g

b

g b g

g

the object and the second lens 17.7 cm toward the object. Alternatively, the first lens can be moved 0.927 cm toward the object and the second lens 4.44 cm toward the object.

372 P36.62

Image Formation

1 1 1 1 1 = − = − q1 f1 p1 10.0 cm 12.5 cm so

q1 = 50.0 cm (to left of mirror).

This serves as an object for the lens (a virtual object), so 1 1 1 1 1 = − = − and q 2 = −50.3 cm, q2 f2 p 2 −16.7 cm −25.0 cm

a

f a

f

meaning 50.3 cm to the right of the lens. Thus, the final image is located 25.3 cm to right of mirror . M1 = −

q1 50.0 cm =− = −4.00 p1 12.5 cm

M2 = −

−50.3 cm q2 =− = −2.01 p2 −25.0 cm

a a

f f

M = M 1 M 2 = 8.05 Thus, the final image is virtual, upright , 8.05 times the size of object, and 25.3 cm to right of the mirror. P36.63

We first find the focal length of the mirror. 1 1 1 1 1 9 = + = + = f p q 10.0 cm 8.00 cm 40.0 cm

*P36.64

and

f = 4. 44 cm .

Hence, if p = 20.0 cm ,

1 1 1 1 1 15.56 = − = − = . q f p 4. 44 cm 20.0 cm 88.8 cm

Thus,

q = 5.71 cm , real.

A telescope with an eyepiece decreases the diameter of a beam of parallel rays. When light is sent through the same device in the opposite direction, the beam expands. Send the light first through the diverging lens. It will then be diverging from a virtual image found like this: 1 1 1 + = p q f

1 1 1 + = ∞ q −12 cm

FIG. P36.64

q = −12 cm . Use this image as a real object for the converging lens, placing it at the focal point on the object side of the lens, at p = 21 cm . Then 1 1 1 + = p q f

1 1 1 + = 21 cm q 21 cm q=∞.

The exiting rays will be parallel. The lenses must be 21.0 cm − 12.0 cm = 9.00 cm apart. By similar triangles,

d 2 21 cm = = 1.75 times . d 1 12 cm

Chapter 36

P36.65

A hemisphere is too thick to be described as a thin lens. The light is undeviated on entry into the flat face. We next consider the light’s exit from the second surface, for which R = −6.00 cm . The incident rays are parallel, so p = ∞ . Then,

n1 n 2 n 2 − n 1 + = p q R

becomes

0+

1 1.00 − 1.56 = q −6.00 cm

q = 10.7 cm .

and P36.66

FIG. P36.65

(a)

I=

P 4.50 W = 2 4π r 4π 1.60 × 10 −2 m

(b)

I=

P 4.50 W = 2 4π r 4π 7.20 m

(c)

1 1 1 + = : p q f

1 1 1 + = 7.20 m q 0.350 m

so

q = 0.368 m

and

M=

e a

j

f

2

2

= 1.40 kW m 2

= 6.91 mW m 2

q h′ 0.368 m =− =− 3. 20 cm p 7.20 m

h′ = 0.164 cm (d)

e

jLMN π4 a0.150 mf OPQ

e

j a f

The lens intercepts power given by

P = IA = 6.91 × 10 −3 W m 2

and puts it all onto the image where

I=

2

f

6.91 × 10 −3 W m 2 π 15.0 cm P = 2 A π 0.164 cm 4

I = 58.1 W m 2 .

a

2

4

373

374 P36.67

Image Formation

a

fa a

f f

−6.00 cm 12.0 cm f1 p1 = = −4.00 cm . p1 − f1 12.0 cm − −6.00 cm

From the thin lens equation,

q1 =

When we require that q 2 → ∞ ,

the thin lens equation becomes p 2 = f 2 .

In this case,

p 2 = d − −4.00 cm .

Therefore,

d + 4.00 cm = f 2 = 12.0 cm

a

f

and

d = 8.00 cm .

FIG. P36.67 *P36.68

The inverted real image is formed by the lens operating on light directly from the object, on light that has not reflected from the mirror. q p

q = 1.50 p

For this we have

M = −1.50 = −

1 1 1 + = p q f

1 1 1 2.50 + = = p 1.50 p 10 cm 1.50 p

FG 2.5 IJ = 16.7 cm H 1.5 K

p = 10 cm

40.0 cm − 16.7 cm = 23.3 cm .

Then the object is distant from the mirror by

The second image seen by the person is formed by light that first reflects from the mirror and then goes through the lens. For it to be in the same position as the inverted image, the lens must be receiving light from an image formed by the mirror at the same location as the physical object. The formation of this image is described by 1 1 1 + = p q f P36.69

1 1 1 + = 23.3 cm 23.3 cm f

f = 11.7 cm .

R = +1.50 m . In addition, because the distance to the Sun is so much larger than 2 any other distances, we can take p = ∞ .

For the mirror, f =

The mirror equation,

1 1 1 + = , then gives p q f

q = f = 1.50 m . M=−

Now, in

q h′ = p h

the magnification is nearly zero, but we can be more precise: Thus, the image diameter is h′ = −

fFGH

IJ a K

h is the angular diameter of the object. p

hq π rad = −0.533° 1.50 m = −0.014 0 m = −1.40 cm . 180° p

a

f

Chapter 36

P36.70

(a)

For the light the mirror intercepts, P = I 0 A = I 0 π R a2

e

j

350 W = 1 000 W m 2 π R a2

Ra = 0.334 m or larger .

and (b)

In we have so

h′ = − q

so where

1 1 1 2 + = = p q f R p→∞ R q= 2 q h′ M= =− h p

F h I = −FG R IJ L0.533° FG π rad IJ O = −FG R IJ a9.30 m radf GH p JK H 2 K MN H 180° K PQ H 2 K

h is the angle the Sun subtends. The intensity at the image is p

then

I=

4I 0 π R a2 P = = πh ′ 2 4 π h′ 2 R2

4I 0 R a2

b g e9.30 × 10 16e1 000 W m jR = R e9.30 × 10 radj 2

2

120 × 10 3 W m 2 so P36.71

−3

2

rad

j

2

2 a 2

Ra = 0.025 5 or larger . R

In the original situation,

p1 + q1 = 1.50 m .

In the final situation,

p 2 = p1 + 0.900 m

and

q 2 = q1 − 0.900 m = 0.600 m − p1 . 1 1 1 1 1 + = = + . p1 q 1 f p 2 q 2

Our lens equation is

−3

Substituting, we have

1 1 1 1 + = + . p1 1.50 m − p1 p1 + 0.900 0.600 − p1

Adding the fractions,

1.50 m − p1 + p1 0.600 − p1 + p1 + 0.900 = . p1 1.50 m − p1 p1 + 0.900 0.600 − p1

Simplified, this becomes

b g b p b1.50 m − p g = b p

(a)

p1 =

(b) (c)

Thus,

1

1

1

gb g + 0.900gb0.600 − p g . 1

0.540 m = 0.300 m 1.80

1 1 1 = + f 0.300 m 1.50 m − 0.300 m

and

p 2 = p1 + 0.900 = 1.20 m f = 0.240 m

The second image is real, inverted, and diminished with

M=−

q2 = −0.250 . p2

FIG. P36.71

375

376 P36.72

Image Formation

(a)

becomes: (b)

a fFGH

1 1 1 = n−1 + f R1 R 2

The lens makers’ equation,

a fLM N

OP giving n = fQ

1 1 1 = n−1 − 9.00 cm −11.0 cm 5.00 cm

a

IJ K

1.99 .

As the light passes through the lens for the first time, the thin lens equation 1 1 1 + = p 1 q1 f 1 1 1 + = 8.00 cm q1 5.00 cm

becomes: or

q1 = 13.3 cm ,

M1 = −

and

q1 13.3 cm =− = −1.67 . 8.00 cm p1

This image becomes the object for the concave mirror with: p m = 20.0 cm − q1 = 20.0 cm − 13.3 cm = 6.67 cm R = +4.00 cm . 2 1 1 1 + = 6.67 cm qm 4.00 cm f=

and The mirror equation becomes: giving

qm = 10.0 cm

and

M2 = −

qm 10.0 cm =− = −1.50 . 6.67 cm pm

The image formed by the mirror serves as a real object for the lens on the second pass of the light through the lens with: p 3 = 20.0 cm − q m = +10.0 cm . The thin lens equation yields:

1 1 1 + = 10.0 cm q3 5.00 cm

or

q3 = 10.0 cm

and

M3 = −

10.0 cm to the left of the lens .

The final image is a real image located

M total = M1 M 2 M 3 = −2.50 .

The overall magnification is (c) P36.73

q3 10.0 cm =− = −1.00 . 10.0 cm p3

Since the total magnification is negative, this final image is inverted .

For the objective:

1 1 1 + = becomes p q f

1 1 1 + = so q = 25.5 mm . 3.40 mm q 3.00 mm q 25.5 mm =− = −7.50 . 3.40 mm p

The objective produces magnification

M1 = −

For the eyepiece as a simple magnifier,

me =

and overall

M = M1 m e = −75.0 .

25.0 cm 25.0 cm = = 10.0 f 2.50 cm

Chapter 36

P36.74

(a)

377

Start with the second lens: This lens must form a virtual image located 19.0 cm to the left of it (i.e., q 2 = −19.0 cm ). The required object distance for this lens is then p2 =

a

fa

f

−19.0 cm 20.0 cm 380 cm q2 f2 = . = 39.0 q2 − f2 −19.0 cm − 20.0 cm

The image formed by the first lens serves as the object for the second lens. Therefore, the image distance for the first lens is q1 = 50.0 cm − p 2 = 50.0 cm −

380 cm 1 570 cm = . 39.0 39.0

The distance the original object must be located to the left of the first lens is then given by 1 1 1 1 39.0 157 − 39.0 118 = − = − = = p1 f1 q1 10.0 cm 1 570 cm 1 570 cm 1 570 cm (b)

P36.75

P36.76

or

p1 =

1 570 cm = 13.3 cm . 118

F q I F − q I = LMFG 1 570 cm IJ F 118 I OPL a−19.0 cmfa39.0f O = GH p JK GH p JK MNH 39.0 K GH 1 570 cm JK PQMN 380 cm PQ

M = M1 M 2 = −

1

2

1

2

(c)

Since M < 0 , the final image is inverted .

(a)

P=

1 1 1 1 1 = + = + = 44.6 diopters f p q 0.022 4 m ∞

(b)

P=

1 1 1 1 1 = + = + = 3.03 diopters 0.330 m ∞ f p q

b a

−5.90

g

f

The object is located at the focal point of the upper mirror. Thus, the upper mirror creates an image at infinity (i.e., parallel rays leave this mirror). The lower mirror focuses these parallel rays at its focal point, located at the hole in the upper mirror. Thus, the

image is real, inverted, and actual size .

For the upper mirror: 1 1 1 + = : p q f

1 1 1 + = 7.50 cm q1 7.50 cm

q1 = ∞.

For the lower mirror: 1 1 1 + = ∞ q 2 7.50 cm

q 2 = 7.50 cm.

Light directed into the hole in the upper mirror reflects as shown, to behave as if it were reflecting from the hole.

FIG. P36.76

378 P36.77

Image Formation

(a)

For lens one, as shown in the first figure, 1 1 1 + = 40.0 cm q1 30.0 cm q1 = 120 cm M1 = −

q1 120 cm =− = −3.00 p1 40.0 cm

This real image I 1 = O 2 is a virtual object for the second lens. That is, it is behind the lens, as shown in the second figure. The object distance is p 2 = 110 cm − 120 cm = −10.0 cm 1 1 1 + = : −10.0 cm q 2 −20.0 cm q 2 = 20.0 cm M2 = −

q2 20.0 cm =− = +2.00 −10.0 cm p2

a

f

M overall = M 1 M 2 = −6.00 (b)

M overall < 0 , so final image is

inverted . (c)

If lens two is a converging lens (third figure): 1 1 1 + = −10.0 cm q 2 20.0 cm q 2 = 6.67 cm M2 = −

a

6.67 cm = +0.667 −10.0 cm

f

M overall = M 1 M 2 = −2.00 Again, M overall < 0 and the final image is inverted .

FIG. P36.77

Chapter 36

*P36.78

379

The first lens has focal length described by

gFGH

IJ b K

gFGH

IJ b K

gFGH

IJ K

n −1 1 1 1 1 1 = n1 − 1 − = n1 − 1 − =− 1 . ∞ R f1 R11 R12 R

b

For the second lens

gFGH

IJ K

b

g

2 n2 − 1 1 1 1 1 1 = n2 − 1 − =+ = n2 − 1 − . +R −R f2 R 21 R 22 R

b

Let an object be placed at any distance p1 large compared to the thickness of the doublet. The first lens forms an image according to 1 1 1 + = p1 q1 f1 1 −n 1 + 1 1 = − . q1 R p1

b

g

This virtual q1 < 0 image is a real object for the second lens at distance p 2 = − q1 . For the second lens 1 1 1 + = p 2 q2 f2 2n − 2 1 2n 2 − 2 −n1 + 1 1 2n 2 − n1 − 1 1 1 2n 2 − 2 1 = − = 2 + = + − = − . q2 R p2 R q1 R R p1 R p1 Then

1 1 2n 2 − n1 − 1 1 2n − n 1 − 1 + = so the doublet behaves like a single lens with = 2 . p1 q 2 R f R

ANSWERS TO EVEN PROBLEMS P36.2

4.58 m

P36.22

4.82 cm

P36.4

see the solution

P36.24

P36.6

(a) p1 + h ; (b) virtual; (c) upright; (d) +1; (e) No

see the solution; real, inverted, diminished

P36.26

2.00

P36.28

20.0 cm

P36.30

(a) q = 40.0 cm real, inverted, actual size M = −1.00 ; (b) q = ∞ , M = ∞ , no image is formed; (c) q = −20.0 cm upright, virtual , enlarged M = +2.00

P36.8

at q = −0.267 m virtual upright and diminished with M = 0.026 7

P36.10

at 3.33 m from the deepest point of the niche

P36.12

30.0 cm

P36.14

(a) 160 mm; (b) R = −267 mm

P36.32

(a) 6.40 cm; (b) −0. 250 ; (c) converging

P36.16

(a) convex; (b) At the 30.0 cm mark; (c) –20.0 cm

P36.34

(a) 3.40 , upright; (b) see the solution

P36.18

(a) 15.0 cm; (b) 60.0 cm

P36.36

(a) 39.0 mm; (b) 39.5 mm

P36.20

(a) see the solution; (b) at 0.639 s and at 0.782 s

P36.38

1.16 mm s toward the lens

380

Image Formation

P36.40

(a) 13.3 cm; (b) see the solution; a trapezoid; (c) 224 cm 2

P36.60

(a) inside the rod, 47.1 cm from the second surface ; (b) virtual, inverted, and enlarged

P36.42

2.18 mm away from the film

P36.62

P36.44

(a) at q = −34.7 cm virtual, upright and diminshed; (b) at q = −36.1 cm virtual, upright and diminshed

25.3 cm to right of mirror , virtual, upright , enlarged 8.05 times

P36.64

place the lenses 9.00 cm apart and let light pass through the diverging lens first. 1.75 times (a) 1.40 kW m 2 ; (b) 6.91 mW m 2 ;

P36.46

f 1.41

P36.66

P36.48

23. 2 cm

P36.68

11.7 cm

P36.50

(a) at 4.17 cm; (b) 6.00

P36.70

P36.52

2.14 cm

(a) 0.334 m or larger ; R (b) a = 0.025 5 or larger R

P36.54

(a) see the solution; (b) h ′ = −

P36.72

(a) 1.99 ; (b) 10.0 cm to the left of the lens ; −2.50 ; (c) inverted

P36.74

(a) 13.3 cm; (b) −5.90 ; (c) inverted

P36.76

see the solution; real, inverted, and actual size

P36.78

see the solution

(c) 0.164 cm; (d) 58.1 W m 2

hf ; p

(c) −1.07 mm P36.56

3.38 min

P36.58

if M < 1 , f =

− Md

a1 − M f Md if M > 1 , f = a M − 1f

2

2

,

37 Interference of Light Waves CHAPTER OUTLINE 37.1 37.2 37.3

37.4 37.5 37.6 37.7

Conditions for Interference Young’s Double-Slit Experiment Intensity Distribution of the Double-Slit Interference Pattern Phasor Addition of Waves Change of Phase Due to Reflection Interference in Thin Films The Michelson Interferometer

ANSWERS TO QUESTIONS Q37.1

(a)

Two waves interfere constructively if their path difference is zero, or an integral multiple of the wavelength, according to δ = mλ , with m = 0 , 1, 2 , 3 , ….

(b)

Two waves interfere destructively if their path difference is a half wavelength, or an odd multiple of λ 1 , described by δ = m + λ , with m = 0 , 1, 2 , 3 , …. 2 2

FG H

Q37.2

IJ K

The light from the flashlights consists of many different wavelengths (that’s why it’s white) with random time differences between the light waves. There is no coherence between the two sources. The light from the two flashlights does not maintain a constant phase relationship over time. These three equivalent statements mean no possibility of an interference pattern.

λ air . Since the positions of light n water and dark bands are proportional to λ, (according to Equations 37.2 and 37.3), the underwater fringe separations will decrease.

Q37.3

Underwater, the wavelength of the light would decrease, λ water =

Q37.4

Every color produces its own pattern, with a spacing between the maxima that is characteristic of the wavelength. With several colors, the patterns are superimposed and it can be difficult to pick out a single maximum. Using monochromatic light can eliminate this problem.

Q37.5

The threads that are woven together to make the cloth have small meshes between them. These bits of space act as pinholes through which the light diffracts. Since the cloth is a grid of such pinholes, an interference pattern is formed, as when you look through a diffraction grating.

Q37.6

If the oil film is brightest where it is thinnest, then n air < n oil < n water . With this condition, light reflecting from both the top and the bottom surface of the oil film will undergo phase reversal. Then these two beams will be in phase with each other where the film is very thin. This is the condition for constructive interference as the thickness of the oil film decreases toward zero.

381

382

Interference of Light Waves

Q37.7

As water evaporates from the ‘soap’ bubble, the thickness of the bubble wall approaches zero. Since light reflecting from the front of the water surface is phase-shifted 180° and light reflecting from the back of the soap film is phase-shifted 0°, the reflected light meets the conditions for a minimum. Thus the soap film appears black, as in the illustration accompanying textbook Example 37.5, “Interference in a Wedge-Shaped Film.”

Q37.8

If the film is more than a few wavelengths thick, the interference fringes are so close together that you cannot resolve them.

Q37.9

If R is large, light reflecting from the lower surface of the lens can interfere with light reflecting from the upper surface of the flat. The latter undergoes phase reversal on reflection while the former does not. Where there is negligible distance between the surfaces, at the center of the pattern you will see a dark spot because of the destructive interference associated with the 180° phase shift. Colored rings surround the dark spot. If the lens is a perfect sphere the rings are perfect circles. Distorted rings reveal bumps or hollows on the fine scale of the wavelength of visible light.

Q37.10

A camera lens will have more than one element, to correct (at least) for chromatic aberration. It will have several surfaces, each of which would reflect some fraction of the incident light. To maximize light throughput the surfaces need antireflective coatings. The coating thickness is chosen to produce destructive interference for reflected light of some wavelength.

Q37.11

To do Young’s double-slit interference experiment with light from an ordinary source, you must first pass the light through a prism or diffraction grating to disperse different colors into different directions. With a single narrow slit you select a single color and make that light diffract to cover both of the slits for the interference experiment. Thus you may have trouble lining things up and you will generally have low light power reaching the screen. The laser light is already monochromatic and coherent across the width of the beam.

Q37.12

Suppose the coating is intermediate in index of refraction between vacuum and the glass. When the coating is very thin, light reflected from its top and bottom surfaces will interfere constructively, so you see the surface white and brighter. As the thickness reaches one quarter of the wavelength of violet light in the coating, destructive interference for violet will make the surface look red or perhaps orange. Next to interfere destructively are blue, green, yellow, orange, and red, making the surface look red, purple, and then blue. As the coating gets still thicker, we can get constructive interference for violet and then for other colors in spectral order. Still thicker coating will give constructive and destructive interference for several visible wavelengths, so the reflected light will start to look white again.

Q37.13

Assume the film is higher in refractive index than the medium on both sides of it. The condition for

λ . The 2 ray that reflects through the film undergoes phase reversal both at the bottom and at the top surface. λ Then this ray should also travel an extra distance of . Since this ray passes through two extra 2 λ thicknesses of film, the thickness should be . This is different from the condition for destructive 4 interference of light reflected from the film, but it is the same as the condition for constructive interference of reflected light. The energy of the extra reflected light is energy diverted from light otherwise transmitted. destructive interference of the two transmitted beams is that the waves be out of phase by

Chapter 37

Q37.14

383

The metal body of the airplane is reflecting radio waves broadcast by the television station. The reflected wave that your antenna receives has traveled an extra distance compared to the stronger signal that came straight from the transmitter tower. You receive it with a short time delay. On the television screen you see a faint image offset to the side.

SOLUTIONS TO PROBLEMS Section 37.1

Conditions for Interference

Section 37.2

Young’s Double-Slit Experiment

P37.1

∆y bright =

P37.2

y bright =

ja f

e

−9 5.00 λL 632.8 × 10 = m = 1.58 cm −4 d 2.00 × 10

λL m d λ=

For m = 1 , P37.3

e

je

j

3.40 × 10 −3 m 5.00 × 10 −4 m yd = = 515 nm 3.30 m L

Note, with the conditions given, the small angle approximation does not work well. That is, sin θ , tan θ , and θ are significantly different. We treat the interference as a Fraunhofer pattern. (a)

At the m = 2 maximum, tan θ =

400 m = 0.400 1 000 m

400 m

θ = 21.8° λ=

so (b)

300 m

a

f

300 m sin 21.8° d sin θ = = 55.7 m . 2 m

The next minimum encountered is the m = 2 minimum;

FG H

IJ K

1 λ 2

and at that point,

d sin θ = m +

which becomes

d sin θ =

or

sin θ =

and

θ = 27.7°

so

y = 1 000 m tan 27.7° = 524 m .

b

5 λ 2

FG H

IJ K

5 λ 5 55.7 m = = 0.464 2 d 2 300 m

g

Therefore, the car must travel an additional 124 m . If we considered Fresnel interference, we would more precisely find 1 550 2 + 1 000 2 − 250 2 + 1 000 2 = 55.2 m and (b) 123 m. (a) λ = 2

FH

IK

1 000 m

FIG. P37.3

384 P37.4

Interference of Light Waves

λ=

v 354 m s = = 0.177 m f 2 000 s −1

a0.300 mf sinθ = 1a0.177 mf so d sin 36.2° = 1b0.030 0 mg d sin θ = mλ e1.00 × 10 mj sin 36.2° = a1fλ d sin θ = mλ

(a) (b)

−6

(c)

f=

P37.5

c

λ

=

3.00 × 10 8 m s

FG H

d sin θ = m +

The first minimum is described by

m=0

and the tenth by m = 9 :

sin θ =

sin θ ≈ tan θ .

Thus,

d=

λ=

ja

7.26 × 10

m

d = 5.08 cm

so

λ = 590 nm

f = 1.54 × 10

IJ K

y L

IJ K

Source

y L

but for small θ,

−3

and

1 λ. 2

FG H

tan θ =

e

θ = 36.2°

λ 1 9+ . 2 d

Also,

9.5 5 890 × 10 −10 m 2.00 m

and

= 508 THz

5.90 × 10 −7 m

In the equation

d=

P37.6

so

FIG. P37.5

9.5 λ 9.5 λL = sin θ y

−3

m = 1.54 mm .

340 m s = 0.170 m 2 000 Hz

Maxima are at

d sin θ = mλ :

m=0

gives

θ = 0°

m=1

gives

sin θ =

λ 0.170 m = d 0.350 m

θ = 29.1°

m=2

gives

sin θ =

2λ = 0.971 d

θ = 76.3°

m=3

gives

sin θ = 1. 46

Minima are at

d sin θ = m +

m=0

gives

sin θ =

m=1

gives

sin θ =

m=2

gives

sin θ = 1.21

FG H

IJ K

No solution.

1 λ: 2

λ

= 0.243

θ = 14.1°

3λ = 0.729 2d

θ = 46.8°

2d

No solution.

So we have maxima at 0° , 29.1° , and 76.3° ; minima at 14.1° and 46.8° .

d

Chapter 37

P37.7

(a)

For the bright fringe, y bright = y=

y

mλ L where m = 1 d

e546.1 × 10

−9

ja

f = 2.62 × 10

m 1.20 m

0. 250 × 10

−3

m

y 2 − y1

−3

m = 2.62 mm . Source

IJ K λL LF 1 I F 1 I O λL = 1 + J − G0 + JP = a1f G M 2K H 2KQ d d NH e546.1 × 10 mja1.20 mf =

For the dark bands, y dark =

(b)

L = 1.20 m

FG H

λL 1 ; m = 0 , 1, 2 , 3 , … m+ 2 d

d = 0.250 m

bright bright bright dark dark dark

−9

0.250 × 10 −3 m

y1

∆y = 2.62 mm .

y2 FIG. P37.7

P37.8

Taking m = 0 and y = 0.200 mm in Equation 37.6 gives L≈

2dy

=

e

je

Bright

j = 0.362 m

2 0.400 × 10 −3 m 0.200 × 10 −3 m

λ L ≈ 36.2 cm

442 × 10

−9

m

Bright

0.2 mm

Geometric optics incorrectly predicts bright regions opposite the slits and darkness in between. But, as this example shows, interference can produce just the opposite. P37.9

Dark

0.2 mm

Dark L

FIG. P37.7

Location of A = central maximum, Location of B = first minimum.

P37.10

FG IJ H K a3.00 mfa150 mf = =

So,

∆y = y min − y max =

Thus,

d=

a

f

40.0 m

11.3 m .

d sin θ = mλ

At 30.0° ,

e3.20 × 10

λL 2 20.0 m

1 1 λL λL 0+ −0= = 20.0 m . 2 2 d d

−4

j

e

j

m sin 30.0° = m 500 × 10 −9 m

so

m = 320

There are 320 maxima to the right, 320 to the left, and one for m = 0 straight ahead. There are 641 maxima .

Bright

385

386 *P37.11

Interference of Light Waves

Observe that the pilot must not only home in on the airport, but must be headed in the right direction when she arrives at the end of the runway. 8 c 3 × 10 m s = = 10.0 m f 30 × 10 6 s −1

(a)

λ=

(b)

The first side maximum is at an angle given by d sin θ = 1 λ .

af

θ = 14.5° a40 mf sinθ = 10 m y = L tan θ = b 2 000 mg tan 14.5° = 516 m (c)

*P37.12

tan θ =

y L

The signal of 10-m wavelength in parts (a) and (b) would show maxima at 0°, 14.5°, 30.0°, 48.6°, and 90°. A signal of wavelength 11.23-m would show maxima at 0°, 16.3°, 34.2°, and 57.3°. The only value in common is 0°. If λ 1 and λ 2 were related by a ratio of small integers λ n (a just musical consonance!) in 1 = 1 , then the equations d sin θ = n 2 λ 1 and d sin θ = n1 λ 2 λ 2 n2 would both be satisfied for the same nonzero angle. The pilot could come flying in with that inappropriate bearing, and run off the runway immediately after touchdown.

In d sin θ = mλ

d

y = mλ L

e e

y=

mλ L d

j

−9 dy mλ dL 1 633 × 10 m = = 3 m s = 6.33 mm s dt d dt 0.3 × 10 −3 m

P37.13

φ=

2π

λ

d sin θ =

j

2π

d

λ

FG y IJ H LK

2π

1.20 × 10 je

(a)

φ=

(b)

φ=

(c)

If φ = 0.333 rad =

e

2π

e5.00 × 10

−7

e1.20 × 10 mj 2πd sin θ

λ

j a

f

−4

m sin 0.500° = 13.2 rad

−4

m

5.00 × 10 −7 m

× 10 m I = jFGH 5.001.20 m JK

θ = sin −1

−3

F λ φ I = sin GH 2π d JK

6.28 rad

LM e5.00 × 10 mja0.333 radf OP MN 2π e1.20 × 10 mj PQ −7

−1

−4

θ = 1.27 × 10 −2 deg .

(d)

If d sin θ =

λ 4

θ = sin −1

FG λ IJ = sin H 4d K

−1

LM 5 × 10 m OP MN 4e1.20 × 10 mj PQ

θ = 5.97 × 10 −2 deg .

−7

−4

Chapter 37

P37.14

δ = d sin θ 1 − d sin θ 2 .

The path difference between rays 1 and 2 is:

For constructive interference, this path difference must be equal to an integral number of wavelengths: d sin θ 1 − d sin θ 2 = mλ , or

b

g

d sin θ 1 − sin θ 2 = mλ . P37.15

(a)

The path difference δ = d sin θ and when L >> y

δ=

e

je

j

1.80 × 10 −2 m 1.50 × 10 −4 m yd = = 1.93 × 10 −6 m = 1.93 µm . 1.40 m L

(b)

δ 1.93 × 10 −6 m = = 3.00 , or δ = 3.00λ λ 6.43 × 10 −7 m

(c)

Point P will be a maximum since the path difference is an integer multiple of the wavelength.

Section 37.3 P37.16

(a)

Intensity Distribution of the Double-Slit Interference Pattern I I max

= cos 2

FG φ IJ H 2K

(Equation 37.11)

φ = 2 cos −1

Therefore,

(b)

P37.17

δ=

a

fa

I I max

f

486 nm 1.29 rad λφ = = 99.8 nm 2π 2π

I av = I max cos 2

FG π d sinθ IJ H λ K

For small θ,

sin θ =

and

I av = 0.750 I max y=

y L

λL cos −1 πd

I av I max

e6.00 × 10 ja1.20 mf cos π e 2.50 × 10 mj −7

y=

P37.18

= 2 cos −1 0.640 = 1.29 rad .

−1

−3

FG π yd IJ H λL K L π e6.00 × 10 mje1.80 × 10 mj OP = cos M MN e656.3 × 10 mja0.800 mf PQ =

0.750 I max = 48.0 µm I max

I = I max cos 2 I I max

−3

−4

2

−9

0.968

387

388 P37.19

Interference of Light Waves

(a)

From Equation 37.8,

φ=

2π d

λ

sin θ =

e

2π d

λ

y

⋅

y2 + D2

je ja

j

−3 −3 2π yd 2π 0.850 × 10 m 2.50 × 10 m φ≈ = = 7.95 rad λD 600 × 10 −9 m 2.80 m

(b)

I I max I I max

P37.20

(a)

=

e

b

g

cos 2 π d λ sin θ cos

2

= cos 2

=

f

b g

cos 2 φ 2

cos mπ bπ d λ g sinθ φ F 7.95 rad IJ = 0.453 = cos G H 2 K 2 2

max

2

The resultant amplitude is

b

g b

g

Er = E0 sin ω t + E0 sin ω t + φ + E0 ω t + 2φ ,

where

φ=

2π

λ

d sin θ .

b g E = E bsin ω t ge1 + cos φ + 2 cos φ − 1j + E bcos ω t gbsin φ + 2 sin φ cos φ g E = E b1 + 2 cos φ gbsin ω t cos φ + cos ω t sin φ g = E b1 + 2 cos φ g sinbω t + φ g F 1I Then the intensity is I ∝ E = E b1 + 2 cos φ g G J H 2K Er = E0 sin ω t + sin ω t cos φ + cos ω t sin φ + sin ω t cos 2φ + cos ω t sin 2φ r

0

r

0

2

0

0

2 r

b

2

2 0

g

sin 2 ω t + φ is

where the time average of

1 . 2

From one slit alone we would get intensity I max ∝ E02 I = I max (b)

LM1 + 2 cosFG 2π d sinθ IJ OP H λ KQ N

FG 1 IJ so H 2K

2

.

Look at the N = 3 graph in Figure 37.14. Minimum intensity is zero, attained where 1 cos φ = − . One relative maximum occurs at cos φ = −1.00 , where I = I max . 2 The larger local maximum happens where cos φ = +1.00 , giving I = 9.00 I 0 . The ratio of intensities at primary versus secondary maxima is 9.00 .

389

Chapter 37

Section 37.4 P37.21

(a)

Phasor Addition of Waves

FG A + B IJ cosFG A − B IJ to find the sum of the two sine functions H 2 2K H 2 2K

We can use sin A + sin B = 2 sin to be

f b g a = b19.7 kN C g sina15 x − 4.5t + 35.0°f

E1 + E2 = 24.0 kN C sin 15 x − 4.5t + 35.0° cos 35.0° E1 + E2

Thus, the total wave has amplitude 19.7 kN C and has a constant phase difference of

35.0° from the first wave. (b)

In units of kN/C, the resultant phasor is

e j e a16.1f + a11.3f at tan FGH 1116..31 IJK =

kx - ωt

y

j

E R = E 1 + E 2 = 12.0 i + 12.0 cos 70.0° i + 12.0 sin 70.0 j = 16.1 i + 11.3 j 2

ER =

2

−1

ER

19.7 kN C at 35.0°

E2

70.0°

α

x

E1

FIG. P37.21(b) (c)

y

E R = 12.0 cos 70.0° i + 12.0 sin 70.0° j +17.0 cos 160° i + 17.0 sin 160° j

P37.22

(a)

f

e

x

E3

The wave function of the total wave is EP = 9.36 kN C sin 15 x − 4.5t + 169° .

g a

E2

ER

E R = −9.18 i + 1.83 j = 9.36 kN C at 169°

b

FIG. P37.21(c)

j e

E R = E0 i + i cos 20.0°+ j sin 20.0° + i cos 40.0°+ j sin 40.0°

j

y ER

E R = E0 2.71i + 0.985 j = 2.88E0 at 20.0° = 2.88E0 at 0.349 rad

b

EP = 2.88E0 sin ω t + 0.349

kx - ωt

E1

+15.5 cos 80.0° i − 15.5 sin 80.0° j

g

E3 E2

x

E1

FIG. P37.22(a) (b)

e

j e

E R = E0 i + i cos 60.0°+ j sin 60.0° + i cos 120°+ j sin 120° E R = E0 1.00 i + 1.73 j = 2.00E0 at 60.0° = 2.00E0 at

FG H

EP = 2.00E0 sin ω t +

π 3

IJ K

j

y

E3

π rad 3

ER E2 E1

x

FIG. P37.22(b) continued on next page

390

Interference of Light Waves

e

j e

E R = E0 i + i cos 120°+ j sin 120° + i cos 240°+ j sin 240°

(c)

j

y E3

E R = E0 0 i + 0 j = 0

E2

EP = 0

x

E1

FIG. P37.22(c)

LM FG N H

E R = E0 i + i cos

(d)

IJ e K

3π 3π + i cos 3π + j sin 3π + j sin 2 2

jOPQ

y E3

3π rad E R = E0 0 i − 1.00 j = E0 at 270° = E0 at 2 3π EP = E0 sin ω t + 2

FG H

P37.23

E R = 6.00 i + 8.00 j =

E2

IJ K

a6.00f + a8.00f 2

2

at tan −1

FIG. P37.22(d)

FG 8.00 IJ H 6.00 K

y ER

E R = 10.0 at 53.1° = 10.0 at 0.927 rad

b

EP = 10.0 sin 100π t + 0.927

x

E1

g

8.00

α

π2

6.00

x

FIG. P37.23 P37.24

b

g

If E1 = E01 sin ω t and E2 = E02 sin ω t + φ , then by phasor addition, the amplitude of E is E0 =

bE

01

+ E02 cos φ

g + bE 2

02

sin φ

g

2

e

x

FIG. P37.24

j

y

E R = 21.0 i + 15.6 j = 26.2 at 36.6°

b

E 02

φ

θ E 01

E R = 12.0 i + 18.0 cos 60.0° i + 18.0 sin 60.0° j ER = 26.2 sin ω t + 36.6°

E0

2 2 + 2E01 E02 cos φ + E02 E01

=

E sin φ and the phase angle is found from sin θ = 02 . E0 P37.25

y

ER

g

θ

18.0

60.0°

12.0

x

FIG. P37.25 P37.26

Constructive interference occurs where m = 0 , 1, 2 , 3 , … , for

FG 2π x − 2π ft + π IJ − FG 2π x H λ 6K H λ bx − x g + 1 − 1 = m 1

1

2

λ

12

16

2

− 2π ft +

IJ K

π = 2π m 8

b

g + FG π − π IJ = 2π m H 6 8K F 1I = Gm − J λ m = 0 , 1, 2 , 3 , … H 48 K

2π x 1 − x 2

λ x1 − x 2

.

Chapter 37

P37.27

See the figure to the right:

φ=

391

π /2

π . 2

π /2

π /2 ωt

FIG. P37.27 P37.28

ER2 = E12 + E22 − 2E1 E2 cos β

where

ER

β = 180 − φ .

E2 φ = π /4

β

I ∝ E2

Since

E1

I R = I 1 + I 2 + 2 I 1 I 2 cos φ .

FIG. P37.28 P37.29

360° where N defines the N number of coherent sources. Then, Take φ =

ER =

y The N = 6 case

φ=

N

∑ E0 sinbω t + mφ g = 0 .

360° N

=

360° 6

= 60. 0°

m =1

In essence, the set of N electric field components complete a full circle and return to zero.

60.0° x

FIG. P37.29

Section 37.5

Change of Phase Due to Reflection

Section 37.6

Interference in Thin Films

P37.30

Light reflecting from the first surface suffers phase reversal. Light reflecting from the second surface does not, but passes twice through the thickness t of the film. So, for constructive interference, we require

λn + 2t = λ n 2 where

λn =

λ is the wavelength in the material. n

Then

2t =

λn λ = 2 2n

a fa

f

λ = 4nt = 4 1.33 115 nm = 612 nm .

392 P37.31

Interference of Light Waves

(a)

The light reflected from the top of the oil film undergoes phase reversal. Since 1.45 > 1.33 , the light reflected from the bottom undergoes no reversal. For constructive interference of reflected light, we then have

FG H

2nt = m + or

λm =

IJ K

1 λ 2

a fa

f

FIG. P37.31

2 1.45 280 nm 2nt . = m+ 1 2 m+ 1 2

b g

Substituting for m gives:

b g

m=0 ,

λ 0 = 1 620 nm (infrared)

m=1,

λ 1 = 541 nm (green)

m=2,

λ 2 = 325 nm (ultraviolet).

Both infrared and ultraviolet light are invisible to the human eye, so the dominant color in reflected light is green . (b)

The dominant wavelengths in the transmitted light are those that produce destructive interference in the reflected light. The condition for destructive interference upon reflection is 2nt = mλ or

λm =

2nt 812 nm = . m m

Substituting for m gives:

m=1,

λ 1 = 812 nm (near infrared)

m=2,

λ 2 = 406 nm (violet)

m=3 ,

λ 3 = 271 nm (ultraviolet).

Of these, the only wavelength visible to the human eye (and hence the dominate wavelength observed in the transmitted light) is 406 nm. Thus, the dominant color in the transmitted light is violet . P37.32

Since 1 < 1.25 < 1.33 , light reflected both from the top and from the bottom surface of the oil suffers phase reversal. mλ cons n

For constructive interference we require

2t =

and for destructive interference,

2t =

Then

λ cons 1 640 nm =1+ = = 1.25 and m = 2 . 2m 512 nm λ dest

Therefore,

t=

b g

m + 1 2 λ des n

a

f

.

2 640 nm = 512 nm . 2 1.25

a f

Chapter 37

P37.33

393

Treating the anti-reflectance coating like a camera-lens coating,

FG H

2t = m + Let m = 0 :

t=

λ 4n

=

IJ K

1 λ . 2 n

3.00 cm = 0.500 cm . 4 1.50

a f

This anti-reflectance coating could be easily countered by changing the wavelength of the radar—to 1.50 cm—now creating maximum reflection! P37.34

FG H

2nt = m +

IJ K

1 λ 2

so

Minimum P37.35

FG 1 IJ λ H 2 K 2n F 1 I a500 nmf = t=G J H 2 K 2a1.30f t= m+

96.2 nm .

Since the light undergoes a 180° phase change at each surface of the film, the condition for 2nt . The film thickness is constructive interference is 2nt = mλ , or λ = m t = 1.00 × 10 −5 cm = 1.00 × 10 −7 m = 100 nm . Therefore, the wavelengths intensified in the reflected light are

λ=

a fa

f

2 1.38 100 nm 276 nm where m = 1, 2 , 3 , … = m m

or λ 1 = 276 nm , λ 2 = 138 nm , . . . . All reflection maxima are in the ultraviolet and beyond.

No visible wavelengths are intensified. P37.36

(a)

For maximum transmission, we want destructive interference in the light reflected from the front and back surfaces of the film. If the surrounding glass has refractive index greater than 1.378, light reflected from the front surface suffers no phase reversal and light reflected from the back does undergo phase reversal. This effect by itself would produce destructive interference, so we want the distance down and back to be one whole wavelength in the film: 2t = t=

=

656.3 nm = 238 nm 2 1.378

a

f

(b)

The filter will expand. As t increases in 2nt = λ , so does λ increase .

(c)

Destructive interference for reflected light happens also for λ in 2nt = 2 λ , or

P37.37

λ 2n

λ . n

a

f

λ = 1.378 238 nm = 328 nm

anear ultravioletf .

If the path length difference ∆ = λ , the transmitted light will be bright. Since ∆ = 2d = λ , d min =

λ 580 nm = = 290 nm . 2 2

394 P37.38

Interference of Light Waves

The condition for bright fringes is 2t +

λ λ =m 2n n

θ

m = 1, 2 , 3 , … .

R

From the sketch, observe that

f FGH

a

t = R 1 − cos θ ≈ R 1 − 1 +

P37.39

I JK

FG IJ H K

θ2 R r = 2 2 R

2

=

r2 . 2R

t 2

FG H

r

IJ K

The condition for a bright fringe becomes

r 1 λ = m− . R 2 n

Thus, for fixed m and λ,

nr 2 = constant .

Therefore, n liquid r f2 = n air ri2 and

n liquid = 1.00

cmf a f aa11..5031 cm f

FIG. P37.38

2 2

= 1.31 .

