Solutions Manual for Tipler Mosca Physics for Scientists and Engineers [6 ed.]

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Chapter 1 Measurement and Vectors Conceptual Problems 1 • [SSM] Which of the following is not one of the base quantities in the SI system? (a) mass, (b) length, (c) energy, (d) time, (e) All of the above are base quantities.

Determine the Concept The base quantities in the SI system include mass, length, and time. Force is not a base quantity. (c) is correct. In doing a calculation, you end up with m/s in the numerator and m/s2 in the denominator. What are your final units? (a) m2/s3, (b) 1/s, (c) s3/m2, (d) s, (e) m/s.

2

•

Picture the Problem We can express and simplify the ratio of m/s to m/s2 to determine the final units. m 2 s = m ⋅ s = s and (d ) is correct. m m ⋅s 2 s

Express and simplify the ratio of m/s to m/s2:

3

•

The prefix giga means (a) 103, (b) 106, (c) 109, (d) 1012, (e) 1015.

Determine the Concept Consulting Table 1-1 we note that the prefix giga means 109. (c) is correct. 4

•

The prefix mega means (a) 10–9, (b) 10–6, (c) 10–3, (d) 106, (e) 109.

Determine the Concept Consulting Table 1-1 we note that the prefix mega means 106. (d ) is correct. [SSM] Show that there are 30.48 cm per foot. How many centimeters are there in one mile? 5

•

Picture the Problem We can use the facts that there are 2.540 centimeters in 1 inch and 12 inches in 1 foot to show that there are 30.48 cm per ft. We can then use the fact that there are 5280 feet in 1 mile to find the number of centimeters in one mile.

1

2

Chapter 1

Multiply 2.540 cm/in by 12 in/ft to find the number of cm per ft:

cm ⎞ ⎛ in ⎞ ⎛ ⎜ 2.540 ⎟ ⎜12 ⎟ = 30.48 cm/ft in ⎠ ⎝ ft ⎠ ⎝

Multiply 30.48 cm/ft by 5280 ft/mi to find the number of centimeters in one mile: ft ⎞ cm ⎞ ⎛ ⎛ 5 ⎜ 30.48 ⎟ ⎜ 5280 ⎟ = 1.609 × 10 cm/mi mi ⎠ ft ⎠ ⎝ ⎝ Remarks: Because there are exactly 2.54 cm in 1 in and exactly 12 inches in 1 ft, we are justified in reporting four significant figures in these results.

The number 0.000 513 0 has • (c) four, (d) seven, (e) eight.

6

significant figures. (a) one, (b) three,

Determine the Concept Counting from left to right and ignoring zeros to the left of the first nonzero digit, the last significant figure is the first digit that is in doubt. Applying this criterion, the three zeros after the decimal point are not significant figures, but the last zero is significant. Hence, there are four significant figures in this number. (c) is correct. 7 • The number 23.0040 has (c) four, (d) five, (e) six.

significant figures. (a) two, (b) three,

Determine the Concept Counting from left to right, the last significant figure is the first digit that is in doubt. Applying this criterion, there are six significant figures in this number. (e) is correct. 8 • Force has dimensions of mass times acceleration. Acceleration has dimensions of speed divided by time. Pressure is defined as force divided by area. What are the dimensions of pressure? Express pressure in terms of the SI base units kilogram, meter and second. Determine the Concept We can use the definitions of force and pressure, together with the dimensions of mass, acceleration, and length, to find the dimensions of pressure. We can express pressure in terms of the SI base units by substituting the base units for mass, acceleration, and length in the definition of pressure.

Measurement and Vectors Use the definition of pressure and the dimensions of force and area to obtain:

ML 2 [P] = [F ] = T 2 = M 2 [A] L LT

Express pressure in terms of the SI base units to obtain:

m kg ⋅ 2 N s = kg = 2 2 m m m ⋅ s2

3

• True or false: Two quantities must have the same dimensions in order to be multiplied.

9

False. For example, the distance traveled by an object is the product of its speed (length/time) multiplied by its time of travel (time). 10 • A vector has a negative x component and a positive y component. Its angle measured counterclockwise from the positive x axis is (a) between zero and 90 degrees. (b) between 90 and 180 degrees. (c) More than 180 degrees. Determine the Concept Because a vector with a negative x-component and a positive y-component is in the second quadrant, its angle is between 90 and 180 degrees. (b ) is correct.

r [SSM] A vector A points in the +x direction. Show graphically at r r r least three choices for a vector B such that B + A points in the +y direction. 11

•

Determine the Concept The figure r shows a vector A pointing in the positive x direction and three unlabeled r possibilities for vector B. Note that the r choices for B start at the end of vector r A rather than at its initial point. Note further that this configuration could be in any quadrant of the reference system shown.

y

r Several B choices

r A

x

r • A vector A points in the +y direction. Show graphically at least three r r r choices for a vector B such that B − A points in the + x direction. 12

4

Chapter 1

Determine the Concept Let the +x

direction be to the right and the +y direction be upward. The figure shows r the vector − A pointing in the −y direction and three unlabeled r possibilities for vector B. Note that the r choices for B start at the end of vector r − A rather than at its initial point.

y

r −A

r Several B choices

x

13 • [SSM] Is it possible for three equal magnitude vectors to add to zero? If so, sketch a graphical answer. If not, explain why not. Determine the Concept In order for the three equal magnitude vectors to add to zero, the sum of the three vectors must form a triangle. The equilateral triangle shown to the right satisfies this r r condition for the vectors A , B , and r C for which it is true that A = B = C, r r r whereas A + B + C = 0.

r C

r B

r A

Estimation and Approximation The angle subtended by the moon’s diameter at a point on Earth is about 0.524º (Fig. 1-2). Use this and the fact that the moon is about 384 Mm away to find the diameter of the moon. HINT: The angle can be determined from the diameter of the moon and the distance to the moon. 14

•

Picture the Problem Let θ represent the angle subtended by the moon’s diameter, D represent the diameter of the moon, and rm the distance to the moon. Because θ is small, we can approximate it by θ ≈ D/rm where θ is in radian measure. We can solve this relationship for the diameter of the moon.

Express the moon’s diameter D in terms of the angle it subtends at Earth θ and the Earth-moon distance rm:

D = θ rm

Substitute numerical values and evaluate D:

2π rad ⎞ ⎛ D = ⎜ 0.524° × ⎟ (384 Mm ) 360° ⎠ ⎝ = 3.51 × 10 6 m

Measurement and Vectors

5

15 • [SSM] Some good estimates about the human body can be made if it is assumed that we are made mostly of water. The mass of a water molecule is 29.9 ×10−27 kg. If the mass of a person is 60 kg, estimate the number of water molecules in that person. Picture the Problem We can estimate the number of water molecules in a person whose mass is 60 kg by dividing this mass by the mass of a single water molecule.

Letting N represent the number of water molecules in a person of mass mhuman body, express N in terms of mhuman body and the mass of a water molecule mwater molecule: Substitute numerical values and evaluate N:

N=

mhuman body mwater molecule

60 kg

N=

29.9 ×10 −27

kg molecule

= 2.0 ×10 27 molecules •• In 1989, IBM scientists figured out how to move atoms with a scanning tunneling microscope (STM). One of the first STM pictures seen by the general public was of the letters IBM spelled with xenon atoms on a nickel surface. The letters IBM were 15 xenon atoms across. If the space between the centers of adjacent xenon atoms is 5 nm (5 ×10−9 m), estimate how many times could ″IBM″ could be written across this 8.5 inch page. 16

Picture the Problem We can estimate the number of times N that ″IBM″ could be written across this 8.5-inch page by dividing the width w of the page by the distance d required by each writing of ″IBM.″

w d

Express N in terms of the width w of the page and the distance d required by each writing of ″IBM″:

N=

Express d in terms of the separation s of the centers of adjacent xenon atoms and the number n of xenon atoms in each writing of ″IBM″:

d = sn

Substitute for d in the expression for N to obtain:

N=

w sn

6

Chapter 1

Substitute numerical values and evaluate N:

(8.5 in )⎛⎜ 2.540 cm ⎞⎟

N=

in ⎠ ⎝ = 3 × 10 6 −9 ⎛ 10 m ⎞ ⎜⎜ 5 nm × ⎟ (15) nm ⎟⎠ ⎝

There is an environmental debate over the use of cloth versus disposable diapers. (a) If we assume that between birth and 2.5 y of age, a child uses 3 diapers per day, estimate the total number of disposable diapers used in the United States per year. (b) Estimate the total landfill volume due to these diapers, assuming that 1000 kg of waste fills about 1 m3 of landfill volume. (c) How many square miles of landfill area at an average height of 10 m is needed for the disposal of diapers each year?

17

••

Picture the Problem We’ll assume a population of 300 million and a life expectancy of 76 y. We’ll also assume that a diaper has a volume of about half a liter. In (c) we’ll assume the disposal site is a rectangular hole in the ground and use the formula for the volume of such an opening to estimate the surface area required.

(a) Express the total number N of disposable diapers used in the United States per year in terms of the number of children n in diapers and the number of diapers D used by each child per year:

N = nD

Use the estimated daily consumption and the number of days in a year to estimate the number of diapers D required per child per year:

D=

Use the assumed life expectancy to estimate the number of children n in diapers yearly:

⎛ 2.5 y ⎞ ⎟⎟ 300 × 10 6 children n = ⎜⎜ ⎝ 76 y ⎠ ≈ 10 7 children

Substitute numerical values in equation (1) to obtain:

⎛ diapers ⎞ ⎟ N = 10 7 children ⎜⎜ 3 × 10 3 y ⎟⎠ ⎝

(1)

3 diapers 365.24 d × child ⋅ d y

≈1.1× 10 3 diapers/child ⋅ y

(

(

)

≈ 1.1 × 1010 diapers

)

Measurement and Vectors (b) Express the required landfill volume V in terms of the volume of diapers to be buried:

7

V = NVone diaper

Substitute numerical values and evaluate V: ⎛ 0.5 L 10 −3 m 3 ⎞ ⎟⎟ ≈ 5.5 × 10 6 m 3 V = 1.1 × 10 diapers ⎜⎜ × L ⎠ ⎝ diaper

(

10

)

(c) Express the required volume in terms of the volume of a rectangular parallelepiped: Substitute numerical values evaluate A: Use a conversion factor (see Appendix A) to express this area in square miles:

V = Ah ⇒ A =

A=

V h

5.5 × 10 6 m 3 = 5.5 × 10 5 m 2 10 m

A = 5.5 × 10 5 m 2 ×

0.3861 mi 2 1km 2

≈ 0.2 mi 2

•• (a) Estimate the number of gallons of gasoline used per day by automobiles in the United States and the total amount of money spent on it. (b) If 19.4 gal of gasoline can be made from one barrel of crude oil, estimate the total number of barrels of oil imported into the United States per year to make gasoline. How many barrels per day is this? 18

Picture the Problem The population of the United States is roughly 3 × 108 people. Assuming that the average family has four people, with an average of two cars per family, there are about 1.5 × 108 cars in the United States. If we double that number to include trucks, cabs, etc., we have 3 × 108 vehicles. Let’s assume that each vehicle uses, on average, 14 gallons of gasoline per week and that the United States imports half its oil.

(

)

(a) Find the daily consumption of gasoline G:

G = 3×10 8 vehicles (2 gal/d )

Assuming a price per gallon P = $3.00, find the daily cost C of gasoline:

C = GP = 6 × 10 8 gal/d ($3.00 / gal )

= 6 ×10 gal/d 8

(

= $18 × 10 8 / d ≈ 2 billion dollars/d

)

8

Chapter 1

(b) Relate the number of barrels N of crude oil imported annually to the yearly consumption of gasoline Y and the number of gallons of gasoline n that can be made from one barrel of crude oil: Substitute numerical values and estimate N:

Convert barrels/y to barrels/d to obtain:

fY fGΔt = n n where f is the fraction of the oil that is imported. N=

(0.5)⎛⎜ 6 ×108 gallons ⎞⎟ ⎛⎜⎜ 365.24 d ⎞⎟⎟

d ⎠⎝ gallons 19.4 barrel barrels ≈ 6 × 10 9 y

N=

⎝

N = 6 × 10 9

y⎠

barrels 1y × y 365.24 d

≈ 2 × 10 7

barrels d

•• [SSM] A megabyte (MB) is a unit of computer memory storage. A CD has a storage capacity of 700 MB and can store approximately 70 min of high-quality music. (a) If a typical song is 5 min long, how many megabytes are required for each song? (b) If a page of printed text takes approximately 5 kilobytes, estimate the number of novels that could be saved on a CD.

19

Picture the Problem We can set up a proportion to relate the storage capacity of a CD to its playing time, the length of a typical song, and the storage capacity required for each song. In (b) we can relate the number of novels that can be stored on a CD to the number of megabytes required per novel and the storage capacity of the CD.

(a) Set up a proportion relating the ratio of the number of megabytes on a CD to its playing time to the ratio of the number of megabytes N required for each song: Solve this proportion for N to obtain:

700 MB N = 70 min 5 min

⎛ 700 MB ⎞ N =⎜ ⎟ (5 min ) = 50 MB ⎝ 70 min ⎠

Measurement and Vectors (b) Letting n represent the number of megabytes per novel, express the number of novels N novels that can be

N novels =

9

700 MB n

stored on a CD in terms of the storage capacity of the CD: Assuming that a typical page in a novel requires 5 kB of memory, express n in terms of the number of pages p in a typical novel:

⎛ kB ⎞ ⎟⎟ p n = ⎜⎜ 5 ⎝ page ⎠

Substitute for n in the expression for N novels to obtain:

N novels =

Assuming that a typical novel has 200 pages:

700 MB ⎛ kB ⎞ ⎜⎜ 5 ⎟⎟ p ⎝ page ⎠

103 kB MB = ⎛ kB ⎞ ⎛ pages ⎞ ⎜⎜ 5 ⎟⎟ ⎜ 200 ⎟ novel ⎠ ⎝ page ⎠ ⎝ 700 MB ×

N novels

= 7 × 102 novels

Units 20 • Express the following quantities using the prefixes listed in Table 1-1 and the unit abbreviations listed in the table Abbreviations for Units. For example, 10,000 meters = 10 km. (a) 1,000,000 watts, (b) 0.002 gram, (c) 3 × 10–6 meter, (d) 30,000 seconds. Picture the Problem We can use the metric prefixes listed in Table 1-1 and the abbreviations on page EP-1 to express each of these quantities. (a)

(c)

1,000,000 watts = 10 watts = 1 MW

3 × 10 −6 meter = 3 μm

(b)

(d)

6

−3

0.002gram = 2 × 10 g = 2 mg

30,000 seconds = 30 × 103 s = 30 ks

21 • Write each of the following without using prefixes: (a) 40 μW, (b) 4 ns, (c) 3 MW, (d) 25 km.

10

Chapter 1

Picture the Problem We can use the definitions of the metric prefixes listed in Table 1-1 to express each of these quantities without prefixes. (a)

(c)

40 μW = 40 × 10 W = 0.000040 W

3 MW = 3 × 106 W = 3,000,000 W

(b)

(d)

−6

−9

4 ns = 4 × 10 s = 0.000000004 s

25 km = 25 × 103 m = 25,000 m

22 • Write the following (which are not SI units) using prefixes (but not their abbreviations). For example, 103 meters = 1 kilometer: (a) 10–12 boo, (b) 109 low, (c) 10–6 phone, (d) 10–18 boy, (e) 106 phone, (f) 10–9 goat, (g) 1012 bull. Picture the Problem We can use the definitions of the metric prefixes listed in Table 1-1 to express each of these quantities without abbreviations. (a) 10 −12 boo = 1 picoboo

(e) 106 phone = 1megaphone

(b) 10 9 low = 1 gigalow

(f) 10 −9 goat = 1 nanogoat

(c) 10 −6 phone = 1 microphone

(g) 1012 bull = 1 terabull

(d) 10 −18 boy = 1 attoboy

23 •• [SSM] In the following equations, the distance x is in meters, the time t is in seconds, and the velocity v is in meters per second. What are the SI units of the constants C1 and C2? (a) x = C1 + C2t, (b) x = 12 C1t2, (c) v2 = 2C1x, (d) x = C1 cos C2t, (e) v2 = 2C1v – (C2x)2. Picture the Problem We can determine the SI units of each term on the righthand side of the equations from the units of the physical quantity on the left-hand side.

(a) Because x is in meters, C1 and C2t must be in meters:

C1 is in m; C2 is in m/s

(b) Because x is in meters, ½C1t2 must be in meters:

C1 is in m/s 2

Measurement and Vectors (c) Because v2 is in m2/s2, 2C1x must be in m2/s2:

C1 is in m/s 2

(d) The argument of a trigonometric function must be dimensionless; i.e. without units. Therefore, because x is in meters:

C1 is in m; C2 is in s −1

(e) All of the terms in the expression must have the same units. Therefore, because v is in m/s:

C1 is in m/s; C2 is in s −1

11

If x is in feet, t is in milliseconds, and v is in feet per second, what are the units of the constants C1 and C2 in each part of Problem 23? 24

••

Picture the Problem We can determine the US customary units of each term on the right-hand side of the equations from the units of the physical quantity on the left-hand side.

(a) Because x is in feet, C1 and C2t must be in feet:

C1 is in ft; C 2 is in ft/ms

(b) Because x is in feet, 12 C1t 2

C1 is in ft/ (ms )

2

must be in feet: (c) Because v2 is in ft2/(ms)2, 2C1x must be in ft2/s2:

C1 is in ft/ (ms )

(d) The argument of a trigonometric function must be dimensionless; that is, without units. Therefore, because x is in feet:

C1 is in ft; C 2 is in (ms )

(e) The argument of an exponential function must be dimensionless; that is, without units. Therefore, because v is in ft/s:

C1 is in ft/ms; C 2 is in (ms )

Conversion of Units

2

−1

−1

12

Chapter 1

25 • From the original definition of the meter in terms of the distance along a meridian from the equator to the North Pole, find in meters (a) the circumference of Earth and (b) the radius of Earth. (c) Convert your answers for (a) and (b) from meters into miles. Picture the Problem We can use the formula for the circumference of a circle to find the radius of Earth and the conversion factor 1 mi = 1.609 km to convert distances in meters into distances in miles.

(a) The Pole-Equator distance is one-fourth of the circumference:

C = 4 Dpole-equator = 4 × 10 7 m

(b) The formula for the circumference of a circle is:

C = πD = 2πR ⇒ R =

Substitute numerical values and evaluate R:

R=

C 2π

4 × 10 7 m = 6.37 × 10 6 m 2π

= 6 × 10 6 m 1 km 1mi × 3 10 m 1.609 km

(c) Use the conversion factors 1 km = 1000 m and 1 mi = 1.609 km to express C in mi:

C = 4 × 10 7 m ×

Use the conversion factors 1 km = 1000 m and 1 mi = 1.61 km to express R in mi:

R = 6.37 × 10 6 m ×

= 2 × 10 4 mi 1 km 1mi × 3 10 m 1.609 km

= 4 × 10 3 mi

26 • The speed of sound in air is 343 m/s at normal room temperature. What is the speed of a supersonic plane that travels at twice the speed of sound? Give your answer in kilometers per hour and miles per hour. Picture the Problem We can use the conversion factor 1 mi = 1.61 km to convert speeds in km/h into mi/h.

Find the speed of the plane in km/h:

v = 2(343 m/s ) = 686 m/s m ⎞ ⎛ 1 km ⎞ ⎛ s⎞ ⎛ = ⎜ 686 ⎟ ⎜⎜ 3 ⎟⎟ ⎜ 3600 ⎟ s ⎠ ⎝ 10 m ⎠ ⎝ h⎠ ⎝ = 2.47 × 10 3 km/h

Measurement and Vectors Convert v into mi/h:

km ⎞ ⎛ 1 mi ⎞ ⎛ ⎟ v = ⎜ 2.47 × 10 3 ⎟⎜ h ⎠ ⎜⎝ 1.609 km ⎟⎠ ⎝ = 1.53 × 10 3 mi/h

A basketball player is 6 ft 10 12 in tall. What is his height in centimeters? 27

•

Picture the Problem We’ll first express his height in inches and then use the conversion factor 1 in = 2.540 cm.

12 in + 10.5 in = 82.5 in ft

Express the player’s height in inches:

h = 6 ft ×

Convert h into cm:

h = 82.5 in ×

28 • Complete the following: (a) 100 km/h = in, (c) 100 yd = m.

2.540 cm = 210 cm in mi/h, (b) 60 cm =

Picture the Problem We can use the conversion factors 1 mi = 1.609 km, 1 in = 2.540 cm, and 1 m = 1.094 yd to perform these conversions.

(a) Convert 100

mi km : to h h

100

km km 1 mi = 100 × h h 1.609 km = 62.2 mi/h

(b) Convert 60 cm to in:

60 cm = 60 cm ×

1in = 23.6 in 2.540 cm

= 24 in

(c) Convert 100 yd to m:

100 yd = 100 yd ×

1m 1.094 yd

= 91.4 m

The main span of the Golden Gate Bridge is 4200 ft. Express this 29 • distance in kilometers. Picture the Problem We can use the conversion factor 1.609 km = 5280 ft to convert the length of the main span of the Golden Gate Bridge into kilometers.

13

14

Chapter 1

Convert 4200 ft into km by multiplying by 1 in the form 1.609 km : 5280 ft

4200 ft = 4200 ft ×

1.609 km 5280 ft

= 1.28 km

Find the conversion factor to convert from miles per hour into kilometers per hour. 30

•

Picture the Problem Let v be the speed of an object in mi/h. We can use the conversion factor 1 mi = 1.609 km to convert this speed to km/h.

Multiply v mi/h by 1.609 km/mi to convert v to km/h:

v

mi mi 1.609 km =v × = 1.61v km/h h h mi

31 • Complete the following: (a) 1.296 × 105 km/h2 = km/(h⋅s), m/s2, (c) 60 mi/h = ft/s, (d) 60 mi/h = m/s. (b) 1.296 × 105 km/h2 = Picture the Problem Use the conversion factors 1 h = 3600 s, 1.609 km = 1 mi, and 1 mi = 5280 ft to make these conversions. (a) 1.296 × 105

km ⎛ km ⎞ ⎛ 1 h ⎞ ⎟ = 36.00 km/h ⋅ s = ⎜1.296 × 105 2 ⎟ ⎜⎜ 2 h h ⎠ ⎝ 3600 s ⎟⎠ ⎝

km ⎞ ⎛ 1 h ⎞ km ⎛ ⎟ (b) 1.296 × 10 2 = ⎜1.296 × 105 2 ⎟ ⎜⎜ h ⎠ ⎝ 3600 s ⎟⎠ h ⎝

2

5

(c) 60

⎛ 103 m ⎞ ⎜⎜ ⎟⎟ = 10.00 m/s 2 ⎝ km ⎠

mi ⎛ mi ⎞ ⎛ 5280 ft ⎞ ⎛ 1 h ⎞ ⎟⎜ ⎟ = 88 ft/s = ⎜ 60 ⎟ ⎜⎜ h ⎝ h ⎠ ⎝ 1 mi ⎟⎠ ⎜⎝ 3600 s ⎟⎠

mi ⎛ mi ⎞ ⎛ 1.609 km ⎞ ⎛ 103 m ⎞ ⎛ 1 h ⎞ ⎟⎜ ⎟⎜ ⎟ = 26.8 m/s = 27 m/s (d) 60 = ⎜ 60 ⎟ ⎜⎜ h ⎝ h ⎠ ⎝ 1 mi ⎟⎠ ⎜⎝ km ⎟⎠ ⎜⎝ 3600 s ⎟⎠

32 • There are 640 acres in a square mile. How many square meters are there in one acre? Picture the Problem We can use the conversion factor given in the problem statement and the fact that 1 mi = 1.609 km to express the number of square meters in one acre. Note that, because there are exactly 640 acres in a square mile, 640 acres has as many significant figures as we may wish to associate with it.

Measurement and Vectors

Multiply by 1 twice, properly chosen, to convert one acre into square miles, and then into square meters:

⎛ 1 mi 2 ⎞ ⎛ 1609 m ⎞ ⎟⎟ ⎜ 1acre = (1acre)⎜⎜ ⎟ ⎝ 640 acres ⎠ ⎝ mi ⎠

15

2

= 4045 m 2

33 •• [SSM] You are a delivery person for the Fresh Aqua Spring Water Company. Your truck carries 4 pallets. Each pallet carries 60 cases of water. Each case of water has 24 one-liter bottles. You are to deliver 10 cases of water to each convenience store along your route. The dolly you use to carry the water into the stores has a weight limit of 250 lb. (a) If a milliliter of water has a mass of 1 g, and a kilogram has a weight of 2.2 lb, what is the weight, in pounds, of all the water in your truck? (b) How many full cases of water can you carry on the cart? Picture the Problem The weight of the water in the truck is the product of the volume of the water and its weight density of 2.2 lb/L.

(a) Relate the weight w of the water on the truck to its volume V and weight density (weight per unit volume) D:

w = DV

Find the volume V of the water:

V = (4 pallets) (60

cases L ) (24 ) pallet case

= 5760 L

Substitute numerical values for D and L and evaluate w:

lb ⎞ ⎛ w = ⎜ 2.2 ⎟ (5760 L ) = 1.267 × 10 4 lb L⎠ ⎝ = 1.3 × 10 4 lb

(b) Express the number of cases of water in terms of the weight limit of the cart and the weight of each case of water:

N=

Substitute numerical values and evaluate N:

N=

weight limit of the cart weight of each case of water

250 lb = 4.7 cases lb ⎞ ⎛ L ⎞ ⎛ ⎜ 2.2 ⎟ ⎜ 24 ⎟ L ⎠ ⎝ case ⎠ ⎝

You can carry 4 cases.

16

Chapter 1

34 •• A right circular cylinder has a diameter of 6.8 in and a height of 2.0 ft. What is the volume of the cylinder in (a) cubic feet, (b) cubic meters, (c) liters? Picture the Problem The volume of a right circular cylinder is the area of its base multiplied by its height. Let d represent the diameter and h the height of the right circular cylinder; use conversion factors to express the volume V in the given units.

(a) Express the volume of the cylinder:

V = 14 πd 2 h

Substitute numerical values and evaluate V:

V = 14 π (6.8 in ) (2.0 ft ) 2

⎛ 1ft ⎞ ⎟⎟ = π (6.8 in ) (2.0 ft )⎜⎜ ⎝ 12 in ⎠

2

2

1 4

= 0.504 ft 3 = 0.50 ft 3 3

(b) Use the fact that 1 m = 3.281 ft to convert the volume in cubic feet into cubic meters:

⎛ 1m ⎞ ⎟⎟ = 0.0143 m 3 V = 0.504 ft ⎜⎜ ⎝ 3.281ft ⎠

(c) Because 1 L = 10−3 m3:

⎛ 1L ⎞ V = 0.0143m 3 ⎜⎜ −3 3 ⎟⎟ = 14 L ⎝ 10 m ⎠

(

3

)

= 0.014 m 3

(

)

[SSM] In the following, x is in meters, t is in seconds, v is in meters per second, and the acceleration a is in meters per second squared. Find the SI units of each combination: (a) v2/x, (b) x a , (c) 12 at2.

35

••

Picture the Problem We can treat the SI units as though they are algebraic quantities to simplify each of these combinations of physical quantities and constants.

(a) Express and simplify the units of v2/x: (b) Express and simplify the units of x a:

(m s )2 m

=

m2 m = 2 2 m⋅s s

m = s2 = s m/s 2

Measurement and Vectors (c) Noting that the constant factor 1 2 has no units, express and simplify

17

⎛m⎞ 2 ⎜ 2 ⎟(s ) = m ⎝s ⎠

the units of 12 at 2 :

Dimensions of Physical Quantities 36

What are the dimensions of the constants in each part of Problem 23?

•

Picture the Problem We can use the facts that each term in an equation must have the same dimensions and that the arguments of a trigonometric or exponential function must be dimensionless to determine the dimensions of the constants.

(a) x = C1 + C2 t L L L T T

(d) x

(b) x =

(e) v2

1 2

C1

t

L T2 T2

L

(c ) v 2 = 2 C1 2

L T2

37

L T2

•

2

=

L

⎛L⎞ ⎜ ⎟ ⎝T⎠

C1 cos C2 t 1 L T T

= 2 C1 v − (C2)2 x2 2

L L T T

1 2 L T

x L

The law of radioactive decay is N (t ) = N 0 e − λt , where N0 is the number

of radioactive nuclei at t = 0, N(t) is the number remaining at time t, and λ is a quantity known as the decay constant. What is the dimension of λ? Picture the Problem Because the exponent of the exponential function must be dimensionless, the dimension of λ must be T −1 . 38 •• The SI unit of force, the kilogram-meter per second squared (kg⋅m/s2) is called the newton (N). Find the dimensions and the SI units of the constant G in Newton’s law of gravitation F = Gm1m2/r2.

18

Chapter 1

Picture the Problem We can solve Newton’s law of gravitation for G and substitute the dimensions of the variables. Treating them as algebraic quantities will allow us to express the dimensions in their simplest form. Finally, we can substitute the SI units for the dimensions to find the units of G.

Solve Newton’s law of gravitation for G to obtain: Substitute the dimensions of the variables:

Use the SI units for L, M, and T to obtain:

Fr 2 m1m2

G=

ML 2 ×L 2 L3 T [G ] = = M2 MT 2

m3 kg ⋅ s 2

The magnitude of the force (F) that a spring exerts when it is stretched 39 •• a distance x from its unstressed length is governed by Hooke’s law, F = kx. (a) What are the dimensions of the force constant, k? (b) What are the dimensions and SI units of the quantity kx2? Picture the Problem The dimensions of mass and velocity are M and L/T, respectively. We note from Table 1-2 that the dimensions of force are ML/T2.

F x

(a) We know, from Hooke’s law, that:

k=

Write the corresponding dimensional equation:

[k ] = [F ] [x] L T2 = M L T2

Substitute the dimensions of F and x and simplify to obtain:

[k ] =

(b) Substitute the dimensions of k and x2 and simplify to obtain:

[kx ] = TM L

Substitute the units of kx2 to obtain:

M

2

2

kg ⋅ m 2 s2

2

=

ML 2 T2

Measurement and Vectors

19

40 •• Show that the product of mass, acceleration, and speed has the dimensions of power. Picture the Problem We note from Table 1-2 that the dimensions of power are ML2/T3. The dimensions of mass, acceleration, and speed are M, L/T2, and L/T respectively.

Express the dimensions of mav:

[mav] = M ×

From Table 1-2:

[P] = ML3

L L ML2 × = 3 T2 T T

2

T

Comparing these results, we see that the product of mass, acceleration, and speed has the dimensions of power. 41 •• [SSM] The momentum of an object is the product of its velocity and mass. Show that momentum has the dimensions of force multiplied by time. Picture the Problem The dimensions of mass and velocity are M and L/T, respectively. We note from Table 1-2 that the dimensions of force are ML/T2.

Express the dimensions of momentum:

[mv] = M × L = ML

From Table 1-2:

[F ] = ML2

T

T

T

Express the dimensions of force multiplied by time:

[Ft ] = ML2 × T = ML T

T

Comparing these results, we see that momentum has the dimensions of force multiplied by time. 42 •• What combination of force and one other physical quantity has the dimensions of power? Picture the Problem Let X represent the physical quantity of interest. Then we can express the dimensional relationship between F, X, and P and solve this relationship for the dimensions of X.

20

Chapter 1

Express the relationship of X to force and power dimensionally:

[F ][X ] = [P]

Solve for [X ] :

[X ] = [P] [F ]

Substitute the dimensions of force and power and simplify to obtain:

ML2 T3 = L [X ] = ML T 2 T

Because the dimensions of velocity are L/T, we can conclude that:

[P] = [F ][v]

Remarks: While it is true that P = Fv, dimensional analysis does not reveal the presence of dimensionless constants. For example, if P = πFv , the analysis shown above would fail to establish the factor of π. 43 •• [SSM] When an object falls through air, there is a drag force that depends on the product of the cross sectional area of the object and the square of its velocity, that is, Fair = CAv2, where C is a constant. Determine the dimensions of C. Picture the Problem We can find the dimensions of C by solving the drag force equation for C and substituting the dimensions of force, area, and velocity.

Fair Av 2

Solve the drag force equation for the constant C:

C=

Express this equation dimensionally:

[C ] = [Fair ]2 [A][v]

Substitute the dimensions of force, area, and velocity and simplify to obtain:

ML 2 [C ] = T 2 = M3 L ⎛L⎞ L2 ⎜ ⎟ ⎝T⎠

44 •• Kepler’s third law relates the period of a planet to its orbital radius r, the constant G in Newton’s law of gravitation (F = Gm1m2/r2), and the mass of the sun Ms. What combination of these factors gives the correct dimensions for the period of a planet?

Measurement and Vectors

21

Picture the Problem We can express the period of a planet as the product of the factors r, G, and MS (each raised to a power) and then perform dimensional analysis to determine the values of the exponents.

Express the period T of a planet as the product of r a , G b , and M Sc :

Solve the law of gravitation for the constant G: Express this equation dimensionally:

T = Cr a G b M Sc

(1)

where C is a dimensionless constant. G=

Fr 2 m1m2

2 [ F ][r ] [G ] = [m1 ][m2 ]

Substitute the dimensions of F, r, and m:

ML 2 × (L ) 2 L3 [G ] = T = M×M MT 2

Noting that the dimension of time is represented by the same letter as is the period of a planet, substitute the dimensions in equation (1) to obtain:

⎛ L3 ⎞ ⎟ (M )c T = (L ) ⎜⎜ 2 ⎟ ⎝ MT ⎠

Introduce the product of M 0 and L0 in the left hand side of the equation and simplify to obtain:

M 0 L0 T1 = M c −b La +3b T −2b

Equating the exponents on the two sides of the equation yields:

0 = c – b, 0 = a + 3b, and 1 = –2b

Solve these equations simultaneously to obtain:

a = 32 , b = − 12 , and c = − 12

Substitute for a, b, and c in equation (1) and simplify to obtain:

b

a

T = Cr 3 2 G −1 2 M S−1 2 =

C r3 2 GM S

Scientific Notation and Significant Figures 45 • [SSM] Express as a decimal number without using powers of 10 notation: (a) 3 × 104, (b) 6.2 × 10–3, (c) 4 × 10–6, (d) 2.17 × 105.

22

Chapter 1

Picture the Problem We can use the rules governing scientific notation to express each of these numbers as a decimal number.

(a) 3 × 10 4 = 30,000

(c) 4 × 10 −6 = 0.000004

(b) 6.2 × 10 −3 = 0.0062

(d) 2.17 × 105 = 217,000

46 • Write the following in scientific notation: (a) 1345100 m = ____ km, (b) 12340. kW = ____ MW, (c) 54.32 ps = ____ s, (d) 3.0 m = ____ mm Picture the Problem We can use the rules governing scientific notation to express each of these numbers in scientific notation.

(a) 1345100 m = 1.3451×106 m

( c) 54.32 ps = 54.32 × 10 −12 s = 5.432 × 10 −11 s

= 1.3451× 103 km

(b) 12340. kW = 1.2340 × 104 kW = 1.2340 × 107 W = 1.2340 × 101 MW

(d) 103 mm 3.0 m = 3.0 m × m = 3.0 × 103 mm

47 • [SSM] Calculate the following, round off to the correct number of significant figures, and express your result in scientific notation: (a) (1.14)(9.99 × 104), (b) (2.78 × 10–8) – (5.31 × 10–9), (c) 12π /(4.56 × 10–3), (d) 27.6 + (5.99 × 102). Picture the Problem Apply the general rules concerning the multiplication, division, addition, and subtraction of measurements to evaluate each of the given expressions.

(a) The number of significant figures in each factor is three; therefore the result has three significant figures:

(1.14)(9.99 × 104 ) =

1.14 × 105

Measurement and Vectors (b) Express both terms with the same power of 10. Because the first measurement has only two digits after the decimal point, the result can have only two digits after the decimal point:

23

(2.78 × 10 ) − (5.31 × 10 ) −8

−9

= (2.78 − 0.531) × 10−8 = 2.25 × 10−8

(c) We’ll assume that 12 is exact. Hence, the answer will have three significant figures:

12π = 8.27 × 103 −3 4.56 × 10

(d) Proceed as in (b):

27.6 + 5.99 × 10 2 = 27.6 + 599 = 627

(

)

= 6.27 × 10 2

48 • Calculate the following, round off to the correct number of significant figures, and express your result in scientific notation: (a) (200.9)(569.3), (b) (0.000000513)(62.3 × 107), (c) 28 401 + (5.78 × 104), (d) 63.25/(4.17 × 10–3). Picture the Problem Apply the general rules concerning the multiplication, division, addition, and subtraction of measurements to evaluate each of the given expressions.

(a) Note that both factors have four significant figures.

(200.9)(569.3) =

1.144 × 105

(b) Express the first factor in scientific notation and note that both factors have three significant figures.

(0.000000513)(62.3 × 107 ) = (5.13 × 10−7 )(62.3 × 107 ) =

3.20 × 10 2

(c) Express both terms in scientific notation and note that the second has only three significant figures. Hence the result will have only three significant figures. 28401 + (5.78 × 10 4 ) = (2.841 × 10 4 ) + (5.78 × 10 4 ) = (2.841 + 5.78) × 10 4 = 8.62 × 10 4

24

Chapter 1

(d) Because the divisor has three significant figures, the result will have three significant figures.

63.25 = 1.52 × 104 4.17 × 10 −3

49 • [SSM] A cell membrane has a thickness of 7.0 nm. How many cell membranes would it take to make a stack 1.0 in high? Picture the Problem Let N represent the required number of membranes and express N in terms of the thickness of each cell membrane.

Express N in terms of the thickness of a single membrane:

N=

1.0 in 7.0 nm

Convert the units into SI units and simplify to obtain: N=

1.0 in 2.540 cm 1m 1nm × × × −9 = 3.63 × 10 6 = 3.6 × 10 6 7.0 nm in 100 cm 10 m

50 •• A circular hole of radius 8.470 ×10−1 cm must be cut into the front panel of a display unit. The tolerance is 1.0 × 10−3 cm, which means the actual hole cannot differ by more than this much from the desired radius. If the actual hole is larger than the desired radius by the allowed tolerance, what is the difference between the actual area and the desired area of the hole? Picture the Problem Let r0 represent the larger radius and r the desired radius of the hole. We can find the difference between the actual and the desired area of the hole by subtracting the smaller area from the larger area.

(

Express the difference between the two areas in terms of r and r0:

ΔA = πr02 − πr 2 = π r02 − r 2

Factoring r02 − r 2 to obtain:

ΔA = π (r0 − r )(r0 + r )

)

Substitute numerical values and evaluate ΔA:

[

]

ΔA = π (1.0 × 10 −3 cm ) 8.470 × 10 −1 cm + (8.470 × 10 −1 cm + 1.0 × 10 −3 cm ) = 5.3 × 10 −3 cm 2

51 •• [SSM] A square peg must be made to fit through a square hole. If you have a square peg that has an edge length of 42.9 mm, and the square hole has an edge length of 43.2 mm, (a) what is the area of the space available when

Measurement and Vectors

25

the peg is in the hole? (b) If the peg is made rectangular by removing 0.10 mm of material from one side, what is the area available now? Picture the Problem Let sh represent the side of the square hole and sp the side of the square peg. We can find the area of the space available when the peg is in the hole by subtracting the area of the peg from the area of the hole.

(a) Express the difference between the two areas in terms of sh and sp:

ΔA = sh2 − s p2

Substitute numerical values and evaluate ΔA:

ΔA = (43.2 mm ) − (42.9 mm )

(b) Express the difference between

ΔA' = s h2 − s p l p

the area of the square hole and the rectangular peg in terms of sh ,sp and the new length of the peg lp:

2

2

≈ 26 mm 2

Substitute numerical values and evaluate ΔA′: ΔA' = (43.2 mm ) − (42.9 mm )(42.9 mm − 0.10 mm ) ≈ 30 mm 2 2

Vectors and Their Properties 52 • A vector 7.00 units long and a vector 5.50 units long are added. Their sum is a vector 10.0 units long. (a) Show graphically at least one way that this can be accomplished. (b) Using your sketch in Part (a), determine the angle between the original two vectors.

r r Picture the Problem Let A be the vector whose length is 7.00 units and let B be r r r r r the vector whose length is 5.50 units. C = A + B is the sum of A and B. The fact that their sum is 10.0 long tells us that the vectors are not collinear. We can use the r r of the vectors to find the angle between A and B. r r r (a) A graphical representation of vectors A , B and C is shown below. θ is the r r angle between A and B.

26

Chapter 1 y

B

r =5

.50

r = 10.0 C

α

θ

r A = 7.00

(b) The components of the vectors are related as follows:

x

Ax + Bx = C x

and Ay + B y = C y 7.00 + (5.50)cos θ = (10.0 )cos α and (5.50)sin θ = (10.0)sin α

Substituting for the components gives:

Squaring and adding these equations yields:

(100)sin 2 α + (100)cos 2 α = (5.50)sin 2 θ + [7.00 + (5.50)cosθ ]2 or

(100)(sin 2 α + cos 2 α ) = (5.50)sin 2 θ + [7.00 + (5.50)cosθ ]2

Because sin 2 α + cos 2 α = 1 : 100 = (5.50) sin 2 θ + [7.00 + (5.50)cosθ ] 2

Solve this equation for θ (you can (1) use the same trigonometric identity used in the previous step to eliminate either sin2θ or cos2θ in favor of the other and then solve the resulting equation or (2) use your graphing calculator’s SOLVER program) to obtain:

2

θ = 74.4°

Remarks: You could also solve Part (b) of this problem by using the law of cosines. 53 • [SSM] Determine the x and y components of the following three vectors in the xy plane. (a) A 10-m displacement vector that makes an angle of 30° clockwise from the +y direction. (b) A 25-m/s velocity vector that makes an

Measurement and Vectors

27

angle of −40° counterclockwise from the −x direction. (c) A 40-lb force vector that makes an angle of 120° counterclockwise from the −y direction. Picture the Problem The x and y components of these vectors are their projections onto the x and y axes. Note that the components are calculated using the angle each vector makes with the +x axis. y

(a) Sketch the displacement vector r (call it A ) and note that it makes an angle of 60° with the +x axis:

Ay

r A 30°

x

Ax

r Find the x and y components of A :

Ax = (10 m )cos 60° = 5.0 m

and Ay = (10 m )sin 60° = 8.7 m (b) Sketch the velocity vector (call r it v ) and note that it makes an angle of 220° with the +x axis:

y

vx

x

40°

r v

vy r Find the x and y components of v :

v x = (25 m/s )cos 220° = − 19 m/s and

v y = (25 m/s )sin 220° = − 16 m/s

28

Chapter 1

y

(c) Sketch the force vector (call it r F ) and note that it makes an angle of 30° with the +x axis:

r F

x

120°

Find the x and y components of r F:

Fx = (40 lb )cos 30° = 35 lb and

Fy = (40 lb )sin 30° = 20 lb

54 • Rewrite the following vectors in terms of their magnitude and angle (counterclockwise from the + x direction). (a) A displacement vector with an x component of +8.5 m and a y component of −5.5 m. (b) A velocity vector with an x component of −75 m/s and a y component of +35 m/s. (c) A force vector with a magnitude of 50 lb that is in the third quadrant with an x component whose magnitude is 40 lb. Picture the Problem We can use the Pythagorean Theorem to find magnitudes of these vectors from their x and y components and trigonometry to find their direction angles.

Use the Pythagorean Theorem to r find the magnitude of A :

r Find the direction angle of A :

y

Ax = 8.5 m

θ

x

r A

A y = −5.5 m

(a) Sketch the components ofr the displacement vector (call it A ) and show their resultant:

A = Ax2 + Ay2 =

(8.5 m ) 2 + (− 5.5 m ) 2

= 10 m ⎛ Ay ⎝ Ax

θ = tan −1 ⎜⎜

⎞ − 5 .5 m ⎞ ⎟⎟ = tan −1 ⎛⎜ ⎟ ⎝ 8 .5 m ⎠ ⎠

= − 33° = 327°

Measurement and Vectors

vy = 35 m/s

y

(b) Sketch the components of the r velocity vector (call it v ) and show their resultant:

r v

θ

α vx =

Use the Pythagorean Theorem to r find the magnitude of v and trigonometry to find its direction angle:

29

x

− 75 m/s

(− 75 m/s) 2 + (35 m/s) 2

v = v x2 + v y2 = = 83 m/s ⎛ vy ⎝ vx

⎞ 35 m/s ⎞ ⎟⎟ = tan −1 ⎛⎜ ⎟ = 25° ⎝ 75 m/s ⎠ ⎠

α = tan −1 ⎜⎜

and θ = 180° − α = 180° − 25° = 155° r (c) Sketch the force vector (call it F ) and its x-component and show Fy:

y F x = −40 lb

θ

x

α 0l

F

Use trigonometry to find α and, hence, θ :

Use the relationship between a vector and its vertical component to find Fy:

=5

b

r Fy

⎛ Fx ⎞ −1 ⎛ 40 lb ⎞ ⎟ = cos ⎜ ⎟ = 37° ⎝ 50 lb ⎠ ⎝F ⎠ θ = 180° + α = 180° + 37° = 217°

α = cos −1 ⎜

Fy = F sin θ = (50 lb)sin 217° = − 30 lb

Remarks: In Part (c) we could have used the Pythagorean Theorem to find the magnitude of Fy. 55 • You walk 100 m in a straight line on a horizontal plane. If this walk took you 50 m east, what are your possible north or south movements? What are the possible angles that your walk made with respect to due east?

30

Chapter 1

Picture the Problem There are two directions that you could have walked that are consistent with your having walked 50 m to the east. One possibility is walking in the north-east direction and the other is walking in the south-east direction. We can use the Pythagorean Theorem to find the distance you walked either north or south and then use trigonometry to find the two angles that correspond to your having walked in either of these directions.

Letting A represent the magnitude of your displacement on the walk, use the Pythagorean Theorem to relate A to its horizontal (Ax) and vertical (Ay) components:

Ay = A 2 − Ax2

Substitute numerical values and evaluate Ay:

Ay =

(100 m )2 − (50 m )2

= ±87 m

The plus-and-minus-signs mean that you could have gone 87 m north or 87 m south. Use trigonometry to relate the directions you could have walked to the distance you walked and its easterly component:

⎛ 50 m ⎞ ⎟ = ±60° ⎝ 100 m ⎠

θ = cos −1 ⎜

The plus-and-minus-signs mean that you could have walked 60° north of east or 60° south of east. 56 • The final destination of your journey is 300 m due east of your starting point. The first leg of this journey is the walk described in Problem 55, and the second leg is also a walk along a single straight-line path. Estimate graphically the length and heading for the second leg of your journey.

r Picture the Problem Let A be the vector whose length is 100 m and whose r direction is 60° N of E. Let B be the vector whose direction and magnitude we are to determine and assume that you initially walked in a direction north of east. The graphical representation that we can use to estimate these quantities is shown r below. Knowing that the magnitude of A is 100 m, use any convenient scale to r determine the length of B and a protractor to determine the value of θ.

Measurement and Vectors

31

N r B

r A =1

00

m

θ

86.6 m

60°

E

250 m

50 m

The magnitude of the second leg of your journey is about 260 m at an angle of approximately 20° S of E. If you had initially walked into the southeast, the magnitude of the second leg of your journey would still be about 260 m but its direction would be approximately 20° N of E. Remarks: If you use the Pythagorean Theorem and right-triangle trigonometry, you’ll find that the length of the second leg of your journey is 265 m and that θ = 19° S of E. r r A = 3.4iˆ + 4.7 ˆj , B = (− 7.7)iˆ + 3.2 ˆj , and 57 •• Given the following vectors: r r Cr = 5.4r iˆ + (r− 9.1)rˆj . (a) Find the vector D , in unit vector notation, such that D + 2 A − 3C + 4 B = 0 . (b) Express your answer in Part (a) in terms of magnitude and angle with the +x direction.

r Picture the Problem We can find the vector D by solving the equation r r r r r r r r D + 2 A − 3C + 4 B = 0 for D and then substituting for the vectors A, B and C . In r (b) we can use the components of D to find its magnitude and direction. r r r r D = −2 A + 3C − 4 B

(a) Solve the vector equation that gives the condition that must r satisfied for D :

r r r Substitute for A, C and B and simplify to obtain:

(

) (

) (

r D = −2 3.4iˆ + 4.7 ˆj + 3 5.4iˆ − 9.1 ˆj − 4 − 7.7iˆ + 3.2 ˆj = (− 6.8 + 16.2 + 30.8)iˆ + (− 9.4 − 27.3 − 12.8) ˆj = 40.2 iˆ − 49.5 ˆj = 40 iˆ − 50 ˆj

(b) Use the Pythagorean Theorem to r relate the magnitude of D to its components Dx and Dy:

D = Dx2 + D y2

)

32

Chapter 1

(40.2)2 + (− 49.5)2

Substitute numerical values and evaluate D:

D=

Use trigonometry to express and evaluate the angle θ :

θ = tan −1 ⎜⎜

⎛ Dy ⎝ Dx

= 63.8

⎞ − 49.5 ⎞ ⎟⎟ = tan −1 ⎛⎜ ⎟ ⎝ 40.2 ⎠ ⎠

= − 51°

r where the minus sign means that D is in the 4th quadrant. r 58 •• Given the following force vectors: A is 25 lb at an angle of 30° r clockwise from the +x axis, and B is 42 lb at an angle of 50° clockwise from the +y axis.r(a) Make a sketch r r andr visually estimate the magnitude and angle of the vector C such that 2 A + C − B results in a vector with a magnitude of 35 lb pointing in the +x direction. (b) Repeat the calculation in Part (a) using the method of components and compare your result to the estimate in (a).

r r r Picture the Problem A diagram showing the condition that 2 A + C − B = 35iˆ and from which one can scale the values of C and θ is shown below. In (b) we can use the two scalar equations corresponding to this vector equation to check our graphical results.

(a) A diagram showing the conditions r r r imposed by 2 A + C − B = 35iˆ approximately to scale is shown to the right.

y r B 50°

35iˆ

x

30°

r C

r The magnitude of C is approximately 57 lb and the angle θ is

The corresponding scalar equations are:

50°

r −B

approximately 68°.

(b) Express the condition relating the r r r vectors A, B, and C :

r 2A

θ

r r r 2 A + C − B = 35iˆ

2 Ax + C x − Bx = 35 and 2 Ay + C y − B y = 0

Measurement and Vectors Solve these equations for Cx and Cy to obtain:

Substitute for the x and y r r components of A and B to obtain:

33

C x = 35 − 2 Ax + Bx

and C y = −2 Ay + B y C x = 35 lb − 2[(25 lb)cos 330°]

+ (42 lb )cos 40°

= 23.9 lb

and C y = −2[(25 lb)sin 330°] + (42 lb)sin 40° = 52 lb

Use the Pythagorean Theorem to r relate the magnitude of C to its components:

C = C x2 + C y2

Substitute numerical values and evaluate C:

C=

Use trigonometry to find the r direction of C :

θ = tan −1 ⎜⎜

(23.9 lb)2 + (52.0 lb)2 ⎛ Cy ⎝ Cx

= 57 lb

⎞ 52.0 lb ⎞ ⎟⎟ = tan −1 ⎛⎜ ⎟ = 65° ⎝ 23.9 lb ⎠ ⎠

Remarks: The analytical results for C and θ are in excellent agreement with the values determined graphically. [SSM] Calculate the unit vector (in terms of iˆ and ˆj ) in the direction 59 •• opposite to the direction of each of the vectors in Problem 57. Picture the Problem The unit vector in the direction opposite to the direction of a vector is found by taking the negative of the unit vector in the direction of the vector. The unit vector in the direction of a given vector is found by dividing the given vector by its magnitude.

The unit vector in the direction of r A is given by:

r A ˆ A= r = A

r Substitute for A, Ax, and Ay and evaluate Aˆ :

Aˆ =

r A Ax2 + Ay2

3.4iˆ + 4.7 ˆj

(3.4) + (4.7) 2

2

= 0.59iˆ + 0.81 ˆj

34

Chapter 1

The unit vector in the direction r opposite to that of A is given by:

− Aˆ = − 0.59iˆ − 0.81 ˆj

The unit vector in the direction of r B is given by:

r B ˆ B= r = B

r Substitute for B , Bx, and By and evaluate Bˆ :

Bˆ =

r B Bx2 + B y2

− 7.7 iˆ + 3.2 ˆj

(− 7.7 )2 + (3.2)2

= −0.92iˆ + 0.38 ˆj

The unit vector in the direction r opposite to that of B is given by:

− Bˆ = 0.92iˆ − 0.38 ˆj

The unit vector in the direction of r C is given by:

r r C C ˆ C= r = C C x2 + C y2

r Substitute for C , Cx, and Cy and evaluate Cˆ :

Cˆ =

The unit vector in the direction r opposite that of C is given by:

− Cˆ = − 0.51iˆ + 0.86 ˆj

5.4iˆ − 9.1 ˆj

(5.4) + (− 9.1) 2

2

= 0.51iˆ − 0.86 ˆj

Unit vectors iˆ and ˆj are directed east and north, respectively. Calculate the unit vector (in terms of iˆ and ˆj ) in the following directions. (a) northeast,

60

••

(b) 70° clockwise from the −y axis, (c) southwest. Picture the Problem The unit vector in a given direction is a vector pointing in that direction whose magnitude is 1.

(a) The unit vector in the northeast direction is given by:

uˆ NE = (1) cos 45°iˆ + (1) sin 45° ˆj

(b) The unit vector 70° clockwise from the −y axis is given by:

uˆ = (1) cos 200°iˆ + (1) sin 200° ˆj

= 0.707iˆ + 0.707 ˆj

= − 0.940iˆ − 0.342 ˆj

Measurement and Vectors (c) The unit vector in the southwest direction is given by:

35

uˆSW = (1) cos 225°iˆ + (1) sin 225° ˆj = − 0.707iˆ − 0.707 ˆj

Remarks: One can confirm that a given vector is, in fact, a unit vector by checking its magnitude.

General Problems 61 • [SSM] The Apollo trips to the moon in the 1960's and 1970's typically took 3 days to travel the Earth-moon distance once they left Earth orbit. Estimate the spacecraft's average speed in kilometers per hour, miles per hour, and meters per second. Picture the Problem Average speed is defined to be the distance traveled divided by the elapsed time. The Earth-moon distance and the distance and time conversion factors can be found on the inside-front cover of the text. We’ll assume that 3 days means exactly three days.

Express the average speed of Apollo as it travels to the moon:

vav =

distance traveled elapsed time

Substitute numerical values to obtain:

vav =

2.39 × 10 5 mi 3d

Use the fact that there are 24 h in 1 d to convert 3 d into hours:

vav =

2.39 × 10 5 mi 24 h 3d× d = 3.319 × 10 3 mi/h = 3.32 × 10 3 mi/h

Use the fact that 1 mi is equal to 1.609 km to convert the spacecraft’s average speed to km/h:

vav = 3.319 × 10 3

km km mi × 1.609 = 5.340 × 10 3 = 5.34 × 10 3 km/h h mi h

36

Chapter 1

Use the facts that there are 3600 s in 1 h and 1000 m in 1 km to convert the spacecraft’s average speed to m/s:

vav = 5.340 × 10 6

km 10 3 m 1h × × = 1.485 × 10 3 m/s = 1.49 × 10 3 m/s h km 3600 s

Remarks: An alternative to multiplying by 103 m/km in the last step is to replace the metric prefix ″k″ in ″km″by 103. 62 • On many of the roads in Canada the speed limit is 100 km/h. What is this speed limit in miles per hour? Picture the Problem We can use the conversion factor 1 mi = 1.609 km to convert 100 km/h into mi/h.

Multiply 100 km/h by 1 mi/1.609 km to obtain:

100

km 1 mi km = 100 × h 1.609 km h = 62.2 mi/h

63 • If you could count $1.00 per second, how many years would it take to count 1.00 billion dollars? Picture the Problem We can use a series of conversion factors to convert 1 billion seconds into years.

Multiply 1 billion seconds by the appropriate conversion factors to convert into years: 109 s = 109 s ×

1h 1day 1y × × = 31.7 y 3600 s 24 h 365.24 days

(a) The speed of light in vacuum is 186,000 mi/s = 3.00 × 108 m/s. Use this fact to find the number of kilometers in a mile. (b) The weight of 1.00 ft3 of water is 62.4 lb, and 1.00 ft = 30.5 cm. Use this and the fact that 1.00 cm3 of water has a mass of 1.00 g to find the weight in pounds of a 1.00-kg mass. 64

•

Picture the Problem In both the examples cited we can equate expressions for the physical quantities, expressed in different units, and then divide both sides of the equation by one of the expressions to obtain the desired conversion factor.

Measurement and Vectors (a) Divide both sides of the equation expressing the speed of light in the two systems of measurement by 186,000 mi/s to obtain:

37

3.00 × 108 m/s = 1.61 × 10 3 m/mi 1= 5 1.86 × 10 mi/h m ⎞ ⎛ 1 km ⎞ ⎛ ⎟ = ⎜1.61 × 10 3 ⎟⎜ mi ⎠ ⎜⎝ 10 3 m ⎟⎠ ⎝ = 1.61 km/mi

(b) Find the volume of 1.00 kg of water:

Volume of 1.00 kg = 103 g is 103 cm3

Express 103 cm3 in ft3:

⎞ ⎛ (10 cm ) ⎜⎜ 1.00 ft ⎟⎟ = 0.03525 ft 3 ⎝ 30.5 cm ⎠

Relate the weight of 1.00 ft3 of water to the volume occupied by 1.00 kg of water:

lb 1.00 kg = 62.4 3 3 ft 0.03525 ft

Divide both sides of the equation by the left-hand side to obtain:

lb ft 3 = 2.20 lb/kg 1= 1.00 kg 0.03525 ft 3

3

3

62.4

65 • The mass of one uranium atom is 4.0 × 10–26 kg. How many uranium atoms are there in 8.0 g of pure uranium? Picture the Problem We can use the given information to equate the ratios of the number of uranium atoms in 8 g of pure uranium and of 1 atom to its mass.

Express the proportion relating the number of uranium atoms NU in 8.0 g of pure uranium to the mass of 1 atom:

NU 1atom = 8.0 g 4.0 × 10 − 26 kg

Solve for and evaluate NU:

⎞ ⎛ 1 kg ⎞ ⎛ 1atom ⎟ N U = ⎜⎜ 8.0 g × 3 ⎟⎟ ⎜⎜ − 26 10 g ⎠ ⎝ 4.0 × 10 kg ⎟⎠ ⎝ = 2.0 × 10 23

38

Chapter 1

66 •• During a thunderstorm, a total of 1.4 in of rain falls. How much water falls on one acre of land? (1 mi2 = 640 acres.) Express your answer in (a) cubic inches, (b) cubic feet, (c) cubic meters, and (d) kilograms. Note that the density of water is 1000 kg/m3. Picture the Problem Assuming that the water is distributed uniformly over the one acre of land, its volume is the product of the area over which it is distributed and its depth. The mass of the water is the product of its density and volume. The required conversion factors can be found in the front material of the text. V = Ah

(a) Express the volume V of water in terms of its depth h and the area A over which it falls:

Substitute numerical values and evaluate V: 2

2

⎛ 1 mi 2 ⎞ ⎛ 5280 ft ⎞ ⎛ 12 in ⎞ 6 3 ⎟⎟ ⎜ V = (1 acre)⎜⎜ ⎟ ⎜ ⎟ (1.4 in ) = 8.782 × 10 in ⎝ 640 acres ⎠ ⎝ mi ⎠ ⎝ ft ⎠ = 8.8 × 10 6 in 3

(b) Convert in3 into ft3:

⎛ 1 ft ⎞ V = 8.782 × 10 in × ⎜ ⎟ ⎝ 12 in ⎠ 6

3

3

= 5.082 × 10 3 ft 3 = 5.1× 10 3 ft 3

(c) Convert ft3 into m3: 3

3

3

⎛ 12 in ⎞ ⎛ 2.540 cm ⎞ ⎛ 1 m ⎞ 3 V = 5.082 × 10 ft × ⎜ ⎟ = 1.439 × 102 m ⎟ ×⎜ ⎟ ×⎜ ft in 100 cm ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 3

3

= 1.4 × 10 2 m 3

(d) The mass of the water is the product of its density ρ and volume V:

m = ρV

Substitute numerical values and evaluate m:

kg ⎞ ⎛ m = ⎜10 3 3 ⎟ 1.439 × 10 2 m 3 m ⎠ ⎝

(

= 1.4 × 10 5 kg

)

Measurement and Vectors

39

67 •• An iron nucleus has a radius of 5.4 × 10–15 m and a mass of 9.3 × 10–26 kg. (a) What is its mass per unit volume in kg/ m3? (b) If Earth had the same mass per unit volume, what would be its radius? (The mass of Earth is 5.98 × 1024 kg.) Picture the Problem The mass per unit volume of an object is its density.

(a) The density ρ of an object is its mass m per unit volume V:

ρ=

m V

Assuming the iron nucleus to be spherical, its volume as a function of its radius r is given by:

4 V = πr 3 3

Substitute for V and simplify to obtain:

ρ=

Substitute numerical values and evaluate ρ:

ρ=

m 4 3 πr 3

=

3m 4πr 3

( 4π (5.4 × 10

(1)

) m)

3 9.3 × 10 −26 kg −15

3

= 1.410 × 1017 kg/m 3 = 1.4 × 1017 kg/m 3 (b) Solve equation (1) for r to obtain: Substitute numerical values and evaluate r:

r =3

3m 4πρ

(

)

3 5.98 ×10 24 kg r= 3 kg ⎞ ⎛ 4π ⎜1.410 ×1017 3 ⎟ m ⎠ ⎝ = 2.2 ×10 2 m

or about 200 m! 68 •• The Canadian Norman Wells Oil Pipeline extends from Norman Wells, Northwest Territories, to Zama, Alberta. The 8.68 ×105-m-long pipeline has an inside diameter of 12 in and can be supplied with oil at 35 L/s. (a) What is the volume of oil in the pipeline if it is full at some instant in time? (b) How long would it take to fill the pipeline with oil if it is initially empty?

40

Chapter 1

Picture the Problem The volume of a cylinder is the product of its crosssectional area and its length. The time required to fill the pipeline with oil is the ratio of its volume to the flow rate R of the oil. We’ll assume that the pipe has a diameter of exactly 12 in.

V = π r2L

(a) Express the volume V of the cylindrical pipe in terms of its radius r and its length L:

Substitute numerical values and evaluate V: 2

⎛ 2.540 × 10 - 2 m ⎞ ⎟⎟ 8.68 × 10 5 m = 6.3 × 10 4 m 3 V = π ⎜⎜ 6 in × in ⎝ ⎠ (b) Express the time Δt to fill the pipe in terms of its volume V and the flow rate R of the oil: Substitute numerical values and evaluate Δt:

(

)

Δt =

V R

6.3 × 10 4 m 3 Δt = = 1.8 × 10 6 s 3 −3 ⎛ L ⎞ ⎛ 10 m ⎞ ⎟⎟ ⎜ 35 ⎟ ⎜⎜ ⎝ s ⎠⎝ L ⎠ or about 21 days!

69 •• The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the sun, namely 1.496 × 1011 m. The parsec is the radius of a circle for which a central angle of 1 s intercepts an arc of length 1 AU. The light-year is the distance that light travels in 1 y. (a) How many parsecs are there in one astronomical unit? (b) How many meters are in a parsec? (c) How many meters in a light-year? (d) How many astronomical units in a light-year? (e) How many light-years in a parsec? Picture the Problem We can use the relationship between an angle θ, measured in radians, subtended at the center of a circle, the radius R of the circle, and the length L of the arc to answer these questions concerning the astronomical units of measure. We’ll take the speed of light to be 2.998 × 108 m/s.

(a) Relate the angle θ subtended by an arc of length S to the distance R:

θ=

S ⇒ S = Rθ R

(1)

Measurement and Vectors

41

Substitute numerical values and evaluate S: ⎛ 1 min ⎞ ⎛ 1° ⎞ ⎛ 2π rad ⎞ −6 ⎟⎟ ⎜⎜ ⎟⎟ ⎜ S = (1 parsec)(1s )⎜⎜ ⎟ = 4.848 × 10 parsec ⎝ 60 s ⎠ ⎝ 60 min ⎠ ⎝ 360° ⎠

(b) Solving equation (1) for R yields:

R=

Substitute numerical values and evaluate R:

R=

S

θ 1.496 × 1011 m ⎛ ⎞⎛ ⎞ (1s ) ⎜⎜ 1min ⎟⎟ ⎜⎜ 1° ⎟⎟ ⎛⎜ 2π rad ⎞⎟ ⎝ 60 s ⎠ ⎝ 60 min ⎠ ⎝ 360° ⎠

= 3.086 × 1016 m D = cΔt

(c) The distance D light travels in a given interval of time Δt is given by:

Substitute numerical values and evaluate D: s ⎞ m ⎞⎛ d ⎞ ⎛ h ⎞ ⎛ min ⎞ ⎛ ⎛ 15 D = ⎜ 2.998× 108 ⎟ ⎜⎜ 365.24 ⎟⎟ ⎜ 24 ⎟ ⎜ 60 ⎟ ⎜ 60 ⎟ = 9.461 × 10 m s ⎠⎝ y ⎠⎝ d ⎠⎝ h ⎠ ⎝ min ⎠ ⎝ ⎛ ⎞ 1 AU ⎟⎟ 1c ⋅ y = 9.461 × 1015 m ⎜⎜ 11 ⎝ 1.496 × 10 m ⎠

(d) Use the definition of 1 AU and the result from Part (c) to obtain:

(

)

= 6.324 × 10 4 AU

(e) Combine the results of Parts (b) and (c) to obtain: ⎛ ⎞ 1c ⋅ y ⎟⎟ = 3.262 c ⋅ y 1 parsec = 3.086 × 1016 m ⎜⎜ 15 ⎝ 9.461 × 10 m ⎠

(

)

70 •• If the average density of the universe is at least 6 × 10–27 kg/m3, then the universe will eventually stop expanding and begin contracting. (a) How many electrons are needed in each cubic meter to produce the critical density? (b) How many protons per cubic meter would produce the critical density? (me = 9.11 × 10–31 kg; mp = 1.67 × 10–27 kg.)

42

Chapter 1

Picture the Problem Let Ne and Np represent the number of electrons and the number of protons, respectively and ρ the critical average density of the universe. We can relate these quantities to the masses of the electron and proton using the definition of density. N m N e me ρ = ⇒ e = V V V me

(a) Using its definition, relate the required density ρ to the electron density Ne/V:

ρ=

Substitute numerical values and evaluate Ne/V:

Ne 6 × 10 −27 kg/m 3 = V 9.11 × 10 −31 kg/electron

(1)

= 6.586 × 10 3 electrons/m 3 ≈ 7 × 10 3 electrons/m 3

(b) Express and evaluate the ratio of the masses of an electron and a proton:

me 9.11 × 10 −31 kg = = 5.455 × 10 − 4 − 27 mp 1.67 × 10 kg

Rewrite equation (1) in terms of protons:

Np

Divide equation (2) by equation (1) to obtain:

Np

V

=

ρ mp

(2)

V = me or N p = me ⎛ N e ⎞ ⎜ ⎟ Ne V mp ⎝ V ⎠ mp V

Substitute numerical values and use the result from Part (a) to evaluate Np/V:

Np V

= (5.455 × 10 −4 )(6.586 × 10 3 electrons/m 3 ) ≈ 4 protons/m 3

71 ••• You are an astronaut doing physics experiments on the moon. You are interested in the experimental relationship between distance fallen, y, and time elapsed, t, of falling objects dropped from rest. You have taken some data for a falling penny, which is represented in the table below. You expect that a general relationship between distance y and time t is y = Bt C , where B and C are constants to be determined experimentally. To accomplish this, create a log-log plot of the data: (a) graph log(y) vs. log(t), with log(y) the ordinate variable and log(t) the abscissa variable. (b) Show that if you take the log of each side of your equation, you get log(y) = log(B) + Clog(t). (c) By comparing this linear relationship to the graph of the data, estimate the values of B and C. (d) If you

Measurement and Vectors

43

drop a penny, how long should it take to fall 1.0 m? (e) In the next chapter, we will show that the expected relationship between y and t is y = 12 at 2 , where a is the acceleration of the object. What is the acceleration of objects dropped on the moon?

y (m) 10 20 30 40 50 t (s) 3.5 5.2 6.0 7.3 7.9 Picture the Problem We can plot log y versus log t and find the slope of the bestfit line to determine the exponent C. The value of B can be determined from the intercept of this graph. Once we know C and B, we can solve y = Bt C for t as a function of y and use this result to determine the time required for an object to fall a given distance on the surface of the moon.

(a) The following graph of log y versus log t was created using a spreadsheet program. The equation shown on the graph was obtained using Excel’s ″Add Trendline″ function. (Excel’s ″Add Trendline″ function uses regression analysis to generate the trendline.)

1.80 1.70 1.60

log y = 1.9637log t - 0.0762

1.50

log y

1.40 1.30 1.20 1.10 1.00 0.90 0.80 0.50

0.60

0.70

0.80

0.90

1.00

log t

(b) Taking the logarithm of both sides of the equation y = Bt C yields: (c) Note that this result is of the form:

log y = log(Bt C ) = log B + log t C

= log B + C log t Y = b + mX where Y = log y, b = log B, m = C, and X = log t

44

Chapter 1

From the regression analysis (trendline) we have:

log B = −0.076

Solving for B yields:

B = 10 −0.076 = 0.84 m/s 2

where we have inferred the units from tho C given for y = Bt . Also, from the regression analysis we have: (d) Solve y = Bt for t to obtain: C

C = 1.96 ≈ 2.0

1

⎛ y ⎞C t =⎜ ⎟ ⎝B⎠ 1

Substitute numerical values and evaluate t to determine how long it would take a penny to fall 1.0 m:

⎛ ⎞2 ⎜ 1.0 m ⎟ ⎟ ≈ 1.1 s t =⎜ ⎜ 0.84 m ⎟ ⎜ ⎟ s2 ⎠ ⎝

(e) Substituting for B and C in y = Bt C yields:

m⎞ ⎛ y = ⎜ 0.84 2 ⎟ t 2 s ⎠ ⎝

Compare this equation to y = 12 at 2 to obtain:

1 2

a = 0.84

m s2

and m⎞ m ⎛ a = 2⎜ 0.84 2 ⎟ = 1.7 2 s ⎠ s ⎝

Remarks: One could use a graphing calculator to obtain the results in Parts (a) and (c).

A particular company’s stock prices vary with the market and with 72 ••• the company’s type of business, and can be very unpredictable, but people often try to look for mathematical patterns where they may not belong. Corning is a materials-engineering company located in upstate New York. Below is a table of the price of Corning stock on August 3, for every 5 years from 1981 to 2001. Assume that the price follows a power law: price (in $) = BtC where t is expressed in years. (a) Evaluate the constants B and C. (b) According to the power law, what should the price of Corning stock have been on August 3, 2000? (It was actually $82.83!)

Measurement and Vectors Price (dollars) Years since 1980

2.10 1

4.19 6

9.14 11

10.82 16

45

16.85 21

Picture the Problem We can plot log P versus log t and find the slope of the best-fit line to determine the exponent C. The value of B can be determined from the intercept of this graph. Once we know C and B, we can use P = Bt C to predict the price of Corning stock as a function of time.

(a) The following graph of log P versus log was created using a spreadsheet program. The equation shown on the graph was obtained using Excel’s ″Add Trendline″ function. (Excel’s ″Add Trendline″ function uses regression analysis to generate the trendline.) 1.40 1.20

log P = 0.658log t + 0.2614

log P

1.00 0.80 0.60 0.40 0.20 0.00 0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

log t

Taking the logarithm of both sides of the equation P = Bt C yields:

log P = log(Bt C ) = log B + log t C = log B + C log t

Note that this result is of the form:

Y = b + mX where Y = log P, b = log B, m = C, and X = log t

From the regression analysis (trendline) we have:

log B = 0.2614⇒ B = 10 0.2614 = $1.83

Also, from the regression analysis we have:

C = 0.658

(b) Substituting for B and C in P = Bt C yields:

P = ($1.83)t 0.658

46

Chapter 1

Evaluate P(20 y) to obtain:

P(20 y ) = ($1.83)(20 )

0.658

= $13.14

Remarks: One could use a graphing calculator to obtain these results. 73 ••• [SSM] The Super-Kamiokande neutrino detector in Japan is a large transparent cylinder filled with ultra pure water. The height of the cylinder is 41.4 m and the diameter is 39.3 m. Calculate the mass of the water in the cylinder. Does this match the claim posted on the official Super-K Web site that the detector uses 50000 tons of water? Picture the Problem We can use the definition of density to relate the mass of the water in the cylinder to its volume and the formula for the volume of a cylinder to express the volume of water used in the detector’s cylinder. To convert our answer in kg to lb, we can use the fact that 1 kg weighs about 2.205 lb.

Relate the mass of water contained in the cylinder to its density and volume:

m = ρV

π

Express the volume of a cylinder in terms of its diameter d and height h:

V = Abase h =

Substitute in the expression for m to obtain:

m=ρ

Substitute numerical values and evaluate m:

⎛π ⎞ 2 m = (10 3 kg/m 3 ) ⎜ ⎟ (39.3 m ) (41.4 m ) ⎝4⎠ 7 = 5.022 × 10 kg

Convert 5.02 × 107 kg to tons:

π 4

4

d 2h

d 2h

m = 5.022 × 10 7 kg ×

2.205 lb 1 ton × kg 2000 lb

= 55.4 × 10 3 ton The 50,000-ton claim is conservative. The actual weight is closer to 55,000 tons. 74 ••• You and a friend are out hiking across a large flat plain and decide to determine the height of a distant mountain peak, and also the horizontal distance from you to the peak Figure 1-19). In order to do this, you stand in one spot and determine that the sightline to the top of the peak is inclined at 7.5° above the horizontal. You also make note of the heading to the peak at that point: 13° east of north. You stand at the original position, and your friend hikes due west for 1.5 km. He then sights the peak and determines that its sightline has a heading of 15° east of north. How far is the mountain from your position, and how high is its summit above your position?

Measurement and Vectors

47

r Picture the Problem Vector A lies in the plane of the plain and locates the base r or the peak relative to you. Vector B also lies in the plane of the plain and locates the base of the peak relative to your friend when he/she has walked 1.5 km to the west. We can use the geometry of the diagram and the E-W components of the r r r vectors d , B and A to find the distance to the mountain from your position. Once we know the distance A, we can use a trigonometric relationship to find the height of the peak above your position. N

Peak

r B 13° 15°

W

β

r d d = 1.5 km

r A

d

α

Your friend's position

E Your position

Referring to the diagram, note that:

B sin β = h and A sin α = h

Equating these expressions for h gives:

B sin β = A sin α ⇒ B =

Adding the E-W components of the r r r vectors d , B and A yields:

1.5 km = B cos β − A cos α

Substitute for B and simplify to obtain:

1.5 km = A

sin α A sin β

sin α cos β − A cos α sin β

⎛ sin α ⎞ = A⎜⎜ − cos α ⎟⎟ ⎝ tan β ⎠ Solving for A yields:

A=

1.5 km sin α − cos α tan β

48

Chapter 1

Substitute numerical values and evaluate A:

A=

1.5 km = 41.52 km sin 77° − cos 77° tan 75°

= 42 km Referring to the following diagram, we note that: h

7.5°

h = A tan 7.5° = (41.52 km ) tan 7.5° = 5.5 km

A

Remarks: One can also solve this problem using the law of sines. 75 ••• The table below gives the periods T and orbit radii r for the motions of four satellites orbiting a dense, heavy asteroid. (a) These data can be fitted by the formula T = Cr n . Find the values of the constants C and n. (b) A fifth satellite is discovered to have a period of 6.20 y. Find the radius for the orbit of this satellite, which fits the same formula.

Period T, y 0.44 1.61 3.88 7.89 Radius r, Gm 0.088 0.208 0.374 0.600 Picture the Problem We can plot log T versus log r and find the slope of the best-fit line to determine the exponent n. We can then use any of the ordered pairs to evaluate C. Once we know n and C, we can solve T = Cr n for r as a function of T.

(a) Take the logarithm (we’ll arbitrarily use base 10) of both sides of T = Cr n and simplify to obtain:

log(T ) = log(Cr n ) = log C + log r n = n log r + log C Note that this equation is of the form y = mx + b . Hence a graph of log T vs. log r should be linear with a slope of n and a log T -intercept log C.

Measurement and Vectors

49

The following graph of log T versus log r was created using a spreadsheet program. The equation shown on the graph was obtained using Excel’s ″Add Trendline″ function. (Excel’s ″Add Trendline″ function uses regression analysis to generate the trendline.) 1.0 0.8

log T = 1.5036log r + 1.2311

log T

0.6 0.4 0.2 0.0 -0.2 -0.4 -1.1

-1.0

-0.9

-0.8

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

log r

From the regression analysis we note that:

n = 1.50 ,

C = 101.2311 = 17.0 y/(Gm )

32

and

(

T = 17.0 y/(Gm )

32

)r

(b) Solve equation (1) for the radius of the planet’s orbit:

⎞ ⎛ T ⎟ r = ⎜⎜ 32 ⎟ ⎝ 17.0 y / (Gm ) ⎠

Substitute numerical values and evaluate r:

⎛ ⎞ 6.20 y ⎟ r = ⎜⎜ 32 ⎟ ⎝ 17.0 y/(Gm ) ⎠

1.50

,

(1)

23

23

= 0.510 Gm

76 ••• The period T of a simple pendulum depends on the length L of the pendulum and the acceleration of gravity g (dimensions L/T2). (a) Find a simple combination of L and g that has the dimensions of time. (b) Check the dependence of the period T on the length L by measuring the period (time for a complete swing back and forth) of a pendulum for two different values of L. (c) The correct formula relating T to L and g involves a constant that is a multiple of π , and cannot be obtained by the dimensional analysis of Part (a). It can be found by

50

Chapter 1

experiment as in Part (b) if g is known. Using the value g = 9.81 m/s2 and your experimental results from Part (b), find the formula relating T to L and g. Picture the Problem We can express the relationship between the period T of the pendulum, its length L, and the acceleration of gravity g as T = CLa g b and perform dimensional analysis to find the values of a and b and, hence, the function relating these variables. Once we’ve performed the experiment called for in Part (b), we can determine an experimental value for C.

(a) Express T as the product of L and g raised to powers a and b:

T = CLa g b (1) where C is a dimensionless constant.

Write this equation in dimensional form:

[T ] = [L] a [g ] b

Substituting the dimensions of the physical quantities yields:

⎛ L ⎞ T=L ⎜ 2⎟ ⎝T ⎠

Because L does not appear on the left-hand side of the equation, we can write this equation as:

L0 T1 = La +b T −2b

Equate the exponents to obtain:

a + b = 0 and − 2b = 1

Solve these equations simultaneously to find a and b:

a = 12 and b = − 12

Substitute in equation (1) to obtain:

(b) If you use pendulums of lengths 1.0 m and 0.50 m; the periods should be about:

(c) Solving equation (2) for C yields:

b

a

T = CL1 2 g −1 2 = C T (1.0 m ) = 2.0 s

and T (0.50 m ) = 1.4 s

C =T

g L

L g

(2)

Measurement and Vectors Evaluate C with L = 1.0 m and T = 2.0 s:

9.81m/s 2 = 6.26 ≈ 2π 1.0 m

C = (2.0 s )

Substitute in equation (2) to obtain:

T = 2π

51

L g

77 ••• A sled at rest is suddenly pulled in three horizontal directions at the same time but it goes nowhere. Paul pulls to the northeast with a force of 50 lb. Johnny pulls at an angle of 35° south of due west with a force of 65 lb. Connie pulls with a force to be determined. (a) Express the boys' two forces in terms of the usual unit vectors (b) Determine the third force (from Connie), expressing it first in component form and then as a magnitude and angle (direction). Picture the Problem A diagram showing the forces exerted by Paul, Johnny, and Connie is shown below. Once we’ve expressed the forces exerted by Paul and Johnny in vector form we can use them to find the force exerted by Connie. r FPaul

50

lb

N

45°

W

35°

65

r FConnie

θ

E

lb

r FJohnny

S

(a) The force that Paul exerts is: r FPaul = [(50 lb )cos 45°]iˆ + [(50 lb )sin 45°] ˆj = (35.4 lb ) iˆ + (35.4 lb ) ˆj

= (35 lb ) iˆ + (35 lb ) ˆj The force that Johnny exerts is: r FJohnny = [(65 lb )cos 215°]iˆ + [(65 lb )sin 215°] ˆj = (− 53.2 lb ) iˆ − (37.3 lb ) ˆj

=

(− 53 lb) iˆ + (− 37 lb) ˆj

52

Chapter 1

The sum of the forces exerted by Paul and Johnny is: r r FPaul + FJohnny = (35.4 lb ) iˆ + (35.4 lb ) ˆj − (53.2 lb ) iˆ − (37.3 lb ) ˆj

= (− 17.8 lb ) iˆ − (1.9 lb ) ˆj (b) The condition that the three forces must satisfy is:

r r r FPaul + FJohnny + FConnie = 0

(

)

r Solving for FConnie yields:

r r r FConnie = − FPaul + FJohnny

r r Substitute for FPaul + FJohnny to

r FConnie = − (− 17.8 lb ) iˆ − (1.9 lb ) ˆj

obtain: r The magnitude of FConnie is:

The direction that the force exerted by Connie acts is given by:

[

]

= (18 lb ) iˆ + (1.9 lb ) ˆj

(17.8 lb)2 + (1.9 lb)2

FConnie =

= 18 lb

⎛ 1.9 lb ⎞ ⎟ = 6.1° N of E ⎝ 17.8 lb ⎠

θ = tan −1 ⎜

78 ••• You spot a plane that is 1.50 km North, 2.5 km East and at an altitude of 5.0 km above your position. (a) How far from you is the plane? (b) At what angle from due north (in the horizontal plane) are you looking? (c) Determine the plane's position vector (from your location) in terms of the unit vectors, letting iˆ be toward the east direction, ˆj be toward the north direction, and kˆ be vertically upward. (d) At what elevation angle (above the horizontal plane of Earth) is the airplane? Picture the Problem A diagram showing the given information is shown below. We can use the Pythagorean Theorem, trigonometry, and vector algebra to find the distance, angles, and expression called for in the problem statement.

Measurement and Vectors

53

Plane N

h

1.5

km

d

φ

r

θ E

2.5 km You are here

(a) Use the Pythagorean Theorem to express d in terms of h and l:

d = l2 + h2

Substitute numerical values and evaluate d:

d=

(b) Use trigonometry to evaluate the angle from due north at which you are looking at the plane:

θ = tan −1 ⎜

(c) We can use the coordinates of the plane relative to your position to r express the vector d : (d) Express the elevation angle φ in terms of h and l: Substitute numerical values and evaluate φ:

(2.5 km )2 + (1.5 km )2 + (5.0 km )2

= 5.8 km ⎛ 2.5 km ⎞ ⎟ = 59° E of N ⎝ 1.5 km ⎠

r d=

(2.5 km ) iˆ + (1.5 km ) ˆj + (5.0 km ) kˆ

⎛h⎞ ⎝l⎠

φ = tan −1 ⎜ ⎟ ⎛ φ = tan ⎜ ⎜ ⎝ −1

5.0 km

(2.5 km )2 + (1.5 km )2

= 60° above the horizon

⎞ ⎟ ⎟ ⎠

54

Chapter 1

Chapter 2 Motion in One Dimension Conceptual Problems 1 • What is the average velocity over the ″round trip″ of an object that is launched straight up from the ground and falls straight back down to the ground? Determine the Concept The "average velocity" is being requested as opposed to "average speed". The average velocity is defined as the change in position or displacement divided by the change in time. The change in position for any "round trip" is zero by definition. So the average velocity for any round trip must also be zero.

vav =

Δy Δt

vav =

Δy 0 = = 0 Δt Δt

2 • An object thrown straight up falls back and is caught at the same place it is launched from. Its time of flight is T, its maximum height is H. Neglect air resistance. The correct expression for its average speed for the entire flight is (a) H/T, (b) 0, (c) H/(2T), and (d) 2H/T. Determine the Concept The important concept here is that "average speed" is being requested as opposed to "average velocity".

Under all circumstances, including constant acceleration, the definition of the average speed is the ratio of the total distance traveled (H + H) to the elapsed time, in this case 2H/T. (d ) is correct. Remarks: Because this motion involves a round trip, if the question asked for "average velocity," the answer would be zero. 3 • Using the information in the previous question, what is its average speed just for the first half of the trip? What is its average velocity for the second half of the trip? (Answer in terms of H and T.) Determine the Concept Under all circumstances, including constant acceleration, the definition of the average speed is the ratio of the total distance traveled to the elapsed time. The average velocity, on the other hand, is the ratio of the displacement to the elapsed time.

55

56

Chapter 2

The average speed for the first half of the trip is the height to which the object rises divided by one-half its time of flight: The average velocity for the second half of the trip is the distance the object falls divided by one-half its time of flight:

vav, 1st half =

H 2H = T T

1 2

vel av, 2nd half =

−H 2H = − 1 T 2T

Remarks: We could also say that the average velocity for the second half of the trip is −2H/T. 4 • Give an everyday example of one-dimensional motion where (a) the velocity is westward and the acceleration is eastward, and (b) the velocity is northward and the acceleration is northward. Determine the Concept The important concept here is that a = dv/dt, where a is the acceleration and v is the velocity. Thus, the acceleration is positive if dv is positive; the acceleration is negative if dv is negative.

(a) An example of one-dimensional motion where the velocity is westward and acceleration is eastward is a car traveling westward and slowing down. (b)An example of one-dimensional motion where the velocity is northward and the acceleration is northward is a car traveling northward and speeding up. 5 • [SSM] Stand in the center of a large room. Call the direction to your right ″positive,″ and the direction to your left ″negative.″ Walk across the room along a straight line, using a constant acceleration to quickly reach a steady speed along a straight line in the negative direction. After reaching this steady speed, keep your velocity negative but make your acceleration positive. (a) Describe how your speed varied as you walked. (b) Sketch a graph of x versus t for your motion. Assume you started at x = 0. (c) Directly under the graph of Part (b), sketch a graph of vx versus t. Determine the Concept The important concept is that when both the acceleration and the velocity are in the same direction, the speed increases. On the other hand, when the acceleration and the velocity are in opposite directions, the speed decreases.

(a) Your speed increased from zero, stayed constant for a while, and then decreased.

Motion in One Dimension (b) A graph of your position as a function of time is shown to the right. Note that the slope starts out equal to zero, becomes more negative as the speed increases, remains constant while your speed is constant, and becomes less negative as your speed decreases.

x

(c) The graph of v(t) consists of a straight line with negative slope (your acceleration is constant and negative) starting at (0,0), then a flat line for a while (your acceleration is zero), and finally an approximately straight line with a positive slope heading to v = 0.

v

57

t

x

t

6 • True/false: The displacement always equals the product of the average velocity and the time interval. Explain your choice. Determine the Concept True. We can use the definition of average velocity to express the displacement Δx as Δx = vavΔt. Note that, if the acceleration is constant, the average velocity is also given by vav = (vi + vf)/2. 7 • Is the statement ″for an object’s velocity to remain constant, its acceleration must remain zero″ true or false? Explain your choice. Determine the Concept True. Acceleration is the slope of the velocity versus time curve, a = dv/dt; while velocity is the slope of the position versus time curve, v = dx/dt. The speed of an object is the magnitude of its velocity. Zero acceleration implies that the velocity is constant. If the velocity is constant (including zero), the speed must also be constant. 8 • Draw careful graphs of the position and velocity and acceleration over the time period 0 ≤ t ≤ 30 s for a cart that, in succession, has the following motion. The cart is moving at the constant speed of 5.0 m/s in the +x direction. It passes by the origin at t = 0.0 s. It continues on at 5.0 m/s for 5.0 s, after which it gains speed at the constant rate of 0.50 m/s each second for 10.0 s. After gaining speed for 10.0 s, the cart loses speed at the constant rate of 0.50 m/s for the next 15.0 s. Determine the Concept Velocity is the slope of the position versus time curve and acceleration is the slope of the velocity versus time curve. The following graphs were plotted using a spreadsheet program.

Chapter 2 160 140 120

x, m

100 80 60 40 20 0 0

5

10

15

20

25

30

t, s

10 9 8

v, m/s

7 6 5 4 3 2 1 0 0

5

10

15

20

25

30

t, s

0.6 0.4 0.2 a , m/s2

58

0 -0.2 -0.4 -0.6 0

5

10

15 t, s

20

25

30

Motion in One Dimension

59

9 • True/false; Average velocity always equals one-half the sum of the initial and final velocities. Explain your choice. Determine the Concept False. The average velocity is defined (for any acceleration) as the change in position (the displacement) divided by the change in time vav = Δx Δt . It is always valid. If the acceleration remains constant the average velocity is also given by v +v vav = i f 2 Consider an engine piston moving up and down as an example of non-constant velocity. For one complete cycle, vf = vi and xi = xf so vav = Δx/Δt is zero. The formula involving the mean of vf and vi cannot be applied because the acceleration is not constant, and yields an incorrect nonzero value of vi. 10 • Identical twin brothers standing on a horizontal bridge each throw a rock straight down into the water below. They throw rocks at exactly the same time, but one hits the water before the other. How can this be? Explain what they did differently. Ignore any effects due to air resistance. Determine the Concept This can occur if the rocks have different initial speeds. Ignoring air resistance, the acceleration is constant. Choose a coordinate system in which the origin is at the point of release and upward is the positive direction. From the constant-acceleration equation y = y0 + v0t + 12 at 2 we see that the only way two objects can have the same acceleration (–g in this case) and cover the same distance, Δy = y – y0, in different times would be if the initial velocities of the two rocks were different. Actually, the answer would be the same whether or not the acceleration is constant. It is just easier to see for the special case of constant acceleration. 11 •• [SSM] Dr. Josiah S. Carberry stands at the top of the Sears Tower in Chicago. Wanting to emulate Galileo, and ignoring the safety of the pedestrians below, he drops a bowling ball from the top of the tower. One second later, he drops a second bowling ball. While the balls are in the air, does their separation (a) increase over time, (b) decrease, (c) stay the same? Ignore any effects due to air resistance. Determine the Concept Neglecting air resistance, the balls are in free fall, each with the same free-fall acceleration, which is a constant.

At the time the second ball is released, the first ball is already moving. Thus, during any time interval their velocities will increase by exactly the same amount. What can be said about the speeds of the two balls? The first ball will always be moving faster than the second ball. This being the case, what happens to the separation of the two balls while they are both falling? Their separation increases. (a ) is correct.

60

Chapter 2

12 •• Which of the position-versus-time curves in Figure 2-28 best shows the motion of an object (a) with positive acceleration, (b) with constant positive velocity, (c) that is always at rest, and (d) with negative acceleration? (There may be more than one correct answer for each part of the problem.) Determine the Concept The slope of an x(t) curve at any point in time represents the speed at that instant. The way the slope changes as time increases gives the sign of the acceleration. If the slope becomes less negative or more positive as time increases (as you move to the right on the time axis), then the acceleration is positive. If the slope becomes less positive or more negative, then the acceleration is negative. The slope of the slope of an x(t) curve at any point in time represents the acceleration at that instant.

(a) The correct answer is (d ) . The slope of curve (d) is positive and increasing. Therefore the velocity and acceleration are positive. We would need more information to conclude that a is constant. (b) The correct answer is (b) . The slope of curve (b) is positive and constant. Therefore the velocity is positive and constant. (c) The correct answer is (e) . The slope of curve (e) is zero. Therefore, the velocity and acceleration are zero and the object remains at the same position. (d) The correct answers are (a ) and (c ) . The slope of curve (a) is negative and becomes more negative as time increases. Therefore the velocity is negative and the acceleration is negative. The slope of curve (c) is positive and decreasing. Therefore the velocity is positive and the acceleration is negative. Which of the velocity-versus-time curves in figure 2-29 best 13 •• [SSM] describes the motion of an object (a) with constant positive acceleration, (b) with positive acceleration that is decreasing with time, (c) with positive acceleration that is increasing with time, and (d) with no acceleration? (There may be more than one correct answer for each part of the problem.) Determine the Concept The slope of a v(t) curve at any point in time represents the acceleration at that instant.

(a) The correct answer is

(b )

. The slope of curve (b) is constant and positive.

Therefore the acceleration is constant and positive. (b) The correct answer is (c) . The slope of curve (c) is positive and decreasing with time. Therefore the acceleration is positive and decreasing with time.

Motion in One Dimension

61

(c) The correct answer is (d ) . The slope of curve (d) is positive and increasing with time. Therefore the acceleration is positive and increasing with time. (d) The correct answer is (e) . The slope of curve (e) is zero. Therefore the velocity is constant and the acceleration is zero. 14 •• The diagram in Figure 2-30 tracks the location of an object moving in a straight line along the x axis. Assume that the object is at the origin at t = 0. Of the five times shown, which time (or times) represents when the object is (a) farthest from the origin, (b) at rest for an instant, (c) in the midst of being at rest for awhile, and (d) moving away from the origin? Determine the Concept Because this graph is of distance-versus-time we can use its displacement from the time axis to draw conclusions about how far the object is from the origin. We can also use the instantaneous slope of the graph to decide whether the object is at rest and whether it is moving toward or away from the origin.

(a) The correct answer is B . Because the object’s initial position is at x = 0, point B represents the instant that the object is farthest from x = 0. (b) The correct answers are B and D. Because the slope of the graph is zero at points B and D, the velocity of the object is zero and it is momentarily at rest at these points. (c) The correct answer is E . Because the graph is a horizontal line with zero slope, the object remains at rest at the same position (its velocity is zero). (d) The correct answer is A. Because the slope of the graph is positive at point A, the velocity of the object is positive and it is moving away from the origin. 15 •• [SSM] An object moves along a straight line. Its position versus time graph is shown in Figure 2-30. At which time or times is its (a) speed at a minimum, (b) acceleration positive, and (c) velocity negative? Determine the Concept Because this graph is of distance-versus-time we can use its instantaneous slope to describe the object’s speed, velocity, and acceleration.

(a) The minimum speed is at B, D, and E , where it is zero. In the onedimensional motion shown in the figure, the velocity is a minimum when the slope of a position-versus-time plot goes to zero (i.e., the curve becomes horizontal). At these points the velocity is zero and, therefore, the speed is zero.

62

Chapter 2

(b) The acceleration is positive at points A and D. Because the slope of the graph is increasing at these points, the velocity of the object is increasing and its acceleration is positive. (c) The velocity is negative at point C. Because the slope of the graph is negative at point C, the velocity of the object is negative. 16 •• For each of the four graphs of x versus t in Figure 2-31 answer the following questions. (a) Is the velocity at time t2 greater than, less than, or equal to the velocity at time t1? (b) Is the speed at time t2 greater than, less than, or equal to the speed at time t1? Determine the Concept In one-dimensional motion, the velocity is the slope of a position-versus-time plot and can be either positive or negative. On the other hand, the speed is the magnitude of the velocity and can only be positive. We’ll use v to denote velocity and the word ″speed″ for how fast the object is moving.

(a) curve a: v(t 2 ) < v(t1 ) curve b: v(t 2 ) = v(t1 ) curve c: v(t 2 ) > v(t1 ) curve d: v(t 2 ) < v(t1 )

(b) curve a: speed(t 2 ) < speed(t1 ) curve b: speed(t 2 ) = speed(t1 ) curve c: speed(t 2 ) < speed(t1 ) curve d: speed(t 2 ) > speed(t1 )

17 •• True/false: Explain your reasoning for each answer. If the answer is true, give an example.

(a) (b) ( c)

If the acceleration of an object is always zero, then it cannot be moving. If the acceleration of an object is always zero, then its x-versus-t curve must be a straight line. If the acceleration of an object is nonzero at an instant, it may be momentarily at rest at that instant.

Explain your reasoning for each answer. If an answer is true, give an example. (a) False. An object moving in a straight line with constant speed has zero acceleration. (b) True. If the acceleration of the object is zero, then its speed must be constant. The graph of x-versus-t for an object moving with constant speed is a straight line. (c) True. A ball thrown upward is momentarily at rest when it is at the top of its trajectory. Its acceleration, however, is non-zero at this instant. Its value is the same as it was just before it came to rest and after it has started its descent.

Motion in One Dimension

63

18 •• A hard-thrown tennis ball is moving horizontally when it bangs into a vertical concrete wall at perpendicular incidence. The ball rebounds straight back off the wall. Neglect any effects due to gravity for the small time interval described here. Assume that towards the wall is the +x direction. What are the directions of its velocity and acceleration (a) just before hitting the wall, (b) at maximum impact, and (c) just after leaving the wall. Determine the Concept The tennis ball will be moving with constant velocity immediately before and after its collision with the concrete wall. It will be accelerated during the duration of its collision with the wall.

(a) Just before hitting the wall the velocity of the ball is in the +x direction and, because its velocity is constant, its acceleration is zero. (b) At maximum impact, the ball is reversing direction and its velocity is zero. Its acceleration is in the −x direction. (c) Just after leaving the wall, the velocity of the ball is in the −x direction and constant. Because its velocity is constant, its acceleration is zero. 19 •• [SSM] A ball is thrown straight up. Neglect any effects due to air resistance. (a) What is the velocity of the ball at the top of its flight? (b) What is its acceleration at that point? (c) What is different about the velocity and acceleration at the top of the flight if instead the ball impacts a horizontal ceiling very hard and then returns. Determine the Concept In the absence of air resistance, the ball will experience a constant acceleration and the graph of its position as a function of time will be parabolic. In the graphs to the right, a coordinate system was chosen in which the origin is at the point of release and the upward direction is the +y direction. The top graph shows the position of the ball as a function of time and the bottom graph shows the velocity of a ball as a function of time.

y

t v

t

(a) v top of flight = 0 (b) The acceleration of the ball is the same at every point of its trajectory, including the point at which v = 0 (at the top of its flight). Hence a top of = − g . flight

64

Chapter 2

(c) If the ball impacts a horizontal ceiling very hard and then returns, its velocity at the top of its flight is still zero and its acceleration is still downward but greater than g in magnitude. 20 •• An object that is launched straight up from the ground, reaches a maximum height H, and falls straight back down to the ground, hitting it T seconds after launch. Neglect any effects due to air resistance. (a) Express the average speed for the entire trip as a function of H and T. (b) Express the average speed for the same interval of time as a function of the initial launch speed v0. Picture the Problem The average speed is being requested as opposed to average velocity. We can use the definition of average speed as distance traveled divided by the elapsed time and the expression for the average speed of an object when it is experiencing constant acceleration to express vav in terms of v0.

(a) The average speed is defined as the total distance traveled divided by the change in time:

vav =

total distance traveled total time

Substitute for the total distance traveled and the total time and simplify to obtain:

vav =

H +H 2H = T T

(b) The average speed for the upward flight of the object is given by:

vav, up =

The average speed for the same interval of time as a function of the initial launch speed v0 is twice the average speed during the upward portion of the flight:

v0 + 0 H H = 1 ⇒ = 14 v0 2 T 2T

vav = 2vav, up = 2( 14 v0 ) =

1 2

v0

Because v0 ≠ 0 , the average speed is not zero.

Remarks: 1) Because this motion involves a roundtrip, if the question asked for ″average velocity″, the answer would be zero. 2) Another easy way to obtain this result is take the absolute value of the velocity of the object to obtain a graph of its speed as a function of time. A simple geometric argument leads to the result we obtained above. 21 •• A small lead ball is thrown directly upward. Neglect any effects due to air resistance. True or false: (a) The magnitude of its acceleration decreases on the way up. (b) The direction of its acceleration on its way down is opposite to the direction of its acceleration on its way up. (c) The direction of its velocity on its way down is opposite to the direction of its velocity on its way up. Determine the Concept For free fall, the acceleration is the same (g) throughout the entire flight.

Motion in One Dimension

65

(a) False. The velocity of the ball decreases at a steady rate. This means that the acceleration of the ball is constant. (b) False. The velocity of the ball decreases at a steady rate (g) throughout its entire flight. (c) True. On the way up the velocity vector points upward and on the way down it points downward. 22 •• At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a window 10 m below the roof. Air resistance is negligible. During the descent of B to the ground, the distance between the two objects (a) is proportional to t, (b) is proportional to t2, (c) decreases, (d) remains 10 m throughout. Determine the Concept Both objects experience the same constant acceleration. Choose a coordinate system in which downward is the positive direction and use a constant-acceleration equation to express the position of each object as a function of time.

Using constant-acceleration equations, express the positions of both objects as functions of time:

xA = x0, A + v0t + 12 gt 2

and xB = x0, B + v0t + 12 gt 2 where v0 = 0.

Express the separation of the two objects by evaluating xB − xA:

xB − xA = x0,B − x0.A = 10 m (d ) is correct.

23 •• You are driving a Porsche that accelerates uniformly from 80.5 km/h (50 mi/h) at t = 0.00 to 113 km/h (70 mi/h) at t = 9.00 s. (a) Which graph in Figure 2-32 best describes the velocity of your car? (b) Sketch a position-versustime graph showing the location of your car during these nine seconds, assuming we let its position x be zero at t = 0. Determine the Concept Because the Porsche accelerates uniformly, we need to look for a graph that represents constant acceleration.

(a) Because the Porsche has a constant acceleration that is positive (the velocity is increasing), we must look for a velocity-versus-time curve with a positive constant slope and a nonzero intercept. Such a graph is shown in (c ).

66

Chapter 2

(b) Use the data given in the problem statement to determine that the acceleration of the Porsche is 1.00 m/s2 and that its initial speed is 22.4 m/s. The equation describing the position of the car as a function of time is x = 22.4 m/s 2 t + 12 1.00 m/s 2 t 2 .

(

)

(

)

The following graph of this equation was plotted using a spreadsheet program. 250

200

x, m

150

100

50

0 0

1

2

3

4

5

6

7

8

9

t, s

24 •• A small heavy object is dropped from rest and falls a distance D in a time T. After it has fallen for a time 2T, what will be its (a) fall distance from its initial location in terms of D, (b) its speed in terms of D and t, and (c) its acceleration? (Neglect air resistance.) Picture the Problem In the absence of air resistance, the object experiences constant acceleration. Choose a coordinate system in which the downward direction is positive and use the constant-acceleration equation to describe its motion.

(a) Relate the distance D that the object, released from rest, falls in time t:

x(t ) = D = 12 gt 2

(1)

Evaluate x(2t) to obtain:

x(2t ) = 12 g (2t ) = 2 gt 2

(2)

Dividing equation (2) by equation (1) and simplifying yields:

x(2t ) 2 gt 2 = 1 2 = 4 ⇒ x(2t ) = 4 D D 2 gt

2

Motion in One Dimension (b) Express the speed of the object as a function of time:

Solving equation (1) for g yields:

v = v0 + gt or, because v0 = 0, v = gt g=

2x t2

Substitute for g in equation (3) to obtain:

v=

2x 2x t= 2 t t

Evaluating equation (4) at time 2t and simplifying yields:

v(2t ) =

67

(3)

(4)

2 x(2t ) 4D = 2t t

(c) The acceleration of the object is independent of time (that is, it is constant) and is equal to g . 25 •• In a race, at an instant when two horses are running right next to each other and in the same direction (the +x direction), horse A's instantaneous velocity and acceleration are +10 m/s and +2.0 m/s2 respectively, and horse B's instantaneous velocity and acceleration are +12 m/s and –1.0 m/s2 respectively. Which horse is passing the other at this instant? Explain. Determine the Concept The information about the horses’ accelerations is irrelevant to the determination of which horse is passing the other at this instant. The horse running with the greater instantaneous velocity will by passing the slower horse. Hence B is passing A. The accelerations are relevant to the

determination of which horse will be in the lead at some later time. 26 •• True or false: (a) The equation x − x0 = v0 x t + 12 a x t 2 is always valid for all particle motion in one dimension. (b) If the velocity at a given instant is zero, the acceleration at that instant must also be zero. (c) The equation Δx = vavΔt holds for all particle motion in one dimension. Determine the Concept As long as the acceleration remains constant the following constant-acceleration equations hold. If the acceleration is not constant, they do not, in general, give correct results except by coincidence.

x = x0 + v0t + 12 at 2

v = v0 + at

v 2 = v02 + 2aΔx

vav =

vi + vf 2

(a) False. This statement is true if and only if the acceleration is constant.

68

Chapter 2

(b) False. Consider a rock thrown straight up into the air. At the "top" of its flight, the velocity is zero but it is changing (otherwise the velocity would remain zero and the rock would hover); therefore the acceleration is not zero. (c) True. The definition of average velocity, vav = Δx Δt , requires that this always be true. 27 •• If an object is moving in a straight line at constant acceleration, its instantaneous velocity halfway through any time interval is (a) greater than its average velocity, (b) less than its average velocity, (c) equal to its average velocity, (d) half its average velocity, (e) twice its average velocity. Determine the Concept Because the acceleration of the object is constant, the constant-acceleration equations can be used to describe its motion. The special v +v expression for average velocity for constant acceleration is vav = i f . (c ) is 2 correct. 28 •• A turtle, seeing his owner put some fresh lettuce on the opposite side of his terrarium, begins to accelerate (at a constant rate) from rest at time t = 0, heading directly toward the food. Let t1 be the time at which the turtle has covered half the distance to his lunch. Derive an expression for the ratio of t2 to t1, where t2 is the time at which the turtle reaches the lettuce. Picture the Problem We are asked, essentially, to determine the time t2, at which a displacement, Δx, is twice what it was at an earlier time, t1. The turtle is crawling with constant acceleration so we can use the constant-acceleration 2 equation Δx = v0 x Δt + 12 a x (Δt ) to describe the turtle’s displacement as a function of time.

Express the displacement Δt of the turtle at the end of a time interval Δt:

Δx = v0 x Δt + 12 a x (Δt ) or, because v0x = 0, 2 Δx = 12 a x (Δt )

For the two time intervals:

Δx1 = 12 a x t12 and Δx2 = 12 a x t 22

Express the ratio of Δx2 to Δx1 to obtain:

Δx2 12 a x t 22 t 22 = = Δx1 12 a x t12 t12

We’re given that:

Δx2 =2 Δx1

2

(1)

Motion in One Dimension Substitute in equation (1) and simplify to obtain:

t 22 t = 2⇒ 2 = 2 t1 t1

69

2

29 •• [SSM] The positions of two cars in parallel lanes of a straight stretch of highway are plotted as functions of time in the Figure 2-33.Take positive values of x as being to the right of the origin. Qualitatively answer the following: (a) Are the two cars ever side by side? If so, indicate that time (those times) on the axis. (b) Are they always traveling in the same direction, or are they moving in opposite directions for some of the time? If so, when? (c) Are they ever traveling at the same velocity? If so, when? (d) When are the two cars the farthest apart? (e) Sketch (no numbers) the velocity versus time curve for each car. Determine the Concept Given the positions of the two cars as a function of time, we can use the intersections of the curves and their slopes to answer these questions.

(a) The positions of cars A and B are the same at two places where the graphs cross.

x

Car A

Car B 1

9

t (s)

Cars are side by side (b) When the slopes of the curves have opposite signs, the velocities of the cars are oppositely directed. Thus, after approximately 7 s, car A is moving leftward while car B is moving rightward. Before t = 7 s, the two cars are traveling in the same direction. (c) The two cars have the same velocity when their curves have the same slopes. This occurs at about 6 s. (d) The time at which the two cars are farthest apart is roughly 6 s as indicated by the place at which, vertically, the two curves are farthest part.

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Chapter 2

(e) v Ca

rA

Car B

t

30 •• A car driving at constant velocity passes the origin at time t = 0. At that instant, a truck, at rest at the origin, begins to accelerate uniformly from rest. Figure 2-34 shows a qualitative plot of the velocities of truck and car as functions of time. Compare their displacements (from the origin), velocities, and accelerations at the instant that their curves intersect. Determine the Concept The graph is a plot of velocity versus time. Thus, where the two curves cross, the truck and car are, at that instant, moving with equal velocities. The slope of a velocity versus time curve is equal to the instantaneous acceleration – thus, since the curve that represents the truck’s velocity has a positive slope, and the car’s curve has zero slope, the truck is accelerating at a higher rate than the car. Finally, the displacements of the two cars are determined by calculating the areas under the curves. In this instance, the curve representing the truck’s velocity as a function of time encloses a triangular area that is exactly half that of the curve representing the car’s velocity. Thus, at the instant represented by the point where the curves intersect, the truck has gone half as far as has the car. 31 •• Reginald is out for a morning jog, and during the course of his run on a straight track, has a velocity that depends upon time as shown in Figure 2-35. That is, he begins at rest, and ends at rest, peaking at a maximum velocity vmax at an arbitrary time tmax. A second runner, Josie, runs throughout the time interval t = 0 to t = tf at a constant speed vR, so that each has the same displacement during the time interval. Note: tf is NOT twice tmax, but represents an arbitrary time. What is the relationship between vJ and vmax? Determine the Concept In this problem we are presented with curves representing the velocity as a function of time for both runners. The area under each curve represents the displacement for each runner and we are told that Josie and Reginald each have the same displacement during the time interval of length tf. Since this is the case, we can find the relationship between vR and vmax by equating the areas under the two curves.

Express the condition on the displacement of the two runners:

ΔxR = ΔxJ

(1)

Motion in One Dimension Josie runs at a constant velocity v for the whole of the time interval. Express her displacement ΔxJ:

Δx J = v J t f

Reginald has a different velocity profile, one which results in a triangle of height vmax and length tf. Express his displacement ΔxJ:

ΔxR = 12 vmax t f

Substitute for ΔxR and ΔxJ in equation (1) and simplify to obtain:

1 2

vmax t f = vJ t f = ⇒ vJ =

1 2

71

vmax

32 •• Which graph (or graphs), if any, of v versus t in Figure 2-36 best describes the motion of a particle with (a) positive velocity and increasing speed, (b) positive velocity and zero acceleration, (c) constant non-zero acceleration, and (d) a speed decrease? Determine the Concept The velocity of the particle is positive if the curve is above the v = 0 line (the t axis), and the acceleration is positive if the curve has a positive slope. The speed of the particle is the magnitude of its velocity.

(a) Graph

(c )

describes the motion of a particle with positive velocity and

increasing speed because v(t) is above the t axis and has a positive slope. (b) Graph (a )

describes the motion of a particle with positive velocity and zero

acceleration because v(t) is above the t axis and its slope is zero. (c) Graphs

(c ), (d ) and (e )

describe the motion of a particle with constant non-

zero acceleration because v(t) is linear and has a non-zero slope. (d) Graph

(e )

describes the motion of a particle with a speed decrease because

it shows the speed of the particle decreasing with time. 33 •• Which graph (or graphs), if any, of vx versus t in Figure 2-36 best describes the motion of a particle with (a) negative velocity and increasing speed, (b) negative velocity and zero acceleration, (c) variable acceleration, and (d) increasing speed? Determine the Concept The velocity of the particle is positive if the curve is above the v = 0 line (the t axis), and the acceleration is positive if the curve has a positive slope. The speed of the particle is the magnitude of its velocity.

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Chapter 2

(a) Graph

(d )

describes the motion of a particle with negative velocity and

increasing speed because v(t) is below the t axis and has a negative slope. (b) Graph (b )

describes the motion of a particle with negative velocity and zero

acceleration because v(t) is below the t axis and its slope is zero. (c) None of these graphs describe the motion of a particle with a variable acceleration because v(t) is linear. (d) Graphs

(c ) and (d )

describe the motion of a particle with an increasing

speed because they show the speed of the particle increasing with time. 34 •• Sketch a v-versus-t curve for each of the following conditions: (a) Acceleration is zero and constant while velocity is not zero. (b) Acceleration is constant but not zero. (c) Velocity and acceleration are both positive. (d) Velocity and acceleration are both negative. (e) Velocity is positive and acceleration is negative. (f) Velocity is negative and acceleration is positive. (g) Velocity is momentarily zero but the acceleration is not zero. Determine the Concept Acceleration is the slope of a velocity-versus-time curve.

(a) Acceleration is zero and constant while velocity is not zero.

v

t

(b) Acceleration is constant but not zero.

v

t

(c) Velocity and acceleration are both positive.

v

t

(d) Velocity and acceleration are both negative.

v t

Motion in One Dimension (e) Velocity is positive and acceleration is negative.

73

v

t

(f) Velocity is negative and acceleration is positive.

v

(g) Velocity is momentarily zero but the acceleration is not zero.

v

t

t

35 •• Figure 2-37 shows nine graphs of position, velocity, and acceleration for objects in motion along a straight line. Indicate the graphs that meet the following conditions: (a) Velocity is constant, (b) velocity reverses its direction, (c) acceleration is constant, and (d) acceleration is not constant. (e) Which graphs of position, velocity, and acceleration are mutually consistent? Determine the Concept Velocity is the slope and acceleration is the slope of the slope of a position-versus-time curve. Acceleration is the slope of a velocityversus-time curve.

(a) Graphs

(a ), ( f ), and (i )

describe motion at constant velocity. For constant

velocity, x versus t must be a straight line; v-versus-t must be a horizontal straight line; and a versus t must be a straight horizontal line at a = 0. (b) Graphs

(c ) and (d )

describe motion in which the velocity reverses its

direction. For velocity to reverse its direction x-versus-t must have a slope that changes sign and v versus t must cross the time axis. The acceleration cannot remain zero at all times. (c) Graphs

(a ), (d ), (e ), ( f ), (h ), and (i )

describe motion with constant

acceleration. For constant acceleration, x versus t must be a straight horizontal line or a parabola, v versus t must be a straight line, and a versus t must be a horizontal straight line.

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Chapter 2

(d) Graphs

(b ), (c ), and (g )

describe motion with non-constant acceleration. For

non-constant acceleration, x versus t must not be a straight line or a parabola; v versus t must not be a straight line, or a versus t must not be a horizontal straight line. (e) The following pairs of graphs are mutually consistent: (d ) and (h), and ( f ) and (i ).

(a ) and (i ),

For two graphs to be mutually consistent, the

curves must be consistent with the definitions of velocity and acceleration.

Estimation and Approximation 36 • While engrossed in thought about the scintillating lecture just delivered by your physics professor you mistakenly walk directly into the wall (rather than through the open lecture hall door). Estimate the magnitude of your average acceleration as you rapidly come to a halt. Picture the Problem The speed of one’s walk varies from person to person, but 1.0 m/s is reasonable. We also need to estimate a distance within which you would stop in such a case. We’ll assume a fairly short stopping distance of 1.5 cm. We’ll also assume (unrealistically) that you experience constant acceleration and choose a coordinate system in which the direction you are walking is the +x direction.

Using a constant-acceleration equation, relate your final speed to your initial speed, acceleration, and displacement while stopping: Substitute numerical values and evaluate the magnitude of your acceleration:

vf2 = vi2 + 2a x Δx ⇒ a x =

vf2 − vi2 2Δx

2

(0) − ⎛⎜1.0 m ⎞⎟ s⎠ ⎝ ax = = 33 m/s 2 −2 2 1.5 × 10 m 2

(

)

37 • [SSM] Occasionally, people can survive falling large distances if the surface they land on is soft enough. During a traverse of the Eiger’s infamous Nordvand, mountaineer Carlos Ragone’s rock anchor gave way and he plummeted 500 feet to land in snow. Amazingly, he suffered only a few bruises and a wrenched shoulder. Assuming that his impact left a hole in the snow 4.0 ft deep, estimate his average acceleration as he slowed to a stop (that is, while he was impacting the snow).

Motion in One Dimension

75

Picture the Problem In the absence of air resistance, Carlos’ acceleration is constant. Because all the motion is downward, let’s use a coordinate system in which downward is the positive direction and the origin is at the point at which the fall began.

Using a constant-acceleration equation, relate Carlos’ final velocity v2 to his velocity v1 just before his impact, his stopping acceleration as upon impact, and his stopping distance Δy:

v22 = v12 + 2as Δy ⇒ as =

v22 − v12 2Δy

or, because v2 = 0, v2 as = − 1 2Δy

(1)

Using a constant-acceleration equation, relate Carlos’ speed just before impact to his acceleration during free-fall and the distance he fell h:

v12 = v02 + 2afree-fall h or, because v0 = 0 and afree- fall = g ,

Substituting for v12 in equation (1) yields:

as = −

Substitute numerical values and evaluate as:

a=−

v12 = 2 gh 2 gh 2Δy

(

)

2 9.81 m/s 2 (500 ft ) 2(4.0 ft )

= − 1.2 × 10 3 m/s 2 Remarks: The magnitude of this acceleration is about 125g! 38 •• When we solve free-fall problems near Earth, it’s important to remember that air resistance may play a significant role. If its effects are significant, we may get answers that are wrong by orders of magnitude if we ignore it. How can we tell when it is valid to ignore the affects of air resistance? One way is to realize that air resistance increases with increasing speed. Thus, as an object falls and its speed increases, its downward acceleration decreases. Under these circumstances, the object's speed will approach, as a limit, a value called its terminal speed. This terminal speed depends upon such things as the mass and cross-sectional area of the body. Upon reaching its terminal speed, its acceleration is zero. For a ″typical″ skydiver falling through the air, a typical the terminal speed is about 50 m/s (roughly 120 mph). At half its terminal speed, the skydiver’s acceleration will be about 34 g . Let’s take half the terminal speed as a reasonable ″upper bound″ beyond which we shouldn’t use our constant acceleration free-fall relationships. Assuming the skydiver started from rest, (a) estimate how far, and for how long, the skydiver falls before we can no longer neglect air resistance. (b) Repeat the analysis for a ping-pong ball, which has a terminal speed of about 5.0 m/s. (c) What can you conclude by comparing your answers for Parts (a) and (b)?

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Chapter 2

Picture the Problem Because we’re assuming that the accelerations of the skydiver and a ping-pong ball are constant to one-half their terminal velocities, we can use constant-acceleration equations to find the times required for them to reach their ″upper-bound″ velocities and their distances of fall. Let’s use a coordinate system in which downward is the +y direction.

(a) Using a constant-acceleration equation, relate the upper-bound velocity to the free-fall acceleration and the time required to reach this velocity: Substitute numerical values and evaluate Δt:

vupper bound = v0 + gΔt or, because v0 = 0, vupper bound = gΔt ⇒ Δt =

Δt =

vupper bound g

25 m/s = 2.55 s ≈ 2.6 s 9.81 m/s 2

( 12 vt )2 = v02 + 2 gΔy

Using a constant-acceleration equation, relate the skydiver’s terminal speed to his/her acceleration and distance of fall:

or, because v0 = 0,

Substitute numerical values and evaluate Δy:

Δy =

(b) Proceed as in (a) with vupper bound = 5.0 m/s to obtain:

Δt =

2 1 ( 2 vt ) ( vt ) = 2 gΔy ⇒ Δy = 1 2

and Δy =

2

2g

[12 (50 m/s)]2

(

2 9.81 m/s 2 1 2

(5.0 m/s)

9.81 m/s 2

32 m

= 0.255 s ≈ 0.26 s

[12 (5 m/s)]2

(

)≈

2 9.81 m/s 2

)≈

32 cm

(c) The analysis of the motion of a ping-pong ball requires the inclusion of air resistance for almost any situation, whereas the analysis of the motion of the sky diver doesn’t require it until the fall distances and times are considerably longer. 39 •• On June 14, 2005 Asafa Powell of the Jamaica set a world’s record for the 100-m dash with a time t = 9.77 s. Assuming he reached his maximum speed in 3.00 s, and then maintained that speed until the finish, estimate his acceleration during the first 3.00 s. Picture the Problem This is a constant-acceleration problem. Choose a coordinate system in which the direction Powell is running is the +x direction. During the first 3 s of the race his acceleration is positive and during the rest of the race it is zero. The pictorial representation summarizes what we know about Powell’s race.

Motion in One Dimension

t0 = 0

t1 = 3.00 s

t 2 = 9.77 s

x0 = 0

x1

x 2 = 100 m

v0 = 0

v1 = v max

v 2 = v max

77

Express the total distance covered by Powell in terms of the distances covered in the two phases of his race:

100 m = Δx01 + Δx12

Express the distance he runs getting to his maximum velocity:

Δx01 = v0 Δt01 + 12 a01 (Δt01 ) = 12 a(3 s )

The distance covered during the rest of the race at the constant maximum velocity is given by:

Δx12 = vmax Δt12 + 12 a12 (Δt12 )

Substitute for these displacements in equation (1) to obtain:

100 m = 12 a(3.00 s ) + a (3.00 s )(6.77 s )

Solving for a yields:

(1)

2

2

2

= (aΔt 01 )Δt12

= a(3.00 s )(6.77 s ) 2

a=

1 2

100 m (3.00 s ) + (3.00 s )(6.77 s ) 2

= 4.03 m/s 2 40 •• The photograph in Figure 2-38 is a short-time exposure (1/30 s) of a juggler with two tennis balls in the air. (a) The tennis ball near the top of its trajectory is less blurred than the lower one. Why is that? (b) Estimate the speed of the ball that he is just releasing from his right hand. (c) Determine how high the ball should have gone above the launch point and compare it to an estimate from the picture. (Hint: You have a built-in distance scale if you assume some reasonable value for the height of the juggler.) Determine the Concept This is a constant-acceleration problem with a = −g if we take upward to be the positive direction. At the maximum height the ball will reach, its speed will be near zero and when the ball has just been tossed in the air its speed is near its maximum value.

78

Chapter 2

(a) Because the ball is moving slowly its blur is relatively short (i.e., there is less blurring). (b) The average speed of the ball is given by:

vav =

Estimating how far the ball has traveled in 1/30 s yields:

vav =

The diameter of a tennis ball is 6.5 cm:

vav =

(c) Use a constant-acceleration equation to relate the initial and final speeds of the ball to its maximum height h:

Substitute numerical values and evaluate h:

distance traveled elapsed time 2 ball diameters 1 s 30

2 (6.5 cm ) ≈ 3.9 m/s 1 s 30

v 2 − v02 2a y or, because v = 0 and ay = g, v2 h=− 0 2g v 2 = v02 + 2a y h ⇒ h =

h=−

(3.9 m/s)2

(

2 − 9.81 m/s 2

)≈

78 cm

If we assume that the juggler is approximately 6 ft (1.8 m) tall, then our calculated value for h seems to be a good approximation to the height shown in the photograph. 41 •• A rough rule of thumb for determining the distance between you and a lightning strike is to start counting the seconds that elapse ("one-Mississippi, two-Mississippi, …") until you hear the thunder (sound emitted by the lightning as it rapidly heats the air around it). Assuming the speed of sound is about 750 mi/h, (a) estimate how far away is a lightning strike if you counted about 5 s until you heard the thunder. (b) Estimate the uncertainty in the distance to the strike in Part (a). Be sure to explain your assumptions and reasoning. (Hint: The speed of sound depends on the air temperature and your counting is far from exact!) Picture the Problem We can use the relationship between distance, speed, and time to estimate the distance to the lightning strike.

(a) Relate the distance Δd to the lightning strike to the speed of sound in air v and the elapsed time Δt:

Δd = vΔt

Motion in One Dimension Substitute numerical values and evaluate Δd:

79

m ⎞ ⎛ 0.3048 ⎟ ⎜ mi ⎞ ⎛ s ⎟ (5 s ) Δd = ⎜ 750 ⎟ ⎜ mi h ⎜ ⎟ ⎠ 0.6818 ⎝ ⎜ ⎟ h ⎠ ⎝ ≈ 1.7 km ≈ 1 mi

(b) You are probably lucky if the uncertainty in your time estimate is less than 1 s (±20%), so the uncertainty in the distance estimate is about 20% of 1.7 km or approximately 300 m. This is probably much greater than the error made by assuming v is constant.

Speed, Displacement, and Velocity 42 • (a) An electron in a television tube travels the 16-cm distance from the grid to the screen at an average speed of 4.0 × 107 m/s. How long does the trip take? (b) An electron in a current-carrying wire travels at an average speed of 4.0 × 10–5 m/s. How long does it take to travel 16 cm? Picture the Problem Think of the electron as traveling in a straight line at constant speed and use the definition of average speed.

(a) Using its definition, express the average speed of the electron:

Average speed =

Solve for and evaluate the time of flight:

Δt =

distance traveled time of flight Δs = Δt

0.16 m Δs = Average speed 4.0 × 10 7 m s

= 4.0 × 10 −9 s = 4.0 ns (b) Calculate the time of flight for an electron in a 16-cm long current carrying wire similarly.

0.16 m Δs = Average speed 4.0 × 10 −5 m s 1 min = 4.0 × 10 3 s × = 67 min 60 s

Δt =

43 • [SSM] A runner runs 2.5 km, in a straight line, in 9.0 min and then takes 30 min to walk back to the starting point. (a) What is the runner’s average velocity for the first 9.0 min? (b) What is the average velocity for the time spent walking? (c) What is the average velocity for the whole trip? (d) What is the average speed for the whole trip?

80

Chapter 2

Picture the Problem In this problem the runner is traveling in a straight line but not at constant speed - first she runs, then she walks. Let’s choose a coordinate system in which her initial direction of motion is taken as the +x direction.

(a) Using the definition of average velocity, calculate the average velocity for the first 9 min: (b) Using the definition of average velocity, calculate her average velocity for the 30 min spent walking:

vav =

Δx 2.5 km = = 0.28 km / min Δt 9.0 min

vav =

Δx − 2.5 km = Δt 30 min

= − 83 m / min

(c) Express her average velocity for the whole trip:

vav =

(d) Finally, express her average speed for the whole trip:

speed av =

Δxround trip Δt

=

0 = 0 Δt

distance traveled elapsed time 2(2.5 km) = 30 min + 9.0 min = 0.13 km / min

44 • A car travels in a straight line with an average velocity of 80 km/h for 2.5 h and then with an average velocity of 40 km/h for 1.5 h. (a) What is the total displacement for the 4.0-h trip? (b) What is the average velocity for the total trip? Picture the Problem The car is traveling in a straight line but not at constant speed. Let the direction of motion be the +x direction.

(a) The total displacement of the car for the entire trip is the sum of the displacements for the two legs of the trip:

Δx total = Δx1 + Δx2

Find the displacement for each leg of the trip:

Δx1 = vav ,1Δt1 = (80 km/h )(2.5 h ) = 200 km and Δx2 = vav , 2 Δt2 = (40 km/h )(1.5 h ) = 60.0 km

Add the individual displacements to get the total displacement:

Δxtotal = Δx1 + Δx2 = 200 km + 60.0 km = 2.6 × 105 m

Motion in One Dimension (b) As long as the car continues to move in the same direction, the average velocity for the total trip is given by: Substitute numerical values and evaluate vav:

vav =

Δxtotal Δt total

vav =

2.6 × 10 5 m = 65 km h 2.5 h + 1.5 h

81

45 • One busy air route across the Atlantic Ocean is about 5500 km. The now-retired Concord, a supersonic jet capable of flying at twice the speed of sound was used to travel such routes. (a) Roughly how long did it take for a oneway flight? (Use 343 m/s for the speed of sound.) (b) Compare this time to the time taken by a subsonic jet flying at 0.90 times the speed of sound. Picture the Problem However unlikely it may seem, imagine that both jets are flying in a straight line at constant speed and use the definition of average speed to find the flight times.

(a) The time of flight is the ratio of the distance traveled to the speed of the supersonic jet.

t supersonic = =

s Atlantic vsupersonic 5500 km 2(343 m/s)(3600 s/h )

= 2.23 h = 2.2 h

(b) Express the ratio of the time for the trip at supersonic speed to the time for the trip at subsonic speed and simplify to obtain:

t supersonic

Substitute numerical values and evaluate the ratio of the flight times:

t supersonic

t subsonic

t subsonic

s Atlantic vsupersonic v = = subsonic s Atlantic vsupersonic vsubsonic =

(0.90)(343 m/s) = (2)(343 m/s)

0.45

46 •• The speed of light, designated by the universally recognized symbol c, has a value, to two significant figures, of 3.0 × 108 m/s. (a) How long does it take for light to travel from the Sun to Earth, a distance of 1.5 × 1011 m? (b) How long does it take light to travel from the Moon to Earth, a distance of 3.8 × 108 m? Picture the Problem In free space, light travels in a straight line at constant speed, c.

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Chapter 2

(a) Using the definition of average speed, express the time Δt required for light to travel from the Sun to Earth:

Δs average speed where Δs is the distance from the Sun to Earth.

Substitute numerical values and evaluate Δt:

Δt =

Δt =

1.5 × 1011 m = 5.0 × 10 2 s 8 3.0 × 10 m/s

≈ 8.3 min (b) Proceed as in (a) this time using the Moon-Earth distance:

3.8 ×10 8 m Δt = ≈ 1.3 s 3.0 ×108 m/s

47 • [SSM] Proxima Centauri, the closest star to us besides our own sun, 13 is 4.1 × 10 km from Earth. From Zorg, a planet orbiting this star, a Gregor places an order at Tony’s Pizza in Hoboken, New Jersey, communicating via light signals. Tony’s fastest delivery craft travels at 1.00 × 10–4c (see Problem 46). (a) How long does it take Gregor’s order to reach Tony’s Pizza? (b) How long does Gregor wait between sending the signal and receiving the pizza? If Tony’s has a "1000-years-or-it’s-free" delivery policy, does Gregor have to pay for the pizza? Picture the Problem In free space, light travels in a straight line at constant speed, c. We can use the definition of average speed to find the elapsed times called for in this problem.

(a) Using the definition of average speed (equal here to the assumed constant speed of light), solve for the time Δt required to travel the distance to Proxima Centauri: Substitute numerical values and evaluate Δt:

Δt =

distance traveled speed of light

4.1×1016 m Δt = = 1.37 ×108 s 8 3.0 ×10 m s = 4.3 y

Motion in One Dimension

83

(b) The delivery time (Δttotal) is the sum of the time for the order to reach Hoboken and the travel time for the delivery craft to travel to Proxima Centauri: Δt total = Δt order to be

sent to Hoboken

= 4.33 y +

+ Δt order to

be delivered

4.1×1013 km 1y ⎛ ⎞ ⎜ 8 7 ⎟ −4 1.00 × 10 3.0 × 10 m s ⎝ 3.156 × 10 s ⎠

(

)(

)

= 4.3 y + 4.3×10 4 y ≈ 4.3×10 4 y Because 4.3 × 104 y >> 1000 y, Gregor does not have to pay. 48 • A car making a 100-km journey travels 40 km/h for the first 50 km. How fast must it go during the second 50 km to average 50 km/h? Picture the Problem The time for the second 50 km is equal to the time for the entire journey less the time for the first 50 km. We can use this time to determine the average speed for the second 50 km interval from the definition of average speed.

Using the definition of average speed, find the time required for the total journey:

Δt total =

Find the time required for the first 50 km:

Δt1st 50 km =

Find the time remaining to travel the last 50 km: Finally, use the time remaining to travel the last 50 km to determine the average speed over this distance:

Δx 100 km = = 2.0 h vav 50 km h 50 km = 1.25 h 40 km h

Δt2nd 50 km = t total − t1st 50 km = 2.0 h − 1.25 h = 0.75 h

vav, 2nd 50 km =

Δx 2nd 50 km 50 km = Δt 2nd 50 km 0.75 h

= 67 km h

49 •• Late in ice hockey games, the team that is losing sometimes "pulls" their goalkeeper off the ice to add an additional offensive player and increase their chances of scoring. In such cases, the goalie on the opposing team might have an opportunity to score into the unguarded net that is 55.0 m away. Suppose you are the goaltender for your university team and are in just such a situation. You launch a shot (in hopes of getting your first career goal) on the frictionless ice. You hear a disappointing ″clang″ as the puck strikes a goalpost (instead of going in!) exactly 2.50 s later. In this case, how fast did the puck travel? You should assume 343 m/s for the speed of sound.

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Chapter 2

Picture the Problem The distance over which both the puck and the sound from the puck hitting the goalpost must travel is 55.0 m. The time between the shot being released and the sound reaching the goalie’s ear can be found by expressing the total elapsed time as the sum of the elapsed times for the shot to travel 55.0 m and for the sound to travel back to you.

The total elapsed time as the sum of the elapsed times for the shot and for the sound to travel back to you :

Δt total = Δt shot + Δt sound

Express the time for the shot to travel to the other net a distance Δx away:

Δt shot =

Express the time for the sound to travel a distance Δx back to you:

Δt sound =

Substitute in the expression for Δt total to obtain:

Δt total =

Solving this equation for vshot yields:

Substitute numerical values and evaluate vshot :

Δx vshot Δx vsound

Δx Δx + vshot vsound

vshot =

vsound Δx vsound Δttotal − Δx

vshot =

(343 m/s)(55.0 m ) (343 m/s)(2.50 s ) − 55.0 m

= 23.5 m/s 50 •• Cosmonaut Andrei, your co-worker at the International Space Station, tosses a banana at you with a speed of 15 m/s. At exactly the same instant, you fling a scoop of ice cream at Andrei along exactly the same path. The collision between banana and ice cream produces a banana split 7.2 m from your location 1.2 s after the banana and ice cream were launched. (a) How fast did you toss the ice cream? (b) How far were you from Andrei when you tossed the ice cream? (Neglect any effects due to gravity.) Picture the Problem Let the subscript b refer to the banana and the subscript ic refer to the ice cream. Then the distance covered by the ice cream before collision is given by Δxic = vicΔt and the distance covered by the banana is Δxb = vbΔt. The distance between you and Andrei is then the sum of these distances: Δxtot = Δxic + Δxb.

(a) The speed of the ice cream is given by:

Δxic Δt where Δt is the time-to-collision. vic =

Motion in One Dimension Substitute numerical values and evaluate vic:

vic =

85

7.2 m = 6.0 m/s 1.2 s

(b) Express the distance between yourself and Andrei as the sum of the distances the ice cream and the banana travel:

Δx total = Δxic + Δxb

Because Δxb = vb Δt :

Δx total = Δxic + vb Δt

Substitute numerical values and evaluate Δxtotal :

Δxtotal = 7.2 m + (15 m/s )(1.2 s ) = 25 m

51 •• Figure 2-39 shows the position of a particle as a function of time. Find the average velocities for the time intervals a, b, c, and d indicated in the figure. Picture the Problem The average velocity in a time interval is defined as the displacement divided by the time elapsed; that is vav = Δx / Δt .

(a) Δxa = 0

vav = 0

(b) Δxb = 1 m and Δtb = 3 s

vav = 0.3 m/s

(c) Δxc = –6 m and Δtc = 3 s

vav = − 2 m/s

(d) Δxd = 3 m and Δtd = 3 s

vav = 1 m/s

52 •• It has been found that, on average, galaxies are moving away from Earth at a speed that is proportional to their distance from Earth. This discovery is known as Hubble’s law, named for its discoverer, astrophysicist Sir Edwin Hubble. He found that the recessional speed v of a galaxy a distance r from Earth is given by v = Hr, where H = 1.58 × 10–18 s–1 is called the Hubble constant. What are the expected recessional speeds of galaxies (a) 5.00 × 1022 m from Earth and (b) 2.00 × 1025 m from Earth? (c) If the galaxies at each of these distances had traveled at their expected recessional speeds, how long ago would they have been at our location? Picture the Problem In free space, light travels in a straight line at constant speed c. We can use Hubble’s law to find the speed of the two planets.

(a) Using Hubble’s law, calculate the speed of the first galaxy:

(

)(

va = 5.00 × 10 22 m 1.58 × 10 −18 s −1 = 7.90 × 10 4 m/s

)

86

Chapter 2

(b) Using Hubble’s law, calculate the speed of the second galaxy:

(c) Using the relationship between distance, speed, and time for both galaxies, express how long ago Δt they were both located at the same place as Earth: Substitute numerical values and evaluate Δt:

(

)(

vb = 2.00 × 10 25 m 1.58 × 10 −18 s −1

)

= 3.16 × 10 7 m/s

Δt =

r r 1 = = v rH H

Δt = 6.33 × 1017 s ≈ 20 × 109 y

53 •• [SSM] The cheetah can run as fast as 113 km/h, the falcon can fly as fast as 161 km/h, and the sailfish can swim as fast as 105 km/h. The three of them run a relay with each covering a distance L at maximum speed. What is the average speed of this relay team for the entire relay? Compare this average speed with the numerical average of the three individual speeds. Explain carefully why the average speed of the relay team is not equal to the numerical average of the three individual speeds. Picture the Problem We can find the average speed of the relay team from the definition of average speed.

Using its definition, relate the average speed to the total distance traveled and the elapsed time:

vav =

Express the time required for each animal to travel a distance L:

t cheetah =

distance traveled elapsed time L vcheetah

, t falcon =

L vfalcon

and t sailfish = Express the total time Δt:

L vsailfish

⎛ 1 1 1 ⎞ ⎟⎟ Δt = L⎜⎜ + + ⎝ vcheetah vfalcon vsailfish ⎠

Use the total distance traveled by the relay team and the elapsed time to calculate the average speed: vav =

3L = 122.03 km/h = 122 km/h ⎛ ⎞ 1 1 1 ⎟⎟ L⎜⎜ + + ⎝ 113 km/h 161 km/h 105 km/h ⎠

Motion in One Dimension

87

Calculating the average of the three speeds yields: Averagethree speeds =

113 km/h + 161 km/h + 105 km/h = 126.33 km/h = 126 km/h 3

= 1.04vav

The average speed of the relay team is not equal to the numerical average of the three individual speeds because the ″runners″ did not run for the same interval of time. The average speed would be equal to one-third the sum of the three speeds if the three speeds were each maintained for the same length of time instead of for the same distance. 54 •• Two cars are traveling along a straight road. Car A maintains a constant speed of 80 km/h and car B maintains a constant speed of 110 km/h. At t = 0, car B is 45 km behind car A. (a) How much farther will car A travel before car B overtakes it? (b) How much ahead of A will B be 30 s after it overtakes A? Picture the Problem Let the position of car A at t = 0 be the origin of our coordinate system. Then we can use a constant-acceleration equation to express the positions of both cars as functions of time and equate these expressions to determine the time at which car A is overtaken by car B.

(a) Car B overtakes car A when their x coordinates are the same:

xA (t ) = xB (t )

Using a constant-acceleration equation with a = 0, express the position of car A as a function of time:

xA (t ) = x0A + vA t where x0A is the position of car A at t = 0.

Because we’ve let car A be at the origin at t = 0:

xA (t ) = vA t

Using a constant-acceleration equation, express the position of car B as a function of time:

xB (t ) = x0 B + vBt where x0B is the position of car B at t = 0.

Substitute for xA(t) and xB(t) in equation (1) to obtain:

v A t = x0 B + v B t ⇒ t =

Substitute numerical values and evaluate the time t at which car B overtakes car A:

t=

(1)

(2)

x0 B vA − vB

− 45 km = 1.50 h 80 km/h − 110 km/h

88

Chapter 2

Now we can evaluate equation (2) at t = 1.50 h to obtain:

⎛ km ⎞ xA (1.50 h ) = ⎜ 80 ⎟ (1.50 h ) = 120 km h ⎠ ⎝ = 1.2 × 10 5 m Δx(t ) = xB (t ) − xA (t ) = x0B − vB t − vA t

(b) The separation of the cars as a function of time is given by:

1 Substitute numerical values and evaluate Δx(1.50 h + 30 s ) = Δx(1.50 h + 120 h ) to obtain:

km km ⎞ ⎛ 1 1 Δx(1.50 h + 120 h ) = −45 km + ⎜110 − 80 h ) = 0.25 km ⎟ (1.50 h + 120 h h ⎠ ⎝ Remarks: One can use a graphing calculator or a spreadsheet program to solve this problem. A spreadsheet program was used to plot the following graph: 300 250

Car A Car B

x, km

200 150 100 50 0 0

0.5

1

1.5

2

2.5

3

t, h

Note that this graph confirms our result that the cars are at the same location at t = 1.5 h. 55 •• [SSM] A car traveling at a constant speed of 20 m/s passes an intersection at time t = 0. A second car traveling at a constant speed of 30 m/s in the same direction passes the same intersection 5.0 s later. (a) Sketch the position functions x1(t) and x2(t) for the two cars for the interval 0 ≤ t ≤ 20 s. (b) Determine when the second car will overtake the first. (c) How far from the intersection will the two cars be when they pull even? (d) Where is the first car when the second car passes the intersection? Picture the Problem One way to solve this problem is by using a graphing calculator to plot the positions of each car as a function of time. Plotting these positions as functions of time allows us to visualize the motion of the two cars

Motion in One Dimension

89

relative to the (fixed) ground. More importantly, it allows us to see the motion of the two cars relative to each other. We can, for example, tell how far apart the cars are at any given time by determining the length of a vertical line segment from one curve to the other. (a) Letting the origin of our coordinate system be at the intersection, the position of the slower car, x1(t), is given by:

x1(t) = 20t where x1 is in meters if t is in seconds.

Because the faster car is also moving at a constant speed, we know that the position of this car is given by a function of the form:

x2(t) = 30t + b

We know that when t = 5.0 s, this second car is at the intersection (that is, x2(5.0 s) = 0). Using this information, you can convince yourself that:

b = −150 m

Thus, the position of the faster car is given by:

x2 (t ) = 30t − 150

One can use a graphing calculator, graphing paper, or a spreadsheet to obtain the following graphs of x1(t) (the solid line) and x2(t) (the dashed line): 450 400 350

x, m

300 250 200 150 100 50 0 0

5

10

15

20

t, s

(b) Use the time coordinate of the intersection of the two lines to determine the time at which the second car overtakes the first:

From the intersection of the two lines, one can see that the second car will "overtake" (catch up to) the first car at t = 15 s.

90

Chapter 2

(c) Use the position coordinate of the intersection of the two lines to determine the distance from the intersection at which the second car catches up to the first car:

From the intersection of the two lines, one can see that the distance from the intersection is 300 m.

(d) Draw a vertical line from t = 5 s to the solid line and then read the position coordinate of the intersection of the vertical line and the solid line to determine the position of the first car when the second car went through the intersection. From the graph, when the second car passes the intersection, the first car was 100 m ahead. 56 •• Bats use echolocation to determine their distance from objects they cannot easily see in the dark. The time between the emission of high-frequency sound pulse (a click) and the detection of its echo is used to determine such distances. A bat, flying at a constant speed of 19.5 m/s in a straight line toward a vertical cave wall, makes a single clicking noise and hears the echo 0.15 s later. Assuming that she continued flying at her original speed, how close was she to the wall when she received the echo? Assume a speed of 343 m/s for the speed of sound. Picture the Problem The sound emitted by the bat travels at vsound and during the

time interval Δt during which the sound travels to the wall (a distance L from the bat’s initial position) and back to the bat, the bat travels a distance of Δxbat = vbat Δt , where vbat is the bat’s flying speed. The distance of the bat from the wall when she received the echo of her click is Δxaway = L − vbat Δt where Δt = Δt to wall + Δt back . to bat

vbat Δt

L − vbat Δt

L

Express the distance of the bat from the wall of the cave when it hears the echo of its click:

Δxaway = L − vbat Δt

The elapsed time between the bat clicking and hearing the sound is:

Δt = Δt to wall + Δt back

to bat

(1)

Motion in One Dimension Substituting for Δt to wall and Δt back gives:

Δt =

Solving for L yields:

L=

Substitute for L in equation (1) and simplify to obtain:

Δxaway =

1 2

Substitute numerical values and evaluate Δxaway :

Δxaway =

1 2

L vsound

+

91

L − vbat Δt vsound

to bat

1 2

(vsound + vbat )Δt (vsound + vbat )Δt − vbat Δt = 12 (vsound − vbat )Δt

(343 m/s − 19.5 m/s)(0.15 s )

= 24 m

A submarine can use sonar (sound traveling through water) to determine its distance from other objects. The time between the emission of a sound pulse (a ″ping″) and the detection of its echo can be used to determine such distances. Alternatively, by measuring the time between successive echo receptions of a regularly timed set of pings, the submarine's speed may be determined by comparing the time between echoes to the time between pings. Assume you are the sonar operator in a submarine traveling at a constant velocity underwater. Your boat is in the eastern Mediterranean Sea, where the speed of sound is known to be 1522 m/s. If you send out pings every 2.000 s, and your apparatus receives echoes reflected from an undersea cliff every 1.980 s, how fast is your submarine traveling? 57

•••

Picture the Problem Both the pulses sent out by the submarine and the pulses returning from the sea-wall are traveling at 1522 m/s. Consequently, we can determine the distance in water between two successive echo (or emitted) pulses of sound which were emitted with a time interval Δtemitted between them. The actual distance in the seawater between the echoed pulses is given by Δx = vsoundΔtemitted. We need to find the time Δtreceived between successive pulses received by the submarine. We start our ″clock″, as it were, when the submarine passes one of two successive pulses that approach it, separated by the distance Δx. After passing the first pulse, the next sound pulse moves toward the submarine at vsound and the submarine moves toward the pulse at speed vsub. The distance between successive pulses Δx may be divided into Δxsub and Δxsound, which are equal to vsubΔtreceived and vsubΔtreceived, respectively.

The distance between successive pulses is given by:

Δx = Δxsub + Δxreceived

Substituting for all three terms in this equation yields:

vsound Δt emitted = vsub Δt received + vsound Δt received

92

Chapter 2

Solve for vsub to obtain:

Substitute numerical values and evaluate vsub :

vsub =

vsub =

vsound (Δtemitted − Δtreceived ) Δtreceived

(1522 m/s )(2.000 s − 1.980 s ) 2.000 s

= 15 m/s

Acceleration 58 • A sports car accelerates in third gear from 48.3 km/h (about 30 mi/h) to 80.5 km/h (about 50 mi/h) in 3.70 s. (a) What is the average acceleration of this car in m/s2? (b) If the car maintained this acceleration, how fast would it be moving one second later? Picture the Problem In Part (a), we can apply the definition of average acceleration to find aav. In Part (b), we can find the change in the car’s velocity in one second and add this change to its velocity at the beginning of the interval to find its speed one second later.

(a) The definition of average acceleration is:

aav =

Δv Δt

Substitute numerical values and evaluate aav:

aav =

80.5 km/h − 48.3 km/h km = 8.70 3.70 s h ⋅s

Convert aav to m/s2:

m ⎞⎛ 1h ⎞ ⎛ ⎟ aav = ⎜ 8.70 × 103 ⎟⎜ h ⋅ s ⎠⎜⎝ 3600 s ⎟⎠ ⎝ = 2.42 m/s 2

(b) Express the speed of the car at the end of 4.7 s:

v(4.7 s ) = v(3.70 s ) + Δv1s

Find the change in the speed of the car in 1.00 s:

km ⎞ ⎛ Δv = aav Δt = ⎜ 8.70 ⎟(1.00 s ) h ⋅s ⎠ ⎝ = 8.70 km/h

Substitute and evaluate v(4.7 s):

v(4.7 s ) = 80.5 km/h + 8.7 km/h

= 80.5 km/h + Δv1s

= 89.2 km/h

Motion in One Dimension

93

59 • [SSM] An object is moving along the x axis. At t = 5.0 s, the object is at x = +3.0 m and has a velocity of +5.0 m/s. At t = 8.0 s, it is at x = +9.0 m and its velocity is –1.0 m/s. Find its average acceleration during the time interval 5.0 s ≤ t ≤ 8.0 s. Picture the Problem We can find the change in velocity and the elapsed time from the given information and then use the definition of average acceleration.

The average acceleration is defined as the change in velocity divided by the change in time: Substitute numerical values and evaluate aav:

aav =

Δv Δt

aav =

(− 1.0 m/s) − (5.0 m/s) (8.0 s ) − (5.0 s )

= − 2.0 m/s 2 60 •• A particle moves along the x axis with velocity vx = (8.0 m/s2) t – 7.0 m/s. (a) Find the average acceleration for two different one-second intervals, one beginning at t = 3.0 s and the other beginning at t = 4.0 s. (b) Sketch vx versus t over the interval 0 < t < 10 s. (c) How do the instantaneous accelerations at the middle of each of the two time intervals specified in Part (a) compare to the average accelerations found in Part (a)? Explain. Picture the Problem The important concept here is the difference between average acceleration and instantaneous acceleration.

(a) The average acceleration is defined as the change in velocity divided by the change in time: Determine vx at t = 3.0 s, t = 4.0 s, and t = 5.0 s:

aav =

Δv Δt

(

)

v x (3.0 s ) = 8.0 m/s 2 (3.0 s ) − 7.0 m/s = 17 m/s v x (4.0 s ) = 8.0 m/s 2 (4.0 s ) − 7.0 m/s

(

)

= 25 m/s v x (5.0 s ) = 8.0 m/s 2 (5.0 s ) − 7.0 m/s

(

= 33 m/s

)

94

Chapter 2 25 m/s − 17 m/s 1.0 s = 8.0 m/s 2

Find aav for the two 1-s intervals:

aav, 3.0 s to 4.0 s = and

33 m/s − 25 m/s 1.0 s = 8.0 m/s 2

aav, 4.0 s to 5.0 s =

The instantaneous acceleration is defined as the time derivative of the velocity or the slope of the velocity- versus-time curve:

ax =

[(

)

dv x d = 8.0 m/s 2 t − 7.0 m/s dt dt

]

= 8.0 m/s 2

(b) The given function and a spreadsheet program were used to plot the following graph of v-versus-t: 35 30 25

v, m/s

20 15 10 5 0 -5 -10 0

1

2

3

4

5

t, s

(c) Because the particle’s speed varies linearly with time, these accelerations are the same. [SSM] The position of a certain particle depends on time according 61 •• to the equation x(t) = t2 – 5.0t + 1.0, where x is in meters if t is in seconds. (a) Find the displacement and average velocity for the interval 3.0 s ≤ t ≤ 4.0 s. (b) Find the general formula for the displacement for the time interval from t to t + Δt. (c) Use the limiting process to obtain the instantaneous velocity for any time t. Picture the Problem We can closely approximate the instantaneous velocity by the average velocity in the limit as the time interval of the average becomes small. This is important because all we can ever obtain from any measurement is the average velocity, vav, which we use to approximate the instantaneous velocity v.

Motion in One Dimension (a) The displacement of the particle during the interval 3.0 s ≤ t ≤ 4.0 s is given by: The average velocity is given by:

Δx = x(4.0 s ) − x(3.0 s )

vav =

95

(1)

Δx Δt

(2)

Find x(4.0 s) and x(3.0 s):

x(4.0 s) = (4.0)2 – 5(4.0) + 1 = –3.0 m and x(3.0 s) = (3.0)2 – 5(3.0) + 1 = −5.0 m

Substitute numerical values in equation (1) and evaluate Δx:

Δx = (− 3.0 m ) − (− 5.0 m ) = 2.0 m

Substitute numerical values in equation (2) and evaluate vav:

vav =

2.0 m = 2.0 m/s 1.0 s

(b) Find x(t + Δt):

x(t + Δt) = (t + Δt)2 − 5(t + Δt) + 1 = (t2 + 2tΔt + (Δt)2) – 5(t + Δt) + 1

Express x(t + Δt) – x(t) = Δx:

Δx =

(2t − 5)Δt + (Δt )2

where Δx is in meters if t is in seconds. (c) From (b) find Δx/Δt as Δt → 0:

Δx (2t − 5)Δt + (Δt ) = Δt Δt = 2t − 5 + Δt and v = lim Δt →0 (Δx / Δt ) = 2t − 5 2

where v is in m/s if t is in seconds. Alternatively, we can take the derivative of x(t) with respect to time to obtain the instantaneous velocity.

dx(t ) d = at 2 + bt + 1 dt dt = 2at + b

v(t ) =

(

)

= 2t − 5

62 •• The position of an object as a function of time is given by 2 x = At – Bt + C, where A = 8.0 m/s2, B = 6.0 m/s, and C = 4.0 m. Find the instantaneous velocity and acceleration as functions of time. Picture the Problem The instantaneous velocity is dx dt and the acceleration is dv dt .

96

Chapter 2

Using the definitions of instantaneous velocity and acceleration, determine v and a:

v=

]

and a=

Substitute numerical values for A and B and evaluate v and a:

[

dx d = At 2 − Bt + C = 2 At − B dt dt dv d = [2 At − B ] = 2 A dt dt

v = 2(8.0 m/s 2 ) t − 6.0 m/s = (16 m/s 2 ) t − 6.0 m/s

and

a = 2(8.0 m/s 2 ) = 16 m/s 2

••• The one-dimensional motion of a particle is plotted in Figure 2-40. (a) What is the average acceleration in each of the intervals AB, BC, and CE? (b) How far is the particle from its starting point after 10 s? (c) Sketch the displacement of the particle as a function of time; label the instants A, B, C, D, and E on your graph. (d) At what time is the particle traveling most slowly? 63

Picture the Problem We can use the definition of average acceleration (aav = Δv/Δt) to find aav for the three intervals of constant acceleration shown on the graph.

(a) Using the definition of average acceleration, find aav for the interval AB: Find aav for the interval BC:

Find aav for the interval CE:

aav, AB =

15.0 m/s − 5.0 m/s = 3.3 m/s 2 3.0 s

aav, BC =

15.0 m/s − 15.0 m/s = 0 3.0 s

aav, CE =

− 15.0 m/s − 15.0 m/s 4.0 s

= − 7.5 m/s 2 (b) Use the formulas for the areas of trapezoids and triangles to find the area under the graph of v as a function of t. Δx = (Δx )A→B + (Δx )B→C + (Δx )C→D + (Δx )D→E =

1 2

(5.0 m/s + 15.0 m/s)(3.0 s ) + (15.0 m/s)(3.0 s) + 12 (15.0 m/s)(2.0 s) + 12 (−15.0 m/s)(2.0 s)

= 75 m

Motion in One Dimension

97

(c) The graph of displacement, x, as a function of time, t, is shown in the following figure. In the region from B to C the velocity is constant so the x- versus-t curve is a straight line. 100

80

x, m

60

40

20

0 0

2

4

6

8

10

t, s

(d) Reading directly from the figure, we can find the time when the particle is moving the slowest. At point D, t = 8 s, the graph crosses the time axis; therefore v = 0.

Constant Acceleration and Free-Fall 64 • An object projected vertically upward with initial speed v0 attains a maximum height h above its launch point. Another object projected up with initial speed 2v0 from the same height will attain a maximum height of (a) 4h, (b) 3h, (c) 2h, (d) h. (Air resistance is negligible.) Picture the Problem Because the acceleration is constant (–g) we can use a constant-acceleration equation to find the height of the projectile.

Using a constant-acceleration equation, express the height of the object as a function of its initial speed, the acceleration due to gravity, and its displacement: Express the ratio of the maximum height of the second object to that of the first object and simplify to obtain:

v 2 = v02 − 2 gΔy or, because v(h) = 0, 0 = v02 − 2 gh ⇒ h =

v02 2g

(2v0 )2 h2nd object h1st object

=

2g =4 (v0 )2 2g

98

Chapter 2

Solving for h2nd object yields:

h2nd object = 4h ⇒ (a ) is correct.

65 •• A car traveling along the x axis starts from rest at x = 50 m and accelerates at a constant rate of 8.0 m/s2. (a) How fast is it going after 10 s? (b) How far has it gone after 10 s? (c) What is its average velocity for the interval 0 ≤ t ≤ 10 s? Picture the Problem Because the acceleration of the car is constant we can use constant-acceleration equations to describe its motion.

(a) Using a constant-acceleration equation, relate the velocity to the acceleration and the time:

v = v0 + at

Substitute numerical values and evaluate v:

m⎞ ⎛ v = 0 + ⎜ 8.0 2 ⎟ (10 s ) = 80 m s s ⎠ ⎝

(b) Using a constant-acceleration equation, relate the displacement to the acceleration and the time:

a Δx = x − x0 = v0t + t 2 2

Substitute numerical values and evaluate Δx:

Δx =

1⎛ m⎞ 2 ⎜ 8.0 2 ⎟ (10 s ) = 0.40 km 2⎝ s ⎠

vav =

Δx 400 m = = 40 m/s Δt 10 s

(c) Use the definition of vav :

Remarks: Because the area under a velocity-versus-time graph is the displacement of the object, we could solve this problem graphically. 66 • An object traveling along the x axis with an initial velocity of +5.0 m/s has a constant acceleration of +2.0 m/s2. When its speed is 15 m/s, how far has it traveled? Picture the Problem Because the acceleration of the object is constant we can use constant-acceleration equations to describe its motion.

Using a constant-acceleration equation, relate the speed of the object to its acceleration and displacement: Substitute numerical values and evaluate Δx:

v 2 = v02 + 2 a Δx ⇒ Δx =

Δx =

(15

v 2 − v02 2a

− 5.0 2 )m 2 s 2 = 50 m 2 (2.0 m s 2 )

2

Motion in One Dimension

99

67 • [SSM] An object traveling along the x axis at constant acceleration has a velocity of +10 m/s when it is at x = 6.0 m and of +15 m/s when it is at x = 10 m. What is its acceleration? Picture the Problem Because the acceleration of the object is constant we can use constant-acceleration equations to describe its motion.

v 2 − v02 = a v = v + 2 a Δx ⇒ 2 Δx

Using a constant-acceleration equation, relate the velocity to the acceleration and the displacement:

2

Substitute numerical values and evaluate a:

a=

2 0

(15

− 10 2 )m 2 s 2 = 16 m s 2 2(10 m − 6.0 m ) 2

68 • The speed of an object traveling along the x axis increases at the constant rate of +4.0 m/s each second. At t = 0.0 s, its velocity is +1.0 m/s and its position is +7.0 m. How fast is it moving when its position is +8.0 m, and how much time has elapsed from the start at t = 0.0 s? Picture the Problem Because the acceleration of the object is constant we can use constant-acceleration equations to describe its motion.

Using a constant-acceleration equation, relate the velocity to the acceleration and the displacement:

v 2 = v02 + 2 a Δx ⇒ v = v02 + 2 a Δx

Substitute numerical values and evaluate v to obtain: v=

(1.0 m s )2 + 2 (4.0 m

)

s 2 (8.0 m − 7.0 m ) = 3.0 m/s

From the definition of average acceleration we have:

Δt =

Δv aav

Substitute numerical values and evaluate Δt:

Δt =

3.0 m s − 1.0 m s = 0.50 s 4.0 m s 2

69 •• A ball is launched directly upward from ground level with an initial speed of 20 m/s. (Air resistance is negligible.) (a) How long is the ball in the air? (b) What is the greatest height reached by the ball? (c) How many seconds after launch is the ball 15 m above the release point? Picture the Problem In the absence of air resistance, the ball experiences constant acceleration. Choose a coordinate system with the origin at the point of release and the positive direction upward.

100 Chapter 2 (a) Using a constant-acceleration equation, relate the displacement of the ball to the acceleration and the time:

Δy = v0t + 12 at 2

Setting Δy = 0 (the displacement for a round trip), solve for the time required for the ball to return to its starting position:

2 0 = v0 t roundtrip + 12 at round trip

and t round =

2v0 g

t round =

2(20 m/s ) = 4.1s 9.81m/s 2

trip

Substitute numerical values and evaluate t round : trip

(b) Using a constant-acceleration equation, relate the final speed of the ball to its initial speed, the acceleration, and its displacement: Substitute numerical values and evaluate H:

trip

2 vtop = v02 + 2aΔy

or, because vtop = 0 and a = −g, v2 0 = v02 + 2(− g )H ⇒ H = 0 2g H=

(20 m s )2

(

2 9.81m s 2

)=

20 m

(c) Using the same constantacceleration equation with which we began part (a), express the displacement as a function of time:

Δy = v0t + 12 at 2

Substitute numerical values to obtain:

⎛ 9.81 m/s 2 ⎞ 2 ⎟⎟ t 15 m = (20 m/s )t − ⎜⎜ 2 ⎝ ⎠

Use your graphing calculator or the quadratic formula to solve this equation for the times at which the displacement of the ball is 15 m:

The solutions are t = 0.99 s (this corresponds to passing 15 m on the way up) and t = 3.1s (this corresponds to passing 15 m on the way down).

70 •• In the Blackhawk landslide in California, a mass of rock and mud fell 460 m down a mountain and then traveled 8.00 km across a level plain. It has been theorized that the rock and mud moved on a cushion of water vapor. Assume that the mass dropped with the free-fall acceleration and then slid horizontally, losing speed at a constant rate. (a) How long did the mud take to drop the 460 m? (b) How fast was it traveling when it reached the bottom? (c) How long did the mud take to slide the 8.00 km horizontally?

Motion in One Dimension 101 Picture the Problem This is a multipart constant-acceleration problem using two different constant accelerations. We’ll choose a coordinate system in which downward is the positive direction and apply constant-acceleration equations to find the required times.

(a) Using a constant-acceleration equation, relate the time for the slide to the distance of fall and the acceleration:

Δy = y − y0 = h − 0 = v0t1 + 12 at12 or, because v0 = 0, 2h h = 12 at12 ⇒ t1 = g 2(460 m ) = 9.684 s = 9.68 s 9.81 m s 2

Substitute numerical values and evaluate t1:

t1 =

(b) Using a constant-acceleration equation, relate the velocity at the bottom of the mountain to the acceleration and time:

v1 = v0 + a1t1 or, because v0 = 0 and a1 = g, v1 = gt1

Substitute numerical values and evaluate v1:

v1 = 9.81 m s 2 (9.684 s ) = 95.0 m s

(

)

Δt =

Δx vav

vav =

v1 + vf v1 + 0 v1 = = 2 2 2

Substitute for vav to obtain:

Δt =

2Δx v1

Substitute numerical values and evaluate Δt:

Δt =

2(8000 m ) = 168 s 95.0 m s

(c) Using a constant-acceleration equation, relate the time required to stop the mass of rock and mud to its average speed and the distance it slides: Because the acceleration is constant:

71 •• [SSM] A load of bricks is lifted by a crane at a steady velocity of 5.0 m/s when one brick falls off 6.0 m above the ground. (a) Sketch the position of the brick y(t) versus time from the moment it leaves the pallet until it hits the ground. (b) What is the greatest height the brick reaches above the ground? (c) How long does it take to reach the ground? (d) What is its speed just before it hits the ground?

102 Chapter 2 Picture the Problem In the absence of air resistance, the brick experiences constant acceleration and we can use constant-acceleration equations to describe its motion. Constant acceleration implies a parabolic position-versus-time curve.

(a) Using a constant-acceleration equation, relate the position of the brick to its initial position, initial velocity, acceleration, and time into its fall:

y = y0 + v0 t + 12 (− g )t 2

Substitute numerical values to obtain: y = 6.0 m + (5.0 m s )t − (4.91m s 2 )t 2

(1)

The following graph of y = 6.0 m + (5.0 m s )t − (4.91m s 2 )t 2 was plotted using a spreadsheet program: 8 7 6

y, m

5 4 3 2 1 0 0.0

0.5

1.0

1.5

t, s

(b) Relate the greatest height reached by the brick to its height when it falls off the load and the additional height it rises Δymax :

h = y0 + Δymax

Using a constant-acceleration equation, relate the height reached by the brick to its acceleration and initial velocity:

2 vtop = v02 + 2(− g )Δymax

or, because vtop = 0,

Substitute numerical values and evaluate Δymax :

Δy max =

(2)

0 = v + 2(− g )Δymax ⇒ Δymax 2 0

(5.0 m s ) 2

(

2 9.81 m s 2

) = 1.3 m

v02 = 2g

Motion in One Dimension 103 Substitute numerical values in equation (2) and evaluate h:

(c) Setting y = 0 in equation (1) yields: Use the quadratic equation or your graphing calculator to obtain:

h = 6.0 m + 1.3 m = 7.3 m

The graph shown above confirms this result. 0 = 6.0 m + (5.0 m s )t − (4.91 m s 2 )t 2 t = 1.7 s and t = –0.71 s. The

negative value for t yields (from v = v0 − gt ) v = − 12 m/s and so has no physical meaning.

(d) Using a constant-acceleration equation, relate the speed of the brick on impact to its acceleration and displacement:

v 2 = v02 + 2 gh or, because v0 = 0, v 2 = 2 gh ⇒ v = 2 gh

Substitute numerical values and evaluate v:

v = 2(9.81 m/s 2 )(7.3 m ) = 12 m/s

72 •• A bolt comes loose from underneath an elevator that is moving upward at a speed of 6.0 m/s. The bolt reaches the bottom of the elevator shaft in 3.0 s. (a) How high above the bottom of the shaft was the elevator when the bolt came loose? (b) What is the speed of the bolt when it hits the bottom of the shaft? Picture the Problem In the absence of air resistance, the acceleration of the bolt is constant. Choose a coordinate system in which upward is positive and the origin is at the bottom of the shaft (y = 0).

(a) Using a constant-acceleration equation, relate the position of the bolt to its initial position, initial velocity, and fall time:

ybottom = 0

Solve for the position of the bolt when it came loose:

y0 = −v0t + 12 gt 2

= y0 + v0t + 12 (− g )t 2

Substitute numerical values and evaluate y0: m⎞ m⎞ ⎛ ⎛ 2 y 0 = −⎜ 6.0 ⎟ (3.0 s ) + 12 ⎜ 9.81 2 ⎟ (3.0 s ) = 26 m s s ⎠ ⎝ ⎠ ⎝

104 Chapter 2 (b) Using a constant-acceleration equation, relate the speed of the bolt to its initial speed, acceleration, and fall time: Substitute numerical values and evaluate v :

v = v0 + at

v = 6.0

m ⎛ m⎞ − ⎜ 9.81 2 ⎟ (3.0 s ) = −23 m s s ⎝ s ⎠

and v = 23 m s 73 •• An object is dropped from rest at a height of 120 m. Find the distance it falls during its final second in the air. Picture the Problem In the absence of air resistance, the object’s acceleration is constant. Choose a coordinate system in which downward is the +y direction and the origin is at the point of release. In this coordinate system, a = g and y = 120 m at the bottom of the fall.

Express the distance fallen in the last second in terms of the object’s position at impact and its position 1 s before impact: Using a constant-acceleration equation, relate the object’s position upon impact to its initial position, initial velocity, and fall time:

Δylast

second

= y − 12 gt 2

(1)

where t is the time the object has been falling. y = y0 + v0t + 12 gt 2 or, because y0 = 0 and v0 = 0, 2y 2 y = 12 gtfall ⇒ tfall = g

Substitute numerical values and evaluate tfall:

tfall =

2(120 m ) = 4.95 s 9.81 m/s 2

Evaluating equation (1) for t = 3.95 s gives:

Δylast

= 120 m − 12 9.81 m/s 2 (3.95 s )

second

(

)

2

= 44 m 74 •• An object is released from rest at a height h. During the final second of its fall, it traverses a distance of 38 m. Determine h. Picture the Problem In the absence of air resistance, the acceleration of the object is constant. Choose a coordinate system with the origin at the point of release and downward as the positive direction.

Motion in One Dimension 105 Using a constant-acceleration equation, relate the height h of the object to its initial and final velocities and acceleration: Using the definition of average velocity, find the average velocity of the object during its final second of fall:

vf2 = v02 + 2aΔy or, because v0 = 0, a = g, and Δy = h , vf2 = 2 gh ⇒ h =

vav =

vf -1s + vf

2

vf2 2g

=

(1)

Δy 38 m = = 38 m s Δt 1s

Express the sum of the final velocity and the velocity 1 s before impact:

vf -1s + vf = 2(38 m s ) = 76 m s

From the definition of acceleration, we know that the change in velocity of the object, during 1 s of fall, is 9.81 m/s:

Δv = vf − vf -1s = 9.81 m s

Add the equations that express the sum and difference of vf -1s and vf to

vf -1s + vf + vf − vf -1s = 76 m s + 9.81 m s

obtain: Solving for vf yields: Substitute numerical values in equation (1) and evaluate h:

vf =

h=

76 m s + 9.81m s = 43 m s 2

(43 m s )2

2(9.81m s 2 )

= 94 m

75 •• [SSM] A stone is thrown vertically downward from the top of a 200-m cliff. During the last half second of its flight, the stone travels a distance of 45 m. Find the initial speed of the stone. Picture the Problem In the absence of air resistance, the acceleration of the stone is constant. Choose a coordinate system with the origin at the bottom of the trajectory and the upward direction positive. Let vf -1 2 be the speed one-half

second before impact and vf the speed at impact. Using a constant-acceleration equation, express the final speed of the stone in terms of its initial speed, acceleration, and displacement:

vf2 = v02 + 2aΔy ⇒ v0 = vf2 + 2 gΔy (1)

106 Chapter 2 Find the average speed in the last half second:

vav =

vf -1 2 + vf

2 = 90 m s

=

Δxlast half second 45 m = Δt 0.5 s

and vf -1 2 + vf = 2(90 m s ) = 180 m s

Using a constant-acceleration equation, express the change in speed of the stone in the last half second in terms of the acceleration and the elapsed time and solve for the change in its speed: Add the equations that express the sum and difference of vf – ½ and vf and solve for vf: Substitute numerical values in equation (1) and evaluate v0:

Δv = vf − vf -1 2 = gΔt

(

)

= 9.81 m s 2 (0.5 s ) = 4.91 m s

vf =

180 m s + 4.91m s = 92.5 m s 2

v0 =

(92.5 m s )2 + 2(9.81 m

)

s 2 (− 200m )

= 68 m s

Remarks: The stone may be thrown either up or down from the cliff and the results after it passes the cliff on the way down are the same. 76 •• An object is released from rest at a height h. It travels 0.4h during the first second of its descent. Determine the average velocity of the object during its entire descent. Picture the Problem In the absence of air resistance, the acceleration of the object is constant. Choose a coordinate system in which downward is the positive direction and the object starts from rest. Apply constant-acceleration equations to find the average velocity of the object during its descent.

Express the average velocity of the falling object in terms of its initial and final velocities: Using a constant-acceleration equation, express the displacement of the object during the 1st second in terms of its acceleration and the elapsed time:

vav =

v0 + vf 2

Δy1st second =

gt 2 = 4.91 m = 0.4 h 2

Motion in One Dimension 107 Solving for the displacement h gives:

h = 12.3 m

Using a constant-acceleration equation, express the final velocity of the object in terms of its initial velocity, acceleration, and displacement:

vf2 = v02 + 2 gΔy or, because v0 = 0, vf = 2 gΔy

Substitute numerical values and evaluate the final velocity of the object:

vf = 2 9.81m s 2 (12.3 m ) = 15.5 m s

Substitute in the equation for the average velocity to obtain:

(

vav =

)

0 + 15.5 m s = 7.77 m s 2

77 •• A bus accelerates from rest at 1.5 m/s2 for 12 s. It then travels at constant velocity for 25 s, after which it slows to a stop with an acceleration of 1.5 m/s2. (a) What is the total distance that the bus travels? (b) What is its average velocity? Picture the Problem This is a three-part constant-acceleration problem. The bus starts from rest and accelerates for a given period of time, and then it travels at a constant velocity for another period of time, and, finally, decelerates uniformly to a stop. The pictorial representation will help us organize the information in the problem and develop our solution strategy.

(a) Express the total displacement of the bus during the three intervals of time:

Δxtotal = Δx(0 → 12 s ) + Δx(12 s → 37 s )

Using a constant-acceleration equation, express the displacement of the bus during its first 12 s of motion in terms of its initial velocity, acceleration, and the elapsed time; solve for its displacement:

Δx(0 → 12 s ) = v0t + 12 at 2 or, because v0 = 0, Δx(0 → 12 s ) = 12 at 2 = 108 m

+ Δx(37 s → end )

108 Chapter 2

(

)

Using a constant-acceleration equation, express the velocity of the bus after 12 seconds in terms of its initial velocity, acceleration, and the elapsed time; solve for its velocity at the end of 12 s:

v12 s = v0 + a0→12 s Δt = 1.5 m/s 2 (12 s )

During the next 25 s, the bus moves with a constant velocity. Using the definition of average velocity, express the displacement of the bus during this interval in terms of its average (constant) velocity and the elapsed time:

Δx(12 s → 37 s ) = v12 s Δt = (18 m/s )(25 s )

Because the bus slows down at the same rate that its velocity increased during the first 12 s of motion, we can conclude that its displacement during this braking period is the same as during its acceleration period and the time to brake to a stop is equal to the time that was required for the bus to accelerate to its cruising speed of 18 m/s. Hence:

Δx(37 s → 49s ) = 108 m

Add the displacements to find the distance the bus traveled:

Δxtotal = 108 m + 450 m + 108 m

(b) Use the definition of average velocity to calculate the average velocity of the bus during this trip:

= 18 m/s

= 450 m

= 666 m = 0.67 km vav =

Δxtotal 666 m = = 14 m s Δt 49 s

Remarks: One can also solve this problem graphically. Recall that the area under a velocity as a function-of-time graph equals the displacement of the moving object. 78 •• Al and Bert are jogging side-by-side on a trail in the woods at a speed of 0.75 m/s. Suddenly Al sees the end of the trail 35 m ahead and decides to speed up to reach it. He accelerates at a constant rate of 0.50 m/s2, while Bert continues on at a constant speed. (a) How long does it take Al to reach the end of the trail? (b) Once he reaches the end of the trail, he immediately turns around and heads back along the trail with a constant speed of 0.85 m/s. How long does it take him

Motion in One Dimension 109 to meet up with Bert? (c) How far are they from the end of the trail when they meet? Picture the Problem Because the accelerations of both Al and Bert are constant, constant-acceleration equations can be used to describe their motions. Choose the origin of the coordinate system to be where Al decides to begin his sprint.

(a) Using a constant-acceleration equation, relate Al's initial velocity, his acceleration, and the time to reach the end of the trail to his displacement in reaching the end of the trail:

Δx = v0t + 12 at 2

Substitute numerical values to obtain:

35 m = (0.75 m/s)t + 12 (0.50 m/s 2 )t 2

Use your graphing calculator or the quadratic formula to solve for the time required for Al to reach the end of the trail:

t = 10.43 s = 10 s

(b) Using constant-acceleration equations, express the positions of Bert and Al as functions of time. At the instant Al turns around at the end of the trail, t = 0. Also, x = 0 at a point 35 m from the end of the trail:

x Bert = x Bert,0 + (0 .75 m/s ) t and

xAl = xAl,0 − (0.85 m/s ) t

= 35 m − (0.85 m/s ) t

Calculate Bert’s position at t = 0. At that time he has been running for 10.4 s:

xBert,0 = (0.75 m/s )(10.43 s ) = 7.823 m

Because Bert and Al will be at the same location when they meet, equate their position functions and solve for t:

7.823 m + (0.75 m/s )t = 35 m − (0.85 m/s )t and t = 16.99 s

To determine the elapsed time from when Al began his accelerated run, we need to add 10.43 s to this time:

t start = 16.99 s + 10.43 s = 27.42 s

(c) Express Bert’s distance from the end of the trail when he and Al meet:

d end of = 35 m − xBert,0 − d Bert runs until

= 27 s

trail

he meets Al

110 Chapter 2 Substitute numerical values and evaluate d end of : trail

d end of = 35 m − 7.823 m − (16.99 s) (0.75 m/s ) = 14.43 m = 14 m trail

79 •• You have designed a rocket to be used to sample the local atmosphere for pollution. It is fired vertically with a constant upward acceleration of 20 m/s2. After 25 s, the engine shuts off and the rocket continues rising (in freefall) for a while. (Neglect any effects due to air resistance.) The rocket eventually stops rising and then falls back to the ground. You want to get a sample of air that is 20 km above the ground. (a) Did you reach your height goal? If not, what would you change so that the rocket reaches 20 km? (b) Determine the total time the rocket is in the air. (c) Find the speed of the rocket just before it hits the ground. Picture the Problem This is a two-part constant-acceleration problem. Choose a coordinate system in which the upward direction is positive and to the right. The pictorial representation will help us organize the information in the problem and develop our solution strategy.

(a) Express the highest point the rocket reaches, h, as the sum of its displacements during the first two stages of its flight:

h = Δy1st stage + Δy 2nd stage

Using a constant-acceleration equation, express the altitude reached in the first stage in terms of the rocket’s initial velocity, acceleration, and burn time; solve for the first stage altitude:

2 Δy1st stage = v0 y t + 12 a1st stage t1st stage

Using a constant-acceleration equation, express the speed of the rocket at the end of its first stage in terms of its initial speed, acceleration, and displacement:

or, because y0y = 0, 2 Δy1st stage = 12 a1st stage t1st stage

v1st stage = v0 y + a1st stage t1st stage or, because y0y = 0, v1st stage = a1st stage t1st stage

(1)

Motion in One Dimension 111 Using a constant-acceleration equation, express the final speed of the rocket during the remainder of its climb in terms of its shut-off speed, free-fall acceleration, and displacement: Solving for Δy 2 nd stage yields:

Using a constant-acceleration equation, express the speed of the rocket at the end of its first stage in terms is its initial speed, acceleration, and displacement: Substituting for vshutoff yields:

Substitute for Δy1st stage and Δy 2nd stage in equation (1) and simplify to obtain:

Substitute numerical values and evaluate h:

2 2 vhighest = vshutoff + 2a 2 nd stage Δy 2nd stage point

or, because vhighest = 0 and point

a2 nd stage = − g , 2 0 = vshutoff − 2 gΔy 2nd stage

Δy 2 nd stage

2 vshutoff = 2g

v1st stage = v0 + a1st stage t1st stage or, because v0 = 0 and v1st stage = vshutoff , vshutoff = a1st stage t1st stage

Δy 2 nd stage

h= a 1 2

(a =

)

2

t

1st stage 1st stage

2g

2 1st stage 1st stage

t

⎛ 1 a1st stage = ⎜⎜ + 2g ⎝2

+

(a

t

1st stage 1st stage

)

2

2g

⎞ 2 ⎟⎟a1st stage t1st stage ⎠

⎛ m ⎞ ⎜ 20 2 ⎟ 1 s ⎟ ⎛⎜ 20 m ⎞⎟ (25 s )2 h=⎜ + ⎜2 m⎞⎟ s2 ⎠ ⎛ 2⎜ 9.81 2 ⎟ ⎟ ⎝ ⎜ s ⎠⎠ ⎝ ⎝ ≈ 19 km

You did not achieve your goal. To go higher, you can increase the acceleration value or the time of acceleration. (b) Express the total time the rocket is in the air in terms of the three segments of its flight:

Δt total = Δt powered + Δt 2nd climb

segment

= 25 s + Δt 2nd

segment

Express Δt2nd segment in terms of the rocket’s displacement and average speed during the second segment:

Δt 2nd segment =

+ Δt descent

+ Δt descent

Δy 2nd

segment

vav, 2nd segment

112 Chapter 2

(a

Substituting for Δy 2nd segment and

)

2

t

1st stage 1st stage

simplifying yields:

2g

Δt 2nd segment = =

(a

vav, 2nd segment t

1st stage 1st stage

)

2

2 gvav, 2nd segment

Δydescent = v0t + 12 g (Δtdescent ) or, because v0 = 0, 2 Δydescent = 12 g (Δtdescent )

Using a constant-acceleration equation, relate the fall distance to the descent time:

2

Solving for Δtdescent yields:

Δtdescent =

2Δydescent g

Substitute for Δt 2nd segment and Δtdescent in the expression for Δt total to obtain: Δt total = 25 s +

(a

Δt1st stage )

2

1st stage

+

2 gvav, 2nd segment

2Δy descent g

Noting that, because the acceleration is constant, vav, 2nd segment is the average of the initial and final speeds during the second stage, substitute numerical values and evaluate Δt total : 2

Δt total

⎞ ⎛⎛ m ⎞ ⎜⎜ ⎜ 20 2 ⎟ (25 s )⎟⎟ 2 19 × 103 m ⎠ ⎝⎝ s ⎠ + ≈ 1.4 × 10 2 s = 25 s + m m⎞ ⎛ 9.81 2 0 + 500 ⎟ m ⎞⎜ ⎛ s s ⎟ ⎜ 2⎜ 9.81 2 ⎟ s ⎠⎜ 2 ⎟ ⎝ ⎟ ⎜ ⎠ ⎝

(c) Using a constant-acceleration equation, express the impact velocity of the rocket in terms of its initial downward velocity, acceleration under free-fall, and time of descent; solve for its impact velocity: Substituting for Δtdescent yields:

(

)

vimpact = v0 + gΔtdescent and, because v0 = 0, vimpact = gΔtdescent

vimpact = g

2Δydescent = 2 gΔydescent g

Motion in One Dimension 113 Substitute numerical values and evaluate vimpact :

(

)

vimpact = 2 9.81 m/s 2 (19 km ) = 6.1×10 2 m/s

80 •• A flowerpot falls from a windowsill of an apartment that is on the tenth floor of an apartment building. A person in an apartment below, coincidentally in possession of a high-speed high-precision timing system, notices that it takes 0.20 s for the pot to fall past his window, which is 4.0-m from top to bottom. How far above the top of the window is the windowsill from which the pot fell? (Air resistance is negligible.) Picture the Problem In the absence of air resistance, the acceleration of the flowerpot is constant. Choose a coordinate system in which downward is positive and the origin is at the point from which the flowerpot fell. Let t = time when the pot is at the top of the window, and t + Δt the time when the pot is at the bottom of the window. To find the distance from the ledge to the top of the window, first find the time ttop that it takes the pot to fall to the top of the window.

Using a constant-acceleration equation, express the distance y below the ledge from which the pot fell as a function of time: Express the position of the pot as it reaches the top of the window: Express the position of the pot as it reaches the bottom of the window: Subtract ybottom from ytop to obtain an expression for the displacement Δy window of the pot as it passes the window:

y = y0 + v0t + 12 at 2 or, because a = g and v0 = y0 = 0 , y = 12 gt 2 2 ytop = 12 gttop

ybottom = 12 g (t top + Δt window )

2

where Δtwindow = t top − tbottom

[ g [2t

2 Δywindow = 12 g (t top + Δt window ) − t top

=

Solving for t top yields: t top Substitute for t top in equation (1) to obtain:

(1)

y top

1 2

2

top

]

Δt window + (Δt window )

2Δy window 2 − (Δt window ) g = 2Δt window ⎛ 2Δy window 2 ⎞ − (Δt window ) ⎟ ⎜ g ⎟ = 12 g ⎜ ⎟ ⎜ 2Δt window ⎟ ⎜ ⎠ ⎝

2

2

]

114 Chapter 2 Substitute numerical values and evaluate y top : ⎛ 2(4.0 m ) 2 ⎞ − (0.20 s ) ⎟ ⎜ 2 9.81 m/s ⎟ = 18 m 9.81 m/s 2 ⎜ ⎟ ⎜ 2(0.20 s ) ⎟ ⎜ ⎠ ⎝ 2

y top =

1 2

(

)

81 •• [SSM] In a classroom demonstration, a glider moves along an inclined air track with constant acceleration. It is projected from the low end of the track with an initial velocity. After 8.00 s have elapsed, it is 100 cm from the low end and is moving along the track at a velocity of –15 cm/s. Find the initial velocity and the acceleration. Picture the Problem The acceleration of the glider on the air track is constant. Its average acceleration is equal to the instantaneous (constant) acceleration. Choose a coordinate system in which the initial direction of the glider’s motion is the positive direction.

Using the definition of acceleration, express the average acceleration of the glider in terms of the glider’s velocity change and the elapsed time: Using a constant-acceleration equation, express the average velocity of the glider in terms of the displacement of the glider and the elapsed time: Substitute numerical values and evaluate v0:

a = aav =

Δv Δt

vav =

2Δx Δx v0 + v ⇒ v0 = = −v 2 Δt Δt

v0 =

2(100 cm ) − (−15 cm/s) 8.00 s

= 40 cm/s The average acceleration of the glider is:

a=

− 15 cm/s − (40 cm/s) 8.00 s

= − 6.9 cm/s 2

82 •• A rock dropped from a cliff covers one-third of its total distance to the ground in the last second of its fall. Air resistance is negligible. How high is the cliff?

Motion in One Dimension 115 Picture the Problem In the absence of air resistance, the acceleration of the rock is constant and its motion can be described using the constant-acceleration equations. Choose a coordinate system in which the downward direction is positive and let the height of the cliff, which equals the displacement of the rock, be represented by h.

Using a constant-acceleration equation, express the height h of the cliff in terms of the initial velocity of the rock, acceleration, and time of fall:

Δy = v0t + 12 at 2 or, because v0 = 0, a = g, and Δy = h, h = 12 gt 2

Using this equation, express the displacement of the rock during the first two-thirds of its fall:

2 3

Using the same equation, express the displacement of the rock during its complete fall in terms of the time required for it to fall this distance:

h = 12 g (t + 1s )

Substitute equation (2) in equation (1) to obtain:

t2 – (4 s)t – 2 s2 = 0

Use the quadratic formula or your graphing calculator to solve for the positive root:

t = 4.45 s

Evaluate Δt = t + 1 s:

Δt = 4.45 s + 1 s = 5.45 s

Substitute numerical values in equation (2) and evaluate h:

h=

h = 12 gt 2

(1)

2

1 2

(2)

(9.81m/s )(5.45 s) 2

2

= 146 m

83 •• [SSM] A typical automobile under hard braking loses speed at a rate of about 7.0 m/s2; the typical reaction time to engage the brakes is 0.50 s. A local school board sets the speed limit in a school zone such that all cars should be able to stop in 4.0 m. (a) What maximum speed does this imply for an automobile in this zone? (b) What fraction of the 4.0 m is due to the reaction time? Picture the Problem Assume that the acceleration of the car is constant. The total distance the car travels while stopping is the sum of the distances it travels during the driver’s reaction time and the time it travels while braking. Choose a coordinate system in which the positive direction is the direction of motion of the automobile and apply a constant-acceleration equation to obtain a quadratic equation in the car’s initial speed v0.

116 Chapter 2 (a) Using a constant-acceleration equation, relate the speed of the car to its initial speed, acceleration, and displacement during braking:

v 2 = v02 + 2aΔxbrk or, because the final speed of the car is zero, v2 0 = v02 + 2aΔx brk ⇒ Δxbrk = − 0 2a

Express the total distance traveled by the car as the sum of the distance traveled during the reaction time and the distance traveled while slowing down:

Δxtot = Δxreact + Δxbrk

Rearrange this quadratic equation to obtain:

v02 − 2aΔtreact v0 + 2aΔxtot = 0

Substitute numerical values and simplify to obtain:

v02 + (7.0 m/s )v0 − 56 m 2 / s 2 = 0

Use your graphing calculator or the quadratic formula to solve the quadratic equation for its positive root:

v0 = 4.76 m/s

Convert this speed to mi/h:

⎛ 1mi/h ⎞ ⎟⎟ v0 = (4.76 m/s )⎜⎜ ⎝ 0.4470 m/s ⎠

= v0 Δtreact −

v02 2a

= 11 mi/h

(b) Find the reaction-time distance:

Δxreact = v0 Δt react = (4.76 m/s)(0.50 s) = 2.38 m

Express and evaluate the ratio of the reaction distance to the total distance:

Δxreact 2.38 m = = 0.60 Δx tot 4.0 m

84 •• Two trains face each other on adjacent tracks. They are initially at rest, and their front ends are 40 m apart. The train on the left accelerates rightward at 1.0 m/s2. The train on the right accelerates leftward at 1.3 m/s2. (a) How far does the train on the left travel before the front ends of the trains pass? (b) If the trains are each 150 m in length, how long after the start are they completely past one another, assuming their accelerations are constant?

Motion in One Dimension 117 Picture the Problem Assume that the accelerations of the trains are constant. Choose a coordinate system in which the direction of the motion of the train on the left is the positive direction. Take x0 = 0 as the position of the train on the left at t = 0 and note that the acceleration of the train on the right is negative.

(a) Using a constant-acceleration equation, express the position of the front of the train on the left as a function of time: Using a constant-acceleration equation, express the position of the front of the train on the right as a function of time: Equate xL and xR to obtain:

xL = 12 aLt 2

(1)

xR = 40 m + 12 aR t 2

1 2

aL t 2 = 40 m + 12 aR t 2 ⇒ t =

80 m aL − aR

Substituting for t in equation (1) gives:

⎛ ⎜ ⎛ 80 m ⎞ 1 ⎜ 80 m 1 ⎟⎟ = 2 xL = 2 aL ⎜⎜ ⎜ aR ⎝ aL − aR ⎠ ⎜1− a L ⎝

The acceleration of the train on the left is 1.0 m/s2 and the acceleration of the train on the right is −1.3 m/s2. Substitute numerical values and evaluate xL:

⎛ ⎜ 80 m 1⎜ xL = 2 ⎜ − 1.3 m/s 2 ⎜1− 1.0 m/s 2 ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

⎞ ⎟ ⎟ = 17 m ⎟ ⎟ ⎠

(b) Let the rear of the left train be at the origin at t = 0. Then the initial location of the rear end of the train on the right is at x = 340 m (150 m + 40 m + 150 m). Using a constant-acceleration equation, express the position of the rear of the train on the left as a function of time:

xL = 12 aLt 2

Using a constant-acceleration equation, express the position of the rear of the train on the right as a function of time:

xR = 340 m + 12 aR t 2

The trains will be completely past each other when:

xL = xR ⇒ 12 aL t 2 = 340 m + 12 aR t 2

118 Chapter 2 Solving for t yields:

Substitute numerical values and evaluate t:

t=

680 m aL − aR

t=

680 m = 17 s 1.0 m/s − − 1.3 m/s 2 2

(

)

x, m

Remarks: One can also solve this problem by graphing xL and xR as functions of t. The coordinates of the intersection of the two curves give one the time-to-passing and the distance traveled by the train on the left for Part (a).

xR xL 17

t, s

17

85 •• Two stones are dropped from the edge of a 60-m cliff, the second stone 1.6 s after the first. How far below the top of the cliff is the second stone when the separation between the two stones is 36 m? Picture the Problem In the absence of air resistance, the acceleration of the stones is constant. Choose a coordinate system in which the downward direction is positive and the origin is at the point of release of the stones.

Using constant-acceleration equations, relate the positions of the two stones to their initial positions, accelerations, and time-of-fall:

x1 = 12 gt 2

and x2 = 12 g (t − 1.6 s) 2

Express the difference between x1 and x2:

x1 − x2 = 36 m

Substitute for x1 and x2 to obtain:

36 m = 12 gt 2 − 12 g (t − 1.6 s )

Solve this equation for the time t at which the stones will be separated by 36 m:

t = 3.09 s

Substitute this result in the expression for x2 and solve for x2:

x2 =

2

1 2

(9.81m s )(3.09 s − 1.6 s )

= 11 m

2

2

Motion in One Dimension 119 86 •• A motorcycle officer hidden at an intersection observes a car driven by an oblivious driver who ignores a stop sign and continues through the intersection at constant speed. The police officer takes off in pursuit 2.0 s after the car has passed the stop sign. She accelerates at 4.2 m/s2 until her speed is 110 km/h, and then continues at this speed until she catches the car. At that instant, the car is 1.4 km from the intersection. (a) How long did it take for the officer to catch up to the car? (b) How fast was the car traveling? Picture the Problem The acceleration of the police officer’s car is positive and constant and the acceleration of the speeder’s car is zero. Choose a coordinate system such that the direction of motion of the two vehicles is the positive direction and the origin is at the stop sign.

(a) The time traveled by the car is given by:

Convert 110 km/h into m/s:

Express and evaluate t1:

Express and evaluate d1:

t car = 2.0 s + t1 + t 2 (1) where t1 is the time during which the motorcycle was accelerating and t2 is the time during which the motorcycle moved with constant speed. km 10 3 m 1h × × h km 3600 s m = 30.56 s

v1 = 110

t1 =

v1 a motorcycle

=

30.56 m/s = 7.276 s 4.2 m/s 2

d1 = 12 v1t1 = 12 (30.56 m/s)(7.276 s) = 111.2 m

Determine d2:

d 2 = d caught − d1 = 1400 m − 111 m = 1289 m

Express and evaluate t2:

t2 =

d2 1289 m = = 42.18 s v1 30.56 m/s

Substitute in equation (1) and evaluate tcar:

t car = 2.0 s + 7.3 s + 42.2 s = 51.5 s

(b) The speed of the car when it was overtaken is the ratio of the distance it traveled to the elapsed time:

vcar =

= 52 s d caught t car

120 Chapter 2 Substitute numerical values and evaluate vcar:

⎛ 1400 m ⎞ ⎛ 1 mi/h ⎞ ⎟⎟ = 61 mi/h vcar = ⎜ ⎟ ⎜⎜ ⎝ 51.5 s ⎠ ⎝ 0.447 m/s ⎠

87 •• At t = 0, a stone is dropped from the top of a cliff above a lake. Another stone is thrown downward 1.6 s later from the same point with an initial speed of 32 m/s. Both stones hit the water at the same instant. Find the height of the cliff. Picture the Problem In the absence of air resistance, the acceleration of the stone is constant. Choose a coordinate system in which downward is positive and the origin is at the point of release of the stone and apply constant-acceleration equations.

Using a constant-acceleration equation, express the height of the cliff in terms of the initial position of the stones, acceleration due to gravity, and time for the first stone to hit the water:

h = 12 gt12

Express the displacement of the second stone when it hits the water in terms of its initial velocity, acceleration, and time required for it to hit the water.

d 2 = v02t2 + 12 gt22 where t2 = t1 – 1.6 s.

Because the stones will travel the same distances before hitting the water, equate h and d2 to obtain:

gt12 = v02t2 + 12 gt22 or, because t2 = t1 – 1.6 s, 2 2 1 1 2 gt1 = v02t 2 + 2 g (t1 − 1.6 s ) 1 2

Substitute numerical values to obtain: 1 2

(9.81 m/s )t = (32 m/s)(t 2

2 1

1

(

)

− 1.6 s )+ 12 9.81 m/s 2 (t1 − 1.6 s )

2

Solving this quadratic equation for t1 yields:

t1 = 2.370 s

Substitute for t1 and evaluate h:

h = 12 (9.81 m/s 2 )(2.370 s) 2 = 28 m

88 •• A passenger train is traveling at 29 m/s when the engineer sees a freight train 360 m ahead of his train traveling in the same direction on the same track. The freight train is moving at a speed of 6.0 m/s. (a) If the reaction time of the engineer is 0.40 s, what is the minimum (constant) rate at which the passenger train must lose speed if a collision is to be avoided? (b) If the engineer’s reaction

Motion in One Dimension 121 time is 0.80 s and the train loses speed at the minimum rate described in Part (a), at what rate is the passenger train approaching the freight train when the two collide? (c) How far will the passenger train have traveled in the time between the sighting of the freight train and the collision? Picture the Problem Assume that the acceleration of the passenger train is constant. Let xp = 0 be the location of the passenger train engine at the moment of sighting the freight train’s end; let t = 0 be the instant the passenger train begins to slow (0.40 s after the passenger train engineer sees the freight train ahead). Choose a coordinate system in which the direction of motion of the trains is the positive direction and use constant-acceleration equations to express the positions of the trains in terms of their initial positions, speeds, accelerations, and elapsed time.

(a) Using constant-acceleration equations, write expressions for the positions of the front of the passenger train and the rear of the freight train, xp and xf, respectively:

xp = (29 m/s )(t + 0.40 s ) − 12 at 2

(1)

and xf = (360 m ) + (6.0 m/s ) (t + 0.40 s) where xp and xf are in meters if t is in seconds. at 2 − (23 m/s )t + 350.8 m = 0

Equate xf and xp to obtain an equation for t:

1 2

Find the discriminant D = B2 − 4AC of this equation:

⎛a⎞ 2 D = (23 m/s ) − 4⎜ ⎟(350.8 m ) ⎝2⎠

The equation must have real roots if it is to describe a collision. The necessary condition for real roots is that the discriminant be greater than or equal to zero:

If (23 m/s)2 – a (701.6 m) ≥ 0, then

(b) Differentiate equation (1) to express the speed of the passenger train as a function of time:

vp (t ) =

(

)

a ≤ 0.75 m/s 2

dxp (t ) dt

= 29 m/s − at

(2)

Repeat the previous steps with a = 0.75 m/s2 and a 0.80 s reaction time. The quadratic equation that guarantees real roots with the longer reaction time is:

(0.375 m/s )t − (23 m/s)t + 341.6m = 0 2

Use the quadratic formula or your graphing calculator to find the collision times:

2

t = 25.23 s and t = 36.10 s

122 Chapter 2 Because, when t = 36.10 s, the trains have already collided, this root is not a meaningful solution to our problem.

Note: In the graph shown below, you will see why we keep only the smaller of the two solutions.

Evaluate equation (2) for t = 25.23 s and a = 0.75 m/s2:

vp (t ) = 29 m/s − 0.75 m/s 2 (25.23 s )

(

)

= 10 m/s

(c) The position of the passenger train as a function of time is given by: xp (t ) = (29 m/s )(t + 0.80 s ) − 12 (0.75 m/s 2 )t 2

Evaluate xp (25.23 s ) to obtain:

(

)

x p (26 s ) = (29 m/s )(25.23 s + 0.80 s ) − 12 0.75 m/s 2 (25.23 s ) = 0.52 km 2

The following graph shows the positions of the trains as a function of time. The solid straight line is for the constant velocity freight train; the dashed curves are for the passenger train, with reaction times of 0.40 s for the lower curve and 0.80 s for the upper curve. 700 600

x, m

500 400 300 0.40 s reaction time

200

Freight train 100

0.80 s reaction time

0 0

5

10

15

20

25

30

35

40

t, s

Remarks: A collision occurs the first time the curve for the passenger train crosses the curve for the freight train. The smaller of the two solutions will always give the time of the collision. 89 •• The click beetle can project itself vertically with an acceleration of about 400g (an order of magnitude more than a human could survive!). The beetle jumps by ″unfolding″ its 0.60-cm long legs. (a) How high can the click beetle jump? (b) How long is the beetle in the air? (Assume constant acceleration while in contact with the ground and neglect air resistance.)

Motion in One Dimension 123 Picture the Problem Choose a coordinate system in which the upward direction is positive. We can use a constant-acceleration equation to find the beetle’s speed as its feet lose contact with the ground and then use this speed to calculate the height of its jump.

(a) Using a constant-acceleration equation, relate the beetle’s maximum height to its launch speed, speed at the top of its trajectory, and acceleration once it is airborne; solve for its maximum height:

2 2 vhighest = vlaunch + 2aΔy free fall point

2 = vlaunch + 2(− g )h or, because vhighest = 0 , point

2 0 = vlaunch + 2(− g )h ⇒ h =

Now, in order to determine the beetle’s launch speedy, relate its time of contact with the ground to its acceleration and push-off distance:

2 vlaunch = v02 + 2aΔylaunch or, because v0 = 0, 2 vlaunch = 2aΔylaunch

2 Substituting for vlaunch in equation (1) yields:

h=

Substitute numerical values in equation (1) to find the height to which the beetle can jump:

(b) Using a constant-acceleration equation, relate the speed of the beetle at its maximum height to its launch speed, free-fall acceleration while in the air, and time-tomaximum height:

2 vlaunch (1) 2g

2aΔylaunch 2g

2(400 )(9.81m/s 2 )(0.60 × 10−2 m ) h= 2(9.81m/s 2 ) = 2.4 m v = v0 + at or vmax. = vlaunch − gt max height

height

and, because v max = 0 , height

0 = vlaunch − gt max ⇒ t max = height

For zero displacement and constant acceleration, the time-of-flight is twice the time-to-maximum height:

t flight = 2t max =

Substitute numerical values and evaluate tflight :

tflight =

height

height

2vlaunch g

2(6.9 m/s ) = 1.4 s 9.81m/s 2

vlaunch g

124 Chapter 2 90 •• An automobile accelerates from rest at 2.0 m/s2 for 20 s. The speed is then held constant for 20 s, after which there is an acceleration of –3.0 m/s2 until the automobile stops. What is the total distance traveled? Picture the Problem This is a multipart constant-acceleration problem using three different constant accelerations (+2.0 m/s2 for 20 s, then zero for 20 s, and then –3.0 m/s2 until the automobile stops). The final speed is zero. The pictorial representation will help us organize the information in the problem and develop a solution strategy.

Add up all the displacements to get the total:

Δx 03 = Δx01 + Δx12 + Δx 23

Using constant-acceleration formulas, find the first displacement:

Δx01 = v0t1 + 12 a01t12

(1)

or, because v0 =0, Δx01 = 12 a01t12

The speed is constant for the second displacement. The second displacement is given by:

Δx12 = v1 (t2 − t1 ) or, because v1 = v0 + a01t1 = 0 + a01t1 ,

The displacement during the braking interval is related to v3, v2, and a23:

v32 = v22 + 2a23 Δx23 and, because v2 = v1 = a01t1 and v3 = 0, 2 − (a01t1 ) Δx23 = 2a23

Substituting for Δx01 , Δx12 , and Δx23 in equation (1) yields:

Δx12 = a01t1 (t2 − t1 )

Δx03 = a t 1 2

2 01 1

2 ( a01t1 ) + a01t1 (t2 − t1 ) −

2a23

Substitute numerical values and evaluate Δx03 : Δx03 =

1 2

2 [ (2.0 m/s) (20 s)] (2.0 m/s) (20 s ) + (2.0 m/s) (20 s) (40 s − 20 s ) − 2 2

2(− 3.0 m s

)

= 1.5 km Remarks: Because the area under the curve of a velocity-versus-time graph equals the displacement of the object experiencing the acceleration, we could

Motion in One Dimension 125 solve this problem by plotting the velocity as a function of time and finding the area bounded by it and the time axis. 91 •• [SSM] Consider measuring the free-fall motion of a particle (neglect air resistance). Before the advent of computer-driven data-logging software, these experiments typically employed a wax-coated tape placed vertically next to the path of a dropped electrically conductive object. A spark generator would cause an arc to jump between two vertical wires through the falling object and through the tape, thereby marking the tape at fixed time intervals Δt. Show that the change in height during successive time intervals for an object falling from rest follows Galileo’s Rule of Odd Numbers: Δy21 = 3Δy10, Δy32 = 5Δy10, . . . , where Δy10 is the change in y during the first interval of duration Δt, Δy21 is the change in y during the second interval of duration Δt, etc. Picture the Problem In the absence of air resistance, the particle experiences constant acceleration and we can use constant-acceleration equations to describe its position as a function of time. Choose a coordinate system in which downward is positive, the particle starts from rest (vo = 0), and the starting height is zero (y0 = 0).

Using a constant-acceleration equation, relate the position of the falling particle to the acceleration and the time. Evaluate the y-position at successive equal time intervals Δt, 2Δt, 3Δt, etc:

1 2 y1 = − g (Δt ) 2 1 4 2 2 y 2 = − g (2Δt ) = − g (Δt ) 2 2 1 9 2 2 y3 = − (3Δt ) = − g (Δt ) 2 2 1 16 2 2 y 4 = − g (4Δt ) = − g (Δt ) 2 2 etc.

Evaluate the changes in those positions in each time interval:

⎛ 1 ⎞ 2 Δy10 = y1 − 0 = ⎜ − g ⎟ (Δt ) ⎝ 2 ⎠ 4 1 2 2 Δy 21 = y 2 − y1 = − g (Δt ) + g (Δt ) 2 2 3 ⎡ 1 2 2⎤ = − g (Δt ) = 3⎢− g (Δt ) ⎥ 2 ⎣ 2 ⎦ = 3Δy10

126 Chapter 2 9 4 2 2 Δy32 = y3 − y 2 = − g (Δt ) + g (Δt ) 2 2 5 ⎡ 1 2 2⎤ = − g (Δt ) = 5⎢− g (Δt ) ⎥ 2 ⎣ 2 ⎦ = 5Δy10

16 9 2 2 g (Δt ) + g (Δt ) 2 2 ⎡ 1 2⎤ = 7 ⎢− g (Δt ) ⎥ ⎣ 2 ⎦

Δy 43 = y 4 − y3 = − 7 2 = − g (Δt ) 2 = 7 Δy10 etc.

92 •• A particle travels along the x axis with a constant acceleration of +3.0 2 m/s . It is at x = −100 m at time t = 4.0 s. In addition, it has a velocity of +15 m/s at a time 6.0 s later. Find its position at this later time. Picture the Problem Because the particle moves with a constant acceleration we can use constant-acceleration equations to relate its positions and velocities to the time into its motion. A pictorial representation will help us organize the information in the problem and develop our solution strategy.

t1 = 4.0 s

t 2 = 10.0 s

v1 = ?

v2 = +15 m/s

x1 = −100 m

x2 = ?

0

x, m

a x = +3.0 m/s 2 Using a constant-acceleration equation, relate the velocity of the particle at time t2 to its velocity at time t1:

v2 − v1 = a x (t 2 − t1 )

Using a constant-acceleration equation, relate the position of the particle at time t2 to its position at time t1:

x2 − x1 = v1 (t 2 − t1 ) + 12 a x (t 2 − t1 )

Eliminating v1 and solving for x2 gives:

x2 = x1 + v2 (t 2 − t1 ) − 12 a x (t 2 − t1 )

2

2

Motion in One Dimension 127 Substitute numerical values and evaluate x2:

(

)

x2 = −100 m + (+ 15 m/s )(10.0 s − 4.0 s ) − 12 + 3.0 m/s 2 (10.0 s − 4.0 s ) = − 64 m 2

93 •• [SSM] If it were possible for a spacecraft to maintain a constant acceleration indefinitely, trips to the planets of the Solar System could be undertaken in days or weeks, while voyages to the nearer stars would only take a few years. (a) Using data from the tables at the back of the book, find the time it would take for a one-way trip from Earth to Mars (at Mars’ closest approach to Earth). Assume that the spacecraft starts from rest, travels along a straight line, accelerates halfway at 1g, flips around, and decelerates at 1g for the rest of the trip. (b) Repeat the calculation for a 4.1 × 1013-km trip to Proxima Centauri, our nearest stellar neighbor outside of the sun. (See Problem 47.) Picture the Problem Note: No material body can travel at speeds faster than light. When one is dealing with problems of this sort, the kinematic formulae for displacement, velocity and acceleration are no longer valid, and one must invoke the special theory of relativity to answer questions such as these. For now, ignore such subtleties. Although the formulas you are using (i.e., the constantacceleration equations) are not quite correct, your answer to Part (b) will be wrong by about 1%.

(a) Let t1/2 represent the time it takes to reach the halfway point. Then the total trip time is:

t = 2 t1/2

Use a constant- acceleration equation to relate the half-distance to Mars Δx to the initial speed, acceleration, and half-trip time t1/2 :

Δx = v0t + 12 at12 2

Because v0 = 0 and a = g:

Substitute for t1/ 2 in equation (1) to obtain: The distance from Earth to Mars at closest approach is 7.8 × 1010 m. Substitute numerical values and evaluate t : (b) From Problem 47 we have:

t1 / 2 =

t=2

(1)

2Δx a 2Δx a

t round trip = 2

(2)

(

≈ 2.1 d

d Proxima = 4.1× 1013 km Centauri

)

2 3.9 × 1010 m = 18 × 10 4 s 2 9.81 m/s

128 Chapter 2 Substitute numerical values in equation (2) to obtain:

t round trip = 2

(

)

2 4.1× 1013 km = 18 × 107 s 9.81 m/s 2

≈ 5.8 y

Remarks: Our result in Part (a) seems remarkably short, considering how far Mars is and how low the acceleration is. 94 • The Stratosphere Tower in Las Vegas is 1137 ft high. It takes 1 min, 20 s to ascend from the ground floor to the top of the tower using the high-speed elevator. The elevator starts and ends at rest. Assume that it maintains a constant upward acceleration until it reaches its maximum speed, and then maintains a constant acceleration of equal magnitude until it comes to a stop. Find the magnitude of the acceleration of the elevator. Express this acceleration magnitude as a multiple of g (the acceleration due to gravity). Picture the Problem Because the elevator accelerates uniformly for half the distance and uniformly decelerates for the second half, we can use constantacceleration equations to describe its motion

Let t1/2 = 40 s be the time it takes to reach the halfway mark. Use the constant-acceleration equation that relates the acceleration to the known variables to obtain:

Δy = v0t + 12 at 2 or, because v0 = 0, 2 Δy Δy = 12 at 2 ⇒ a = 2 t1 / 2

Substitute numerical values and evaluate a:

a=

2( 12 )(1137 ft )(1 m/3.281ft )

(40 s )

2

= 0.22 m/s 2

= 0.022 g 95 •• A train pulls away from a station with a constant acceleration of 2 0.40 m/s . A passenger arrives at a point next to the track 6.0 s after the end of the train has passed the very same point. What is the slowest constant speed at which she can run and still catch the train? On a single graph plot the position versus time curves for both the train and the passenger. Picture the Problem Because the acceleration is constant, we can describe the motions of the train using constant-acceleration equations. Find expressions for the distances traveled, separately, by the train and the passenger. When are they equal? Note that the train is accelerating and the passenger runs at a constant minimum speed (zero acceleration) such that she can just catch the train.

Motion in One Dimension 129 1. Using the subscripts ″train″ and ″p″ to refer to the train and the passenger and the subscript ″c″ to identify ″critical″ conditions, express the position of the train and the passenger:

x train c (t c ) =

Express the critical conditions that must be satisfied if the passenger is to catch the train:

v train c = vp,c

2. Express the train’s average speed:

a train 2 tc 2

and xp,c (t c ) = vp,c (t c − Δt )

and x train c = xp,c vav (0 to t c ) =

0 + v train c v train c = 2 2

3. Using the definition of average speed, express vav in terms of xp,c and tc.

vav =

Δx 0 + xp,c xp,c = = 0 + tc tc Δt

4. Combine steps 2 and 3 and solve for xp,c.

xp,c =

v train c t c 2

5. Combine steps 1 and 4 and solve for tc.

vp,c (t c − Δt ) =

v train c t c 2

or t c − Δt =

tc 2

and tc = 2 Δt = 2 (6 s) = 12 s 6. Finally, combine steps 1 and 5 and solve for vtrain c.

(

)

vp,c = v train c = a train t c = 0.40 m/s 2 (12 s ) = 4.8 m/s

The following graph shows the location of both the passenger and the train as a function of time. The parabolic solid curve is the graph of xtrain(t) for the accelerating train. The straight dashed line is passenger's position xp(t) if she arrives at Δt = 6.0 s after the train departs. When the passenger catches the train, our graph shows that her speed and that of the train must be equal ( v train c = vp,c ). Do you see why?

130 Chapter 2 50 45 Train

40

Passenger

35

x, m

30 25 20 15 10 5 0 0

2

4

6

8

10

12

14

16

t, s

96 ••• Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what fraction of the height of the building does the collision occur? Picture the Problem Both balls experience constant acceleration once they are in flight. Choose a coordinate system with the origin at the ground and the upward direction positive. When the balls collide they are at the same height above the ground.

Using constant-acceleration equations, express the positions of both balls as functions of time. At the ground y = 0.

yA = h − 12 gt 2 and yB = v0t − 12 gt 2 yA = yB

The conditions at collision are that the heights are equal and the speeds are related:

and

Express the speeds of both balls as functions of time:

vA = − gt and vB = v0 − gt

Substituting the position and speed functions into the conditions at collision gives:

h − 12 gtc2 = v0tc − 12 gtc2

We now have two equations and two unknowns, tc and v0. Solving the equations for the unknowns gives:

vA = −2vB

and − gtc = −2(v0 − gtc ) where tc is the time of collision. tc =

2h 3 gh and v0 = 3g 2

Motion in One Dimension 131 Substitute the expression for tc into the equation for yA to obtain the height at collision:

⎛ 2h ⎞ 2h y A = h − 12 g ⎜⎜ ⎟⎟ = 3 ⎝ 3g ⎠

Remarks: We can also solve this problem graphically by plotting velocityversus-time for both balls. Because ball A starts from rest its velocity is given by v A = − gt . Ball B initially moves with an unknown velocity vB0 and its velocity is given by v B = v B0 − gt . The graphs of these equations are shown below with T representing the time at which they collide. v vBO vB = vBO − gt 2 3

vBO

1 3

vBO

1 3

vBO

T 2 3

t

vBO

vA = − gt

The height of the building is the sum of the sum of the distances traveled by the balls. Each of these distances is equal to the magnitude of the area ″under″ the corresponding v-versus-t curve. Thus, the height of the building equals the area of the parallelogram, which is vB0T. The distance that A falls is the area of the lower triangle, which is 13 vBOT . Therefore, the ratio of the distance fallen by A to the height of the building is 1/3, so the collision takes place at 2/3 the height of the building. 97 ••• Solve Problem 96 if the collision occurs when the balls are moving in the same direction and the speed of A is 4 times that of B. Picture the Problem Both balls are moving with constant acceleration. Take the origin of the coordinate system to be at the ground and the upward direction to be positive. When the balls collide they are at the same height above the ground. The speeds at collision are related by vA = 4vB.

132 Chapter 2 Using constant-acceleration equations, express the positions of both balls as functions of time:

y A = h − 12 gt 2 and y B = v0 t − 12 gt 2

The conditions at collision are that the heights are equal and the speeds are related:

y A = y B and vA = 4vB

Express the speeds of both balls as functions of time:

vA = − gt and vB = v0 − gt

Substitute the position and speed functions into the conditions at collision to obtain:

h − 12 gtc2 = v0t c − 12 gt c2 and

− gtc = 4(v0 − gtc ) where tc is the time of collision.

We now have two equations and two unknowns, tc and v0. Solving the equations for the unknowns gives:

tc =

Substitute the expression for tc into the equation for yA to obtain the height at collision:

⎛ 4h ⎞ h y A = h − 12 g ⎜⎜ ⎟⎟ = 3 ⎝ 3g ⎠

4h 3gh and v 0 = 3g 4

98 ••• Starting at one station, a subway train accelerates from rest at a constant rate of 1.00 m/s2 for half the distance to the next station, then slows down at the same rate for the second half of the journey. The total distance between stations is 900 m. (a) Sketch a graph of the velocity vx as a function of time over the full journey. (b) Sketch a graph of the position as a function of time over the full journey. Place appropriate numerical values on both axes. Picture the Problem The problem describes two intervals of constant acceleration; one when the train’s speed is increasing, and a second when it is decreasing.

(a) Using a constant-acceleration equation, relate the half-distance Δx between stations to the initial speed v0, the acceleration a of the train, and the time-to-midpoint Δt: Substitute numerical values and evaluate the time-to-midpoint Δt:

Δx = v0 Δt + 12 a (Δt ) or, because v0 = 0,

2

Δx = 12 a (Δt ) ⇒ Δt = 2

Δt =

2Δx a

2(450 m ) = 30.0 s 1.00 m/s 2

Motion in One Dimension 133 Because the train accelerates uniformly and from rest, the first part of its velocity graph will be linear, pass through the origin, and last for 30 s. Because it slows down uniformly and at the same rate for the second half of its journey, this part of its graph will also be linear but with a negative slope. A graph of v as a function of t follows. 30 25

v, m/s

20 15 10 5 0 0

10

20

30

40

50

60

t, s

(b) The graph of x as a function of t is obtained from the graph of v as a function of t by finding the area under the velocity curve. Looking at the velocity graph, note that when the train has been in motion for 10 s, it will have traveled a distance of 1 2

(10 s )(10 m/s) = 50 m

and that this distance is plotted above 10 s on the following graph. 900 800 700

x, m

600 500 400 300 200 100 0 0

10

20

30

40

50

60

t, s

Selecting additional points from the velocity graph and calculating the areas under the curve will confirm the graph of x as a function of t.

134 Chapter 2 99 ••• [SSM] A speeder traveling at a constant speed of 125 km/h races past a billboard. A patrol car pursues from rest with constant acceleration of (8.0 km/h)/s until it reaches its maximum speed of 190 km/h, which it maintains until it catches up with the speeder. (a) How long does it take the patrol car to catch the speeder if it starts moving just as the speeder passes? (b) How far does each car travel? (c) Sketch x(t) for each car. Picture the Problem This is a two-stage constant-acceleration problem. Choose a coordinate system in which the direction of the motion of the cars is the positive direction. The pictorial representation summarizes what we know about the motion of the speeder’s car and the patrol car. xS,0 = 0

xS,1 vS,1 = 125 km/h

vS,0 = 125 km/h

t0 = 0 0

xS,2

t1

aS,01 = 0

aP,01 = 2.22 m/s 2

xP,0 = 0 v P,0 = 0

1

aS,12 = 0 aP,12 = 0

xP,1

vS,2 = 125 km/h

t2

x 2

xP,2

v P,1 = 190 km/h

v P,2 = 190 km/h

Convert the speeds of the vehicles and the acceleration of the police car into SI units:

km km 1h = 8.0 × = 2.2 m/s 2 , h ⋅s h ⋅ s 3600 s km km 1h 125 = 125 × = 34.7 m/s , h h 3600 s and km km 1h 190 = 190 × = 52.8 m/s h h 3600 s

(a) Express the condition that determines when the police car catches the speeder; that is, that their displacements will be the same:

ΔxP,02 = ΔxS,02

Using a constant-acceleration equation, relate the displacement of the patrol car to its displacement while accelerating and its displacement once it reaches its maximum velocity:

ΔxP,02 = ΔxP,01 + ΔxP,12

8.0

= ΔxP,01 + vP ,1 (t 2 − t1 )

Motion in One Dimension 135 Using a constant-acceleration equation, relate the displacement of the speeder to its constant velocity and the time it takes the patrol car to catch it: Calculate the time during which the police car is speeding up:

ΔxS,02 = vS,02 Δt02

= (34.7 m/s ) t 2

Δt P,01 = =

Express the displacement of the patrol car:

ΔvP,01 aP,01

=

vP ,1 − vP, 0 aP,01

52.8 m/s − 0 = 24 s 2.2 m/s 2

ΔxP,01 = vP,0 Δt P,01 + 12 aP ,01Δt P2,01

(

)

= 0 + 12 2.2 m/s 2 (24 s )

2

= 630 m

Equate the displacements of the two vehicles:

ΔxP,02 = ΔxP,01 + ΔxP,12

= ΔxP,01 + vP ,1 (t 2 − t1 )

= 630 m + (52.8 m/s )(t 2 − 24 s)

Substitute for ΔxP,02 to obtain:

(34.7 m/s) t2 = 630 m + (52.8 m/s)(t2 – 24 s)

Solving for the time to catch up yields:

t 2 = 35.19 s = 35 s

(b) The distance traveled is the displacement, Δx02,S, of the speeder during the catch:

ΔxS,02 = vS,02 Δt 02 = (35 m/s )(35.19 s ) = 1.2 km

136 Chapter 2 (c) The graphs of xS and xP follow. The straight line (solid) represents xS(t) and the parabola (dashed) represents xP(t). 1400 1200 Speeder

x, m

1000

Officer

800 600 400 200 0 0

5

10

15

20

25

30

35

40

t, s

100 ••• When the patrol car in Problem 99 (traveling at 190 km/h), is 100 m behind the speeder (traveling at 125 km/h), the speeder sees the police car and slams on his brakes, locking the wheels. (a) Assuming that each car can brake at 6.0 m/s2 and that the driver of the police car brakes instantly as she sees the brake lights of the speeder (reaction time = 0.0 s), show that the cars collide. (b) At what time after the speeder applies his brakes do the two cars collide? (c) Discuss how reaction time affects this problem. Picture the Problem The accelerations of both cars are constant and we can use constant-acceleration equations to describe their motions. Choose a coordinate system in which the direction of motion of the cars is the positive direction, and the origin is at the initial position of the police car.

(a) The collision will not occur if, during braking, the displacements of the two cars differ by less than 100 m:

ΔxP − ΔxS < 100 m

Using a constant-acceleration equation, relate the speeder’s initial and final speeds to its displacement and acceleration and solve for the displacement:

vs2 = v02,s + 2as Δxs

or, because vs = 0, − v02,s Δxs = 2as

Substitute numerical values and evaluate Δxs:

− (34.7 m/s ) ΔxS = = 100 m 2 − 6.0 m/s 2 2

(

)

Motion in One Dimension 137 Using a constant-acceleration equation, relate the patrol car’s initial and final speeds to its displacement and acceleration and solve for the displacement:

or, assuming vp = 0, − v02, p Δxp = 2ap

Substitute numerical values and evaluate Δxp:

ΔxP =

Finally, substitute these displacements into the inequality that determines whether a collision occurs:

232 m − 100 m = 132 m Because this difference is greater than 100 m, the cars collide .

(b) Using constant-acceleration equations, relate the positions of both vehicles to their initial positions, initial speeds, accelerations, and time in motion:

xS = 100 m + (34.7 m/s )t − 3.0 m/s 2 t 2 and xP = (52.8 m/s )t − 3.0 m/s 2 t 2

vp2 = v02, p + 2ap Δxp

− (52.8 m/s ) = 232 m 2 − 6.0 m/s 2 2

(

)

(

(

)

)

Equate xS and xP to obtain:

(

)

(

)

100 m + (34.7 m/s )t − 3.0 m/s 2 t 2 = (52.8 m/s )t − 3.0 m/s 2 t 2 Solving this equation for t yields:

t = 5.5 s

(c) If you take the reaction time into account, the collision will occur sooner and be more severe. 101 ••• Leadfoot Lou enters the "Rest-to-Rest" auto competition, in which each contestant’s car begins and ends at rest, covering a fixed distance L in as short a time as possible. The intention is to demonstrate driving skills, and to find which car is the best at the total combination of speeding up and slowing down. The course is designed so that maximum speeds of the cars are never reached. (a) If Lou's car maintains an acceleration (magnitude) of a during speedup, and maintains a deceleration (magnitude) of 2a during braking, at what fraction of L should Lou move his foot from the gas pedal to the brake? (b) What fraction of the total time for the trip has elapsed at that point? (c) What is the fastest speed Lou's car ever reaches? Neglect Lou's reaction time, and answer in terms of a and L.

138 Chapter 2 Picture the Problem Lou’s acceleration is constant during both parts of his trip. Let t1 be the time when the brake is applied; L1 the distance traveled from t = 0 to t = t1. Let tfin be the time when Lou's car comes to rest at a distance L from the starting line. A pictorial representation will help organize the given information and plan the solution.

(a) Express the total length, L, of the course in terms of the distance over which Lou will be accelerating, Δx01, and the distance over which he will be braking, Δx12:

L = Δx01 + Δx12

Express the final speed over the first portion of the course in terms of the initial speed, acceleration, and displacement; solve for the displacement:

v12 = v02 + 2a01Δx01 or, because v0 = 0, Δx01 = L1, and a01 = a, v2 v2 Δx01 = L1 = 1 = max 2a 2a

Express the final speed over the second portion of the course in terms of the initial speed, acceleration, and displacement; solve for the displacement:

v22 = v12 + 2a12 Δx12 or, because v2 = 0 and a12 = −2a, v12 L1 Δx12 = = 4a 2

Substitute for Δx01 and Δx12 to obtain:

L = Δx01 + Δx12 = L1 + 12 L1 = 32 L1 and L1 = 23 L

(b) Using the fact that the acceleration was constant during both legs of the trip, express Lou’s average speed over each leg: Express the time for Lou to reach his maximum speed as a function of L1 and his maximum speed:

vav, 01 = vav,12 =

Δt01 =

(1)

vmax 2

Δx01 2 L1 = vav,01 vmax

and Δt 01 ∝ L1 = 23 L

Motion in One Dimension 139 Having just shown that the time required for the first segment of the trip is proportional to the length of the segment, use this result to express Δt01 (= t1) in terms tfin:

t1 =

(c) Express Lou’s displacement during the speeding up portion of his trip:

Δx01 = 12 at12

2 3 fin

t

The time required for Lou to make this speeding portion of his trip is given by:

t1 =

Express Lou’s displacement during the slowing down portion of his trip:

Δx12 = v1t1 − 12 (2a )t 22 = v1t1 − at 22 where t2 is his slowing down time.

Substitute in equation (1) to obtain:

L = 12 at12 + v1t1 − at 22

Using a constant-acceleration equation, relate Lou’s slowing down time for this portion of his trip to his initial and final speeds: Substituting for t1 and t2 in equation (2) yields: Solve for v1 to obtain:

v1 a

0 = v1 − 2at 2 ⇒ t 2 =

2

(2) v1 2a

⎛v ⎞ ⎛v ⎞ ⎛ v ⎞ L = 12 a⎜ 1 ⎟ + v1 ⎜ 1 ⎟ − a⎜ 1 ⎟ ⎝a⎠ ⎝ a ⎠ ⎝ 2a ⎠ v1 =

2

4aL 5

102 ••• A physics professor, equipped with a rocket backpack, steps out of a helicopter at an altitude of 575 m with zero initial velocity. (Neglect air resistance.) For 8.0 s, she falls freely. At that time she fires her rockets and slows her rate of descent at 15 m/s2 until her rate of descent reaches 5.0 m/s. At this point she adjusts her rocket engine controls to maintain that rate of descent until she reaches the ground. (a) On a single graph, sketch her acceleration and velocity as functions of time. (Take upward to be positive.) (b) What is her speed at the end of the first 8.0 s? (c) What is the duration of her slowing down period? (d) How far does she travel while slowing down? (e) How much time is required for the entire trip from the helicopter to the ground? (f) What is her average velocity for the entire trip?

140 Chapter 2 y

Picture the Problem There are three intervals of constant acceleration described in this problem. Choose a coordinate system in which the upward direction is positive. The pictorial representation summarizes the information we are given. We can apply constant-acceleration equations to find her speeds and distances traveled at the end of each phase of her descent.

y0 = 575 m

t0 = 0

v0 = 0 a01 = − g

t1 = 8 s

y1 v1

a12 = 15 m/s 2

t2

y2 v 2 = 5 m/s

a23 = 0

y3 = 0

t3

v3 = 5 m/s

(a) The graphs of a(t) (dashed lines) and v(t) (solid lines) follow.

v, m/s and a, m/s^2

20

0

-20

-40 Velocity Acceleration

-60

-80 0

2

4

6

8

10

12

14

16

t, s

(b) Use a constant- acceleration equation to express her speed in terms of her acceleration and the elapsed time:

v1 = v0 + a01t1 or, because a01 = −g and v0 = 0, v1 = − gt1

Substitute numerical values and evaluate v1:

v1 = − 9.81 m/s 2 (8.0 s ) = −78.5 m/s

(

)

and her speed is 79 m/s

Motion in One Dimension 141 (c) As in (b), use a constantacceleration equation to express her speed in terms of her acceleration and the elapsed time: Substitute numerical values and evaluate Δt12:

v2 = v1 + a12 Δt12 ⇒ Δ t12 =

Δ t12 =

v2 − v1 a12

− 5.0 m/s − (− 78.5 m/s ) 15 m/s 2

= 4.9 s (d) Find her average speed as she slows from 78.5 m/s to 5 m/s:

Use this value to calculate how far she travels in 4.90 s:

v1 + v 2 78.5 m/s + 5.0 m/s = 2 2 = 41.7 m/s

vav =

Δy12 = vav Δt12 = (41.7 m/s )(4.90 s) = 204.3 m

She travels approximately 204 m while slowing down. (e) Express the total time in terms of the times for each segment of her descent:

Δt total = Δt01 + Δt12 + Δt23

We know the times for the intervals from 0 to 1 and 1 to 2 so we only need to determine the time for the interval from 2 to 3. We can calculate Δt23 from her displacement and constant speed during that segment of her descent.

Δy 23 = Δy total − Δy 01 − Δy12

Add the times to get the total time:

Δt total = Δt 01 + Δt12 + Δt 23

⎛ 78.5 m/s ⎞ = 575 m − ⎜ ⎟(8.0 s ) 2 ⎝ ⎠ − 204.3 m = 56.7 m

= 8.0 s + 4.9 s +

56.7 m 5.0 m/s

= 24.24 s = 24 s (f) Using its definition, calculate her average velocity:

vav =

Δy − 575 m = = − 24 m/s Δt total 24.24 s

142 Chapter 2

Integration of the Equations of Motion 103 • [SSM] The velocity of a particle is given by vx(t) = (6.0 m/s2)t + (3.0 m/s). (a) Sketch v versus t and find the area under the curve for the interval t = 0 to t = 5.0 s. (b) Find the position function x(t). Use it to calculate the displacement during the interval t = 0 to t = 5.0 s. Picture the Problem The integral of a function is equal to the "area" between the curve for that function and the independent-variable axis.

(a) The following graph was plotted using a spreadsheet program: 35 30

v, m/s

25 20 15 10 5 0 0

1

2

3

4

5

t, s

The distance is found by determining the area under the curve. There are approximately 18 blocks each having an area of (5.0 m/s)(1.0 s) = 5.0 m.

Aunder = (18 blocks)(5.0 m/block )

You can confirm this result by using the formula for the area of a trapezoid:

⎛ 33 m/s + 3 m/s ⎞ A=⎜ ⎟(5.0 s − 0 s ) 2 ⎝ ⎠ = 90 m

(b) To find the position function x(t), we integrate the velocity function v(t) over the time interval in question:

curve

= 90 m

t

x(t ) = ∫ v(t ) dt 0 t

[(

)

]

= ∫ 6.0 m/s 2 t + (3.0 m/s ) dt 0

and x(t ) = 3.0 m/s 2 t 2 + (3.0 m/s ) t

(

)

Motion in One Dimension 143 Now evaluate x(t) at 0 s and 5.0 s respectively and subtract to obtain Δx:

Δx = x(5.0 s ) − x(0 s ) = 90 m − 0 m = 90 m

104 • Figure 2-41 shows the velocity of a particle versus time. (a) What is the magnitude in meters represented by the area of the shaded box? (b) Estimate the displacement of the particle for the two 1-s intervals, one beginning at t = 1.0 s and the other at t = 2.0 s. (c) Estimate the average velocity for the interval 1.0 s ≤ t ≤ 3.0 s. (d) The equation of the curve is vx = (0.50 m/s3)t2. Find the displacement of the particle for the interval 1.0 s ≤ t ≤ 3.0 s by integration and compare this answer with your answer for Part (b). Is the average velocity equal to the mean of the initial and final velocities for this case? Picture the Problem The integral of v(t) over a time interval is the displacement (change in position) during that time interval. The integral of a function is equivalent to the "area" between the curve for that function and the independentvariable axis. Count the grid boxes.

(a) Find the area of the shaded grid box:

Area = (1 m/s )(1 s ) = 1 m per box

(b) Find the approximate area under the curve for 1.0 s ≤ t ≤ 2.0 s:

Δx1.0 s to 2.0 s = 1.2 m

Find the approximate area under the curve for 2.0 s ≤ t ≤ 3.0 s:

Δx2.0 s to 3.0 s = 3.2 m

(c) Sum the displacements to obtain the total in the interval 1.0 s ≤ t ≤ 3.0 s:

Δx1.0 s to 3.0 s = 1.2 m + 3.2 m = 4.4 m

Using its definition, express and evaluate vav:

vav =

Δx1.0 s to 3.0 s 4.4 m = = 2.2 m/s Δt 1.0 s to 3.0 s 2.0 s

144 Chapter 2 (d) Because the velocity of the particle is dx/dt, separate the variables and integrate over the interval 1.0 s ≤ t ≤ 3.0 s to determine the displacement in this time interval:

(

)

dx = 0.50 m/s 3 t 2 dt so x

(

Δx1.0 s→3.0 s = ∫ dx = 0.50 m/s 3 x0

3.0 s

) ∫t

2

dt

1.0 s

3.0s

⎡t 3 ⎤ = 0.50 m/s 3 ⎢ ⎥ = 4.33 m ⎣ 3 ⎦ 1.0s

(

)

= 4.3 m

This result is a little smaller than the sum of the displacements found in Part (b). Calculate the average velocity over the 2-s interval from 1.0 s to 3.0 s:

vav (1.0s−3.0s) =

Δx1.0s−3.0s Δt1.0s−3.0s

=

4.33 m = 2.2 m/s 2.0 s

( ) v(3.0 s ) = (0.50 m/s )(3.0 s )

Calculate the initial and final velocities of the particle over the same interval:

v(1.0 s ) = 0.50 m/s 3 (1.0 s ) = 0.50 m/s

Finally, calculate the average value of the velocities at t = 1.0 s and t = 3.0 s:

v(1.0 s) + v(3.0 s) 0.50 m/s + 4.5 m/s = 2 2 = 2.5 m/s This average of 2.5 m/s is not equal to the average velocity calculated above because this is not an example of constant acceleration.

2

3

2

= 4.5 m/s

105 •• The velocity of a particle is given by vx = (7.0 m/s3)t2 – 5.0 m/s, where t is in seconds and v is in meters per second. If the particle is at the origin at t0 = 0, find the position function x(t). Picture the Problem Because the velocity of the particle varies with the square of the time, the acceleration is not constant. The position of the particle is found by integrating the velocity function.

dx(t ) dt

Express the velocity of a particle as the derivative of its position function:

v(t ) =

Separate the variables to obtain:

dx(t ) = v(t )dt

Motion in One Dimension 145 Express the integral of x from xo = 0 to x and t from t0 = 0 to t:

Substitute for v(t) and carry out the integration to obtain:

x(t ) =

x(t ) =

x (t )

t

t0 =0

t0 =0

∫ dx = ∫ v(t )dt

∫ [(7.0 m/s )t − (5.0 m/s)]dt t

3

t0 =0

(

2

)

= 2.3 m/s 3 t 3 − (5.0 m/s )t 106 •• Consider the velocity graph in Figure 2-42. Assuming x = 0 at t = 0, write correct algebraic expressions for x(t), vx(t), and ax(t) with appropriate numerical values inserted for all constants. Picture the Problem The graph is one of constant negative acceleration. Because vx = v(t) is a linear function of t, we can make use of the slope-intercept form of the equation of a straight line to find the relationship between these variables. We can then differentiate v(t) to obtain a(t) and integrate v(t) to obtain x(t).

Find the acceleration (the slope of the graph) and the velocity at time 0 (the v-intercept) and use the slopeintercept form of the equation of a straight line to express vx(t): Find x(t) by integrating v(t):

a x = − 10 m/s 2

and v x (t ) = 50 m/s + (−10 m/s 2 )t

[(

]

)

x(t ) = ∫ − 10 m/s 2 t + 50 m/s dt

(

)

= (50 m/s )t − 5.0 m/s 2 t 2 + C

(

)

Using the fact that x = 0 when t = 0, evaluate C:

0 = (50 m/s )(0) − 5.0 m/s 2 (0) + C and C=0

Substitute to obtain:

x(t ) =

2

(50 m/s) t − (5.0 m/s 2 ) t 2

Note that this expression is quadratic in t and that the coefficient of t2 is negative and equal in magnitude to half the constant acceleration. Remarks: We can check our result for x(t) by evaluating it over the 10-s interval shown and comparing this result with the area bounded by this curve and the time axis.

146 Chapter 2 107 ••• Figure 2-43 shows the acceleration of a particle versus time. (a) What is the magnitude, in m/s, of the area of the shaded box? (b) The particle starts from rest at t = 0. Estimate the velocity at t = 1.0 s, 2.0 s, and 3.0 s by counting the boxes under the curve. (c) Sketch the curve v versus t from your results for Part (b), then estimate how far the particle travels in the interval t = 0 to t = 3.0 s using your curve. Picture the Problem During any time interval, the integral of a(t) is the change in velocity and the integral of v(t) is the displacement. The integral of a function equals the "area" between the curve for that function and the independent-variable axis.

(a) Find the area of the shaded grid box in Figure 2-44:

m⎞ ⎛ Area = ⎜ 0.50 2 ⎟ (0.50 s ) s ⎠ ⎝ = 0.25 m/s per box

(b) We start from rest (vo = 0) at t = 0. For the velocities at the other times, count boxes and multiply by the 0.25 m/s per box that we found in Part (a):

t

# of boxes

v(t)

(s) 1.0

(m/s) 3.7

0.93 m/s

2.0

12

3.0 m/s

3.0

24

6.0 m/s

(c) The graph of v as a function of t follows: 7 6

v, m/s

5 4 3 2 1 0 0

0.5

1

1.5

2

2.5

t, s

Area = hw = (1.0 m/s )(0.50 s ) = 0.50 m per box

3

Motion in One Dimension 147 Count the boxes under the v(t) curve to find the distance traveled:

x(3.0 s ) = Δx(0 → 3.0 s )

= (13 boxes )[(0.50 m ) / box ] = 6.5 m

108 ••• Figure 2-44 is a graph of vx versus t for a particle moving along a straight line. The position of the particle at time t = 0 is x0 = 5.0 m. (a) Find x for various times t by counting boxes, and sketch x as a function of t. (b) Sketch a graph of the acceleration a as a function of the time t. (c) Determine the displacement of the particle between t = 3.0 s and 7.0 s. Picture the Problem The integral of v(t) over a time interval is the displacement (change in position) during that time interval. The integral of a function equals the "area" between the curve for that function and the independent-variable axis. We can estimate this value by counting the number of squares under the curve between given limits and multiplying this count by the "area" of each square. Because acceleration is the slope of a velocity versus time curve, this is a nonconstant-acceleration problem. The derivative of a function is equal to the "slope" of the function at that value of the independent variable. We can approximate the slope of any graph by drawing tangent lines and estimating their slopes.

(a) To obtain the data for x(t), we must estimate the accumulated area under the v(t) curve at each time interval: Find the area of a shaded grid box in Figure 2-44:

Area = hw = (1.0 m/s )(0.50 s ) = 0.50 m per box

We start from rest (vo = 0) at to= 0. For the position at the other times, count boxes and multiply by the 0.50 m per box that we found above. Remember to add the offset from the origin, xo = 5.0 m, and that boxes below the v = 0 line are counted as negative:

Examples: ⎛ 0.50 m ⎞ x(2 s ) = (15.5 boxes )⎜ ⎟ + 5.0 m ⎝ box ⎠ = 13 m ⎛ 0.50 m ⎞ x(5 s ) = (49 boxes )⎜ ⎟ + 5.0 m ⎝ box ⎠ = 30 m ⎛ 0.50 m ⎞ x(10 s ) = (51 boxes )⎜ ⎟ ⎝ box ⎠ ⎛ 0.50 m ⎞ − (36 boxes )⎜ ⎟ + 5.0 m ⎝ box ⎠ = 13 m

148 Chapter 2 A graph of x as a function of t follows: 35 30

x, m

25 20 15 10 5 0 0

2

4

6

8

10

t, s

(b) To obtain the data for a(t), we must estimate the slope (Δv/Δt) of the v(t) curve at several times. A good way to get reasonably reliable readings from the graph is to enlarge it several fold and then estimate the slope of the tangent line at selected points on the graph:

Examples: 8.0 m/s − 4.1 m/s a(2 s ) = = 0.8 m/s 2 4.9 s a (3 s ) = 0 − 6.0 m/s − 8.0 m/s a(5 s ) = = −4.5 m/s 2 7.1 s − 4.0 s − 6.0 m/s − 1.4 m/s a(7 s ) = = −3.0 m/s 2 7.6 s − 5.0 s a (8 s ) = 0

A graph of a as a function of t follows: 6 4

a, m/s^2

2 0 -2 -4 -6 0

2

4

6

t, s

8

10

Motion in One Dimension 149 (c) The displacement of the particle between t = 3.0 s and 7.0 s is given by:

Δx3.0s →7.0 s = x(7.0 s ) − x(3.0 s )

Use the graph in (a) to obtain:

x(7.0 s ) = 29 m and x(3.0 s ) = 18 m

Substitute for Δx(7.0 s) and Δx(3.0 s) to obtain:

Δx3.0s →7.0 s = 29 m − 18 m = 11 m

109 ••• [SSM] Figure 2-45 shows a plot of x versus t for a body moving along a straight line. For this motion, sketch graphs (using the same t axis) of (a) vx as a function of t, and (b) ax as a function of t. (c) Use your sketches to qualitatively compare the time(s) when the object is at its largest distance from the origin to the time(s) when its speed is greatest. Explain why the times are not the same. (d) Use your sketches to qualitatively compare the time(s) when the object is moving fastest to the time(s) when its acceleration is the largest. Explain why the times are not the same. Picture the Problem Because the position of the body is not described by a parabolic function, the acceleration is not constant.

(a) Select a series of points on the graph of x(t) (e.g., at the extreme values and where the graph crosses the t axis), draw tangent lines at those points, and measure their slopes. In doing this, you are evaluating v = dx/dt at these points. Plot these slopes above the times at which you measured the slopes. Your graph should closely resemble the following graph. 8 6 4 2

v 0 -2 -4 -6 -8 0

0.2

0.4

0.6

0.8

1

1.2

1.4

t

(b) Select a series of points on the graph of v(t) (e.g., at the extreme values and where the graph crosses the t axis), draw tangent lines at those points, and measure their slopes. In doing this, you are evaluating a = dv/dt at these points. Plot these slopes above the times at which you measured the slopes. Your graph should closely resemble the following graph.

150 Chapter 2 15 10 5

a

0 -5 -10 -15 0

0.5

1

1.5

t

(c) The points at the greatest distances from the time axis correspond to turnaround points. The velocity of the body is zero at these points. (d) The body is moving fastest as it goes through the origin. At these times the velocity is not changing and hence the acceleration is zero. The maximum acceleration occurs at the maximum distances where the velocity is zero but changing direction rapidly. 110 ••• The acceleration of a certain rocket is given by a = bt, where b is a positive constant. (a) Find the position function x(t) if x = x0 and vx = v0x at t = 0. (b) Find the position and velocity at t = 5.0 s if x0 = 0, v0x = 0 and b = 3.0 m/s3. (c) Compute the average velocity of the rocket between t = 4.5 s and 5.5 s at t = 5.0 s if x0 = 0, v0x = 0 and b = 3.0 m/s3. Compare this average velocity with the instantaneous velocity at t = 5.0 s. Picture the Problem Because the acceleration of the rocket varies with time, it is not constant and integration of this function is required to determine the rocket’s velocity and position as functions of time. The conditions on x and v at t = 0 are known as initial conditions.

(a) Integrate a(t) to find v(t):

v(t ) = ∫ a(t ) dt = b ∫ t dt = 12 bt 2 + C where C, the constant of integration, can be determined from the initial conditions.

Integrate v(t) to find x(t):

[

]

x(t ) = ∫ v(t ) dt = ∫ 12 bt 2 + C dt = 16 bt 3 + Ct + D where D is a second constant of integration.

Motion in One Dimension 151 Using the initial conditions, find the constants C and D:

v x (0 ) = v0 x ⇒ C = v0 x and x(0) = x0 ⇒ D = x0 and so x(t ) = 16 bt 3 + v0 x t + x0

(b) Evaluate v(5.0 s) and x(5.0 s) with C = D = 0 and b = 3.0 m/s3:

v(5.0 s ) =

1 2

(3.0 m/s )(5.0 s)

and x(5.0 s ) =

1 6

(3.0 m/s )(5.0 s )

2

3

3

3

= 38 m/s = 63 m

(c) The average value of v(t) over the interval Δt is given by:

v=

Substitute for Δt and v(t) and evaluate the integral to obtain:

1 21 2 b 3 t2 b t 23 − t13 v = ∫ 2 bt dt = t = Δt t1 6Δt t1 6Δt

Substitute numerical values and evaluate v :

(3.0 m/s )[(5.5 s ) − (4.5 s ) ] v=

t

1 2 v(t )dt Δt ∫t1

(

t

3

3

)

3

6(5.5 s − 4.5 s )

= 38 m/s ... a result in good agreement with the value found in (b). 111 ••• [SSM] In the time interval from 0.0 s to 10.0 s, the acceleration of a particle traveling in a straight line is given by ax = (0.20 m/s3)t. Let to the right be the +x direction. The particle initially has a velocity to the right of 9.5 m/s and is located 5.0 m to the left of the origin. (a) Determine the velocity as a function of time during the interval, (b) determine the position as a function of time during the interval, (c) determine the average velocity between t = 0.0 s and10.0 s, and compare it to the average of the instantaneous velocities at the start and ending times. Are these two averages equal? Explain. Picture the Problem The acceleration is a function of time; therefore it is not constant. The instantaneous velocity can be determined by integration of the acceleration function and the average velocity from the general expression for the average value of a non-linear function.

(a) The instantaneous velocity function v(t) is the time-integral of the acceleration function:

b v(t ) = ∫ a x dt = ∫ b t dt = t 2 + C1 2 where b = 0.20 m/s3

152 Chapter 2 The initial conditions are:

1) x(0 ) = −5.0 m and 2) v(0 ) = 9.5 m/s

Use initial condition 2) to obtain:

v(0) = 9.5 m/s = C1

Substituting in v(t) for b and C1 yields:

v(t ) = 0.10 m/s 3 t 2 + 9.5 m/s

(b) The instantaneous position function x(t) is the time-integral of the velocity function:

x(t ) = ∫ v(t )dt = ∫ ct 2 + C1 dt

(

)

(

(1)

)

c = t 3 + C1t + C 2 3 where c = 0.10 m/s3. x(0 ) = −5.0 m = C 2

Using initial condition 1) yields: Substituting in x(t) for C1 and C2 yields: x(t ) =

1 3

(0.20 m/s )t + (9.5 m/s)t − 5.0 m/s 3

(c) The average value of v(t) over the interval Δt is given by:

3

t

v=

1 2 v(t )dt Δt ∫t1

Substitute for v(t) and evaluate the integral to obtain: 2 1 2⎛b 1 ⎡b 1 ⎡b ⎞⎤ ⎛b ⎤ ⎞ v = ∫ ⎜ t 2 + C1 ⎟dt = ⎢ t 3 + C1t ⎥ = ⎢ t 23 + C1t 2 − ⎜ t13 + C1t1 ⎟⎥ Δt t1 ⎝ 2 Δt ⎣ 6 ⎠⎦ ⎝6 ⎦ t1 Δt ⎣ 6 ⎠

t

t

Simplifying this expression yields:

Because t1 = 0:

(

)

v=

1 ⎡b 3 3 ⎤ t 2 − t1 + C1 (t 2 − t1 )⎥ ⎢ Δt ⎣ 6 ⎦

v=

1 ⎡b 3 ⎤ t 2 + C1t 2 ⎥ ⎢ Δt ⎣ 6 ⎦

Substitute numerical values and simplify to obtain: v=

⎤ 1 ⎡⎛ 0.20 m/s 3 ⎞ ⎟⎟ (10.0 s )3 + (9.5 m/s )(10.0 s )⎥ = 13 m/s ⎢⎜⎜ 10.0 s ⎣⎝ 6 ⎠ ⎦

Motion in One Dimension 153 The average of the initial instantaneous and final instantaneous velocities is given by: Using equation (1), evaluate v(0) and v(10 s):

Substitute in equation (2) to obtain:

vav =

v(0 ) + v(10.0 s ) 2

(2)

v(0) = 9.5 m/s and 2 v(10 s ) = 0.10 m/s 3 (10.0 s ) + 9.5 m/s = 19.5 m/s

(

vav =

)

9.5 m/s + 19.5 m/s = 15 m/s 2

vav is not the same as v because the velocity does not change linearly with time. The velocity does not change linearly with time because the acceleration is not constant. 112 ••• Consider the motion of a particle that experiences a variable acceleration given by a x = a0 x + bt , where a0x and b are constants and x = x0 and vx = v0x at t = 0. (a) Find the instantaneous velocity as a function of time. (b) Find the position as a function of time. (c) Find the average velocity for the time interval with an initial time of zero and arbitrary final time t. (d) Compare the average of the initial and final velocities to your answer to Part (c). Are these two averages equal? Explain. Determine the Concept Because the acceleration is a function of time, it is not constant. Hence we’ll need to integrate the acceleration function to find the velocity of the particle as a function of time and integrate the velocity function to find the position of the particle as a function of time. We can use the general expression for the average value of a non-linear function to find the average velocity over an arbitrary interval of time.

(a) From the definition of acceleration we have:

vx

t

v0 x

0

∫ dv = ∫ adt

Integrating the left-hand side of the equation and substituting for a on the right-hand side yields:

v x − v0 x = ∫ (a0 + bt )dt

Integrate the right-hand side to obtain:

v x − v0 x = a0 t + 12 bt 2

Solving this equation for v x yields:

vx = v0 x + a0t + 12 bt 2

t

0

(1)

154 Chapter 2 (b) The position of the particle as a function of time is the integral of the velocity function:

x

t

x0

0

∫ dx = ∫ v (t )dt x

Integrating the left-hand side of the equation and substituting for a on the right-hand side yields:

b ⎞ ⎛ x − x0 = ∫ ⎜ v0 x + a0 t + t 2 ⎟dt 2 ⎠ 0⎝

Integrate the right-hand side to obtain:

x − x0 = v0 x t + 12 a0 t 2 + 16 bt 3

Solving this equation for x yields:

x = x0 x + v0 t + 12 a0 t 2 + 16 bt 3

(c) The average value of v(t) over an interval Δt is given by:

1 2 v x = ∫ v x (t )dt Δt t1

t

t

or, because t1 = 0 and t2 = t, t 1 v x = ∫ v x (t )dt t0 Substituting for vx(t) yields:

Carry out the details of the integration to obtain:

t

vx =

(

)

1 v0 x + a0 t + 12 bt 2 dt t ∫0

[

]

[

]

1 v x = v0 x t + 12 a0 t 2 + 16 bt 3 t 1 = v0 x t + 12 a0 t 2 + 16 bt 3 t

t

0

= v0 x + 12 a0 t + 16 bt 2 (d) The average of the initial instantaneous and final instantaneous velocities is given by: Using equation (1), substitute to obtain:

vav, x =

vav, x

v0 x + v x 2

v0 x + v0 x + a0 t + 12 bt 2 = 2 2v + a0 t + 12 bt 2 = 0x 2 = v0 x + 12 a0 t + 14 bt 2

vav, x is not the same as v x because the acceleration is not constant.

Motion in One Dimension 155

General Problems 113 ••• You are a student in a science class that is using the following apparatus to determine the value of g. Two photogates are used. (Note: You may be familiar with photogates in everyday living. You see them in the doorways of some stores. They are designed to ring a bell when someone interrupts the beam while walking through the door.) One photogate is located at the edge of a table that is 1.00 m above the floor, and the second photogate is located directly below the first, at a height 0.500 m above the floor. You are told to drop a marble through these gates, releasing it from rest a negligible distance above the upper gate. The upper gate starts a timer as the marble passes through its beam. The second photogate stops the timer when the marble passes through its beam. (a) Prove that the experimental magnitude of free-fall acceleration is given by gexp = (2Δy)/(Δt)2, where Δy is the vertical distance between the photogates and Δt is the fall time. (b) For your setup, what value of Δt would you expect to measure, assuming gexp is the value (9.81 m/s2)? (c) During the experiment, a slight error is made. Instead of locating the first photogate even with the top of the table, your not-so-careful lab partner locates it 0.50 cm lower than the top of the table. However, she releases the marble from the same height that it was released from when the photogate was 1.00 m above the floor. What value of gexp will you and your partner determine? What percentage difference does this represent from the standard value at sea level? Picture the Problem The acceleration of the marble is constant. Because the motion is downward, choose a coordinate system with downward as the positive direction and use constant-acceleration equations to describe the motion of the marble.

Δy = v0 Δt + 12 g exp (Δt )

(a) Use a constant-acceleration equation to relate the vertical distance the marble falls to its time of fall and its free-fall acceleration:

or, because v0 = 0,

(b) Solve equation (1) for Δt to obtain:

Δt =

2Δy g exp

Substitute numerical values and evaluate Δt

Δt =

2(1.00 m ) = 0.452 s 9.81 m/s 2

(c) Using a constant-acceleration equation, express the speed of the marble in terms of its initial speed, acceleration, and displacement:

vf2 = v02 + 2aΔy or, because v0 = 0 and a = g, vf2 = 2 gΔy ⇒ vf = 2 gΔy

2

Δy = 12 g exp (Δt ) ⇒ g exp = 2

2Δy (Δt )2

(1)

156 Chapter 2 Let v1 be the speed the ball has reached when it has fallen 0.50 cm, and v2 be the speed the ball has reached when it has fallen 0.500 m to obtain.

Using a constant-acceleration equation, express v2 in terms of v1, g and Δt: Substitute numerical values and evaluate Δt: The distance the marble falls during this interval is: Use equation (1) to calculate the experimental value of the acceleration due to gravity: Finally, calculate the percent difference between this experimental result and the value accepted for g at sea level.

(

)

v1 = 2 9.81 m/s 2 (0.0050 m ) = 0.313 m/s

(

and

)

v2 = 2 9.81m/s 2 (0.500 m ) = 3.13 m/s v 2 = v1 + gΔt ⇒ Δt =

Δt =

v2 − v1 g

3.13 m/s − 0.313 m/s = 0.2872 s 9.81 m/s 2

Δy = 1.00 m − 0.5000 m − 0.0050 m = 0.50 m

g exp =

2(0.50 m ) = 12 m/s 2 2 (0.2872 s )

% difference =

9.81 m/s 2 − 12 m/s 2 9.81 m/s 2

= 22%

114 ••• The position of a body oscillating on a spring is given by x = A sin ωt, where A and ω are constants, A = 5.0 cm. and ω = 0.175 s–1. (a) Plot x as a function of t for 0 ≤ t ≤ 36 s. (b) Measure the slope of your graph at t = 0 to find the velocity at this time. (c) Calculate the average velocity for a series of intervals, beginning at t = 0 and ending at t = 6.0, 3.0, 2.0, 1.0, 0.50, and 0.25 s. (d) Compute dx/dt to find the velocity at time t = 0. (e) Compare your results in Parts (c) and (d) and explain why your Part (c) results approach your Part (d) results. Picture the Problem We can obtain an average velocity, vav = Δx/Δt, over fixed time intervals. The instantaneous velocity, v = dx/dt can only be obtained by differentiation.

Motion in One Dimension 157 (a) The following graph of x versus t was plotted using a spreadsheet program: 8 6

x, m

4 2 0 -2 -4 -6 0

5

10

15

20

25

30

35

t, s

(b) Draw a tangent line at the origin and measure its rise and run. Use this ratio to obtain an approximate value for the slope at the origin:

The tangent line appears to, at least approximately, pass through the point (5, 4). Using the origin as the second point, Δx = 4 cm – 0 = 4 cm and Δt = 5 s – 0 = 5 s

Therefore, the slope of the tangent line and the velocity of the body as it passes through the origin is approximately:

v(0) =

rise Δx 4 cm = = = 0.8 cm/s run Δt 5s

(c) Calculate the average velocity for the series of time intervals given by completing the table shown below: x0 t0 t x Δt Δx vav=Δx/Δt (s) (s) (s) (cm) (cm) (cm) (m/s) 0 6 6 0 4.34 4.34 0.723 0 3 3 0 2.51 2.51 0.835 0 2 2 0 1.71 1.71 0.857 0 1 1 0 0.871 0.871 0.871 0 0.5 0.5 0 0.437 0.437 0.874 0 0.25 0.25 0 0.219 0.219 0.875 (d) Express the time derivative of the position function:

dx = Aω cos ωt dt

158 Chapter 2 Substitute numerical values and dx evaluate at t = 0: dt

dx = Aω cos 0 = Aω dt = (0.050 m ) 0.175 s −1

(

)

= 0.88 cm/s

(e) Compare the average velocities from Part (c) with the instantaneous velocity from Part (d):

As Δt, and thus Δx, becomes small, the value for the average velocity approaches that for the instantaneous velocity obtained in Part (d). For Δt = 0.25 s, they agree to three significant figures.

115 ••• [SSM] Consider an object that is attached to a horizontally oscillating piston. The object moves with a velocity given by v = B sin(ωt), where B and ω are constants and ω is in s−1. (a) Explain why B is equal to the maximum speed vmax. (b) Determine the acceleration of the object as a function of time. Is the acceleration constant? (c) What is the maximum acceleration (magnitude) in terms of ω and vmax. (d) At t = 0, the object's position is known to be x0. Determine the position as a function of time in terms of in terms of t, ω, x0 and vmax. Determine the Concept Because the velocity varies nonlinearly with time, the acceleration of the object is not constant. We can find the acceleration of the object by differentiating its velocity with respect to time and its position function by integrating the velocity function.

(a) The maximum value of the sine function (as in v = vmax sin(ωt)) is 1. Hence the coefficient B represents the maximum possible speed vmax. (b) The acceleration of the object is the derivative of its velocity with respect to time:

a=

dv d = [vmax sin (ωt )] dt dt

= ω vmax cos (ωt )

Because a varies sinusoidally with time it is not constant. (c) Examination of the coefficient of the cosine function in the expression for a leads one to the conclusion that a max = ω vmax . (d) The position of the object as a function of time is the integral of the velocity function:

∫ dx = ∫ v(t )dt

Motion in One Dimension 159 Integrating the left-hand side of the equation and substituting for v on the right-hand side yields:

x = ∫ vmax sin (ωt )dt + C

− vmax

Integrate the right-hand side to obtain:

x=

Use the initial condition x(0) = x0 to obtain:

x0 =

Solving for C yields:

Substitute for C in equation (1) to obtain: Solving this equation for x yields:

ω

− vmax

ω

C = x0 + x=

cos (ωt ) + C

(1)

+C

vmax

ω

− vmax

ω

x = x0 +

cos (ωt ) + x0 + vmax

ω

vmax

ω

[1 − cos(ωt )]

116 ••• Suppose the acceleration of a particle is a function of x, where ax(x) = (2.0 s–2)x. (a) If the velocity is zero when x = 1.0 m, what is the speed when x = 3.0 m? (b) How long does it take the particle to travel from x = 1.0 m to x = 3.0 m? Picture the Problem Because the acceleration of the particle is a function of its position, it is not constant. Changing the variable of integration in the definition of acceleration will allow us to determine its velocity and position as functions of position.

(a) The acceleration of the particle is given by:

a=

Separating variables yields:

vdv = adx

Substitute for a to obtain:

vdv = bxdx where b = 2.0 s−2

The limits of integration are vo = 0 and v on the left-hand side and xo and x on the right-hand: Integrating both sides of the equation yields:

dv dv dv dx = =v dt dx dt dx

v

x

v0 = 0

x0

∫ vdv = ∫ bxdx

(

v 2 − v02 = b x 2 − x02

)

160 Chapter 2

(

Solve for v to obtain:

v = v02 + b x 2 − x02

)

Now set vo = 0, xo = 1.0 m, x = 3 m, and b =2.0 s–2 and evaluate the speed: v(3.0 m ) =

(2.0 s )[(3.0 m ) − (1.0 m ) ] = 2

−2

(b) From the definition of instantaneous velocity we have: Substitute the expression for v(x) derived in (a) to obtain:

2

t

t=

t=

x0

x

dx

∫ b(x 1

x

∫ b

x0

=

dx

∫ v( x )

0

x0

Evaluate the integral using the formula found in standard integral tables to obtain:

x

t = ∫ dt =

4.0 m/s

2

− x02

)

dx x 2 − x02

⎛ x + x2 − x2 0 ln⎜ ⎜ x0 b ⎝

1

⎞ ⎟ ⎟ ⎠

Substitute numerical values and evaluate t:

⎛ 3.0 m + t= ln⎜ 2.0 s -2 ⎜⎝ 1

(3.0 m )2 − (1.0 m )2 ⎞⎟ 1.0 m

⎟ ⎠

= 1.2 s

117 ••• [SSM] A rock falls through water with a continuously decreasing acceleration. Assume that the rock’s acceleration as a function of velocity has the form a y = g − bv y where b is a positive constant. (The +y direction is directly

downward.) (a) What are the SI units of b? (b) Prove mathematically that if the rock is released from rest at time t = 0, the acceleration will depend exponentially on time according to a y (t ) = ge − bt . (c) What is the terminal speed for the rock in terms of g and b? (See Problem 38 for an explanation of the phenomenon of terminal speed.) Picture the Problem Because the acceleration of the rock is a function of its velocity, it is not constant and we will have to integrate the acceleration function in order to find the velocity function. Choose a coordinate system in which downward is positive and the origin is at the point of release of the rock.

(a) All three terms in a y = g − bv y must have the same units in order for the equation to be valid. Hence the units of bvy must be acceleration units. Because the SI units of vy are m/s, b must have units of s −1.

Motion in One Dimension 161 (b) Rewrite a y = g − bv y explicitly as a differential equation: Separate the variables, vy on the left, t on the right: Integrate the left-hand side of this equation from 0 to vy and the righthand side from 0 to t: Integrating this equation yields:

Solve this expression for vy to obtain: Differentiate this expression with respect to time to obtain an expression for the acceleration and complete the proof: (c) Take the limit, as t →∞, of both sides of equation (1):

dv y

= g − bv y

dt

dv y g − bv y vy

= dt

t

dv y

∫ g − bv 0

y

= ∫ dt 0

1 ⎛ g − bv y − ln⎜⎜ b ⎝ g

vy = ay =

⎞ ⎟⎟ = t ⎠

g ( 1 − e −bt ) b dv y dt

=

(1)

d ⎛g −bt ⎞ −bt ⎜ (1 − e )⎟ = ge dt ⎝ b ⎠

(

)

⎡g ⎤ lim t →∞ v y = lim t →∞ ⎢ 1 − e −bt ⎥ ⎣b ⎦ and vt =

g b

Notice that this result depends only on b (inversely so). Thus b must include geometric factors like the shape and cross-sectional area of the falling object, as well as properties of the liquid such as density and temperature. 118 ••• A small rock sinking through water (see Problem 117) experiences an exponentially decreasing acceleration given by ay(t) = ge–bt, where b is a positive constant that depends on the shape and size of the rock and the physical properties of the water. Based upon this, find expressions for the velocity and position of the rock as functions of time. Assume that its initial position and velocity are both zero and that the +y direction is directly downward.

162 Chapter 2 Picture the Problem Because the acceleration of the rock is a function of time, it is not constant and we will have to integrate the acceleration function in order to find the velocity of the rock as a function of time. Similarly, we will have to integrate the velocity function in order to find the position of the rock as a function of time. Choose a coordinate system in which downward is positive and the origin at the point of release of the rock.

Separate variables in a y (t ) = dv y dt = ge −bt to obtain:

dv y = ge −bt dt

Integrating from t0 = 0, v0y = 0 to some later time t and velocity vy yields: v

t

v = ∫ dv = ∫ ge −bt dt = 0

0

[ ]

g −bt e −b

t

0

=

where vt =

g b

dy = v t (1 − e −bt )dt

Separate variables in dy v= = v t 1 − e −bt to obtain: dt

(

g (1 − e −bt ) = vt (1 − e −bt ) b

)

Integrate from to = 0, yo = 0 to some later time t and position y: y

t

(

y = ∫ dy = ∫ v t 1 − e 0

0

−bt

)

t

(

v ⎡ 1 ⎤ dt = v term ⎢t + e −bt ⎥ = v t t − t 1 − e −bt b ⎣ b ⎦0

)

Remarks: This last result is very interesting. It says that throughout its freefall, the object experiences drag; therefore it has not fallen as far at any given time as it would have if it were falling at the constant velocity, vt. 119 ••• The acceleration of a skydiver jumping from an airplane is given by a y = g − bv y2 where b is a positive constant depending on the skydiver’s cross-

sectional area and the density of the surrounding atmosphere she is diving through. The +y direction is directly downward. (a) If her initial speed is zero when stepping from a hovering helicopter, show that her speed as a function of time is given by v y (t ) = vt tanh (t T ) where vt is the terminal speed (see Problem 38) given by v t = g b and T = vt g is a time-scale parameter (b) What fraction of the terminal speed is the speed at t = T? (c) Use a spreadsheet program to graph vy(t) as a function of time, using a terminal speed of 56 m/s (use this value to calculate b and T). Does the resulting curve make sense?

Motion in One Dimension 163 Picture the Problem The skydiver’s acceleration is a function of her speed; therefore it is not constant. Expressing her acceleration as the derivative of her speed, separating the variables, and then integrating will give her speed as a function of time.

(a) Rewrite a y = g − bv y2 explicitly

dv y

as a differential equation:

dt

Separate the variables, with vy on the left, and t on the right:

Eliminate b by using b =

g : vt2

= g − bv y2

dv y

= dt

g − bv y2

dv y dv y = dt = 2 g 2 ⎤ ⎡ v ⎛ ⎞ y g − 2 vy g ⎢1 − ⎜⎜ ⎟⎟ ⎥ vt ⎢⎣ ⎝ v t ⎠ ⎥⎦

or, separating variables,

dv y ⎛ vy 1 − ⎜⎜ ⎝ vt Integrate the left-hand side of this equation from 0 to vy and the righthand side from 0 to t:

vy

∫ 0

2

⎞ ⎟⎟ ⎠

= gdt

t

dv y ⎛ vy 1 − ⎜⎜ ⎝ vt

⎞ ⎟⎟ ⎠

2

= g ∫ dt = gt 0

The integral on the left-hand side can be found in integral tables:

⎛ vy v t tanh −1 ⎜⎜ ⎝ vt

Solving this equation for vy yields:

⎛g v y = v t tanh ⎜⎜ ⎝ vt

Because c has units of m−1, and g has units of m/s2, (bg)−1/2 will have units of time. Let’s represent this expression with the time-scale factor T:

T = (bg )

−1 2

⎞ ⎟⎟ = gt ⎠ ⎞ ⎟⎟ t ⎠

164 Chapter 2 The skydiver falls with her terminal speed when a = 0. Using this definition, relate her terminal speed to the acceleration due to gravity and the constant b in the acceleration equation:

0 = g − bvt2 ⇒ vt =

g b

Convince yourself that T is also equal to vt/g and use this relationship to eliminate g and vt in the solution to the differential equation:

⎛t⎞ v y (t ) = v t tanh ⎜ ⎟ ⎝T ⎠

(b) The ratio of the speed at any time to the terminal speed is given by:

⎛t⎞ v t tanh⎜ ⎟ v y (t ) ⎝ T ⎠ = tanh⎛ t ⎞ = ⎜ ⎟ vt vt ⎝T ⎠

Evaluate this ratio for t = T to obtain:

v y (T ) vt

⎛T ⎞ = tanh⎜ ⎟ = tanh (1) = 0.762 ⎝T ⎠

(c) The following table was generated using a spreadsheet and the equation we derived in Part (a) for v(t). The cell formulas and their algebraic forms are: Cell Content/Formula Algebraic Form D1 56 vT D2 5.71 T B7 B6 + 0.25 t + 0.25 C7 $B$1*TANH(B7/$B$2) ⎛t⎞ vT tanh⎜ ⎟ ⎝T ⎠ A 1 2 3 4 5 6 7 8 9 56 57 58 59

B

vT = 56 T=5.71

C m/s s

t (s) 0.00 0.25 0.50 0.75

v (m/s) 0.00 2.45 4.89 7.32

12.50 12.75 13.00 13.25

54.61 54.73 54.83 54.93

Motion in One Dimension 165 A graph of v as a function of t follows: 60 50

v, m/s

40 30 20 10 0 0

2

4

6

8

10

12

14

t, s

Note that the speed increases linearly over time (that is, with constant acceleration) for about time T, but then it approaches the terminal speed as the acceleration decreases. 120 ••• Imagine that you are standing at a wishing well, wishing that you knew how deep the surface of the water was. Cleverly, you make your wish. Then, you take a penny from your pocket and drop it into the well. Exactly three seconds after you dropped the penny, you hear the sound it made when it struck the water. If the speed of sound is 343 m/s, how deep is the well? Neglect any effects due to air resistance. Picture the Problem We know that the sound was heard exactly 3.00 s after the penny was dropped. This total time may be broken up into the time required for the penny to drop from your hand to the water’s surface, and the time required for the sound to bounce back up to your ears. The time required for the penny to drop is related to the depth of the well. The use of a constant-acceleration equation in expressing the total fall time will lead to a quadratic equation one of whose roots will be the depth of the well

Express the time required for the sound to reach your ear as the sum of the drop time for the penny and the time for the sound to travel to your ear from the bottom of the well:

Δt tot = Δt drop + Δt sound

Use a constant-acceleration equation to express the depth of the well:

Δy well = 12 g (Δt drop ) ⇒ Δt drop = 2

(1)

2Δy well g

166 Chapter 2 From the definition of average speed, the time Δtsound required for the sound to travel from the bottom of the well to your ear is given by: Substituting for Δtsound and Δtdrop in equation (1) yields: Isolate the radical term and square both sides of the equation to obtain:

Δtsound =

Δywell vsound

Δt tot =

2Δy well Δy well + g vsound 2

⎛ Δy ⎞ 2Δy well ⎜⎜ Δt tot − well ⎟⎟ = vsound ⎠ g ⎝

Expanding the left side of this equation yields:

(Δttot )

2

2

⎛ Δy ⎞ Δy 2Δywell − 2Δt tot well + ⎜⎜ well ⎟⎟ = vsound ⎝ vsound ⎠ g

Rewrite this equation explicitly as a quadratic equation in Δywell :

⎛ 1 ⎜⎜ ⎝ vsound

2

⎛ 1 Δt ⎞ ⎟⎟ (Δy well )2 − 2⎜⎜ + tot ⎝ g vsound ⎠

⎞ ⎟⎟Δy well + (Δt tot )2 = 0 ⎠

Simplify further to obtain: ⎛

2

(Δy well )2 − 2⎜⎜ vsound ⎝ g

⎞ 2 + vsound Δt tot ⎟⎟Δy well + (vsound Δt tot ) = 0 ⎠

Substituting numerical values yields:

(Δy well )

2

or

⎞ ⎛ (343 m/s )2 ⎟Δy well + [(343 m/s )(3.00 s )]2 = 0 ( )( ) 343 m/s 3 . 00 s − 2⎜⎜ + 2 ⎟ ⎠ ⎝ 9.81 m/s

(Δy well )2 − (2.604 ×10 4 m )Δy well + 1.059 × 10 6 m 2 = 0

Use the quadratic formula or your graphing calculator to solve for the smaller root of this equation (the larger root is approximately 26 km and is too large to be physically meaningful):

Δy well = 40.73 m = 41 m

Motion in One Dimension 167 121 ••• You are driving a car at the speed limit of 25-mi/h speed limit when you observe the light at the intersection 65 m in front of you turn yellow. You know that at that particular intersection the light remains yellow for exactly 5.0 s before turning red. After you think for 1.0 s, you then accelerate the car at a constant rate. You somehow manage to pass your 4.5-m-long car completely through the 15.0-m-wide intersection just as the light turns red, thus narrowly avoiding a ticket for being in an intersection when the light is red. Immediately after passing through the intersection, you take your foot off the accelerator, relieved. However, down the road, you are pulled over for speeding. You assume that you were ticketed for the speed of your car as it exited the intersection. Determine this speed and decide whether you should fight this ticket in court. Explain. Picture the Problem First, we should find the total distance covered by the car between the time that the light turned yellow (and you began your acceleration) and the time that the back end of the car left the intersection. This distance is Δxtot = 65.0 m + 15.0 m + 4.5 m = 84.5 m. Part of this distance, Δxc is covered at constant speed, as you think about trying to make it through the intersection in time – this can be subtracted from the full distance Δxtot above to yield Δxacc, the displacement during the constant acceleration. With this information, we then can utilize a constant-acceleration equation to obtain an expression for your velocity as you exited the intersection. The following pictorial representation will help organize the information given in this problem. t0 = 0 v0 = 11.2 m/s

x0 = 0

t1 = 1 s v1 = 11.2 m/s

x1 = 11.2 m

Using a constant-acceleration equation relate the distance over which you accelerated to your initial velocity v1, your acceleration a, and the time during which you accelerated aacc: Relate the final velocity v3 of the car to its velocity v1 when it begins to accelerate, its acceleration, and the time during which it accelerates:

t2

t3 = 5 s

v2

v3 = ?

x 2 = 65 m

x3 = 80 m

Δxacc = Δxtot − Δxc = v1Δtacc + 12 a (Δtacc )

2

v3 = v1 + aΔt acc ⇒ a =

v3 − v1 Δt acc

Substitute for a in the expression for Δxacc to obtain: ⎛v −v ⎞ 2 Δx tot − Δxc = v1Δt acc + 12 ⎜⎜ 3 1 ⎟⎟(Δt acc ) = v1Δt acc + 12 (v3 − v1 )(Δt acc ) ⎝ Δt acc ⎠

168 Chapter 2 Solving for v3 yields:

v3 =

2(Δxtot − Δxc ) − v1 Δt acc

Substitute numerical values and evaluate v3:

v3 =

2(84.5 m − 11.2 m ) − 11.2 m/s = 25 m/s 4.0 s

Because 25 m/s is approximately 57 mi/h, you were approximately 32 mi/h over the speed limit! You would be foolish to contest your ticket. 122 ••• For a spherical celestial object of radius R, the acceleration due to gravity g at a distance x from the center of the object is g = g 0 R 2 x 2 , where g0 is the acceleration due to gravity at the object’s surface and x > R. For the moon, take g0 = 1.63 m/s2 and R = 3200 km. If a rock is released from rest at a height of 4R above the lunar surface, with what speed does the rock impact the moon? Hint: Its acceleration is a function of position and increases as the object falls. So do not use constant acceleration free-fall equations, but go back to basics. Picture the Problem Let the origin be at the center of the moon and the +x direction be radially outward. Because the acceleration of the rock is a function of its distance from the center of the moon, we’ll need to change the variables of integration in the definition of acceleration to v and x in order to relate the rock’s acceleration to its speed. Separating variables and integrating will yield an expression for the speed of the rock as a function of its distance from the center of the moon.