For destructive interference in the air, 2t = mλ . For 30 dark fringes, including the one where the plates meet, t=

a

f

29 600 nm = 8.70 × 10 −6 m . 2

Therefore, the radius of the wire is r=

t 8.70 µm = = 4.35 µm . 2 2

FIG. P37.39 P37.40

For total darkness, we want destructive interference for reflected light for both 400 nm and 600 nm. With phase reversal at just one reflecting surface, the condition for destructive interference is 2n air t = mλ

m = 0 , 1, 2 , … .

The least common multiple of these two wavelengths is 1 200 nm, so we get no reflected light at 2 1.00 t = 3 400 nm = 2 600 nm = 1 200 nm , so t = 600 nm at this second dark fringe.

a f a

f a

f

By similar triangles,

600 nm 0.050 0 mm = , x 10.0 cm

or the distance from the contact point is

x = 600 × 10 −9 m

e

m I J= jFGH 5.000.100 × 10 m K −5

1.20 mm .

Chapter 37

Section 37.7 P37.41

The Michelson Interferometer

When the mirror on one arm is displaced by ∆ , the path difference changes by 2∆ . A shift resulting in the reversal between dark and bright fringes requires a path length change of one-half mλ , where in this case, m = 250 . wavelength. Therefore, 2 ∆ = 2 ∆ =m

P37.42

395

a fe

j

250 6.328 × 10 −7 m λ = = 39.6 µm 4 4

e

j

Distance = 2 3.82 × 10 −4 m = 1 700 λ

λ = 4.49 × 10 −7 m = 449 nm

The light is blue . P37.43

Counting light going both directions, the number of wavelengths originally in the cylinder is 2L 2L 2nL = m1 = . It changes to m 2 = as the cylinder is filled with gas. If N is the number of bright λ λn λ 2L fringes passing, N = m 2 − m1 = n − 1 , or the index of refraction of the gas is

λ

n = 1+

a f

Nλ 2L

Additional Problems *P37.44

(a)

Where fringes of the two colors coincide we have d sin θ = mλ = m ′λ ′ , requiring

(b)

λ = 430 nm , λ ′ = 510 nm ∴

λ m′ . = λ′ m

m ′ 430 nm 43 = = , which cannot be reduced any further. Then m = 51, m ′ = 43 . m 510 nm 51

F mλ IJ = sin LM a51fe430 × 10 mj OP = 61.3° = sin G HdK MN 0.025 × 10 m PQ = L tan θ = a1.5 mf tan 61.3° = 2.74 m −9

θm ym P37.45

−1

−1

−3

m

The wavelength is λ =

c 3.00 × 10 8 m s = = 5.00 m. f 60.0 × 10 6 s −1

Along the line AB the two traveling waves going in opposite directions add to give a standing wave. The two transmitters are exactly 2.00 wavelengths apart and the signal from B, when it arrives at A, will always be in phase with transmitter B. Since B is 180° out of phase with A, the two signals always interfere destructively at the position of A. The first antinode (point of constructive interference) is located at distance

λ 5.00 m = = 1.25 m from the node at A. 4 4

396 *P37.46

Interference of Light Waves

Along the line of length d joining the source, two identical waves moving in opposite directions add to give a standing wave. An d λ antinode is halfway between the sources. If > , there is space for 2 2 d two more antinodes for a total of three. If > λ , there will be at least 2 d five antinodes, and so on. To repeat, if > 0 , the number of antinodes

s

N

A

N

s

λ 4 FIG. P37.46

λ

d > 1, the number of antinodes is 3 or more. If > 2 , the λ λ number of antinodes is 5 or more. In general, is 1 or more. If

d

d . λ

The number of antinodes is 1 plus 2 times the greatest integer less than or equal to

d λ d λ < , there will be no nodes. If > , there will be space for at least two nodes, as shown in the 2 4 2 4 d 3λ d 5λ picture. If > , there will be at least four nodes. If > six or more nodes will fit in, and so on. 2 4 2 4 To repeat, if 2d < λ the number of nodes is 0. If 2d > λ the number of nodes is 2 or more. If 2d > 3 λ d 1 + > 1, the number of nodes is 4 or more. If 2d > 5 λ the number of nodes is 6 or more. Again, if λ 2 d 1 d 1 the number of nodes is at least 2. If + > 2 , the number of nodes is at least 4. If + > 3 , the λ 2 λ 2 number of nodes is at least 6. In general, If

FG H

IJ K

FG H

FG H

the number of nodes is 2 times the greatest nonzero integer less than

IJ K

IJ K

FG d + 1 IJ . H λ 2K

Next, we enumerate the zones of constructive interference. They are described by d sin θ = mλ , m = 0 , 1, 2 , … with θ counted as positive both left and right of the maximum at θ = 0 in the center. d The number of side maxima on each side is the greatest integer satisfying sin θ ≤ 1, d1 ≥ mλ , m ≤ .

λ

So the total number of bright fringes is one plus 2 times the greatest integer less than or equal to It is equal to the number of antinodes on the line joining the sources.

FG H

The interference minima are to the left and right at angles described by d sin θ = m +

FG H

IJ K

d . λ

IJ K

1 λ, 2

1 d 1 d 1 λ , m max < − or m max + 1 < + . Let n = 1, 2 , 3 , …. λ 2 λ 2 2 d 1 Then the number of side minima is the greatest integer n less than + . Counting both left and λ 2 d 1 right, the number of dark fringes is two times the greatest positive integer less than + . It is λ 2 equal to the number of nodes in the standing wave between the sources. m = 0 , 1, 2 , …. With sin θ < 1, d1 > m max +

FG H

IJ K

Chapter 37

P37.47

397

My middle finger has width d = 2 cm . (a)

Two adjacent directions of constructive interference for 600-nm light are described by d sin θ = mλ

θ0 = 0

e2 × 10

(b)

−2

j

e

m sin θ 1 = 1 6 × 10 −7 m

Thus,

θ 1 = 2 × 10 −3 degree

and

θ 1 − θ 0 ~ 10 −3 degree .

Choose

θ 1 = 20°

e2 × 10

−2

j

af

j

m sin 20° = 1 λ

λ = 7 mm Millimeter waves are microwaves . f= P37.48

λ

f=

:

3 × 10 8 m s 7 × 10 −3 m

~ 10 11 Hz

If the center point on the screen is to be a dark spot rather than bright, passage through the plastic must delay the light by one-half wavelength. Calling the thickness of the plastic t. t

λ P37.49

c

+

t nt 1 = = 2 λn λ

t=

or

λ 2 n −1

a f

where n is the index of refraction for the plastic.

No phase shift upon reflection from the upper surface (glass to air) of the film, but there will be a

λ due to the reflection at the lower surface of the film (air to metal). The total phase 2 difference in the two reflected beams is

shift of

then

δ = 2nt +

λ . 2

For constructive interference, δ = mλ or

a f

2 1.00 t +

λ = mλ . 2

Thus, the film thickness for the m th order bright fringe is

FG H

tm = m −

FG IJ H K

IJ K

λ λ 1 λ =m − 2 2 2 4

and the thickness for the m − 1 bright fringe is:

a

t m−1 = m − 1

fFGH λ2 IJK − λ4 .

Therefore, the change in thickness required to go from one bright fringe to the next is ∆t = t m − t m −1 = continued on next page

λ . 2

398

Interference of Light Waves

To go through 200 bright fringes, the change in thickness of the air film must be: 200

FG λ IJ = 100λ . H 2K

Thus, the increase in the length of the rod is

e

j

∆L = 100 λ = 100 5.00 × 10 −7 m = 5.00 × 10 −5 m .

P37.50

From

∆L = Liα∆t

we have:

α=

∆L 5.00 × 10 −5 m = = 20.0 × 10 −6 ° C −1 . Li ∆T 0.100 m 25.0° C

a

fa

f

Since 1 < 1.25 < 1.34 , light reflected from top and bottom surfaces of the oil undergoes phase reversal. The path difference is then 2t, which must be equal to mλ n =

mλ n

for maximum reflection, with m = 1 for the given first-order condition and n = 1.25 . So t=

a

f

mλ 1 500 nm = = 200 nm . 2n 2 1.25

a f

a

A=

P37.51

1.00 m3

e

200 10

−9

m

j

f

1.00 m3 = 200 nm A

The volume we assume to be constant: = 5.00 × 10 6 m 2 = 5.00 km 2 .

One radio wave reaches the receiver R directly from the distant source at an angle θ above the horizontal. The other wave undergoes phase reversal as it reflects from the water at P. Constructive interference first occurs for a path difference of d=

λ 2

(1)

It is equally far from P to R as from P to R ′ , the mirror image of the telescope. The angles θ in the figure are equal because they each form part of a right triangle with a shared angle at R ′ .

a

f

FIG. P37.51

a

f

So the path difference is

d = 2 20.0 m sin θ = 40.0 m sin θ .

The wavelength is

λ=

Substituting for d and λ in Equation (1),

a40.0 mf sinθ = 5.002 m .

Solving for the angle θ, sin θ =

8 c 3.00 × 10 m s = = 5.00 m . f 60.0 × 10 6 Hz

5.00 m and θ = 3.58° . 80.0 m

Chapter 37

P37.52

For destructive interference, the path length must differ by mλ . We may treat this problem as a

π -phase shift at the mirror. The 2 second slit is the mirror image of the source, 1.00 cm below the mirror plane. Modifying Equation 37.5, double slit experiment if we remember the light undergoes a

y dark

P37.53

e

ja

f

−7 mλL 1 5.00 × 10 m 100 m = = = 2.50 mm . d 2.00 × 10 −2 m

e

j

a15.0 kmf + h = 30.175 km a15.0 kmf + h = 227.63 2

2

2

2

2

h = 1.62 km P37.54

FIG. P37.53 2nt = mλ

For dark fringes,

and at the edge of the wedge, t =

so

P37.55

a fa fa

f

2nt = mλ

When submerged in water, m=

a

84 500 nm . 2

f

2 1.33 42 500 nm 500 nm

FIG. P37.54

m + 1 = 113 dark fringes . I

From Equation 37.13,

I max I

Let λ 2 equal the wavelength for which

I max

λ2 =

Then

But

FG H

π yd I1 = λ 1 cos −1 L I max

IJ K

12

a

FG I IJ HI K max

→

F π yd I . GH λ L JK

I2

= 0.640 .

I max

π yd L cos

−1

a

bI

2

I max

g

12

.

f

= 600 nm cos −1 0.900 = 271 nm .

Substituting this value into the expression for λ 2 ,

Note that in this problem, cos −1

f

= cos 2

λ2 =

271 nm cos

−1

e0.640 j

12

must be expressed in radians.

12

= 421 nm .

399

400 P37.56

Interference of Light Waves

At entrance, 1.00 sin 30.0° = 1.38 sin θ 2 θ 2 = 21.2° Call t the unknown thickness. Then t t cos 21.2° = a= a cos 21.2° c tan 21.2° = c = t tan 21.2° t b sin θ 1 = b = 2t tan 21.2° sin 30.0° 2c The net shift for the second ray, including the phase reversal on reflection of the first, is

FIG. P37.56

λ 2 where the factor n accounts for the shorter wavelength in the film. For constructive interference, we require λ 2 an − b − = mλ . 2 λ 2 an − b − = 0 . The minimum thickness will be given by 2 λ nt = 2 an − b = 2 − 2t tan 21.2° sin 30.0° 2 cos 21. 2° 590 nm 2 × 1.38 t = 115 nm = − 2 tan 21.2° sin 30.0° t = 2.57t cos 21.2° 2 2 an − b −

a

f

FG H

P37.57

IJ K

λ 2 where a and b are as shown in the ray diagram, n is the index of λ is due to phase reversal at the top refraction, and the term 2 surface. For constructive interference, δ = mλ where m has integer values. This condition becomes 1 2na − b = m + λ (1) 2 t From the figure’s geometry, a = cos θ 2 t sin θ 2 c = a sin θ 2 = cos θ 2 2t sin θ 2 b = 2 c sin φ 1 = sin φ 1 cos θ 2 Also, from Snell’s law, sin φ 1 = n sin θ 2 . The shift between the two reflected waves is δ = 2na − b −

FG H

Thus,

b=

IJ K

FIG. P37.57

2nt sin 2 θ 2 . cos θ 2

With these results, the condition for constructive interference given in Equation (1) becomes:

FG t IJ − 2nt sin θ = 2nt e1 − sin θ j = FG m + 1 IJ λ H 2K H cos θ K cos θ cosθ F 1I 2nt cos θ = G m + J λ . H 2K 2

2n

2

or

2

2

2

2

2

2

Chapter 37

P37.58

(a)

Minimum:

2nt = mλ 2

Maximum:

2nt = m ′ +

for λ 1 > λ 2 ,

FG m′ + 1 IJ < m H 2K

so

m′ = m − 1 .

Then

2nt = mλ 2 = m −

FG H

401

for m = 0 , 1, 2 , …

IJ K

1 λ1 2

FG H

for m ′ = 0 , 1, 2 , …

IJ K

1 λ1 2

2mλ 2 = 2mλ 1 − λ 1 m=

so

(b)

m=

λ1 2 λ1 − λ 2

b

g

.

500 = 1.92 → 2 (wavelengths measured to ±5 nm ) 2 500 − 370

a

f

Minimum:

2nt = mλ 2

a f a f 1I F 2nt = G m − 1 + J λ = 1.5 λ H 2K 2a1.40ft = 1.5a500 nmf 2 1.40 t = 2 370 nm

Maximum:

t = 264 nm

t = 268 nm

Film thickness = 266 nm . P37.59

From the sketch, observe that x = h2 +

FG d IJ H 2K

2

=

4h 2 + d 2 . 2

d x

h

x

Including the phase reversal due to reflection from the ground, the total shift between the two waves is δ = 2 x − d −

λ . 2

d/2

FIG. P37.59 (a)

For constructive interference, the total shift must be an integral number of wavelengths, or δ = mλ where m = 0 , 1, 2 , 3 , … . Thus,

FG H

2x − d = m +

IJ K

1 λ 2

or

λ=

4x − 2d . 2m + 1

For the longest wavelength, m = 0 , giving λ = 4x − 2d = 2 4h 2 + d 2 − 2d .

(b)

For destructive interference, Thus,

FG H

IJ K

1 λ where m = 0 , 1, 2 , 3 , … . 2 2x − d λ= . or 2x − d = mλ m

δ = m−

For the longest wavelength, m = 1 giving λ = 2 x − d =

4h 2 + d 2 − d .

402 P37.60

Interference of Light Waves

FG IJ H K FλI 2t = G J m . H nK F hI t = G Jx . HK 2t =

Bright fringes occur when and dark fringes occur when The thickness of the film at x is Therefore, x bright =

P37.61

FG H

λ 1 m+ 2 hn 2

IJ K

λ 1 m+ n 2

and x dark =

λ m . 2 hn

FIG. P37.60

Call t the thickness of the film. The central maximum corresponds to zero phase difference. Thus, the added distance ∆r traveled by the light from the lower slit must introduce a phase difference equal to that introduced by the plastic film. The phase difference φ is

φ = 2π

FG t IJ an − 1f . Hλ K

L Thin film y' d

θ

α

Screen

∆r

a

The corresponding difference in path length ∆r is

FIG. P37.61

Fλ I F t I Fλ I ∆r = φ G J = 2π G J an − 1fG J = tan − 1f . H 2π K H λ K H 2π K a

Zero order m=0

a

a

Note that the wavelength of the light does not appear in this equation. In the figure, the two rays from the slits are essentially parallel.

P37.62

∆r y ′ = . d L

Thus the angle θ may be expressed as

tan θ =

Eliminating ∆r by substitution,

t n −1 L y′ t n − 1 = gives y ′ = . L d d

a f

a f

The shift between the waves reflecting from the top and bottom surfaces of the film at the point where the film has thickness t is

θ

λ λ , with the factor of being due to a phase reversal 2 2 at one of the surfaces. δ = 2tn film +

For the dark rings (destructive interference), the total shift should 1 be δ = m + λ with m = 0 , 1, 2 , 3 , … . This requires that 2 mλ t= . 2n film

FG H

IJ K

a f

To find t in terms of r and R,

R2 = r 2 + R − t

Since t is much smaller than R,

t 2 1. a

or if

a < λ = 632.8 nm .

x = 1.22

F GH

Ie JK

λ 5.00 × 10 −7 m D = 1.22 250 × 10 3 m = 30.5 m d 5.00 × 10 −3 m

j

D = 250 × 10 3 m

λ = 5.00 × 10 −7 m d = 5.00 × 10 −3 m

424 P38.53

Diffraction Patterns and Polarization

d=

1 400 mm

= 2.50 × 10 −6 m

−1

(a)

d sin θ = mλ

(b)

λ=

(c)

d sin θ a = 2 λ

θ a = sin −1

541 × 10 −9 m = 4.07 × 10 −7 m 1.33

θ b = sin −1

d sin θ b =

F 2 × 541 × 10 m I = 25.6° GH 2.50 × 10 m JK F 2 × 4.07 × 10 m I = 19.0° GH 2.50 × 10 m JK −9

−6

−7

−6

2λ n

af

n sin θ b = 1 sin θ a P38.54

(a)

λ=

v : f

λ=

θ min = 1.22

3.00 × 10 8 m s 1.40 × 10 9 s −1

= 0.214 m

FG H

IJ K

λ 0.214 m = 7.26 µrad : θ min = 1.22 D 3.60 × 10 4 m

FG 180 × 60 × 60 s IJ = H π K

θ min = 7.26 µrad

P38.55

g e jb F 500 × 10 m I = 50.8 µrad a10.5 seconds of arcf λ θ θ = 1.22 = 1.22G D H 12.0 × 10 m JK d = θ L = e50.8 × 10 radja30.0 mf = 1.52 × 10 m = 1.52 mm

(b)

θ min =

(c)

min

(d)

1.50 arc seconds

d : L

d = θ min L = 7.26 × 10 −6 rad 26 000 ly = 0.189 ly −9

min

−3

−6

min

−3

With a grazing angle of 36.0°, the angle of incidence is 54.0°

tan θ p = n = tan 54.0° = 1.38 . In the liquid, λ n = *P38.56

(a)

λ n

=

750 nm = 545 nm . 1.38

Bragg’s law applies to the space lattice of melanin rods. Consider the planes d = 0.25 µm apart. For light at near-normal incidence, strong reflection happens for the wavelength given by 2d sin θ = mλ . The longest wavelength reflected strongly corresponds to m = 1:

e

j

2 0.25 × 10 −6 m sin 90° = 1λ

λ = 500 nm . This is the blue-green color.

(b)

For light incident at grazing angle 60°, 2d sin θ = mλ gives 1λ = 2 0.25 × 10 −6 m sin 60° = 433 nm . This is violet.

(c)

Your two eyes receive light reflected from the feather at different angles, so they receive light incident at different angles and containing different colors reinforced by constructive interference.

e

continued on next page

j

Chapter 38

P38.57

425

(d)

The longest wavelength that can be reflected with extra strength by these melanin rods is the one we computed first, 500 nm blue-green.

(e)

If the melanin rods were farther apart (say 0.32 µm ) they could reflect red with constructive interference.

(a)

d sin θ = mλ or d =

e

j

3 500 × 10 −9 m mλ = = 2.83 µm sin θ sin 32.0°

Therefore, lines per unit length =

1 1 = d 2.83 × 10 −6 m

or lines per unit length = 3.53 × 10 5 m −1 = 3.53 × 10 3 cm −1 .

sin θ =

(b)

P38.58

e

j a

−9 mλ m 500 × 10 m = = m 0.177 d 2.83 × 10 −6 m

f a

f

For sin θ ≤ 1.00 , we must have

m 0.177 ≤ 1.00

or

m ≤ 5.66 .

Therefore, the highest order observed is

m=5 .

Total number of primary maxima observed is

2m + 1 = 11 .

For the air-to-water interface, n water 1.33 = 1.00 n air

tan θ p =

θp

θ p = 53.1°

θ

a1.00f sinθ = a1.33f sinθ F sin 53.1° IJ = 36.9° . θ = sin G H 1.33 K

and

p

2

θ2 θ3

2

−1

For the water-to-glass interface, tan θ p = tan θ 3 =

n glass n water

=

1.50 so 1.33

FIG. P38.58

θ 3 = 48.4° . The angle between surfaces is θ = θ 3 − θ 2 = 11.5° . *P38.59

A central maximum and side maxima in seven orders of interference appear. If the seventh order is just at 90°,

d sin θ = mλ

e

j

d1 = 7 654 × 10 −9 m

d = 4.58 µm .

If the seventh order is at less than 90°, the eighth order might be nearly ready to appear according to

e

j

d1 = 8 654 × 10 −9 m

d = 5.23 µm .

Thus 4.58 µm < d < 5.23 µm .

FIG. P38.59

Air Water

426 P38.60

Diffraction Patterns and Polarization

(a)

We require

D

=

radius of diffraction disk D = . L 2L

D 2 = 2.44λL .

Then

e

ja

f

D = 2.44 500 × 10 −9 m 0.150 m = 428 µm

(b)

P38.61

λ

θ min = 1.22

The limiting resolution between lines θ min

e e

j j

550 × 10 −9 m λ = 1.22 = 1.22 = 1.34 × 10 −4 rad . D 5.00 × 10 −3 m

Assuming a picture screen with vertical dimension , the minimum viewing distance for no visible 485 . The desired ratio is then lines is found from θ min = L L

P38.62

(a)

=

1 1 = = 15.4 . 485θ min 485 1.34 × 10 −4 rad

e

j

Applying Snell’s law gives n 2 sin φ = n1 sin θ . From the sketch, we also see that:

b

g

θ + φ + β = π , or φ = π − θ + β .

which reduces to:

b g sin φ = sinbθ + β g .

Applying the identity again:

sin φ = sin θ cos β + cos θ sin β .

Snell’s law then becomes:

n 2 sin θ cos β + cos θ sin β = n1 sin θ

or (after dividing by cos θ ): Solving for tan θ gives: (b)

(a)

b g n btan θ cos β + sin β g = n tan θ . 2

1

tan θ =

n 2 sin β . n1 − n 2 cos β

If β = 90.0° , n1 = 1.00 , and n 2 = n , the above result becomes: tan θ =

P38.63

b

a f

n 1.00 , or n = tan θ , which is Brewster’s law. 1.00 − 0

From Equation 38.1,

θ = sin −1

In this case m = 1 and

λ=

Thus,

θ = sin −1

continued on next page

g

sin φ = sin π cos θ + β − cos π sin θ + β ,

Using the given identity:

FG mλ IJ . HaK

8 c 3.00 × 10 m s = = 4.00 × 10 −2 m . 9 f 7.50 × 10 Hz

F 4.00 × 10 GH 6.00 × 10

−2 −2

I= J mK m

41.8° .

FIG. P38.62(a)

Chapter 38

(b)

427

L sinbβ 2g OP where β = 2π a sinθ . =M λ I MN β 2 PQ 2π b0.060 0 mg sin 15.0° β= = 2.44 rad 2

I

From Equation 38.4,

max

When θ = 15.0° ,

0.040 0 m

I

and

(c)

sin θ =

I max

L sina1.22 radf OP =M N 1.22 rad Q

2

= 0.593 .

λ so θ = 41.8° : a

This is the minimum angle subtended by the two sources at the slit. Let α be the half angle between the sources, each a distance = 0.100 m from the center line and a distance L from the slit plane. Then, 41.8° = 0.262 m . L = cot α = 0.100 m cot 2

a

f FGH

IJ K

P38.64

I 1 1 = cos 2 45.0° cos 2 45.0° = I max 2 8

P38.65

d sin θ = mλ

e

je

j

a

d cos θ dθ = mdλ

or

d 1 − sin 2 θ ∆θ ≈ m∆λ d 1−

*P38.66

f

and, differentiating,

∆θ ≈

so

FIG. P38.63(c)

m 2 λ2 ∆θ ≈ m∆λ d2 ∆λ

e

d

2

.

j

m 2 − λ2

af

(a)

The angles of bright beams diffracted from the grating are given by d sin θ = mλ . The dθ dθ m dθ =m = angular dispersion is defined as the derivative : d cos θ dλ dλ d cos θ dλ

(b)

For the average wavelength 578 nm,

af

d sin θ = mλ

θ = sin −1

0.02 m sin θ = 2 578 × 10 −9 m 8 000

e

j

2 × 578 × 10 −9 m = 27.5° 2.5 × 10 −6 m

The separation angle between the lines is dθ m 2 ∆λ = ∆λ = 2.11 × 10 −9 m −6 dλ d cos θ 2.5 × 10 m cos 27.5° 180° = 0.109° = 0.001 90 = 0.001 90 rad = 0.001 90 rad π rad

∆θ =

FG H

IJ K

428 *P38.67

Diffraction Patterns and Polarization

(a)

Constructive interference of light of wavelength λ on the screen is described by d sin θ = mλ −1 2 y y = mλ . Differentiating with where tan θ = so sin θ = . Then d y L2 + y 2 2 2 L L +y

afe

j

respect to y gives

a f FGH 12 IJK eL + y j b0 + 2yg = m ddyλ adfy = m dλ = adfL + adfy − adfy d − dy eL + y j e L + y j eL + y j adfL dλ = dy m L + y e j e

d1 L2 + y 2

j

−1 2

2

+ d y −

2 −3 2

2

2

2 12

2

2

2 3 2

2

2

2 32

2

2

2 3 2

2

(b)

Here d sin θ = mλ gives

I JK

F GH

10 −2 m 0.55 × 10 −6 m sin θ = 1 550 × 10 −9 m , θ = sin −1 = 26.1° 8 000 1.25 × 10 −6 m

e

j

y = L tan θ = 2. 40 m tan 26.1° = 1.18 m Now

P38.68

dλ dL2 = dy m L2 + y 2

e

j

32

=

a

f 1ea 2.4 mf + a1.18 mf j 1.25 × 10 −6 m 2.40 m

2

2 32

2

= 3.77 × 10 −7 = 3.77 nm cm .

For a diffraction grating, the locations of the principal maxima for wavelength λ are given by a mλ y where a is the width of the grating ≈ . The grating spacing may be expressed as d = sin θ = N d L NLmλ and N is the number of slits. Thus, the screen locations of the maxima become y = . If two a nearly equal wavelengths are present, the difference in the screen locations of corresponding maxima is ∆y =

b g.

NLm ∆λ a

For a single slit of width a, the location of the first diffraction minimum is sin θ = y=

λ y ≈ , or a L

FG L IJ λ . If the two wavelengths are to be just resolved by Rayleigh’s criterion, y = ∆y from above. H aK

Therefore,

FG L IJ λ = NLmb∆λ g H aK a

or the resolving power of the grating is

R≡

λ = Nm . ∆λ

Chapter 38

P38.69

(a)

429

The E and O rays, in phase at the surface of the plate, will have a phase difference

θ=

FG 2π IJδ HλK

after traveling distance d through the plate. Here δ is the difference in the optical path lengths of these rays. The optical path length between two points is the product of the actual path length d and the index of refraction. Therefore,

δ = dnO − dn E . The absolute value is used since

θ=

(b)

P38.70

(a)

(b)

FG 2π IJ dn HλK

O

− dn E =

FG 2π IJ d n HλK

O

nO may be more or less than unity. Therefore, nE − nE .

jb g

e

550 × 10 −9 m π 2 λθ d= = = 1.53 × 10 −5 m = 15.3 µm 2π nO − n E 2π 1.544 − 1.553 I

From Equation 38.4,

I max

If we define

φ≡

β 2

this becomes

Therefore, when

1 I = I max 2

we must have

Let y1 = sin φ and y 2 =

I I max sin φ

φ

the right.

2π a sin θ

λ

gives If

.

2

=

1 2

, or sin φ =

φ . 2

π is shown to 2

The solution to the transcendental equation is found to be φ = 1.39 rad .

β=

2

φ . 2

A plot of y1 and y 2 in the range 1.00 ≤ φ ≤

(c)

L sinbβ 2g OP =M MN β 2 PQ L sin φ OP . =M Nφ Q

FIG. P38.70(b)

= 2φ sin θ =

FG φ IJ λ = 0.443 λ . Hπ K a a

λ λ is small, then θ ≈ 0. 443 . a a

This gives the half-width, measured away from the maximum at θ = 0 . The pattern is symmetric, so the full width is given by ∆θ = 0.443

FG H

IJ K

λ λ 0.886 λ − −0.443 = . a a a

430 P38.71

Diffraction Patterns and Polarization

φ 1 2 1.5 1.4 1.39 1.395 1.392 1.391 5 1.391 52 1.391 6 1.391 58 1.391 57 1.391 56 1.391 559 1.391 558 1.391 557 1.391 557 4

2 sin φ 1.19 1.29 1.41 1.394 1.391 1.392 1.391 7 1.391 54 1.391 55 1.391 568 1.391 563 1.391 561 1.391 558 1.391 557 8 1.391 557 5 1.391 557 3 1.391 557 4

bigger than φ smaller than φ smaller bigger smaller bigger bigger smaller

We get the answer to seven digits after 17 steps. Clever guessing, like using the value of the next guess for φ, could reduce this to around 13 steps.

P38.72

2 sin φ as

b g I LM bβ 2g cosbβ 2gb1 2g − sinbβ 2gb1 2g OP JK M PQ bβ 2g N FβI and require that it be zero. The possibility sinG J = 0 locates all of the minima and the central H 2K In I = I max

LM sinbβ 2g OP MN β 2 PQ

2

find

F GH

2 sin β 2 dI = I max β 2 dβ

2

maximum, according to 2π a sin θ

β = 0 , π , 2π , … ; 2

β=

The side maxima are found from

β β β β β − sin = 0 , or tan = . cos 2 2 2 2 2

This has solutions

λ

FG IJ H K

= 0 , 2π , 4π , … ; a sin θ = 0 , λ , 2 λ , … .

FG IJ H K

FG IJ H K

β β = 4.493 4 , = 7.725 3 , and others, giving 2 2

(a)

π a sin θ = 4.493 4λ

a sin θ = 1.430 3 λ

(b)

π a sin θ = 7.725 3 λ

a sin θ = 2.459 0λ

431

Chapter 38

P38.73

The first minimum in the single-slit diffraction pattern occurs at sin θ =

λ y min ≈ . a L

Thus, the slit width is given by a=

λL . y min

For a minimum located at y min = 6.36 mm ± 0.08 mm, the width is a=

e632.8 × 10

−9

ja

f=

m 1.00 m

6.36 × 10

−3

m

FIG. P38.73 99.5 µm ± 1% .

ANSWERS TO EVEN PROBLEMS P38.2

547 nm

P38.34

(a) 0.738 mm; (b) see the solution

P38.4

91.2 cm

P38.36

0.455 nm

P38.6

(a) 1.09 m ; (b) 1.70 mm

P38.38

3

P38.8

see the solution

P38.40

P38.10

(a) 0° , 10.3°, 21.0° , 32.5° , 45.8°, 63.6° ; (b) nine bright fringes at 0° and on either side at 10.3°, 21.0° , 32.5° , and 63.6° ; (c) 1.00 , 0.811 , 0.405 , 0.090 1 , 0.032 4

3 8

P38.42

(a) 6.89 units ; (b) 5.63 units

P38.44

(a) see the solution; (b) For light confined to a plane, yes. tan −1

FG n IJ − tan FG n IJ Hn K Hn K 3

−1

P38.12

2.61 µm

P38.14

869 m

P38.46

see the solution

P38.16

0.512 m

P38.48

see the solution

P38.18

6.10 cm

P38.50

see the solution

P38.20

105 m

P38.52

30.5 m

P38.22

(a) 2.40 µ rad ; (b) 213 km

P38.54

P38.24

514 nm

(a) 1.50 sec; (b) 0.189 ly; (c) 10.5 sec; (d) 1.52 mm

P38.26

1.81 µm

P38.56

see the solution

P38.28

see the solution

P38.58

11.5°

P38.30

74.2 grooves/mm

P38.60

(a) see the solution; (b) 428 µm

P38.32

2

P38.62

see the solution

2

1

2

432

Diffraction Patterns and Polarization

P38.64

1 8

P38.70

(a) see the solution; (b) φ = 1.39 rad; (c) see the solution

P38.66

(a) see the solution; (b) 0.109°

P38.72

(a) a sin θ = 1.430 3 λ ; (b) a sin θ = 2.459 0λ

P38.68

see the solution

39 Relativity CHAPTER OUTLINE The Principle of Galilean Relativity 39.2 The Michelson-Morley Experiment 39.3 Einstein’s Principle of Relativity 39.4 Consequences of the Special Theory of Relativity 39.5 The Lorentz Transformation Equations 39.6 The Lorentz Velocity Transformation Equations 39.7 Relativistic Linear Momentum and the Relativistic Form of Newton’s Laws 39.8 Relativistic Energy 39.9 Mass and Energy 39.10 The General Theory of Relativity

ANSWERS TO QUESTIONS

39.1

Q39.1

The speed of light c and the speed v of their relative motion.

Q39.2

An ellipsoid. The dimension in the direction of motion would be measured to be scrunched in.

Q39.3

No. The principle of relativity implies that nothing can travel faster than the speed of light in a vacuum, which is 300 Mm/s. The electron would emit light in a conical shock wave of Cerenkov radiation.

Q39.4

The clock in orbit runs slower. No, they are not synchronized. Although they both tick at the same rate after return, a time difference has developed between the two clocks.

Q39.5

Suppose a railroad train is moving past you. One way to measure its length is this: You mark the tracks at the cowcatcher forming the front of the moving engine at 9:00:00 AM, while your assistant marks the tracks at the back of the caboose at the same time. Then you find the distance between the marks on the tracks with a tape measure. You and your assistant must make the marks simultaneously in your frame of reference, for otherwise the motion of the train would make its length different from the distance between marks.

Q39.6

(a)

Yours does.

(b)

His does.

(c)

If the velocity of relative motion is constant, both observers have equally valid views.

Q39.7

Get a Mr. Tompkins book by George Gamow for a wonderful fictional exploration of this question. Driving home in a hurry, you push on the gas pedal not to increase your speed by very much, but rather to make the blocks get shorter. Big Doppler shifts in wave frequencies make red lights look green as you approach them and make car horns and car radios useless. High-speed transportation is very expensive, requiring huge fuel purchases. And it is dangerous, as a speeding car can knock down a building. Having had breakfast at home, you return hungry for lunch, but you find you have missed dinner. There is a five-day delay in transmission when you watch the Olympics in Australia on live television. It takes ninety-five years for sunlight to reach Earth. We cannot see the Milky Way; the fireball of the Big Bang surrounds us at the distance of Rigel or Deneb.

Q39.8

Nothing physically unusual. An observer riding on the clock does not think that you are really strange, either. 433

434

Relativity

Q39.9

By a curved line. This can be seen in the middle of Speedo’s world-line in Figure 39.12, where he turns around and begins his trip home.

Q39.10

According to p = γmu , doubling the speed u will make the momentum of an object increase by the factor 2

LM c − u OP N c − 4u Q 2

2

2

2

12

.

Q39.11

As the object approaches the speed of light, its kinetic energy grows without limit. It would take an infinite investment of work to accelerate the object to the speed of light.

Q39.12

There is no upper limit on the momentum of an electron. As more energy E is fed into the object E without limit, its speed approaches the speed of light and its momentum approaches . c

Q39.13

Recall that when a spring of force constant k is compressed or stretched from its relaxed position a 1 distance x, it stores elastic potential energy U = kx 2 . According to the special theory of relativity, 2 any change in the total energy of the system is equivalent to a change in the mass of the system. Therefore, the mass of a compressed or stretched spring is greater than the mass of a relaxed spring U by an amount 2 . The fractional change is typically unobservably small for a mechanical spring. c

Q39.14

You see no change in your reflection at any speed you can attain. You cannot attain the speed of light, for that would take an infinite amount of energy.

Q39.15

Quasar light moves at three hundred million meters per second, just like the light from a firefly at rest.

Q39.16

A photon transports energy. The relativistic equivalence of mass and energy means that is enough to give it momentum.

Q39.17

Any physical theory must agree with experimental measurements within some domain. Newtonian mechanics agrees with experiment for objects moving slowly compared to the speed of light. Relativistic mechanics agrees with experiment for objects at all speeds. Thus the two theories must and do agree with each other for ordinary nonrelativistic objects. Both statements given in the question are formally correct, but the first is clumsily phrased. It seems to suggest that relativistic mechanics applies only to fast-moving objects.

Q39.18

The point of intersection moves to the right. To state the problem precisely, let us assume that each of the two cards moves toward the other parallel to the long dimension of the picture, with velocity 2v = 2 v cot φ , where φ is the of magnitude v. The point of intersection moves to the right at speed tan φ small angle between the cards. As φ approaches zero, cot φ approaches infinity. Thus the point of intersection can move with a speed faster than c if v is sufficiently large and φ sufficiently small. For example, take v = 500 m s and φ = 0.000 19° . If you are worried about holding the cards steady enough to be sure of the angle, cut the edge of one card along a curve so that the angle will necessarily be sufficiently small at some place along the edge. Let us assume the spinning flashlight is at the center of a grain elevator, forming a circular screen of radius R. The linear speed of the spot on the screen is given by v = ω R , where ω is the angular speed of rotation of the flashlight. With sufficiently large ω and R, the speed of the spot moving on the screen can exceed c. continued on next page

Chapter 39

435

Neither of these examples violates the principle of relativity. Both cases are describing a point of intersection: in the first case, the intersection of two cards and in the second case, the intersection of a light beam with a screen. A point of intersection is not made of matter so it has no mass, and hence no energy. A bug momentarily at the intersection point could yelp, take a bite out of one card, or reflect the light. None of these actions would result in communication reaching another bug so soon as the intersection point reaches him. The second bug would have to wait for sound or light to travel across the distance between the first bug and himself, to get the message. As a child, the author used an Erector set to build a superluminal speed generator using the intersecting-cards method. Can you get a visible dot to run across a computer screen faster than light? Want’a see it again? Q39.19

In this case, both the relativistic and Galilean treatments would yield the same result: it is that the experimentally observed speed of one car with respect to the other is the sum of the speeds of the cars.

Q39.20

The hotter object has more energy per molecule than the cooler one. The equivalence of energy and mass predicts that each molecule of the hotter object will, on average, have a larger mass than those in the cooler object. This implies that given the same net applied force, the cooler object would have a larger acceleration than the hotter object would experience. In a controlled experiment, the difference will likely be too small to notice.

Q39.21

Special relativity describes inertial reference frames: that is, reference frames that are not accelerating. General relativity describes all reference frames.

Q39.22

The downstairs clock runs more slowly because it is closer to the Earth and hence in a stronger gravitational field than the upstairs clock.

Q39.23

The ants notice that they have a stronger sense of being pushed outward when they venture closer to the rim of the merry-go-round. If they wish, they can call this the effect of a stronger gravitational field produced by some mass concentration toward the edge of the disk. An ant named Albert figures out that the strong gravitational field makes measuring rods contract when they are near the rim of the disk. He shows that this effect precisely accounts for the discrepancy.

SOLUTIONS TO PROBLEMS Section 39.1 P39.1

The Principle of Galilean Relativity

In the rest frame, pi = m1 v1i + m 2 v 2i = 2 000 kg 20.0 m s + 1 500 kg 0 m s = 4.00 × 10 4 kg ⋅ m s

b

g

b

b

gb

g

g b

gb

g

p f = m1 + m 2 v f = 2 000 kg + 1 500 kg v f Since pi = p f ,

vf =

4.00 × 10 4 kg ⋅ m s = 11.429 m s . 2 000 kg + 1 500 kg

In the moving frame, these velocities are all reduced by +10.0 m/s.

b

g

v1′ i = v1i − v ′ = 20.0 m s − +10.0 m s = 10.0 m s

b

g

v ′2i = v 2i − v ′ = 0 m s − +10.0 m s = −10.0 m s

b

g

v ′f = 11.429 m s − +10.0 m s = 1.429 m s Our initial momentum is then

b

gb

g b

b

gb

gb

g

pi′ = m1 v1′ i + m 2 v ′2i = 2 000 kg 10.0 m s + 1 500 kg −10.0 m s = 5 000 kg ⋅ m s and our final momentum is

b

g

g

p ′f = 2 000 kg + 1 500 kg v ′f = 3 500 kg 1.429 m s = 5 000 kg ⋅ m s .

436 P39.2

P39.3

Relativity

(a)

v = vT + v B = 60.0 m s

(b)

v = vT − v B = 20.0 m s

(c)

v = vT2 + v B2 = 20 2 + 40 2 = 44.7 m s

The first observer watches some object accelerate under applied forces. Call the instantaneous velocity of the object v 1 . The second observer has constant velocity v 21 relative to the first, and measures the object to have velocity v 2 = v 1 − v 21 . dv 2 dv 1 = . dt dt This is the same as that measured by the first observer. In this nonrelativistic case, they measure the same forces as well. Thus, the second observer also confirms that ∑ F = ma . a2 =

The second observer measures an acceleration of

P39.4

The laboratory observer notes Newton’s second law to hold: F1 = ma 1 (where the subscript 1 implies the measurement was made in the laboratory frame of reference). The observer in the accelerating frame measures the acceleration of the mass as a 2 = a 1 − a ′ (where the subscript 2 implies the measurement was made in the accelerating frame of reference, and the primed acceleration term is the acceleration of the accelerated frame with respect to the laboratory frame of reference). If Newton’s second law held for the accelerating frame, that observer would then find valid the relation F2 = ma 2

or

F1 = ma 2

(since F1 = F2 and the mass is unchanged in each). But, instead, the accelerating frame observer will find that F2 = ma 2 − ma ′ which is not Newton’s second law.

Section 39.2

The Michelson-Morley Experiment

Section 39.3

Einstein’s Principle of Relativity

Section 39.4

Consequences of the Special Theory of Relativity

P39.5

L = Lp

Taking L =

P39.6

∆t =

v=c

p

Lp 2

where L p = 1.00 m gives

∆t p

b g

1− v c

2 12

so

v=c

∆t = 2 ∆t p

p

p

L F ∆t I OP v = c M1 − G MN H ∆t JK PQ L F ∆t I OP v = c M1 − G MN H 2 ∆t JK PQ

2

= c 1−

1 = 0.866 c . 4

2 12

p

.

2 12

For

FLI 1−G J HL K F L 2I 1−G H L JK 2

v2 1− 2 c

p

p

LM N

= c 1−

1 4

OP Q

12

= 0.866 c .