The acceleration of the object is given by:

a=

Separating variables yields:

vdv = adx

Because a = −g:

Expressing the integral of v from 0 to v and x from 4R to R yields: Carry out the integration to obtain:

Solve for v to obtain an expression for the speed of the object upon impact:

dv dv dx dv = =v dt dx dt dx

vdv = −

g0 R 2 dx x2

v

R

0

4R

∫ vdv = ∫ −

g0 R 2 dx dx = − g 0 R 2 ∫ 2 2 x x 4R

⎡1 R ⎤ 3 v2 = g0 R 2 ⎢ ⎥ = g0 R 2 ⎢⎣ x 4 R ⎥⎦ 4 v=

3 g0 R 2

R

Motion in One Dimension 169 Substitute numerical values and evaluate v:

v=

3 ( 1.63 m/s 2 )(1.738 × 10 6 m ) 2

= 2.06 km/s

170 Chapter 2

Chapter 3 Motion in Two and Three Dimensions Conceptual Problems 1 • [SSM] Can the magnitude of the displacement of a particle be less than the distance traveled by the particle along its path? Can its magnitude be more than the distance traveled? Explain. Determine the Concept The distance traveled along a path can be represented as a sequence of displacements.

Suppose we take a trip along some path and consider the trip as a sequence of many very small displacements. The net displacement is the vector sum of the very small displacements, and the total distance traveled is the sum of the magnitudes of the very small displacements. That is,

r r r r total distance = Δr0,1 + Δr1, 2 + Δr2,3 + ... + ΔrN −1, N where N is the number of very small displacements. (For this to be exactly true we have to take the limit as N goes to infinity and each displacement magnitude goes to zero.) Now, using ″the shortest distance between two points is a straight line,″ we have

r r r r r Δr0, N ≤ Δr0,1 + Δr1, 2 + Δr2,3 + ... + ΔrN −1, N , r where Δr0, N is the magnitude of the net displacement. Hence, we have shown that the magnitude of the displacement of a particle is less than or equal to the distance it travels along its path. 2 • Give an example in which the distance traveled is a significant amount, yet the corresponding displacement is zero. Can the reverse be true? If so, give an example.

171

172 Chapter 3 Determine the Concept The displacement of an object is its final position vector r r r minus its initial position vector ( Δr = rf − ri ). The displacement can be less but never more than the distance traveled. Suppose the path is one complete trip around Earth at the equator. Then, the displacement is 0 but the distance traveled is 2πRe. No, the reverse cannot be true. 3 • What is the average velocity of a batter who hits a home run (from when he hits the ball to when he touches home plate after rounding the bases)? Determine the Concept The important distinction here is that average velocity is being requested, as opposed to average speed.

The average velocity is defined as the displacement divided by the elapsed time.

r r Δr 0 vav = = =0 Δt Δt

The displacement for any trip around the bases is zero. Thus we see that no matter how fast the runner travels, the average velocity is always zero at the end of each complete circuit of the bases. What is the correct answer if we were asked for average speed? The average speed is defined as the distance traveled divided by the elapsed time.

vav =

total distance Δt

For one complete circuit of any track, the total distance traveled will be greater than zero and so the average speed is not zero. 4 • A baseball is hit so its initial velocity upon leaving the bat makes an angle of 30° above the horizontal. It leaves that bat at a height of 1.0 m above the ground and lands untouched for a single. During its flight, from just after it leaves the bat to just before it hits the ground, describe how the angle between its velocity and acceleration vectors changes. Neglect any effects due to air resistance. Determine the Concept The angle between its velocity and acceleration vectors starts at 30° + 90° or 120° because the acceleration of the ball is straight down. At the peak of the flight of the ball the angle reduces to 90° because the ball’s velocity vector is horizontal. When the ball reaches the same elevation that it started from the angle is 90° − 30° or 60°. 5 • If an object is moving toward the west at some instant, in what direction is its acceleration? (a) north, (b) east, (c) west, (d) south, (e) may be any direction.

Motion in One and Two Dimensions 173 Determine the Concept The instantaneous acceleration is the limiting value, as r r Δt approaches zero, of Δv Δt and is in the same direction as Δv . r Other than through the definition of a, the instantaneous velocity and acceleration vectors are unrelated. Knowing the direction of the velocity at one instant tells one nothing about how the velocity is changing at that instant. (e) is correct. 6 • Two astronauts are working on the lunar surface to install a new telescope. The acceleration due to gravity on the Moon is only 1.64 m/s2. One astronaut tosses a wrench to the other astronaut but the speed of throw is excessive and the wrench goes over her colleague’s head. When the wrench is at the highest point of its trajectory (a) its velocity and acceleration are both zero, (b) its velocity is zero but its acceleration is nonzero, (c) its velocity is nonzero but its acceleration is zero, (d) its velocity and acceleration are both nonzero, (e) insufficient information is given to choose between any of the previous choices. Determine the Concept When the wrench reaches its maximum height, it is still moving horizontally but its acceleration is downward. (d ) is correct.

The velocity of a particle is directed toward the east while the 7 • acceleration is directed toward the northwest as shown in Figure 3-27. The particle is (a) speeding up and turning toward the north, (b) speeding up and turning toward the south, (c) slowing down and turning toward the north, (d) slowing down and turning toward the south, (e) maintaining constant speed and turning toward the south. Determine the Concept The change in the velocity is in the same direction as the acceleration. Choose an x-y coordinate system with east being the positive x direction and north the positive y direction. Given our choice of coordinate r r system, the x component of a is negative and so v will decrease. The y r r component of a is positive and so v will increase toward the north. (c) is

correct. 8 • Assume you know the position vectors of a particle at two points on its path, one earlier and one later. You also know the time it took the particle to move from one point to the other. Then you can then compute the particle’s (a) average velocity, (b) average acceleration, (c) instantaneous velocity, (d) instantaneous acceleration. Determine the Concept All you can compute is the average velocity, since no instantaneous quantities can be computed and you need two instantaneous velocities to compute the average acceleration. (a ) is correct.

174 Chapter 3 9 • Consider the path of a moving particle. (a) How is the velocity vector related geometrically to the path of the particle? (b) Sketch a curved path and draw the velocity vector for the particle for several positions along the path. Determine the Concept (a) The velocity vector, as a consequence of always being in the direction of motion, is tangent to the path.

(b) A sketch showing two velocity vectors for a particle moving along a curved path is shown to the right.

10 • The acceleration of a car is zero when it is (a) turning right at a constant speed, (b) driving up a long straight incline at constant speed, (c) traveling over the crest of a hill at constant speed, (d) bottoming out at the lowest point of a valley at constant speed, (e) speeding up as it descends a long straight decline. Determine the Concept An object experiences acceleration whenever either its speed changes or it changes direction.

The acceleration of a car moving in a straight path at constant speed is zero. In the other examples, either the magnitude or the direction of the velocity vector is changing and, hence, the car is accelerated. (b) is correct. 11 • [SSM] Give examples of motion in which the directions of the velocity and acceleration vectors are (a) opposite, (b) the same, and (c) mutually perpendicular.

r r Determine the Concept The velocity vector is defined by v = dr / dt , while the r r acceleration vector is defined by a = dv / dt. (a) A car moving along a straight road while braking. (b) A car moving along a straight road while speeding up. (c) A particle moving around a circular track at constant speed. How is it possible for a particle moving at constant speed to be 12 • accelerating? Can a particle with constant velocity be accelerating at the same time?

Motion in One and Two Dimensions 175 Determine the Concept A particle experiences accelerated motion when either its speed or direction of motion changes.

A particle moving at constant speed in a circular path is accelerating because the direction of its velocity vector is changing. If a particle is moving at constant velocity, it is not accelerating. 13 • [SSM] Imagine throwing a dart straight upward so that it sticks into the ceiling. After it leaves your hand, it steadily slows down as it rises before it sticks. (a) Draw the dart’s velocity vector at times t1 and t2, where t1 and t2 occur after it leaves your hand but before it impacts the ceiling, and Δt =r t2 −rt1 is rsmall. From your drawing find the direction of the change in velocity Δv = v 2 – v1 , and thus the direction of the acceleration vector. (b) After it has stuck in the ceiling for a few seconds, the dart falls down to the floor. As it falls it speeds up, of course, until it hits the floor. Repeat Part (a) to find the direction of its acceleration vector as it falls. (c) Now imagine tossing the dart horizontally. What is the direction of its acceleration vector after it leaves your hand, but before it strikes the floor? Determine the Concept The acceleration vector is in the same direction as the r change in velocity vector, Δv .

(a) The sketch for the dart thrown upward is shown below. The acceleration vector is in the direction of the change in the velocity r vector Δv .

r v2 r v1

r − v1

r r r Δv = v 2 − v1

r v2

(b) The sketch for the falling dart is shown below. Again, the acceleration vector is in the direction of the change in the velocity r vector Δv .

r v2

r v1 rr r Δv = v 2 − v1

(c) The acceleration vector is in the direction of the change in the velocity vector … and hence is downward as shown below:

r v1 r v2

r r r Δv = v 2 − v1

14 • As a bungee jumper approaches the lowest point in her descent, the rubber cord holding her stretches and she loses speed as she continues to move downward. Assuming that she is dropping straight down, make a motion diagram to find the direction of her acceleration vector as she slows down by drawing her velocity vectors at times t1 and t2, where t1 and t2 are two instants during the portion of her descent that she is losing speed and t2 − t1 is small. From your

176 Chapter 3 r r r drawing find the direction of the change in velocity Δv = v 2 − v1 , and thus the direction of the acceleration vector.

Determine the Concept The acceleration vector is in the same direction as the change in velocity r vector, Δv . The drawing is shown to the right.

r v1 r − v1

r r r Δv = v 2 − v1

r v2

r v2

15 • After reaching the lowest point in her jump at time tlow, a bungee jumper moves upward, gaining speed for a short time until gravity again dominates her motion. Draw her velocity vectors at times t1 and t2, where Δt = t2 – t1 is small and t1 < tlow < t2. From your drawing find the direction of the r r r change in velocity Δv = v 2 − v1 , and thus the direction of the acceleration vector at time tlow. Determine the Concept The acceleration vector is in the same direction as the change in the r velocity vector, Δv . The drawing is shown to the right.

r v2

r v1

r r r Δv = v 2 − v1

r − v1

16 • A river is 0.76 km wide. The banks are straight and parallel (Figure 328). The current is 4.0 km/h and is parallel to the banks. A boat has a maximum speed of 4.0 km/h in still water. The pilot of the boat wishes to go on a straight line from A to B, where the line AB is perpendicular to the banks. The pilot should (a) head directly across the river, (b) head 53º upstream from the line AB, (c) head 37º upstream from the line AB, (d) give up—the trip from A to B is not possible with a boat of this limited speed, (e) do none of the above. Determine the Concept We can decide what the pilot should do by considering the speeds of the boat and of the current. The speed of the stream is equal to the maximum speed of the boat in still water. The best the boat can do is, while facing directly upstream, maintain its position relative to the bank. So, the pilot should give up. (d ) is correct.

Motion in One and Two Dimensions 177 17 • [SSM] During a heavy rain, the drops are falling at a constant velocity and at an angle of 10° west of the vertical. You are walking in the rain and notice that only the top surfaces of your clothes are getting wet. In what direction are you walking? Explain. Determine the Concept You must be walking west to make it appear to you that the rain is exactly vertical. 18 • In Problem 17, what is your walking speed if the speed of the drops relative to the ground is 5.2 m/s? r Determine the Concept Let v YG represent your velocity relative to the r ground and v RG represent the velocity of the rain relative to the ground. Then your speed relative to the ground is given by vYG = (5.2 m/s )sin 10° = 0.90 m/s

r vRG

10°

r vYG

True or false (Ignore any effects due to air resistance):

19

•

(a)

When a projectile is fired horizontally, it takes the same amount of time to reach the ground as an identical projectile dropped from rest from the same height. When a projectile is fired from a certain height at an upward angle, it takes longer to reach the ground than does an identical projectile dropped from rest from the same height. When a projectile is fired horizontally from a certain height, it has a higher speed upon reaching the ground than does an identical projectile dropped from rest from the same height.

(b) (c)

(a) True. In the absence of air resistance, both projectiles experience the same downward acceleration. Because both projectiles have initial vertical velocities of zero, their vertical motions must be identical. (b) True. When a projectile is fired from a certain height at an upward angle, its time in the air is larger because it first rises to a greater height before beginning to fall with zero initial vertical velocity.

178 Chapter 3 (c) True. When a projectile is fired horizontally, its velocity upon reaching the ground has a horizontal component in addition to the vertical component it has when it is dropped from rest. The magnitude of this velocity is related to its horizontal and vertical components through the Pythagorean Theorem. 20 • A projectile is fired at 35º above the horizontal. Any effects due to air resistance are negligible. At the highest point in its trajectory, its speed is 20 m/s. The initial velocity had a horizontal component of (a) 0, (b) (20 m/s) cos 35º, (c) (20 m/s) sin 35º, (d) (20 m/s)/cos 35º, (e) 20 m/s. Determine the Concept In the absence of air resistance, the horizontal component of the projectile’s velocity is constant for the duration of its flight. At the highest point, the speed is the horizontal component of the initial velocity. The vertical component is zero at the highest point. (e) is correct. 21 • [SSM] A projectile is fired at 35º above the horizontal. Any effects due to air resistance are negligible. The initial velocity of the projectile in Problem 20 has a vertical component that is (a) less than 20 m/s, (b) greater than 20 m/s, (c) equal to 20 m/s, (d) cannot be determined from the data given. Determine the Concept (a ) is correct. Because the initial horizontal velocity

is 20 m/s and the launch angle is less than 45 degrees, the initial vertical velocity must be less than 20 m/s. 22 • A projectile is fired at 35º above the horizontal. Any effects due to air resistance are negligible. The projectile lands at the same elevation of launch, so the vertical component of the impact velocity is (a) the same as the vertical component of its initial velocity in both magnitude and direction, (b) the same as the vertical component of its initial velocity, (c) less than the vertical component of its initial velocity, (d) less than the vertical component of its initial velocity. Determine the Concept (b) is correct. The landing speed is the same as the

launch speed. Because the horizontal component of its initial velocity does not change, the vertical component of the velocity at landing must be the same magnitude but oppositely directed its vertical component at launch. 23 • Figure 3-29 represents the trajectory of a projectile going from A to E. Air resistance is negligible. What is the direction of the acceleration at point B? (a) up and to the right, (b) down and to the left, (c) straight up, (d) straight down, (e) The acceleration of the ball is zero.

Motion in One and Two Dimensions 179 Determine the Concept

(d ) is correct. In the absence of air resistance, the

acceleration of the ball depends only on the change in its velocity and is independent of its velocity. As the ball moves along its trajectory between points A and C, the vertical component of its velocity decreases and the change in its velocity is a downward pointing vector. Between points C and E, the vertical component of its velocity increases and the change in its velocity is also a downward pointing vector. There is no change in the horizontal component of the velocity. 24 • Figure 3-29 represents the trajectory of a projectile going from A to E. Air resistance is negligible. (a) At which point(s) is the speed the greatest? (b) At which point(s) is the speed the least? (c) At which two points is the speed the same? Is the velocity also the same at these points? Determine the Concept In the absence of air resistance, the horizontal component of the velocity remains constant throughout the flight. The vertical component has its maximum values at launch and impact.

(a) The speed is greatest at A and E. (b) The speed is least at point C. (c) The speed is the same at A and E. No. The horizontal components are equal at these points but the vertical components are oppositely directed, so the velocity is not the same at A and E. [SSM]

True or false:

25

•

(a) (b) (c) (d) (e)

If an object's speed is constant, then its acceleration must be zero. If an object's acceleration is zero, then its speed must be constant. If an object's acceleration is zero, its velocity must be constant. If an object's speed is constant, then its velocity must be constant. If an object's velocity is constant, then its speed must be constant.

Determine the Concept Speed is a scalar quantity, whereas acceleration, equal to the rate of change of velocity, is a vector quantity.

(a) False. Consider a ball moving in a horizontal circle on the end of a string. The ball can move with constant speed (a scalar) even though its acceleration (a vector) is always changing direction. (b) True. From its definition, if the acceleration is zero, the velocity must be constant and so, therefore, must be the speed. (c) True. An object’s velocity must change in order for the object to have acceleration other than zero.

180 Chapter 3

(d) False. Consider an object moving at constant speed along a circular path. Its velocity changes continuously along such a path. (e) True. If the velocity of an object is constant, then both its direction and magnitude (speed) must be constant. 26 • The initial and final velocities of a particle are as shown in Figure 330. Find the direction of the average acceleration. Determine the Concept The average acceleration vector is defined by r r r aav = Δv / Δt. The direction of aav is that r r r of Δv = vf − vi , as shown to the right.

r vf r r r

Δ v = vf − vi

r − vi 27 •• The automobile path shown in Figure 3-31 is made up of straight lines and arcs of circles. The automobile starts from rest at point A. After it reaches point B, it travels at constant speed until it reaches point E. It comes to rest at point F. (a) At the middle of each segment (AB, BC, CD, DE, and EF), what is the direction of the velocity vector? (b) At which of these points does the automobile have a nonzero acceleration? In those cases, what is the direction of the acceleration? (c) How do the magnitudes of the acceleration compare for segments BC and DE? Determine the Concept The velocity vector is in the same direction as the change in the position vector while the acceleration vector is in the same direction as the change in the velocity vector. Choose a coordinate system in which the y direction is north and the x direction is east.

(b)

(a) Path AB BC CD DE EF

Direction of velocity vector north northeast east southeast south

Path AB BC CD DE EF

Direction of acceleration vector north southeast 0 southwest north

(c) The magnitudes are comparable, but larger for DE because the radius of the path is smaller there.

Motion in One and Two Dimensions 181 28 •• Two cannons are pointed directly toward each other as shown in Figure 3-32. When fired, the cannonballs will follow the trajectories shown—P is the point where the trajectories cross each other. If we want the cannonballs to hit each other, should the gun crews fire cannon A first, cannon B first, or should they fire simultaneously? Ignore any effects due to air resistance. Determine the Concept We’ll assume that the cannons are identical and use a constant-acceleration equation to express the displacement of each cannonball as a function of time. Having done so, we can then establish the condition under which they will have the same vertical position at a given time and, hence, collide. The modified diagram shown below shows the displacements of both cannonballs.

Express the displacement of the cannonball from cannon A at any time t after being fired and before any collision:

r r r Δr = v0t + 12 gt 2

Express the displacement of the cannonball from cannon A at any time t′ after being fired and before any collision:

r r r Δr ′ = v0′t ′ + 12 gt ′2

If the guns are fired simultaneously, t = t′ and the balls are the same distance 2 1 2 gt below the line of sight at all times. Also, because the cannons are identical, the cannonballs have the same horizontal component of velocity and will reach the horizontal midpoint at the same time. Therefore, they should fire the guns simultaneously. Remarks: This is the ″monkey and hunter″ problem in disguise. If you imagine a monkey in the position shown below, and the two guns are fired simultaneously, and the monkey begins to fall when the guns are fired, then the monkey and the two cannonballs will all reach point P at the same time.

182 Chapter 3

29 •• Galileo wrote the following in his Dialogue concerning the two world systems: ″Shut yourself up . . . in the main cabin below decks on some large ship, and . . . hang up a bottle that empties drop by drop into a wide vessel beneath it. When you have observed [this] carefully . . . have the ship proceed with any speed you like, so long as the motion is uniform and not fluctuating this way and that. The droplets will fall as before into the vessel beneath without dropping towards the stern, although while the drops are in the air the ship runs many spans.″ Explain this quotation. Determine the Concept The droplet leaving the bottle has the same horizontal velocity as the ship. During the time the droplet is in the air, it is also moving horizontally with the same velocity as the rest of the ship. Because of this, it falls into the vessel, which has the same horizontal velocity. Because you have the same horizontal velocity as the ship does, you see the same thing as if the ship were standing still. 30 • A man swings a stone attached to a rope in a horizontal circle at constant speed. Figure 3-33 represents the path of the rock looking down from above. (a) Which of the vectors could represent the velocity of the stone? (b) Which could represent the acceleration?

r r Determine the Concept (a) Because A and D are tangent to the path of the stone, either of them could represent the velocity of the stone.

r r (b) Let the vectors A(t ) and A(t + Δt ) be of equal length but point in slightly different directions as the stone moves around the circle. These two vectors and r r ΔA are shown in the diagram above. Note that ΔA is nearly perpendicular to r r r A(t ) . For very small time intervals, ΔA and A(t ) are perpendicular to one r r r another. Therefore, dA/dt is perpendicular to A and only the vector E could represent the acceleration of the stone.

Motion in One and Two Dimensions 183 True or false:

31

••

(a) (b) (c) (d)

An object cannot move in a circle unless it has centripetal acceleration. An object cannot move in a circle unless it has tangential acceleration. An object moving in a circle cannot have a variable speed. An object moving in a circle cannot have a constant velocity.

(a) True. An object accelerates when its velocity changes; that is, when either its speed or its direction changes. When an object moves in a circle the direction of its motion is continually changing. (b) False. An object moving in a circular path at constant speed has a tangential acceleration of zero. (c) False. A good example is a rock wedged in the tread of an automobile tire. Its speed changes whenever the car’s speed changes. (d) True. The velocity vector of any object moving in a circle is continually changing direction. 32 •• Using a motion diagram, find the direction of the acceleration of the bob of a pendulum when the bob is at a point where it is just reversing its direction. Picture the Problem In the diagram, (a) shows the pendulum just before it reverses direction and (b) shows the pendulum just after it has reversed its direction. The acceleration of the bob is in the direction of the change in the r r r velocity Δv = v f − v i and is tangent to the pendulum trajectory at the point of reversal of direction. This makes sense because, at an extremum of motion, v = 0, so there is no centripetal acceleration. However, because the velocity is reversing direction, the tangential acceleration is nonzero. (a)

(b)

r vi r vi r vf

r

r vf

r r

Δ v = vf − vi

33 •• [SSM] During your rookie bungee jump, your friend records your fall using a camcorder. By analyzing it frame by frame, he finds that the ycomponent of your velocity is (recorded every 1/20th of a second) as follows:

184 Chapter 3 12.05 12.10 12.15 12.20 12.25 12.30 12.35 12.40 12.45 t (s) vy −0.78 −0.69 −0.55 −0.35 −0.10 0.15 0.35 0.49 0.53 (m/s) (a) Draw a motion diagram. Use it to find the direction and relative magnitude of your average acceleration for each of the eight successive 0.050 s time intervals in the table. (b) Comment on how the y component of your acceleration does or does not vary in sign and magnitude as you reverse your direction of motion. Determine the Concept (a) The motion diagram shown below was constructed using the data in the table shown below the motion diagram. The column for Δv in the table was calculated using Δv = vi − vi −1 and the column for a was calculated using a = (vi − vi −1 ) Δt .

+y

r v1 1 r a12

r a 23 r a34 r a 45

i 1 2 3 4 5 6

v (m/s) −0.78 −0.69 −0.55 −0.35 −0.10 0.15

r v2

9

r v9

8

r v8

7

r v7

2

r v3 3

r v4

4

r v5

5

Δv (m/s)

aave (m/s2)

0.09 0.14 0.20 0.25 0.25

1.8 2.8 4.0 5.0 5.0

r v r 6 a56 6

r a89

r a78

r a67

Motion in One and Two Dimensions 185 7 8 9

0.35 0.49 0.53

0.20 0.14 0.04

4.0 2.8 0.8

(b) The acceleration vector always points upward and so the sign of its y component does not change. The magnitude of the acceleration vector is greatest when the bungee cord has its maximum extension (your speed, the magnitude of your velocity, is least at this time and times near it) and is less than this maximum value when the bungee cord has less extension.

Estimation and Approximation 34 •• Estimate the speed in mph with which water comes out of a garden hose using your past observations of water coming out of garden hoses and your knowledge of projectile motion. Picture the Problem Based on your experience with garden hoses, you probably know that the maximum range of the water is achieved when the hose is inclined at about 45° with the vertical. A reasonable estimate of the range of such a stream is about 4.0 m when the initial height of the stream is 1.0 m. Use constantacceleration equations to obtain expressions for the x and y coordinates of a droplet of water in the stream and then eliminate time between these equations to obtain an equation that you can solve for v0. y r v0 (x,y)

y0

θ0

R x Use constant-acceleration equations to express the x and y components of a molecule of water in the stream:

x = x0 + v0 x t + 12 a x t 2 and y = y0 + v0 y t + 12 a y t 2

Because x0 = 0, x = R, ay = −g, and ax = 0:

x = v0 x t and y = y0 + v0 y t − 12 gt 2

(1) (2)

186 Chapter 3 Express v0x and v0y in terms of v0 and θ :

Substitute in equations (1) and (2) to obtain:

v0 x = v0 cosθ 0 and v0 y = v0 sin θ 0 where θ0 is the angle the stream makes with the horizontal.

x = (v0 cosθ 0 )t and y = y0 + (v0 sin θ 0 )t − 12 gt 2

(3)

(4)

Eliminating t from equations (3) and (4) yields:

y = y0 + (tan θ 0 )x −

g x2 2v cos 2 θ 0

When the stream of water hits the ground, y = 0 and x = R:

0 = y0 + (tan θ 0 )R −

g R2 2 2v cos θ 0

Solving this equation for v0 gives:

v0 = R

2 0

2 0

g 2 cos θ [R tan θ + y0 ] 2

Substitute numerical values and evaluate v0: v0 = (4.0 m )

9.81 m/s 2 = 5.603 m/s 2 cos 2 45°[(4.0 m ) tan 45° + 1.0 m]

Use a conversion factor, found in Appendix A, to convert m/s to mi/h:

v0 = 5.603 m/s ×

1 mi/h 0.4470 m/s

≈ 13 mi/h

35 •• [SSM] You won a contest to spend a day with all team during their spring training camp. You are allowed to try to hit some balls thrown by a pitcher. Estimate the acceleration during the hit of a fastball thrown by a major league pitcher when you hit the ball squarely-straight back at the pitcher. You will need to make reasonable choices for ball speeds, both just before and just after the ball is hit, and of the contact time the ball has with the bat. Determine the Concept The magnitude of the acceleration of the ball is given by r r Δv r a= a = . Let vafter represent the velocity of the ball just after its collision with Δt r the bat and v before its velocity just before this collision. Most major league pitchers can throw a fastball at least 90 mi/h and some occasionally throw as fast as 100 mi/h. Let’s assume that the pitcher throws you an 80 mph fastball.

Motion in One and Two Dimensions 187 The magnitude of the acceleration of the ball is: Assuming that vafter and vbefore are both 80 mi/h and that the ball is in contact with the bat for 1 ms: Converting a to m/s2 yields:

a=

r r v after − v before . Δt

a=

80 mi/h − (− 80 mi/h ) 160 mi/h = 1 ms 1 ms

a=

160 mi/h 0.447 m/s × 1 ms 1 mi/h

≈ 7 × 10 4 m/s 2

Estimate how far you can throw a ball if you throw it (a) horizontally 36 •• while standing on level ground, (b) at θ = 45º above horizontal while standing on level ground, (c) horizontally from the top of a building 12 m high, (d) at θ = 45º above horizontal from the top of a building 12 m high. Ignore any effects due to air resistance. Picture the Problem During the flight of the ball the acceleration is constant and equal to 9.81 m/s2 directed downward. We can find the flight time from the vertical part of the motion, and then use the horizontal part of the motion to find the horizontal distance. We’ll assume that the release point of the ball is 2.0 m above your feet. A sketch of the motion that includes coordinate axes, the initial and final positions of the ball, the launch angle, and the initial velocity follows. y r v0 (x,y)

y0

θ

R x Obviously, how far you throw the ball will depend on how fast you can throw it. A major league baseball pitcher can throw a fastball at 90 mi/h or so. Assume that you can throw a ball at two-thirds that speed to obtain:

v0 = 60 mi/h ×

0.447 m/s ≈ 27 m/s 1mi/h

188 Chapter 3 There is no acceleration in the x direction, so the horizontal motion is one of constant velocity. Express the horizontal position of the ball as a function of time:

x = v0 x t = (v0 cosθ 0 )t

Assuming that the release point of the ball is a distance y0 above the ground, express the vertical position of the ball as a function of time:

y = y0 + v0 y t + 12 a y t 2 = y0 + (v0 sin θ 0 )t + 12 a y t 2

Eliminating t between equations (1) and (2) yields:

ay ⎛ ⎞ 2 ⎟⎟ x y = y0 + (tan θ 0 )x + ⎜⎜ 2 2 ⎝ 2v0 cos θ 0 ⎠

(a) If you throw the ball horizontally our equation becomes:

⎛ ay ⎞ y = y0 + ⎜⎜ 2 ⎟⎟ x 2 ⎝ 2v0 ⎠

Substitute numerical values to obtain:

⎛ − 9.81 m/s 2 ⎞ 2 ⎟x y = 2.0 m + ⎜⎜ 2 ⎟ ⎝ 2(27 m/s ) ⎠

At impact, y = 0 and x = R:

⎛ − 9.81 m/s 2 ⎞ 2 ⎟R 0 = 2.0 m + ⎜⎜ 2 ⎟ ⎝ 2(27 m/s ) ⎠

Solving for R yields:

R ≈ 17 m

(b) For θ = 45° we have: ⎛ ⎞ 2 − 9.81 m/s 2 ⎟x y = 2.0 m + (tan 45°)x + ⎜⎜ 2 2 ⎟ ⎝ 2(27 m/s ) cos 45° ⎠

At impact, y =0 and x = R: ⎛ ⎞ 2 − 9.81 m/s 2 ⎜ ⎟R 0 = 2.0 m + (tan 45°)R + ⎜ 2 2 ⎟ ( ) 2 27 m/s cos 45 ° ⎝ ⎠

Use the quadratic formula or your graphing calculator to obtain:

R = 75 m

(c) If you throw the ball horizontally from the top of a building that is 12 m high our equation becomes:

⎛ − 9.81 m/s 2 ⎞ 2 ⎟x y = 14.0 m + ⎜⎜ 2 ⎟ ( ) 2 27 m/s ⎝ ⎠

(1)

(2)

Motion in One and Two Dimensions 189 At impact, y = 0 and x = R:

⎛ − 9.81 m/s 2 ⎞ 2 ⎟R 0 = 14.0 m + ⎜⎜ 2 ⎟ ⎝ 2(27 m/s ) ⎠

Solve for R to obtain:

R = 45 m

(d) If you throw the ball at an angle of 45° from the top of a building that is 12 m high our equation becomes: ⎛ ⎞ 2 − 9.81 m/s 2 ⎜ ⎟x y = 14 m + (tan 45°)x + ⎜ 2 2 ⎟ ( ) 2 27 m/s cos 45 ° ⎝ ⎠

At impact, y = 0 and x = R: ⎛ ⎞ 2 − 9.81 m/s 2 ⎟R 0 = 14 m + (tan 45°)R + ⎜⎜ 2 2 ⎟ ( ) 2 27 m/s cos 45 ° ⎝ ⎠

Use the quadratic formula or your graphing calculator to obtain:

R = 85 m

37 •• In 1978, Geoff Capes of Great Britain threw a heavy brick a horizontal distance of 44.5 m. Find the approximate speed of the brick at the highest point of its flight, neglecting any effects due to air resistance. Assume the brick landed at the same height it was launched. Picture the Problem We’ll ignore the height of Geoff’s release point above the ground and assume that he launched the brick at an angle of 45°. Because the velocity of the brick at the highest point of its flight is equal to the horizontal component of its initial velocity, we can use constant-acceleration equations to relate this velocity to the brick’s x and y coordinates at impact. The diagram shows an appropriate coordinate system and the brick when it is at point P with coordinates (x, y). y

r v0

P (x , y )

θ0

R = 44.5 m

Using a constant-acceleration equation, express the x and y coordinates of the brick as a function of time:

x = x0 + v0 xt + 12 axt 2 and y = y0 + v0 y t + 12 a y t 2

x

190 Chapter 3 Because x0 = 0, y0 = 0, ay = −g, and ax = 0:

x = v0 xt and y = v0 y t − 12 gt 2

Eliminate t between these equations to obtain:

y = (tan θ 0 )x −

g 2 x 2v02x

where we have used tan θ 0 = When the brick strikes the ground y = 0 and x = R:

Solve for v0x to obtain:

Substitute numerical values and evaluate v0x:

v0 y v0 x

.

g 2 R 2v02x where R is the range of the brick. 0 = (tan θ 0 )R −

v0 x =

v0 x =

gR 2 tan θ 0

(9.81m/s )(44.5 m) ≈ 2

15 m/s 2 tan 45° Note that, at the brick’s highest point, vy = 0.

Position, Displacement, Velocity and Acceleration Vectors 38 • A wall clock has a minute hand with a length of 0.50 m and an hour hand with a length of 0.25 m. Take the center of the clock as the origin, and use a Cartesian coordinate system with the positive x axis pointing to 3 o'clock and the positive y axis pointing to 12 o’clock. Using unit vectors iˆ and jˆ , express the r position vectors of the tip of the hour hand ( A ) and the tip of the minute hand r ( B ) when the clock reads (a) 12:00, (b) 3:00, (c) 6:00, (d) 9:00. Picture the Problem Let the +y direction be straight up and the +x direction be to the right.

(a) At 12:00, both hands are positioned along the +y axis. The position vector for the tip of the hour hand at12:00 is:

r A =

The position vector for the tip of the minute hand at 12:00 is:

r B=

(0.25 m ) ˆj (0.50 m ) ˆj

Motion in One and Two Dimensions 191 (b) At 3:00, the minute hand is positioned along the +y axis, while the hour hand is positioned along the +x axis. The position vector for the tip of the hour hand at 3:00 is:

r A=

(0.25 m )iˆ

The position vector for the tip of the minute hand at 3:00 is:

r B=

(0.50 m ) ˆj

(c) At 6:00, the minute hand is positioned along the −y axis, while the hour hand is positioned along the +y axis. The position vector for the tip of the hour hand at 6:00 is:

r A = − (0.25 m ) ˆj

The position vector for the tip of the minute hand at 6:00 is:

r B=

(0.50 m ) ˆj

(d) At 9:00, the minute hand is positioned along the +y axis, while the hour hand is positioned along the −x axis. The position vector for the tip of the hour hand at 9:00 is:

r A = − (0.25 m ) iˆ

The position vector for the tip of the minute hand at 9:00 is:

r B=

(0.50 m ) ˆj

39 • [SSM] In Problem 38, find the displacements of the tip of each hand r r (that is, ΔA and ΔB ) when the time advances from 3:00 P.M. to 6:00 P.M. Picture the Problem Let the +y direction be straight up, the +x direction be to the right, and use the vectors describing the ends of the hour and minute hands in r r Problem 38 to find the displacements ΔA and ΔB.

The displacement of the minute hand as time advances from 3:00 P.M. to 6:00 P.M. is given by:

r r r ΔB = B6 − B3

From Problem 38:

r r B6 = (0.50 m ) ˆj and B3 = (0.50 m ) ˆj

Substitute and simplify to obtain:

r ΔB = (0.50 m ) ˆj − (0.50 m ) ˆj = 0

192 Chapter 3 r r r ΔA = A6 − A3

The displacement of the hour hand as time advances from 3:00 P.M. to 6:00 P.M. is given by: From Problem 38:

r r A6 = −(0.25 m ) ˆj and A3 = (0.25 m )iˆ

Substitute and simplify to obtain:

r ΔA = − (0.25 m ) ˆj − (0.25 m )iˆ

40 • In Problem 38, write the vector that describes the displacement of a fly if it quickly goes from the tip of the minute hand to the tip of the hour hand at 3:00 P.M. Picture the Problem Let the positive y direction be straight up, the positive x direction be to the right, and use the vectors describing the ends of the hour and r minute hands in Problem 38 to find the displacement D of the fly as it goes from the tip of the minute hand to the tip of the hour hand at 3:00 P.M. r r r D = A3 − B3

The displacement of the fly is given by: From Problem 38:

r r A3 = (0.25 m ) iˆ and B3 = (0.50 m ) ˆj

r r Substitute for A3 and B3 to obtain:

r D=

(0.25 m ) iˆ − (0.50 m ) ˆj

41 • A bear, awakening from winter hibernation, staggers directly northeast for 12 m and then due east for 12 m. Show each displacement graphically and graphically determine the single displacement that will take the bear back to her cave, to continue her hibernation. Picture the Problem The single displacement for the bear to make it back to its r cave is the vector D. Its magnitude D and direction θ can be determined by drawing the bear’s displacement vectors to scale. Ν 45°

12 m θ

12 m

r D

Ε

Motion in One and Two Dimensions 193 D ≈ 22 m and θ ≈ 23°

r Remarks: The direction of D is 180° + θ ≈ 203° 42 • A scout walks 2.4 km due East from camp, then turns left and walks 2.4 km along the arc of a circle centered at the campsite, and finally walks 1.5 km directly toward the camp. (a) How far is the scout from camp at the end of his walk? (b) In what direction is the scout’s position relative to the campsite? (c) What is the ratio of the final magnitude of the displacement to the total distance walked? Picture the Problem The figure shows the paths walked by the Scout. The length of path A is 2.4 km; the length of path B is 2.4 km; and the length of path C is 1.5 km:

North C

B

θ A

East

(a) Express the distance from the campsite to the end of path C:

2.4 km – 1.5 km = 0.9 km

(b) Determine the angle θ subtended by the arc at the origin (campsite):

arc length 2.4 km = radius 2.4 km = 1 rad = 57.3° His direction from camp is 1 rad North of East.

(c) Express the total distance as the sum of the three parts of his walk:

d tot = d east + d arc + d toward camp

Substitute the given distances to find the total:

dtot = 2.4 km + 2.4 km + 1.5 km = 6.3 km

θradians =

Express the ratio of the magnitude of his displacement to the total distance he walked and substitute to obtain a numerical value for this ratio: Magnitude of his displacement 0.9 km 1 = = Total distance walked 6.3 km 7

194 Chapter 3 43 •• [SSM] The faces of a cubical storage cabinet in your garage has 3.0-m-long edges that are parallel to the xyz coordinate planes. The cube has one corner at the origin. A cockroach, on the hunt for crumbs of food, begins at that corner and walks along three edges until it is at the far corner. (a) Write the roach's displacement using the set of ˆi , jˆ , and kˆ unit vectors, and (b) find the magnitude of its displacement. Picture the Problem While there are several walking routes the cockroach could take to get from the origin to point C, its displacement will be the same for all of them. One possible route is shown in the figure.

r (a) The roach’s displacement D during its trip from the origin to point C is:

r r r r D = A+ B +C

(b) The magnitude of the roach’s displacement is given by:

D = Dx2 + D y2 + Dz2

Substitute numerical values for Dx, Dy, and Dz and evaluate D to obtain:

D=

=

(3.0 m )iˆ + (3.0 m ) ˆj + (3.0 m )kˆ

(3.0 m )2 + (3.0 m )2 + (3.0 m )2

= 5.2 m

44 • You are the navigator of a ship at sea. You receive radio signals from two transmitters A and B, which are 100 km apart, one due south of the other. The direction finder shows you that transmitter A is at a heading of 30º south of east from the ship, while transmitter B is due east. Calculate the distance between your ship and transmitter B. Picture the Problem The diagram shows the locations of the transmitters relative to the ship and defines the distances separating the transmitters from each other and from the ship. We can find the distance between the ship and transmitter B using trigonometry.

Motion in One and Two Dimensions 195

S

DSB

A

θ

DAB

DSA

B

Relate the distance between A and B to the distance from the ship to A and the angle θ : Substitute numerical values and evaluate DSB:

tan θ =

DSB =

D DAB ⇒ DSB = AB DSB tan θ

100 km = 1.7 × 10 5 m tan 30°

Velocity and Acceleration Vectors 45 • A stationary radar operator determines that a ship is 10 km due south of him. An hour later the same ship is 20 km due southeast. If the ship moved at constant speed and always in the same direction, what was its velocity during this time? Picture the Problem For constant speed and direction, the instantaneous velocity is identical to the average velocity. Take the origin to be the location of the stationary radar and let the +x direction be to the East and the +y direction be to the North.

Express the average velocity: r Determine the position vectors r1 r and r2 :

N (y ) E (x )

r r1 1

r r2

r Δr

2

r r Δr v av = Δt

r r1 = (− 10 km ) ˆj and r r2 = (14.1km ) iˆ + (− 14.1km ) ˆj

(1)

196 Chapter 3

r Find the displacement vector Δr :

r r r Δr = r2 − r1 = (14.1 km ) iˆ + (− 4.1 km ) ˆj

r Substitute for Δr and Δt in equation (1) to find the average velocity.

r (14.1km ) iˆ + (− 4.1km ) ˆj vav = 1.0 h = (14 km/h )iˆ + (− 4.1km/h ) ˆj

46 • A particle’s position coordinates (x, y) are (2.0 m, 3.0 m) at t = 0; (6.0 m, 7.0 m) at t = 2.0 s; and (13 m, 14 m) at t = 5.0 s. (a) Find the magnitude of the average velocity from t = 0 to t = 2.0 s. (b) Find the magnitude of the average velocity from t = 0 to t = 5.0 s. Picture the Problem The average velocity is the change in position divided by the elapsed time.

(a) The magnitude of the average velocity is given by:

r Δr r vav = Δt

Find the position vectors and the displacement vector:

r r0 = (2.0 m )iˆ + (3.0 m ) ˆj r r2 = (6.0 m )iˆ + (7.0 m ) ˆj and r r r Δr = r2 − r1 = (4.0 m )iˆ + (4.0 m ) ˆj

Find the magnitude of the displacement vector for the interval between t = 0 and t = 2.0 s:

r Δr02 =

r Substitute to determine v av :

r 5.66 m vav = = 2.8 m/s 2.0 s

θ is given by:

(4.0 m )2 + (4.0 m )2

= 5.66 m

⎛ Δry , 02 ⎞ ⎟ = tan −1 ⎛⎜ 4.0 m ⎞⎟ ⎟ ⎝ 4.0 m ⎠ ⎝ Δrx , 02 ⎠

θ = tan −1 ⎜⎜ = 45°

measured from the positive x axis.

Motion in One and Two Dimensions 197 (b) Repeat (a), this time using the displacement between t = 0 and t = 5.0 s to obtain:

r r5 = (13 m )iˆ + (14 m ) ˆj  r r r ˆ Δr05 = r5 − r0 = (11 m )i + (11 m ) ˆj  r 2 2 Δr05 = (11 m ) + (11 m ) = 15.6 m

r 15.6 m vav = = 3.1 m/s  5.0 s and ⎛ 11 m ⎞ ⎟⎟ = 45° θ = tan −1 ⎜⎜ ⎝ 11 m ⎠ measured from the +x axis. 47 • [SSM] A particle moving at a velocity of 4.0 m/s in the +x direction is given an acceleration of 3.0 m/s2 in the +y direction for 2.0 s. Find the final speed of the particle. Picture the Problem The magnitude of the velocity vector at the end of the 2 s of acceleration will give us its speed at that instant. This is a constant-acceleration problem.

Find the final velocity vector of the particle:

r v = v x iˆ + v y ˆj = v x 0 iˆ + a y tˆj

r The magnitude of v is:

r v = v x2 + v y2

Substitute for vx and vy and evaluate r v:

r v =

= (4.0 m/s ) iˆ + (3.0 m/s 2 )(2.0 s ) ˆj = (4.0 m/s ) iˆ + (6.0 m/s ) ˆj

(4.0 m/s)2 + (6.0 m/s )2

= 7.2 m/s

48 • Initially, a swift-moving hawk is moving due west with a speed of 30 m/s; 5.0 s later it is moving due north with a speed of 20 m/s. (a) What are the r magnitude and direction of Δvav during this 5.0-s interval? (b) What are the r magnitude and direction of a av during this 5.0-s interval? Picture the Problem Choose a coordinate system in which north coincides with the +y direction and east with the +x direction. Expressing the hawk’s velocity r r vectors is the first step in determining Δv and a av .

198 Chapter 3 (a) The change in the hawk’s velocity during this interval is:

r r r Δvav = v N − v W

r r v W and v N are given by:

r r v W = −(30 m/s )iˆ and v N = (20 m/s ) ˆj

r r Substitute for v W and v N and r evaluate Δv :

r Δvav = (20 m/s ) ˆj − − (30 m/s )iˆ = (30 m/s ) iˆ + (20 m/s ) ˆj

r The magnitude of Δvav is given by:

r Δv av = Δv x2 + Δv y2

Substitute numerical values and evaluate Δv:

r Δvav =

The direction of Δv is given by: r − vW

θ = tan −1 ⎜⎜

θ

r vN

[

(30 m/s)2 + (20 m/s)2

= 36 m/s ⎛ Δv y ⎞ ⎟⎟ ⎝ Δvx ⎠

where θ is measured from the +x axis.

r Δvav θ

Substitute numerical values and evaluate θ :

θ = tan −1 ⎜

(b) The hawk’s average acceleration during this interval is:

r r Δvav aav = Δt

r Substitute for Δv and Δt to obtain:

]

⎛ 20 m/s ⎞ ⎟ = 34° ⎝ 30 m/s ⎠

r (30 m/s) iˆ + (20 m/s ) ˆj aav = 5.0 s 2 ˆ = 6.0 m/s i + 4.0 m/s 2 ˆj

(

) (

r The magnitude of aav is given by:

r a av = a x2 + a y2

Substitute numerical values and r evaluate aav :

r a av =

)

(6.0 m/s ) + (4.0 m/s )

= 7.2 m/s 2

2

2

Motion in One and Two Dimensions 199 r The direction of aav is given by:

Substitute numerical values and evaluate θ :

⎛ ay ⎝ ax

θ = tan −1 ⎜⎜

⎞ ⎟⎟ ⎠

⎛ 4.0 m/s 2 ⎞ ⎟ = 34° 2 ⎟ ⎝ 6.0 m/s ⎠

θ = tan −1 ⎜⎜

where θ is measured from the + x axis. r r Remarks: Because aav is in the same direction as Δv av , the calculation of θ in Part (b) was not necessary. 49 • At t = 0, a particle located at the origin has a velocity of 40 m/s at θ = 45º. At t = 3.0 s, the particle is at x = 100 m and y = 80 m and has a velocity of 30 m/s at θ = 50º. Calculate (a) the average velocity and (b) the average acceleration of the particle during this 3.0-s interval. Picture the Problem The initial and final positions and velocities of the particle are given. We can find the average velocity and average acceleration using their definitions by first calculating the given displacement and velocities using unit vectors iˆ and ˆj.

(a) The average velocity of the particle is the ratio of its displacement to the elapsed time:

r r Δr v av = Δt

The displacement of the particle during this interval of time is:

r Δr = (100 m ) iˆ + (80 m ) ˆj

Substitute to find the average velocity:

r (100 m ) iˆ + (80 m ) ˆj vav = 3.0 s = (33.3 m/s ) iˆ + (26.7 m/s ) ˆj =

(b) The average acceleration is: r r v1 and v 2 are given by:

(33 m/s) iˆ + (27 m/s) ˆj

r r r r Δv v 2 − v1 = a av = Δt Δt r v1 = [(40 m/s )cos 45°]iˆ + [(40 m/s )sin 45°] ˆj = (28.28 m/s ) iˆ + (28.28 m/s ) ˆj

and r v 2 = [(30 m/s) cos 50°]iˆ + [(30 m/s)sin 50°] ˆj = (19.28 m/s ) iˆ + (22.98 m/s ) ˆj

200 Chapter 3 r r r Substitute for v1 and v 2 and evaluate aav :

[

]

r (19.28 m/s) iˆ + (22.98 m/s) ˆj − (28.28 m/s) iˆ + (28.28 m/s) ˆj aav = 3.0 s (− 9.00 m/s ) iˆ + (− 5.30 m/s) ˆj = − 3.0 m/s 2 iˆ + − 1.8 m/s 2 ˆj = 3.0 s

(

) (

)

50 •• At time zero, a particle is at x = 4.0 m and y = 3.0 m and has velocity r v = (2.0 m/s) ˆi + (–9.0 m/s) jˆ . The acceleration of the particle is constant and is r given by a = (4.0 m/s2) ˆi + (3.0 m/s2) jˆ . (a) Find the velocity at t = 2.0 s. (b) Express the position vector at t = 4.0 s in terms of iˆ and jˆ . In addition, give

the magnitude and direction of the position vector at this time. Picture the Problem The acceleration is constant so we can use the constantacceleration equations in vector form to find the velocity at t = 2.0 s and the position vector at t = 4.0 s.

r r r v = v0 + at

(a) The velocity of the particle, as a function of time, is given by: r r r Substitute for v 0 and a to find v (2.0 s ) :

[

]

r v (2.0 s ) = (2.0 m/s) iˆ + (−9.0 m/s) ˆj + (4.0 m/s 2 ) iˆ + (3.0 m/s 2 ) ˆj (2.0 s ) = (10 m/s) iˆ + (−3.0 m/s) ˆj r r r r r = r0 + v0t + 12 at 2

(b) Express the position vector as a function of time:

r Substitute numerical values and evaluate r :

[

]

r r = (4.0 m) iˆ + (3.0 m) ˆj + (2.0 m/s) iˆ + (−9.0 m/s) ˆj (4.0 s ) 2 + 1 (4.0 m/s 2 ) iˆ + (3.0 m/s 2 ) ˆj (4.0 s ) 2

[

]

= (44.0 m) iˆ + (−9.0 m) ˆj = (44 m) iˆ + (−9.0 m) ˆj

r The magnitude of r is given by r = rx2 + ry2 . Substitute numerical

values and evaluate r(4.0 s):

r (4.0 s) =

(44.0 m )2 + (− 9.0 m )2

= 45 m

Motion in One and Two Dimensions 201 r The direction of r is given by ⎛r ⎞ r θ = tan −1 ⎜⎜ y ⎟⎟ . Because r is in the ⎝ rx ⎠ th 4 quadrant, its direction measured from the +x axis is:

⎛ − 9.0 m ⎞ ⎟⎟ = −11.6° = − 12° ⎝ 44.0 m ⎠

θ = tan −1 ⎜⎜

r 51 •• [SSM] A particle has a position vector given by r = (30t) ˆi + (40t – 5t2) jˆ , where r is in meters and t is in seconds. Find the instantaneous-velocity and instantaneous-acceleration vectors as functions of time t. Picture the Problem The velocity vector is the time-derivative of the position vector and the acceleration vector is the time-derivative of the velocity vector.

r Differentiate r with respect to time:

r r dr d (30t )iˆ + 40t − 5t 2 ˆj v= = dt dt = 30iˆ + (40 − 10t ) ˆj

[

)]

(

r where v has units of m/s if t is in seconds. r r dv d a= = 30iˆ + (40 − 10t ) ˆj dt dt = (− 10 m/s 2 ) ˆj

r Differentiate v with respect to time:

[

]

r 52 •• A particle has a constant acceleration of a = (6.0 m/s2) ˆi + (4 .0 m/s2) jˆ . At time t = 0, the velocity is zero and the position vector is r r 0 = (10 m) ˆi . (a) Find the velocity and position vectors as functions of time t. (b) Find the equation of the particle’s path in the xy plane and sketch the path. Picture the Problem We can use constant-acceleration equations in vector form to find the velocity and position vectors as functions of time t. In (b), we can eliminate t from the equations giving the x and y components of the particle to find an expression for y as a function of x.

r r r r (a) Use v = v0 + at with v0 = 0 to r find v :

r v=

[(6.0 m/s ) iˆ + (4.0 m/s ) ˆj ]t 2

2

r r r r r r Use r = r0 + v0t + 12 at 2 with r0 = (10 m ) iˆ to find r :

r r=

[(10 m) + (3.0 m/s )t ] iˆ + [(2.0 m/s ) t ]ˆj 2

2

2

2

202 Chapter 3 (b) Obtain the x and y components of the path from the vector equation in (a): Eliminate t from these equations and solve for y to obtain:

x = 10 m + (3.0 m/s 2 ) t 2 and y = (2.0 m/s 2 ) t 2 y = 23 x − 203 m

The graph of y = 23 x − 203 m is shown below. Note that the path in the xy plane is a straight line. 20 18 16 14 y, m

12 10 8 6 4 2 0 0

10

20

30

40

x, m

53 •• Starting from rest at a dock, a motor boat on a lake heads north while gaining speed at a constant 3.0 m/s2 for 20 s. The boat then heads west and continues for 10 s at the speed that it had at 20 s. (a) What is the average velocity of the boat during the 30-s trip? (b) What is the average acceleration of the boat during the 30-s trip? (c) What is the displacement of the boat during the 30-s trip? Picture the Problem The displacements of the boat are shown in the following figure. Let the +x direction be to the east and the +y direction be to the north. We need to determine each of the displacements in order to calculate the average velocity of the boat during the 30-s trip.

Motion in One and Two Dimensions 203 (a) The average velocity of the boat is given by:

r r Δr vav = Δt

(1)

The total displacement of the boat is given by:

r r r Δr = ΔrN + ΔrW

To calculate the displacement we first have to find the speed after the first 20 s:

v W = v N, f = a N Δt N

Substitute numerical values and evaluate vW:

v W = 3.0 m/s 2 (20 s ) = 60 m/s

( )

2 = 12 a N (Δt N ) ˆj + v W Δt W − iˆ

(

(2)

)

r Substitute numerical values in equation (2) and evaluate Δr (30 s ) :

(

)

r 2 Δr (30 s ) = 12 3.0 m/s 2 (20 s ) ˆj − (60 m/s )(10 s ) iˆ = (600 m ) ˆj − (600 m ) iˆ

Substitute numerical values in equation (1) to find the boat’s average velocity:

(

r r Δr (600 m ) − iˆ + ˆj v av = = Δt 30 s =

)

(20 m/s)(− iˆ + ˆj )

(b) The average acceleration of the boat is given by:

r r r r Δv v f − v i = a av = Δt Δt

Substitute numerical values and r evaluate aav :

r (− 60 m/s ) iˆ − 0 = aav = 30 s

(c) The displacement of the boat from the dock at the end of the 30-s trip was one of the intermediate results we obtained in Part (a).

r Δr = (600 m ) ˆj + (− 600 m ) iˆ =

(− 2.0 m/s ) iˆ 2

(600 m ) (− iˆ + ˆj )

54 •• Starting from rest at point A, you ride your motorcycle north to point B 75.0 m away, increasing speed at steady rate of 2.00 m/s2. You then gradually turn toward the east along a circular path of radius 50.0 m at constant speed from B to point C until your direction of motion is due east at C. You then continue eastward, slowing at a steady rate of 1.00 m/s2 until you come to rest at point D. (a) What is your average velocity and acceleration for the trip from A to D? (b) What is your displacement during your trip from A to C? (c) What distance did you travel for the entire trip from A to D?

204 Chapter 3 Picture the Problem The following diagram summarizes the information given in the problem statement. Let the +x direction be to the east and the +y direction be to ther north. To find your average velocity you’ll need to find your displacement r Δr and the total time required for you to make this trip. You can express Δr as the r r r sum of your displacements ΔrAB , ΔrBC , and ΔrCD . The total time for your trip is the sum of the times required for each of the segments. Because the acceleration is constant (but different) along each of the segments of your trip, you can use constant-acceleration equations to find each of these quantities.

North

D

50 m

C

B

50 m

75 m

r

Δr

East

A r r Δr vav = Δt

(1)

The total displacement for your trip is the sum of the displacements along the three segments:

r r r r Δr = ΔrAB + ΔrBC + ΔrCD

(2)

The total time for your trip is the sum of the times for the three segments:

Δt = Δt AB + Δt BC + Δt CD

(3)

The displacements of the three segments of the trip are:

r ΔrAB = 0iˆ + (75.0 m ) ˆj , r ΔrBC = (50.0 m )iˆ + (50.0 m ) ˆj , and r ΔrCD = ΔrCD iˆ + 0 ˆj

(a) The average velocity for your trip is given by:

Motion in One and Two Dimensions 205 In order to find ΔxCD , you need to find the time for the C to D segment of the trip. Use a constantacceleration equation to express ΔxCD :

ΔxCD = vC Δt CD + 12 aCD (Δt CD ) (4)

Because vC = vB:

vC = vB = vA + aABΔt AB or, because vA = 0, vC = aAB Δt AB

Use a constant-acceleration equation to relate ΔtAB to ΔyAB:

2

(5)

Δy AB = vA Δt AB + 12 a AB (Δt AB ) or, because vA = 0,

2

Δy AB = 12 a AB (Δt AB ) ⇒ Δt AB = 2

Substitute numerical values and evaluate ΔtAB:

Δt AB =

Substitute for ΔtAB in equation (5) to obtain:

vC = a AB

Substitute numerical values and evaluate vC:

vC = 2(2.00 m/s 2 ) (75.0 m )

The time required for the circular segment BC is given by:

Δt BC =

Substitute numerical values and evaluate ΔtBC:

Δt BC =

The time to travel from C to D is given by:

Substitute numerical values and evaluate ΔtCD :

2(75.0 m) = 8.660 s 2.00 m/s 2 2Δy AB = 2a AB Δy AB a AB

= 17.32 m/s

πr

1 2

vB 1 2

π (50.0 m)

17.32 m/s

= 4.535 s

ΔvCD vD − vC = aCD aCD or, because vD = 0, −v Δt CD = C aCD Δt CD =

Δt CD =

− 17.32 m/s = 17.32 s − 1.00 m/s 2

2Δy AB a AB

206 Chapter 3 Now we can use equation (4) to evaluate ΔtCD :

(

)

ΔxCD = (17.32 m/s )(17.32 s ) + 12 − 1.00 m/s 2 (17.32 s ) = 150 m

The three displacements thatryou need to add in order to get Δr are:

2

r ΔrAB = 0iˆ + (75.0 m ) ˆj , r ΔrBC = (50.0 m ) iˆ + (50.0 m ) ˆj , and r ΔrCD = (150 m ) iˆ + 0 ˆj

Substitute in equation (2) to obtain: r Δr = 0iˆ + (75.0 m ) ˆj + (50.0 m )iˆ + (50.0 m ) ˆj + (150 m )iˆ + 0 ˆj = (200 m )iˆ + (125 m ) ˆj

Substituting in equation (3) for Δt AB , Δt BC , and Δt CD yields:

Δt = 8.660 s + 4.535 s + 17.32 s = 30.51 s

r Use equation (1) to find vav :

r (200 m )iˆ + (125 m ) ˆj v av = 30.51 s = (6.56 m/s) iˆ + (4.10 m/s ) ˆj

Because the final and initial velocities are zero, the average acceleration is zero. (b) Express your displacement from A to C as the sum of the displacements from A to B and from B to C:

r r r ΔrAC = ΔrAB + ΔrBC

From (a) you have:

r ΔrAB = 0iˆ + (75.0 m ) ˆj and r ΔrBC = (50.0 m ) iˆ + (50.0 m ) ˆj

Substitute and simplify to obtain:

r ΔrAC = 0iˆ + (75.0 m ) ˆj + (50.0 m )iˆ + (50.0 m ) ˆj = (50.0 m ) iˆ + (125 m ) ˆj

(c) Express the total distance you traveled as the sum of the distances traveled along the three segments:

d tot = d AB + d BC + d CD where d BC = 12 π r and r is the radius of the circular arc.

Motion in One and Two Dimensions 207 Substituting for d BC yields:

d tot = d AB + 12 π r + d CD

Substitute numerical values and evaluate dtot:

d tot = 75.0 m + 12 π (50.0 m ) + 150 m

= 304 m

Relative Velocity 55 •• A plane flies at an airspeed of 250 km/h. A wind is blowing at 80 km/h toward the direction 60º east of north. (a) In what direction should the plane head in order to fly due north relative to the ground? (b) What is the speed of the plane relative to the ground? Picture the Problem Choose a coordinate system in which north is the +y direction and east is the +x direction. Let θ be the angle between north and the direction of the plane’s heading. The velocity of the plane relative to the r ground, v PG , is the sum of the velocity r of the plane relative to the air, v PA , and the velocity of the air relative to the r r r r ground, vAG . That is, v PG = v PA + v AG . The pilot must head in such a direction r that the east-west component of v PG is zero in order to make the plane fly due north.

(a) From the diagram one can see that: Solving for θ yields:

Substitute numerical values and evaluate θ :

r v AG

N

30°

r v PG

r v PA

θ W S vAG cos 30° = v PA sin θ ⎡ vAG cos 30° ⎤ ⎥ v PA ⎣ ⎦

θ = sin −1 ⎢

⎡ (80 km/h)cos30° ⎤ ⎥⎦ 250 km/h ⎣

θ = sin −1 ⎢

= 16.1° ≈ 16° west of north (b) Because the plane is headed due north, add the north components of r r v PA and v AG to determine the plane’s ground speed: r v PG = (250 km/h)cos16.1° + (80 km/h)sin30° = 280 km/s

E

208 Chapter 3 56 •• A swimmer heads directly across a river, swimming at 1.6 m/s relative to the water. She arrives at a point 40 m downstream from the point directly across the river, which is 80 m wide. (a) What is the speed of the river current? (b) What is the swimmer’s speed relative to the shore? (c) In what direction should the swimmer head in order to arrive at the point directly opposite her starting point?

r Picture the Problem Let vSB represent the velocity of the swimmer relative to r the shore; vSW the velocity of the swimmer relative to the water; and r v WB the velocity of the water relative to the shore; i.e., r r r vSB = vSW + v WB The current of the river causes the swimmer to drift downstream. (a) The triangles shown in the figure are similar right triangles. Set up a proportion between their sides and solve for the speed of the water relative to the bank:

vWB 40 m = vSW 80 m and v WB = 12 (1.6 m/s ) = 0.80 m/s

(b) Use the Pythagorean Theorem to solve for the swimmer’s speed relative to the shore:

2 2 vSB = vSW + v WB

=

(1.6 m/s)2 + (0.80 m/s)2

= 1.8 m/s r vWB

(c) The swimmer should head in a direction such that the upstream component of her velocity relative to r the shore ( vSB ) is equal to the speed of the water relative to the shore r ( v WB ):

r vSW

r vSB θ

Referring to the diagram, relate r r sinθ to v WB and vSB :

sin θ =

Substitute numerical values and evaluate θ:

θ = sin −1 ⎜⎜

⎛v ⎞ v WB ⇒ θ = sin −1 ⎜⎜ WB ⎟⎟ vSB ⎝ vSB ⎠ ⎛ 0.80 m/s ⎞ ⎟⎟ = 26° ⎝ 1.8 m/s ⎠

Motion in One and Two Dimensions 209 57 •• [SSM] A small plane departs from point A heading for an airport 520 km due north at point B. The airspeed of the plane is 240 km/h and there is a steady wind of 50 km/h blowing directly toward the southeast. Determine the proper heading for the plane and the time of flight. Picture the Problem Let the velocity of the plane relative to the ground be r represented by v PG ; the velocity of the plane relative to the air by r v PA , and the velocity of the air r relative to the ground by v AG . Then r r r v PG = v PA + v AG (1) Choose a coordinate system with the origin at point A, the +x direction to the east, and the +y direction to the north. θ is the angle between north and the direction of the plane’s heading. The pilot must head so that r the east-west component of v PG is zero in order to make the plane fly due north.

Use the diagram to express the condition relating the eastward r component of v AG and the westward r component of v PA . This must be satisfied if the plane is to stay on its northerly course. [Note: this is equivalent to equating the xcomponents of equation (1).] Now solve for θ to obtain:

Substitute numerical values and evaluate θ :

vAG cos 45° = vPA sin θ

⎡ vAG cos 45° ⎤ ⎥ vPA ⎣ ⎦

θ = sin −1 ⎢

⎡ (50 km/h ) cos45° ⎤ ⎥⎦ = 8.47° 240 km/h ⎣

θ = sin −1 ⎢ = 8.5°

r Add the north components of v PA r and v AG to find the velocity of the plane relative to the ground:

vPG + v AG sin 45° = vPA cos 8.47°

210 Chapter 3 Solving for vPG yields:

vPG = vPA cos 8.47° − v AG sin 45°

Substitute numerical values and evaluate vPG to obtain: vPG = (240 km/h) cos 8.47° − (50 km/h) sin 45° = 202.0 km/h The time of flight is given by:

Substitute numerical values and evaluate tflight:

t flight =

distance travelled vPG

t flight =

520 km = 2.57 h 202.0 km/h

58 •• Two boat landings are 2.0 km apart on the same bank of a stream that flows at 1.4 km/h. A motorboat makes the round trip between the two landings in 50 min. What is the speed of the boat relative to the water?

r Picture the Problem Let v BS be the velocity of the boat relative to the shore; r v BW be the velocity of the boat relative r to the water; and v WS represent the velocity of the water relative to the shore. Independently of whether the boat is going upstream or downstream: r r r v BS = v BW + v WS Going upstream, the speed of the boat relative to the shore is reduced by the speed of the water relative to the shore. Going downstream, the speed of the boat relative to the shore is increased by the same amount. For the upstream leg of the trip:

vBS = vBW − v WS

For the downstream leg of the trip:

vBS = vBW + v WS

Express the total time for the trip in terms of the times for its upstream and downstream legs:

t total = t upstream + t downstream =

L L + vBW − v WS vBW + vWS

Motion in One and Two Dimensions 211 Multiply both sides of the equation by (v BW − v WS )(v BW + v WS ) (the product of the denominators) and rearrange the terms to obtain:

2 vBW −

2L 2 vBW − vWS =0 t total

Substituting numerical values gives: 2 vBW −

or

2(2.0 km) vBW − (1.4 km/h) 2 = 0 5 6 h

2 vBW − (4.80 km/h )v BW − 1.96 (km ) /h 2 = 0 2

Use the quadratic formula or your graphing calculator to solve the quadratic equation for vBW (Note that only the positive root is physically meaningful.):

vBW = 5.2 km/h

59 •• During a radio-controlled model-airplane competition, each plane must fly from the center of a 1.0-km-radius circle to any point on the circle and back to the center. The winner is the plane that has the shortest round-trip time. The contestants are free to fly their planes along any route as long as the plane begins at the center, travels to the circle, and then returns to the center. On the day of the race, a steady wind blows out of the north at 5.0 m/s. Your plane can maintain an air speed of 15 m/s. Should you fly your plane upwind on the first leg and downwind on the trip back, or across the wind flying east and then west? Optimize your chances by calculating the round-trip time for both routes using your knowledge of vectors and relative velocities. With this pre-race calculation, you can determine the best route and have a major advantage over the competition!

r Picture the Problem Let v pg be the velocity of the plane relative to the ground; r r v ag be the velocity of the air relative to the ground; and v pa the velocity of the r r r plane relative to the air. Then, v pg = v pa + v ag . The wind will affect the flight times differently along these two paths.

r vpa

r vag W

r vpg

N

θ

E S

212 Chapter 3 The speed of the plane, relative to the ground, on its eastbound leg is equal to its speed on its westbound leg. Using the diagram, express the speed of the plane relative to the ground for both directions:

vpg = vpa2 − vag2

Substitute numerical values and evaluate vpg:

vpg =

(15 m/s)2 − (5.0 m/s)2

= 14.1m/s

Express the time for the east-west roundtrip in terms of the distances and velocities for the two legs: t round trip,EW = t eastbound + t westbound =

radius of the circle radius of the circle + vpg,eastbound vpg,westbound

Substitute numerical values and evaluate t roundtrip,EW :

t round trip,EW =

2 ×1.0 km = 142 s 14.1 m/s

Use the distances and speeds for the two legs to express the time for the northsouth roundtrip: t round trip,NS = t northbound + t southbound =

radius of the circle radius of the circle + vpg,northbound v pg,southbound

Substitute numerical values and evaluate t roundtrip,NS : t round trip, NS =

1.0 k m 1.0 k m + (15 m/s) − (5.0 m/s) (15 m/s) + (5.0 m/s)

= 150 s

Because t roundtrip,EW < t round trip, NS , you should fly your plane across the wind. 60 • You are piloting a small plane that can maintain an air speed of 150 kt (knots, or nautical miles per hour) and you want to fly due north (azimuth = 000º) relative to the ground. (a) If a wind of 30 kt is blowing from the east (azimuth = 090º), calculate the heading (azimuth) you must ask your co-pilot to maintain. (b) At that heading, what will be your groundspeed?

Motion in One and Two Dimensions 213 Picture the Problem This is a relative velocity problem. The given quantities are the direction of the velocity of the plane relative to the ground and the velocity (magnitude and direction) of the air relative to the ground. Asked for is the direction of the velocity of the air relative to the ground. Using r r r v PG = v PA + v AG , draw a vector addition diagram and solve for the unknown quantities.

(a) Referring to the diagram, relate the heading you must take to the wind speed and the speed of your plane relative to the air: Substitute numerical values and evaluate θ :

⎛ vAG ⎞ ⎟⎟ ⎝ vPA ⎠

θ = sin −1 ⎜⎜

⎛ 30 kts ⎞ ⎟⎟ = 11.5° ⎝ 150 kts ⎠

θ = sin −1 ⎜⎜

Because this is also the angle of the plane's heading clockwise from north, it is also its azimuth or the required true heading:

Az = (011.5°) = (012°)

(b) Referring to the diagram, relate your heading to your plane’s speeds relative to the ground and relative to the air:

cosθ =

Solve for the speed of your plane relative to the ground to obtain:

vPG = vPA cosθ

Substitute numerical values and evaluate vPG:

vPG = (150 kt )cos11.5° = 147 kt

vPG vPA

= 1.5 × 10 2 kt

61 •• [SSM] Car A is traveling east at 20 m/s toward an intersection. As car A crosses the intersection, car B starts from rest 40 m north of the intersection and moves south, steadily gaining speed at 2.0 m/s2. Six seconds after A crosses the intersection find (a) the position of B relative to A, (b) the velocity of B relative to A, and (c) the acceleration of B relative to A. (Hint: Let the unit vectors iˆ and ˆj be toward the east and north, respectively, and express your answers using iˆ and ˆj .)

214 Chapter 3 y, m

Picture the Problem The position of B relative to A is the vector from A to B; that is, r r r rBA = rB − rA

40 B

The velocity of B relative to A is r r v BA = drBA dt and the acceleration of B relative to A is r r aBA = dv BA dt

r rBA

r rB

Choose a coordinate system with the origin at the intersection, the +x direction to the east, and the +y direction to the north.

0

[

r r (a) Find rB and rA :

x, m

r rA'

A

) ]

(

r rB = 40 m − 12 2.0 m/s 2 t 2 ˆj and r rA = [(20 m/s )t ]iˆ

r r r r Use rBA = rB − rA to find rBA :

[

) ]

(

r rBA = [(− 20 m/s ) t ]iˆ + 40 m − 12 2.0 m/s 2 t 2 ˆj r Evaluate rBA at t = 6.0 s:

[

(

]

)

r 2 rBA (6.0 s ) = [(− 20 m/s )(6.0 s )] iˆ + 40 m − 12 2.0 m/s 2 (6.0 s ) ˆj = (−1.2 ×10 2 m) iˆ + (4.0 m) ˆj r r (b) Find v BA = drBA dt :

r r drBA d {(− 20 m/s)t}iˆ + 40 m − 12 2.0 m/s 2 t 2 ˆj v BA = = dt dt = ( −20 m/s) iˆ + ( −2.0 m/s 2 ) t ˆj

[

{

(

) }]

r Evaluate vBA at t = 6.0 s:

r v BA (6.0 s ) = ( −20 m/s) iˆ + ( −2.0 m/s 2 ) (6.0 s ) ˆj =

(− 20 m/s) iˆ − (12 m/s) ˆj

Motion in One and Two Dimensions 215 r r (c) Find aBA = dv BA dt :

[

r d aBA = (−20 m/s) iˆ + ( −2.0 m/s 2 ) t ˆj dt = − 2.0 m/s 2 ˆj r Note that aBA is independent of time.

(

]

)

62 •• While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at a constant speed of 2.5 m/s relative to the surface of the walkway, you decide to try to determine the speed of the walkway itself. You watch the child run on the entire 21-m walkway in one direction, immediately turn around, and run back to his starting point. The entire trip taking a total elapsed time of 22 s. Given this information, what is the speed of the moving walkway relative to the airport terminal? Picture the Problem We don’t need to know which direction the child ran first, and which he ran second. Because we have the total length of the walkway, the elapsed time for the round trip journey, and the child’s walking speed relative to the walkway, we are given sufficient information to determine the moving walkway’s speed. The distance covered in the airport is 21 m, and this is covered in a total time of 22 s. When the child walks in the direction of the walkway, his r velocity in the airport is the sum of his walking velocity, vchild , and the walkway r velocity, v ww . On the return trip, the velocity in the airport is the difference of these two velocities.

Express the total time for the child’s run in terms of his running times with and against the moving walkway: In terms of the length L of the walkway and the speeds, relative to the airport, of the child walking with and against the moving walkway: The speeds of the child relative to the airport, then, in each case, are:

Substitute for vwith and vagainst in equation (1) to obtain: Solving this equation for vww yields:

Δt tot = Δt with + Δt against

Δt tot =

L v with

+

L vagainst

(1)

r r v with = v child + v ww = vchild + v ww

and r r vagainst = v child − v ww = vchild − v ww Δt tot =

L L + vchild + v ww vchild + v ww

2 v ww = vchild −

2vchild L Δt tot

216 Chapter 3 Substitute numerical values and evaluate vww : v ww = (2.5 m/s) 2 −

2(2.5 m/s) (21 m) = 1.2 m/s 22 s

63 •• [SSM] Ben and Jack are shopping in a department store. Ben leaves Jack at the bottom of the escalator and walks east at a speed of 2.4 m/s. Jack then stands on the escalator, which is inclined at an angle of 37° above the horizontal and travels eastward and upward at a speed of 2.0 m/s. (a) What is the velocity of Ben relative to Jack? (b) At what speed should Jack walk up the escalator so that he is always directly above Ben (until he reaches the top)?

r Picture the Problem The velocity of Ben relative to Jack ( v BJ ) is the difference r r between the vectors v B and vescalator . Choose the coordinate system shown and express these vectors using the unit vectors iˆ and ˆj . r vB

θ y

r vescalator

r v BJ

37°

x

(a) The velocity of Ben relative to Jack is given by:

r r r v BJ = v B − vescalator

The velocities of the floor walker and the escalator are:

r v B = (2.4 m/s) iˆ and r vescalator = (2.0 m/s)cos37°iˆ + (2.0 m/s)sin37° ˆj

r r Substitute for v B and vescalator and simplify to obtain:

[

r v BJ = (2.4 m/s) iˆ − (2.0 m/s)cos37°iˆ + (2.0 m/s)sin37° ˆj = (0.803 m/s) iˆ − (1.20 m/s) ˆj = (0.80 m/s) iˆ − (1.2 m/s) ˆj

]

Motion in One and Two Dimensions 217

r The magnitude and direction of v BJ are given by:

r 2 2 v BJ = vBJ, x + vBJ, y and ⎛ vBJ, y ⎞ ⎟ ⎟ ⎝ vBJ, x ⎠

θ = tan −1 ⎜⎜ Substitute numerical values and r evaluate v BJ and θ :

r v BJ = (0.803 m/s) 2 + (−1.20 m/s) 2 = 1.4 m/s and ⎛ − 1.20 m/s ⎞ ⎟ = −56.2° ⎝ 0.803 m/s ⎠

θ = tan −1 ⎜

= 56° below the horizontal

64 ••• A juggler traveling in a train on level track throws a ball straight up, relative to the train, with a speed of 4.90 m/s. The train has a velocity of 20.0 m/s due east. As observed by the juggler, (a) what is the ball’s total time of flight, and (b) what is the displacement of the ball during its rise? According to a friend standing on the ground next to the tracks, (c) what is the ball’s initial speed, (d) what is the angle of the launch, and (e) what is the displacement of the ball during its rise? Picture the Problem The vector diagram shows the relationship between the velocity of the ball relative r to the train v BT , the velocity of the train r relative to the ground vTG , and the velocity of the ball relative to the r ground v BG . Choose a coordinate system in which the +x direction is to the east and the +y direction is to the north.

N

r v BG

θ

(a) The ball’s time-of-flight is twice the time it takes it to reach its maximum height:

t flight = 2t max height

Use a constant-acceleration equation to relate the ball’s speed to its initial speed and the time into its flight:

v y = v0 y + a y t

r v BT

r v TG

E

(1)

218 Chapter 3 At its maximum height, vy = 0:

0 = v0 y + a y t max height or, because ay = −g, 0 = v0 y − gt max height

Solving for t max height yields:

t max height =

Substitute for t max height in equation (1) to obtain:

t flight =

Substitute numerical values and evaluate tflight :

t flight =

v0 y g

2v 0 y g

2(4.90 m/s ) = 1.00 s 9.81 m/s 2

(b) The displacement of the ball during its rise is given by:

r ΔyBT = ΔyBT ˆj

Use a constant-acceleration equation to relate the displacement of the ball during its ascent to its initial and final speeds:

vx2 = v02x − 2 gΔyBT or, because vx = 0,

Substituting for ΔyBT gives:

r v02x ˆ ΔyBT = j 2g

Substitute numerical values and r evaluate ΔyBT :

r (4.90 m/s )2 ˆj = (1.22 m ) ˆj ΔyBT = 2 9.81 m/s 2

r r r (c) v BG is the sum of v BT and v TG :

r r r v BG = v BT + vTG

r r Express v BT and v TG in terms of the unit vectors iˆ and ˆj :

r v BT = 0 iˆ + (4.90 m/s) ˆj and r vTG = (20.0 m/s ) iˆ + 0 ˆj

r r Substitute for v BT and v TG in equation (1) to obtain:

r v BG = (20.0 m/s ) iˆ + (4.90 m/s) ˆj

r The magnitude of v BG is given by:

0 = v02x − 2 gΔyBT ⇒ ΔyBT =

(

vBG =

v02x 2g

)

(20.0 m/s)2 + (4.90 m/s)2

= 20.6 m/s

(3)

Motion in One and Two Dimensions 219 (d) Referring to the vector diagram, express the angle of the launch θ in terms of vBT and vTG:

θ = tan −1 ⎜⎜

Substitute numerical values and evaluate θ :

θ = tan −1 ⎜

(e) When the ball is at its peak it is 20.0 m east and 1.22 m above the friend observing from the ground. Hence its displacement is:

⎛ vBT ⎝ vTG

⎞ ⎟⎟ ⎠

⎛ 4.90 m/s ⎞ ⎟ = 13.8° ⎝ 20.0 m/s ⎠

r d=

(20.0 m ) iˆ + (1.22 m ) ˆj

Circular Motion and Centripetal Acceleration 65 • What is the magnitude of the acceleration of the tip of the minute hand of the clock in Problem 38? Express it as a fraction of the magnitude of free-fall acceleration g. Picture the Problem We can use the definition of centripetal acceleration to express ac in terms of the speed of the tip of the minute hand. We can find the tangential speed of the tip of the minute hand by using the distance it travels each revolution and the time it takes to complete each revolution.

Express the magnitude of the acceleration of the tip of the minute hand of the clock as a function of the length of the hand and the speed of its tip: Use the distance the minute hand travels every hour to express its speed:

ac =

v=

v2 R

(1)

2πR T

Substituting for v in equation (1) yields:

4π 2 R ac = T2

Substitute numerical values and evaluate ac:

ac =

4π 2 (0.50 m ) = 1.52 × 10 −6 m/s 2 2 (3600 s )

= 1.5 × 10 −6 m/s 2 Express the ratio of ac to g:

ac 1.52 × 10 −6 m/s 2 = = 1.55 × 10 −7 2 g 9.81m/s

220 Chapter 3 66 • You are designing a centrifuge to spins at a rate of 15,000 rev/min. (a) Calculate the maximum centripetal acceleration that a test-tube sample held in the centrifuge arm 15 cm from the rotation axis must withstand. (b) It takes 1 min, 15 s for the centrifuge to spin up to its maximum rate of revolution from rest. Calculate the magnitude of the tangential acceleration of the centrifuge while it is spinning up, assuming that the tangential acceleration is constant. Picture the Problem The following diagram shows the centripetal and tangential accelerations experienced by the test tube. The tangential acceleration will be zero when the centrifuge reaches its maximum speed. The centripetal acceleration increases as the tangential speed of the centrifuge increases. We can use the definition of centripetal acceleration to express ac in terms of the speed of the test tube. We can find the tangential speed of the test tube by using the distance it travels each revolution and the time it takes to complete each revolution. The tangential acceleration can be found from the change in the tangential speed as the centrifuge is spinning up.

(a) Express the maximum centripetal acceleration of the centrifuge arm as a function of the length of its arm and the speed of the test tube:

ac, max

v2 = R

(1)

2πR T

Use the distance the test tube travels every revolution to express its speed:

v=

Substituting for v in equation (1) yields:

ac, max =

Substitute numerical values and evaluate ac:

ac, max =

4π 2 R T2 4π 2 (0.15 m )

⎛ 1 min 60 s ⎞ ⎜⎜ ⎟⎟ × ⎝ 15000 rev min ⎠

= 3.7 × 10 5 m/s 2

2

Motion in One and Two Dimensions 221 (b) Express the tangential acceleration in terms of the difference between the final and initial tangential speeds:

2πR −0 vf − vi 2πR T at = = = Δt Δt TΔt

Substitute numerical values and evaluate aT:

at =

2π (0.15 m ) ⎛ 1min 60 s ⎞ ⎜⎜ ⎟⎟ (75 s ) × ⎝ 15000 rev min ⎠

= 3.1m/s 2 67 ••• [SSM] Earth rotates on its axis once every 24 hours, so that objects on its surface that are stationary with respect to the surface execute uniform circular motion about the axis with a period of 24 hours. Consider only the effect of this rotation on the person on the surface. (Ignore Earth’s orbital motion about the Sun.) (a) What is the speed and what is the magnitude of the acceleration of a person standing on the equator? (Express the magnitude of this acceleration as a percentage of g.) (b) What is the direction of the acceleration vector? (c) What is the speed and what is the magnitude of the centripetal acceleration of a person standing on the surface at 35°N latitude? (d) What is the angle between the direction of the acceleration of the person at 35°N latitude and the direction of the acceleration of the person at the equator if both persons are at the same longitude? Picture the Problem The radius of Earth is 6370 km. Thus at the equator, a person undergoes circular motion with radius equal to Earth’s radius, and a period of 24 h = 86400 s. At 35o N latitude, the person undergoes circular motion having radius r cos35o = 5220 km, and the same period.

The centripetal acceleration experienced by a person traveling with a speed v in a circular path of radius r is given by: The speed of the person is the distance the person travels in one revolution divided by the elapsed time (the period T): Substitute for v in equation (1) to obtain:

(a) Substitute numerical values and evaluate v for the person at the equator:

a=

v2 r

v=

2π r T

(1)

2

⎛ 2π r ⎞ ⎜ ⎟ 4π 2 r T ⎠ ⎝ a= = r T2 v=

2π (6370 km) = 463 m/s 86400 s

222 Chapter 3 Substitute numerical values and evaluate a for the person at the equator:

4π 2 (6370 km) = 3.369 × 10 −2 m/s 2 a= 2 (86400 s )

The ratio of a to g is:

a 3.369 × 10 −2 m/s 2 = = 0.343% g 9.81 m/s 2

= 3.37 cm/s2

(b) The acceleration vector points directly at the center of Earth. (c) Substitute numerical values and evaluate v for the person at 35oN latitude: Substitute numerical values and evaluate v for the person at 35oN latitude:

v=

2π (5220 km) = 380 m/s 86400 s

4π 2 (5220 km) a= = 2.76 cm/s 2 2 (86400 s )

(d) The plane of the person’s path is parallel to the plane of the equator – the acceleration vector is in the plane – so that it is perpendicular to Earth’s axis, pointing at the center of the person’s revolution, rather than the center of Earth. 68 •• Determine the acceleration of the Moon toward Earth, using values for its mean distance and orbital period from the Terrestrial and Astronomical Data table in this book. Assume a circular orbit. Express the acceleration as a fraction of the magnitude of free-fall acceleration g. Picture the Problem We can relate the acceleration of the Moon toward Earth to its orbital speed and distance from Earth. Its orbital speed can be expressed in terms of its distance from Earth and its orbital period. From the Terrestrial and Astronomical Data table, we find that the sidereal period of the Moon is 27.3 d and that its mean distance from Earth is 3.84×108 m.

Express the centripetal acceleration of the Moon:

ac =

Express the orbital speed of the Moon:

v=

Substituting for v in equation (1) yields:

ac =

v2 r

2πr T 4π 2 r T2

(1)

Motion in One and Two Dimensions 223 Substitute numerical values and evaluate ac:

(

4π 2 3.84 × 108 m

ac =

)

24 h 3600 s ⎞ ⎛ × ⎜ 27.3 d × ⎟ d h ⎠ ⎝ = 2.72 × 10 −3 m/s 2

2

= 2.78 × 10 − 4 g

ac radius of Earth = (ac is just the g distance from Earth to Moon acceleration due to Earth’s gravity evaluated at the moon’s position). This is Newton’s famous ″falling apple″ observation.

Remarks: Note that

69 •• (a) What are the period and speed of the motion of a person on a carousel if the person has an acceleration magnitude of 0.80 m/s2 when she is standing 4.0 m from the axis? (b) What are her acceleration magnitude and speed if she then moves in to a distance of 2.0 m from the carousel center and the carousel keeps rotating with the same period? Picture the Problem The person riding on this carousel experiences a centripetal acceleration due to the fact that her velocity is continuously changing. Use the expression for centripetal acceleration to relate her speed to her centripetal acceleration and the relationship between distance, speed, and time to find the period of her motion.

In general, the acceleration and period of any object moving in a circular path at constant speed are given by:

v2 r

(1)

2π r v

(2)

Solving equation (1) for v yields:

v = ac r

(3)

Substituting for v in equation (2) yields:

T=

2π r r = 2π ac ac r

v=

(0.80 m/s ) (4.0 m ) = 1.79 m/s

(a) Substitute numerical values in equation (3) and evaluate v for a person standing 4.0 m from the axis: Substitute numerical values in equation (4) and evaluate T for a person standing 4.0 m from the axis:

ac = and T=

(4)

2

= 1.8 m/s

T = 2π

4.0 m = 14.05 s = 14 s 0.80 m/s 2

224 Chapter 3 (b) Solve equation (2) for v to obtain:

v=

2π r T

For a person standing 2.0 m from the axis:

v=

2π (2.0 m ) = 0.894 m/s = 0.89 m/s 14.05 s

From equation (1) we have, for her acceleration:

ac =

(0.894 m/s)2 2.0 m

= 0.40 m/s 2

70 ••• Pulsars are neutron stars that emit X rays and other radiation in such a way that we on Earth receive pulses of radiation from the pulsars at regular intervals equal to the period that they rotate. Some of these pulsars rotate with periods as short as 1 ms! The Crab Pulsar, located inside the Crab Nebula in the constellation Orion, has a period currently of length 33.085 ms. It is estimated to have an equatorial radius of 15 km, which is an average radius for a neutron star. (a) What is the value of the centripetal acceleration of an object on the surface and at the equator of the pulsar? (b) Many pulsars are observed to have periods that lengthen slightly with time, a phenomenon called "spin down." The rate of slowing of the Crab Pulsar is 3.5 × 10−13 s per second, which implies that if this rate remains constant, the Crab Pulsar will stop spinning in 9.5 × 1010 s (about 3000 years). What is the tangential acceleration of an object on the equator of this neutron star? Picture the Problem We need to know the speed of a point on the equator where the object sits. The circular path that the object at the equator undergoes has a radius equal to that of the star, and therefore is simply the equatorial circumference – and is traversed in a time interval of one period, 33.085 ms. In (b) we need only consider the change in speed of an object on the surface over the course of this time period.

(a) The centripetal acceleration of the object is given by: The speed of the object is equal to related to the radius of its path and the period of its motion: Substituting for v in equation (1) yields:

Substitute numerical values and evaluate ac:

ac = v=

v2 r

(1)

2π r T 2

⎛ 2π r ⎞ ⎜ ⎟ 4π 2 r T ⎠ ⎝ ac = = r T2 ac =

4π 2 (15 km ) = 5.4 × 10 8 m/s 2 2 (33.085 ms)

Motion in One and Two Dimensions 225 (b) The tangential acceleration of an object on the equator of the neutron star is given by: Because v r,final = 0 :

We can obtain the initial speed of the object from equation (1):

at =

at =

Δt spin down

at =

=

vt, final − vt, initial Δt spin down

− v t, initial Δt spin down

vt, initial =

Substitute for v r,final = 0 to obtain:

Substitute numerical values and evaluate a t :

Δv t

−

2π r T

2π r T =−

Δt spin down

at = −

2π r Δt spin downT

2π (15 km ) (9.5 ×10 s)(33.085 ×10 −3 s) 10

= − 3.0 × 10 −5 m/s 2 71 ••• Human blood contains plasma, platelets and blood cells. To separate the plasma from other components, centrifugation is used. Effective centrifugation requires subjecting blood to an acceleration of 2000g or more. In this situation, assume that blood is contained in test tubes that are 15 cm long and are full of blood. These tubes ride in the centrifuge tilted at an angle of 45.0° above the horizontal (See Figure 3-34.) (a) What is the distance of a sample of blood from the rotation axis of a centrifuge rotating at 3500 rpm, if it has an acceleration of 2000g? (b) If the blood at the center of the tubes revolves around the rotation axis at the radius calculated in Part (a), calculate the centripetal accelerations experienced by the blood at each end of the test tube. Express all accelerations as multiples of g. Picture the Problem The equations that describe the centripetal acceleration and speed of objects moving in circular paths at constant speed can be used to find the distance of the sample of blood from the rotation axis as well as the accelerations experienced by the blood at each end of the test tube under the conditions described in the problem statement.

(a) The centripetal acceleration of the blood sample is given by:

v2 ac = r

The speed of the blood’s revolution is given by:

2π r = 2π rf T where f is the frequency of revolution. v=

226 Chapter 3 Substitute for v in the expression for ac to obtain:

ac

Solving for r yields:

2 ( 2π rf ) =

r=

r

= 4π 2 f 2 r

(1)

ac

4π 2 f 2 or, because ac = g, 2000 g r= 4π 2 f 2 Substitute numerical values and evaluate r:

r=

2000(9.81 m/s 2 ) rev 1 min ⎞ ⎛ 4π ⎜ 3500 × ⎟ min 60 s ⎠ ⎝

2

= 0.146 m

2

= 15 cm

(b) The center of the tube undergoes uniform circular motion with a radius of 15cm. Because the tubes are tilted at a 45 degree angle, the top surface of the blood is undergoing uniform circular motion with a radius of:

rtop = 14.6 cm − (7.5 cm )cos 45°

The bottom surface is moving in a circle of radius:

rbottom = 14.6 cm + (7.5 cm )cos 45°

The range of accelerations can be found from equation (1):

ac = 4π 2 f 2 r

The minimum acceleration corresponds to rtop:

amin = 4π 2 f 2 rtop

= 9.30 cm

= 19.90 cm

Substitute numerical values and evaluate amin: 2

a min

rev 1 min ⎞ ⎛ 4 2 = 4π ⎜ 3500 × ⎟ (9.30 cm ) = 1.249 × 10 m/s ≈ 1300 g min 60 s ⎠ ⎝ 2

The maximum acceleration corresponds to rbottom:

amax = 4π 2 f 2 rbottom

Motion in One and Two Dimensions 227 Substitute numerical values and evaluate amax: 2

a max

rev 1 min ⎞ ⎛ 4 2 = 4π ⎜ 3500 × ⎟ (19.90 cm ) = 2.673 × 10 m/s ≈ 2700 g min 60 s ⎠ ⎝ 2

The range of accelerations is 1300g to 2700g.

Projectile Motion and Projectile Range 72 • While trying out for the position of pitcher on your high school baseball team, you throw a fastball at 87 mi/h toward home plate, which is 18.4 m away. How far does the ball drop due to effects of gravity by the time it reaches home plate? (Ignore any effects due to air resistance.) Picture the Problem Neglecting air resistance, the accelerations of the ball are constant and the horizontal and vertical motions of the ball are independent of each other. We can use the horizontal motion to determine the time-of-flight and then use this information to determine the distance the ball drops. Choose a coordinate system in which the origin is at the point of release of the ball, downward is the positive y direction, and the horizontal direction is the positive x direction.

Express the vertical displacement of the ball:

Use vx = Δx/Δt to express the time of flight:

Δy = v0 y Δt + 12 a y (Δt )

2

or, because v0y = 0 and ay = g, 2 Δy = 12 g (Δt ) (1) Δt =

Δx vx

Substitute for Δt in equation (1) to obtain:

2 ⎛ Δx ⎞ g (Δx ) Δy = g ⎜⎜ ⎟⎟ = 2 2(v x ) ⎝ vx ⎠

Substitute numerical values and evaluate Δy:

m⎞ ⎛ 2 ⎜ 9.81 2 ⎟(18.4 m ) s ⎠ = 1.1 m Δy = ⎝ 2 m⎞ ⎛ ⎜ mi 0.4470 ⎟ s ⎟ ⋅ 2 ⎜ 87 mi h ⎜ ⎟ 1 ⎜ ⎟ h ⎝ ⎠

2

1 2

228 Chapter 3 73 • A projectile is launched with speed v0 at an angle of θ0 above the horizontal. Find an expression for the maximum height it reaches above its starting point in terms of v0, θ0, and g. (Ignore any effects due to air resistance.) Picture the Problem In the absence of air resistance, the maximum height achieved by a projectile depends on the vertical component of its initial velocity.

Use a constant-acceleration equation to relate the displacement of a projectile to its initial and final speeds and its acceleration: Because vy = 0 and ay = −g:

The vertical component of the projectile’s initial velocity is: Setting Δy = h and substituting for v0y yields:

v 2y = v02 y + 2 a yΔ y ⇒ Δy =

Δy =

v y2 − v02y 2a y

v02y 2g

v0 y = v0 sin θ 0

h=

(v0 sin θ 0 )2 2g

74 •• A cannonball is fired with initial speed v0 at an angle 30º above the horizontal from a height of 40 m above the ground. The projectile strikes the ground with a speed of 1.2v0. Find v0. (Ignore any effects due to air resistance.) Picture the Problem Choose the coordinate system shown to the right. Because, in the absence of air resistance, the horizontal and vertical speeds are independent of each other, we can use constant-acceleration equations to relate the impact speed of the projectile to its components.

y, m

h

r v0

θ0

x, m

r v = 1.2v0

The horizontal and vertical velocity components are:

v0 x = v x = v0 cosθ 0 and v0 y = v0 sin θ 0

Motion in One and Two Dimensions 229 Using a constant-acceleration equation, relate the vertical component of the velocity to the vertical displacement of the projectile:

v y2 = v02y + 2a y Δy

or, because ay = −g and Δy = −h, 2 v y2 = (v0 sin θ 0 ) + 2 gh

Express the relationship between the magnitude of a velocity vector and its components, substitute for the components, and simplify to obtain:

v 2 = v x2 + v y2 = (v0 cosθ 0 ) + v y2

Substituting for v and solving for v0 gives:

(1.2v0 )2 = v02 + 2 gh ⇒ v0 =

Substitute numerical values and evaluate v0:

v0 =

2

= v02 (sin 2 θ 0 + cos 2 θ 0 ) + 2 gh

= v02 + 2 gh 2 gh 0.44

2(9.81 m/s 2 )(40 m ) = 42 m/s 0.44

Remarks: Note that v is independent of θ0. This will be more obvious once conservation of energy has been studied. 75 •• In Figure 3.35, what is the minimum initial speed of the dart if it is to hit the monkey before the monkey hits the ground, which is 11.2 m below the initial position of the monkey, if x is 50 m and h = 10 m? (Ignore any effects due to air resistance.) Picture the Problem Example 3-12 shows that the dart will hit the monkey unless the dart hits the ground before reaching the monkey’s line of fall. What initial speed does the dart need in order to just reach the monkey’s line of fall? First, we will calculate the fall time of the monkey, and then we will calculate the horizontal component of the dart’s velocity.

Relate the horizontal speed of the dart to its launch angle and initial velocity v0:

v x = v0 cosθ ⇒ v0 =

Use the definition of vx and the fact that, in the absence of air resistance, it is constant to obtain:

vx =

Δx Δt

Substituting in the expression for v0 yields:

v0 =

Δx (cosθ )Δt

vx cosθ

(1)

230 Chapter 3 Using a constant-acceleration equation, relate the monkey’s fall distance to the fall time:

Δh = 12 g (Δt ) ⇒ Δt =

Substituting for Δt in equation (1) yields:

v0 =

Let θ be the angle the barrel of the dart gun makes with the horizontal. Then:

θ = tan −1 ⎜⎜

Substitute numerical values in equation (2) and evaluate v0:

v0 =

2

Δx cosθ

2Δh g

g 2Δh

(2)

⎛ 10 m ⎞ ⎟⎟ = 11.3° ⎝ 50 m ⎠

50 m 9.81 m/s 2 = 34 m/s cos11.3° 2(11.2 m )

76 •• A projectile is launched from ground level with an initial speed of 53 m/s. Find the launch angle (the angle the initial velocity vector is above the horizontal) such that the maximum height of the projectile is equal to its horizontal range. (Ignore any effects due to air resistance.) Picture the Problem Choose the coordinate system shown in the figure to the right. In the absence of air resistance, the projectile experiences constant acceleration in both the x and y directions. We can use the constant-acceleration equations to express the x and y coordinates of the projectile along its trajectory as functions of time. The elimination of the parameter t will yield an expression for y as a function of x that we can evaluate at (R, 0) and (R/2, h). Solving these equations simultaneously will yield an expression for θ.

y

(12 R, h) r v0 θ0

(x, y )

x

(R,0)

Express the position coordinates of the projectile along its flight path in terms of the parameter t:

x = (v0 cosθ )t and y = (v0 sin θ )t − 12 gt 2

Eliminate t from these equations to obtain:

y = (tan θ )x −

g x2 2 2v cos θ 2 0

(1)

Motion in One and Two Dimensions 231 Evaluate equation (1) at (R, 0) to obtain:

2v02 sin θ cos θ R= g

Evaluate equation (1) at (R/2, h) to obtain:

h=

(v0 sin θ )2 2g

Equating R and h yields:

2v02 sin θ cosθ (v0 sin θ ) = g 2g

Solving for θ gives:

θ = tan −1 (4 ) = 76°

2

Note that this result is independent of v0. 77 •• [SSM] A ball launched from ground level lands 2.44 s later 40.0-m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.) Picture the Problem In the absence of air resistance, the motion of the ball is uniformly accelerated and its horizontal and vertical motions are independent of each other. Choose the coordinate system shown in the figure to the right and use constant-acceleration equations to relate the x and y components of the ball’s initial velocity.

Express θ0 in terms of v0x and v0y:

y, m

r v0

θ0 40

⎛ v0 y ⎝ v0 x

θ 0 = tan −1 ⎜⎜

⎞ ⎟⎟ ⎠

Use the Pythagorean relationship between the velocity and its components to express v0:

v0 = v02x + v02y

Using a constant-acceleration equation, express the vertical speed of the projectile as a function of its initial upward speed and time into the flight:

v y = v0 y + a y t ⇒ v0 y = v y − a y t

Because vy = 0 halfway through the flight (at maximum elevation) and ay = − g:

v0 y = gt max elevation

(1)

(2)

x, m

232 Chapter 3 Substitute numerical values and evaluate v0y:

(

)

v0 y = 9.81 m/s 2 (1.22 s ) = 11.97 m/s

Because there is no acceleration in the horizontal direction, v0x can be found from:

v0x =

Δx 40.0 m = = 16.39 m/s Δt 2.44 s

Substitute for v0x and v0y in equation (2) and evaluate v0:

v0 =

(16.39 m/s)2 + (11.97 m/s)2

Substitute for v0x and v0y in equation (1) and evaluate θ0 :

θ 0 = tan −1 ⎜⎜

= 20.3 m/s ⎛ 11.97 m/s ⎞ ⎟⎟ = 36.1° ⎝ 16.39 m/s ⎠

78 •• Consider a ball that is launched from ground level with initial speed v0 at an angle θ0 above the horizontal. If we consider its speed to be v at height h above the ground, show that for a given value of h, v is independent of θ0. (Ignore any effects due to air resistance.) Picture the Problem In the absence of friction, the acceleration of the ball is constant and we can use the constantacceleration equations to describe its motion. The figure shows the launch conditions and an appropriate coordinate system. The speeds v, vx, and vy are related by the Pythagorean Theorem.

The squares of the vertical and horizontal components of the object’s velocity are:

y h

r v

v y ˆj r v0

v x iˆ

θ0

x

v y2 = v02 sin 2 θ 0 − 2 gh

and v x2 = v02 cos 2 θ 0

The relationship between these variables is:

v 2 = vx2 + v y2

Substitute for vx and vy and simplify to obtain:

v 2 = v02 cos 2 θ 0 + v02 sin 2 θ 0 − 2 gh

= v02 (cos 2 θ 0 + sin 2 θ 0 ) − 2 gh = v02 − 2 gh

79 ••• At 12 of its maximum height, the speed of a projectile is 34 of its initial speed. What was its launch angle? (Ignore any effects due to air resistance.)

Motion in One and Two Dimensions 233 Picture the Problem In the absence of air resistance, the projectile experiences constant acceleration during its flight and we can use constant-acceleration equations to relate the speeds at half the maximum height and at the maximum height to the launch angle θ0 of the projectile.

y h

1 2

h

r v0 θ0

The angle θ the initial velocity makes with the horizontal is related to the initial velocity components:

x

θ 0 = tan −1 ⎜⎜

⎛ v0 y ⎞ ⎟⎟ ⎝ v0 x ⎠

(1)

Using a constant-acceleration equation, relate the vertical component of the speed to the projectile to its acceleration and displacement:

v y2 = v02y + 2aΔy

(2)

When the projectile is at the top of its trajectory ∆y = h and vy = 0. Hence:

0 = v02y − 2 gh ⇒ v02y = 2 gh

When Δy = 12 h equation (2) becomes:

v y2 = v02y − gh

Substituting for v02y and simplifying

v y2 = 2 gh − gh = gh

(3)

yields: We are given that v = 34 v0 when Δy = 12 h . Square both sides of v = 34 v0 and express this using the components of the velocity.

v02x + v y2 = ( 34 ) (v02x + v02y ) 2

(4)

where we have used v0x for vx because the x component of the velocity remains constant.

Solving equation (2) for v02y gives: Substitute for v02y in equation (3) to

v y2 = 2 gh − gh = gh

obtain: Substituting for v y2 and v02y in equation (4) yields:

(

)

v02x + gh = 169 v02x + 2 gh ⇒ v02x = 72 gh

234 Chapter 3 Finally, substituting for v0y and v0x in equation (1) and simplifying gives:

⎛ 2 gh ⎞ ⎟ = tan −1 ⎜ 2 gh ⎟ ⎝ 7 ⎠

θ 0 = tan −1 ⎜

( 7)

= 69.3°

80 •• A cargo plane is flying horizontally at an altitude of 12 km with a speed of 900 km/h when a large crate falls out of the rear-loading ramp. (Ignore any effects due to air resistance.) (a) How long does it take the crate to hit the ground? (b) How far horizontally is the crate from the point where it fell off when it hits the ground? (c) How far is the crate from the aircraft when the crate hits the ground, assuming that the plane continues to fly with the same velocity? Picture the Problem The horizontal speed of the crate, in the absence of air resistance, is constant and equal to the speed of the cargo plane. Choose a coordinate system in which the direction the plane is moving is the positive x direction and downward is the positive y direction and apply the constant-acceleration equations to describe the crate’s displacements at any time during its flight.

(a) Using a constant-acceleration equation, relate the vertical displacement of the crate Δy to the time of fall Δt: Substitute numerical values and evaluate Δt:

y h

r v0

R

x

Δy = v0 y Δt + 12 g (Δt )

2

or, because v0y = 0 and Δy = h, 2h 2 h = 12 g (Δt ) ⇒ Δt = g Δt =

2(12 × 10 3 m ) = 49.46 s = 49 s 9.81 m/s 2

(b) The horizontal distance traveled in time Δt is:

R = Δx = v0 x Δt

Substitute numerical values and evaluate R:

⎛ 1h ⎞ ⎟⎟(49.46 s ) R = (900 km/h )⎜⎜ ⎝ 3600 s ⎠ = 12 km

(c) Because the velocity of the plane is constant, it will be directly over the crate when it hits the ground; that is, the distance to the aircraft will be the elevation of the aircraft. Therefore Δy = 12 km .

Motion in One and Two Dimensions 235 81 •• [SSM] Wile E. Coyote (Carnivorous hungribilous) is chasing the Roadrunner (Speedibus cantcatchmi) yet again. While running down the road, they come to a deep gorge, 15.0 m straight across and 100 m deep. The Roadrunner launches himself across the gorge at a launch angle of 15º above the horizontal, and lands with 1.5 m to spare. (a) What was the Roadrunner’s launch speed? (b) Wile E. Coyote launches himself across the gorge with the same initial speed, but at a different launch angle. To his horror, he is short the other lip by 0.50 m. What was his launch angle? (Assume that it was less than 15º.) Picture the Problem In the absence of air resistance, the accelerations of both Wiley Coyote and the Roadrunner are constant and we can use constantacceleration equations to express their coordinates at any time during their leaps across the gorge. By eliminating the parameter t between these equations, we can obtain an expression that relates their y coordinates to their x coordinates and that we can solve for their launch angles.

(a) Using constant-acceleration equations, express the x coordinate of the Roadrunner while it is in flight across the gorge:

x = x0 + v0 x t + 12 a xt 2 or, because x0 = 0, ax = 0 and v0x = v0 cosθ0, x = (v0 cos θ 0 ) t (1)

Using constant-acceleration equations, express the y coordinate of the Roadrunner while it is in flight across the gorge:

y = y0 + v0 y t + 12 a y t 2

or, because y0 = 0, ay =−g and v0y = v0 sinθ0, y = (v0 sin θ 0 ) t − 12 gt 2 (2)

Eliminate the variable t between equations (1) and (2) to obtain:

y = (tan θ 0 )x −

g x2 2 2v cos θ 0

When y = 0, x = R and equation (3) becomes:

0 = (tan θ 0 )R −

g R2 2 2v cos θ 0

Using the trigonometric identity sin2θ = 2sinθ cosθ, solve for v0:

v0 =

Substitute numerical values and evaluate v0:

v0 =

(3)

2 0

2 0

Rg sin 2θ 0

(16.5 m )(9.81m/s 2 ) = sin 30°

18 m/s

236 Chapter 3 (b) Letting R represent Wiley’s range, solve equation (1) for his launch angle:

θ 0 = sin −1 ⎜⎜ 2 ⎟⎟ 2 ⎝ v0 ⎠

Substitute numerical values and evaluate θ0:

1 −1 ⎡ (14.5 m ) 9.81 m/s 2 ⎤ θ 0 = sin ⎢ 2 (18.0 m/s )2 ⎥⎦ ⎣

⎛ Rg ⎞

1

(

)

= 13°

82 •• A cannon barrel is elevated 45º above the horizontal. It fires a ball with a speed of 300 m/s. (a) What height does the ball reach? (b) How long is the ball in the air? (c) What is the horizontal range of the cannon ball? (Ignore any effects due to air resistance.) Picture the Problem Because, in the absence of air resistance, the vertical and horizontal accelerations of the cannonball are constant, we can use constant-acceleration equations to express the ball’s position and velocity as functions of time and acceleration. The maximum height of the ball and its time-of-flight are related to the components of its launch velocity.

(a) Using a constant-acceleration equation, relate h to the initial and final speeds of the cannon ball:

The vertical component of the firing speed is given by: Substituting for v0y gives:

y h r v0

(x, y )

θ0 R

x

v 2 = v02y + 2a y Δy

or, because v = 0, ay = −g, and Δy = h, v02y 0 = v02y − 2 gh ⇒ h = 2g v0 y = v0 sin θ 0

h=

v02 sin 2 θ 0 2g

Substitute numerical values and evaluate h:

h=

(300 m/s)2 sin 2 45° =

(b) The total flight time of the cannon ball is given by:

Δt = t up + t dn = 2t up = 2 =

2(9.81 m/s 2 )

2v0 sin θ 0 g

v0 y g

2.3 km

Motion in One and Two Dimensions 237 Substitute numerical values and evaluate Δt:

Δt =

2(300 m/s )sin 45° = 43.2 s 9.81 m/s 2

= 43 s

(c) Express the x coordinate of the cannon ball as a function of time:

x = v0 x Δt = (v0 cosθ 0 )Δt

Evaluate x (= R) when Δt = 43.2 s:

x = [(300 m/s )cos45°](43.2 s )

= 9.2 km 83 •• [SSM] A stone thrown horizontally from the top of a 24-m tower hits the ground at a point 18 m from the base of the tower. (Ignore any effects due to air resistance.) (a) Find the speed with which the stone was thrown. (b) Find the speed of the stone just before it hits the ground. Picture the Problem Choose a coordinate system in which the origin is at the base of the tower and the x- and y-axes are as shown in the figure to the right. In the absence of air resistance, the horizontal speed of the stone will remain constant during its fall and a constant-acceleration equation can be used to determine the time of fall. The final velocity of the stone will be the vector sum of its x and y components.

Because the stone is thrown horizontally: (a) Using a constant-acceleration equation, express the vertical displacement of the stone as a function of the fall time:

y, m r v0

24

18

v x = v0 x =

Δx Δt

x, m

(1)

Δy = v0 y Δt + 12 a y (Δt )

2

or, because v0y = 0 and a = −g, 2Δy 2 Δy = − 12 g (Δt ) ⇒ Δt = − g

Substituting for Δt in equation (1) and simplifying yields:

v x = Δx

Substitute numerical values and evaluate vx:

vx = (18 m )

g − 2Δy 9.81 m/s 2 = 8.1 m/s − 2(− 24 m )

238 Chapter 3 (b) The speed with which the stone hits the ground is related to the x and y components of its speed:

v = v x2 + v y2

The y component of the stone’s velocity at time t is:

v y = v0 y − gt

(2)

or, because v0y = 0, v y = − gt

Substitute for vx and vy in equation (2) and simplify to obtain:

2

⎛ g ⎞ ⎟ + (− gt )2 v = ⎜⎜ Δx − 2Δy ⎟⎠ ⎝ g (Δx ) + g 2t 2 − 2Δy 2

=

Substitute numerical values and evaluate v: v=

(9.81 m/s )(18 m ) + (9.81 m/s ) (2.21 s ) − 2(− 24 m ) 2

2

2 2

2

= 23 m/s

84 •• A projectile is fired into the air from the top of a 200-m cliff above a valley (Figure 3-36). Its initial velocity is 60 m/s at 60º above the horizontal. Where does the projectile land? (Ignore any effects due to air resistance.) Picture the Problem In the absence of air resistance, the acceleration of the projectile is constant and its horizontal and vertical motions are independent of each other. We can use constant-acceleration equations to express the horizontal and vertical displacements of the projectile in terms of its time-of-flight.

Using a constant-acceleration equation, express the horizontal displacement of the projectile as a function of time: Using a constant-acceleration equation, express the vertical displacement of the projectile as a function of time: Substitute numerical values to obtain the quadratic equation:

Δx = v0 x Δt + 12 a x (Δt )

2

or, because v0x = v0cosθ0 and ax = 0, Δx = (v0 cosθ 0 )Δt Δy = v0 y Δt + 12 a y (Δt )

2

or, because v0y = v0sinθ0 and ay = −g, 2 Δy = (v0 cosθ 0 )Δt − 12 g (Δt ) − 200 m = (60m/s )(sin 60°)Δt

(

)

− 12 9.81 m/s 2 (Δt )

2

Motion in One and Two Dimensions 239 Use the quadratic formula or your graphing calculator to obtain:

Δt = 13.6 s

Substitute for Δt and evaluate the horizontal distance traveled by the projectile:

Δx = (60 m/s)(cos60°)(13.6 s) 0.41 km =

85 •• The range of a cannonball fired horizontally from a cliff is equal to the height of the cliff. What is the direction of the velocity vector of the projectile as it strikes the ground? (Ignore any effects due to air resistance.) Picture the Problem In the absence of air resistance, the acceleration of the cannonball is constant and its horizontal and vertical motions are independent of each other. Choose the origin of the coordinate system to be at the base of the cliff and the axes directed as shown and use constantacceleration equations to describe both the horizontal and vertical displacements of the cannonball.

Express the direction of the velocity vector when the projectile strikes the ground: Express the vertical displacement using a constant-acceleration equation:

y

r v0

R

x

θ

r v

⎛ vy ⎞ ⎟⎟ ⎝ vx ⎠

θ = tan −1 ⎜⎜

(1)

Δy = v0 y Δt + 12 a y (Δt )

2

or, because v0y = 0 and ay = −g, 2 Δy = − 12 g (Δt ) Δx 1 = gΔt Δt 2

Set Δx = −Δy (R = −h) to obtain:

Δx = vx Δt = 12 g (Δt ) ⇒ vx =

Find the y component of the projectile as it hits the ground:

v y = v0 y + aΔt = − gΔt = −2vx

Substituting for vy in equation (1) yields:

θ = tan −1 ⎜⎜

2

⎛ − 2v x ⎝ vx

⎞ ⎟⎟ = tan −1 (− 2) ⎠

= − 63.4°

An archerfish launches a droplet of water from the surface of a small 86 •• lake at an angle of 60° above the horizontal. He is aiming at a juicy spider sitting on a leaf 50 cm to the east and on a branch 25 cm above the water surface. The

240 Chapter 3 fish is trying to knock the spider into the water so that the fish may eat the spider. (a) What must the speed of the water droplet be for the fish to be successful? (b) When it hits the spider, is the droplet rising or falling? Picture the Problem The diagram to the right shows the trajectory of the water droplet launched by the archerfish. We can use constantacceleration equations to derive expressions for the required speed of the droplet and for the vertical velocity of the droplet as a function of its position.

y, cm 25

r v0 θ

50

(a) Use a constant-acceleration equation to express the x- and y-coordinates of the droplet:

x = x0 + v0 x t + 12 a x t 2 and y = y0 + v0 y t + 12 a y t 2

Because x0 = y0 = 0, v0x = v0cosθ, v0y = v0sinθ, and, in the absence of air resistance, ax = 0 :

x = (v0 cosθ )t and y = (v0 sin θ )t − 12 gt 2

Solve equation (1) for t to obtain:

(x,y)

t=

x, cm

(1) (2)

x v0 cosθ

Substituting for t in equation (2) yields:

⎛ x ⎞ 1 ⎛ x ⎞ ⎟⎟ ⎟⎟ − 2 g ⎜⎜ y = (v0 sin θ )⎜⎜ v cos θ v cos θ ⎠ ⎝ 0 ⎠ ⎝ 0

Simplify this equation to obtain:

⎛ ⎞ 2 g ⎟⎟ x y = (tan θ )x − ⎜⎜ 2 2 2 cos θ v 0 ⎝ ⎠

Solving for v0 yields:

v0 =

g x 2( x tan θ − y )cos 2 θ

When the droplet hits the spider, its coordinates must be x = 0.50 m and y = 0.25 m. Substitute numerical values and evaluate v0: v0 =

9.81 m/s 2 (0.50 m ) = 2.822 m/s = 2.8 m/s 2((0.50 m ) tan 60° − 0.25 m )cos 2 60°

2

Motion in One and Two Dimensions 241 v y (t ) = v0 y + a y t

(b) Use a constant-acceleration equation to express the y component of the speed of the droplet as a function of time:

or, because v0y = v0sinθ and ay = −g, (3) v y (t ) = v0 sin θ − gt

From equation (1), the time-to-thetarget, tto target, is given by:

t to target =

Substitute in equation (3) to obtain:

xtarget

=

xtarget

xtarget

=

v0 x v0 cosθ where xtarget is the x-coordinate of the target. vx

v y (t to target ) = v0 sin θ − gt to target

= v0 sin θ −

gx target v0 cosθ

Substitute numerical values and evaluate vy(tto target): v y (t to target ) = (2.822 m/s )sin 60° −

(9.81 m/s )(0.50 m ) = −1.032 m/s 2

(2.822 m/s)cos 60°

Because the y-component of the velocity of the droplet is negative at the location of the spider, the droplet is falling and moving horizontally when it hits the spider. [SSM] You are trying out for the position of place-kicker on a 87 •• professional football team. With the ball teed up 50.0 m from the goalposts with a crossbar 3.05 m off the ground, you kick the ball at 25.0 m/s and 30° above the horizontal. (a) Is the field goal attempt good? (b) If so, by how much does it clear the bar? If not, by how much does it go under the bar? (c) How far behind the plane of the goalposts does the ball land? Picture the Problem We can use constant-acceleration equations to express the x and y coordinates of the ball along its flight path. Eliminating t between the equations will leave us with an equation for y as a function of x that we can use to find the height of the ball when it has reached the cross bar. We can use this same equation to find the range of the ball and, hence, how far behind the plane of the goalposts the ball lands.

y, m 3.05

cross bar

(x, y ) r v0

θ0 50.0

R

x, m

242 Chapter 3 (a) Use a constant-acceleration equation to express the x coordinate of the ball as a function of time:

x(t ) = x0 + v0 x t + 12 a x t 2 or, because x0 = 0, v0x = v0cosθ0, and ax = 0, x(t ) x(t ) = (v0 cosθ 0 )t ⇒ t = v0 cosθ 0

Use a constant-acceleration equation to express the y coordinate of the ball as a function of time:

y (t ) = y0 + v0 y t + 12 a y t 2

or, because y0 = 0, v0y = v0sinθ0, and ay = −g, y (t ) = (v0 sin θ 0 )t − 12 g t 2

Substituting for t yields: ⎛ x(t ) ⎞ x(t ) ⎟⎟ y ( x ) = (v0 sin θ 0 ) − 12 g ⎜⎜ v0 cosθ 0 ⎝ v0 cosθ 0 ⎠

2

Simplify to obtain: y ( x ) = (tan θ 0 ) x(t ) −

g (x(t ))2 2 2v cos θ 0 2 0

Substitute numerical values and evaluate y(50.0 m): y (50.0 m ) = (tan 30°)(50.0 m ) −

9.81 m/s 2 (50.0 m )2 = 2.71 m 2 2 2(25.0 m/s ) cos 30°

Because 2.71 m < 3.05 m, the ball goes under the crossbar and the kick is no good. (b) The ball goes under the bar by:

d under = 3.05 m − 2.71 m = 0.34 m

(c) The distance the ball lands behind the goalposts is given by:

d behind the = R − 50.0 m

Evaluate the equation derived in (a) for y = 0 and x(t) = R:

0 = (tan θ 0 ) R −

Solving for v0 yields:

goal posts

R=

v02 sin 2θ 0 g

g R2 2 2v cos θ 0 2 0

(3)

Motion in One and Two Dimensions 243 Substitute for R in equation (3) to obtain:

d behind the goal posts

v02 sin 2θ 0 = − 50.0 m g

Substitute numerical values and evaluate d behind the : goal posts

d behind the =

(25.0 m/s)2 sin 2(30°) − 50.0 m = 9.81 m/s 2

goal posts

5.2 m

88 •• The speed of an arrow fired from a compound bow is about 45.0 m/s. (a) A Tartar archer sits astride his horse and launches an arrow into the air, elevating the bow at an angle of 10º above the horizontal. If the arrow is 2.25 m above the ground at launch, what is the arrow’s horizontal range? Assume that the ground is level, and ignore any effects due to air resistance. (b) Now assume that his horse is at full gallop and moving in the same direction as the direction the archer will fire the arrow. Also assume that the archer elevates the bow at the same elevation angle as in Part (a) and fires. If the horse’s speed is 12.0 m/s, what is the arrow’s horizontal range now? Picture the Problem Choose a coordinate system in which the origin is at ground level. Let the positive x direction be to the right and the positive y direction be upward. We can apply constant-acceleration equations to obtain equations in time that relate the range to the initial horizontal speed and the height h to which the initial upward speed. Eliminating time from these equations will leave us with a quadratic equation in R, the solution to which will give us the range of the arrow. In (b), we’ll find the launch speed and angle as viewed by an observer who is at rest on the ground and then use these results to find the arrow’s range when the horse is moving at 12.0 m/s.

y

h

r v0

θ0

(x,y)

R

(a) Use constant-acceleration equations to express the horizontal and vertical coordinates of the arrow’s motion:

x

Δx = x − x0 = v0 x t and y = h + v0 y t + 12 (− g )t 2 where v0 x = v0 cosθ 0 and v0 y = v0 sin θ 0

244 Chapter 3 Solve the x-component equation for time:

t=

Δx Δx = v0 x v0 cosθ 0

Eliminating time from the y-component equation yields:

Δx 1 ⎛ Δx ⎞ ⎟ y = h + v0 y − g⎜ v0 x 2 ⎜⎝ v0 x ⎟⎠

When y = 0, Δx = R and:

0 = h + (tan θ 0 )R −

Solve for the range R to obtain:

2

g R2 2 2v cos θ 0 2 0

⎛ v02 2 gh R= sin 2θ 0 ⎜⎜1 + 1 + 2 2 2g v0 sin θ 0 ⎝

Substitute numerical values and evaluate R: R=

2 ⎞ ⎛ ⎜1 + 1 + 2(9.81 m/s )(2.25 m ) ⎟ = 82 m sin 20 ° ⎜ 2(9.81 m/s 2 ) (45.0 m/s)2 (sin 2 10°) ⎟⎠ ⎝

(45.0 m/s)2

(b) Express the speed of the arrow in the horizontal direction:

v x = varrow + varcher

= (45.0 m/s )cos10° + 12.0 m/s = 56.32 m/s

Express the vertical speed of the arrow:

v y = (45.0 m/s )sin10° = 7.814 m/s

Express the angle of elevation from the perspective of someone on the ground:

θ 0 = tan −1 ⎜⎜

Substitute numerical values and evaluate θ0 :

θ 0 = tan −1 ⎜⎜

The arrow’s speed relative to the ground is given by:

v0 = v x2 + v y2

Substitute numerical values and evaluate v0:

v0 =

⎛ vy ⎝ vx

⎞ ⎟⎟ ⎠

⎛ 7.814 m/s ⎞ ⎟⎟ = 7.899° ⎝ 56.32 m/s ⎠

(56.32 m/s)2 + (7.814 m/s)2

= 56.86 m/s

⎞ ⎟ ⎟ ⎠

Motion in One and Two Dimensions 245 Substitute numerical values and evaluate R: 2 ⎛ ( 56.86 m/s ) sin (15.8°)⎜1 + R=

2(9.81m/s 2 )

⎜ ⎝

1+

2(9.81m/s 2 )(2.25 m ) ⎞⎟ = 0.10 km (56.86 m/s)2 (sin 2 7.899°) ⎟⎠

Remarks: An alternative solution for part (b) is to solve for the range in the reference frame of the archer and then add to it the distance the frame travels, relative to Earth, during the time of flight. 89 •• [SSM] The roof of a two-story house makes an angle of 30° with the horizontal. A ball rolling down the roof rolls off the edge at a speed of 5.0 m/s. The distance to the ground from that point is about two stories or 7.0 m. (a) How long is the ball in the air? (b) How far from the base of the house does it land? (c) What is its speed and direction just before landing? Picture the Problem Choosing the coordinate system shown in the figure to the right, we can use a constant acceleration equation to express the y coordinate of the ball’s position as a function of time. When the ball hits the ground, this position coordinate is zero and we can solve the resulting quadratic equation for the time-to-the-ground. Because, in the absence of air resistance, there is no acceleration in the horizontal direction we can find R, v, and φ using constantacceleration equations. y, m 7.0

θ0 r v0

(x,y)

R

x, m

φ

r v

(a) Use a constant-acceleration equation to relate the y coordinate of the ball to its time-in-flight: y(tground) = 0 when the ball hits the ground and:

y (t ) = y0 + v0 y t + 12 a y t 2

or, because v0y = v0sinθ0 and ay = −g, y (t ) = y0 + (v0 sin θ 0 )t − 12 g t 2 2 0 = y 0 + (v0 sin θ 0 )t ground − 12 g t ground

246 Chapter 3 Substitute numerical values to obtain:

(

)

2 0 = 7.0 m + (− 5.0 m/s )sin 30° t ground − 12 9.81 m/s 2 t ground

Simplifying yields:

(4.91 m/s )t 2

2 ground

+ (2.50 m/s )t ground − 7.0 m = 0

Use the quadratic formula or your graphing calculator to obtain:

t ground = 0.966 s or − 1.48 s

Because only the positive root has physical meaning:

t ground = 0.97 s

(b) The distance from the house R is related to the ball’s horizontal speed and the time-to-ground:

R = v0 x t ground = (v0 cosθ 0 )t ground

Substitute numerical values and evaluate R:

R = (5.0 m/s )(cos 30°)(0.966 s )

(c) The speed and direction of the ball just before landing are given by:

v = v x2 + v y2

= 4.2 m (1)

and ⎛ vy ⎝ vx

φ = tan −1 ⎜⎜

⎞ ⎟⎟ ⎠

(2)

Express vx and vy in terms of v0:

v x = v0 cosθ 0 and v y = v0 y + a y t = v0 sin θ 0 + gt

Substitute numerical values and evaluate vx and vy:

v x = (5.0 m/s) cos 30° = 4.330 m/s and v y = (5.0 m/s)sin30° + (9.81 m/s 2 )(0.966 s) = 11.98 m/s

Motion in One and Two Dimensions 247 Substituting numerical values in equations (1) and (2) yields:

v = (4.330 m/s) 2 + (11.98 m/s) 2 = 13 m/s and ⎛ 11.98 m/s ⎞ ⎟ ⎝ 4.330 m/s ⎠

φ = tan −1 ⎜

= 70° below the horizontal 90 •• Compute dR/dθ0 from R = (v 20 / g)sin( 2θ0 ) and show that setting dR/dθ0 = 0 gives θ0 = 45º for the maximum range. Picture the Problem An extreme value (i.e., a maximum or a minimum) of a function is determined by setting the appropriate derivative equal to zero. Whether the extremum is a maximum or a minimum can be determined by evaluating the second derivative at the point determined by the first derivative.

Evaluate dR/dθ0:

2v02 dR v02 d [sin (2θ 0 )] = cos(2θ 0 ) = dθ 0 g dθ 0 g

Set dR/dθ0= 0 for extrema:

2v02 cos(2θ 0 ) = 0 g

Solve for θ0 to obtain:

θ 0 = 12 cos −1 (0) = 45°

Determine whether 45° corresponds to a maximum or a minimum value of R:

d 2R 2 dθ 0

[ (

)

= − 4 v02 g sin 2θ 0

]

θ0 =45°

θ0 =45°

0? Repeat the question for the region x < 0. (c) Does the potential energy U increase or decrease as x increases in the region x > 0? (d) Answer Parts (b) and (c) where C is a negative constant. Picture the Problem Fx is defined to be the negative of the derivative of the potential-energy function with respect to x, that is Fx = − dU dx . Consequently,

given U as a function of x, we can find Fx by differentiating U with respect to x. (a) Evaluate Fx = −

dU : dx

Fx = −

d ⎛C ⎞ C ⎜ ⎟= 2 dx ⎝ x ⎠ x

r (b) Because C > 0, if x > 0, Fx is positive and F points away from the origin. If r x < 0, Fx is still positive and F points toward the origin. (c) Because U is inversely proportional to x and C > 0, U(x) decreases with increasing x.

r (d) When C < 0, if x > 0, Fx is negative and F points toward the origin. If x < 0, r Fx is negative and F points away from the origin. Because U is inversely proportional to x and C < 0, U(x) becomes less negative as x increases and U(x) increases with increasing x.

622 Chapter 7 26 •• The force Fy is associated with the potential-energy function U(y). On the potential-energy curve for U versus y, shown in Figure 7-37, the segments AB and CD are straight lines. Plot Fy versus y. Include numerical values, with units, on both axes. These values can be obtained from the U versus y plot. Picture the Problem Fy is defined to be the negative of the derivative of the potential-energy function with respect to y; that is, Fy = − dU dy . Consequently,

we can obtain Fy by examining the slopes of the graph of U as a function of y. The table to the right summarizes the information we can obtain from Figure 7-37:

Interval A→B B→C C→D

Slope (N) −2 transitional 1.4

Fy (N) 2 2 → −1.4 −1.4

The following graph shows F as a function of y: 2.5 2.0 1.5

F, N

1.0 0.5 0.0 -0.5 -1.0 -1.5 0

1

2

3

4

5

6

y, m

27 •• The force acting on an object is given by Fx = a/x2. At x = 5.0 m, the force is known to point in the –x direction and have a magnitude of 25.0 N. Determine the potential energy associated with this force as a function of x, assuming we assign a reference value of –10.0 J at x = 2.0 m for the potential energy. Picture the Problem Fx is defined to be the negative of the derivative of the potential-energy function with respect to x, i.e. Fx = − dU dx . Consequently, given F as a function of x, we can find U by integrating Fx with respect to x. Applying the condition on Fx x will allow us to determine the value of a and

using the fact that the potential energy is –10.0 J at x = 2.00 m will give us the value of U0.

Conservation of Energy 623 Evaluate the integral of Fx with respect to x:

U ( x ) = − ∫ Fx dx = − ∫

a dx x2

a = +U0 x

Because Fx (5.0 m ) = −25.0 N :

(1)

a = −25.0 N (5.00 m )2

Solving for a yields:

a = −625 N ⋅ m 2

Substitute for a in equation (1) to obtain:

U (x ) =

Applying the condition U(2.00 m) = −10.0 J yields:

− 10.0 J =

Solve for U0 to obtain:

U 0 = 303 J

Substituting for U0 in equation (2) yields:

U (x ) =

− 625 N ⋅ m 2 +U0 x

(2)

− 625 N ⋅ m 2 +U0 2.00 m

− 625 N ⋅ m 2 + 303 J x

28 •• The potential energy of an object constrained to the x axis is given by U(x) = 3x2 – 2x3, where U is in joules and x is in meters. (a) Determine the force Fx associated with this potential energy function. (b) Assuming no other forces act on the object, at what positions is this object in equilibrium? (c) Which of these equilibrium positions are stable and which are unstable? Picture the Problem Fx is defined to be the negative of the derivative of the potential-energy function with respect to x, that is, Fx = − dU dx . Consequently, given U as a function of x, we can find Fx by differentiating U with respect to x.

To determine whether the object is in stable or unstable equilibrium at a given point, we’ll evaluate d 2U dx 2 at the point of interest. (a) Evaluate Fx = −

dU : dx

(b) We know that, at equilibrium, Fx = 0:

Fx = −

(

)

d 3 x 2 − 2 x 3 = 6 x( x − 1) dx

When Fx = 0, 6x(x – 1) = 0. Therefore, the object is in equilibrium at x = 0 and x = 1 m.

624 Chapter 7 (c) To decide whether the equilibrium at a particular point is stable or unstable, evaluate the 2nd derivative of the potential energy function at the point of interest: Evaluate

d 2U at x = 0: dx 2

(

)

dU d 3x 2 − 2 x 3 = 6 x − 6 x 2 = dx dx and d 2U = 6 − 12 x dx 2 d 2U dx 2

=6>0 x =0

⇒ stable equilibrium at x = 0

Evaluate

d 2U at x = 1 m: dx 2

d 2U dx 2

= 6 − 12 < 0 x =1 m

⇒ unstable equilibrium at x = 1 m

29 •• [SSM] The potential energy of an object constrained to the x axis is given by U(x) = 8x2 – x4, where U is in joules and x is in meters. (a) Determine the force Fx associated with this potential energy function. (b) Assuming no other forces act on the object, at what positions is this object in equilibrium? (c) Which of these equilibrium positions are stable and which are unstable? Picture the Problem Fx is defined to be the negative of the derivative of the potential-energy function with respect to x, that is Fx = − dU dx . Consequently, given U as a function of x, we can find Fx by differentiating U with respect to x.

To determine whether the object is in stable or unstable equilibrium at a given point, we’ll evaluate d 2U dx 2 at the point of interest. (a) Evaluate the negative of the derivative of U with respect to x:

Fx = −

(

dU d =− 8x 2 − x 4 dx dx

)

= 4 x 3 − 16 x = 4 x( x + 2 )( x − 2 )

(b) The object is in equilibrium wherever Fnet = Fx = 0 :

4 x( x + 2 )( x − 2 ) = 0 ⇒ the equilibrium points are x = −2 m, 0, and 2 m.

Conservation of Energy 625 (c) To decide whether the equilibrium at a particular point is stable or unstable, evaluate the 2nd derivative of the potential energy function at the point of interest:

(

)

d 2U d 16 x − 4 x 3 = 16 − 12 x 2 = dx 2 dx

Evaluating d 2U dx 2 at x = −2 m, 0 and x = 2 m yields the following results: x, m

d 2U dx 2

Equilibrium

−2 0 2

−32 16 −32

Unstable Stable Unstable

Remarks: You could also decide whether the equilibrium positions are stable or unstable by plotting F(x) and examining the curve at the equilibrium positions. 30 •• The net force acting on an object constrained to the x axis is given by Fx(x) = x3 – 4x. (The force is in newtons and x in meters.) Locate the positions of unstable and stable equilibrium. Show that each of these positions is either stable or unstable by calculating the force one millimeter on either side of the locations. Picture the Problem The equilibrium positions are those values of x for which F(x) = 0. Whether the equilibrium positions are stable or unstable depends on whether the signs of the force either side of the equilibrium position are the same (unstable equilibrium) of opposite (stable equilibrium).

Determine the equilibrium locations by setting Fnet = F ( x ) = 0 :

F(x) = x3 – 4x = x(x2 – 4) = 0 and the positions of stable and unstable equilibrium are at x = −2 m, 0 and 2 m .

626 Chapter 7 Noting that we need only determine whether each value of F(x) is positive or negative, evaluate F(x) at x = −201 mm and x = −199 mm to determine the stability at x = −200 mm … and repeat these calculations at x = −1 mm, 1 mm and x = 199 mm, 201 mm to complete the following table: x, mm

Fx −1 mm

Fx +1 mm

Equilibrium

−200 0 200

0 >0

0:

U decreases as x increases

(b) As x → ∞,

1 A → 0. Hence: 2 x2

U0 = 0 and U (x ) = =

1 A 1 ⎛ 8.0 N ⋅ m 3 ⎞ ⎟⎟ = ⎜ 2 x 2 2 ⎜⎝ x2 ⎠ 4.0 N ⋅ m3 2 x

(c) The graph of U(x) follows: 400 350 300

U (J)

250 200 150 100 50 0 0.0

0.5

1.0

1.5

2.0

x (m)

33 •• [SSM] A straight rod of negligible mass is mounted on a frictionless pivot, as shown in Figure 7-38. Blocks have masses m1 and m2 are attached to the rod at distances l1 and l 2 . (a) Write an expression for the gravitational potential energy of the blocks-Earth system as a function of the angle θ made by the rod and the horizontal. (b) For what angle θ is this potential energy a minimum? Is the statement ″systems tend to move toward a configuration of minimum potential energy″ consistent with your result? (c) Show that if m1l1 = m2 l 2 , the potential energy is the same for all values of θ. (When this holds, the system will balance at any angle θ. This result is known as Archimedes’ law of the lever.)

Conservation of Energy 629 Picture the Problem The gravitational potential energy of this system of two objects is the sum of their individual potential energies and is dependent on an arbitrary choice of where, or under what condition(s), the gravitational potential energy is zero. The best choice is one that simplifies the mathematical details of the expression for U. In this problem let’s choose U = 0 where θ = 0.

(a) Express U for the 2-object system as the sum of their gravitational potential energies; noting that because the object whose mass is m2 is above the position we have chosen for U = 0, its potential energy is positive while that of the object whose mass is m1 is negative:

U (θ ) = U1 + U 2 = m2 gl 2 sin θ − m1 gl 1 sin θ

(m2l 2 − m1l 1 )g sin θ

=

(b) Differentiate U with respect to θ and set this derivative equal to zero to identify extreme values:

dU = (m2 l 2 − m1l 1 )g cos θ = 0 dθ from which we can conclude that cosθ = 0 and θ = cos−10.

To be physically meaningful, − π 2 ≤ θ ≤ π 2 . Hence:

θ = ±π 2

Express the 2nd derivative of U with respect to θ and evaluate this derivative at θ = ± π 2 :

d 2U = −(m2l 2 − m1l 1 )g sin θ dθ 2

If we assume, in the expression for U that we derived in (a), that m2l2 – m1l1 > 0, then U(θ) is a sine function and, in the interval of interest, −π 2 ≤θ ≤ π 2, takes on its minimum value when θ = −π/2:

d 2U dθ 2

(c) If m2l2 = m1l1, then:

m1l 1 − m2 l 2 = 0 and U = 0 independent of θ .

> 0 and −π 2

U is a minimum at θ = − π 2 d 2U dθ 2

< 0 and π 2

U is a maximum at θ = π 2

630 Chapter 7 Remarks: An alternative approach to establishing that U is a maximum at θ = π/2 is to plot its graph and note that, in the interval of interest, U is concave downward with its maximum value at θ = π/2. Similarly, it can be shown that U is a minimum at θ = −π/2 (Part (b)). 34 •• An Atwood’s machine (Figure 7-39) consists of masses m1 and m2, and a pulley of negligible mass and friction. Starting from rest, the speed of the two masses is 4.0 m/s at the end of 3.0 s. At that time, the kinetic energy of the system is 80 J and each mass has moved a distance of 6.0 m. Determine the values of m1 and m2. Picture the Problem In a simple Atwood’s machine, the only effect of the pulley is to connect the motions of the two objects on either side of it; that is, it could be replaced by a piece of polished pipe. We can relate the kinetic energy of the rising and falling objects to the mass of the system and to their common speed and relate their accelerations to the sum and difference of their masses … leading to simultaneous equations in m1 and m2.

(m1 + m2 )v 2 ⇒ m1 + m2 = 2 K2

Relate the kinetic energy of the system to the total mass being accelerated:

K=

Substitute numerical values and evaluate m1 + m2:

m1 + m2 =

In Chapter 4, the acceleration of the masses was shown to be:

a=

Because v(t) = at, we can eliminate a in the previous equation to obtain:

v(t ) =

Solving for m1 − m2 yields:

Substitute numerical values and evaluate m1 − m2 :

1 2

v

2(80 J ) (4.0 m/s )2 = 10.0 kg

(1)

m1 − m2 g m1 + m2 m1 − m2 gt m1 + m2

m1 − m2 =

m1 − m2 =

(m1 + m2 )v(t ) gt

(10 kg )(4.0 m/s) (9.81m/s 2 )(3.0 s)

(2)

= 1.36 kg Solve equations (1) and (2) simultaneously to obtain:

m1 = 5.7 kg and m2 = 4.3 kg

Conservation of Energy 631 35 ••• You have designed a novelty desk clock, as shown in Figure 7-40. You are worried that it is not ready for market because the clock itself might be in an unstable equilibrium configuration. You decide to apply your knowledge of potential energies and equilibrium conditions and analyze the situation. The clock (mass m) is supported by two light cables running over the two frictionless pulleys of negligible diameter, which are attached to counterweights that each have mass M. (a) Find the potential energy of the system as a function of the distance y. (b) Find the value of y for which the potential energy of the system is a minimum. (c) If the potential energy is a minimum, then the system is in equilibrium. Apply Newton’s second law to the clock and show that it is in equilibrium (the forces on it sum to zero) for the value of y obtained for Part (b). (d) Finally, determine whether you are going to be able to market this gadget: is this a point of stable or unstable equilibrium? Picture the Problem Let L be the total length of one cable and the zero of gravitational potential energy be at the top of the pulleys. We can find the value of y for which the potential energy of the system is an extremum by differentiating U(y) with respect to y and setting this derivative equal to zero. We can establish that this value corresponds to a minimum by evaluating the second derivative of U(y) at the point identified by the first derivative. We can apply Newton’s second law to the clock to confirm the result we obtain by examining the derivatives of U(y).

(a) Express the potential energy of the system as the sum of the potential energies of the clock and counterweights:

U ( y ) = U clock ( y ) + U weights ( y )

(

Substitute for U clock ( y ) and U weights ( y ) to obtain:

U ( y ) = − mgy − 2 Mg L − y 2 + d 2

(b) Differentiate U(y) with respect to y:

[

(

)]

⎡ dU ( y ) d mgy + 2Mg L − y 2 + d 2 = − ⎢mg − 2Mg =− dy dy ⎢⎣

For extreme values (relative maxima and minima): Solve for y′ to obtain:

mg − 2 Mg

y' = d

y' y' 2 + d 2

m2 4M 2 − m 2

⎤ ⎥ y 2 + d 2 ⎥⎦ y

=0

)

632 Chapter 7 d 2U ( y ) : Find dy 2

d 2U ( y ) d ⎡ = − ⎢mg − 2Mg dy 2 dy ⎢ ⎣ 2Mgd 2 = 32 y2 + d 2

(

d 2U ( y ) at y = y′: Evaluate dy 2

⎤ ⎥ y 2 + d 2 ⎦⎥ y

)

d 2U ( y ) 2Mgd 2 = 32 dy 2 y' y2 + d 2

(

)

y'

2Mgd

=

32

⎛ ⎞ m2 ⎜⎜ + 1⎟⎟ 2 2 ⎝ 4M − m ⎠ >0 and the potential energy is a minimum at y= d

m2 4M 2 − m 2

(c) The free-body diagram, showing the magnitudes of the forces acting on the support point just above the clock, is shown to the right:

y Mg

Mg

θ

θ

x

mg

Apply

∑F

y

= 0 to this point to

obtain: Express sinθ in terms of y and d:

Equate the two expressions for sinθ to obtain:

2Mg sin θ − mg = 0 ⇒ sin θ =

sin θ =

m = 2M

m 2M

y y + d2 2

y y +d2 2

which is equivalent to the first equation in Part (b).

Conservation of Energy 633 (d) This is a point of stable equilibrium. If the clock is displaced downward, θ increases, leading to a larger upward force on the clock. Similarly, if the clock is displaced upward, the net force from the cables decreases. Because of this, the clock will be pulled back toward the equilibrium point if it is displaced away from it. Remarks: Because we’ve shown that the potential energy of the system is a minimum at y = y′ (i.e., U(y) is concave upward at that point), we can conclude that this point is one of stable equilibrium.

The Conservation of Mechanical Energy 36 • A block of mass m on a horizontal frictionless tabletop is pushed against a horizontal spring, compressing it a distance x, and the block is then released. The spring propels the block along the tabletop, giving a speed v. The same spring is then used to propel a second block of mass 4m, giving it a speed 3v. What distance was the spring compressed in the second case? Express your answer in terms of x. Picture the Problem The work done in compressing the spring is stored in the spring as potential energy. When the block is released, the energy stored in the spring is transformed into the kinetic energy of the block. Equating these energies will give us a relationship between the compressions of the spring and the speeds of the blocks. kx22 = 12 m2 v22

Let the numeral 1 refer to the first case and the numeral 2 to the second case. Relate the compression of the spring in the second case to its potential energy, which equals its initial kinetic energy when released:

1 2

Relate the compression of the spring in the first case to its potential energy, which equals its initial kinetic energy when released:

1 2

kx12 = 12 m1v12 ⇒ m1v12 = kx12

1 2

kx22 = 18kx12 ⇒ x 2 = 6x1

Substitute for m1v12 to obtain:

=

1 2

(4m1 )(3v1 )2

= 18m1v12

37 • A simple pendulum of length L with a bob of mass m is pulled aside until the bob is at a height L/4 above its equilibrium position. The bob is then released. Find the speed of the bob as it passes through the equilibrium position. Neglect any effects due to air resistance.

634 Chapter 7 Picture the Problem The diagram shows the pendulum bob in its initial position. Let the zero of gravitational potential energy be at the low point of the pendulum’s swing, the equilibrium position, and let the system include the pendulum and Earth. We can find the speed of the bob at it passes through the equilibrium position by applying conservation of mechanical energy to the system.

Apply conservation of mechanical energy to the system to obtain:

L

Δh = 14 L

Ug = 0 Wext = ΔK + ΔU or, because Wext = 0, ΔK + ΔU = 0

Because K i −U f = 0 :

K f −U i = 0

Substituting for K f and U i yields:

1 2

Express Δh in terms of the length L of the pendulum:

Δh =

Substitute for Δh in the expression for vf and simplify to obtain:

vf =

mvf2 − mgΔh = 0 ⇒ vf = 2 gΔh L 4 gL 2

38 • A 3.0-kg block slides along a frictionless horizontal surface with a speed of 7.0 m/s (Figure 7-41). After sliding a distance of 2.0 m, the block makes a smooth transition to a frictionless ramp inclined at an angle of 40° to the horizontal. What distance along the ramp does the block slide before coming momentarily to rest? Picture the Problem The pictorial representation shows the block in its initial, intermediate, and final states. It also shows a choice for Ug = 0. Let the system consist of the block, ramp, and Earth. Because the surfaces are frictionless, the initial kinetic energy of the system is equal to its final gravitational potential energy when the block has come to rest on the incline.

Conservation of Energy 635 3

x

x3 = 2.0 m + r v3 = 0 h

1

2 x0

=0 0

θ

Ug = 0

x1 = 0 x 2 = 2.0 m v1 = 7.0 m/s v 2

Apply conservation of mechanical energy to the system to obtain:

Wext = ΔK + ΔU or, because Wext = 0, ΔK + ΔU = 0

Because K3 = U1 = 0:

− K1 + U 3 = 0

Substituting for K1 and U3 yields:

v12 2g where h is the change in elevation of the block as it slides to a momentary stop on the ramp.

Relate the height h to the displacement l of the block along the ramp and the angle the ramp makes with the horizontal:

h = l sin θ

Equate the two expressions for h and solve for l to obtain: Substitute numerical values and evaluate l:

− 12 mv12 + mgh = 0 ⇒ h =

l sin θ =

l=

(

v12 v12 ⇒l = 2g 2 g sin θ

(7.0 m/s) 2

)

2 9.81 m/s 2 sin40°

= 3.9 m

39 • The 3.00-kg object in Figure 7-42 is released from rest at a height of 5.00 m on a curved frictionless ramp. At the foot of the ramp is a spring of force constant 400 N/m. The object slides down the ramp and into the spring, compressing it a distance x before coming momentarily to rest. (a) Find x. (b) Describe the motion object (if any) after the block momentarily comes to rest? Picture the Problem Let the system consist of Earth, the block, and the spring. With this choice there are no external forces doing work to change the energy of the system. Let Ug = 0 at the elevation of the spring. Then the initial gravitational

636 Chapter 7 potential energy of the 3.00-kg object is transformed into kinetic energy as it slides down the ramp and then, as it compresses the spring, into potential energy stored in the spring. (a) Apply conservation of mechanical energy to the system to relate the distance the spring is compressed to the initial potential energy of the block: Substitute numerical values and evaluate x:

Wext = ΔK + ΔU = 0 and, because ΔK = 0, − mgh + 12 kx 2 = 0 ⇒ x =

x=

(

2mgh k

)

2(3.00 kg ) 9.81 m/s 2 (5.00 m ) 400 N/m

= 0.858 m

(b) The energy stored in the compressed spring will accelerate the block, launching it back up the incline and the block will retrace its path, rising to a height of 5.00 m. 40 • You are designing a game for small children and want to see if the ball’s maximum speed is sufficient to require the use of goggles. In your game, a 15.0-g ball is to be shot from a spring gun whose spring has a force constant of 600 N/m. The spring will be compressed 5.00 cm when in use. How fast will the ball be moving as it leaves the gun and how high will the ball go if the gun is aimed vertically upward? What would be your recommendation on the use of goggles? Picture the Problem With Ug chosen to be zero at the uncompressed level of the spring, the ball’s initial gravitational potential energy is negative. Let the system consist of the ball, the spring and gun, and Earth. The difference between the initial potential energy of the spring and the gravitational potential energy of the ball is first converted into the kinetic energy of the ball and then into gravitational potential energy as the ball rises and slows … eventually coming momentarily to rest.

Apply conservation of mechanical energy to the system as the ball leaves the gun to obtain: Solving for vf yields:

Wext = ΔK + ΔU g + ΔU s = 0 or, because Ki = Us,f = Ug,f = 0, 2 2 1 1 2 mvf + mgx − 2 kx = 0 ⎛k ⎞ vf = ⎜ x − 2 g ⎟ x ⎝m ⎠

Conservation of Energy 637 Substitute numerical values and evaluate vf: ⎡⎛ 600 N/m ⎞ ⎤ ⎟⎟ (0.0500 m ) − 2(9.81 m/s 2 )⎥ (0.0500 m ) = 9.95 m/s vf = ⎢⎜⎜ ⎣⎝ 0.0150 kg ⎠ ⎦ This initial speed of the ball is fast enough to warrant the use of goggles. Apply conservation of mechanical energy to the system as the ball rises to its maximum height to obtain:

Wext = ΔK + ΔU g + ΔU s = 0 or, because ΔK = Us,f = 0, mgh + mgx − 12 kx 2 = 0 where h is the maximum height of the ball.

Solving for h gives:

Substitute numerical values and evaluate h:

h=

kx 2 −x 2mg

2 ( 600 N/m )(0.0500 m ) h= − 0.0500 m 2(0.0150 kg )(9.81m/s 2 )

= 5.05 m 41 • A 16-kg child on a 6.0-m-long playground swing moves with a speed of 3.4 m/s when the swing seat passes through its lowest point. What is the angle that the swing makes with the vertical when the swing is at its highest point? Assume that the effects due to air resistance are negligible, and assume that the child is not pumping the swing. Picture the Problem Let the system consist of Earth and the child. Then Wext = 0. Choose Ug = 0 at the child’s lowest point as shown in the diagram to the right. Then the child’s initial energy is entirely kinetic and its energy when it is at its highest point is entirely gravitational potential. We can determine h from conservation of mechanical energy and then use trigonometry to determineθ.

Using the diagram, relate θ to h and L:

θ L

L−h

r vi Ug = 0

h

h⎞ ⎛ L−h⎞ −1 ⎛ ⎟ = cos ⎜1 − ⎟ (1) ⎝ L ⎠ ⎝ L⎠

θ = cos −1 ⎜

638 Chapter 7 Apply conservation of mechanical energy to the system to obtain:

Wext = ΔK + ΔU = 0 or, because Kf = Ug,i = 0, − K i + U g,f = 0

Substituting for Ki and Ug,f yields:

vi2 − mv + mgh = 0 ⇒ h = 2g

Substitute for h in equation (1) to obtain:

⎛ vi2 ⎞ ⎜ ⎟⎟ θ = cos ⎜1 − 2 gL ⎝ ⎠

Substitute numerical values and evaluate θ :

θ = cos −1 ⎜⎜1 −

1 2

2 i

−1

⎛ ⎝

⎞ (3.4 m/s) 2 ⎟ 2 2(9.81 m/s )(6.0 m ) ⎟⎠

= 26° 42 •• The system shown in Figure 7-44 is initially at rest when the lower string is cut. Find the speed of the objects when they are momentarily at the same height. The frictionless pulley has negligible mass. Picture the Problem Let the system include the two objects and Earth. Then Wext = 0. Choose Ug = 0 at the elevation at which the two objects meet. With this choice, the initial potential energy of the 3.0-kg object is positive and that of the 2.0-kg object is negative. Their sum, however, is positive. Given our choice for Ug = 0, this initial potential energy is transformed entirely into kinetic energy.

Apply conservation of mechanical energy to the system to obtain:

Wext = ΔK + ΔU g = 0 or, because Wext = 0, ΔK = −ΔUg mvf2 − 12 mvi2 = − ΔU g

Noting that m represents the sum of the masses of the objects as they are both moving in the final state, substitute for ΔK:

or, because vi = 0,

ΔUg is given by:

ΔU g = U g,f − U g,i = 0 − (m3 − m2 )gh

Substitute for ΔUg to obtain:

1 2

1 2

mvf2 = −ΔU g ⇒ vf =

vf =

2(m3 − m2 )gh m

− 2ΔU g m

Conservation of Energy 639 Substitute numerical values and evaluate vf :

vf =

(

)

2(3.0 kg − 2.0 kg )(0.50 m ) 9.81m/s 2 = 1.4 m/s 3.0 kg + 2.0 kg

43 •• A block of mass m rests on an inclined plane (Figure 7-44). The coefficient of static friction between the block and the plane is μs. A gradually increasing force is pulling down on the spring (force constant k). Find the potential energy U of the spring (in terms of the given symbols) at the moment the block begins to move. y

Picture the Problem Fs is the force exerted by the spring and, because the block is on the verge of sliding, fs = fs,max. We can use Newton’s second law, under equilibrium conditions, to express the elongation of the spring as a function of m, k and θ and then substitute in the expression for the potential energy stored in a stretched or compressed spring.

Express the potential energy of the spring when the block is about to move: Apply

r

r

∑ F = ma, under equilibrium

conditions, to the block:

Using f s, max = μ s Fn and Fs = kx , eliminate f s, max and Fs from the x

r Fn

r f s, max

r Fs

x

θ r Fg

U = 12 kx 2

∑F

x

=Fs − f s,max − mg sin θ = 0

and ∑ Fy = Fn − mg cosθ = 0

x=

mg (sin θ + μs cos θ ) k

equation and solve for x: Substitute for x in the expression for U and simplify to obtain:

⎡ mg (sin θ + μs cosθ ) ⎤ U = k⎢ ⎥⎦ k ⎣ 1 2

=

[mg (sin θ + μs cosθ )] 2 2k

2

640 Chapter 7 44 •• A 2.40-kg block is dropped onto a spring and platform (Figure 7-45) of negligible mass. The block is released a distance of 5.00 m above the platform. When the block is momentary at rest, the spring is compressed by 25.0 cm. Find the speed of the block when the compression of the spring is only 15.0 cm Picture the Problem Let the system include the block, the spring, and Earth. Let Ug = 0 where the spring is compressed 15.0 cm. Then the mechanical energy when the compression of the spring is 15.0 cm will be partially kinetic and partially stored in the spring. We can use conservation of mechanical energy to relate the initial potential energy of the system to the energy stored in the spring and the kinetic energy of block when it has compressed the spring 15.0 cm.

Apply conservation of mechanical energy to the system to obtain:

m

h = 5.00 m

x = 15.0 cm

m

Ug = 0

Wext = ΔU + ΔK = 0 or U g,f − U g,i + U s,f − U s,i + K f − K i = 0

Because U g,f = U s,i = K i = 0 :

− U g,i + U s,f + K f = 0

Substitute to obtain:

− mg (h + x ) + 12 kx 2 + 12 mv 2 = 0

Solving for v yields:

v = 2 g (h + x ) −

The spring’s force constant is given by:

wblock mg = Δx Δx where ∆x is the compression of the spring when the block is momentarily at rest.

Substituting for k and simplifying gives:

v = 2 g (h + x ) −

kx 2 m

k=

mgx 2 mΔx

⎛ x2 ⎞ = g ⎜⎜ 2(h + x ) − ⎟⎟ Δx ⎠ ⎝

Conservation of Energy 641 Substitute numerical values and evaluate v: m) ⎤ (9.81m/s )⎡⎢2 (5.00 m + 0.150 m) − (0.150 ⎥= 0.250 m 2

v=

2

⎣

⎦

10.0 m/s

45 •• [SSM] A ball at the end of a string moves in a vertical circle with constant mechanical energy E. What is the difference between the tension at the bottom of the circle and the tension at the top? Picture the Problem The diagram represents the ball traveling in a circular path with constant energy. Ug has been chosen to be zero at the lowest point on the circle and the superimposed free-body diagrams show the forces acting on the ball at the top (T) and bottom (B) of the circular path. We’ll apply Newton’s second law to the ball at the top and bottom of its path to obtain a relationship between TT and TB and conservation of mechanical energy to relate the speeds of the ball at these two locations.

Apply

∑F

radial

= maradial to the ball at

the bottom of the circle and solve for TB:

F = maradial Apply ∑ radial to the ball at the top of the circle and solve for TT:

m

r mg r TT

R

r TB m

TB − mg = m

r mg

r v

m

Ug = 0

vB2 R

and

TB = mg + m

vB2 R

TT + mg = m

vT2 R

(1)

and

TT = −mg + m

vT2 R

(2)

642 Chapter 7 Subtract equation (2) from equation (1) to obtain:

TB − TT = mg + m

vB2 R

⎛ vT2 ⎞ ⎜ − ⎜ − mg + m ⎟⎟ R⎠ ⎝ 2 2 v v = m B − m T + 2mg R R

Using conservation of mechanical energy, relate the energy of the ball at the bottom of its path to its mechanical energy at the top of the circle: Substituting in equation (3) yields:

1 2

(3)

mvB2 = 12 mvT2 + mg (2 R )

or m

vB2 v2 − m T = 4mg R R

TB − TT = 6mg

46 •• A girl of mass m is taking a picnic lunch to her grandmother. She ties a rope of length R to a tree branch over a creek and starts to swing from rest at a point that is a distance R/2 lower than the branch. What is the minimum breaking tension for the rope if it is not to break and drop the girl into the creek? Picture the Problem Let the system consist of the girl and Earth and let Ug = 0 at the lowest point in the girl’s swing. We can apply conservation of mechanical energy to the system to relate the girl’s speed v to R. The force diagram shows the forces acting on the girl at the low point of her swing. Applying Newton’s second law to her will allow us to establish the relationship between the tension T and her speed.

Apply

∑F

radial

= maradial to the girl

at her lowest point and solve for T:

R

R

r T

=0 r Ug r Fg = m g

T − mg = m

v2 R

and v2 T = mg + m R

Apply conservation of mechanical energy to the system to obtain:

1 2

Wext = ΔK + ΔU = 0

or, because Ki = Uf = 0,

(1)

Conservation of Energy 643 K f −U i = 0

Substituting for Kf and Ui yields:

Substitute for v2/R in equation (1) and simplify to obtain:

1 2

mv 2 − mg

R v2 =0⇒ = g R 2

T = mg + mg = 2mg

47 •• A 1500-kg roller coaster car starts from rest a height H = 23.0 m (Figure 7-46) above the bottom of a 15.0-m-diameter loop. If friction is negligible, determine the downward force of the rails on the car when the upsidedown car is at the top of the loop. Picture the Problem Let the system include the car, the track, and Earth. The pictorial representation shows the forces acting on the car when it is upside down at the top of the loop. Choose Ug = 0 at the bottom of the loop. We can express Fn in terms of v and R by apply Newton’s second law to the car and then obtain a second expression in these same variables by applying conservation of mechanical energy to the system. The simultaneous solution of these equations will yield an expression for Fn in terms of known quantities.

Apply

∑F

radial

= maradial to the car at

the top of the circle and solve for Fn:

m

r Fg

r Fn

Ug = 0

Fn + mg = m

Substitute for Kf, Uf, and Ui to obtain:

v2 R

and Fn = m

Using conservation of mechanical energy, relate the energy of the car at the beginning of its motion to its energy when it is at the top of the loop:

R

v2 − mg R

Wext = ΔK + ΔU = 0

or, because Ki = 0, Kf +U f −Ui = 0

1 2

mv 2 + mg (2 R ) − mgH = 0

(1)

644 Chapter 7 v2 Solving for m yields: R

v2 ⎞ ⎛H m = 2mg ⎜ − 2 ⎟ R ⎠ ⎝R

Substitute equation (2) in equation (1) to obtain:

⎞ ⎛H Fn = 2mg ⎜ − 2 ⎟ − mg ⎠ ⎝R ⎛ 2H ⎞ = mg ⎜ − 5⎟ ⎝ R ⎠

(2)

Substitute numerical values and evaluate Fn: ⎡ 2(23.0 m ) ⎤ Fn = (1500 kg ) 9.81 m/s 2 ⎢ − 5⎥ = 16.7 kN ⎣ 7.50 m ⎦

(

)

48 •• A single roller-coaster car is moving with speed v0 on the first section of track when it descends a 5.0-m-deep valley, then climbs to the top of a hill that is 4.5 m above the first section of track. Assume any effects of friction or of air resistance are negligible. (a) What is the minimum speed v0 required if the car is to travel beyond the top of the hill? (b) Can we affect this speed by changing the depth of the valley to make the coaster pick up more speed at the bottom? Explain. Picture the Problem Let the system include the roller coaster, the track, and Earth and denote the starting position with the numeral 0 and the top of the second hill with the numeral 1. We can use the work-energy theorem to relate the energies of the coaster at its initial and final positions. Let m be the mass of the roller coaster. 1 m 0 m

h1 = 9.5 m

h0 = 5.0 m Ug = 0

(a) Use conservation of mechanical energy to relate the work done by external forces to the change in the energy of the system:

Wext = ΔEsys = ΔK + ΔU

Because the track is frictionless, Wext = 0:

ΔK + ΔU = 0 and K1 − K 0 + U1 − U 0 = 0

Substitute to obtain:

1 2

mv12 − 12 mv02 + mgh1 − mgh0 = 0

Conservation of Energy 645 Solving for v0 yields:

v0 = v12 + 2 g (h1 − h0 )

If the coaster just makes it to the top of the second hill, v1 = 0 and:

v0 = 2 g (h1 − h0 )

Substitute numerical values and evaluate v0:

v0 = 2 9.81m/s 2 (9.5 m − 5.0 m )

(

)

= 9.4 m/s

(b) No. Note that the required speed depends only on the difference in the heights of the two hills. 49 •• The Gravitron single-car roller coaster consists of a single loop-theloop. The car is initially pushed, giving it just the right mechanical energy so the riders on the coaster will feel ″weightless″ when they pass through the top of the circular arc. How heavy will they feel when they pass through the bottom of the arc (that is, what is the normal force pressing up on them when they are at the bottom of the loop)? Express the answer as a multiple of mg (their actual weight). Assume any effects of friction or of air resistance are negligible. Picture the Problem Let the radius of the loop be R and the mass of one of the riders be m. At the top of the loop, the centripetal force on her is her weight (the force of gravity). The two forces acting on her at the bottom of the loop are the normal force exerted by the seat of the car, pushing up, and the force of gravity, pulling down. We can apply Newton’s second law to her at both the top and bottom of the loop to relate the speeds at those locations to m and R and, at b, to F, and then use conservation of mechanical energy to relate vt and vb. t r mg

R

r Fn

b r mg

Apply

∑F

radial

= ma radial to the rider

at the bottom of the circular arc:

Ug = 0

Fn − mg = m

vb2 v2 ⇒ Fn = mg + m b (1) R R

646 Chapter 7 Apply

∑F

radial

= ma radial to the rider

at the top of the circular arc: Apply conservation of mechanical energy to the system to obtain:

vt2 mg = m ⇒ vt2 = gR R Kb − Kt + U b − U t = 0 or, because Ub = 0, Kb − Kt − U t = 0

Substitute for Kb, Kt, and Ut to obtain:

1 2

Solving for vb2 yields:

vb2 = 5 gR

Substitute for vb2 in equation (1) and simplify to obtain:

5 gR = 6mg R That is, the rider will feel six times heavier than her normal weight.

mvb2 − 12 mvt2 − 2mgR = 0

Fn = mg + m

50 •• A stone is thrown upward at an angle of 53° above the horizontal. Its maximum height above the release point is 24 m. What was the stone’s initial speed? Assume any effects of air resistance are negligible. Picture the Problem Let the system consist of the stone and Earth and ignore the influence of air resistance. Then Wext = 0. Choose Ug = 0 as shown in the figure. Apply conservation of mechanical energy to describe the energy transformations as the stone rises to the highest point of its trajectory.

y h

r vx

r v0 θ0

Apply conservation of mechanical energy to the system:

Ug = 0

Wext = ΔK + ΔU = 0 and K1 − K 0 + U1 − U 0 = 0

Because U0 = 0:

K1 − K 0 + U1 = 0

Substitute for the kinetic and potential energies yields:

1 2

mvx2 − 12 mv02 + mgh = 0

x

Conservation of Energy 647 In the absence of air resistance, the r horizontal component of v is constant and equal to vx = v0 cosθ : Solving for v0 gives:

Substitute numerical values and evaluate v0:

1 2

m(v0 cos θ ) − 12 mv02 + mgh = 0 2

v0 =

2 gh 1 − cos 2 θ

v0 =

2 9.81m/s 2 (24 m ) = 27 m/s 1 − cos 2 53°

(

)

51 •• A 0.17-kg baseball is launched from the roof of a building 12 m above the ground. Its initial velocity is 30 m/s at 40° above the horizontal. Assume any effects of air resistance are negligible. (a) What is the maximum height above the ground the ball reaches? (b) What is the speed of the ball as it strikes the ground? Picture the Problem The figure shows the ball being thrown from the roof of the building. Let the system consist of the ball and Earth. Then Wext = 0. Choose Ug = 0 at ground level. We can use conservation of mechanical energy to determine the maximum height of the ball and its speed at impact with the ground.

(a) Apply conservation of mechanical energy to obtain:

r v1

y, m H 12

θ h2

h1

Ug = 0

h3 = 0 x

Wext = ΔK + ΔU = 0 or K 2 − K1 + U 2 − U 1 = 0 mv 22 − 12 mv12 + mgh2 − mgh1 = 0

Substitute for the energies to obtain:

1 2

Note that, at point 2, the ball is moving horizontally and:

v 2 = v1 cosθ

Substitute for v2 and h2 to obtain:

1 2

m(v1 cosθ ) − 12 mv12 + mgH 2

− mgh1 = 0

Solving for H gives:

H = h1 −

(

)

v2 cos 2 θ − 1 2g

648 Chapter 7 Substitute numerical values and evaluate H:

H = 12 m −

(30 m/s)2

2(9.81m/s

2

) (cos

2

40° − 1)

= 31 m (b) Apply conservation of mechanical energy to the system to relate the initial mechanical energy of the ball to its just-before-impact energy:

Wext = ΔK + ΔU = 0 or, because U3 = 0, K 3 − K1 − U1 = 0

mv32 − 12 mv12 − mgh1 = 0

Substituting for K3, K1, and U1 yields:

1 2

Solve for v 3 to obtain:

v3 = v12 + 2gh1

Substitute numerical values and evaluate v3 :

v3 =

(30 m/s) 2 + 2(9.81m/s2 )(12 m )

= 34 m/s

52 •• An 80-cm-long pendulum with a 0.60-kg bob is released from rest at an initial angle of θ0 with the vertical. At the bottom of the swing, the speed of the bob is 2.8 m/s. (a) What is θ0? (b) What angle does the pendulum make with the vertical when the speed of the bob is 1.4 m/s? Is this angle equal to 12 θ 0 ? Explain why or why not. Picture the Problem The figure shows the pendulum bob in its release position and in the two positions in which it is in motion with the given speeds. Let the system consist of the pendulum and Earth and choose Ug = 0 at the low point of the swing. We can apply conservation of mechanical energy to relate the two angles of interest to the speeds of the bob at the intermediate and low points of its trajectory.

θ

θ0

Lcosθ0 L cosθ

L

m

r vf

m

r v ' h'

h

Ug = 0

Wext = ΔK + ΔU = 0

(a) Apply conservation of mechanical energy to the system to obtain:

or K f − K i + U f − U i = 0.

Because Ki = Uf = 0:

K f −U i = 0

(1)

Conservation of Energy 649 Refer to the pictorial representation to see that Ui is given by: Substitute for Kf and Ui in equation (1) to obtain: Solving for θ0 yields:

U i = mgh = mgL(1 − cos θ 0 ) mvf2 − mgL(1 − cos θ 0 ) = 0

1 2

⎛

θ 0 = cos −1 ⎜⎜1 − ⎝

Substitute numerical values and evaluate θ0:

v2 ⎞ ⎟ 2 gL ⎟⎠

2 ⎡ ⎤ ( 2.8 m/s ) θ 0 = cos ⎢1 − ⎥ 2 ⎣ 2 9.81 m/s (0.80 m ) ⎦ −1

(

)

= 60° (b) Letting primed quantities describe the indicated location, use conservation of mechanical energy to obtain :

K f' − Ki + U f' − U i = 0

Because Ki = 0:

K f' + U f' − U i = 0

Refer to the pictorial representation to see that U f' is given by:

U f' = mgh' = mgL(1 − cosθ )

Substitute for K f' , U f' and U i :

1 2

m(vf' ) + mgL(1 − cos θ ) 2

− mgL(1 − cos θ 0 ) = 0

⎡ (vf ')2

Solving for θ yields :

⎤ + cosθ 0 ⎥ ⎣ 2 gL ⎦

θ = cos −1 ⎢

Substitute numerical values and evaluate θ : ⎤ (1.4 m/s) 2 + cos 60°⎥ = 2 ⎣ 2(9.81m/s )(0.80 m ) ⎦ ⎡

θ = cos −1 ⎢

51°

No. The change in gravitational potential energy is linearly dependent on the cosine of the angle rather than on the angle itself. 53 •• The Royal Gorge bridge over the Arkansas River is 310 m above the river. A 60-kg bungee jumper has an elastic cord with an unstressed length of 50 m attached to her feet. Assume that, like an ideal spring, the cord is massless and

650 Chapter 7 provides a linear restoring force when stretched. The jumper leaps, and at her lowest point she barely touches the water. After numerous ascents and descents, she comes to rest at a height h above the water. Model the jumper as a point particle and assume that any effects of air resistance are negligible. (a) Find h. (b) Find the maximum speed of the jumper. Picture the Problem Choose Ug = 0 at the bridge and let the system be Earth, the jumper and the bungee cord. Then Wext = 0. We can use conservation of mechanical energy to relate her initial and final gravitational potential energies to the energy stored in the stretched bungee cord Us. In Part (b), we’ll use a similar strategy but include a kinetic energy term because we are interested in finding her maximum speed.

(a) Express her final height h above the water in terms of L, d and the distance x the bungee cord has stretched:

h=L–d−x

Use conservation of mechanical energy to relate her gravitational potential energy as she just touches the water to the energy stored in the stretched bungee cord:

Wext = ΔK + ΔU = 0

Because ΔK = 0 and ΔU = ΔUg + ΔUs:

2mgL s2 where s is the maximum distance the bungee cord has stretched.

Find the maximum distance the bungee cord stretches:

s = 310 m – 50 m = 260 m.

Substitute numerical values and evaluate k:

− mgL + 12 ks 2 = 0 ⇒ k =

2(60 kg ) (9.81 m/s 2 )(310 m ) (260 m )2 = 5.40 N/m

k=

(1)

Conservation of Energy 651 Express the relationship between the forces acting on her when she has finally come to rest x: Substitute numerical values and evaluate x:

Fnet = kx − mg = 0 ⇒ x =

mg k

( 60 kg ) (9.81 m/s 2 ) x= = 109 m 5.40 N/m

Substitute in equation (1) and evaluate h:

h = 310 m − 50 m − 109 m = 151m

(b) Using conservation of mechanical energy, express her total energy E:

E = K + U g + U s = Ei = 0

Because v is a maximum when K is a maximum, solve for K to obtain:

K = −U g − U s

Use the condition for an extreme value to obtain:

mg dK = mg − kx = 0 ⇒ x = k dx

Substitute numerical values and evaluate x:

x=

From equation (2) we have:

1 2

Solve for v to obtain:

= 0.15 km

= mg (d + x ) − 12 kx 2

(2)

(60 kg )(9.81m/s2 ) = 109 m 5.40 N/m

mv 2 = mg (d + x ) − 12 kx 2

v = 2 g (d + x ) −

kx 2 m

Substitute numerical values and evaluate v for x = 109 m: v = 2(9.81 m/s 2 )(50 m + 109 m ) −

(5.4 N/m )(109 m )2 60 kg

= 45 m/s

d 2K = −k < 0, x = 109 m corresponds to Kmax and so v is a maximum. Because dx 2 54 •• A pendulum consists of a 2.0-kg bob attached to a light 3.0-m-long string. While hanging at rest with the string vertical, the bob is struck a sharp horizontal blow, giving it a horizontal velocity of 4.5 m/s. At the instant the string

652 Chapter 7 makes an angle of 30° with the vertical, what is (a) the speed, (b) the gravitational potential energy (relative to its value is at the lowest point), and (c) the tension in the string? (d) What is the angle of the string with the vertical when the bob reaches its greatest height? Picture the Problem Let the system be Earth and the pendulum bob. Then Wext = 0. Choose Ug = 0 at the low point of the bob’s swing and apply conservation of mechanical energy to the system. When the bob reaches the 30° position its energy will be partially kinetic and partially potential. When it reaches its maximum height, its energy will be entirely potential. Applying Newton’s second law will allow us to express the tension in the string as a function of the bob’s speed and its angular position.

(a) Apply conservation of mechanical energy to relate the energies of the bob at points 1 and 2: Because U1 = 0: The potential energy of the system when the bob is at point 2 is given by: Substitute for U2 in equation (1) to obtain: Solving for v2 yields:

θ

L

L cos θ

m

2

Ug = 0

m 1

Wext = ΔK + ΔU = 0 or K 2 − K1 + U 2 − U 1 = 0 1 2

mv22 − 12 mv12 + U 2 = 0

(1)

U 2 = mgL(1 − cos θ )

1 2

mv22 − 12 mv12 + mgL(1 − cosθ ) = 0

v2 = v12 − 2 gL(1 − cos θ )

Substitute numerical values and evaluate v2: v2 =

(4.5 m/s)2 − 2(9.81m/s 2 )(3.0 m )(1 − cos30°) = 3.52 m/s =

(b) From (a) we have:

U 2 = mgL(1 − cos θ )

3.5 m/s

Conservation of Energy 653 Substitute numerical values and evaluate U2: U 2 = (2.0 kg )(9.81 m/s 2 )(3.0 m )(1 − cos30°) = 7.9 J

(c) Apply

∑F

radial

= maradial to the

T − mg cos θ = m

bob to obtain:

v22 L

⎛ v2 ⎞ T = m⎜⎜ g cos θ + 2 ⎟⎟ L⎠ ⎝

Solving for T yields:

Substitute numerical values and evaluate T: 2 ⎡ ( 3.52 m/s ) ⎤ 2 T = (2.0 kg )⎢ 9.81 m/s cos30° + ⎥ = 25 N 3.0 m ⎦ ⎣

(

)

(d) When the bob reaches its greatest height:

Substitute for K1 and U max : Solve for θmax to obtain:

U = U max = mgL(1 − cosθ max ) and − K1 + U max = 0 − 12 mv12 + mgL(1 − cos θ max ) = 0 ⎛

θ max = cos −1 ⎜⎜1 − ⎝

Substitute numerical values and evaluate θmax:

θ max

v12 ⎞ ⎟ 2 gL ⎟⎠

2 ⎡ ⎤ ( 4.5 m/s ) = cos ⎢1 − ⎥ 2 ⎣ 2 9.81m/s (3.0 m )⎦

−1

(

)

= 49° 55 •• [SSM] A pendulum consists of a string of length L and a bob of mass m. The bob is rotated until the string is horizontal. The bob is then projected downward with the minimum initial speed needed to enable the bob to make a full revolution in the vertical plane. (a) What is the maximum kinetic energy of the bob? (b) What is the tension in the string when the kinetic energy is maximum? Picture the Problem Let the system consist of Earth and the pendulum bob. Then Wext = 0. Choose Ug = 0 at the bottom of the circle and let points 1, 2 and 3 represent the bob’s initial point, lowest point and highest point, respectively. The bob will gain speed and kinetic energy until it reaches point 2 and slow down

654 Chapter 7 until it reaches point 3; so it has its maximum kinetic energy when it is at point 2. We can use Newton’s second law at points 2 and 3 in conjunction with conservation of mechanical energy to find the maximum kinetic energy of the bob and the tension in the string when the bob has its maximum kinetic energy. 3

m

r mg

L

1 m r v1

r T2 2 m

(a) Apply

∑F

radial

= maradial to the

bob at the top of the circle and solve for v32 :

r mg

Ug = 0

mg = m

v32 ⇒ v32 = gL L

K3 − K2 + U 3 −U 2 = 0

Apply conservation of mechanical energy to the system to express the relationship between K2, K3 and U3:

or, because U2 = 0, K3 − K2 + U 3 = 0

Solving for K2 yields:

K 2 = K max = K 3 + U 3

Substituting for K3 and U3 yields:

K max = 12 mv32 + mg (2 L )

Substitute for v32 and simplify to

K max = 12 m( gL ) + 2mgL =

5 2

mgL

obtain: (b) Apply ∑ Fradial = mac to the bob at the bottom of the circle and solve for T2:

Fnet = T2 − mg = m

v22 L

and T2 = mg + m

v22 L

(1)

Conservation of Energy 655 Use conservation of mechanical energy to relate the energies of the bob at points 2 and 3 and solve for K2:

K 3 − K 2 + U 3 − U 2 = 0 where U 2 = 0 and K 2 = K 3 + U 3 = 12 mv32 + mg (2 L ) mv22 = 12 m( gL ) + mg (2 L ) ⇒ v22 = 5 gL

Substitute for v32 and K2 to obtain:

1 2

Substitute for v22 in equation (1) and simplify to obtain:

T2 = mg + m

5 gL = 6mg L

56 •• A child whose weight is 360 N swings out over a pool of water using a rope attached to the branch of a tree at the edge of the pool. The branch is 12 m above ground level and the surface of the water is 1.8 m below ground level. The child holds onto the rope at a point 10.6 m from the branch and moves back until the angle between the rope and the vertical is 23°. When the rope is in the vertical position, the child lets go and drops into the pool. Find the speed of the child just as he impacts the surface of the water. (Model the child as a point particle attached to the rope 10.6 m from the branch.) Picture the Problem Let the system consist of Earth and the child. Then Wext = 0. In the figure, the child’s initial position is designated with the numeral 1; the point at which the child releases the rope and begins to fall with a 2, and its point of impact with the water is identified with a 3. Choose Ug = 0 at the water level. While one could apply conservation of mechanical energy between points 1 and 2 and then between points 2 and 3, it is more direct to consider the energy transformations between points 1 and 3.

Apply conservation of mechanical energy between points 1 and 3:

θ L

1 2

L(1 − cos θ ) h

Ug = 0

3

Wext = ΔK + ΔU = 0 K 3 − K1 + U 3 − U1 = 0 where U 3 and K1are zero. mv32 − mg [h + L(1 − cos θ )] = 0

Substitute for K3 and U1;

1 2

Solving for v3 yields:

v3 = 2 g [h + L(1 − cos θ )]

656 Chapter 7 Substitute numerical values and evaluate v3:

(

)

v3 = 2 9.81 m/s 2 [3.2 m + (10.6 m )(1 − cos23°)] = 8.9 m/s 57 •• Walking by a pond, you find a rope attached to a stout tree limb that is 5.2 m above ground level. You decide to use the rope to swing out over the pond. The rope is a bit frayed, but supports your weight. You estimate that the rope might break if the tension is 80 N greater than your weight. You grab the rope at a point 4.6 m from the limb and move back to swing out over the pond. (Model yourself as a point particle attached to the rope 4.6 m from the limb.) (a) What is the maximum safe initial angle between the rope and the vertical at which it will not break during the swing? (b) If you begin at this maximum angle, and the surface of the pond is 1.2 m below the level of the ground, with what speed will you enter the water if you let go of the rope when the rope is vertical? Picture the Problem Let the system consist of you and Earth. Then there are no external forces to do work on the system and Wext = 0. In the figure, your initial position is designated with the numeral 1, the point at which you release the rope and begin to fall with a 2, and your point of impact with the water is identified with a 3. Choose Ug = 0 at the water level. We can apply Newton’s second law to the forces acting on you at point 2 and apply conservation of mechanical energy between points 1 and 2 to determine the maximum angle at which you can begin your swing and then between points 1 and 3 to determine the speed with which you will hit the water.

(a) Use conservation of mechanical energy to relate your speed at point 2 to your potential energy there and at point 1: Because K1 = 0:

θ L r T

L(1 − cos θ ) 2 r r h Fg = m g

1

Ug = 0

3

Wext = ΔK + ΔU = 0 or K 2 − K1 + U 2 − U 1 = 0

1 2

mv22 + mgh

− [mgL(1 − cos θ ) + mgh] = 0

Conservation of Energy 657 Solve this equation for θ to obtain:

⎡

θ = cos −1 ⎢1 − ⎣

Apply

∑F

radial

= maradial to yourself

at point 2 and solve for T: Because you’ve estimated that the rope might break if the tension in it exceeds your weight by 80 N, it must be that:

T − mg = m

m

v22 ⎤ ⎥ 2 gL ⎦

(1)

v22 v2 and T = mg + m 2 L L

(80 N )L v22 = 80 N ⇒ v22 = L m

(80 N )(4.6 m ) = 5.26 m 2 /s 2

Let’s assume that your mass is 70 kg. Then:

v22 =

Substitute numerical values in equation (1) to obtain:

θ = cos −1 ⎢1 −

70 kg ⎡ ⎣

⎤ 5.26 m 2 /s 2 ⎥ 2 2 9.81 m/s (4.6 m ) ⎦

(

)

= 19.65° = 20°

(b) Apply conservation of mechanical energy between points 1 and 3: Substitute for K3 and U1 to obtain: Solving for v3 yields:

Wext = ΔK + ΔU = 0 or, because U3 = K1 = 0, K 3 −U 1 = 0 1 2

mv32 − mg [h + L(1 − cos θ )] = 0

v3 = 2 g [h + L(1 − cos θ )]

Substitute numerical values and evaluate v3: v3 = 2(9.81 m/s 2 )[1.2 m + (4.6 m )(1 − cos19.65°)] = 5.4 m/s 58 ••• A pendulum bob of mass m is attached to a light string of length L and is also attached to a spring of force constant k. With the pendulum in the position shown in Figure 7-47, the spring is at its unstressed length. If the bob is now pulled aside so that the string makes a small angle θ with the vertical and released, what is the speed of the bob as it passes through the equilibrium position? Hint: Recall the small-angle approximations: if θ is expressed in radians, and if θ < 1 , then sin θ ≈ tan θ ≈ θ and cos θ ≈ 1 – 12 θ 2 .

658 Chapter 7 Picture the Problem Choose Ug = 0 at point 2, the lowest point of the bob’s trajectory and let the system consist of the bob, spring, string and Earth. Given this choice, there are no external forces doing work on the system. Because θ >1 → k

(b) F is given by:

For r T), the maximum value of T′ that can be maintained without the rope slipping is Tmax ′ = Teμs Δθ , where μs is the coefficient of static friction. Consider the Atwood’s machine in Figure 9-77: the pulley has a radius R = 0.15 m, the moment of inertia is I = 0.35 kg⋅m2, and the coefficient of static friction between the wheel and the string is μs = 0.30. (a) If the tension on one side of the pulley is 10 N, what is the maximum tension on the other side that will prevent the rope from slipping on the pulley? (b) What is the acceleration of the blocks in this case? (c) If the mass of one of the hanging blocks is 1.0 kg, what is the maximum mass of the other block if, after the blocks are released, the pulley is to rotate without slipping? Picture the Problem Free-body diagrams for the pulley and the two blocks are shown to the right. Choose a coordinate system in which the direction of motion of the block whose mass is M (downward) is the +y direction. We can use the given relationship T 'max = Te μsΔθ to relate the tensions in the rope on either side of the pulley and apply Newton’s second law in both rotational form (to the pulley) and translational form (to the blocks) to obtain a system of equations that we can solve simultaneously for a, T1, T2, and M.

R

T1

T2

T2

T1 m

M mg

Mg

(a) Use T 'max = Te μsΔθ to evaluate the maximum tension required to prevent the rope from slipping on the pulley:

T ' max = (10 N )e (0.30 )π = 25.66 N

(b) Given that the angle of wrap is π radians, express T2 in terms of T1:

T2 = T1e (0.30 )π

Because the rope doesn’t slip, we can relate the angular acceleration, α, of the pulley to the acceleration, a, of the hanging masses by:

α=

= 26 N

a R

(1)

964 Chapter 9 Apply

∑τ = Iα to the pulley to

obtain: Substitute for T2 from equation (1) in equation (2) to obtain: Solving for T1 yields:

∑F

(T2 − T1 ) R = I a

(T e (

0.30 )π

1

T1 =

(e

)

− T1 R = I

I

(0.30 )π

)

−1 R 2

whose mass is m to obtain:

T1 − mg = ma and T1 = ma + mg

Equating these two expressions for T1 and solving for a yields:

a=

Apply

y

= ma y to the block

Substitute numerical values and evaluate a:

(2)

R a R

a

(3)

g

(e

(0.30 )π

a=

(e (

0.30 )π

I −1 − 1 mR 2

)

9.81 m/s 2 0.35 kg ⋅ m 2 −1 2 − 1 (1.0 kg )(0.15 m )

)

= 1.098 m/s 2 = 1.1 m/s 2 (c) Apply

∑F

y

= ma y to the block

whose mass is M to obtain: Substitute for T2 (from equation (1)) and T1 (from equation (3)) yields:

Mg − T2 = Ma ⇒ M =

M=

T2 g −a

m(a + g )e (0.30 )π g −a

Substitute numerical values and evaluate M:

M=

(1.0 kg )(1.098 m/s 2 + 9.81 m/s 2 )e (0.30 )π 9.81 m/s 2 − 1.098 m/s 2

= 3.2 kg

128 ••• A massive, uniform cylinder rhas a mass m and a radius R (Figure 977). It is accelerated by a tension force T that is applied through a rope wound around a light drum of radius r that is attached to the cylinder. The coefficient of static friction is sufficient for the cylinder to roll without slipping. (a) Find the frictional force. (b) Find the acceleration a of the center of the cylinder. (c) Show that it is possible to choose r so that a is greater than T/m. (d) What is the direction of the frictional force in the circumstances of Part (c)?

Rotation 965 Picture the Problem When the tension is horizontal, the cylinder will roll r forward and the friction force will be in the direction of T . We can use Newton’s second law to obtain equations that we can solve simultaneously for a and f.

(a) Apply Newton’s second law to the cylinder:

∑F

x

= T + f = ma

and

∑ τ = Tr − fR = Iα

Substitute for I and α in equation (2) to obtain:

Tr − fR = 12 mR 2

Solve equation (3) for f:

f =

Tr 1 − 2 ma R

Substitute equation (4) in equation (1) to obtain:

T+

Tr 1 − 2 ma = ma R

Solving for a gives:

a=

r⎞ 2T ⎛ ⎜1 + ⎟ 3m ⎝ R ⎠

f =

Tr 1 ⎡ 2T ⎛ r ⎞⎤ − m ⎢ ⎜1 + ⎟ ⎥ R 2 ⎣ 3m ⎝ R ⎠⎦

Substitute equation (5) in equation (4) to obtain: Simplifying yields:

(1)

f =

a 1 = mRa R 2

T ⎛ 2r ⎞ ⎜ − 1⎟ 3⎝ R ⎠

(b) Solve equation (4) for a:

Tr ⎞ ⎛ 2⎜ f − ⎟ R⎠ a= ⎝ m

Substituting for f yields:

⎛ T ⎛ 2r ⎞ Tr ⎞ 2⎜⎜ ⎜ − 1⎟ − ⎟⎟ 3⎝ R ⎠ R⎠ a= ⎝ m = −

r⎞ 2T ⎛ ⎜1 + ⎟ 3m ⎝ R ⎠

(2) (3)

(4)

(5)

966 Chapter 9

r⎞ T r⎞ 2⎛ 2T ⎛ ⎜1 + ⎟ > ⇒ ⎜ 1 + ⎟ > 1 3⎝ R⎠ 3m ⎝ R ⎠ m or r > 12 R

(c) Express the condition that T a> : m

r f > 0, i.e., in the direction of T .

(d) If r > 12 R :

129 ••• A uniform rod that has a length L and a mass M is free to rotate about a horizontal axis through one end, as shown in Figure 9-78. The rod is released from rest at θ = θ0. Show that the parallel and perpendicular components of the force exerted by the axis on the rod are given by F|| = 12 Mg (5 cosθ − 3 cosθ 0 ) and

F⊥ = 14 Mg sin θ , where F|| is the component parallel with the rod and F⊥ is the component perpendicular to the rod. Picture the Problem The system is shown in the drawing in two positions, with angles θ0 and θ with the vertical. The drawing also shows all the forces that act on the rod. These forces result in a rotation of the rod—and its center of mass—about the pivot, and a tangential acceleration of the center of mass. We’ll apply the conservation of mechanical energy and Newton’s second law to relate the radial and tangential forces acting on the stick.

Apply conservation of mechanical energy to relate the kinetic energy of the stick when it makes an angle θ with the vertical to its initial potential energy: Substituting for Kf, Uf and Ui gives:

θ0

r F⊥

L

θ

Mg

r F⏐⏐

Kf − Ki + U f − U i = 0 or, because Ki = 0, Kf +Uf −Ui = 0

1 2

Iω 2 + Mg

L L cosθ − Mg cosθ 0 = 0 2 2

Substitute for I to obtain:

(

1 1 2 3

)

ML2 ω 2 + Mg

θ

L L cosθ − Mg cosθ 0 = 0 2 2

Rotation 967 Solving for ω2 yields:

ω2 =

3g (cosθ 0 − cosθ ) L

Applying Newton’s second law in the centripetal direction gives:

L 2 ω 2 where toward the pivot is the positive centripetal direction.

Substituting for the gravitational force component yields:

F|| + Mg cosθ = M

F|| + Fg c = M

L 2 ω 2

or, solving for F|| ,

F|| = M Substitute for ω 2 and simplify to obtain:

L 2 ω − Mg cosθ 2

L ⎡ 3g (cosθ 0 − cosθ )⎤⎥ − Mg cosθ ⎢ 2⎣ L ⎦ 1 = 2 Mg [(3 cos θ 0 − 3 cos θ )] − Mg cos θ

F|| = M

= 12 Mg [(3 cosθ 0 − 3 cosθ − 2 cosθ )] = − 12 Mg [(5 cosθ − 3 cosθ 0 )]

where the minus sign is a consequence of our choice of the positive direction for the centripetal force. Relate the tangential acceleration of the center of mass to its angular acceleration: Use Newton’s second law to relate the angular acceleration of the stick to the net torque acting on it: Express a⊥ in terms of α:⊥ Solving for F⊥ yields:

a⊥= 12 Lα

α=

τ net I

=

L sin θ 3g sin θ 2 = 2 1 2L 3 ML

Mg

a⊥ = 12 Lα = 34 g sin θ = g sin θ +

F⊥ M

F⊥ = − 14 Mg sin θ where the minus sign indicates that the force is directed oppositely to the tangential component r of Mg.

968 Chapter 9

Chapter 10 Angular Momentum Conceptual Problems 1

•

True or false:

(a)

If two vectors are exactly opposite in direction, their vector product must be zero. (b) The magnitude of the vector product of two vectors is at a minimum when the two vectors are perpendicular. (c) Knowing the magnitude of the vector product of two nonzero vectors and the vectors′ individual magnitudes uniquely determines the angle between them.

r r Determine the Concept The vector product of A and B is defined to be r r r A × B = AB sin φ nˆ where nˆ is a unit vector normal to the plane defined by A and r r r B and φ is the angle between A and B .

r r (a) True. If A and B are in opposite direction, then sinφ = sin 180° = 0. r r (b) False. If rA and B are perpendicular, then sinφ = sin 90° = 1 and the vector r product of A and B is a maximum. (c) False. Knowing the magnitude of the vector product and the vectors′ individual magnitudes only gives the magnitude of the sine of the angle between the vectors. It does not determine the angle uniquely, nor does this knowledge tell us if the sine of the angle is positive or negative. r r 2 • Consider two r vectors A and B . Their vector product has the r nonzero greatest magnitude if A and B are (a) parallel, (b) perpendicular, (c) antiparallel, (d) at an angle of 45° to each other.

r r Determine the Concept The vector product of the vectors A and B is defined to r r be A × B = AB sin φ nˆ where nˆ is a unit vector normal to the plane defined by r r r r A and B and φ is the angle between A and B . Hence, the vector product of r r r A and B is a maximum when sinφ = 1. This condition is satisfied provided A and r B are perpendicular. (b) is correct.

969

970

Chapter 10

r r 3 • What is the angle between a force vector F and a torque vector τ r produced by F ? r r r Determine the Concept Because τ = r × F = rF sin φ nˆ , where nˆ is a unit vector r r r r normal to the plane defined by r and F , the angle between F and τ is 90°. 4 • A point particle of mass m is moving with a constant speed v along a straight line that passes through point P. What can you say about the angular momentum of the particle relative to point P? (a) Its magnitude is mv. (b) Its magnitude is zero. (c) Its magnitude changes sign as the particle passes through point P. (d) It varies in magnitude as the particle approaches point P.

r r r r r Determine the Concept L and p are related according to L = r × p and the r r r magnitude of L is L = rp sin φ where φ is the angle between r and p . Because

the motion is along a line that passes through point P, r = 0 and so is L. (b) is correct. 5 • [SSM] A particle travels in a circular path and point P is at the r center of the circle. (a) If the particle’s linear momentum p is doubled without changing the radius of the circle, how is the magnitude of its angular momentum about P affected? (b) If the radius of the circle is doubled but the speed of the particle is unchanged, how is the magnitude of its angular momentum about P affected?

r r r r r Determine the Concept L and p are related according to L = r × p and the r r r magnitude of L is L = rp sin φ where φ is the angle between r and p . r r (a) Because L is directly proportional to p , L is doubled. r r (b) Because L is directly proportional to r , L is doubled. 6 • A particle moves along a straight line at constant speed. How does its angular momentum about any fixed point vary with time? Determine the Concept We can determine how the angular momentum of the particle about any fixed point varies with time by examining the derivative of the r r cross product of r and p .

The angular momentum of the particle is given by:

r r r L=r× p

Angular Momentum

r Differentiate L with respect to time to obtain:

r r r dL ⎛ r dp ⎞ ⎛ dr r ⎞ = ⎜ r × ⎟ + ⎜ × p⎟ dt ⎝ dt ⎠ ⎝ dt ⎠

r r r dp r = Fnet , and Because p = mv , dt r dr r =v: dt

r r r dL r r = r × Fnet + (v × p ) dt

(

Because the particle moves along a straight line at constant speed:

971

(1)

)

r r Fnet = 0 ⇒ r × Fnet = 0

r r r Because v and p(= mv ) are parallel:

r r v× p=0

Substitute in equation (1) to obtain:

r r dL = 0 ⇒ L = constant dt

An alternate solution The following diagram shows a particle whose mass is m moving along a straight line r at constant speed. The point identified as P is any fixed point and the vector r is the position vector, relative to this fixed point, of the particle moving with constant velocity.

φ

m r r

r v

d

φ P From the definition of angular momentum, the magnitude of the angular momentum of the particle, relative to the fixed point at P, is given by:

r L = rmv sin φ = mv(r sin φ )

Because d = r sin φ :

r L = mvd

or, because m, v and d are constants, r L = constant 7 •• True or false: If the net torque on a rotating system is zero, the angular velocity of the system cannot change. If your answer is false, give an example of such a situation.

972

Chapter 10

False. The net torque acting on a rotating system equals the change in the system’s angular momentum; that is, τ net = dL dt where L = Iω. Hence, if τ net is zero, all we can say for sure is that the angular momentum (the product of I and ω) is constant. If I changes, so must ω. An example is a high diver going from a tucked to a layout position. 8 •• You are standing on the edge of a turntable with frictionless bearings that is initially rotating when you catch a ball that is moving in the same direction but faster than you are moving and on a line tangent to the edge of the turntable. Assume you do not move relative to the turntable. (a) Does the angular speed of the turntable increase, decrease, or remain the same during the catch? (b) Does the magnitude of your angular momentum (about the rotation axis of the table) increase, decrease, or remain the same after the catch? (c) How does the ball’s angular momentum (about the rotation axis of the table) change after the catch? (d) How does the total angular momentum of the system, you-table-ball (about the rotation axis of the table) change after the catch? Determine the Concept You can apply conservation of angular momentum to the you-table-ball system to answer each of these questions.

(a) Because the ball is moving in the same direction that you are moving, your angular speed will increase when you catch it. Your Newton’s 3rd law interaction with the ball causes a torque that acts on the you-table-ball system to increase ω. (b) The ball has angular momentum relative to the rotation axis of the table before you catch it and so catching it increases your angular momentum relative to the rotation axis of the table. (c) The ball will slow down as a result of your catch and so its angular momentum relative to the center of the table will decrease. (d) Because there is zero net torque on the you-table-ball system, its angular momentum remains the same. 9 •• If the angular momentum of a system about a fixed point P is constant, which one of the following statements must be true?

(a) (b) ( c) (d) ( e)

No torque about P acts on any part of the system. A constant torque about P acts on each part of the system. Zero net torque about P acts on each part of the system. A constant external torque about P acts on the system. Zero net external torque about P acts on the system.

Angular Momentum

973

Determine the Concept If L is constant, we know that the net torque acting on the system is zero. There may be multiple constant or time-dependent torques acting on the system as long as the net torque is zero. (e) is correct.

A block sliding on a frictionless table is attached to a string that passes 10 •• through a narrow hole through the tabletop. Initially, the block is sliding with speed v0 in a circle of radius r0. A student under the table pulls slowly on the string. What happens as the block spirals inward? Give supporting arguments for your choice. (The term angular momentum refers to the angular momentum about a vertical axis through the hole.) (a) Its energy and angular momentum are conserved. (b) Its angular momentum is conserved and its energy increases. (c) Its angular momentum is conserved and its energy decreases. (d) Its energy is conserved and its angular momentum increases. (e) Its energy is conserved and its angular momentum decreases. Determine the Concept The pull that the student exerts on the block is at right r r r angles to its motion and exerts no torque (recall that τ = r × F and τ = rF sin φ ). Therefore, we can conclude that the angular momentum of the block is conserved. The student does, however, do work in displacing the block in the direction of the radial force and so the block’s energy increases. (b) is correct. 11 •• [SSM] One way to tell if an egg is hardboiled or uncooked without breaking the egg is to lay the egg flat on a hard surface and try to spin it. A hardboiled egg will spin easily, while an uncooked egg will not. However, once spinning, the uncooked egg will do something unusual; if you stop it with your finger, it may start spinning again. Explain the difference in the behavior of the two types of eggs. Determine the Concept The hardboiled egg is solid inside, so everything rotates with a uniform angular speed. By contrast, when you start an uncooked egg spinning, the yolk will not immediately spin with the shell, and when you stop it from spinning the yolk will continue to spin for a while. 12 •• Explain why a helicopter with just one main rotor has a second smaller rotor mounted on a horizontal axis at the rear as in Figure 10-40. Describe the resultant motion of the helicopter if this rear rotor fails during flight. Determine the Concept The purpose of the second smaller rotor is to prevent the body of the helicopter from rotating. If the rear rotor fails, the body of the helicopter will tend to rotate on the main axis due to angular momentum being conserved.

974

Chapter 10

13 •• The spin angular-momentum vector for a spinning wheel is parallel with its axle and is pointed east. To cause this vector to rotate toward the south, in which direction must a force be exerted on the east end of the axle? (a) up, (b) down, (c) north, (d) south, (e) east. r r r Determine the Concept The vector ΔL = Lf − Li (and the torque that is responsible for this change in the direction of the angular momentum vector) initially points to the south. One can use a right-hand rule to determine the direction of this torque, and hence the force exerted on the east end of the axle, required to cause the angular-momentum vector to rotate toward the south. Letting the fingers of your right hand point east, rotate your wrist until your thumb points south. Note that your curled fingers, which point in the direction of the force that must be exerted on the east end of the axle, point upward. (a ) is

correct. 14 •• You are walking toward the north and in your left hand you are carrying a suitcase that contains a massive spinning wheel mounted on an axle attached to the front and back of the case. The angular velocity of the gyroscope points north. You now begin to turn to walk toward the east. As a result, the front end of the suitcase will (a) resist your attempt to turn it and will try to maintain its original orientation, (b) resist your attempt to turn and will pull to the west, (c) rise upward, (d) dip downward, (e) show no effect whatsoever. Determine the Concept In turning toward the east, you redirect the angular momentum vector from north to east by exerting a torque on the spinning wheel. The force that you must exert to produce this torque (use a right-hand rule with your thumb pointing either east of north and note that your fingers point upward) is upward. That is, the force you exert on the front end of the suitcase is upward and the force the suitcase exerts on you is downward. Consequently, the front end of the suitcase will dip downward. (d ) is correct. 15 •• [SSM] The angular momentum of the propeller of a small singleengine airplane points forward. The propeller rotates clockwise if viewed from behind. (a) Just after liftoff, as the nose of the plane tilts upward, the airplane tends to veer to one side. To which side does it tend to veer and why? (b) If the plane is flying horizontally and suddenly turns to the right, does the nose of the plane tend to veer upward or downward? Why? r (a) The plane tends to veer to the right. The change in angular momentum ΔLprop r for the propeller is upward, so the net torque τ on the propeller is upward as well. The propeller must exert an equal but opposite torque on the plane. This downward torque exerted on the plane by the propeller tends to cause a downward

Angular Momentum

975

change in the angular momentum of the plane. This means the plane tends to rotate clockwise as viewed from above. (b) The nose of the plane tends to veer downward. The change in angular r r momentum ΔLprop for the propeller is to the right, so the net torque τ on the propeller is toward the right as well. The propeller must exert an equal but opposite torque on the plane. This leftward directed torque exerted by the propeller on the plane tends to cause a leftward-directed change in angular momentum for the plane. This means the plane tends to rotate clockwise as viewed from the right. 16 •• You have designed a car that is powered by the energy stored in a r single flywheel with a spin angular momentum L . In the morning, you plug the car into an electrical outlet and a motor spins the flywheel up to speed, adding a huge amount of rotational kinetic energy to it—energy that will be changed into translational kinetic energy of the car during the day. Having taken a physics course involving angular momentum and torques, you realize that problems would arise during various maneuvers of the car. Discuss some of these problems. r For example, suppose the flywheel is mounted so L points vertically upward when the car is on a horizontal road. What would happen as the car travels over a r hilltop? Through a valley? Suppose the flywheel is mounted so L points forward, or to one side, when the car is on a horizontal road. Then what would happen as the car attempts to turn to the left or right? In each case that you examine, consider the direction of the torque exerted on the car by the road. r Determine the Concept If L points upward and the car travels over a hill or through a valley, the force the road exerts on the wheels on one side (or the other) r will increase and car will tend to tip. If L points forward and the car turns left or right, the front (or rear) of the car will tend to lift. These problems can be averted by having two identical flywheels that rotate on the same shaft in opposite directions. 17 •• [SSM] You are sitting on a spinning piano stool with your arms folded. (a) When you extend your arms out to the side, what happens to your kinetic energy? What is the cause of this change? (b) Explain what happens to your moment of inertia, angular speed and angular momentum as you extend your arms. Determine the Concept The rotational kinetic energy of the you-stool system is L2 given by K rot = 12 Iω 2 = . Because the net torque acting on the you-stool system 2I r is zero, its angular momentum L is conserved.

976

Chapter 10

(a) Your kinetic energy decreases. Increasing your moment of inertia I while L2 . conserving your angular momentum L decreases you kinetic energy K = 2I (b) Extending your arms out to the side increases your moment of inertia, decreases your angular speed. The angular momentum of the system is unchanged. 18 •• A uniform rod of mass M and length L rests on a horizontal frictionless table. A blob of putty of mass m = M/4 moves along a line perpendicular to the rod, strikes the rod near its end, and sticks to the rod. Describe qualitatively the subsequent motion of the rod and putty. Determine the Concept The center of mass of the rod-and-putty system moves in a straight line, and the system rotates about its center of mass.

Estimation and Approximation 19 •• [SSM] An ice-skater starts her pirouette with arms outstretched, rotating at 1.5 rev/s. Estimate her rotational speed (in revolutions per second) when she brings her arms flat against her body. Picture the Problem Because we have no information regarding the mass of the skater, we’ll assume that her body mass (not including her arms) is 50 kg and that each arm has a mass of 4.0 kg. Let’s also assume that her arms are 1.0 m long and that her body is cylindrical with a radius of 20 cm. Because the net external torque acting on her is zero, her angular momentum will remain constant during her pirouette.

Because the net external torque acting on her is zero:

ΔL = Lf − Li = 0 or I arms in ω arms in − I arms out ω arms out = 0 (1)

Express her total moment of inertia with her arms outstretched:

I arms out = I body + I arms

Treating her body as though it is cylindrical, calculate the moment of inertia of her body, minus her arms:

I body = 12 mr 2 =

1 2

(50 kg )(0.20 m )2

= 1.00 kg ⋅ m 2

Angular Momentum

[

]

Modeling her arms as though they are rods, calculate their moment of inertia when they are outstretched:

I arms = 2 13 (4 kg )(1.0 m )

Substitute to determine her total moment of inertia with her arms outstretched:

I arm s out = 1.00 kg ⋅ m 2 + 2.67 kg ⋅ m 2

Express her total moment of inertia with her arms flat against her body:

I arms in = I body + I arms

2

= 2.67 kg ⋅ m 2

= 3.67 kg ⋅ m 2

= 1.00 kg ⋅ m 2

[

+ 2 (4.0 kg )(0.20 m )

2

= 1.32 kg ⋅ m Solve equation (1) for ωarms in to obtain: Substitute numerical values and evaluate ωarms in :

977

ωarms in =

ωarms in =

I arms out I arms in

]

2

ωarms out

3.67 kg ⋅ m 2 (1.5 rev/s) 1.32 kg ⋅ m 2

≈ 4 rev/s 20 •• Estimate the ratio of angular velocities for the rotation of a diver between the full tuck position and the full-layout position. Picture the Problem Because the net external torque acting on the diver is zero, the diver’s angular momentum will remain constant as she rotates from the full tuck to the full layout position. Assume that, in layout position, the diver is a thin rod of length 2.5 m and that, in the full tuck position, the diver is a sphere of radius 0.50 m.

Because the net external torque acting on the diver is zero:

Solving for the ratio of the angular velocities gives:

ΔL = Llayout − Ltuck = 0 or I layoutωlayout − I tuckω tuck = 0

ω tuck I layout = ω layout I tuck

978

Chapter 10

Substituting for the moment of inertia of a thin rod relative to an axis through its center of mass and the moment of inertia of a sphere relative to its center of mass and simplifying yields:

ω tuck 121 ml 2 5l 2 = = ω layout 52 mr 2 24r 2

Substitute numerical values and evaluate ω tuck ω layout :

2 ω tuck 5(2.5 m ) = ≈ 5 ω layout 24(0.50 m )2

21 •• The days on Mars and Earth are of nearly identical length. Earth’s mass is 9.35 times Mars’s mass, Earth’s radius is 1.88 times Mars’s radius, and Mars is on average 1.52 times farther away from the Sun than Earth is. The Martian year is 1.88 times longer than Earth’s year. Assume that they are both uniform spheres and that their orbits about the Sun are circles. Estimate the ratio (Earth to Mars) of (a) their spin angular momenta, (b) their spin kinetic energies, (c) their orbital angular momenta, and (d) their orbital kinetic energies. Picture the Problem We can use the definitions of spin angular momentum, spin kinetic energy, orbital angular momentum, and orbital kinetic energy to evaluate these ratios.

(a) The ratio of the spin angular momenta of Earth and Mars is:

⎛ LE ⎜⎜ ⎝ LM

⎞ I ω ⎟⎟ = E E ⎠ spin I Mω M

Because Mars and Earth have nearly identical lengths of days, ωE ≈ ωM:

⎛ LE ⎜⎜ ⎝ LM

⎞ I ⎟⎟ ≈ E ⎠ spin I M

Substituting for the moments of inertia and simplifying yields:

⎛ LE ⎜⎜ ⎝ LM

2 ⎞ M R2 M ⎟⎟ ≈ 5 E E2 = E 2 ⎠ spin 5 M M RM M M

Substitute numerical values for the ⎛L ⎞ ratios and evaluate ⎜⎜ E ⎟⎟ : ⎝ LM ⎠ spin

⎛ LE ⎜⎜ ⎝ LM

⎞ ⎟⎟ ≈ 9.35(1.88)2 ≈ 33 ⎠ spin

(b) The ratio of the spin kinetic energies of Earth and Mars is:

⎛ KE ⎜⎜ ⎝ KM

⎞ ⎟⎟ ⎠ spin

L2E 2I E L2E I M = 2 = 2 LM I E LM 2I M

⎛ RE ⎜⎜ ⎝ RM

⎞ ⎟⎟ ⎠

2

Angular Momentum

979

⎛L Because ⎜⎜ E ⎝ LM

⎞ I ⎟⎟ ≈ E : ⎠ spin I M

⎛ KE ⎜⎜ ⎝ KM

⎞ ⎛L ⎞ L2 L ⎟⎟ = 2E M = ⎜⎜ E ⎟⎟ ⎠ spin LM LE ⎝ LM ⎠ spin

⎛L From (a) ⎜⎜ E ⎝ LM

⎞ ⎟⎟ ≈ 33 . Hence: ⎠ spin

⎛ KE ⎜⎜ ⎝ KM

⎞ ⎟⎟ ≈ 33 ⎠ spin

(c) Treating Earth and Mars as point objects, the ratio of their orbital angular momenta is:

⎛ LE ⎜⎜ ⎝ LM

I ω ⎞ ⎟⎟ = E orb, E ⎠ orb I Mω orb, M

Substituting for the moments of inertia and orbital angular speeds yields:

⎛ 2π ⎞ ⎟ M E rE2 ⎜⎜ TE ⎟⎠ ⎛ LE ⎞ ⎝ ⎜⎜ ⎟⎟ = ⎝ LM ⎠ orb M r 2 ⎛⎜ 2π ⎞⎟ M M⎜ ⎟ ⎝ TM ⎠ where rE and rM are the radii of the orbits of Earth and Mars, respectively.

Simplify to obtain:

⎛ LE ⎜⎜ ⎝ LM

⎛ M ⎞⎛ r ⎞ ⎟⎟ = ⎜⎜ E ⎟⎟ ⎜⎜ E ⎠ orb ⎝ M M ⎠ ⎝ rM

Substitute numerical values for the ⎛L ⎞ three ratios and evaluate ⎜⎜ E ⎟⎟ : ⎝ LM ⎠ orb

⎛ LE ⎜⎜ ⎝ LM

⎞ ⎛ 1 ⎞ ⎟⎟ = (9.35)⎜ ⎟ (1.88) ≈ 8 ⎝ 1.52 ⎠ ⎠ orb

(d) The ratio of the orbital kinetic energies of Earth and Mars is:

⎛ KE ⎜⎜ ⎝ KM

⎞ ⎟⎟ ⎠

2

⎛ TM ⎜⎜ ⎝ TE

⎞ ⎟⎟ ⎠

2

1 I ω2 E ⎞ ⎟⎟ = 12 E orb, 2 ⎠ orb 2 I Mω orb, M

Substituting for the moments of inertia and angular speeds and simplifying gives: 2

⎛ KE ⎜⎜ ⎝ KM

⎞ ⎟⎟ ⎠ orb

⎛ 2π ⎞ ⎟ M r ⎜⎜ TE ⎟⎠ ⎛ M ⎞⎛ r ⎝ = = ⎜⎜ E ⎟⎟ ⎜⎜ E 2 ⎛ 2π ⎞ ⎝ M M ⎠ ⎝ rM ⎟⎟ M M rM2 ⎜⎜ ⎝ TM ⎠ 2 E E

Substitute numerical values for the ⎛K ⎞ ratios and evaluate ⎜⎜ E ⎟⎟ : ⎝ K M ⎠ orb

⎛ KE ⎜⎜ ⎝ KM

⎞ ⎟⎟ ⎠

2

⎛ TM ⎜⎜ ⎝ TE

⎞ ⎟⎟ ⎠

2

2

⎞ 1 ⎞ 2 ⎟⎟ = (9.35)⎛⎜ ⎟ (1.88) ≈ 14 ⎝ 1.52 ⎠ ⎠ orb

980

Chapter 10

22 •• The polar ice caps contain about 2.3 × 1019 kg of ice. This mass contributes negligibly to the moment of inertia of Earth because it is located at the poles, close to the axis of rotation. Estimate the change in the length of the day that would be expected if the polar ice caps were to melt and the water were distributed uniformly over the surface of Earth. Picture the Problem The change in the length of the day is the difference between its length when the ice caps have melted and the water has been distributed over the surface of Earth and the length of the day before the ice caps melt. Because the net torque acting on Earth during this process is zero, angular momentum is conserved and we can relate the angular speed (which are related to the length of the day) of Earth before and after the ice caps melt to the moments of inertia of the Earth-plus-spherical shell the ice caps melt.

Express the change in the length of a day as:

ΔT = Tafter − Tbefore

Because the net torque acting on Earth during this process is zero, angular momentum is conserved:

ΔL = Lafter − Lbefore = 0

Substituting for Lafter and Lbefore yields: Because ω = 2π T :

(I

sphere

+ I shell )ωafter − I sphereω before = 0

(I

sphere

+ I shell )

2π 2π − I sphere =0 Tafter Tbefore

or, simplifying, I sphere + I shell I sphere − =0 Tafter Tbefore Solve for Tafter to obtain:

⎛ ⎞ I Tafter = ⎜1 + shell ⎟Tbefore ⎜ I ⎟ sphere ⎠ ⎝

Substituting for Tafter in equation (1)

⎛ ⎞ I ΔT = ⎜1 + shell ⎟Tbefore − Tbefore ⎜ I ⎟ sphere ⎠ ⎝ I = shell Tbefore I sphere

and simplifying yields:

(1)

Angular Momentum

981

mr 2 5m ΔT = 2 Tbefore T = 2 before 3M E 5 M E RE

Substitute for Ishell and Isphere and simplify to obtain:

2 3

Substitute numerical values and evaluate ΔT: ΔT =

( (

) ⎛⎜1d × 24 h × 3600 s ⎞⎟ = d h ⎠ )⎝

5 2.3 × 1019 kg 3 5.98 × 10 24 kg

0.55 s

23 •• [SSM] A 2.0-g particle moves at a constant speed of 3.0 mm/s around a circle of radius 4.0 mm. (a) Find the magnitude of the angular momentum of the particle. (b) If L = l(l + 1)h , where l is an integer, find the

value of l(l + 1) and the approximate value of l . (c) By how much does l change if the particle’s speed increases by one-millionth of a percent, and nothing else changing? Use your result to explain why the quantization of angular momentum is not noticed in macroscopic physics. Picture the Problem We can use L = mvr to find the angular momentum of the particle. In (b) we can solve L = l(l + 1)h for l(l + 1) and the approximate value

of l . (a) Use the definition of angular momentum to obtain:

(

)(

)(

)

L = mvr = 2.0 × 10 −3 kg 3.0 × 10 −3 m/s 4.0 × 10 −3 m = 2.40 × 10 −8 kg ⋅ m 2 /s = 2.4 × 10 −8 kg ⋅ m 2 /s (b) Solve the equation L = l(l + 1)h for l(l + 1) :

L2 l(l + 1) = 2 h

Substitute numerical values and evaluate l(l + 1) :

⎛ 2.40 × 10 −8 kg ⋅ m 2 /s ⎞ ⎟⎟ l(l + 1) = ⎜⎜ −34 × ⋅ 1.05 1 0 J s ⎝ ⎠

(1) 2

= 5.2 × 10 52 Because l >>1, approximate its value with the square root of l(l + 1) :

l ≈ 2.3 × 10 26

(c) The change in l is:

Δl = l new − l

(2)

982

Chapter 10

(

)

If the particle’s speed increases by one-millionth of a percent while nothing else changes:

v → v + 10 −8 v = 1 + 10 −8 v and L → L + 10 −8 L = 1 + 10 −8 L

Equation (1) becomes:

l new (l new + 1) =

(

and

)

[(1 + 10 )L] −8

2

h2

(1 + 10 )L ≈ −8

l new

Substituting in equation (2) yields:

Substitute numerical values and evaluate Δl :

Δl = l new

h

(1 + 10 )L − L = 10 −l ≈ −8

h

−8

h

L h

⎛ 2.40 × 10 −8 kg ⋅ m 2 /s ⎞ ⎟⎟ Δl = 10 −8 ⎜⎜ −34 ⎝ 1.05 × 01 J ⋅ s ⎠ = 2.3 × 1018

and Δl 2.3 × 1018 ≈ 10 −6 % = 26 l 2.3 × 10 The quantization of angular momentum is not noticed in macroscopic physics because no experiment can detect a fractional change in l of 10 −6 % . 24 ••• Astrophysicists in the 1960s tried to explain the existence and structure of pulsars—extremely regular astronomical sources of radio pulses whose periods ranged from seconds to milliseconds. At one point, these radio sources were given the acronym LGM (Little Green Men), a reference to the idea that they might be signals of extraterrestrial civilizations. The explanation given today is no less interesting. Consider the following. Our Sun, which is a fairly typical star, has a mass of 1.99 × 1030 kg and a radius of 6.96 × 108 m. Although it does not rotate uniformly, because it is not a solid body, its average rate of rotation is about 1 rev/25 d. Stars larger than the Sun can end their life in spectacular explosions called supernovae, leaving behind a collapsed remnant of the star called a neutron star. These neutron-stars have masses comparable to the original masses of the stars but radii of only a few kilometers! The high rotation rates are due to the conservation of angular momentum during the collapses. These stars emit beams of radio waves. Because of the rapid angular speed of the stars, the beam sweeps past Earth at regular, very short, intervals. To produce the observed radio-wave pulses, the star has to rotate at rates that range from about 1 rev/s to 1000 rev/s. (a) Using data from the textbook, estimate the rotation rate of the Sun if it were to collapse into a neutron star of radius 10 km. The Sun is not a uniform sphere of

Angular Momentum

983

gas, and its moment of inertia is given by I = 0.059MR2. Assume that the neutron star is spherical and has a uniform mass distribution. (b) Is the rotational kinetic energy of the Sun greater or smaller after the collapse? By what factor does it change, and where does the energy go to or come from? Picture the Problem We can use conservation of angular momentum in Part (a) to relate the before-and-after collapse rotation rates of the Sun. In Part (b), we can express the fractional change in the rotational kinetic energy of the Sun as it collapses into a neutron star to decide whether its rotational kinetic energy is greater initially or after the collapse.

(a) Use conservation of angular momentum to relate the angular momenta of the Sun before and after its collapse:

I bωb = I aωa ⇒ ωa =

Ib ωb Ia

(1)

Using the given formula, approximate the moment of inertia Ib of the Sun before collapse:

(

)(

)

2

2 I b = 0.059MRsun = 0.059 1.99 × 10 30 kg 6.96 × 10 5 km = 5.69 × 10 46 kg ⋅ m 2

Find the moment of inertia Ia of the Sun when it has collapsed into a spherical neutron star of radius 10 km and uniform mass distribution: Substitute numerical values in equation (1) and simplify to obtain:

I a = 52 MR 2 =

2 5

(1.99 ×10

30

)

kg (10 km )

2

= 7.96 ×1037 kg ⋅ m 2

ωa =

5.69 × 10 46 kg ⋅ m 2 ωb 7.96 × 1037 kg ⋅ m 2

= 7.15 × 108 ωb

Given that ωb = 1 rev/25 d, evaluate ωa:

⎛ 1rev ⎞ ⎟⎟ = 2.86 rev/d ⎝ 25 d ⎠

ωa = 7.15 × 108 ⎜⎜

= 2.9 × 10 7 rev/d

Note that the rotational period decreases by the same factor of Ib/Ia and becomes: Ta =

2π

ωa

=

2π = 3.0 × 10 −3 s rev 2π rad 1d 1h × × × 2.86 × 10 7 d rev 24 h 3600 s

984

Chapter 10

(b) Express the fractional change in the Sun’s rotational kinetic energy as a consequence of its collapse:

ΔK K a − K b K a = = −1 Kb Kb Kb

Substituting for the kinetic energies and simplifying yields:

I a ωa2 ΔK 12 I a ωa2 =1 − 1 = −1 K b 2 I bω b2 I bω b2

Substitute numerical values and evaluate ΔK/Kb: 2

7 ΔK ⎛ 1 ⎞⎛ 2.86 × 10 rev/d ⎞ ⎟⎟ − 1 = 7.1× 108 ⎜ =⎜ ⎟ 8 ⎜ K b ⎝ 7.15 × 10 ⎠⎝ 1rev/25 d ⎠

That is, the rotational kinetic energy increases by a factor of approximately 7 × 108. The additional rotational kinetic energy comes at the expense of gravitational potential energy, which decreases as the Sun gets smaller. 25 •• The moment of inertia of Earth about its spin axis is approximately 8.03 × 1037 kg⋅m2. (a) Because Earth is nearly spherical, assume that the moment of inertia can be written as I = CMR2, where C is a dimensionless constant, M = 5.98 × 1024 kg is the mass of Earth, and R = 6370 km is its radius. Determine C. (b) If Earth’s mass were distributed uniformly, C would equal 2/5. From the value of C calculated in Part (a), is Earth’s density greater near its center or near its surface? Explain your reasoning. Picture the Problem We can solve I = CMR 2 for C and substitute numerical values in order to determine an experimental value of C for Earth. We can then compare this value to those for a spherical shell and a sphere in which the mass is uniformly distributed to decide whether Earth’s mass density is greatest near its core or near its crust.

(a) Express the moment of inertia of Earth in terms of the constant C:

I = CMR 2 ⇒ C =

Substitute numerical values and evaluate C:

8.03 ×1037 kg ⋅ m 2 C= 2 5.98 × 1024 kg (6370 km )

(

I MR 2

)

= 0.331 (b) Because experimentally C < 0.4, the mass density must be greater near the center of Earth. 26 ••• Estimate Timothy Goebel’s initial takeoff speed, rotational velocity, and angular momentum when he performs a quadruple Lutz (Figure 10-41). Make any assumptions you think reasonable, but justify them. Goebel’s mass is about 60 kg and the height of the jump is about 0.60 m. Note that his angular speed will

Angular Momentum

985

change quite a bit during the jump, as he begins with arms outstretched and pulls them in. Your answer should be accurate to within a factor of 2, if you’re careful. Picture the Problem We’ll assume that he launches himself at an angle of 45° with the horizontal with his arms spread wide, and then pulls them in to increase his rotational speed during the jump. We’ll also assume that we can model him as a 2.0-m long cylinder with an average radius of 0.15 m and a mass of 60 kg. We can then find his take-off speed and ″air time″ using constant-acceleration equations, and use the latter, together with the definition of rotational velocity, to find his initial rotational velocity. Finally, we can apply conservation of angular momentum to find his initial angular momentum.

Using a constant-acceleration equation, relate his takeoff speed v0 to his maximum elevation Δy:

v 2 = v02y + 2a y Δy

or, because v0y = v0sin(45°), v = 0, and ay = − g, 0 = v02 sin 2 45° − 2 gΔy

Solving for v0 and simplifying yields:

v0 =

2 gΔy 2 gΔy = 2 sin 45° sin 45°

Substitute numerical values and evaluate v0:

v0 =

2 9.81m/s 2 (0.60 m ) sin45°

(

)

= 4.9 m/s Δθ Δt

Use its definition to express Goebel’s angular velocity:

ω=

Use a constant-acceleration equation to express Goebel’s ″air time″ Δt:

Δt = 2Δtrise 0.6 m = 2

Substitute numerical values and evaluate Δt:

Δt = 2

Substitute numerical values and evaluate ω:

ω=

Use conservation of angular momentum to relate his take-off angular velocity ω0 to his average angular velocity ω as he performs a quadruple Lutz:

I 0ω0 = Iω

2Δy g

2(0.60 m ) = 0.699 s 9.81 m/s 2

4 rev 2π rad × = 36 rad/s 0.699 s rev

986

Chapter 10

Assuming that he can change his moment of inertia by a factor of 2 by pulling his arms in, solve for and evaluate ω0:

ω0 =

Express his take-off angular momentum:

L0 = I 0ω0

Assuming that we can model him as a solid cylinder of length l with an average radius r and mass m, express his moment of inertia with arms outstretched (his take-off configuration):

I 0 = 2 12 mr 2 = mr 2 where the factor of 2 represents our assumption that he can double his moment of inertia by extending his arms.

Substituting for I0 gives:

L0 = mr 2ω0

Substitute numerical values and evaluate L0:

L0 = (60 kg )(0.15 m ) (18 rad/s )

I 1 ω = (36 rad/s ) = 18 rad/s I0 2

(

)

2

= 24 kg ⋅ m 2 /s

The Vector Product and the Vector Nature of Torque and Rotation 27 • [SSM] A force of magnitude F is applied horizontally in the negative r x direction to the rim of a disk of radius R as shown in Figure 10-42. Write F and r r in terms of the unit vectors ˆi , jˆ , and kˆ , and compute the torque produced by this force about the origin at the center of the disk.

r r Picture the Problem We can express F and r in terms of the unit vectors iˆ and ˆj and then use the definition of the vector product to find τr . r Express F in terms of F and the unit vector iˆ :

r F = − Fiˆ

r Express r in terms of R and the unit vector ˆj :

r r = Rˆj

Calculate the vector product of r r r and F :

r r τ = r × F = FR ˆj × −iˆ = FR iˆ × ˆj

r

= FR kˆ

(

)

( )

Angular Momentum

987

28 Compute the torque about the origin of the gravitational force r • r ˆ F = −mgj acting on a particle of mass m located at r = x iˆ + y jˆ and show that this torque is independent of the y coordinate. r r Picture the Problem We can find the torque from the vector product of r and F .

Compute the vector product of r r r and F :

) = −mgx(iˆ × ˆj ) − mgy ( ˆj × ˆj ) r

(

)(

r r τ = r × F = xiˆ + yˆj − mgˆj = − mgxkˆ

r r r r Find A × B for the following choices: (a) A = 4ˆi and B = 6ˆi + 6 jˆ , r r r r (b) A = 4ˆi and B = 6ˆi + 6 kˆ , and (c) A = 2ˆi + 3 jˆ and B = 3iˆ + 2 jˆ .

29

•

Picture the Problem We can user therdefinitions of the vector products of the unit vectors iˆ , ˆj , and kˆ to evaluate A × B in each case.

r r r (a) Evaluate A × B for A = 4 iˆ and r B = 6 iˆ + 6 ˆj :

(

) ( ) ( )

r r A × B = 4iˆ × 6iˆ + 6 ˆj = 24 iˆ × iˆ + 24 iˆ × ˆj = 24(0) + 24kˆ = 24kˆ

r r r (b) Evaluate A × B for A = 4 iˆ and r B = 6 iˆ + 6 kˆ :

(

) ( ) ( ) = 24(0) + 24(− ˆj )

r r A × B = 4iˆ × 6iˆ + 6kˆ = 24 iˆ × iˆ + 24 iˆ × kˆ = − 24 ˆj

r r (c) Evaluate A × B for r r A = 2 iˆ + 3 ˆj and B =3 iˆ + 2 ˆj :

(

)( ) ( ) ( ) ( ) ( ) = 6(0) + 4(kˆ )+ 9(− kˆ ) + 6(0)

r r A × B = 2iˆ + 3 ˆj × 3iˆ + 2 ˆj = 6 iˆ × iˆ + 4 iˆ × ˆj + 9 ˆj × iˆ + 6 ˆj × ˆj = − 5kˆ

30

••

r r r r For each case in Problem 31, compute A × B . Compare it to A B to

estimate which of the pairs of vectors are closest to being perpendicular. Verify your answers by calculating the angle using the dot product.

988

Chapter 10

r r r r r r Picture the Problem Because A × B = A B sin φ , if vectors A and B are r r A× B r r r r perpendicular, then A × B = A B or r r = 1 . The scalar product of vectors AB r r r r r r A and B is A ⋅ B = A B cos φ . We can verify our estimations using this definition to calculate φ for each pair of vectors. r r (a) For A = 4 iˆ and B = 6 iˆ + 6 ˆj :

( ) ( )

r r 24kˆ A × B 4iˆ × 6iˆ + 6 ˆj 1 = = r r = (4) 6 2 24 2 2 AB ≈ 0.707 r r and the vectors A and B are not perpendicular.

r r The angle between A and B is:

r r (b) For A = 4 iˆ and B = 6 iˆ + 6 kˆ :

(

)

r r ˆ ⎞ ⎛ ˆ ˆ A ⋅ B ⎞⎟ −1 4 i ⋅ 6i + 6 j ⎟ r r ⎟ = cos ⎜⎜ ⎟ 24 2 A B⎟ ⎝ ⎠ ⎠ 24 = cos −1 = 45°, 24 2 a result confirming that obtained above. ⎛ φ = cos ⎜⎜ ⎜ ⎝ −1

( ) ( )

r r − 24 ˆj A × B 4iˆ × 6iˆ + 6kˆ 1 = = = r r (4) 6 2 24 2 2 AB ≈ 0.707 r r and the vectors A and B are not perpendicular.

r r The angle between A and B is:

(

)

r r ⎛ 4iˆ ⋅ 6iˆ + 6kˆ ⎞ A ⋅ B ⎞⎟ ⎟ r r ⎟ = cos −1 ⎜⎜ ⎟ 24 2 AB⎟ ⎠ ⎝ ⎠ 24 = cos −1 = 45°, 24 2 a result confirming that obtained above. ⎛ φ = cos −1 ⎜⎜ ⎜ ⎝

Angular Momentum r (c) For A = 2 iˆ + 3 ˆj and r B =3 iˆ + 2 ˆj :

(

)(

989

)

r r 2iˆ + 3 ˆj × 3iˆ + 2 ˆj − 5kˆ A× B = r r = 13 13 13 AB 5 ≈ 0.385 13 r r and the vectors A and B are not perpendicular. =

r r The angle between A and B is:

r r⎞ ⎛A ⋅B ⎜ φ = cos ⎜ r r ⎟⎟ ⎜ AB⎟ ⎝ ⎠ ⎛ 2iˆ + 3 ˆj ⋅ 3iˆ + 2 ˆj ⎞ ⎟ = cos −1 ⎜⎜ ⎟ 13 13 ⎝ ⎠ ⎛ 12 ⎞ = cos −1 ⎜ ⎟ = 23°, ⎝ 13 ⎠ a result confirming that obtained above. −1

(

)(

)

While none of these sets of vectors are perpendicular, those in (a) and (b) are the closest, with φ = 45°, to being perpendicular. 31 •• A particle moves in a circle that is centered at the origin. The particle has r r r r r position r and angular velocity ω . (a) Show that its velocity is given by v = ω × r . r r r r r r (b) Show that its centripetal acceleration is given by a c = ω × v = ω × (ω × r ) .

r Picture the Problem Let r be in the xy plane and point in the +x r direction. Then ω points in the +z direction. We can establish the results called for in this problem by forming the appropriate vector r products and by differentiating v . r (a) Express ω using unit vector notation:

r ω = ω kˆ

r Express r using unit vector notation:

r r = riˆ

r r Form the vector product of ω and r :

r r ω × r = ω kˆ × r iˆ = rω kˆ × iˆ = rω ˆj = vˆj r r r and v = ω × r

( )

990

Chapter 10

r r (b) Differentiate v with respect to t to express a : r r r r r d v d r r dω r r dr dω r r r r r r r = (ω × r ) = × r + ω× = × r + ω × v = a t + ω × (ω × r ) a= dt dt dt dt dt r r = a t + ac r r r r r r where a c = ω × (ω × r ) and a t and a c are the tangential and centripetal

accelerations, respectively. 32 You are given r •• r three vectors and theirr components in the form: ˆ ˆ ˆ A = ax i + ay j + az k , B = bx ˆi + by jˆ + bz kˆ , and C = cx iˆ + cy jˆ + c z kˆ . Show that the r r r r r r r r r following equalities hold: A ⋅ B × C = C ⋅ A × B = B ⋅ C × A

(

)

(

)

(

)

Picture the Problem We can establish these equalities by carrying out the details of the vector- and scalar-products and comparing the results of these operations. r r Evaluate the vector product of B and C to obtain:

r r B × C = (b y c z − bz c y )iˆ + (bz c x − bx c z ) ˆj + (bx c y − b y c x )kˆ r r r Form the scalar product of A with B × C to obtain:

(

)

r r r A ⋅ B × C = a x b y c z − a x b z c y + a y b z c x − a y bx c z + a z b x c y − a z b y c x

(1)

r r Evaluate the vector product of A and B to obtain: r r A × B = (a y bz − a z b y )iˆ + (a z bx − a x bz ) ˆj + (a x b y − a y bx )kˆ r r r Form the scalar product of C with A × B to obtain:

(

)

r r r C ⋅ A × B = c x a y b z − c x a z b y + c y a z bx − c y a x b z + c z a x b y − c z a y b x

(2)

r r Evaluate the vector product of C and A to obtain: r r C × A = (c y a z − a z a y )iˆ + (c z a x − a x a z ) ˆj + (c x a y − a y a x )kˆ r r r Form the scalar product of B with C × A to obtain:

(

)

r r r B ⋅ C × A = b x c y a z − bx c z a y + b y c z a x − b y c x a z + b z c x a y − bz c y a x

(3)

Angular Momentum

991

The equality of equations (1), (2), and (3) establishes the equalities. 33

••

r r r r r r If A = 3jˆ , A × B = 9iˆ , and A ⋅ B = 12, find B .

r r Picture the Problem We can write B in the form B = B x iˆ + B y ˆj + B z kˆ and use r r the scalar product of A and B to find By and their vector product to find Bx and Bz. r Express B in terms of its components:

r B = B x iˆ + B y ˆj + B z kˆ

r r Evaluate A ⋅ B :

r r A ⋅ B = 3B y = 12 ⇒By = 4

r r Evaluate A × B :

r r A × B = 3 ˆj × B x iˆ + 4 ˆj + B z kˆ = −3B kˆ + 3B iˆ

(

x

r r Because A × B = 9 iˆ : Substitute for By and Bz in equation (1) to obtain: 34

••

(1)

)

z

Bx = 0 and Bz = 3.

r B = 4 ˆj + 3kˆ

r r r r r If A = 4iˆ , Bz = 0, B = 5 , and A × B = 12kˆ , determine B .

r r Picture the Problem Because Bz = 0, we can express B as B = B x iˆ + B y ˆj and r form its vector product with A to determine Bx and By. r Express B in terms of its components:

r B = B x iˆ + B y ˆj

r r Express A × B :

r r A × B = 4iˆ × Bx iˆ + B y ˆj = 4 B y kˆ = 12kˆ

Solving for By yields:

By = 3

Relate B to Bx and By:

B 2 = B x2 + B y2

Solve for and evaluate Bx:

Bx = B 2 − B y2 = 5 2 − 32 = 4

Substitute for Bx and By in equation (1) to obtain:

(

r B = 4iˆ + 3 ˆj

(1)

)

992 35

Chapter 10 •••

r r r r r r Given three noncoplanar vectors A , B , and C , show that A ⋅ B × C

(

)

is the volume of the parallelepiped formed by the three vectors. r Picture the Problem Let, without loss of generality, the vector C lie along the x r axis and the vector B lie in the xy plane as shown below to the left. The diagram to the right shows the parallelepiped spanned by the three vectors. We can apply the r r r definitions of the vector- and scalar-products to show that A ⋅ B × C is the volume of the parallelepiped.

(

)

r r B×C

r A r C

A cos θ

θ φ

Bsin φ

r B

The magnitude of the vector product r r of B and C is:

r r B × C = BC sin φ and r r B × C = (B sin φ )C = area of the base parallelogram

Forming the scalar-product of r r A with the vector-product of B and r C gives:

(

)

r r r A ⋅ B × C = A(B sin φ )C cosθ = (BC sin φ )( A cosθ )

= (area of base )(height ) = Vparallelepiped

36 ••• Using the cross product, prove the law of sines for the triangle shown in Figure 10-43. That is, if A, B, and C are the lengths of each side of the triangle, show that A/sin a = B/sin b = C/sin c. Picture the Problem Draw the triangle using the three vectors as shown below. r r r r Note that A + B = C . We can find the magnitude of the vector product of A and r r r r r B and of A and C and then use the vector product of A and C , using

Angular Momentum

993

r r r B C A + B = C, to show that AC sin b = AB sin c or = . Proceeding similarly, sin b sin c we can extend the law of sines to the third side of the triangle and the angle opposite it. c r A

r B

a

b

r C

Express the magnitude of the vector r r product of A and B :

r r A × B = AB sin (180° − c ) = AB sin c

Express the magnitude of the vector r r product of A and C :

r r A × C = AC sin b

r Form the vector product of A with r C to obtain:

r r r r Because A × C = A × B :

(

)

r r r r r A× C = A× A + B r r r r = A× A + A× B r r = A× B r r because A × A = 0 . r r r r A× C = A× B

and AC sin b = AB sin c Simplify and rewrite this expression to obtain:

B C = sin b sin c

Proceed similarly to extend this result to the law of sines:

A B C = = sin a sin b sin c

Torque and Angular Momentum 37 • [SSM] A 2.0-kg particle moves directly eastward at a constant speed of 4.5 m/s along an east-west line. (a) What is its angular momentum (including direction) about a point that lies 6.0 m north of the line? (b) What is its angular momentum (including direction) about a point that lies 6.0 m south of the line?

994

Chapter 10

(c) What is its angular momentum (including direction) about a point that lies 6.0 m directly east of the particle? r r r Picture the Problem The angular momentum of the particle is L = r × p where r r r is the vector locating the particle relative to the reference point and p is the particle’s linear momentum.

(a) The magnitude of the particle’s angular momentum is given by: Substitute numerical values and evaluate L:

L = rp sin φ = rmv sin φ = mv(r sin φ )

L = (2.0 kg )(4.5 m/s )(6.0 m ) = 54 kg ⋅ m 2 /s

Use a right-hand rule to establish r the direction of L :

L = 54 kg ⋅ m 2 /s, upward

(b) Because the distance to the line along which the particle is moving is the same, only the direction of r L differs:

L = 54 kg ⋅ m 2 /s, downward

r r (c) Because r × p = 0 for a point on the line along which the particle is moving:

r L= 0

38 • You observe a 2.0-kg particle moving at a constant speed of 3.5 m/s in a clockwise direction around a circle of radius 4.0 m. (a) What is its angular momentum (including direction) about the center of the circle? (b) What is its moment of inertia about an axis through the center of the circle and perpendicular to the plane of the motion? (c) What is the angular velocity of the particle?

r r r Picture the Problem The angular momentum of the particle is L = r × p where r r r is the vector locating the particle relative to the reference point and p is the particle’s linear momentum.

(a) The magnitude of the particle’s angular momentum is given by: Substitute numerical values and evaluate the magnitude of L:

L = rp sin φ = rmv sin φ = mv(r sin φ )

L = (2.0 kg )(3.5 m/s )(4.0 m ) = 28 kg ⋅ m 2 /s

Angular Momentum Use a right-hand rule to establish the r direction of L :

L = 28 kg ⋅ m 2 /s, away from you

(b) Treat the 2.0-kg particle as a point particle to obtain:

I = mr 2

Substitute numerical values and evaluate I:

I = (2.0 kg )(4.0 m ) = 32 kg ⋅ m 2

(c) Because L = Iω, the angular speed of the particle is the ratio of its angular momentum and its moment of inertia: Substitute numerical values and evaluate ω:

995

2

ω=

L I

ω=

28 kg ⋅ m 2 /s = 0.88 rad/s 2 2 32 kg ⋅ m

39 •• (a) A particle moving at constant velocity has zero angular momentum about a particular point. Use the definition of angular momentum to show that under this condition the particle is moving either directly toward or directly away from the point. (b) You are a right-handed batter and let a waist-high fastball go past you without swinging. What is the direction of its angular momentum relative to your navel? (Assume the ball travels in a straight horizontal line as it passes you.)

r r r r r r Picture the Problem L and p are related according to L = r × p. If L = 0, then r r examination of the magnitude of r × p will allow us to conclude that sin φ = 0 and that the particle is moving either directly toward the point, directly away from the point, or through the point. r (a) Because L = 0:

r r r r r r r × p = r × mv = mr × v = 0 or r r r ×v = 0

r r Express the magnitude of r × v :

r r r × v = rv sin φ = 0

Because neither r nor v is zero:

sin φ = 0 r r where φ is the angle between r and v .

Solving for φ yields:

φ = sin −1 (0) = 0° or 180°

996

Chapter 10

(b) Use the right-hand rule to establish that the ball’s angular momentum is downward. r 40 •• A particle that has a mass m is traveling with a constant velocity v along a straight line that is a distance b from the origin O (Figure 10-44). Let dA be the area swept out by the position vector from O to the particle during a time interval dt. Show that dA/dt is constant and is equal to L 2m , where L is the magnitude of the angular momentum of the particle about the origin. Picture the Problem The area swept out by the position vector (the shaded area in Figure 10-44) is the difference between the area of a trapezoid and the area of a r r triangle. Let x1 be the x component of r1 and x1 + Δx be the x component of r2 . Use the formulas for the areas of a trapezoid and a triangle to express ΔA and then r take the limit as Δt → 0 to express dA/dt. Letting θ be the angle between r1 and r the horizontal axis, we can express b as a function of r1 and θ.

The area swept out by the position vector (the shaded area in Figure 1044) is given by:

ΔA = Atrap − Atriangle

In the limit as Δt → 0:

dA 1 dx 1 = 2b = 2 bv = constant dt dt

r Because b = r1 sin θ :

r ( r1 sin θ )mv r dA 1 1 = 2 bv = 2 ( r1 sin θ )v = 2m dt 1 r ( r1 p sin θ ) = L = 2m 2m

= 12 b[Δx + ( x1 + Δx )] − 12 b( x1 + Δx ) = 12 bΔx

41 •• A 15-g coin that has a diameter of 1.5 cm is spinning at 10 rev/s about a fixed vertical axis. The coin is spinning on edge with its center directly above the point of contact with the tabletop. As you look down on the tabletop, the coin spins clockwise. (a) What is the angular momentum (including direction) of the coin about its center of mass? (To find the moment of inertia about the axis, see Table 9-1.) Model the coin as a cylinder of length L and take the limit as L approaches zero. (b) What is the coin’s angular momentum (including direction) about a point on the tabletop 10 cm from the axis? (c) Now the coin’s center of mass travels at 5.0 cm/s in a straight line east across the tabletop, while spinning the same way as in Part (a). What is the angular momentum (including direction) of the coin about a point on the line of motion of the center of mass? (d) When it is both spinning and sliding, what is the angular momentum of the coin (including direction) about a point 10 cm north of the line of motion of the center of mass?

Angular Momentum

997

Picture the Problem We can find the total angular momentum of the coin from the sum of its spin and orbital angular momenta.

(a) The spin angular momentum of the coin is:

Lspin = Iωspin

From Table 9-1, for L negligible compared to R:

I = 14 MR 2

Substitute for I to obtain:

Lspin = 14 MR 2ωspin

Substitute numerical values and evaluate Lspin: Lspin =

1 4

(0.015 kg )(0.0075 m )2 ⎛⎜10 rev × 2π rad ⎞⎟ = 1.33 × 10 −5 kg ⋅ m 2 /s ⎝

s

rev ⎠

r Use a right-hand rule to establish the direction of Lspin : Lspin = 1.3 × 10 −5 kg ⋅ m 2 /s, away from you.

(b)The total angular momentum of the coin is the sum of its orbital and spin angular momenta:

Ltotal = Lorbital + Lspin

Substitute numerical values and evaluate Ltotal:

Ltotal = 0 + Lspin = 1.3 × 10 −5 kg ⋅ m 2 /s

r Use a right-hand rule to establish the direction of Ltotal :

Ltotal = 1.3 ×10 −5 kg ⋅ m 2 /s, away from you

(c) Because Lorbital = 0 : Ltotal = 1.3 ×10 −5 kg ⋅ m 2 /s, away from you

(d) When it is both spinning and sliding, the total angular momentum of the coin is:

Ltotal = Lorbital + Lspin

998

Chapter 10

The orbital angular momentum of the coin is:

Lorbital = MvR

The spin angular momentum of the coin is:

Lspin = I spin ωspin = 14 MR 2ωspin

Substituting for Lorbital and Lspin yields:

Ltotal = MvR + 14 MR 2ωspin

Substitute numerical values and evaluate Ltotal : Ltotal = (0.015 kg )(0.050 m/s )(0.10 m )

rev 2π rad ⎞ 2⎛ + 14 (0.015 kg )(0.0075 m ) ⎜10 × ⎟ s rev ⎠ ⎝ = 8.8 × 10 −5 kg ⋅ m 2 /s, pointing toward you

r r 42 •• (a) Two stars of masses m1 and m2 are located at r1 and r2 relative to some origin O, as shown in Figure 10-45. They exert equal and opposite attractive gravitational forces on each other. For this two-star system, calculate the net torque exerted by these internal forces about the origin O and show that it is zero only if both forces lie along the line joining the particles. (b)The fact that the Newton’s third-law pair of forces are not only equal and oppositely directed but also lie along the line connecting the two objects is sometimes called the strong form of Newton’s third law. Why is it important to add that last phrase? Hint: Consider what would happen to these two objects if the forces were offset from each other. Picture the Problem Both the forces acting on the particles exert torques with respect to an axis perpendicular to the page and through point O and the net torque about this axis is their vector sum.

(a) The net torque about an axis perpendicular to the page and through point O is given by:

r r r Because r1 − r2 points along − F1 :

r

r

r

r

r

r

τ net = ∑τ i = r1 × F1 + r2 × F2 i

r r or, because F2 = − F1 , r r r r τ net = (r1 − r2 ) × F1 r

r

r

r

τ net = (r1 − r2 )× F1 = 0

Angular Momentum

999

(b) If the forces are not along the same line, there will be a net torque (but still no net force) acting on the system. This net torque would cause the system to accelerate angularly, contrary to observation, and hence makes no sense physically. 43 •• A 1.8-kg particle moves in a circle of radius 3.4 m. As you look down on the plane of its orbit, it is initially moving clockwise. If we call the clockwise direction positive, its angular momentum relative to the center of the circle varies with time according to L(t ) = 10 N ⋅ m ⋅ s − (4.0 N ⋅ m )t . (a) Find the magnitude and direction of the torque acting on the particle. (b) Find the angular velocity of the particle as a function of time. Picture the Problem The angular momentum of the particle changes because a net torque acts on it. Because we know how the angular momentum depends on time, we can find the net torque acting on the particle by differentiating its angular momentum. We can use a constant-acceleration equation and Newton’s second law to relate the angular speed of the particle to its angular acceleration.

(a) The magnitude of the torque acting on the particle is the rate at which its angular momentum changes:

τ net =

dL dt

Evaluate dL/dt to obtain:

τ net =

d [10 N ⋅ m ⋅ s − (4.0 N ⋅ m )t ] dt

= − 4.0 N ⋅ m

Note that, because L decreases as the particle rotates clockwise, the angular acceleration and the net torque are both upward.

Lorbital I orbital

(b) The angular speed of the particle is given by:

ωorbital =

Treating the 1.8-kg particle as a point particle, express its moment of inertia relative to an axis through the center of the circle and normal to it:

I orbital = MR 2

Substitute for I orbital and Lorbital to

ω orbital =

obtain:

10 N ⋅ m ⋅ s − (4.0 N ⋅ m )t MR 2

1000 Chapter 10 Substitute numerical values and evaluate ωorbital:

ω orbital =

10 N ⋅ m ⋅ s − (4.0 N ⋅ m )t = 0.48 rad/s − (0.19 rad/s 2 )t (1.8 kg )(3.4 m )2

Note that the direction of the angular velocity is downward. 44 •• You are designing a lathe motor and part of it consists of a uniform cylinder whose mass is 90 kg and radius is 0.40 m that is mounted so that it turns without friction on its axis, which is fixed. The cylinder is driven by a belt that wraps around its perimeter and exerts a constant torque. At t = 0, the cylinder’s angular velocity is zero. At t = 25 s, its angular speed is 500 rev/min. (a) What is the magnitude of its angular momentum at t = 25 s? (b) At what rate is the angular momentum increasing? (c) What is the magnitude of the torque acting on the cylinder? (d) What is the magnitude of the frictional force acting on the rim of the cylinder? Picture the Problem The angular momentum of the cylinder changes because a net torque acts on it. We can find the angular momentum at t = 25 s from its definition and the magnitude of the net torque acting on the cylinder from the rate at which the angular momentum is changing. The magnitude of the frictional force acting on the rim can be found using the definition of torque.

L = Iω = 12 mr 2ω

(a) The angular momentum of the cylinder is given by:

Substitute numerical values and evaluate L: L=

1 2

⎛

⎞

(90 kg )(0.40 m )2 ⎜⎜ 500 rev × 2π rad × 1min ⎟⎟ ⎝

min

rev

60 s ⎠

= 377 kg ⋅ m 2 /s = 3.8 × 10 2 kg ⋅ m 2 /s

(b) The rate at which the angular momentum of the cylinder is increasing is given by:

dL (377 kg ⋅ m 2 /s ) = = 15 kg ⋅ m 2 /s 2 dt 25 s = 15 kg ⋅ m 2 /s 2

Angular Momentum 1001 (c) Because the torque acting on the uniform cylinder is constant, the rate of change of the angular momentum is constant and hence the instantaneous rate of change of the angular momentum at any instant is equal to the average rate of change over the time during which the torque acts: (d) The magnitude of the frictional force f acting on the rim is:

dL = 15 kg ⋅ m 2 /s 2 dt

τ=

f =

τ l

=

15.1kg ⋅ m 2 /s 2 = 38 N 0.40 m

45 •• [SSM] In Figure 10-46, the incline is frictionless and the string passes through the center of mass of each block. The pulley has a moment of inertia I and radius R. (a) Find the net torque acting on the system (the two masses, string, and pulley) about the center of the pulley. (b)Write an expression for the total angular momentum of the system about the center of the pulley. Assume the masses are moving with a speed v. (c) Find the acceleration of the masses by using your results for Parts (a) and (b) and by setting the net torque equal to the rate of change of the system’s angular momentum. Picture the Problem Let the system include the pulley, string, and the blocks and assume that the mass of the string is negligible. The angular momentum of this system changes because a net torque acts on it. We’ll take clockwise to be positive to be consistent with a positive upward velocity of the block whose mass is m1 as indicated in the figure.

(a) Express the net torque about the center of mass of the pulley:

τ net = −m1 gR + T1 R − T1 R + T2 R + (m2 g sin θ )R − Rm1 g − T2 R = Rg (m2 sin θ − m1 ) (b) Express the total angular momentum of the system about an axis through the center of the pulley:

L = Iω + m1vR + m2 vR ⎛ I ⎞ = vR⎜ 2 + m1 + m2 ⎟ ⎝R ⎠

1002 Chapter 10 (c) Express τ as the time derivative of the angular momentum:

τ=

Equate this result to that of Part (a) and solve for a to obtain:

a=

dL d ⎡ ⎛ I ⎞⎤ = ⎢vR⎜ 2 + m1 + m2 ⎟⎥ dt dt ⎣ ⎝ R ⎠⎦ ⎛ I ⎞ = aR⎜ 2 + m1 + m2 ⎟ ⎝R ⎠ g (m2 sin θ − m1 ) I + m1 + m2 R2

46 •• Figure 10-47 shows the rear view of a space capsule that was left rotating rapidly about its longitudinal axis at 30 rev/min after a collision with another capsule. You are the flight controller and have just moments to tell the crew how to stop this rotation before they become ill from the rotation and the situation becomes dangerous. You know that they have access to two small jets mounted tangentially at a distance of 3.0 m from the axis, as indicated in the figure. These jets can each eject 10 g/s of gas with a nozzle speed of 800 m/s. Determine the length of time these jets must run to stop the rotation. In flight, the moment of inertia of the ship about its axis (assumed constant) is known to be 4000 kg⋅m2. Picture the Problem The forces resulting from the release of gas from the jets will exert a torque on the spaceship that will slow and eventually stop its rotation. We can relate this net torque to the angular momentum of the spaceship and to the time the jets must fire.

ΔL

Relate the firing time of the jets to the desired change in angular momentum:

Δt =

Express the magnitude of the net torque exerted by the jets:

τ net = 2 FR

Letting Δm/Δt′ represent the mass of gas per unit time exhausted from the jets, relate the force exerted by the gas on the spaceship to the rate at which the gas escapes: Substituting for F yields:

F=

τ net

=

IΔω

τ net

Δm v Δt '

τ net = 2vR

Δm Δt '

(1)

Angular Momentum 1003 Substitute for τ net in equation (1) to obtain:

Δt =

IΔω Δm 2vR Δt '

Substitute numerical values and evaluate Δt:

rev 2π rad 1 min ⎞ ⎟ (4000 kg ⋅ m )⎛⎜⎜ 30 min × × rev 60 s ⎟⎠ ⎝ = Δt = 2(10 kg/s )(800 m/s )(3.0 m ) 2

−2

2.6 × 10 2 s

47 •• A projectile (mass M) is launched at an angle θ with an initial speed v0. Considering the torque and angular momentum about the launch point, explicitly show that dL/dt = τ. Ignore the effects of air resistance. (The equations for projectile motion are found in Chapter 3.) Picture the Problem We can use constant-acceleration equations to express the projectile’s position and velocity coordinates as functions of time. We rcan use r these coordinates to express the particle’s position and velocity vectors r and v . r Using its definition, we can express the projectile’s angular momentum L as a r function of time and then differentiate this expression to obtain dL dt . Finally, we can use the definition of the torque, relative to an origin located at the launch r position, the gravitational force exerts on the projectile to express τ and complete r r the demonstration that dL dt = τ .

Using its definition, express the r angular momentum vector L of the projectile:

r r r L = r × mv

(1)

Using constant-acceleration equations, express the position coordinates of the projectile as a function of time:

x = v0 x t = (v0 cosθ )t and y = y 0 + v0 y t + 12 a y t 2

Express the projectile’s position r vector r :

r r = [(v0 cosθ )t ]iˆ + (v0 sin θ )t − 12 gt 2 ˆj

Using constant-acceleration equations, express the velocity of the projectile as a function of time:

v x = v0 x = v0 cosθ and v y = v0 y + a y t = v0 sin θ − gt

Express the projectile’s velocity r vector v :

r v = [v0 cosθ ]iˆ + [v0 sin θ − gt ] ˆj

= (v0 sin θ )t − 12 gt 2

[

]

1004 Chapter 10 Substituting in equation (1) and simplifying yields:

{

[

]} {

}

r L = [(V cosθ )t ]iˆ + (V sin θ )t − 12 gt 2 ˆj × m [V cosθ ]iˆ + [V sin θ − gt ] ˆj = (− 1 mgt 2V cosθ )kˆ 2

r Differentiate L with respect to t to obtain:

r dL d 1 = − 2 mgt 2V cosθ kˆ dt dt = (− mgtV cosθ ) kˆ

(

)

(2)

Using its definition, express the torque acting on the projectile:

τ = r × (− mg ) ˆj = [(v0 cosθ )t ]iˆ + [(v0 sin θ )t − 12 gt 2 ] ˆj × (− mg ) ˆj r

r

= (− mgtV cosθ )kˆ

(3)

r dL r =τ dt

Comparing equations (2) and (3) we see that:

Conservation of Angular Momentum 48 • A planet moves in an elliptical orbit about the Sun, with the Sun at one focus of the ellipse, as in Figure 10-48. (a) What is the torque about the center of the Sun due to the gravitational force of attraction of the Sun on the planet? (b) At position A, the planet has an orbital radius r1 and is moving with a speed v1 perpendicular to the line from the sun to the planet. At position B, the planet has an orbital radius r2 and is moving with speed v2, again perpendicular to the line from the sun to the planet. What is the ratio of v1 to v2 in terms of r1 and r2? Picture the Problem Let m represent the mass of the planet and apply the definition of torque to find the torque produced by the gravitational force of r r attraction. We can use Newton’s second law of motion in the form τ = dL dt to r show that L is constant and apply conservation of angular momentum to the motion of the planet at points A and B. r

r

r

r

(a) Express the torque produced by the gravitational force of attraction of the sun for the planet:

τ = r × F = 0 because F acts along

r (b) Because τ = 0 :

r r r r dL = 0 ⇒ L = r × mv = constant dt

r the direction of r .

Angular Momentum 1005 Noting that at points A and B r r r × v = rv , express the relationship

r1v1 = r2 v2 ⇒

v1 r = 2 v2 r1

between the distances from the sun and the speeds of the planets: 49 •• [SSM] You stand on a frictionless platform that is rotating at an angular speed of 1.5 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 6.0 kg⋅m2. When you pull the weights in toward your body, the moment of inertia decreases to 1.8 kg⋅m2. (a) What is the resulting angular speed of the platform? (b) What is the change in kinetic energy of the system? (c) Where did this increase in energy come from? Picture the Problem Let the system consist of you, the extended weights, and the platform. Because the net external torque acting on this system is zero, its angular momentum remains constant during the pulling in of the weights.

Ii ωi If

(a) Using conservation of angular momentum, relate the initial and final angular speeds of the system to its initial and final moments of inertia:

I i ω i − I f ω f = 0 ⇒ ωf =

Substitute numerical values and evaluate ω f :

6.0 kg ⋅ m 2 (1.5 rev/s) = 5.0 rev/s ωf = 1.8 kg ⋅ m 2

(b) Express the change in the kinetic energy of the system:

ΔK = K f − K i = 12 I f ωf2 − 12 I iωi2

Substitute numerical values and evaluate ΔK:

(

)

2

(

)

rev 2π rad ⎞ 1 rev 2π rad ⎞ ⎛ 2 ⎛ ΔK = 1.8 kg ⋅ m ⎜ 5.0 × × ⎟ ⎟ − 2 6.0 kg ⋅ m ⎜1.5 s s rev ⎠ rev ⎠ ⎝ ⎝ 1 2

2

2

= 0.62 kJ (c) Because no external agent does work on the system, the energy comes from your internal energy. 50 •• A small blob of putty of mass m falls from the ceiling and lands on the outer rim of a turntable of radius R and moment of inertia I0 that is rotating freely with angular speed ω0 about its vertical fixed-symmetry axis. (a) What is the postcollision angular speed of the turntable-putty system? (b) After several turns, the

1006 Chapter 10 blob flies off the edge of the turntable. What is the angular speed of the turntable after the blob’s departure? Picture the Problem Let the system consist of the blob of putty and the turntable. Because the net external torque acting on this system is zero, its angular momentum remains constant when the blob of putty falls onto the turntable.

I0 ω 0 (1) If

(a) Using conservation of angular momentum, relate the initial and final angular speeds of the turntable to its initial and final moments of inertia and solve for ωf:

I 0ω 0 − I f ω f = 0 ⇒ ω f =

Express the final rotational inertia of the turntable-plus-blob:

I f = I 0 + I blob = I 0 + mR 2

Substitute for If in equation (1) and simplify to obtain:

ωf =

I0 ω0 = I 0 + mR 2

1 ω0 mR 2 1+ I0

(b) If the blob flies off tangentially to the turntable, its angular momentum doesn’t change (with respect to an axis through the center of turntable). Because there is no external torque acting on the blob-turntable system, the total angular momentum of the system will remain constant and the angular momentum of the turntable will not change. The turntable will continue to spin at ω' = ωf . 51 •• [SSM] A lazy Susan consists of a heavy plastic disk mounted on a frictionless bearing resting on a vertical shaft through its center. The cylinder has a radius R = 15 cm and mass M = 0.25 kg. A cockroach (mass m = 0.015 kg) is on the lazy Susan, at a distance of 8.0 cm from the center. Both the cockroach and the lazy Susan are initially at rest. The cockroach then walks along a circular path concentric with the center of the Lazy Susan at a constant distance of 8.0 cm from the axis of the shaft. If the speed of the cockroach with respect to the lazy Susan is 0.010 m/s, what is the speed of the cockroach with respect to the room? Picture the Problem Because the net external torque acting on the lazy Susancockroach system is zero, the net angular momentum of the system is constant (equal to zero because the lazy Susan is initially at rest) and we can use conservation of angular momentum to find the angular velocity ω of the lazy Susan. The speed of the cockroach relative to the floor vf is the difference between its speed with respect to the lazy Susan and the speed of the lazy Susan at the location of the cockroach with respect to the floor.

Angular Momentum 1007 Relate the speed of the cockroach with respect to the floor vf to the speed of the lazy Susan at the location of the cockroach:

vf = v − ωr

(1)

Use conservation of angular momentum to obtain:

LLS − LC = 0

(2)

Express the angular momentum of the lazy Susan:

LLS = I LSω = 12 MR 2ω

Express the angular momentum of the cockroach:

⎞ ⎛v LC = I CωC = mr 2 ⎜ − ω ⎟ ⎠ ⎝r

Substitute for LLS and LC in equation (2) to obtain:

1 2

Solving for ω yields: Substitute for ω in equation (1) to obtain:

⎞ ⎛v MR 2ω − mr 2 ⎜ − ω ⎟ = 0 ⎠ ⎝r

ω=

2mrv MR 2 + 2mr 2

2mr 2 v vf = v − MR 2 + 2mr 2

Substitute numerical values and evaluate vf: 2(0.015 kg )(0.080 m ) (0.010 m/s ) vf = 0.010 m/s − = 10 mm/s (0.25 m )(0.15 m )2 + 2(0.015 kg )(0.080 m )2 2

Remarks: Because the moment of inertia of the lazy Susan is so much larger than the moment of inertia of the cockroach, after the cockroach begins moving, the angular speed of the lazy Susan is very small. Therefore, the speed of the cockroach relative to the floor is almost the same as the speed relative to the lazy Susan.

Two disks of identical mass but different radii (r and 2r) are spinning 52 •• on frictionless bearings at the same angular speed ω0 but in opposite directions (Figure 10-49). The two disks are brought slowly together. The resulting frictional force between the surfaces eventually brings them to a common angular velocity. (a) What is the magnitude of that final angular velocity in terms of ω0? (b) What is the change in rotational kinetic energy of the system? Explain.

1008 Chapter 10 Picture the Problem The net external torque acting on this system is zero and so we know that angular momentum is conserved as these disks are brought together. Let the numeral 1 refer to the disk to the left and the numeral 2 to the disk to the right. Let the angular momentum of the disk with the larger radius be positive.

(a) Using conservation of angular momentum, relate the initial angular speeds of the disks to their common final speed and to their moments of inertia: Solving for ωf yields:

Express I1 and I2:

I iω i = I f ω f or I 1ω 0 − I 2ω 0 = (I 1 + I 2 )ω f

I1 − I 2 ω0 I1 + I 2

ωf =

(1)

I 1 = 12 m(2r ) = 2mr 2 2

and I 2 = 12 mr 2 Substitute for I1 and I2 in equation (1) and simplify to obtain:

2mr 2 − 12 mr 2 ωf = ω0 = 2mr 2 + 12 mr 2

(b) The change in kinetic energy of the system is given by:

ΔK = K f − K i

The initial kinetic energy of the system is the sum of the kinetic energies of the two disks:

K i = K1 + K 2 = 12 I1ω02 + 12 I 2ω02

Substituting for Kf and Ki in equation (2) yields:

ΔK = 12 (I1 + I 2 )ωf2 − 12 (I1 + I 2 )ω 02

Substitute for ωf from part (a) and simplify to obtain:

ΔK =

Noting that the quantity in brackets is Ki, substitute to obtain:

ΔK = − 16 25 K i

=

1 2

3 5

ω0 (2)

(I1 + I 2 )ω02

(I1 + I 2 )( 53 ω0 )2 − 12 (I1 + I 2 )ω02 2 1 = − 16 25 [2 (I 1 + I 2 )ω 0 ] 1 2

The frictional force between the surfaces is responsible for some of the initial kinetic energy being converted to thermal energy as the two disks come together.

Angular Momentum 1009 53 •• A block of mass m sliding on a frictionless table is attached to a string that passes through a narrow hole through the center of the table. The block is sliding with speed v0 in a circle of radius r0. Find (a) the angular momentum of the block, (b) the kinetic energy of the block, and (c) the tension in the string. (d) A student under the table now slowly pulls the string downward. How much work is required to reduce the radius of the circle from r0 to r0/2? Picture the Problem (a) and (b) We can express the angular momentum and kinetic energy of the block directly from their definitions. (c) The tension in the string provides the centripetal force required for the uniform circular motion and can be expressed using Newton’s second law. (d) Finally, we can use the workkinetic energy theorem to express the work required to reduce the radius of the circle by a factor of two.

(a) Express the initial angular momentum of the block:

L0 = r0 mv0

(b) Express the initial kinetic energy of the block:

K0 =

(c) Using Newton’s second law, relate the tension in the string to the centripetal force required for the circular motion:

v02 T = Fc = m r0

(d) Use the work-kinetic energy theorem to relate the required work to the change in the kinetic energy of the block:

L20 L2f W = ΔK = K f − K 0 = − 2I f 2I 0

Substituting for L0 from Part (a) and simplifying gives:

1 2

mv02

L20 L2 L2 ⎛ 1 1 ⎞ − 0 = 0 ⎜⎜ − ⎟⎟ 2I f 2I 0 2 ⎝ If I0 ⎠ L2 ⎛ 1 1 ⎞ = 0⎜ − 2⎟ 2 2 ⎜⎝ m( 12 r0 ) mr0 ⎟⎠ L2 ⎛ 4 1 ⎞ L2 ⎛ 3 ⎞ = 0 ⎜⎜ 2 − 2 ⎟⎟ = 0 ⎜⎜ 2 ⎟⎟ 2 ⎝ mr0 mr0 ⎠ 2 ⎝ mr0 ⎠

=

W

2 ( r0 mv0 ) ⎛ ⎜ =

2

3 ⎞ ⎜ mr 2 ⎟⎟ = ⎝ 0 ⎠

3 2

mv02

54 ••• A 0.20-kg point mass moving on a frictionless horizontal surface is attached to a rubber band whose other end is fixed at point P. The rubber band exerts a force whose magnitude is F = bx, where x is the length of the rubber band

1010 Chapter 10 and b is an unknown constant. The rubber band force points inward towards P. The mass moves along the dotted line in Figure 10-50. When it passes point A, its velocity is 4.0 m/s, directed as shown. The distance AP is 0.60 m and BP is 1.0 m. (a) Find the speed of the mass at points B and C. (b) Find b. Picture the Problem Because the force exerted by the rubber band is parallel to the position vector of the point mass, the net external torque acting on it is zero and we can use the conservation of angular momentum to determine the speeds of the ball at points B and C. We’ll use mechanical energy conservation to find b by relating the kinetic and elastic potential energies at A and B.

(a) Use conservation of momentum to relate the angular momenta at points A, B and C:

L A = LB = LC

or mv A rA = mv B rB = mvC rC

(1)

rA rB

Solve for vB in terms of v A :

vB = v A

Substitute numerical values and evaluate vB :

v B = (4.0 m/s )

Solve equation (1) for vC in terms of vA :

vC = v A

Substitute numerical values and evaluate vC :

vC = (4.0 m/s )

(b) Use conservation of mechanical energy between points A and B to relate the kinetic energy of the point mass and the energy stored in the stretched rubber band:

ΔE = E A − E B = 0 or 2 2 2 2 1 1 1 1 2 mv A + 2 brA − 2 mv B − 2 brB = 0

Solving for b yields:

Substitute numerical values and evaluate b:

0.60 m = 2.4 m/s 1.0 m

rA rC

(

0.60 m = 4.0 m/s 0.60 m

)

b=

m vB2 − v A2 rA2 − rB2

b=

(0.20 kg )([ 2.4 m/s)2 − (4.0 m/s)2 ] (0.60 m )2 − (1.0 m )2

= 3 N/m

Angular Momentum 1011

*Quantization of Angular Momentum 55

••

[SSM]

The z component of the spin of an electron is − 12 h , but the

magnitude of the spin vector is 0.75h . What is the angle between the electron’s spin angular momentum vector and the positive z-axis? z

Picture the Problem The electron’s spin angular momentum vector is related to its z component as shown in the diagram. The angle between r s and the +z-axis is φ.

φ 5r 0.7

θ

r s

− 12 r

Express φ in terms of θ to obtain:

φ = 180° − θ

Using trigonometry, relate the r magnitude of s to its −z component:

θ = cos −1 ⎜⎜

Substitute for θ in the expression for φ to obtain:

θ = 180° − cos −1 ⎜⎜

⎞ ⎟⎟ ⎝ 0.75h ⎠ ⎛

1 2

h

⎞ ⎟⎟ = 125° ⎝ 0.75h ⎠ ⎛

1 2

h

56 •• Show that the energy difference between one rotational state of a molecule and the next higher state is proportional to l + 1. Picture the Problem Equation 10-29a describes the quantization of rotational energy. We can show that the energy difference between a given state and the next higher state is proportional to l + 1 by using Equation 10-27a to express the energy difference.

From Equation 10-29a we have: Using this equation, express the difference between one rotational state and the next higher state:

K l = l(l + 1)E 0 r ΔE = (l + 1)(l + 2)E0 r − l(l + 1)E0 r = 2(l + 1)E0 r

57 •• [SSM] You work in a bio-chemical research lab, where you are investigating the rotational energy levels of the HBr molecule. After consulting the periodic chart, you know that the mass of the bromine atom is 80 times that of the hydrogen atom. Consequently, in calculating the rotational motion of the molecule, you assume, to a good approximation, that the Br nucleus remains stationary as the H atom (mass 1.67 × 10–27 kg) revolves around it. You also know

1012 Chapter 10 that the separation between the H atom and bromine nucleus is 0.144 nm. Calculate (a) the moment of inertia of the HBr molecule about the bromine nucleus, and (b) the rotational energies for the bromine nucleus’s ground state (lowest energy) l = 0, and the next two states of higher energy (called the first and second excited states) described by l = 1, and l = 2. Picture the Problem The rotational energies of HBr molecule are related to l and E 0 r according to K l = l(l + 1)E 0 r where E0 r = h 2 2 I .

(a) Neglecting the motion of the bromine molecule:

I HBr ≈ mp r 2 = mH r 2

Substitute numerical values and evaluate IHBr:

I HBr ≈ 1.67 × 10 −27 kg 0.144 × 10 −9 m

(

)(

)

2

= 3.463 × 10 − 47 kg ⋅ m 2 = 3.46 × 10 − 47 kg ⋅ m 2

(b) Relate the rotational energies to l and E 0 r :

K l = l(l + 1)E 0 r where E0 r =

Substitute numerical values and evaluate E 0 r :

E0 r =

h2 2 I HBr

( (

)

2

1.055 × 10 −34 J ⋅ s h2 = 2 I 2 3.463 × 10 − 47 kg ⋅ m 2

= 1.607 × 10 −22 J ×

)

1eV 1.602 × 10 −19 J

= 1.003 meV Evaluate E0 to obtain:

E0 = K 0 = 1.00 meV

Evaluate E1 to obtain:

E1 = K1 = (1 + 1)(1.003 meV ) = 2.01meV

Evaluate E2 to obtain:

E 2 = K 2 = 2(2 + 1)(1.003 meV ) = 6.02 meV

58 ••• The equilibrium separation between the nuclei of the nitrogen molecule (N2) is 0.110 nm and the mass of each nitrogen nucleus is 14.0 u, where u = 1.66 × 10–27 kg. For rotational energies, the total energy is due to rotational kinetic energy. (a) Approximate the nitrogen molecule as a rigid dumbbell of two equal point masses and calculate the moment of inertia about its center of mass.

Angular Momentum 1013 (b) Find the energy El of the lowest three energy levels using El = K l = l(l + 1)h 2 (2 I ) . (c) Molecules emit a particle (or quantum) of light called a photon when they make a transition from a higher energy state to a lower one. Determine the energy of a photon emitted when a nitrogen molecule drops from the l = 2 to the l = 1 state. Visible light photons each have between 2 and 3 eV of energy. Are these photons in the visible region? Picture the Problem We can use the definition of the moment of inertia of point particles to calculate the rotational inertia of the nitrogen molecule. The rotational energies of nitrogen molecule are related to l and E 0 r according to

El = K l = l(l + 1)E0 r where E0 r = h 2 2 I .

(a) Using a rigid dumbbell model, express and evaluate the moment of inertia of the nitrogen molecule about its center of mass:

I N 2 = ∑ mi ri 2 = mN r 2 + mN r 2

Substitute numerical values and evaluate I N 2 :

⎛ 0.110 nm ⎞ I N 2 = 2(14) 1.66 × 10 −27 kg ⎜ ⎟ 2 ⎝ ⎠ 2 − 46 = 1.406 × 10 kg ⋅ m

i

= 2m N r 2

(

)

= 1.41× 10 −46 kg ⋅ m 2 (b) Relate the rotational energies to l and E 0 r :

El = K l = l(l + 1)E0 r where E0 r =

h2 2 I N2

(1.055 ×10 2(1.406 × 10

Substitute numerical values and evaluate E 0 r :

E0 r =

Evaluate E0 to obtain:

E0 = 0.247 meV

Evaluate E1 to obtain:

E1 = (1 + 1)(0.2474 meV )

−34

)

2

J ⋅s − 46 kg ⋅ m 2 1eV = 3.958 × 10 − 23 J × 1.60 × 10 −19 J = 0.2474 meV

= 0.495 meV

)

2

1014 Chapter 10 Evaluate E2 to obtain:

E 2 = 2(2 + 1)(0.2474 meV ) = 1.48 meV

(c) The energy of a photon emitted when a nitrogen molecule drops from the l = 2 to the l = 1 state is:

ΔEl = 2→l =1 = E 2 − E1

= 1.48 meV − 0.495 meV = 0.99 meV

No. This energy is too low to produce radiation in the visible portion of the spectrum. 59 ••• Consider a transition from a lower energy state to a higher one. That is, the absorption of a quantum of energy resulting in an increase in the rotational energy of an N2 molecule (see Problem 64). Suppose such a molecule, initially in its ground rotational state, was exposed to photons each with energy equal to the three times the energy of its first excited state. (a) Would the molecule be able to absorb this photon energy? Explain why or why not and if it can, determine the energy level to which it goes. (b) To make a transition from its ground state to its second excited state requires how many times the energy of the first excited state? Picture the Problem The rotational energies of a nitrogen molecule depend on the quantum number l according to El = L2 / 2 I = l(l + 1)h 2 / 2 I .