Chapter 39

P39.7

(a)

1

γ =

b g

1− v c

2

=

a

1

1 − 0.500

f

2

=

437

2 3

The time interval between pulses as measured by the Earth observer is 2 60.0 s ∆t = γ∆t p = = 0.924 s . 3 75.0 60.0 s min = 64.9 min . Thus, the Earth observer records a pulse rate of 0.924 s

FG H

(b)

IJ K

At a relative speed v = 0.990 c , the relativistic factor γ increases to 7.09 and the pulse rate recorded by the Earth observer decreases to 10.6 min . That is, the life span of the astronaut (reckoned by the duration of the total number of his heartbeats) is much longer as measured by an Earth clock than by a clock aboard the space vehicle.

*P39.8

(a)

The 0.8 c and the 20 ly are measured in the Earth frame, x 20 ly 20 ly 1c ∆t = = = = 25.0 yr . so in this frame, v 0.8 c 0.8 c 1 ly yr

(b)

We see a clock on the meteoroid moving, so we do not measure proper time; that clock measures proper time. 25.0 yr ∆t ∆t p = = = 25.0 yr 1 − 0.8 2 = 25.0 yr 0.6 = 15.0 yr ∆t = γ∆t p : γ 1 1 - v2 c2

a f

(c)

Method one: We measure the 20 ly on a stick stationary in our frame, so it is proper length. The tourist measures it to be contracted to Lp 20 ly 20 ly = = = 12.0 ly . L= γ 1 1 − 0.8 2 1.667 Method two: The tourist sees the Earth approaching at 0.8 c 0.8 ly yr 15 yr = 12.0 ly .

b

gb

g

Not only do distances and times differ between Earth and meteoroid reference frames, but within the Earth frame apparent distances differ from actual distances. As we have interpreted it, the 20-lightyear actual distance from the Earth to the meteoroid at the time of discovery must be a calculated result, different from the distance measured directly. Because of the finite maximum speed of information transfer, the astronomer sees the meteoroid as it was years previously, when it was much farther away. Call its apparent distance d. The time d required for light to reach us from the newly-visible meteoroid is the lookback time t = . c The astronomer calculates that the meteoroid has approached to be 20 ly away as it moved with constant velocity throughout the lookback time. We can work backwards to reconstruct her calculation: 0.8 cd d = 20 ly + 0.8 ct = 20 ly + c 0.2d = 20 ly d = 100 ly Thus in terms of direct observation, the meteoroid we see covers 100 ly in only 25 years. Such an apparent superluminal velocity is actually observed for some jets of material emanating from quasars, because they happen to be pointed nearly toward the Earth. If we can watch events unfold on the meteoroid, we see them slowed by relativistic time dilation, but also greatly speeded up by the Doppler effect.

438 P39.9

P39.10

Relativity

∆t p

∆t = γ∆t p =

∆t p =

so

1 − v2 c2

F GG H

I F v I J ∆t ≅ GH1 − 2 c JK ∆t c JK F v I ∆t . =G H 2c JK

1−

v2

2

2

2

2

and

∆t − ∆t p

If

v = 1 000 km h =

then

v = 9.26 × 10 −7 c

and

e∆t − ∆t j = e4.28 × 10 jb3 600 sg = 1.54 × 10

For

2

1.00 × 10 6 m = 277.8 m s 3 600 s

−13

p

−9

v = 0.990 , γ = 7.09 c

(a)

The muon’s lifetime as measured in the Earth’s rest frame is ∆t =

4.60 km 0.990 c

and the lifetime measured in the muon’s rest frame is ∆t p =

(b) P39.11

∆t

γ

=

L = Lp 1 −

LM MN

1 4.60 × 10 3 m 7.09 0.990 3.00 × 10 8 m s

FG v IJ H cK

2

e

=

Lp

γ

=

OP = j PQ

2.18 µs .

4.60 × 10 3 m = 649 m 7.09

The spaceship is measured by the Earth observer to be length-contracted to L = Lp 1 −

v2 c2

or

F GH

L2 = L2p 1 −

v2 c

2

I. JK

Also, the contracted length is related to the time required to pass overhead by:

L = vt

or

v2

L2 = v 2 t 2 = v2

c

2

af

Equating these two expressions gives

L2p − L2p

or

L2p + ct

Using the given values:

L p = 300 m

this becomes

e1.41 × 10

giving

v = 0.800 c .

c2

af

= ct v2

2

c

5

actf

2

m2

2

2

.

v2 c2

= L2p . and

j vc

2 2

t = 7.50 × 10 −7 s

= 9.00 × 10 4 m 2

s = 1.54 ns .

Chapter 39

P39.12

(a)

439

The spaceship is measured by Earth observers to be of length L, where L = Lp 1 −

v2 c2

v∆ t = L p 1 −

F v G ∆t H 2

Solving for v,

2

L2p

+

c

2

v2 c2

and

L = v∆t

and

v 2 ∆t 2 = L2p 1 −

F GH

I =L JK

2 p

The tanks move nonrelativistically, so we have v =

(c)

For the data in problem 11, v=

a

f

c 300 m

e3 × 10

8

ms

j e0.75 × 10 sj + a300 mf 2

−6

cL p

v=

(b)

2

2

c 2 ∆t 2 + L2p

a

f

c 300 m

=

a f m sj a75 sf + a300 mf c 300 m

e3 × 10

8

2

2

2

225 2 + 300 2 m

a

c 300 m

=

.

300 m = 4.00 m s . 75 s

in agreement with problem 11. For the data in part (b), v=

I JK

v2 . c2

e2.25 × 10 j

10 2

f

= 0.800 c

= 1.33 × 10 −8 c = 4.00 m s

+ 300 2 m

in agreement with part (b). P39.13

GMm

We find Cooper’s speed:

r2

=

mv 2 . r

L GM OP = LM e6.67 × 10 je5.98 × 10 j OP v=M N aR + hf Q MN e6.37 × 10 + 0.160 × 10 j PQ 2π a R + hf 2π e6.53 × 10 j T= = = 5. 25 × 10 s . 12

Solving,

−11

24

6

6

12

= 7.82 km s .

6

Then the time period of one orbit,

(a)

v

7.82 × 10

LF v I The time difference for 22 orbits is ∆t − ∆t = bγ − 1g∆t = MG 1 − J MNH c K F 1 v − 1I a22T f = 1 F 7.82 × 10 m s I 22e5.25 × 10 sj = ∆t − ∆t ≈ G 1 + 2 GH 3 × 10 m s JK H 2 c JK 2

p

p

P39.14

γ =

2

8

For one orbit, ∆t − ∆t p = 1

e

1 − v2 c2

j

= 1.01

p

3

2

(b)

3

3

2

−1 2

OP PQa f

− 1 22T

2

3

39.2 µs .

39.2 µs = 1.78 µs . The press report is accurate to one digit . 22

so

v = 0.140 c

440 P39.15

*P39.16

Relativity

(a)

Since your ship is identical to his, and you are at rest with respect to your own ship, its length is 20.0 m .

(b)

His ship is in motion relative to you, so you measure its length contracted to 19.0 m .

(c)

We have

L = Lp 1 −

from which

L 19.0 m v2 = = 0.950 = 1 − 2 and v = 0.312 c . L p 20.0 m c

v2 c2

In the Earth frame, Speedo’s trip lasts for a time ∆t =

20.0 ly ∆x = = 21.05 yr . v 0.950 ly yr

Speedo’s age advances only by the proper time interval ∆t p =

∆t

γ

= 21.05 yr 1 − 0.95 2 = 6.574 yr during his trip.

Similarly for Goslo, ∆t p =

20.0 ly ∆x v2 1− 2 = 1 − 0.75 2 = 17.64 yr . v 0.750 ly yr c

While Speedo has landed on Planet X and is waiting for his brother, he ages by 20.0 ly 20.0 ly − = 5.614 yr . 0.750 ly yr 0.950 ly yr

b

g

Then Goslo ends up older by 17.64 yr − 6.574 yr + 5.614 yr = 5.45 yr . P39.17

(a)

∆t = γ∆t p =

∆t p

=

15.0 yr

= 21.0 yr

(b)

b g 1 − a0.700f d = va ∆t f = 0.700 c b 21.0 yr g = a0.700fb1.00 ly yr g b 21.0 yr g =

(c)

The astronauts see Earth flying out the back window at 0.700 c :

e j

1− v c

2

b

2

g a

fb

gb

14.7 ly

g

d = v ∆t p = 0.700 c 15.0 yr = 0.700 1.00 ly yr 15.0 yr = 10.5 ly (d)

Mission control gets signals for 21.0 yr while the battery is operating, and then for 14.7 years after the battery stops powering the transmitter, 14.7 ly away: 21.0 yr + 14.7 yr = 35.7 yr

Chapter 39

P39.18

The orbital speed of the Earth is as described by

v=

GmS = r

e6.67 × 10

−11

∑ F = ma :

je

N ⋅ m 2 kg 2 1.99 × 10 30 kg 1.496 × 10

11

m

GmS m E r2

j = 2.98 × 10

4

=

441

mE v 2 r

m s.

The maximum frequency received by the extraterrestrials is f obs = fsource

1+v c = 57.0 × 10 6 Hz 1−v c

e

j 1 − ee2.98 × 10

j e3.00 × 10 m sj e3.00 × 10

1 + 2.98 × 10 4 m s 4

8 8

j = 57.005 66 × 10 m sj

6

Hz .

j = 56.994 34 × 10 m sj

6

Hz .

ms

The minimum frequency received is f obs = fsource

1−v c = 57.0 × 10 6 Hz 1+v c

e

j 1 + ee2.98 × 10

j e3.00 × 10 m sj e3.00 × 10

1 − 2.98 × 10 4 m s 4

8 8

ms

The difference, which lets them figure out the speed of our planet, is

b57.005 66 − 56.994 34g × 10 P39.19

(a)

6

Hz = 1.13 × 10 4 Hz .

Let f c be the frequency as seen by the car. Thus,

f c = f source

and, if f is the frequency of the reflected wave,

f = fc

(c)

which gives

a c + vf . a c − vf f a c − vf = f a c + vf b f − f gc = b f + f g v ≈ 2 f

The beat frequency is then

f beat = f − fsource =

Using the above result,

f beat

λ=

(d)

c+v . c−v

f = f source

Combining gives (b)

c+v c−v

v=

source

source

a2fb30.0 m sge10.0 × 10 = 8

3.00 × 10 m s

c fsource f beat λ 2

=

3.00 × 10 8 m s 10.0 × 10 9 Hz so

∆v =

9

Hz

j = a2fb30.0 m sg = 2 000 Hz = b0.030 0 mg

source

2 f source v 2v . = λ c

2.00 kHz

= 3.00 cm

a

fb

g

source v .

5 Hz 0.030 0 m ∆f beat λ = = 0.075 0 m s ≈ 0.2 mi h 2 2

442 P39.20

Relativity

(a)

When the source moves away from an observer, the observed frequency is f obs = fsource

FG c − v IJ Hc+v K

12

s

where v s = v source .

s

When v s 0 for x < 0 .

and

0

(a)

ψ x

(b)

Prob x < 0 =

af

2

a

= 0, x < 0

f

z 0

af

ψ x

2

dx =

−∞

Normalization

2 −2 x a e , x>0 a

FIG. P41.49

za f 0

0 dx = 0

−∞

z z z z FGH IJK

∞

(c)

af

ψ2 x =

and

af

ψ x

2

dx =

−∞ 0

0

z

∞

2

−∞

0dx +

−∞

∞ 0

a

2

ψ dx + ψ dx = 1 0

2 −2 x a e dx = 0 − e −2 x a a

f

z a

2

Prob 0 < x < a = ψ dx = 0

z FGH IJK a

0

∞ 0

e

j

= − e −∞ − 1 = 1

2 −2 x a e dx = − e −2 x a a

a 0

= 1 − e −2 = 0.865

509

510 P41.50

Quantum Mechanics

(a)

The requirement that E=

bpcg + emc j

2 2

2

K n = En − mc 2 =

(b)

nλ h nh = L so p = = is still valid. λ 2L 2

⇒ En =

FG nhc IJ + emc j H 2L K 2

FG nhc IJ + emc j H 2L K 2

2 2

2 2

− mc 2

Taking L = 1.00 × 10 −12 m, m = 9.11 × 10 −31 kg , and n = 1, we find K 1 = 4.69 × 10 −14 J .

e

j

2

6.626 × 10 −34 J ⋅ s h2 = Nonrelativistic, E1 = 8mL2 8 9.11 × 10 −31 kg 1.00 × 10 −12 m

e

je

j

2

= 6.02 × 10 −14 J .

Comparing this to K 1 , we see that this value is too large by 28.6% . P41.51

LM N

FG H

U=

(b)

From Equation 41.12,

(c)

E = U + K and

d=

g

a7fe18k e m j 2

K = 2E1 = 7kee2

dE = 0 for a minimum: dd

3h2 e

(d)

IJ a fOP b K Q

− 7 3 e2 7k e 2 e2 1 1 1 −1 + − + −1 + + −1 = = − e 4π ∈0 d 2 3 2 4π ∈0 d 3d

(a)

e

3d 2

e

−

2h2

e j

8m e 9d 2

j je

2

6.626 × 10 −34 h2 = = 42m e k e e 2 42 9.11 × 10 −31 8.99 × 10 9 1.60 × 10 −19 C

a fe

je

of one atom, we get: 13

h2 . 36m e d 2

h2 =0 18m e d 3

Since the lithium spacing is a, where Na 3 = V , and the density is

F Vm IJ a=G H Nm K

=

F m I =G H density JK

13

(5.62 times larger than c).

F 1.66 × 10 kg × 7 I =G H 530 kg JK −27

j

2

= 0.049 9 nm .

Nm , where m is the mass V

13

m = 2.80 × 10 −10 m = 0. 280 nm

511

Chapter 41

P41.52

(a)

ψ = Bxe − bmω

g

2= x2

dψ − mω = Be b dx

g

2= x 2

FG H

IJ K

mω − mω 2 xe b 2=

+ Bx −

FG H

IJ K d ψ F mω IJ xe b = −3BG H=K dx

d 2ψ mω − mω = Bx − xe b = dx 2 2

g

2= x 2

g

− mω 2 = x 2

2

g

2= x2

= Be

b

FG mω IJ 2 xe b H=K F mω IJ x e b + BG H=K

g

− mω 2 = x 2

−B

2

FG mω IJ x e b g H=K F mω IJ x FG − mω IJ xe b − BG H=K H =K g

− mω 2 = x 2

2 2 − mω 2 = x

−B

g

− mω 2 = x 2

2

g

2 3 − mω 2 = x

Substituting into the Schrödinger Equation (41.13), we have

FG mω IJ xe b H=K

g

− mω 2 = x 2

−3B

This is true if −3ω = −

FG mω IJ H=K

+B

2

x3 e

b

g

− mω 2 = x 2

=−

2mE =

2

Bxe

We never find the particle at x = 0 because ψ = 0 there.

(c)

ψ is maximized if

(d)

We require

z

g

+

FG mω IJ H=K

2

x 2 Bxe

b

g

− mω 2 = x 2

.

2E 3 =ω ; it is true if E = . = 2

(b)

∞

b

− mω 2 = x 2

FG IJ H K

= dψ mω = 0 = 1 − x2 , which is true at x = ± . = dx mω

2

ψ dx = 1 :

−∞

1=

z

∞

B2 x2 e

b

g

− mω = x 2

z

dx = 2B 2 x 2 e

−∞

FG IJ H K

2 1 2 mω Then B = 1 4 = π

3 4

b

g

− mω = x 2

F 4m ω I GH π = JK 3

=

3

dx = 2B 2

π

1 4

b mω = g

3

=

B2 π 1 2=3 2 . 2 mω 3 2

a f

14

.

3

FG IJ H K

= 1 1 4= = 2 =ω . This is larger than the , the potential energy is mω 2 x 2 = mω 2 mω 2 2 mω 3 =ω , so there is zero classical probability of finding the particle here. total energy 2

(e)

At x = 2

(f)

Probability = ψ dx = Bxe

2

Probability = δ

2

π1 2

FH

b

g

− mω 2 = x 2

IK

2

FG mω IJ FG 4= IJ e b H = K H mω K 32

δ = δB 2 x 2 e −b mω

gb

− mω = 4 = mω

g=

g

= x2

8δ

FG mω IJ H =π K

12

e −4

512

Quantum Mechanics

z

L

P41.53

(a)

FG π x IJ + 16 sin FG 2π x IJ + 8 sinFG π x IJ sinFG 2π x IJ OPdx = 1 HLK H L K H L K H L KQ LMFG L IJ + 16FG L IJ + 8 sinFG π x IJ sinFG 2π x IJ dxOP = 1 MNH 2 K H 2 K z H L K H L K PQ LM 17L + 16 sin FG π x IJ cosFG π x IJ dxOP = A LM 17L + 16L sin FG π x IJ OP = 1 MN 2 z H L K H L K PQ MN 2 3π H L K PQ 2

A2

ψ dx = 1 :

z LMN L

sin 2

2

0

0

A2

L 0

A2

x=L

L

2

2

3

x=0

0

A2 =

z a

(b)

2 2 , so the normalization constant is A = . 17L 17L

z LMN a

2

ψ dx = 1 :

2

A cos 2

−a

−a

FG π x IJ + B H 2a K

2

2

sin 2

FG π x IJ + 2 A B cosFG π x IJ sinFG π x IJ OPdx = 1 HaK H 2a K H a K Q

2

The first two terms are A a and B a . The third term is:

z a

2 A B cos −a

FG π x IJ LM2 sinFG π x IJ cosFG π x IJ OPdx = 4 A B z cos FG π x IJ sinFG π x IJ dx H 2a K N H 2a K H 2a K Q H 2a K H 2a K 8a A B F π x IJ = 0 = cos G H 2a K 3π a

2

−a

a

3

−a

e

2

so that a A + B

*P41.54

(a)

x

z

∞

=

0

x

−∞

FG a IJ HπK

F 4a I = z xG H π JK ∞

(b)

(c)

x

x

1

12

01

3

2

j = 1 , giving

2

2

A +B =

1 . a

2

e − ax dx = 0 , since the integrand is an odd function of x.

12

2

x 2 e − ax dx = 0 , since the integrand is an odd function of x.

−∞

=

z

∞

x

−∞

b

1 ψ 0 +ψ 1 2

g

2

dx =

1 x 2

0

+

1 x 2

+ 1

z

∞

af af

xψ 0 x ψ 1 x dx

−∞

The first two terms are zero, from (a) and (b). Thus:

F aI = z xG J HπK F 2a I = 2G H π JK ∞

x

01

14

−∞

2

=

1 2a

12

F 4a I xe dx = 2F 2a I e GH π JK GH π JK 1F π I G J , from Table B.6 4Ha K 3

− ax 2 2

12

3

14

− ax 2 2

2

1 2∞

z 0

2

x 2 e − ax dx

Chapter 41

P41.55

2

2

With one slit open

P1 = ψ 1

With both slits open,

P = ψ 1 +ψ 2 .

At a maximum, the wave functions are in phase

Pmax = ψ 1 + ψ 2

At a minimum, the wave functions are out of phase

Pmin = ψ 1 − ψ 2

ψ1 P Now 1 = P2 ψ 2

c c

h = c5.00 ψ h c5.00 ψ

c

h.

c

h.

2

2

ψ1 = 5.00 ψ2

= 25.0 , so

ψ1 + ψ2 P and max = Pmin ψ1 − ψ2

or P2 = ψ 2 . 2

2 2

513

2

2

2

2

+ψ2 −ψ2

h = a6.00f h a4.00f 2

2

2

2

=

36.0 = 2.25 . 16.0

ANSWERS TO EVEN PROBLEMS P41.2

1 2

P41.4

(a) 4; (b) 6.03 eV

P41.6

0.517 MeV , 3.31 × 10 −20 kg ⋅ m s

P41.8

FG 3hλ IJ H 8m c K

P41.22

af af af af af

12

e

P41.10

(a) 5.13 meV ; (b) 9.41 eV ; (c) The much smaller mass of the electron requires it to have much more energy to have the same momentum.

P41.12

(a)

FG 15hλ IJ H 8m c K

12

; (b) 1.25λ

P41.16

P41.18 P41.20

see the solution;

=2k2 2m

L ; (b) 5.26 × 10 −5 ; (c) 3.99 × 10 −2 ; 2 (d) see the solution (a)

F GH

I JK

=2 4x 2 − 6 ; (b) see the solution 2mL2 L2

P41.24

(a)

P41.26

see the solution

P41.28

1.03 × 10 −3

P41.30

85.9

P41.32

3.92%

P41.34

(a) see the solution; b =

e

P41.14

FG IJ H K FG IJ H K FG IJ H K FG IJ H K FG IJ H K FG IJ H K

πx 2 cos ; L L πx 2 P1 x = cos 2 ; L L 2π x 2 ψ2 x = sin ; L L 2π x 2 ; P2 x = sin 2 L L 3π x 2 ψ3 x = cos ; L L 3π x 2 P3 x = cos 2 ; L L (b) see the solution

af

(a) ψ 1 x =

0.250

FG H

IJ K

2π A A 1 sin ; (b) see the solution; − L 2π L (c) 0.585L

(a)

(c) first excited state

mω 3 ; (b) E = =ω ; 2 2=

514

Quantum Mechanics

P41.36

F mω I (a) B = G H π = JK

P41.38

see the solution

P41.40

(a) 2.00 × 10 −10 m; (b) 3.31 × 10 −24 kg ⋅ m s ; (c) 0.172 eV

P41.42

(a) see the solution; (b) 0.092 0 , 0.908

14

F mω I ; (b) δ G H π = JK

12

P41.44

(a) see the solution; (b) 0.200 ; (c) 0.351 ; (d) 0.377 eV , 1.51 eV

P41.46

(a)

h2 5h2 h2 5h2 , , , ; 4m e L2 8m e L2 m e L2 4m e L2 3h2 (b) see the solution, 4m e L2

P41.48

(a)

P41.50

(a)

2 L

; (b) 0.409

FG nhc IJ H 2L K

2

+ m 2 c 4 − mc 2 ;

(b) 46.9 fJ ; 28.6% P41.52

(a)

= 3 =ω ; (b) x = 0 ; (c) ± ; mω 2

F 4m ω I (d) G H π = JK 3

3

3

P41.54

14

; (e) 0; (f) 8δ

a f

(a) 0; (b) 0; (c) 2 a

−1 2

FG mω IJ H =π K

12

e −4

42 Atomic Physics CHAPTER OUTLINE 42.1 42.2 42.3

Atomic Spectra of Gases Early Models of the Atom Bohr’s Model of the Hydrogen Atom 42.4 The Quantum Model of the Hydrogen Atom 42.5 The Wave Functions of Hydrogen 42.6 Physical Interpretation of the Quantum Numbers 42.7 The Exclusion Principle and the Periodic Table 42.8 More on Atomic Spectra: Visible and X-ray 42.9 Spontaneous and Stimulated Transitions 42.10 Lasers

ANSWERS TO QUESTIONS Q42.1

Neon signs emit light in a bright-line spectrum, rather than in a continuous spectrum. There are many discrete wavelengths which correspond to transitions among the various energy levels of the neon atom. This also accounts for the particular color of the light emitted from a neon sign. You can see the separate colors if you look at a section of the sign through a diffraction grating, or at its reflection in a compact disk. A spectroscope lets you read their wavelengths.

Q42.2

One assumption is natural from the standpoint of classical physics: The electron feels an electric force of attraction to the nucleus, causing the centripetal acceleration to hold it in orbit. The other assumptions are in sharp contrast to the behavior of ordinary-size objects: The electron’s angular momentum must be one of a set of certain special allowed values. During the time when it is in one of these quantized orbits, the electron emits no electromagnetic radiation. The atom radiates a photon when the electron makes a quantum jump from one orbit to a lower one.

Q42.3

If an electron moved like a hockey puck, it could have any arbitrary frequency of revolution around an atomic nucleus. If it behaved like a charge in a radio antenna, it would radiate light with frequency equal to its own frequency of oscillation. Thus, the electron in hydrogen atoms would emit a continuous spectrum, electromagnetic waves of all frequencies smeared together.

Q42.4

(a)

Yes—provided that the energy of the photon is precisely enough to put the electron into one of the allowed energy states. Strangely—more precisely non-classically—enough, if the energy of the photon is not sufficient to put the electron into a particular excited energy level, the photon will not interact with the atom at all!

(b)

Yes—a photon of any energy greater than 13.6 eV will ionize the atom. Any “extra” energy will go into kinetic energy of the newly liberated electron.

Q42.5

An atomic electron does not possess enough kinetic energy to escape from its electrical attraction to the nucleus. Positive ionization energy must be injected to pull the electron out to a very large separation from the nucleus, a condition for which we define the energy of the atom to be zero. The atom is a bound system. All this is summarized by saying that the total energy of an atom is negative. 515

516 Q42.6

Atomic Physics

From Equations 42.7, 42.8 and 42.9, we have − E = −

kee2 k e2 k e2 = + e − e = K + U e . Then K = E and r 2r 2r

U e = −2 E . Q42.7

Bohr modeled the electron as moving in a perfect circle, with zero uncertainity in its radial coordinate. Then its radial velocity is always zero with zero uncertainty. Bohr’s theory violates the uncertainty principle by making the uncertainty product ∆r∆p r be zero, less than the minimum allowable

2

.

Q42.8

Fundamentally, three quantum numbers describe an orbital wave function because we live in threedimensional space. They arise mathematically from boundary conditions on the wave function, expressed as a product of a function of r, a function of θ, and a function of φ.

Q42.9

Bohr’s theory pictures the electron as moving in a flat circle like a classical particle described by ∑ F = ma . Schrödinger’s theory pictures the electron as a cloud of probability amplitude in the three-dimensional space around the hydrogen nucleus, with its motion described by a wave equation. In the Bohr model, the ground-state angular momentum is 1 ; in the Schrödinger model the ground-state angular momentum is zero. Both models predict that the electron’s energy is −13.606 eV with n = 1, 2 , 3 . limited to discrete energy levels, given by n2

Q42.10

The term electron cloud refers to the unpredictable location of an electron around an atomic nucleus. It is a cloud of probability amplitude. An electron in an s subshell has a spherically symmetric probability distribution. Electrons in p, d, and f subshells have directionality to their distribution. The shape of these electron clouds influences how atoms form molecules and chemical compounds.

Q42.11

The direction of the magnetic moment due to an orbiting charge is given by the right hand rule, but assumes a positive charge. Since the electron is negatively charged, its magnetic moment is in the opposite direction to its angular momentum.

Q42.12

Practically speaking, no. Ions have a net charge and the magnetic force q v × B would deflect the beam, making it difficult to separate the atoms with different orientations of magnetic moments.

Q42.13

The deflecting force on an atom with a magnetic moment is proportional to the gradient of the magnetic field. Thus, atoms with oppositely directed magnetic moments would be deflected in opposite directions in an inhomogeneous magnetic field.

Q42.14

If the exclusion principle were not valid, the elements and their chemical behavior would be grossly different because every electron would end up in the lowest energy level of the atom. All matter would be nearly alike in its chemistry and composition, since the shell structures of all elements would be identical. Most materials would have a much higher density. The spectra of atoms and molecules would be very simple, and there would be very little color in the world.

Q42.15

The Stern-Gerlach experiment with hydrogen atoms shows that the component of an electron’s spin angular momentum along an applied magnetic field can have only one of two allowed values. So does electron spin resonance on atoms with one unpaired electron.

a

f

Chapter 42

517

Q42.16

The three elements have similar electronic configurations. Each has filled inner shells plus one electron in an s orbital. Their single outer electrons largely determine their chemical interactions with other atoms.

Q42.17

When a photon interacts with an atom, the atom’s orbital angular momentum changes, thus the photon must carry orbital angular momentum. Since the allowed transitions of an atom are restricted to a change in angular momentum of ∆ = ±1 , the photon must have spin 1.

Q42.18

In a neutral helium atom, one electron can be modeled as moving in an electric field created by the nucleus and the other electron. According to Gauss’s law, if the electron is above the ground state it moves in the electric field of a net charge of +2 e − 1e = +1e We say the nuclear charge is screened by the inner electron. The electron in a He + ion moves in the field of the unscreened nuclear charge of 2 protons. Then the potential energy function for the electron is about double that of one electron in the neutral atom.

Q42.19

At low density, the gas consists of essentially separate atoms. As the density increases, the atoms interact with each other. This has the effect of giving different atoms levels at slightly different energies, at any one instant. The collection of atoms can then emit photons in lines or bands, narrower or wider, depending on the density.

Q42.20

An atom is a quantum system described by a wave function. The electric force of attraction to the nucleus imposes a constraint on the electrons. The physical constraint implies mathematical boundary conditions on the wave functions, with consequent quantization so that only certain wave functions are allowed to exist. The Schrödinger equation assigns a definite energy to each allowed wave function. Each wave function is spread out in space, describing an electron with no definite position. If you like analogies, think of a classical standing wave on a string fixed at both ends. Its position is spread out to fill the whole string, but its frequency is one of a certain set of quantized values.

Q42.21

Each of the electrons must have at least one quantum number different from the quantum numbers of each of the other electrons. They can differ (in m s ) by being spin-up or spin-down. They can also differ (in ) in angular momentum and in the general shape of the wave function. Those electrons with = 1 can differ (in m ) in orientation of angular momentum—look at Figure Q42.21. FIG. Q42.21

Q42.22

The Mosely graph shows that the reciprocal square root of the wavelength of K α characteristic xrays is a linear function of atomic number. Then measuring this wavelength for a new chemical element reveals its location on the graph, including its atomic number.

Q42.23

No. Laser light is collimated. The energy generally travels in the same direction. The intensity of a laser beam stays remarkably constant, independent of the distance it has traveled.

Q42.24

Stimulated emission coerces atoms to emit photons along a specific axis, rather than in the random directions of spontaneously emitted photons. The photons that are emitted through stimulation can be made to accumulate over time. The fraction allowed to escape constitutes the intense, collimated, and coherent laser beam. If this process relied solely on spontaneous emission, the emitted photons would not exit the laser tube or crystal in the same direction. Neither would they be coherent with one another.

518 Q42.25

Atomic Physics

(a)

The terms “I define” and “this part of the universe” seem vague, in contrast to the precision of the rest of the statement. But the statement is true in the sense of being experimentally verifiable. The way to test the orientation of the magnetic moment of an electron is to apply a magnetic field to it. When that is done for any electron, it has precisely a 50% chance of being either spin-up or spin-down. Its spin magnetic moment vector must make one of two 12 S and allowed angles with the applied magnetic field. They are given by cos θ = z = S 3 2 cos θ =

−1 2 3 2

. You can calculate as many digits of the two angles allowed by “space

quantization” as you wish. (b)

This statement may be true. There is no reason to suppose that an ant can comprehend the cosmos, and no reason to suppose that a human can comprehend all of it. Our experience with macroscopic objects does not prepare us to understand quantum particles. On the other hand, what seems strange to us now may be the common knowledge of tomorrow. Looking back at the past 150 years of physics, great strides in understanding the Universe— from the quantum to the galactic scale—have been made. Think of trying to explain the photoelectric effect using Newtonian mechanics. What seems strange sometimes just has an underlying structure that has not yet been described fully. On the other hand still, it has been demonstrated that a “hidden-variable” theory, that would model quantum uncertainty as caused by some determinate but fluctuating quantity, cannot agree with experiment.

SOLUTIONS TO PROBLEMS Section 42.1 P42.1

(a)

Atomic Spectra of Gases Lyman series

1

λ 1

λ (b)

Paschen series:

1

λ

F GH

= R 1− =

1 ni2

I JK

ni = 2 , 3 , 4, …

jFGH

1 1 = 1.097 × 10 7 1 − 2 −9 ni 94.96 × 10

=R

e

F1 − 1I GH 3 n JK 2

I JK

ni = 5

ni = 4, 5 , 6 , …

2 i

The shortest wavelength for this series corresponds to ni = ∞ for ionization

F GH

1 1 1 = 1.097 × 10 7 − 9 ni2 λ

I JK

For ni = ∞ , this gives λ = 820 nm This is larger than 94.96 nm, so this wave length cannot be associated with the Paschen series . Balmer series:

F1 − 1I GH 2 n JK λ F1 1 I 1 = 1.097 × 10 G − J λ H4 n K 1

=R

2

2 i 7

2 i

ni = 3 , 4, 5 , … with ni = ∞ for ionization, λ min = 365 nm

Once again the shorter given wavelength cannot be associated with the Balmer series .

Chapter 42

P42.2

(a)

(b)

Section 42.2 P42.3

(a)

hc Emax

λ min =

Lyman (n f = 1 ):

λ min =

hc 1 240 eV ⋅ nm = = 91.2 nm E1 13.6 eV

(Ultraviolet)

Balmer (n f = 2 ):

λ min =

hc 1 240 eV ⋅ nm = = 365 nm E2 1 4 13.6 eV

(UV)

Paschen (n f = 3 ):

λ min

Bracket (n f = 4 ):

λ min

Emax =

2

821 nm

(Infrared)

2

1 460 nm

(IR)

hc

λ min

c= E h 3.40 eV c = E h 1.51 eV c = E h 0.850 eV c= E h

Lyman:

Emax = 13.6 eV

1

Balmer:

Emax =

2

Paschen:

Emax =

Brackett:

Emax =

3

4

Early Models of the Atom For a classical atom, the centripetal acceleration is a=

1 v2 e2 = 4π ∈0 r 2 m e r

E=−

so

m v2 e2 e2 + e =− 4π ∈0 r 2 8π ∈0 r

F GH

−1 e 2 a 2 −e2 dE e2 dr e2 = = = dt 8π ∈0 r 2 dt 6π ∈0 c 3 6π ∈0 c 3 4π ∈0 r 2 m e

−

z 0

12π

I JK

2

.

dr e4 =− . 2 2 2 2 3 dt 12π ∈0 r m e c

Therefore,

(b)

b g = … = 3 a91.2 nmf = = … = 4 a91.2 nmf =

2

2.00 ×10 −10 m

∈02

r

2

m e2 c 3 dr

=e

4

z

T 0

dt

12π 2 ∈02 m e2 c 3 r 3 3 e4

FIG. P42.3

2.00 ×10 −10

= T = 8.46 × 10 −10 s 0

Since atoms last a lot longer than 0.8 ns, the classical laws (fortunately!) do not hold for systems of atomic size.

519

520 P42.4

Atomic Physics

(a)

The point of closest approach is found when k q q E = K + U = 0 + e α Au r k e 2 e 79 e or rmin = E

a fa f

e8.99 × 10 N ⋅ m C ja158fe1.60 × 10 Cj = a4.00 MeVfe1.60 × 10 J MeV j 9

rmin (b)

2

−19

2

2

−13

= 5.68 × 10 −14 m .

The maximum force exerted on the alpha particle is Fmax =

k e qα q Au 2 rmin

e8.99 × 10 =

9

ja fe

N ⋅ m 2 C 2 158 1.60 × 10 −19 C

e5.68 × 10

−14

j

m

2

j

2

= 11.3 N away from the

nucleus.

Section 42.3 P42.5

Bohr’s Model of the Hydrogen Atom kee2 m e r1

v1 =

(a)

af e8.99 × 10 N ⋅ m C je1.60 × 10 Cj = e9.11 × 10 kg je5.29 × 10 mj 2

r1 = 1 a 0 = 0.005 29 nm = 5.29 × 10 −11 m

where

9

v1

P42.6

2

−19

2

−31

−11

1 1 m e v12 = 9.11 × 10 −31 kg 2.19 × 10 6 m s 2 2

e

je

j

2

K1 =

(c)

8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C kee2 =− U1 = − r1 5. 29 × 10 −11 m

a

∆E = 13.6 eV

= 2.19 × 10 6 m s

= 2.18 × 10 −18 J = 13.6 eV

(b)

e

2

je

j

2

= −4.35 × 10 −18 J = −27.2 eV

fFG n1 − n1 IJ H K 2 i

2 f

Where for ∆E > 0 we have absorption and for ∆E < 0 we have emission. (i)

for ni = 2 and n f = 5 , ∆E = 2.86 eV (absorption)

(ii)

for ni = 5 and n f = 3 , ∆E = −0.967 eV (emission)

(iii)

for ni = 7 and n f = 4 , ∆E = −0.572 eV (emission)

(iv)

for ni = 4 and n f = 7 , ∆E = 0.572 eV (absorption)

(a)

E=

(b)

The atom gains most energy in transition i .

(c)

The atom loses energy in transitions ii and iii .

hc

λ

so the shortest wavelength is emitted in transition ii .

Chapter 42

P42.7

ga f

b

(a)

r22 = 0.052 9 nm 2

(b)

meke e2 me v2 = = r2

2

= 0.212 nm

e9.11 × 10

−31

je

je

kg 8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C

j

2

0.212 × 10 −9 m

m e v 2 = 9.95 × 10 −25 kg ⋅ m s

P42.8

e

je

j

(c)

L 2 = m e v 2 r2 = 9.95 × 10 −25 kg ⋅ m s 0.212 × 10 −9 m = 2.11 × 10 −34 kg ⋅ m 2 s

(d)

me v2 1 K 2 = m e v 22 = 2 2m e

(e)

8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C kee2 =− U2 = − r2 0.212 × 10 −9 m

(f)

E2 = K 2 + U 2 = 3.40 eV − 6.80 eV = −3.40 eV

b

g = e9.95 × 10 kg ⋅ m sj kg j 2e9.11 × 10 −25

2

2

−31

e

= 5.43 × 10 −19 J = 3.40 eV

je

j

En =

We use

2

= −1.09 × 10 −18 J = −6.80 eV

−13.6 eV . n2

To ionize the atom when the electron is in the n th level, it is necessary to add an amount of energy given by

P42.9

E = − En =

13.6 eV . n2

(a)

Thus, in the ground state where n = 1, we have E = 13.6 eV .

(b)

In the n = 3 level,

E=

13.6 eV = 1.51 eV . 9

(a)

F 1 − 1 I = 1.097 × 10 m F 1 − 1 I so λ = 410 nm jGH 2 6 JK GH n n JK e λ J ⋅ sje3.00 × 10 m sj hc e6.626 × 10 J = 3.03 eV E= = = 4.85 × 10

(c)

f=

(b)

1

=R

2 f

7

2 i

−34

λ

2

=

2

8

410 × 10 −9 m

λ

c

−1

3.00 × 10 8 = 7.32 × 10 14 Hz 410 × 10 −9

−19

521

522 P42.10

P42.11

Atomic Physics

Starting with

k e2 1 mev2 = e 2 2r

we have

v2 =

kee2 mer

and using

rn =

n2 2 me ke e2

gives

vn2 =

or

vn =

ke e2

e

me n 2

2

me ke e2

j

kee2 . n

Each atom gives up its kinetic energy in emitting a photon, so

e

je

j

6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s 1 hc 2 mv = = = 1.63 × 10 −18 J λ 2 1.216 × 10 −7 m

e

j

v = 4.42 × 10 4 m s . P42.12

The batch of excited atoms must make these six transitions to get back to state one: 2 → 1 , and also 3 → 2 and 3 → 1 , and also 4 → 3 and 4 → 2 and 4 → 1 . Thus, the incoming light must have just enough energy to produce the 1 → 4 transition. It must be the third line of the Lyman series in the absorption spectrum of hydrogen. The absorbing atom changes from energy Ei = −

13.6 eV 13.6 eV = −13.6 eV to E f = − = −0.850 eV , 2 1 42

so the incoming photons have wavelength

e

je a

j F 1.00 eV I = 9.75 × 10 f GH 1.60 × 10 J JK

6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s hc λ= = −0.850 eV − −13.6 eV E f − Ei P42.13

(a)

−19

−8

m = 97.5 nm .

The energy levels of a hydrogen-like ion whose charge number is Z are given by

a

En = −13.6 eV

f Zn

2 2

a

.

f

Thus for Helium Z = 2 , the energy levels are En = − (b)

54.4 eV n2

n = 1, 2 , 3 , … .

For He + , Z = 2 , so we see that the ionization energy (the energy required to take the electron from the n = 1 to the n = ∞ state) is E = E∞ − E1 = 0 −

a−13.6 eVfa2f a1f 2

2

= 54.4 eV .

FIG. P42.13

Chapter 42

*P42.14

(a)

1

= Z 2 RH

λ

523

F 1 − 1 I . The shortest wavelength, λ , corresponds to n = ∞ , and the longest GH n n JK 2 f

s

2 i

i

wavelength, λ , to ni = n f + 1 . 1

λs

=

Z 2 RH

(1)

n 2f

L 1 1 OP Z R L F n I =Z R M − MM n dn + 1i PP = n MM1 − GH n + 1 JK λ N N Q F n I λ =1−G Divide (1) and (2): λ H n + 1 JK 1

2

2

H

2 f

2

2 f

f

f

H

f

2

OP PQ

(2)

2

f

s

f

∴

nf nf +1

= 1−

λs 22.8 nm = 1− = 0.800 63.3 nm λ n 2f

From (1): Z =

=

λ s RH

e22.8 × 10

∴n f = 4

42

−9

je

m 1.097 × 10 7 m −1

j

= 8.00 .

Hence the ion is O 7+ .

(b)

R| λ = Se7.020 8 × 10 |T

8

m

−1

L jMM 41 − a4 +1kf N 2

2

OPU| PQV|W

−1

, k = 1, 2 , 3 , …

Setting k = 2 , 3 , 4 gives λ = 41.0 nm, 33.8 nm, 30.4 nm .

P42.15

(a)

The speed of the moon in its orbit is v = So,

(b)

j

je

je

j

L = mvr = 7.36 × 10 22 kg 1.02 × 10 3 m s 3.84 × 10 8 m = 2.89 × 10 34 kg ⋅ m 2 s .