(a) No. None of the allowed values of El are equal to E0r. (b) The upward transition from the ground state to the second excited state requires energy given by:

ΔEl =0→l =2 = E2 − E0

Set this energy difference equal to a constant n times the energy of the 1st excited state:

E2 − E0 = nE1 ⇒ n =

Substitute numerical values and evaluate n:

n=

E 2 − E0 E1

2(2 + 1)E0 r − E0 r = 2.5 (1 + 1)E0r

Collisions with Rotations 60 •• A 16.0-kg, 2.40-m-long rod is supported at its midpoint on a knife edge. A 3.20-kg ball of clay drops from rest from a height of 1.20 m and makes a perfectly inelastic collision with the rod 0.90 m from the point of support (Figure 10-51). Find the angular momentum of the rod and clay system about the point of support immediately after the inelastic collision.

Angular Momentum 1015 Picture the Problem Let the zero of gravitational potential energy be at the elevation of the rod. Because the net external torque acting on this system is zero, we know that angular momentum is conserved in the collision. We’ll use the definition of angular momentum to express the angular momentum just after the collision and conservation of mechanical energy to determine the speed of the ball just before it makes its perfectly inelastic collision with the rod.

Use conservation of angular momentum to relate the angular momentum before the collision to the angular momentum just after the perfectly inelastic collision:

Lf = Li = mvr

Use conservation of mechanical energy to relate the kinetic energy of the ball just before impact to its initial potential energy:

Kf − Ki + U f − U i = 0 or, because Ki = Uf = 0, Kf −Ui = 0

(1)

mv 2 − mgh = 0 ⇒ v = 2 gh

Letting h represent the distance the ball falls, substitute for K f and U i to obtain:

1 2

Substituting for v in equation (1) yields:

Lf = mr 2 gh

Substitute numerical values and evaluate Lf:

(

)

Lf = (3.20 kg )(0.90 m ) 2 9.81m/s 2 (1.20 m ) = 14 J ⋅ s

61 •• [SSM] Figure 10-52 shows a thin uniform bar of length L and mass M and a small blob of putty of mass m. The system is supported r by a frictionless horizontal surface. The putty moves to the right with velocity v , strikes the bar at a distance d from the center of the bar, and sticks to the bar at the point of contact. Obtain expressions for the velocity of the system’s center of mass and for the angular speed following the collision. Picture the Problem The velocity of the center of mass of the bar-blob system does not change during the collision and so we can calculate it before the collision using its definition. Because there are no external forces or torques acting on the bar-blob system, both linear and angular momentum are conserved in the collision. Let the direction the blob of putty is moving initially be the +x direction. Let lower-case letters refer to the blob of putty and upper-case letters refer to the bar.

1016 Chapter 10 The diagram to the left shows the blob of putty approaching the bar and the diagram to the right shows the bar-blob system rotating about its center of mass and translating after the perfectly inelastic collision. M

M

cm d m

ycm

vcm

d

ω

r v

The velocity of the center of mass before the collision is given by:

Using its definition, express the location of the center of mass relative to the center of the bar:

Express the angular momentum, relative to the center of mass, of the bar-blob system: Express the angular momentum about the center of mass:

Using the parallel axis theorem, express the moment of inertia of the system relative to its center of mass:

r

(M + m )vrcm = mvr + MV

r or, because V = 0 , r m r v cm = v M +m

(M + m ) ycm = md

⇒ ycm =

md M +m

below the center of the bar. Lcm = I cmω ⇒ ω =

Lcm I cm

(1)

Lcm = mv(d − ycm ) md ⎞ mMvd ⎛ = mv⎜ d − ⎟= M +m⎠ M +m ⎝ 2 I cm = 121 ML2 + Mycm + m(d − ycm )

Substitute for ycm and simplify to obtain: 2 2 2 md ⎞ ⎛ md ⎞ ⎛ 2 + mMd 1 1 I cm = 12 ML2 + M ⎜ + m d − = ML ⎟ ⎜ ⎟ 12 M +m⎠ M +m ⎝M +m⎠ ⎝

2

Angular Momentum 1017 Substitute for Icm and Lcm in equation (1) and simplify to obtain:

ω=

1 12

mMvd ML (M + m ) + Mmd 2 2

Remarks: You can verify the expression for Icm by letting m → 0 to obtain I cm = 121 ML2 and letting M → 0 to obtain Icm = 0. 62 •• Figure 10-52 shows a thin uniform bar whose length is L and mass is M and a compact hard sphere whose mass is m. The system is supported by a frictionless horizontal surface. The sphere moves to the right r with velocity v , strikes the bar at a distance 14 L from the center of the bar. The collision is elastic, and following the collision the sphere is at rest. Find the value of the ratio m/M. Picture the Problem Because there are no external forces or torques acting on the bar-sphere system, both linear and angular momentum are conserved in the collision. Kinetic energy is also conserved in the elastic collision of the hard sphere with the bar. Let the direction the sphere is moving initially be the +x direction Let lower-case letters refer to the compact hard sphere and upper-case characters refer to the bar. Let unprimed characters refer to before the collision and primed characters to after the collision. The diagram to the left shows the path of the sphere before its collision with the bar and the diagram to the right shows the sphere at rest after the collision and the bar rotating about its center of mass and translating to the right. M

M

cm

cm

r V'

d= 14 L m

m

r v

Apply conservation of linear momentum to the collision to obtain:

mv = 0 + MV ' ⇒ V ' =

Apply conservation of angular momentum to the collision to obtain:

mvd = 0 + I cmω

ω

m v M

(1) (2)

1018 Chapter 10 mv 2 = 0 + 12 MV ' 2 + 12 I cmω 2

Apply conservation of mechanical energy to the elastic collision to obtain:

1 2

Use Table 9-1 to find the moment of inertia of a thin bar about an axis through its center:

I cm = 121 ML2

Substitute for I cm in equation (2) and

⎛ 12vd ⎞ m mvd = 121 ML2ω ⇒ ω = ⎜ 2 ⎟ ⎝ L ⎠M

simplify to obtain: Substitute for I cm and V ' in equation (3) and simplify to obtain:

(3)

2

⎛m⎞ mv = M ⎜ ⎟ v 2 + 121 ML2ω 2 ⎝M ⎠ 2

Substituting for ω yields: 2

⎡⎛ 12vd ⎞ m ⎤ ⎛m⎞ mv = M ⎜ ⎟ v 2 + 121 ML2 ⎢⎜ 2 ⎟ ⎥ ⎝M ⎠ ⎣⎝ L ⎠ M ⎦

2

2

Solve this equation for

Because d = L/4:

m to obtain: M

m = M

m = M

1 ⎛d ⎞ 1 + 12⎜ ⎟ ⎝ L⎠

2

1 ⎛1⎞ 1 + 12⎜ ⎟ ⎝4⎠

2

=

4 7

63 •• Figure 10-53 shows a uniform rod whose length is L and whose mass is M pivoted at the top. The rod, which is initially at rest, is struck by a particle whose mass is m at a point x = 0.8L below the pivot. Assume that the particle sticks to the rod. What must be the speed v of the particle so that following the collision the maximum angle between the rod and the vertical is 90°? Picture the Problem Let the zero of gravitational potential energy be a distance x below the pivot and ignore friction between the rod and the pivot. Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. We can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing.

Angular Momentum 1019

Using conservation of mechanical energy, relate the rotational kinetic energy of the system just after the collision to its gravitational potential energy when it has swung through an angle θ :

ΔK + ΔU = 0 or, because Kf = Ui = 0, − Ki + Uf = 0

Substitute for Ki and Uf to obtain: L ⎞ ⎛ − 12 Iω 2 + ⎜ Mg + mgx ⎟(1 − cosθ ) = 0 2 ⎠ ⎝

(1)

Applying conservation of angular momentum to the collision gives:

ΔL = Lf − Li = 0

(2)

The moment of inertia of the system about the pivot is given by:

I f = m(0.8 L ) + 13 ML2

(3)

Substituting for If and Ii in equation (2) yields: Solving for ω yields:

2

[ ML + (0.8L) m]ω − 0.8Lmv = 0 1 3

ω=

2

1 3

2

0.8Lmv ML2 + 0.64mL2

Substitute equations (3) and (4) in equation (1) and simplify to obtain: L 0.32(Lmv ) ⎞ ⎛ − ⎜ Mg + mg (0.8L )⎟(1 − cosθ ) = 0 2 2 1 2 ⎠ ⎝ 3 ML + 0.64mL 2

(4)

1020 Chapter 10 Solving for v gives: v=

(0.5 M + 0.8m ) (13 ML2 + 0.64mL2 )g (1 − cosθ ) 0.32 Lm 2

Evaluate v for θ = 90° to obtain: v=

(0.5 M + 0.8m )(13 ML2 + 0.64mL2 )g 0.32 Lm 2

64 •• If, for the system of Problem 63, L = 1.2 m, M = 0.80 kg, m = 0.30 kg, and the maximum angle between the rod and the vertical following the collision is 60°, find the speed of the particle before impact. Picture the Problem Let the zero of gravitational potential energy be a distance x below the pivot and ignore friction between the rod and pivot. Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. We can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing.

Using conservation of mechanical energy, relate the rotational kinetic energy of the system just after the collision to its gravitational potential energy when it has swung through an angle θ :

Kf − Ki + U f − U i = 0 or, because Kf = Ui = 0, − Ki + Uf = 0

Substitute for Ki and Uf to obtain: L ⎞ ⎛ − 12 Iω 2 + ⎜ Mg + mgx ⎟(1 − cosθ ) = 0 2 ⎠ ⎝

(1)

Angular Momentum 1021 Apply conservation of momentum to the collision:

ΔL = Lf − Li = 0

The moment of inertia of the system about the pivot is:

I = m(0.80 L ) + 13 ML2

Substituting for Lf and Li in equation (2) gives:

(13 M + 0.64m ) L2ω − 0.80 Lmv = 0

Solving for ω yields:

(2)

2

= (0.64m + 13 M ) L2

ω=

1 2

0.80 Lmv ML2 + 0.64mL2

Substitute for ω in equation (1) and simplify to obtain: L 0.32(Lmv ) ⎛ ⎞ − + ⎜ Mg + 0.80 Lmg ⎟ (1 − cosθ ) = 0 I 2 ⎠ ⎝ 2

Solving for v yields:

g (0.50 M + 0.80m )(1 − cosθ )I 0.32 Lm 2

v=

Substitute numerical values and evaluate I:

I = [0.64(0.30 kg ) + 13 (0.80 kg )](1.2 m )

2

= 0.660 kg ⋅ m 2

Substitute numerical values and evaluate v for θ = 60° to obtain: v=

(9.81m/s ) [0.50 (0.80 kg ) + (0.80)(0.30 kg )](0.50)(0.660 kg ⋅ m ) = 2

2

0.32 (1.2 m )(0.30 kg )

2

7.7 m/s

65 •• A uniform rod is resting on a frictionless table when it is suddenly struck at one end by a sharp horizontal blow in a direction perpendicular to the rod. The mass of the rod is M and the magnitude of the impulse applied by the blow is J. Immediately after the rod is struck, (a) what is the velocity of the center of mass of the rod, (b) what is the velocity of the end that is struck, (c) and what is the velocity of the other end of the rod? (d) Is there a point on the rod that remains motionless?

1022 Chapter 10 Picture the Problem Let the length of the uniform stick be l. We can use the impulse-change in momentum theorem to express the velocity of the center of mass of the stick. By expressing the velocity V of the end of the stick in terms of the velocity of the center of mass and applying the angular impulse-change in angular momentum theorem we can find the angular velocity of the stick and, hence, the velocity of the end of the stick.

(a) Apply the impulse-change in momentum theorem to obtain:

J = Δp = p − p 0 = p or, because p0 = 0 and p = Mvcm, J J = Mvcm ⇒ vcm = M

(b) Relate the velocity V of the end of the stick to the velocity of the center of mass vcm :

V = vcm + vrel to cm = vcm + ω ( 12 l )

(1)

Relate the angular impulse to the change in the angular momentum of the stick:

J ( 12 l ) = ΔL = L − L0 = I cmω or, because L0 = 0, J ( 12 l ) = I cmω

(2)

Refer to Table 9-1 to find the moment of inertia of the stick with respect to its center of mass:

I cm = 121 Ml 2

Substitute for Icm in equation (2) to obtain:

J ( 12 l ) = 121 Ml 2ω ⇒ ω =

Substituting for ω in equation (1) yields:

V=

(c) Relate the velocity V′ of the other end of the stick to the velocity of the center of mass vcm :

6J Ml

J ⎛ 6J ⎞ l 4J +⎜ ⎟ = M ⎝ Ml ⎠ 2 M

V = vcm − vrel to cm = vcm − ω ( 12 l ) =

J ⎛ 6J ⎞ l 2J −⎜ ⎟ = − M ⎝ Ml ⎠ 2 M

(d) Yes, one point remains motionless, but only for a very brief time. r 66 •• A projectile of mass mp is traveling at a constant velocity v 0 toward a stationary disk of mass M and radius R that is free to rotate about its axis O (Figure 10-54). Before impact, the projectile is traveling along a line displaced a distance b below the axis. The projectile strikes the disk and sticks to point B. Model the projectile as a point mass. (a) Before impact, what is the total angular momentum L0 of the disk-projectile system about the axis? Answer the following questions in terms of the symbols given at the start of this problem. (b) What is

Angular Momentum 1023 the angular speed ω of the disk-projectile system just after the impact? (c) What is the kinetic energy of the disk-projectile system after impact? (d) How much mechanical energy is lost in this collision? Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision.

(a) Use its definition to express the total angular momentum of the disk and projectile just before impact:

L0 = mp v0 b

(b) Use conservation of angular momentum to relate the angular momenta just before and just after the collision:

L0 = L = Iω ⇒ ω =

L0 I

The moment of inertia of the diskprojectile after the impact is:

I = 12 MR 2 + mp R 2 =

1 2

Substitute for I in the expression for ω to obtain:

ω=

(c) Express the kinetic energy of the system after impact in terms of its angular momentum:

L2 = Kf = 2I 2

(M + 2m )R

2

p

2mp v0b

(M + 2m )R

2

p

(m v b)

2

p 0

[ (M + 2m )R ] 2

1 2

p

(m v b)

=

2

p 0

(M + 2m )R

2

p

(d) Express the difference between the initial and final kinetic energies, substitute, and simplify to obtain:

ΔE = K i − K f

(m v b) − (M + 2m )R 2

= mv

2 p 0

1 2

p 0

2

p

=

1 2

⎡ ⎤ 2mp b 2 m v ⎢1 − 2⎥ ⎣⎢ (M + 2mp )R ⎦⎥ 2 p 0

67 •• [SSM] A uniform rod of length L1 and mass M equal to 0.75 kg is attached to a hinge of negligible mass at one end and is free to rotate in the vertical plane (Figure 10-55). The rod is released from rest in the position shown. A particle of mass m = 0.50 kg is supported by a thin string of length L2 from the hinge. The particle sticks to the rod on contact. What should the ratio L2/L1 be so that θmax = 60° after the collision?

1024 Chapter 10 Picture the Problem Assume that there is no friction between the rod and the hinge. Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the perfectly inelastic collision with the particle and the rotational kinetic of the after-collision system is then transformed into gravitational potential energy as the rod-plus-particle swing upward. Let the zero of gravitational potential energy be at a distance L1 below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L1 and L2 and the masses M and m.

Use conservation of energy to relate the initial and final potential energy of the rod to its rotational kinetic energy just before it collides with the particle: Substitute for Kf, Uf, and Ui to obtain: Solving for ω yields:

Letting ω′ represent the angular speed of the rod-and-particle system just after impact, use conservation of angular momentum to relate the angular momenta before and after the collision: Solve for ω′ to obtain:

Use conservation of energy to relate the rotational kinetic energy of the rod-plus-particle just after their collision to their potential energy when they have swung through an angle θmax:

Kf − Ki + U f − U i = 0 or, because Ki = 0, Kf + U f − Ui = 0

(

1 1 2 3

)

ML12 ω 2 + Mg

ω=

L1 − MgL1 = 0 2

3g L1

ΔL = Lf − Li = 0 or 2 2 2 1 1 3 ML1 + mL2 ω '− 3 ML1 ω = 0

(

ω'=

) (

ML12 ω 2 2 1 3 ML1 + mL2 1 3

Kf − Ki + U f − U i = 0

)

Angular Momentum 1025 Because Kf = 0: − 12 Iω '2 + Mg ( 12 L1 )(1 − cosθ max ) + mgL2 (1 − cosθ max ) = 0 Express the moment of inertia of the system with respect to the pivot: Substitute for θmax, I and ω′ in equation (1):

Simplify to obtain:

(1)

I = 13 ML12 + mL22

(

)

g 1 2 2 3 ML1 L1 = Mg ( 12 L1 ) + mgL2 2 2 1 + ML mL 1 2 3

3

L13 = 2

m 2 m m L1 L2 + 3 L22 L1 + 6 L32 M M M

Dividing both sides of the equation by L13 yields: ⎛m 1 = 2⎜ ⎝M

⎞ ⎛ L2 ⎞ ⎛ m ⎟ ⎜⎜ ⎟⎟ + 3⎜ ⎠ ⎝ L1 ⎠ ⎝ M

2

2

⎞ ⎛ L2 ⎞ ⎛ m ⎞ ⎛L ⎞ ⎟ ⎜⎜ ⎟⎟ + 6⎜ ⎟ ⎜⎜ 2 ⎟⎟ ⎠ ⎝ L1 ⎠ ⎝ M ⎠ ⎝ L1 ⎠

3

Let α = m/M and β = L2/L1 to obtain:

6α 2 β 3 + 3αβ 2 + 2αβ − 1 = 0

Substitute for α and simplify to obtain the cubic equation in β:

8β 3 + 6 β 2 + 4 β − 3 = 0

Use the solver function of your calculator to find the only real value of β:

β = 0.39

68 •• A uniform rod that has a length L1 equal to 1.2 m and a mass M equal to 2.0 kg is supported by a hinge at one end and is free to rotate in the vertical plane (Figure 10-55). The rod is released from rest in the position shown. A particle whose mass is m is supported by a thin string that has a length L2 equal to 0.80 m from the hinge. The particle sticks to the rod on contact, and after the collision the rod continues to rotate until θmax = 37°. (a) Find m. (b) How much energy is dissipated during the collision? Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the perfectly inelastic collision with the particle and the rotational

1026 Chapter 10 kinetic energy of the after-collision system is then transformed into gravitational potential energy as the rod-plus-particle swing upward. Let the zero of gravitational potential energy be at a distance L1 below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L1 and L2 and the mass M to m. (a) Use conservation of energy to relate the initial and final potential energy of the rod to its rotational kinetic energy just before it collides with the particle: Substitute for Kf, Uf, and Ui to obtain: Solving for ω yields:

Kf − Ki + U f − U i = 0 or, because Ki = 0, Kf + U f − Ui = 0

(

1 1 2 3

)

ML12 ω 2 + Mg

ω=

L1 − MgL1 = 0 2

3g L1

Letting ω′ represent the angular speed of the system after impact, use conservation of angular momentum to relate the angular momenta before and after the collision:

ΔL = Lf − Li = 0 or 2 2 2 1 1 3 ML1 + mL2 ω '− 3 ML1 ω = 0 (1)

Solving for ω′ and simplifying yields:

ML12 ω' = 1 2 ω 2 3 ML1 + mL2

(

) (

)

1 3

=

ML12 2 2 1 3 ML1 + mL2 1 3

3g L1

Substitute numerical values and simplify to obtain:

ω' =

(2.0 kg )(1.2 m )2 2 2 1 3 (2.0 kg )(1.2 m ) + m(0.80 m ) 1 3

3(9.81m/s 2 ) 4.75 kg / s = 1.2 m 0.960 kg + 0.64m

Angular Momentum 1027 Use conservation of energy to relate the rotational kinetic energy of the rod-plus-particle just after their collision to their potential energy when they have swung through an angle θmax:

Kf − Ki + U f − U i = 0 or, because Kf = 0, − Ki + U f − U i = 0

Substitute for Ki, Uf, and Ui to obtain:

− 12 Iω '2 + Mg ( 12 L1 )(1 − cos θ max )

Express the moment of inertia of the system with respect to the pivot:

I = 13 ML12 + mL22

Substitute for θmax, I and ω′ in equation (1) and simplify to obtain: Substitute for M, L1 and L2 and simplify to obtain:

+ mgL2 (1 − cos θ max ) = 0

1 2

(4.75 kg/s )2

0.960 kg + 0.64m 1 2

= 0.2 g (ML1 + mL2 )

(4.75 kg/s )2

0.960 kg + 0.64m = 0.2 g (2.4 kg ⋅ m + (0.80 m )m )

Solve for m to obtain:

m = 1.18 kg = 1.2 kg

(b) The energy dissipated in the inelastic collision is:

ΔE = U i − U f

Express Ui:

U i = Mg

Express Uf:

⎛ L ⎞ U f = (1 − cosθ max )g ⎜ M 1 + mL2 ⎟ ⎝ 2 ⎠

L1 2

Substitute for Ui and Uf in equation (2) to obtain: ΔE = Mg

(2)

L1 ⎛ L ⎞ − (1 − cosθ max )g ⎜ M 1 + mL2 ⎟ 2 2 ⎝ ⎠

1028 Chapter 10 Substitute numerical values and evaluate ΔE: ΔE =

(2.0 kg )(9.81m/s 2 )(1.2 m ) 2

⎛ (2.0 kg )(1.2 m ) ⎞ − (1 − cos37°) 9.81 m/s 2 ⎜ + (1.18 kg )(0.80 m )⎟ 2 ⎝ ⎠

(

)

= 7.5 J

Precession 69 •• [SSM] A bicycle wheel that has a radius equal to 28 cm is mounted at the middle of an axle 50 cm long. The tire and rim weigh 30 N. The wheel is spun at 12 rev/s, and the axle is then placed in a horizontal position with one end resting on a pivot. (a) What is the angular momentum due to the spinning of the wheel? (Treat the wheel as a hoop.) (b) What is the angular velocity of precession? (c) How long does it take for the axle to swing through 360° around the pivot? (d) What is the angular momentum associated with the motion of the center of mass, that is, due to the precession? In what direction is this angular momentum? Picture the Problem We can determine the angular momentum of the wheel and the angular velocity of its precession from their definitions. The period of the precessional motion can be found from its angular velocity and the angular momentum associated with the motion of the center of mass from its definition.

w 2 Rω g

(a) Using the definition of angular momentum, express the angular momentum of the spinning wheel:

L = Iω = MR 2ω =

Substitute numerical values and evaluate L:

⎛ 30 N ⎞ ⎟(0.28 m )2 L = ⎜⎜ 2 ⎟ ⎝ 9.81m/s ⎠ ⎛ rev 2π rad ⎞ × ⎜12 × ⎟ s rev ⎠ ⎝ = 18.1J ⋅ s = 18 J ⋅ s

(b) Using its definition, express the angular velocity of precession:

ωp =

dφ MgD = dt L

Angular Momentum 1029 Substitute numerical values and evaluate ωp:

ωp =

(30 N )(0.25 m ) = 0.414 rad/s 18.1 J ⋅ s

= 0.41rad/s 2π

2π = 15 s 0.414 rad/s

(c) Express the period of the precessional motion as a function of the angular velocity of precession:

T=

(d) Express the angular momentum of the center of mass due to the precession:

Lp = I cmωp = MD 2ωp

Substitute numerical values and evaluate Lp :

⎛ 30 N ⎞ ⎟(0.25 m )2 (0.414 rad/s ) Lp = ⎜⎜ 2 ⎟ ⎝ 9.81 m/s ⎠

ωp

=

= 0.079 J ⋅ s

The direction of Lp is either up or down, depending on the direction of L. 70 •• A uniform disk, whose mass is 2.50 kg and radius is 6.40 cm is mounted at the center of a 10.0-cm-long axle and spun at 700 rev/min. The axle is then placed in a horizontal position with one end resting on a pivot. The other end is given an initial horizontal velocity such that the precession is smooth with no nutation. (a) What is the angular velocity of precession? (b) What is the speed of the center of mass during the precession? (c) What is the acceleration (magnitude and direction) of the center of mass? (d) What are the vertical and horizontal components of the force exerted by the pivot on the axle? Picture the Problem The angular velocity of precession can be found from its definition. Both the speed and the magnitude of the acceleration of the center of mass during precession are related to the angular velocity of precession. We can use Newton’s second law to find the vertical and horizontal components of the force exerted by the pivot on the axle.

(a) The angular velocity of precession is given by:

ωp =

Substituting for Is and simplifying yields:

ωp =

dφ MgD = dt I sω s

1 2

MgD 2 gD = 2 2 MR ωs R ωs

1030 Chapter 10 Substitute numerical values and evaluate ωp:

ωp =

(

)

2 9.81m/s 2 (0.050 m ) = 3.27 rad/s = 3.3 rad/s rev 2π rad 1min ⎞ 2⎛ ⎟ (0.064 m ) ⎜⎜ 700 × × min rev 60 s ⎟⎠ ⎝

(b) Express the speed of the center of mass in terms of its angular speed of precession:

vcm = Dω p = (0.050 m )(3.27 rad/s ) = 16 cm/s

(c) Relate the acceleration of the center of mass to its angular speed of precession:

acm = Dω p2 = (0.050 m )(3.27 rad/s )

(d) Use Newton’s second law to relate the vertical component of the force exerted by the pivot to the weight of the disk:

Fv = Mg = (2.5 kg ) 9.81m/s 2

Relate the horizontal component of the force exerted by the pivot on the axle to the acceleration of the center of mass:

FH = Macm = (2.5 kg ) 0.535 m/s 2

2

= 0.535 m/s 2 = 54 cm/s 2

(

)

= 25 N

(

)

= 1.3 N

General Problems 71 • [SSM] A particle whose mass is 3.0 kg moves in the xy plane with r v = (3.0 m / s)iˆ along the line y = 5.3 m. (a) Find the angular momentum velocity r Lr about the origin when the particle is at (12 m, 5.3 m). (b) A force F = (–3.9 N) ˆi is applied to the particle. Find the torque about the origin due to this force as the particle passes through the point (12 m, 5.3 m). Picture the Problem While the 3-kg particle is moving in a straight line, it has r r r r r angular momentum given by L = r × p where r is its position vector and p is its r r r linear momentum. The torque due to the applied force is given by τ = r × F .

(a) The angular momentum of the particle is given by:

r r r L=r×p

Angular Momentum 1031 r r Express the vectors r and p :

r r = (12 m ) iˆ + (5.3 m ) ˆj and r p = mvˆi = (3.0 kg )(3.0 m/s ) iˆ = (9.0 kg ⋅ m/s ) iˆ

r r Substitute for r and p :and simplify r to find L :

r L = (12 m ) iˆ + (5.3 m ) ˆj × (9.0 kg ⋅ m/s ) iˆ = (47.7 kg ⋅ m 2 /s ) ˆj × iˆ

[

]

( )

(

)

= − 48 kg ⋅ m 2 /s kˆ (b) Using its definition, express the torque due to the force: r r Substitute for r and F and simplify to r find τ :

r r r τ = r ×F

[

] ( )

r τ = (12 m ) iˆ + (5.3 m ) ˆj × (− 3.9 N ) iˆ = −(15.9 N ⋅ m ) ˆj × iˆ

=

(21 N ⋅ m ) kˆ

72 • The position vector of a particle whose mass is 3.0 kg is given by r r ˆ r = 4.0 i + 3.0t2 jˆ , where r is in meters and t is in seconds. Determine the angular momentum and net torque, about the origin, acting on the particle. Picture the Problem The angular momentum of the particle is given by r r r r r L = r × p where r is its position vector and p is its linear momentum. The torque r r acting on the particle is given by τ = dL dt . r r r r r r r L = r × p = r × mv = mr × v r r dr = mr × dt

The angular momentum of the particle is given by:

r dr yields: Evaluating dt

r dr d = 4.0iˆ + 3.0t 2 ˆj = (6.0t ) ˆj dt dt

[

]

r r r dr Substitute for mr and and simplify to find L : dt

[

{

) }]

r L = (3.0 kg ) (4.0 m ) iˆ + 3.0t 2 m/s 2 ˆj × (6.0t m/s ) ˆj =

(

(72t J ⋅ s ) kˆ

1032 Chapter 10 Find the net torque due to the force:

r r dL d (72t J ⋅ s ) kˆ τ net = = dt dt = (72 N ⋅ m ) kˆ

[

]

73 •• Two ice skaters, whose masses are 55 kg and 85 kg, hold hands and rotate about a vertical axis that passes between them, making one revolution in 2.5 s. Their centers of mass are separated by 1.7 m and their center of mass is stationary. Model each skater as a point particle and find (a) the angular momentum of the system about their center of mass and (b) the total kinetic energy of the system. Picture the Problem The ice skaters rotate about their center of mass; a point we can locate using its definition. Knowing the location of the center of mass we can determine their moment of inertia with respect to an axis through this point. The angular momentum of the system is then given by L = I cmω and its kinetic energy can be found from K = L2 (2 I cm ).

(a) Express the angular momentum of the system about the center of mass of the skaters:

L = I cmω

Using its definition, locate the center of mass, relative to the 85-kg skater, of the system:

xcm =

Calculate I cm :

I cm = (55 kg )(1.7 m − 0.668 m )

(55 kg )(1.7 m ) + (85 kg )(0)

55 kg + 85 kg = 0.668 m 2

+ (85 kg )(0.668 m )

2

= 96.5 kg ⋅ m 2 Substitute to determine L:

⎛ 1 rev 2π rad ⎞ ⎟ × L = 96.5 kg ⋅ m 2 ⎜⎜ rev ⎟⎠ ⎝ 2.5 s

(

)

= 243 J ⋅ s = 0.24 kJ ⋅ s

(b) Relate the total kinetic energy of the system to its angular momentum and evaluate K: Substitute numerical values and evaluate K:

K=

K=

L2 2I cm

(

(243 J ⋅ s )2

2 96.5 kg ⋅ m 2

)=

0.31kJ

Angular Momentum 1033

74 •• A 2.0-kg ball attached to a string whose length is 1.5 m moves counterclockwise (as viewed from above) in a horizontal circle (Figure 10-56). The string makes an angle θ = 30° with the vertical. (a) Determine both the r horizontal and vertical components of the angular momentum L of the ball about r the point of support P. (b) Find the magnitude of dL dt and verify that it equals the magnitude of the torque exerted by gravity about the point of support. y

Picture the Problem Let the origin of the coordinate system be at the pivot. The diagram shows the forces acting on the ball. We’ll apply Newton’s second law to the ball to determine its speed. We’ll then use the derivative of its position vector to express its velocity and the definition of angular r momentum to show that L has both horizontal and vertical components. We r can use the derivative of L with respect to time to show that the rate at which the angular momentum of the ball changes is equal to the torque, relative to the pivot point, acting on it. (a) Express the angular momentum of the ball about the point of support: Apply Newton’s second law to the ball:

P

x

θ z r

θ

r T

m r mg

r r r r r L = r × p = mr × v

∑ Fx = T sin θ = m

(1)

v2 r sin θ

and ∑ Fz = T cosθ − mg = 0 Eliminate T between these equations and solve for v to obtain:

v = rg sin θ tan θ

Substitute numerical values and evaluate v:

v=

Express the position vector of the ball:

r r = (1.5 m )sin 30° cos ωt iˆ + sin ωtˆj − (1.5 m )cos 30°kˆ

(1.5 m )(9.81m/s2 ) sin30°tan30°

= 2.06 m/s

(

)

1034 Chapter 10 r r dr v= dt = (0.75ω m/s ) − sin ωt iˆ + cos ωt ˆj

The velocity of the ball is:

(

Evaluating ω yields:

ω=

2.06 m/s = 2.75 rad/s (1.5 m )sin 30°

(

r v = (2.06 m/s ) − sin ωt iˆ + cos ωtˆj

Substitute for ω to obtain:

)

)

r Substitute in equation (1) and evaluate L :

[

(

)

r L = (2.0 kg )(1.5 m )sin 30° cos ωt iˆ + sin ωt ˆj − (1.5 m )cos 30°kˆ × [(2.06 m/s ) − sin ωt iˆ + cos ωt ˆj

(

= (5.35 J ⋅ s )cos ωt iˆ + (5.35 J ⋅ s )sin ωt ˆj + (3.09 J ⋅ s ) kˆ

)]

]

r The horizontal component of L is the component in the xy plane: r Lhor = (5.4 J ⋅ s )cos ωt iˆ + (5.4 J ⋅ s )sin ωt ˆj

r The vertical component of L is its z component:

r Lvertical = (3.1J ⋅ s )kˆ

r dL (b) Evaluate : dt

r dL = 5.36ω − sin ωt iˆ + cos ωtˆj J dt

r dL Evaluate the magnitude of : dt

r dL = (5.36 N ⋅ m ⋅ s )(2.75 rad/s ) dt

[

(

)]

= 15 N ⋅ m

Express the magnitude of the torque exerted by gravity about the point of support: Substitute numerical values and evaluate τ :

τ = mgr sin θ

τ = (2.0 kg )(9.81m/s 2 )(1.5 m )sin 30° = 15 N ⋅ m

Angular Momentum 1035

75 •• A compact object whose mass is m resting on a horizontal, frictionless surface is attached to a string that wraps around a vertical cylindrical post attached to the surface. Thus, when the object is set into motion, it follows a path that spirals inward. (a) Is the angular momentum of the object about the axis of the post conserved? Explain your answer. (b) Is the energy of the object conserved? Explain your answer. (c) If the speed of the object is v0 when the unwrapped length of the string is r, what is its speed when the unwrapped length has shortened to r/2? Picture the Problem The pictorial representation depicts the object rotating counterclockwise around the cylindrical post. Let the system be the object. In Part (a) we need to decide whether a net torque acts on the object and in Part (b) the issue is whether any external forces act on the object. In Part (c) we can apply the definition of kinetic energy to find the speed of the object when the unwrapped length has shortened to r/2. (a) The net torque acting on the object is given by:

R

r

r T

m

τ net =

dL = RT dt

Because τnet ≠ 0, angular momentum is not conserved. (b) Because, in this frictionless environment, the net external force acting on the object is the tension force and it acts at right angles to the object’s velocity, the energy of the object is conserved. (c) Apply conservation of mechanical energy to the object to obtain:

Substituting for the kinetic energies yields:

ΔE = ΔK + ΔU = 0 or, because ΔU = 0, ΔK rot = 0 1 2

I'ω' 2 − 12 Iω02 = 0

or I'ω' 2 − Iω 02 = 0

1036 Chapter 10 Substitute for I, I′, ω′, and ω0 to obtain:

⎛ ⎞ 2 2⎜ ⎟ ⎛ r ⎞ ⎜ v' ⎟ 1 2 ⎛ v0 ⎞ 1 − 2 mr ⎜ ⎟ = 0 2 m⎜ ⎟ ⎝2⎠ ⎜ r ⎟ ⎝ r ⎠ ⎜ ⎟ ⎝2⎠

Solving for v′ yields:

v' = v0

2

76 •• Figure 10-57 shows a hollow cylindrical tube that has a mass M, a length L, and a moment of inertia ML2/10. Inside the cylinder are two disks each of mass m, separated by a distance l, and tied to a central post by a thin string. The system can rotate about a vertical axis through the center of the cylinder. You are designing this cylinder-disk apparatus to shut down the rotations when the strings break by triggering an electronic ″shutoff″ signal (sent to the rotating motor) when the disks hit the ends of the cylinder. During development, you notice that with the system rotating at some critical angular speed ω, the string suddenly breaks. When the disks reach the ends of the cylinder, they stick. Obtain expressions for the final angular speed and the initial and final kinetic energies of the system. Assume that the inside walls of the cylinder are frictionless. Picture the Problem Because the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular velocities of the system. See Table 9-1 for the moment of inertia of a disk. Using conservation of angular momentum, relate the initial and final angular speeds to the initial and final moments of inertia: Solving for ωf yields:

ΔL = Lf − Li = 0 or I f ω f − I iωi = 0

ωf =

Ii I ωi = i ω If If

(1)

Use the parallel-axis theorem to express the moment of inertia of each of the disks with respect to the axis of rotation:

I i, each disk = I cm + m( 12 l )

Express the initial moment of inertia Ii of the cylindrical tube plus disks system:

I i = I cylindrical + 2 I i, each disk

2

= 14 mr 2 + 14 ml 2

(

= 14 m r 2 + l 2

tube

)

[ ( )] m(r + l )

= 101 ML2 + 2 14 m r 2 + l 2 = 101 ML + 12 2

2

2

Angular Momentum 1037

(

)

( (

) )

I f = 101 ML2 + 12 m r 2 + L2

When the disks have moved out to the end of the cylindrical tube: Substitute for Ii and If in equation (1) and simplify to obtain:

ωf = =

ML2 + 12 m r 2 + l 2 ω ML2 + 12 m r 2 + L2

1 10 1 10

( (

) )

ML2 + 5m r 2 + l 2 ω ML2 + 5m r 2 + L2

The initial kinetic energy of the system is:

K i = 12 I iω 2

Substituting for Ii and simplifying yields:

Ki =

1 1 2 10

=

1 20

The final kinetic energy of the system is:

K f = 12 I f ωf2

[ [

( m(r

)] )]ω

ML2 + 12 m r 2 + l 2 ω 2 ML2 + 14

2

+ l2

2

Substitute for If and ωf and simplify to obtain: Kf =

=

[

1 1 2 10

1 20

( (

) )

⎛ ML2 + 5m r 2 + l 2 ⎞ ω ⎟⎟ ML + m r + L ⎜⎜ 2 2 2 ⎠ ⎝ ML + 5m r + L 2

[

1 2

(

2

( (

2

)]

2

)] )

⎡ ML2 + 5m r 2 + l 2 2 ⎤ 2 ⎢ ⎥ω 2 2 2 ⎣⎢ ML + 5m r + L ⎥⎦

77 •• [SSM] Repeat Problem 76, this time friction between the disks and the walls of the cylinder is not negligible. However, the coefficient of friction is not great enough to prevent the disks from reaching the ends of the cylinder. Can the final kinetic energy of the system be determined without knowing the coefficient of kinetic friction? Determine the Concept Yes. The solution depends only upon conservation of angular momentum of the system, so it depends only upon the initial and final moments of inertia. 78 •• Suppose that in Figure 10-57 l = 0.60 m, L = 2.0 m, M = 0.80 kg, and m = 0.40 kg. The string breaks when the system’s angular speed approaches the critical angular speed ωi, at which time the tension in the string is 108 N. The masses then move radially outward until they undergo perfectly inelastic collisions with the ends of the cylinder. Determine the critical angular speed and the angular speed of the system after the inelastic collisions. Find the total kinetic

1038 Chapter 10 energy of the system at the critical angular speed, and again after the inelastic collisions. Assume that the inside walls of the cylinder are frictionless.

Picture the Problem Because the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular speeds of the system. Using conservation of angular momentum, relate the initial and final angular speeds to the initial and final moments of inertia:

ΔL = Lf − Li = 0 or

Express the tension in the string as a function of the critical angular speed of the system:

l 2T T = mrωi2 = m ωi2 ⇒ ωi = ml 2

Substitute numerical values and evaluate ωi :

ωi =

I f ω f − I iωi = 0 ⇒ ω f =

Ii ωi (1) If

2(108 N ) = 30.0 rad/s (0.40 kg )(0.60 m )

= 30 rad/s

(

)

Express Ii:

I i = 101 ML2 + 2 14 ml 2

Substitute numerical values and evaluate Ii:

I i = 101 (0.80 kg )(2.0 m )

2

+ 12 (0.40 kg )(0.60 m )

2

= 0.392 kg ⋅ m 2

(

)

Express If:

I f = 101 ML2 + 2 14 mL2

Substitute numerical values and evaluate If:

I f = 101 (0.80 kg )(2.0 m )

2

+ 12 (0.40 kg )(2.0 m )

2

= 1.12 kg ⋅ m 2

Substitute numerical values in equation (1) and evaluate ωf:

ωf =

0.392 kg ⋅ m 2 (30.0 rad/s ) 1.12 kg ⋅ m 2

= 10.5 rad/s = 11 rad/s

Angular Momentum 1039 The total kinetic energy of the system at the critical angular speed is:

K i = 12 I iωi2

Substitute numerical values and evaluate K i :

Ki =

The total kinetic energy of the system after the inelastic collisions is:

K f = 12 I f ω f2

Substitute numerical values and evaluate K f :

K f = 12 1.12 kg ⋅ m 2 (10.5 rad/s )

1 2

(0.392 kg ⋅ m )(30.0 rad/s)

2

2

= 176 J = 0.18 kJ

(

)

2

= 62 J

79 •• [SSM] Kepler’s second law states: The line from the center of the Sun to the center of a planet sweeps out equal areas in equal times. Show that this law follows directly from the law of conservation of angular momentum and the fact that the force of gravitational attraction between a planet and the Sun acts along the line joining the centers of the two celestial objects. Picture the Problem The pictorial representation shows an elliptical orbit. The triangular element of the area is dA = 12 r (rdθ ) = 12 r 2 dθ .

r dθ

dA θ

Differentiate dA with respect to t to obtain:

dA 1 2 dθ 1 2 = r = r ω dt 2 dt 2

(1)

Because the gravitational force acts along the line joining the two objects, τ = 0. Hence:

L = mr 2ω = constant

(2)

1040 Chapter 10 Eliminate r2ω between equations (1) and (2) to obtain:

dA L = = constant 2m dt

80 •• Consider a cylindrical turntable whose mass is M and radius is R, turning with an initial angular speed ω1 . (a) A parakeet of mass m, hovering in flight above the outer edge of the turntable, gently lands on it and stays in one place on it as shown in Figure 10-58. What is the angular speed of the turntable off (not flies off) after the parakeet r lands? (b) Becoming dizzy, the parakeet jumps r with a velocity v relative to the turntable. The direction of v is tangent to the edge of the turntable, and in the direction of its rotation. What will be the angular speed of the turntable afterwards? Express your answer in terms of the two masses m and M, the radius R, the parakeet speed v and the initial angular speed ω1 . Picture the Problem The angular momentum of the turntable-parakeet is conserved in both parts of this problem. (a) Apply conservation of angular momentum to the turntable-parakeet system as the parakeet lands to obtain:

ΔL = Lf − Li = 0

The final angular momentum of the system is given by:

Lf = Lturntable + Lparakeet r r = I turntableωf + r × pparakeet

Because I turntable = 12 MR 2 and r r r × pparakeet = Rmvparakeet :

Lf = 12 MR 2ω f + Rmvparakeet

(1)

= 12 MR 2ω f + Rm(Rωf ) = 12 MR 2ω f + mR 2ωf

The initial angular momentum of the system is given by:

Li = I turntableωi = 12 MR 2ωi

Substituting for Lf and Li in equation (1) yields:

1 2

Solve for ωf to obtain:

MR 2ωf + mR 2ωf − 12 MR 2ωi = 0

ωf =

M ωi M + 2m

Angular Momentum 1041 (b) Apply conservation of angular momentum to the turntable-parakeet system as the parakeet jumps off to obtain:

ΔL = Lf − Li = 0

The final angular momentum of the system is given by:

Lf = Lturntable + Lparakeet r r = I turntableωf + r × pparakeet

Because I turntable = 12 MR 2 and r r r × pparakeet = Rmvparakeet :

Lf = 12 MR 2ω f + Rmvparakeet

Express the speed of the parakeet relative to the ground:

vparakeet = v turntable + v = Rωf + v

Using the expression derived in (a), substitute for ωf to obtain:

vparakeet =

Substituting for vparakeet in equation (3) and simplifying yields:

⎛ M ⎞ Lf = 12 MR 2ωf + mR⎜ Rωi + v ⎟ ⎝ M + 2m ⎠

The initial angular momentum of the system is the same as the final angular momentum in (a):

Li = 12 MR 2ωi

(2)

(3)

M Rωi + v M + 2m

Substituting for Lf and Li in equation (2) yields: 1 2

⎡⎛ M ⎞ ⎤ 2 MR 2ωf + mR ⎢⎜ ⎟ Rωi + v ⎥ − 12 MR ωi = 0 + M m 2 ⎠ ⎣⎝ ⎦

Solving for ωf gives:

ωf =

M ⎛m ωi − 2⎜ M + 2m ⎝M

⎞v ⎟ ⎠R

81 •• You are given a heavy but thin metal disk (like a coin, but larger; Figure 10-59). (Objects like this are called Euler disks.) Placing the disk on a turntable, you spin the disk, on edge, about a vertical axis through a diameter of the disk and the center of the turntable. As you do this, you hold the turntable still with your other hand, letting it go immediately after you spin the disk. The turntable is a uniform solid cylinder with a radius equal to 0.250 m and a mass equal to 0.735 kg and rotates on a frictionless bearing. The disk has an initial angular speed of 30.0 rev/min. (a) The disk spins down and falls over, finally

1042 Chapter 10 coming to rest on the turntable with its symmetry axis coinciding with the turntable’s. What is the final angular speed of the turntable? (b) What will be the final angular speed if the disk’s symmetry axis ends up 0.100 m from the axis of the turntable?

Picture the Problem Let the letters d, m, and r denote the disk and the letters t, M, and R the turntable. We can use conservation of angular momentum to relate the final angular speed of the turntable to the initial angular speed of the Euler disk and the moments of inertia of the turntable and the disk. In part (b) we’ll need to use the parallel-axis theorem to express the moment of inertia of the disk with respect to the rotational axis of the turntable. You can find the moments of inertia of the disk in its two orientations and that of the turntable in Table 9-1. (a) Use conservation of angular momentum to relate the initial and final angular momenta of the system:

I df ω df + I tf ω tf − I diωdi = 0

Because ωtf = ωdf:

I df ω tf + I tf ω tf − I diωdi = 0

Solving for ωtf yields:

ωtf =

I di ωdi I df + I tf

Ignoring the negligible thickness of the disk, express its initial moment of inertia:

I di = 14 mr 2

Express the final moment of inertia of the disk:

I df = 12 mr 2

Express the final moment of inertia of the turntable:

I tf = 12 MR 2

Substitute in equation (1) and simplify to obtain:

ω tf = =

Express ωdi in rad/s:

mr 2 ωdi 2 1 1 2 mr + 2 MR

(1)

1 4 2

1 ωdi MR 2 2+2 mr 2

ωdi = 30.0

rev 2π rad 1min × × min rev 60 s

= π rad/s

(2)

Angular Momentum 1043 Substitute numerical values in equation (2) and evaluate ωtf:

ω tf =

π rad/s

2 ( 0.735 kg )(0.250 m ) 2+2 (0.500 kg )(0.125 m )2

= 0.228 rad/s (b) Use the parallel-axis theorem to express the final moment of inertia of the disk when it is a distance L from the center of the turntable: Substitute in equation (1) to obtain:

(

I df = 12 mr 2 + mL2 = m 12 r 2 + L2

ωtf = =

)

mr 2 ωdi m 12 r 2 + L2 + 12 MR 2 1 4

(

)

1 2

L MR 2 2+4 2 +2 r mr 2

ωdi

Substitute numerical values and evaluate ωtf:

ω tf =

π rad/s

(0.100 m ) + 2 (0.735 kg )(0.250 m )2 2+4 (0.125 m )2 (0.500 kg )(0.125 m )2 2

= 0.192 rad/s

82 •• (a) Assuming Earth to be a homogeneous sphere that has a radius r and a mass m, show that the period T (time for one daily rotation) of Earth’s rotation about its axis is related to its radius by T = br2, where b = (4/5)π m/L. Here L is the magnitude of the spin angular momentum of Earth. (b) Suppose that the radius r changes by a very small amount Δr due to some internal cause such as thermal expansion. Show that the fractional change in the period ΔT is given approximately by ΔT/T = 2Δr/r. (c) By how many kilometers would r need to increase for the period to change by 0.25 d/y (so that leap years would no longer be necessary)? Picture the Problem We can express the period of Earth’s rotation in terms of its angular velocity of rotation and relate its angular velocity to its angular momentum and moment of inertia with respect to an axis through its center. We can differentiate this expression with respect to T and then use differentials to approximate the changes in r and T.

(a) Express the period of Earth’s rotation in terms of its angular velocity of rotation:

T=

2π

ω

1044 Chapter 10 Relate Earth’s angular speed to its angular momentum and moment of inertia: Substitute for ω and simplify to obtain: (b) Find dT/dr:

ω=

T=

L L = 2 2 I 5 mr 2π

(

)

mr 2 4π m 2 = r L 5L

2 5

dT d ⎛ 4π m 2 ⎞ ⎛ 4π m ⎞ r ⎟ = 2⎜ = ⎜ ⎟r dr dr ⎝ 5L ⎠ ⎝ 5L ⎠

2T ⎛T ⎞ = 2⎜ 2 ⎟r = r ⎝r ⎠ Solving for dT/T yields:

dT dr ΔT Δr ≈ 2 =2 ⇒ T r T r

(c) Using the equation we just derived, substitute for the change in the period of Earth:

1y 1 ΔT 14 d Δr = × = =2 y 365.24 d 1460 T r

Solving for Δr yields:

Substitute numerical values and evaluate Δr:

Δr =

r 2(1460)

6.37 × 10 3 km Δr = = 2.18 km 2(1460 )

83 •• [SSM] The term precession of the equinoxes refers to the fact that Earth’s spin axis does not stay fixed but sweeps out a cone once every 26,000 y. (This explains why our pole star, Polaris, will not remain the pole star forever.) The reason for this instability is that Earth is a giant gyroscope. The spin axis of Earth precesses because of the torques exerted on it by the gravitational forces of the Sun and moon. The angle between the direction of Earth’s spin axis and the normal to the ecliptic plane (the plane of Earth’s orbit) is 22.5 degrees. Calculate an approximate value for this torque, given that the period of rotation of Earth is 1.00 d and its moment of inertia is 8.03 × 1037 kg⋅m2. Picture the Problem Let ωP be the angular velocity of precession of Earth-asgyroscope, ωs its angular velocity about its spin axis, and I its moment of inertia with respect to an axis through its poles, and relate ωP to ωs and I using its definition.

Angular Momentum 1045

τ

Use its definition to express the precession rate of Earth as a giant gyroscope:

ωP =

Substitute for I and solve for τ to obtain:

τ = LωP = IωωP

The angular velocity ωs of Earth about its spin axis is given by:

2π where T is the period of T rotation of Earth.

L

ω=

Substitute for ω to obtain:

τ=

2π Iω P T

Substitute numerical values and evaluateτ:

τ=

(

)(

)

2π 8.03 × 1037 kg ⋅ m 2 7.66 × 10−12 s −1 = 4.47 × 1022 N ⋅ m 24 h 3600 s 1d × × d h

84 •• As indicated in the text, according to the Standard Model of Particle Physics, electrons are point-like particles having no spatial extent. (This assumption has been confirmed experimentally, and the radius of the electron has been shown to be less than 10−18 m.) The intrinsic spin of an electron could in principle be due to its rotation. Let’s check to see if this conclusion is feasible. (a) Assuming that the electron is a uniform sphere whose radius is 1.00 × 10−18 m, what angular speed would be necessary to produce the observed intrinsic angular momentum of h/2? (b) Using this value of angular speed, show that the speed of a point on the ″equator″ of a ″spinning″ electron would be moving faster than the speed of light. What is your conclusion about the spin angular momentum being analogous to a spinning sphere with spatial extent? Picture the Problem We can use the definition of the angular momentum of a spinning sphere, together with the expression for its moment of inertia, to find the angular speed of a point on the surface of a spinning electron. The speed of such a point is directly proportional to the angular speed of the sphere.

(a) Express the angular momentum of the spinning electron:

L = Iω = 12 h

Assuming a spherical electron of radius R, its moment of inertia, relative to its spin axis, is:

I = 52 MR 2

1046 Chapter 10 Substituting for I yields:

Substitute numerical values and evaluate ω:

2 5

MR 2ω = 12 h ⇒ ω =

ω=

(

(

5h 4 MR 2

5 1.05 × 10 −34 J ⋅ s

)(

)

4 9.11× 10 −31 kg 10 −18 m

)

2

= 1.44 × 10 32 rad/s

(b) The speed of a point on the ″equator″ of a spinning electron of radius R is given by: Substitute numerical values and evaluate v:

v = Rω

v = (10 −18 m )(1.44 ×10 32 rad/s ) = 1.44 ×1014 m/s > c

Given that our model predicts a value for the speed of a point on the ″equator″ of a spinning electron that is greater than the speed of light, the idea that the spin angular momentum of an electron is analogous to that of a spinning sphere with spatial extent lacks credibility. 85 •• An interesting phenomenon occurring in certain pulsars (see Problem 24) is an event known as a ″spin glitch,″ that is, a quick change in the spin rate of the pulsar due to a shift in mass location and a resulting rotational inertia change. Imagine a pulsar whose radius is 10.0 km and whose period of rotation is 25.032 ms. The rotation period is observed to suddenly decrease from 25.032 ms to 25.028 ms. If that decrease was related to a contraction of the star, by what amount would the pulsar radius have had to change? Picture the Problem We can apply the conservation of angular momentum to the shrinking pulsar to relate its radii to the observed periods.

The change in the radius of the pulsar is:

ΔR = Rf − Ri

Apply conservation of angular momentum to the shrinking pulsar to obtain:

ΔL = Lf − Li = 0 or I f ω f − I iωi = 0

Substituting for If and Ii yields:

2 5

MRf2ωf − 52 MR12ωi = 0

(1)

Angular Momentum 1047 Solve for ωf to obtain:

Ri2 ωf = 2 ωi Rf

Because ω = 2π T , where T is the rotation period:

T 2π Ri2 2π = 2 ⇒ Rf = f Ri Ti Tf Rf Ti

Substitute for Rf in equation (1) and simplify to obtain:

ΔR =

Substitute numerical values and evaluate ΔR:

⎛ 25.028 ms ⎞ ΔR = ⎜⎜ − 1⎟⎟ (10.0 km ) 25 . 032 ms ⎠ ⎝

⎛ T ⎞ Tf Ri − Ri = ⎜⎜ f − 1⎟⎟ Ri Ti ⎝ Ti ⎠

= − 79.9 cm

a 0.008% decrease in radius. 86 ••• Figure 10.60 shows a pulley in the form of a uniform disk with a rope hanging over it. The circumference of the pulley is 1.2 m and its mass is 2.2 kg. The rope is 8.0 m long and its mass is 4.8 kg. At the instant shown in the figure, the system is at rest and the difference in height of the two ends of the rope is 0.60 m. (a) What is the angular speed of the pulley when the difference in height between the two ends of the rope is 7.2 m? (b) Obtain an expression for the angular momentum of the system as a function of time while neither end of the rope is above the center of the pulley. There is no slippage between rope and pulley wheel. Picture the Problem Let the origin of the coordinate system be at the center of the pulley with the upward direction positive. Let λ be the linear density (mass per unit length) of the rope and L1 and L2 the lengths of the hanging parts of the rope. We can use conservation of mechanical energy to find the angular speed of the pulley when the difference in height between the two ends of the rope is 7.2 m. ΔK + ΔU = 0 or, because Ki = 0, K + ΔU = 0

(a) Apply conservation of energy to relate the final kinetic energy of the system to the change in potential energy:

(1)

Express the change in potential energy of the system: ΔU = U f − U i = − 12 L1f (L1f λ )g − 12 L2f (L2f λ )g − [− 12 L1i (L1i λ )g − 12 L2i (L2i λ )g ]

(

)

(

)

[(

) (

= − 12 L1f2 + L22f λ g + 12 L1i2 + L22i λ g = − 12 λ g L1f2 + L22f − L1i2 + L22i

)]

1048 Chapter 10 Because L1 + L2 = 7.4 m, L2i – L1i = 0.6 m, and L2f – L1f = 7.2 m, we obtain:

L1i = 3.4 m, L2i = 4.0 m, L1f = 0.1 m, and L2f = 7.3 m.

Substitute numerical values and evaluate ΔU:

(

)[

ΔU = − 12 (0.60 kg/m ) 9.81 m/s 2 (0.10 m ) + (7.3 m ) − (3.4 m ) − (4.0 m ) 2

2

2

2

]

= −75.75 J

Express the kinetic energy of the system when the difference in height between the two ends of the rope is 7.2 m:

K = 12 I pω 2 + 12 Mv 2

Substitute numerical values and simplify:

⎛ 1.2 m ⎞ 2 K = [ (2.2 kg ) + 4.8 kg ] ⎜ ⎟ω ⎝ 2π ⎠ = 0.1076 kg ⋅ m 2 ω 2

( (

=

1 1 2 2

=

1 1 2 2

)

M p R 2 ω 2 + 12 MR 2ω 2 M p + M )R 2ω 2

2

1 1 2 2

(

Substitute in equation (1) and solve for ω:

)

(0.1076 kg ⋅ m )ω 2

2

− 75.75 J = 0

and 75.75 J = 27 rad/s 0.1076 kg ⋅ m 2

ω= (b) Noting that the moment arm of each portion of the rope is the same, express the total angular momentum of the system:

L = Lp + Lr = I pω + M r R 2ω

Letting θ be the angle through which the pulley has turned, express U(θ):

U (θ ) = − 12 (L1i − Rθ ) + (L2i + Rθ ) λ g

Express ΔU and simplify to obtain:

ΔU = U f − U i = U (θ ) − U (0)

( =( =

1 2 1 2

)

M p R2 + M r R2 ω M p + M r )R 2ω

[

[

(2)

2

2

]

]

= − 12 (L1i − Rθ ) + (L2i + Rθ ) λ g 2

(

)

+ 12 L1i2 + L22i λ g

2

= − R θ λ g + (L1i − L2i )Rθλ g 2

Assuming that, at t = 0, L1i ≈ L2i:

2

ΔU ≈ − R 2θ 2λ g

Angular Momentum 1049 Substitute for K and ΔU in equation (1) to obtain: Solving for ω yields:

(0.1076 kg ⋅ m )ω 2

− R 2θ 2λ g = 0

R 2θ 2λ g 0.1076 kg ⋅ m 2

ω=

Substitute numerical values to obtain:

2

2

(

)

⎛ 1.2 m ⎞ 2 ⎜ ⎟ (0.6 kg/m ) 9.81m/s 2π ⎠ ω= ⎝ θ 0.1076 kg ⋅ m 2

(

)

= 1.41s -1 θ

Express ω as the rate of change of θ :

dθ dθ = 1.41s −1 θ ⇒ = 1.41s −1 dt θ dt

Integrate θ from 0 to θ to obtain:

ln θ = (1.41s −1 )t

(

(

)

Transform from logarithmic to exponential form to obtain:

θ (t ) = e (1.41s )t

Differentiate to express ω as a function of time:

ω (t ) =

)

−1

−1 dθ = 1.41s −1 e (1.41 s )t dt

(

)

(

)

−1 L = (12 M p + M r )R 2 1.41s −1 e (1.41 s )t

Substitute for ω in equation (2) to obtain:

Substitute numerical values and evaluate L: L=[

2

1 2

[(

)

] (

)

−1 −1 (2.2 kg ) + (4.8 kg )]⎛⎜ 1.2 m ⎞⎟ 1.41s −1 e (1.41s )t = 0.30 kg ⋅ m 2 / s e (1.41s )t ⎝ 2π ⎠

1050 Chapter 10

Chapter R Relativity Conceptual Problems 1 • You are standing on a corner and a friend is driving past in an automobile. Each of you is wearing a wristwatch. Both of you note the times when the car passes two different intersections and determine from your watch readings the time that elapses between the two events. Have either of you determined the proper time interval? Explain your answer. Determine the Concept In the reference frame of the car both events occur at the same location (the location of the car). Thus, your friend’s watch measures the proper time between the two events. 2 • In Problem 1, suppose your friend in the car measures the width of the car door to be 90 cm. You also measure the width as he goes by you. (a) Does either of you measures the proper width of the door? Explain your answer. (b) How will your value for the door width compare to his? (1) Yours will be smaller, (2) yours will be larger, (3) yours will be the same, (4) you can’t compare the widths, as the answer depends on the car’s speed. Determine the Concept The proper length of an object is the length of the object in the rest frame of the object. The proper length of a meter stick is one meter. (a) Because the door is at rest in the reference frame of the car, its width in that frame is its proper width. If your friend measures this width, say by placing a meter stick against the door, then he will measure the proper width of the door. (b) In the reference frame in which you are at rest, the door is moving, so its width is less than its proper width. To measure this width would be challenging. (You could measure the width by measuring the time for the door to go by. The width of the door is the product of the speed of the car and the time.) 3 • [SSM] If event A occurs at a different location than event B in some reference frame, might it be possible for there to be a second reference frame in which they occur at the same location? If so, give an example. If not, explain why not. Determine the Concept Yes. Let the initial frame of reference be frame 1. In frame 1 let L be the distance between the events, let T be the time between the events, and let the +x direction be the direction of event B relative to event A. Next, calculate the value of L/T. If L/T is less than c, then consider the two events in a reference frame 2, a frame moving at speed v = L/T in the +x direction. In frame 2 both events occur at the same location.

1051

1052 Chapter R 4 • [SSM] If event A occurs prior to event B in some frame, might it be possible for there to be a second reference frame in which event B occurs prior to event A? If so, give an example. If not, explain why not. Determine the Concept Yes. Let L and T be the distance and time between the two events in reference frame 1. If L ≤ cT, then something moving at a speed less than or equal to c could travel from the location of event A to the location of event B in a time less than T. Thus, it is possible that event A could cause event B. For events like these, causality demands that event A must precede event B in all reference frames. However, if L > cT then event A cannot be the cause of event B. For events like these, event B does precede event A in certain reference frames. 5 • [SSM] Two events are simultaneous in a frame in which they also occur at the same location. Are they simultaneous in all other reference frames? Determine the Concept Yes. If two events occur at the same time and place in one reference frame they occur at the same time and place in all reference frames. (Any pair of events that occur at the same time and at the same place in one reference frame are called a spacetime coincidence.) 6 • Two inertial observers are in relative motion. Under what circumstances can they agree on the simultaneity of two different events? Determine the Concept We will refer to the two events as event A and event B. Assume that in the reference frame of the first observer there is a stationary clock at the location of each event, with clock A at the location of event A and clock B at the location of event B, and that the two clocks are synchronized. Because the two events are simultaneous in this frame, the readings of the two clocks at the time the events occur are the same. Also, event A and the reading of clock A at the time of event A are a spacetime coincidence, so all observers must agree with that clock reading. In like manner, event B and the reading of clock B at the time of event B are a spacetime coincidence. If observer B is moving parallel with the line joining the two clocks then the clocks readings will differ by Lv/c2 in the reference frame of B, where L is the distance between the clocks in the reference frame of observer A. This means that observer B will agree that the two clock readings at the times of the events are the same, but will not agree that the events occurred at the same time unless L = 0. 7 • The approximate total energy of a particle of mass m moving at speed v > mc2 to obtain the second result. In Parts (b) and (c) we can use the first expression obtained in (a), with E = E0 + K, to find the speeds of electrons with the given kinetic energies. See Table 39-1 for the rest energy of an electron.

(a) The relativistic energy of a particle is given by Equation R-15:

E=

mc 2 1−

v2 c2

( )

Solving for v/c yields:

2 ⎡ v mc 2 ⎤ = ⎢1 − ⎥ c E 2 ⎦⎥ ⎢⎣

12

Expand the radical expression binomially to obtain:

(

v mc 2 = 1− c E2

)

2

= 1−

(

1 mc 2 2 E2

)

2

+ higher - order terms

(1)

1086 Chapter R

( )

Because the higher-order terms are much smaller than the 2nd-degree term when E >> mc2: (b) Solve equation (1) for v:

Because E = E0 + K:

v mc 2 ≈ 1− c 2E 2

v = c 1−

v = c 1−

For an electron whose kinetic energy is 0.510 MeV:

2

(mc )

2 2

E2

E02 1 = c 1− 2 2 (E0 + K ) ⎛ K⎞ ⎜⎜1 + ⎟⎟ ⎝ E0 ⎠

v(0.510 MeV ) = c 1 −

1 ⎛ 0.510 MeV ⎞ ⎜⎜1 + ⎟⎟ ⎝ 0.511 MeV ⎠

2

= 0.866c

(c) For an electron whose kinetic energy is 10.0 MeV:

v(10.0 MeV ) = c 1 −

1 ⎛ 10.0 MeV ⎞ ⎜⎜1 + ⎟⎟ ⎝ 0.511 MeV ⎠

2

= 0.999c

46 •• Use the binomial expansion and Equation R-17 to show that when 2 pc R2.

Gravity For r < R1:

g (r < R1 ) = 0

For r > R2, g(r) is the field due to the thick spherical shell of mass M centered at the origin:

g (r > R2 ) =

For R1 < r < R2, g(r) is determined by the mass within the shell of radius r:

Gm r2 where m = 43 πρ r 3 − R13

Express the density of the spherical shell:

ρ=

M M = 4 3 3 V 3 π R2 − R1

Substitute for ρ in equation (2) and simplify to obtain:

m=

M r 3 − R13 R23 − R13

Substitute for m in equation (1) to obtain:

gr =

GM r2

g (R1 < r < R2 ) =

(

(

(

1185

(1)

)

(2)

)

)

(

) )

GM r 3 − R13 r 2 R23 − R13

(

A graph of gr with R1 = 2, R2 = 3, and GM = 1 follows: 0.12 0.10 0.08 g r 0.06 0.04 0.02 0.00 0

1

2

3

4

5

6

7

8

r

100 •• (a) A thin uniform ring of mass M and radius R lies in the x = 0 plane and is centered at the origin. Sketch a plot of the gravitational field gx versus x for all points on the x axis. (b) At what point, or points, on the axis is the magnitude of gx a maximum?

1186

Chapter 11

Picture the Problem A ring of radius R is shown to the right. Choose a coordinate system in which the origin is at the center of the ring and x axis is as shown. An element of length dL and mass dm is responsible for the field dg at a distance x from the center of the ring. We can express the x component of dg and then integrate over the circumference of the ring to find the total field as a function of x. Gdm R2 + x2

(a) Express the differential gravitational field at a distance x from the center of the ring in terms of the mass of an elemental segment of length dL:

dg =

Relate the mass of the element to its length:

dm = λ dL where λ is the linear density of the ring.

Substitute for dm to obtain:

dg =

By symmetry, the y and z components of g vanish. The x component of dg is:

dg x = dg cosθ =

Refer to the figure to obtain:

cosθ =

Substituting for cosθ yields:

Because λ =

M : 2π R

Gλ dL R2 + x2

dg x =

dg x =

Gλ dL cosθ R2 + x2

x R2 + x2

Gλ dL R2 + x2

x R +x 2

GM xdL

(

2π R R 2 + x 2

)

2

3/ 2

=

Gλ xdL

(R

2

+ x2

)

3/ 2

Gravity Integrate to find gx:

gx =

2πR

GM x

(

2π R R 2 + x 2

=

(R

GM 2

+ x2

1187

)

3/ 2

dL ) ∫ 3/ 2

0

x

A graph of gx follows. The curve is normalized with R = 1 and GM = 1. 0.40 0.35 0.30 0.25

g x 0.20 0.15 0.10 0.05 0.00 0

1

2

3

4

x

(b) Differentiate g(x) with respect to x and set the derivative equal to zero to identify extreme values:

(

)

⎡ x 2 + R 2 3 / 2 − x( 3 ) 2 dg 2 = GM ⎢ x + R2 3 2 2 dx R +x ⎣⎢ Simplify to obtain: Solving for x yields:

(

)

(

(x

2

⎤

) (2 x )⎥ = 0 for extrema 1/ 2

+ R2

x= ±

⎦⎥

)

3/ 2

(

− 3x 2 x 2 + R 2

)

1/ 2

=0

R 2

Because the curve is concave downward, we can conclude that this result corresponds to a maximum. Note that this result agrees with our graphical maximum.

1188

Chapter 11

101 ••• Find the magnitude of the gravitational field a distance r from an infinitely long uniform thin rod whose mass per unit length is λ. Picture the Problem The diagram shows a segment of the wire of length dx and mass dm = λdx at a distance x from the origin of our coordinate system. We can find the magnitude of the gravitational field at a distance r from the wire from the resultant gravitational force acting on a particle of mass m located at point P and then integrating over the length of the wire.

0

dx

x

dm = λ dx

r dF

r

θ P

ρ =

The magnitude of the gravitational field at P is given by: Express the gravitational force acting on a particle of mass m at a distance r from the wire due to the segment of the wire of length dx: Substituting for F and dm gives:

2

2

x

+r

g=

F m 90°

F =2∫

Gmdm cos θ

ρ2

0

90°

g =2∫

Gλdx cos θ

ρ2

0

Because tan θ =

x : r

Note that ρ = r cos θ and substitute for dx and ρ in the expression for x and simplify to obtain:

dx = r sec 2 θdθ = 90°

g =2∫ 0

=

rdθ cos 2 θ

Gλrdθ cos θ ⎛ r2 ⎞ ⎟⎟ cos 2 θ ⎜⎜ 2 ⎝ cos θ ⎠

2Gλ r

90°

∫ cosθ dθ = 0

2Gλ r

102 ••• One question in early planetary science was whether each of the rings of Saturn were solid or were, instead, composed of individual chunks, each in its own orbit. The issue could be resolved by an observation in which astronomers would measure the speed of the inner and outer portions of the ring. If the inner portion of the ring moved more slowly than the outer portion, then the ring was solid; if the opposite was true, then it was actually composed of separate chunks. Let us see how this results from a theoretical viewpoint. Let the radial

Gravity

1189

width of a given ring (there are many) be Δr, the average distance of that ring from the center of Saturn be represented by R, and the average speed of that ring be vavg . (a) If the ring is solid, show that the difference in speed between its outermost and innermost portions, Δv, is given by the approximate expression Δr . Here, vout is the speed of the outermost portion of the Δv = vout − vin ≈ vavg R ring, and vin is the speed of the innermost portion. (b) If, however, the ring is composed of many small chunks, show that Δv ≈ − 12 (vavg Δr R ) . (Assume that Δr ρ B , (b) ρ A < ρ B , (c) ρ A = ρ B , (d) not enough information is given to compare their densities. Determine the Concept The density of an object is its mass per unit volume. The pictorial representation shown to the right summarizes the information concerning the radii and masses of the two spheres. We can determine the relationship between the densities of A and B by examining their ratio.

Express the densities of the two spheres:

A B 2r

ρA =

mA m = 4 A3 VA 3 π rA

and

ρB =

1271

mB m = 4 B3 VB 3 π rB

8m

r

m

1272 Chapter 13 Divide the first of these equations by the second and simplify to obtain:

mA π rA3 mA rB3 ρA = = mB ρB mB rA3 3 4 3 π rB

Substituting for the masses and radii and simplifying yields:

ρ A 8m r 3 = = 1 ⇒ ρA = ρB ρ B m (2r )3

4 3

and

(c )

is correct.

3 • [SSM] Two objects differ in density and mass. Object A has a mass that is eight times the mass of object B. The density of object A is four times the density of object B. How do their volumes compare? (a) VA = 12 VB , (b) VA = VB , (c) VA = 2VB, (d) not enough information is given to compare their volumes. Determine the Concept The density of an object is its mass per unit volume. We can determine the relationship between the volumes of A and B by examining their ratio.

Express the volumes of the two objects: Divide the first of these equations by the second and simplify to obtain:

VA =

mA

ρA

and VB =

mB

ρB

mA

VA ρ A ρ B mA = = VB m B ρ A m B

ρB

Substituting for the masses and densities and simplifying yields:

VA ρ 8mB = B = 2 ⇒ VA = 2VB VB 4 ρ B m B

and

(c )

is correct.

4 • A sphere is constructed by gluing together two hemispheres of different density materials. The density of each hemisphere is uniform, but the density of one is greater than the density of the other. True or false: The average density of the sphere is the numerical average of the two different densities. Clearly explain your reasoning. Determine the Concept True. This is a special case. Because the volumes are equal, the average density is the numerical average of the two densities. This is not a general result.

Fluids 1273 5 • In several jungle adventure movies, the hero and heroine escape the bad guys by hiding underwater for extended periods of time. To do this, they breathe through long vertical hollow reeds. Imagine that in one movie, the water is so clear that to be safely hidden the two are at a depth of 15 m. As a science consultant to the movie producers, you tell them that this is not a realistic depth and the knowledgeable viewer will laugh during this scene. Explain why this is so. Determine the Concept Pressure increases approximately 1 atm every 10 m in fresh water. To breathe requires creating a pressure of less than 1 atm in your lungs. At the surface you can do this easily, but not at a depth of 10 m. 6 •• Two objects are balanced as in Figure 13-28. The objects have identical volumes but different masses. Assume all the objects in the figure are denser than water and thus none will float. Will the equilibrium be disturbed if the entire system is completely immersed in water? Explain your reasoning. Determine the Concept Yes. Because the volumes of the two objects are equal, the downward force on each side is reduced by the same amount (the buoyant force acting on them) when they are submerged. The buoyant force is independent of their masses. That is, if m1L1 = m2L2 and L1 ≠ L2, then (m1 − c)L1 ≠ (m2 − c)L2. 7 •• [SSM] A solid 200-g block of lead and a solid 200-g block of copper are completely submerged in an aquarium filled with water. Each block is suspended just above the bottom of the aquarium by a thread. Which of the following is true?

(a) (b) ( c) (d)

The buoyant force on the lead block is greater than the buoyant force on the copper block. The buoyant force on the copper block is greater than the buoyant force on the lead block. The buoyant force is the same on both blocks. More information is needed to choose the correct answer.

Determine the Concept The buoyant forces acting on these submerged objects are equal to the weight of the water each displaces. The weight of the displaced water, in turn, is directly proportional to the volume of the submerged object. Because ρPb > ρCu, the volume of the copper must be greater than that of the lead and, hence, the buoyant force on the copper is greater than that on the lead. (b) is correct. 8 •• A 20-cm3 block of lead and a 20-cm3 block of copper are completely submerged in an aquarium filled with water. Each is suspended just above the bottom of the aquarium by a thread. Which of the following is true?

1274 Chapter 13 (a) The buoyant force on the lead block is greater than the buoyant force on the copper block.. (b) The buoyant force on the copper block is greater than the buoyant force on the lead block. (c) The buoyant force is the same on both blocks. (d) More information is needed to choose the correct answer. Determine the Concept The buoyant forces acting on these submerged objects are equal to the weight of the water each displaces. The weight of the displaced water, in turn, is directly proportional to the volume of the submerged object. Because their volumes are the same, the buoyant forces on them must be the same. (c) is correct. 9 •• Two bricks are completely submerged in water. Brick 1 is made of lead and has rectangular dimensions of 2″ × 4″ × 8″. Brick 2 is made of wood and has rectangular dimensions of 1″ × 8″ × 8″. True or false: The buoyant force on brick 2 is larger than the buoyant force on brick 1. Determine the Concept False. The buoyant force on a submerged object depends on the weight of the displaced fluid which, in turn, depends on the volume of the displaced fluid. Because the bricks have the same volume, they will displace the same volume of water and the buoyant force will be the same on both of them. 10 •• Figure 13-29 shows an object called a ″Cartesian diver.″ The diver consists of a small tube, open at the bottom, with an air bubble at the top, inside a closed plastic soda bottle that is partly filled with water. The diver normally floats, but sinks when the bottle is squeezed hard. (a) Explain why this happens. (b) Explain the physics behind how a submarine can ″silently″ sink vertically simply by allowing water to flow into empty tanks near its keel. (c) Explain why a floating person will oscillate up and down on the water surface as he or she breathes in and out. Determine the Concept

(a) When the bottle is squeezed, the force is transmitted equally through the fluid, leading to a pressure increase on the air bubble in the diver. The air bubble shrinks, and the loss in buoyancy is enough to sink the diver. (b) As water enters its tanks, the weight of the submarine increases. When the submarine is completely submerged, the volume of the displaced water and, hence, the buoyant force acting on the submarine become constant. Because the weight of the submarine is now greater than the buoyant force acting on it, the submarine will start to sink.

Fluids 1275 (c) Breathing in lowers one’s average density and breathing out increases one’s average density. Because denser objects float lower on the surface than do less dense objects, a floating person will oscillate up and down on the water surface as he or she breathes in and out. 11 •• A certain object has a density just slightly less than that of water so that it floats almost completely submerged. However, the object is more compressible than water. What happens if the floating object is given a slight downward push? Explain. Determine the Concept Because the pressure increases with depth, the object will be compressed and its density will increase as its volume decreases. Thus, the object will sink to the bottom. 12 •• In Example 13.11 the fluid is accelerated to a greater speed as it enters the narrow part of the pipe. Identify the forces that act on the fluid at the entrance to the narrow region to produce this acceleration. Determine the Concept The acceleration-producing force acting on the fluid is the product of the difference in pressure between the wide and narrow parts of the pipe and the area of the narrow part of the pipe. 13 •• [SSM] An upright glass of water is accelerating to the right along a flat, horizontal surface. What is the origin of the force that produces the acceleration on a small element of water in the middle of the glass? Explain by using a diagram. Hint: The water surface will not remain level as long as the glass of water is accelerating. Draw a free body diagram of the small element of water. Determine the Concept The pictorial representation shows the glass and an element of water in the middle of the glass. As is readily established by a simple demonstration, the surface of the water is not level while the glass is accelerated, showing that there is a pressure gradient (a difference in pressure) due to the differing depths (h1 > h2 and hence F1 > F2) of water on the two sides of the element of water. This pressure gradient results in a net force on the element as shown in the figure. The upward buoyant force is equal in magnitude to the downward gravitational force.

1276 Chapter 13

h1 Fr g

h2 r F2

r F1 r Fb

14 •• You are sitting in a boat floating on a very small pond. You take the anchor out of the boat and drop it into the water. Does the water level in the pond rise, fall, or remain the same? Explain your answer. Determine the Concept The water level in the pond will fall slightly. When the anchor is in the boat, the boat displaces enough water so that the buoyant force on it equals the sum of the weight of the boat, your weight, and the weight of the anchor. When you drop the anchor into the water, it displaces just its volume of water (rather than its weight as it did while in the boat). The total weight of the boat becomes less and the boat displaces less water as a consequence. 15 •• A horizontal pipe narrows from a diameter of 10 cm at location A to 5.0 cm at location B. For a nonviscous incompressible fluid flowing without turbulence from location A to location B, how do the flow speeds v (in m/s) at the two locations compare? (a) vA = vB , (b) vA = 12 vB , (c) vA = 14 vB , (d) vA = 2vB , (e) vA = 4vB Determine the Concept We can use the equation of continuity to compare the flow rates at the two locations.

Apply the equation of continuity at locations A and B to obtain: Substitute for AB and AA and simplify to obtain:

AA vA = AB vB ⇒ vA =

AB vB AA 2

⎛d ⎞ πd B2 vA = v = ⎜⎜ B ⎟⎟ vB 2 B πd A ⎝ dA ⎠ 1 4 1 4

Fluids 1277 Substitute numerical values and evaluate vA:

2

⎛ 5 cm ⎞ vA = ⎜ ⎟ vB = 14 vB ⎝ 10 cm ⎠ and (c) is correct.

16 •• A horizontal pipe narrows from a diameter of 10 cm at location A to 5.0 cm at location B. For a nonviscous incompressible fluid flowing without turbulence from location A to location B, how do the pressures P (in N/m2) at the two locations compare? (a) PA = PB , (b) PA = 12 PB , (c) PA = 14 PB , (d) PA = 2PB , (e) PA = 4PB , (f) There is not enough information to compare the pressures quantitatively. Determine the Concept We can use the equation of continuity to compare the flow rates at the two locations.

Apply Bernoulli’s equation for constant elevations at locations A and B to obtain: Apply the equation of continuity at locations A and B to obtain:

PA + 12 ρvA2 = PB + 12 ρvB2

AA vA = AB vB ⇒ vA =

(1)

AB vB AA

Substitute for AB and AA and simplify to obtain:

⎛d ⎞ πd B2 v = ⎜⎜ B ⎟⎟ vB vA = 2 B πd A ⎝ dA ⎠

Substituting for vA in equation (1) yields:

⎛⎛ d PA + 12 ρ ⎜ ⎜⎜ B ⎜⎝ dA ⎝

2

1 4 1 4

2

2 ⎞ ⎞ ⎟⎟ vB ⎟ = PB + 12 ρvB2 ⎟ ⎠ ⎠

While the values of dB, dA, and ρ are known to us, we need a value for vB (or vA) in order to compare PA and PB. Hence ( f ) is correct. 17 •• [SSM] Figure 13-30 is a diagram of a prairie dog tunnel. The geometry of the two entrances are such that entrance 1 is surrounded by a mound and entrance 2 is surrounded by flat ground. Explain how the tunnel remains ventilated, and indicate in which direction air will flow through the tunnel.

1278 Chapter 13 Determine the Concept The mounding around entrance 1 will cause the streamlines to curve concave downward over the entrance. An upward pressure gradient produces the downward centripetal force. This means there is a lowering of the pressure at entrance 1. No such lowering occurs aver entrance 2, so the pressure there is higher than the pressure at entrance 1. The air circulates in entrance 2 and out entrance 1. It has been demonstrated that enough air will circulate inside the tunnel even with slightest breeze outside.

Estimation and Approximation 18 •• Your undergraduate research project involves atmospheric sampling. The sampling device has a mass of 25.0 kg. Estimate the diameter of a heliumfilled balloon required to lift the device off the ground. Neglect the mass of the balloon ″skin″ and the small buoyancy force on the device itself. Picture the Problem We can use Archimedes’ principle and the condition for vertical equilibrium to estimate the diameter of the helium-filled balloon that would just lift the sampling device.

Express the equilibrium condition that must be satisfied if the balloonpayload is to ″just lift off″:

Fbouyant = B = Fgravitational = mg

Using Archimedes’ principle, express the buoyant force B:

B = wdisplaced fluid = mdisplaced fluid g

Substituting for B in equation (1) yields:

ρ airVballoon g = mg

= ρ airVballoon g

or

ρ airVballoon = m

The total mass m to be lifted is the sum of the mass of the payload mp and the mass of the helium mHe:

m = mp + mHe = mp + ρ HeVballoon

Substitute for m in equation (2) to obtain:

ρ airVballoon = mp + ρ HeVballoon

Solving for Vballoon yields:

The volume of the balloon is given by:

(1)

Vballoon =

mp

ρ air − ρ He

Vballoon = 43 π r 3 = 16 π d 3

(2)

Fluids Substituting for Vballoon yields:

Substitute numerical values and evaluate d:

1 6

π d3 =

d =3

mp

ρ air − ρ He

⇒d = 3

1279

6mp

π (ρ air − ρ He )

6(25.0 kg ) π 1.293 kg/m 3 − 0.179 kg/m 3

(

)

= 3.50 m

Your friend wants to start a business giving hot-air balloon rides. The 19 •• empty balloon, the basket and the occupants have a total maximum mass of 1000 kg. If the balloon has a diameter of 22.0 m when fully inflated with hot air, estimate the required density of the hot air. Neglect the buoyancy force on the basket and people. Picture the Problem We can use Archimedes’ principle and the condition for vertical equilibrium to estimate the density of the hot air that would enable the balloon and its payload to lift off.

Express the equilibrium condition that must be satisfied if the balloonpayload is to ″just lift off″:

Fbouyant = B = Fgravitational = mg

Using Archimedes’ principle, express the buoyant force B:

B = wdisplaced fluid = mdisplaced fluid g

Substituting for B in equation (1) yields:

ρ airVballoon g = mg ⇒ ρ airVballoon = m (2)

The total mass m to be lifted is the sum of the mass of the payload mp and the mass of the hot air:

m = mp + mhot air = mp + ρ hot airVballoon

Substitute for m in equation (2) to obtain:

ρ airVballoon = mp + ρ hot airVballoon

Solving for ρhot air and simplifying yields:

ρ hot air =

The volume of the balloon is given by:

Vballoon = 43 π r 3 = 16 π d 3

Substituting for Vballoon yields:

(1)

= ρ airVballoon g

ρ airVballoon − mp Vballoon

ρ hot air = ρ air −

mp 1 6

πd

3

= ρ air −

= ρ air −

mp Vballoon

6mp

π d3

1280 Chapter 13 Substitute numerical values and evaluate ρhot air:

ρ hot air = 1.293 kg/m 3 −

6(1000 kg ) π (22.0 m )3

= 1.11 kg/m 3 Remarks: As expected, the density of the hot air is considerably less than the density of the surrounding cooler air.

Density 20

•

Find the mass of a solid lead sphere with a radius equal to 2.00 cm.

Picture the Problem The mass of the sphere is the product of its density and volume. The density of lead can be found in Figure 13-1.

(

)

Using the definition of density, express the mass of the sphere:

m = ρV = ρ 43 π R 3

Substitute numerical values and evaluate m:

m = 43 π 11.3 × 10 3 kg/m 3 2.00 × 10 − 2 m

(

)(

)

3

= 0.379 kg

21 • [SSM] Consider a room measuring 4.0 m × 5.0 m × 4.0 m. Under normal atmospheric conditions at Earth’s surface, what would be the mass of the air in the room? Picture the Problem The mass of the air in the room is the product of its density and volume. The density of air can be found in Figure 13-1.

Use the definition of density to express the mass of the air in the room: Substitute numerical values and evaluate m:

m = ρV = ρLWH

m = (1.293 kg/m 3 )(4.0 m )(5.0 m )(4.0 m ) = 1.0 × 10 2 kg

22 •• An average neutron star has approximately the same mass as the Sun, but is compressed into a sphere of radius roughly 10 km. What would be the approximate mass of a teaspoonful of matter that dense? Picture the Problem We can use the definition of density to find the approximate mass of a teaspoonful of matter from a neutron star. Assume that the volume of a teaspoon is about 5 mL.

Fluids Use the definition of density to express the mass of a teaspoonful of matter whose density is ρ :

m = ρVteaspoon

The density of the neutron star is given by:

ρ NS =

mNS VNS

Substituting for mNS and VNS yields:

ρ NS =

mSun 3 4 3 π rNS

Let ρ = ρNS in equation (1) to obtain:

Substitute numerical values and evaluate m:

m=

m=

1281

(1)

3mSunVteaspoon 3 4π rNS

(

)(

3 1.99 × 10 30 kg 5 × 10 −6 m 3

(

4π 10 × 10 3 m

)

)

3

≈ 2 Tg

23 •• A 50.0-g ball consists of a plastic spherical shell and a water-filled core. The shell has an outside diameter equal to 50.0 mm and an inside diameter equal to 20.0 mm. What is the density of the plastic? Picture the Problem We can use the definition of density to find the density of the plastic of which the spherical shell is constructed.

The density of the plastic is given by:

ρ plastic =

mplastic Vplastic

The mass of the plastic is the difference between the mass of the ball and the mass of its water-filled core:

mplastic = mball − mwater

Use the definition of density to express the mass of the water:

mwater = ρ waterVwater

Substituting for mwater yields:

mplastic = mball − ρ waterVwater

Substitute for mplastic in equation (1) to obtain:

ρ plastic =

mball − ρ waterVwater Vplastic

(1)

1282 Chapter 13 3 and Because Vwater = 43 πRinside

(

ρ plastic =

)

3 3 Vball = 43 π Routside − Rinside :

3 ) mball − ρ water ( 43 πRinside 3 3 4 3 π (Routside − Rinside )

Substitute numerical values and evaluate ρplsstic:

ρ plastic

(

)

g ⎞4 ⎛ π (10.0 mm )3 50.0 g − ⎜1.00 3 ⎟ 3 cm ⎠ ⎝ = = 0.748 g/cm3 3 3 4 ( ) ( ) π 25 . 0 mm − 10 . 0 mm 3

(

)

24 •• A 60.0-mL flask is filled with mercury at 0ºC (Figure 13-31). When the temperature rises to 80ºC, 1.47 g of mercury spills out of the flask. Assuming that the volume of the flask stays constant, find the change in density of mercury at 80ºC if its density at 0ºC is 13 645 kg/m3. Picture the Problem We can use the definition of density to relate the change in the density of the mercury to the amount spilled during the heating process.

The change in the density of the mercury as it is warmed is given by:

The density of the mercury before it is warmed is the ratio of its mass to the volume it occupies: The volume of the mercury that spills is the ratio of its mass to its density at the higher temperature: The density ρ of the mercury at the higher temperature is given by:

Solving for ρ yields:

Substituting equations (2) and (3) in equation (1) yields:

Δρ = ρ 0 − ρ

(1)

where ρ0 is the density of the mercury before it is warmed.

ρ0 =

m0 V0

Vspilled =

ρ=

ρ=

(2)

mspilled

ρ

m0 = V0 + Vspilled

m0 − mspilled V0

m0 mspilled V0 +

= ρ0 −

ρ

mspilled V0

(3)

mspilled ⎞ mspilled ⎛ ⎟= Δρ = ρ 0 − ⎜⎜ ρ 0 − V0 ⎟⎠ V0 ⎝

Fluids 1283 Substitute numerical values and evaluate Δρ:

Δρ =

1.47 ×10 −3 kg = 24.5 kg/m 3 60.0 ×10 −6 m 3

25 •• One sphere is made of gold and has a radius rAu and another sphere is made of copper and has a radius rCu. If the spheres have equal mass, what is the ratio of the radii, rAu/ rCu? Picture the Problem We can use the definition of density to find the ratio of the radii of the two spheres. See Table 13-1 for the densities of gold and copper.

Use the definition of density to express the mass of the gold sphere:

3 mAu = ρ AuVAu = 43 πρ Au rAu

The mass of the copper sphere is given by:

3 mCu = ρ CuVCu = 43 πρ Cu rCu

Dividing the first of these equations by the second and simplifying yields:

3 ρ mAu 43 πρ Au rAu = 4 = Au 3 mCu 3 πρ Cu rCu ρ Cu

Solve for rAu/rCu to obtain:

rAu m ρ = 3 Au Cu rCu mCu ρ Au

Because the spheres have the same mass:

rAu ρ = 3 Cu rCu ρ Au

Substitute numerical values and evaluate rAu/rCu:

rAu 8.93 kg/m 3 =3 = 0.773 rCu 19.3 kg/m 3

⎛ rAu ⎜⎜ ⎝ rCu

⎞ ⎟⎟ ⎠

3

26 ••• Since 1983, the US Mint has coined pennies that are made out of zinc with a copper cladding. The mass of this type of penny is 2.50 g. Model the penny as a uniform cylinder of height 1.23 mm and radius 9.50 mm. Assume the copper cladding is uniformly thick on all surfaces. If the density of zinc is 7140 kg/m3 and that of copper is 8930 kg/m3, what is the thickness of the copper cladding? Picture the Problem The pictorial representation shows a zinc penny with its copper cladding. We can use the definition of density to relate the difference between the mass of an all-copper penny and the mass of a copper-zinc penny to the thickness d of the copper cladding.

1284 Chapter 13

d

Cu

h

Zn

2r

Express the mass 0f the cladded penny:

m = mCu + mZn

In terms of the densities of copper and zinc, our equation becomes:

m = ρ CuVCu + ρ Zn VZn

Substituting for the volumes of zinc and copper gives:

[

]

m = ρ Cu 2πr 2 d + 2πr (h − 2d )d + ρ Zn π (r − d ) (h − 2d ) 2

Factor h and r from (h − 2d) and (r − d)2 to obtain: ⎡ d⎞ ⎛ 2d ⎞ ⎤ 2⎛ m = ρ Cu ⎢2πr 2 d + 2πrh⎜1 − ⎟d ⎥ + ρ Zn πr ⎜1 − ⎟ h ⎠ ⎦ ⎝ r⎠ ⎝ ⎣

Assuming that d 0, the speed is v2. A sinusoidal wave is incident on the knot from the left (x < 0); part of the wave is reflected and part is transmitted. For x < 0, the displacement of the wave is describabed by y(x,t) = A sin (k1x – ωt) + B sin(k1x + ωt), while for x > 0, y(x,t) = C sin (k2x – ωt), where ω/k1 = v1 and ω/k2 = v2. (a) If we assume that both the wave function y and its first spatial derivative ∂y/∂x must be continuous at x = 0, show that C/A = 2v2/( v1 + v2), and that B/A = (v1 – v2)/( v1 + v2). (b) Show that B2 + (v1/v2)C2 = A2. Picture the Problem We can use the assumption that both the wave function and its first spatial derivative are continuous at x = 0 to establish equations relating A, B, C, k1, and k2. Then, we can solve these simultaneous equations to obtain expressions for B and C in terms of A, v1, and v2.

(a) Let y1(x, t) represent the wave function in the region x < 0, and y2(x, t) represent the wave function in the region x > 0. Express the continuity of the two wave functions at x = 0:

y1 (0, t ) = y2 (0, t ) and A sin [k1 (0) − ωt ] + B sin[k1 (0) + ωt ]

= C sin[k 2 (0) − ωt ]

or A sin (− ωt ) + B sin ωt = C sin (− ωt )

Because the sine function is odd; that is, sin (− θ ) = − sin θ :

− A sin ωt + B sin ωt = −C sin ωt and A− B =C (1)

Differentiate the wave functions with respect to x to obtain:

∂y1 = Ak1 cos(k1 x − ωt ) ∂x + Bk1 cos(k1 x + ωt ) and ∂y2 = Ck 2 cos(k 2 x − ωt ) ∂x

Traveling Waves 1479 Express the continuity of the slopes of the two wave functions at x = 0:

∂y 1 ∂y = 2 ∂x x =0 ∂x x =0 and Ak1 cos[k1 (0 ) − ωt ] + Bk1 cos[k1 (0) + ωt ] = Ck 2 cos[k 2 (0) − ωt ]

or Ak1 cos(− ωt ) + Bk1 cos ωt

= Ck 2 cos(− ωt )

Because the cosine function is even; that is, cos(− θ ) = cos θ :

Ak1 cos ωt + Bk1 cos ωt = Ck 2 cos ωt and k1 A + k1 B = k 2C (2)

Multiply equation (1) by k1 and add it to equation (2) to obtain:

2k1 A = (k1 + k 2 )C

Solving for C yields:

C=

2k1 2 A= A k1 + k 2 1 + k 2 k1

Solve for C/A and substitute ω/v1 for k1 and ω/v2 for k2 to obtain:

2v2 C 2 2 = = = A 1 + k 2 k1 1 + v1 v2 v1 + v2

Substitute in equation (1) to obtain:

⎛ 2v2 ⎞ ⎟⎟ A A − B = ⎜⎜ ⎝ v2 + v1 ⎠

Solving for B/A yields:

v −v B = 1 2 A v1 + v2

1480 Chapter 15 (b) We wish to show that B2 + (v1/v2)C2 = A2 Use the results of (a) to obtain the expressions B = −[(1 − α)/(1 + α)] A and C = 2A/(1 + α), where α = v1/v2.

B2 + 2

2

⎛1−α ⎞ 2 ⎛ 2 ⎞ 2 2 ⎜ ⎟ A + α⎜ ⎟ A =A 1 + 1 + α α ⎝ ⎠ ⎝ ⎠ 2

and check to see if the resulting equation is an identity:

2

⎛ 2 ⎞ ⎛1−α ⎞ ⎟ =1 ⎜ ⎟ + α⎜ ⎝1+ α ⎠ ⎝1+ α ⎠

(1 − α )2 + 4α (1 + α )2

Substitute these expressions into B2 + (v1/v2)C2 = A2

v1 2 C = A2 v2

=1

1 − 2α + α 2 + 4α =1 (1 + α )2 1 + 2α + α 2 =1 (1 + α )2

(1 + α )2 (1 + α )2

=1

1=1 The equation is an identity:

Therefore, B 2 +

v1 2 C = A2 v2

Remarks: Our result in (a) can be checked by considering the limit of B/A as v2/v1 → 0. This limit gives B/A = +1, telling us that the transmitted wave has zero amplitude and the incident and reflected waves superpose to give a standing wave with a node at x = 0.

Harmonic Sound Waves 47

•

A sound wave in air produces a pressure variation given by

p( x, t ) = 0.75 cos

π

(x − 343t ) , where p is in pascals, x is in meters, and t is in 2 seconds. Find (a) the pressure amplitude, (b) the wavelength, (c) the frequency, and (d) the wave speed. Picture the Problem The pressure variation is of the form p( x,t ) = p0 cos k ( x − vt ) where k is the wave number, p0 is the pressure amplitude,

and v is the wave speed. We can find λ from k and f from ω and k. (a) By inspection of the equation:

p0 = 0.75 Pa

Traveling Waves 1481 (b) Because k =

(c) Solve v =

ω k

2π

λ =

=

π 2

:

2πf for f to obtain: k

λ = 4.00 m

f =

kv 2π

π

Substitute numerical values and evaluate f:

f = 2

(d) By inspection of the equation:

v = 343 m/s

(343 m/s ) 2π

= 85.8 Hz

48 • (a) Middle C on the musical scale has a frequency of 262 Hz. What is the wavelength of this note in air? (b) The frequency of the C an octave above middle C is twice that of middle C. What is the wavelength of this note in air? Picture the Problem The frequency, wavelength, and speed of the sound waves are related by v = fλ.

(a) The wavelength of middle C is given by:

λ=

v 340 m/s = = 1.30 m f 262 s −1

(b) Evaluate λ for a frequency twice that of middle C:

λ=

v 340 m/s = = 0.649 m f 2 262 s −1

(

)

49 • [SSM] The density of air is 1.29 kg/m3. (a) What is the displacement amplitude for a sound wave with a frequency of 100 Hz and a pressure amplitude of 1.00 × 10–4 atm? (b) The displacement amplitude of a sound wave of frequency 300 Hz is 1.00 ×10–7 m. What is the pressure amplitude of this wave? Picture the Problem The pressure amplitude depends on the density of the medium ρ, the angular frequency of the sound wave μ, the speed of the wave v, and the displacement amplitude s0 according to p0 = ρωvs0 .

(a) Solve p0 = ρωvs0 for s0:

s0 =

p0 ρωv

1482

Chapter 15

Substitute numerical values and evaluate s0: s0 =

(1.00 ×10 atm)(1.01325 ×10 Pa/atm) = 3.64 ×10 2π (1.29 kg/m )(100 s )(343 m/s ) −4

5

−5

−1

3

m = 36.4 μm

(b) Use p0 = ρωvs0 to find p0:

(

)(

)

(

)

p0 = 2π 1.29 kg/m 3 300 s −1 (343 m/s ) 1.00 ×10 −7 m = 83.4 mPa

50 • The density of air is 1.29 kg/m3. (a) What is the displacement amplitude of a sound wave that has a frequency of 500 Hz at the pain-threshold pressure amplitude of 29.0 Pa? (b) What is the displacement amplitude of a sound wave that has the same pressure amplitude as the wave in Part (a), but has a frequency of 1.00 kHz? Picture the Problem The pressure amplitude depends on the density of the medium ρ, the angular frequency of the sound wave μ, the wave speed v, and the displacement amplitude s0 according to p0 = ρωvs0 .

(a) Solve p0 = ρωvs0 for s0:

Substitute numerical values and evaluate s0:

s0 =

p0 ρωv

s0 =

29.0 Pa 2π 1.29 kg/m 3 (500 Hz )(343 m/s )

(

)

= 20.9 μm (b) Proceed as in (a) with f = 1.00 kHz:

s0 =

29.0 Pa 2π 1.29 kg/m (1.00 kHz )(343 m/s )

(

3

)

= 10.4 μm 51 • A typical loud sound wave that has a frequency of 1.00 kHz has a pressure amplitude of about 1.00 × 10–4 atm. (a) At t = 0, the pressure is a maximum at some point x1. What is the displacement at that point at t = 0? (b) Assuming the density of air is 1.29 kg/m3, what is the maximum value of the displacement at any time and place? Picture the Problem The pressure or density wave is 90° out of phase with the displacement wave. When the displacement is zero, the pressure and density changes are either a maximum or a minimum. When the displacement is a

Traveling Waves 1483 maximum or minimum, the pressure and density changes are zero. We can use p0 = ρωvs0 to find the maximum value of the displacement at any time and place. (a) If the pressure is a maximum at x1 when t = 0, the displacement s is zero. (b) Solve p0 = ρωvs0 for s0:

s0 =

p0 ρωv

Substitute numerical values and evaluate s0:

(1.00 ×10 atm )(1.01325 ×10 Pa/atm) = = 2π (1.29 kg/m )(1.00 kHz )(343 m/s ) −4

s0

5

3

3.64 μm

52 • An octave represents a change in frequency by a factor of two. Over how many octaves can a typical person hear? Picture the Problem A human can hear sounds between roughly 20 Hz and 20 kHz; a factor of 1000. An octave represents a change in frequency by a factor of 2. We can evaluate 2N = 1000 to find the number of octaves heard by a person who can hear this range of frequencies.

Relate the number of octaves to the difference between 20 kHz and 20 Hz:

2 N = 1000

Take the logarithm of both sides of the equation to obtain:

log 2 N = log103 ⇒ N log 2 = 3

Solving for N yields:

N=

3 = 9.97 ≈ 10 log 2

53 •• In the oceans, whales communicate by sound transmission through the water. A whale emits a sound of 50.0 Hz to tell a wayward calf to catch up to the pod. The speed of sound in water is about 1500 m/s. (a) How long does it take the sound to reach the calf if he is 1.20 km away? (b) What is the wavelength of this sound in the water? (c) If the whales are close to the surface, some of the sound energy might refract out into the air. What would be the frequency and wavelength of the sound in the air?

1484 Chapter 15 Picture the Problem (a) We can use the definition of average speed to find the time required for the sound to travel to the calf. (b) We can use the relationship between wavelength, frequency, and speed to find the wavelength of the sound in water. (c) The frequency of the sound does not change as it travels from water to air, but its wavelength changes because of the difference in the speed of sound in water and in air.

(a) Relate the time it takes the sound to reach the calf to the distance from the whale to the calf and the speed of sound in water:

d 1.20 km = = 0.80 s v 1500 m/s

Δt =

(b) The wavelength of this sound in water is the ratio of its speed in water to its frequency:

λwater =

(c) Because the frequency does not change as it travels from water to air:

f = 50.0 Hz

The wavelength of this sound in air is the ratio of its speed in air to its frequency:

λair =

vwater 1500 m/s = = 30 m f 50.0 Hz

vair 343 m/s = = 6.86 m f 50.0 Hz

Waves in Three Dimensions: Intensity 54 • A spherical sinusoidal source radiates sound uniformly in all directions. At a distance of 10.0 m, the sound intensity level is 1.00 × 10–4 W/m2. (a) At what distance from the source is the intensity 1.00 × 10–6 W/m2? (b) What power is radiated by this source? Picture the Problem The intensity of the sound from the spherical sinusoidal source varies inversely with the square of the distance from the source. The power radiated by the source is the product of the intensity of the radiation and the surface area over which it is distributed.

(a) Relate the intensity I1 at a distance R1 from the source to the energy per unit time (power) arriving at the point of interest:

I1 =

Pav,1 ⇒ Pav,1 = 4πR12 I1 4πR12

Traveling Waves 1485 At a distance R2 from the source:

Because Pav,1 = Pav, 2 :

Substituting numerical values and evaluating R2 gives:

I2 =

Pav, 2 ⇒ Pav, 2 = 4πR22 I 2 4πR22

4πR12 I1 = 4πR22 I 2 ⇒ R2 = R1

I1 I2

1.00 × 10 −4 W/m 2 R2 = (10.0 m ) 1.00 × 10 −6 W/m 2 = 100 m

(b) Solve I =

Pav for Pav: 4π r 2

Substitute numerical values and evaluate Pav:

Pav = 4π r 2 I

(

Pav = 4π (10.0 m ) 1.00 × 10 −4 W/m 2 2

)

= 126 mW

55 • [SSM] A loudspeaker at a rock concert generates a sound that has an intensity level equal to 1.00 × 10–2 W/m2 at 20.0 m and has a frequency of 1.00 kHz. Assume that the speaker spreads its energy uniformly in three dimensions. (a) What is the total acoustic power output of the speaker? (b) At what distance will the sound intensity be at the pain threshold of 1.00 W/m2? (c) What is the sound intensity at 30.0 m? Picture the Problem Because the power radiated by the loudspeaker is the product of the intensity of the sound and the area over which it is distributed, we can use this relationship to find the average power, the intensity of the radiation, or the distance to the speaker for a given intensity or average power.

(a) Use Pav = 4πr 2 I to find the total acoustic power output of the speaker: (b) Relate the intensity of the sound at 20 m to the distance from the speaker:

(

Pav = 4π (20.0 m ) 1.00 × 10 −2 W/m 2 2

= 50.27 W = 50.3 W 1.00 ×10 − 2 W/m 2 =

Pav

4π (20.0 m )

2

)

1486 Chapter 15 Relate the threshold-of-pain intensity to the distance from the speaker:

1.00 W/m 2 =

Divide the first of these equations by the second and solve for r:

r=

(c) Use I =

Pav to find the intensity 4πr 2

Pav 4πr 2

(1.00 ×10 )(20.0 m )

2

−2

I (30.0 m ) =

at 30.0 m:

= 2.00 m

50.3 W 2 4π (30.0 m )

= 4.45 × 10 −3 W/m 2

56 •• When a pin of mass 0.100 g is dropped from a height of 1.00 m, 0.050 percent of its energy is converted into a sound pulse that has a duration of 0.100 s. (a) Estimate how far away the dropped pin can be heard if the minimum audible intensity is 1.00 × 10–11 W/m2. (b) Your result in Part (a) is much too large in practice because of background noise. If you assume that the intensity must be at least 1.00 × 10–8 W/m2 for the sound to be heard, estimate how far away the dropped pin can be heard. (In both parts, assume that the intensity is P/4πr2.) Picture the Problem We can use conservation of energy to find the acoustical energy resulting from the dropping of the pin. The power developed can then be found from the given time during which the energy was transformed from mechanical to acoustical form. We can find the range at which the dropped pin can be heard from I = P/4π r2.

(a) Assuming that I = P/4π r2, express the distance at which one can hear the dropped pin: Use conservation of energy to determine the sound energy generated when the pin falls:

r=

P 4π I

E = ε mgh

( ) × (9.81 m/s )(1.00 m )

= (0.00050) 0.100 × 10 −3 kg 2

= 4.905 × 10 −7 J

The power of the sound pulse is given by:

P=

E 4.905 ×10 −7 J = Δt 0.100 s

= 4.905 ×10 −6 W

Traveling Waves 1487 Substitute numerical values and evaluate r:

r=

4.905 × 10 −6 W 4π 1.00 × 10 −11 W/m 2

(

)

= 0.20 km

(b) Repeat the last step in (a) with I = 1.00 × 10–8 W/m2:

4.905 × 10 −6 W r= = 6.2 m 4π (1.00 × 10 −8 W/m 2 )

*Intensity Level 57 • [SSM] What is the intensity level in decibels of a sound wave that has an intensity of (a) 1.00 × 10–10 W/m2 and (b) 1.00 × 10–2 W/m2? Picture the Problem The intensity level β of a sound wave, measured in decibels, is given by β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to be

the threshold of hearing. ⎛ 1.00 × 10 −10 W/m 2 ⎞ ⎟⎟ −12 2 ⎝ 10 W/m ⎠

(a) Using its definition, calculate the intensity level of a sound wave whose intensity is 1.00 × 10–10 W/m2:

β = (10 dB)log⎜⎜

(b) Proceed as in (a) with I = 1.00 × 10–2 W/m2:

⎛ 1.00 × 10 −2 W/m 2 ⎞ ⎟⎟ β = (10 dB)log⎜⎜ 2 −12 10 W/m ⎝ ⎠

= 10 log10 2 = 20.0 dB

= 10 log1010 = 100 dB

What is the intensity of a sound wave if, at a particular location, the 58 • intensity level is (a) β = 10 dB and (b) β = 3 dB? Picture the Problem (a) and (b) The intensity level β of a sound wave, measured in decibels, is given by β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to

be the threshold of hearing. (a) Solve β = (10 dB) log(I I 0 ) for I

I = 10 β

(10 dB )

I0

to obtain: Evaluate I for β = 10 dB:

I = 10 (10 dB ) (10 dB ) I 0 = 10 I 0

(

)

= 10 10 −12 W/m 2 = 10 −11 W/m 2

1488 Chapter 15 (b) Proceed as in (a) with β = 3 dB:

I = 10 (3 dB ) (10 dB ) I 0 = 2 I 0

(

)

= 2 10 −12 W/m 2 = 2 × 10 −12 W/m 2

59 • At a certain distance, the sound intensity level of a dog’s bark is 50 dB. At that same distance, the sound intensity of a rock concert is 10,000 times that of the dog’s bark. What is the sound intensity level of the rock concert? Picture the Problem The intensity level β of a sound wave, measured in decibels, is given by β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to be

the threshold of hearing. ⎛ I concert ⎝ I0

⎞ ⎟⎟ ⎠

Express the sound intensity level of the rock concert:

β concert = (10 dB)log⎜⎜

Express the sound intensity level of the dog’s bark:

⎛ I dog ⎞ ⎟⎟ 50 dB = (10 dB) log⎜⎜ ⎝ I0 ⎠

Solve for the intensity of the dog’s bark:

I dog = 10 5 I 0 = 10 5 10 −12 W/m 2

Express the intensity of the rock concert in terms of the intensity of the dog’s bark:

I concert = 10 4 I dog = 10 4 10 −7 W/m 2

Substitute in equation (1) and evaluate βconcert:

β concert = (10 dB)log⎜⎜

(

(1)

)

= 10 −7 W/m 2

(

)

= 10 −3 W/m 2 ⎛ 10 −3 W/m 2 ⎞ ⎟ −12 2 ⎟ ⎝ 10 W/m ⎠

= (10 dB)log10 9 = 90 dB

60 • What fraction of the acoustic power of a noise would have to be eliminated to lower its sound intensity level from 90 to 70 dB? Picture the Problem We can express the intensity levels at both 90 dB and 70 dB in terms of the intensities of the sound at those levels. By subtracting the two expressions, we can solve for the ratio of the intensities at the two levels and then find the fractional change in the intensity that corresponds to a decrease in intensity level from 90 dB to 70 dB.

Traveling Waves 1489 Express the intensity level at 90 dB:

⎛I ⎞ 90 dB = (10 dB) log⎜⎜ 90 ⎟⎟ ⎝ I0 ⎠

Express the intensity level at 70 dB:

⎛I ⎞ 70 dB = (10 dB)log⎜⎜ 70 ⎟⎟ ⎝ I0 ⎠ Δβ = 20 dB

Express Δβ = β90 − β70:

⎛I ⎞ ⎛I ⎞ = (10 dB) log⎜⎜ 90 ⎟⎟ − (10 dB) log⎜⎜ 70 ⎟⎟ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎛I ⎞ = (10 dB) log⎜⎜ 90 ⎟⎟ ⎝ I 70 ⎠ Solving for I90 yields:

I 90 = 100I 70

Express the fractional change in the intensity from 90 dB to 70 dB:

I 90 − I 70 100 I 70 − I 70 = = 99% I 90 100 I 70

Because P ∝ I , the fractional change in power is 99% . 61 •• A spherical source radiates sound uniformly in all directions. At a distance of 10 m, the sound intensity level is 80 dB. (a) At what distance from the source is the intensity level 60 dB? (b) What power is radiated by this source? Picture the Problem The intensity at a distance r from a spherical source varies with distance from the source according to I = Pav 4πr 2 . We can use this

relationship to relate the intensities corresponding to an 80-dB intensity level (I80) and the intensity corresponding to a 60-dB intensity level (I60) to their distances from the source. We can relate the intensities to the intensity levels through β = (10 dB) log(I I 0 ) . (a) Express the intensity of the sound where the intensity level is 80 dB:

I 80 =

Pav 4π r802

Express the intensity of the sound where the intensity level is 60 dB:

I 60 =

Pav 4π r602

1490 Chapter 15 Divide the first of these equations by the second to obtain:

I 80 I 60

Pav 2 r602 4π (10 m ) = = Pav 100 m 2 4π r602

Solving for r60 yields:

r60 = (10 m )

Find the intensity of the 80-dB sound level radiation:

⎛I ⎞ 80 dB = (10 dB) log⎜⎜ 80 ⎟⎟ ⎝ I0 ⎠ and I 80 = 108 I 0 = 10 −4 W/m 2

Find the intensity of the 60-dB sound level radiation:

⎛I ⎞ 60 dB = (10 dB) log⎜⎜ 60 ⎟⎟ ⎝ I0 ⎠ and I 60 = 10 6 I 0 = 10 −6 W/m 2

Substitute numerical values for I80 and I60 and evaluate r60:

r60 = (10 m )

(b) Using the intensity corresponding to an intensity level of 80 dB, express and evaluate the power radiated by this source:

P = I 80 A = 10 − 4 W/m 2 4π(10 m )

(

I 80 I 60

10 −4 W/m 2 = 0.10 km 10 −6 W/m 2

)[

2

]

= 0.13 W

62 •• Harry and Sally are sitting on opposite sides of a circus tent when an elephant trumpets a loud blast. If Harry experiences a sound intensity level of 65 dB and Sally experiences only 55 dB, what is the ratio of the distance between Sally and the elephant to the distance between Harry and the elephant? Picture the Problem The intensity of the sound heard by Harry and Sally depends inversely on the square of the distance between the elephant and each of them. We can use the definition of sound intensity (decibel) level ( β = (10 dB) log(I I 0 ) ) and the definition of intensity ( I = Pav A ) to find the ratio

of these distances.

Traveling Waves 1491 Express the sound intensity level at Harry’s location:

⎛ IH ⎝ I0

β H = (10 dB)log⎜⎜

⎞ ⎟⎟ ⎠

⎛ Pav ⎞ ⎟⎟ = (10 dB)log⎜⎜ 2 4 π r I H 0 ⎠ ⎝ ⎛

Pav ⎞ ⎟⎟ 2 ⎝ 4π rS I 0 ⎠

Similarly, the sound intensity level at Sally’s location is:

β S = (10 dB)log⎜⎜

The difference in decibel level’s at the two locations is given by:

Δβ = β H − β S

Substituting for βH, βS, and Δβ yields: ⎛ Pav ⎞ ⎛ Pav ⎞ ⎟ ⎜⎜ ⎟⎟ = 65 dB − 55 dB = 10 dB ( ) − Δβ = (10 dB)log⎜⎜ 10 dB log 2 2 ⎟ r I r I π π 4 4 H 0 ⎠ S 0 ⎠ ⎝ ⎝

Simplifying this expression yields:

⎛ r2 ⎞ r2 log⎜⎜ S2 ⎟⎟ = 1 ⇒ S2 = 101 rH ⎝ rH ⎠

Solving for the ratio rS rH yields:

rS = 10 = 3.2 rH

Three noise sources produce intensity levels of 70, 73, and 80 dB 63 •• when acting separately. When the sources act together, the resultant intensity is the sum of the individual intensities. (a) Find the sound intensity level in decibels when the three sources act at the same time. (b) Discuss the effectiveness of eliminating the two least intense sources in reducing the intensity level of the noise. Picture the Problem We can find the intensities of the three sources from their intensity levels and, because their intensities are additive, find the intensity level when all three sources are acting.

(a) Express the sound intensity level when the three sources act at the same time:

⎛ I 3 sources ⎞ ⎟⎟ I 0 ⎝ ⎠

β 3 sources = (10 dB)log⎜⎜

⎛I +I +I ⎞ = (10 dB)log⎜⎜ 70 73 80 ⎟⎟ I0 ⎠ ⎝

1492 Chapter 15 Find the intensities of each of the three sources:

⎛I ⎞ 70 dB = (10 dB)log⎜⎜ 70 ⎟⎟ ⇒ I 70 = 10 7 I 0 ⎝ I0 ⎠ ⎛I ⎞ 73 dB = (10 dB)log⎜⎜ 73 ⎟⎟ ⇒ I 73 = 10 7.3 I 0 ⎝ I0 ⎠

and ⎛I ⎞ 80 dB = (10 dB)log⎜⎜ 80 ⎟⎟ ⇒ I 80 = 10 8 I 0 ⎝ I0 ⎠

Substituting in the expression for β3 sources yields: ⎛ 10 7 I 0 + 10 7.3 I 0 + 10 8 I 0 ⎞ ⎟⎟ = (10 dB)log 10 7 + 10 7.3 + 10 8 I 0 ⎝ ⎠

(

β 3 sources = (10 dB)log⎜⎜

)

= 81dB

(b) Find the intensity level with the two least intense sources eliminated:

⎛ 10 8 I 0 ⎞ ⎟⎟ = 80 dB ⎝ I0 ⎠

β 80 = (10 dB)log⎜⎜

Eliminating the 70-dB and 73-dB sources does not reduce the intensity level significantly. 64 •• Show that if two people are different distances away from a sound source, the difference Δβ between the intensity levels reaching the people, in decibels, will always be the same, no matter the power radiated by the source. Picture the Problem Let the two people be identified by the numerals 1 and 2 and use the definition of the intensity level, in decibels, to express Δβ as a function of the distances r1 and r2 of the two people from the source.

Express the difference in the sound intensity level heard by the two people:

Δβ = β 2 − β1

The sound level intensity (decibel level) heard by the second person is given by:

β 2 = (10 dB)log⎜⎜

⎛ I2 ⎞ ⎟⎟ I ⎝ 0⎠

⎛ P ⎞ = (10 dB)log⎜⎜ av ⎟⎟ ⎝ A2 I 0 ⎠ ⎛ Pav ⎞ ⎟⎟ = (10 dB)log⎜⎜ 2 ⎝ 4π r2 I 0 ⎠

Traveling Waves 1493 ⎛

Pav ⎞ ⎟⎟ 2 ⎝ 4π r1 I 0 ⎠

In like manner:

β1 = (10 dB)log⎜⎜

Substitute in the expression for Δβ to obtain: ⎛ Pav ⎞ ⎛ Pav ⎞ ⎟⎟ − (10 dB)log⎜⎜ ⎟⎟ Δβ = (10 dB)log⎜⎜ 2 2 4 π r I 4 π r I 2 0 ⎠ 1 0 ⎠ ⎝ ⎝

Simplifying yields:

⎛ Pav ⎜ 4π r22 I 0 Δβ = (10 dB) log⎜ ⎜ Pav ⎜ 2 ⎝ 4π r1 I 0

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

⎛ r2 ⎞ = (10 dB) log⎜⎜ 1 2 ⎟⎟ ⎝ r2 ⎠ This result shows that the difference in sound level intensities depends only on the distances to the source and not on the source’s power output. 65 ••• Everyone at a party is talking equally loudly. One person is talking to you and the sound intensity level at your location is 72 dB. Assuming that all 38 people at the party are at the same distance from you as the person who you are talking to, find the sound intensity level at your location. Picture the Problem The sound intensity level can be found from the intensity of the sound due to the 38 people. When 38 people are talking, the intensities add.

Express the sound level when all 38 people are talking:

⎛ 38 I1 ⎞ ⎟⎟ I ⎝ 0 ⎠

β 38 = (10 dB) log⎜⎜

⎛I ⎞ = (10 dB) log 38 + (10 dB) log⎜⎜ 1 ⎟⎟ ⎝ I0 ⎠ = (10 dB) log 38 + 72 dB = 88 dB An equivalent but longer solution:

Express the sound intensity level when all 38 people are talking:

⎛ 38 I1 ⎞ ⎟⎟ ⎝ I0 ⎠

β 38 = (10 dB) log⎜⎜

1494 Chapter 15 ⎛ I1 ⎞ ⎟⎟ ⎝ I0 ⎠

Express the sound intensity level when only one person is talking:

β1 = 72 dB = (10 dB) log⎜⎜

Solving for I1 yields:

I1 = 10 7.2 I 0 = 107.2 10 −12 W/m 2

(

)

= 1.58 × 10 −5 W/m 2

Express the sound intensity when all 38 people are talking:

I 38 = 38I1

The sound intensity level is:

β 38 = (10 dB)log ⎢

(

)

⎡ 38 1.58 × 10 −5 W/m 2 ⎤ ⎥ 10 −12 W/m 2 ⎣ ⎦

= 88 dB

66 ••• When a violinist pulls the bow across a string, the force with which the bow is pulled is fairly small, about 0.60 N. Suppose the bow travels across the A string, which vibrates at 440 Hz, at 0.50 m/s. A listener 35 m from the performer hears a sound of 60-dB intensity. Assuming that the sound radiates uniformly in all directions, with what efficiency is the mechanical energy of bowing converted to sound energy? Picture the Problem Let η represent the efficiency with which mechanical energy is converted to sound energy. Because we’re given information regarding the rate at which mechanical energy is delivered to the string and the rate at which sound energy arrives at the location of the listener, we’ll take the efficiency to be the ratio of the sound power delivered to the listener divided by the power delivered to the string. We can calculate the power input directly from the given data. We’ll calculate the intensity of the sound at 35 m from its intensity level at that distance and use this result to find the power output.

Express the efficiency of the conversion of mechanical energy to sound energy: Find the power delivered by the bow to the string: Using β = (10 dB) log(I I 0 ) , find the intensity of the sound at 35 m:

η=

Pout Pin

Pin = Fv = (0.60 N )(0.50 m/s ) = 0.30 W

⎛ I 35 m ⎞ ⎟⎟ 60 dB = (10 dB) log⎜⎜ I ⎝ 0 ⎠ and I 35 m = 10 6 I 0 = 1.00 × 10 −6 W/m 2

Traveling Waves 1495

(

)

The rate at which sound energy is emitted is given by:

Pout = IA = 4π 1.00 × 10 − 6 W/m 2 (35 m )

Substitute numerical values and evaluate η:

η=

2

= 0.0154 W

0.0154 W = 5.1% 0.30 W

67 ••• [SSM] The noise intensity level at some location in an empty examination hall is 40 dB. When 100 students are writing an exam, the noise level at that location increases to 60 dB. Assuming that the noise produced by each student contributes an equal amount of acoustic power, find the noise intensity level at that location when 50 students have left. Picture the Problem Because the sound intensities are additive, we’ll find the noise intensity level due to one student by subtracting the background noise intensity from the intensity due to the students and dividing by 100. Then, we’ll use this result to calculate the intensity level due to 50 students. ⎛ 50 I1 ⎞ ⎟⎟ ⎝ I0 ⎠

Express the intensity level due to 50 students:

β 50 = (10 dB) log⎜⎜

Find the sound intensity when 100 students are writing the exam:

⎛I ⎞ 60 dB = (10 dB) log⎜⎜ 100 ⎟⎟ ⎝ I0 ⎠ and I100 = 10 6 I 0 = 10 −6 W/m 2

Find the sound intensity due to the background noise:

⎛ I background ⎞ ⎟⎟ 40 dB = (10 dB) log⎜⎜ I 0 ⎝ ⎠ and I background = 10 4 I 0 = 10 −8 W/m 2

Express the sound intensity due to the 100 students:

I100 − I background = 10 −6 W/m 2 − 10 −8 W/m 2

Find the sound intensity due to 1 student:

I 100 − I background

Substitute numerical values and evaluate the noise intensity level due to 50 students:

β 50 = (10 dB)log

≈ 10 − 6 W/m 2

100

= 57 dB

= 10 −8 W/m 2

(

50 1.00 × 10 −8 W/m 2 10 −12 W/m 2

)

1496 Chapter 15

String Waves Experiencing Speed Changes 68 • A 3.00-m-long piece of string with a mass of 25.0 g is tied to 4.00 m of heavy twine with a mass of 75.0 g and the combination is put under a tension of 100 N. If a transverse pulse is sent down the less dense string, determine the reflection and transmission coefficients at the junction point. Picture the Problem Let the subscript ″s″ refer to the string and the subscript ″t″ to the heavy twine. We can use the definitions of the reflection and transmission coefficients and the expression for the speed of waves on a string (Equation 15-3) to find the speeds of the pulse on the string and the heavy twine.

Use their definitions to express the reflection and transmission coefficients:

vs v −v vt r= t s = vt + vs 1 + vs vt 1−

(1)

and 2v t = v t + vs

τ=

Use Equation 15-3 to express vt and vs : Dividing the expression for vs by the expression for vt and simplifying yields:

vt =

FT

2 v 1+ s vt

and vs =

μt

(2)

FT

μs

FT vs = vt

μs

FT

=

μt μs

μt

Substitute for vs v t in equation (1) to obtain:

Express the ratio of μt to μs:

1−

r=

μt μs

μt 1+ μs

mt μ t l t mt l s = = μ s ms ms l t ls

(3)

Traveling Waves 1497 Substituting in equation (3) yields:

1−

mt l s ms l t

1+

mt l s ms l t

r=

Substitute numerical values and evaluate r:

(75.0 ×10 (25.0 ×10 (75.0 ×10 (25.0 ×10

1− r= 1+

−3 −3 −3 −3

) ) kg )(3.00 m ) kg )(4.00 m )

kg (3.00 m ) kg (4.00 m )

= − 0.20 Substitute for vs v t in equation (2) to obtain:

Substitute numerical values and evaluateτ:

2

τ= 1+

μt μs

2

=

mt l s ms l t

1+

2

τ= 1+

(75.0 ×10 (25.0 ×10

−3 −3

kg )(3.00 m ) kg )(4.00 m )

= 0.80 Remarks: Because r + τ = 1 , we could have used this relationship to find τ. 69 • [SSM] Consider a taut string with a mass per unit length μ1, carrying transverse wave pulses that are incident upon a point where the string connects to a second string with a mass per unit length μ2. (a) Show that if μ2 = μ1, then the reflection coefficient r equals zero and the transmission coefficient τ equals +1. (b) Show that if μ2 >> μ1, then r ≈ –1 and τ ≈ 0; and (c) if μ2 > μ1 then v1 >> v2:

r=

v2 − v1 − v1 ≈ = −1 v2 + v1 v1

and

τ=

(c) If μ2 > v1:

2v2 = v2 + v1

2 ≈ 0 v 1+ 1 v2

v1 v −v v2 r= 2 1 = ≈ 1 v1 v2 + v1 1+ v2 1−

and

τ=

2v 2 = v2 + v1

2 ≈ 2 v1 1+ v2

70 •• Verify the validity of 1 = r 2 + (v1 v2 )τ 2 (Equation 15-36) by substituting the expressions for r and τ into it. Picture the Problem Making the indicated substitutions will lead us to the identity 1 = 1.

2v2 v2 − v1 and τ = v2 + v1 v2 + v1

The reflection and transmission coefficients are given by:

r=

Substituting into the equation given in the problem statement yields:

⎛ v − v ⎞ v ⎛ 2v2 ⎞ ⎟⎟ 1 = ⎜⎜ 2 1 ⎟⎟ + 1 ⎜⎜ ⎝ v2 + v1 ⎠ v2 ⎝ v2 + v1 ⎠

2

2

Simplify this expression algebraically to obtain: ⎛

1=

⎞

(v2 − v1 )2 + 4v22 ⎜⎜ v1 ⎟⎟

2 2 2 2 ⎝ v2 ⎠ = v2 − 2v2 v1 + v1 + 4v 2 v1 = v 2 + 2v2 v1 + v1 = (v2 + v1 ) (v2 + v1 )2 (v2 + v1 )2 (v2 + v1 )2 (v2 + v1 )2

2

= 1

71 ••• Consider a taut string that has a mass per unit length μ1 carrying transverse wave pulses of the form y = f(x – v1t) that are incident upon a point P where the string connects to a second string with mass per unit length μ2. Derive 1 = r 2 + (v1 v2 )τ 2 by equating the power incident on point P to the power reflected at P plus the power transmitted at P.

1500 Chapter 15 Picture the Problem Choose the direction of propagation of the incident pulse as the +x direction and let x = 0 at point P. Energy is conserved as the incident pulse is partially reflected and partially transmitted at point P.

From the conservation of energy we have:

Pin + Pr = Pt

From Equation 15-20, the power transmitted in the direction of increasing x is given by:

P = − FT

(1)

∂y ∂y ∂x ∂x

Substituting in equation (1) yields: ∂y ∂y ⎞ ∂y ∂y ⎞ ⎛ ∂y ∂y ⎞ ⎛ ⎛ ⎜ − FT in in ⎟ + ⎜ − FT r r ⎟ = ⎜ − FT t t ⎟ ∂x ∂x ⎠ ∂x ∂x ⎠ ⎝ ∂x ∂x ⎠ ⎝ ⎝ or, upon simplification, ∂yin ∂yin ∂y r ∂y r ∂y t ∂y t = + (2) ∂x ∂x ∂x ∂x ∂x ∂x Because the incident pulse is given by yin = f ( x − v1t ) , the reflected and transmitted pulses are given by:

y r = rf (− x − v1t ) and ⎛v ⎞ y t = τ f ⎜⎜ 1 [x − v2 t ]⎟⎟ ⎝ v2 ⎠

∂y ∂y and using the ∂x ∂t chain rule yields:

∂y df ∂η ∂y df ∂η = and = ∂x dη ∂x ∂t dη ∂t where η is the argument of the wave function.

For the transmitted pulse:

⎞ ∂y t df ∂ ⎛ v1 df v1 ⎜⎜ [x − v2 t ]⎟⎟ = τ =τ ∂x dη ∂x ⎝ v2 dη v2 ⎠ and ⎞ ∂y t df df ∂ ⎛ v1 ⎜⎜ [x − v2 t ]⎟⎟ = τ (− v1 ) =τ dη dη ∂t ⎝ v2 ∂t ⎠

Evaluating

= −τ v1

df dη

Traveling Waves 1501 For the reflected pulse:

∂y r df ∂ (− x − v1t ) = −r df =r dη ∂x dη ∂x and ∂yr df ∂ (− x − v1t ) = r df (− v1 ) =r dη ∂t dη ∂t df = −rv1 dη

For the incident pulse:

∂yin df ∂ (x − v1t ) = df = dη ∂x dη ∂x and ∂yin df ∂ (x − v1t ) = df (− v1 ) = dη ∂t dη ∂t df = −v1 dη

Substitute in equation (2) to obtain: df ⎛ df ⎞ ⎛ df ⎞ ⎛ df ⎞ ⎛ df v1 ⎞ ⎛ df ⎞ ⎟⎟ ⎜⎜ − τ v1 ⎜⎜ − v1 ⎟⎟ + ⎜⎜ − r ⎟⎟ ⎜⎜ − rv1 ⎟⎟ = ⎜⎜τ ⎟ dη ⎝ dη ⎠ ⎝ dη ⎠ ⎝ dη ⎠ ⎝ dη v2 ⎠ ⎝ dη ⎟⎠

Simplifying and rearranging terms yields:

1= r2 +

v1 2 τ v2

The Doppler Effect 72 • A sound source is moving at 80 m/s toward a stationary listener that is standing in still air. (a) Find the wavelength of the sound in the region between the source and the listener. (b) Find the frequency heard by the listener. Picture the Problem We can use Equation 15-38 ( λ =

v ± us ) to find the fs

wavelength of the sound between the source and the listener and Equation15-41a v ± ur ( fr = f s ) to find the frequency heard by the listener. v ± us (a) Apply Equation 15-38 to find λ:

λ=

v ± us v − us 343 m/s − 80 m/s = = fs fs 200 s −1

= 1.32 m

1502 Chapter 15 (b) Apply Equation 15-41a to obtain fr:

fr = =

v ± ur v±0 fs fs = v − us v ± us

(

343 m/s 200 s −1 343 m/s − 80 m/s

)

= 261 Hz 73 • Consider the situation described in Problem 72 from the reference frame of the source. In this frame, the listener and the air are moving toward the source at 80 m/s and the source is at rest. (a) At what speed, relative to the source, is the sound traveling in the region between the source and the listener? (b) Find the wavelength of the sound in the region between the source and the listener. (c) Find the frequency heard by the listener. Picture the Problem (a) In the reference frame of the source, the speed of sound from the source to the listener is reduced by the speed of the air. (b) We can find the wavelength of the sound in the region between the source and the listener from v = fλ. (c) Because the sound waves in the region between the source and the listener will be compressed by the motion of the listener, the frequency of the sound heard by the listener will be higher than the frequency emitted by the v ± ur source and can be calculated using f r = f s (Equation 15-41a). v ± us

(a) The speed of sound in the reference frame of the source is:

v' = v − u wind = 343 m/s − 80 m/s

(b) Noting that the frequency is unchanged, express the wavelength of the sound:

λ=

v' 263 m/s = = 1.32 m f 200 s −1

fr =

v' ± u r ⎛ v' + u r ⎞ fs = ⎜ ⎟ fs v' ± u s ⎝ v' ± 0 ⎠

(c) Apply Equation 15-41a to obtain:

= 263 m/s

⎛ 263 m/s + 80 m/s ⎞ ⎟⎟ 200 s −1 = ⎜⎜ 263 m/s ⎝ ⎠

(

)

= 261 Hz

A sound source is moving away from the stationary listener at 74 • 80 m/s. (a) Find the wavelength of the sound waves in the region between the source and the listener. (b) Find the frequency heard by the listener.

Traveling Waves 1503 Picture the Problem We can use λ = (v ± us ) f s ( Equation 15-38) to find the

wavelength of the sound in the region between the source and the listener v ± ur and f r = f s (Equation 15-41a) to find the frequency heard by the listener. v ± us Because the sound waves in the region between the source and the listener will be spread out by the motion of the listener, the frequency of the sound heard by the listener will be lower than the frequency emitted by the source. (a) Because the source is moving away from the listener, use the positive sign in the numerator of Equation 15-38 to find the wavelength of the sound between the source and the listener:

λ=

(b) Because the listener is at rest and the source is receding, ur = 0 and the denominator of Equation 15-41a is the sum of the two speeds:

fr =

=

v ± us v + us = fs fs 343 m/s + 80 m/s 200 s −1

= 2.12 m

=

v ± ur v±0 fs = fs v ± us v + us

(

343 m/s 200 s −1 343 m/s + 80 m/s

)

= 162 Hz 75 • The listener is moving 80 m/s from the stationary source that is in rest relative to the air. Find the frequency heard by the listener. Picture the Problem Because the listener is moving away from the source, we know that the frequency he/she will hear will be less than the frequency emitted v ± ur by the source. We can use f r = f s (Equation 15-41a), with us = 0 and the v ± us

minus sign in the numerator, to determine its value. Relate the frequency heard by the listener to that of the source:

fr =

v ± ur ⎛ v − ur fs = ⎜ v ± us ⎝ v±0

⎞ ⎟ fs ⎠

⎛ 343 m/s − 80 m/s ⎞ ⎟⎟ 200 s −1 = ⎜⎜ 343 m/s ⎝ ⎠

(

)

= 153 Hz

76 •• You have made the trek to observe a Space Shuttle landing. Near the end of its descent, the ship is traveling at Mach 2.50 at an altitude of 5000 m. (a) What is the angle that the shock wave makes with the line of flight of the shuttle? (b) How far are you from the shuttle by the time you hear its shock wave,

1504 Chapter 15 assuming the shuttle maintains both a constant heading and a constant 5000-m altitude after flying directly over your head? Picture the Problem The diagram shows the position of the shuttle at time t after it was directly over your head (located at point P) Let u represent the speed of the shuttle and v the speed of sound. We can use trigonometry to determine the angle of the shock wave as well as the location of the shuttle x when you hear the shock wave.

(a) Referring to the diagram, express θ in terms of v, u, and t: Solving for θ yields:

(b) Using the diagram, relate θ to the altitude h of the shuttle and the distance x: Substitute numerical values and evaluate x:

sin θ =

vt 1 1 = = ut u v 2.5 ⎛ 1 ⎞ ⎟ = 23.58° = 23.6° ⎝ 2.50 ⎠

θ = sin −1 ⎜

tan θ =

x=

h h ⇒ x= tan θ x

5000 m = 11.5 km tan23.58°

77 •• The SuperKamiokande neutrino detector in Japan is a water tank the size of a 14-story building. When neutrinos collide with electrons in water, most of their energy is transferred to the electrons. As a consequence, the electrons then fly off at speeds that approach c. The neutrino is counted by detecting the shock wave, called Cerenkov radiation, that is produced when the high-speed electrons travel through the water at speeds greater than the speed of light in water. If the maximum angle of the Cerenkov shock-wave cone is 48.75°, what is the speed of light in water? Picture the Problem The angle θ of the Cerenkov shock wave is related to the speed of light in water v and the speed of light in a vacuum c according to sin θ = v c .

Traveling Waves 1505 Relate the speed of light in water v to the angle of the Cerenkov cone: Substitute numerical values and evaluate v:

sin θ =

v ⇒ v = c sin θ c

v = (2.998 × 108 m/s )sin 48.75° = 2.254 × 108 m/s

78 •• You are in charge of calibrating the radar guns for a local police department. One such device emits microwaves at a frequency of 2.00 GHz. During the trials, you have it arranged so that these waves are reflected from a car moving directly away from the stationary emitter. In this situation, you detect a frequency difference (between the received microwaves and the ones sent out) of 293 Hz. Find the speed of the car. Picture the Problem Because the car is moving away from the stationary emitter at a speed ur, the frequency fr it receives will be less than the frequency emitted by the emitter. The microwaves reflected from the car, moving away from a stationary detector, will be of a still lower frequency fr′. We can use the Doppler shift equations to derive an expression for the speed of the car in terms of difference of these frequencies.

Express the frequency fr received by the moving car in terms of fs, ur, and c: The waves reflected by the car are like waves re-emitted by a source moving away from the radar gun: Substitute equation (1) in equation (2) to eliminate fr:

−1

u ⎛ u ⎞ Because ur TE, c'V ≈ 3R . In Part (b), we can use the result of Problem 94 to obtain values for c'V every 100 K between 300 K and 600 K and use this data to find ΔU numerically. 2

(a) From Problem 94 we have:

eTE T ⎛T ⎞ c' V = 3R⎜ E ⎟ ⎝ T ⎠ e TE T − 1

(

Divide the numerator and denominator by eTE T to obtain:

)

2

⎛T ⎞ c' V = 3R⎜ E ⎟ 2TE ⎝T ⎠ e 2

⎛T ⎞ = 3R⎜ E ⎟ TE ⎝T ⎠ e

T

T

2

1 − 2eTE eTE T

T

1 − 2 + e −TE

+1

T

Express the exponential terms in their power series to obtain: 2

eTE T − 2 + e −TE

T

= 1+

2

TE 1 ⎛ TE ⎞ T 1⎛T ⎞ + ⎜ ⎟ + ... − 2 + 1 − E + ⎜ E ⎟ + ... T 2⎝ T ⎠ T 2⎝ T ⎠ 2

⎛T ⎞ ≈ ⎜ E ⎟ for T >> TE ⎝T ⎠ Substitute for eTE T − 2 + e −TE T to obtain:

2

1 ⎛T ⎞ = 3R c' V ≈ 3R⎜ E ⎟ 2 ⎝ T ⎠ ⎛ TE ⎞ ⎜ ⎟ ⎝T ⎠

(b) Use the result of Problem 94 to verify the following table:

T cV

(K) 300 400 500 600 (J/mol⋅K) 9.65 14.33 17.38 19.35

Heat and the First Law of Thermodynamics 1775 The following graph of specific heat as a function of temperature was plotted using a spreadsheet program: 21 19

C V (J/mol-K)

17 15 13 11 9 7 5 300

350

400

450

500

550

600

T (K)

Integrate numerically, using the formula for the area of a trapezoid, to obtain: ΔU =

1 2

(1.00 mol)(100 K )(9.65 + 14.33)

J mol ⋅ K

+ 12 (1.00 mol)(100 K )(14.33 + 17.38)

J mol ⋅ K

+ 12 (1.00 mol)(100 K )(17.38 + 19.35)

J mol ⋅ K

= 4.62 kJ

96 ••• Use the results of the Einstein model in Problem 94 to determine the molar internal energy of diamond (TE = 1060 K) at 300 K and 600 K, and thereby the increase in internal energy as diamond is heated from 300 K to 600 K. Compare your result to that of Problem 95. Picture the Problem We can simplify our calculations by relating Avogadro’s number NA, Boltzmann’s constant k, the number of moles n, and the number of molecules N in the gas and solving for NAk. We can then calculate U300 K, U600 K, and their difference.

1776 Chapter 18 Express the increase in internal energy per mole resulting from the heating of diamond:

ΔU = U 600 K − U 300 K

Express the relationship between Avogadro’s number NA, Boltzmann’s constant k, the number of moles n, and the number of molecules N in the gas:

nR = Nk ⇒ R =

Substitute in the given equation to obtain:

U=

(1)

N k = NAk n

3RTE eTE T − 1

Determine U300 K:

U 300 K

J ⎞ ⎛ 3⎜ 8.314 ⎟ (1060 K ) mol ⋅ K ⎠ ⎝ = e1060 K 300 K − 1

= 795.4 J = 795 J Determine U600 K:

U 600 K

J ⎞ ⎛ 3⎜ 8.314 ⎟ (1060 K ) mol ⋅ K ⎠ ⎝ = e1060 K 600 K − 1

= 5.4498 kJ = 5.45 kJ Substituting in equation (1) yields:

ΔU = U 600 K − U 300 K

= 5.4498 kJ − 795.4 J = 4.65 kJ This result agrees with the result of Problem 95 to within 1%.

97 ••• During an isothermal expansion, an ideal gas at an initial pressure P0 expands until its volume is twice its initial volume V0. (a) Find its pressure after the expansion. (b) The gas is then compressed adiabatically and quasi-statically until its volume is V0 and its pressure is 1.32P0. Is the gas monatomic, diatomic, or polyatomic? (c) How does the translational kinetic energy of the gas change in each stage of this process? Picture the Problem The isothermal expansion followed by an adiabatic compression is shown on the PV diagram. The path 1→2 is isothermal and the path 2→3 is adiabatic. We can apply the ideal-gas law for a fixed amount of gas and an isothermal process to find the pressure at point 2 and the pressure-volume relationship for a quasi-static adiabatic process to determineγ.

Heat and the First Law of Thermodynamics 1777

(a) Relate the initial and final pressures and volumes for the isothermal expansion and solve for and evaluate the final pressure:

P1V1 = P2V2 and V V P2 = P1 1 = P0 1 = 2V1 V2

(b) Relate the initial and final pressures and volumes for the adiabatic compression:

P2V2γ = P3V3γ

1 2

P0

or γ γ 1 2 P0 (2V0 ) = 1.32 P0V0 which simplifies to 2γ = 2.64

Take the natural logarithm of both sides of this equation and solve for and evaluate γ :

ln 2.64 = 1.40 ln 2 and the gas is diatomic .

γ ln 2 = ln 2.64 ⇒ γ =

(c) During the isothermal process, T is constant and the translational kinetic energy is unchanged. During the adiabatic process, the translational kinetic energy increases by a factor of 1.32.

98 ••• If a hole is punctured in a tire, the gas inside will gradually leak out. Assume the following: the area of the hole is A; the tire volume is V; and the time, τ, it takes for most of the air to leak out of the tire can be expressed in terms of the ratio A/V, the temperature T, the Boltzmann constant k, and the initial mass m of the gas inside the tire. (a) Based on these assumptions, use dimensional analysis to find an estimate for τ. (b) Use the result of Part (a) to estimate the time it takes for a car tire with a nail hole punched in it to go flat. Picture the Problem In (a) we’ll assume that τ = f (A/V, T, k, m) with the factors dependent on constants a, b, c, and d that we’ll find using dimensional analysis. In (b) we’ll use our result from (a) and assume that the diameter of the puncture is about 2 mm, that the tire volume is 0.1 m3, and that the air temperature is 20°C.

1778 Chapter 18 (a) Express τ = f (A/V, T, k, m):

⎛ A⎞ τ = ⎜ ⎟ (T ) b (k ) c (m ) d ⎝V ⎠

Rewrite this equation in terms of the dimensions of the physical quantities to obtain:

⎛ ML2 ⎞ d T = (L ) (K ) ⎜⎜ 2 ⎟⎟ (M ) ⎝T K⎠ where K represents the dimension of temperature.

Simplify this dimensional equation to obtain:

T1 = L− a K b M c L2c K − c T -2c M d or T1 = L2 c−a K b−c M c+d T −2c

Equate exponents to obtain:

T : − 2c = 1 , L : 2c − a = 0 , K : b − c = 0, and M: c+d = 0

Solve these equations simultaneously to obtain:

a = −1 , b = − 12 , c = − 12 , and d =

Substituting in equation (1) yields:

τ = ⎜ ⎟ (T )− (k )− (m ) =

a

(1) c

−a

1

⎛ A⎞ ⎝V ⎠

b

−1

1 2

1 2

1 2

(b) Substitute numerical values and evaluate τ:

τ=

π 4

0.1 m 3

(2 ×10

−3

(1.293 kg/m )(0.1m ) = 569 s ≈ 3

m

)

2

3

(1.381J/K )(293 K )

9 min

1 2

V m A kT

Chapter 19 The Second Law of Thermodynamics Conceptual Problems 1 • Modern automobile gasoline engines have efficiencies of about 25%. About what percentage of the heat of combustion is not used for work but released as heat? (a) 25%, (b) 50%, (c) 75%, (d) 100%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. The percentage of the heat of combustion (heat absorbed from the hightemperature reservoir) is the ratio of Qc to Qh. We can use the relationship between W, Qh, and Qc ( W = Qh − Qc ) to find Qc/ Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain:

ε=

Q W Qh − Qc = = 1− c Qh Qh Qh

Solving for Qc/ Qh yields:

Qc = 1− ε Qh

Substitute for ε to obtain:

Qc = 1 − 0.25 = 0.75 Qh

and

(c )

is correct.

2 • If a heat engine does 100 kJ of work per cycle while releasing 400 kJ of heat, what is its efficiency? (a) 20%, (b) 25%, (c) 80%, (d) 400%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( W = Qh − Qc ) to express the efficiency of the heat engine in terms of Qc and W. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain:

ε=

1779

W W = = Qh W + Qc

1 Q 1+ c W

1780 Chapter 19 Substitute for Qc and W to obtain:

1 = 0.2 400 kJ 1+ 100 kJ and (a ) is correct.

ε=

3 • If the heat absorbed by a heat engine is 600 kJ per cycle, and it releases 480 kJ of heat each cycle, what is its efficiency? (a) 20%, (b) 80%, (c) 100%, (d) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( W = Qh − Qc ) to express the

efficiency of the heat engine in terms of Qc and Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: Substitute for Qc and Qh to obtain:

4

•

ε=

Q W Qh − Qc = = 1− c Qh Qh Qh

480 kJ = 0.2 600 kJ and (a ) is correct.

ε = 1−

Explain what distinguishes a refrigerator from a ″heat pump.″

Determine the Concept The job of a refrigerator is to move heat from its cold interior to the warmer kitchen environment. This process moves heat in a direction that is opposite its ″natural″ direction of flow, analogous to the use of a water pump to pump water out of a boat. The term heat pumps is used to describe devices, such as air conditioners, that are used to cool living and working spaces in the summer and warm them in the winter.

An air conditioner’s COP is mathematically identical to that Q of a refrigerator, that is, COPAC = COPref = c . However a heat pump’s COP is W Q defined differently, as COPhp = h . Explain clearly why the two COPs are W defined differently. Hint: Think of the end use of the three different devices. 5

•

[SSM]

Determine the Concept The COP is defined so as to be a measure of the

The Second Law of Thermodynamics 1781 effectiveness of the device. For a refrigerator or air conditioner, the important quantity is the heat drawn from the already colder interior, Qc. For a heat pump, the ideas is to focus on the heat drawn into the warm interior of the house, Qh. • Explain why you cannot cool your kitchen by leaving your refrigerator 6 door open on a hot day. (Why does turning on a room air conditioner cool down the room, but opening a refrigerator door does not?) Determine the Concept As described by the second law of thermodynamics, more heat must be transmitted to the outside world than is removed by a refrigerator or air conditioner. The heating coils on a refrigerator are inside the room, so the refrigerator actually heats the room in which it is located. The heating coils on an air conditioner are outside one’s living space, so the waste heat is vented to the outside. 7 • Why do steam-power-plant designers try to increase the temperature of the steam as much as possible? Determine the Concept Increasing the temperature of the steam increases its energy content. In addition, it increases the Carnot efficiency, and generally increases the efficiency of any heat engine. 8 • To increase the efficiency of a Carnot engine, you should (a) decrease the temperature of the hot reservoir, (b) increase the temperature of the cold reservoir, (c) increase the temperature of the hot reservoir, (d) change the ratio of maximum volume to minimum volume. Determine the Concept Because the efficiency of a Carnot cycle engine is given T by ε C = 1 − c , you should increase the temperature of the hot reservoir. (c ) is Th

correct. 9 •• [SSM] Explain why the following statement is true: To increase the efficiency of a Carnot engine, you should make the difference between the two operating temperatures as large as possible; but to increase the efficiency of a Carnot cycle refrigerator, you should make the difference between the two operating temperatures as small as possible. Determine the Concept A Carnot-cycle refrigerator is more efficient when the temperatures are close together because it requires less work to extract heat from an already cold interior if the temperature of the exterior is close to the temperature of the interior of the refrigerator. A Carnot-cycle heat engine is more efficient when the temperature difference is large because then more work is done by the engine for each unit of heat absorbed from the hot reservoir. A Carnot engine operates between a cold temperature reservoir of 10 ••

1782 Chapter 19 27°C and a high temperature reservoir of 127°C. Its efficiency is (a) 21%, (b) 25%, (c) 75%, (d) 79%. Determine the Concept The efficiency of a Carnot cycle engine is given by ε C = 1 − Tc Th where Tc and Th (in kelvins) are the temperatures of the cold and hot

reservoirs, respectively. Substituting numerical values for Tc and Th yields:

300 K = 0.25 400 K (b ) is correct.

εC = 1−

11 •• The Carnot engine in Problem 10 is run in reverse as a refrigerator. Its COP is (a) 0.33, (b) 1.3, (c) 3.0 (d) 4.7. Determine the Concept The coefficient of performance of a Carnot cycle engine Q run in reverse as refrigerator is given by COPref = c . We can use the relationship W between W, Qc, and Qh to eliminate W from this expression and then use the Q T relationship, applicable only to a device operating in a Carnot cycle, c = c to Qh Th

express the refrigerator’s COP in terms of Tc and Th. The coefficient of performance of a refrigerator is given by:

Qc W or, because W = Qh − Qc , COPref =

COPref =

Qc Qh − Qc

Dividing the numerator and denominator of this fraction by Qc yields:

COPref =

For a device operating in a Carnot cycle:

Qc Tc = Qh Th

Substitute in the expression for COPref to obtain:

COPref, C =

1 Qh −1 Qc

1 Th −1 Tc

The Second Law of Thermodynamics 1783 Substitute numerical values and evaluate COPref, C:

1 = 3.0 400 K −1 300 K (c ) is correct.

COPref, C =

12 •• On a humid day, water vapor condenses on a cold surface. During condensation, the entropy of the water (a) increases, (b) remains constant, (c) decreases, (d) may decrease or remain unchanged. Explain your answer. Determine the Concept When water vapor condenses, its entropy decreases (the liquid state is a more ordered state than is the vapor state). (c) is correct. 13 •• An ideal gas is taken reversibly from an initial state Pi, Vi, Ti to the final state Pf, Vf, Tf. Two possible paths are (A) an isothermal expansion followed by an adiabatic compression and (B) an adiabatic compression followed by an isothermal expansion. For these two paths, (a) ΔEint A > ΔEint B, (b) ΔSA > ΔSB, (c) ΔSA < ΔSB, (d) None of the above. Determine the Concept The two paths are shown on the PV diagram to the right. We can use the concept of a state function to choose from among the alternatives given as possible answers to the problem.

P

B

B

Ti

f

Pf Pi

i

A

Tf

A Vi

Vf

V

(a) Because Eint is a state function and the initial and final states are the same for the two paths, ΔEint, A = ΔEint, B . (b) and (c) S, like Eint, is a state function and its change when the system moves from one state to another depends only on the system’s initial and final states. It is not dependent on the process by which the change occurs. Thus ΔS A = ΔS B . (d) (d ) is correct. Figure 19-12 shows a thermodynamic cycle for an ideal gas on an ST 14 •• diagram. Identify this cycle and sketch it on a PV diagram.

1784 Chapter 19 Determine the Concept The processes A→B and C→D are adiabatic and the processes B→C and D→A are isothermal. Therefore, the cycle is the Carnot cycle shown in the adjacent PV diagram.

P B

C A D V

15 •• Figure 19-13 shows a thermodynamic cycle for an ideal gas on an SV diagram. Identify the type of engine represented by this diagram. Determine the Concept Note that A→B is an adiabatic expansion, B→C is a constant-volume process in which the entropy decreases, C→D is an adiabatic compression and D→A is a constant-volume process that returns the gas to its original state. The cycle is that of the Otto engine (see Figure 19-3). The points A, B, C, and D in Figure 19-13 correspond to points c, d, a, and b, respectively, in Figure 19-3.

Sketch an ST diagram of the Otto cycle. (The Otto cycle is discussed 16 •• in Section 19-1.) Determine the Concept The Otto cycle consists of four quasi-static steps. Refer to Figure 19-3. There a→b is an adiabatic compression, b→c is a constant volume heating, c→d is an adiabatic expansion and d→a is a constant-volume cooling. So, from a to b, S is constant and T increases, from b to c, heat is added to the system and both S and T increase, from c→d S is constant while T decreases, and from d to a both S and T decrease.

To determine how S depends on T along b→c and d→a, consider the entropy change of the gas from point b to an arbitrary point on the path b→c where the entropy and temperature of the gas are S and T, respectively:

Q T where, because heat is entering the system, Q is positive.

Because Won = 0 for this constantvolume process:

ΔEint = Qin = Q = C V ΔT = C V (T − Tb )

Substituting for Q yields:

ΔS =

ΔS =

C V (T − Tb ) ⎛ T ⎞ = C V ⎜1 − b ⎟ T ⎝ T ⎠

The Second Law of Thermodynamics 1785 On path b→c the entropy is given by:

⎛ T ⎞ S = S b + ΔS = S b + C V ⎜1 − b ⎟ ⎝ T ⎠

The first and second derivatives, dS dT and d 2 S dT 2 , give the slope and concavity of the path. Calculate these derivatives assuming CV is constant. (For an ideal gas CV is a positive constant.):

T dS = C V b2 dT T 2 T d S = −2C V b3 2 dT T

These results tell us that, along path b→c, the slope of the path is positive and the slope decreases as T increases. The concavity of the path is negative for all T. Following the same procedure on path d→a gives:

⎛ T ⎞ S = S d + C V ⎜1 − d ⎟ ⎝ T ⎠ T dS = C V d2 dT T 2 T d S = −2C V d3 2 dT T

These results tell us that, along path d→a, the slope of the path is positive and the slope decreases as T increases. The concavity of the path is negative for all T. An ST diagram for the Otto cycle is shown to the right.

S d

a

c

b T

17 ••

[SSM]

Sketch an SV diagram of the Carnot cycle for an ideal gas.

Determine the Concept Referring to Figure 19-8, process 1→2 is an isothermal expansion. In this process heat is added to the system and the entropy and volume increase. Process 2→3 is adiabatic, so S is constant as V increases. Process 3→4 is an isothermal compression in which S decreases and V also decreases. Finally, process 4→1 is adiabatic, that is, isentropic, and S is constant while V decreases.

1786 Chapter 19 During the isothermal expansion (from point 1 to point 2) the work done by the gas equals the heat added to the gas. The change in entropy of the gas from point 1 (where the temperature is T1) to an arbitrary point on the curve is given by:

ΔS =

For an isothermal expansion, the work done by the gas, and thus the heat added to the gas, are given by:

⎛V ⎞ Q = W = nRT1 ln⎜⎜ ⎟⎟ ⎝ V1 ⎠

Substituting for Q yields:

⎛V ⎞ ΔS = nR ln⎜⎜ ⎟⎟ ⎝ V1 ⎠

Since S = S1 + ΔS , we have:

⎛V ⎞ S = S1 + nR ln⎜⎜ ⎟⎟ ⎝ V1 ⎠

S

The graph of S as a function of V for an isothermal expansion shown to the right was plotted using a spreadsheet program. This graph establishes the curvature of the 1→2 and 3→4 paths for the SV graph.

Q T1

V

An SV graph for the Carnot cycle (see Figure 19-8) is shown to the right.

S 2

1

3

4 V

18 •• Sketch an SV diagram of the Otto cycle. (The Otto cycle is discussed in Section 19-1.)

The Second Law of Thermodynamics 1787 Determine the Concept The Otto cycle is shown in Figure 19-3. Process a→b takes place adiabatically and so both Q = 0 and ΔS = 0 along this path. Process b→c takes place at constant volume. Qin, however, is positive and so, while ΔV = 0 along this path, Q > 0 and, therefore ΔS > 0. Process c→d also takes place adiabatically and so, again, both Q = 0 and ΔS = 0 along this path. Finally, process d→a is a constant-volume process, this time with heat leaving the system and ΔS < 0. A sketch of the SV diagram for the Otto cycle follows: S c

d

b

a V

19 •• Figure 19-14 shows a thermodynamic cycle for an ideal gas on an SP diagram. Make a sketch of this cycle on a PV diagram. Determine the Concept Process A→B is at constant entropy; that is, it is an adiabatic process in which the pressure increases. Process B→C is one in which P is constant and S decreases; heat is exhausted from the system and the volume decreases. Process C→D is an adiabatic compression. Process D→A returns the system to its original state at constant pressure. The cycle is shown in the adjacent PV diagram.

P

C

B

D

A

V

20 •• One afternoon, the mother of one of your friends walks into his room and finds a mess. She asks your friend how the room came to be in such a state, and your friend replies, ″Well, it is the natural destiny of any closed system to degenerate toward greater and greater levels of entropy. That’s all, Mom.″ Her reply is a sharp ″Nevertheless, you’d better clean your room!″ Your friend retorts, ″But that can’t happen. It would violate the second law of thermodynamics.″ Critique your friend’s response. Is his mother correct to ground him for not cleaning his room, or is cleaning the room really impossible? Determine the Concept The son is out of line, here, but besides that, he’s also wrong. While it is true that systems tend to degenerate to greater levels of disorder, it is not true that order cannot be brought forth from disorder. What is

1788 Chapter 19 required is an agent doing work – for example, your friend – on the system in order to reduce the level of chaos and bring about order. His cleanup efforts will be rewarded with an orderly system after a sufficient time for him to complete the task. It is true that order will not come about from the disordered chaos of his room – unless he applies some elbow grease.

Estimation and Approximation Estimate the change in COP of your electric food freezer when it is 21 • removed from your kitchen to its new location in your basement, which is 8°C cooler than your kitchen. Picture the Problem We can use the definition of the coefficient of performance to express the ratio of the coefficient of performance in your basement to the coefficient of performance in the kitchen. If we further assume that the freezer operates in a Carnot cycle, then we can use the proportion Qh Qc = Th Tc to

express the ratio of the coefficients of performance in terms of the temperatures in the kitchen, basement, and freezer. The ratio of the coefficients of performance in the basement and kitchen is given by:

COPbasement COPkit

Qc,basement W = c,basement Qc,kit Wc,kit

COPbasement COPkit

Qc,basement Q − Qc,basement = h,basement Qc,kit Qh,kit − Qc,kit

Because W = Qh − Qc for a heat engine or refrigerator:

Divide the numerators and denominators by Qc,basement and Qc,kit and simplify to obtain:

1 COPbasement COPkit

Qh,basement −1 Qc,basement = 1 Qh,kit −1 Qc,kit =

Qh,kit −1 Qc,kit Qh,basement −1 Qc,basement

The Second Law of Thermodynamics 1789 If we assume that the freezer unit operates in a Carnot cycle, then Qh Th = and our expression for the Qc Tc ratio of the COPs becomes: Assuming that the temperature in your kitchen is 20°C and that the temperature of the interior of your freezer is −5°C, substitute numerical values and evaluate the ratio of the coefficients of performance:

Th,kit −1 Tc,kit

COPbasement = Th,basement COPkit −1 Tc,basement 293 K −1 COPbasement 268 K = 1.47 = 285 K COPkit −1 268 K or an increase of 47% in the performance of the freezer!

22 •• Estimate the probability that all the molecules in your bedroom are located in the (open) closet which accounts for about 10% of the total volume of the room. Picture the Problem The probability that all the molecules in your bedroom are N

⎛V ⎞ located in the (open) closet is given by p = ⎜⎜ 2 ⎟⎟ where N is the number of air ⎝ V1 ⎠ molecules in your bedroom and V1 and V2 are the volumes of your bedroom and closet, respectively. We can use the ideal-gas law to find the number of molecules N. We’ll assume that the volume of your room is about 50 m3 and that the temperature of the air is 20°C. If the original volume of the air in your bedroom is V1, the probability p of finding the N molecules, normally in your bedroom, confined to your closet whose volume is V2 is given by:

⎛V ⎞ p = ⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠ or, because V2 = 101 V1 ,

Use the ideal-gas law to express N:

N=

Substitute numerical values and evaluate N:

( 101.325 kPa )(50 m 3 ) N= (1.381×10 −23 J/K )(293 K )

N

⎛1⎞ p=⎜ ⎟ ⎝ 10 ⎠

N

PV kT

= 1.252 × 10 27 molecules

(1)

1790 Chapter 19 Substitute for N in equation (1) and evaluate p:

⎛1⎞ p=⎜ ⎟ ⎝ 10 ⎠

1.252×10 27

=

≈ 10 −10

1 101.252×10

27

= 10 −1.252×10

27

27

23 •• [SSM] Estimate the maximum efficiency of an automobile engine that has a compression ratio of 8.0:1.0. Assume the engine operates according to the Otto cycle and assume γ = 1.4. (The Otto cycle is discussed in Section 19-1. ) Picture the Problem The maximum efficiency of an automobile engine is given by the efficiency of a Carnot engine operating between the same two temperatures. We can use the expression for the Carnot efficiency and the equation relating V and T for a quasi-static adiabatic expansion to express the Carnot efficiency of the engine in terms of its compression ratio.

Express the Carnot efficiency of an engine operating between the temperatures Tc and Th: Relate the temperatures Tc and Th to the volumes Vc and Vh for a quasistatic adiabatic compression from Vc to Vh: Substitute for

Tc to obtain: Th

Express the compression ratio r:

Substituting for r yields:

Substitute numerical values for r and γ (1.4 for diatomic gases) and evaluate εC:

εC = 1 −

γ −1

TcVc

Tc Th

γ −1

= ThVh

⎛V ⎞ ε C = 1 − ⎜⎜ h ⎟⎟ ⎝ Vc ⎠

r=

T V γ −1 ⎛ V ⎞ ⇒ c = hγ −1 = ⎜⎜ h ⎟⎟ Th Vc ⎝ Vc ⎠

γ −1

γ −1

Vc Vh

εC = 1 − ε C = 1−

1

r

γ −1

1 ≈ 56% (8.0)1.4−1

24 •• You are working as an appliance salesperson during the summer. One day, your physics professor comes into your store to buy a new refrigerator. Wanting to buy the most efficient refrigerator possible, she asks you about the efficiencies of the available models. She decides to return the next day to buy the most efficient refrigerator. To make the sale, you need to provide her with the following estimates: (a) the highest COP possible for a household refrigerator, and (b) and the highest rate possible for the heat to be released by the refrigerator if the refrigerator uses 600 W of electrical power.

The Second Law of Thermodynamics 1791 Picture the Problem If we assume that the temperature on the inside of the refrigerator is 0°C (273 K) and the room temperature to be 20°C (293 K), then the refrigerator must be able to maintain a temperature difference of 20 K. We can use the definition of the COP of a refrigerator and the relationship between the temperatures of the hot and cold reservoir and Qh and Qc to find an upper limit on the COP of a household refrigerator. In (b) we can solve the definition of COP for Qc and differentiate the resulting equation with respect to time to estimate the rate at which heat is being drawn from the refrigerator compartment.

(a) Using its definition, express the COP of a household refrigerator:

COP =

Apply conservation of energy to the refrigerator to obtain:

W + Qc = Qh ⇒ W = Qh − Qc

Substitute for W and simplify to obtain:

COP =

Assume, for the sake of finding the upper limit on the COP, that the refrigerator is a Carnot refrigerator and relate the temperatures of the hot and cold reservoirs to Qh and Qc: Substitute for

Qh to obtain: Qc

Qc W

(1)

Qc 1 = Qh − Qc Qh −1 Qc

Qh Th = Qc Tc

COPmax =

1 Th −1 Tc

Substitute numerical values and evaluate COPmax:

COPmax =

(b) Solve equation (1) for Qc:

Qc = W (COP )

Differentiate equation (2) with respect to time to obtain:

dQc dW = (COP ) dt dt

Substitute numerical values and dQc evaluate : dt

dQc = (13.65)(600 J/s ) = 8.2 kW dt

1 = 13.65 ≈ 14 293 K −1 273 K (2)

1792 Chapter 19 25 •• [SSM] The average temperature of the surface of the Sun is about 5400 K, the average temperature of the surface of Earth is about 290 K. The solar constant (the intensity of sunlight reaching Earth’s atmosphere) is about 1.37 kW/m2. (a) Estimate the total power of the sunlight hitting Earth. (b) Estimate the net rate at which Earth’s entropy is increasing due to this solar radiation. Picture the Problem We can use the definition of intensity to find the total power of sunlight hitting Earth and the definition of the change in entropy to find the changes in the entropy of Earth and the Sun resulting from the radiation from the Sun.

(a) Using its definition, express the intensity of the Sun’s radiation on Earth in terms of the power P delivered to Earth and Earth’s cross sectional area A:

I=

P A

Solve for P and substitute for A to obtain:

P = IA = IπR 2 where R is the radius of Earth.

Substitute numerical values and evaluate P:

P = π 1.37 kW/m 2 6.37 × 10 6 m

(b) Express the rate at which Earth’s entropy SEarth changes due to the flow of solar radiation:

dS Earth P = dt TEarth

Substitute numerical values and dS Earth evaluate : dt

dS Earth 1.746 ×1017 W = dt 290 K

(

)(

)

2

= 1.746 × 1017 W = 1.75 × 1017 W

= 6.02 ×1014 J/K ⋅ s

26 •• A 1.0-L box contains N molecules of an ideal gas, and the positions of the molecules are observed 100 times per second. Calculate the average time it should take before we observe all N molecules in the left half of the box if N is equal to (a) 10, (b) 100, (c) 1000, and (d) 1.0 mole. (e) The best vacuums that have been created to date have pressures of about 10–12 torr. If a vacuum chamber has the same volume as the box, how long will a physicist have to wait before all of the gas molecules in the vacuum chamber occupy only the left half of it? Compare that to the expected lifetime of the universe, which is about 1010 years. Picture the Problem If you had one molecule in a box, it would have a 50% chance of being on one side or the other. We don’t care which side the molecules are on as long as they all are on one side, so with one molecule you have a 100%

The Second Law of Thermodynamics 1793 chance of it being on one side or the other. With two molecules, there are four possible combinations (both on one side, both on the other, one on one side and one on the other, and the reverse), so there is a 25% (1 in 4) chance of them both being on a particular side, or a 50% chance of them both being on either side. Extending this logic, the probability of N molecules all being on one side of the box is P = 2/2N, which means that, if the molecules shuffle 100 times a second, the time it would take them to cover all the combinations and all get on one side 2N . In (e) we can apply the ideal gas law to find the number or the other is t = 2(100 ) of molecules in 1.0 L of air at a pressure of 10−12 torr and an assumed temperature of 300 K. (a) Evaluate t for N = 10 molecules:

(b) Evaluate t for N = 100 molecules:

t=

210 = 5.12 s ≈ 5 s 2 100 s −1

t=

2100 2 100 s −1

(

)

(

)

= 6.34 × 10 27 s ×

1y 3.156 × 10 7 s

≈ 2 × 10 20 y

(c) Evaluate t for N = 1000 molecules:

To evaluate 21000 let 10 x = 21000 and take the logarithm of both sides of the equation to obtain:

Substitute to obtain:

t=

21000 2 100 s −1

(

)

(1000)ln 2 = x ln10 ⇒ x = 301

t=

10 301 2 100 s −1

(

)

= 0.5 × 10 299 s ×

1y 3.156 × 10 7 s

≈ 2 × 10 291 y

(d) Evaluate t for N = 1.0 mol =6.022 ×1023 molecules: 23

To evaluate 26.022×10 let 23 10 x = 26.022×10 and take the logarithm of both sides of the equation to obtain:

23

26.022×10 t= 2(100 s −1 )

(6.022 ×10 )ln 2 = x ln10 ⇒ x ≈ 10 23

23

1794 Chapter 19 Substituting for x yields:

23

1010 1y ⎛ ⎞ t≈ 7 ⎟ −1 ⎜ 2 100 s ⎝ 3.156 ×10 s ⎠

(

)

23

≈ 1010 y

(e) Solve the ideal gas law for the number of molecules N in the gas: Assuming the gas to be at room temperature (300 K), substitute numerical values and evaluate N: Evaluate t for N = 3.22×107 molecules: 7

To evaluate 23.22×10 let 7 10 x = 23.22×10 and take the logarithm of both sides of the equation to obtain: Substituting for x yields:

N=

PV kT

(10 N=

torr )(133.32 Pa/torr )(1.0 L ) (1.381×10−23 J/K )(300 K )

−12

= 3.22 × 107 molecules 7

2 3.22×10 t= 2(100 s −1 )

(3.22 × 10 )ln 2 = x ln10 ⇒ x ≈ 10 7

7

7

1010 1y × t= −1 2 100 s 3.156 × 10 7 s

(

)

7

≈ 1010 y

Express the ratio of this waiting time to the lifetime of the universe tuniverse:

t tuniverse or

7

7 1010 y = 10 ≈ 1010 10 y

7

t ≈ 1010 tuniverse

Heat Engines and Refrigerators 27 • [SSM] A heat engine with 20.0% efficiency does 0.100 kJ of work during each cycle. (a) How much heat is absorbed from the hot reservoir during each cycle? (b) How much heat is released to the cold reservoir during each cycle?

The Second Law of Thermodynamics 1795 Picture the Problem (a) The efficiency of the engine is defined to be ε = W Qh where W is the work done per cycle and Qh is the heat absorbed from the hot reservoir during each cycle. (b) Because, from conservation of energy, Qh = W + Qc , we can express the efficiency of the engine in terms of the heat Qc

released to the cold reservoir during each cycle. 100 J = 500 J 0.200

(a) Qh absorbed from the hot reservoir during each cycle is given by:

Qh =

(b) Use Qh = W + Qc to obtain:

Qc = Qh − W = 500 J − 100 J = 400 J

W

ε

=

28 • A heat engine absorbs 0.400 kJ of heat from the hot reservoir and does 0.120 kJ of work during each cycle. (a) What is its efficiency? (b) How much heat is released to the cold reservoir during each cycle? Picture the Problem (a) The efficiency of the engine is defined to be ε = W Qh where W is the work done per cycle and Qh is the heat absorbed from the hot reservoir during each cycle. (b) We can apply conservation of energy to the engine to obtain Qh = W + Qc and solve this equation for the heat Qc released

to the cold reservoir during each cycle. W 120 J = = 30% Qh 400 J

(a) The efficiency of the heat engine is given by:

ε=

(b) Apply conservation of energy to the engine to obtain:

Qh = W + Qc ⇒ Qc = Qh − W

Substitute numerical values and evaluate Qc:

Qc = 400 J − 120 J = 280 J

29 • A heat engine absorbs 100 J of heat from the hot reservoir and releases 60 J of heat to the cold reservoir during each cycle. (a) What is its efficiency? (b) If each cycle takes 0.50 s, find the power output of this engine. Picture the Problem We can use its definition to find the efficiency of the engine and the definition of power to find its power output.

(a) The efficiency of the heat engine is given by:

ε=

Q W Qh − Qc = = 1− c Qh Qh Qh

1796 Chapter 19 60 J = 40% 100 J

Substitute numerical values and evaluate ε:

ε = 1−

(b) The power output P of this engine is the rate at which it does work:

P=

Substitute numerical values and evaluate P:

⎛ 100 J ⎞ ⎟⎟ = 80 W P = (0.40 )⎜⎜ ⎝ 0.500 s ⎠

dQh dW d = ε Qh = ε dt dt dt

30 • A refrigerator absorbs 5.0 kJ of heat from a cold reservoir and releases 8.0 kJ to a hot reservoir. (a) Find the coefficient of performance of the refrigerator. (b) The refrigerator is reversible. If it is run backward as a heat engine between the same two reservoirs, what is its efficiency? Picture the Problem We can apply their definitions to find the COP of the refrigerator and the efficiency of the heat engine.

(a) The COP of a refrigerator is defined to be:

COP =

Apply conservation of energy to relate the work done per cycle to Qh and Qc:

W = Qh − Qc

Substitute for W to obtain:

COP =

Qc Qh − Qc

Substitute numerical values and evaluate COP:

COP =

5.0 kJ = 1.7 8.0 kJ − 5.0 kJ

(b) The efficiency of a heat pump is defined to be:

ε=

W Qh

Apply conservation of energy to the heat pump to obtain:

ε=

Qh − Qc Q = 1− c Qh Qh

Substitute numerical values and evaluate ε :

ε = 1−

Qc W

5.0 kJ = 38% 8.0 kJ

The Second Law of Thermodynamics 1797 31 •• [SSM] The working substance of an engine is 1.00 mol of a monatomic ideal gas. The cycle begins at P1 = 1.00 atm and V1 = 24.6 L. The gas is heated at constant volume to P2 = 2.00 atm. It then expands at constant pressure until its volume is 49.2 L. The gas is then cooled at constant volume until its pressure is again 1.00 atm. It is then compressed at constant pressure to its original state. All the steps are quasi-static and reversible. (a) Show this cycle on a PV diagram. For each step of the cycle, find the work done by the gas, the heat absorbed by the gas, and the change in the internal energy of the gas. (b) Find the efficiency of the cycle. Picture the Problem To find the heat added during each step we need to find the temperatures in states 1, 2, 3, and 4. We can then find the work done on the gas during each process from the area under each straight-line segment and the heat that enters the system from Q = CV ΔT and Q = CP ΔT . We can use the 1st law of

thermodynamics to find the change in internal energy for each step of the cycle. Finally, we can find the efficiency of the cycle from the work done each cycle and the heat that enters the system each cycle. (a) The cycle is shown to the right:

Apply the ideal-gas law to state 1 to find T1: T1 =

P1V1 = nR

(1.00 atm )(24.6 L ) (1.00 mol)⎛⎜ 8.206 ×10 − 2 L ⋅ atm ⎞⎟ mol ⋅ K ⎠ ⎝

The pressure doubles while the volume remains constant between states 1 and 2. Hence:

T2 = 2T1 = 600 K

The volume doubles while the pressure remains constant between states 2 and 3. Hence:

T3 = 2T2 = 1200 K

= 300 K

1798 Chapter 19 T4 = 12 T3 = 600 K

The pressure is halved while the volume remains constant between states 3 and 4. Hence: For path 1→2:

W12 = PΔV12 = 0

and

J ⎞ ⎛ Q12 = CV ΔT12 = 32 RΔT12 = 32 ⎜ 8.314 ⎟ (600 K − 300 K ) = 3.74 kJ mol ⋅ K ⎠ ⎝ The change in the internal energy of the system as it goes from state 1 to state 2 is given by the 1st law of thermodynamics:

ΔEint = Qin + Won

Because W12 = 0 :

ΔEint,12 = Q12 = 3.74 kJ

For path 2→3:

⎛ 101.325 J ⎞ Won = −W23 = − PΔV23 = −(2.00 atm )(49.2 L − 24.6 L )⎜ ⎟ = − 4.99 kJ ⎝ L ⋅ atm ⎠

J ⎞ ⎛ Q23 = CP ΔT23 = 52 RΔT23 = 52 ⎜ 8.314 ⎟ (1200 K − 600 K ) = 12.5 kJ mol ⋅ K ⎠ ⎝ Apply ΔEint = Qin + Won to obtain:

ΔEint, 23 = 12.5 kJ − 4.99 kJ = 7.5 kJ

For path 3→4: W34 = PΔV34 = 0

and J ⎞ ⎛ Q34 = ΔEint, 34 = CV ΔT34 = 32 RΔT34 = 32 ⎜ 8.314 ⎟ (600 K − 1200 K ) = − 7.48 kJ mol ⋅ K ⎠ ⎝ Apply ΔEint = Qin + Won to obtain:

ΔEint, 34 = −7.48 kJ + 0 = − 7.48 kJ

The Second Law of Thermodynamics 1799 For path 4→1:

⎛ 101.325 J ⎞ Won = −W41 = − PΔV41 = −(1.00 atm )(24.6 L − 49.2 L )⎜ ⎟ = 2.49 kJ ⎝ L ⋅ atm ⎠

and J ⎞ ⎛ Q41 = CP ΔT41 = 52 RΔT41 = 52 ⎜ 8.314 ⎟ (300 K − 600 K ) = − 6.24 kJ mol ⋅ K ⎠ ⎝ Apply ΔEint = Qin + Won to obtain:

ΔEint, 41 = −6.24 kJ + 2.49 kJ = − 3.75 kJ

For easy reference, the results of the preceding calculations are summarized in the following table: Process Won , kJ Qin , kJ ΔEint (= Qin + Won ) , kJ 1→2 2→3 3→4 4→1

0 −4.99 0 2.49

3.74 12.5 −7.48 −6.24

3.74 7.5 −7.48 −3.75

(b) The efficiency of the cycle is given by:

ε=

Substitute numerical values and evaluate ε:

ε=

Wby Qin

=

− W23 + (− W41 ) Q12 + Q23

4.99 kJ − 2.49 kJ ≈ 15% 3.74 kJ + 12.5 kJ

Remarks: Note that the work done per cycle is the area bounded by the rectangular path. Note also that, as expected because the system returns to its initial state, the sum of the changes in the internal energy for the cycle is zero. 32 •• The working substance of an engine is 1.00 mol of a diatomic ideal gas. The engine operates in a cycle consisting of three steps: (1) an adiabatic expansion from an initial volume of 10.0 L to a pressure of 1.00 atm and a volume of 20.0 L, (2) a compression at constant pressure to its original volume of 10.0 L, and (3) heating at constant volume to its original pressure. Find the efficiency of this cycle.

1800 Chapter 19 P (atm)

Picture the Problem The three steps in the process are shown on the PV diagram. We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle.

1

2.639 2

1

2

3

0 0

20.0

10.0

V(L)

γ

The pressures and volumes at the end points of the adiabatic expansion are related according to:

⎛V ⎞ P1V1 = P2V2 ⇒ P1 = ⎜⎜ 2 ⎟⎟ P2 ⎝ V1 ⎠

Substitute numerical values and evaluate P1:

⎛ 20.0 L ⎞ P1 = ⎜ ⎟ ⎝ 10.0 L ⎠

Express the efficiency of the cycle:

ε=

No heat enters or leaves the system during the adiabatic expansion:

Q12 = 0

Find the heat entering or leaving the system during the isobaric compression:

Q23 = CP ΔT23 = 72 RΔT23 = 72 PΔV23

Find the heat entering or leaving the system during the constantvolume process:

Q31 = C V ΔT31 = 52 RΔT31 = 52 ΔPV31

Apply the 1st law of thermodynamics to the cycle ( ΔEint, cycle = 0 ) to obtain:

Won = ΔEint − Qin = −Qin

γ

γ

1.4

(1.00 atm) = 2.639 atm

W Qh

=

7 2

(1)

(1.00 atm )(10.0 L − 20.0 L )

= −35.0 atm ⋅ L

=

5 2

(2.639 atm − 1.00 atm )(10.0 L )

= 41.0 atm ⋅ L

= Q12 + Q23 + Q31 = 0 − 35.0 atm ⋅ L + 41.0 atm ⋅ L = 6.0 atm ⋅ L

Substitute numerical values in equation (1) and evaluate ε :

ε=

6.0 atm ⋅ L = 15% 41atm ⋅ L

The Second Law of Thermodynamics 1801 33 •• An engine using 1.00 mol of an ideal gas initially at a volume of 24.6 L and a temperature of 400 K performs a cycle consisting of four steps: (1) an isothermal expansion at 400 K to twice its initial volume, (2) cooling at constant volume to a temperature of 300 K (3) an isothermal compression to its original volume, and (4) heating at constant volume to its original temperature of 400 K. Assume that Cv = 21.0 J/K. Sketch the cycle on a PV diagram and find its efficiency. Picture the Problem We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle. 2 P (atm)

The PV diagram of the cycle is shown to the right. A, B, C, and D identify the four states of the gas and the numerals 1, 2, 3, and 4 represent the four steps through which the gas is taken.

A

1.5 4

1

3

0.5 0

1

D

C

0

10

20

30 40 V (L)

B 2

50

Express the efficiency of the cycle:

ε=

W1 + W2 + W3 + W4 W = Qh Qh,1 + Qh, 2 + Qh,3 + Qh, 4

Because steps 2 and 4 are constantvolume processes, W2 = W4 = 0:

ε=

W1 + 0 + W3 + 0 W = Qh Qh,1 + Qh, 2 + Qh,3 + Qh, 4

Because the internal energy of the gas increases in step 4 while no work is done, and because the internal energy does not change during step 1 while work is done by the gas, heat enters the system only during these processes:

ε=

W + W3 W = 1 Qh Qh,1 + Qh, 4

The work done during the isothermal expansion (1) is given by:

⎛V W1 = nRT ln⎜⎜ B ⎝ VA

The work done during the isothermal compression (3) is given by:

⎛V ⎞ W3 = nRTc ln⎜⎜ D ⎟⎟ ⎝ VC ⎠

⎞ ⎟⎟ ⎠

(1)

400 K 300 K

60

1802 Chapter 19 Because there is no change in the internal energy of the system during step 1, the heat that enters the system during this isothermal expansion is given by: The heat that enters the system during the constant-volume step 4 is given by: Substituting in equation (1) yields:

Noting the

⎛V Q1 = W1 = nRTh ln⎜⎜ B ⎝ VA

⎞ ⎟⎟ ⎠

Q4 = C V ΔT = C V (Th − Tc )

⎛V ⎞ ⎛V ⎞ nRTh ln⎜⎜ B ⎟⎟ + nRTc ln⎜⎜ D ⎟⎟ ⎝ VA ⎠ ⎝ VC ⎠ ε= ⎛V ⎞ nRTh ln⎜⎜ B ⎟⎟ + C V (Th − Tc ) ⎝ VA ⎠

VB V 1 = 2 and D = , substitute and simplify to obtain: VA VC 2

⎛1⎞ Th ln (2 ) + Tc ln⎜ ⎟ Th − Tc ⎝ 2 ⎠ = Th ln (2) − Tc ln (2) = ε= C CV C (Th − Tc ) Th ln (2) + V (Th − Tc ) Th ln (2) + V (Th − Tc ) Th + nR nR nR ln (2) Substitute numerical values and evaluate ε:

ε=

400 K − 300 K = 13.1% J 21.0 K (400 K − 300 K ) 400 K + J ⎞ ⎛ (1.00 mol)⎜ 8.314 ⎟ ln (2) mol ⋅ K ⎠ ⎝

34 •• Figure 19-15 shows the cycle followed by 1.00 mol of an ideal monatomic gas initially at a volume of 25.0 L. All the processes are quasi-static. Determine (a) the temperature of each numbered state of the cycle, (b) the heat transfer for each part of the cycle, and (c) the efficiency of the cycle. Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas and the heat capacities at constant volume and constant pressure to find the heat flow for the constant-volume and isobaric processes. Because the change in internal energy is zero for the isothermal process, we can use the expression for the work done on or by a gas during an isothermal process

The Second Law of Thermodynamics 1803 to find the heat flow during such a process. Finally, we can find the efficiency of the cycle from its definition. (a) Use the ideal-gas law to find the temperature at point 1:

T1 =

P1V1 = nR

(100 kPa )(25.0 L ) (1.00 mol)⎛⎜ 8.314 ⎝

J ⎞ ⎟ mol ⋅ K ⎠

= 301 K

Use the ideal-gas law to find the temperatures at points 2 and 3:

P2V2 nR (200 kPa )(25.0 L ) = (1.00 mol)⎛⎜ 8.314 J ⎞⎟ mol ⋅ K ⎠ ⎝

T2 = T3 =

= 601 K (b) Find the heat entering the system for the constant-volume process from 1 → 2: J ⎞ ⎛ Q12 = C V ΔT12 = 32 RΔT12 = 32 ⎜ 8.314 ⎟ (601 K − 301 K ) = 3.74 kJ mol ⋅ K ⎠ ⎝ Find the heat entering or leaving the system for the isothermal process from 2 → 3: ⎛V ⎞ ⎛ 50.0 L ⎞ J ⎞ ⎛ ⎟⎟ = 3.46 kJ Q23 = nRT2 ln⎜⎜ 3 ⎟⎟ = (1.00 mol)⎜ 8.314 ⎟ (601 K )ln⎜⎜ mol ⋅ K ⎠ ⎝ ⎝ 25.0 L ⎠ ⎝ V2 ⎠

Find the heat leaving the system during the isobaric compression from 3 → 1: J ⎞ ⎛ Q31 = C P ΔT31 = 52 RΔT31 = 52 ⎜ 8.314 ⎟ (301 K − 601 K ) = − 6.24 kJ mol ⋅ K ⎠ ⎝ (c) Express the efficiency of the cycle:

ε=

Apply the 1st law of thermodynamics to the cycle:

W = ∑ Q = Q12 + Q23 + Q31

W W = Qin Q12 + Q23

(1)

= 3.74 kJ + 3.46 kJ − 6.24 kJ = 0.96 kJ because, for the cycle, ΔEint = 0 .

1804 Chapter 19 Substitute numerical values in equation (1) and evaluate ε :

ε=

0.96 kJ = 13% 3.74 kJ + 3.46 kJ

35 •• An ideal diatomic gas follows the cycle shown in Figure 19-16. The temperature of state 1 is 200 K. Determine (a) the temperatures of the other three numbered states of the cycle and (b) the efficiency of the cycle. Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas. We can find the efficiency of the cycle from its definition; using the area enclosed by the cycle to find the work done per cycle and the heat entering the system between states 1 and 2 and 2 and 3 to determine Qin.

(a) Use the ideal-gas law for a fixed amount of gas to find the temperature in state 2 to the temperature in state 1:

PV P P1V1 P2V2 = ⇒ T2 = T1 2 2 = T1 2 T1 T2 P1V1 P1

Substitute numerical values and evaluate T2:

T2 = (200 K )

Apply the ideal-gas law for a fixed amount of gas to states 2 and 3 to obtain:

T3 = T2

Substitute numerical values and evaluate T3:

T3 = (600 K )

Apply the ideal-gas law for a fixed amount of gas to states 3 and 4 to obtain:

T4 = T3

Substitute numerical values and evaluate T4:

T4 = (1800 K )

(b) The efficiency of the cycle is:

ε=

Use the area of the rectangle to find the work done each cycle:

W = ΔPΔV

(3.0 atm ) = (1.0 atm )

600 K

P3V3 V = T2 3 P2V2 V2

(300 L ) = (100 L )

1800 K

P4V4 P = T3 4 P3V3 P3

W Qin

(1.0 atm ) = (3.0 atm )

600 K

(1)

= (300 L − 100 L )(3.0 atm − 1.0 atm ) = 400 atm ⋅ L

The Second Law of Thermodynamics 1805 Apply the ideal-gas law to state 1 to find the product of n and R:

nR =

P1V1 (1.0 atm )(100 L ) = T1 200 K

= 0.50 L ⋅ atm/K Qin = Q12 + Q23 = C V ΔT12 + C P ΔT23

Noting that heat enters the system between states 1 and 2 and states 2 and 3, express Qin:

= 52 nRΔT12 + 72 nRΔT23 = ( 52 ΔT12 + 72 ΔT23 )nR

Substitute numerical values and evaluate Qin: L ⋅ atm ⎞ ⎛ Qin = [52 (600 K − 200 K ) + 72 (1800 K − 600 K )]⎜ 0.50 ⎟ = 2600 atm ⋅ L K ⎠ ⎝ Substitute numerical values in equation (1) and evaluate ε :

ε=

400 atm ⋅ L = 15% 2600 atm ⋅ L

36 ••• Recently, an old design for a heat engine, known as the Stirling engine has been promoted as a means of producing power from solar energy. The cycle of a Stirling engine is as follows: (1) isothermal compression of the working gas (2) heating of the gas at constant volume, (3) an isothermal expansion of the gas, and (4) cooling of the gas at constant volume. (a) Sketch PV and ST diagrams for the Stirling cycle. (b) Find the entropy change of the gas for each step of the cycle and show that the sum of these entropy changes is equal to zero. Picture the Problem (a) The PV and ST cycles are shown below. (b) We can show that the entropy change during one Stirling cycle is zero by adding up the entropy changes for the four processes. P

S 2 ΔV = 0

(2)

3 (3)

Th Th

Tc

1

(4)

(1) Tc

4 V

ΔV = 0

T

1806 Chapter 19 (b) The change in entropy for one Stirling cycle is the sum of the entropy changes during the cycle:

ΔS cycle = ΔS12 + ΔS 23 + ΔS 34 + ΔS 41 (1)

Express the entropy change for the isothermal process from state 1 to state 2:

⎛V ⎞ ΔS12 = nR ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠

Similarly, the entropy change for the isothermal process from state 3 to state 4 is:

⎛V ⎞ ΔS 34 = nR ln⎜⎜ 4 ⎟⎟ ⎝ V3 ⎠ or, because V2 = V3 and V1 = V4, ⎛V ⎞ ⎛V ⎞ ΔS 34 = nR ln⎜⎜ 1 ⎟⎟ = − nR ln⎜⎜ 2 ⎟⎟ ⎝ V2 ⎠ ⎝ V1 ⎠

The change in entropy for a constantvolume process is given by:

T

ΔS isochoric

dQ f nCV dT =∫ =∫ T T Ti ⎛T = nC V ln⎜⎜ f ⎝ Ti

⎞ ⎟⎟ ⎠

For the constant-volume process from state 2 to state 3:

⎛T ΔS 23 = C V ln⎜⎜ c ⎝ Th

⎞ ⎟⎟ ⎠

For the constant-volume process from state 4 to state 1:

⎛T ΔS 41 = C V ln⎜⎜ h ⎝ Tc

⎞ ⎛T ⎟⎟ = −C V ln⎜⎜ c ⎝ Th ⎠

⎞ ⎟⎟ ⎠

Substituting in equation (1) yields: ⎛V ⎞ ⎛T ⎞ ⎛V ⎞ ⎛T ⎞ ΔS cycle = nR ln⎜⎜ 2 ⎟⎟ + CV ln⎜⎜ c ⎟⎟ − nR ln⎜⎜ 2 ⎟⎟ − CV ln⎜⎜ c ⎟⎟ = 0 ⎝ V1 ⎠ ⎝ Th ⎠ ⎝ V1 ⎠ ⎝ Th ⎠

37 •• ″As far as we know, Nature has never evolved a heat engine″—Steven Vogel, Life’s Devices, Princeton University Press (1988). (a) Calculate the efficiency of a heat engine operating between body temperature (98.6ºF) and a typical outdoor temperature (70ºF), and compare this to the human body’s efficiency for converting chemical energy into work (approximately 20%). Does this efficiency comparison contradict the second law of thermodynamics? (b) From the result of Part (a), and a general knowledge of the conditions under which most warm-blooded organisms exist, give a reason why no warm-blooded organisms have evolved heat engines to increase their internal energies.

The Second Law of Thermodynamics 1807 Picture the Problem We can use the efficiency of a Carnot engine operating between reservoirs at body temperature and typical outdoor temperatures to find an upper limit on the efficiency of an engine operating between these temperatures.

(a) Express the maximum efficiency of an engine operating between body temperature and 70°F: Use T =

5 9

(tF − 32) + 273 to obtain:

Substitute numerical values and evaluate ε C :

εC = 1 −

Tc Th

Tbody = 310 K and Troom = 294 K

ε C = 1−

294 K = 5.16% 310 K

The fact that this efficiency is considerably less than the actual efficiency of a human body does not contradict the second law of thermodynamics. The application of the second law to chemical reactions such as the ones that supply the body with energy have not been discussed in the text but we can note that we don’t get our energy from heat swapping between our body and the environment. Rather, we eat food to get the energy that we need. (b) Most warm-blooded animals survive under roughly the same conditions as humans. To make a heat engine work with appreciable efficiency, internal body temperatures would have to be maintained at an unreasonably high level. 38 ••• The diesel cycle shown in Figure 19-17 approximates the behavior of a diesel engine. Process ab is an adiabatic compression, process bc is an expansion at constant pressure, process cd is an adiabatic expansion, and process da is cooling at constant volume. Find the efficiency of this cycle in terms of the volumes Va, Vb and Vc. Picture the Problem The working fluid will be modeled as an ideal gas and the process will be modeled as quasistatic. To find the efficiency of the diesel cycle we can find the heat that enters the system and the heat that leaves the system and use the expression that gives the efficiency in terms of these quantities. Note that no heat enters or leaves the system during the adiabatic processes ab and cd. Heat enters the system during the isobaric process bc and leaves the system during the isovolumetric process da.

Express the efficiency of the cycle in terms of Qc and Qh: Express Q for the isobaric warming process bc:

ε=

Q W Qh − Qc = = 1− c Qh Qh Qh

Qbc = Qh = C P (Tc − Tb )

1808 Chapter 19 Qda = Qc = C V (Td − Ta )

Because CV is independent of T, Qda (the constant-volume cooling process) is given by: Substitute for Qh and Qc and simplify using γ = C P C V to obtain:

ε = 1−

Using an equation for a quasistatic adiabatic process, relate the temperatures Ta and Tb to the volumes Va and Vb: Proceeding similarly, relate the temperatures Tc and Td to the volumes Vc and Vd: Use equations (1) and (2) to eliminate Ta and Td:

(T − Ta ) C V (Td − Ta ) = 1− d C P (Tc − Tb ) γ (Tc − Tb )

TaVaγ −1 = TbVbγ −1 ⇒ Ta = Tb

Vbγ −1 (1) Vaγ −1

TcVcγ −1 = TdVdγ −1 ⇒ Td = Tc

Vcγ −1 (2) Vdγ −1

⎛ Vcγ −1 V γ −1 ⎞ ⎜⎜ Tc γ −1 − Tb bγ −1 ⎟⎟ Vd Va ⎠ ε = 1− ⎝ γ (Tc − Tb )

Because Va = Vd:

ε = 1−

⎛⎛ V ⎜⎜ c ⎜ ⎜⎝ Va ⎝

⎞ ⎟⎟ ⎠

γ −1

T ⎛V − b ⎜⎜ b Tc ⎝ Va

⎛

γ ⎜⎜1 − ⎝

Noting that Pb = Pc, apply the idealgas law to relate Tb and Tc:

⎞ ⎟⎟ ⎠

γ −1

Tb ⎞ ⎟ Tc ⎟⎠

Tb Vb = Tc Vc

Substitute for the ratio of Tb to Tc and simplify to obtain: ⎛ Vc ⎜⎜ V ε = 1− ⎝ a

⎞ ⎟⎟ ⎠

γ −1

⎛ Vb ⎜⎜ ⎝ Va ⎛ V ⎞ γ ⎜⎜1 − b ⎟⎟ ⎝ Vc ⎠

γ

V − b Vc

γ

⎞ ⎟⎟ ⎠

γ −1

⎛ Vc Vc ⎜⎜ V Va ⋅ = 1− ⎝ a Vc Va

⎞ ⎟⎟ ⎠

γ −1

⎛ Vb ⎜⎜ ⎝ Va ⎛V V ⎞ γ ⎜⎜ c − b ⎟⎟ ⎝ Va Va ⎠

⎛ Vc ⎞ ⎛ Vb ⎞ ⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ Va ⎠ ⎝ Va ⎠ Vcγ − Vbγ ⎝ = 1− = 1 − γ −1 γVa (Vc − Vb ) ⎛ Vc Vb ⎞ γ ⎜⎜ − ⎟⎟ ⎝ Va Va ⎠

V − b Vc

⎞ ⎟⎟ ⎠

γ −1

⎞ ⎟ ⎟ ⎠

The Second Law of Thermodynamics

1809

Second Law of Thermodynamics 39 •• [SSM] A refrigerator absorbs 500 J of heat from a cold reservoir and releases 800 J to a hot reservoir. Assume that the heat-engine statement of the second law of thermodynamics is false, and show how a perfect engine working with this refrigerator can violate the refrigerator statement of the second law of thermodynamics. Determine the Concept The following diagram shows an ordinary refrigerator that uses 300 J of work to remove 500 J of heat from a cold reservoir and releases 800 J of heat to a hot reservoir (see (a) in the diagram). Suppose the heat-engine statement of the second law is false. Then a ″perfect″ heat engine could remove energy from the hot reservoir and convert it completely into work with 100 percent efficiency. We could use this perfect heat engine to remove 300 J of energy from the hot reservoir and do 300 J of work on the ordinary refrigerator (see (b) in the diagram). Then, the combination of the perfect heat engine and the ordinary refrigerator would be a perfect refrigerator; transferring 500 J of heat from the cold reservoir to the hot reservoir without requiring any work (see (c) in the diagram).This violates the refrigerator statement of the second law. Hot reservoir at temperature Th

⇓ ⇓

Ordinary refrigerator

300 J

300 J

Perfect heat engine

⇓ 500 J

Perfect refrigerator

⇓

⇓

⇓

500 J

⇓

300 J

800 J

500 J

Cold reservoir at temperature Tc (a)

(b)

(c )

If two curves that represent quasi-static adiabatic processes could 40 •• intersect on a PV diagram, a cycle could be completed by an isothermal path between the two adiabatic curves shown in Figure 19-18. Show that such a cycle violates the second law of thermodynamics. Determine the Concept The work done by the system is the area enclosed by the cycle, where we assume that we start with the isothermal expansion. It is only in this expansion that heat is extracted from a reservoir. There is no heat transfer in the adiabatic expansion or compression. Thus, we would completely convert heat to mechanical energy, without exhausting any heat to a cold reservoir, in violation of the second law of thermodynamics.

1810 Chapter 19

Carnot Cycles 41 • [SSM] A Carnot engine works between two heat reservoirs at temperatures Th = 300 K and Tc = 200 K. (a) What is its efficiency? (b) If it absorbs 100 J of heat from the hot reservoir during each cycle, how much work does it do each cycle? (c) How much heat does it release during each cycle? (d) What is the COP of this engine when it works as a refrigerator between the same two reservoirs? Picture the Problem We can find the efficiency of the Carnot engine using ε = 1 − Tc / Th and the work done per cycle from ε = W / Qh . We can apply

conservation of energy to find the heat rejected each cycle from the heat absorbed and the work done each cycle. We can find the COP of the engine working as a refrigerator from its definition. Tc 200 K = 1− = 33.3% Th 300 K

(a) The efficiency of the Carnot engine depends on the temperatures of the hot and cold reservoirs:

ε C = 1−

(b) Using the definition of efficiency, relate the work done each cycle to the heat absorbed from the hot reservoir:

W = ε C Qh = (0.333)(100 J ) = 33.3 J

(c) Apply conservation of energy to relate the heat given off each cycle to the heat absorbed and the work done:

Qc = Qh − W = 100 J − 33.3 J = 66.7 J

(d) Using its definition, express and evaluate the refrigerator’s coefficient of performance:

COP =

= 67 J

Qc 66.7 J = = 2.0 W 33.3 J

An engine absorbs 250 J of heat per cycle from a reservoir at 300 K 42 • and releases 200 J of heat per cycle to a reservoir at 200 K. (a) What is its efficiency? (b) How much additional work per cycle could be done if the engine were reversible? Picture the Problem We can find the efficiency of the engine from its definition and the additional work done if the engine were reversible from W = ε CQh , where

εC is the Carnot efficiency.

The Second Law of Thermodynamics

1811

Q W Qh − Qc = = 1− c Qh Qh Qh

(a) Express the efficiency of the engine in terms of the heat absorbed from the high-temperature reservoir and the heat exhausted to the lowtemperature reservoir:

ε=

(b) Express the additional work done if the engine is reversible:

ΔW = WCarnot − WPart (a )

Relate the work done by a reversible engine to its Carnot efficiency:

⎛ T W = ε C Qh = ⎜⎜1 − c ⎝ Th

Substitute numerical values and evaluate W:

⎛ 200 K ⎞ ⎟⎟ (250 J ) = 83.3 J W = ⎜⎜1 − ⎝ 300 K ⎠

Substitute numerical values in equation (1) and evaluate ΔW:

ΔW = 83.3 J − 50 J = 33 J

= 1−

200 J = 20.0% 250 J (1)

⎞ ⎟⎟Qh ⎠

43 •• A reversible engine working between two reservoirs at temperatures Th and Tc has an efficiency of 30%. Working as a heat engine, it releases 140 J per cycle of heat to the cold reservoir. A second engine working between the same two reservoirs also releases 140 J per cycle to the cold reservoir. Show that if the second engine has an efficiency greater than 30%, the two engines working together would violate the heat-engine statement of the second law. Determine the Concept Let the first engine be run as a refrigerator. Then it will remove 140 J from the cold reservoir, deliver 200 J to the hot reservoir, and require 60 J of energy to operate. Now take the second engine and run it between the same reservoirs, and let it eject 140 J into the cold reservoir, thus replacing the heat removed by the refrigerator. If ε2, the efficiency of this engine, is greater than 30%, then Qh2, the heat removed from the hot reservoir by this engine, is 140 J/(1 − ε2) > 200 J, and the work done by this engine is W = ε2Qh2 > 60 J. The end result of all this is that the second engine can run the refrigerator, replacing the heat taken from the cold reservoir, and do additional mechanical work. The two systems working together then convert heat into mechanical energy without rejecting any heat to a cold reservoir, in violation of the second law. 44 •• A reversible engine working between two reservoirs at temperatures Th and Tc has an efficiency of 20%. Working as a heat engine, it does 100 J of work per cycle. A second engine working between the same two reservoirs also does 100 J of work per cycle. Show that if the efficiency of the second engine is greater

1812 Chapter 19 than 20%, the two engines working together would violate the refrigerator statement of the second law. Determine the Concept If the reversible engine is run as a refrigerator, it will require 100 J of mechanical energy to take 400 J of heat from the cold reservoir and deliver 500 J to the hot reservoir. Now let the second engine, with ε2 > 0.2, operate between the same two heat reservoirs and use it to drive the refrigerator. Because ε2 > 0.2, this engine will remove less than 500 J from the hot reservoir in the process of doing 100 J of work. The net result is then that no net work is done by the two systems working together, but a finite amount of heat is transferred from the cold reservoir to the hot reservoir, in violation of the refrigerator statement of the second law. 45 •• A Carnot engine works between two heat reservoirs as a refrigerator. During each cycle, 100 J of heat are absorbed and 150 J are released to the hot reservoir. (a) What is the efficiency of the Carnot engine when it works as a heat engine between the same two reservoirs? (b) Show that no other engine working as a refrigerator between the same two reservoirs can have a COP greater than 2.00. Picture the Problem We can use the definition of efficiency to find the efficiency of the Carnot engine operating between the two reservoirs.

(a) The efficiency of the Carnot engine is given by:

εC =

W 50 J = = 33% Qh 150 J

(b) If the COP > 2, then 50 J of work will remove more than 100 J of heat from the cold reservoir and put more than 150 J of heat into the hot reservoir. So running the engine described in Part (a) to operate the refrigerator with a COP > 2 will result in the transfer of heat from the cold to the hot reservoir without doing any net mechanical work in violation of the second law. 46 •• A Carnot engine works between two heat reservoirs at temperatures Th = 300 K and Tc = 77.0 K. (a) What is its efficiency? (b) If it absorbs 100 J of heat from the hot reservoir during each cycle, how much work does it do? (c) How much heat does it release to the low-temperature reservoir during each cycle? (d) What is the coefficient of performance of this engine when it works as a refrigerator between these two reservoirs?

The Second Law of Thermodynamics

1813

Picture the Problem We can use the definitions of the efficiency of a Carnot engine and the coefficient of performance of a refrigerator to find these quantities. The work done each cycle by the Carnot engine is given by W = ε CQh and we can

use the conservation of energy to find the heat rejected to the low-temperature reservoir. Tc 77.0 K = 1− = 74.3% Th 300 K

(a) The efficiency of a Carnot engine depends on the temperatures of the hot and cold reservoirs:

ε C = 1−

(b) Express the work done each cycle in terms of the efficiency of the engine and the heat absorbed from the high-temperature reservoir:

W = ε C Qh = (0.743)(100 J ) = 74.3 J

(c) Apply conservation of energy to obtain:

Qc = Qh − W = 100 J − 74.3 J = 26 J

(d) Using its definition, express and evaluate the refrigerator’s coefficient of performance:

COP =

Qc 26 J = = 0.35 W 74.3 J

47 •• [SSM] In the cycle shown in Figure 19-19, 1.00 mol of an ideal diatomic gas is initially at a pressure of 1.00 atm and a temperature of 0.0ºC. The gas is heated at constant volume to T2 = 150ºC and is then expanded adiabatically until its pressure is again 1.00 atm. It is then compressed at constant pressure back to its original state. Find (a) the temperature after the adiabatic expansion, (b) the heat absorbed or released by the system during each step, (c) the efficiency of this cycle, and (d) the efficiency of a Carnot cycle operating between the temperature extremes of this cycle. Picture the Problem We can use the ideal-gas law for a fixed amount of gas and the equations of state for an adiabatic process to find the temperatures, volumes, and pressures at the end points of each process in the given cycle. We can use Q = C V ΔT and Q = C P ΔT to find the heat entering and leaving during the

constant-volume and isobaric processes and the first law of thermodynamics to find the work done each cycle. Once we’ve calculated these quantities, we can use its definition to find the efficiency of the cycle and the definition of the Carnot efficiency to find the efficiency of a Carnot engine operating between the extreme temperatures.

1814 Chapter 19 (a) Apply the ideal-gas law for a fixed amount of gas to relate the temperature at point 3 to the temperature at point 1:

P1V1 P3V3 = T1 T3 or, because P1 = P3, V T3 = T1 3 V1

(1)

Apply the ideal-gas law for a fixed amount of gas to relate the pressure at point 2 to the temperatures at points 1 and 2 and the pressure at 1:

PV T P1V1 P2V2 = ⇒ P2 = 1 1 2 T1 T2 V2T1

Because V1 = V2:

P2 = P1

Apply an equation for an adiabatic process to relate the pressures and volumes at points 2 and 3:

⎛ P ⎞γ P1V1γ = P3V3γ ⇒ V3 = V1 ⎜⎜ 1 ⎟⎟ ⎝ P3 ⎠

Noting that V1 = 22.4 L, evaluate V3:

⎛ 1.55 atm ⎞ 1.4 ⎟⎟ = 30.6 L V3 = (22.4 L )⎜⎜ ⎝ 1atm ⎠

Substitute numerical values in equation (1) and evaluate T3 and t3:

T3 = (273 K )

T2 423 K = (1.00 atm ) = 1.55 atm T1 273 K 1

1

30.6 L = 373 K 22.4 L

and t 3 = T3 − 273 = 100°C (b) Process 1→2 takes place at constant volume (note that γ = 1.4 corresponds to a diatomic gas and that CP – CV = R):

Q12 = CV ΔT12 = 52 RΔT12 J ⎞ ⎛ = 52 ⎜ 8.314 ⎟ (423 K − 273 K ) mol ⋅ K ⎠ ⎝ = 3.12 kJ

Process 2→3 takes place adiabatically:

Q23 = 0

Process 3→1 is isobaric (note that CP = CV + R):

Q31 = C P ΔT31 = 72 RΔT12 J ⎞ ⎛ = 72 ⎜ 8.314 ⎟ (273 K − 373 K ) mol ⋅ K ⎠ ⎝ = − 2.91 kJ

The Second Law of Thermodynamics

1815

(c) The efficiency of the cycle is given by:

ε=

Apply the first law of thermodynamics to the cycle:

ΔEint = Qin + Won or, because ΔEint, cycle = 0 (the system

W Qin

(2)

begins and ends in the same state) and Won = −Wby the gas = Qin . Evaluating W yields:

W = ∑ Q = Q12 + Q23 + Q31 = 3.12 kJ + 0 − 2.91 kJ = 0.21 kJ

0.21kJ = 6.7% 3.12 kJ

Substitute numerical values in equation (2) and evaluate ε :

ε=

(d) Express and evaluate the efficiency of a Carnot cycle operating between 423 K and 273 K:

εC = 1−

Tc 273 K = 1− = 35.5% Th 423 K

48 •• You are part of a team that is completing a mechanical-engineering project. Your team built a steam engine that takes in superheated steam at 270ºC and discharges condensed steam from its cylinder at 50.0ºC. Your team has measured its efficiency to be 30.0%. (a) How does this efficiency compare with the maximum possible efficiency for your engine? (b) If the useful power output of the engine is known to be 200 kW, how much heat does the engine release to its surroundings in 1.00 h? Picture the Problem We can find the maximum efficiency of the steam engine by calculating the Carnot efficiency of an engine operating between the given temperatures. We can apply the definition of efficiency to find the heat discharged to the engine’s surroundings in 1.00 h.

(a) The efficiency of the steam engine as a percentage of the maximum possible efficiency is given by:

ε steam engine 0.300 = ε max ε max

The efficiency of a Carnot engine operating between temperatures Fc and Th is:

ε max = 1 −

Tc 323 K = 1− = 40.52% Th 543 K

1816 Chapter 19 Substituting for εmax yields:

ε steam engine 0.300 = = 74.05% ε max 0.4052 or

ε steam engine = 0.740ε max W Qh − Qc = ⇒ Qc = (1 − ε )Qh Qh Qh

(b) Relate the heat Qc discharged to the engine’s surroundings to Qh and the efficiency of the engine:

ε=

Using its definition, relate the efficiency of the engine to the heat intake of the engine and the work it does each cycle:

Qh =

Substitute for Qh in the expression for Qc and simplify to obtain:

Qc = (1 − ε )

W

ε

=

PΔt

ε

PΔt

ε

⎛1 ⎞ = ⎜ − 1⎟ PΔt ⎝ε ⎠

Substitute numerical values and evaluate Qc (1.00 h ) : kJ ⎞ ⎞⎛ ⎛ 1 Qc (1.00 h ) = ⎜ − 1⎟ ⎜ 200 ⎟ (3600 s ) = 1.68 GJ s ⎠ ⎝ 0.300 ⎠ ⎝

*Heat Pumps 49 • [SSM] As an engineer, you are designing a heat pump that is capable of delivering heat at the rate of 20 kW to a house. The house is located where, in January, the average outside temperature is –10ºC. The temperature of the air in the air handler inside the house is to be 40ºC. (a) What is maximum possible COP for a heat pump operating between these temperatures? (b) What must the minimum power of the electric motor driving the heat pump be? (c) In reality, the COP of the heat pump will be only 60 percent of the ideal value. What is the minimum power of the electric motor when the COP is 60 percent of the ideal value? Picture the Problem We can use the definition of the COPHP and the Carnot efficiency of an engine to express the maximum efficiency of the refrigerator in terms of the reservoir temperatures. We can apply the definition of power to find the minimum power needed to run the heat pump.

The Second Law of Thermodynamics (a) Express the COPHP in terms of Th and Tc:

COPHP = =

Substitute numerical values and evaluate COPHP:

COPHP =

1817

Qh Qh = W Qh − Qc Th 1 1 = = Q T Th − Tc 1− c 1− c Qh Th

313 K = 6.26 313 K − 263 K

= 6.3 (b) The COPHP is also given by:

COPHP =

Substitute numerical values and evaluate Pmotor:

Pmotor =

(c) The minimum power of the electric motor is given by:

Pmin

Pout Pout ⇒ Pmotor = COPHP Pmotor

20 kW = 3.2 kW 6.26

dQc = dt =

dQc dt ε (COPHP,max )

ε HP where ε HP is the efficiency of the heat pump.

Substitute numerical values and evaluate Pmin:

Pmin =

20 kW = 5.3 kW (0.60)(6.26)

50 • A refrigerator is rated at 370 W. (a) What is the maximum amount of heat it can absorb from the food compartment in 1.00 min if the foodcompartment temperature of the refrigerator is 0.0ºC and it releases heat into a room at 20.0ºC? (b) If the COP of the refrigerator is 70% of that of a reversible refrigerator, how much heat can it absorb from the food compartment in 1.00 min under these conditions? Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc, Th, P, and Δt.

(a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP:

Qc = (COP )W

= (COP )PΔt

1818 Chapter 19 Express the COP in terms of Th and Tc and simplify to obtain:

COP = =

=

Substituting for COP yields:

Qc Q Q −W = c = h W ε Qh ε Qh 1− ε

ε

=

1

ε

−1 =

1 −1 Tc 1− Th

Tc Th − Tc

⎛ Tc ⎞ ⎟⎟ PΔt Qc = ⎜⎜ ⎝ Th − Tc ⎠

Substitute numerical values and evaluate Qc: 273 K 60 s ⎞ ⎛ ⎞ ⎛ Qc = ⎜ ⎟ (370 W )⎜1.00 min × ⎟ = 303 kJ = 0.30 MJ min ⎠ ⎝ 293 K − 273 K ⎠ ⎝ (b) If the COP is 70% of the efficiency of an ideal pump:

Qc' = (0.70 )(303 kJ ) = 0.21 MJ

51 • A refrigerator is rated at 370 W. (a) What is the maximum amount of heat it can absorb for the food compartment in 1.00 min if the temperature in the compartment is 0.0ºC and it releases heat into a room at 35ºC? (b) If the COP of the refrigerator is 70% of that of a reversible pump, how much heat can it absorb from the food compartment in 1.00 min? Is the COP for the refrigerator greater when the temperature of the room is 35ºC or 20ºC? Explain. Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc, Th, P, and Δt.

(a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP:

Qc = (COP )W

= (COP )PΔt

The Second Law of Thermodynamics 1819 Express the COP in terms of Th and Tc and simplify to obtain:

COP = = =

Substituting for COP yields:

Qc Q Q −W = c = h W εQh εQh 1− ε

ε

=

1

ε

−1

Tc 1 −1 = T Th − Tc 1− c Th

⎛ Tc ⎞ ⎟⎟ PΔt Qc = ⎜⎜ ⎝ Th − Tc ⎠

Substitute numerical values and evaluate Qc: 273 K 60 s ⎞ ⎛ ⎞ ⎛ Qc = ⎜ ⎟ = 173 kJ = 0.17 MJ ⎟ (370 W )⎜1.00 min × min ⎠ ⎝ 308 K − 273 K ⎠ ⎝ (b) If the COP is 70% of the efficiency of an ideal pump:

Qc' = (0.70 )(173 kJ ) = 0.12 MJ

Because the temperature difference increases when the room is warmer, the COP decreases. 52 ••• You are installing a heat pump, whose COP is half the COP of a reversible heat pump. You will use the pump on chilly winter nights to increase the air temperature in your bedroom. Your bedroom’s dimensions are 5.00 m × 3.50 m × 2.50 m. The air temperature should increase from 63°F to 68°F. The outside temperature is 35°F , and the temperature at the air handler in the room is 112°F. If the pump’s electric power consumption is 750 W, how long will you have to wait in order for the room’s air to warm (take the specific heat of air to be 1.005 kJ/(kg·°C)? Assume you have good window draperies and good wall insulation so that you can neglect the release of heat through windows, walls, ceilings and floors. Also assume that the heat capacity of the floor, ceiling, walls and furniture are negligible. Picture the Problem We can use the definition of the coefficient of performance of a heat pump and the relationship between the work done per cycle and the pump’s power consumption to find your waiting time.

1820 Chapter 19 The coefficient of performance of the heat pump is defined as:

We’re given that the coefficient of performance of the heat pump is half the coefficient of performance of an ideal heat pump:

Substituting for COPHP yields:

The heat required to warm the room is related to the volume of the room, the density of air, and the desired increase in temperature: Substitute for Qh to obtain:

Qh Q Qh = h ⇒ Δt = (COPHP )P W PΔt where Qh is the heat required to raise the temperature of your bedroom, P is the power consumption of the heat pump, and Δt is the time required to warm the bedroom. COPHP =

Qh 1 = COPmax W 2 ⎛ Th ⎞ ⎟⎟ = 12 ⎜⎜ − T T ⎝ h c⎠

COPHP =

Δt =

2Qh ⎛ Th ⎞ ⎜⎜ ⎟⎟ P ⎝ Th − Tc ⎠

Qh = mcΔT = ρVcΔT where ρ is the density of air and c is its specific heat capacity.

Δt =

2 ρVcΔT ⎛ Th ⎞ ⎜⎜ ⎟⎟ P ⎝ Th − Tc ⎠

Substitute numerical values and evaluate Δt: ⎛ kg ⎞ J ⎞⎛ 5 C° ⎞ ⎛ ⎟⎟ ⎜ 5 F° × 2⎜1.293 3 ⎟ (5.00 m × 3.50 m × 2.50 m )⎜⎜1005 ⎟ kg ⋅ C° ⎠ ⎝ 9 F° ⎠ m ⎠ ⎝ ⎝ Δt = = 56 s 317 K ⎛ ⎞ ⎜ ⎟ (750 W ) ⎝ 317 K − 275 K ⎠

Entropy Changes 53 • [SSM] You inadvertently leave a pan of water boiling away on the hot stove. You return just in time to see the last drop converted into steam. The pan originally held 1.00 L of boiling water. What is the change in entropy of the water associated with its change of state from liquid to gas?

The Second Law of Thermodynamics 1821 Picture the Problem Because the water absorbed heat in the vaporization process Qabsorbed by H 2 O its change in entropy is positive and given by ΔS H 2 O = . See Table 18-2 T for the latent heat of vaporization of water.

The change in entropy of the water is given by:

The heat absorbed by the water as it vaporizes is the product of its mass and latent heat of vaporization: Substituting for Qabsorbed yields: by H 2 O

Substitute numerical values and evaluate ΔS H 2O :

ΔS H 2 O =

Qabsorbed by H 2 O

T

Qabsorbed = mLv = ρVLv by H 2 O

ΔS H 2 O =

ΔS H 2 O

ρVLv T

⎛ kg ⎞ kJ ⎞ ⎛ ⎜1.00 ⎟ (1.00 L )⎜⎜ 2257 ⎟⎟ L⎠ kg ⎠ ⎝ ⎝ = 373 K kJ = 6.05 K

54 • What is the change in entropy of 1.00 mol of liquid water at 0.0ºC that freezes to ice at 0.0°C? Picture the Problem We can use the definition of entropy change to find the change in entropy of the liquid water as it freezes. Because heat is removed from liquid water when it freezes, the change in entropy of the liquid water is negative. See Appendix C for the molar mass of water and Table 18-2 for the latent heat of fusion of water.

The change in entropy of the water is given by:: The heat removed from the water as it freezes is the product of its mass and latent heat of fusion:

ΔS H 2 O =

Qremoved

from H 2 O

T

Qremoved = −mLf from H 2 O

or, because m = nM H 2 O , Qremoved = − nM H 2 O Lf from H 2 O

1822 Chapter 19 Substitute numerical values and evaluate ΔS H 2O :

ΔS H 2O

g ⎞⎛ J⎞ ⎛ − (1.00 mol)⎜18.015 ⎟ ⎜⎜ 333.5 ⎟⎟ mol ⎠ ⎝ g⎠ J ⎝ = = − 22.0 K 273 K

55 •• Consider the freezing of 50.0 g of water once it is placed in the freezer compartment of a refrigerator. Assume the walls of the freezer are maintained at –10ºC. The water, initially liquid at 0.0ºC, is frozen into ice and cooled to –10ºC. Show that even though the entropy of the water decreases, the net entropy of the universe increases. Picture the Problem The change in the entropy of the universe resulting from the freezing of this water and the cooling of the ice formed is the sum of the entropy changes of the water-ice and the freezer. Note that, while the entropy of the water decreases, the entropy of the freezer increases.

The change in entropy of the universe resulting from this freezing and cooling process is given by:

ΔS u = ΔS water + ΔS freezer

(1)

Express ΔS water :

ΔS water = ΔS freezing + ΔS cooling

(2)

Express ΔS freezing :

ΔS freezing =

− Qfreezing

(3)

Tfreezing

where the minus sign is a consequence of the fact that energy is leaving the water as it freezes. Relate Qfreezing to the latent heat of

Qfreezing = mLf

fusion and the mass of the water: − mLf Tfreezing

Substitute in equation (3) to obtain:

ΔS freezing =

Express ΔS cooling :

⎛T ΔS cooling = mC p ln⎜⎜ f ⎝ Ti

Substitute in equation (2) to obtain:

ΔS water =

⎞ ⎟⎟ ⎠

⎛T − mLf + mCp ln⎜⎜ f Tfreezing ⎝ Ti

⎞ ⎟⎟ ⎠

The Second Law of Thermodynamics 1823 Noting that the freezer gains heat (at 263 K) from the freezing water and cooling ice, express ΔS freezer :

ΔS freezer = =

ΔQice ΔQcooling ice + Tfreezer Tfreezer mCp ΔT mLf + Tfreezer Tfreezer

Substitute for ΔS water and ΔS freezer in equation (1): ΔS u =

⎛T − mLf + mCp ln⎜⎜ f Tfreezing ⎝ Ti

mCp ΔT ⎞ mLf ⎟⎟ + + Tfreezer ⎠ Tfreezer

⎡ − Lf ⎛T = m⎢ + C p ln⎜⎜ f ⎝ Ti ⎣⎢ Tfreezing

⎞ Lf + C p ΔT ⎤ ⎟⎟ + ⎥ Tfreezer ⎦⎥ ⎠

Substitute numerical values and evaluate ΔSu: ⎡ 3 J ⎢ 333.5 ×10 kg ⎛ J ⎞ ⎛ 263 K ⎞ ⎟ ⎟ ln⎜ + ⎜⎜ 2100 ΔS u = (0.0500 kg ) ⎢− kg ⋅ K ⎟⎠ ⎜⎝ 273 K ⎟⎠ 273 K ⎢ ⎝ ⎢⎣ ⎤ J ⎞ J ⎛ ⎟⎟ (273 K − 263 K ) ⎥ + ⎜⎜ 2100 kg ⋅ K ⎠ kg ⎝ ⎥ = 2.40 J/K + ⎥ 263 K ⎥ ⎦ and, because ΔSu > 0, the entropy of the universe increases. 333.5 ×10 3

56 • In this problem, 2.00 mol of an ideal gas at 400 K expand quasistatically and isothermally from an initial volume of 40.0 L to a final volume of 80.0 L. (a) What is the entropy change of the gas? (b) What is the entropy change of the universe for this process? Picture the Problem We can use the definition of entropy change and the 1st law of thermodynamics to express ΔS for the ideal gas as a function of its initial and final volumes.

(a) The entropy change of the gas is given by:

ΔS gas =

Q T

(1)

1824 Chapter 19 Apply the first law of thermodynamics to the isothermal process to express Q in terms of Won:

Q = ΔEint − Won or, because ΔEint = 0 for an isothermal expansion of a gas, Q = −Won

The work done on the gas is given by:

⎛V Won = nRT ln⎜⎜ i ⎝ Vf

⎛V ⎞ ⎟⎟ ⇒ Q = −nRT ln⎜⎜ i ⎝ Vf ⎠

Substitute for Q in equation (1) to obtain:

⎛V ΔS gas = −nR ln⎜⎜ i ⎝ Vf

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ ⎠

Substitute numerical values and evaluate ΔS: J J ⎞ ⎛ 40.0 L ⎞ ⎛ ⎟⎟ = 11.5 ΔS gas = −(2.00 mol)⎜ 8.314 ⎟ ln⎜⎜ K mol ⋅ K ⎠ ⎝ 80.0 L ⎠ ⎝

(b) Because the process is reversible:

ΔS u = 0

Remarks: The entropy change of the environment of the gas is −11.5 J/K. 57 •• [SSM] A system completes a cycle consisting of six quasi-static steps, during which the total work done by the system is 100 J. During step 1 the system absorbs 300 J of heat from a reservoir at 300 K, during step 3 the system absorbs 200 J of heat from a reservoir at 400 K, and during step 5 it absorbs heat from a reservoir at temperature T3. (During steps 2, 4 and 6 the system undergoes adiabatic processes in which the temperature of the system changes from one reservoir’s temperature to that of the next.) (a) What is the entropy change of the system for the complete cycle? (b) If the cycle is reversible, what is the temperature T3? Picture the Problem We can use the fact that the system returns to its original state to find the entropy change for the complete cycle. Because the entropy change for the complete cycle is the sum of the entropy changes for each process, we can find the temperature T3 from the entropy changes during the 1st two processes and the heat released during the third.

(a) Because S is a state function of the system, and because the system’s final state is identical to its initial state:

ΔS system

1 complete cycle

= 0

The Second Law of Thermodynamics 1825 (b) Relate the entropy changes for each of the three heat reservoirs and the system for one complete cycle of the system:

ΔS1 + ΔS 2 + ΔS 3 + ΔS system = 0 or Q1 Q2 Q3 + + +0 = 0 T1 T2 T3

Substitute numerical values. Heat is rejected by the two high-temperature reservoirs and absorbed by the cold reservoir:

− 300 J − 200 J 400 J + + =0 300 K 400 K T3

Solving for T3 yields:

T3 = 267 K

58 •• In this problem, 2.00 mol of an ideal gas initially has a temperature of 400 K and a volume of 40.0 L. The gas undergoes a free adiabatic expansion to twice its initial volume. What is (a) the entropy change of the gas and (b) the entropy change of the universe? Picture the Problem The initial and final temperatures are the same for a free expansion of an ideal gas. Thus, the entropy change ΔS for a free expansion from Vi to Vf is the same as ΔS for an isothermal process from Vi to Vf. We can use the definition of entropy change and the 1st law of thermodynamics to express ΔS for the ideal gas as a function of its initial and final volumes.

(a) The entropy change of the gas is given by:

ΔS gas =

Apply the first law of thermodynamics to the isothermal process to express Q:

Q = ΔEint − Won

Q T

(1)

or, because ΔEint = 0 for a free expansion of a gas, Q = −Won ⎞ ⎛V ⎟⎟ ⇒ Q = −nRT ln⎜⎜ i ⎠ ⎝ Vf

The work done on the gas is given by:

⎛V Won = nRT ln⎜⎜ i ⎝ Vf

Substitute for Q in equation (1) to obtain:

⎛V ΔS gas = −nR ln⎜⎜ i ⎝ Vf

⎞ ⎟⎟ ⎠

Substitute numerical values and evaluate ΔS: J ⎞ ⎛ 40.0 L ⎞ J ⎛ ⎟⎟ = 11.5 ΔS gas = −(2.00 mol)⎜ 8.314 ⎟ ln⎜⎜ mol ⋅ K ⎠ ⎝ 80.0 L ⎠ K ⎝

⎞ ⎟⎟ ⎠

1826 Chapter 19 (b) The change in entropy of the universe is the sum of the entropy changes of the gas and the surroundings:

ΔS u = ΔS gas + ΔS surroundings

For the change in entropy of the surroundings we use the fact that, during the free expansion, the surroundings are unaffected:

ΔS surroundings =

The change in entropy of the universe is the change in entropy of the gas:

ΔS u = 11.5

Qrev 0 = =0 T T

J K

59 •• A 200-kg block of ice at 0.0ºC is placed in a large lake. The temperature of the lake is just slightly higher than 0.0ºC, and the ice melts very slowly. (a) What is the entropy change of the ice? (b) What is the entropy change of the lake? (c) What is the entropy change of the universe (the ice plus the lake)? Picture the Problem Because the ice gains heat as it melts, its entropy change is positive and can be calculated from its definition. Because the temperature of the lake is just slightly greater than 0°C and the mass of water is so much greater than that of the block of ice, the absolute value of the entropy change of the lake will be approximately equal to the entropy change of the ice as it melts.

(a) The entropy change of the ice is given by: Substitute numerical values and evaluate ΔSice :

(b) Relate the entropy change of the lake to the entropy change of the ice: (c) The entropy change of the universe due to this melting process is the sum of the entropy changes of the ice and the lake:

ΔS ice =

ΔS ice =

mLf T

(200 kg )⎛⎜⎜ 333.5 kJ ⎞⎟⎟ ⎝ 273 K

kg ⎠

ΔS lake ≈ −ΔS ice = − 244

ΔS u = ΔS ice + ΔS lake

kJ K

= 244

kJ K

The Second Law of Thermodynamics 1827 Because the temperature of the lake is slightly greater than that of the ice, the magnitude of the entropy change of the lake is less than 244 kJ/K and the entropy change of the universe is just slightly greater than zero. The melting of the ice is an irreversible process and ΔS u > 0 . 60 •• A 100-g piece of ice at 0.0ºC is placed in an insulated calorimeter with negligible heat capacity containing 100 g of water at 100ºC. (a) What is the final temperature of the water once thermal equilibrium is established? (b) Find the entropy change of the universe for this process. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a melting process and for constant-pressure processes to find the entropy change of the universe (the entropy change of the piece of ice plus the entropy change of the water in the insulated container).

∑Q

(a) Apply conservation of energy to obtain:

i

=0

i

or Qmelting + Qwarming − Qcooling = 0 ice

water

water

Substitute to relate the masses of the ice and water to their temperatures, specific heats, and the final temperature of the water:

(100 g )⎛⎜⎜ 333.5 kJ ⎞⎟⎟ + (100 g )⎛⎜⎜ 4.18 ⎝

kg ⎠

⎝

kJ ⎞ ⎟t kg ⋅ C° ⎟⎠

⎛ kJ ⎞ ⎟ (100°C − t ) = 0 − (100 g )⎜⎜ 4.18 kg ⋅ C° ⎟⎠ ⎝

Solving for t yields:

t = 10.1°C

(b) The entropy change of the universe is the sum of the entropy changes of the ice and the water:

ΔS u = ΔSice + ΔS water

Using the expression for the entropy change for a constant-pressure process, express the entropy change of the melting ice and warming icewater:

ΔSice = ΔS melting ice + ΔS warming water =

⎛T mLf + mcP ln⎜⎜ f Tf ⎝ Ti

⎞ ⎟⎟ ⎠

1828 Chapter 19 Substitute numerical values to obtain:

ΔS ice =

(0.100 kg )⎛⎜⎜ 333.5 kJ ⎞⎟⎟ ⎝ 273 K

kg ⎠

⎛ J kJ ⎞ ⎛ 283 K ⎞ ⎟⎟ = 137 ⎟⎟ ln⎜⎜ + (0.100 kg )⎜⎜ 4.18 K kg ⋅ K ⎠ ⎝ 273 K ⎠ ⎝

Find the entropy change of the cooling water: ⎛ J kJ ⎞ ⎛ 283 K ⎞ ⎟⎟ = −115 ⎟⎟ ln⎜⎜ ΔS water = (0.100 kg )⎜⎜ 4.18 K kg ⋅ K ⎠ ⎝ 373 K ⎠ ⎝

Substitute for ΔSice and ΔSwater and evaluate the entropy change of the universe:

ΔS u = 137

J J J − 115 = 22 K K K

Remarks: The result that ΔSu > 0 tells us that this process is irreversible. 61 •• [SSM] A 1.00-kg block of copper at 100ºC is placed in an insulated calorimeter of negligible heat capacity containing 4.00 L of liquid water at 0.0ºC. Find the entropy change of (a) the copper block, (b) the water, and (c) the universe. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a constant pressure process to find the entropy changes of the copper block, the water, and the universe.

(a) Use the equation for the entropy change during a constant-pressure process to express the entropy change of the copper block:

⎛T ΔS Cu = mCu cCu ln⎜⎜ f ⎝ Ti

Apply conservation of energy to obtain:

∑Q

i

=0

i

or Qcopper + Qwarming = 0 block

water

⎞ ⎟⎟ ⎠

(1)

The Second Law of Thermodynamics 1829 Substitute to relate the masses of the block and water to their temperatures, specific heats, and the final temperature Tf of the water:

(1.00 kg )⎛⎜⎜ 0.386 ⎝

kJ ⎞ ⎟ (Tf − 373 K ) kg ⋅ K ⎟⎠ kg ⎞ ⎛ kJ ⎞ ⎛ ⎟ (Tf − 273 K ) = 0 + (4.00 L )⎜1.00 ⎟ ⎜⎜ 4.18 L ⎠⎝ kg ⋅ K ⎟⎠ ⎝

Solve for Tf to obtain:

Tf = 275.26 K

Substitute numerical values in equation (1) and evaluate ΔS Cu : ⎛ J kJ ⎞ ⎛ 275.26 K ⎞ ⎟⎟ = − 117 ⎟⎟ln⎜⎜ ΔS Cu = (1.00 kg )⎜⎜ 0.386 K kg ⋅ K ⎠ ⎝ 373 K ⎠ ⎝

(b) The entropy change of the water is given by:

⎛T ΔS water = mwater c water ln⎜⎜ f ⎝ Ti

⎞ ⎟⎟ ⎠

Substitute numerical values and evaluate ΔS water : ⎛ J kJ ⎞ ⎛ 275.26 K ⎞ ⎟⎟ = 138 ⎟⎟ ln⎜⎜ ΔS water = (4.00 kg )⎜⎜ 4.18 K kg ⋅ K ⎠ ⎝ 273 K ⎠ ⎝

(c) Substitute for ΔS Cu and ΔS water and evaluate the entropy change of the universe:

ΔS u = ΔS Cu + ΔS water = −117 = 21

J J + 138 K K

J K

Remarks: The result that ΔSu > 0 tells us that this process is irreversible. 62 •• If a 2.00-kg piece of lead at 100ºC is dropped into a lake at 10ºC, find the entropy change of the universe. Picture the Problem Because the mass of the water in the lake is so much greater than the mass of the piece of lead, the temperature of the lake will increase only slightly and we can reasonably assume that its final temperature is 10°C. We can apply the equation for the entropy change during a constant pressure process to find the entropy changes of the piece of lead, the water in the lake, and the universe.

1830 Chapter 19 Express the entropy change of the universe in terms of the entropy changes of the lead and the water in the lake:

ΔS u = ΔS Pb + ΔS w

(1)

Using the equation for the entropy change during a constant-pressure process, express and evaluate the entropy change of the lead: ⎛T ΔS Pb = mPb cPb ln⎜⎜ f ⎝ Ti

⎞ ⎛ J kJ ⎞ ⎛ 283 K ⎞ ⎟⎟ = (2.00 kg )⎜⎜ 0.128 ⎟⎟ = −70.69 ⎟⎟ ln⎜⎜ K kg ⋅ K ⎠ ⎝ 373 K ⎠ ⎝ ⎠

The entropy change of the water in the lake is given by: Substitute numerical values and evaluate ΔSw:

ΔS w =

ΔS w =

Qw QPb mPb cPb ΔTPb = = Tw Tw Tw

kJ ⎞ ⎟ (90 K ) kg ⋅ K ⎟⎠ ⎝ 283 K

(2.00 kg )⎛⎜⎜ 0.128

= 81.41 J/K Substitute numerical values in equation (1) and evaluate ΔSu:

ΔS u = −70.69

J J J + 81.41 = 11 K K K

Entropy and ″Lost″ Work 63 •• [SSM] A a reservoir at 300 K absorbs 500 J of heat from a second reservoir at 400 K. (a) What is the change in entropy of the universe, and (b) how much work is lost during the process? Picture the Problem We can find the entropy change of the universe from the entropy changes of the high- and low-temperature reservoirs. The maximum amount of the 500 J of heat that could be converted into work can be found from the maximum efficiency of an engine operating between the two reservoirs.

(a) The entropy change of the universe is the sum of the entropy changes of the two reservoirs:

ΔS u = ΔS h + ΔSc = − ⎛1 1⎞ = −Q⎜⎜ − ⎟⎟ ⎝ Th Tc ⎠

Q Q + Th Tc

The Second Law of Thermodynamics 1831 Substitute numerical values and evaluate ΔSu:

⎛ 1 1 ⎞ ⎟⎟ − ΔS u = (− 500 J ) ⎜⎜ ⎝ 400 K 300 K ⎠ = 0.42 J/K

(b) Relate the heat that could have been converted into work to the maximum efficiency of an engine operating between the two reservoirs:

W = ε max Qh

The maximum efficiency of an engine operating between the two reservoir temperatures is the efficiency of a Carnot device operating between the reservoir temperatures:

ε max = ε C = 1 −

Substitute for εmax to obtain:

⎛ T ⎞ W = ⎜⎜1 − c ⎟⎟Qh ⎝ Th ⎠

Substitute numerical values and evaluate W:

⎛ 300 K ⎞ ⎟⎟ (500 J ) = 125 J W = ⎜⎜1 − ⎝ 400 K ⎠

Tc Th

64 •• In this problem, 1.00 mol of an ideal gas at 300 K undergoes a free adiabatic expansion from V1 = 12.3 L to V2 = 24.6 L. It is then compressed isothermally and reversibly back to its original state. (a) What is the entropy change of the universe for the complete cycle? (b) How much work is lost in this cycle? (c) Show that the work lost is TΔSu. Picture the Problem Although no energy is lost by the gas in the adiabatic free expansion, the process is irreversible and the entropy of the gas (and the universe) increases. In the isothermal reversible process that returns the gas to its original state, the gas releases energy to the surroundings. However, because the process is reversible, the entropy change of the universe is zero. Consequently, the net entropy change is the negative of that of the gas in the isothermal compression.

1832 Chapter 19 (a) Relate the entropy change of the universe to the entropy changes of the gas during 1 complete cycle:

ΔS u = ΔS gas during

free expansion

+ ΔS gas during

isothermal compresion

= 0,

or, because ΔSgas during

isothermal compression

ΔS u = ΔS gas during

=

free expansion

Q T

The work done by the gas during its isothermal compression is given by:

⎛V Wby = −Won = −Q = −nRT ln⎜⎜ f ⎝ Vi

Substituting for Q in the expression for ΔS u and simplifying yields:

⎛V ΔS u = − nR ln⎜⎜ f ⎝ Vi

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ ⎠

(1)

Substitute numerical values and evaluate ΔSu: J J J ⎞ ⎛ 12.3 L ⎞ ⎛ ⎟⎟ = 5.763 = 5.76 ΔS u = −(1.00 mol)⎜ 8.314 ⎟ln⎜⎜ K K mol ⋅ K ⎠ ⎝ 24.6 L ⎠ ⎝

(b) Use Equation 19-22 to find the amount of energy that becomes unavailable for doing work during this process:

J⎞ ⎛ Wlost = TΔS u = (300 K )⎜ 5.763 ⎟ K⎠ ⎝

(c) No work is done in the free expansion. In the adiabatic compression, the work done on the gas is:

⎛V ⎞ Wby gas, f →i = −Won gas, f →i = −nRT ln⎜⎜ i ⎟⎟ ⎝ Vf ⎠ ⎛ ⎛V ⎞⎞ ⎛V ⎞ = nRT ln⎜⎜ f ⎟⎟ = T ⎜⎜ nR ln⎜⎜ f ⎟⎟ ⎟⎟ ⎝ Vi ⎠ ⎠ ⎝ Vi ⎠ ⎝

= 1.73 kJ

= TΔS u

General Problems 65 • A heat engine with an output of 200 W has an efficiency of 30%. It operates at 10.0 cycles/s. (a) How much work is done by the engine during each cycle? (b) How much heat is absorbed from the hot reservoir and how much is released to the cold reservoir during each cycle? Picture the Problem We can use the definition of power to find the work done each cycle and the definition of efficiency to find the heat that is absorbed each cycle. Application of the first law of thermodynamics will yield the heat given off each cycle.

The Second Law of Thermodynamics 1833

(a) Use the definition of power to relate the work done in each cycle to the frequency of each cycle: Substitute numerical values and evaluate Wcycle: (b) Express the heat absorbed in each cycle in terms of the work done and the efficiency of the engine: Apply the 1st law of thermodynamics to find the heat given off in each cycle:

Wcycle = PΔt =

P f

where f is the frequency of the engine. Wcycle =

200 W = 20.0 J 10.0 s −1

Qh,cycle =

Wcycle

ε

=

20.0 J = 67 J 0.30

Qc,cycle = Qh,cycle − W = 67 J − 20 J = 47 J

66 • During each cycle, a heat engine operating between two heat reservoirs absorbs 150 J from the reservoir at 100ºC and releases 125 J to the reservoir at 20ºC. (a) What is the efficiency of this engine? (b) What is the ratio of its efficiency to that of a Carnot engine working between the same reservoirs? (This ratio is called the second law efficiency.) Picture the Problem We can use their definitions to find the efficiency of the engine and that of a Carnot engine operating between the same reservoirs. Q W Qh − Qc = = 1− c Qh Qh Qh

(a) The efficiency of the engine is given by:

ε=

Substitute numerical values and evaluate ε:

ε = 1−

(b) Find the efficiency of a Carnot engine operating between the same reservoirs:

εC = 1−

Express the ratio of the two efficiencies:

ε 16.67% = = 0.777 ε C 21.45%

125 J = 16.67% = 16.7% 150 J Tc 293 K = 1− = 21.45% Th 373 K

67 • [SSM] An engine absorbs 200 kJ of heat per cycle from a reservoir at 500 K and releases heat to a reservoir at 200 K. Its efficiency is 85 percent of that of a Carnot engine working between the same reservoirs. (a) What is the

1834 Chapter 19 efficiency of this engine? (b) How much work is done in each cycle? (c) How much heat is released to the low-temperature reservoir during each cycle? Picture the Problem We can use the definition of efficiency to find the work done by the engine during each cycle and the first law of thermodynamics to find the heat released to the low-temperature reservoir during each cycle.

(a) Express the efficiency of the engine in terms of the efficiency of a Carnot engine working between the same reservoirs: Substitute numerical values and evaluate ε :

⎛

ε = 0.85ε C = 0.85⎜⎜1 − ⎝

⎛

ε = 0.85⎜⎜1 − ⎝

Tc Th

⎞ ⎟⎟ ⎠

200 K ⎞ ⎟ = 0.510 = 51% 500 K ⎟⎠

(b) Use the definition of efficiency to find the work done in each cycle:

W = ε Qh = (0.510)(200 kJ ) = 102 kJ

(c) Apply the first law of thermodynamics to the cycle to obtain:

Qc,cycle = Qh,cycle − W = 200 kJ − 102 kJ

= 0.10 MJ

= 98 kJ

68 • Estimate the change in entropy of the universe associated with an Olympic diver diving into the water from the 10-m platform. Picture the Problem Assume that the mass of the diver is 75 kg and that the temperature of the water in the pool is 25°C. The energy added to the water in the pool is the change in the gravitational potential energy of the diver during the dive.

The change in entropy of the universe associated with a dive is given by:

Qadded to water Twater where Qadded to water is the energy entering

ΔS u = ΔS water =

the water as a result of the kinetic energy of the diver as he enters the water. The energy added to the water is the change in the gravitational potential energy of the diver:

ΔS u =

mgh Twater

The Second Law of Thermodynamics 1835 Substitute numerical values and evaluate ΔS u :

( 75 kg ) (9.81 m/s 2 )(10 m ) ΔS u = (25 + 273)K ≈ 25

J K

69 • To maintain the temperature inside a house at 20ºC, the electric power consumption of the electric baseboard heaters is 30.0 kW on a day when the outside temperature is –7ºC. At what rate does this house contribute to the increase in the entropy of the universe? Picture the Problem The change in entropy of the universe is the change in entropy of the house plus the change in entropy of the environment. We can find the change in entropy of the house by exploiting the given information that the temperature inside the house is maintained at a constant temperature. We can find the change in entropy of the surrounding by dividing the heat added by the temperature.

Entropy is a state function, and the state of the house does not change. Therefore the entropy of the house does not change: Heat is absorbed by the surroundings at the same rate R that energy is delivered to the house: Substitute for ΔSsurroundings yields:

Substitute numerical values and evaluate ΔSu/Δt:

ΔS u = ΔS house + ΔS surroundings or, because ΔS house = 0 , ΔS u = ΔS surroundings

ΔS surroundings =

ΔS u =

Q Tsurroundings

RΔt Tsurroundings

⇒

=

RΔt Tsurroundings

ΔS u R = Δt Tsurroundings

ΔS u 30.0 kW W = = 113 K Δt 266 K

70 •• Calvin Cliffs Nuclear Power Plant, located on the Hobbes River, generates 1.00 GW of power. In this plant, liquid sodium circulates between the reactor core and a heat exchanger located in the superheated steam that drives the turbine. Heat is absorbed by the liquid sodium in the core, and released by the liquid sodium (and into the superheated steam) in the heat exchanger. The temperature of the superheated steam is 500 K. Heat is released into the river, and the water in the river flows by at a temperature of 25ºC. (a) What is the highest efficiency that this plant can have? (b) How much heat is released into the river every second? (c) How much heat must be released by the core to supply

1836 Chapter 19 1.00 GW of electrical power? (d) Assume that new environmental laws have been passed to preserve the unique wildlife of the river. Because of these laws, the plant is not allowed to heat the river by more than 0.50ºC. What is the minimum flow rate that the water in the Hobbes River must have? Picture the Problem We can use the expression for the Carnot efficiency of the plant to find the highest efficiency this plant can have. We can then use this efficiency to find the power that must be supplied to the plant to generate 1.00 GW of power and, from this value, the power that is wasted. The rate at which heat is being released to the river is related to the requisite flow rate of the river by dQ dt = cΔTρ dV dt .

(a) The Carnot efficiency of a plant operating between temperatures Tc and Th is given by:

ε max = ε C = 1 −

Substitute numerical values and evaluate εC:

ε max = 1 −

298 K = 0.404 500 K

Psupplied =

Poutput

(c) The power that must be supplied, at 40.4% efficiency, to produce an output of 1.00 GW is given by:

ε max

Tc Th

=

1.00 GW 0.404

= 2.48 GW

(b) Relate the wasted power to the power generated and the power supplied:

Pwasted = Psupplied − Pgenerated

Substitute numerical values and evaluate Pwasted :

Pwasted = 2.48 GW − 1.00 GW

(d) Express the rate at which heat is being dumped into the river:

dQ dm d = cΔT = cΔT (ρV ) dt dt dt dV = cΔTρ dt

Solve for the flow rate dV/dt of the river:

dV dQ dt = dt cΔTρ

= 1.48 GW

The Second Law of Thermodynamics 1837 Substitute numerical values (see Table 19-1 for the specific heat of water) and evaluate dV/dt:

J dV s = dt ⎛ J ⎞ kg ⎞ ⎛ ⎜⎜ 4180 ⎟⎟ (0.50 K )⎜10 3 3 ⎟ kg ⎠ m ⎠ ⎝ ⎝ 1.48 × 10 9

= 7.1× 10 5 L/s 71 •• An inventor comes to you to explain his new invention. It is a novel heat engine using water vapor as the working substance. He claims that the water vapor absorbs heat at 100°C, does work at the rate of 125 W, and releases heat to the air at the rate of only 25.0 W, when the air temperature is 25°C. (a) Explain to him why he cannot be correct. (b) After careful analysis of the data in his prospectus folder, you decide he has made an error in the measurement of his exhausted-heat value. What is the minimum rate of exhausting heat that would make you consider believing him? Picture the Problem We can use the inventor’s data to calculate the thermal efficiency of his steam engine and then compare this value to the efficiency of a Carnot engine operating between the same temperatures.

(a) The Carnot efficiency of an engine operating between these temperatures is: The thermal efficiency of the inventor’s device, in terms of the rate at which it expels heat to the air and does work is:

ε C = 1−

Tc 298 K = 1− = 20.1% Th 373 K

dW dW dt ε = dt = dQh dW dQc + dt dt dt 125 W = = 83.3% 125 W + 25.0 W

You should explain to him that, because the efficiency he claims for his invention is greater than the efficiency of a Carnot engine operating between the same two temperatures, his data is not consistent with what is known about the thermodynamics of engines. He must have made a mistake in his analysis of his data−or he is a con man looking for suckers to swindle. (b) The maximum efficiency of a steam engine that has ever been achieved is about 50% of the Carnot efficiency of an engine operating between the same temperatures. Setting the efficiency of his steam engine equal to half the Carnot efficiency of the engine yields:

dW dW dt 1 ε = dt = 2 C dQh dW dQc + dt dt dt

1838 Chapter 19 Solve for dQc/dt to obtain:

⎞ dW dQc ⎛ 2 = ⎜⎜ − 1⎟⎟ dt ⎝ ε C ⎠ dt

Assuming that the inventor has measured the work done per cycle by his invention correctly:

dQc ⎛ 2 ⎞ =⎜ − 1⎟ (125 W ) ≈ 1100 W dt ⎝ 0.201 ⎠ a value totally inconsistent with the inventor’s claims for his engine.

Ignoring his claim that 125.0 W of work are done per cycle, let’s assume that his device does take in energy at the rate of 150 W each cycle and find how much work it would do with an efficiency half that of a Carnot engine:

dW dW 1 dQh 1 ε = dt ⇒ = 2 εC 2 C dQh dt dt dt

Substituting numerical values yields:

dW = dt

dQc dQh dW = − ,a dt dt dt reasonable value for dQc/dt is:

Because

1 2

(0.201)(150 W ) ≈ 15 W

dQc = 150 W − 15 W = 135 W dt

72 •• The cycle represented in Figure 19-12 (next to Problem 19-14) is for 1.00 mol of an ideal monatomic gas. The temperatures at points A and B are 300 and 750 K, respectively. What is the efficiency of the cyclic process ABCDA? Picture the Problem Because the cycle represented in Figure 19-12 is a Carnot cycle, its efficiency is that of a Carnot engine operating between the temperatures of its isotherms.

The Carnot efficiency of the cycle is given by:

εC = 1−

Tc Th

Substitute numerical values and evaluate εC:

εC = 1−

300 K = 60.0% 750 K

73 •• [SSM] (a) Which of these two processes is more wasteful? (1) A block moving with 500 J of kinetic energy being slowed to rest by sliding (kinetic) friction when the temperature of the environment is 300 K, or (2) A reservoir at 400 K releasing 1.00 kJ of heat to a reservoir at 300 K? Explain your choice. Hint: How much of the 1.00 kJ of heat could be converted into work by an ideal cyclic process? (b) What is the change in entropy of the universe for each process?

The Second Law of Thermodynamics 1839 Picture the Problem All 500 J of mechanical energy are lost, i.e., transformed into heat in process (1). For process (2), we can find the heat that would be converted to work by a Carnot engine operating between the given temperatures and subtract that amount of work from 1.00 kJ to find the energy that is lost. In Part (b) we can use its definition to find the change in entropy for each process.

(a) For process (2):

W2,max = Wrecovered = ε CQin

The efficiency of a Carnot engine operating between temperatures Th and Tc is given by:

ε C = 1−

Tc Th

and hence ⎛ T Wrecovered = ⎜⎜1 − c ⎝ Th

Substitute for ε C to obtain:

⎞ ⎟⎟Qin ⎠

⎛ 300 K ⎞ Wrecovered = ⎜1 − ⎟(1.00 kJ ) = 250 J ⎝ 400 K ⎠ or 750 J are lost.

Process (1) produces more waste heat. Process (2) is more wasteful of available work. ΔQ 500 J = = 1.67 J/K T 300 K

(b) Find the change in entropy of the universe for process (1):

ΔS1 =

Express the change in entropy of the universe for process (2):

ΔS 2 = ΔS h + ΔSc = −

ΔQ ΔQ + Th Tc

⎛1 1⎞ = ΔQ ⎜⎜ − ⎟⎟ ⎝ Tc Th ⎠ Substitute numerical values and evaluate ΔS2:

⎛ 1 1 ⎞ ⎟⎟ − ΔS 2 = (1.00 kJ )⎜⎜ ⎝ 300 K 400 K ⎠ = 0.833 J/K

Helium, a monatomic gas, is initially at a pressure of 16 atm, a volume 74 •• of 1.0 L, and a temperature of 600 K. It is quasi-statically expanded at constant temperature until its volume is 4.0 L and is then quasi-statically compressed at constant pressure until its volume and temperature are such that a quasi-static adiabatic compression will return the gas to its original state. (a) Sketch this

1840 Chapter 19 cycle on a PV diagram. (b) Find the volume and temperature after the compression at constant pressure. (c) Find the work done during each step of the cycle. (d) Find the efficiency of the cycle. Picture the Problem Denote the three states of the gas as 1, 2, and 3 with 1 being the initial state. We can use the ideal-gas law and the equation of state for an adiabatic process to find the temperatures, volumes, and pressures at points 1, 2, and 3. To find the work done during each cycle, we can use the equations for the work done during isothermal, isobaric, and adiabatic processes. Finally, we find the efficiency of the cycle from the work done each cycle and the heat that enters the system during the isothermal expansion.

(a) The PV diagram of the cycle is shown to the right.

(b) Apply the ideal-gas law to the isothermal expansion 1→2 to find P2:

P2 = P1

⎛ 1.0 L ⎞ V1 ⎟⎟ = 4.0 atm = (16 atm )⎜⎜ V2 ⎝ 4.0 L ⎠ 1γ

Apply an equation for an adiabatic process to relate the pressures and volumes at 1 and 3:

⎛P⎞ P1V1 = P3V3 ⇒ V3 = V1 ⎜⎜ 1 ⎟⎟ ⎝ P3 ⎠

Substitute numerical values and evaluate V3:

⎛ 16 atm ⎞ ⎟⎟ V3 = (1.0 L )⎜⎜ ⎝ 4.0 atm ⎠

γ

γ

1 1.67

= 2.294 L

= 2.3 L Apply an equation for an adiabatic process (γ =1.67) to relate the temperatures and volumes at 1 and 3:

γ −1

T3V3

γ −1

= T1V1

⎛V ⎞ ⇒ T3 = T1 ⎜⎜ 1 ⎟⎟ ⎝ V3 ⎠

γ −1

The Second Law of Thermodynamics 1841 Substitute numerical values and evaluate T3:

1.67 −1

⎛ 1.0 L ⎞ ⎟⎟ T3 = (600 K )⎜⎜ ⎝ 2.294 L ⎠

= 344 K

= 3.4 ×10 2 K (c) Express the work done each cycle:

W = W12 + W23 + W31

For the process 1→2:

⎛V ⎞ ⎛V ⎞ W12 = nRT1 ln⎜⎜ 2 ⎟⎟ = P1V1 ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠ ⎝ V1 ⎠ ⎛ 4.0 L ⎞ ⎟⎟ = (16 atm )(1.0 L )ln⎜⎜ ⎝ 1.0 L ⎠

(1)

= 22.18 atm ⋅ L For the process 2→3:

W23 = P2 ΔV23

= (4.0 atm )(2.294 L − 4.00 L ) = −6.824 atm ⋅ L

For the process 3→1: W31 = −C V ΔT31 = − 32 nR(T1 − T3 ) = − 32 (P1V1 − P3V3 ) = − 32 [(16 atm )(1.0 L ) − (4.0 atm )(2.294 L )] = −10.24 atm ⋅ L Substitute numerical values in equation (1) and evaluate W:

W = 22.18 atm ⋅ L − 6.824 atm ⋅ L − 10.24 atm ⋅ L = 5.116 atm ⋅ L = 5 atm ⋅ L

(d) Use its definition to express the efficiency of the cycle:

ε=

W W W = = Qin Q12 W12

Substitute numerical values and evaluate ε:

ε=

5.116 atm ⋅ L ≈ 20% 22.18 atm ⋅ L

75 •• [SSM] A heat engine that does the work of blowing up a balloon at a pressure of 1.00 atm absorbs 4.00 kJ from a reservoir at 120ºC. The volume of the balloon increases by 4.00 L, and heat is released to a reservoir at a temperature Tc, where Tc < 120ºC. If the efficiency of the heat engine is 50% of the efficiency of a Carnot engine working between the same two reservoirs, find the temperature Tc.

1842 Chapter 19 Picture the Problem We can express the temperature of the cold reservoir as a function of the Carnot efficiency of an ideal engine and, given that the efficiency of the heat engine is half that of a Carnot engine, relate Tc to the work done by and the heat input to the real heat engine.

Using its definition, relate the efficiency of a Carnot engine working between the same reservoirs to the temperature of the cold reservoir:

εC = 1−

Relate the efficiency of the heat engine to that of a Carnot engine working between the same temperatures:

ε=

Substitute for ε C to obtain:

⎛ 2W Tc = Th ⎜⎜1 − ⎝ Qin

The work done by the gas in expanding the balloon is:

Tc ⇒ Tc = Th (1 − ε C ) Th

2W W = 12 ε C ⇒ ε C = Qin Qin

⎞ ⎟⎟ ⎠

W = PΔV = (1.00 atm )(4.00 L ) = 4.00 atm ⋅ L

Substitute numerical values and evaluate Tc: ⎛ 101.325 J ⎞ ⎞ ⎛ ⎜ 2 ⎜ 4.00 atm ⋅ L × ⎟⎟ atm ⋅ L ⎠ ⎟ ⎝ ⎜ Tc = (393 K ) 1 − = 313 K ⎜ ⎟ 4.00 kJ ⎜ ⎟ ⎝ ⎠ Show that the coefficient of performance of a Carnot engine run as a 76 •• refrigerator is related to the efficiency of a Carnot engine operating between the same two temperatures by ε C × COPC = Tc Th . Picture the Problem We can use the definitions of the COP and εC to show that their relationship is ε C × COPC = TC Th .

Using the definition of the COP, relate the heat removed from the cold reservoir to the work done each cycle:

COP =

Qc W

The Second Law of Thermodynamics 1843 Apply energy conservation to relate Qc, Qh, and W:

Qc = Qh − W

Substitute for Qc to obtain:

COP =

Divide the numerator and denominator by Qh and simplify to obtain:

W Q −W Qh COP = h = W W Qh

Because ε C =

T W = 1− c : Qh Th

Qh − W W 1−

⎛ T 1 − ⎜⎜1 − c Th 1− ε C = ⎝ COPc =

εC

εC

⎞ Tc ⎟⎟ ⎠ = Th

εC

and

ε C × COPc =

Tc Th

77 •• A freezer has a temperature Tc = –23ºC. The air in the kitchen has a temperature Th = 27ºC. The freezer is not perfectly insulated and some heat leaks through the walls of the freezer at a rate of 50 W. Find the power of the motor that is needed to maintain the temperature in the freezer. Picture the Problem We can use the definition of the COP to express the work the motor must do to maintain the temperature of the freezer in terms of the rate at which heat flows into the freezer. Differentiation of this expression with respect to time will yield an expression for the power of the motor that is needed to maintain the temperature in the freezer.

Using the definition of the COP, relate the heat that must be removed from the freezer to the work done by the motor:

COP =

Differentiate this expression with respect to time to express the power of the motor:

P=

Express the maximum COP of the motor:

COPmax =

Q Qc ⇒W = c W COP

dW dQc dt = dt COP

Tc ΔT

1844 Chapter 19

dQc ΔT dt Tc

Substitute for COPmax to obtain:

P=

Substitute numerical values and evaluate P:

⎛ 50 K ⎞ ⎟⎟ = 10 W P = (50 W )⎜⎜ ⎝ 250 K ⎠

78 •• In a heat engine, 2.00 mol of a diatomic gas are taken through the cycle ABCA as shown in Figure 19-20. (The PV diagram is not drawn to scale.) At A the pressure and temperature are 5.00 atm and 600 K. The volume at B is twice the volume at A. The segment BC is an adiabatic expansion and the segment CA is an isothermal compression. (a) What is the volume of the gas at A? (b) What are the volume and temperature of the gas at B? (c) What is the temperature of the gas at C? (d) What is the volume of the gas at C? (e) How much work is done by the gas in each of the three segments of the cycle? (f) How much heat is absorbed or released by the gas in each segment of this cycle? Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points A, B, and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the constantpressure, adiabatic, and isothermal processes of the cycle.

(a) Apply the ideal-gas law to find the volume of the gas at A:

VA =

nRTA PA

(2.00 mol)⎛⎜ 8.314

J ⎞ ⎟ (600 K ) mol ⋅ K ⎠ ⎝ = 101.325 kPa 5.00 atm × atm 1L = 1.969 ×10 − 2 m 3 × −3 3 = 19.69 L 10 m = 19.7 L (b) We’re given that VB = 2VA . Hence:

VB = 2(19.69 L ) = 39.38 L = 39.4 L

Apply the ideal-gas law to this constant-pressure process to obtain:

TB = TA

VB 2V = (600 K ) A VA VA

= 1200 K

The Second Law of Thermodynamics 1845 (c) Because the process C→A is isothermal:

TC = TA = 600 K

(d) Apply an equation for an adiabatic process (γ = 1.4) to find the volume of the gas at C:

⎛ T ⎞ γ −1 TBVBγ −1 = TCVCγ −1 ⇒ VC = VB ⎜⎜ B ⎟⎟ ⎝ TC ⎠

Substitute numerical values and evaluate VC:

⎛ 1200 K ⎞ 1.4 −1 ⎟⎟ = 222.77 L VC = (39.38 L )⎜⎜ ⎝ 600 K ⎠

1

1

= 223 L (e) The work done by the gas during the constant-pressure process AB is given by: Substitute numerical values and evaluate WAB:

WAB = PA (VB − VA ) = PA (2VA − VA ) = PAVA

WAB = (5.00 atm )(19.69 L ) = 98.45 atm ⋅ L ×

101.325 J atm ⋅ L

= 9.9754 ×10 3 J = 9.98 kJ Apply the first law of thermodynamics to express the work done on the gas during the adiabatic expansion BC:

WBC = ΔEint, BC − Qin, BC = ΔEint, BC − 0 = ΔEint, BC = −ncV ΔTBC = − 52 nRΔTBC

Substitute numerical values and evaluate WBC: J ⎞ ⎛ 4 WBC = − 52 (2.00 mol)⎜ 8.314 ⎟ (600 K − 1200 K ) = 2.494 ×10 J mol ⋅ K ⎠ ⎝ = 24.9 kJ The work done by the gas during the isothermal compression CA is: ⎛V WCA = nRTC ln⎜⎜ A ⎝ VC

⎞ ⎛ 19.69 L ⎞ J ⎞ ⎟⎟ = (2.00 mol)⎛⎜ 8.314 ⎟⎟ ⎟ (600 K ) ln⎜⎜ mol ⋅ K ⎠ ⎝ ⎝ 222.77 L ⎠ ⎠

= −24.20 kJ = − 24.2 kJ

1846 Chapter 19 (f) The heat absorbed during the constant-pressure expansion AB is:

QAB = ncP ΔTA − B = 72 nRΔTA − B =

7 2

(2.00 mol)⎛⎜ 8.314 ⎝

J ⎞ ⎟ (1200 K − 600 K ) mol ⋅ K ⎠

= 34.92 kJ = 34.9 kJ The heat absorbed during the adiabatic expansion BC is:

QBC = 0

Use the first law of thermodynamics to find the heat absorbed during the isothermal compression CA:

QCA = WCA + ΔEint, CA = WCA = − 24.2 kJ because ΔEint, CA = 0 for an isothermal process.

79 •• [SSM] In a heat engine, 2.00 mol of a diatomic gas are carried through the cycle ABCDA shown in Figure 19-21. (The PV diagram is not drawn to scale.) The segment AB represents an isothermal expansion, the segment BC an adiabatic expansion. The pressure and temperature at A are 5.00 atm and 600 K. The volume at B is twice the volume at A. The pressure at D is 1.00 atm. (a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total work done by the gas in one cycle. Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D. We can then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle.

(a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB to find the pressure at B:

PB = PA

VA V = (5.00 atm ) A VB 2VA

= 2.50 atm ×

101.325 kPa = 253.3 kPa 1atm

= 253 kPa (b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC to express the temperature at C:

TC = TB

PCVC PBVB

(1)

The Second Law of Thermodynamics 1847 Use the ideal-gas law to find the volume of the gas at B:

VB =

nRTB PB

(2.00 mol)⎛⎜ 8.314

J ⎞ ⎟ (600 K ) mol ⋅ K ⎠ ⎝ 253.3 kPa

= = 39.39 L 1γ

Use the equation of state for an adiabatic process and γ = 1.4 to find the volume occupied by the gas at C:

⎛P ⎞ VC = VB ⎜⎜ B ⎟⎟ ⎝ PC ⎠ = 75.78 L

Substitute numerical values in equation (1) and evaluate TC:

TC = (600 K )

1 1.4

⎛ 2.50 atm ⎞ ⎟⎟ = (39.39 L )⎜⎜ 1.00 atm ⎝ ⎠

(1.00 atm )(75.78 L) (2.50 atm )(39.39 L )

= 462 K (c) The work done by the gas in one cycle is given by:

W = WAB + WBC + WCD + WDA

The work done during the isothermal expansion AB is: ⎛V WAB = nRTA ln⎜⎜ B ⎝ VA

⎞ ⎛ 2V J ⎞ ⎛ ⎟⎟ = (2.00 mol)⎜ 8.314 ⎟ (600 K ) ln⎜⎜ A mol ⋅ K ⎠ ⎝ ⎠ ⎝ VA

⎞ ⎟⎟ = 6.915 kJ ⎠

The work done during the adiabatic expansion BC is:

J ⎞ ⎛ WBC = −C V ΔTBC = − 52 nRΔTBC = − 52 (2.00 mol)⎜ 8.314 ⎟ (462 K − 600 K ) mol ⋅ K ⎠ ⎝ = 5.737 kJ The work done during the isobaric compression CD is: WCD = PC (VD − VC ) = (1.00 atm )(19.7 L − 75.78 L ) = −56.09 atm ⋅ L × = −5.680 kJ

Express and evaluate the work done during the constant-volume process DA:

WDA = 0

101.325 J atm ⋅ L

1848 Chapter 19 Substitute numerical values and evaluate W:

W = 6.915 kJ + 5.737 kJ − 5.680 kJ + 0 = 6.972 kJ = 6.97 kJ

80 •• In a heat engine, 2.00 mol of a monatomic gas are taken through the cycle ABCA as shown in Figure 19-20. (The PV diagram is not drawn to scale.) At A the pressure and temperature are 5.00 atm and 600 K. The volume at B is twice the volume at A. The segment BC is an adiabatic expansion and the segment CA is an isothermal compression. (a) What is the volume of the gas at A? (b) What are the volume and temperature of the gas at B? (c) What is the temperature of the gas at C? (d) What is the volume of the gas at C? (e) How much work is done by the gas in each of the three segments of the cycle? (f) How much heat is absorbed by the gas in each segment of the cycle? Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points A, B, and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the isobaric, adiabatic, and isothermal processes of the cycle.

(a) Apply the ideal-gas law to find the volume of the gas at A:

VA =

nRTA PA

(2.00 mol)⎛⎜ 8.314

J ⎞ ⎟ (600 K ) mol ⋅ K ⎠ ⎝ 101.325 kPa 5.00 atm × atm

=

= 19.69 L = 19.7 L (b) We’re given that:

VB = 2VA = 2(19.69 L ) = 39.39 L = 39.4 L VB 2V = (600 K ) A VA VA

Apply the ideal-gas law to this isobaric process to find the temperature at B:

TB = TA

(c) Because the process CA is isothermal:

TC = TA = 600 K

(d) Apply an equation for an adiabatic process (γ = 5/3) to express the volume of the gas at C:

= 1200 K

1

γ −1

TBVB

γ −1

= TCVC

⎛ T ⎞ γ −1 ⇒ VC = VB ⎜⎜ B ⎟⎟ ⎝ TC ⎠

The Second Law of Thermodynamics 1849 3

Substitute numerical values and evaluate VC:

⎛ 1200 K ⎞ 2 ⎟⎟ VC = (39.39 L )⎜⎜ 600 K ⎝ ⎠ = 111.4 L = 111 L WAB = PA (VB − VA ) = PA (2VA − VA )

(e) The work done by the gas during the isobaric process AB is given by:

= PAVA WAB = (5.00 atm )(19.69 L )

Substitute numerical values and evaluate WAB:

= 98.45 atm ⋅ L ×

101.325 J atm ⋅ L

= 9.975 kJ = 9.98 kJ Apply the first law of thermodynamics to express the work done by the gas during the adiabatic expansion BC:

WBC = ΔEint, BC − Qin, BC = ΔEint, BC − 0

= ΔEint, BC = −(ncV ΔTBC ) = − 32 nRΔTBC

Substitute numerical values and evaluate WBC: J ⎞ ⎛ WBC = − 32 (2.00 mol)⎜ 8.314 ⎟ (600 K − 1200 K ) = 14.97 kJ mol ⋅ K ⎠ ⎝ = 15.0 kJ The work done by the gas during the isothermal compression CA is: ⎛V WCA = nRTC ln⎜⎜ A ⎝ VC

⎞ ⎛ 19.69 L ⎞ J ⎞ ⎟⎟ = (2.00 mol)⎛⎜ 8.314 ⎟⎟ ⎟ (600 K ) ln⎜⎜ mol ⋅ K ⎠ ⎝ ⎝ 111.4 L ⎠ ⎠

= −17.29 kJ − 17.3 kJ

(f) The heat absorbed during the isobaric expansion AB is: Qin, AB = ncP ΔTAB = 52 nRΔTAB = = 24.9 kJ

5 2

(2.00 mol)⎛⎜ 8.314 ⎝

J ⎞ ⎟ (1200 K − 600 K ) mol ⋅ K ⎠

1850 Chapter 19 Express and evaluate the heat absorbed during the adiabatic expansion BC:

QBC = 0

Use the first law of thermodynamics to express and evaluate the heat absorbed during the isothermal compression CA:

QCA = WCA + ΔEint, CA = WCA = − 17.3 kJ because ΔEint = 0 for an isothermal process.

81 •• In a heat engine, 2.00 mol of a monatomic gas are carried through the cycle ABCDA shown in Figure 19-21. (The PV diagram is not drawn to scale.) The segment AB represents an isothermal expansion, the segment BC an adiabatic expansion. The pressure and temperature at A are 5.00 atm and 600 K. The volume at B is twice the volume at A. The pressure at D is 1.00 atm. (a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total work done by the gas in one cycle. Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle.

(a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB:

PB = PA

VA V = (5.00 atm ) A 2VA VB

= 2.50 atm ×

101.325 kPa 1atm

= 253.3 kPa = 253 kPa (b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC:

TC = TB

Use the ideal-gas law to find the volume at B:

VB =

PCVC PBVB

(1)

nRTB PB

(2.00 mol)⎛⎜ 8.314 = = 39.39 L

J ⎞ ⎟ (600 K ) mol ⋅ K ⎠ ⎝ 253.3 kPa

The Second Law of Thermodynamics 1851 1γ

Use the equation of state for an adiabatic process and γ = 5/3 to find the volume occupied by the gas at C:

⎛P ⎞ VC = VB ⎜⎜ B ⎟⎟ ⎝ PC ⎠ = 68.26 L

Substitute numerical values in equation (1) and evaluate TC:

TC = (600 K )

⎛ 2.50 atm ⎞ ⎟⎟ = (39.39 L )⎜⎜ 1.00 atm ⎝ ⎠

(1.00 atm )(68.26 L ) (2.50 atm )(39.39 L )

= 415.9 K = 416 K (c) The work done by the gas in one cycle is given by:

W = WAB + WBC + WCD + WDA

(2)

The work done during the isothermal expansion AB is: ⎛V WAB = nRTA ln⎜⎜ B ⎝ VA

⎞ ⎛ 2V J ⎞ ⎛ ⎟⎟ = (2.00 mol)⎜ 8.314 ⎟ (600 K ) ln⎜⎜ A mol ⋅ K ⎠ ⎝ ⎠ ⎝ VA

⎞ ⎟⎟ = 6.915 kJ ⎠

The work done during the adiabatic expansion BC is: WBC = −C V ΔTBC = − 32 nRΔTBC J ⎞ ⎛ = − 32 (2.00 mol)⎜ 8.314 ⎟ (415.9 K − 600 K ) mol ⋅ K ⎠ ⎝ = 4.592 kJ The work done during the isobaric compression CD is: WCD = PC (VD − VC ) = (1.00 atm )(19.7 L − 68.26 L ) = −48.56 atm ⋅ L ×

101.325 J atm ⋅ L

= −4.920 kJ

The work done during the constantvolume process DA is:

WDA = 0

Substitute numerical values in equation (2) to obtain:

W = 6.915 kJ + 4.592 kJ − 4.920 kJ + 0 = 6.59 kJ

82 •• Compare the efficiency of the Otto cycle to the efficiency of the Carnot cycle operating between the same maximum and minimum temperatures. (The Otto cycle is discussed in Section 19-1.)

35

1852 Chapter 19 Picture the Problem We can express the efficiency of the Otto cycle using the result from Example 19-2. We can apply the relation TV γ −1 = constant to the adiabatic processes of the Otto cycle to relate the end-point temperatures to the volumes occupied by the gas at these points and eliminate the temperatures at c and d. We can use the ideal-gas law to find the highest temperature of the gas during its cycle and use this temperature to express the efficiency of a Carnot engine. Finally, we can compare the efficiencies by examining their ratio.