We have L = n or

(c)

e

e

8 2π r 2π 3.84 × 10 m = = 1.02 × 10 3 m s . T 2.36 × 10 6 s

n=

L

=

2.89 × 10 34 kg ⋅ m 2 s 1.055 × 10 −34 J ⋅ s

We have n = L = mvr = m so

r=

2

m 2 GM e

FG GM IJ H r K e

12

n 2 = Rn 2 and

which is approximately equal to

= 2.74 × 10 68 .

r,

a f

2

n + 1 R − n 2 R 2n + 1 ∆r = = r n2R n2

2 = 7.30 × 10 −69 . n

524

Atomic Physics

Section 42.4 P42.16

The Quantum Model of the Hydrogen Atom

The reduced mass of positronium is less than hydrogen, so the photon energy will be less for positronium than for hydrogen. This means that the wavelength of the emitted photon will be longer than 656.3 nm. On the other hand, helium has about the same reduced mass but more charge than hydrogen, so its transition energy will be larger, corresponding to a wavelength shorter than 656.3 nm. All the factors in the given equation are constant for this problem except for the reduced mass and the nuclear charge. Therefore, the wavelength corresponding to the energy difference for the transition can be found simply from the ratio of mass and charge variables. mpme

For hydrogen,

µ=

Its wavelength is

λ = 656.3 nm ,

(a)

mp + me

µ=

For positronium,

≈ me .

The photon energy is

∆E = E3 − E2 .

where

λ=

c hc = . f ∆E

meme m = e 2 me + me

so the energy of each level is one half as large as in hydrogen, which we could call “protonium”. The photon energy is inversely proportional to its wavelength , so for positronium,

a

f

λ 32 = 2 656.3 nm = 1.31 µm (in the infrared region). (b)

For He + ,

µ ≈ m e , q1 = e , and q 2 = 2 e ,

so the transition energy is 2 2 = 4 times larger than hydrogen.

λ 32 =

Then,

P42.17

(a)

∆x∆p ≥

(b)

Choosing ∆p ≈

U= (c)

2

so if ∆x = r , ∆p ≥

r

,K=

FG 656 IJ nm = H4K

2r

164 nm (in the ultraviolet region).

.

b g

2

2 ∆p p2 ≈ = 2m e 2m e 2m e r 2

2 −kee2 ke e2 − , so E = K + U ≈ . r r 2m e r 2

To minimize E, 2 2 k e2 dE =− + e2 = 0 → r = = a0 3 dr mer r meke e2

Then, E =

2

2m e

Fm k e I GH JK e e 2

2

2

− kee2

(the Bohr radius).

Fm k e I =−m k e GH JK 2 e e 2

2

2 4 e e 2

= −13.6 eV .

Chapter 42

Section 42.5 P42.18

The Wave Functions of Hydrogen

af

1

ψ 1s r =

e−r

a 03

π

4r 2 −2 r e a 03

af

P1s r =

a0

(Eq. 42.22)

a0

(Eq. 42.25) FIG. P42.18

P42.19

z

(a)

z

∞

2

2

ψ dV = 4π ψ r 2 dr = 4π 0

z

Using integral tables,

F 1Ire GH π a JK z 2 L dV = − Me a MN ∞

3 0

2 −2 r a 0

2

ψ

dr

0

2 0

−2 r a 0

Fr GH

a2 + a0 r + 0 2

2

I OP = F − 2 I F − a I = JK PQ GH a JK GH 2 JK ∞

2 0

2 0

0

1

so the wave function as given is normalized. Pa0

(b)

2 → 3 a0 2

= 4π

z

3 a0 2

2

ψ r 2 dr = 4π

a0 2

F 1I GH π a JK 3 0

Again, using integral tables, Pa0 2 → 3 a0 2 = −

P42.20

ψ=

1

1

b g

3 2 a0

3 2

LMe MN

2 a 02

r −r e a0

−2 r a 0

Fr GH

2

z

3 a0 2

a2 + a0 r + 0 2

I OP JK PQ

r 2 −r e 24a 05

.

r 2 e −2 r

a0

dr

a0 2 3 a0 2

=− a0 2

2 a 02

LMe F 17a I − e F 5 a I OP = MN GH 4 JK GH 4 JK PQ −3

2 a0

so

Pr = 4π r 2 ψ 2 = 4π r 2

Set

dP 4π 4r 3 e − r = dr 24a 05

LM MN

a0

a0

FG 1 IJ e H aK

+ r4 −

− r a0

0

OP = 0 . PQ

Solving for r, this is a maximum at r = 4a 0 . P42.21

ψ= d 2ψ dr

2

1

π =

a 03

e −r 1

π a 07

−2 2 dψ 2 ψ = e − r a0 = − 5 r dr r π a ra 0 0

a0

e−r

a0

2

But so or

a0 = −

=

1 a 02

ψ

−

2

2m e

F 1 − 2 Iψ − e ψ = Eψ GH a ra JK 4π ∈ r 2

2 0

0

b4π ∈ g 0

mee

2

2

e =E 8π ∈0 a 0

E=−

kee2 . 2 a0

This is true, so the Schrödinger equation is satisfied.

0

2 0

−1

2 0

0.497 .

525

526 P42.22

Atomic Physics

The hydrogen ground-state radial probability density is

af

P r = 4π r 2 ψ 1s

2

=

4r 2 a 03

FG 2r IJ . H aK

exp −

0

The number of observations at 2 a 0 is, by proportion

b g = 1 000 b2 a g N = 1 000 Pb a 2g b a 2g P 2 a0

0

0

Section 42.6

2 2

0

e −4 a 0 a 0 = 1 000 16 e −3 = 797 times . − a0 a0 e

a f

Physical Interpretation of the Quantum Numbers

Note: Problems 17 and 25 in Chapter 29 and Problem 68 in Chapter 30 can be assigned with this section. P42.23

(a)

In the 3d subshell, n = 3 and we have n

= 2,

m

3 2 +2

3 2 +2

3 2 +1

3 2 +1

3 2 0

3 2 0

3 2 –1

3 2 –1

3 2 –2

3 2 –2

ms

+1/2

–1/2

+1/2

–1/2

+1/2

–1/2

+1/2

–1/2

+1/2

–1/2

(A total of 10 states) (b)

In the 3p subshell, n = 3 and we have n

=1,

m

3 1 +1

3 1 +1

3 1 +0

3 1 +0

3 1 –1

3 1 –1

ms

+1/2

–1/2

+1/2

–1/2

+1/2

–1/2

(A total of 6 states) P42.24

P42.25

= 2.58 × 10 −34 J ⋅ s .

(a)

For the d state,

= 2,

L=

6

(b)

For the f state,

=3,

L=

a + 1f

L=

a + 1f

:

=

= 3.65 × 10 −34 J ⋅ s .

a + 1f FGH 6.6262×π10 IJK a + 1f = e4.714 × 10 j a2π f = 1.998 × 10 e6.626 × 10 j −34

4.714 × 10 −34 =

−4 2

−34 2

so

12

=4 .

2

1

a f

≈ 20 = 4 4 + 1

Chapter 42

P42.26

The 5th excited state has n = 6 , energy The atom loses this much energy:

P42.27

−13.6 eV = −0.378 eV . 36

e6.626 × 10 J ⋅ sje3.00 × 10 = λ e1 090 × 10 mje1.60 × 10 −34

hc

−9

−0.378 eV − 1.14 eV = −1.52 eV

which is the energy in state 3:

−

(a)

33

n =1:

For n = 1 ,

= 0 , m = 0 , ms = ±

1 1

ms

0 0

–1/2 +1/2

0 0

af

Yields 2 sets; 2n 2 = 2 1

n=2:

m

2

j = 1.14 eV J eV j ms

= −1.51 eV .

a + 1f

can be as large as 2, giving angular momentum

n

(b)

13.6 eV

8

−19

to end up with energy

While n = 3 ,

527

=

6

.

1 2

= 2

For n = 2 ,

we have n 2 2 2 2

0 1 1 1

m

ms

0 –1 0 1

±1/2 ±1/2 ±1/2 ±1/2

af

2n 2 = 2 2

yields 8 sets;

2

= 8

Note that the number is twice the number of m values. Also, for each different m values. Finally,

can take on values ranging from 0 to n − 1 .

So the general expression is

number =

n −1

∑ 2a 2

f

+1 .

0

The series is an arithmetic progression:

2 + 6 + 10 + 14…

the sum of which is

number =

n 2a + n − 1 d 2

where a = 2 , d = 4 :

number =

n 4 + n − 1 4 = 2n 2 . 2

(c)

n=3:

(d)

n=4:

(e)

n=5:

af af af 2a1f + 2a3f + 2a5f + 2a7 f = 32 32 + 2a9f = 32 + 18 = 50

2 1 + 2 3 + 2 5 = 2 + 6 + 10 = 18

af = 2a 4f = 2a 5 f

a f

a f

2n 2 = 2 3

2

= 18

2n 2

2

= 32

2

= 50

2n 2

a

f

there are 2 + 1

528 P42.28

Atomic Physics

For a 3d state,

n = 3 and

Therefore,

L=

m can have the values

–2, –1, 0, 1, and 2

we find the possible values of θ

1.67 × 10 −27 kg m = V 4 3 π 1.00 × 10 −15 m

ρ=

(b)

Size of model electron:

F 3m I r =G H 4π ρ JK

(c)

Moment of inertia:

I=

b ge

v=

P42.30

.

F 3e9.11 × 10 =G GH 4π e3.99 × 10

13

= 3.99 × 10 17 kg m3 .

j IJ kg m j JK

−31

17

kg

13

= 8.17 × 10 −17 m .

3

e

e

2

=

je

j

2

= 2.43 × 10 −63 kg ⋅ m 2

Iv . r

je

j

6.626 × 10 −34 J ⋅ s 8.17 × 10 −17 m r = = 1.77 × 10 12 m s . −63 2 2I 2π 2 × 2. 43 × 10 kg ⋅ m

e

j

This is 5.91 × 10 3 times larger than the speed of light.

In the N shell, n = 4 . For n = 4 , can take on values of 0, 1, 2, and 3. For each value of , m can be − to in integral steps. Thus, the maximum value for m is 3. Since L z = m , the maximum value for L z is L z = 3

P42.31

j

3

2 2 mr 2 = 9.11 × 10 −31 kg 8.17 × 10 −17 m 5 5

L z = Iω = Therefore,

and 2

145° , 114° , 90.0° , 65.9° , and 35.3° .

Density of a proton:

(d)

= 2.58 × 10 −34 J ⋅ s

6

Lz L

cos θ =

Using the relation

(a)

=

L z can have the values − 2 , − , 0 ,

so

P42.29

a + 1f

= 2.

The 3d subshell has

.

= 2 , and n = 3 . Also, we have s = 1 .

Therefore, we can have n = 3 ,

= 2 ; m = −2 , − 1, 0 , 1, 2 ; s = 1; and m s = −1, 0 , 1

leading to the following table: n m s ms

3 2 –2

3 2 –2

3 2 –2

3 2 –1

3 2 –1

3 2 –1

3 2 0

3 2 0

3 2 0

3 2 1

3 2 1

3 2 1

3 2 2

3 2 2

3 2 2

1 –1

1 0

1 1

1 –1

1 0

1 1

1 –1

1 0

1 1

1 –1

1 0

1 1

1 –1

1 0

1 1

529

Chapter 42

Section 42.7 P42.32

The Exclusion Principle and the Periodic Table 1s 2 2 s 2 2 p 4

(a) (b)

P42.33

For the 1s electrons,

n =1,

=0, m =0,

ms = +

1 2

and

−

1 . 2

For the two 2s electrons,

n=2,

=0, m =0,

ms = +

1 2

and

−

1 . 2

For the four 2p electrons,

n=2;

= 1 ; m = −1 , 0, or 1; and

ms = +

1 2

or

−

1 . 2

The 4s subshell fills first , for potassium and calcium, before the 3d subshell starts to fill for scandium through zinc. Thus, we would first suppose that Ar 3d 4 4s 2 would have lower energy than Ar 3d 5 4s 1 . But the latter has more unpaired spins, six instead of four, and Hund’s rule suggests that this could give the latter configuration lower energy. In fact it must, for Ar 3d 5 4s 1 is the ground state for chromium.

P42.34

Electronic configuration: 2

2

1s 2s 2 p

6

→

Na 11

+3s 2

→

Mg 12

+3s 2 3 p 1

→

Al 13

+3s 2 3 p 2

→

Si 14

+3s 2 3 p 3

→

P 15

+3s 2 3 p 4

→

S 16

+3s 2 3 p 5

→

Cl 17

+3s 2 3 p 6

→

Ar 18

→

K 19

+3s

1s 2 2s 2 2 p 6 3s 2 3 p 6 4s 1 *P42.35

Sodium to Argon

1

In the table of electronic configurations in the text, or on a periodic table, we look for the element whose last electron is in a 3p state and which has three electrons outside a closed shell. Its electron configuration then ends in 3s 2 3 p 1 . The element is aluminum .

530 P42.36

Atomic Physics

(a)

For electron one and also for electron two, n = 3 and here in columns giving the other quantum numbers:

= 1 . The possible states are listed

electron m

1

1

1

1

1

1

1

1

1

1

0

0

0

0

0

one

1 2 1

1 2 0

1 2 0

1 2 –1

1 2 –1

1 − 2 1

1 − 2 0

1 − 2 0

1 − 2 –1

1 − 2 –1

1 2 1

1 2 1

1 2 0

1 2 –1

1 2 –1

1 2

−

1 2

1 2

−

1 2

1 2

−

1 2

1 2

−

1 2

1 2

−

0

–1

–1

–1

–1

–1

–1

–1

–1

–1

–1

1 2 –1

1 2 1

1 2 1

1 2 0

1 2 0

1 2 –1

1 2

1 2

ms

electron m two

ms

electron m

−

1 2

0

electron m

1 2 1

two

1 2

one

ms

ms

−

1 2

−

0

0

1 2 1

−

−

1 2

1 2

1 2 0

−

0 1 2 –1

−

1 2

1 2

1 2

−

−

−

1 2

1 2

−

1 2

−

1 2

1 2 1

−

1 2

1 2

1 2 1

−

−

1 2

−

1 2 0

−

1 2

1 2 0

−

−

1 2

1 2

1 2 –1

−

1 2

There are thirty allowed states, since electron one can have any of three possible values for m for both spin up and spin down, amounting to six states, and the second electron can have any of the other five states. (b)

Were it not for the exclusion principle, there would be 36 possible states, six for each electron independently.

P42.37

(a)

n+ subshell

(b)

Z = 15 :

Z = 47 :

Z = 86 :

1 1s

2 2s

3 2p, 3s

5 3d, 4p, 5s

6 4d, 5p, 6s

7 4f, 5d, 6p, 7s

Valence subshell: Prediction: Element is phosphorus,

1s , 2s , 2 p , 3s (12 electrons) 3 electrons in 3p subshell Valence = +3 or – 5 Valence = +3 or – 5 (Prediction correct)

Filled subshells:

1s , 2s , 2 p , 3s , 3 p , 4s , 3 d , 4 p , 5 s

Outer subshell: Prediction: Element is silver,

(38 electrons) 9 electrons in 4d subshell Valence = −1 (Prediction fails) Valence is +1

Filled subshells:

Filled subshells:

Prediction Element is radon, inert P42.38

4 3p, 4s

1s , 2s , 2 p , 3s , 3 p , 4s , 3 d , 4 p , 5 s , 4d , 5 p , 6 s , 4 f , 5d , 6 p (86 electrons) Outer subshell is full: inert gas (Prediction correct)

Listing subshells in the order of filling, we have for element 110, 1s 2 2s 2 2 p 6 3s 2 3 p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5 p 6 6s 2 4 f 14 5 d 10 6 p 6 7 s 2 5 f 14 6 d 8 . In order of increasing principal quantum number, this is 1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4s 2 4p 6 4d 10 4 f 14 5s 2 5 p 6 5d 10 5 f 14 6s 2 6 p 6 6d 8 7 s 2 .

Chapter 42

*P42.39

In the ground state of sodium, the outermost electron is in an s state. This state is spherically symmetric, so it generates no magnetic field by orbital motion, and has the same energy no matter whether the electron is spin-up or spin-down. The energies of the states 3p and 3p above 3s are hc hc hf1 = and hf 2 = .

A

λ

2 µ B B = hc

B=

so

B

λ2

The energy difference is

FG 1 Hλ

1

−

1

λ2

IJ K

IJ e K

FG H

je

6.63 × 10 −34 J ⋅ s 3 × 10 8 m s hc 1 1 − = 2µ B λ 1 λ 2 2 9.27 × 10 −24 J T

e

j

j FG 1 H 588.995 × 10

−9

m

−

1 589.592 × 10

−9

IJ mK

B = 18.4 T .

Section 42.8 P42.40

More on Atomic Spectra: Visible and X-ray

(a)

n = 3,

= 0, m = 0

n = 3,

= 1 , m = −1 , 0 , 1

For n = 3 ,

= 2 , m = −2 , − 1 , 0 , 1 , 2

ψ 300 corresponds to E300 = −

(b)

Z 2 E0 n

2

=−

a f=

2 2 13.6 32

−6.05 eV .

ψ 31 −1 , ψ 310 , ψ 311 have the same energy since n is the same. ψ 32 − 2 , ψ 32 −1 , ψ 320 , ψ 321 , ψ 322 have the same energy since n is the same. All states are degenerate. P42.41

E=

hc

λ

e6.626 × 10 J ⋅ sje3.00 × 10 e10.0 × 10 mj −34

= e∆ V :

8

ms

−9

j = e1.60 × 10 j∆V −19

∆V = 124 V P42.42

531

Some electrons can give all their kinetic energy K e = e∆V to the creation of a single photon of xradiation, with hf =

λ=

hc

λ

= e∆ V

e

je

j

6.626 1 × 10 −34 J ⋅ s 2.997 9 × 10 8 m s 1 240 nm ⋅ V hc = = −19 ∆V e∆V 1.602 2 × 10 C ∆V

e

j

532 P42.43

Atomic Physics

Eγ =

Following Example 42.9

a

3 42 − 1 4

f a13.6 eVf = 1.71 × 10 2

4

eV = 2.74 × 10 −15 J

f = 4.14 × 10 18 Hz

λ = 0.072 5 nm .

and

P42.44

E= For

hc

λ

=

1 240 eV ⋅ nm

=

1.240 keV ⋅ nm

λ λ 1 = 0.018 5 nm ,

λ E = 67.11 keV

λ 2 = 0.020 9 nm ,

E = 59.4 keV

λ 3 = 0.021 5 nm ,

E = 57.7 keV

The ionization energy for the K shell is 69.5 keV, so the ionization energies for the other shells are:

L shell = 11.8 keV P42.45

M shell = 10.1 keV

FIG. P42.44

N shell = 2.39 keV .

The K β x-rays are emitted when there is a vacancy in the ( n = 1 ) K shell and an electron from the ( n = 3 ) M shell falls down to fill it. Then this electron is shielded by nine electrons originally and by one in its final state. hc

λ

=−

a

13.6 Z − 9 32

f

2

eV +

a

13.6 Z − 1

f

2

12

eV

e6.626 × 10 J ⋅ sje3.00 × 10 m sj = a13.6 eVfF − Z GH 9 e0.152 × 10 mje1.60 × 10 J eVj F 8 Z − 8I 8.17 × 10 eV = a13.6 eV fG H 9 JK −34

8

−9

2

−19

8Z 2 −8 9

so

601 =

and

Z = 26

Iron .

Section 42.9

Spontaneous and Stimulated Transitions

Section 42.10

Lasers

a

f

E4 − E3 = 20.66 − 18.70 eV = 1.96 eV =

The photon energy is

λ=

P42.47

I JK

18Z 81 − + Z 2 − 2Z + 1 9 9

2

3

P42.46

+

F GH

e6.626 × 10

−34

je

8

J ⋅ s 3.00 × 10 m s

e

j

1.96 1.60 × 10 −19 J

I FG 1 J IJ = JK H 1 V ⋅ C K

f=

0.117 eV 1.60 × 10 −19 C E = h 6.630 × 10 −34 J ⋅ s e

λ=

c 3.00 × 10 8 m s = = 10.6 µm , infrared f 2.82 × 10 13 s −1

hc

λ

j=

2.82 × 10 13 s −1

633 nm .

Chapter 42

(a)

e3.00 × 10 Jj I= e1.00 × 10 sjLMNπ e15.0 × 10

(b)

e3.00 × 10

533

−3

P42.48

P42.49

−9

e0.600 × 10 mj Jj e30.0 × 10 mj −9

−3

e

2

2

−6

je

4.24 × 10 15 W m 2

= mj O PQ 2

= 1.20 × 10 −12 J = 7.50 MeV

j

E = P ∆t = 1.00 × 10 6 W 1.00 × 10 −8 s = 0.010 0 J Eγ = hf = N=

*P42.50

−6

(a)

hc

λ

e6.626 × 10 je3.00 × 10 j J = 2.86 × 10 −34

=

694.3 × 10

8

−9

−19

J

0.010 0 E = = 3.49 × 10 16 photons Eγ 2.86 × 10 −19 − E3 N3 N g e = N 2 N g e − E2

b k ⋅300 K g b k ⋅300 K g B B

=e

b

− E3 − E 2

g b k ⋅300 K g = e − hc λ b k ⋅300 K g B

B

where λ is the wavelength of light radiated in the 3 → 2 transition. N3 − 6.63 × 10 −34 J⋅s je 3 ×10 8 m s j =e e N2 N3 = e −75 .9 = 1.07 × 10 −33 N2 (b)

Nu − E −E =e b u N

e632.8 ×10

−9

ja

je

m 1.38 × 10 −23 J K 300 K

f

gkT B

where the subscript u refers to an upper energy state and the subscript state. Since Eu − E = Ephoton =

hc

λ

to a lower energy

Nu = e − hc λk BT . N 1.02 = e − hc λk BT

Thus, we require

a f

ln 1.02 = −

or

T=−

e6.63 × 10

e632.8 × 10

−34

−9

je

j J K jT

J ⋅ s 3 × 10 8 m s

je

m 1.38 × 10

23

2.28 × 10 4 = −1.15 × 10 6 K . ln 1.02

a f

A negative-temperature state is not achieved by cooling the system below 0 K, but by heating it above T = ∞ , for as T → ∞ the populations of upper and lower states approach equality. (c)

Because Eu − E > 0 , and in any real equilibrium state T > 0 ,

e − bEu − E

g k T EF , we can take Eav =

z

EF

1 ne

CE 3 2 dE =

0

z

EF

C ne

2C 5 2 EF . 5n e

E 3 2 dE =

0

But from Equation 43.24, P43.34

af N aEf = CE

N E = 0 for E > EF ;.

At T = 0 ,

C 3 −3 2 = EF , so that ne 2

Eav =

12

=

8 2π m e3 2 h3

FG 2 IJ FG 3 E IJ E H 5KH 2 K −3 2 F

5 2 F

E1 2

3 EF . 5

=

Consider first the wave function in x. At x = 0 and x = L , ψ = 0. Therefore,

sin k x L = 0

and

k x L = π , 2π , 3π , ….

Similarly,

sin k y L = 0

and

k y L = π , 2π , 3π , …

sin k z L = 0

and

k z L = π , 2π , 3π , …

FG n π x IJ sinFG n π y IJ sinFG n π z IJ . H L K H L K H L K ∂ ψ ∂ ψ ∂ ψ 2m + + = From aU − Efψ , ∂x ∂y ∂z F − n π − n π − n π I ψ = 2m −E ψ a f GH L L L JK

ψ = A sin

y

x

2

2

2

2 x

2

2

2

2 y

2

2

2

z

2

2 z

2

e

2

2

2

we have inside the box, where U = 0,

e

E=

2

π2 2 n x + n y2 + n z2 2 2m e L

e

j

n x , n y , n z = 1, 2 , 3 , … .

Outside the box we require ψ = 0. The minimum energy state inside the box is P43.35

(a)

n x = n y = n z = 1, with E =

The density of states at energy E is Hence, the required ratio is

(b)

3 2π 2 2m e L2

af g a8.50 eV f C a8.50f = g a7.00 eV f C a7.00f

g E = CE 1 2 .

12 12

= 1.10 .

From Eq. 43.22, the number of occupied states having energy E is

Hence, the required ratio is At T = 300 K , k BT = 4.14 × 10 −21 J = 0.025 9 eV , And

12

a f eb CEg + 1 . N a8.50 eV f a8.50f L e a = M N a7.00 eV f a7.00f MN e a N a8.50 eV f a8.50f L = N a7.00 eV f a7.00f MN e a N a8.50 eV f = 1.55 × 10 N a7.00 eV f

NE =

E − EF k B T

f f

OP . + 1 PQ OP . + 1Q

12

7.00 − 7 .00 k BT

12

8 .50 − 7.00 kB T

12 12

2.00

f

1.50 0 .025 9

−25

+1

.

Comparing this result with that from part (a), we conclude that very few states with E > EF are occupied.

Chapter 43

Section 43.6 P43.36

559

Electrical Conduction in Metals, Insulators, and Semiconductors E g = 1.14 eV for Si

(a)

a

fe

j

hf = 1.14 eV = 1.14 eV 1.60 × 10 −19 J eV = 1.82 × 10 −19 J so f ≥ 2.75 × 10 14 Hz c = λf ; λ =

(b) P43.37

c 3.00 × 10 8 m s = = 1.09 × 10 −6 m = 1.09 µm (in the infrared region) f 2.75 × 10 14 Hz

Photons of energy greater than 2.42 eV will be absorbed. This means wavelength shorter than

λ=

e

je

j

6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s hc = = 514 nm . E 2.42 × 1.60 × 10 −19 J

All the hydrogen Balmer lines except for the red line at 656 nm will be absorbed. hc

−34

je

J ⋅ s 3.00 × 10 8 m s

j J≈

Eg =

P43.39

If λ ≤ 1.00 × 10 −6 m, then photons of sunlight have energy E≥

λ

=

e6.626 × 10

P43.38

650 × 10

e6.626 × 10 =

hc

λ max

−34

−9

m

je

J ⋅ s 3.00 × 10 8 m s

1.00 × 10

−6

1.91 eV

j F 1 eV I = 1.24 eV . GH 1.60 × 10 J JK −19

m

Thus, the energy gap for the collector material should be E g ≤ 1.24 eV . Since Si has an energy gap E g ≈ 1.14 eV , it will absorb radiation of this energy and greater. Therefore, Si is acceptable as a material for a solar collector. P43.40

If the photon energy is 5.5 eV or higher, the diamond window will absorb. Here,

bhf g

max

=

hc

λ min

λ min = 2.26 × 10 *P43.41

λ min =

= 5.5 eV : −7

a′ =

2

me ke e2 2

je

a

fe

j

j

m = 226 nm .

In the Bohr model we replace k e by a0 =

e

6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s hc = 5.5 eV 5.5 eV 1.60 × 10 −19 J eV

ke

κ

and m e by m* . Then the radius of the first Bohr orbit,

in hydrogen, changes to

FG IJ H K

FG H

FG IJ H K

IJ b K

2 me me me κ κ κ a0 = = = 11.7 0.052 9 nm = 2.81 nm . 0. 220m e m∗ke e2 m∗ me ke e2 m∗

The energy levels are in hydrogen En = − En′ = −

ke e2 1 and here 2 a0 n 2

F I GH JK

ke e2 1 ke e2 m ∗ En =− = − 2 κ 2a ′ n me κ 2 κ 2 m e m ∗ κ a0

e

For n = 1 , E1′ = −0.220

j

13.6 eV 11.7 2

= −0.021 9 eV .

g

560

Molecules and Solids

Section 43.7 P43.42

Semiconductor Devices

I = I 0 e ea ∆V f

Thus,

e ea ∆V f

and

∆V =

k BT I ln 1 + . e I0

At T = 300 K ,

∆V =

e

k BT

j

−1 .

k BT

IJ K

FG H

e1.38 × 10

=1+

−23

I I0

ja

J K 300 K

1.60 × 10

−19

C

f lnF 1 + I I = a25.9 mVf lnF 1 + I I . GH I JK GH I JK 0

a f a f ∆V = a 25.9 mV f lna0.100f =

0

∆V = 25.9 mV ln 10.0 = 59.5 mV .

(a)

If I = 9.00 I 0 ,

(b)

If I = −0.900 I 0 ,

−59.5 mV .

The basic idea behind a semiconductor device is that a large current or charge can be controlled by a small control voltage. P43.43

P43.44

The voltage across the diode is about 0.6 V. The voltage drop across the resistor is 0.025 A 150 Ω = 3.75 V . Thus, ε − 0.6 V − 3.8 V = 0 and ε = 4.4 V .

a

fa

f

First, we evaluate I 0 in I = I 0 e ea ∆V f

ja

a f e e

e

k BT

j

− 1 , given that I = 200 mA when ∆V = 100 mV and T = 300 K .

f f

1.60 × 10 −19 C 0.100 V e ∆V 200 mA I = 3.86 = 4.28 mA = = 3.86 so I 0 = ea ∆V f k T − 23 B k BT −1 −1 e 1.38 × 10 J K 300 K e If ∆V = −100 mV , I = I 0 e ea ∆V f

e

*P43.45

(a)

k BT

a f

ja

e ∆V = −3.86 ; and the current will be k BT

j a

fe

j

− 1 = 4.28 mA e −3.86 − 1 = −4.19 mA .

The currents to be plotted are

e

je

I D = 10 −6 A e ∆V IW =

0.025 V

j

−1 ,

20

2.42 V − ∆V 745 Ω

je

Diode Wire

10

The two graphs intersect at ∆V = 0.200 V . The currents are then

e

Diode and Wire Currents

0

j

I D = 10 −6 A e 0. 200 V 0.025 V − 1

0

0.1

∆V(volts)

= 2.98 mA FIG. P43.45 2.42 V − 0.200 V = 2.98 mA . They agree to three digits. IW = 745 Ω ∴ I D = I W = 2.98 mA (b)

∆V 0.200 V = = 67.1 Ω ID 2.98 × 10 −3 A

(c)

d ∆V dI D = dI D d ∆V

a f LM OP = LM 10 A e N a f Q N 0.025 V −1

−6

0 . 200 V 0.025 V

OP Q

−1

= 8.39 Ω

0.2

0.3

Chapter 43

Section 43.8 P43.46

(a) (b)

Superconductivity See the figure at right. For a surface current around the outside of the cylinder as shown, B=

Nµ 0 I

or NI =

B

µ0

a0.540 Tfe2.50 × 10 mj = = e4π × 10 j T ⋅ m A −2

−7

10.7 kA .

FIG. P43.46 P43.47

By Faraday’s law (Equation 32.1),

∆Φ B ∆I ∆B =L =A . ∆t ∆t ∆t

Thus,

A ∆B π 0.010 0 m 0.020 0 T ∆I = = = 203 A . L 3.10 × 10 −8 H

a f b

gb

g

2

The direction of the induced current is such as to maintain the B– field through the ring. P43.48

(a)

∆V = IR If R = 0 , then ∆V = 0 , even when I ≠ 0 .

(b)

The graph shows a direct proportionality. Slope =

a

f f

155 − 57.8 mA ∆I 1 = = = 43.1 Ω −1 ∆ R V 3.61 − 1.356 mV

a

R = 0.023 2 Ω (c)

FIG. P43.48

Expulsion of magnetic flux and therefore fewer current-carrying paths could explain the decrease in current.

Additional Problems P43.49

(a)

Since the interatomic potential is the same for both molecules, the spring constant is the same. Then f =

1 2π

k

µ

where µ 12 =

a12 ufa16 uf = 6.86 u and µ = a14 ufa16 uf = 7.47 u . 14

12 u + 16 u

14 u + 16 u

Therefore, f14 =

1 2π

k

µ 14

continued on next page

=

1 2π

k

µ 12

FG µ IJ = f Hµ K 12

14

12

µ 12 = 6.42 × 10 13 Hz µ 14

e

j

6.86 u = 6.15 × 10 13 Hz . 7.47 u

561

562

Molecules and Solids

(b)

The equilibrium distance is the same for both molecules.

FG µ IJ µ r = FG µ IJ I Hµ K Hµ K F 7.47 u IJ e1.46 × 10 kg ⋅ m j = 1.59 × 10 kg ⋅ m I =G H 6.86 u K The molecule can move to the b v = 1, J = 9g state or to the b v = 1, J = 11g state. The energy it I 14 = µ 14 r 2 =

14

12

12

14

12

12

−46

14

(c)

2

−46

2

2

can absorb is either

LMFG 1 + 1 IJ hf 2K λ NH hc LF 1I = MG 1 + J hf ∆E = H λ N 2K ∆E =

or

hc

=

a f 2 I OP − LMFGH 0 − 12 IJK hf + 10a10 + 1f 2 I OP , Q N Q OP − LMFG 0 + 1 IJ hf + 10a10 + 1f OP . + 11a11 + 1f 2 I Q NH 2K 2I Q 2

14 + 9 9 + 1

2

14

14

14

2

14

2

14

14

14

The wavelengths it can absorb are then

λ= These are: λ =

λ=

and

P43.50

c f14 − 10

b2π I g

or λ =

14

c f14 + 11

b2π I g . 14

3.00 × 10 8 m s

e

6.15 × 10 13 Hz − 10 1.055 × 10 −34 J ⋅ s

j

3.00 × 10 8 m s

e

6.15 × 10 13 Hz + 11 1.055 × 10 −34 J ⋅ s

e

j

e

j

2π 1.59 × 10 −46 kg ⋅ m 2

j

2π 1.59 × 10 −46 kg ⋅ m 2

For the N 2 molecule, k = 2 297 N m, m = 2.32 × 10 −26 kg , r = 1.20 × 10 −10 m , µ = k

ω=

µ

e

je

j

I = µ r 2 = 1.16 × 10 −26 kg 1.20 × 10 −10 m

= 4.45 × 10 14 rad s,

2

= 4.96 µm

= 4.79 µm .

m 2

= 1.67 × 10 −46 kg ⋅ m 2 .

For a rotational state sufficient to allow a transition to the first exited vibrational state, 2

2I

a f

a f

J J + 1 = ω so J J + 1 =

2 Iω

=

e

je

2 1.67 × 10 −46 4.45 × 10 14 1.055 × 10

−34

j = 1 410 .

Thus J = 37 .

P43.51

FG H

∆Emax = 4.5 eV = v + 8.25 > 7.5

P43.52

1 2

IJ K

a4.5 eVfe1.6 × 10 J eVj ≥ F v + 1 I G J e1.055 × 10 J ⋅ sje8.28 × 10 s j H 2 K −19

ω

so

−34

14

−1

v=7

With 4 van der Waal bonds per atom pair or 2 electrons per atom, the total energy of the solid is

e

10 atoms I jFGH 6.02 ×4.00 JK = g

E = 2 1.74 × 10 −23 J atom

23

5. 23 J g .

563

Chapter 43

P43.53

The total potential energy is given by Equation 43.17: U total = −α

ke e2 B + m. r r

The total potential energy has its minimum value U 0 at the equilibrium spacing, r = r0 . At this point, dU = 0, dr r = r0 dU dr

or

P43.54

= r = r0

F GH

k e e 2 m −1 r0 . m

Thus,

B=α

Substituting this value of B into U total ,

U 0 = −α

I JK

k e2 d B −α e + m dr r r

=α r = r0

kee2 r02

F I GH JK

−

mB = 0. r0m +1

FG H

kee2 k e2 k e2 1 1 1− + α e r0m −1 m = −α e r0 m r0 m r0

IJ K

1 3 hf + 4. 48 eV = hf + 3.96 eV is the depth 2 2 of the well below the dissociation point. We see hf = 0.520 eV , so the depth of the well is

Suppose it is a harmonic-oscillator potential well. Then,

a

f

1 1 hf + 4.48 eV = 0.520 eV + 4.48 eV = 4.74 eV . 2 2 P43.55

.

(a)

For equilibrium,

dU = 0: dx

d Ax −3 − Bx −1 = −3 Ax −4 + Bx −2 = 0 dx

e

j

x → ∞ describes one equilibrium position, but the stable equilibrium position is at 3 Ax 0−2 = B . x0 =

(b)

e

The depth of the well is given by U0 = U x= x = − 0

(c)

j

3 0.150 eV ⋅ nm3 3A = = 0.350 nm 3.68 eV ⋅ nm B

Fx = −

U0 = U x= x = 0

a e

f

A B AB 3 2 BB 1 2 − = − x 03 x 0 3 3 2 A 3 2 3 1 2 A 1 2

3 2

2 3.68 eV ⋅ nm 2B 3 2 =− 32 12 3 A 3 3 2 0.150 eV ⋅ nm3

j

12

= −7.02 eV .

dU = 3 Ax −4 − Bx −2 dx

To find the maximum force, we determine finite x m such that

dF dx

= 0. x = xm

FG 6 A IJ . HBK F B IJ − BFG B IJ = − B = − a3.68 eV ⋅ nmf = 3 AG H 6 A K H 6 A K 12 A 12e0.150 eV ⋅ nm j F 1.60 × 10 J I FG 1 nm IJ = −1.20 × 10 = −7.52 eV nm G H 1 eV JK H 10 m K

Thus,

−12 Ax −5 + 2Bx −3

Then

Fmax

or

Fmax

2

x = xm

12

= 0 so that x m =

2

2

3

−19

−9

−9

N = −1.20 nN .

564 P43.56

Molecules and Solids

(a)

For equilibrium,

dU = 0: dx

d Ax −3 − Bx −1 = −3 Ax −4 + Bx −2 = 0 dx

e

j

x → ∞ describes one equilibrium position, but the stable equilibrium position is at 3 Ax 0−2 = B or x 0 =

3A . B

(b)

The depth of the well is given by U 0 = U x = x =

(c)

Fx = −

0

A B AB 3 2 BB 1 2 B3 − = 3 2 3 2 − 1 2 1 2 = −2 . 3 27 A 3 A x0 x0 3 A

dU = 3 Ax −4 − Bx −2 dx

To find the maximum force, we determine finite x m such that dFx dx P43.57

(a)

= −12 Ax −5 + 2Bx −3 x = xm

x = x0

= 0 then Fmax = 3 A dU dr

At equilibrium separation, r = re ,

FG B IJ H 6AK

2

−B

FG B IJ = H 6AK

−

−a r −r − a r −r = −2 aB e b e 0 g − 1 e b e 0 g = 0 . r = re

We have neutral equilibrium as re → ∞ and stable equilibrium at

e − abre − r0 g = 1 ,

or

re = r0 .

(b)

At r = r0 , U = 0. As r → ∞ , U → B . The depth of the well is B .

(c)

We expand the potential in a Taylor series about the equilibrium point:

af b g

2

br − r g + 12 ddrU br − r g 1 U ar f ≈ 0 + 0 + a −2BafL− ae b g − ae b g e e b MN 2 U r ≈ U r0 +

dU dr

r = r0

0

−2 r − r0

0

2

2

r = r0

− r − r0

This is of the form

(d)

B2 . 12 A

−2 r − r0

g − 1 jO

QP br − r g ≈ Ba br − r g 1 1 kx = k br − r g 2 2 0

r = r0

2

2

2

0

for a simple harmonic oscillator with

k = 2Ba 2 .

Then the molecule vibrates with frequency

f=

The zero-point energy is

ha 1 1 ω = hf = π 2 2

1 2π

k

µ

=

a 2π

0

2

2

2B

µ

=

a

π

B . 2µ

B . 8µ

Therefore, to dissociate the molecule in its ground state requires energy B −

ha

π

B . 8µ

Chapter 43

T=0

P43.58 E EF 0 0.500 0.600 0.700 0.800 0.900 1.00 1.10 1.20 1.30 1.40 1.50

T = 0.1TF

af

bE E g−1 bT T g

e e −∞ e −∞ e −∞ e −∞ e −∞ e −∞ e0 e +∞ e +∞ e +∞ e +∞ e +∞

F

bE E g−1 bT T g

f E

F

T = 0.2TF

F e −10 .0 e e −5.00 e −4.00 e −3.00 e −2.00 e −1.00 e0 e 1.00 e 2.00 e 3 .00 e 4.00 e 5 .00

1.00 1.00 1.00 1.00 1.00 1.00 0.500 0.00 0.00 0.00 0.00 0.00

F

af

f E

1.000 0.993 0.982 0.953 0.881 0.731 0.500 0.269 0.119 0.047 4 0.018 0 0.006 69

T = 0.5TF

bE E g−1 bT T g

F e −5.00 e e −2.50 e −2.00 e −1.50 e −1.00 e −0.500 e0 e 0 .500 e 1.00 e 1.50 e 2.00 e 2.50

F

af

f E

bE E g−1 bT T g

F e −2 . 00 0.993 e 0.924 e −1.00 0.881 e −0.800 0.818 e −0.600 0.731 e −0. 400 0.622 e −0. 200 0.500 e0 0.378 e 0. 200 0.269 e 0. 400 0.182 e 0 .600 0.119 e 0 .800 0.075 9 e 1.00

FIG. P43.58 P43.59

(a)

There are 6 Cl − ions at distance r = r0 . The contribution of these ions to the electrostatic −6 k e e 2 potential energy is . r0

There are 12 Na + ions at distance r = 2 r0 . Their contribution to the electrostatic potential +12 k e e 2 energy is . Next, there are 8 Cl − ions 2 r0 at distance r = 3 r0 . These contribute a term of −8 k e e 2 3 r0

to the electrostatic potential energy.

FIG. P43.59

To three terms, the electrostatic potential energy is:

FG H

U = −6 +

12 2

continued on next page

−

IJ k e 3K r

8

e

0

2

= −2.13

565

k e2 kee2 with α = 2.13 . or U = −α e r0 r0

F

af

f E

0.881 0.731 0.690 0.646 0.599 0.550 0.500 0.450 0.401 0.354 0.310 0.269

566

Molecules and Solids

(b)

The fourth term consists of 6 Na + at distance r = 2r0 . Thus, to four terms,

a

U = −2.13 + 3

f k re e

2

= 0.866

0

kee2 . r0

So we see that the electrostatic potential energy is not even attractive to 4 terms, and that the infinite series does not converge rapidly when groups of atoms corresponding to nearest neighbors, next-nearest neighbors, etc. are added together.

ANSWERS TO EVEN PROBLEMS P43.2

4.3 eV

P43.30

2%

P43.4

(a) 1.28 eV; (b) σ = 0.272 nm, ∈= 4.65 eV ; (c) 6.55 nN; (d) 576 N m

P43.32

3.40 × 10 17 electrons

P43.6

(a) 40.0 µeV , 9.66 GHz; (b) If r is 10% too small, f is 20% too large.