The efficiency of the Otto engine is given in Example 19-2:

ε Otto = 1 −

Td − Ta Tc − Tb

(1)

where the subscripts refer to the various points of the cycle as shown in Figure 19-3. Apply the relation TV γ −1 = constant to the adiabatic process a→b to obtain:

⎛V ⎞ Tb = Ta ⎜⎜ a ⎟⎟ ⎝ Vb ⎠

γ −1

Apply the relation TV γ −1 = constant to the adiabatic process c→d to obtain:

⎛V Tc = Td ⎜⎜ d ⎝ Vc

γ −1

Subtract the first of these equations from the second to obtain:

⎛V Tc − Tb = Td ⎜⎜ d ⎝ Vc

In the Otto cycle, Va = Vd and Vc = Vb. Substitute to obtain:

⎛V ⎞ Tc − Tb = Td ⎜⎜ a ⎟⎟ ⎝ Vb ⎠

⎞ ⎟⎟ ⎠

γ −1

⎛V ⎞ − Ta ⎜⎜ a ⎟⎟ ⎝ Vb ⎠

γ −1

⎛V ⎞ − Ta ⎜⎜ a ⎟⎟ ⎝ Vb ⎠

⎞ ⎟⎟ ⎠

⎛V ⎞ = (Td − Ta )⎜⎜ a ⎟⎟ ⎝ Vb ⎠ Substitute in equation (1) and simplify to obtain:

ε Otto = 1 −

γ −1

γ −1

γ −1

Td − Ta ⎛ (Td − Ta )⎜⎜ Va ⎝ Vb

γ −1

⎞ ⎟⎟ ⎠

γ −1

⎛V ⎞ T = 1 − ⎜⎜ b ⎟⎟ = 1 − a Tb ⎝ Va ⎠ Note that, while Ta is the lowest temperature of the cycle, Tb is not the highest temperature.

The Second Law of Thermodynamics 1853 Apply the ideal-gas law to c and b to obtain an expression for the cycle’s highest temperature Tc:

P Pc Pb = ⇒ Tc = Tb c > Tb Pb Tc Tb

The efficiency of a Carnot engine operating between the maximum and minimum temperatures of the Otto cycle is given by:

ε Carnot = 1 −

Express the ratio of the efficiency of a Carnot engine to the efficiency of an Otto engine operating between the same temperatures:

ε Carnot ε Otto

Ta Tc

Ta Tc > 1 because Tc > Tb. = Ta 1− Tb 1−

Hence, ε Carnot > ε Otto 83 ••• [SSM] A common practical cycle, often used in refrigeration, is the Brayton cycle, which involves (1) an adiabatic compression, (2) an isobaric (constant pressure) expansion,(3) an adiabatic expansion, and (4) an isobaric compression back to the original state. Assume the system begins the adiabatic compression at temperature T1, and transitions to temperatures T2, T3 and T4 after each leg of the cycle. (a) Sketch this cycle on a PV diagram. (b) Show that the (T − T ) efficiency of the overall cycle is given by ε = 1 − 4 1 . (c) Show that this (T3 − T2 )

efficiency, can be written as ε = 1 − r (1− γ ) γ , where r is the pressure ratio Phigh/Plow of the maximum and minimum pressures in the cycle. Picture the Problem The efficiency of the cycle is the ratio of the work done to the heat that flows into the engine. Because the adiabatic transitions in the cycle do not have heat flow associated with them, all we must do is consider the heat flow in and out of the engine during the isobaric transitions.

(a) The Brayton heat engine cycle is shown to the right. The paths 1→2 and 3→4 are adiabatic. Heat Qh enters the gas during the isobaric transition from state 2 to state 3 and heat Qc leaves the gas during the isobaric transition from state 4 to state 1.

Qh

P 2

⇓

3

1

⇓

Qc

4 V

1854 Chapter 19 (b) The efficiency of a heat engine is given by: During the constant-pressure expansion from state 1 to state 2 heat enters the system: During the constant-pressure compression from state 3 to state 4 heat enters the system: Substituting in equation (1) and simplifying yields:

ε=

Q − Qc W = h Qin Qin

(1)

Q23 = Qh = nC P ΔT = nC P (T3 − T2 )

Q41 = −Qc = −nC P ΔT = −nC P (T1 − T4 )

ε=

nC P (T3 − T2 ) − (− nC P (T1 − T4 )) nC P (T3 − T2 )

(T3 − T2 ) + (T1 − T4 ) (T3 − T2 ) (T − T ) = 1− 4 1 (T3 − T2 ) =

(c) Given that, for an adiabatic transition, TV γ −1 = constant , use the ideal-gas law to eliminate V and obtain:

Tγ = constant P γ −1

Let the pressure for the transition from state 1 to state 2 be Plow and the pressure for the transition from state 3 to state 4 be Phigh. Then for the adiabatic transition from state 1 to state 2:

⎛P T1γ T2γ ⇒ T1 = ⎜ low = γ −1 γ −1 ⎜P Plow Phigh ⎝ high

Similarly, for the adiabatic transition from state 3 to state 4:

⎛P T4 = ⎜ low ⎜P ⎝ high

Subtract T1 from T4 and simplify to obtain:

⎛P T4 − T1 = ⎜ low ⎜P ⎝ high

⎞ ⎟ ⎟ ⎠

⎛P = ⎜ low ⎜P ⎝ high

⎞ ⎟ ⎟ ⎠

⎞ ⎟ ⎟ ⎠

⎞ ⎟ ⎟ ⎠

γ −1 γ

T2

γ −1 γ

T3 γ −1 γ

γ −1 γ

⎛P T3 − ⎜ low ⎜P ⎝ high

(T3 − T2 )

⎞ ⎟ ⎟ ⎠

γ −1 γ

T2

The Second Law of Thermodynamics 1855 Dividing both sides of the equation by T3 − T2 yields:

Substitute in the result of Part (b) and simplify to obtain:

T4 − T1 ⎛⎜ Plow = T3 − T2 ⎜⎝ Phigh

⎛P ε = 1 − ⎜ low ⎜P ⎝ high

⎞ ⎟ ⎟ ⎠

⎞ ⎟ ⎟ ⎠

γ −1 γ

γ −1 γ

⎛ Phigh = 1 − ⎜⎜ ⎝ Plow

⎞ ⎟⎟ ⎠

1−γ

γ

1−γ

= 1 − (r ) γ where r =

Phigh Plow

84 ••• Suppose the Brayton cycle engine (see Problem 83) is run in reverse as a refrigerator in your kitchen. In this case, the cycle begins at temperature T1 and expands at constant pressure until its temperature T4. Then the gas is adiabatically compressed, until its temperature is T3. And then it is compressed at constant pressure until its temperature T2. Finally, it adiabatically expands until it returns to its initial state at temperature T1. (a) Sketch this cycle on a PV diagram. (T4 − T1 ) . (b) Show that the coefficient of performance is COPB = (T3 − T2 − T4 + T1 )

(c) Suppose your ″Brayton cycle refrigerator″ is run as follows. The cylinder containing the refrigerant (a monatomic gas) has an initial volume and pressure of 60 mL and 1.0 atm. After the expansion at constant pressure, the volume and temperature are 75 mL and –25°C. The pressure ratio r = Phigh/Plow for the cycle is 5.0. What is the coefficient of performance for your refrigerator? (d) To absorb heat from the food compartment at the rate of 120 W, what is the rate at which electrical energy must be supplied to the motor of this refrigerator? (e) Assuming the refrigerator motor is actually running for only 4.0 h each day, how much does it add to your monthly electric bill. Assume 15 cents per kWh of electric energy and thirty days in a month. Picture the Problem The efficiency of the Brayton refrigerator cycle is the ratio of the heat that enters the system to the work done to operate the refrigerator. Because the adiabatic transitions in the cycle do not have heat flow associated with them, all we must do is consider the heat flow in and out of the refrigerator during the isobaric transitions.

1856 Chapter 19 Qh

2

3

1

⇓

P

⇓

(a) The Brayton refrigerator cycle is shown to the right. The paths 1→2 and 3→4 are adiabatic. Heat Qc enters the gas during the constantpressure transition from state 1 to state 4 and heat Qh leaves the gas during the constant-pressure transition from state 3 to state 2.

4

Qc

(b) The coefficient of performance of the Brayton cycle refrigerator is given by: During the constant-pressure compression from state 3 to state 2 heat leaves the system: During the constant-pressure expansion from state 1 to state 4 heat enters the system: Substituting in equation (1) and simplifying yields:

Qc W where W = Q32 − Q14 COPB =

V (1)

Q32 = −Qh = −nC P ΔT = − nC P (T2 − T3 )

Q14 = Qc = nC P ΔT = nC P (T4 − T1 )

COPB =

= =

nCP (T4 − T1 ) − nCP (T2 − T3 ) − nCP (T4 − T1 )

(T4 − T1 ) − (T4 − T1 ) − (T2 − T3 ) T4 − T1 T3 − T2 − T4 + T1

(c) The COPB requires the temperatures corresponding to states 1, 2, 3, and 4. We’re given that the temperature in state 4 is:

T4 = −25°C + 273 K = 248 K

For the constant-pressure transition from state 1 to state 4, the quotient T/V is constant:

⎛V ⎞ T1 T4 = ⇒ T1 = ⎜⎜ 1 ⎟⎟T4 V1 V4 ⎝ V4 ⎠

Substitute numerical values and evaluate T1:

⎛ 60 mL ⎞ T1 = ⎜ ⎟ (248 K ) = 198 K ⎝ 75 mL ⎠

The Second Law of Thermodynamics 1857 Given that, for an adiabatic transition, TV γ −1 = constant , use the ideal-gas law to eliminate V and obtain: For the adiabatic transition from state 4 to state 3:

Substitute numerical values and evaluate T3: Similarly, for the adiabatic transition from state 2 to state 1:

Tγ = constant P γ −1

⎛P ⎞ T3γ T4γ ⇒ T3 = ⎜⎜ 3 ⎟⎟ = γ −1 γ −1 P3 P4 ⎝ P4 ⎠

T4

1.67 −1

T3 = (5) 1.67 (248 K ) = 473 K γ −1

1.67 −1 ⎛P ⎞ γ T2 = ⎜⎜ 2 ⎟⎟ T1 = (5) 1.67 (198 K ) ⎝ P1 ⎠ = 378 K

248 K − 198 K 473 K − 378 K − 248 K + 198 K

Substitute numerical values in the expression derived in Part (a) and evaluate COPB:

COPB =

(d) From the definition of COPB:

W=

The rate at which energy must be supplied to this refrigerator is given by:

1 dQc dW = dt COPB dt

Express the heat Qc that is drawn from the cold reservoir:

γ −1 γ

= 1.1 Qc COPB

or, if the frequency of the AC power input is f, fQc dW = dt COPB

Qc = nC P ΔT = nC P (T4 − T1 )

Substituting for Qc yields:

fnC P (T4 − T1 ) dW = COPB dt

Use the ideal-gas law to express the number of moles of the gas:

n=

P4V4 RT4

1858 Chapter 19 Because the gas is monatomic, C P = 52 R . Substitute for n and CP to obtain:

dW = dt =

5 2

5 2

f

P4V4 R(T4 − T1 ) RT4 COPB

fP4V4 (T4 − T1 ) (COPB )T4

Substitute numerical values and evaluate dW/dt:

dW = dt

5 2

⎛ 10 −3 m 3 ⎞ ⎜ ⎟⎟(248 K − 198 K ) (101.325 kPa )⎜ 75 mL × L ⎝ ⎠ = 207 W (1.11)(248 K )

(60 s ) −1

= 0.21 kW (e) The monthly cost of operation is given by Monthly Cost = Cost Per Unit of Power × Power Consumption

= rate × daily consumption × number of days per month Substitute numerical values and evaluate the monthly cost of operation: Monthly Cost =

$0.15 4.0 h × 0.207 kW × × 30 d ≈ $4 kWh d

85 ••• Using ΔS = CV ln (T2 T1 ) + nR ln (V2 V1 ) (Equation 19-16) for the entropy change of an ideal gas, show explicitly that the entropy change is zero for a quasi-static adiabatic expansion from state (V1, T1) to state (V2, T2). Picture the Problem We can use nR = CP − CV , γ = CP CV , and

TV γ −1 = a constant to show that the entropy change for a quasi-static adiabatic expansion that proceeds from state (V1,T1) to state (V2,T2) is zero. Express the entropy change for a general process that proceeds from state 1 to state 2:

⎛T ⎞ ⎛V ⎞ ΔS = CV ln⎜⎜ 2 ⎟⎟ + nR ln⎜⎜ 2 ⎟⎟ ⎝ T1 ⎠ ⎝ V1 ⎠

For an adiabatic process:

T2 ⎛ V1 ⎞ =⎜ ⎟ T1 ⎜⎝ V2 ⎟⎠

γ −1

The Second Law of Thermodynamics 1859

Substitute for

T2 and simplify to obtain: T1

γ −1 ⎡ ⎛ V1 ⎞ ⎤ ⎢ CV ln⎜⎜ ⎟⎟ ⎥ γ −1 ⎛ V1 ⎞ ⎛ V2 ⎞ ⎛ V2 ⎞ ⎢ ⎝ V2 ⎠ ⎥ ΔS = CV ln⎜⎜ ⎟⎟ + nR ln⎜⎜ ⎟⎟ = ln⎜⎜ ⎟⎟ ⎢nR + ⎥ V ⎝ V2 ⎠ ⎝ V1 ⎠ ⎝ V1 ⎠ ⎢ ln 2 ⎥ V1 ⎢⎣ ⎥⎦

⎡ ⎛ ⎞⎤ (γ − 1)CV ln⎜⎜ V1 ⎟⎟ ⎥ ⎢ ⎛V ⎞ ⎝ V2 ⎠ ⎥ = ln⎛⎜ V2 ⎞⎟ [nR − (γ − 1)C ] = ln⎜⎜ 2 ⎟⎟ ⎢nR + V ⎜V ⎟ V1 ⎥ ⎝ 1⎠ ⎝ V1 ⎠ ⎢ − ln ⎢ ⎥ V2 ⎢⎣ ⎥⎦ Use the relationship between CP and CV to obtain:

nR = CP − CV

Substituting for nR and γ and simplifying yields:

⎛C ⎞ ⎤ ⎛ V ⎞⎡ ΔS = ln⎜⎜ 2 ⎟⎟ ⎢CP − CV − ⎜⎜ p − 1⎟⎟CV ⎥ ⎝ V1 ⎠ ⎣ ⎝ CV ⎠ ⎦ = 0

86 ••• (a) Show that if the refrigerator statement of the second law of thermodynamics were not true, then the entropy of the universe could decrease. (b) Show that if the heat-engine statement of the second law were not true, then the entropy of the universe could decrease. (c) A third statement of the second law is that the entropy of the universe cannot decrease. Have you just proved that this statement is equivalent to the refrigerator and heat-engine statements? Picture the Problem (a) Suppose the refrigerator statement of the second law is violated in the sense that heat Qc is taken from the cold reservoir and an equal amount of heat is transferred to the hot reservoir and W = 0. The entropy change of the universe is then ΔSu = Qc/Th − Qc/Tc. Because Th > Tc, Su < 0, i.e., the entropy of the universe would decrease.

(b) In this case, heat Qh is taken from the hot reservoir and no heat is rejected to the cold reservoir; that is, Qc = 0, then the entropy change of the universe is ΔSu = −Qh/Th + 0, which is negative. Again, the entropy of the universe would decrease.

1860 Chapter 19 (c) The heat-engine and refrigerator statements of the second law only state that some heat must be rejected to a cold reservoir and some work must be done to transfer heat from the cold to the hot reservoir, but these statements do not specify the minimum amount of heat rejected or work that must be done. The statement ΔSu ≥ 0 is more restrictive. The heat-engine and refrigerator statements in conjunction with the Carnot efficiency are equivalent to ΔSu ≥ 0. 87 ••• Suppose that two heat engines are connected in series, such that the heat released by the first engine is used as the heat absorbed by the second engine as shown in Figure 19-22. The efficiencies of the engines are ε1 and ε2, respectively. Show that the net efficiency of the combination is given by ε net = ε 1 + ε 2 − ε 1ε 2 . Picture the Problem We can express the net efficiency of the two engines in terms of W1, W2, and Qh and then use ε1 = W1/Qh and ε2 = W2/Qm to eliminate W1, W2, Qh, and Qm.

W1 + W2 Qh

Express the net efficiency of the two heat engines connected in series:

ε net =

Express the efficiencies of engines 1 and 2:

ε1 =

Solve for W1 and W2 and substitute to obtain:

ε net =

Express the efficiency of engine 1 in terms of Qm and Qh:

ε1 = 1 −

Substitute for Qm/Qh and simplify to obtain:

ε net = ε 1 + (1 − ε 1 )ε 2 = ε 1 + ε 2 − ε 1ε 2

W1 W and ε 2 = 2 Qh Qm

ε1Qh + ε 2Qm Qh

= ε1 +

Qm ε2 Qh

Qm Q ⇒ m = 1 − ε1 Qh Qh

88 ••• Suppose that two heat engines are connected in series, such that the heat released by the first engine is used as the heat absorbed by the second engine, as shown in Figure 19-22. Suppose that each engine is an ideal reversible heat engine. Engine 1 operates between temperatures Th and Tm and Engine 2 operates between Tm and Tc, where Th > Tm > Tc. Show that the net efficiency of T the combination is given by ε net = 1 − c . (Note that this result means that two Th

The Second Law of Thermodynamics

1861

reversible heat engines operating ″in series″ are equivalent to one reversible heat engine operating between the hottest and coldest reservoirs.) Picture the Problem We can express the net efficiency of the two engines in terms of W1, W2, and Qh and then use ε1 = W1/Qh and ε2 = W2/Qm to eliminate W1, W2, Qh, and Qm. Finally, we can substitute the expressions for the efficiencies of the ideal reversible engines to obtain ε net = 1 − Tc Th .

Express the efficiencies of ideal reversible engines 1 and 2:

Tm Th

(1)

Tc Tm

(2)

W1 + W2 Qh

(3)

ε1 = 1 − and

ε2 = 1− The net efficiency of the two engines connected in series is given by:

ε net =

Express the efficiencies of engines 1 and 2:

ε1 =

Solve for W1 and W2 and substitute in equation (3) to obtain:

ε net =

Express the efficiency of engine 1 in terms of Qm and Qh:

ε1 = 1 −

Substitute for

Qm to obtain: Qh

Substitute for ε1 and ε2 and simplify to obtain:

W1 W and ε 2 = 2 Qh Qm

ε1Qh + ε 2Qm Qh

= ε1 +

Qm ε2 Qh

Q Qm ⇒ m = 1 − ε1 Qh Qh

ε net = ε1 + (1 − ε1 )ε 2

Tm ⎛ Tm ⎞ ⎛ Tc ⎞ + ⎜ ⎟ ⎜1 − ⎟ Th ⎜⎝ Th ⎟⎠ ⎜⎝ Tm ⎟⎠ T T T T = 1− m + m − c = 1− c Th Th Th Th

ε net = 1 −

89 ••• [SSM] The English mathematician and philosopher Bertrand Russell (1872-1970) once said that if a million monkeys were given a million typewriters and typed away at random for a million years, they would produce all of Shakespeare’s works. Let us limit ourselves to the following fragment of Shakespeare (Julius Caesar III:ii):

1862 Chapter 19 Friends, Romans, countrymen! Lend me your ears. I come to bury Caesar, not to praise him. The evil that men do lives on after them, The good is oft interred with the bones. So let it be with Caesar. The noble Brutus hath told you that Caesar was ambitious, And, if so, it were a grievous fault, And grievously hath Caesar answered it . . . Even with this small fragment, it will take a lot longer than a million years! By what factor (roughly speaking) was Russell in error? Make any reasonable assumptions you want. (You can even assume that the monkeys are immortal.) Picture the Problem There are 26 letters and four punctuation marks (space, comma, period, and exclamation point) used in the English language, disregarding capitalization, so we have a grand total of 30 characters to choose from. This fragment is 330 characters (including spaces) long; there are then 30330 different possible arrangements of the character set to form a fragment this long. We can use this number of possible arrangements to express the probability that one monkey will write out this passage and then an estimate of a monkey’s typing speed to approximate the time required for one million monkeys to type the passage from Shakespeare.

Assuming the monkeys type at random, express the probability P that one monkey will write out this passage: Use the approximation 30 ≈ 1000 = 101.5 to obtain: Assuming the monkeys can type at a rate of 1 character per second, it would take about 330 s to write a passage of length equal to the quotation from Shakespeare. Find the time T required for a million monkeys to type this particular passage by accident: Express the ratio of T to Russell’s estimate:

P=

P=

T=

1 30330

1

10

(1.5 )(330 )

=

1 = 10 −495 495 10

(330 s )(10495 ) 106

⎛ ⎞ 1y ⎟ = 3.30 × 10 491 s ⎜⎜ 7 ⎟ ⎝ 3.16 × 10 s ⎠

(

)

≈ 10484 y

T

=

10484 y = 10478 6 10 y

TRussell or T ≈ 10 478 TRussell

Chapter 20 Thermal Properties and Processes Conceptual Problems 1 • Why does the mercury level of a thermometer first decrease slightly when the thermometer is first placed in warm water? Determine the Concept The glass bulb warms and expands first, before the mercury warms and expands. 2 • A large sheet of metal has a hole cut in the middle of it. When the sheet is heated, the area of the hole will (a) not change, (b) always increase, (c) always decrease, (d) increase if the hole is not in the exact center of the sheet, (e) decrease only if the hole is in the exact center of the sheet. Determine the Concept The heating of the sheet causes the average separation of its molecules to increase. The consequence of this increased separation is that the area of the hole always increases. (b) is correct. 3 • [SSM] Why is it a bad idea to place a tightly sealed glass bottle that is completely full of water, into your kitchen freezer in order to make ice? Determine the Concept Water expands greatly as it freezes. If a sealed glass bottle full of water is placed in a freezer, as the water freezes there will be no room for the expansion to take place. The bottle will be broken. 4 • The windows of your physics laboratory are left open on a night when the temperature of the outside dropped well below freezing. A steel ruler and a wooden ruler were left on the window sill, and when you arrive in the morning they are both very cold. The coefficient of linear expansion of wood is about 5 × 10−6 K−1. Which ruler should you use to make the most accurate length measurements? Explain your answer. Determine the Concept You should use the wooden ruler. Because the coefficient of expansion for wood is about half that for metal, the metal ruler will have shrunk considerably more than will have the wooden ruler. 5 • Bimetallic strips are used both for thermostats and for electrical circuit breakers. A bimetallic strip consists of a pair of thin strips of metal that have different coefficients of linear expansion and are bonded together to form one doubly thick strip. Suppose a bimetallic strip is constructed out of one steel strip and one copper strip, and suppose the bimetallic strip is curled in the shape of a circular arc with the steel strip on the outside. If the temperature of the strip is decreased, will the strip straighten out or curl more tightly?

1863

1864

Chapter 20

Determine the Concept The strip will curl more tightly. Because the coefficient of linear expansion for copper (17 × 10−6 K−1) is greater than the coefficient of linear expansion for steel (11 × 10−6 K−1), the length of the copper strip will decrease more than the length of the steel strip−resulting in a tighter curl. 6 • Metal A has a coefficient of linear expansion that is three times the coefficient of linear expansion of metal B. How do their coefficients of volume expansion β compare? (a) β A = β B , (b) β A = 3β B , (c) β A = 6β B , (d) β A = 9β B , (e) You cannot tell from the data given. Determine the Concept We know that the coefficient of volume expansion is three times the coefficient of linear expansion and so can use this fact to express the ratio of β A to β B . Express the coefficient of volume expansion of metal A in terms of its coefficient of linear expansion:

β A = 3α A

Express the coefficient of volume expansion of metal B in terms of its coefficient of linear expansion:

β B = 3α B

Dividing the first of these equations by the second yields:

β A 3α A α A = = β B 3α B α B

Because α A = 3α B :

β A 3α B = = 3 ⇒ (b ) is correct. βB αB

7 • The summit of Mount Rainier is 14 410 ft above sea level. Mountaineers say that you cannot hard boil an egg at the summit. This statement is true because at the summit of Mount Rainier (a) the air temperature is too low to boil water, (b) the air pressure is too low for alcohol fuel to burn, (c) the temperature of boiling water is not hot enough to hard boil the egg, (d) the oxygen content of the air is too low to support combustion, (e) eggs always break in climbers′ backpacks. Determine the Concept Actually, an egg can be hard boiled, but it takes quite a bit longer than at sea level. (c) is the best response. 8 • Which gases in Table 20-3 cannot be condensed by applying pressure at 20ºC? Explain your answer. Determine the Concept Gases that cannot be liquefied by applying pressure at 20°C are those for which Tc < 293 K. These are He, Ar, Ne, H2, O2, NO.

Thermal Properties and Processes 1865 9 •• [SSM] The phase diagram in Figure 20-15 can be interpreted to yield information on how the boiling and melting points of water change with altitude. (a) Explain how this information can be obtained. (b) How might this information affect cooking procedures in the mountains? Determine the Concept (a) With increasing altitude, decreases; from curve OC, the temperature of the liquid-gas interface decreases as the pressure decreases, so the boiling temperature decreases. Likewise, from curve OB, the melting temperature increases with increasing altitude. (b) Boiling at a lower temperature means that the cooking time will have to be increased. 10 •• Sketch a phase diagram for carbon dioxide using information from Section 20-3. Determine the Concept The following phase diagram for carbon dioxide was constructed using information in Section 20-3. P , atm Gas

Critical point Liquid Gas

5.1 Solid

Vapor

216.6

304.2

T, K

11 •• Explain why the carbon dioxide on Mars is found in the solid state in the polar regions even though the atmospheric pressure at the surface of Mars is only about 1 percent of the atmospheric pressure at the surface of Earth. Determine the Concept At very low pressures and temperatures, carbon dioxide can exist only as a solid or gas (or vapor above the gas). The atmosphere of Mars is 95 percent carbon dioxide. Mars, on average, is warm enough so that the atmosphere is mostly gaseous carbon dioxide. The polar regions are cold enough to enable solid carbon dioxide (dry ice) to exist, even at the low pressure. 12 •• Explain why decreasing the temperature of your house at night in winter can save money on heating costs. Why doesn’t the cost of the fuel consumed to heat the house back to the daytime temperature in the morning equal

1866

Chapter 20

the savings realized by cooling it down in the evening and keeping it cool throughout the night? Determine the Concept The amount of heat lost by the house is proportional to the difference between the temperature inside the house and that of the outside air. Hence, the rate at which the house loses heat (that must be replaced by the furnace) is greater at night when the temperature of the house is kept high than when it is allowed to cool down. 13 •• [SSM] Two solid cylinders made of materials A and B have the same lengths; their diameters are related by dA = 2dB. When the same temperature difference is maintained between the ends of the cylinders, they conduct heat at the same rate. Their thermal conductivities are therefore related by which of the following equations? (a) kA = kB/4, (b) kA = kB/2, (c) kA = kB, (d) kA = 2kB, (e) kA = 4kB B

Picture the Problem The rate at which heat is conducted through a cylinder is given by I = dQ / dt = kAΔT / Δx (see Equation 20-7) where A is the crosssectional area of the cylinder. The heat current in cylinder A is the same as the heat current in cylinder B: Substituting for the heat currents yields: Because dA = 2dB:

IA = IB

k A AA

A ΔT ΔT ⇒ kA = kB B = k B AB L L AA

⎛ A ⎞ k A = k B ⎜⎜ B ⎟⎟ ⇒ k A = 14 k B ⎝ 4AB ⎠

(a) is correct. 14 •• Two solid cylinders made of materials A and B have the same diameter; their lengths are related by LA = 2LB. When the same temperature difference is maintained between the ends of the cylinders, they conduct heat at the same rate. Their thermal conductivities are therefore related by which of the following equations? (a) kA = kB/4, (b) kA = kB/2, (c) kA = kB, (d) kA = 2kB, (e) kA = 4kB. B

Determine the Concept We can use the expression for the heat current in a conductor, Equation 20-7, to relate the heat current in each cylinder to its thermal conductivity, cross-sectional area, temperature difference, and length.

Thermal Properties and Processes 1867 The heat current in cylinder A is the same as the heat current in cylinder B: Substituting for the heat currents yields: Because LA = 2LB:

IA = IB

kA A

ΔT ΔT L = kB A ⇒ kA = kB A LA LB LB

kA = kB

2 LB = 2k B ⇒ (d ) is correct. LB

15 •• If you feel the inside of a single pane window during a very cold day, it is cold, even though the room temperature can be quite comfortable. Assuming the room temperature is 20.0°C and the outside temperature is 5.0°C, Construct a plot of temperature versus position starting from a point 5.0 m in behind the window (inside the room) and ending at a point 5.0 m in front of the window. Explain the heat transfer mechanisms that occur along this path. Determine the Concept The temperatures on both sides of the glass are almost the same. Because glass is an excellent conductor of heat, there need not be a huge temperature difference. Thus the temperature must drop quickly as you near the pane on the warm side and the same on the outside. This is sketched qualitatively in the following diagram. Convection and radiation are primarily responsible for heat transfer on the inside and outside, and it is mainly conduction through the glass. Conduction through the interior and exterior air is minimal. T , °C

20

5 Inside

Outside

0

5

x, m

16 •• During the thermal retrofitting of many older homes in California, it was found that the 3.5-in-deep spaces between the wallboards and the outer sheathing were filled with just air (no insulation). Filling the space with insulating material certainly reduces heating and cooling costs; although, the insulating material is a better conductor of heat than air is. Explain why adding the insulation is a good idea. Determine the Concept The tradeoff is the reduction of convection cells between the walls by putting in the insulating material, versus a slight increase in conductivity. The net reduction in convection results in a higher R value.

1868 Chapter 20

Estimation and Approximation 17 •• [SSM] You are using a cooking pot to boil water for a pasta dish. The recipe calls for at least 4.0 L of water to be used. You fill the pot with 4.0 L of room temperature water and note that this amount of water filled the pot to the brim. Knowing some physics, you are counting on the volume expansion of the steel pot to keep all of the water in the pot while the water is heated to a boil. Is your assumption correct? Explain. If your assumption is not correct, how much water runs over the sides of the pot due to the thermal expansion of the water? Determine the Concept The volume of water overflowing is the difference between the change in volume of the water and the change in volume of the pot. See Table 20-1 for the coefficient of volume expansion of water and the coefficient of linear expansion of steel.

Express the volume of water that overflows when the pot and the water are heated:

Vovefrlow = ΔVH 2O − ΔVpot

Because the coefficient of volume expansion of steel is three times its coefficient of linear expansion:

β steel = 3α steel

Substituting for β H 2O and β steel yields:

Voverflow = α H2O − 3α steel V0ΔT

= β H 2OV0ΔT − β steelV0ΔT

(

)

= β H 2O − β steel V0 ΔT

(

)

Substitute numerical values and evaluate Voverflow :

(

(

))

Voverflow = 0.207 ×10 −3 K −1 − 3 11×10 − 6 K −1 (4.0 L )(100°C − 20°C) = 56 mL Your assumption was not correct and 56 mL of water overflowed. 18 •• Liquid helium is stored in containers fitted with 7.00-cm-thick ″superinsulation″ consisting of numerous layers of very thin aluminized Mylar sheets. The rate of evaporation of liquid helium in a 200-L container is about 0.700 L per day when the container is stored at room temperature (20ºC). The density of liquid helium is 0.125 kg/L and the latent heat of vaporization is 21.0 kJ/kg. Estimate the thermal conductivity of the superinsulation. Picture the Problem We can express the heat current through the insulation in terms of the rate of evaporation of the liquid helium and in terms of the temperature gradient across the superinsulation. Equating these equations will lead to an expression for the thermal conductivity k of the superinsulation. Note that the boiling temperature of liquid helium is 4.2 K.

Thermal Properties and Processes 1869 Express the heat current in terms of the rate of evaporation of the liquid helium:

I = Lv

dm dt

Express the heat current in terms of the temperature gradient across the superinsulation and the conductivity of the superinsulation:

I = kA

ΔT Δx

Equate these expressions and solve for k to obtain:

k=

dm dV =ρ dt dt

Using the definition of density, express the rate of loss of liquid helium: Substitute for

dm dt AΔT

Lv Δx

dm to obtain: dt

k=

dV dt AΔT

Lv Δxρ

Express the ratio of the area of the spherical container to its volume:

A 4πr 2 = 4 3 ⇒ A = 3 36πV 2 V 3 πr

Substituting for A yields:

dV dt k=3 2 36πV ΔT Lv Δxρ

Substitute numerical values and evaluate k:

⎛ kJ ⎞ kg ⎞ ⎛ 0.700 × 10 −3 m 3 ⎞ ⎛ ⎟⎟ ⎜⎜ 21.0 ⎟⎟ 7.00 × 10 −2 m ⎜125 3 ⎟ ⎜⎜ kg ⎠ 86400 s m ⎠⎝ ⎝ ⎝ ⎠ = 3.12 × 10 −6 W k= 2 m⋅K 3 36π 200 × 10 −3 m 3 (289 K )

(

)

(

19

••

[SSM]

)

Estimate the thermal conductivity of human skin.

Picture the Problem We can use the thermal current equation for the thermal conductivity of the skin. If we model a human body as a rectangular parallelepiped that is 1.5 m high × 7 cm thick × 50 cm wide, then its surface area is about 1.8 m2. We’ll also assume that a typical human, while resting, produces energy at the rate of 120 W, that normal internal and external temperatures are 33°C and 37°C, respectively, and that an average skin thickness is 1.0 mm.

1870 Chapter 20 Use the thermal current equation to express the rate of conduction of thermal energy:

I = kA

Substitute numerical values and evaluate k:

k=

ΔT I ⇒k = ΔT Δx A Δx

120 W mW = 17 37°C − 33°C m⋅K 1.8 m 2 −3 1.0 × 10 m

(

)

20 •• You are visiting Finland with a college friend and have met some Finnish friends. They talk you into taking part in a traditional Finnish after-sauna exercise which consists of leaving the sauna, wearing only a bathing suit, and running out into the mid-winter Finnish air. Estimate the rate at which you initially lose energy to the cold air. Compare this rate of initial energy loss to the resting metabolic rate of a typical human under normal temperature conditions. Explain the difference. Picture the Problem We can use the Stefan-Boltzmann law to estimate the rate at which you lose energy when you first step out of the sauna. If we model a human body as a rectangular parallelepiped that is 1.5 m high × 7 cm thick × 50 cm wide, then its surface area is about 1.8 m2. Assume that your skin temperature is initially 37°C (310 K), that the mid-winter outside temperature is −10°C (263 K), and that the emissivity of your skin is 1.

(

4 Pnet = εσA Tskin − Tair4

Use the Stefan-Boltzmann law to express the net rate at which you radiate energy to the cold air:

)

Substitute numerical values and evaluate Pnet:

(

)(

)

W ⎞ ⎛ 4 4 Pnet = (1)⎜ 5.670 × 10 −8 2 4 ⎟ 1.8 m 2 (310 K ) − (263 K ) = 450 W m ⋅K ⎠ ⎝

This result is almost four times greater than the basal metabolic rate of 120 W. We can understand the difference in terms of the temperature of your skin when you first step out of the sauna and the fact that radiation loses are dependent on the fourth power of the absolute temperature. Remarks: The emissivity of your skin is, in fact, very close to 1. 21 •• Estimate the rate of heat conduction through a 2.0-in-thick wooden door during a cold winter day in Minnesota. Include the brass doorknob. What is the ratio of the heat that escapes through the doorknob to the heat that escapes through the whole door? What is the total overall R-factor for the door, including the knob? The thermal conductivity of brass is 85 W (m ⋅ K ) .

Thermal Properties and Processes 1871 Picture the Problem We can use the thermal current equation (Equation 20-7) to estimate the rate of heat loss through the door and its knob. We’ll assume that the door is made of oak with an area of 2.0 m2, that the knob is made of brass, that the conducting path has a diameter of 3.0 cm, and that the inside temperature is 20°C. Take the outside temperature to be −20°C. The reciprocal of the equivalent Rfactor is the sum of the reciprocals of the R-values of the door and the knob.

The total thermal current through the door is the sum of the thermal currents through the door and the knob:

I tot = I door + I knob

Assume that Δxdoor = Δxknob = Δx to obtain:

I tot = (k door Adoor + k knob Aknob )

=

k door Adoor ΔT k knob Aknob ΔT + Δxdoor Δxknob ΔT Δx

Substitute numerical values and evaluate Itot: 2 ⎤ 20°C − (− 20°C ) ⎡⎛ W ⎞π W ⎞ ⎛ 2 −2 I tot = ⎢⎜ 0.15 ⎟ 2.0 m + ⎜ 85 ⎟ 3.0 × 10 m ⎥ m⋅K ⎠ ⎝ m⋅K ⎠ 4 ⎣⎝ ⎦ (2.0 in )⎛⎜ 1 m ⎞⎟ ⎝ 39.37 in ⎠

(

)

(

)

= 0.28 kW Express the ratio of the thermal currents through the knob and the door:

Because the temperature difference across the door and the knob are the same and we’ve assumed that the thickness of the door and length of the knob are the same: Substitute numerical values to obtain:

I knob I door

⎛ ΔT ⎞ k door Adoor ⎜ ⎟ ⎝ Δx ⎠ door = ⎛ ΔT ⎞ k knob Aknob ⎜ ⎟ ⎝ Δx ⎠ knob

I knob k door Adoor = I door k knob Aknob

I knob I door

(

(

= 5.0 Relate the equivalent R-factor to the R-factors of the door and knob;

)

W ⎞ ⎛ 2 ⎜ 0.15 ⎟ 2.0 m ⋅ m K ⎝ ⎠ = W ⎞π ⎛ −2 ⎜ 85 ⎟ 3.0 × 10 m m ⋅ K 4 ⎝ ⎠

1 1 1 = + Req Rdoor Rknob

)

2

1872 Chapter 20 Substitute for Rdoor and Rknob to obtain:

1 = Req

Solving for Req yields:

1 1 + Δx Δx k door Adoor k knob Aknob

=

k door Adoor + k knob Aknob Δx

Req =

Δx k door Adoor + k knob Aknob

Substitute numerical values and evaluate Req:

(2.0 in )⎛⎜

1m ⎞ ⎟ K 39.37 in ⎠ ⎝ Req = = 0.14 2 W ⎞ W ⎞π W ⎛ ⎛ 2 −2 ⎜ 0.15 ⎟ 2.0 m + ⎜ 85 ⎟ 3.0 × 10 m m⋅K ⎠ ⎝ ⎝ m⋅K ⎠ 4

(

)

(

)

22 •• Estimate the effective emissivity of Earth, given the following information. The solar constant, which is the intensity of radiation incident on Earth from the Sun, is about 1.37 kW/m2. Seventy percent of this energy is absorbed by Earth, and Earth’s average surface temperature is 288 K. (Assume that the effective area that is absorbing the light is πR2, where R is Earth’s radius, while the blackbody-emission area is 4πR2.) Picture the Problem The amount of heat radiated by Earth must equal the solar flux from the Sun, or else the temperature on Earth would continually increase. The emissivity of Earth is related to the rate at which it radiates energy into space by the Stefan-Boltzmann law Pr = eσAT 4 .

Using the Stefan-Boltzmann law, express the rate at which Earth radiates energy as a function of its emissivity e and temperature T:

Pr σA'T 4 where A′ is the surface area of Earth.

Use its definition to express the intensity of the radiation received by Earth:

Pabsorbed A where A is the cross-sectional area of Earth.

Pr = eσA'T 4 ⇒ e =

I=

For 70% absorption of the Sun’s radiation incident on Earth:

I=

(0.70)Pr

Substitute for Pr and A in the expression for e and simplify to

e=

(0.70)AI = (0.70)π R 2 I = (0.70)I

A

σA'T 4

4π R 2σT 4

4σT 4

Thermal Properties and Processes 1873 obtain: Substitute numerical values and evaluate e:

e=

(0.70) (1.37 kW/m 2 ) 4 4(5.670 ×10 −8 W/m 2 ⋅ K 4 )(288 K )

= 0.61 23 •• Black holes are highly condensed remnants of stars. Some black holes, together with a normal star, form binary systems. In such systems the black hole and the normal star orbit about the center of mass of the system. One way black holes can be detected from Earth is by observing the frictional heating of the atmospheric gases from the normal star that fall into the black hole. These gases can reach temperatures greater than 1.0 × 106 K. Assuming that the falling gas can be modeled as a blackbody radiator, estimate λmax for use in an astronomical detection of a black hole. (Remark: This is in the X-ray region of the electromagnetic spectrum.) Picture the Problem The wavelength at which maximum power is radiated by the gas falling into a black hole is related to its temperature by Wien’s displacement law.

Wien’s displacement law relates the wavelength at which maximum power is radiated by the gas to its temperature: Substitute for T and evaluate λmax:

λmax =

2.898 mm ⋅ K T

λmax =

2.898 mm ⋅ K = 2.9 nm 1.0 × 10 6 K

24 ••• Your cabin in northern Michigan has walls that consist of pine logs that have average thicknesses of about 20 cm. You decide to finish the interior of the cabin to improve the look and to increase the insulation of the exterior walls. You choose to buy insulation with an R-factor of 31 to cover the walls. In addition, you cover the insulation with 1.0-in-thick gypsum wallboard. Assuming heat transfer is only due to conduction, estimate the ratio of thermal current through the walls during a cold winter night before the renovation to the thermal current through the walls following the renovation. Picture the Problem We can use the thermal current equation to find the thermal current per square meter through the walls of the cabin both before and after the walls have been insulated. See Table 20-5 for the R-factor of gypsum wallboard and Table 20-4 for the thermal conductivity of white pine.

The rate at which heat is conducted through the walls is given by the thermal current equation:

I before = k before A

ΔT ΔT = Δx Rbefore

1874 Chapter 20 With the insulation in place:

I after = k after A

Divide the second of these equations by the first to obtain:

I after I before

ΔT ΔT = Δx Rafter

ΔT R R = after = before ΔT Rafter Rbefore

Rbefore is the R-factor for pine and

Rbefore = Rpine

Rafter is the sum of the R-factors for pine, the insulating material, and the gypsum board:

and Rafter = Req = Rpine + R31 + Rgypsum

Substitute for Rbefore and Rafter to obtain:

Rpine I after = I before Rpine + R31 + Rgypsum

Because Rpine =

Δx k pine

Δx : k pine

I after = Δx I before + R31 + Rgypsum k pine

Substitute numerical values and evaluate the ratio I after / I before :

1 in 2.54 cm Btu ⋅ in 0.78 h ⋅ ft 2 ⋅ F°

20 cm × I after = I before 20 cm ×

1 in 2 2 2.54 cm + 31 h ⋅ ft ⋅ F° + 1 in × 0.32 h ⋅ ft ⋅ F° Btu ⋅ in Btu 0.375 in Btu 0.78 h ⋅ ft 2 ⋅ F°

= 24%

25 ••• [SSM] You are in charge of transporting a liver from New York, New York to Los Angeles, California for a transplant surgery. The liver is kept cold in a Styrofoam ice chest initially filled with 1.0 kg of ice. It is crucial that the liver temperature is never warmer than 5.0°C. Assuming the trip from the hospital in New York to the hospital in Los Angeles takes 7.0 h, estimate the R-value the Styrofoam walls of the ice chest must have. Picture the Problem The R factor is the thermal resistance per unit area of a slab of material. We can use the thermal current equation to express the thermal resistance of the styrofoam in terms of the maximum amount of heat that can enter the chest in 7.0 h without raising the temperature above 5.0°C. We’ll

Thermal Properties and Processes 1875 assume that the surface area of the ice chest is 1.0 m2 and that the ambient temperature is 25°C The R-factor needed for the Rf = RA (1) styrofoam walls of the ice chest is the product of their thermal resistance and area: Use the thermal current equation to express R:

R=

Substitute for R in equation (1) to obtain:

Rf =

The total heat entering the chest in 7 h is given by:

Qtot = Qmelt + Qwarm

Substitute for Qtot and simplify to obtain:

Rf =

ΔT ΔT ΔTΔt = = Qtot I Qtot Δt A ΔT Δt Qtot

ice

ice water

= mice Lf + mice cH 2 O ΔTH 2 O

mice

AΔTΔt Lf + cH 2 O ΔTH 2O

(

)

Substitute numerical values and evaluate Rf:

⎞⎛ ⎛ 1 ft 2 9 F° ⎞ ⎛ 1054.35 J ⎞ ⎟ 20 C° × ⎜⎜1.0 m 2 × ⎟ (7 h )⎜ ⎟ 2 ⎟⎜ −2 5 C° ⎠ 9.29 × 10 m ⎠ ⎝ F° ⋅ h ⋅ ft 2 ⎝ Btu ⎠ ⎝ ≈ 8 Rf = Btu ⎛ ⎞ (1 kg )⎜⎜ 333.5 kJ + ⎛⎜⎜ 4.18 kJ ⎞⎟⎟ (5 C°)⎟⎟ kg ⎝ kg ⋅ K ⎠ ⎝ ⎠

Thermal Expansion 26 •• You have inherited your grandfather’s grandfather clock that was calibrated when the temperature of the room was 20ºC. Assume that the pendulum consists of a thin brass rod of negligible mass with a compact heavy bob at its end. (a) During a hot day, the temperature is 30ºC, does the clock run fast or slow? Explain. (b) How much time does it gain or lose during this day? Picture the Problem We can determine whether the clock runs fast or slow from the expression for the period of a simple pendulum and the dependence of its length on the temperature. We can use the expression for the period of a simple pendulum and the equation describing its length as a function of temperature to find the time gained or lost in a 24-h period.

1876 Chapter 20 (a) Express the period of the pendulum in terms of its length:

TP = 2π

L g

Because TP ∝ L and L is temperature dependent, the clock runs slow. (b) The period of the pendulum when the temperature is 20°C is given by: When the temperature increases to 30°C, the period of the pendulum increases due to the increase in its length:

T20 = 2π

L20 g

(1)

L (1 + αΔt C ) L = 2π 20 g g

T = 2π

= T20 1 + αΔt C

The daily fractional gain or loss is given by :

Δτ

Substituting for T and simplifying yields:

Δτ

τ

τ

=

=

ΔT T − T20 = T20 T20

T20 1 + αΔt C − T20 T20

= 1 + αΔt C − 1 Solve for Δτ to obtain:

Δτ =

( 1 + α Δt

C

)

−1τ

Substitute numerical values and Δτ: Δτ =

( 1 + (19 ×10

−6

)

)

3600 s ⎞ ⎛ K −1 (30°C − 20°C ) − 1 ⎜ 24 h × ⎟ = 8.2 s h ⎠ ⎝

[SSM] You need to fit a copper collar tightly around a steel shaft that 27 •• has a diameter of 6.0000 cm at 20ºC. The inside diameter of the collar at that temperature is 5.9800 cm. What temperature must the copper collar have so that it will just slip on the shaft, assuming the shaft itself remains at 20ºC? Picture the Problem Because the temperature of the steel shaft does not change, we need consider just the expansion of the copper collar. We can express the required temperature in terms of the initial temperature and the change in temperature that will produce the necessary increase in the diameter D of the copper collar. This increase in the diameter is related to the diameter at 20°C and the increase in temperature through the definition of the coefficient of linear expansion.

Express the temperature to which the

T = Ti + ΔT

Thermal Properties and Processes 1877 copper collar must be raised in terms of its initial temperature and the increase in its temperature: Apply the definition of the coefficient of linear expansion to express the change in temperature required for the collar to fit on the shaft: Substitute for ΔT to obtain:

Substitute numerical values and evaluate T:

⎛ ΔD ⎞ ⎜ ⎟ D ⎠ ΔD ⎝ ΔT = = α αD

T = Ti +

ΔD αD

T = 20°C +

6.0000 cm − 5.9800 cm 17 × 10 −6 K −1 (5.9800 cm )

(

)

= 220°C

28 •• You have a copper collar and a steel shaft. At 20°C, the collar has an inside diameter of 5.9800 cm and the steel shaft has diameter of 6.0000 cm. The copper collar was heated. When its inside diameter exceeded 6.0000 cm is was slipped on the shaft. The collar fitted tightly on the shaft after they cooled to room temperature. Now, several years later, you need to remove the collar from the shaft. To do this you heat them both until you can just slip the collar off the shaft. What temperature must the collar have so that the collar will just slip off the shaft? Picture the Problem Because the temperatures of both the steel shaft and the copper collar change together, we can find the temperature change required for the collar to fit the shaft by equating their diameters for a temperature increase ΔT. These diameters are related to their diameters at 20°C and the increase in temperature through the definition of the coefficient of linear expansion.

Express the temperature to which the collar and the shaft must be raised in terms of their initial temperature and the increase in their temperature: Express the diameter of the steel shaft when its temperature has been increased by ΔT: Express the diameter of the copper collar when its temperature has been

T = Ti + ΔT

Dsteel = Dsteel,20°C (1 + α steelΔT )

DCu = DCu,20° C (1 + α Cu ΔT )

(1)

1878 Chapter 20 increased by ΔT: DCu,20°C (1 + α Cu ΔT )

If the collar is to fit over the shaft when the temperature of both has been increased by ΔT:

= Dsteel,20°C (1 + α steel ΔT )

Solving for ΔT yields:

ΔT =

Substitute in equation (1) to obtain:

Dsteel,20°C − DCu,20°C DCu,20°Cα Cu − Dsteel,20°Cα steel

T = Ti +

Dsteel,20°C − DCu,20°C DCu,20°Cα Cu − Dsteel,20°Cα steel

Substitute numerical values and evaluate T: T = 20°C +

6.0000 cm − 5.9800 cm = 580°C (5.9800 cm ) 17 × 10 −6 /K − (6.0000 cm ) 11× 10 −6 /K

(

)

(

)

29 •• A container is filled to the brim with 1.4 L of mercury at 20ºC. As the temperature of container and mercury is increased to 60ºC, a total of 7.5 mL of mercury spill over the brim of the container. Determine the linear expansion coefficient of the material that makes up the container. Picture the Problem The linear expansion coefficient of the container is onethird its coefficient of volume expansion. We can relate the changes in volume of the mercury and the container to their initial volumes, temperature change, and coefficients of volume expansion, and, because we know the amount of spillage, obtain an equation that we can solve for βc.

Relate the linear expansion coefficient of the container to its coefficient of volume expansion:

α c = 13 β c

(1)

Express the difference in the change in the volume of the mercury and the container in terms of the spillage:

ΔVHg − ΔVc = 7.5 mL

(2)

Express ΔVHg using the definition of

ΔVHg = β HgVHg ΔT

the coefficient of volume expansion: Express ΔVc using the definition of the coefficient of volume expansion:

ΔVc = β cVc ΔT

Thermal Properties and Processes 1879

1880 Chapter 20 Substitute for ΔVHg and ΔVc in

β HgVHg ΔT − β cVc ΔT = 7.5 mL

equation (2) to obtain: Solving for βc yields:

βc =

β HgVHg ΔT − 7.5 mL Vc ΔT

or, because V = VHg = Vc, β VΔT − 7.5 mL β c = Hg VΔT 7.5 mL = β Hg − VΔT Substitute for β c in equation (1) to obtain: Substitute numerical values and evaluate αc:

α c = 13 β Hg −

7.5 mL 7.5 mL = α Hg − 3VΔT 3VΔT

α c = 13 (0.18 × 10 −3 K −1 ) −

7.5 mL 3(1.4 L )(40 K )

= 15 × 10 −6 K −1

30 •• A car has a 60.0-L steel gas tank filled to the brim with 60.0-L of gasoline when the temperature of the outside is 10ºC. The coefficient of volume expansion for gasoline at 20°C is 0.950 × 10−3 K−1. How much gasoline spills out of the tank when the outside temperature increases to 25ºC? Take the expansion of the steel tank into account. Picture the Problem The amount of gas that spills is the difference between the change in the volume of the gasoline and the change in volume of the tank. We can find this difference by expressing the changes in volume of the gasoline and the tank in terms of their common volume at 10°C, their coefficients of volume expansion, and the change in the temperature.

Express the spill in terms of the change in volume of the gasoline and the change in volume of the tank:

Vspill = ΔVgasoline − ΔVtank

Relate ΔVgasoline to the coefficient

ΔVgas = β gasolineVΔT

of volume expansion for gasoline:

Thermal Properties and Processes 1881 Relate ΔVtank to the coefficient of

ΔVtank = β tankVΔT

linear expansion for steel:

or, because βsteel = 3αsteel, ΔVtank = 3α steelVΔT

Substitute for ΔVgasoline and ΔVtank

Vspill = β gasVΔT − 3α steelVΔT

= VΔT (β gasoline − 3α steel )

and simplify to obtain: Substitute numerical values and evaluate Vspill :

[

(

)]

Vspill = (60.0 L )(25°C − 10°C ) 0.950 × 10 −3 K −1 − 3 11× 10 −6 K −1 ≈ 0.8 L 31 ••• What is the tensile stress in the copper collar of Problem 27 when its temperature returns to 20ºC? Picture the Problem We can use the definition of Young’s modulus to express the tensile stress in the copper in terms of the strain it undergoes as its temperature returns to 20°C. We can show that ΔL/L for the circumference of the collar is the same as Δd/d for its diameter.

Using Young’s modulus, relate the stress in the collar to its strain:

Stress = Y × Strain = Y

ΔL L20°C

where L20°C is the circumference of the collar at 20°C. Express the circumference of the collar at the temperature at which it fits over the shaft:

LT = πdT

Express the circumference of the collar at 20 °C:

L20°C = πd 20°C

Substitute for LT and L20°C and simplify to obtain:

Stress = Y =Y

πd T − πd 20°C πd 20°C d T − d 20°C d 20°C

1882 Chapter 20 Substitute numerical values and evaluate the stress:

(

Stress = 11× 10 −10 N/m 2

cm − 5.9800 cm ) 6.00005.9800 = cm

3.7 × 10 −12 N/m 2

The van der Waals Equation, Liquid-Vapor Isotherms, and Phase Diagrams 32 • (a) Calculate the volume of 1.00 mol of an ideal gas at a temperature of 100ºC and a pressure of 1.00 atm. (b) Calculate the temperature at which 1.00 mol of steam at a pressure of 1.00 atm has the volume calculated in Part (a). Use a = 0.550 Pa⋅m6/mol2 and b = 30.0 cm3/mol. Picture the Problem We can apply the ideal-gas law to find the volume of 1.00 mol of steam at 100°C and a pressure of 1.00 atm and then use the van der Waals equation to find the temperature at which the steam will this volume.

(a) Solving the ideal-gas law for the volume gives:

V=

Substitute numerical values and evaluate V:

V=

nRT P

(1.00mol)(8.314 J/mol ⋅ K )(373 K )

101.325 kPa atm 1 L = 3.06 × 10 − 2 m 3 × −3 3 10 m 1.00 atm ×

= 30.6 L ⎛ an 2 ⎞ ⎜⎜ P + 2 ⎟⎟(V − bn ) V ⎠ T=⎝ nR

(b) Solve van der Waals equation for T to obtain:

Substitute numerical values and evaluate T:

(

)

2 ⎛ 0.550 Pa ⋅ m 6 /mol 2 (1.00 mol) ⎞⎟ ⎜ T = 101.325 kPa + 2 ⎟ ⎜ 3.06 × 10 − 2 m 3 ⎠ ⎝ −2 3 −6 3 3.06 × 10 m − 30.0 × 10 m /mol (1.00 mol) × (1.00 mol) (8.314 J/mol ⋅ K )

(

= 375 K

(

) )

Thermal Properties and Processes 1883 33 •• [SSM] Using Figure 20-16, find the following quantities. (a) The temperature at which water boils on a mountain where the atmospheric pressure is 70.0 kPa, (b) the temperature at which water boils in a container where the pressure inside the container is 0.500 atm, and (c) the pressure at which water boils at 115ºC. Picture the Problem Consulting Figure 20-16, we see that:

(a) At 70.0 kPa, water boils at approximately 90°C . (b) At 0.500 atm (about 51 kPa), water boils at approximately 78°C . (c) The pressure at which water boils at 115°C is approximately 127 kPa . 34 •• The van der Waals constants for helium are a = 0.03412 L2⋅atm/mol2 and b = 0.0237 L/mol. Use these data to find the volume in cubic centimeters occupied by one helium atom. Then, estimate the radius of the helium atom. Picture the Problem Assume that a helium atom is spherical. Then we can find its volume from the van der Waals equation and its radius from V = 43 π r 3 .

In the van der Waals equation, b is the volume of 1 mol of molecules. For He, 1 molecule = 1 atom. Use Avogadro’s number to express b in cm3/atom:

L ⎞ ⎛ 3 cm 3 ⎞ ⎛ ⎟ ⎜ 0.0237 ⎟ ⎜10 mol ⎠ ⎜⎝ L ⎟⎠ ⎝ b= atoms 6.022 × 10 23 mol = 3.94 × 10 −23

The volume of a spherical helium atom is given by:

V = 43 πr 3 ⇒ r = 3

Substitute numerical values and evaluate r:

r=3

(

cm 3 atom 3V 3 3b = 4π 4π

)

3 3.94 ×10 − 23 cm 3 = 0.211nm 4π

Conduction 35 • [SSM] A 20-ft × 30-ft slab of insulation has an R factor of 11. At what rate is heat conducted through the slab if the temperature on one side is a constant 68ºF and the temperature of the other side is a constant 30ºF?

1884 Chapter 20 Picture the Problem We can use its definition to express the thermal current in the slab in terms of the temperature differential across it and its thermal resistance and use the definition of the R factor to express I as a function of ΔT, the crosssectional area of the slab, and Rf.

Express the thermal current through the slab in terms of the temperature difference across it and its thermal resistance:

I=

ΔT R

Substitute to express R in terms of the insulation’s R factor:

I=

AΔT ΔT = Rf / A Rf

Substitute numerical values and evaluate I:

I=

(20 ft )(30 ft )(68°F − 30°F) 11

= 2.1

h ⋅ ft 2 ⋅ F° Btu

kBtu h

36 •• A copper cube and an aluminum, cube each with 3.00-cm-long edges, are arranged as shown in Figure 20-17. Find (a) the thermal resistance of each cube, (b) the thermal resistance of the two-cube combination, (c) the thermal current I, and (d) the temperature at the interface of the two cubes. Picture the Problem We can use R = Δx kA to find the thermal resistance of each cube and the fact that they are in series to find the thermal resistance of the twocube system. We can use I = ΔT R to find the thermal current through the cubes and the temperature at their interface. See Table 20-4 for the thermal conductivities of copper and aluminum. Δx kA

(a) Using its definition, express the thermal resistance of each cube:

R=

Substitute numerical values and evaluate the thermal resistance of the copper cube:

RCu =

3.00 cm W ⎞ ⎛ 2 ⎟ (3.00 cm ) ⎜ 401 m⋅K ⎠ ⎝

= 0.08313 K/W = 0.0831 K/W

Thermal Properties and Processes 1885 Substitute numerical values and evaluate the thermal resistance of the aluminum cube:

RAl =

3.00 cm W ⎞ ⎛ 2 ⎜ 237 ⎟ (3.00 cm ) m⋅K ⎠ ⎝

= 0.1406 K/W = 0.141K/W

(b) Because the cubes are in series, their thermal resistances are additive:

R = RCu + RAl = 0.08313 K/W + 0.1406 K/W = 0.2237 K/W = 0.224 K/W

(c) Using its definition, find the thermal current:

I=

ΔT 100°C − 20°C = 0.2237 K/W R

= 357.6 W = 0.36 kW

(d) Express the temperature at the interface between the two cubes:

Tinterface = 100°C − ΔTCu

Express the temperature differential across the copper cube:

ΔTCu = I Cu RCu = IRCu

Substitute for ΔTCu to obtain:

Tinterface = 100°C − IRCu

Substitute numerical values and evaluate Tinterface:

Tinterface = 100°C

− (357.6 W )(0.08313 K/W )

≈ 70°C

37 •• Two metal cubes, one copper and one aluminum, with 3.00-cm-long edges, are arranged in parallel, as shown in Figure 20-18. Find (a) the thermal current in each cube, (b) the total thermal current, and (c) the thermal resistance of the two-cube combination. Picture the Problem We can use Equation 20-9 to find the thermal current in each cube. Because the currents are additive, we can find the equivalent resistance of the two-cube system from Req = ΔT I total .

(a) The thermal current in each cube is given by Equation 20-9:

I=

ΔT ΔT = kA R Δx

1886 Chapter 20 Substitute numerical values and evaluate the thermal current in the copper cube: W ⎞ ⎛ 2 ⎛ 100°C − 20°C ⎞ ⎟⎟ = 962.4 W = 0.96 kW I Cu = ⎜ 401 ⎟ (3.00 cm ) ⎜⎜ m⋅K ⎠ ⎝ ⎝ 3.00 cm ⎠ Substitute numerical values and evaluate the thermal current in the aluminum cube: W ⎞ ⎛ 2 ⎛ 100°C − 20°C ⎞ ⎟⎟ = 568.8 W = 0.57 kW I Al = ⎜ 237 ⎟ (3.00 cm ) ⎜⎜ m⋅K ⎠ ⎝ ⎝ 3.00 cm ⎠ (b) Because the cubes are in parallel, their total thermal currents are additive:

I = I Cu + I Al = 962.4 W + 568.8 W

(c) Use the relationship between the total thermal current, temperature differential and thermal resistance to find Req:

Req =

= 1.531 kW = 1.5 kW ΔT 100°C − 20°C = 1.531 kW I total

= 0.052 K/W

38 •• The cost of air conditioning a house is approximately proportional to the rate at which heat is absorbed by the house from its surroundings divided by the coefficient of performance (COP) of the air conditioner. Let us denote the temperature difference between the inside temperature and the outside temperature as ΔT. Assuming that the rate at which heat is absorbed by a house is proportional to ΔT and that the air conditioner is operating ideally, show that the cost of air conditioning is proportional to (ΔT)2 divided by the temperature inside the house. Picture the Problem The cost of operating the air conditioner is proportional to the energy used in its operation. We can use the definition of the COP to relate the rate at which the air conditioner removes heat from the house to rate at which it must do work to maintain a constant temperature differential between the interior and the exterior of the house. To obtain an expression for the minimum rate at which the air conditioner must do work, we’ll assume that it is operating with the maximum efficiency possible. Doing so will allow us to derive an expression for the rate at which energy is used by the air conditioner that we can integrate to obtain the energy (and hence the cost of operation) required.

Relate the cost C of air conditioning the energy W required to operate the air conditioner:

C = uW (1) where u is the unit cost of the energy.

Thermal Properties and Processes 1887 Express the rate dQ/dt at which heat flows into a house provided the house is maintained at a constant temperature:

Use the definition of the COP to relate the rate at which the air conditioner must remove heat dW/dt to maintain a constant temperature: Express the maximum value of the COP:

dQ = kΔT dt where ΔT is the temperature difference between the interior and exterior of the house. P=

COP =

dQ dt dW 1 dQ ⇒ = dW dt dt COP dt

Tc ΔT where Tc = Tinside the is the temperature COPmax =

house

of the cold reservoir. Letting COP = COPmax, substitute to obtain an expression for the minimum rate at which the air conditioner must do work in order to maintain a constant temperature:

dQ dW = dt ΔT dt Tc

Substituting for dQ/dt gives:

dW kΔT k 2 = ΔT = (ΔT ) dt Tc Tc

Separate variables and integrate this equation to obtain: Substitute in equation (1) to obtain:

Δt

W =

k (ΔT )2 ∫ dt' = k (ΔT )2 Δt Tc Tc 0

k 2 C = u (ΔT ) Δt ⇒ C ∝ Tc

(ΔT )2 Tc

39 •• A spherical shell of thermal conductivity k has inside radius r1 and outside radius r2 (Figure 20-19). The inside of the shell is held at a temperature T1, and the outside of the shell is held at temperature T2, with T1 < T2. In this problem, you are to show that the thermal current through the shell is given by 4π kr1r2 I=− (T − T ) r2 − r1 2 1 where I is positive if heat is transferred in the +r direction. Here is a suggested procedure for obtaining this result: (1) obtain an expression for the thermal current I through a thin spherical shell of radius r and thickness dr when there is a temperature difference dT across the thickness of the shell; (2) explain why the thermal current is the same through each such thin shell; (3) express the thermal current I through such a shell element in terms of the area A = 4πr2, the thickness

1888 Chapter 20

dr, and the temperature difference dT across the element; and (4) separate variables (solve for dT in terms of r and dr) and integrate. Picture the Problem We can follow the step-by-step instructions given in the problem statement to obtain the differential equation describing the variation of T with r. Integrating this equation will yield an equation that we can solve for the current I.

(1) The thermal current through a thin spherical shell surface of area A and thickness dr due to a dT temperature gradient is given by: dr

I = kA

dT dr

(2) Conservation of energy requires that the thermal current through each shell be the same. (3) For a shell of area A = 4π r2:

(4) Separating the variables yields:

Integrate from r = r1 to r = r2 and simplify to obtain:

I = 4π kr 2

dT =

dT dr

dr 4π k r 2

T2

∫ dT =

T1

I

I

r2

dr

4π k ∫ r

2

r1

and r

T2 − T1 = −

Solving for I gives:

I= −

I ⎡1 ⎤ 2 I ⎛1 1⎞ ⎜ − ⎟ =− ⎢ ⎥ 4π k ⎣ r ⎦ r1 4π k ⎜⎝ r1 r2 ⎟⎠

4π kr1r2 (T2 − T1 ) r2 − r1

Radiation 40 • Calculate λmax (the wavelength at which the emitted power is maximum) for a human skin. Assume the human skin is a blackbody emitter with a temperature of 33ºC. Picture the Problem We can apply Wein’s displacement law to find the wavelength at which the power is a maximum.

Thermal Properties and Processes 1889 Wein’s law relates the maximum wavelength of the radiation to the temperature of its source:

λmax =

2.898 mm ⋅ K T

Substitute numerical values and evaluate λmax:

λmax =

2.898 mm ⋅ K = 9.47 μm 273 K + 33 K

41 • [SSM] The universe is filled with radiation that is believed to be remaining from the Big Bang. If the entire universe is considered to be a blackbody with a temperature equal to 2.3 K, what is the λmax (the wavelength at which the power of the radiation is maximum) of this radiation? Picture the Problem We can use Wein’s law to find the peak wavelength of this radiation.

Wein’s law relates the maximum wavelength of the background radiation to the temperature of the universe:

λmax =

2.898 mm ⋅ K T

Substituting for T gives:

λmax =

2.898 mm ⋅ K = 1.3 mm 2 .3 K

42 • What is the range of temperatures for star surfaces for which λmax (the wavelength at which the power of the emitted radiation is maximum) is in the visible range? Picture the Problem The visible portion of the electromagnetic spectrum extends from approximately 400 nm to 700 nm. We can use Wein’s law to find the range of temperatures corresponding to these wavelengths.

Wein’s law relates the maximum wavelength of the radiation to the temperature of its source: Solving for T yields:

For λmax = 400 nm: For λmax = 700 nm:

λmax =

T=

2.898 mm ⋅ K T

2.898 mm ⋅ K

λmax

T=

2.898 mm ⋅ K = 7250 K 400 nm

T=

2.898 mm ⋅ K = 4140 K 700 nm

1890 Chapter 20 The range of temperatures is 4140 K ≤ T ≤ 7250 K . 43 • The heating wires of a 1.00-kW electric heater are red hot at a temperature of 900ºC. Assuming that 100 percent of the heat released is due to radiation and that the wires act as blackbody emitters, what is the effective area of the radiating surface? (Assume a room temperature of 20ºC.) Picture the Problem We can apply the Stefan-Boltzmann law to find the net power radiated by the wires of its heater to the room.

(

Relate the net power radiated to the surface area of the heating wires, their temperature, and the room temperature:

)

Pnet = eσA T 4 − T04 ⇒ A =

Pnet eσ T 4 − T04

(

)

Substitute numerical values and evaluate A: A=

(1)⎛⎜ 5.6703 × 10 −8 ⎝

1.00 kW = 93.5 cm 2 W ⎞ (1173 K )4 − (293 K )4 2 4 ⎟ m ⋅K ⎠

[

]

44 •• A blackened, solid copper sphere that has a radius equal to 4.0 cm hangs in an evacuated enclosure whose walls have a temperature of 20ºC. If the sphere is initially at 0ºC, find the initial rate at which its temperature changes, assuming that heat is transferred by radiation only. (Assume the sphere is a blackbody emitter.) Picture the Problem The rate at which the copper sphere absorbs radiant energy is given by dQ / dt = mcdT / dt and, from the Stephan-Boltzmann law,

(

Pnet = eσA T 4 − T04

)

where A is the surface area of the sphere, T0 is its

temperature, and T is the temperature of the walls. We can solve the first equation for dT/dt and substitute Pnet for dQ/dt in order to find the initial rate at which the temperature of the sphere changes. dQ dT = mc dt dt

Relate the rate at which the sphere absorbs radiant energy to the rate at which its temperature changes:

Pnet =

dT and substituting dt for mc gives:

P P dT Pnet = = net = 4 net3 dt mc ρVc 3 π r ρ c

Solving for

Thermal Properties and Processes 1891

Pnet = eσA(T 4 − T04 )

Apply the Stefan-Boltzmann law to relate the net power radiated to the sphere to the difference in temperature of the walls and the blackened copper sphere:

= 4π r 2 eσ (T 4 − T04 )

dT 4π r 2eσ (T 4 − T04 ) = 3 4 dt 3π r ρc

Substitute for Pnet to obtain:

=

(

3eσ T 4 − T04 rρ c

)

Substitute numerical values and evaluate dT/dt:

dT =− dt

[

]

W ⎞ ⎛ 4 4 3(1) ⎜ 5.6703 × 10−8 2 4 ⎟ (293 K ) − (273 K ) m ⋅K ⎠ ⎝ = 2.2 × 10− 3 K/s kg ⎞ ⎛ kJ ⎞ ⎛ ⎟ 4.0 × 10− 2 m ⎜ 8.93 × 103 3 ⎟ ⎜⎜ 0.386 m ⎠⎝ kg ⋅ K ⎟⎠ ⎝

(

)

45 •• The surface temperature of the filament of an incandescent lamp is 1300ºC. If the electric power input is doubled, what will the new temperature be? Hint: Show that you can neglect the temperature of the surroundings. Picture the Problem We can apply the Stephan-Boltzmann law to express the net power radiated by the incandescent lamp to its surroundings.

Express the rate at which energy is radiated to the surroundings:

Pnet = eσA(T 4 − T04 ) ⎛ ⎛ T0 ⎞ 4 ⎞ = eσAT ⎜1 − ⎜ ⎟ ⎟ ⎜ ⎝T ⎠ ⎟ ⎝ ⎠ 4

4

4

⎛ T0 ⎞ ⎛ 293 K ⎞ ⎟⎟ ≈ 1× 10 −3 ⎜ ⎟ = ⎜⎜ T 1573 K ⎝ ⎠ ⎝ ⎠

⎛T ⎞ Because ⎜ 0 ⎟ is so small, we can ⎝T ⎠ neglect the temperature of the surroundings. Hence:

4

⎛P ⎞ Pnet ≈ eσAT ⇒ T = ⎜ net ⎟ ⎝ eσA ⎠

Express the temperature T ′when the electric power input is doubled:

⎛ 2P ⎞ T' = ⎜ net ⎟ ⎝ eσA ⎠

⎛T ⎞ Evaluate ⎜ 0 ⎟ : ⎝T ⎠

4

14

4

14

1892 Chapter 20 Dividing the second of these equations by the first and solving for T gives:

T' 14 14 = (2 ) ⇒ T' = (2 ) T T

Substitute numerical values and evaluate T ′

T' = (2 ) (1573 K ) = 1871 K 14

= 1598°C

46 •• Liquid helium is stored at its boiling point (4.2 K) in a spherical can that is separated by an evacuated region of space from a surrounding shield that is maintained at the temperature of liquid nitrogen (77 K). If the can is 30 cm in diameter and is blackened on the outside so that it acts as a blackbody emitter, how much helium boils away per hour? Picture the Problem We can differentiate Q = mL, where L is the latent heat of boiling for helium, with respect to time to obtain an expression for the rate at which the helium boils away.

(

)

Express the rate at which the helium boils away in terms of the rate at which its container absorbs radiant energy:

dm Pnet eσA T 4 − T04 = = dt L L 2 4 eσπ d T − T04 = L 4 2 eσπ d 4 ⎛⎜ ⎛ T0 ⎞ ⎞⎟ = T 1− ⎜ ⎟ ⎜ ⎝T ⎠ ⎟ L ⎝ ⎠

If T0 R2. (b) What should the ratio of the charges q1/q2 and the relative signs for q1 and q2 be for the electric field to be zero throughout the region r > R2? (c) Sketch the electric field lines for the situation in Part (b) when q1 is positive. Picture the Problem To find En in these three regions we can choose Gaussian surfaces of appropriate radii and apply Gauss’s law. On each of these surfaces, Er is constant and Gauss’s law relates Er to the total charge inside the surface.

(a) Use Gauss’s law to find the electric field in the region r < R1:

∫

S

En dA =

1

∈0

Qinside

and r Q E r < R1 = inside rˆ where rˆ is a unit radial ∈0 A vector. Because Qinside = 0:

r E r R2 = 1 rˆ = rˆ 2 ∈ 0 4πr r2

(b) Set Er >R2 = 0 to obtain:

q1 + q2 = 0 ⇒

(

)

q1 = −1 q2

(c) The electric field lines for the situation in (b) with q1 positive is shown to the right.

38 • A spherical shell of radius 6.00 cm carries a uniform surface charge density of A non-conducting thin spherical shell of radius 6.00 cm has a uniform surface charge density of 9.00 nC/m2. (a) What is the total charge on the shell? Find the electric field at the following distances from the sphere’s center: (b) 2.00 cm, (c) 5.90 cm, (d) 6.10 cm, and (e) 10.0 cm. Picture the Problem We can use the definition of surface charge density and the formula for the area of a sphere to find the total charge on the shell. Because the charge is distributed uniformly over a spherical shell, we can choose a spherical Gaussian surface and apply Gauss’s law to find the electric field as a function of the distance from the center of the spherical shell.

(a) Using the definition of surface charge density, relate the charge on the sphere to its area:

Q = σA = 4πσ r 2

Substitute numerical values and evaluate Q:

Q = 4π 9.00 nC/m 2 (0.0600 m )

Apply Gauss’s law to a spherical surface of radius r that is concentric the spherical shell to obtain:

∫

(

)

2

= 0.4072 nC = 0.407 nC

S

En dA =

1

∈0

Qinside ⇒ 4π r 2 En =

Qinside

∈0

128

Chapter 22 Qinside 1 kQinside = 4π ∈0 r 2 r2

Noting that, due to symmetry, Er = En , solve for Er to obtain:

Er =

(b) The charge inside a sphere whose radius is 2.00 cm is zero and hence:

Er = 2.00 cm = 0

(c) The charge inside a sphere whose radius is 5.90 cm is zero and hence:

Er = 5.90 cm = 0

(d) The charge inside a sphere whose radius is 6.10 cm is 0.4072 nC and hence: Er = 6.10 cm =

(8.988 ×10

)

N ⋅ m 2 /C 2 (0.4072 nC ) = 983 N/C (0.0610 m )2

9

(e) The charge inside a sphere whose radius is 10.0 cm is 0.4072 nC and hence: Er = 10 cm =

(8.988 ×10

9

)

N ⋅ m 2 /C 2 (0.4072 nC ) = 366 N/C (0.100 m )2

39 •• [SSM] A non-conducting sphere of radius 6.00 cm has a uniform volume charge density of 450 nC/m3. (a) What is the total charge on the sphere? Find the electric field at the following distances from the sphere’s center: (b) 2.00 cm, (c) 5.90 cm, (d) 6.10 cm, and (e) 10.0 cm. Picture the Problem We can use the definition of volume charge density and the formula for the volume of a sphere to find the total charge of the sphere. Because the charge is distributed uniformly throughout the sphere, we can choose a spherical Gaussian surface and apply Gauss’s law to find the electric field as a function of the distance from the center of the sphere.

(a) Using the definition of volume charge density, relate the charge on the sphere to its volume:

Q = ρV = 43 πρr 3

Substitute numerical values and evaluate Q:

Q = 43 π 450 nC/m 3 (0.0600 m )

(

)

= 0.4072 nC = 0.407 nC

3

The Electric Field 2: Continuous Charge Distributions 1

129

Qinside

Apply Gauss’s law to a spherical surface of radius r < R that is concentric with the spherical shell to obtain:

∫

Noting that, due to symmetry, En = Er , solve for Er to obtain:

Er =

Because the charge distribution is uniform, we can find the charge inside the Gaussian surface by using the definition of volume charge density to establish the proportion:

Q Qinside = V V' where V′ is the volume of the Gaussian surface.

S

Solve for Qinside to obtain:

En dA =

∈0

Qinside 1 kQinside = 4π ∈0 r 2 r2

Qinside = Q

Substitute for Qinside to obtain:

∈0

Qinside ⇒ 4π r 2 En =

Er < R =

V' r3 =Q 3 V R

Qinside 1 kQ = r 4π ∈0 r 2 R 3

(b) Evaluate Er = 2.00 cm: Er = 2.00 cm =

(8.988 ×10

)

N ⋅ m 2 /C 2 (0.4072 nC ) (0.0200 m ) = 339 N/C (0.0600 m )3

9

(c) Evaluate Er = 5.90 cm: Er = 5.90 cm =

(8.988 ×10

N ⋅ m 2 /C 2 )(0.4072 nC ) (0.0590 m ) = 1.00 kN/C (0.0600 m )3

9

Apply Gauss’s law to the Gaussian surface with r > R:

4π r 2 Er =

Qinside

∈0

⇒ Er =

kQinside kQ = 2 r2 r

(d) Evaluate Er = 6.10 cm: Er = 6.10 cm =

(8.988 ×10

N ⋅ m 2 /C 2 )(0.4072 nC ) = 983 N/C (0.0610 m )2

9

130

Chapter 22

(e) Evaluate Er = 10.0 cm: Er = 10.0 cm =

(8.988 ×10

9

N ⋅ m 2 /C 2 )(0.4072 nC ) = 366 N/C (0.100 m )2

40 •• Consider the solid conducting sphere and the concentric conducting spherical shell in Figure 22-41. The spherical shell has a charge –7Q. The solid sphere has a charge +2Q. (a) How much charge is on the outer surface and how much charge is on the inner surface of the spherical shell? (b) Suppose a metal wire is now connected between the solid sphere and the shell. After electrostatic equilibrium is re-established, how much charge is on the solid sphere and on each surface of the spherical shell? Does the electric field at the surface of the solid sphere change when the wire is connected? If so, in what way? (c) Suppose we return to the conditions in Part (a), with +2Q on the solid sphere and –7Q on the spherical shell. We next connect the solid sphere to ground with a metal wire, and then disconnect it. Then how much total charge is on the solid sphere and on each surface of the spherical shell? Determine the Concept The charges on a conducting sphere, in response to the repulsive Coulomb forces each experiences, will separate until electrostatic equilibrium conditions exit. The use of a wire to connect the two spheres or to ground the outer sphere will cause additional redistribution of charge.

(a) Because the outer sphere is conducting, the field in the thin shell must vanish. Therefore, −2Q, uniformly distributed, resides on the inner surface, and −5Q, uniformly distributed, resides on the outer surface. (b) Now there is no charge on the inner surface and −5Q on the outer surface of the spherical shell. The electric field just outside the surface of the inner sphere changes from a finite value to zero. (c) In this case, the −5Q is drained off, leaving no charge on the outer surface and −2Q on the inner surface. The total charge on the outer sphere is then −2Q. 41 •• A non-conducting solid sphere of radius 10.0 cm has a uniform volume charge density. The magnitude of the electric field at 20.0 cm from the sphere’s center is 1.88 × 103 N/C. (a) What is the sphere’s volume charge density? (b) Find the magnitude of the electric field at a distance of 5.00 cm from the sphere’s center.

The Electric Field 2: Continuous Charge Distributions

131

Picture the Problem (a) We can use the definition of volume charge density, in conjunction with Equation 22-18a, to find the sphere’s volume charge density. (b) We can use Equation 22-18b, in conjunction with our result from Part (a), to find the electric field at a distance of 5.00 cm from the solid sphere’s center.

(a) The solid sphere’s volume charge density is the ratio of its charge to its volume:

ρ=

For r ≥ R, Equation 22-18a gives the electric field at a distance r from the center of the sphere:

Qinside Qinside = 4 3 V 3 πR

(1)

Qinside 4π ∈ 0 r 2

(2)

Er =

1

Qinside = 4π ∈ 0 Er r 2

Solving for Qinside yields: Substitute for Qinside in equation

ρ=

(1) and simplify to obtain:

4π ∈ 0 E r r 2 3∈ 0 Er r 2 = 3 4 R3 3 πR

Substitute numerical values and evaluate ρ:

ρ=

(

)(

)

3 8.854 × 10 −12 C 2 /N ⋅ m 2 1.88 × 10 3 N/C (20.0 cm ) = 1.997 μC/m 3 3 (10.0 cm ) 2

= 2.00 μC/m 3

(b) For r ≤ R, the electric field at a distance r from the center of the sphere is given by:

Er =

Express Qinside for r ≤ R:

1

Qinside r 4π ∈0 R 3

(3)

Qinside = ρVsphere whose = 43 π r 3 ρ radius is r

Substituting for Qinside in equation (3)

Er =

and simplifying yields:

1 4π ∈0

4 3

π r 3ρ R3

r=

ρr 4 3 ∈0 R 3

Substitute numerical values and evaluate Er = 5.00 cm:

(1.997μC/m )(5.00 cm) = 3(8.854 × 10 C /N ⋅ m )(10.0 cm ) 4

3

Er = 5.00 cm

−12

2

2

3

= 470 N/C

132

Chapter 22

42 •• A non-conducting solid sphere of radius R has a volume charge density that is proportional to the distance from the center. That is, ρ = Ar for r ≤ R, where A is a constant. (a) Find the total charge on the sphere. (b) Find the expressions for the electric field inside the sphere (r < R) and outside the sphere (r > R). (c) Sketch the magnitude of the electric field as a function of the distance r from the sphere’s center. Picture the Problem We can find the total charge on the sphere by expressing the charge dq in a spherical shell and integrating this expression between r = 0 and r = R. By symmetry, the electric fields must be radial. To find Er inside the charged sphere we choose a spherical Gaussian surface of radius r < R. To find Er outside the charged sphere we choose a spherical Gaussian surface of radius r > R. On each of these surfaces, Er is constant. Gauss’s law then relates Er to the total charge inside the surface.

(a) Express the charge dq in a shell of thickness dr and volume 4πr2 dr: Integrate this expression from r = 0 to R to find the total charge on the sphere:

dq = 4π r 2 ρdr = 4π r 2 ( Ar ) dr = 4πAr 3dr

[

R

Q = 4πA∫ r 3 dr = πAr 4

]

R 0

= πAR 4

0

(b) Apply Gauss’s law to a spherical surface of radius r > R that is concentric with the nonconducting sphere to obtain:

∫

Solving for Er yields:

Er > R =

S

Er dA =

1

∈0

Qinside ⇒ 4π r 2 Er =

Qinside

∈0

Qinside 1 kQinside = 4π ∈0 r 2 r2

kAπR 4 AR 4 = = r2 4 ∈0 r 2

Apply Gauss’s law to a spherical surface of radius r < R that is concentric with the nonconducting sphere to obtain: Solve for Er and simplify to obtain:

∫

S

Er dA =

Er < R =

1

∈0

Qinside ⇒ 4π r 2 Er =

Qinside

∈0

Qinside Ar 2 πAr 4 = = 4πr 2 ∈0 4πr 2 ∈0 4 ∈0

The Electric Field 2: Continuous Charge Distributions

133

(c) The following graph of Er versus r/R, with Er in units of A/(4∈0), was plotted using a spreadsheet program. 1.0

0.8

0.6

Er 0.4

0.2

0.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

r/R

Remarks: Note that the results for (a) and (b) agree at r = R. 43 •• [SSM] A sphere of radius R has volume charge density ρ = B/r for r < R , where B is a constant and ρ = 0 for r > R. (a) Find the total charge on the sphere. (b) Find the expressions for the electric field inside and outside the charge distribution (c) Sketch the magnitude of the electric field as a function of the distance r from the sphere’s center. Picture the Problem We can find the total charge on the sphere by expressing the charge dq in a spherical shell and integrating this expression between r = 0 and r = R. By symmetry, the electric fields must be radial. To find Er inside the charged sphere we choose a spherical Gaussian surface of radius r < R. To find Er outside the charged sphere we choose a spherical Gaussian surface of radius r > R. On each of these surfaces, Er is constant. Gauss’s law then relates Er to the total charge inside the surface.

B dr r

(a) Express the charge dq in a shell of thickness dr and volume 4πr2 dr:

dq = 4π r 2 ρdr = 4π r 2

Integrate this expression from r = 0 to R to find the total charge on the sphere:

Q == 4πB ∫ rdr = 2πBr 2

(b) Apply Gauss’s law to a spherical surface of radius r > R that is concentric with the nonconducting sphere to obtain:

∫

= 4πBrdr R

[

]

R 0

0

= 2πBR 2

S

Er dA =

1

∈0

Qinside or 4π r 2 Er =

Qinside

∈0

134

Chapter 22

Solving for Er yields:

Er > R = =

Apply Gauss’s law to a spherical surface of radius r < R that is concentric with the nonconducting sphere to obtain: Solving for Er yields:

∫

S

Qinside 1 kQinside = 4π ∈0 r 2 r2 k 2πBR 2 BR 2 = r2 2 ∈0 r 2

Er dA =

Er < R =

1

∈0

Qinside ⇒ 4π r 2 Er =

Qinside

∈0

Qinside B 2πBr 2 = = 2 2 4π r ∈0 4π r ∈0 2 ∈0

(c) The following graph of Er versus r/R, with Er in units of B/(2∈0), was plotted using a spreadsheet program. 1.2 1.0 0.8 E r 0.6 0.4 0.2 0.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

r /R

Remarks: Note that our results for (a) and (b) agree at r = R. 44 •• A sphere of radius R has volume charge density ρ = C/r2 for r < R, where C is a constant and ρ = 0 for r > R. (a) Find the total charge on the sphere. (b) Find the expressions for the electric field inside and outside the charge distribution (c) Sketch the magnitude of the electric field as a function of the distance r from the sphere’s center. Picture the Problem We can find the total charge on the sphere by expressing the charge dq in a spherical shell and integrating this expression between r = 0 and r = R. By symmetry, the electric fields must be radial. To find Er inside the charged sphere we choose a spherical Gaussian surface of radius r < R. To find Er outside the charged sphere we choose a spherical Gaussian surface of radius r > R.

The Electric Field 2: Continuous Charge Distributions

135

On each of these surfaces, Er is constant. Gauss’s law then relates Er to the total charge inside the surface. (a) Express the charge dq in a shell of thickness dr and volume 4πr2 dr:

dq = 4π r 2 ρdr = 4π r 2 R

C dr = 4πCdr r2

Integrate this expression from r = 0 to R to find the total charge on the sphere:

Q = 4πC ∫ dr = [4πCr ] 0 = 4πCR

(b) Apply Gauss’s law to a spherical surface of radius r > R that is concentric with the nonconducting sphere to obtain:

∫

Solving for Er yields:

Er > R =

0

S

Er dA =

=

Apply Gauss’s law to a spherical surface of radius r < R that is concentric with the nonconducting sphere to obtain: Solving for Er yields:

R

∫

S

∈0

Qinside ⇒ 4π r 2 Er =

Qinside

∈0

Qinside 1 kQinside = 4π ∈0 r 2 r2 k 4πCR CR = 2 r ∈0 r 2

Er dA =

Er < R =

1

1

∈0

Qinside or 4π r 2 Er =

Qinside

∈0

Qinside 4πCr C = = 2 2 4πr ∈0 4πr ∈0 ∈0 r

136

Chapter 22

(c) The following graph of Er versus r/R, with Er in units of C / (∈0 R ) , was plotted using a spreadsheet program. 10

8

6

Er 4

2

0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

r/R

45 ••• A non-conducting spherical shell of inner radius R1 and outer radius R2 has a uniform volume charge density ρ. (a) Find the total charge on the shell. (b) Find expressions for the electric field everywhere. Picture the Problem By symmetry, the electric fields resulting from this charge distribution must be radial. To find Er for r < R1 we choose a spherical Gaussian surface of radius r < R1. To find Er for R1 < r < R2 we choose a spherical Gaussian surface of radius R1 < r < R2. To find Er for r > R2 we choose a spherical Gaussian surface of radius r > R2. On each of these surfaces, Er is constant. Gauss’s law then relates Er to the total charge inside the surface.

(a) The charge in an infinitesimal spherical shell of radius r and thickness dr is: Integrate dQ from r = R1 to R2 to find the total charge in the spherical shell in the interval R1 < r < R2:

dQ = ρdV = 4πρr 2 dr

R2

Qtotal

R2

⎡ 4πCr 3 ⎤ = 4πρ ∫ r dr = ⎢ ⎥ ⎣ 3 ⎦ R1 R1 =

2

4πρ 3 (R2 − R13 ) 3

The Electric Field 2: Continuous Charge Distributions 1

(b) Apply Gauss’s law to a spherical surface of radius r that is concentric with the nonconducting spherical shell to obtain:

∫

Solving for Er yields:

Er =

Evaluate Er < R1 :

Qinside 1 kQinside = = 0 4π ∈0 r 2 r2 because ρ r R2 = =

kQinside 4πkρ 3 = r − R13 2 2 3r r

(

)

ρ (r 3 − R13 ) 3 ∈0 r 2

4πρ 3 R2 − R13 3

(

)

4πkρ 3 R2 − R13 2 3r

(

)

ρ (R23 − R13 ) 3 ∈0 r 2

Remarks: Note that E is continuous at r = R2.

Gauss’s Law Applications in Cylindrical Symmetry Situations 46 • For your senior project you are in charge of designing a Geiger tube for detecting radiation in the nuclear physics laboratory. This instrument will consist of a long metal cylindrical tube that has a long straight metal wire running down its central axis. The diameter of the wire is to be 0.500 mm and the inside diameter of the tube will be 4.00 cm. The tube is to be filled with a dilute gas in which electrical discharge (breakdown) occurs when the electric field reaches 5.50 × 106 N/C. Determine the maximum linear charge density on the wire if breakdown of the gas is not to happen. Assume that the tube and the wire are infinitely long.

138

Chapter 22

Picture the Problem The electric field of a line charge of infinite length is given 1 λ by E r = , where r is the distance from the center of the line of charge and 2π ∈ 0 r λ is the linear charge density of the wire.

The electric field of a line charge of infinite length is given by: Because Er varies inversely with r, its maximum value occurs at the surface of the wire where r = R, the radius of the wire: Solving for λ yields:

Er =

λ

1

2π ∈ 0 r

E max =

λ

1

2π ∈ 0 R

λ = 2π ∈ 0 REmax

Substitute numerical values and evaluate λ: ⎛

λ = 2π ⎜⎜ 8.854 × 10 −12 ⎝

C2 ⎞ N⎞ ⎛ ⎟ (0.250 mm )⎜ 5.50 × 10 6 ⎟ = 76.5 nC/m 2 ⎟ C⎠ N⋅m ⎠ ⎝

47 ••• In Problem 54, suppose ionizing radiation produces an ion and an electron at a distance of 2.00 cm from the long axis of the central wire of the Geiger tube. Suppose that the central wire is positively charged and has a linear charge density equal to 76.5 pC/m. (a) In this case, what will be the electron’s speed as it impacts the wire? (b) Qualitatively, how will the electron’s speed compare to that of the ion’s final speed when it impacts the outside cylinder? Explain your reasoning. Picture the Problem Because the inward force on the electron increases as its distance from the wire decreases, we’ll need to integrate the net electric force acting on the electron to obtain an expression for its speed as a function of its distance from the wire in the Geiger tube.

(a) The force the electron experiences is the radial component of the force on the electron and is the product of its charge and the radial component of the electric field due to the positively charged central wire: The radial electric field due to the charged wire is given by:

Fe , r = eEr

Er =

1

λ

2π ∈ 0 r

The Electric Field 2: Continuous Charge Distributions

139

Substituting for Er yields:

⎛ eλ ⎞ 1 ⎟⎟ where the minus Fe , r = −⎜⎜ ⎝ 2π ∈ 0 ⎠ r sign indicates that the force acting on the electron is radially inward.

Apply Newton’s second law to the electron to obtain:

⎛ eλ − ⎜⎜ ⎝ 2π ∈ 0

Separating variables yields:

⎛ eλ ⎞ dr ⎟⎟ vdv = −⎜⎜ ⎝ 2πm ∈ 0 ⎠ r

Express the integral of this equation to obtain:

vf

Integrating yields:

r

⎛ eλ ⎞ 2 dr ⎜⎜ ⎟⎟ ∫ where the = − vdv ∫0 ⎝ 2πm ∈ 0 ⎠ r1 r lower limit on the left-hand side is zero because the electron is initially at rest. 1 2

Solve for vf to obtain:

⎞1 dv dv dr ⎟⎟ = m =m dt dt dr ⎠r dv dr dv =m = mv dr dr dt

⎛ eλ vf2 = −⎜⎜ ⎝ 2πm ∈ 0

⎛ eλ vf = ⎜⎜ ⎝ πm ∈ 0

⎞ ⎛ r2 ⎟⎟ ln⎜⎜ ⎠ ⎝ r1

⎞ ⎛ r1 ⎟⎟ ln⎜⎜ ⎠ ⎝ r2

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ ⎠

Substitute numerical values and evaluate vf: ⎛ ⎞ pC ⎞ ⎛ ⎜ ⎟ 1.602 ×10 −19 C ⎜ 76.5 ⎟ ⎜ ⎟ ⎛ 0.0200 m − 0.0025 m ⎞ m ⎝ ⎠ ln⎜ vf = ⎜ ⎟ 2 0.250 mm ⎛ ⎞⎟ ⎝ C ⎠ 31 12 − − ⎜ π 9.109 ×10 kg ⎜ 8.854 ×10 ⎟⎟ 2 ⎟⎟ ⎜ ⎜ N⋅m ⎠ ⎠ ⎝ ⎝

(

(

)

)

= 1.43 × 106 m/s (b) The positive ion is accelerated radially outward and will impact the tube instead of the wire. Because of its much larger mass, the impact speed of the ion will be much less than the impact speed of the electron.

140

Chapter 22

48 •• Show that the electric field due to an infinitely long, uniformly charged thin cylindrical shell of radius a having a surface charge density σ is given by the following expressions: E = 0 for 0 ≤ R < a and ER = σ a (∈0 R ) for

R > a. Picture the Problem From symmetry, the field in the tangential direction must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the infinitely long, uniformly charged cylindrical shell.

Apply Gauss’s law to the cylindrical surface of radius r and length L that is concentric with the infinitely long, uniformly charged cylindrical shell:

∫

S

En dA =

1

∈0

Qinside

or 2πrLE R =

Qinside

∈0

where we’ve neglected the end areas because no there is no flux through them and, due to symmetry, En = ER . Qinside 2kQinside = Lr 2πrL ∈ 0

Solve for E R :

ER =

For R < a, Qinside = 0 and:

E R a, Qinside = λL and:

2kλL 2kλ 2k (2πaσ ) = = Lr r r aσ = ∈0 r

E R >a =

49 •• A thin cylindrical shell of length 200 m and radius 6.00 cm has a uniform surface charge density of 9.00 nC/m2. (a) What is the total charge on the shell? Find the electric field at the following radial distances from the long axis of the cylinder. (b) 2.00 cm, (c) 5.90 cm, (d) 6.10 cm, and (e) 10.0 cm. (Use the results of Problem 48.)

The Electric Field 2: Continuous Charge Distributions

141

Picture the Problem We can use the definition of surface charge density to find the total charge on the shell. From symmetry, the electric field in the tangential direction must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the uniformly charged cylindrical shell.

Q = σA = 2πRLσ

(a) Using its definition, relate the surface charge density to the total charge on the shell:

Q = 2π (0.0600 m )(200 m ) (9.00 nC/m 2 )

Substitute numerical values and evaluate Q:

= 679 nC

(b) From Problem 48 we have, for r = 2.00 cm:

Er = 2.00 cm = 0

(c) From Problem 48 we have, for r = 5.90 cm:

Er = 5.90 cm = 0

(d) From Problem 48 we have, for r = 6.10 cm:

Er =

σR ∈0 r

Substitute numerical values and evaluate Er = 6.10 cm :

(9.00 nC/m )(0.0600 m) = = (8.854 ×10 C /N ⋅ m )(0.0610 m) 2

Er = 6.10 cm

−12

2

2

1.00 kN/C

(e) From Problem 48 we have, for r = 10.0 cm: Er = 10.0 cm =

(9.00 nC/m )(0.0600 m ) = (8.854 ×10 C /N ⋅ m )(0.100 m ) 2

−12

2

2

610 N/C

50 •• An infinitely long non-conducting solid cylinder of radius a has a uniform volume charge density of ρ0. Show that the electric field is given by the following expressions: Ea = ρ 0 a (2 ∈0 ) for 0 ≤ r < a and Ea = ρ 0 a 2 (2 ∈0 r ) for r > a , where r is the distance from the long axis of the cylinder.

142

Chapter 22

Picture the Problem From symmetry, the field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the infinitely long nonconducting cylinder.

Apply Gauss’s law to a cylindrical surface of radius r and length L that is concentric with the infinitely long nonconducting cylinder:

∫

S

En dA =

1

∈0

Qinside

or 2πrLEn =

Qinside

∈0

where we’ve neglected the end areas because there is no flux through them. Due to symmetry, En = Er . Solving for Er yields:

Er =

Express Qinside for r < a:

Qinside = ρ (r )V = ρ 0 πr 2 L

Substitute to obtain:

Qinside 2kQinside = Lr 2πrL ∈0

(

Er < R =

(

)

)

2k πρ0 Lr 2 ρ0 = r 2 ∈0 Lr

or, because λ = ρπa 2 , Er < R =

Express Qinside for r > a: Substitute for Qinside to obtain:

λ r 2π ∈0 a 2

(

Qinside = ρ (r )V = ρ 0 πa 2 L Er > a =

(

)

)

2k πρ0 La 2 ρ0 a 2 = Lr 2 ∈0 r

2 or, because λ = ρπa

Er > a =

λ 2π ∈0 r

51 •• [SSM] A solid cylinder of length 200 m and radius 6.00 cm has a uniform volume charge density of 300 nC/m3. (a) What is the total charge of the cylinder? Use the formulas given in Problem 50 to calculate the electric field at a point equidistant from the ends at the following radial distances from the cylindrical axis: (b) 2.00 cm, (c) 5.90 cm, (d) 6.10 cm, and (e) 10.0 cm.

The Electric Field 2: Continuous Charge Distributions

143

Picture the Problem We can use the definition of volume charge density to find the total charge on the cylinder. From symmetry, the electric field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the uniformly charged cylinder.

(

)

(a) Use the definition of volume charge density to express the total charge of the cylinder:

Qtotal = ρV = ρ πR 2 L

Substitute numerical values to obtain:

Qtotal = π 300 nC/m 3 (0.0600 m ) (200 m )

(b) From Problem 50, for r < R, we have:

Er < R =

(

)

= 679 nC

ρ r 2 ∈0

For r = 2.00 cm: Er = 2.00 cm =

(300 nC/m )(0.0200 m) = 2(8.854 × 10 C /N ⋅ m ) 3

−12

2

2

339 N/C

(c) For r = 5.90 cm:

(300 nC/m )(0.0590 m) = = 2(8.854 × 10 C /N ⋅ m ) 3

Er = 5.90 cm

−12

From Problem 50, for r > R, we have:

2

2

Er > R =

1.00 kN/C

ρR 2 2 ∈0 r

(d) For r = 6.10 cm:

(300 nC/m )(0.0600 m ) = = 2(8.854 × 10 C /N ⋅ m )(0.0610 m ) 2

3

Er = 6.10 cm

−12

2

2

1.00 kN/C

(e) For r = 10.0 cm:

(300 nC/m )(0.0600 m) = = 2(8.854 × 10 C /N ⋅ m )(0.100 m ) 2

3

Er = 10.0 cm

−12

2

2

610 N/C

2

144

Chapter 22

52 •• Consider two infinitely long, coaxial thin cylindrical shells. The inner shell has a radius a1 and has a uniform surface charge density of σ1, and the outer shell has a radius a2 and has a uniform surface charge density of σ2. (a) Use Gauss’s law to find expressions for the electric field in the three regions: 0 ≤ r < a1 , a1 < r < a2 , and r > a2 , where r is the distance from the axis. (b) What is the ratio of the surface charge densities σ2/σ1 and their relative signs if the electric field is to be zero everywhere outside the largest cylinder? (c) For the case in Part (b), what would be the electric field between the shells? (d) Sketch the electric field lines for the situation in Part (b) if σ1 is positive. Picture the Problem From symmetry; the field tangent to the surfaces of the shells must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the infinitely long, uniformly charged cylindrical shells.

(a) Apply Gauss’s law to the cylindrical surface of radius r and length L that is concentric with the infinitely long, uniformly charged cylindrical shell:

∫

S

En dA =

1

∈0

Qinside ⇒ 2πrLEn =

2kQinside Lr

Er =

For r < R1, Qinside = 0 and:

Er a2:

Ea1 a2 to obtain:

(c) Because the electric field is determined by the charge inside the Gaussian surface, the field under these conditions would be as given above:

145

σ a σ 1a1 + σ 2 a2 =0⇒ 2 = − 1 σ1 a2 ∈0 r Ea1 b:

Vr >b = − ∫ Er >b dr

∫

S

r Q E r ⋅ nˆ dA = enclosed = 0

∈0

and Er>b = 0 because Qenclosed = 0 for r > b. Substitute for Er>b to obtain:

Vr >b = − ∫ (0 )dr = 0

(b) Express the potential of the metal sphere:

Va = VQ at its center + Vsurface

Express the potential at the surface of the metal sphere:

Vsurface =

Substitute and simplify to obtain:

Va =

k (− Q ) kQ =− b b

kQ kQ ⎛1 1⎞ − = kQ⎜ − ⎟ a b ⎝a b⎠

45 •• [SSM] Two coaxial conducting cylindrical shells have equal and opposite charges. The inner shell has charge +q and an outer radius a, and the outer shell has charge –q and an inner radius b. The length of each cylindrical shell is L, and L is very long compared with b. Find the potential difference, Va – Vb between the shells.

232

Chapter 23

Picture the Problem The diagram is a cross-sectional view showing the charges on the inner and outer conducting shells. A portion of the Gaussian surface over which we’ll integrate E in order to find V in the region a < r < b is also shown. Once we’ve determined how E varies with r, we can find Va – Vb from Vb − Va = − ∫ Er dr .

Express the potential difference Vb – Va:

Vb − Va = − ∫ Er dr ⇒ Va − Vb = ∫ Er dr

Apply Gauss’s law to cylindrical Gaussian surface of radius r and length L:

∫ E ⋅ nˆ dA = E (2πrL ) = ∈

Solving for Er yields:

Er =

Substitute for Er and integrate from r = a to b:

r

q

r

S

0

q 2π ∈0 rL

Va − Vb = =

q 2π ∈ 0

b

dr r a

L∫

2kq ⎛ b ⎞ ln⎜ ⎟ L ⎝a⎠

46 •• Positive charge is placed on two conducting spheres that are very far apart and connected by a long very-thin conducting wire. The radius of the smaller sphere is 5.00 cm and the radius of the larger sphere is 12.0 cm. The electric field strength at the surface of the larger sphere is 200 kV/m. Estimate the surface charge density on each sphere. Picture the Problem Let L and S refer to the larger and smaller spheres, respectively. We can use the fact that both spheres are at the same potential to estimate the electric fields near their surfaces. Knowing the electric fields, we can use σ =∈0 E to estimate the surface charge density of each sphere.

Express the electric fields at the surfaces of the two spheres:

ES =

kQS kQ and EL = 2L 2 RS RL

Electric Potential Divide the first of these equations by the second to obtain:

kQS ES R2 Q R2 = S = S L2 EL kQL QL RS RL2

Because the potentials are equal at the surfaces of the spheres:

Q R kQL kQS = and S = S RL RS QL RL

Substitute for

QS to obtain: QL

233

ES RS RL2 RL R = = ⇒ ES = L EL 2 EL RL RS RS RS

Substitute numerical values and evaluate ES:

ES =

12.0 cm (200 kV/m ) = 480 kV/m 5.00 cm

Use σ =∈0 E to estimate the surface charge density of each sphere:

σ 12 cm =∈ 0 E12 cm = (8.854 × 10 −12 C 2 /N ⋅ m 2 )(200 kV/m ) = 1.77 μC/m 2 and

σ 5 cm =∈ 0 E5 cm = (8.854 × 10 −12 C 2 /N ⋅ m 2 )(480 kV/m ) = 4.25 μC/m 2

47 •• Two concentric conducting spherical shells have equal and opposite charges. The inner shell has outer radius a and charge +q; the outer shell has inner radius b and charge –q. Find the potential difference, Va – Vb between the shells. Picture the Problem The diagram is a cross-sectional view showing the charges on the concentric spherical shells. The Gaussian surface over which we’ll integrate E in order to find V in the region r ≥ b is also shown. We’ll also find E in the region for which a < r < b. We can then use the relationship V = − ∫ Edr to

find Va and Vb and their difference. −q

+q

Gaussian Surface

a

b

234

Chapter 23

Express Vb :

b

Vb = − ∫ E r ≥b dr ∞

Apply Gauss’s law for r ≥ b:

∫

S

r Q E r ⋅ nˆ dA = enclosed = 0

∈0

and E r ≥ b = 0 because Qenclosed = 0 for r ≥ b. Substitute for Er ≥ b to obtain:

b

Vb = − ∫ (0)dr = 0 ∞

Express Va :

a

Va = − ∫ Er ≥ a dr b

Apply Gauss’s law for r ≥ a:

(

)

Er ≥a 4π r 2 =

q

∈0

and E r ≥a = Substitute for Er ≥ a to obtain:

The potential difference between the shells is given by:

q 4π ∈ 0 r

2

=

kq r2

a

dr kq kq = − r2 a b b

Va = − kq ∫

⎛1 1⎞ Va − Vb = Va = kq⎜ − ⎟ ⎝a b⎠

48 •• The electric potential at the surface of a uniformly charged sphere is 450 V. At a point outside the sphere at a (radial) distance of 20.0 cm from its surface, the electric potential is 150 V. (The potential is zero very far from the sphere.) What is the radius of the sphere, and what is the charge of the sphere? Picture the Problem Let R be the radius of the sphere and Q its charge. We can express the potential at the two locations given and solve the resulting equations simultaneously for R and Q.

Relate the potential of the sphere at its surface to its radius:

kQ = 450 V R

(1)

Electric Potential Express the potential at a distance of 20.0 cm from its surface:

kQ = 150 V R + 0.200 m

Divide equation (1) by equation (2) to obtain:

kQ 450 V R = kQ 150 V R + 0.200 m

235

(2)

or R + 0.200 m = 3 ⇒ R = 10.0 cm R Solving equation (1) for Q yields:

Q = (450 V )

Substitute numerical values and evaluate Q:

Q = (450 V )

R k

(0.100 m )

(8.988 ×10

9

N ⋅ m 2 /C 2

)

= 5.01 nC 49 •• Consider two infinite parallel thin sheets of charge, one in the x = 0 plane and the other in the x = a plane. The potential is zero at the origin. (a) Find the electric potential everywhere in space if the planes have equal positive charge densities +σ. (b) Find the electric potential everywhere in space if the sheet in the x = 0 plane has a charge density +σ and the sheet in the x = a plane has a charge density –σ . Picture the Problem Let the charge density on the infinite plane at x = a be σ1 and that on the infinite plane at x = 0 be σ2. Call that region in space for which x < 0, region I, the region for which 0 < x < a, region II, and the region for which a < x, region III. We can integrate E due to the planes of charge to find the electric potential in each of these regions.

236

Chapter 23 x

(a) Express the potential in region I in terms of the electric field in that region:

r r VI = − ∫ E I ⋅ dx

Express the electric field in region I as the sum of the fields due to the charge densities σ1 and σ2:

r σ σ E I = − 1 iˆ − 2 iˆ 2 ∈0 2 ∈0

r Substitute for E I and evaluate VI:

0

⎛ σ ⎞ σ VI = − ∫ ⎜⎜ − ⎟⎟dx = x + V (0) ∈0 ⎠ ∈0 0⎝ x

=

Express the potential in region II in terms of the electric field in that region: Express the electric field in region II as the sum of the fields due to the charge densities σ1 and σ2:

r Substitute for E II and evaluate VII:

σ ˆ σ ˆ σ i− i = − iˆ 2 ∈0 2 ∈0 ∈0

=−

σ σ x+0= x ∈0 ∈0

r r VII = − ∫ E II ⋅ dx + V (0)

r σ σ E II = − 1 iˆ + 2 iˆ 2 ∈0 2 ∈0

σ ˆ σ ˆ i+ i =0 2 ∈0 2 ∈0

=−

x

VII = − ∫ (0)dx = 0 + V (0) = 0 0

Express the potential in region III in terms of the electric field in that region:

x r r VIII = − ∫ E III ⋅ dx

Express the electric field in region III as the sum of the fields due to the charge densities σ1 and σ2:

r σ σ E III = 1 iˆ + 2 iˆ 2 ∈0 2 ∈0

a

=

σ ˆ σ ˆ σ ˆ i+ i= i ∈0 2 ∈0 2 ∈0

Electric Potential r Substitute for E III and evaluate VIII:

x ⎛σ ⎞ σ σ VIII = − ∫ ⎜⎜ ⎟⎟dx = − x + a ∈ ∈0 ∈0 a⎝ 0 ⎠

= − (b) Proceed as in (a) with σ1 = −σ and σ2 = σ to obtain:

237

σ (x − a ) ∈0

VI = 0 , VII = −

σ x , and ∈0

VIII = 0

These results are summarized in the following table: Region x ≤ 0 0 ≤ x ≤ a 0 Part (a) σ x

∈0

Part (b)

0

−

σ x ∈0

x≥a

−

σ (x − a ) ∈0 0

The expression for the potential along the axis of a thin uniformly ⎛ ⎞ R2 charge disk is given by V = 2πkσ z ⎜ 1 + 2 − 1⎟ (Equation 23-20), where R and ⎜ ⎟ z ⎝ ⎠ σ are the radius and the charge per unit area of the disk, respectively. Show that this expression reduces to V = kQ z for z >> R , where Q = σπ R 2 is the total 50

•••

charge on the disk. Explain why this result is expected. Hint: Use the binomial theorem to expand the radical. Picture the Problem

Expand the radical expression binomially to obtain: 2 2 R2 ⎛ 1 ⎞ ⎛ R ⎞ ( 12 )(− 12 ) ⎛ R ⎞ ⎜ ⎟ + higher order terms 1 + 2 = 1 + ⎜ ⎟ ⎜⎜ 2 ⎟⎟ + 2! ⎜⎝ z 2 ⎟⎠ z ⎝ 2 ⎠⎝ z ⎠ 2

For z >> R :

1+

2 R2 ⎛ 1 ⎞⎛ R ⎞ ⎜ 1 ≈ + ⎜ ⎟ ⎜ 2 ⎟⎟ z2 ⎝ 2 ⎠⎝ z ⎠

and 1+

R2 R2 1 − ≈ z2 2z 2

238

Chapter 23

Substituting in Equation 23-20 yields:

⎛ R2 ⎞ ⎛ R2 ⎞ V = 2πkσ z ⎜⎜ 2 ⎟⎟ = πkσ z ⎜⎜ 2 ⎟⎟ ⎝z ⎠ ⎝ 2z ⎠

The total charge on the disk is given by:

Q = σπ R 2 ⇒ σ =

Substitute for σ and simplify to obtain:

⎛ Q ⎞ V = πk ⎜ 2 ⎟ ⎝ πR ⎠

If z ≥ 0 , then z = z and:

If z < 0 , then z = − z and:

Thus, for z >> R , Equation 23-20 reduces to:

Q πR 2

⎛ R2 ⎞ k z Q z ⎜⎜ 2 ⎟⎟ = 2 z ⎝z ⎠

V=

kzQ kQ kQ = = z2 z z

V=

k (− z )Q kQ kQ = = −z z2 z

V=

kQ z

51 •• [SSM] A rod of length L has a total charge Q uniformly distributed along its length. The rod lies along the y-axis with its center at the origin. (a) Find an expression for the electric potential as a function of position along the x-axis. (b) Show that the result obtained in Part (a) reduces to V = kQ x for x >> L .

Explain why this result is expected. Picture the Problem Let the charge per unit length be λ = Q/L and dy be a line element with charge λdy. We can express the potential dV at any point on the x axis due to the charge element λdy and integrate to find V(x, 0).

(a) Express the element of potential dV due to the line element dy:

dV =

kλ dy r

where r = x 2 + y 2 and λ = Substituting for r and λ yields:

dV =

kQ L

dy x + y2 2

Q . L

Electric Potential Use a table of integrals to integrate dV from y = −L/2 to y = L/2:

V ( x,0) =

L2

kQ dy ∫ L −L 2 x2 + y 2

2 2 kQ ⎛⎜ x + 14 L + 12 L ⎞⎟ = ln L ⎜ x 2 + 14 L2 − 14 L2 ⎟ ⎝ ⎠

(b) Factor x from the numerator and denominator within the parentheses to obtain:

2 ⎞ ⎛ ⎜ 1+ L + L ⎟ 2 kQ ⎜ 2x ⎟ 4x ln⎜ V ( x,0 ) = ⎟ 2 L ⎜ 1+ L − L ⎟ ⎟ ⎜ 4x 2 2x ⎠ ⎝

⎛a⎞ Use ln⎜ ⎟ = ln a − ln b to obtain: ⎝b⎠ V (x,0 ) =

⎛ kQ ⎧⎪ ⎛⎜ L2 L ⎞⎟ L2 L ⎞⎟⎫⎪ ⎜ ln 1 ln 1 + + − + − ⎬ ⎨ ⎜ L ⎪ ⎜⎝ 4 x 2 2 x ⎟⎠ 4 x 2 2 x ⎟⎠⎪ ⎝ ⎭ ⎩

L2 L2 12 1 1 2 Let ε = 2 and use (1 + ε ) = 1 + 2 ε − 8 ε + ... to expand 1 + 2 : 4x 4x 12

⎛ L2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ 4x ⎠

2

1 L2 1 ⎛ L2 ⎞ ⎟ + ... ≈ 1 for x >> L . = 1+ − ⎜ 2 4 x 2 8 ⎜⎝ 4 x 2 ⎟⎠

12

⎛ L2 ⎞ Substitute for ⎜⎜1 + 2 ⎟⎟ to obtain: ⎝ 4x ⎠ V (x,0 ) =

Let δ =

kQ ⎧ ⎛ L⎞ L ⎞⎫ ⎛ ⎨ln⎜1 + ⎟ − ln⎜1 − ⎟⎬ L ⎩ ⎝ 2x ⎠ ⎝ 2 x ⎠⎭

L L⎞ ⎛ and use ln (1 + δ ) = δ − 12 δ 2 + ... to expand ln⎜1 ± ⎟ : 2x ⎝ 2x ⎠ L⎞ L L2 L⎞ L L2 ⎛ ⎛ ln⎜1 + ⎟ ≈ − 2 and ln⎜1 − ⎟ ≈ − − 2 for x >> L. 2x 4x ⎝ 2x ⎠ 2x 4x ⎝ 2x ⎠

239

240

Chapter 23

L⎞ L⎞ ⎛ ⎛ Substitute for ln⎜1 + ⎟ and ln⎜1 − ⎟ and simplify to obtain: ⎝ 2x ⎠ ⎝ 2x ⎠ kQ ⎧ L L2 ⎛ L L2 ⎞⎫ kQ ⎜ − 2 ⎟⎟⎬ = V ( x,0) = ⎨ − 2 − ⎜− L ⎩ 2 x 4 x ⎝ 2 x 4 x ⎠⎭ x Because, for x >> L , the charge carried by the rod is far enough away from the point of interest to look like a point charge, this result is what we would expect. 52 •• A rod of length L has a charge Q uniformly distributed along its length. The rod lies along the y-axis with one end at the origin. (a) Find an expression for the electric potential as a function of position along the x-axis. (b) Show that the result obtained in Part (a) reduces to V = kQ x for x >> L .

Explain why this result is expected. Picture the Problem Let the charge per unit length be λ = Q/L and dy be a line element with charge λdy. We can express the potential dV at any point on the x axis due to λdy and integrate to find V(x, 0).

(a) Express the element of potential dV due to the line element dy:

dV =

kλ dy r

where r = x 2 + y 2 and λ = Substituting for r and λ yields:

Use a table of integrals to integrate dV from y = 0 to y = L:

dV =

kQ L

V ( x,0) =

Q . L

dy x2 + y2 L

kQ dy ∫ L 0 x2 + y2

( [(

) )

L kQ ln y + x 2 + y 2 0 L kQ = ln L + x 2 + L2 − ln ( x ) L

=

⎛a⎞ Because ln a − ln b = ln⎜ ⎟ : ⎝b⎠

V ( x,0) =

kQ ⎡ ⎛⎜ L + x 2 + L2 ⎢ln L ⎢ ⎜⎝ x ⎣

]

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

(1)

Electric Potential

241

(b) Factor x2 under the radical to obtain: 2 ⎞ ⎛ ⎛ ⎜ L + x 2 ⎛⎜1 + ⎛ L ⎞ ⎞⎟ ⎟ ⎜ L + x2 ⎜ ⎟ ⎟⎟ ⎜ ⎜ ⎜ 2 2 ⎛L+ x +L ⎞ ⎝ ⎝x⎠ ⎠⎟ ⎜ ⎟ ⎜ = ln = ln⎜ ln ⎟ ⎜ ⎜ ⎟ ⎜ x x ⎠ ⎝ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎝ ⎠ ⎝

⎛ ⎛ L ⎞ 2 ⎞ ⎞⎟ ⎜1 + ⎜ ⎟ ⎟ ⎜ ⎝x⎠ ⎟⎟ ⎝ ⎠⎟ ⎟ x ⎟ ⎟ ⎠

Because x >> L :

⎛ L + x 2 + L2 ln⎜ ⎜ x ⎝

⎛ L⎞ Expanding ln⎜1 + ⎟ binomially ⎝ x⎠ yields:

⎛ L⎞ L 1⎛ L⎞ ln⎜1 + ⎟ = − ⎜ ⎟ x ⎠ x 2⎝ x ⎠ ⎝ + higher order terms

Again, because x >> L :

⎛ L⎞ L ln⎜1 + ⎟ ≈ x⎠ x ⎝

Substitute in equation (2) to obtain:

⎛ L + x 2 + L2 ln⎜ ⎜ x ⎝

Finally, substituting in equation (1) yields:

V (x,0 ) =

2 ⎛ ⎞ ⎟ ≈ ln⎜ L + x ⎜ ⎟ x ⎝ ⎠ ⎛ L⎞ = ln⎜1 + ⎟ x⎠ ⎝

⎞ ⎟ ⎟ ⎠

2

⎞ L ⎟≈ ⎟ x ⎠

kQ ⎡ L ⎤ kQ = ⎢ ⎥ L ⎣x⎦ x

Because, for x >> L , the charge carried by the rod is far enough away from the point of interest to look like a point charge, this result is what we would expect. 53 •• [SSM] A disk of radius R has a surface charge distribution given by 2 2 σ = σ0r /R where σ0 is a constant and r is the distance from the center of the disk. (a) Find the total charge on the disk. (b) Find an expression for the electric potential at a distance z from the center of the disk on the axis that passes through the disk′s center and is perpendicular to its plane.

242

Chapter 23

Picture the Problem We can find Q by integrating the charge on a ring of radius r and thickness dr from r = 0 to r = R and the potential on the axis of the disk by integrating the expression for the potential on the axis of a ring of charge between the same limits.

σ r z R dr

(a) Express the charge dq on a ring of radius r and thickness dr:

⎛ r2 ⎞ dq = 2πrσdr = 2πr ⎜⎜ σ 0 2 ⎟⎟dr ⎝ R ⎠ 2πσ 0 3 = r dr R2

Integrate from r = 0 to r = R to obtain:

2πσ 0 R 3 Q= r dr = R 2 ∫0

(b)Express the potential on the axis of the disk due to a circular element 2πσ 0 3 of charge dq = r dr : R2

dV =

kdq 2πkσ 0 = r' R2

[(

)

1 2

πσ 0 R 2 r3 x2 + r 2

dr

Integrate from r = 0 to r = R to obtain: 2πkσ 0 V = R2

R

∫ 0

r 3dr x +r 2

2

=

2πkσ 0 2 R − 2x2 2 3R

x 2 + R 2 + 2 x3

]

54 ••• A disk of radius R has a surface charge distribution given by σ = σ0R/r where σ0 is a constant and r is the distance from the center of the disk. (a) Find the total charge on the disk. (b) Find an expression for the electric potential at a distance x from the center of the disk on the axis that passes through the disk′s center and is perpendicular to its plane.

Electric Potential Picture the Problem We can find Q by integrating the charge on a ring of radius r and thickness dr from r = 0 to r = R and the potential on the axis of the disk by integrating the expression for the potential on the axis of a ring of charge between the same limits.

243

σ r z R dr

(a) Express the charge dq on a ring of radius r and thickness dr:

Integrate from r = 0 to r = R to obtain:

⎛ R⎞ dq = 2πrσdr = 2πr ⎜ σ 0 ⎟dr ⎝ r⎠ = 2πσ 0 Rdr R

Q = 2πσ 0 R ∫ dr = 2πσ 0 R 2 0

kdq 2πkσ 0 Rdr = r' x2 + r 2

(b) Express the potential on the axis of the disk due to a circular element of charge dq = 2πrσdr :

dV =

Integrate from r = 0 to r = R to obtain:

V = 2πkσ 0 R ∫

R

0

dr x2 + r 2

⎛ R + x2 + R2 = 2πkσ 0 R ln⎜ ⎜ x ⎝

⎞ ⎟ ⎟ ⎠

55 •• A rod of length L has a total charge Q uniformly distributed along its length. The rod lies along the x-axis with its center at the origin. (a) What is the electric potential as a function of position along the x-axis for x > L/2? (b) Show that for x >> L/2, your result reduces to that due to a point charge Q. Picture the Problem We can express the electric potential dV at x due to an elemental charge dq on the rod and then integrate over the length of the rod to find V(x). In the second part of the problem we use a binomial expansion to show that, for x >> L/2, our result reduces to that due to a point charge Q.

244

Chapter 23

(a) Express the potential at x due to the element of charge dq located at u:

kdq kλdu = r x−u or, because λ = Q/L, kQ du dV = L x−u

Integrate V from u = −L/2 to u = L/2 and simplify to obtain:

V (x ) =

dV =

L2

kQ du ∫ L −L 2 x − u

kQ L2 ln (x − u ) − L 2 L kQ [− ln(x − 12 L ) + ln(x + 12 L )] = L =−

=

(b) From Part (a):

kQ ⎛ x + 12 L ⎞ ⎟ ln⎜ L ⎜⎝ x − 12 L ⎟⎠

kQ [ln(x + 12 L ) − ln(x − 12 L )] L kQ ⎡ ⎛ L⎞ L ⎞⎤ ⎛ = ln x⎜1 + ⎟ − ln x⎜1 − ⎟⎥ ⎢ L ⎣ ⎝ 2x ⎠ ⎝ 2 x ⎠⎦

V (x ) =

Expanding the expression in the square brackets and simplifying gives: V (x ) =

For

kQ ⎡ L⎞ L ⎞⎤ kQ ⎡ ⎛ L⎞ L ⎞⎤ ⎛ ⎛ ⎛ ln x + ln⎜1 + ⎟ − ln x + ln⎜1 − ⎟⎥ = ln⎜1 + ⎟ + ln⎜1 − ⎟⎥ ⎢ ⎢ L ⎣ ⎝ 2 x ⎠⎦ ⎝ 2 x ⎠⎦ L ⎣ ⎝ 2 x ⎠ ⎝ 2x ⎠

L > b the electric potential on the axis of the uniformly charged disk with cutout approaches kQ/x, where Q = σπ(b2– a2) is the total charge on the disk.

Electric Potential

245

Picture the Problem The potential on the axis of the uniformly charged disk is the sum of the potential Vb due to the disk of radius b and the potential Va due to

the disk of radius a that has been removed. We can think of the charged disk that has been removed as having a negative charge density −σ. Note that if x >> b, then it is also true that x >> a. V (x ) = Vb ( x ) + Va ( x )

(a)The potential on the axis of the circular disk is:

where Vb ( x ) = 2πkσ x 2 + b 2

[(

)

12

−x

and Va ( x ) = 2πk (− σ ) x 2 + a 2

[(

)

]

12

−x

]

Substitute for Vb ( x ) and Va (x ) and simplify to obtain:

[( = 2πkσ [(x

V ( x ) = 2πkσ x 2 + b 2 = 2πkσ

⎛ b2 ⎞ (b) Expanding ⎜⎜1 + 2 ⎟⎟ ⎝ x ⎠ yields:

(

2

+ b2

) − x]+ 2πk (− σ )[(x ) − (x + a ) ] 12

12

2

binomially

+ a2

12

12

−x

]

)

⎛ b2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ x ⎠

12

= 1+

b2 2x 2

+ higher order terms ≈ 1+

⎛ a2 ⎞ Expanding ⎜⎜1 + 2 ⎟⎟ binomially ⎝ x ⎠ yields:

)

2 12

x2 + b2 − x2 + a2

12

2

12

⎛ a2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ x ⎠

b2 2x 2

a2 ≈ 1+ 2 2x

Substituting in the expression for V ( x ) and simplifying yields:

(

⎡ ⎡ b2 b2 ⎛ a 2 ⎞⎤ a 2 ⎤ πkσ b 2 − a 2 V ( x ) ≈ 2πkσx ⎢1 + 2 − ⎜⎜1 + 2 ⎟⎟⎥ = 2πkσx ⎢ 2 − 2 ⎥ = 2x ⎦ x ⎣ 2x ⎣ 2 x ⎝ 2 x ⎠⎦ The total charge on the disk is:

(

)

Q = σπ b 2 − a 2 ⇒ σ =

Q π b − a2

(

2

)

)

246

Chapter 23

Substituting for σ yields:

⎛

V (x ) ≈

Q 2 ⎝π b −a x

πk ⎜⎜

(

2

⎞ 2 ⎟⎟ b − a 2 kQ ⎠ = x

)(

)

The expression for the electric potential inside a uniformly charged kQ ⎛ r2 ⎞ solid sphere is given by V ( r ) = − 3 ⎜ ⎟ , where R is the radius of the sphere 2R ⎝ R2 ⎠ and r is the distance from the center. This expression was obtained in Example 23-12 by first finding the electric field. In this problem, you derive the same expression by modeling the sphere as a nested collection of thin spherical shells, and then adding the potentials of these shells at a field point inside the sphere. The potential dV that is a distance r′ from the center of a uniformly charged thin spherical shell that has a radius r′ and a charge dQ is given by dV = kdQ/r′ for r′ ≥ r and dV = kdQ/r for r′ ≤ r (Equation 23-22). Consider a sphere of radius R containing a charge Q that is uniformly distributed and you want to find V at some point inside the sphere (that is for r < R). (a) Find an expression for the charge dQ on a spherical shell of radius r′ and thickness dr′. (b) Find an expression for the potential dV at r due to the charge in a shell of radius r′ and thickness dr′, where r ≤ r′ ≤ R. (c) Integrate your expression in Part (b) from r′ = r to r′ = R to find the potential at r due to all the charge in the region farther from r than the center of the sphere. (d) Find an expression for the potential dV at r due to the charge in a shell of radius r′ and thickness dr′, where r′ ≤ r. (e) Integrate your expression in Part (d) from r′ = 0 to r′ = r to find the potential at r due to all the charge in the region closer than r to the center of the sphere. (f) Find the total potential V at r by adding your Part (c) and Part (e) results. 57

•••

Picture the Problem The diagram shows a uniformly charged sphere of radius R and the field point P at which we wish to find the total potential. We can use the definition of charge density to find the charge inside a sphere of radius r and the potential at r due to this charge. We can express the potential at r due to the charge in a shell of radius r′ and thickness dr′ at r′ ≥ r using dV = kdq' r and then integrate this expression from r′ = r to r′ = R to find V. Q P r' r

R

Electric Potential (a) The charge dQ in a shell of radius r′ and thickness dr′ at r′ > r is given by:

dQ = ρdV ' = ρA' dr '

Because A' = 4πr '2 :

dQ = 4πr '2 ρdr '

Because the sphere is uniformly charged:

ρ=

Substituting for ρ and simplifying yields:

3Q 2 ⎛ 3Q ⎞ dQ = 4πr' 2 ⎜ dr' = r' dr' 3 ⎟ R3 ⎝ 4πR ⎠

(b) Express the potential dV in the interval r ≤ r ' ≤ R due to dQ:

dV =

kdQ r'

Substituting for dQ and simplifying yields:

dV =

k ⎛ 3Q 2 ⎞ 3kQ r ' dr ' ⎜ 3 r' dr' ⎟ = r' ⎝ R R3 ⎠

(c) Integrate dV from r′ = r to r′ = R to find V:

V=

(d) The potential dV at r due to the charge in a shell of radius r′ and thickness dr′, where r′ ≤ r, is given by: Substituting for ρ and simplifying yields:

Q Q 3Q = 4 3 = V 3 πR 4πR 3

R

(

3kQ 3kQ 2 r'dr' = R − r2 3 ∫ 3 2R R r

kdQ kρdV ' kρA' dr ' = = r r r 2 kρ 4π r ' dr ' 4π kρ 2 = = r ' dr ' r r

dV =

(

dV =

)

4πk ⎛ Q ⎞ 2 ⎟ r ' dr ' ⎜ r ⎜⎝ 43 πR 3 ⎟⎠

⎛ 3kQ ⎞ = ⎜ 3 ⎟ r '2 dr ' ⎝R r⎠ (e) Integrate from r′ = 0 to r′ = r to find the potential at r due to all the charge in the region closer than r to the center of the sphere:

)

r

kQ 2 ⎛ 3kQ ⎞ V = ⎜ 3 ⎟ ∫ r '2 dr ' = r R3 ⎝ R r ⎠0

247

248

Chapter 23

(f) The sum of our results from Part (c) and Part (e) is:

V= =

(

kQ 2 3kQ 2 r + R − r2 3 3 R 2R

)

kQ ⎛ r2 ⎞ ⎜⎜ 3 − 2 ⎟⎟ 2R ⎝ R ⎠

58 •• Calculate the electric potential at the point a distance R/2 from the center of a uniformly charged thin spherical shell of radius R and charge Q. (Assume the potential is zero far from the shell.) Picture the Problem We can find the potential relative to infinity at a distance R/2 from the center of the spherical shell by integrating the electric field for 0 to ∞. We can apply Gauss’s law to find the electric field both inside and outside the spherical shell.

The potential relative to infinity at a distance R/2 from the center of the spherical shell is given by:

R2

R2 R r r V = − ∫ E ⋅ dr = − ∫ E r > R dr − ∫ E r < R dr

Because there is no charge inside the spherical shell, Er < R = 0 and:

R2

R

R

∞

Apply Gauss’s law to a spherical surface of radius r > R to obtain:

∫

Solving for Er>R yields:

Er > R =

∞

R

∫ Er R dr (1) S

Substitute for Er>R in equation (1) to obtain:

∞

(

)

En dA = Er > R 4πr 2 = Q

4π ∈0 r

2

=

Qinside

∈0

=

Q

∈0

kQ r2

R

dr r2 ∞

V = −kQ ∫

R

Evaluating this integral yields:

kQ ⎡ 1⎤ V = − kQ ⎢− ⎥ = R ⎣ r ⎦∞

59 •• [SSM] A circle of radius a is removed from the center of a uniformly charged thin circular disk of radius b. Show that the potential at a point on the central axis of the disk a distance z from its geometrical center is given by

V ( z ) = 2π kσ

(

)

z 2 + b 2 − z 2 + a 2 , where σ is the charge density of the disk.

Picture the Problem We can find the electrostatic potential of the conducting washer by treating it as two disks with equal but opposite charge densities.

Electric Potential

249

⎛ ⎞ R2 V ( x ) = 2πkσ x ⎜ 1 + 2 − 1⎟ ⎜ ⎟ x ⎝ ⎠

The electric potential due to a charged disk of radius R is given by:

Superimpose the electrostatic potentials of the two disks with opposite charge densities and simplify to obtain: ⎛ ⎞ ⎛ ⎞ b2 a2 V ( x ) = 2πkσ x ⎜ 1 + 2 − 1⎟ − 2πkσ x ⎜ 1 + 2 − 1⎟ ⎜ ⎟ ⎜ ⎟ x x ⎝ ⎠ ⎝ ⎠ ⎛⎛ ⎞ ⎛ ⎞⎞ b2 a2 = 2πkσ x ⎜ ⎜ 1 + 2 − 1⎟ − ⎜ 1 + 2 − 1⎟ ⎟ ⎟ ⎜ ⎟⎟ ⎜⎜ x x ⎠ ⎝ ⎠⎠ ⎝⎝ ⎛ ⎛ x2 + b2 b2 a 2 ⎞⎟ x 2 + a 2 ⎞⎟ ⎜ ⎜ = 2πkσ x 1 + 2 − 1 + 2 = 2πkσ x − ⎜ ⎜ x x ⎟⎠ x2 x 2 ⎟⎠ ⎝ ⎝ ⎛ x2 + b2 ⎛ x2 + b2 x 2 + a 2 ⎞⎟ x2 + a2 = 2πkσ x ⎜ − = 2πkσ x ⎜ − ⎜ ⎜ x x x2 x 2 ⎟⎠ ⎝ ⎝ = 2πkσ

(x

2

+ b2 − x2 + a2

The charge density σ is given by:

)

σ=

⎞ ⎟ ⎟ ⎠

Q π (b − a 2 ) 2

Substituting for σ yields: V (x ) =

2kQ (b − a 2 ) 2

(x

2

+ b2 − x2 + a2

)

Equipotential Surfaces 60 • An infinite flat sheet of charge has a uniform surface charge density equal to 3.50 μC/m2. How far apart are the equipotential surfaces whose potentials differ by 100 V? Picture the Problem We can equate the expression for the electric field due to an infinite plane of charge and −ΔV/Δx and solve the resulting equation for the separation of the equipotential surfaces.

Express the electric field due to the infinite plane of charge:

E=

σ 2 ∈0

250

Chapter 23 ΔV Δx

Relate the electric field to the potential:

E=−

Equate these expressions and solve for Δx to obtain:

Δx =

Substitute numerical values and evaluate Δx :

⎛ C2 ⎞ ⎟ (100 V ) 2⎜⎜ 8.854 × 10−12 N ⋅ m 2 ⎟⎠ ⎝ Δx = 3.50 μC/m 2

2 ∈0 ΔV

σ

= 0.506 mm 61 •• [SSM] Consider two parallel uniformly charged infinite planes that are equal but oppositely charged. (a) What is (are) the shape(s) of the equipotentials in the region between them? Explain your answer. (b) What is (are) the shape(s) of the equipotentials in the regions not between them? Explain your answer. Picture the Problem The two parallel planes, with their opposite charges, are shown in the pictorial representation. z

+Q

−Q

y

x (a) Because the electric field between the charged plates is uniform and perpendicular to the plates, the equipotential surfaces are planes parallel to the charged planes.

(b) The regions to either side of the two charged planes are equipotential regions, so any surface in either of these regions is an equipotential surface. 62 •• A Geiger tube consists of two elements, a long metal cylindrical shell and a long straight metal wire running down its central axis. Model the tube as if both the wire and cylinder are infinitely long. The central wire is positively charged and the outer cylinder is negatively charge. The potential difference between the wire and the cylinder is 1.00 kV. (a) What is the direction of the electric field inside the tube? (b) Which element is at a higher electric potential?

Electric Potential

251

(c) What is (are) the shape(s) of the equipotentials inside the tube? (d) Consider two equipotentials described in Part (c). Suppose they differ in electric potential by 10 V. Do two such equipotentials near the central wire have the same spacing as they would near the outer cylinder? If not, where in the tube are the equipotentials that are more widely spaced? Explain your answer. Determine the Concept (a) The direction of the electric field inside the tube is the direction of the force the electric field exerts on a positively charged object. Because the wire is positively charged and the tube is negatively charged, and because of the cylindrical geometry of the Geiger tube, the electric field is directed radially away from the central wire.