P43.34

see the solution

P43.36

(a) 275 THz; (b) 1.09 µm

P43.8

1.46 × 10 −46 kg ⋅ m 2

P43.38

1.91 eV

P43.10

(a) 1.81 × 10 −45 kg ⋅ m 2 ; (b) 1.62 cm

P43.40

226 nm

P43.12

(a) 11.8 pm; (b) 7.72 pm; HI is more loosely bound

P43.42

(a) 59.5 mV; (b) –59.5 mV

P43.44

4.19 mA

P43.46

(a) see the solution; (b) 10.7 kA

P43.48

see the solution

P43.50

37

P43.52

5.23 J g

P43.54

4.74 eV

P43.56

(a) x 0 =

P43.58

see the solution

P43.14

(a) 0, 364 µeV , 1.09 meV; (b) 98.2 meV, 295 meV, 491 meV

P43.16

(a) 472 µm ; (b) 473 µm ; (c) 0.715 µm

P43.18

2.9 × 10 −47 kg ⋅ m 2

P43.20

only 64.1 THz

P43.22

(a) ~ 10 17 ; (b) ~ 10 5 m 3

P43.24

(a) 0.444 nm, 0.628 nm, 0.769 nm

P43.26

see the solution

P43.28

(a) 1.57 Mm s; (b) larger by 10 orders of magnitude

B2 3A B3 ; (b) −2 ; (c) − B 27 A 12 A

44 Nuclear Structure ANSWERS TO QUESTIONS

CHAPTER OUTLINE 44.1 44.2 44.3 44.4 44.5 44.6 44.7 44.8

Some Properties of Nuclei Nuclear Binding Energy Nuclear Models Radioactivity The Decay Processes Natural Radioactivity Nuclear Reactions Nuclear Magnetic Resonance and Magnetic Resonance Imagining

Q44.1

Because of electrostatic repulsion between the positivelycharged nucleus and the +2e alpha particle. To drive the αparticle into the nucleus would require extremely high kinetic energy.

Q44.2

There are 86 protons and 136 neutrons in the nucleus 222 86 Rn. For the atom to be neutral, there must be 86 electrons orbiting the nucleus—the same as the number of protons.

Q44.3

All of these isotopes have the same number of protons in the nucleus. Neutral atoms have the same number of electrons. Isotopes only differ in the number of neutrons in the nucleus.

Q44.4

Nuclei with more nucleons than bismuth-209 are unstable because the electrical repulsion forces among all of the protons is stronger than the nuclear attractive force between nucleons.

Q44.5

The nuclear force favors the formation of neutron-proton pairs, so a stable nucleus cannot be too far away from having equal numbers of protons and neutrons. This effect sets the upper boundary of the zone of stability on the neutron-proton diagram. All of the protons repel one another electrically, so a stable nucleus cannot have too many protons. This effect sets the lower boundary of the zone of stability.

Q44.6

Nucleus Y will be more unstable. The nucleus with the higher binding energy requires more energy to be disassembled into its constituent parts.

Q44.7

Extra neutrons are required to overcome the increasing electrostatic repulsion of the protons. The neutrons participate in the net attractive effect of the nuclear force, but feel no Coulomb repulsion.

Q44.8

In the liquid-drop model the nucleus is modeled as a drop of liquid. The nucleus is treated as a whole to determine its binding energy and behavior. The shell model differs completely from the liquid-drop model, as it utilizes quantum states of the individual nucleons to describe the structure and behavior of the nucleus. Like the electrons that orbit the nucleus, each nucleon has a spin state to which the Pauli exclusion principle applies. Unlike the electrons, for protons and neutrons the spin and orbital motions are strongly coupled.

Q44.9

The liquid drop model gives a simpler account of a nuclear fission reaction, including the energy released and the probable fission product nuclei. The shell model predicts magnetic moments by necessarily describing the spin and orbital angular momentum states of the nucleons. Again, the shell model wins when it comes to predicting the spectrum of an excited nucleus, as the quantum model allows only quantized energy states, and thus only specific transitions. 567

568

Nuclear Structure

Q44.10

4

Q44.11

If one half the number of radioactive nuclei decay in one year, then one half the remaining number will decay in the second year. Three quarters of the original nuclei will be gone, and one quarter will remain.

Q44.12

The statement is false. Both patterns show monotonic decrease over time, but with very different shapes. For radioactive decay, maximum activity occurs at time zero. Cohorts of people now living will be dying most rapidly perhaps forty years from now. Everyone now living will be dead within less than two centuries, while the mathematical model of radioactive decay tails off exponentially forever. A radioactive nucleus never gets old. It has constant probability of decay however long it has existed.

Q44.13

Since the samples are of the same radioactive isotope, their half-lives are the same. When prepared, sample A has twice the activity (number of radioactive decays per second) of sample B. After 5 halflives, the activity of sample A is decreased by a factor of 2 5 , and after 5 half-lives the activity of sample B is decreased by a factor of 2 5 . So after 5 half-lives, the ratio of activities is still 2:1.

Q44.14

After one half-life, one half the radioactive atoms have decayed. After the second half-life, one half 1 1 3 of the remaining atoms have decayed. Therefore + = of the original radioactive atoms have 2 4 4 decayed after two half-lives.

Q44.15

The motion of a molecule through space does not affect anything inside the nucleus of an atom of the molecule. The half-life of a nucleus is based on nuclear stability which, as discussed in Questions 44.4 and Q44.5, is predominantly determined by Coulomb repulsion and nuclear forces, not molecular motion.

Q44.16

Long-lived progenitors at the top of each of the three natural radioactive series are the sources of our radium. As an example, thorium-232 with a half-life of 14 Gyr produces radium-228 and radium-224 at stages in its series of decays, shown in Figure 44.17.

Q44.17

A free neutron decays into a proton plus an electron and an antineutrino. This implies that a proton is more stable than a neutron, and in particular the proton has lower mass. Therefore a proton cannot decay into a neutron plus a positron and a neutrino. This reaction satisfies every conservation law except for energy conservation.

Q44.18

A neutrino has spin

Q44.19

Let us consider the carbon-14 decay used in carbon dating. 146 C→ 147 N + e − + ν . The carbon-14 atom has 6 protons in the nucleus. The nitrogen-14 atom has 7 protons in the nucleus, but the additional + charge from the extra proton is canceled by the – charge of the ejected electron. Since charge is conserved in this (and every) reaction, the antineutrino must have zero charge. Similarly, when nitrogen-12 decays into carbon-12, the nucleus of the carbon atom has one fewer protons, but the change in charge of the nucleus is balanced by the positive charge of the ejected positron. Again according to charge conservation, the neutrino must have no charge.

He ,

16

O,

40

Ca , and

208

Pb.

1 while a photon has spin 1. A neutrino interacts by the weak interaction while 2 a photon is a quantum of the electromagnetic interaction.

Chapter 44

Q44.20

569

1 . When a nucleus undergoes beta decay, an electron and 2 antineutrino are ejected. With all nucleons paired, in their ground states the carbon-14, nitrogen-14, nitrogen-12, and carbon-12 nuclei have zero net angular momentum. Angular momentum is conserved in any process in an isolated system and in particular in the beta-decays of carbon-14 and 1 nitrogen-12. Conclusion: the neutrino must have spin quantum number , so that its z-component 2 = = of angular momentum can be just or − . A proton and a neutron have spin quantum number 1. 2 2 For conservation of angular momentum in the beta-decay of a free neutron, an antineutrino must 1 have spin quantum number . 2 An electron has spin quantum number

Q44.21

The alpha particle and the daughter nucleus carry equal amounts of momentum in opposite p2 directions. Since kinetic energy can be written as , the small-mass alpha particle has much more 2m of the decay energy than the recoiling nucleus.

Q44.22

Bullet and rifle carry equal amounts of momentum p. With a much smaller mass m, the bullet has p2 much more kinetic energy K = . The daughter nucleus and alpha particle have equal momenta 2m and the massive daughter nucleus, like the rifle, has a very small share of the energy released.

Q44.23

Yes. The daughter nucleus can be left in its ground state or sometimes in one of a set of excited states. If the energy carried by the alpha particle is mysteriously low, the daughter nucleus can quickly emit the missing energy in a gamma ray.

Q44.24

In a heavy nucleus each nucleon is strongly bound to its momentary neighbors. Even if the nucleus could step down in energy by shedding an individual proton or neutron, one individual nucleon is never free to escape. Instead, the nucleus can decay when two protons and two neutrons, strongly bound to one another but not to their neighbors, happen momentarily to have a lot of kinetic energy, to lie at the surface of the nucleus, to be headed outward, and to tunnel successfully through the potential energy barrier they encounter.

Q44.25

mv 2 , or qBr = mv , a charged particle fired into a magnetic field is deflected r into a path with radius proportional to its momentum. If they have equal kinetic energies K, the much greater mass m of the alpha particle gives it more momentum mv = 2mK than an electron. Thus the electron undergoes greater deflection. This conclusion remains true if one or both particles are moving relativistically. From

∑ F = ma , or qvB =

Q44.26

The alpha particle stops in the wood, while many beta particles can make it through to deposit some or all of their energy in the film emulsion.

Q44.27

The reaction energy is the amount of energy released as a result of a nuclear reaction. Equation 44.28 in the text implies that the reaction energy is (initial mass – final mass) c 2 . The Q-value is taken as positive for an exothermic reaction.

570

Nuclear Structure

Q44.28

Carbon-14 is produced when carbon-12 is bombarded by cosmic rays. Both carbon-12 and carbon-14 combine with oxygen to form the atmospheric CO 2 that plants absorb in respiration. When the plant dies, the carbon-14 is no longer replenished and decays at a known rate. Since carbon-14 is a beta-emitter, one only needs to compare the activity of a living plant to the activity of the sample to determine its age, since the activity of a radioactive sample exponentially decreases in time.

Q44.29

The samples would have started with more carbon-14 than we first thought. We would increase our estimates of their ages.

Q44.30

There are two factors that determine the uncertainty on dating an old sample. The first is the fact that the activity level decreases exponentially in time. After a long enough period of time, the activity will approach background radiation levels, making precise dating difficult. Secondly, the ratio of carbon-12 to carbon-14 in the atomsphere can vary over long periods of time, and this effect contributes additional uncertainty.

Q44.31

An α-particle is a helium nucleus: 24 He

A β-particle is an electron or a positron: either e − or e + . A γ-ray is a high-energy photon emitted when a nucleus makes a downward transition between two states. Q44.32

I z may have 6 values for I = I = 3.

1 3 5 5 5 3 1 , namely , , , − , − , and − . Seven I z values are possible for 2 2 2 2 2 2 2

Q44.33

The frequency increases linearly with the magnetic field strength.

Q44.34

The decay of a radioactive nucleus at one particular moment instead of at another instant cannot be predicted and has no cause. Natural events are not just like a perfect clockworks. In history, the idea of a determinate mechanical Universe arose temporarily from an unwarranted wild extrapolation of Isaac Newton’s account of planetary motion. Before Newton’s time [really you can blame Pierre Simon de Laplace] and again now, no one thought of natural events as just like a perfect row of falling dominos. We can and do use the word “cause” more loosely to describe antecedent enabling events. One gear turning another is intelligible. So is the process of a hot dog getting toasted over a campfire, even though random molecular motion is at the essence of that process. In summary, we say that the future is not determinate. All natural events have causes in the ordinary sense of the word, but not necessarily in the contrived sense of a cause operating infallibly and predictably in a way that can be calculated. We have better reason now than ever before to think of the Universe as intelligible. First describing natural events, and second determining their causes form the basis of science, including physics but also scientific medicine and scientific bread-baking. The evidence alone of the past hundred years of discoveries in physics, finding causes of natural events from the photoelectric effect to x-rays and jets emitted by black holes, suggests that human intelligence is a good tool for figuring out how things go. Even without organized science, humans have always been searching for the causes of natural events, with explanations ranging from “the will of the gods” to Schrödinger’s equation. We depend on the principle that things are intelligible as we make significant strides towards understanding the Universe. To hope that our search is not futile is the best part of human nature.

Chapter 44

571

SOLUTIONS TO PROBLEMS Section 44.1 P44.1

Some Properties of Nuclei

An iron nucleus (in hemoglobin) has a few more neutrons than protons, but in a typical water molecule there are eight neutrons and ten protons. So protons and neutrons are nearly equally numerous in your body, each contributing mass (say) 35 kg: 35 kg

P44.2

F 1 nucleon I GH 1.67 × 10 kg JK −27

~ 10 28 protons

and

~ 10 28 neutrons .

The electron number is precisely equal to the proton number,

~ 10 28 electrons .

1 mv 2 = q∆V 2

and

2m∆V = qr 2 B 2 :

mv 2 = qvB r 2 m∆ V

r=

qB

2

=

e

r = 5.59 × 10 11 (a)

b

2 1 000 V

g

2

m

−27

kg

j

−27

kg

j

e1.60 × 10 Cja0.200 Tf m kg j m −19

e

For 12 C , m = 12 u and r = 5.59 × 10 11 m

kg

j 12e1.66 × 10

r = 0.078 9 m = 7.89 cm . For 13 C :

e

r = 5.59 × 10 11 m

kg

j 13e1.66 × 10

r = 0.082 1 m = 8.21 cm .

(b)

With

the ratio gives

r1 = r1 = r2

2 m 1 ∆V qB

2

and r2 =

m1 m2

r1 7.89 cm = = 0.961 r2 8.21 cm and so they do agree.

m1 12 u = = 0.961 13 u m2

2 m 2 ∆V qB 2

572 P44.3

P44.4

Nuclear Structure

Q 1Q 2

a2fa6fe1.60 × 10 Cj C j e1.00 × 10 mj −19

e

9

2

(a)

F = ke

(b)

a=

(c)

QQ U = k e 1 2 = 8.99 × 10 9 N ⋅ m 2 C 2 r

r2

= 8.99 × 10 N ⋅ m

2

e

= 27.6 N

2

−14

27.6 N F = = 4.17 × 10 27 m s 2 m 6.64 × 10 −27 kg

2

away from the nucleus.

a2fa6f 1.60 × 10 C j e1.e00 × 10 mj j −19

2

= 2.76 × 10 −13 J = 1.73 MeV

−14

Eα = 7.70 MeV 4k e Ze 2

e

ja fe

−19 9 2 k e Ze 2 2 8.99 × 10 79 1.60 × 10 = = Eα 7.70 1.60 × 10 −13

(a)

d min =

(b)

The de Broglie wavelength of the α is

λ=

(c)

mv 2

h = mα vα

h 2mα Eα

=

e

j

j

2

= 29.5 × 10 −15 m = 29.5 fm

6.626 × 10 −34

e

2 6.64 × 10

−27

j7.70e1.60 × 10 j −13

= 5.18 × 10 −15 m = 5.18 fm .

Since λ is much less than the distance of closest approach , the α may be considered a particle.

P44.5

(a)

The initial kinetic energy of the alpha particle must equal the electrostatic potential energy at the distance of closest approach. Ki = U f = rmin

(b)

e

ja fa fe

8.99 × 10 9 N ⋅ m 2 C 2 2 79 1.60 × 10 −19 C k qQ = e = Ki 0.500 MeV 1.60 × 10 −13 J MeV

Since K i =

vi =

k e qQ rmin

a

fe

j

j

2

= 4.55 × 10 −13 m

k qQ 1 mα vi2 = e , 2 rmin

2 k e qQ = mα rmin

e

ja fa fe kg uje3.00 × 10

2 8.99 × 10 9 N ⋅ m 2 C 2 2 79 1.60 × 10 −19 C

a4.00 ufe1.66 × 10

−27

−13

m

j

j

2

= 6.04 × 10 6 m s .

Chapter 44

P44.6

It must start with kinetic energy equal to K i = U f = the 24 He and

197 79 Au

13

k e qQ . Here r f stands for the sum of the radii of rf

nuclei, computed as

e

13

je

j

r f = r0 A1 + r0 A 2 = 1.20 × 10 −15 m 41 3 + 197 1 3 = 8.89 × 10 −15 m . Thus, K i = U f P44.7

P44.8

e8.99 × 10 =

e

ja f

e

ja f

13

r = r0 A 1 3 = 1.20 × 10 −15 m 4

(b)

r = r0 A 1 3 = 1.20 × 10 −15 m 238

2

= 4.09 × 10 −12 J = 25.6 MeV .

= 1.90 × 10 −15 m 13

= 7.44 × 10 −15 m

a f

From r = r0 A 1 3 , the radius of uranium is rU = r0 238

a f

1 1 rU then r0 A 1 3 = r0 238 2 2

13

.

13

A = 30 .

from which

The number of nucleons in a star of two solar masses is A= Therefore

P44.10

j

8.89 × 10 −15 m

(a)

Thus, if r =

P44.9

ja fa fe

N ⋅ m 2 C 2 2 79 1.60 × 10 −19 C

9

V=

e

2 1.99 × 10 30 kg 1.67 × 10

−27

j

kg nucleon

e

= 2.38 × 10 57 nucleons .

je

r = r0 A 1 3 = 1.20 × 10 −15 m 2.38 × 10 57

b

g

4 4 4 π r = π 0.021 5 m 3 3

3

j

13

= 16.0 km .

= 4.16 × 10 −5 m3

We take the nuclear density from Example 44.2

e

je

j

m = ρV = 2.3 × 10 17 kg m3 4.16 × 10 −5 m3 = 9.57 × 10 12 kg and

F =G

m1 m 2 r2

9.57 × 10 kg j e a1.00 mf j 12

e

= 6.67 × 10 −11 N ⋅ m 2 kg 2

2

2

F = 6.11 × 10 15 N toward the other ball. P44.11

The stable nuclei that correspond to magic numbers are: Z magic:

2 He

8O

20 Ca

28 Ni

50 Sn

82 Pb

An artificially produced nucleus with Z = 126 might be more stable than other nuclei with lower values for Z, since this number of protons is magic. N magic:

3 1T ,

4 2 He,

15 7 N,

16 8O,

37 17 Cl ,

39 19 K ,

40 20 Ca ,

51 23 V ,

89 39 Y ,

90 40 Zr ,

136 54 Xe ,

138 56 Ba ,

139 57 La,

140 58 Ce ,

141 59 Pr ,

142 208 60 Nd , 82 Pb,

210 84 Po

573

52 24 Cr ,

88 38 Sr , 209 83 Bi ,

574 *P44.12

Nuclear Structure

(a)

For even Z, even N, even A, the list begins 24 He, 126 C , and ends

194 196 202 208 78 Pt , 78 Pt , 80 Hg , 82 Pb,

containing 48 isotopes. (b)

195 The whole even Z, odd N, odd A list is 94 Be , 129 54 Xe , 78 Pt , with 3 entries.

(c)

The odd Z, even N, odd A list has 46 entries, represented as 11 H , 73 Li , …,

203 205 81Tl , 81Tl ,

209 83 Bi .

(d)

The odd Z, odd N, even A list has 1 entry, 147 N . Do not be misled into thinking that nature favors nuclei with even numbers of neutrons. The form of the question here forces a count with essentially equal numbers of odd-Z and even-Z nuclei. If we counted all of the stable nuclei we would find many even-even isotopes but also lots of even-Z odd-N nuclei and odd-Z even-N nuclei; we would find roughly equal numbers of these two kinds of odd-A nuclei. A nucleus with one odd neutron is no more likely to be unstable than a nucleus with one odd proton. With the arbitrary 25% abundance standard, we can note that most elements have a single predominant isotope. Ni, Cu, Zn, Ga, Ge, Pd, Ag, Os, Ir, and Pt form a compact patch on the periodic table and have two common isotopes, as do some others. Tungsten is the only element with three isotopes over 25% abundance.

*P44.13

(a)

Z1 = 8Z 2

N1 = 5 N 2

b

g

N 1 + Z1 = 6 N 1 + Z 2 and N 1 = Z1 + 4 Thus:

N 1 + Z1 = 6

FG 1 N H5

1

+

1 Z1 8

IJ K

5 Z1 4 5 ∴ Z1 + 4 = Z1 4 ∴ Z1 = 16 ∴ N1 =

N 1 = Z1 + 4 = 20 , A1 = Z1 + N 1 = 36 N2 = Hence: (b)

6 2 He

36 16 S

1 1 N 1 = 4, Z 2 = Z1 = 2 , A 2 = Z 2 + N 2 = 6 5 8

and 62 He.

is unstable. Two neutrons must be removed to make it stable

e Hej . 4 2

Chapter 44

Section 44.2 P44.14

Nuclear Binding Energy

Using atomic masses as given in Table A.3,

b

g b

g

−2.014 102 + 1 1.008 665 + 1 1.007 825

For 21 H:

(a)

2 Eb 931.5 MeV = 0.001 194 u = 1.11 MeV nucleon . u A

gFGH

b

b

g b

IJ K

g

2 1.008 665 + 2 1.007 825 − 4.002 603

(b)

For 24 He:

(c)

For

56 26 Fe :

30 1.008 665 + 26 1.007 825 − 55.934 942 = 0.528 u Eb 0.528 = = 0.009 44 uc 2 = 8.79 MeV nucleon . 56 A

(d)

For

238 92 U :

146 1.008 665 + 92 1.007 825 − 238.050 783 = 1.934 2 u Eb 1.934 2 = = 0.008 13 uc 2 = 7.57 MeV nucleon . A 238

4 Eb 2 = 0.007 59 uc = 7.07 MeV nucleon . A

b

g b

b

g b

g

g

a

∆M = Zm H + Nm n − M

BE ∆M 931.5 = A A

Nuclei

Z

N

M in u

∆M in u

55

25 26 27

30 30 32

54.938 050 55.934 942 58.933 200

0.517 5 0.528 46 0.555 35

P44.15

Mn Fe 59 Co 56

∴56 Fe has a greater P44.16

575

f BE in MeV A 8.765 8.790 8.768

BE than its neighbors. This tells us finer detail than is shown in Figure 44.5. A

Use Equation 44.2. The

23 11 Na ,

and for

23 12 Mg ,

Eb = 8.11 MeV nucleon A Eb = 7.90 MeV nucleon . A

The binding energy per nucleon is greater for

23 11 Na

by 0.210 MeV . (There is less proton repulsion

23

in Na .) P44.17

(a) (b) (c)

The neutron-to-proton ratio 139

A−Z is greatest for Z

139 55 Cs

and is equal to 1.53.

La has the largest binding energy per nucleon of 8.378 MeV.

139

Cs with a mass of 138.913 u. We locate the nuclei carefully on Figure 44.4, the neutron–proton plot of stable nuclei. Cesium appears to be farther from the center of the zone of stability. Its instability means extra energy and extra mass.

576 P44.18

Nuclear Structure

(a)

40

The radius of the

ja f

e

R = r0 A 1 3 = 1.20 × 10 −15 m 40

Ca nucleus is:

13

= 4.10 × 10 −15 m .

The energy required to overcome electrostatic repulsion is

e

j e

−19 2 2 9 C 3 k Q 2 3 8.99 × 10 N ⋅ m C 20 1.60 × 10 = U= e 5R 5 4.10 × 10 −15 m

(b)

e

The binding energy of

b

40 20 Ca

j

j

2

= 1.35 × 10 −11 J = 84.1 MeV .

is

g b

g

b

g

Eb = 20 1.007 825 u + 20 1.008 665 u − 39.962 591 u 931.5 MeV u = 342 MeV . (c)

P44.19

The nuclear force is so strong that the binding energy greatly exceeds the minimum energy needed to overcome electrostatic repulsion.

f a f e Xj b931.494 MeV ug . = 8b1.007 825 ug + 7b1.008 665 ug − 15.003 065 u b931.494 MeV ug = 111.96 MeV . = 7b1.007 825 ug + 8b1.008 665 ug − 15.000 109 u b931.494 MeV ug = 115.49 MeV .

For 158 O :

Eb

For 157 N:

Eb

Therefore, the binding energy of P44.20

15 7N

Sn

A Z

is larger by 3.54 MeV .

Removal of a neutron from 43 20 Ca would result in the residual nucleus, separation energy is Sn , the overall process can be described by

42 20 Ca .

j + massanf = b 41.958 618 + 1.008 665 − 42.958 767 g u = b0.008 516 ugb931.5 MeV ug =

mass

Section 44.3 P44.21

a

Eb MeV = ZM H + Nmn − M

The binding energy of a nucleus is

e

43 20 Ca

j+S

n

= mass

e

If the required

42 20 Ca

7.93 MeV .

Nuclear Models

∆Eb = Ebf − Ebi Eb = 7.4 MeV A

For

A = 200,

so

Ebi = 200 7.4 MeV = 1 480 MeV .

For

A ≈ 100 ,

so

Ebf = 2 100 8.4 MeV = 1 680 MeV .

a

f

Eb = 8.4 MeV A

a fa

f

∆Eb = Ebf − Ebi : Eb = 1 680 MeV − 1 480 MeV = 200 MeV FIG. P44.21

Chapter 44

P44.22

P44.23

(a)

The first term overstates the importance of volume and the second term subtracts this overstatement.

(b)

For spherical volume

(a)

“Volume” term:

b4 3gπ R

3

R R3 R . For cubical volume . = 2 2 3 6 4π R 6R The maximum binding energy or lowest state of energy is achieved by building “nearly” spherical nuclei. =

a

fa f

E1 = C1 A = 15.7 MeV 56 = 879 MeV .

a fa f = −260 MeV . ZaZ − 1f a26fa25f = −121 MeV . = −C = −a0.71 MeV f A a56f a A − 2Zf = −a23.6 MeVf a56 − 52f = −6.74 MeV . =C 56 A

“Surface” term:

E2 = −C 2 A

“Coulomb” term:

E3

“Asymmetry” term:

E4

23

3

= − 17.8 MeV 56

23

13

13

2

4

Eb = 491 MeV (b)

Section 44.4

E1 E E E = 179%; 2 = −53.0%, 3 = −24.6% ; 4 = −1.37% Eb Eb Eb Eb

Radioactivity

a

f

FG H

IJ e K

P44.24

R = R0 e − λ t = 6. 40 mCi e

P44.25

dN = − λN dt so

λ= T1 2

P44.26

b

ga

− ln 2 8.04 d 40. 2 d

f = a6.40 mCifee − ln 2 j5 = a6.40 mCifFG

1I H 2 JK =

je

0.200 mCi

j

a

λ

f

F ln 2 I F 1.00 g I e6.02 × 10 j GH 5.27 yr JK GH 59.93 g mol JK F 1 yr IJ = 4.18 × 10 R = e1.32 × 10 decays yr jG H 3.16 × 10 s K 23

R = λN =

(a)

5

1 dN − = 1.00 × 10 −15 6.00 × 10 11 = 6.00 × 10 −4 s −1 N dt ln 2 = = 1.16 × 10 3 s = 19.3 min

21

P44.27

2

7

13

Bq

R = R0 e − λ t , R 1 1 10.0 ln λ = ln 0 = = 5.58 × 10 −2 h −1 = 1.55 × 10 −5 s −1 4.00 h 8.00 t R ln 2 T1 2 = = 12.4 h

From

FG IJ FG H K H

IJ FG IJ K H K

λ

R0

F GH

I JK

N0 =

(c)

R = R0 e − λ t = 10.0 mCi exp −5.58 × 10 −2 × 30.0 = 1.88 mCi

λ

=

10.0 × 10 −3 Ci 3.70 × 10 10 s = 2.39 × 10 13 atoms 1 Ci 1.55 × 10 −5 s

(b)

a

f e

j

577

578 P44.28

Nuclear Structure

where

λ=

R = 0.100 = e − λ t R0

so

ln 0.100 = − λ t

2.30 = P44.29

ln 2 = 0.026 6 h −1 26.0 h

R = R0 e − λ t

a

FG 0.026 6 IJ t H h K

f

t = 86.4 h

e

j

The number of nuclei which decay during the interval will be N 1 − N 2 = N 0 e − λ t1 − e − λ t 2 . ln 2 0.693 = = 0.010 7 h −1 = 2.97 × 10 −6 s −1 T1 2 64.8 h

First we find l :

λ=

and

N0 =

Substituting these values,

N 1 − N 2 = 4.98 × 10 11 e

R0

λ

b40.0 µCige3.70 × 10

=

2.97 × 10

jLMN

e

−6

4

s

s −1 µCi −1

ja

e

− 0.010 7 h −1 10.0 h

j = 4.98 × 10

11

nuclei .

f − e −e0.010 7 h ja12.0 hf O . −1

PQ

Hence, the number of nuclei decaying during the interval is N1 − N 2 = 9.47 × 10 9 nuclei . P44.30

*P44.31

e

j

The number of nuclei which decay during the interval will be N 1 − N 2 = N 0 e − λ t1 − e − λ t 2 . ln 2 T1 2

First we find λ:

λ=

so

e −λ t = e

and

N0 =

Substituting in these values

N1 − N 2 =

e

ln 2 − t T1 2

R0

λ

=

j = 2 −t T

R0 T1 2 ln 2 R0 T1 2

af

ln 2

12

.

ee

af

We have all this information: N x 0 = 2.50 N y 0

a f N a0 fe

a f

N x 3d = 4.20 N y 3d x

−λ x 3d

af

= 4.20 N y 0 e

−λ y 3d

= 4.20

2.5 3 dλ y e 4.2 2.5 + 3 dλ y 3dλ x = ln 4.2 0.693 2.5 0.693 = ln + 3d = 0.781 3d T1 2 x 4.2 1.60 d e 3 dλ x =

FG IJ H K FG IJ H K

T1 2 x = 2.66 d

af

N x 0 −λ y 3d e 2.50

− λ t1

j

− e − λ t2 =

R0 T1 2 ln 2

e2

− t1 T1 2

−2

− t 2 T1 2

j.

Chapter 44

*P44.32

(a)

579

dN 2 = rate of change of N 2 dt = rate of production of N 2 – rate of decay of N 2 = rate of decay of N 1 – rate of decay of N 2 = λ 1 N1 − λ 2 N 2

(b)

From the trial solution

af

N2 t = ∴ ∴

e

j

dN 2 N λ = 10 1 − λ 2 e − λ 2 t + λ 1 e − λ 1t λ1 − λ 2 dt

e

j

(1)

dN 2 N λ + λ 2 N 2 = 10 1 − λ 2 e − λ 2 t + λ 1 e − λ 1t + λ 2 e − λ 2 t − λ 2 e − λ 1t λ1 − λ 2 dt N λ = 10 1 λ 1 − λ 2 e − λ 1t λ1 − λ 2 = λ 1 N1

e b

So (c)

N10 λ 1 − λ 2 t − e − λ 1t e λ1 − λ 2

j

g

dN 2 = λ 1 N 1 − λ 2 N 2 as required. dt

The functions to be plotted are

e af L N at f = 1 130.8 M e e N N 1 t = 1 000 e

Decay of

j

j

2

−e

Po and

214

Pb

1 200

− 0. 223 6 min −1 t − 0 . 223 6 min −1 t

218

e

jO

− 0 .025 9 min −1 t

PQ

From the graph: t m ≈ 10.9 min

1 000

Po Pb

800 600 400 200 0

0

10

20 time (min)

30

FIG. P44.32(c) (d)

From (1),

b

ln λ 1 λ 2 λ dN 2 = 0 if λ 2 e − λ 2 t = λ 1 e − λ 1t . ∴ e b λ 1 − λ 2 gt = 1 . Thus, t = t m = λ1 − λ 2 λ2 dt

g

.

With λ 1 = 0.223 6 min −1 , λ 2 = 0.025 9 min −1 , this formula gives t m = 10.9 min , in agreement with the result of part (c).

Section 44.5 P44.33

The Decay Processes

b gb g Q = b 238.050 783 − 234.043 596 − 4.002 603 g ub931.5 MeV ug = Q = M U- 238 − M Th- 234 − M He- 4 931.5 MeV u

4.27 MeV

40

580 P44.34

P44.35

Nuclear Structure

(a)

A gamma ray has zero charge and it contains no protons or neutrons. So for a gamma ray Z = 0 and A = 0. Keeping the total values of Z and A for the system conserved then requires Z = 28 and A = 65 for X. With this atomic number it must be nickel, and the nucleus must be * in an exited state, so it is 65 28 Ni .

(b)

α = 24 He has Z = 2 so for X

and

A=4

we require

Z = 84 − 2 = 82

for Pb

and

A = 215 − 4 = 211, X= 211 82 Pb.

(c)

A positron e + = 01 e has charge the same as a nucleus with Z = 1 . A neutrino 00 ν has no charge. Neither contains any protons or neutrons. So X must have by conservation Z = 26 + 1 = 27 . It is Co. And A = 55 + 0 = 55 . It is 55 27 Co. Similar reasoning about balancing the sums of Z and A across the reaction reveals:

(d)

0 −1 e

(e)

1 1H

(or p). Note that this process is a nuclear reaction, rather than radioactive decay. We can solve it from the same principles, which are fundamentally conservation of charge and conservation of baryon number.

NC =

F 0.021 0 g I e6.02 × 10 GH 12.0 g mol JK

23

molecules mol

j

eN = 1.05 × 10 carbon atomsj of which 1 in 7.70 × 10 is a C atom ln 2 = 1.21 × 10 yr = 3.83 × 10 s bN g = 1.37 × 10 , λ = 5 730 yr 21

C

0 C-14

11

9

−4

C-14

−1

−12

R = λN = λN 0 e − λ t

e

837 = 951 decays week . 0.88 R −1 R = −λ t ln so t= ln R0 λ R0

Taking logarithms,

t=

N = N0 e −λ t R R0

L 400 sg OP = 3.17 × 10 jMM 7b186week PQ N

3

decays week .

FG H

IJ K

FG IJ H K

951 −1 ln = 9.96 × 10 3 yr . −4 −1 3 1.21 × 10 yr 3.17 × 10

dN = R = − λN 0 e − λ t = R0 e − λ t dt

FG R IJ = ln 2 t H RK T lnb0.25 0.13g If R = 0.13 Bq , t = 5 730 yr = 5 406 yr . 0.693 lnb0.25 0.11g = 6 787 yr . If R = 0.11 Bq , t = 5 730 yr e −λ t =

−1

R=

At time t,

*P44.36

je

R0 = λN 0 = 3.83 × 10 −12 s −1 1.37 × 10 9

At t = 0 ,

14

eλ t =

R0 R

λ t = ln

0

12

t = T1 2

b

ln R0 R

g

ln 2

0.693 The range is most clearly written as between 5 400 yr and 6 800 yr , without understatement.

Chapter 44

P44.37

3 1H

nucleus→ 32 He nucleus + e − + ν 3 1H

becomes

nucleus + e − → 32 He nucleus + 2e − + ν .

Ignoring the slight difference in ionization energies, 3 1H

we have

atom→ 32 He atom + ν

3.016 049 u = 3.016 029 u + 0 +

Q

c2 Q = 3.016 049 u − 3.016 029 u 931.5 MeV u = 0.018 6 MeV = 18.6 keV

b

P44.38

(a)

gb

g

For e + decay,

b

g

b

gb

Q = M X − M Y − 2m e c 2 = 39.962 591 u − 39.963 999 u − 2 0.000 549 u 931.5 MeV u Q = −2.33 MeV Since Q < 0 , the decay cannot occur spontaneously. (b)

For alpha decay,

b

g

b

Q = M X − Mα − M Y c 2 = 91.905 287 u − 4.002 603 u − 93.905 088 u 931.5 MeV u

g

Q = −2.24 MeV Since Q < 0 , the decay cannot occur spontaneously. (c)

For alpha decay,

b

g

b

Q = M X − Mα − M Y c 2 = 143.910 083 u − 4.002 603 u − 139.905 434 u 931.5 MeV u Q = 1.91 MeV Since Q > 0 , the decay can occur spontaneously. P44.39

(a) (b)

e− + p → n + ν For nuclei,

15

O + e − → 15 N + ν . 15 8O

Add seven electrons to both sides to obtain (c)

From Table A.3,

m

e Oj = me Nj + cQ 15

15

2

∆m = 15.003 065 u − 15.000 109 u = 0.002 956 u

b

gb

g

Q = 931.5 MeV u 0.002 956 u = 2.75 MeV

atom→ 157 N atom + ν .

g

g

581

582

Nuclear Structure

Section 44.6 P44.40

(a)

Natural Radioactivity Let N be the number of

238

U nuclei and N ′ be

206

a

Pb nuclei.

f

Then N = N 0 e − λ t and N 0 = N + N ′ so N = N + N ′ e − λ t or e λ t = 1 +

FG H

Taking logarithms,

λ t = ln 1 +

Thus,

t=

If

N = 1.164 for the N′

238

From above, e λ t = 1 +

IJ K

where λ =

ln 2 . T1 2

F T I lnFG 1 + N ′ IJ . GH ln 2 JK H N K 12

U → 206 Pb chain with T1 2 = 4.47 × 10 9 yr , the age is:

t=

(b)

N′ N

N′ . N

F 4.47 × 10 yr I lnFG 1 + 1 IJ = GH ln 2 JK H 1.164 K 9

4.00 × 10 9 yr .

N N e −λ t N′ . Solving for = gives . N′ N ′ 1 − e−λ t N

With t = 4.00 × 10 9 yr and T1 2 = 7.04 × 10 8 yr for the

235

U → 207 Pb chain,

F ln 2 I t = aln 2fe4.00 × 10 yrj = 3.938 and GH T JK 7.04 × 10 yr 9

λt=

8

12

With t = 4.00 × 10 9 yr and T1 2 = 1.41 × 10 10 yr for the

Th→ 208 Pb chain,

aln 2fe4.00 × 10 yrj = 0.196 6 and 9

λt=

232

1.41 × 10 10 yr

P44.41

FIG. P44.41

N = 0.019 9 . N′

N = 4.60 . N′

Chapter 44

P44.42

(a)

(b)

(c)

F 4.00 × 10 Ci I F 3.70 × 10 Bq I F 1.00 × 10 L I = 148 Bq m GH 1 L JK GH 1 Ci JK GH 1 m JK F T I = e148 Bq m jFG 3.82 d IJ FG 86 400 s IJ = 7.05 × 10 atoms m R N = = RG H ln2 K H 1 d K λ H ln 2 JK F 1 mol IJ FG 222 g IJ = 2.60 × 10 g m mass = e7.05 × 10 atoms m jG H 6.02 × 10 atoms K H 1 mol K 10

−12

4.00 pCi L =

3

3

3

12

7

3

7

3

583

3

−14

23

3

Since air has a density of 1.20 kg m3 , the fraction consisting of radon is fraction = *P44.43

(a)

2.60 × 10 −14 g m 3 1 200 g m

3

= 2.17 × 10 −17 .

Let x, y denote the half-lives of the nuclei X, Y. R X R0 e − λ X t − a 0.685 h fa ln 2 fb1 x −1 y g = =e = 1.04 , which gives RY R0 e − λ Y t 1 1 − = −0.082 603 69 h −1 . x y

(1)

From the data: x − y = 77.2 h .

(2)

1 1 − = −0.082 603 69 h −1 . x x − 77.2 h

Substitute (2) into (1):

This reduces to the quadratic equation x 2 − 77.2 x − 934.6 = 0 x = 87.84 h or −10.64 h .

which has solutions:

Thus: x = T1 2 , X = 87.84 h = 3.66 days is the only physical root. From (2): y = T1 2 , Y = 87.84 h − 77.2 h = 10.6 h .

P44.44

(b)

From Table A.3, X is

(c)

From Figure 44.18,

224

224

212

Ra and Y is

Ra decays to

212

Pb .

Pb by three successive alpha-decays.

a f

− ln 2 t T1 2

Number remaining:

N = N0 e

Fraction remaining:

N − a ln 2 ft T1 2 = e−λ t = e . N0

.

(a)

With T1 2 = 3.82 d and t = 7.00 d ,

N = e − aln 2 fa7.00 f N0

(b)

When t = 1.00 yr = 365.25 d ,

N = e − aln 2 fa365. 25 f N0

(c)

a3.82 f = a3.82 f =

0. 281 . 1.65 × 10 −29 .

Radon is continuously created as one daughter in the series of decays starting from the long-lived isotope

238

U.

584

Nuclear Structure

Section 44.7 P44.45

Nuclear Reactions

Q = M 27 Al + Mα − M 30 P − mn c 2

b

g

Q = 26.981 539 + 4.002 603 − 29.978 314 − 1.008 665 u 931.5 MeV u = −2.64 MeV P44.46

(a)

For X,

A = 24 + 1 − 4 = 21

and

Z = 12 + 0 − 2 = 10 ,

21 10 Ne

.

so X is

144 54 Xe

.

A = 235 + 1 − 90 − 2 = 144

(b)

and

Z = 92 + 0 − 38 − 0 = 54 ,

A=2−2=0

(c)

and

Z = 2 − 1 = +1,

so X must be a positron.

X = 01 e +

As it is ejected, so is a neutrino: P44.47

so X is

(a)

* 197 1 198 198 79 Au + 0 n → 79 Au → 80 Hg

(b)

Consider adding 79 electrons: 197 79 Au

+

and X ′= 00 ν .

0 −1 e + ν

atom+ 01 n→ 198 80 Hg atom + ν + Q

Q = M 197 Au + mn − M 198 Hg c 2

b

g

Q = 196.966 552 + 1.008 665 − 197.966 752 u 931.5 MeV u = 7.89 MeV P44.48

Neglect recoil of product nucleus, (i.e., do not require momentum conservation for the system of colliding particles). The energy balance gives K emerging = K incident + Q . To find Q:

b g b g Q = b1.007 825 + 26.981 539g − b 26.986 705 + 1.008 665 g ub931.5 MeV ug = −5.59 MeV Q = M H + M Al − M Si + mn c 2

Thus, P44.49

9 4 Be + 1.665

K emerging = 6.61 MeV − 5.59 MeV = 1.02 MeV . MeV → 84 Be+ 01 n , so M 8 Be = M 9 Be −

M 8 Be = 9.012 182 u − 4

4

Q − mn c2

a−1.665 MeVf − 1.008 665 u = 931.5 MeV u

9 1 10 4 Be+ 0 n→ 4 Be + 6.812

8.005 3 u

MeV , so M 10 Be = M 9 Be + mn −

M 10 Be = 9.012 182 u + 1.008 665 u − 4

4

4

4

Q c2

6.812 MeV = 10.013 5 u 931.5 MeV u

Chapter 44

P44.50

10 5B

(a)

+ 24 He→ 136 C + 11 H

The product nucleus is 13 6C

(b)

236 92 U

→

.

10 5B

.

+ 11 H → 105 B + 24 He

The product nucleus is P44.51

13 6C

90 143 37 Rb + 55 Cs

+ 3 01 n ,

so Q = M 236 U − M 90 Rb − M 143 Cs − 3mn c 2 92

37

55

From Table A.3,

b

g b

g

Q = 236.045 562 − 89.914 809 − 142.927 330 − 3 1.008 665 u 931.5 MeV u = 165 MeV .

Section 44.8

Nuclear Magnetic Resonance and Magnetic Resonance Imagining

P44.52

FIG. P44.52 P44.53

2 µB

b

ja

ge

(a)

fn =

(b)

fp =

(c)

In the Earth’s magnetic field, fp =

h

b

=

2 1.913 5 5.05 × 10 −27 J T 1.00 T 6.626 × 10 −34 J ⋅ s

ja

ge

2 2.792 8 5.05 × 10 −27 J T 1.00 T 6.626 × 10

b

−34

ge

J⋅s

je

6.626 × 10

29.2 MHz

f=

42.6 MHz

j=

2.13 kHz .

2 2.792 8 5.05 × 10 −27 50.0 × 10 −6 −34

f=

585

586

Nuclear Structure

Additional Problems *P44.54

(a)

With mn and vn as the mass and speed of the neutrons, Eq. 9.23 of the text becomes, after making appropriate notational changes, for the two collisions v1 =

FG 2m IJ v Hm +m K ∴ bm + m g v = bm + m g v ∴ m bv − v g = m v − m v n

v2 =

n

2

n

∴ mn =

P44.55

n

n

n

1

n

2

2

n

FG 2m IJ v , Hm +m K

2

1

n

1

1 1

1

= 2 m n vn

2 2

m 1 v1 − m 2 v 2 v 2 − v1

a1 ufe3.30 × 10

j a fe

7

m s − 14 u 4.70 × 10 6 m s

(b)

mn =

(a)

Q = M 9 Be + M 4 He − M 12 C − mn c 2

6

7

4.70 × 10 m s − 3.30 × 10 m s

j=

1.16 u

b

g

Q = 9.012 182 u + 4.002 603 u − 12.000 000 u − 1.008 665 u 931.5 MeV u = 5.70 MeV (b)

Q = 2 M 2 H − M 3 He − mn

b

g

b

a

g

Q = 2 2.014 102 − 3.016 029 − 1.008 665 u 931.5 MeV u = 3.27 MeV exothermic P44.56

(a)

f

At threshold, the particles have no kinetic energy relative to each other. That is, they move like two particles that have suffered a perfectly inelastic collision. Therefore, in order to calculate the reaction threshold energy, we can use the results of a perfectly inelastic collision. Initially, the projectile M a moves with velocity v a while the target M X is at rest. We have from momentum conservation for the projectile-target system: Ma va = M a + M X vc . 1 The initial energy is: Ei = M a v a2 . 2 The final kinetic energy is:

b

Ef =

g

b

g

b

1 1 M a + M X v c2 = M a + M X 2 2

gLMN MM+ vM OPQ = LMN M M+ M OPQE . 2

a a

a

a

X

a

i

X

From this, we see that E f is always less than Ei and the change in energy, E f − Ei , is given by E f − Ei =

LM M NM + M a

a

X

OP Q

− 1 Ei = −

LM M OPE . NM + M Q X

a

X

i

This loss of kinetic energy in the isolated system corresponds to an increase in mass-energy during the reaction. Thus, the absolute value of this kinetic energy change is equal to –Q (remember that Q is negative in an endothermic reaction). The initial kinetic energy Ei is the threshold energy Eth . Therefore,

LM M OPE NM + M Q L M + M OP = = −Q M N M Q X

−Q =

a

or continued on next page

Eth

X

X

th

a

X

LM N

−Q 1 +

Ma MX

OP Q

.

Chapter 44

(b)

First, calculate the Q-value for the reaction:

587

Q = M N-14 + M He- 4 − M O-17 − M H-1 c 2

b OP = −a−1.19 MeVfLM1 + 4.002 603 OP = Q N 14.003 074 Q

g

Q = 14.003 074 + 4.002 603 − 16.999 132 − 1.007 825 u 931.5 MeV u = −1.19 MeV . Eth = −Q

Then, P44.57

1 1H

LM M + M N M X

a

X

1.53 MeV .

+ 73 Li → 74 Be + 01 n

b g b g b931.5 MeV ug Q = b1.007 825 u + 7.016 004 ug − b7.016 929 u + 1.008 665 ug b931.5 MeV ug Q = e −1.765 × 10 ujb931.5 MeV ug = −1.644 MeV F m I Q = F 1 + 1.007 825 I a1.644 MeVf = 1.88 MeV = G1 + Thus, KE JK GH 7.016 004 JK H m Q = M H + M Li − M Be + M n −3

incident projectile

min

P44.58

.

target nucleus

(a)

N0 =

(b)

λ=

1.00 kg mass = = 2.52 × 10 24 mass per atom 239.05 u 1.66 × 10 −27 kg u

a

fe

j

ln 2 ln 2 = = 9.106 × 10 −13 s −1 4 T1 2 2.412 × 10 yr 3.156 × 10 7 s yr

e

je

e

j

je

j

R0 = λN 0 = 9.106 × 10 −13 s −1 2.52 × 10 24 = 2.29 × 10 12 Bq (c)

t=

P44.59

(a)

FG R IJ = 1 lnFG R IJ λ HR K λ H R K F 2.29 × 10 Bq I = 3.38 × 10 sFG 1 yr IJ = lnG H 3.156 × 10 s K s H 0.100 Bq JK

R = R0 e − λ t , so t =

−1

0

ln

0

12

1 9.106 × 10 −13

13

−1

7

1.07 × 10 6 yr

57 57 0 0 27 Co → 26 Fe + +1 e + 0 ν

The Q-value for this positron emission is Q = M 57 Co − M 57 Fe − 2m e c 2 .

b

g b

g

Q = 56.936 296 − 56.935 399 − 2 0.000 549 u 931.5 MeV u = −0.187 MeV Since Q < 0 , this reaction cannot spontaneously occur . (b)

14 6C

→

14 7N

+

0 0 −1 e + 0 ν

The Q-value for this e − decay is Q = M 14 C − M 14 N c 2 .

b

g

Q = 14.003 242 − 14.003 074 u 931.5 MeV u = 0.156 MeV = 156 keV Since Q > 0 , the decay can spontaneously occur . (c)

The energy released in the reaction of (b) is shared by the electron and neutrino. Thus, K e can range from zero to 156 keV .

588 P44.60

Nuclear Structure

(a)

r = r0 A 1 3 = 1.20 × 10 −15 A 1 3 m.

r = 2.75 × 10 −15 m .

When A = 12 ,

(b)

(c)

(d)

P44.61

(a)

F=

a

f

ke Z − 1 e2

r2 When Z = 6

e8.99 × 10 =

9

ja

r2 r = 2.75 × 10 −15 m,

and

a

f

fe

N ⋅ m 2 C 2 Z − 1 1.60 × 10 −19 C

ja

e

fe

j

2

F = 152 N .

8.99 × 10 9 Z − 1 1.6 × 10 −19 k e q1 q 2 k e Z − 1 e 2 = = U= r r r

j

2

When Z = 6

and

r = 2.75 × 10 −15 m,

U = 4.19 × 10 −13 J = 2.62 MeV .

A = 238;

Z = 92,

r = 7.44 × 10 −15 m

F = 379 N

and

U = 2.82 × 10 −12 J = 17.6 MeV .

Because the reaction p → n + e + + ν would violate the law of conservation of energy m p = 1.007 276 u

me + = 5.49 × 10 −4 u .

mn = 1.008 665 u

mn + m e+ > m p .

Note that (b)

The required energy can come from the electrostatic repulsion of protons in the nucleus.

(c)

Add seven electrons to both sides of the reaction for nuclei to obtain the reaction for neutral atoms

13 7N

atom →

13 6C

13 13 7 N → 6C

+ e+ + ν

atom + e + + e − + ν

e N j − me C j − m − m − m Q = b931.5 MeV ug 13.005 739 − 13.003 355 − 2e5.49 × 10 j − 0 u Q = b931.5 MeV uge1.286 × 10 uj = 1.20 MeV Q = c2 m

13

13

e+

ν

e−

−4

−3

P44.62

(a)

A least-square fit to the graph yields:

e

j

λ = − slope = − −0.250 h −1 = 0.250 h −1 and

b g = intercept = 8.30 . F 1 h IJ = 4.17 × 10 λ = 0.250 h G H 60.0 min K ln cpm

(b)

t= 0

−1

T1 2 =

ln 2

=

min −1

ln 2

λ 4.17 × 10 −3 min −1 = 166 min = 2.77 h

continued on next page

−3

FIG. P44.62

589

Chapter 44

(c)

0

= 8.30 .

bcpmg = e counts min = 4.02 × 10 counts min . 4.02 × 10 counts min 1 bcpmg = = = 9.65 × 10 atoms λ Eff e4.17 × 10 min ja0.100f 8.30

Thus,

P44.63

b g

intercept = ln cpm

From (a),

3

0

R0

3

0

6

(d)

N0 =

(a)

The reaction is 145 61 Pm →

(b)

Q = M Pm − Mα − M Pr 931.5 = 144.912 744 − 4.002 603 − 140.907 648 931.5 = 2.32 MeV

(c)

The alpha and daughter have equal and opposite momenta

λ

−3

b

Eα =

141 59 Pr + α

g

pα2 2 mα

−1

b

Ed =

g

pα = p d

p d2 2m d

pα2 2mα 1 2 mα Eα Eα md 141 = = 2 = = = = 97. 2% or 2 + +4 Etot Eα + Ed m m 141 + m m 1 2 1 2 pα 2mα + pα 2m d α d α d

e

j e

j b

g b

g

2.26 MeV. This is carried away by the alpha. P44.64

(a)

If ∆E is the energy difference between the excited and ground states of the nucleus of mass M, and hf is the energy of the emitted photon, conservation of energy for the nucleusphoton system gives ∆E = hf + Er .

(1)

Where Er is the recoil energy of the nucleus, which can be expressed as Er =

a f

2

Mv Mv 2 = . 2 2M

(2)

Since system momentum must also be conserved, we have Mv =

(b)

hf . c 2 hf

b g

Hence, Er can be expresses as

Er =

When

hf 2.44 MeV Electron capture is allowed For positron emission,

to all specified excited states in 93 93 0 43 Tc → 42 Mo + +1 e + γ

93 42 Mo.

.

The disintegration energy is Q ′ = M 93 Tc − M 93 Mo − 2m e c 2 .

b

g b

g

Q ′ = 92.910 2 − 92.906 8 − 2 0.000 549 u 931.5 MeV u = 2.14 MeV Positron emission can reach the 1.35, 1.48, and 2.03 MeV states but there is insufficient energy to reach the 2.44 MeV state. (b)

The daughter nucleus in both forms of decay is

93 42 Mo

.

FIG. P44.72

Chapter 44

P44.73

K=

1 mv 2 , 2

so

v=

2K = m

b

ge

2 0.040 0 eV 1.60 × 10 −19 J eV 1.67 × 10

The time for the trip is t =

−27

kg

j = 2.77 × 10

3

m s.

1.00 × 10 4 m x = = 3.61 s . v 2.77 × 10 3 m s

The number of neutrons finishing the trip is given by N = N 0 e − λ t . N − a ln 2 ft T1 2 − a ln 2 fb 3 .61 s 624 s g =1−e =1−e = 0.004 00 = 0.400% . N0

The fraction decaying is 1 − P44.74

(a)

If we assume all the then

N = N0 e

yields

t=

P44.75

λ

ln

Sr came from

87

Rb,

−λ t

FG N IJ = T lnFG N IJ H N K ln 2 H N K 12

0

0

where

N = N Rb-87

and

N 0 = N Sr-87 + N Rb-87 t=

(b)

−1

87

e4.75 × 10

10

ln 2

yr

j lnF 1.82 × 10 + 1.07 × 10 I = GH 1.82 × 10 JK 10

9

10

It could be no older . The rock could be younger if some

b g

R = R0 exp − λ t lets us write

ln R = ln R0 − λ t

which is the equation of a straight line with

slope = λ .

3.91 × 10 9 yr . 87

Sr were originally present.

The logarithmic plot shown in Figure P44.75 is fitted by ln R = 8.44 − 0.262 t . If t is measured in minutes, then decay constant λ is 0.262 per minute. The half–life is T1 2 =

ln 2

λ

=

ln 2 = 2.64 min . 0.262 min

The reported half–life of 137 Ba is 2.55 min. The difference reflects experimental uncertainties. FIG. P44.75

ANSWERS TO EVEN PROBLEMS P44.2

(a) 7.89 cm and 8.21 cm ; (b) see the solution

P44.4

(a) 29.5 fm; (b) 5.18 fm; (c) see the solution

P44.6

25.6 MeV

30

Si with A = 30

P44.8

a nucleus such as

P44.10

6.11 PN toward the other ball

P44.12

(a) 48; (b) 3; (c) 46; (d) 1

593

594 P44.14

P44.16 P44.18

Nuclear Structure

(a) 1.11 MeV nucleon; (b) 7.07 MeV nucleon ; (c) 8.79 MeV nucleon ; (d) 7.57 MeV nucleon 0.210 MeV greater for less proton repulsion

23

Na because it has

(a) 84.1 MeV ; (b) 342 MeV ; (c) The nuclear force of attraction dominates over electrical repulsion

P44.20

7.93 MeV

P44.22

(a) see the solution; R R and ; see the solution (b) 3 6

P44.24

0.200 mCi

P44.26

41.8 TBq

P44.28

86.4 h

P44.30 P44.32

R0 T1 2 ln 2

e2

− t1 T1 2

(a) (e)

* 65 28 Ni ; 1 1H

(b)

j

g

211 82 Pb ;

(c)

55 27 Co;

(d)

0 −1 e;

P44.36

between 5 400 yr and 6 800 yr

P44.38

(a) cannot occur ; (b) cannot occur ; (c) can occur

P44.40

(a) 4.00 Gyr ; (b) 0.019 9 and 4.60

P44.42

(a) 148 Bq m3 ; (b) 7.05 × 10 7 atoms m 3 ; (c) 2.17 × 10 −17

(a) 0.281 ; (b) 1.65 × 10 −29 ; (c) see the solution

P44.46

(a)

P44.48

1.02 MeV

P44.50

(a)

P44.52

see the solution

P44.54

(a) see the solution; (b) 1.16 u

P44.56

(a) see the solution; (b) 1.53 MeV

P44.58

(a) 2.52 × 10 24 ; (b) 2.29 TBq ; (c) 1.07 Myr

P44.60

(a) 2.75 fm; (b) 152 N ; (c) 2.62 MeV ; (d) 7.44 fm , 379 N , 17.6 MeV

P44.62

(a) see the solution; (b) 4.17 × 10 −3 min −1 ; 2.77 h ; (c) 4.02 × 10 3 counts min ;

21 10 Ne;

13 6C ;

0 + 0 (b) 144 54 Xe; (c) 1 e and 0 ν

(b)

10 5B

(d) 9.65 × 10 6 atoms

(a) see the solution; (b) see the solution; (c) see the solution; 10.9 min ; ln λ 1 λ 2 ; yes (d) t m = λ1 − λ 2

b

P44.34

−2

− t 2 T1 2

P44.44

P44.64

(a) see the solution; (b) 1.94 meV

P44.66

(a) ~ 10 −1 353 ; (b) 0.892

P44.68

(a) 4.28 pJ; (b) 1.19 × 10 57 atoms ; (c) 107 Gyr

P44.70

(a) 12.3 mg ; (b) 0.166 W

P44.72

(a) electron capture to all; positron emission to the 1.35 MeV , 1.48 MeV , and 2.03 MeV states ; (b) 93 42 Mo; see the solution

P44.74

(a) 3.91 Gyr ; (b) No older; it could be younger if some 87 Sr were originally present, contrary to our assumption.

45 Applications of Nuclear Physics CHAPTER OUTLINE 45.1 45.2 45.3 45.4 45.5 45.6 45.7

Interactions Involving Neutrons Nuclear Fission Nuclear Reactors Nuclear Fusion Radiation Damage Radiation Detectors Uses of Radiation

ANSWERS TO QUESTIONS Q45.1

A moderator is used to slow down neutrons released in the fission of one nucleus, so that they are likely to be absorbed by another nucleus to make it fission.

Q45.2

The hydrogen nuclei in water molecules have mass similar to that of a neutron, so that they can efficiently rob a fast-moving neutron of kinetic energy as they scatter it. Once the neutron is slowed down, a hydrogen nucleus can absorb it in the reaction n + 11 H → 21 H .

Q45.3

The excitation energy comes from the binding energy of the extra nucleon.

Q45.4

The advantage of a fission reaction is that it can generate much more electrical energy per gram of fuel compared to fossil fuels. Also, fission reactors do not emit greenhouse gasses as combustion byproducts like fossil fuels—the only necessary environmental discharge is heat. The cost involved in producing fissile material is comparable to the cost of pumping, transporting and refining fossil fuel. The disadvantage is that some of the products of a fission reaction are radioactive—and some of those have long half-lives. The other problem is that there will be a point at which enough fuel is spent that the fuel rods do not supply power economically and need to be replaced. The fuel rods are still radioactive after removal. Both the waste and the “spent” fuel rods present serious health and environmental hazards that can last for tens of thousands of years. Accidents and sabotage involving nuclear reactors can be very serious, as can accidents and sabotage involving fossil fuels.

Q45.5

The products of fusion reactors are generally not themselves unstable, while fission reactions result in a chain of reactions which almost all have some unstable products.

Q45.6

For the deuterium nuclei to fuse, they must be close enough to each other for the nuclear forces to overcome the Coulomb repulsion of the protons—this is why the ion density is a factor. The more time that the nuclei in a sample spend in close proximity, the more nuclei will fuse—hence the confinement time is a factor.

595

596

Applications of Nuclear Physics

Q45.7

In a fusion reaction, the main idea is to get the nuclear forces, which act over very short distances, to overcome the Coulomb repulsion of the protons. Tritium has one more neutron in the nucleus, and thus increases the nuclear force, decreasing the necessary kinetic energy to obtain D–T fusion as compared to D–D fusion.

Q45.8

The biggest obstacle is power loss due to radiation. Remember that a high temperature must be maintained to keep the fuel in a reactive plasma state. If this kinetic energy is lost due to bremsstrahlung radiation, then the probability of nuclear fusion will decrease significantly. Additionally, each of the confinement techniques requires power input, thus raising the bar for sustaining a reaction in which the power output is greater than the power input.

Q45.9

Fusion of light nuclei to a heavier nucleus releases energy. Fission of a heavy nucleus to lighter nuclei releases energy. Both processes are steps towards greater stability on the curve of binding energy, Figure 44.5. The energy release per nucleon is typically greater for fusion, and this process is harder to control.

Q45.10

Advantages of fusion: high energy yield, no emission of greenhouse gases, fuel very easy to obtain, reactor can not go supercritical like a fission reactor, low amounts of radioactive waste. Disadvantages: requires high energy input to sustain reaction, lithium and helium are scarce, neutrons released by reaction cause structural damage to reactor housing.

Q45.11

The fusion fuel must be heated to a very high temperature. It must be contained at a sufficiently high density for a sufficiently long time to achieve a reasonable energy output.

Q45.12

The first method uses magnetic fields to contain the plasma, reducing its contact with the walls of the container. This way, there is a reduction in heat loss to the environment, so that the reaction may be sustained over seconds. The second method involves striking the fuel with high intensity, focused lasers from multiple directions, effectively imploding the fuel. This increases the internal pressure and temperature of the fuel to the point of ignition.

Q45.13

No. What is critical in radiation safety is the type of radiation encountered. The curie is a measure of the rate of decay, not the products of the decay or of their energies.

Q45.14

X-ray radiation can cause genetic damage in the developing fetus. If the damaged cells survive the radiation and reproduce, then the genetic errors will be replicated, potentially causing severe birth defects or death of the child.

Q45.15

For each additional dynode, a larger applied voltage is needed, and hence a larger output from a power supply—“infinite” amplification would not be practical. Nor would it be desirable: the goal is to connect the tube output to a simple counter, so a massive pulse amplitude is not needed. If you made the detector sensitive to weaker and weaker signals, you would make it more and more sensitive to background noise.

Chapter 45

Q45.16

597

Sometimes the references are oblique indeed. Some must serve for more than one form of energy or mode of transfer. Here is one list: kinetic: ocean currents rotational kinetic: Earth turning gravitational: water lifted up elastic: Elastic energy is necessary for sound, listed below. internal: by contrast to a chilly night; or in forging a chain chemical: flames sound: thunder electrical transmission: lightning electromagnetic radiation: heavens blazing; lightning atomic electronic: In the blazing heavens, stars have different colors because of different predominant energy losses by atoms at their surfaces. nuclear: The blaze of the heavens is produced by nuclear reactions in the cores of stars. Remarkably, the word “energy” in this translation is an anachronism. Goethe wrote the song a few years before Thomas Young coined the term.

SOLUTIONS TO PROBLEMS Section 45.1

Interactions Involving Neutrons

Section 45.2

Nuclear Fission

*P45.1

The energy is

FG 1 eV IJ FG 1 U - 235 nucleus IJ FG 235 g IJ FG M IJ = H 1.60 × 10 J K H 208 MeV K H 6.02 × 10 nucleus K H 10 K ∆m = bm + M g − b M + M + 3m g ∆m = b1.008 665 u + 235.043 923 ug − d97.912 7 u + 134.916 5 u + 3b1.008 665 ugi

3.30 × 10 10 J P45.2

−19

n

U

Zr

23

Te

6

0.387 g of U - 235 .

n

∆m = 0.197 39 u = 3.28 × 10 −28 kg

Q = ∆mc 2 = 2.95 × 10 −11 J = 184 MeV

so

Three different fission reactions are possible:

1 235 0 n + 92 U

→

90 144 38 Sr + 54 Xe +

2 01 n

144 54 Xe

1 235 0 n + 92 U

→

90 143 38 Sr + 54 Xe +

1 235 0 n + 92 U

→

90 142 38 Sr + 54 Xe +

4 01 n

142 54 Xe

P45.4

1 238 0 n + 92 U

→

239 92 U

P45.5

− 1 232 233 233 0 n + 90Th → 90Th → 91 Pa + e

P45.6

(a)

Q = ∆m c 2 = mn + M U235 − M Ba141 − M Kr92 − 3mn c 2

(b)

f=

P45.3

→

3 01 n

− 239 93 Np + e

143 54 Xe

+ν +ν

− 239 239 93 Np → 94 Pu + e 233 233 91 Pa → 92 U

+ν

+ e− + ν

a f ∆m = b1.008 665 + 235.043 923 g − b140.914 4 + 91.926 2 + 3 × 1.008 665 g u = 0.185 993 u Q = b0.185 993 ugb931.5 MeV ug = 173 MeV ∆m 0.185993 u = = 7.88 × 10 −4 = 0.078 8% 236.05 u mi

598 *P45.7

Applications of Nuclear Physics

(a)

(b) P45.8

The initial mass is 1.007 825 u + 11.009 306 u = 12.017 131 u . The final mass is 3 4.002 603 u = 12.007 809 u . The rest mass annihilated is ∆m = 0.009 322 u. The energy 931.5 MeV = 8.68 MeV . created is Q = ∆mc 2 = 0.009 322 u 1u

b

g

FG H

IJ K

The proton and the boron nucleus have positive charges. The colliding particles must have enough kinetic energy to approach very closely in spite of their electric repulsion.

If the electrical power output of 1 000 MW is 40.0% of the power derived from fission reactions, the power output of the fission process is 1 000 MW = 2.50 × 10 9 J s 8.64 × 10 4 s d = 2.16 × 10 14 J d . 0.400

e

je

j

e

The number of fissions per day is 2.16 × 10 14 J d This also is the number of

e6.74 × 10

24

nuclei d

235

IJ FG 1 eV IJ = 6.74 × 10 jFGH 2001 fission × 10 eV K H 1.60 × 10 JK −19

6

235

U nuclei used, so the mass of

F I = 2.63 × 10 g mol jGH 6.02 × 235 J 10 nuclei mol K 23

3

24

d −1 .

U used per day is

g d = 2.63 kg d .

In contrast, a coal-burning steam plant producing the same electrical power uses more than 6 × 10 6 kg d of coal. P45.9

The available energy to do work is 0.200 times the energy content of the fuel.

b1.00 kg fuelge0.034 0 U fueljFGH 1 1000kgg IJK FGH 1235molg IJK e6.02 × 10 e2.90 × 10 Jja0.200f = 5.80 × 10 J = e1.00 × 10 Nj∆r 235

12

11

F a208fe1.60 × 10 JjI moljG GH fission JJK = 2.90 × 10 −13

23

12

J

5

∆r = 5.80 × 10 6 m = 5.80 Mm

Section 45.3 P45.10

Nuclear Reactors 4 3 πr 3

13

For a sphere:

V=

(b)

For a cube:

V=

(c)

For a parallelepiped: V = 2 a 3

(d)

Therefore, the sphere has the least leakage and the

3

and

and

and

r=

FG 3V IJ H 4π K

(a)

= V1 3

a=

FG V IJ H 2K

so

4π r 2 A = = 4.84V −1 3 . V 4 3 π r3

so

A 6 2 = 3 = 6V −1 3 . V

so

2a 2 + 8a 2 A = = 6.30V −1 3 . V 2a3

13

parallelepiped has the greatest leakage for a given volume.

b g e

j

Chapter 45

P45.11

235

mass of

a

fe

U available ≈ 0.007 10 9 metric tons

number of nuclei ≈

F 7 × 10 g I 6.02 × 10 GH 235 g mol JK e 12

23

F 10 g I = 7 × 10 jGH 1 metric ton JK 6

12

g

j

nuclei mol = 1.8 × 10 34 nuclei

The energy available from fission (at 208 MeV/event) is

jb

e

ge

j

E ≈ 1.8 × 10 34 events 208 MeV event 1.60 × 10 −13 J MeV = 6.0 × 10 23 J . This would last for a time interval of ∆t =

P45.12

E

P

≈

jFGH

e

In one minute there are

60.0 s = 5.00 × 10 4 fissions . 1.20 ms

b

So the rate increases by a factor of 1.000 25 P45.13

IJ K

1 yr 6.0 × 10 23 J = 8.6 × 10 10 s ≈ 3 000 yr . 7.0 × 10 12 J s 3.16 × 10 7 s

g

50 000

= 2.68 × 10 5 .

P = 10.0 MW = 1.00 × 10 7 J s If each decay delivers 1.00 MeV = 1.60 × 10 −13 J , then the number of decays/s = 6.25 × 10 19 Bq .

Section 45.4 P45.14

(a)

Nuclear Fusion The Q value for the D-T reaction is 17.59 MeV. Specific energy content in fuel for D-T reaction:

a17.59 MeVfe1.60 × 10 J MeVj = 3.39 × 10 J kg a5 ufe1.66 × 10 kg uj e3.00 × 10 J sjb3 600 s hrg = 31.9 g h burning of D and T r = e3.39 × 10 J kg je10 kg g j −13

14

−27

9

DT

(b)

−3

14

Specific energy content in fuel for D-D reaction: Q = two Q values

a

f

1 3.27 + 4.03 = 3.65 MeV average of 2

a3.65 MeVfe1.60 × 10 J MeV j = 8.80 × 10 J kg a4 ufe1.66 × 10 kg uj e3.00 × 10 J sjb3 600 s hrg = 122 g h burning of D r = e8.80 × 10 J kg je10 kg g j −13

13

−27

9

DD

13

−3

.

.

599

600 P45.15

Applications of Nuclear Physics

(a)

At closest approach, the electrostatic potential energy equals the total energy E. Uf =

b gb g = E :

k e Z1 e Z 2 e rmin

e8.99 × 10 E= (b)

9

je

j

2

N ⋅ m 2 C 2 1.6 × 10 −19 C Z1 Z 2 1.00 × 10 −14 m

e2.30 × 10 JjZ Z −14

=

1

.

2

For both the D-D and the D-T reactions, Z1 = Z 2 = 1 . Thus, the minimum energy required in both cases is

IJ = jFGH 1.601 ×MeV 10 JK

e

E = 2.30 × 10 −14 J

−13

144 keV .

Section 45.4 in the text gives more accurate values for the critical ignition temperatures, of about 52 keV for D-D fusion and 6 keV for D-T fusion. The nuclei can fuse by tunneling. A triton moves more slowly than a deuteron at a given temperature. Then D-T collisions last longer than D-D collisions and have much greater tunneling probabilities. P45.16

e

j a f + a3 f

e

je

13

13

(a)

r f = rD + rT = 1.20 × 10 −15 m 2

= 3.24 × 10 −15 m

(b)

8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C k e2 Uf = e = rf 3.24 × 10 −15 m

(c)

Conserving momentum,

j

2

= 7.10 × 10 −14 J = 444 keV

FG m IJ v = 2 v Hm +m K 5 F m IJ v 1 1 K + 0 = bm + m gv + U = bm + m gG 2 2 Hm +m K F m IJ FG 1 m v IJ + U = FG m IJ K + U K +0=G H m + m KH 2 K Hm +m K F m + m IJ = 5 a444 keVf = 740 keV K =U G H m K 3 b

g

D

m D vi = m D + m T v f , or v f =

D

T

i

i

2

(d)

K i + Ui = K f + U f :

D

i

D

D

P45.17

T

i

f

(e)

Possibly by tunneling.

(a)

Average KE per particle is

Therefore, v rms =

(b)

t=

T

D

i

f

T

D

D

f

2 D i

D

i

FG 1 − m IJ K = U : H m +m K

2 f

f

D

T

D

D

D

T

T

i

T

T

3 1 k B T = mv 2 . 2 2

3 k BT = m

x 0.1 m = ~ 10 −7 s ~ 6 v 10 m s

e

je

3 1.38 × 10 −23 J K 4.00 × 10 8 K

e

2 1.67 × 10

−27

kg

j

j=

2.23 × 10 6 m s .

f

2 i

+U f

Chapter 45

P45.18

(a)

mI J = 1.32 × 10 m e jFGH 1 609 1 mi K = ρV = e10 kg m je1.32 × 10 m j = 1.32 × 10 kg F M I m = FG 2.016 IJ 1.32 × 10 kg = 1.48 × 10 kg =G j H 18.015 K e H M JK = b0.030 0%gm = e0.030 0 × 10 je1.48 × 10 kg j = 4.43 × 10 3

V = 317 × 10 6 mi 3 3

m water

H2

mH 2

H 2O

18

3

18

3

3

21

21

H 2O

m Deuterium

20

−2

H2

601

20

16

kg

The number of deuterium nuclei in this mass is 4.43 × 10 16 kg m Deuterium = = 1.33 × 10 43 . −27 m Deuteron 2.014 u 1.66 × 10 kg u

N=

a

fe

j

Since two deuterium nuclei are used per fusion, 21 H + 21 H → 24 He + Q , the number of events N is = 6.63 × 10 42 . 2 The energy released per event is

b

g

b

g

Q = M 2 H + M 2 H − M 4 He c 2 = 2 2.014 102 − 4.002 603 u 931.5 MeV u = 23.8 MeV . The total energy available is then E= (b)

FG N IJ Q = e6.63 × 10 ja23.8 MeVfFG 1.60 × 10 J IJ = H 2K H 1 MeV K

(a)

2.53 × 10 31 J .

The time this energy could possibly meet world requirements is ∆t =

P45.19

−13

42

E

P

=

2.53 × 10 31 J

e

100 7.00 × 10

12

e J sj

= 3.61 × 10 16 s

jFGH 3.161×yr10 s IJK = 7

1.14 × 10 9 yr ~ 1 billion years .

Including both ions and electrons, the number of particles in the plasma is N = 2nV where n is the ion density and V is the volume of the container. Application of Equation 21.6 gives the total energy as E=

3 Nk BT = 3nVk B T = 3 2.0 × 10 13 cm −3 2

e

L jMMe50 m jFGH 101 mcm N 3

6

3

3

I OPe1.38 × 10 JK PQ

−23

je

J K 4.0 × 10 8 K

j

E = 1.7 × 10 7 J (b)

From Table 20.2, the heat of vaporization of water is L v = 2.26 × 10 6 J kg . The mass of water that could be boiled away is m=

1.7 × 10 7 J E = = 7.3 kg . L v 2.26 × 10 6 J kg

602 P45.20

Applications of Nuclear Physics

(a)

Lawson’s criterion for the D-T reaction is nτ ≥ 10 14 s cm 3 . For a confinement time of

τ = 1.00 s, this requires a minimum ion density of n = 10 14 cm −3 . (b)

At the ignition temperature of T = 4.5 × 10 7 K and the ion density found above, the plasma pressure is

LM MNe

P = 2nk B T = 2 10 14 cm −3 (c)

O jFGH 101 mcm IJK PPe1.38 × 10 Q 6

3

3

je

−23

j

J K 4.5 × 10 7 K = 1.24 × 10 5 J m 3 .

The required magnetic energy density is then uB =

B2 ≥ 10 P = 10 1.24 × 10 5 J m3 = 1.24 × 10 6 J m 3 , 2µ 0

e

j

e

je

j

B ≥ 2 4π × 10 −7 N A 2 1.24 × 10 6 J m 3 = 1.77 T . P45.21

Let the number of 6 Li atoms, each having mass 6.015 u, be N 6 while the number of 7 Li atoms, each with mass 7.016 u, is N 7 .

F 0.925 I N . GH 0.075 0 JK total mass = N a6.015 uf + N a7.016 uf e1.66 × 10 kg uj = 2.00 kg , L F 0.925 I a7.016 ufOPe1.66 × 10 kg uj = 2.00 kg . N Ma6.015 uf + G H 0.075 0 JK MN PQ

b

Also,

6

or

6

−27

7

−27

6

N 6 = 1.30 × 10 25 as the number of 6 Li atoms and

This yields

N7 = P45.22

g

N 6 = 7.50% of N total = 0.075 0 N 6 + N 7 , or N 7 =

Then,

F 0.925 I e1.30 × 10 j = GH 0.075 0 JK 25

1.61 × 10 26

as the number of 7 Li atoms.

The number of nuclei in 1.00 metric ton of trash is

b

N = 1 000 kg 1 000 g kg

23

mol = 1.08 × 10 g 6.02 ×5610.0 gnuclei mol

e

28

nuclei .

ja fe

j

At an average charge of 26.0 e/nucleus,

q = 1.08 × 10 28 26.0 1.60 × 10 −19 = 4.47 × 10 10 C .

Therefore

t=

q 4. 47 × 10 10 = = 4.47 × 10 4 s = 12.4 h . I 1.00 × 10 6

Chapter 45

Section 45.5 P45.23

N0 =

603

Radiation Damage mass present 5.00 kg = = 3.35 × 10 25 nuclei − 27 mass of nucleus 89.907 7 u 1.66 × 10 kg u

b

λ=

ge

j

ln 2 ln 2 = = 2.38 × 10 −2 yr −1 = 4.52 × 10 −8 min −1 T1 2 29.1 yr

e

je

j

R0 = λN 0 = 4.52 × 10 −8 min −1 3.35 × 10 25 = 1.52 × 10 18 counts min 10.0 counts min R = e −λ t = = 6.60 × 10 −18 R0 1.52 × 10 18 counts min

P45.24

e

j

and

λ t = − ln 6.60 × 10 −18 = 39.6

giving

t=

39.6

λ

=

39.6 = 1.66 × 10 3 yr . 2.38 × 10 −2 yr −1

Source: 100 mrad of 2-MeV γ -rays/h at a 1.00-m distance. (a)

For γ -rays, dose in rem = dose in rad. Thus a person would have to stand 10.0 hours to receive 1.00 rem from a 100-mrad/h source.

(b)

1 . r2

If the γ -radiation is emitted isotropically, the dosage rate falls off as

Thus a dosage 10.0 mrad/h would be received at a distance r = 10.0 m = 3.16 m . P45.25

(a)

The number of x-rays taken per year is

b

gb

gb

g

n = 8 x - ray d 5 d wk 50 wk yr = 2.0 × 10 3 x - ray yr . The average dose per photograph is (b)

5.0 rem yr 2.0 × 10 3 x - ray yr

= 2.5 × 10 −3 rem x - ray .

The technician receives low-level background radiation at a rate of 0.13 rem yr . The dose of 5.0 rem yr received as a result of the job is 5.0 rem yr = 38 times background levels . 0.13 rem yr

P45.26

(a)

I = I 0 e − µ x , so With µ = 1.59 cm −1 , the thickness when I =

(b)

When

I0 = 1.00 × 10 4 , I

I0 is 2

FG IJ H K

x=

I 1 ln 0 µ I

x=

1 ln 2 = 0.436 cm . 1.59 cm −1

x=

1 ln 1.00 × 10 4 = 5.79 cm . −1 1.59 cm

af e

j

604 P45.27

Applications of Nuclear Physics

1 rad = 10 −2 J kg ∆t =

mc∆T

=

P

P ∆ t = mc∆T

Q = mc∆T

b a10fe10

ga

m 4 186 J kg ⋅° C 50.0° C −2

ja f

J kg ⋅ s m

f=

2.09 × 10 6 s ≈ 24 days!

Note that power is the product of dose rate and mass. P45.28

10 −2 J kg Q absorbed energy = = 1 000 rad = 10.0 J kg m unit mass 1 rad

b

g

The rise in body temperature is calculated from Q = mc∆T where c = 4 186 J kg for water and the human body ∆T = P45.29

b

g

Q 1 = 10.0 J kg = 2.39 × 10 −3 ° C (Negligible). mc 4 186 J kg ⋅° C

If half of the 0.140-MeV gamma rays are absorbed by the patient, the total energy absorbed is

a0.140 MeVf LMF 1.00 × 10 g I F 6.02 × 10 nuclei I OP = 4.26 × 10 MeV 2 MNGH 98.9 g mol JK GH 1 mol JK PQ J MeV j = 0.682 J E = e 4.26 × 10 MeV je1.60 × 10 0.682 J F 1 rad I = 1.14 rad . Thus, the dose received is Dose = 60.0 kg GH 10 J kg JK F 6.02 × 10 nuclei mol I = 6.70 × 10 The nuclei initially absorbed are N = e1.00 × 10 g jG H 89.9 g mol JK a f ∆N = N − N = N e1 − e j = N e1 − e The number of decays in time t is j. −8

E=

23

12

−13

12

−2

P45.30

0

0

At the end of 1 year,

The energy deposited is Thus, the dose received is

P45.31

T1 2

=

−λ t

0

12

.

− ln 2 t T1 2

0

1.00 yr = 0.034 4 29.1 yr

e je j E = e1.58 × 10 ja1.10 MeV fe1.60 × 10 J MeV j = 0.027 7 J . F 0.027 7 J I = 3.96 × 10 J kg = 0.039 6 rad. Dose = G H 70.0 kg JK ∆N = N 0 − N = 6.70 × 10 12 1 − e −0.023 8 = 1.58 × 10 11 .

and

Section 45.6

t

23

−9

−13

11

−4

Radiation Detectors

b g a f b ge a

fe

a f e

je

je

j j

2

(a)

2 1 2 5.00 × 10 −12 F 1.00 × 10 3 V 1 2 C ∆V E = = = 3.12 × 10 7 Eβ 0.500 MeV 0.500 MeV 1.60 × 10 −13 J MeV

(b)

N=

j

5.00 × 10 −12 F 1.00 × 10 3 V Q C ∆V = = = 3.12 × 10 10 electrons e e 1.60 × 10 −19 C

Chapter 45

P45.32

(a)

(b)

605

EI = 10.0 eV is the energy required to liberate an electron from a dynode. Let ni be the number of electrons incident upon a dynode, each having gained energy e ∆V as it was accelerated to this dynode. The number of electrons that will be freed from this dynode is ∆V : N i = ni e EI

a f

a1fea100 V f =

10 1 electrons .

At the first dynode, ni = 1

and

N1 =

For the second dynode, ni = N 1 = 10 1 ,

so

N2 =

(10 1 ) e 100 V = 10 2 . 10.0 eV

At the third dynode, ni = N 2 = 10 2

and

N3 =

(10 2 ) e 100 V = 10 3 . 10.0 eV

10.0 eV

a

f

a

f

Observing the developing pattern, we see that the number of electrons incident on the seventh and last dynode is n 7 = N 6 = 10 6 . (c)

The number of electrons incident on the last dynode is n 7 = 10 6 . The total energy these electrons deliver to that dynode is given by

a f

a

f

E = ni e ∆V = 10 6 e 700 V − 600 V = 10 8 eV . P45.33

(a)

The average time between slams is 60 min 38 = 1.6 min . Sometimes, the actual interval is nearly zero. Perhaps about equally as often, it is 2 × 1.6 min. Perhaps about half as often, it is 4 × 1.6 min . Somewhere around 5 × 1.6 min = 8.0 min , the chances of randomness producing so long a wait get slim, so such a long wait might likely be due to mischief.

(b)

The midpoints of the time intervals are separated by 5.00 minutes. We use R = R0 e − λ t . Subtracting the background counts,

a f a f e ja F 262 IJ = ln a0.882f = − 3.47 min T ln G H 297 K 337 − 5 15 = 372 − 5 15 e

or (c)

− ln 2 T1

2

5 .00 min

1 2

f

which yields T1 2 = 27.6 min .

As in the random events in part (a), we imagine a ±5 count counting uncertainty. The smallest likely value for the half-life is then given by ln

FG 262 − 5 IJ = − 3.47 min T H 297 + 5 K

1 2

e j

, or T1

2

min

= 21.1 min .

The largest credible value is found from

FG 262 + 5 IJ = − 3.47 min T , yielding eT j = 38.8 min . H 297 − 5 K F 38.8 + 21.1 IJ ± FG 38.8 − 21.1 IJ min = a30 ± 9f min = 30 min ± 30% . T =G H 2 K H 2 K ln

Thus,

1 2

1 2

1 2

max

606

Applications of Nuclear Physics

Section 45.7 P45.34

Uses of Radiation

The initial specific activity of

b R mg

0

=

59

Fe in the steel,

F I GH JK R F RI e =G J e = e3.70 × 10 Bq kg je K H m m F 800 Bq literIJ (6.50 liters) = 86.7 Bq . R =G H 60.0 K

20.0 µCi 100 µCi 3.70 × 10 4 Bq = 3.70 × 10 6 Bq kg . = 0.200 kg kg 1 µCi −λ t

After 1 000 h,

jb

− 6. 40 ×10 −4 h −1 1 000 h

6

g = 1.95 × 10 6

Bq kg .

0

The activity of the oil,

P45.35

oil

86.7 Bq Roil = = 4.45 × 10 −5 kg . Rm 1.95 × 10 6 Bq kg

Therefore,

m in oil =

So that wear rate is

4. 45 × 10 −5 kg = 4.45 × 10 −8 kg h . 1 000 h

The half-life of The

14

14

b g

ln 2 ln 2 = = 0.009 82 s −1 . T1 2 70.6 s

O is 70.6 s, so the decay constant is λ =

e j

O nuclei remaining after five min is N = N 0 e − λ t = 10 10 e

ja

e

− 0.009 82 s −1 300 s

f = 5.26 × 10 8 .

The number of these in one cubic centimeter of blood is

F 1.00 cm I = e5.26 × 10 jF 1.00 cm I = 2 .63 × 10 GH total vol. of blood JK GH 2000 cm JK R = λ N ′ = e0.009 82 s je 2 .63 × 10 j = 2 .58 × 10 Bq ~ 10 and their activity is 3

N′ = N

3

8

−1

P45.36

5

3

5

3

3

Bq .

10 4 MeV = 9.62 × 10 3 . Since only 50% of the photons are 1.04 MeV detected, the number of 65 Cu nuclei decaying is twice this value, or 1.92 × 10 4 . In two half3 lives, three-fourths of the original nuclei decay, so N 0 = 1.92 × 10 4 and N 0 = 2.56 × 10 4 . This 4 is 1% of the 65 Cu , so the number of 65 Cu is 2.56 × 10 6 ~ 10 6 .

(a)

The number of photons is

(b)

Natural copper is 69.17% 63 Cu and 30.83% 65 Cu . Thus, if the sample contains N Cu copper atoms, the number of atoms of each isotope is N 63 = 0.691 7 NCu and N 65 = 0.308 3 N Cu .

FG H

IJ K

FG H

IJ e K

N 63 0.691 7 0.6917 0.6917 = 2 .56 × 10 6 = 5.75 × 10 6 . N 65 = or N 63 = 0.3083 0.3083 N 65 0.308 3

j The total mass of copper present is then m = b62 .93 ugN + a64.93 ufN m = b62 .93ge5.75 × 10 j + a64.93fe 2 .56 × 10 j u e1.66 × 10 g uj

Therefore,

Cu

Cu

6

= 8.77 × 10 −16 g ~ 10 −15 g

6

63

−24

65 :

Chapter 45

P45.37

(a)

Starting with N = 0 radioactive atoms at t = 0 , the rate of increase is (production – decay) dN = R−λN so dN = R − λ N dt . dt The variables are separable.

b

z

N 0

(b)

607

FG R − λ N IJ = t λ H R K FG R − λN IJ = e . H R K

z t

dN = dt : R − λN 0

−

FG R − λN IJ = −λ t H R K

so

ln

Therefore,

1−

λ R

g

1

ln

−λ t

and

N = e−λ t

N=

R

λ

e1 − e j . −λ t

R

The maximum number of radioactive nuclei would be

λ

.

Additional Problems P45.38

(a)

Suppose each 235 U fission releases 208 MeV of energy. Then, the number of nuclei that must have undergone fission is total release 5 × 10 13 J N= = = 1.5 × 10 24 nuclei . −13 energy per nuclei 208 MeV 1.60 × 10 J MeV

a

P45.39

fe

j

F 1.5 × 10 nuclei I b235 g molg ≈ GH 6.02 × 10 nuclei mol JK 24

(b)

mass =

(a)

At 6 × 10 8 K , the average kinetic energy of a carbon atom is 3 k B T = 1.5 8.62 × 10 −5 eV K 6 × 10 8 K = 8 × 10 4 eV 2 Note that 6 × 10 8 K is about 6 2 = 36 times larger than 1.5 × 10 7 K , the core temperature of the Sun. This factor corresponds to the higher potential-energy barrier to carbon fusion compared to hydrogen fusion. It could be misleading to compare it to the temperature ~ 10 8 K required for fusion in a low-density plasma in a fusion reactor.

23

a fe

(b)

je

0.6 kg

j

The energy released is

e j e

j e j

E = 2m C 12 − m Ne 20 − m He 4 c 2

b

ga

f

E = 24.000 000 − 19.992 440 − 4.002 603 931.5 MeV = 4.62 MeV In the second reaction,

j a931.5f MeV u E = b 24.000 000 − 23.985 042 ga931.5 f MeV = e j e

E = 2m C 12 − m Mg 24

continued on next page

13.9 MeV

608

Applications of Nuclear Physics

(c)

The energy released is the energy of reaction of the number of carbon nuclei in a 2.00-kg sample, which corresponds to

F mol I F 4.62 MeV fusion event I F 1 kWh I G e jGH 6.02 ×1210.0 gatoms G J J mol K H 2 nuclei fusion event K H 2.25 × 10 MeV JK e1.00 × 10 ja4.62f kWh = 1.03 × 10 kWh ∆E = 2e 2.25 × 10 j 23

∆E = 2.00 × 10 3 g

19

26

7

19

P45.40

To conserve momentum, the two fragments must move in opposite directions with speeds v1 and v 2 such that m1 v1 = m 2 v 2

v2 =

or

FG m IJ v . Hm K 1

1

2

The kinetic energies after the break-up are then K 1 = 1 m1 v12 2

*P45.41

FG IJ H K

m1 K 2 = 1 m 2 v 22 = 1 m 2 2 2 m2

and

2

v12 =

FG m IJ K . Hm K 1

2

1

The fraction of the total kinetic energy carried off by m1 is

K1 K1 m2 = = K 1 + K 2 K 1 + m1 m 2 K 1 m1 + m 2

and the fraction carried off by m 2 is

1−

(a)

b

g

m2 m1 = . m1 + m 2 m1 + m 2

Q = 236.045 562u c 2 − 86.920 711u c 2 − 148.934 370 u c 2 = 0.190 481u c 2 = 177 MeV Immediately after fission, this Q-value is the total kinetic energy of the fission products.

(b)

FG m IJ Q , from Problem 45.40. Hm +m K F 149 u IJ a177.4 MeVf = 112 MeV =G H 87 u + 149 u K

K Br =

La

Br

La

K La = Q − K Br = 177.4 MeV − 112.0 MeV = 65.4 MeV

(c)

v Br =

2K Br = m Br

v La =

2K La = m La

e

je

2 112 × 10 6 eV 1.6 × 10 −19 J eV

j=

7

1.58 × 10 m s a87 ufe1.66 × 10 kg uj J eV j 2e65.4 × 10 eV je1.6 × 10 a149 ufe1.66 × 10 kg uj = 9.20 × 10 m s −27

−19

6

6

−27

Chapter 45

P45.42

For a typical

235

609

U , Q = 208 MeV ; and the initial mass is 235 u. Thus, the fractional energy loss is Q 208 MeV = = 9.50 × 10 − 4 = 0.095 0% . 235 u 931.5 MeV u mc 2

a

fb

g

For the D-T fusion reaction,

Q = 17.6 MeV .

The initial mass is

m = 2.014 u + 3.016 u = 5.03 u .

The fractional loss in this reaction is

Q 17.6 MeV = = 3.75 × 10 − 3 = 0.375% 2 5.03 u 931.5 MeV u mc

a

f a

a

f

fb

g

0.375% = 3.95 or the fractional loss in D - T is about 4 times that in 0.0950% P45.43

The decay constant is λ =

235

U fission .

ln 2 ln 2 = = 1.78 × 10 −9 s −1 . 7 T1 2 12.3 yr 3.16 × 10 s yr

b

ge

j

The tritium in the plasma decays at a rate of

e

L O j MMFGH 2.00cm× 10 IJK FGH 101 mcm IJK e50.0 m jPP N Q F 1 Ci I = 482 Ci Bq = e1.78 × 10 Bq jG H 3.70 × 10 Bq JK 14

R = λ N = 1.78 × 10 − 9 s −1 R = 1.78 × 10 13

3

3

3

13

The fission inventory is P45.44

6

3

10

4 × 10 10 Ci ~ 10 8 times greater than this amount. 482 Ci

Momentum conservation: 0 = m Li v Li + mα v α , or, m Li v Li = mα vα . K Li

Thus,

g = bm v g = F m I v GH 2m JK 2m F b4.002 6 ug I F I F I =G GH 2 b7.016 0 ug JJK H 9.25 × 10 m sK = a1.14 ufH 9.25 × 10 m sK = 1.14 e1.66 × 10 kg jFH 9.25 × 10 m sIK = 1.62 × 10 J = 1.01 MeV b

1 1 m Li v Li 2 = m Li v Li = 2 2 m Li

2

K Li K Li

2

α α

2 α

Li

2

P45.45

.

Li

2 α

2

6

− 27

2

6

2

6

−13

The complete fissioning of 1.00 gram of U 235 releases Q=

b1.00 g g e6.02 × 10 235 grams mol

23

jb

ge

j

atoms mol 200 MeV fission 1.60 × 10 −13 J MeV = 8.20 × 10 10 J .

If all this energy could be utilized to convert m kilograms of 20.0°C water to 400°C steam (see Chapter 20 of text for values), then

Q = mc w ∆T + mL v + mc s ∆T

b

ga

f

b

ga

f

Q = m 4186 J kg ° C 80.0 ° C + 2.26 × 10 6 J kg + 2010 J kg ° C 300 ° C . Therefore

m=

10

8.20 × 10 J = 2.56 × 10 4 kg . 3.20 × 10 6 J kg

.

610 P45.46

Applications of Nuclear Physics

When mass m of 235 U undergoes complete fission, releasing 200 MeV per fission event, the total energy released is: Q=

F m I N a200 MeVf where N GH 235 g mol JK A

A

is Avogadro’s number.

If all this energy could be utilized to convert a mass m w of liquid water at Tc into steam at Th , then,

b

g

b

Q = m w c w 100° C − Tc + L v + c s Th − 100° C

g

where c w is the specific heat of liquid water, L v is the latent heat of vaporization, and c s is the specific heat of steam. Solving for the mass of water converted gives mw =

P45.47

(a)

b

Q

g

b

c w 100° C − Tc + L v + c s Th − 100° C

a

mN A 200 MeV

=

g b235 g molg c b100° C − T g + L w

c

f

v

b

+ c s Th − 100° C

g

.

The number of molecules in 1.00 liter of water (mass = 1 000 g) is N=

F 1.00 × 10 g I 6.02 × 10 GH 18.0 g mol JK e 3

23

j

molecules mol = 3.34 × 10 25 molecules .

The number of deuterium nuclei contained in these molecules is

e

N ′ = 3.34 × 10 25 molecules

1 deuteron I J = 1.01 × 10 j FGH 3300 molecules K

22

deuterons .

Since 2 deuterons are consumed per fusion event, the number of events possible is N′ = 5.07 × 10 21 reactions, and the energy released is 2

e = e1.66 × 10

jb

g

Efusion = 5.07 × 10 21 reactions 3.27 MeV reaction = 1.66 × 10 22 MeV

Efusion (b)

22

je

j

MeV 1.60 × 10 −13 J MeV = 2.65 × 10 9 J .

In comparison to burning 1.00 liter of gasoline, the energy from the fusion of deuterium is Efusion 2.65 × 10 9 J = == 78.0 times larger . Egasoline 3.40 × 10 7 J

P45.48

j a0.05 mf = 1.23 × 10

e

2

(a)

∆V = 4π r 2 ∆ r = 4π 14.0 × 10 3 m

(b)

The force on the next layer is determined by atmospheric pressure.

e

je

8

m 3 ~ 10 8 m3

j

W = P∆V = 1.013 × 10 5 N / m 2 1.23 × 10 8 m3 = 1.25 × 10 13 J ~ 10 13 J

b

g

1 yield , so yield = 1.25 × 10 14 J ~ 10 14 J 10

(c)

1.25 × 10 13 J =

(d)

1.25 × 10 14 J = 2.97 × 10 4 ton TNT ~ 10 4 ton TNT 4.2 × 10 9 J ton TNT or ~ 10 kilotons

P45.49

a

The thermal power transferred to the water is Pw = 0.970 waste heat

(a)

b

g

f

Chapter 45

611

Pw = 0.970 3 065 − 1 000 MW = 2 .00 × 10 9 J s rw is the mass of water heated per hour: rw = The volume used per hour is

(b) P45.50

The

235

N0 =

λ=

210

F 155 g I e6.02 × 10 GH 209.98 g mol JK 210

The half-life of

4.91 × 10 8 kg h 1.00 × 10 3 kg m3

U fuel is consumed at a rate r f =

The number of nuclei in 0.155 kg of

23

jb ga

e a f b

g f

2 .00 × 10 9 J s 3600 s h Pw = = 4.91 × 10 8 kg h . c ∆T 4186 J kg ⋅° C 3.50 ° C = 4.91 × 10 5 m3 h .

F 3 065 × 10 GH 7.80 × 10

6

10

I F 1 kg I FG 3 600 s IJ = J JG J g K H 1 000 g K H 1 h K Js

0.141 kg h .

Po is

j

nuclei mol = 4.44 × 10 23 nuclei .

Po is 138.38 days, so the decay constant is given by

ln 2 ln 2 = = 5.80 × 10 −8 s −1 . T1 2 138.38 d 8.64 × 10 4 s d

a

fe

j

The initial activity is

e

je

j

R0 = λN 0 = 5.80 × 10 −8 s −1 4.44 × 10 23 nuclei = 2.58 × 10 16 Bq . The energy released in each

210 206 4 84 Po → 82 Pb + 2 He

reaction is

Q = M 210 Po − M 206 Pb − M 4 He c 2 : 84

82

2

b

g

Q = 209.982 857 − 205.974 449 − 4.002 603 u 931.5 MeV u = 5.41 MeV . Thus, assuming a conversion efficiency of 1.00%, the initial power output of the battery is

b

g

b

ge F mI =G J H ρK

jb

ge

j

P = 0.010 0 R0Q = 0.010 0 2.58 × 10 16 decays s 5.41 MeV decay 1.60 × 10 −13 J MeV = 223 W . P45.51

3

=

m

1 3

F 70.0 kg I =G H 18.7 × 10 kg m JK

13

= 0.155 m

(a)

V=

(b)

Add 92 electrons to both sides of the given nuclear reaction. Then it becomes 238 4 206 92 U atom → 8 2 He atom + 82 Pb atom + Q net .

ρ

, so

3

3

b

g

b

Q net = M 238 U − 8 M 4 He − M 206 Pb c 2 = 238.050 783 − 8 4.002 603 − 205.974 449 u 931.5 MeV u 92

2

82

Q net = 51.7 MeV (c)

If there is a single step of decay, the number of decays per time is the decay rate R and the energy released in each decay is Q. Then the energy released per time is P = QR . If there is a series of decays in steady state, the equation is still true, with Q representing the net decay energy.

continued on next page

g

612

Applications of Nuclear Physics

(d)

The decay rate for all steps in the radioactive series in steady state is set by the parent uranium: N=

F 7.00 × 10 g I 6.02 × 10 GH 238 g mol JK e

λ=

ln 2 ln 2 1 = = 1.55 × 10 −10 9 T1 2 4.47 × 10 yr yr

4

F GH

j

nuclei mol = 1.77 × 10 26 nuclei

Ie JK F P = QR = a51.7 MeV fG 2 .75 × 10 H

R = λ N = 1.55 × 10 − 10 so

23

1 1.77 × 10 26 nuclei = 2 .75 × 10 16 decays yr , yr

j

Ie JK

1 1.60 × 10 −13 J MeV = 2.27 × 10 5 J yr . yr

16

dose in rem = dose in rad × RBE

(e)

b

g

j

b

g

5.00 rem yr = dose in rad yr 1.10 , giving dose in rad yr = 4.55 rad yr

b

gb

The allowed whole-body dose is then 70.0 kg 4.55 rad yr

P45.52

a

f

ET ≡ E thermal = ET =

FG 1 IJ H 2K

n

3 k BT = 0.039 eV 2

E where n ≡ number of collisions, and 0.039 =

gFGH 10 1 radJ kg IJK = −2

3.18 J yr .

FG 1 IJ e2.0 × 10 j . H 2K n

6

Therefore, n = 25.6 = 26 collisions . P45.53

Conservation of linear momentum and energy can be applied to find the kinetic energy of the neutron. We first suppose the particles are moving nonrelativistically. The momentum of the alpha particle and that of the neutron must add to zero, so their velocities must be in opposite directions with magnitudes related by m n v n + mα v α = 0

b1.008 7 ugv = b4.002 6 ugv

or

n

α.

At the same time, their kinetic energies must add to 17.6 MeV E=

b

g

b

g

1 1 1 1 mn vn2 + mα vα2 = 1.008 7 u vn2 + 4.002 6 vα2 = 17.6 MeV . 2 2 2 2

Substitute vα = 0.252 0 vn :

b

g b

g

E = 0.504 35 u vn2 + 0.127 10 u vn2 = 17.6 MeV vn =

F I u GH 931.4941 MeV J c K 2

0.018 9 c 2 = 0.173 c = 5.19 × 10 7 m s. 0.631 45

Since this speed is not too much greater than 0.1c, we can get a reasonable estimate of the kinetic energy of the neutron from the classical equation, K=

continued on next page

b

ga

1 1 mv 2 = 1.008 7 u 0.173 c 2 2

f FGH 931.494 uMeV c IJK = 14.1 MeV . 2

2

Chapter 45

For a more accurate calculation of the kinetic energy, we should use relativistic expressions. Conservation of momentum gives

γ n m n v n + γ α mα v α = 0

vα2

yielding

P45.54

vn

1.008 7

c

=

2

1−

vn2 2

c vn2

15.746 c − 14.746 vn2

.

bγ

and

vn = 0.171 c , implying that γ n − 1 mn c 2 = 14.0 MeV .

From Table A.3, the half-life of

32

λ=

b

1 − vα2 c 2

Then

n

g

vα

= 4.002 6

2

g

− 1 mn c 2 + γ α − 1 mα c 2 = 17.6 MeV

b

g

P is 14.26 d. Thus, the decay constant is

ln 2 ln 2 = = 0.048 6 d −1 = 5.63 × 10 −7 s −1 . T1 2 14. 26 d

N0 =

R0

λ

=

5.22 × 10 6 decay s 5.63 × 10 −7 s −1

= 9.28 × 10 12 nuclei

At t = 10.0 days , the number remaining is

e

j

N = N 0 e − λ t = 9.28 × 10 12 nuclei e

jb

e

− 0 .048 6 d −1 10 . 0 d

g = 5.71 × 10 12 nuclei

so the number of decays has been N 0 − N = 3.57 × 10 12 and the energy released is

ja

e

fe

j

E = 3.57 × 10 12 700 keV 1.60 × 10 −16 J keV = 0.400 J . If this energy is absorbed by 100 g of tissue, the absorbed dose is Dose =

P45.55

(a)

F 0.400 J I F 1 rad I = GH 0.100 kg JK GH 10 J kg JK −2

400 rad .

The number of Pu nuclei in 1.00 kg =

6.02 × 10 23 nuclei mol 1 000 g . 239.05 g mol

b

ja

e

g

f

The total energy = 25.2 × 10 23 nuclei 200 MeV = 5.04 × 10 26 MeV

e

je

j

E = 5.04 × 10 26 MeV 4.44 × 10 −20 kWh MeV = 2.24 × 10 7 kWh or (b)

22 million kWh.

b

gb

E = ∆m c 2 = 3.016 049 u + 2.014 102 u − 4.002 603 u − 1.008 665 u 931.5 MeV u E = 17.6 MeV for each D - T fusion

(c)

a

fa fe

En = Total number of D nuclei 17.6 4.44 × 10 −20

e

En = 6.02 × 10 23 continued on next page

IJ a17.6fe4.44 × 10 j = jFGH 12.000 014 K −20

j

2.34 × 10 8 kWh

g

613

614

Applications of Nuclear Physics

En = the number of C atoms in 1.00 kg × 4.20 eV

(d)

En = (e)

P45.56

F 6.02 × 10 I e4.20 × 10 GH 12 JK 26

−6

je

j

MeV 4.44 × 10 −20 = 9.36 kWh

Coal is cheap at this moment in human history. We hope that safety and waste disposal problems can be solved so that nuclear energy can be affordable before scarcity drives up the price of fossil fuels.

Add two electrons to both sides of the given reaction. Then 4 11 H atom→ 24 He atom + Q

a f b Q = a 26.7 MeV fe1.60 × 10

or

g

b

g

Q = ∆m c 2 = 4 1.007 825 − 4.002603 u 931.5 MeV u = 26.7 MeV

where

−13

j

J MeV = 4.28 × 10 −12 J .

The proton fusion rate is then rate =

P45.57

(a)

power output 3.77 × 10 26 J s = = 3.53 × 10 38 protons s . energy per proton 4. 28 × 10 −12 J 4 protons

jb

e

QI = M A + M B − M C − M E c 2 , and

g

QII = M C + M D − M F − M G c 2

Q net = QI + QII = M A + M B − M C − M E + M C + M D − M F − M G c 2 Q net = Q I + QII = M A + M B + M D − M E − M F − M G c 2 Thus, reactions may be added. Any product like C used in a subsequent reaction does not contribute to the energy balance. (b)

Adding all five reactions gives 1 1H

or

+ 11 H+ −01 e+ 11 H+ 11 H+

4 11 H + 2

0 4 −1 e → 2 He +

2ν + Q net

2ν + Q net .

Adding two electrons to each side Thus,

0 4 −1 e → 2 He +

4 11 H atom → 24 He atom + Q net .

b

g

b

g

Q net = 4M 1 H − M 4 He c 2 = 4 1.007 825 − 4.002 603 u 931.5 MeV u = 26.7 MeV . 1

2

Chapter 45

P45.58

(a)

L 4π F 1.50 × 10 The mass of the pellet is m = ρV = e0.200 g cm jM G 2 MN 3 H 3

−2

cm

I JK

3

OP PQ = 3.53 × 10

−7

615

g.

The pellet consists of equal numbers of 2 H and 3 H atoms, so the average molar mass is 2.50 and the total number of atoms is N=

F 3.53 × 10 g I 6.02 × 10 GH 2 .50 g mol JK e −7

23

j

atoms mol = 8.51 × 10 16 atoms .

When the pellet is vaporized, the plasma will consist of 2Nparticles (N nuclei and N electrons). The total energy delivered to the plasma is 1.00% of 200 kJ or 2.00 kJ. The temperature of the plasma is found from E = 2 N 3 k B T as 2

a fe

T= (b)

P45.59

(a)

2 .00 × 10 3 J E = 5.68 × 10 8 K . = 3 Nk B 3 8.51 × 10 16 1.38 × 10 − 23 J K

e

je

j

Each fusion event uses 2 nuclei, so E=

j

FG N IJ Q = FG 8.51 × 10 H 2K H 2

16

N events will occur. The energy released will be 2

I a17.59 MeVfe1.60 × 10 JK

−13

j

J MeV = 1.20 × 10 5 J = 120 kJ .

The solar-core temperature of 15 MK gives particles enough kinetic energy to overcome the k e 2e . The Coulomb-repulsion barrier to 11 H + 32 He → 24 He + e + + ν , estimated as e r k e 7e 7 , larger by times, so Coulomb barrier to Bethe’s fifth and eight reactions is like e 2 r 7 6 7 the required temperature can be estimated as 15 × 10 K ≈ 5 × 10 K . 2

a fa f

a fa f

e

(b)

For 12 C + 1 H →

13

j

N + Q,

b

f

ga

Q1 = 12.000 000 + 1.007 825 − 13.005 739 931.5 MeV = 1.94 MeV For the second step, add seven electrons to both sides to have: 13 N atom → 13 C atom + e + + e − + Q

ga

b

f

Q 2 = 13.005 739 − 13.003 355 − 2 0.000 549 931.5 MeV = 1.20 MeV

b

ga

f

Q3 = Q7 = 2 0.000 549 931.5 MeV = 1.02 MeV

a f = 14.003 074 + 1.007 825 − 15.003 065 a931.5 MeV f = 7.30 MeV = 15.003 065 − 15.000 109 − 2b0.000 549 g a931.5 MeV f = 1.73 MeV = 15.000 109 + 1.007 825 − 12 − 4.002 603 a931.5 MeV f = 4.97 MeV

Q 4 = 13.003 355 + 1.007 825 − 14.003 074 931.5 MeV = 7.55 MeV Q5 Q6 Q8

The sum is 26.7 MeV , the same as for the proton-proton cycle. (c)

Not all of the energy released appears as internal energy in the star. When a neutrino is created, it will likely fly directly out of the star without interacting with any other particle.

616 P45.60

Applications of Nuclear Physics

(a)

I 2 I0 e− µ 2x = = e −b µ 2 − µ 1 gx I1 I 0 e − µ 1 x

(b)

I 50 = e − a5. 40 − 41.0 fa0.100 f = e 3.56 = 35.2 I 100

(c)

I 50 = e − a5. 40 − 41.0 fa1.00 f = e 35.6 = 2.89 × 10 15 I 100 Thus, a 1.00-cm aluminum plate has essentially removed the long-wavelength x-rays from the beam.

*P45.61

(a)

The number of fissions ocurring in the zeroth, first, second, … nth generation is N0 , N0 K , N0 K 2 , … , N0 K n . The total number of fissions that have ocurred up to and including the nth generation is

e

j − 1 = a a + 1fa a − 1f , can be

N = N0 + N0K + N0K 2 +…+ N0K n = N0 1 + K + K 2 +…+ K n . 2

Note that the factoring of the difference of two squares, a generalized to a difference of two quantities to any power,

e

ja f

a3 − 1 = a2 + a + 1 a − 1

e

ja f

a n +1 − 1 = a n + a n − 1 + … + a 2 + a + 1 a − 1 . K n + K n −1 + … + K 2 + K + 1 =

Thus

N = N0

and (b)

K n +1 − 1 . K −1

The number of U-235 nuclei is N = 5.50 kg

K n +1 − 1 K −1

FG 1 atomIJ FG 1 u H 235 u K H 1.66 × 10

−27

I = 1.41 × 10 J kg K

25

nuclei .

We solve the equation from part (a) for n, the number of generations:

a

f

N K − 1 = K n +1 − 1 N0

a

f af F N aK − 1f N + 1 IJ = lnFG NaK − 1f + 1IJ − ln K n ln K = lnG K H K H N K lne1.41 × 10 a0.1f 10 + 1j lnc N aK − 1f N + 1h −1= − 1 = 99.2 n= N K − 1 + 1 = Kn K N0

0

0

25

0

20

ln 1.1

ln K

Therefore time must be alotted for 100 generations:

e

j

∆tb = 100 10 × 10 −9 s = 1.00 × 10 −6 s . continued on next page

617

Chapter 45

(c)

v=

(d)

V=

B

ρ

=

18.7 × 10

3

kg m

3

= 2.83 × 10 3 m s

4 3 m πr = ρ 3

F 3m IJ r =G H 4πρ K ∆t d = (e)

150 × 10 9 N m 2

13

F 3b5.5 kg g I =G GH 4π e18.7 × 10 kg m j JJK 3

13

= 4.13 × 10 −2 m

3

r 4.13 × 10 −2 m = = 1.46 × 10 −5 s v 2.83 × 10 3 m s

14.6 µs is greater than 1 µs , so the entire bomb can fission. The destructive energy released is 1.41 × 10 25 nuclei

F 200 × 10 eV I F 1.6 × 10 J I = 4.51 × 10 GH fissioning nucleus JK GH 1 eV JK −19

6

14

FG 1 ton TNT IJ H 4.2 × 10 J K

J = 4.51 × 10 14 J

9

= 1.07 × 10 5 ton TNT = 107 kilotons of TNT What if? If the bomb did not have an “initiator” to inject 10 20 neutrons at the moment when the critical mass is assembled, the number of generations would be n=

e

a f

ln 1. 41 × 10 25 0.1 1 + 1 ln 1.1

j − 1 = 582 requiring 583e10 × 10 sj = 5.83 µs . −9

This time is not very short compared with 14.6 µs , so this bomb would likely release much less energy.

ANSWERS TO EVEN PROBLEMS P45.2

184 MeV

P45.18

(a) 2.53 × 10 31 J ; (b) 1.14 × 10 9 yr

P45.4

see the solution

P45.20

P45.6

(a) 173 MeV ; (b) 0.078 8%

(a) 10 14 cm −3 ; (b) 1.24 × 10 5 J m3 ; (c) 1.77 T

P45.8

2.63 kg d

P45.22

12.4 h

P45.10

(a) 4.84V −1 3 ; (b) 6V −1 3 ; (c) 6.30V −1 3 ; (d) the sphere has minimum loss and the parallelepiped maximum

P45.24

(a) 10.0 h; (b) 3.16 m

P45.26

(a) 0.436 cm; (b) 5.79 cm

P45.28

2.39 × 10 −3 ° C

P45.30

3.96 × 10 −4 J kg

P45.32

(a) 10; (b) 10 6 ; (c) 10 8 eV

P45.34

4.45 × 10 −8 kg h

5

P45.12

2.68 × 10

P45.14

(a) 31.9 g h ; (b) 122 g h

P45.16

2 (a) 3.24 fm ; (b) 444 keV ; (c) vi ; 5 (d) 740 keV ; (e) possibly by tunneling

618

Applications of Nuclear Physics

P45.36

(a) ~ 10 6 ; (b) ~ 10 −15 g

P45.38

(a) 1.5 × 10 24 ; (b) 0.6 kg

P45.40

see the solution

P45.42

The fractional loss in D - T is about 4 times that in

P45.44 P45.46

235

U fission

1.01 MeV

a

mN A 200 MeV

f

°C − T g + L O P b235 g molgLMM+c c b100 N bT − 100° Cg PQ w

c

s

h

v

P45.48

(a) ~ 10 8 m 3 ; (b) ~ 10 13 J ; (c) ~ 10 14 J ; (d) ~ 10 kilotons

P45.50

223 W

P45.52

26 collisions

P45.54

400 rad

P45.56

3.53 × 10 38 protons s

P45.58

(a) 5.68 × 10 8 K ; (b) 120 kJ

P45.60

(a) see the solution; (b) 35.2 ; (c) 2.89 × 10 15

46 Particle Physics and Cosmology CHAPTER OUTLINE 46.1 46.2 46.3 46.4 46.5 46.6 46.7

46.8 46.9 46.10 46.11 46.12 46.13

The Fundamental Forces in Nature Positrons and Other Antiparticles Mesons and the Beginning of Particle Physics Classification of Particles Conservation Laws Strange Particles and Strangeness Making Elementary Particles and Measuring Their Properties Finding Patterns in the Particles Quarks Multicolored Quarks The Standard Model The Cosmic Connection Problems and Perspectives

ANSWERS TO QUESTIONS Q46.1

Strong Force—Mediated by gluons. Electromagnetic Force—Mediated by photons. Weak Force—Mediated by W + , W − , and Z 0 bosons. Gravitational Force—Mediated by gravitons.

Q46.2

The production of a single gamma ray could not satisfy the law of conservation of momentum, which must hold true in this—and every—interaction.

Q46.3

In the quark model, all hadrons are composed of smaller units called quarks. Quarks have a fractional electric charge and a 1 baryon number of . There are 6 types of quarks: up, down, 3 strange, charmed, top, and bottom. Further, all baryons contain 3 quarks, and all mesons contain one quark and one anti-quark. Leptons are thought to be fundamental particles.

Q46.4

Hadrons are massive particles with structure and size. There are two classes of hadron: mesons and baryons. Hadrons are composed of quarks. Hadrons interact via the strong force. Leptons are light particles with no structure or size. It is believed that leptons are fundamental particles. Leptons interact via the weak force.

Q46.5

Baryons are heavy hadrons with spin

Q46.6

Resonances are hadrons. They decay into strongly interacting particles such as protons, neutrons, and pions, all of which are hadrons.

Q46.7

The baryon number of a proton or neutron is one. Since baryon number is conserved, the baryon number of the kaon must be zero.

Q46.8

Decays by the weak interaction typically take 10 −10 s or longer to occur. This is slow in particle physics.

1 3 or , are composed of three quarks, and have long 2 2 lifetimes. Mesons are light hadrons with spin 0 or 1, are composed of a quark and an antiquark, and have short lifetimes.

619

620

Particle Physics and Cosmology

Q46.9

The decays of the muon, tau, charged pion, kaons, neutron, lambda, charged sigmas, xis, and omega occur by the weak interaction. All have lifetimes longer than 10 −13 s. Several produce neutrinos; none produce photons. Several violate strangeness conservation.

Q46.10

The decays of the neutral pion, eta, and neutral sigma occur by the electromagnetic interaction. These are three of the shortest lifetimes in Table 46.2. All produce photons, which are the quanta of the electromagnetic force. All conserve strangeness.

Q46.11

Yes, protons interact via the weak interaction; but the strong interaction predominates.

Q46.12

You can think of a conservation law as a superficial regularity which we happen to notice, as a person who does not know the rules of chess might observe that one player’s two bishops are always on squares of opposite colors. Alternatively, you can think of a conservation law as identifying some stuff of which the universe is made. In classical physics one can think of both matter and energy as fundamental constituents of the world. We buy and sell both of them. In classical physics you can also think of linear momentum, angular momentum, and electric charge as basic stuffs of which the universe is made. In relativity we learn that matter and energy are not conserved separately, but are both aspects of the conserved quantity relativistic total energy. Discovered more recently, four conservation laws appear equally general and thus equally fundamental: Conservation of baryon number, conservation of electron-lepton number, conservation of tau-lepton number, and conservation of muon-lepton number. Processes involving the strong force and the electromagnetic force follow conservation of strangeness, charm, bottomness, and topness, while the weak interaction can alter the total S, C, B and T quantum numbers of an isolated system.

Q46.13

No. Antibaryons have baryon number –1, mesons have baryon number 0, and baryons have baryon number +1. The reaction cannot occur because it would not conserve baryon number, unless so much energy is available that a baryon-antibaryon pair is produced.

Q46.14

The Standard Model consists of quantum chromodynamics (to describe the strong interaction) and the electroweak theory (to describe the electromagnetic and weak interactions). The Standard Model is our most comprehensive description of nature. It fails to unify the two theories it includes, and fails to include the gravitational force. It pictures matter as made of six quarks and six leptons, interacting by exchanging gluons, photons, and W and Z bosons.

Q46.15

All baryons and antibaryons consist of three quarks. All mesons and antimesons consist of two 1 quarks. Since quarks have spin quantum number and can be spin-up or spin-down, it follows that 2 the three-quark baryons must have a half-integer spin, while the two-quark mesons must have spin 0 or 1.

Q46.16

Each flavor of quark can have colors, designated as red, green and blue. Antiquarks are colored antired, antigreen, and antiblue. A baryon consists of three quarks, each having a different color. By analogy to additive color mixing we call it colorless. A meson consists of a quark of one color and antiquark with the corresponding anticolor, making it colorless as a whole.

Q46.17

In 1961 Gell-Mann predicted the omega-minus particle, with quark composition sss. Its discovery in 1964 confirmed the quark theory.

Chapter 46

Q46.18

621

1 , B = 1, L e = L µ = Lτ = 0, and strangeness –2. 2 All of these are described by its quark composition dss (Table 46.5). The properties of the quarks 1 1 1 1 1 1 1 from Table 46.3 let us add up charge: − e − e − e = − e; spin + − + = , supposing one of the 3 3 3 2 2 2 2 1 1 1 quarks is spin-down relative to the other two; baryon number + + = 1 ; lepton numbers, charm, 3 3 3 bottomness, and topness zero; and strangeness 0 − 1 − 1 = −2 . The Ξ − particle has, from Table 46.2, charge –e, spin

Q46.19

The electroweak theory of Glashow, Salam, and Weinberg predicted the W + , W − , and Z particles. Their discovery in 1983 confirmed the electroweak theory.

Q46.20

Hubble determined experimentally that all galaxies outside the Local Group are moving away from us, with speed directly proportional to the distance of the galaxy from us.

Q46.21

Before that time, the Universe was too hot for the electrons to remain in any sort of stable orbit around protons. The thermal motion of both protons and electrons was too rapid for them to be in close enough proximity for the Coulomb force to dominate.

Q46.22

The Universe is vast and could on its own terms get along very well without us. But as the cosmos is immense, life appears to be immensely scarce, and therefore precious. We must do our work, growing corn to feed the hungry while preserving our planet for future generations. One person has singular abilities and opportunities for effort, faithfulness, generosity, honor, curiosity, understanding, and wonder. His or her place is to use those abilities and opportunities, unique in all the Universe.

SOLUTIONS TO PROBLEMS Section 46.1

The Fundamental Forces in Nature

Section 46.2

Positrons and Other Antiparticles

P46.1

Assuming that the proton and antiproton are left nearly at rest after they are produced, the energy E of the photon must be

a

f

E = 2E0 = 2 938.3 MeV = 1 876.6 MeV = 3.00 × 10 −10 J . Thus,

E = hf = 3.00 × 10 −10 J f=

3.00 × 10 −10 J = 4.53 × 10 23 Hz 6.626 × 10 −34 J ⋅ s

λ=

c 3.00 × 10 8 m s = = 6.62 × 10 −16 m . f 4.53 × 10 23 Hz

622 P46.2

Particle Physics and Cosmology

The minimum energy is released, and hence the minimum frequency photons are produced, when the proton and antiproton are at rest when they annihilate. That is, E = E0 and K = 0 . To conserve momentum, each photon must carry away one-half the energy. Thus,

Emin =

2 E0 = E0 = 938.3 MeV = hf min . 2

Thus,

fmin =

a938.3 MeVfe1.60 × 10 J MeVj = e6.626 × 10 J ⋅ sj −13

λ= P46.3

c fmin

−34

=

3.00 × 10 8 m s 2.27 × 10 23 Hz

2.27 × 10 23 Hz

= 1.32 × 10 −15 m .

In γ → p + + p − , we start with energy

2.09 GeV

we end with energy

938.3 MeV + 938.3 MeV + 95.0 MeV + K 2

where K 2 is the kinetic energy of the second proton. Conservation of energy for the creation process gives

Section 46.3 P46.4

K 2 = 118 MeV .

Mesons and the Beginning of Particle Physics

µ + + e− → ν + ν

The reaction is muon-lepton number before reaction: electron-lepton number before reaction:

a−1f + a0f = −1 a0f + a1f = 1 .

Therefore, after the reaction, the muon-lepton number must be –1. Thus, one of the neutrinos must be the anti-neutrino associated with muons, and one of the neutrinos must be the neutrino associated with electrons:

νµ

νe .

µ + + e− → ν µ + ν e .

Then P46.5

and

The creation of a virtual Z 0 boson is an energy fluctuation ∆E = 93 × 10 9 eV . It can last no longer = and move no farther than than ∆t = 2 ∆E

a f

c ∆t =

e

je

6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s hc = 4π ∆E 4π 93 × 10 9 eV

e

j

j F 1 eV I = 1.06 × 10 GH 1.60 × 10 J JK −19

−18

m = ~ 10 −18 m .

623

Chapter 46

P46.6

A proton has rest energy 938.3 MeV. The time interval during which a virtual proton could exist is at = most ∆t in ∆E∆t = . The distance it could move is at most 2 c∆ t =

e

je

j

1.055 × 10 −34 J ⋅ s 3 × 10 8 m s =c = ~ 10 −16 m . 2 ∆E 2 938.3 1.6 × 10 −13 J

a

fe

j

According to Yukawa’s line of reasoning, this distance is the range of a force that could be associated with the exchange of virtual protons between high-energy particles. P46.7

By Table 46.2,

Mπ 0 = 135 MeV c 2 .

Therefore,

Eγ = 67.5 MeV for each photon p= f=

and P46.8

c Eγ h

= 67.5 MeV c = 1.63 × 10 22 Hz .

The time interval for a particle traveling with the speed of light to travel a distance of 3 × 10 −15 m is ∆t =

P46.9

Eγ

(a)

d 3 × 10 −15 m = = ~ 10 −23 s . v 3 × 10 8 m s

e

j

∆E = m n − m p − m e c 2 From Table A-3,

(b)

b

f

ga

∆E = 1.008 665 − 1.007 825 931.5 = 0.782 MeV .

Assuming the neutron at rest, momentum conservation for the decay process implies p p = p e . Relativistic energy for the system is conserved

em c j a938.3f p

Since p p = p e , Solving the algebra, If p e c = γ m e v e c = 1.19 MeV , then Solving,

2 2

2

em c j + p c = m c . + a0.511f + ( p c ) = 939.6 MeV .

+ p p2 c 2 + + ( p c) 2

e

2 2

2 2 e

2

n

2

2

pc = 1.19 MeV .

γ v e 1.19 MeV v x = = = 2.33 where x = e . 2 0.511 MeV c c 1− x v x 2 = 1 − x 2 5. 43 and x = e = 0.919 c

e

j

v e = 0.919 c . Then m p v p = γ e m e v e :

a

fe

−13 J MeV γ e m e v e c 1.19 MeV 1.60 × 10 = vp = − 27 8 mp c 1.67 × 10 3.00 × 10 m s

e

je

v p = 3.80 × 10 5 m s = 380 km s . (c)

The electron is relativistic, the proton is not.

j

j

624

Particle Physics and Cosmology

Section 46.4 P46.10

Classification of Particles

In ? + p + → n + µ + , charge conservation requires the unknown particle to be neutral. Baryon number conservation requires baryon number = 0. The muon-lepton number of ? must be –1. So the unknown particle must be ν µ .

P46.11

Ω + → Λ0 + K + Λ0 → p + π +

Section 46.5 P46.12

P46.13

K S0 → π + + π −

(or π 0 + π 0 )

n → p + e+ + ν e

Conservation Laws p + p → µ + + e−

Le

0+0→ 0+1

and

Lµ

0 + 0 → −1 + 0

(b)

π− +p→ p+π+

charge

−1 + 1 → +1 + 1

(c)

p+p→ p+π+

baryon number :

1+1→1+0

(d)

p+p→p+p+n

baryon number :

1+1→1+1+1

(e)

γ +p→ n+π0

charge

0+1→ 0+0

(a)

Baryon number and charge are conserved, with values of

0 +1 = 0 +1

and

1 + 1 = 1 + 1 in both reactions.

(a)

(b)

Strangeness is not conserved in the second reaction.

P46.14

Baryon number conservation allows the first and forbids the second .

P46.15

(a)

π − → µ− + νµ

Lµ :

0→1−1

(b)

K+ → µ+ + ν µ

Lµ :

0 → −1 + 1

(c)

ν e + p + → n + e+

Le :

−1 + 0 → 0 − 1

(d)

ν e + n → p + + e−

Le :

1+0→ 0+1

(e)

ν µ + n → p+ + µ −

Lµ :

1+0→ 0+1

(f)

µ − → e− + ν e + ν µ

Lµ :

1→ 0+0+1

and

Le :

0→1−1+0

Chapter 46

P46.16

Momentum conservation for the decay requires the pions to have equal speeds. The total energy of each is

497.7 MeV 2

e j gives a248.8 MeVf = bpcg + a139.6 MeVf . mc FG v IJ pc = 206 MeV = γ mvc = H cK 1 − bv cg pc 206 MeV 1 FG v IJ = 1.48 = = H cK 139.6 MeV mc 1 − bv cg E 2 = p 2 c 2 + mc 2

so

2

2

2

2

2

Solving,

2

2

2

FG IJ H K FG v IJ = 2.18LM1 − FG v IJ OP = 2.18 − 2.18FG v IJ H cK H cK MN H c K PQ F vI 3.18G J = 2.18 H cK v v = 1.48 1 − c c

2

2

and

2

2

2

v = c

so

v = 0.828 c .

and P46.17

2.18 = 0.828 3.18

(a)

p+ → π + + π 0

(b)

p+ + p+ → p+ + p+ + π 0

(c)

p+ + p+ → p+ + π +

(d)

π + → µ+ +ν µ

This reaction can occur .

(e)

n0 → p+ + e− + ν e

This reaction can occur .

(f)

π + → µ+ + n

Violates baryon number :

0→ 0+1

Violates muon - lepton number :

0 → −1 + 0

Baryon number :

1→0+0

This reaction can occur . Baryon number is violated:

1+1→1+0

625

626 P46.18

Particle Physics and Cosmology

(a)

p → e+ + γ +1 → 0 + 0

Baryon number: ∆B ≠ 0 , so baryon number conservation is violated. (b)

(c)

P46.19

From conservation of momentum for the decay:

p e = pγ .

Then, for the positron,

Ee2 = p e c

2

2 0, e

becomes

Ee2

2

2 0, e

From conservation of energy for the system:

E0, p = Ee + Eγ

or

Ee = E0, p − Eγ

so

Ee2 = E02, p − 2E0 , p Eγ + Eγ2 .

Equating this to the result from above gives

Eγ2 + E02, e = E02, p − 2E0 , p Eγ + Eγ2

E02, p − E02, e

b g +E = d p ci + E γ

a938.3 MeVf − a0.511 MeVf 2a938.3 MeV f 2

2

or

Eγ =

Thus,

Ee = E0 , p − Eγ = 938.3 MeV − 469 MeV = 469 MeV .

Also,

pγ =

and

p e = pγ =

2 E0 , p

Eγ c

=

= Eγ2 + E02, e .

469 MeV c

=

469 MeV . c

The total energy of the positron is

Ee = 469 MeV .

But,

E e = γ E0 , e =

FG v IJ H cK

2

=

E0 , e

b g

1− v c E0 , e

so

1−

which yields:

v = 0.999 999 4c .

The relevant conservation laws are:

= 469 MeV .

Ee

=

2

0.511 MeV = 1.09 × 10 −3 469 MeV

∆L e = 0 ∆L µ = 0 ∆Lτ = 0 .

and (a)

π + → π 0 + e+ + ?

Le : 0 → 0 − 1 + Le

implies L e = 1

and we have a ν e

(b)

?+ p → µ − + p + π +

L µ : L µ + 0 → +1 + 0 + 0

implies L µ = 1

and we have a ν µ

continued on next page

Chapter 46

(c)

Λ0 → ρ + µ − + ?

Lµ : 0 → 0 + 1 + Lµ

implies L µ = −1

and we have a ν µ

(d)

τ + → µ + + ?+ ?

L µ : 0 → −1 + L µ

implies L µ = 1

and we have a ν µ

Lτ : −1 → 0 + Lτ

implies L τ = −1

and we have a ν τ

L µ = 1 for one particle, and Lτ = −1 for the other particle.

Conclusion for (d):

Section 46.6 P46.20

627

We have

νµ

and

ντ .

Strange Particles and Strangeness

The ρ 0 → π + + π − decay must occur via the strong interaction. The K S0 → π + + π − decay must occur via the weak interaction.

P46.21

P46.22

(a)

Λ0 → p + π −

Strangeness: −1 → 0 + 0

(strangeness is not conserved )

(b)

π − + p → Λ0 + K 0

Strangeness: 0 + 0 → −1 + 1

(0 = 0 and strangeness is conserved )

(c)

p + p → Λ0 + Λ0

Strangeness: 0 + 0 → +1 − 1

(0 = 0 and strangeness is conserved )

(d)

π − + p → π − + Σ+

Strangeness: 0 + 0 → 0 − 1

(0 ≠ −1 : strangeness is not conserved )

(e)

Ξ − → Λ0 + π −

Strangeness: −2 → −1 + 0

(−2 ≠ −1 so strangeness is not conserved )

(f)

Ξ0 → p + π −

Strangeness: −2 → 0 + 0

(−2 ≠ 0 so strangeness is not conserved )

(a)

µ − → e− + γ

Le :

0 → 1 + 0,

and

Lµ :

1→0

(b)

n → p + e− + ν e

Le :

0→ 0+1+1

(c)

Λ0 → p + π 0

Strangeness:

−1 → 0 + 0 ,

and

charge:

0 → +1 + 0

(d)

p → e+ + π 0

Baryon number:

+1 → 0 + 0

(e)

Ξ0 → n + π 0

Strangeness:

−2 → 0 + 0

628 P46.23

Particle Physics and Cosmology

(a)

π − + p → 2η violates conservation of baryon number as 0 + 1 → 0 , not allowed .

(b)

K − + n → Λ0 + π − Baryon number,

0+1→1+0

Charge,

−1 + 0 → 0 − 1

Strangeness,

−1 + 0 → −1 + 0

Lepton number,

0→0

The interaction may occur via the strong interaction since all are conserved. (c)

K− → π − + π 0 Strangeness,

−1 → 0 + 0

Baryon number,

0→0

Lepton number,

0→0

Charge,

−1 → −1 + 0

Strangeness is violated by one unit, but everything else is conserved. Thus, the reaction can occur via the weak interaction , but not the strong or electromagnetic interaction. (d)

Ω− → Ξ− + π 0 Baryon number,

1→1+0

Lepton number,

0→0

Charge,

−1 → −1 + 0

Strangeness,

−3 → −2 + 0

May occur by weak interaction , but not by strong or electromagnetic. (e)

η → 2γ Baryon number,

0→0

Lepton number,

0→0

Charge,

0→0

Strangeness,

0→0

No conservation laws are violated, but photons are the mediators of the electromagnetic interaction. Also, the lifetime of the η is consistent with the electromagnetic interaction .

Chapter 46

P46.24

(a)

Ξ − → Λ0 + µ − + ν µ Baryon number:

+1 → +1 + 0 + 0

Charge:

−1 → 0 − 1 + 0

Le :

0→0+0+0

Lµ :

0→ 0+1+1

Lτ :

0→0+0+0

Strangeness:

−2 → −1 + 0 + 0 B , charge, L e , and Lτ

Conserved quantities are: (b)

K S0 → 2π 0 Baryon number:

0→0

Charge:

0→0

Le :

0→0

Lµ :

0→0

Lτ :

0→0

Strangeness:

+1 → 0

Conserved quantities are: (c)

B , charge, L e , L µ , and Lτ

K − + p → Σ0 + n Baryon number:

0+1→1+1

Charge:

−1 + 1 → 0 + 0

Le :

0+0→0+0

Lµ :

0+0→0+0

Lτ :

0+0→0+0

Strangeness:

−1 + 0 → −1 + 0 S , charge, L e , L µ , and Lτ

Conserved quantities are: (d)

Σ 0 + Λ0 + γ Baryon number:

+1 → 1 + 0

Charge:

0→0

Le :

0→0+0

Lµ :

0→0+0

Lτ :

0→0+0

Strangeness:

−1 → −1 + 0 B , S , charge, L e , L µ , and Lτ

Conserved quantities are: (e)

e+ + e− → µ + + µ − Baryon number:

0+0→0+0

Charge:

+1 − 1 → +1 − 1

Le :

−1 + 1 → 0 + 0

Lµ :

0 + 0 → +1 − 1

Lτ :

0+0→0+0

Strangeness:

0+0→0+0

Conserved quantities are: (f)

629

B , S , charge, L e , L µ , and Lτ

p + n → Λ0 + Σ − Baryon number:

−1 + 1 → −1 + 1

Charge:

−1 + 0 → 0 − 1

Le :

0+0→0+0

Lµ :

0+0→0+0

Lτ :

0+0→0+0

Strangeness:

0 + 0 → +1 − 1

Conserved quantities are:

B , S , charge, L e , L µ , and Lτ

630 P46.25

Particle Physics and Cosmology

(a)

K+ + p → ? + p The strong interaction conserves everything. Baryon number,

0+1→B+1

so

B=0

Charge,

+1 + 1 → Q + 1

so

Q = +1

Lepton numbers,

0+0→L+0

so

L e = L µ = Lτ = 0

Strangeness,

+1 + 0 → S + 0

so

S=1

The conclusion is that the particle must be positively charged, a non-baryon, with strangeness of +1. Of particles in Table 46.2, it can only be the K + . Thus, this is an elastic scattering process. The weak interaction conserves all but strangeness, and ∆S = ±1. (b)

Ω− → ? + π − Baryon number,

+1 → B + 0

so

B=1

Charge,

−1 → Q − 1

so

Q=0

Lepton numbers,

0→L+0

so

L e = L µ = Lτ = 0

Strangeness,

−3 → S + 0

so

∆S = 1: S = −2

The particle must be a neutral baryon with strangeness of –2. Thus, it is the Ξ 0 . (c)

K+ → ? + µ+ +ν µ Baryon number,

0→B+0+0

so

B=0

Charge,

+1 → Q + 1 + 0

so

Q=0

Lepton numbers,

Le , 0 → Le + 0 + 0

so

Le = 0

Lµ , 0 → Lµ − 1 + 1

so

Lµ = 0

Lτ , 0 → Lτ + 0 + 0

so

Lτ = 0

1→S+0+0

so

∆S = ±1

Strangeness,

(for weak interaction): S = 0 The particle must be a neutral meson with strangeness = 0 ⇒ π 0 .

631

Chapter 46

Section 46.7 *P46.26

(a)

Making Elementary Particles and Measuring Their Properties p Σ + = eBrΣ +

e1.602 177 × 10 = 5.344 288 × 10

(b)

−22

e1.602 177 × 10

pπ + = eBrπ + =

ja

fa f = 686 MeV c bkg ⋅ m sg bMeV cg C ja1.15 Tfa0.580 mf 200 MeV = c bkg ⋅ m sg bMeV cg

−19

−19

5.344 288 × 10 −22

C 1.15 T 1.99 m

Let ϕ be the angle made by the neutron’s path with the path of the Σ + at the moment of decay. By conservation of momentum:

b

g

pn cos ϕ + 199.961 581 MeV c cos 64.5° = 686.075 081 MeV c ∴ pn cos ϕ = 599.989 401 MeV c

b

(1)

g b599.989 401 MeV cg + b180.482 380 MeV cg

pn sin ϕ = 199.961 581 MeV c sin 64.5° = 180.482 380 MeV c

(c)

Eπ + = En =

e p c j + em c j

2 2

2

π+

π+

b p c g + em c j 2

n

2

pn =

From (1) and (2):

n

2 2

=

=

b199.961 581 MeVg + a139.6 MeVf 2

b626.547 022 MeVg + a939.6 MeVf 2

2

2

(2) 2

= 627 MeV c

= 244 MeV

= 1 130 MeV

EΣ + = Eπ + + En = 243.870 445 MeV + 1 129.340 219 MeV = 1 370 MeV (d)

d i

m Σ + c 2 = EΣ2 + − p Σ + c

2

b1 373.210 664 MeVg − b686.075 081 MeVg 2

=

∴ m Σ + = 1 190 MeV c 2

F GH

EΣ + = γ m Σ + c 2 , where γ = 1 −

v2 c2

I JK

−1 2

=

1 373.210 664 MeV = 1.154 4 1 189.541 303 MeV

Solving for v, v = 0.500 c . P46.27

Time-dilated lifetime: T = γ T0 =

0.900 × 10 −10 s 1−v

e

2

c

2

=

0.900 × 10 −10 s 1 − ( 0.960)

je

2

= 3.214 × 10 −10 s

j

distance = 0.960 3.00 × 10 8 m s 3.214 × 10 −10 s = 9.26 cm .

2

= 1 190 MeV

632 *P46.28

Particle Physics and Cosmology

(a)

Let Emin be the minimum total energy of the bombarding particle that is needed to induce the reaction. At this energy the product particles all move with the same velocity. The product particles are then equivalent to a single particle having mass equal to the total mass of the product particles, moving with the same velocity as each product particle. By conservation of energy:

em c j + b p c g

Emin + m 2 c 2 =

2 2

3

2

.

3

p 3 = p1

By conservation of momentum:

b g = bp cg 2

e

2

1

2 = Emin − m1 c 2

Substitute (2) in (1):

Emin + m 2 c 2 =

∴ p3 c

(1)

j. 2

(2)

em c j 3

2 2

e

2 + Emin − m1 c 2

j

2

.

Square both sides:

e

2 + 2Emin m 2 c 2 + m 2 c 2 Emin

∴ Emin

em =

2 3

j

j = em c j 2

2 2

3

e

2 + Emin − m1 c 2

j

2

− m12 − m 22 c 2 2m 2

∴ K min = Emin − m1 c

2

em =

2 3

j

− m12 − m 22 − 2m1 m 2 c 2 2m 2

=

b

m 32 − m1 + m 2

g

2

2m 2

Refer to Table 46.2 for the particle masses. (b)

K min =

a

4 938.3

f

2

e

2 938.3 MeV c 2

b497.7 + 1 115.6g =

(c)

K min

(d)

K min =

a

MeV 2 c 2 − 2 938.3

a

f

2 938.3 + 135

2

2

K min

2

MeV 2 c 2

a

MeV 2 c 2 − 139.6 + 938.3

a

f

f

2

= 5.63 GeV

MeV 2 c 2

2 938.3 MeV c 2

a

MeV 2 c 2 − 2 938.3

a

f

f

2

MeV 2 c 2

2 938.3 MeV c 2

LMe91.2 × 10 j − a938.3 + 938.3f =N 2a938.3f MeV c 3 2

(e)

j

f

2

2

MeV 2 c 2

OP Q=

= 768 MeV

= 280 MeV

4.43 TeV

c2

633

Chapter 46

Section 46.8

Finding Patterns in the Particles

Section 46.9

Quarks

Section 46.10

Multicolored Quarks

Section 46.11

The Standard Model

P46.29

(a)

The number of protons

F 6.02 × 10 molecules I FG 10 protons IJ = 3.34 × 10 JK H molecule K GH 18.0 g F 6.02 × 10 molecules I FG 8 neutrons IJ = 2.68 × 10 = 1 000 g G JK H molecule K 18.0 g H 23

N p = 1 000 g and there are

Nn

23

(b)

26

neutrons .

e j 2 e 2.68 × 10 j + 3.34 × 10

2 3.34 × 10 26 + 2.68 × 10 26 = 9.36 × 10 26 up quarks 26

and there are

protons

3.34 × 10 26 electrons .

So there are for electric neutrality The up quarks have number

26

26

= 8.70 × 10 26 down quarks .

Model yourself as 65 kg of water. Then you contain:

e j ~ 10 65 e9.36 × 10 j ~ 10 65 e8.70 × 10 j ~ 10 65 3.34 × 10 26

28

electrons

26

29

up quarks

26

29

down quarks .

Only these fundamental particles form your body. You have no strangeness, charm, topness or bottomness. P46.30

(a) strangeness baryon number charge

proton 0 1 e

u 0 1/3 2e/3

u 0 1/3 2e/3

d 0 1/3 –e/3

total 0 1 e

strangeness baryon number charge

neutron 0 1 0

u 0 1/3 2e/3

d 0 1/3 –e/3

d 0 1/3 –e/3

total 0 1 0

(b)

P46.31

Quark composition of proton = uud and of neutron = udd. Thus, if we neglect binding energies, we may write mp = 2 m u + md and

m n = m u + 2 md .

(2)

Solving simultaneously, we find

mu =

1 1 2 m p − m n = 2 938 MeV c 2 − 939.6 meV c 2 = 312 MeV c 2 3 3

e

(1)

j

e

and from either (1) or (2), md = 314 MeV c 2 .

j

634 P46.32

Particle Physics and Cosmology

d 0 1/3 –e/3

1 –1/3 e/3

total 1 0 0

strangeness baryon number charge

Λ0 –1 1 0

u 0 1/3 2e/3

d 0 1/3 –e/3

s –1 1/3 –e/3

(b)

P46.33

(a)

(c)

(d)

total –1 1 0

π − + p → K 0 + Λ0 In terms of constituent quarks:

(b)

s

strangeness baryon number charge

K0 1 0 0

(a)

ud + uud → ds + uds

up quarks:

−1 + 2 → 0 + 1,

or

1→1

down quarks:

1 + 1 → 1 + 1,

or

2→2

strange quarks:

0 + 0 → −1 + 1 ,

or

0→0

π + + p → K+ + Σ+

du + uud → us + uus

up quarks:

1 + 2 → 1 + 2,

or

3→3

down quarks:

−1 + 1 → 0 + 0 ,

or

0→0

strange quarks:

0 + 0 → −1 + 1 ,

or

0→0

K − + p → K + + K 0 + Ω−

us + uud → us + ds + sss

up quarks:

−1 + 2 → 1 + 0 + 0 ,

or

1→1

down quarks:

0 +1→ 0 +1+0,

or

1→1

strange quarks:

1 + 0 → −1 − 1 + 3 ,

or

1→1

p + p → K0 + p + π + + ?

uud + uud → ds + uud + ud + ?

The quark combination of ? must be such as to balance the last equation for up, down, and strange quarks. up quarks:

2+ 2 = 0+ 2+1+?

(has 1 u quark)

down quarks:

1+1 =1+1−1+?

(has 1 d quark)

strange quarks:

0 + 0 = −1 + 0 + 0 + ?

(has 1 s quark)

quark composition = uds = Λ0 or Σ 0 P46.34

In the first reaction, π − + p → K 0 + Λ0 , the quarks in the particles are: ud + uud → ds + uds . There is a net of 1 up quark both before and after the reaction, a net of 2 down quarks both before and after, and a net of zero strange quarks both before and after. Thus, the reaction conserves the net number of each type of quark. In the second reaction, π − + p → K 0 + n , the quarks in the particles are: ud + uud → ds + uds . In this case, there is a net of 1 up and 2 down quarks before the reaction but a net of 1 up, 3 down, and 1 anti-strange quark after the reaction. Thus, the reaction does not conserve the net number of each type of quark.

Chapter 46

P46.35

Σ0 + p → Σ+ + γ + X

dds + uud → uds + 0 + ? The left side has a net 3d, 2u and 1s. The right-hand side has 1d, 1u, and 1s leaving 2d and 1u missing. The unknown particle is a neutron, udd. Baryon and strangeness numbers are conserved. P46.36

P46.37

Compare the given quark states to the entries in Tables 46.4 and 46.5: (a)

suu = Σ +

(b)

ud = π −

(c)

sd = K 0

(d)

ssd = Ξ −

(a)

uud :

(b)

udd :

Section 46.12 P46.38

FG 2 eIJ + FG − 2 eIJ + FG 1 eIJ = − e . This is the antiproton . H 3 K H 3 K H3 K F 2 I F1 I F1 I charge = G − eJ + G eJ + G eJ = 0 . This is the antineutron . H 3 K H3 K H3 K

charge = −

The Cosmic Connection

Section 39.4 says

fobserver = fsource

1 + va c . 1 − va c

The velocity of approach, v a , is the negative of the velocity of mutual recession: v a = − v . c

Then,

P46.39

(a)

λ′

c

λ

λ′ = λ

1+ v c 1− v c

1.18 2 =

1+v c = 1.381 1−v c

1+

v = HR :

1−v c 1+v c

and

510 nm = 434 nm

v v = 1.381 − 1.381 c c

v = 0.160 c (b)

=

or

2.38

λ′ = λ

1+ v c . 1− v c

1+v c 1−v c

v = 0.381 c

v = 0.160 c = 4.80 × 10 7 m s

R=

4.80 × 10 7 m s v = = 2.82 × 10 9 ly H 17 × 10 −3 m s ⋅ ly

635

636 P46.40

Particle Physics and Cosmology

(a)

λ n′ = λ n 1+

v=

(b)

P46.41

P46.42

a

1+v c = Z+1 1−v c

f

f − FGH vc IJK aZ + 1f F Z + 2Z I cG H Z + 2Z + 2 JK a

v = Z+1 c

2

a

FG v IJ eZ H cK

2

f

2

j

2

+ 2 Z + 2 = Z2 + 2 Z

2

2

F GH

v c Z 2 + 2Z = H H Z 2 + 2Z + 2

I JK e1.7 × 10 H=

v = HR

−2

ms

j

ly 1+ v c = 590 1.000 113 3 = 590.07 nm 1− v c

b

g

e

j

λ′ = λ

e

j

λ ′ = 590

1 + 0.011 33 = 597 nm 1 − 0.011 33

e

j

λ ′ = 590

1 + 0.113 3 = 661 nm 1 − 0.113 3

(a)

v 2.00 × 10 6 ly = 3.4 × 10 4 m s

(b)

v 2.00 × 10 8 ly = 3.4 × 10 6 m s

(c)

v 2.00 × 10 9 ly = 3.4 × 10 7 m s

(a)

Wien’s law:

λ maxT = 2.898 × 10 −3 m ⋅ K .

Thus,

λ max =

(b) *P46.43

R=

1+ v c = Z + 1 λn 1− v c

2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K = = 1.06 × 10 −3 m = 1.06 mm T 2.73 K .

This is a microwave .

We suppose that the fireball of the Big Bang is a black body.

ja

e

I = eσT 4 = (1) 5.67 × 10 −8 W m 2 ⋅ K 4 2.73 K

f

4

= 3.15 × 10 −6 W m 2

As a bonus, we can find the current power of direct radiation from the Big Bang in the section of the universe observable to us. If it is fifteen billion years old, the fireball is a perfect sphere of radius fifteen billion light years, centered at the point halfway between your eyes:

e

ja fe

P = IA = I ( 4π r 2 ) = 3.15 × 10 −6 W m 2 4π 15 × 10 9 ly P = 7.98 × 10 47 W .

F I j GH 3 ×110ly yrm s JK e3.156 × 10 2

8

2

7

s yr

j

2

Chapter 46

P46.44

637

The density of the Universe is

ρ = 1.20 ρ c = 1.20

F 3H I . GH 8π G JK 2

Consider a remote galaxy at distance r. The mass interior to the sphere below it is M=ρ

FG 4 π r IJ = 1.20FG 3H IJ FG 4 π r IJ = 0.600 H r H 3 K H 8π G K H 3 K G 2

3

2 3

3

both now and in the future when it has slowed to rest from its current speed v = H r . The energy of this galaxy-sphere system is constant as the galaxy moves to apogee distance R: 1 GmM GmM =0− mv 2 − 2 r R −0.100 = −0.600

r R

I JK

F GH

F GH

so

1 Gm 0.600 H 2 r 3 Gm 0.600 H 2 r 3 =0− mH 2 r 2 − 2 r G R G

so

R = 6.00r .

I JK

The Universe will expand by a factor of 6.00 from its current dimensions. P46.45

(a)

k B T ≈ 2m p c 2 so

(b)

(a)

T≈

kB

=

a

f F 1.60 × 10 J I G J J K j H 1 MeV K

~ 10 13 K

f F 1.60 × 10 J I j GH 1 MeV JK

~ 10 10 K

−13

2 938.3 MeV

e1.38 × 10

−23

a

−13

2 0.511 MeV 2m e c 2 = kB 1.38 × 10 −23 J K

e

The Hubble constant is defined in v = HR . The distance R between any two far-separated objects opens at constant speed according to R = v t . Then the time t since the Big Bang is found from v = H vt

(b)

2m p c 2

k B T ≈ 2m e c 2 so

*P46.46

T≈

1= Ht

F GH

t=

I JK

1 . H

3 × 10 8 m s 1 1 = = 1.76 × 10 10 yr = 17.6 billion years H 17 × 10 −3 m s ⋅ ly 1 ly yr

638 *P46.47

Particle Physics and Cosmology

(a)

Consider a sphere around us of radius R large compared to the size of galaxy clusters. If the matter M inside the sphere has the critical density, then a galaxy of mass m at the surface of the sphere is moving just at escape speed v according to 1 GMm = 0. mv 2 − 2 R

K +Ug = 0

The energy of the galaxy-sphere system is conserved, so this equation is true throughout the dR . Then history of the Universe after the Big Bang, where v = dt

F dR I GH d t JK R3 / 2 32

2

=

2 GM R

R

= 2 GM t 0

0

*P46.48

2 R3 2 2 = 3 2 GM 3

z

T

0

dt

R 2GM R

.

2R . 3 v

so

T=

Now Hubble’s law says

v = HR .

So

T=

e

R dR = 2 GM

2GM =v R

From above,

T=

R

0

2 3/2 = 2 GM T R 3

T

T=

(b)

z

dR = R −1 / 2 2 GM dt

2

3 17 × 10 −3

2 R 2 = . 3 HR 3 H

F 3 × 10 m s I = G J m s ⋅ ly j H 1 ly yr K 8

1.18 × 10 10 yr = 11.8 billion years

In our frame of reference, Hubble’s law is exemplified by v 1 = HR 1 and v 2 = HR 2 . From these we may form the equations − v 1 = − HR 1 and v 2 − v 1 = H R 2 − R 1 . These equations express Hubble’s

b

g

b g

law as seen by the observer in the first galaxy cluster, as she looks at us to find − v 1 = H − R 1 and as she looks at cluster two to find v 2 − v 1 = H R 2 − R 1 .

b

Section 46.13

P46.49

g

Problems and Perspectives =G = c3

e1.055 × 10

− 34

(a)

L=

(b)

This time is given as T =

je

J ⋅ s 6.67 × 10 −11 N ⋅ m 2 kg 2

e3.00 × 10

8

ms

j

3

Yes.

1.61 × 10 −35 m

L 1.61 × 10 − 35 m = = 5.38 × 10 −44 s , c 3.00 × 10 8 m s

which is approximately equal to the ultra-hot epoch. (c)

j=

Chapter 46

Additional Problems P46.50

We find the number N of neutrinos:

a

f e

10 46 J = N 6 MeV = N 6 × 1.60 × 10 −13 J

j

N = 1.0 × 10 58 neutrinos The intensity at our location is N N 1.0 × 10 58 = = A 4π r 2 4π 1.7 × 10 5 ly

e

F G j GH e3.00 × 10 2

1 ly

je

8

m s 3.16 × 10 7

I J sj JK

2

= 3.1 × 10 14 m −2 .

The number passing through a body presenting 5 000 cm 2 = 0.50 m 2

FG 3.1 × 10 H

is then

1 m2

IJ e0.50 m j = 1.5 × 10 K 2

14

~ 10 14 .

or *P46.51

14

A photon travels the distance from the Large Magellanic Cloud to us in 170 000 years. The hypothetical massive neutrino travels the same distance in 170 000 years plus 10 seconds:

b

g b

c 170 000 yr = v 170 000 yr + 10 s 170 000 yr v = = c 170 000 yr + 10 s 1 + 10 s

{

g

1

e1.7 × 10 yrje3.156 × 10 5

7

s yr

=

j}

1 1 + 1.86 × 10 −12

For the neutrino we want to evaluate mc 2 in E = γ mc 2 : v2 1 mc = = E 1 − 2 = 10 MeV 1 − γ c 1 + 1.86 × 10 −12 2

e

2

mc ≈ 10 MeV

e

2 1.86 × 10 −12 1

e1 + 1.86 × 10 j − 1 e1 + 1.86 × 10 j −12 2

E

j

2

= 10 MeV

j = 10 MeVe1.93 × 10 j = 19 eV . −6

Then the upper limit on the mass is m= m=

P46.52

19 eV c2 19 eV c

2

F GH 931.5 × 10u

6

eV c

2

I = 2.1 × 10 JK

−8

u.

(a)

π − + p → Σ+ + π 0

is forbidden by charge conservation .

(b)

µ− → π − +νe

is forbidden by energy conservation .

(c)

p→π+ +π+ +π−

is forbidden by baryon number conservation .

−12 2

639

640 P46.53

Particle Physics and Cosmology

The total energy in neutrinos emitted per second by the Sun is:

a0.4fLMN4π e1.5 × 10

j OPQW = 1.1 × 10

11 2

23

W.

Over 10 9 years, the Sun emits 3.6 × 10 39 J in neutrinos. This represents an annihilated mass m c 2 = 3.6 × 10 39 J m = 4.0 × 10 22 kg . About 1 part in 50 000 000 of the Sun’s mass, over 10 9 years, has been lost to neutrinos. P46.54

p+p→ p+π+ + X We suppose the protons each have 70.4 MeV of kinetic energy. From conservation of momentum for the collision, particle X has zero momentum and thus zero kinetic energy. Conservation of system energy then requires

e

j e

M p c 2 + Mπ c 2 + M X c 2 = M p c 2 + K p + M p c 2 + K p

a

j

f

M X c 2 = M p c 2 + 2K p − Mπ c 2 = 938.3 MeV + 2 70.4 MeV − 139.6 MeV = 939.5 MeV X must be a neutral baryon of rest energy 939.5 MeV. Thus X is a neutron . *P46.55

(a)

If 2N particles are annihilated, the energy released is 2 Nmc 2 . The resulting photon E 2 Nmc 2 = 2 Nmc . Since the momentum of the system is conserved, the momentum is p = = c c rocket will have momentum 2Nmc directed opposite the photon momentum. p = 2 Nmc

(b)

Consider a particle that is annihilated and gives up its rest energy mc 2 to another particle which also has initial rest energy mc 2 (but no momentum initially).

e j Thus e 2mc j = p c + emc j . E 2 = p 2 c 2 + mc 2 2 2

2

2 2

2 2

Where p is the momentum the second particle acquires as a result of the annihilation of the

e j

first particle. Thus 4 mc 2

2

e j ,p

= p 2 c 2 + mc 2

2

2

e j . So p =

= 3 mc 2

2

3 mc .

N N protons and antiprotons). Thus the total 2 2 momentum acquired by the ejected particles is 3Nmc , and this momentum is imparted to the rocket. This process is repeated N times (annihilate

p = 3 Nmc (c)

Method (a) produces greater speed since 2 Nmc > 3 Nmc .

641

Chapter 46

P46.56

(a)

∆E∆t ≈ = , and

∆t =

r 1.4 × 10 −15 m = = 4.7 × 10 −24 s c 3 × 10 8 m s

∆E ≈

= 1.055 × 10 −34 J ⋅ s 1 MeV = = 2.3 × 10 −11 J = 1.4 × 10 2 MeV ∆t 4.7 × 10 −24 s 1.60 × 10 −13 J

m= (b) P46.57

IJ K

jFGH

e

∆E ≈ 1.4 × 10 2 MeV c 2 ~ 10 2 MeV c 2 c2

From Table 46.2, mπ c 2 = 139.6 MeV a pi - meson .

m Λ c 2 = 1115.6 MeV

Λ0 → p + π −

m p c 2 = 938.3 MeV

mπ c 2 = 139.6 MeV

The difference between starting rest energy and final rest energy is the kinetic energy of the products. K p + K π = 37.7 MeV

and

p p = pπ = p

Applying conservation of relativistic energy to the decay process, we have

LM a938.3f N

2

OP LM a139.6f Q N

+ p 2 c 2 − 938.3 +

2

OP Q

+ p 2 c 2 − 139.6 = 37.7 MeV .

Solving the algebra yields pπ c = p p c = 100.4 MeV . Then,

Kp = Kπ =

P46.58

em c j + a100.4f a139.6f + a100.4f p

2 2

2

− m p c 2 = 5.35 MeV

2

2

− 139.6 = 32.3 MeV .

By relativistic energy conservation in the reaction,

By relativistic momentum conservation for the system,

Eγ c

=

X=

Subtracting (2) from (1),

me c 2 =

3 − 3X 1− X

2

and X =

4 so Eγ = 4m e c 2 = 2.04 MeV . 5

1 − v2 c2

3m e v 1 − v2 c2 Eγ

Dividing (2) by (1),

Solving, 1 =

3m e c 2

Eγ + m e c 2 =

Eγ + m e c

2

=

3m e c 2 1− X2

.

(1)

.

(2)

v . c −

3m e c 2 X 1− X2

.

642 P46.59

Particle Physics and Cosmology

jb

e

ga

f

Momentum of proton is

qBr = 1.60 × 10 −19 C 0.250 kg C ⋅ s 1.33 m

p p = 5.32 × 10 −20 kg ⋅ m s

c p p = 1.60 × 10 −11 kg ⋅ m 2 s 2 = 1.60 × 10 −11 J = 99.8 MeV .

Therefore,

p p = 99.8 MeV c .

The total energy of the proton is

Ep = E02 + cp

b g

2

=

a983.3f + a99.8f 2

2

= 944 MeV .

For pion, the momentum qBr is the same (as it must be from conservation of momentum in a 2particle decay). pπ = 99.8 MeV c

E0π = 139.6 MeV

b g

Eπ = E02 + cp

2

=

a139.6f + a99.8f 2

2

= 172 MeV

Etotal after = Etotal before = Rest energy .

Thus,

Rest Energy of unknown particle = 944 MeV + 172 MeV = 1 116 MeV

(This is a Λ0 particle!)

Mass = 1 116 MeV c 2 . P46.60

Σ 0 → Λ0 + γ From Table 46.2, m Σ = 1 192.5 MeV c 2

m Λ = 1115.6 MeV c 2 .

and

Conservation of energy in the decay requires

e

j

E0, Σ = Eo , Λ + K Λ + Eγ

F GH

1 192.5 MeV = 1 115.6 MeV +

or

I JK

pΛ 2 + Eγ . 2m Λ

System momentum conservation gives p Λ = pγ , so the last result may be written as

F p I 1 192.5 MeV = G 1 115.6 MeV + J +E 2m K H F p c I 1 192.5 MeV = G 1 115.6 MeV + J +E . 2m c K H γ

2

γ

Λ

γ

or

2 2 Λ

m Λ c 2 = 1 115.6 MeV

Recognizing that

P46.61

1 192.5 MeV = 1 115.6 MeV +

Solving this quadratic equation,

Eγ = 74.4 MeV .

γ

pγ c = Eγ

and

we now have

2

b

Eγ 2

2 1 115.6 MeV

g +E . γ

p+p→ p+ n+π+ The total momentum is zero before the reaction. Thus, all three particles present after the reaction may be at rest and still conserve system momentum. This will be the case when the incident protons have minimum kinetic energy. Under these conditions, conservation of energy for the reaction gives

e

j

2 m p c 2 + K p = m p c 2 + mn c 2 + mπ c 2 so the kinetic energy of each of the incident protons is Kp =

mn c 2 + mπ c 2 − m p c 2 2

=

a939.6 + 139.6 − 938.3f MeV = 2

70.4 MeV

.

Chapter 46

P46.62

π − → µ− +νµ :

From the conservation laws for the decay,

mπ c 2 = 139.6 MeV = Eµ + Eν

P46.63

[1]

d i + a105.7 MeVf = bp cg + a105.7 MeVf − E = a105.7 MeV f . 2

2

and p µ = pν , Eν = pν c :

Eµ2 = p µ c

or

Eµ2

Since

Eµ + Eν = 139.6 MeV

and

dE

µ

2

ν

2

2

2 ν

i a

id

a105.7 MeVf =

Eµ − Eν

Subtracting [3] from [1],

2Eν = 59.6 MeV

[2] [1]

+ Eν Eµ − Eν = 105.7 MeV

then

f

2

[2]

2

139.6 MeV

= 80.0 . and

[3] Eν = 29.8 MeV .

The expression e − E k BT dE gives the fraction of the photons that have energy between E and E + dE . The fraction that have energy between E and infinity is

z z

∞ E ∞

z z

∞

e − E k BT dE = e − E k BT dE

0

E ∞

b

e − E k BT − dE k B T

b

e − E k BT − dE k B T

g g

=

e − E k BT e

∞

E − E k BT ∞

= e − E k BT .

0

0

We require T when this fraction has a value of 0.0100 (i.e., 1.00%) and

E = 1.00 eV = 1.60 × 10 −19 J .

Thus,

0.010 0 = e

b

g

or ln 0.010 0 = −

P46.64

643

(a)

e

− 1.60 × 10 −19 J

1.60 × 10 −19 J

e1.38 × 10

−23

j

j e1.38 ×10

JKT

=−

−23

j

JK T

1.16 × 10 4 K giving T = 2.52 × 10 3 K . T

This diagram represents the annihilation of an electron and an antielectron. From charge and lepton-number conservation at either vertex, the

γ

γ

exchanged particle must be an electron, e − . (b)

This is the tough one. A neutrino collides with a neutron, changing it into a proton with release of a muon. This is a weak interaction. The exchanged particle has charge +e and is a W + .

(a)

(b) FIG. P46.64

644 P46.65

Particle Physics and Cosmology

(a)

The mediator of this weak interaction is a

Z 0 boson . (b)

The Feynman diagram shows a down quark and its antiparticle annihilating each other. They can produce a particle carrying energy, momentum, and angular momentum, but zero charge, zero baryon number, and, it may be, no color charge. In this case the product particle is a photon .

FIG. P46.65

For conservation of both energy and momentum in the collision we would expect two photons; but momentum need not be strictly conserved, according to the uncertainty principle, if the photon travels a sufficiently short distance before producing another matterantimatter pair of particles, as shown in Figure P46.65. Depending on the color charges of the d and d quarks, the ephemeral particle could also be a gluon , as suggested in the discussion of Figure 46.14(b). *P46.66

(a)

At threshold, we consider a photon and a proton colliding head-on to produce a proton and a pion at rest, according to p + γ → p + π 0 . Energy conservation gives mp c 2 2

1−u c

2

+ Eγ = m p c 2 + mπ c 2 . m pu

Momentum conservation gives

1−u

2

c

2

−

Eγ c

= 0.

Combining the equations, we have mp c 2 1−u

2

c

2

+

b

mp c 2

u = m p c 2 + mπ c 2 c 1−u c 2

938.3 MeV 1 + u c

b1 − u cgb1 + u cg

(b)

so

u = 0.134 c

and

Eγ = 127 MeV .

g = 938.3 MeV + 135.0 MeV

λ maxT = 2.898 mm ⋅ K λ max =

(c)

2

2.898 mm ⋅ K = 1.06 mm 2.73 K

Eγ = hf =

hc

λ

=

continued on next page

1 240 eV ⋅ 10 −9 m 1.06 × 10 -3 m

= 1.17 × 10 −3 eV

Chapter 46

(d)

645

u′ = 0.134 and the c photon is moving to the left with hf ′ = 1.27 × 10 8 eV . In the unprimed frame,

In the primed reference frame, the proton is moving to the right at

hf = 1.17 × 10 −3 eV . Using the Doppler effect equation from Section 39.4, we have for the speed of the primed frame 1+ν c 1.17 × 10 −3 1 −ν c

1.27 × 10 8 =

v = 1 − 1.71 × 10 −22 c Then the speed of the proton is given by 0.134 + 1 − 1.71 × 10 −22 u u′ c + ν c = = = 1 − 1.30 × 10 −22 . 2 −22 c 1 + u ′ν c 1 + 0.134 1 − 1.71 × 10

e

j

And the energy of the proton is mp c 2 1−u

2

c

2

=

938.3 MeV

e

1 − 1 − 1.30 × 10

j

−22 2

= 6.19 × 10 10 × 938.3 × 10 6 eV = 5.81 × 10 19 eV .

ANSWERS TO EVEN PROBLEMS P46.2

2.27 × 10 23 Hz ; 1.32 fm

P46.4

ν µ and ν e

P46.6

~ 10 −16 m

P46.8

~ 10 −23 s

P46.10

νµ

P46.12

(a) electron lepton number and muon lepton number; (b) charge; (c) baryon number; (d) baryon number; (e) charge

P46.14

the second violates conservation of baryon number

P46.16

0.828 c

P46.18

(a) see the solution; 469 MeV for both; (b) 469 MeV ; c (c) 0.999 999 4c

P46.20

see the solution

P46.22

(a) electron lepton number and muon lepton number; (b) electron lepton number; (c) strangeness and charge; (d) baryon number; (e) strangeness

P46.24

see the solution

P46.26

686 MeV 200 MeV and ; c c (b) 627 MeV c ; (c) 244 MeV , 1 130 MeV , 1 370 MeV ; (a)

(d) 1 190 MeV c 2 , 0.500 c P46.28

(a) see the solution; (b) 5.63 GeV ; (c) 768 MeV ; (d) 280 MeV ; (e) 4.43 TeV

P46.30

see the solution

P46.32

see the solution

P46.34

see the solution

P46.36

(a) Σ + ; (b) π − ; (c) K 0 ; (d) Ξ −

P46.38

see the solution

646 P46.40

Particle Physics and Cosmology

(a) v = c

F Z + 2Z I ; (b) c F Z + 2Z I GH Z + 2Z + 2 JK H GH Z + 2Z + 2 JK 2

2

2

P46.56

(a) ~ 10 2 MeV c 2 ; (b) a pi - meson

P46.58

2.04 MeV

P46.60

74.4 MeV

P46.62

29.8 MeV

P46.64

(a) electron-position annihilation; e − ; (b) a neutrino collides with a neutron, producing a proton and a muon; W +

P46.66

(a) 127 MeV ; (b) 1.06 mm; (c) 1.17 meV ; (d) 5.81 × 10 19 eV

2

P46.42

(a) 1.06 mm; (b) microwave

P46.44

6.00

P46.46

(a) see the solution; (b) 17.6 Gyr

P46.48

see the solution

P46.50

~ 10 14

P46.52

(a) charge; (b) energy; (c) baryon number

P46.54

neutron