(b) Because it would require more work to bring a positively charged object from infinity to the surface of the wire than it would to bring this test object to the surface of the cylindrical tube, the central wire is at the higher electric potential. (c) Because of the cylindrical geometry of the Geiger tube, the equipotential surfaces are cylinders concentric with the central wire. (d) No. Because the magnitude of the electric field, which is the rate of change with distance (also known as the gradient) of the potential decreases with distance from the wire, the spacing between adjacent equipotential surfaces having the same potential difference between them decreases as you get farther from the central wire. 63 •• [SSM] Suppose the cylinder in the Geiger tube in Problem 62 has an inside diameter of 4.00 cm and the wire has a diameter of 0.500 mm. The cylinder is grounded so its potential is equal to zero. (a) What is the radius of the equipotential surface that has a potential equal to 500 V? Is this surface closer to the wire or to the cylinder? (b) How far apart are the equipotential surfaces that have potentials of 200 and 225 V? (c) Compare your result in Part (b) to the distance between the two surfaces that have potentials of 700 and 725 V respectively. What does this comparison tell you about the electric field strength as a function of the distance from the central wire? Picture the Problem If we let the electric potential of the cylinder be zero, then the surface of the central wire is at +1000 V and we can use Equation 23-23 to find the electric potential at any point between the outer cylinder and the central wire.

252

Chapter 23 ⎛R ⎞ V (r ) = 2kλ ln⎜ ref ⎟ (1) ⎝ r ⎠ where Rref is the radius of the outer cylinder and r is the distance from the center of the central wire and r < Rref.

(a) From Equation 23-23 we have:

Solving for 2kλ yields:

2kλ =

At the surface of the wire, V = 1000 V and r = 0.250 mm. Hence:

2kλ =

V (r ) ⎛R ⎞ ln⎜ ref ⎟ ⎝ r ⎠ 1000 V = 228.2 V ⎛ 2.00 cm ⎞ ln⎜ ⎟ ⎝ 0.250 mm ⎠

and ⎛ 2.00 cm ⎞ V (r ) = (228.2 V )ln⎜ ⎟ r ⎝ ⎠ Setting V = 500 V yields:

⎛ 2.00 cm ⎞ 500 V = (228.2 V )ln⎜ ⎟ r ⎝ ⎠ or 2.00 cm ⎛ 2.00 cm ⎞ = e 2.191 ln⎜ ⎟ = 2.191 ⇒ r r ⎝ ⎠

Solve for r to obtain:

2.00 cm = 0.224 cm , closer to e 2.191 the wire.

(b) The separation of the equipotential surfaces that have potential values of 200 and 225 V is:

Δr = r225 V − r200 V

r=

Solving equation (1) for r yields:

r = Rref e

−

V 2 kλ

(2)

= (2.00 cm )e

−

V 228.2 V

Substitute for the radii in equation (2), simplify, and evaluate Δr to obtain: Δr = (2.00 cm )e

−

= 0.864 mm

225 V 228.2 V

− (2.00 cm )e

−

200 V 228.2 V

= (2.00 cm ) e

−

225 V 228.2 V

−e

−

200 V 228.2 V

Electric Potential (c) The distance between the 700 V and the 725 V equipotentials is:

Δr = (2.00 cm ) e

−

725 V 228.2 V

−e

−

253

700 V 228.2 V

= 0.0966 mm This closer spacing of these two equipotential surfaces was to be expected. Close to the central wire, two equipotential surfaces with the same difference in potential should be closer together to reflect the fact that the higher electric field strength is greater closer to the wire. 64 •• A point particle that has a charge of +11.1 nC is at the origin. (a) What is (are) the shapes of the equipotential surfaces in the region around this charge? (b) Assuming the potential to be zero at r = ∞, calculate the radii of the five surfaces that have potentials equal to 20.0 V, 40.0 V, 60.0 V, 80.0 V and 100.0 V, and sketch them to scale centered on the charge. (c) Are these surfaces equally spaced? Explain your answer. (d) Estimate the electric field strength between the 40.0-V and 60.0-V equipotential surfaces by dividing the difference between the two potentials by the difference between the two radii. Compare this estimate to the exact value at the location midway between these two surfaces. Picture the Problem We can integrate the expression for the electric field due to a point charge to find an expression for the electric potential of the point particle.

(a) The equipotential surfaces are spheres centered on the charge. (b) From the relationship between the electric potential due to the point charge and the electric field of the point charge we have:

b

rb

rb r r −2 ∫ dV = − ∫ E ⋅ dr = −kQ ∫ r dr a

ra

ra

or ⎛1 1⎞ Vb − Va = kQ⎜⎜ − ⎟⎟ ⎝ rb ra ⎠ ⎛1⎞ kQ kQ ⇒r = Vb − 0 = kQ⎜⎜ ⎟⎟ ⇒ V = r V ⎝ rb ⎠

Taking the potential to be zero at ra = ∞ yields: Because Q = +1.11 × 10–8 C:

2 ⎛ N ⋅ m2 ⎞ ⎜⎜ 8.988 × 10 9 ⎟⎟ 1.11× 10 −8 C 99.77 N ⋅ m 2 C ⎠ C r=⎝ = V V

(

)

(1)

254

Chapter 23

Use equation (1) to complete the following table: V (V) 20.0 40.0 60.0 80.0 100.0 r (m) 4.99 2.49 1.66 1.25 1.00 The equipotential surfaces are shown in cross-section to the right:

20.0 V

40.0 V 60.0 V

80.0 V 100.0 V

point charge

(c) No. The equipotential surfaces are closest together where the electric field strength is greatest. (d) The average value of the magnitude of the electric field between the 40.0-V and 60.0-V equipotential surfaces is given by:

E=−

Drop perpendiculars to the r axis from 40.0 V and 60.0 V to approximate the radii corresponding to each of these potential surfaces:

ΔV 40 V − 60 V =− Δr Δr

Eest ≈ −

V 40 V − 60 V = 29 2.4 m − 1.7 m m

The exact value of the electric field at the location midway between these two surfaces is given by E = kQ r 2 , where r is the average of the radii of the 40.0-V and 60.0-V equipotential surfaces. Substitute numerical values and evaluate Eexact. ⎛ N ⋅ m2 ⎞ ⎜⎜ 8.988 ×10 9 ⎟⎟ 1.11×10 −8 C 2 C V ⎠ =⎝ = 23 2 m ⎛ 1.66 m + 2.49 m ⎞ ⎜ ⎟ 2 ⎝ ⎠

(

Eexact

)

The estimated value for E differs by about 21% from the exact value.

Electric Potential

255

Electrostatic Potential Energy 65 • Three point charges are on the x-axis: q1 is at the origin, q2 is at x = +3.00 m, and q3 is at x = +6.00 m. Find the electrostatic potential energy of this system of charges for the following charge values: (a) q1 = q2 = q3 = +2.00 μC; (b) q1 = q2 = +2.00 μC and q3 = –2.00 μC; and (c) q1 = q3 = +2.00 μC and q2 = –2.00 μC. (Assume the potential energy is zero when the charges are very far from each other.)

The electrostatic potential energy of this system of three point charges is the work needed to bring the charges from an infinite separation to the final positions shown in the diagram. Express the work required to assemble this system of charges:

q

1

q3

3

6

x, m

0

U=

q2

kq1q2 kq1q3 kq2 q3 + + r1, 2 r1,3 r2,3

⎛qq qq q q ⎞ = k ⎜⎜ 1 2 + 1 3 + 2 3 ⎟⎟ r1,3 r2,3 ⎠ ⎝ r1, 2 Find the distances r1,2, r1,3, and r2,3:

r1, 2 = 3 m, r2,3 = 3 m, and r1,3 = 6 m

(a) Evaluate U for q1 = q2 = q3 = 2.00 μC:

⎛ N ⋅ m 2 ⎞ ⎛ (2.00 μC )(2.00 μC ) (2.00 μC )(2.00 μC ) ⎟⎜ + U = ⎜⎜ 8.988 ×10 9 C 2 ⎟⎠ ⎜⎝ 3.00 m 6.00 m ⎝ +

(2.00 μC)(2.00 μC) ⎞⎟ 3.00 m

⎟ ⎠

= 30.0 mJ (b) Evaluate U for q1 = q2 = 2.00 μC and q3 = −2.00 μC: ⎛ N ⋅ m 2 ⎞ ⎛ (2.00 μC )(2.00 μC ) (2.00 μC )(− 2.00 μC ) ⎟⎜ + U = ⎜⎜ 8.988 ×10 9 3.00 m 6.00 m C 2 ⎟⎠ ⎜⎝ ⎝ (2.00 μC )(− 2.00 μC) ⎞⎟ + ⎟ 3.00 m ⎠ = − 5.99 mJ

256

Chapter 23

(c) Evaluate U for q1 = q3 = 2.00 μC and q2 = −2.00 μC: ⎛ N ⋅ m 2 ⎞ ⎛ (2.00 μC )(− 2.00 μC ) (2.00 μC )(2.00 μC ) ⎟⎟ ⎜⎜ + U = ⎜⎜ 8.988 ×10 9 2 3 . 00 m 6.00 m C ⎝ ⎝ ⎠ (− 2.00 μC)(2.00 μC) ⎞⎟ + ⎟ 3.00 m ⎠ = − 18.0 mJ 66 • Point charges q1, q2, and q3 are fixed at the vertices of an equilateral triangle whose sides are 2.50 m-long. Find the electrostatic potential energy of this system of charges for the following charge values: (a) q1 = q2 = q3 = +4.20 μC, (b) q1 = q2 = +4.20 μC and q3 = –4.20 μC; and (c) q1 = q2 = –4.20 μC and q3 = +4.20 μC. (Assume the potential energy is zero when the charges are very far from each other.) q2 0m 2.5

0m

Express the work required to assemble this system of charges:

2.5

Picture the Problem The electrostatic potential energy of this system of three point charges is the work needed to bring the charges from an infinite separation to the final positions shown in the diagram.

q1

U=

2.50 m

q3

kq1q2 kq1q3 kq2 q3 + + r1, 2 r1,3 r2,3

⎛qq qq q q ⎞ = k ⎜⎜ 1 2 + 1 3 + 2 3 ⎟⎟ r1,3 r2,3 ⎠ ⎝ r1, 2 Find the distances r1,2, r1,3, and r2,3:

r1, 2 = r2,3 = r1,3 = 2.50 m

(a) Evaluate U for q1 = q2 = q3 = 4.20 μC: ⎛ N ⋅ m 2 ⎞ ⎡ (4.20 μC )(4.20 μC ) (4.20 μC )(4.20 μC ) ⎟ U = ⎜⎜ 8.988 × 10 9 + 2.50 m 2.50 m C 2 ⎟⎠ ⎢⎣ ⎝ (4.20 μC )(4.20 μC)⎤ + ⎥ 2.50 m ⎦ = 190 mJ

Electric Potential

257

(b) Evaluate U for q1 = q2 = 4.20 μC and q3 = −4.20 μC: ⎛ N ⋅ m 2 ⎞ ⎡ (4.20 μC )(4.20 μC ) (4.20 μC )(− 4.20 μC ) ⎟ U = ⎜⎜ 8.988 ×10 9 + 2.50 m 2.50 m C 2 ⎟⎠ ⎢⎣ ⎝ (4.20 μC)(− 4.20 μC)⎤ + ⎥ 2.50 m ⎦ = − 63.4 mJ (c) Evaluate U for q1 = q2 = −4.20 μC and q3 = +4.20 μC: 2 ⎛ (− 4.20 μC )(4.20 μC) 9 N ⋅ m ⎞ ⎡ (− 4.20 μC )(− 4.20 μC ) ⎜ ⎟⎟ ⎢ + U = ⎜ 8.988 ×10 2 2.50 m 2.50 m C ⎠⎣ ⎝ (− 4.20 μC)(4.20 μC)⎤ + ⎥ 2.50 m ⎦

= − 63.4 mJ 67 •• [SSM] (a) How much charge is on the surface of an isolated spherical conductor that has a 10.0-cm radius and is charged to 2.00 kV? (b)What is the electrostatic potential energy of this conductor? (Assume the potential is zero far from the sphere.) Picture the Problem The potential of an isolated spherical conductor is given by V = kQ r , where Q is its charge and r its radius, and its electrostatic potential energy by U = 12 QV . We can combine these relationships to find the sphere’s

electrostatic potential energy. (a) The potential of the isolated spherical conductor at its surface is related to its radius:

RV kQ ⇒Q = R k where R is the radius of the spherical conductor.

Substitute numerical values and evaluate Q:

Q=

V =

(10.0 cm )(2.00 kV ) 8.988 ×10 9

N ⋅ m2 C2

= 22.25 nC = 22.3 nC

258

Chapter 23

(b) Express the electrostatic potential energy of the isolated spherical conductor as a function of its charge Q and potential V:

U = 12 QV

Substitute numerical values and evaluate U:

U=

1 2

(22.25 nC)(2.00 kV ) =

22.3 μJ

68 ••• Four point charges, each having a charge with a magnitude of 2.00 μC, are at the corners of a square whose sides are 4.00 m-long. Find the electrostatic potential energy of this system under the following conditions: (a) all of the charges are negative, (b) three of the charges are positive and one of the charges is negative, and (c) the charges at two adjacent corners are positive and the other two charges are negative. (d) the charges at two opposite corners are positive and the other two charges are negative. (Assume the potential energy is zero when the point charges are very far from each other.) Picture the Problem The electrostatic potential energy of this system of four point charges is the work needed to bring the charges from an infinite separation to the final positions shown in the diagram.

q1

q2

0 4.

q4

0

2

m

4.00 m

4.00 m

The work required to assemble this system of charges equals the potential energy of the assembled system: U=

kq1q2 kq1q3 kq1q4 kq2 q3 kq2 q4 kq3 q4 + + + + + r1, 2 r1,3 r1, 4 r2,3 r2, 4 r3, 4

⎛q q q q q q q q q q q q ⎞ = k ⎜⎜ 1 2 + 1 3 + 1 4 + 2 3 + 2 4 + 3 4 ⎟⎟ r1,3 r1, 4 r2,3 r2, 4 r3, 4 ⎠ ⎝ r1, 2 Find the distances r1,2, r1,3, r1,4, r2,3, r2,4, and r3,4,:

r1, 2 = r2,3 = r3, 4 = r1, 4 = 4.00 m and r1,3 = r2, 4 = 4.00 2 m

q3

Electric Potential

259

(a) Evaluate U for q1 = q2 = q3 = q4 = −2.00 μC: ⎛ N ⋅ m 2 ⎞ ⎡ (− 2.00 μC )(− 2.00 μC ) (− 2.00 μC )(− 2.00 μC ) ⎟⎢ U = ⎜⎜ 8.988 × 109 + C 2 ⎟⎠ ⎣ 4.00 m 4.00 2 m ⎝ (− 2.00 μC )(− 2.00 μC ) + (− 2.00 μC )(− 2.00 μC) + 4.00 m 4.00 m +

(− 2.00 μC )(− 2.00 μC) + (− 2.00 μC)(− 2.00 μC)⎤ 4.00 2 m

⎥ ⎦

4.00 m

= 48.7 mJ (b) Evaluate U for q1 = q2 = q3 = 2 μC and q4 = −2.00 μC: ⎛ N ⋅ m 2 ⎞ ⎡ (2.00 μC )(2.00 μC ) (2.00 μC )(2.00 μC ) ⎟⎢ + U = ⎜⎜ 8.988 × 109 C 2 ⎟⎠ ⎣ 4.00 m 4.00 2 m ⎝ (2.00 μC)(− 2.00 μC ) + (2.00 μC)(2.00 μC ) + 4.00 m 4.00 m +

(2.00 μC)(− 2.00 μC ) + (2.00 μC)(− 2.00 μC)⎤ 4.00 2 m

4.00 m

⎥ ⎦

= 0 (c) Let q1 = q2 = 2.00 μC and q3 = q4 = −2.00 μC: ⎛ N ⋅ m 2 ⎞ ⎡ (2.00 μC )(2.00 μC ) (2.00 μC )(− 2.00 μC ) ⎟⎟ ⎢ + U = ⎜⎜ 8.988 ×10 9 2 C 4 . 00 m 4.00 2 m ⎝ ⎠⎣ (2.00 μC)(− 2.00 μC) + (2.00 μC )(− 2.00 μC ) + 4.00 m 4.00 m + = − 12.7 mJ

(2.00 μC )(− 2.00 μC) + (− 2.00 μC)(− 2.00 μC)⎤ 4.00 2 m

4.00 m

⎥ ⎦

260

Chapter 23

(d) Let q1 = q3 = 2.00 μC and q2 = q4 = −2.00 μC: 2 ⎛ (2.00 μC )(2.00 μC ) 9 N ⋅ m ⎞ ⎡ (2.00 μC )(− 2.00 μC ) ⎟⎟ ⎢ + U = ⎜⎜ 8.988 × 10 2 C ⎠⎣ 4.00 m 4.00 2 m ⎝ (2.00 μC )(− 2.00 μC) + (− 2.00 μC )(2.00 μC) + 4.00 m 4.00 m

+

(− 2.00 μC )(− 2.00 μC) + (2.00 μC)(− 2.00 μC)⎤ ⎥ ⎦

4.00 m

4.00 2 m

= − 23.2 mJ 69 •• [SSM] Four point charges are fixed at the corners of a square centered at the origin. The length of each side of the square is 2a. The charges are located as follows: +q is at (–a, +a), +2q is at (+a, +a), –3q is at (+a, –a), and +6q is at (–a, –a). A fifth particle that has a mass m and a charge +q is placed at the origin and released from rest. Find its speed when it is a very far from the origin. Picture the Problem The diagram shows the four point charges fixed at the corners of the square and the fifth charged particle that is released from rest at the origin. We can use conservation of energy to relate the initial potential energy of the particle to its kinetic energy when it is at a great distance from the origin and the electrostatic potential at the origin to express Ui.

Use conservation of energy to relate the initial potential energy of the particle to its kinetic energy when it is at a great distance from the origin: Express the initial potential energy of the particle to its charge and the electrostatic potential at the origin: Substitute for Kf and Ui to obtain:

y

q

a

2q

2a m, q

a

6q

x

− 3q

ΔK + ΔU = 0 or, because Ki = Uf = 0, K f −U i = 0 U i = qV (0 )

1 2

mv 2 − qV (0 ) = 0 ⇒ v =

2qV (0 ) m

Electric Potential

261

2kq − 3kq 6kq kq + + + 2a 2a 2a 2a 6kq = 2a

Express the electrostatic potential at the origin:

V (0 ) =

Substitute for V(0) and simplify to obtain:

v=

2q ⎛ 6kq ⎞ 6 2k ⎜ ⎟= q m ⎝ 2a ⎠ ma

70 •• Consider two point particles that each have charge +e, are at rest, and are separated by 1.50 × 10–15 m. (a) How much work was required to bring them together from a very large separation distance? (b) If they are released, how much kinetic energy will each have when they are separated by twice their separation at release? (c) The mass of each particle is 1.00 u (1.00 AMU). What speed will each have when they are very far from each other? Picture the Problem (a) In the absence of other charged bodies, no work is required to bring the first proton from infinity to its initial position. We can use the work- energy theorem to find the work required to bring the second proton to a position 1.50 × 10−15 m away from the first proton. (b) and (c) We can apply conservation of mechanical energy to the two-proton system to find the kinetic energy of each proton when they are separated by twice their separation at release and when they are separated by a large distance.

(a) Apply the work-energy theorem to the second proton to obtain:

Wext = ΔK + ΔU = 0 +

ke 2 ke 2 = r r

Substitute numerical values and evaluate Wext : ⎛ N ⋅ m2 ⎞ ⎜⎜ 8.988 ×10 9 ⎟⎟ 1.602 ×10 −19 C 2 C ⎠ =⎝ 1.50 ×10 −15 m

(

Wext

)

2

= 1.538 ×10 −13 J ×

1 eV 1.602 ×10 −19 J

= 960 keV (b) Apply conservation of mechanical energy to the separating protons to obtain:

ΔK + ΔU = 0 ⇒ K f − K i + U f − U i = 0 or, because Ki = 0, Kf +U f −Ui = 0

Substituting for Ui and Uf and simplifying yields:

ke 2 ke 2 ke 2 ke 2 Kf = Ui −U f = − = − ri rf r 2r =

ke 2 2r

262

Chapter 23

Remembering that Kf is the kinetic energy of both protons, substitute numerical values and evaluate Kf, each proton: ⎛ N ⋅ m2 ⎞ ⎜⎜ 8.988 × 10 9 ⎟⎟ 1.602 × 10 −19 C 2 C ⎠ 1 = ⎝ 2 2 1.50 × 10 −15 m 1 eV = 3.844 × 10 −14 J × 1.602 × 10 −19 J

(

K f, each proton

(

)

2

)

= 240 keV

(c) Apply conservation of mechanical energy to the separating protons to obtain: Solving for v∞ yields:

ΔK + ΔU = 0 ⇒ K f − K i + U f − U i = 0 or, because Ki = Uf = 0, K f −U i = 0 ⇒ 12 mp v∞2 − U i = 0

v∞ =

2U i where Ui is half the initial mp

potential energy of the two-proton system. Substitute numerical values and evaluate v∞:

⎛ 1.602 × 10 −19 2⎜⎜ 480 keV × eV v∞ = ⎝ − 27 1.673 × 10 kg

J⎞ ⎟⎟ ⎠

= 9.59 × 10 6 m/s 71 ••• Consider an electron and a proton that are initially at rest and are separated by 2.00 nm. Neglecting any motion of the much more massive proton, what is the minimum (a) kinetic energy and (b) speed that the electron must be projected at so it reaches a point a distance of 12.0 nm from the proton? Assume the electron’s velocity is directed radially away from the proton. (c) How far will the electron travel away from the proton if it has twice that initial kinetic energy? Picture the Problem We can apply the conservation of mechanical energy to the electron-proton system to find the minimum initial kinetic energy (that is, the initial kinetic energy that corresponds to a final kinetic energy of zero) required in Part (a) and, in Part (c), to find how far the electron will travel away from the proton if it has twice the initial kinetic energy found in Part (a).

Electric Potential

263

(a) Apply conservation of mechanical energy to the electronstationary proton system with K f = 0 to obtain:

ΔK + ΔU = 0 ⇒ − K i, min + U f − U i = 0

Solving for K i, min and substituting for

ke 2 ⎛ ke 2 ⎞ ⎟ −⎜− rf ⎜⎝ ri ⎟⎠ or, simplifying further, ⎛1 1⎞ K i, min = ke 2 ⎜⎜ − ⎟⎟ ⎝ ri rf ⎠

U f and U i yields:

K i, min = U f − U i = −

Substitute numerical values and evaluate K i, min : ⎛ 2⎛ N ⋅ m2 ⎞ 1 1 ⎞ ⎟⎟ 1.602 × 10 −19 C ⎜ K i, min = ⎜⎜ 8.988 × 10 9 − ⎟ 2 C ⎠ ⎝ 2.00 nm 12.0 nm ⎠ ⎝

(

)

= 9.611× 10 − 20 J = 9.61× 10 − 20 J

(b) Relate the minimum initial kinetic energy of the electron to its initial speed: Substitute numerical values and evaluate vi:

2 K i, min

K i, min = 12 me vi2 ⇒ vi =

vi =

(

2 9.611× 10 − 20 J 9.109 × 10 −31 kg

me

)

= 4.59 ×10 5 m/s (c) Apply conservation of mechanical energy to the electron- proton system to obtain:

ΔK + ΔU = 0 ⇒ K f − K i + U f − U i = 0

Assuming, as we did in Part (a), that K f = 0 yields:

− 2 K i, min + U f − U i = 0

Substituting for U f and U i yields:

Solve for rf to obtain:

− 2 K i, min −

rf =

ke 2 ke 2 + =0 rf ri

ri 2ri K i, min 1− ke 2

264

Chapter 23

Substituting numerical values and evaluating rf yields a negative value; a result that is not physical and suggests that, contrary to our assumption, K f ≠ 0 . To confirm that this is the case, assume that the electron escapes from the proton (it’s final electrostatic potential will then be equal to zero) and find the initial kinetic energy required for this to occur. If the electron is to escape the influence of the proton, its final electrostatic potential energy will be zero and:

K i, escape

⎛ ke 2 ⎞ ke 2 ⎟⎟ = = −U i = −⎜⎜ − r ri i ⎠ ⎝

Substitute numerical values and evaluate K i, escape : ⎛ N ⋅ m2 ⎞ ⎜⎜ 8.988 ×10 9 ⎟⎟ 1.602 ×10 −19 C 2 C ⎠ =⎝ 2.00 nm

(

K i, escape

)

2

= 1.15 ×10 −19 J

Because 2 K i, min > K i, escape , the electron escapes from the proton with residual kinetic energy.

General Problems A positive point charge equal to 4.80 × 10–19 C is separated from a 72 • negative point charge of the same magnitude by 6.40 × 10–10 m. What is the electric potential at a point 9.20 × 10–10 m from each of the two charges? Picture the Problem Because the charges are point charges, we can use the expression for the Coulomb potential to find the field at any distance from them.

kq+ kq− + r r k = (q+ + q− ) r

Using the expression for the potential due to a system of point charges, express the potential at the point 9.20 ×10−10 m from each of the two charges:

V=

Because q+ = −q−:

V= 0

73 • [SSM] Two positive point charges, each have a charge of +q, and are fixed on the y-axis at y = +a and y = –a. (a) Find the electric potential at any point on the x-axis. (b) Use your result in Part (a) to find the electric field at any point on the x-axis.

Electric Potential Picture the Problem The potential V at any point on the x axis is the sum of the Coulomb potentials due to the two point charges. Once we have found V, r we can use E = −(∂Vx ∂x )iˆ to find the

265

y a

+q x

electric field at any point on the x axis.

r −a +q

kqi ri

(a) Express the potential due to a system of point charges:

V =∑

Substitute to obtain:

V ( x ) = Vcharge at + a + Vcharge at -a

i

= =

(b) The electric field at any point on the x axis is given by:

kq x2 + a2

+

kq x2 + a2

2kq x2 + a2

r ∂V d ⎡ 2kq ⎤ ˆ E ( x ) = − x iˆ = − ⎢ ⎥i dx ⎣ x 2 + a 2 ⎦ ∂x =

(x

2kqx 2

+a

)

2 32

iˆ

74 • If a conducting sphere is to be charged to a potential of 10.0 kV, what is the smallest possible radius of the sphere so that the electric field near the surface of the sphere will not exceed the dielectric strength of air? Picture the Problem The radius of the sphere is related to the electric field and the potential at its surface. The dielectric strength of air is about 3 MV/m.

V (r ) V (r ) ⇒r = r Er

Relate the electric field at the surface of a conducting sphere to the potential at the surface of the sphere:

Er =

When E is a maximum, r is a minimum:

rmin =

V (r ) Emax

Substitute numerical values and evaluate rmin :

rmin =

10.0 kV ≈ 3 mm 3 MV/m

266

Chapter 23

75 •• [SSM] Two infinitely long parallel wires have a uniform charge per unit length λ and –λ respectively. The wires are parallel with the z-axis. The positively charged wire intersects the x-axis at x = –a, and the negatively charged wire intersects the x-axis at x = +a. (a) Choose the origin as the reference point where the potential is zero, and express the potential at an arbitrary point (x, y) in the xy plane in terms of x, y, λ, and a. Use this expression to solve for the potential everywhere on the y axis. (b) Using a = 5.00 cm and λ = 5.00 nC/m, obtain the equation for the equipotential in the xy plane that passes through the point x = 14 a, y = 0. (c) Use a spreadsheet program to plot the equipotential found in part (b). Picture the Problem The geometry of the wires is shown below. The potential at the point whose coordinates are (x, y) is the sum of the potentials due to the charge distributions on the wires.

(x,y) λ

r1

y r2 −λ

−a (a) Express the potential at the point whose coordinates are (x, y):

a

x

V (x, y ) = Vwire at −a + Vwire at a ⎛r ⎞ ⎛r = 2kλ ln⎜⎜ ref ⎟⎟ + 2k (− λ ) ln⎜⎜ ref ⎝ r1 ⎠ ⎝ r2 ⎡ ⎛r ⎞ ⎛ r ⎞⎤ = 2kλ ⎢ln⎜⎜ ref ⎟⎟ − ln⎜⎜ ref ⎟⎟⎥ ⎝ r2 ⎠⎦ ⎣ ⎝ r1 ⎠ =

⎛r ⎞ λ ln⎜⎜ 2 ⎟⎟ 2π ∈0 ⎝ r1 ⎠

where V(0) = 0. Because r1 = r2 =

(x + a )2 + y 2 and

(x − a )2 + y 2 :

On the y-axis, x = 0 and:

⎛ λ V ( x, y ) = ln⎜ 2π ∈0 ⎜ ⎝

(x − a )2 + y 2 ⎞⎟ (x + a )2 + y 2 ⎟⎠

⎛ a2 + y2 λ V (0, y ) = ln⎜ 2π ∈0 ⎜⎝ a 2 + y 2 =

λ ln(1) = 0 2π ∈0

⎞ ⎟ ⎟ ⎠

⎞ ⎟⎟ ⎠

Electric Potential (b) Evaluate the potential at ( 14 a,0) = (1.25 cm, 0) :

V ( 14 a,0) =

⎛ λ ln⎜ 2π ∈0 ⎜

267

( 14 a − a )2 ⎞⎟ ( 14 a + a )2 ⎟⎠

⎝ λ ⎛3⎞ ln⎜ ⎟ = 2π ∈0 ⎝ 5 ⎠

Equate V(x,y) and V ( 14 a,0) :

3 = 5

Solve for y to obtain:

(x − 5)2 + y 2 (x + 5)2 + y 2

y = ± 21.25 x − x 2 − 25

(c) A spreadsheet program to plot y = ± 21.25 x − x 2 − 25 is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell A2

Content/Formula 1.25

Algebraic Form 1 4 a

A3 B2

A2 + 0.05 SQRT(21.25*A2 − A2^2 − 25)

x + Δx y = 21.25 x − x 2 − 25

B4

−B2

y = − 21.25 x − x 2 − 25

1 2 3 4 5

A x 1.25 1.30 1.35 1.40

B ypos 0.00 0.97 1.37 1.67

C yneg 0.00 −0.97 −1.37 −1.67

373 374 375 376

19.80 19.85 19.90 19.95

1.93 1.67 1.37 0.97

−1.93 −1.67 −1.37 −0.97

268

Chapter 23

The following graph shows the equipotential curve in the xy plane for λ ⎛3⎞ V ( 14 a,0) = ln⎜ ⎟ . 2π ∈0 ⎝ 5 ⎠ 10 8 6 4

y (cm)

2 0 -2 -4 -6 -8 -10 0

5

10

15

20

x (cm)

76 •• The equipotential curve graphed in Problem 75 should be a circle. (a) Show mathematically that it is a circle. (b) The equipotential circle in the xy plane is the intersection of a three-dimensional equipotential surface and the xy plane. Describe the three-dimensional surface using one or two sentences. Picture the Problem We can use the expression for the potential at any point in the xy plane to show that the equipotential curve is a circle.

(a) Equipotential surfaces must satisfy the condition: Solving for r2/r1 yields:

V=

⎛r ⎞ λ ln⎜⎜ 2 ⎟⎟ 2π ∈ 0 ⎝ r1 ⎠ 2π ∈ V

0 r2 = e λ = C or r2 = Cr1 r1 where C is a constant.

Substitute for r1 and r2 to obtain:

(x − a )2 + y 2 = C 2 [(x + a )2 + y 2 ]

Expand this expression, combine like terms, and simplify to obtain:

x 2 + 2a

C 2 +1 x + y 2 = −a 2 2 C −1

Electric Potential

269

⎡ ⎛ C 2 + 1 ⎞2 ⎤ ⎟⎟ ⎥ to both sides of the equation: Complete the square by adding ⎢a 2 ⎜⎜ 2 ⎢⎣ ⎝ C − 1 ⎠ ⎥⎦ ⎡ ⎛ C 2 +1⎞2 ⎤ ⎡ ⎛ C 2 +1⎞2 ⎤ C 2 +1 4a 2 C 2 2 2 2 2 ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ ⎢ ⎥ x + 2a 2 x+ a ⎜ 2 ⎟ + y = a ⎜ 2 ⎟ −a = 2 C −1 ⎢⎣ ⎝ C − 1 ⎠ ⎥⎦ ⎢⎣ ⎝ C − 1 ⎠ ⎥⎦ C 2 −1 2

Letting α = 2a

(

C 2 +1 and C 2 −1

C β = 2a 2 yields: C −1

(x + α )2 + y 2 = β 2 ,

)

the equation of

circle in the xy plane with its center at (−α,0).

(b) The three-dimensional surfaces are cylinders whose axes are parallel to the wires and are in the y = 0 plane. 77 ••• The hydrogen atom in its ground state can be modeled as a positive point charge of magnitude +e (the proton) surrounded by a negative charge distribution that has a charge density (the electron) that varies with the distance from the center of the proton r as: ρ (r ) = − ρ 0 e −2 r a (a result obtained from quantum mechanics), where a = 0.523 nm is the most probable distance of the electron from the proton. (a) Calculate the value of ρ0 needed for the hydrogen atom to be neutral. (b) Calculate the electrostatic potential (relative to infinity) of this system as a function of the distance r from the proton. Picture the Problem (a) Expressing the charge dq in a spherical shell of volume 4πr2dr within a distance r of the proton and setting the integral of this expression equal to the charge of the electron will allow us to solve for the value of ρ 0 needed for charge neutrality. (b) The electrostatic potential of this system is the sum of the electrostatic potentials due to the proton and electron’s charge density. The potential due to the proton is ke/r. We can use the given charge density to express the potential function due to the electron’s charge distribution and then integrate this function to find the potential due to the electron.

(a) Express the charge dq in a spherical shell of volume dV = 4πr2dr at a distance r from the proton: Express the condition for charge neutrality:

(

)(

dq = ρdV = − ρ 0 e − 2 r a 4π r 2 dr = −4πρ 0 r 2 e − 2 r a dr

∞

e = −4πρ 0 ∫ r 2 e −2 r a dr 0

From a table of integrals we have:

2 bx ∫ x e dx =

(

e bx 2 2 b x − 2bx + 2 b3

)

)

270

Chapter 23

Using this result yields:

∞

2 −2r a ∫ r e dr = 0

a3 4

Substitute in the expression for e to obtain:

e a3 = −πρ 0 a 3 ⇒ ρ 0 = − e = −4πρ 0 4 π a3

Substitute numerical values and evaluate ρ0:

1.602 ×10 −19 C ρ0 = − π (0.523 nm )3 = − 3.56 ×108 C/m 3

(b) The electrostatic potential of this proton-electron system is the sum of the electrostatic potentials due to the proton and the electron’s charge density:

V = V1 + V2 where ke kQ V1 = + 1 , r r ∞ kρ (r ')4π r '2 dr ' V2 = ∫ r' r and

(1)

(2)

r

Q1 = ∫ ρ (r ')4π r ' 2 dr ' 0

Substituting for ρ (r ') in the expression for Q1 yields:

r

Q1 = 4πρ 0 ∫ r ' 2 e − 2 r ' a dr ' 0

From a table of integrals we have:

2 bx ∫ x e dx =

(

e bx 2 2 b x − 2bx + 2 b3

r

Using this result to evaluate ∫ r '2 e − 2 r ' a dr ' yields: 0

r

∫x 0

2

e

−2 x a

a 3e −2 x a dr ' = − 8

r

⎛ 4 2 ⎞ ⎛2⎞ ⎜⎜ 2 x + 2⎜ ⎟ x + 2 ⎟⎟ ⎝a⎠ ⎝a ⎠0

3 ⎞ a 3 e −2 r a ⎛ 4 2 4 ⎞ ⎛ a ⎜ 2 r + r + 2 ⎟ − ⎜⎜ − (2 )⎟⎟ a 8 ⎝a ⎠ ⎝ 8 ⎠ 3 −2 r a 3 a e ⎛ 4 2 4 ⎞ a =− ⎜ 2 r + r + 2⎟ + a 8 ⎝a ⎠ 4

=−

and 3 ⎡ a 3 e −2 r a ⎛ 4 2 4 ⎞ a ⎤ Q1 = 4πρ 0 ⎢− ⎜ 2 r + r + 2⎟ + ⎥ a 8 ⎝a ⎠ 4⎦ ⎣

)

Electric Potential

271

Substituting for Q1 in the expression for V1 yields: V1 =

ke 4πkρ 0 + r r

⎡ a 3 e −2 r a ⎢− 8 ⎣

3 ⎛ 4 2 4 ⎞ a ⎤ ⎜ 2 r + r + 2⎟ + ⎥ a ⎝a ⎠ 4⎦

Substitute for ρ 0 from Part (a) and simplify to obtain: ⎛ e ⎞ ⎟ 4πk ⎜⎜ − 3 ⎟ 3 −2 r a a π ke ⎝ ⎠ ⎡− a e V1 = + ⎢ r r 8 ⎣ =

3 ⎛ 4 2 4 ⎞ a ⎤ ⎜ 2 r + r + 2⎟ + ⎥ a ⎝a ⎠ 4⎦

ke −2 r a ⎛ 2 2 2 ⎞ e ⎜ 2 r + r + 1⎟ r a ⎝a ⎠

Substituting for ρ (r ') in equation (2) and simplifying yields:

∞

V2 = ∫ r

kρ 0 e −2 r a 4π r ' 2 dr ' r' ∞

= 4πkρ 0 ∫ e − 2 r a r ' dr ' r

From a table of integrals we have:

Using this result to evaluate ∞

−2r a ∫ e r ' dr ' yields:

bx ∫ xe dx =

∞

∫e r

r

e bx (bx − 1) b2 ∞

−2 x a

a2 ⎛2 ⎞ xdx = e −2 x a ⎜ x + 1⎟ 4 ⎝a ⎠r =−

∞

Substitute for ∫ e

−2r a

r ' dr ' and ρ 0 in the

r

expression for V2 to obtain:

a 2 −2 x a ⎛ 2 ⎞ e ⎜ r + 1⎟ 4 ⎝a ⎠

2 ⎛ e ⎞ −2 x a ⎡ a ⎛ 2 ⎞⎤ V2 = 4πk ⎜ − 3 ⎟e ⎢− ⎜ r + 1⎟⎥ ⎝ πa ⎠ ⎠⎦ ⎣ 4 ⎝a

⎛1⎞ ⎛2 ⎞ = ke⎜ ⎟e − 2 r a ⎜ r + 1⎟ ⎝a⎠ ⎝a ⎠

Substituting for V1 and V2 in equation (1) and simplifying yields: V=

ke − 2 r a ⎛ 2 2 2 ⎞ ke −2 r a ⎛ 2 ⎞ ⎛ 1 1 ⎞ −2r a e ⎜ 2 r + r + 1⎟ + e ⎜ r + 1⎟ = ke ⎜ + ⎟e r a ⎝a ⎠ a ⎝a ⎠ ⎝a r⎠

272

Chapter 23

78 •• Charge is supplied to the metal dome of a Van de Graaff generator by the belt at the rate of 200 μC/s when the potential difference between the belt and the dome is 1.25 MV. The dome transfers charge to the atmosphere at the same rate, so the 1.25 MV potential difference is maintained. What minimum power is needed to drive the moving belt and maintain the 1.25 MV potential difference? Picture the Problem We can use the definition of power and the expression for the work done in moving a charge through a potential difference to find the minimum power needed to drive the moving belt.

dW dt

Relate the power needed to drive the moving belt to the rate at which the generator is doing work:

P=

Express the work done in moving a charge q through a potential difference ΔV:

W = qΔV

Substitute for W to obtain:

P=

Substitute numerical values and evaluate P:

P = (1.25 MV )(200 μC/s ) = 250 W

d [qΔV ] = ΔV dq dt dt

79 •• A positive point charge +Q is located on the x axis at x = –a. (a) How much work is required to bring an identical point charge from infinity to the point on the x axis at x = +a? (b) With the two identical point charges in place at x = –a and x = +a, how much work is required to bring a third point charge –Q from infinity to the origin? (c) How much work is required to move the charge –Q from the origin to the point on the x axis at x = 2a along the semicircular path shown (Figure 23-35)? Picture the Problem We can use Wq→final position = QΔVi→ f to find the work

required to move these charges between the given points. (a) Express the required work in terms of the charge being moved and the potential due to the charge at x = +a and simplify to obtain:

W+ Q → + a = QΔV∞ → + a = Q[V (a ) − V (∞ )] kQ 2 ⎛ kQ ⎞ = QV (a ) = Q⎜ = ⎟ 2a ⎝ 2a ⎠

Electric Potential (b) Express the required work in terms of the charge being moved and the potentials due to the charges at x = +a and x = −a and simplify to obtain:

(c) Express the required work in terms of the charge being moved and the potentials due to the charges at x = +a and x = −a and simplify to obtain:

273

W−Q →0 = −QΔV∞ →0 = −Q[V (0) − V (∞ )] ⎡ ⎤ = −QV (0) = −Q ⎢Vcharge + Vcharge ⎥ at + a ⎦ ⎣ at - a − 2kQ 2 ⎛ kQ kQ ⎞ = −Q ⎜ + ⎟= a ⎠ a ⎝ a W−Q → 2 a = −QΔV0→ 2 a

= −Q[V (2a ) − V (0)] ⎡ ⎤ = −Q ⎢Vcharge + Vcharge − V (0)⎥ at + a ⎣ at - a ⎦ ⎛ kQ kQ 2kQ ⎞ = −Q⎜ + − ⎟ a a ⎠ ⎝ 3a

2kQ 2 = 3a A charge of +2.00 nC is uniformly distributed on a ring of radius 80 •• 10.0 cm that lies in the x = 0 plane and is centered at the origin. A point charge of +1.00 nC is initially located on the x axis at x = 50.0 cm. Find the work required to move the point charge to the origin. Picture the Problem Let q represent the charge being moved from x = 50.0 cm to the origin, Q the ring charge, and a the radius of the ring. We can use Wq →final position = qΔVi → f , where V is the expression for the axial field due to a ring

charge, to find the work required to move q from x = 50.0 cm to the origin. Express the required work in terms of the charge being moved and the potential due to the ring charge at x = 50.0 cm and x = 0:

W = qΔV = q[V (0 ) − V (0.500 m )]

kQ

The potential on the axis of a uniformly charged ring is given by:

V (x ) =

At x = 50.0 cm:

V (0.500 m ) =

At x = 0:

V (0 ) =

x2 + a2

kQ a

2

=

kQ

(0.500 m )2 + a 2 kQ a

274

Chapter 23

Substituting for V(0) and V(0.500 m) yields: ⎡ kQ − W = q⎢ ⎢⎣ a

⎤ ⎡1 ⎥ = kQq ⎢ − ⎢⎣ a (0.500 m )2 + a 2 ⎥⎦ kQ

⎤ ⎥ 2 2 ⎥ (0.500 m ) + a ⎦ 1

Substitute numerical values and evaluate W: 2 ⎛ 9 N⋅m ⎞ ⎟⎟ (2.00 nC )(1.00 nC ) ⎜ W = ⎜ 8.988 ×10 2 C ⎠ ⎝ ⎡ 1 ⎤ 1 ⎥ ×⎢ − 2 2 ⎢ 0.100 m (0.500 m ) + (0.100 m ) ⎥⎦ ⎣ 1eV = 1.445 ×10 − 7 J = 1.4 ×10 − 7 J = 1.445 × 10 − 7 J × 1.602 × 10 −19 J

= 9.0 ×1011 eV 81 •• Two metal spheres each have a radius of 10.0 cm. The centers of the two spheres are 50.0 cm apart. The spheres are initially neutral, but a charge Q is transferred from one sphere to the other, creating a potential difference between the spheres of 100 V. A proton is released from rest at the surface of the positively charged sphere and travels to the negatively charged sphere. (a) What is the proton′s kinetic energy just as it strikes the negatively charged sphere? (b) At what speed does it strike the sphere? Picture the Problem The proton′s kinetic energy just as it strikes the negatively charged sphere is the product of its charge and the potential difference through which it has been accelerated. We can find the speed of the proton as it strikes the negatively charged sphere from its kinetic energy and, in turn, its kinetic energy from the potential difference through which it is accelerated.

(a) Apply the work-kinetic energy theorem to the proton to obtain:

Wnet = ΔK = K f − K i or, because Ki = 0, Wnet = K f = K p

The net work done on the proton is given by:

Wnet = qpV

Equating Wnet and K p yields:

K p = qpV = e(100 V ) = 100 eV

Electric Potential (b) Use the definition of kinetic energy to express the speed of the proton when it strikes the negatively charged sphere:

K p = 12 mp v 2 ⇒ v =

Substitute numerical values and evaluate v:

(

2K p mp

=

275

2ΔV mp

)

2 1.602 × 10 −19 C (100 V ) 1.673 × 10 −27 kg

v=

= 1.38 × 10 5 m/s 82 •• (a) Using a spreadsheet program, graph V(z) versus z for a uniformly charged ring in the z = 0 plane and centered at the origin. The potential on the z

axis is given by V ( z ) = kQ a 2 + z 2 (Equation 23-19). (b) Use your graph to estimate the points on the z axis where the electric field strength is greatest. Picture the Problem (b) The electric field strength is greatest where the magnitude of the slope of the graph of electric potential is greatest.

(a) A spreadsheet solution is shown below for kQ = a = 1. The formulas used to calculate the quantities in the columns are as follows: Cell A4 B3

Content/Formula A3 + 0.1 1/(1+A3^2)^(1/2)

Algebraic Form z + Δz kQ a2 + z2

A

B

1 2 3 4 5 6 7 8 9

z/a −5.0 −4.8 −4.6 −4.4 −4.2 −4.0 −3.8

V(z/a) 0.196 0.204 0.212 0.222 0.232 0.243 0.254

49 50 51 52 53

4.2 4.4 4.6 4.8 5.0

0.232 0.222 0.212 0.204 0.196

276

Chapter 23

The following graph, plotted using a spreadsheet program, shows V as a function of z/a: 1.0

V (V)

0.8

0.6

0.4

0.2 -5

-4

-3

-2

-1

0

z/a

1

2

3

4

5

Examining the graph we see that the magnitude of the slope is maximum at z a ≈ 0.7 and at z a ≈ − 0.7 . 83 •• A spherical conductor of radius R1 is charged to 20 kV. When it is connected by a long very-thin conducting wire to a second conducting sphere far away, its potential drops to 12 kV. What is the radius of the second sphere? Picture the Problem Let R2 be the radius of the second sphere and Q1 and Q2 the charges on the spheres when they have been connected by the wire. When the spheres are connected, the charge initially on the sphere of radius R1 will redistribute until the spheres are at the same potential.

Express the common potential of the spheres when they are connected:

12 kV =

kQ1 R1

(1)

kQ2 R2

(2)

and 12 kV =

Electric Potential Express the potential of the first sphere before it is connected to the second sphere:

20 kV =

Solve equation (1) for Q1:

Q1 =

Solve equation (2) for Q2:

Substitute for Q1 and Q2 in equation (3) and simplify to obtain:

k (Q1 + Q2 ) R1

277

(3)

(12 kV )R1 k

Q2 =

(12 kV )R2 k

⎛ (12 kV )R1 (12 kV )R2 ⎞ k⎜ + ⎟ k k ⎠ ⎝ 20 kV = R1 ⎛R ⎞ = 12 kV + 12 kV⎜⎜ 2 ⎟⎟ ⎝ R1 ⎠ or ⎛R ⎞ 2 R1 8 = 12⎜⎜ 2 ⎟⎟ ⇒ R2 = 3 ⎝ R1 ⎠

84 •• A metal sphere centered at the origin has a surface charge density that has a magnitude of 24.6 nC/m2 and a radius less than 2.00 m. A distance of 2.00 m from the origin, the electric potential is 500 V and the electric field strength is 250 V/m. (Assume the potential is zero very far from the sphere.) (a) What is the radius of the metal sphere? (b) What is the sign of the charge on the sphere? Explain your answer. Picture the Problem We can use the definition of surface charge density to relate the radius R of the sphere to its charge Q and the potential function V (r ) = kQ r to relate Q to the potential at r = 2.00 m.

(a) Use its definition, relate the surface charge density σ to the charge Q on the sphere and the radius R of the sphere: Relate the potential to the charge on the sphere:

σ=

Q ⇒R = 4πR 2

V (r ) =

Q 4πσ

kQ rV (r ) ⇒Q = r k

278

Chapter 23

Substitute for Q in the expression for R and simplify to obtain:

R= =

4π ∈0 rV (r ) 4πσ

rV (r ) = 4πkσ

∈0 rV (r ) σ

Substitute numerical values and evaluate R: R=

(8.854 ×10

−12

)

C 2 /N ⋅ m 2 (2.00 m )(500 V ) = 60.0 cm 24.6 nC/m 2

(b) The charge on the sphere is positive. The formula for the electric potential outside a uniformly charged spherical shell is V = kQ r . If V is positive, then so is Q. 85 •• Along the central axis of a uniformly charged disk, at a point 0.60 m from the center of the disk, the potential is 80 V and the magnitude of the electric field is 80 V/m. At a distance of 1.5 m, the potential is 40 V and the magnitude of the electric field is 23.5 V/m. (Assume the potential is zero very far from the sphere.) Find the total charge on the disk. Picture the Problem We can use the definition of surface charge density to relate the radius R of the disk to its charge Q and the potential function V (r ) = kQ r to relate Q to the potential at r = 1.5 m.

Q ⇒ Q = πσR 2 2 πR

Use its definition, relate the surface charge density σ to the charge Q on the disk and the radius R of the disk:

σ=

Relate the potential at r to the charge on the disk:

V (r ) = 2πkσ

Substitute V(0.60 m) = 80 V:

80 V = 2πkσ ⎛⎜ ⎝

Substitute V(1.5 m) = 40 V:

2 40 V = 2πkσ ⎛⎜ (1.5 m ) + R 2 − 1.5 m ⎞⎟ ⎝ ⎠

(x

2

+ R2 − x

(1)

)

(0.60 m )2 + R 2 − 0.60 m ⎞⎟

(0.60 m )2 + R 2 − 0.60 m (1.5 m )2 + R 2 − 1.5 m

Divide the first of these equations by the second to obtain:

2=

Solving for R yields:

R = 0.80 m

⎠

Electric Potential Express the electric field on the axis of a disk charge: Solving for σ yields:

⎛ x E x = 2πkσ ⎜⎜1 − 2 x + R2 ⎝

σ=

Q=

⎞ ⎟ ⎟ ⎠

Ex

⎛ x 2πk ⎜⎜1 − x2 + R2 ⎝ 2∈0 E x = x 1− x2 + R2

Substitute for σ in equation (1) to obtain:

279

⎞ ⎟ ⎟ ⎠

2π ∈ 0 R 2 E x x 1− x2 + R2

Substitute numerical values and evaluate Q: ⎛ C2 ⎞ ⎟ (0.80 m )2 (23.5 V/m ) 2π ⎜⎜ 8.854 × 10 −12 2 ⎟ N⋅m ⎠ Q= ⎝ = 7.1 nC 1.5 m 1− (1.5 m )2 + (0.80 m )2 86 •• A radioactive 210Po nucleus emits an α-particle that has a charge +2e. When the α-particle is a large distance from the nucleus it has a kinetic energy of 5.30 MeV. Assume that the α-particle had negligible kinetic energy as it left at the surface of the nucleus. The ″daughter″ (or residual) nucleus 206Pb has a charge +82e. Determine the radius of the 206Pb nucleus. (Neglect the radius of the α particle and assume the 206Pb nucleus remains at rest.) Picture the Problem We can use U = kq1q2/r to relate the electrostatic potential energy of the particles to their separation.

Express the electrostatic potential energy of the two particles in terms of their charge and separation:

U=

kq q kq1q2 ⇒r = 1 2 r U

Substitute numerical values and evaluate r:

(8.988 ×10 r=

)

(

N ⋅ m 2 /C 2 (2)(82 ) 1.602 × 10 −19 C 1.602 × 10 −19 C 5.30 MeV × eV 9

)

2

= 44.6 fm

280

Chapter 23

87 ••• [SSM] Configuration A consists of two point particles; one particle has a charge of +q and is on the x axis at x = +d and the other particle has a charge of –q and is at x = –d (Figure 23-36a). (a) Assuming the potential is zero at large distances from these charged particles, show that the potential is also zero everywhere on the x = 0 plane. (b) Configuration B consists of a flat metal plate of infinite extent and a point particle located a distance d from the plate (Figure 23-36b) The point particle has a charge equal to +q and the plate is grounded. (Grounding the plate forces its potential to equal zero.) Choose the line perpendicular to the plate and through the point charge as the x axis, and choose the origin at the surface of the plate nearest the particle. (These choices put the particle on the x axis at x = +d.) For configuration B, the electric potential is zero both at all points in the half-space x ≥ 0 that are very far from the particle and at all points on the x = 0 plane—just as was the case for configuration A. (c) A theorem, called the uniqueness theorem, implies that throughout the half-space r x ≥ 0 the potential function V—and thus the electric field E —for the two r configurations are identical. Using this result, obtain the electric field E at every point in the x = 0 plane in the configuration B. (The uniqueness theorem tells us that in configuration B the electric field at each point in the x = 0 plane is the same as it is in configuration A.) Use this result to find the surface charge density σ at each point in the conducting plane (in configuration B). Picture the Problem We can use the relationship between the potential and the electric field to show that this arrangement is equivalent to replacing the plane by a point charge of magnitude −q located a distance d beneath the plane. In (b) we can first find the field at the plane surface and then use σ = ∈0E to find the surface charge density. In (c) the work needed to move the charge to a point 2d away from the plane is the product of the potential difference between the points at distances 2d and 3d from −q multiplied by the separation Δx of these points.

(a) The potential anywhere on the plane is 0 and the electric field is perpendicular to the plane in both configurations, so they must give the same potential everywhere in the xy plane. Also, because the net charge is zero, the potential at infinity is zero. (b) The surface charge density is given by:

σ =∈0 E

(1)

Electric Potential At any point on the plane, the electric field points in the negative x direction and has magnitude:

Because cos θ =

d d 2 + r2

:

kq cos θ d + r2 where θ is the angle between the horizontal and a vector pointing from the positive charge to the point of interest on the xz plane and r is the distance along the plane from the origin (that is, directly to the left of the charge). E=

E= =

Substitute for E in equation (1) and simplify to obtain:

281

σ=

2

kq d + r2 2

d d 2 + r2 qd

(

4π ∈ 0 d 2 + r 2

(

qd

4π d 2 + r 2

=

(d

kqd 2

+ r2

)

32

)

32

)

3/ 2

88 ••• A particle that has a mass m and a positive charge q is constrained to move along the x-axis. At x = –L and x = L are two ring charges of radius L (Figure 23-38). Each ring is centered on the x-axis and lies in a plane perpendicular to it. Each ring has a total positive charge Q uniformly distributed on it. (a) Obtain an expression for the potential V(x) on the x axis due to the charge on the rings. (b) Show that V(x) has a minimum at x = 0. (c) Show that for x > R2, then the reciprocal of R1 is = + Req R1 R2 very small and

1 1 . Hence Req ≈ R2 and (b) is correct. ≈ Req R2

15 • One resistor has a resistance R1 and another resistor has a resistance R2. The resistors are connected in series. If R1 >> R2, the equivalent resistance of the combination is approximately (a) R1, (b) R2, (c) 0, (d) infinity. Determine the Concept The equivalent resistance of the two resistors connected ⎛ R ⎞ in series is given by Req = R1 + R2 = R1 ⎜⎜1 + 2 ⎟⎟ . If R1 >> R2, then R2/R1 is very R1 ⎠ ⎝ small and Req ≈ R1 . Hence (a ) is correct. 16 • A parallel combination consisting of resistors A and B is connected across the terminals of a battery. The resistor A has twice the resistance of resistor B. If the current carried by resistor A is I, then what is the current carried by resistor B? (a) I, (b) 2I, (c) I/2, (d) 4I, (e) I/4

`Electric Current and Direct-Current Circuits 389

Picture the Problem Ohm’s law states that the current in a resistor is proportional to the potential drop across the resistor. Because the potential difference across resistors A and connected in parallel is the same for each resistor, the resistor with half the resistance of the other resistor will carry twice the current carried by the resistor with the larger resistance. Because resistor A has twice the resistance of resistor B, resistor B will carry twice the current of resistor A. (b) is correct. 17 • A series combination consisting of resistors A and B is connected across the terminals of a battery. The resistor A has twice the resistance of resistor B. If the current carried by resistor A is I, then what is the current carried by resistor B? (a) I, (b) 2I, (c) I/2, (d) 4I, (e) I/4 Determine the Concept In a series circuit, because there are no alternative pathways, all resistors carry the same current. (a ) is correct. 18 • Kirchhoff’s junction rule is considered to be a consequence of (a) conservation of charge, (b) conservation of energy, (c) Newton’s laws, (d) Coulomb’s law, (e) quantization of charge. Determine the Concept While Kirchhoff’s junction rule is a statement about current, recall that current is the rate at which charge passes some point in space. Hence, the junction rule is actually a statement that charge is conserved. (a ) is correct. True or False:

19

•

(a) (b) (c)

An ideal voltmeter has a zero internal resistance. An ideal ammeter has a zero internal resistance. An ideal voltage source has a zero internal resistance

(a) False. An ideal voltmeter would have infinite resistance. A voltmeter consists of a galvanometer movement connected in series with a large resistance. The large resistor accomplishes two purposes; 1) it protects the galvanometer movement by limiting the current drawn by it, and 2) minimizes the loading of the circuit by the voltmeter by placing a large resistance in parallel with the circuit element across which the potential difference is being measured.

Chapter 25

390

(b) True. An ideal ammeter would have zero resistance. An ammeter consists of a very small resistance in parallel with a galvanometer movement. The small resistance accomplishes two purposes: 1) It protects the galvanometer movement by shunting most of the current in the circuit around the galvanometer movement, and 2) It minimizes the loading of the circuit by the ammeter by minimizing the resistance of the ammeter. (c) True. An ideal voltage source would have zero internal resistance. The

terminal potential difference of a voltage source is given by V = ε − Ir, where ε is the emf of the source, I is the current drawn from the source, and r is the internal resistance of the source. 20 • Before you and your classmates run an experiment, your professor lectures about safety. She reminds you that to measure the voltage across a resistor you connect a voltmeter in parallel with the resistor, and to measure the current in a resistor you connect an ammeter in series with the resistor. She also states that connecting a voltmeter in series with a resistor will not measure the voltage across the resistor, but also cannot do any damage to the circuit or the instrument. In addition, connecting an ammeter in parallel with a resistor will not measure the current in the resistor, but could cause significant damage to the circuit and the instrument. Explain why connecting a voltmeter in series with a resistor causes no damage while connecting an ammeter in parallel with a resistor can cause significant damage. Determine the Concept Because of the voltmeter’s high resistance, if you connect a voltmeter in series with a circuit element, the current, both in the voltmeter and in the rest of the circuit, will be very small. This means that there is little chance of heating the voltmeter and causing damage. However, because of the ammeter’s low resistance, if you connect an ammeter in parallel with a circuit element, the current, both in the ammeter and in the entire circuit, excluding any elements in parallel with the ammeter, will be very large. This means that there is a good chance of overheating and causing damage, maybe even a fire. For this reason, ammeters are often equipped with fuses or circuit breakers. 21

•

The capacitor in Figure 25-49 is initially uncharged. Just after the

switch S is closed, (a) the voltage across C equals ε, (b) the voltage across R

equals ε, (c) the current in the circuit is zero, (d) both (a) and (c) are correct.

Determine the Concept If we apply Kirchhoff’s loop rule with the switch closed, we obtain ε − IR – VC = 0. Immediately after the switch is closed,

I = I max and VC = 0 . Hence

ε = I max R .

(b) is correct.

`Electric Current and Direct-Current Circuits 391

22 •• A capacitor is discharging through a resistor. If it takes a time T for the charge on a capacitor to drop to half its initial value, how long (in terms of T) does it take for the stored energy to drop to half its initial value? Picture the Problem We can express the variation of charge on the discharging capacitor as a function of time to find the time T it takes for the charge on the capacitor to drop to half its initial value. We can also express the energy remaining in the electric field of the discharging capacitor as a function of time and find the time t′ for the energy to drop to half its initial value in terms of T. Express the time-dependence of the charge stored on a capacitor: For Q(t ) = 12 Q0 : Take the natural logarithm of both sides of the equation and solve for T to obtain:

Q(t ) = Q0e −t τ where τ = RC. 1 2

Q0 = Q0 e −T τ or

U (t ) = 12 CVC2

Express the potential difference across a discharging capacitor as a function of time:

VC = V0e −t RC

Substitute for VC in equation (1) and simplify to obtain:

U (t ) = 12 C V0e −t RC

Take the natural logarithm of both sides of the equation and solve for t′ to obtain:

= e −T τ

T = τ ln (2 )

Express the dependence of the energy stored in a capacitor on the potential difference VC across its terminals:

For U (t ) = 12 U 0 :

1 2

(1)

(

)

2

= 12 CV02e −2t RC

= U 0e −2t RC 1 2

U 0 = U 0 e −2t' RC ⇒

t' = 12 τ ln (2 ) =

1 2

1 2

= e −2t ' RC

T

23 •• [SSM] In Figure 25-50, the values of the resistances are related as follows: R2 = R3 = 2R1. If power P is delivered to R1, what is the power delivered to R2 and R3?

392

Chapter 25

Determine the Concept The power delivered to a resistor varies with the square of the current in the resistor and is directly proportional to the resistance of the resistor ( P = I 2 R ). The power delivered to R1 is:

P = I 12 R1

The power delivered to R2 is:

P2 = I 22 R2 = I 22 (2 R1 )

Because R2 = R3 :

I 2 = I 3 = 12 I 1

Substituting for I2 and simplifying gives:

P2 = ( 12 I 1 ) (2 R1 ) = 12 I 12 R1 =

Similarly:

P3 = I 32 R3 = ( 12 I1 ) (2 R1 ) = 12 I12 R1

2

1 2

P

2

=

1 2

P

24 •• The capacitor in Figure 25-49 is initially uncharged. The switch S is closed and remains closed for a very long time. During this time, (a) the energy supplied by the battery is 12 Cε 2 , (b) the energy dissipated in the resistor is 12 Cε 2 , (c) energy in the resistor is dissipated at a constant rate, (d) the total charge passing through the resistor is 12 Cε . Determine the Concept The energy stored in the fully charged capacitor is

U = 12 Cε 2 . During the charging process, a total charge Qf = εC flows through the

battery. The battery therefore does work W = Qfε = Cε 2. The energy dissipated in the resistor is the difference between W and U. (b) is correct.

Estimation and Approximation 25 •• It is not a good idea to stick the ends of a paper clip into the two rectangular slots of a household electrical wall outlet in the United States. Explain why by estimating the current that a paper clip would carry until either the fuse blows or the breaker trips. Picture the Problem The current drawn by the paper clip is the ratio of the potential difference (120 V) between the two holes and the resistance of the paper clip. We can use R = ρL A to find the resistance of the paper clip. Paper clips are made of steel, which has a resistivity in the range 10 to 100 × 10−8 Ω⋅ m, From the definition of resistance:

I=

ε R

`Electric Current and Direct-Current Circuits 393

The resistance of the paper clip is given by:

R=ρ

Substituting for R in the expression for I and simplifying yields

I=

Assuming that the length of a paper clip is 10 cm and that its diameter is 1.0 mm, substitute numerical values and evaluate I:

I=

L A

ε ρ

L A

=

εA = επr 2 = επd 2 ρL

ρL

4 ρL

(120 V )π (1.0 mm)2 = 4(50 × 10 −8 Ω ⋅ m )(10 cm )

1.9 kA

26 •• (a) Estimate the resistance of an automobile jumper cable. (b) Look up the current required to start a typical car. At that current, what is the potential drop that occurs across the jumper cable? (c) How much power is dissipated in the jumper cable when it carries this current? Picture the Problem (a) We can use the definition of resistivity to find the resistance of the jumper cable. In Part (b), the application of Ohm’s law will yield the potential difference across the jumper cable when it is starting a car, and, in Part (c), we can use the expression for the power dissipated in a conductor to find the power dissipation in the jumper cable.

L A

(a) Noting that a jumper cable has two leads, express the resistance of the cable in terms of the wire’s resistivity and the cable’s length, and cross-sectional area:

R=ρ

Assuming the length of the jumper cables to be 3.0 m and its crosssectional area to be 10 mm2, substitute numerical values (see Table 25-1 for the resistivity of copper) and evaluate R:

R = 1.7 × 10 −8 Ω ⋅ m

(b) Apply Ohm’s law to the cable with a starting current of 90 A to obtain:

(

0m )103.mm

2

= 5.1 mΩ

V = IR = (90 A )(5.1 mΩ ) = 459 mV = 0.46 V

394

Chapter 25

(c) Use the expression for the power dissipated in a conductor to obtain:

P = IV = (90 A )(459 mV ) = 41 W

27 •• Your manager wants you to design a new super-insulated water heater for the residential market. A coil of Nichrome wire is to be used as the heating element. Estimate the length of wire required. HINT: You will need to determine the size of a typical hot water heater and a reasonable time period for creating hot water. Picture the Problem We can combine the expression for the rate at which energy

is delivered to the water to warm it (P = ε 2/R) and the expression for the resistance of a conductor ( R = ρL A ) to obtain an expression for the required length L of Nichrome wire. From Equation 25-10, the length of Nichrome wire required is given by:

L=

RA

ρ Ni

where A is the cross-sectional

area of the wire. Use an expression for the power dissipated in a resistor to relate the required resistance to rate at which energy is delivered to generate the warm water: Substituting for R and simplifying yields: The power required to heat the water is the rate at which the wire will deliver energy to the water: Substitute for P to obtain:

The mass of water to be heated is the product of its density and the volume of the water tank:

P=

L= P=

L=

ε2 ⇒R = ε2 R

P

ε 2A ρ Ni P ΔE Δ(mcΔT ) ΔT = = mc Δt Δt Δt

ε 2A ρ Ni mc

ΔT Δt

m = ρ H 2OV = ρ H 2O A'd

where A′ is the cross-sectional area of the water tank.

`Electric Current and Direct-Current Circuits 395

Finally, substituting for m and simplifying yields:

L=

ε 2A

ΔT Δt 2 2 ε π rwire Δt = 2 ρ Ni ρ H 2Oπ rtank dcΔT

ρ Ni ρ H O A'dc 2

=

2 ε 2 rwire Δt

2 ρ Ni ρ H O rtank dcΔT 2

The diameter of a typical 40-gal water heater is about 50 cm and its height is approximately 1.3 m. Assume that the diameter of the Nichrome wire is 2.0 mm and that the water, initially at 20°C, is to be heated to 80°C in 1.0 h. Substitute numerical values (see Table 25-1 for the resistivity of Nichrome and Table 18-1 for the heat capacity of water) and evaluate L:

(120 V )2 (2.0 ×10 −3 m )2 ⎛⎜1.0 h × 3600 s ⎞⎟

L=

h ⎠ = 26 m ⎛ kg ⎞ kJ ⎞ ⎛ 2 −8 ⎟⎟(60C°) 100 × 10 Ω ⋅ m ⎜1000 3 ⎟ (0.25 m ) (0.50 m )⎜⎜ 4.18 kg ⋅ C ° m ⎠ ⎝ ⎝ ⎠ ⎝

(

)

28 •• A compact fluorescent light bulb costs about $6.00 each and has a typical lifetime of 10 000 h. These bulbs use 20 W of power, but produce illumination equivalent to that of 75-W incandescent bulbs. An incandescent bulb costs about $1.50 and has a typical lifetime of 1000 h. Your family wonders whether it should buy fluorescent light bulbs. Estimate the amount of money your household would save each year by using compact fluorescent light bulbs instead of the incandescent bulbs. Picture the Problem We can find the annual savings by taking into account the costs of the two types of bulbs, the rate at which they consume energy and the cost of that energy, and their expected lifetimes. We’ll assume that the household lights are on for an average of 8 hours per day.

Express the yearly savings:

Δ$ = Cost incandescent − Cost fluorescent

Express the annual costs with the incandescent and fluorescent bulbs:

Cost incandescent = Cost bulbs + Cost energy and Cost fluorescent = Cost bulbs + Cost energy

(1)

396

Chapter 25

The annual cost of the incandescent bulbs is the product of the number of bulbs in use, the annual consumption of bulbs, and the cost per bulb:

Cost bulbs

8h ⎞ ⎛ ⎜ 365.24 d × ⎟ d ⎜ ⎟ ($1.50) = $26.30 = (6 ) 1000 h ⎜ ⎟ ⎜ ⎟ ⎝ ⎠

The cost of operating the incandescent bulbs for one year is the product of the energy consumed and the cost per unit of energy: ⎛ d ⎞ ⎛ 8 h ⎞ ⎛ $0.115 ⎞ Cost energy = 6(75 W )⎜⎜ 365.24 ⎟⎟ ⎜ ⎟ = $151.21 ⎟⎜ y ⎠ ⎝ d ⎠ ⎝ kW ⋅ h ⎠ ⎝ The annual cost of the fluorescent bulbs is the product of the number of bulbs in use, the annual consumption of bulbs, and the cost per bulb:

Cost bulbs

8h ⎞ ⎛ ⎜ 365.24 d × ⎟ d ⎜ ⎟ ($6) = $10.52 = (6 ) ⎜ 10000 h ⎟ ⎜ ⎟ ⎝ ⎠

The cost of operating the fluorescent bulbs for one year is the product of the energy consumed and the cost per unit of energy:

8 h ⎞ ⎛ $0.115 ⎞ ⎛ Cost energy = 6(20 W )⎜ 365.24 d × ⎟⎜ ⎟ = $40.32 d ⎠ ⎝ kW ⋅ h ⎠ ⎝ Substitute in equation (1) and evaluate the cost savings Δ$: Δ$ = Cost incandescent − Cost fluorescent = ($26.30 + $151.20) − ($10.52 + $40.32) = $127 29 •• The wires in a house must be large enough in diameter so that they do not get hot enough to start a fire. While working for a building contractor during the summer, you are involved in remodeling a house. The local building code states that the joule heating of the wire used in houses should not exceed 2.0 W/m. Estimate the maximum gauge of the copper wire that you can use during the rewiring of the house with 20-A circuits.

`Electric Current and Direct-Current Circuits

397

Picture the Problem We can use an expression for the power dissipated in a resistor to relate the Joule heating in the wire to its resistance and the definition of resistivity to relate the resistance to the length and cross-sectional area of the wire. We can find the diameter of the wire from its cross-sectional area and then use Table 25-2 to find the maximum gauge you can use during the rewiring of the house.

Express the power the wires must dissipate in terms of the current they carry and their resistance:

P = I 2R

Divide both sides of the equation by L to express the power dissipation per unit length:

P I 2R = L L

The resistance of the wire is given by:

R=ρ

Substitute for R to obtain:

P 4 ρI 2 ρ = ⇒ d = 2I 2 π (P / L ) L πd

Substitute numerical values (see Table 25-1 for the resistivity of copper wire) and evaluate d:

d = 2(20 A )

L L 4 ρL =ρπ 2 = A π d2 4 d

1.7 ×10 −8 Ω ⋅ m = 2.1mm π (2.0 W/m )

Consulting Table 25-2, we note that the maximum gauge you can use is 12 . 30 •• A laser diode used in making a laser pointer is a highly nonlinear circuit element. Its behavior is as follows: for any voltage drop across it that is less than about 2.30 V, it behaves as if it has an infinite internal resistance, but for voltages higher than 2.30 V it has a very low internal resistance—effectively zero. (a) A laser pointer is made by putting two 1.55 V watch batteries in series across the laser diode. If the batteries each have an internal resistance between 1.00 Ω and 1.50 Ω, estimate the current in the laser beam. (b) About half of the power delivered to the laser diode goes into radiant energy. Using this fact, estimate the power of the laser beam, and compare this value to typical quoted values of about 3.00 mW. (c) If the batteries each have a capacity of 20.0-mA·h (i.e., they can deliver a constant current of 20.0 mA for approximately one hour before discharging), estimate how long one can continuously operate the laser pointer before replacing the batteries.

398

Chapter 25

Picture the Problem Let r be the internal resistance of each battery and apply Kirchhoff’s loop rule to the circuit consisting of the two batteries and the laser diode to relate the current in laser diode to r and the potential differences across the batteries and the diode. We can find the power of the laser diode from the product of the potential difference across the internal resistance of the batteries and the current I delivered by them and the time-to-discharge from the combined capacities of the two batteries and I.

(a) Apply Kirchhoff’s loop rule to the circuit consisting of the two batteries and the laser diode to obtain:

2ε − 2 Ir − Vlaser = 0 diode

Solving for I yields: I= Assuming that r = 125 Ω, substitute numerical values and evaluate I:

I=

ε − Vlaser

diode

2r 2(1.55) − 2.30 V = 3.20 mA 2(125 Ω )

= 3.2 mA (b) The power delivered by the batteries is given by:

P = IV = (3.20 mA )(2.30 V ) = 7.36 mW

(7.36 mW ) = 3.68 mW

The power of the laser is half this value:

Plaser = 12 P =

Express the ratio of Plaser to Pquoted:

Plaser 3.68 mW = = 1.23 Pquoted 3.00 mW

1 2

= 3.7 mW

or Plaser = 1.23Pquoted (c) Express the time-to-discharge:

Δt discharge =

Capacity I

Because each battery has a capacity of 20.0 mA⋅h, the series combination has a capacity of 40.0 mA⋅h and:

Δt discharge =

40.0 mA ⋅ h ≈ 12.5 h 3.20 mA

`Electric Current and Direct-Current Circuits 399

Current, Current Density, Drift Speed and the Motion of Charges 31 • [SSM] A 10-gauge copper wire carries a current equal to 20 A. Assuming copper has one free electron per atom, calculate the drift speed of the free electrons in the wire. Picture the Problem We can relate the drift velocity of the electrons to the current density using I = nevd A . We can find the number density of charge

carriers n using n = ρN A M , where ρ is the mass density, NA Avogadro’s number, and M the molar mass. We can find the cross-sectional area of 10-gauge wire in Table 25-2. I neA

Use the relation between current and drift velocity to relate I and n:

I = nevd A ⇒ vd =

The number density of charge carriers n is related to the mass density ρ, Avogadro’s number NA, and the molar mass M:

n=

ρN A

n=

(8.93 g/cm )(6.022 ×10

For copper, ρ = 8.93 g/cm3 and M = 63.55 g/mol. Substitute and evaluate n: Using Table 25-2, find the crosssectional area of 10-gauge wire:

M

3

23

atoms/mol

)

63.55 g/mol

= 8.459 × 10 28 atoms/m 3 A = 5.261 mm 2

Substitute numerical values and evaluate vd:

vd =

(8.459 ×10

28

m

−3

20 A = 0.28 mm/s 1.602 × 10 −19 C 5.261mm 2

)(

)(

)

32 •• A thin nonconducting ring that has a radius a and a linear charge density λ rotates with angular speed ω about an axis through its center and perpendicular to the plane of the ring. Find the current of the ring. Picture the Problem We can use the definition of current, the definition of charge density, and the relationship between period and frequency to derive an expression for the current as a function of a, λ, and ω.

400

Chapter 25

Use the definition of current to relate the charge ΔQ associated with a segment of the ring to the time Δt it takes the segment to pass a given point:

I=

ΔQ Δt

Because each segment carries a charge ΔQ and the time for one revolution is T:

I=

ΔQ = ΔQf T

Use the definition of the charge density λ to relate the charge ΔQ to the radius a of the ring:

λ=

ΔQ ⇒ ΔQ = 2πaλ 2πa

Substitute for ΔQ in equation (1) to obtain:

I = 2πaλf

Because ω = 2πf :

I = aλω

(1)

33 •• [SSM] A length of 10-gauge copper wire and a length of 14-gauge copper wire are welded together end to end. The wires carry a current of 15 A. (a) If there is one free electron for each copper atom in each wire, find the drift speed of the electrons in each wire. (b) What is the ratio of the magnitude of the current density in the length of10-gauge wire to the magnitude of the current density in the length of 14-gauge wire? Picture the Problem (a) The current will be the same in the two wires and we can relate the drift velocity of the electrons in each wire to their current densities and the cross-sectional areas of the wires. We can find the number density of charge carriers n using n = ρN A M , where ρ is the mass density, NA Avogadro’s number, and M the molar mass. We can find the cross-sectional area of 10- and 14-gauge wires in Table 25-2. In Part (b) we can use the definition of current density to find the ratio of the magnitudes of the current densities in the 10-gauge and 14-gauge wires.

(a) Relate the current density to the drift velocity of the electrons in the 10-gauge wire:

I10 gauge A10 gauge

= nevd, 10 ⇒ vd, 10 =

I10 gauge neA10 gauge

`Electric Current and Direct-Current Circuits 401

ρN A

The number density of charge carriers n is related to the mass density ρ, Avogadro’s number NA, and the molar mass M:

n=

For copper, ρ = 8.93 g/cm3 and M = 63.55 g/mol. Substitute numerical values and evaluate n:

g ⎞⎛ ⎛ 23 atoms ⎞ ⎟ ⎜ 8.93 3 ⎟ ⎜ 6.022 ×10 mol ⎠ cm ⎠ ⎝ ⎝ n= g 63.55 mol atoms = 8.462 ×10 28 m3

Use Table 25-2 to find the crosssectional area of 10-gauge wire:

A10 = 5.261 mm 2

M

Substitute numerical values and evaluate vd.10: vd, 10 =

(8.462 ×10

28

m

−3

15 A = 0.210 mm/s = 0.21 mm/s 1.602 ×10 −19 C 5.261 mm 2

)(

Express the continuity of the current in the two wires:

Solve for vd,14 to obtain:

Use Table 25-2 to find the crosssectional area of 14-gauge wire: Substitute numerical values and evaluate vd,14:

)(

)

I10 gauge = I14 gauge

or nevd,10 A10 gauge = nevd,14 A14 gauge vd,14 = vd,10

A10 gauge A14 gauge

A14 = 2.081 mm 2

vd,14

5.261mm 2 = (0.210 mm/s) 2.081mm 2 = 0.53 mm/s

(b) The ratio of current density, 10gauge wire to 14-gauge wire, is given by:

J 10 J 14

I 10 A I A = 10 = 10 14 I 14 I 14 A10 A14

402

Chapter 25

Because I 10 = I14 :

J 10 A14 = J 14 A10

Substitute numerical values and J evaluate 10 : J14

J 10 2.081 mm 2 = = 0.396 J 14 5.261 mm 2

34 •• An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA. The current density is uniformly distributed through the beam. The kinetic energy of each proton is 20 MeV. The beam strikes a metal target and is absorbed by the target. (a) What is the number density of the protons in the beam? (b) How many protons strike the target each minute? (c) What is the magnitude of the current density in this beam? Picture the Problem We can use I = neAv to relate the number n of protons per unit volume in the beam to current I. We can find the speed of the particles in the beam from their kinetic energy. In Part (b) we can express the number of protons N striking the target per unit time as the product of the number of protons per unit volume n in the beam and the volume of the cylinder containing those protons that will strike the target in an elapsed time Δt and solve for N. Finally, we can use the definition of current to express the charge arriving at the target as a function of time.

(a) Use the relation between current and drift velocity (Equation 25-3) to relate I and n: The kinetic energy of the protons is given by: Relate the cross-sectional area A of the beam to its diameter D: Substitute for v and A and simplify to obtain:

I = neAv ⇒ n =

I eAv

K = 12 mp v 2 ⇒ v =

2K mp

A = 14 πD 2

I

n= 1 4

πeD 2

2K mp

=

4I πeD 2

mp

2K

`Electric Current and Direct-Current Circuits

403

Substitute numerical values and evaluate n: n=

4(1.0 mA ) π (1.602 ×10 −19 C )(2 mm)2

1.673 ×10 −27 kg 2(20 MeV )(1.602 ×10 −19 J/eV )

= 3.21×1013 m −3 = 3.2 ×1013 m −3

(b) Express the number of protons N striking the target per unit time as the product of the number n of protons per unit volume in the beam and the volume of the cylinder containing those protons that will strike the target in an elapsed time Δt and solve for N: Substitute for v and A to obtain:

N = n(vA) ⇒ N = nvAΔt Δt

N = 14 πD 2 nΔt

2K mp

Substitute numerical values and evaluate N:

(

2(20 MeV ) (1.602 × 10 −19 J/eV ) 1.673 × 10 − 27 kg

)

N = 14 π (2 mm) 3.21 × 1013 m −3 (1 min ) 2

= 3.7 × 1017

(c) The magnitude of the current density in this beam is given by:

J=

I 1.0 mA = A π 1.0 × 10 −3 m

(

)

2

= 0.32 kA/m 2 35 •• [SSM] In one of the colliding beams of a planned proton supercollider, the protons are moving at nearly the speed of light and the beam current is 5.00-mA. The current density is uniformly distributed throughout the beam. (a) How many protons are there per meter of length of the beam? (b) If the cross-sectional area of the beam is 1.00 x 10–6 m2, what is the number density of protons? (c) What is the magnitude of the current density in this beam?

404

Chapter 25

Picture the Problem We can relate the number of protons per meter N to the number n of free charge-carrying particles per unit volume in a beam of crosssectional area A and then use the relation between current and drift velocity to relate n to I.

(a) Express the number of protons per meter N in terms of the number n of free charge-carrying particles per unit volume in a beam of crosssectional area A:

N = nA

Use the relation between current and drift velocity to relate I and n:

I = enAv ⇒ n =

Substitute for n and simplify to obtain:

N=

IA I = eAv ev

Substitute numerical values and evaluate N:

N=

5.00 mA (1.602 × 10 C)(2.998 × 108 m/s)

(1)

I eAv

−19

= 1.041 × 10 8 m −1 = 1.04 × 10 8 m −1

(b) From equation (1) we have:

n=

N 1.041× 10 8 m −1 = A 1.00 × 10 −6 m 2

= 1.04 × 1014 m −3

(c) The magnitude of the current density in this beam is given by:

J=

I 5.00 mA = = 5.00 kA/m 2 −6 2 A 1.00 × 10 m

36 •• The solar wind consists of protons from the Sun moving toward Earth ( the wind actually consists of about 95% protons). The number density of protons at a distance from the Sun equal to the orbital radius of Earth is about 7.0 protons per cubic centimeter. Your research team monitors a satellite that is in orbit around the Sun at a distance from the Sun equal to equal to Earth’s orbital radius. You are in charge of the satellite’s mass spectrometer, an instrument used to measure the composition and intensity of the solar wind. The aperture of your spectrometer is a circle of radius 25 cm. The rate of collection of protons by the spectrometer is such that they constitute a measured current of 85 nA. What is the speed of the protons in the solar wind? (Assume the protons enter the aperture at normal incidence.)

`Electric Current and Direct-Current Circuits

405

Picture the Problem We can use Equation 25-3 to express the drift speed of the protons in the solar wind in terms of the proton current, the number density of the protons, and the cross-sectional area of the aperture of the spectrometer.

From Equation 25-3, the drift speed of the protons in the solar winds is given by:

vd =

I qnA

Substitute numerical values and evaluate vd: vd =

(1.602 ×10

−19

85 nA = 3.9 × 10 5 m/s ⎛ 7.0 ⎞ 2 C ⎜ 3 ⎟π (25 cm ) ⎝ cm ⎠

)

37 •• A gold wire has a 0.10-mm-diameter cross section. Opposite ends of this wire are connected to the terminals of a 1.5-V battery. If the length of the wire is 7.5 cm, how much time, on average, is required for electrons leaving the negative terminal of the battery to reach the positive terminal? Assume the resistivity of gold is 2.44 ×10−8  Ω ⋅ m . Picture the Problem We can use the definition of average speed, Equation 25-3 ( I = qnAvd ), Equation 25-10 ( R = ρL A ) and the relationship between the number density of electrons, Avogadro’s number, and the molar mass of gold to derive an expression for the average travel time of the electrons.

The average transit time for the electrons is the ratio of the distance they travel to their drift speed:

Δt =

L where L is the length of the vd

gold wire. I qnA

From Equation 25-3:

vd =

Substituting for vd and I and simplifying yields:

Δt =

Use Equation 25-10 to express the resistance of the gold wire:

R = ρ Au

Substitute for R and simplify to obtain:

Δt =

qnAL qnALR L = = V I V R qnA L A

qnALρ Au V

L 2 A = qnL ρ Au V

406

Chapter 25

The number density of free electrons is given by:

⎛ mass ⎞ ⎛ 1 ⎞ n=⎜ ⎟⎜ ⎟NA ⎝ molar mass ⎠ ⎝ volume ⎠ ⎛ ρ density, Au ⎞ ⎛ ⎞ ⎟ N A = ⎜ N A ⎟ ρ density, Au =⎜ ⎜ M ⎟ ⎜M ⎟ ⎝ mol, Au ⎠ ⎝ mol, Au ⎠

Finally, substituting for n yields:

⎛ NA q⎜⎜ M mol, Au Δt = ⎝

⎞ ⎟ ρ density, Au L2 ρ Au ⎟ ⎠ V

Substitute numerical values and evaluate Δt: ⎛ ⎞ ⎜ 6.022 × 10 23 mol -1 ⎟ ⎟ ⎛⎜19.3 × 10 3 kg3 ⎞⎟ (7.5 cm )2 Δt = 1.602 × 10 −19 C ⎜ m ⎠ ⎜ 0.19697 kg ⎟ ⎝ ⎜ ⎟ mol ⎠ ⎝ −8 2.44 × 10 Ω ⋅ m × = 0.86 s 1.5 V

(

)

(

)

Resistance, Resistivity and Ohm’s Law 38 • A 10-m-long wire has a resistance equal to 0.20 Ω and carries a current equal to 5.0 A. (a) What is the potential difference across the entire length of the wire? (b) What is the electric-field strength in the wire? Picture the Problem We can use Ohm’s law to find the potential difference between the ends of the wire and V = EL to find the magnitude of the electric field in the wire.

(a) Apply Ohm’s law to obtain:

V = RI = (0.20 Ω )(5.0 A ) = 1.0 V

(b) Relate the electric field to the potential difference across the wire and the length of the wire:

E=

V 1.0 V = = 0.10 V/m L 10 m

39 • [SSM] A potential difference of 100 V across the terminals of a resistor produces a current of 3.00 A in the resistor. (a) What is the resistance of the resistor? (b) What is the current in the resistor when the potential difference is only 25.0 V? (Assume the resistance of the resistor remains constant.) Picture the Problem We can apply Ohm’s law to both parts of this problem, solving first for R and then for I.

`Electric Current and Direct-Current Circuits 407

(a) Apply Ohm’s law to obtain:

R=

V 100 V = = 33.3 Ω I 3.00 A

(b) Apply Ohm’s law a second time to obtain:

I=

V 25.0 V = = 0.750 A R 33.3 Ω

40 • A block of carbon is 3.0 cm long and has a square cross-section whose sides are 0.50-cm long. A potential difference of 8.4 V is maintained across its length. (a) What is the resistance of the block? (b) What is the current in this resistor? Picture the Problem We can use R = ρL A to find the resistance of the block and Ohm’s law to find the current in it for the given potential difference across its length.

L A

(a) Relate the resistance of the block to its resistivity ρ, cross-sectional area A, and length L:

R=ρ

Substitute numerical values (see Table 25-1 for the resistivity of carbon) and evaluate R:

R = 3500 × 10 −8 Ω ⋅ m

(b) Apply Ohm’s law to obtain:

I=

(

)

3.0 cm (0.50 cm )2

= 42.0 mΩ = 42 mΩ V 8.4 V = = 0.20 kA R 42.0 mΩ

41 • [SSM] An extension cord consists of a pair of 30-m-long 16-gauge copper wires. What is the potential difference that must be applied across one of the wires if it is to carry a current of 5.0 A? Picture the Problem We can use Ohm’s law in conjunction with R = ρL A to find the potential difference across one wire of the extension cord.

Using Ohm’s law, express the potential difference across one wire of the extension cord:

V = IR

Relate the resistance of the wire to its resistivity ρ, cross-sectional area A, and length L:

R=ρ

L A

408

Chapter 25 LI A

Substitute for R to obtain:

V =ρ

Substitute numerical values (see Table 25-1 for the resistivity of copper and Table 25-2 for the crosssectional area of 16-gauge wire) and evaluate V:

V = 1.7 × 10 −8 Ω ⋅ m

(

m )(5.0 A ) ) (30 1.309 mm 2

= 1.9 V

42 • (a) How long is a 14-gauge copper wire that has a resistance of 12.0 Ω? (b) How much current will it carry if a 120-V potential difference is applied across its length? Picture the Problem We can use R = ρL A to find the length of a 14-gauge copper wire that has a resistance of 12.0 Ω and Ohm’s law to find the current in the wire.

(a) Relate the resistance of the wire to its resistivity ρ, cross-sectional area A, and length L: Substitute numerical values (see Table 25-1 for the resistivity of copper and Table 25-2 for the crosssectional area of 14-gauge wire) and evaluate L: (b) Apply Ohm’s law to find the current in the wire:

R=ρ

L RA ⇒L = ρ A

( 12.0 Ω )(2.081mm 2 ) L= = 1.7 ×10 −8 Ω ⋅ m

I=

1.5 km

V 120 V = = 10.0 A R 12.0 Ω

A cylinder of glass is 1.00 cm long and has a resistivity of 1.01×1012 43 • Ω⋅m. What length of copper wire that has the same cross-sectional area will have the same resistance as the glass cylinder? Picture the Problem We can use R = ρL A to express the resistances of the glass cylinder and the copper wire. Expressing their ratio will eliminate the common cross-sectional areas and leave us with an expression we can solve for the length of the copper wire.

Relate the resistance of the glass cylinder to its resistivity, crosssectional area, and length:

Rglass = ρ glass

Lglass Aglass

`Electric Current and Direct-Current Circuits 409

Relate the resistance of the copper wire to its resistivity, cross-sectional area, and length: Divide the second of these equations by the first to obtain:

Because Aglass = ACu and RCu = Rglass:

Substitute numerical values (see Table 25-1 for the resistivities of glass and copper) and evaluate LCu:

RCu = ρ Cu

RCu = Rglass

1=

LCu ACu

ρ Cu ρ glass

LCu ρ L ACu = Cu ⋅ Cu Lglass ρ glass Lglass Aglass

ρ ρ Cu LCu ⋅ ⇒ LCu = glass Lglass ρ glass Lglass ρ Cu 1.01× 1012 Ω ⋅ m (1.00 cm ) 1.7 × 10 −8 Ω ⋅ m 1c ⋅ y = 5.94 ×1017 m × 9.461×1015 m

LCu =

= 63 c ⋅ y

44 •• While remodeling your garage, you need to temporarily splice, end to end, an 80-m-long copper wire that is 1.00 mm in diameter with a 49-m-long aluminum wire that has the same diameter. The maximum current in the wires is 2.00 A. (a) Find the potential drop across each wire of this system when the current is 2.00 A. (b) Find the electric field in each wire when the current is 2.00 A. Picture the Problem We can use Ohm’s law to relate the potential differences across the two wires to their resistances and R = ρL A to relate their resistances to their lengths, resistivities, and cross-sectional areas. Once we’ve found the potential differences across each wire, we can use E = V L to find the electric field in each wire.

(a) Apply Ohm’s law to express the potential drop across each wire:

VCu = IRCu and VFe = IRFe

Relate the resistances of the wires to their resistivity, cross-sectional area, and length:

RCu = ρ Cu

LCu L and RFe = ρ Cu Fe ACu AFe

410

Chapter 25

Substitute for RCu and RFe to obtain:

Substitute numerical values (see Table 25-1 for the resistivities of copper and iron) and evaluate the potential differences:

VCu =

VCu =

ρ Cu I Cu ACu

I and VFe =

(1.7 ×10 1 4

ρ Fe I Fe AFe

I

)

Ω ⋅ m (80 m ) (2.00 A ) π (1.00 mm )2 −8

= 3.463 V = 3.5 V and VFe =

(10 ×10 1 4

)

Ω ⋅ m (49 m ) (2.00 A ) π (1.00 mm )2 −8

= 12.48 V = 12 V (b) Express the electric field in each conductor in terms of its length and the potential difference across it:

ECu =

VCu V and EFe = Fe I Fe I Cu

Substitute numerical values and evaluate the electric fields:

ECu =

3.463 V = 43 mV/m 80 m

and E Fe =

12.48 V = 0.25 V/m 49 m

45 •• [SSM] A 1.00-m-long wire has a resistance equal to 0.300 Ω. A second wire made of identical material has a length of 2.00 m and a mass equal to the mass of the first wire. What is the resistance of the second wire? Picture the Problem We can use R = ρL A to relate the resistance of the wires to their lengths, resistivities, and cross-sectional areas. To find the resistance of the second wire, we can use the fact that the volumes of the two wires are the same to relate the cross-sectional area of the first wire to the cross-sectional area of the second wire.

Relate the resistance of the first wire to its resistivity, cross-sectional area, and length:

R=ρ

L A

Relate the resistance of the second wire to its resistivity, cross-sectional area, and length:

R' = ρ

L' A'

`Electric Current and Direct-Current Circuits 411

Divide the second of these equations by the first to obtain:

L' A R' L' A = A' = ⇒ R' = 2 R (1) L R L A' A' ρ A

Express the relationship between the volume V of the first wire and the volume V′ of the second wire:

V = V' or LA = L'A' ⇒

Substituting for

A in equation (1) A'

ρ

A L' = =2 A' L

R' = 2(2 )R = 4 R

yields: Because R = 3.00 Ω:

R' = 4(0.300 Ω ) = 1.20 Ω

46 •• A 10-gauge copper wire can safely carry currents up to 30.0 A. (a) What is the resistance of a 100-m length of the wire? (b) What is the electric field in the wire when the current is 30.0 A? (c) How long does it take for an electron to travel 100 m in the wire when the current is 30.0 A? Picture the Problem We can use R = ρL A to find the resistance of the wire from its length, resistivity, and cross-sectional area. The electric field can be found using E = V/L and Ohm’s law to eliminate V. The time for an electron to travel the length of the wire can be found from L = vdΔt, with vd expressed in term of I using I = neAvd .

L A

(a) Relate the resistance of the unstretched wire to its resistivity, cross-sectional area, and length:

R=ρ

Substitute numerical values (see Table 25-1 for the resistivity of copper and Table 25-2 for the cross-sectional area of 10-gauge wire) and evaluate R:

R = 1.7 ×10 −8 Ω ⋅ m

(b) Relate the electric field in the wire to the potential difference between its ends:

E=

V L

Use Ohm’s law to substitute for V:

E=

IR L

(

100 m )5.261 mm

= 0.323 Ω = 0.32 Ω

2

412

Chapter 25

Substitute numerical values and evaluate E: (c) Express the time Δt for an electron to travel a distance L in the wire in terms of its drift speed vd:

E=

(30.0 A )(0.323 Ω ) =

Δt =

L vd

100 m

Relate the current in the wire to the drift speed of the charge carriers:

I = neAvd ⇒ vd =

Substitute for vd to obtain:

Δt =

97 mV/m

I neA

neAL I

Substitute numerical values (in Example 25-1 it is shown that n = 8.47×1028 m−3) and evaluate Δt: Δt =

(8.47 × 10

28

)(

)(

)

m −3 1.602 × 10 −19 C 5.261 mm 2 (100 m ) 30.0 A

= 2.38 × 10 5 s = 2.75 d 47 •• A cube of copper has edges that are 2.00-cm long. If copper in the cube is drawn to form a length of 14-gauge wire, what will the resistance of the length of wire be? Assume the density of the copper does not change. Picture the Problem We can use R = ρL A to find express the resistance of the wire in terms of its length, resistivity, and cross-sectional area. The fact that the volume of the copper does not change as the cube is drawn out to form the wire will allow us to eliminate either the length or the cross-sectional area of the wire and solve for its resistance.

L A

Express the resistance of the wire in terms of its resistivity, crosssectional area, and length:

R=ρ

Relate the volume V of the wire to its length and cross-sectional area:

V = AL ⇒ L =

Substitute for L to obtain:

R=ρ

V A2

V A

`Electric Current and Direct-Current Circuits 413

Substitute numerical values (see Table 25-1 for the resistivity of copper and Table 25-2 for the crosssectional area of 14-gauge wire) and evaluate R:

(

−8

R = 1.7 × 10 Ω ⋅ m

)

(2.00 cm )3

(2.081mm )

2 2

= 31 mΩ

48 ••• Find an expression for the resistance between the ends of the half ring shown in Figure 25-51. The resistivity of the material constituting the half-ring is ρ. Hint: Model the half ring as a parallel combination of a large number of thin half-rings. Assume the current is uniformly distributed on a cross section of the half ring. Picture the Problem We can use, as our element of resistance, the semicircular strip of height t, radius r, and thickness dr shown below. Then dR = (πrρ t )dr . Because the strips are in parallel, integrating over them will give us the reciprocal of the resistance of half ring.

t r

dr

πr

The resistance dR between the ends of the infinitesimal strip is given by:

dR = ρ

Because the infinitesimal strips are in parallel, their equivalent resistance is:

1 t dr = R πρ ∫a r

Integrating dReq from r = a to r = b yields:

1 t ⎛b⎞ = ln⎜ ⎟ ⇒ R = R πρ ⎝ a ⎠

tdr b

ρπ ⎛b⎞ t ln⎜ ⎟ ⎝a⎠

49 ••• [SSM] Consider a wire of length L in the shape of a truncated cone. The radius of the wire varies with distance x from the narrow end according to r = a + [(b – a)/L]x, where 0 < x < L. Derive an expression for the resistance of this wire in terms of its length L, radius a, radius b and resistivity ρ. Hint: Model the wire as a series combination of a large number of thin disks. Assume the current is uniformly distributed on a cross section of the cone.

414

Chapter 25

Picture the Problem The element of resistance we use is a segment of length dx and cross-sectional area π[a + (b − a)x/L]2. Because these resistance elements are in series, integrating over them will yield the resistance of the wire.

dx ρ = dx A π [a + (b − a )( x L )] 2

Express the resistance of the chosen element of resistance:

dR = ρ

Integrate dR from x = 0 to x = L and simplify to obtain:

ρL dx R= ∫ π 0 [a + (b − a )( x L )] 2 ρL

⎞ ⎛1 1 ⎟ ⎜⎜ − π (b − a ) ⎝ a a + (b − a ) ⎟⎠ ρL = πab =

50 ••• The space between two metallic concentric spherical shells is filled with a material that has a resistivity of 3.50 × 10–5 Ω⋅m. If the inner metal shell has an outer radius of 1.50 cm and the outer metal shell has an inner radius of 5.00 cm, what is the resistance between the conductors? Hint: Model the material as a series combination of a large number of thin spherical shells. Picture the Problem The diagram shows a cross-sectional view of the concentric spheres of radii a and b as well as a spherical-shell element of radius r. We can express the resistance dR of the spherical-shell element and then integrate over the volume filled with the material whose resistivity ρ is given to find the resistance between the conductors. Note that the elements of resistance are in series.

Express the element of resistance dR:

dR = ρ

dr dr =ρ A 4πr 2

`Electric Current and Direct-Current Circuits 415

Integrate dR from r = a to r = b to obtain:

ρ b dr ρ ⎛ 1 1 ⎞ = R= ⎜ − ⎟ 4π ∫a r 2 4π ⎝ a b ⎠

Substitute numerical values and evaluate R: R=

3.50 × 10 −5 Ω ⋅ m ⎛ 1 1 ⎞ ⎟⎟ = 130 μΩ ⎜⎜ − 4π ⎝ 1.50 cm 5.00 cm ⎠

51 ••• The space between two metallic coaxial cylinders that have the same length L is completely filled with a nonmetalic material having a resistivity ρ. The inner metal shell has an outer radius a and the outer metal shell has an inner radius b. (a) What is the resistance between the two cylinders? Hint: Model the material as a series combination of a large number of thin cylindrical shells. (b) Find the current between the two metallic cylinders if ρ = 30.0 Ω⋅m, a = 1.50 cm, b = 2.50 cm, L = 50.0 cm, and a potential difference of 10.0 V is maintained between the two cylinders. Picture the Problem The diagram shows a cross-sectional view of the coaxial cylinders of radii a and b as well as a cylindrical-shell element of radius r. We can express the resistance dR of the cylindrical-shell element and then integrate over the volume filled with the material whose resistivity ρ is given to find the resistance between the two cylinders. Note that the elements of resistance are in series.

dr dr =ρ A 2πrL

(a) Express the element of resistance dR:

dR = ρ

Integrate dR from r = a to r = b to obtain:

ρ b dr ρ ⎛b⎞ = ln⎜ ⎟ R= ∫ 2πL a r 2πL ⎝ a ⎠

416

Chapter 25

(b) Apply Ohm’s law to obtain:

I=

Substitute numerical values and evaluate I:

I=

V V 2πLV = = ρ ⎛b⎞ R ⎛b⎞ ln⎜ ⎟ ρ ln⎜ ⎟ 2πL ⎝ a ⎠ ⎝a⎠

2π (50.0 cm )(10.0 V ) = 2.05 A ⎛ 2.50 cm ⎞ ⎟⎟ (30.0 Ω ⋅ m )ln⎜⎜ ⎝ 1.50 cm ⎠

Temperature Dependence of Resistance 52 • A tungsten rod is 50 cm long and has a square cross-section that has 1.0-mm-long edges. (a) What is its resistance at 20°C? (b) What is its resistance at 40°C? Picture the Problem We can use R = ρL/A to find the resistance of the rod at 20°C. Ignoring the effects of thermal expansion, we can we apply the equation defining the temperature coefficient of resistivity, α, to relate the resistance at 40°C to the resistance at 20°C.

L A

(a) Express the resistance of the rod at 20°C as a function of its resistivity, length, and cross-sectional area:

R20 = ρ 20

Substitute numerical values and evaluate R20:

R20 = 5.5 × 10 −8 Ω ⋅ m

(

)

0.50 m (1.0 mm)2

= 27.5 mΩ = 28 mΩ (b) Express the resistance of the rod at 40°C as a function of its resistance at 20°C and the temperature coefficient of resistivity α:

R40 = ρ 40

L A

= ρ 20 [1 + α (tC − 20C°)]

L A

L L + ρ 20 α (tC − 20C°) A A = R20 [1 + α (tC − 20C°)] = ρ 20

Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of tungsten) and evaluate R40:

[ (

)

]

R40 = (27.5 mΩ ) 1 + 4.5 × 10 −3 K −1 (20 C°) = 30 mΩ

`Electric Current and Direct-Current Circuits 417

53 • [SSM] At what temperature will the resistance of a copper wire be 10 percent greater than its resistance at 20°C? Picture the Problem The resistance of the copper wire increases with temperature according to RtC = R20 [1 + α (tC − 20C°)] . We can replace RtC by 1.1R20 and solve for

tC to find the temperature at which the resistance of the wire will be 110% of its value at 20°C. Express the resistance of the wire at 1.10R20: Simplifying this expression yields:

1.10 R20 = R20 [1 + α (t C − 20C°)] 0.10 = α (t C − 20C°)

Solve to tC to obtain:

tC =

Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of copper) and evaluate tC:

tC =

0.10

α

+ 20C°

0.10 + 20C° = 46°C 3.9 × 10 −3 K −1

54 •• You have a toaster that uses a Nichrome wire as a heating element. You need to determine the temperature of the Nichrome wire under operating conditions. First, you measure the resistance of the heating element at 20°C and find it to be 80.0 Ω. Then you measure the current immediately after you plug the toaster into a wall outlet—before the temperature of the Nichrome wire increases significantly. You find this startup current to be 8.70 A. When the heating element reaches its operating temperature, you measure the current to be 7.00 A. Use your data to determine the maximum operating temperature of the heating element. Picture the Problem Let the primed quantities denote the current and resistance at the final temperature of the heating element. We can express R′ in terms of R20 and the final temperature of the wire tC using R' = R20 [1 + α (t C − 20°C )] and relate

I′, R′, I20, and R20 using Ohm’s law. Express the resistance of the heating element at its final temperature as a function of its resistance at 20°C and the temperature coefficient of resistivity for Nichrome:

R' = R20 [1 + α (t C − 20°C )]

(1)

418

Chapter 25

Apply Ohm’s law to the heating element when it is first turned on:

V = I 20 R20

Apply Ohm’s law to the heating element when it has reached its final temperature:

V = I'R'

Because the voltage is constant, we have:

I'R' = I 20 R20 or R' =

Substitute in equation (1) to obtain:

I 20 R20 = R20 [1 + α (t C − 20°C )] I' or I 20 = 1 + α (t C − 20°C ) I'

Solve for tC to obtain:

I 20 −1 + 20°C t C = I'

I 20 R20 I'

α

Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of Nichrome) and evaluate tC:

8.70 A −1 7.00 A tC = + 20°C ≈ 6 × 10 2 °C 0.4 × 10 −3 K −1 where the answer has only one significant figure because the temperature coefficient is only given to one significant figure.

Remarks: At 600°C the wire glows red. 55 •• Your electric space heater has a Nichrome heating element that has a resistance of 8.00 Ω at 20.0°C. When 120 V are applied, the electric current heats the Nichrome wire to 1000°C. (a) What is the initial current in the heating element at 20.0°C? (b) What is the resistance of the heating element at 1000°C? (c) What is the operating power of this heater? Picture the Problem We can apply Ohm’s law to find the initial current drawn by the cold heating element. The resistance of the wire at 1000°C can be found using R1000 = R20 [1 + α (t C − 20.0°C )] and the power consumption of the heater at this

temperature from P = V2/R1000.

`Electric Current and Direct-Current Circuits 419

(a) Apply Ohm’s law to find the initial current I20 drawn by the heating element: (b) Express the resistance of the heating element at 1000°C as a function of its resistance at 20.0°C and the temperature coefficient of resistivity for Nichrome: Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of Nichrome) and evaluate R1000: (c) The operating power of this heater at 1000°C is:

I=

V 120 V = = 15.0 A R20 8.00 Ω

R1000 = R20 [1 + α (t C − 20.0°C )]

[

R1000 = (8.00 Ω ) 1 + (0.4 × 10 −3 K −1 ) × (1000°C − 20.0°C )]

= 11.1Ω

(120 V ) = 1.30 kW V2 P= = R1000 11.1Ω 2

56 •• A 10.0-Ω Nichrome resistor is wired into an electronic circuit using copper leads (wires) that have diameters equal to 0.600 mm. The copper leads have a total length of 50.0 cm. (a) What additional resistance is due to the copper leads? (b) What percentage error in the total resistance is produced by neglecting the resistance of the copper leads? (c) What change in temperature would produce a change in resistance of the Nichrome wire equal to the resistance of the copper leads? Assume that the Nichrome section is the only one whose temperature is changed. Picture the Problem We can find the resistance of the copper leads using RCu = ρ Cu L A and express the percentage error in neglecting the resistance of

the leads as the ratio of RCu to RNichrome. In Part (c) we can express the change in resistance in the Nichrome wire corresponding to a change ΔtC in its temperature and then find ΔtC by substitution of the resistance of the copper wires in this equation. L A

(a) Relate the resistance of the copper leads to their resistivity, length, and cross-sectional area:

RCu = ρ Cu

Substitute numerical values (see Table 25-1 for the resistivity of copper) and evaluate RCu:

RCu = (1.7 × 10 −8 Ω ⋅ m )

50.0 cm 2 1 4 π (0.600 mm )

= 30.1mΩ = 30 mΩ

420

Chapter 25

(b) Express the percentage error as the ratio of RCu to RNichrome:

% error =

RCu RNichrome

=

30.1mΩ 10.0 Ω

= 0.30% (c) Express the change in the resistance of the Nichrome wire as its temperature changes from tC to tC′: Solve for ΔtC to obtain:

Set ΔR equal to the resistance of the copper wires (see Table 25-1 for the temperature coefficient of resistivity of Nichrome wire) and evaluate ΔtC:

ΔR = R' − R = R20 [1 + α (t C' − 20°C )]

− R20 [1 + α (t C − 20°C )]

= R20αΔt C

ΔtC =

ΔR R20α

Δt C =

30.1mΩ = 8°C (10.0 Ω ) 0.4 ×10 −3 K −1

(

)

57 ••• [SSM] A wire that has a cross-sectional area A, a length L1, a resistivity ρ1, and a temperature coefficient α1 is connected end to end to a second wire that has the same cross-sectional area, a length L2, a resistivity ρ2, and a temperature coefficient α2, so that the wires carry the same current. (a) Show that if ρ1L1α1 + ρ2L2α2 = 0, then the total resistance is independent of temperature for small temperature changes. (b) If one wire is made of carbon and the other wire is made of copper, find the ratio of their lengths for which the total resistance is approximately independent of temperature. Picture the Problem Expressing the total resistance of the two current-carrying (and hence warming) wires connected in series in terms of their resistivities, temperature coefficients of resistivity, lengths and temperature change will lead us to an expression in which, if ρ1L1α1 + ρ2L2α2 = 0, the total resistance is temperature independent. In Part (b) we can apply the condition that ρ1L1α1 + ρ2L2α2 = 0 to find the ratio of the lengths of the carbon and copper wires.

(a) Express the total resistance of these two wires connected in series:

R = R1 + R2 = ρ1 =

L1 (1 + α1ΔT ) + ρ 2 L2 (1 + α 2 ΔT ) A A

1 [ρ1 L1 (1 + α1ΔT ) + ρ 2 L2 (1 + α 2 ΔT )] A

`Electric Current and Direct-Current Circuits 421

Expand and simplify this expression to obtain:

R=

1 [ρ1L1 + ρ 2 L2 + (ρ1L1α1 + ρ1L1α 2 )ΔT ] A

If ρ1L1α1 + ρ2L2α2 = 0, then:

1 [ρ1 L1 + ρ 2 L2 ] independently of A the temperature.

R=

(b) Apply the condition for temperature independence obtained in (a) to the carbon and copper wires:

ρ C LCα C + ρ Cu LCuα Cu = 0

Solve for the ratio of LCu to LC:

ρα LCu =− C C ρ Cuα Cu LC

Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of carbon and copper) and evaluate the ratio of LCu to LC:

LCu ( 3500 × 10 −8 Ω ⋅ m )(− 0.5 × 10 −3 K −1 ) =− ≈ 3 × 10 2 −8 −3 −1 LC (1.7 ×10 Ω ⋅ m )(3.9 ×10 K ) 58 ••• The resistivity of tungsten increases approximately linearly with temperature from 56.0 nΩ⋅m at 293 K to 1.10 μΩ⋅m at 3500 K. A light bulb is powered by a a 100-V dc power supply. Under these operating conditions the temperature of the tungsten filament is 2500 K, the length of the filament is equal to 5.00 cm and the power delivered to the filament is 40-W. Estimate (a) the resistance of the filament and (b) the diameter of the filament. Picture the Problem We can use the relationship between the rate at which energy is transformed into heat and light in the filament and the resistance of and potential difference across the filament to estimate the resistance of the filament. The linear dependence of the resistivity on temperature will allow us to find the resistivity of the filament at 2500 K. We can then use the relationship between the resistance of the filament, its resistivity, and cross-sectional area to find its diameter.

(a) Express the wattage of the lightbulb as a function of its resistance R and the voltage V supplied by the source:

P=

V2 V2 ⇒R = R P

422

Chapter 25

Substitute numerical values and evaluate R: (b) Relate the resistance R of the filament to its resistivity ρ, radius r, and length l: The diameter d of the filament is:

Because the resistivity varies linearly with temperature, we can use a proportion to find its value at 2500 K: Solve for ρ2500 K to obtain:

2 ( 100 V ) R=

40 W

R=

= 0.25 kΩ

ρl ρl ⇒r = 2 πr πR

d =2

ρl πR

(1)

ρ 2500 K − ρ 293 K 2500 K − 293 K = ρ3500 K − ρ 293 K 3500 K − 293 K =

ρ 2500 K =

2207 3207

2207 (ρ3500 K − ρ293 K ) + ρ293 K 3207

Substitute numerical values and evaluate ρ2500 K:

ρ 2500 K =

2207 (1.10 μΩ ⋅ m − 56.0 nΩ ⋅ m ) + 56.0 nΩ ⋅ m = 774.5 μΩ ⋅ m 3207

Substitute numerical values in equation (1) and evaluate d:

d =2

(774.5 nΩ ⋅ m )(5.00 cm ) π (250 Ω )

= 14.0 μm

59 ••• A 5.00-V light bulb used in an electronics class has a carbon filament that has a length of 3.00 cm and a diameter of 40.0 μm. At temperatures between 500 K and 700 K, the resistivity of the carbon used in making small light bulb filaments is about 3.00 × 10–5 Ω⋅m. (a) Assuming that the bulb is a perfect blackbody radiator, calculate the temperature of the filament under operating conditions. (b) One concern about carbon filament bulbs, unlike tungsten filament bulbs, is that the resistivity of carbon decreases with increasing temperature. Explain why this decrease in resistivity is a concern. Picture the Problem We can use the relationship between the rate at which an object radiates and its temperature to find the temperature of the bulb.

(a) At a temperature T, the power emitted by a perfect blackbody is:

P = σAT 4 where σ = 5.67×10−8 W/m2⋅K4 is the Stefan-Boltzmann constant.

`Electric Current and Direct-Current Circuits 423

Solve for T and substitute for P to obtain:

T =4

Relate the resistance R of the filament to its resistivity ρ:

R=

Substitute for R in the expression for T to obtain:

T=

P 4 P V2 = =4 σA σπdL σπdLR

ρL A

4

=

4 ρL πd 2

V2 V 2d =4 4 ρL 4σL2 ρ σπdL 2 πd

Substitute numerical values and evaluate T: T =4

(5.00 V )2 (40.0 ×10 −6 m ) 2 4(5.67 × 10 −8 W/m 2 ⋅ K 4 )(0.0300 m ) (3.00 × 10 −5 Ω ⋅ m )

= 636 K

(b) As the filament heats up, its resistance decreases. This results in more power being dissipated, further heat, higher temperature, etc. If not controlled, this thermal runaway can burn out the filament.

Energy in Electric Circuits 60 • A 1.00-kW heater is designed to operate at 240 V. (a) What is the heater’s resistance and what is the current in the wires that supply power to the heater? (b) What is the power delivered to the heater if it operates at 120 V? Assume that its resistance remains the same. Picture the Problem We can use P = V 2 R to find the resistance of the heater and Ohm’s law to find the current it draws.

(a) Express the power output of the heater in terms of its resistance and its operating voltage: Substitute numerical values and evaluate R:

P=

V2 V2 ⇒R = R P

2 ( 240 V ) R=

1.00 kW

Apply Ohm’s law to find the current drawn by the heater:

I=

(b) The power delivered to the heater operating at 120 V is:

P=

= 57.60 Ω = 57.6 Ω

240 V V = = 4.17 A R 57.60 Ω

(120 V )2 57.60 Ω

= 250 W

424

Chapter 25

61 • A battery has an emf of 12 V. How much work does it do in 5.0 s if it delivers a current of 3.0 A? Picture the Problem We can use the definition of power and the relationship between the battery’s power output and its emf to find the work done by it under the given conditions.

Use the definition of power to relate the work done by the battery to the time current is drawn from it: Express the power output of the battery as a function of the battery’s emf: Substituting for P yields: Substitute numerical values and evaluate ΔW:

P=

ΔW ⇒ ΔW = PΔt Δt

P = εI

ΔW = εIΔt ΔW = (12 V )(3.0 A )(5.0 s ) = 0.18 kJ

62 • An automotive battery has an emf of 12.0 V. When supplying power to the starter motor, the current in the battery is 20.0 A and the terminal voltage of the battery is 11.4 V. What is the internal resistance of the battery? Picture the Problem We can relate the terminal voltage of the battery to its emf, internal resistance, and the current delivered by it and then solve this relationship for the internal resistance.

Express the terminal potential difference of the battery in terms of its emf and internal resistance:

Va − Vb = ε − Ir ⇒ r =

Substitute numerical values and evaluate r:

r=

ε − (Va − V )b I

12.0 V − 11.4 V = 0.03 Ω 20.0 A

63 • [SSM] (a) How much power is delivered by the battery in Problem 62 due to the chemical reactions within the battery when the current in the battery is 20 A? (b) How much of this power is delivered to the starter when the current in the battery is 20 A? (c) By how much does the chemical energy of the battery decrease if the current in the starter is 20 A for 7.0 s? (d) How much energy is dissipated in the battery during these 7.0 seconds?

`Electric Current and Direct-Current Circuits

425

Picture the Problem We can find the power delivered by the battery from the product of its emf and the current it delivers. The power delivered to the battery can be found from the product of the potential difference across the terminals of the starter (or across the battery when current is being drawn from it) and the current being delivered to it. In Part (c) we can use the definition of power to relate the decrease in the chemical energy of the battery to the power it is delivering and the time during which current is drawn from it. In Part (d) we can use conservation of energy to relate the energy delivered by the battery to the heat developed in the battery and the energy delivered to the starter

(a) Express the power delivered by the battery as a function of its emf and the current it delivers: (b) Relate the power delivered to the starter to the potential difference across its terminals:

P = εI = (12.0 V )(20 A ) = 240 W = 0.24 kW Pstarter = Vstarter I = (11.4 V )(20 A ) = 228 W = 0.23 kW

(c) Use the definition of power to express the decrease in the chemical energy of the battery as it delivers current to the starter:

ΔE = PΔt

(d) Use conservation of energy to relate the energy delivered by the battery to the heat developed in the battery and the energy delivered to the starter:

Edelivered = E transformed + Edelivered

Express the energy delivered by the battery and the energy delivered to the starter in terms of the rate at which this energy is delivered: Substitute numerical values and evaluate Q:

= (240 W )(7.0 s ) = 1680 J = 1.7 kJ

by battery

into heat

to starter

= Q + Edelivered

to starter

PΔt = Q + Ps Δt ⇒ Q = (P − Ps )Δt

Q = (240 W − 228 W )(7.0 s ) = 84 J

64 • A battery that has an emf of 6.0 V and an internal resistance of 0.30 Ω is connected to a variable resistor with resistance R. Find the current and power delivered by the battery when R is (a) 0, (b) 5.0 Ω, (c) 10 Ω, and (d) infinite.

426

Chapter 25

Picture the Problem We can use conservation of energy to relate the emf of the battery to the potential differences across the variable resistor and the energy converted to heat within the battery. Solving this equation for I will allow us to find the current for the four values of R and we can use P = I 2R to find the power delivered by the battery for the four values of R.

Apply conservation of energy (Kirchhoff’s loop rule) to the circuit:

ε = IR + Ir

Express the power delivered by the battery as a function of the current drawn from it:

P = I 2R

(a) For R = 0:

I=

ε R+r

=

⇒I =

ε R+r

6.0 V = 20 A 0.30 Ω

and 2 P = (20 A ) (0 ) = 0 W (b) For R = 5.0 Ω:

I=

ε R+r

=

6.0 V = 1.13 A 5.0 Ω + 0.30 Ω

= 1.1A

and 2 P = (1.13 A ) (5.0 Ω ) = 6.4 W (c) For R = 10 Ω:

I=

ε R+r

=

6.0 V = 0.583 A 10 Ω + 0.30 Ω

= 0.58 A

and 2 P = (0.583A ) (10 Ω ) = 3.4 W (d) For R = ∞:

I=

ε R+r

and P = 0W

= lim R →∞

6.0 V = 0A R + 0.30 Ω

`Electric Current and Direct-Current Circuits

427

65 •• A 12.0-V automobile battery that has a negligible internal resistance can deliver a total charge of 160 A⋅h. (a) What is the amount of energy stored in the battery? (b) After studying all night for a calculus test, you try to drive to class to take the test. However, you find that the car’s battery is ″dead″ because you had left the headlights on! Assuming the battery was able to produce current at a constant rate until it died, how long were your lights on? Assume the pair of headlights together operates at a power of 150 W. Picture the Problem We can express the total stored energy ΔU in the battery in terms of its emf and the product IΔt of the current it can deliver for a period of time Δt. We can apply the definition of power to relate the lifetime of the battery to the rate at which it is providing energy to the pair of headlights.

(a) Express ΔU in terms of ε and the product IΔt: Substitute numerical values and evaluate ΔU:

ΔU = εIΔt

ΔU = (12.0 V )(160 A ⋅ h ) = 1.92 kW ⋅ h 3.6 MJ = 1.92 kW ⋅ h × kW ⋅ h = 6.9 MJ

(b) Use the definition of power to relate the lifetime of the battery to the rate at which it is providing energy to the pair of headlights:

Δt =

ΔU P

Substitute numerical values and evaluate Δt:

Δt =

1.92 kW ⋅ h = 12.8 h 150 W

66 •• The measured current in a circuit in your uncle’s house is 12.5 A. In this circuit, the only appliance that is on is a space heater that is being used to heat the bathroom. A pair of 12-gauge copper wires carries the current from the supply panel in your basement to the wall outlet in the bathroom, a distance of 30.0 m. You measure the voltage at the supply panel to be exactly 120 V. What is the voltage at the wall outlet in the bathroom that the space heater is connected to? Picture the Problem We can use conservation of energy (Kirchhoff’s loop rule) to relate the emf at the fuse box and the voltage drop in the wires to the voltage at the wall outlet in the bathroom.

428

Chapter 25

ε − Vwires − Voutlet = 0

Apply Kirchhoff’s loop rule to the circuit to obtain:

or

Solve for Voutlet to obtain:

Voutlet = ε − IRwires

ε − IRwires − Voutlet = 0

Relate the resistance of the copper wires to the resistivity of copper, the length of the wires, and the crosssectional area of 12-gauge wire:

Rwires = ρ Cu

L A

Substituting for Rwires yields:

Voutlet = ε −

Iρ Cu L A

Substitute numerical values (see Table 25 -1 for the resistivity of copper and Table 25-2 for the cross-sectional area of 12-gauge wire) and evaluate Voutlet: Voutlet

( 12.5 A ) (1.7 × 10 −8 Ω ⋅ m )(60.0 m ) = 120 V − = 3.309 mm 2

116 V

67 •• [SSM] A lightweight electric car is powered by a series combination of ten 12.0-V batteries, each having negligible internal resistance. Each battery can deliver a charge of 160 A⋅h before needing to be recharged. At a speed of 80.0 km/h, the average force due to air drag and rolling friction is 1.20 kN. (a) What must be the minimum power delivered by the electric motor if the car is to travel at a speed of 80.0 km/h? (b) What is the total charge, in coulombs, that can be delivered by the series combination of ten batteries before recharging is required? (c) What is the total electrical energy delivered by the ten batteries before recharging? (d) How far can the car travel (at 80.0 km/h) before the batteries must be recharged? (e) What is the cost per kilometer if the cost of recharging the batteries is 9.00 cents per kilowatt-hour? Picture the Problem We can use P = fv to find the power the electric motor must develop to move the car at 80 km/h against a frictional force of 1200 N. We can find the total charge that can be delivered by the 10 batteries using ΔQ = NIΔt . The total electrical energy delivered by the 10 batteries before recharging can be found using the definition of emf. We can find the distance the car can travel from the definition of work and the cost per kilometer of driving the car this distance by dividing the cost of the required energy by the distance the car has traveled.

`Electric Current and Direct-Current Circuits

(a) Express the power the electric motor must develop in terms of the speed of the car and the friction force: (b) Because the batteries are in series, the total charge that can be delivered before charging is the same as the charge from a single battery: (c) Use the definition of emf to express the total electrical energy available in the batteries:

429

P = fv = (1.20 kN )(80.0 km/h ) = 26.7 kW

⎛ 3600 s ⎞ ΔQ = IΔt = (160 A ⋅ h )⎜ ⎟ ⎝ h ⎠ = 576 kC

W = Qε = 10 (576 kC )(12.0 V ) = 69.12 MJ = 69.1 MJ W f

(d) Relate the amount of work the batteries can do to the work required to overcome friction:

W = fd ⇒ d =

Substitute numerical values and evaluate d:

d=

(e) The cost per kilometer is the ratio of the cost of the energy to the distance traveled before recharging:

⎛ $0.0900 ⎞ ⎜ ⎟ε I t kW ⋅ h ⎠ ⎝ Cost/km = d

69.12 MJ = 57.6 km 1.20 kN

Substitute numerical values and calculate the cost per kilometer: ⎛ $0.0900 ⎞ ⎜ ⎟(120 V )(160 A ⋅ h ) kW ⋅ h ⎠ ⎝ Cost/km = = $0.300 / km 5.76 km 68 ••• A 100-W heater is designed to operate with an applied voltage of 120 V. (a) What is the heater’s resistance, and what current does the heater carry? (b) Show that if the potential difference V across the heater changes by a small amount ΔV, the power P changes by a small amount ΔP, where ΔP/P ≈ 2ΔV/V. Hint: Approximate the changes by modeling them as differentials, and assume the resistance is constant. (c) Using the Part-(b) result, find the approximate power dissipated in the heater, if the potential difference is decreased to 115 V. Compare your result to the exact answer.

430

Chapter 25

Picture the Problem We can use the definition of power to find the current drawn by the heater and Ohm’s law to find its resistance. In Part (b) we’ll use the hint to show that ΔP/P ≈ 2ΔV/V and in Part (c) use this result to find the approximate power dissipated in the heater if the potential difference is decreased to 115 V.

P V

(a) Use the definition of power to relate the current I drawn by the heater to its power rating P and the potential difference across it V:

P = IV ⇒ I =

Substitute numerical values and evaluate I:

I=

100 W = 0.833 A 120 V

Apply Ohm’s law to relate the resistance of the heater to the voltage across it and the current it draws:

R=

120 V V = = 144 Ω I 0.833 A

(b) Approximating dP/dV as a differential yields:

dP ΔP dP ≈ ⇒ ΔP ≈ ΔV dV ΔV dV

Express the dependence of P on V:

Assuming R to be constant, evaluate dP/dV: Substitute to obtain:

P=

V2 R

dP d ⎡V 2 ⎤ 2V = ⎢ ⎥= dV dV ⎣ R ⎦ R

ΔP ≈

2V V 2 ΔV ΔV ΔV = 2 = 2P R R V V

Divide both sides of the equation by P to obtain:

ΔP ΔV = 2 P V

(c) Express the approximate power dissipated in the heater as the sum of its power consumption and the change in its power dissipation when the voltage is decreased by ΔV:

P ≈ P0 + ΔP ΔV V ΔV ⎞ ⎛ = P0 ⎜1 + 2 ⎟ V ⎠ ⎝ = P0 + 2 P0

(1)

`Electric Current and Direct-Current Circuits

Assuming that the difference between 120 V and 115 V is good to three significant figures, substitute numerical values and evaluate P: The exact power dissipated in the heater is:

431

⎛ ⎛ − 5.00 V ⎞ ⎞ ⎟⎟ ⎟⎟ P ≈ (100 W )⎜⎜1 + 2⎜⎜ ⎝ 120 V ⎠ ⎠ ⎝ = 91.7 W V' 2 (115 V ) = = = 91.8 W 144 Ω R 2

Pexact

The power calculated using the approximation is 0.1 percent less than the power calculated exactly.

Combinations of Resistors 69 • [SSM] If the potential drop from point a to point b (Figure 25-52) is 12.0 V, find the current in each resistor. Picture the Problem We can apply Ohm’s law to find the current through each resistor.

Apply Ohm’s law to each of the resistors to find the current flowing through each:

12.0 V V = = 3.00 A 4.00 Ω 4.00 Ω 12.0 V V I3 = = = 4.00 A 3.00 Ω 3.00 Ω

I4 =

and I6 =

12.0 V V = = 2.00 A 6.00 Ω 6.00 Ω

Remarks: You would find it instructive to use Kirchhoff’s junction rule (conservation of charge) to confirm our values for the currents through the three resistors.

If the potential drop between point a and point b (Figure 25-53) is 70 • 12.0 V, find the current in each resistor. Picture the Problem We can simplify the network by first replacing the resistors that are in parallel by their equivalent resistance. We can then add that resistance and the 3.00-Ω resistance to find the equivalent resistance between points a and b. Denoting the currents through each resistor with subscripts corresponding to the resistance through which the current flows, we can apply Ohm’s law to find those currents.

432

Chapter 25

Express the equivalent resistance of the two resistors in parallel and solve for Req,1:

1 1 1 1 1 = + = + Req,1 R6 R2 6.00 Ω 2.00 Ω

and Req,1 = 1.50 Ω

Because the 3.00-Ω resistor is in series with Req,1:

Req = R3 + Req,1 = 3.00 Ω + 1.50 Ω

Apply Ohm’s law to the network to find I3:

I3 =

= 4.50 Ω Vab 12.0 V = = 2.667 A Req 4.50 Ω

= 2.67 A

Find the potential difference across the parallel resistors:

V6 & 2 = Vab − V3

Use the common potential difference across the resistors in parallel to find the current through each of them:

I6 =

= 12.0 V − (2.667 A )(3.00 Ω ) = 3.999 V

V6 3.999 V = = 0.667 A 6.00Ω R6

and I2 =

V6 & 2 R2

=

3.999 V = 2.00 A 2.00Ω

Remarks: We could have found the currents through the 6.00-Ω and 2.00-Ω resistors by using the fact that the current at the junction between the 3.00-Ω resistor and the parallel branch divides inversely with the resistance in each branch of the parallel network.

(a) Show that the equivalent resistance between point a and point b in 71 • Figure 25-54 is R. (b) How would adding a fifth resistor that has resistance R between point c and point d effect the equivalent resistance between point a and point b? Picture the Problem Note that the resistors between a and c and between c and b are in series as are the resistors between a and d and between d and b. Hence, we have two branches in parallel, each branch consisting of two resistors R in series. In Part (b) it will be important to note that the potential difference between point c and point d is zero.

`Electric Current and Direct-Current Circuits

(a) Express the equivalent resistance between points a and b in terms of the equivalent resistances between acb and adb:

1 1 1 1 1 = + = + Req Racb Radb 2 R 2 R

Solve for Req to obtain:

Req = R

433

(b) It would not affect it. Because the potential difference between points c and d is zero, no current would flow through the resistor connected between these two points, and the addition of that resistor would not change the network. 72 •• The battery in Figure 25-55 has negligible internal resistance. Find (a) the current in each resistor and (b) the power delivered by the battery. Picture the Problem Note that the 2.00-Ω resistors are in parallel with each other and with the 4.00-Ω resistor. We can Apply Kirchhoff’s loop rule to relate the current I3 drawn from the battery to the emf of the battery and equivalent resistance Req of the resistor network. We can find the current through the resistors connected in parallel by applying Kirchhoff’s loop rule a second time. In Part (b) we can find the power delivered by the battery from the product of its emf and the current it delivers to the circuit.

(a) Apply Kirchhoff’s loop rule to obtain: Find the equivalent resistance of the three resistors in parallel:

ε − I 3 Req = 0 ⇒ I 3 = ε

Req

(1)

1 1 1 1 = + + Req,1 R2 R2 R4 =

1 1 1 + + 2.00 Ω 2.00 Ω 4.00 Ω

and Req,1 = 0.800 Ω Find the equivalent resistance of Req,1 and R3 in series:

Req = R3 + Req,1 = 3.00 Ω + 0.800 Ω

Substitute numerical values in equation (1) and evaluate I3:

I3 =

6.00 V = 1.579 V = 1.58 A 3.80 Ω

Express the current through each of the parallel resistors in terms of their common potential difference V:

I2 =

V V and I 4 = R2 R4

= 3.80 Ω

434

Chapter 25

Apply Kirchhoff’s loop rule to obtain:

ε − I 3 R3 − V = 0 ⇒ V = ε − I 3 R3

Substituting for V in the equations for I2 and I4 yields:

I2 =

Substitute numerical values and evaluate I2 and I4:

I2 =

ε − I 3 R3 R2

and I 4 =

ε − I 3 R3 R4

6.00 V − (1.579 A )(3.00 Ω ) 2.00 Ω

= 0.632 A

and I4 =

6.00 V − (1.579 A )(3.00 Ω ) 4.00 Ω

= 0.316 A

(b) P is the product of ε and I3:

P = ε I 3 = (6.00 V )(1.579 A ) = 9.47 W

Remarks: Note that the currents I3, I2, and I4 satisfy Kirchhoff’s junction rule. [SSM] A 5.00-V power supply has an internal resistance of 50.0 Ω. 73 •• What is the smallest resistor that can be put in series with the power supply so that the voltage drop across the resistor is larger than 4.50 V? Picture the Problem Let r represent the resistance of the internal

resistance of the power supply, ε the emf of the power supply, R the resistance of the external resistor to be placed in series with the power supply, and I the current drawn from the power supply. We can use Ohm’s law to express the potential difference across R and apply Kirchhoff’s loop rule to express the current through R in terms of ε, r, and R.

Express the potential difference across the resistor whose resistance is R:

VR = IR

(1)

`Electric Current and Direct-Current Circuits

Apply Kirchhoff’s loop rule to the circuit to obtain:

435

ε − Ir − IR = 0 ⇒ I = ε

r+R

Substitute in equation (1) to obtain:

V r ⎛ ε ⎞ VR = ⎜ ⎟R ⇒ R = R ε − VR ⎝r+R⎠

Substitute numerical values and evaluate R:

R=

(4.50 V )(50.0 Ω ) = 5.00 V − 4.50 V

0.45 kΩ

74 •• You have been handed an unknown battery. Using your multimeter, you determine that when a 5.00-Ω resistor is connected across the battery’s terminals, the current in the battery is 0.500 A. When this resistor is replaced by an 11.0-Ω resistor, the current drops to 0.250 A. From this data, find (a) the emf and (b) internal resistance of your battery. Picture the Problem We can apply Kirchhoff’s loop rule to the two circuits described in the problem statement and solve the resulting equations

simultaneously for r and ε.

(a) and (b) Apply Kirchhoff’s loop rule to the two circuits to obtain: Substitute numerical values and simplify to obtain:

Solve equations (1) and (2) simultaneously to obtain:

ε − I1r − I1R5 = 0 and ε − I 2 r − I 2 R11 = 0

ε − (0.500 A ) r = 2.50 V

(1)

and ε − (0.250 A ) r = 2.75 V

(2)

ε=

3.00 V and r = 1.00 Ω

75 •• (a) Find the equivalent resistance between point a and point b in Figure 25-56. (b) If the potential drop between point a and point b is 12.0 V, find the current in each resistor. Picture the Problem We can simplify the network by first replacing the resistors that are in parallel by their equivalent resistance. We’ll then have a parallel network with two resistors in series in each branch and can use the expressions for resistors in series to simplify the network to two resistors in parallel. The equivalent resistance between points a and b will be the single resistor equivalent to these two resistors. In Part (b) we’ll use the fact that the potential difference across the upper branch is the same as the potential difference across the lower branch, in conjunction with Ohm’s law, to find the currents through each resistor.

436

Chapter 25

(a) The equivalent resistance of the two 6.00-Ω resistors in parallel is given by:

R6 R6 (6.00 Ω )2 = R6 + R6 6.00 Ω + 6.00 Ω

Req,1 =

= 3.00 Ω

Find the equivalent resistance of the 6.00-Ω resistor is in series with Req,1:

Req,2 = R6 + Req,1

Find the equivalent resistance of the 12.0-Ω resistor in series with the 6.00-Ω resistor:

Req,3 = R6 + R12

Finally, find the equivalent resistance of Req,3 in parallel with Req,2:

= 6.00 Ω + 3.00 Ω = 9.00 Ω

= 6.00 Ω + 12.0 Ω = 18.0 Ω

Req =

Req,2 Req,3 Req,2 + Req,3

=

(9.00 Ω )(18.0 Ω ) 9.00 Ω + 18.0 Ω

= 6.00 Ω

Vab 12.0 V = Req,3 18.0 Ω

(b) Apply Ohm’s law to the upper branch to find the current I upper = I 12 = I 6 :

I upper = I 12 = I 6 =

Apply Ohm’s law to the lower branch to find the current I lower = I 6.00-Ω resistor :

I lower = I 6.00-Ω resistor =

Express the current through the 6.00-Ω resistors in parallel:

I 6.00-Ω resistors = 12 I l =

branch

= 667 mA

branch

branch

in series

branch

in series

Vab 12.0 V = Req,2 9.00 Ω

= 1.33 A

in parallel

1 2

(1.33 A )

= 667 mA

In summary, the current in both the 6.00-Ω and the 12.0-Ω resistor in the upper branch is 667 mA. The current in each 6.00-Ω resistor in the parallel combination in the lower branch is 667 mA. The current in the 6.00-Ω resistor on the right in the lower branch is 1.33 A. 76 •• (a) Find the equivalent resistance between point a and point b in Figure 25-57. (b) If the potential drop between point a and point b is 12.0 V, find the current in each resistor.

`Electric Current and Direct-Current Circuits

437

Picture the Problem Assign currents in each of the resistors as shown in the diagram. We can simplify the network by first replacing the resistors that are in parallel by their equivalent resistance. We’ll then have a parallel network with two resistors in series in each branch and can use the expressions for resistors in series to simplify the network to two resistors in parallel. The equivalent resistance between points a and b will be the single resistor equivalent to these two resistors. In Part (b) we’ll use the fact that the potential difference across the upper branch is the same as the potential difference across the lower branch, in conjunction with Ohm’s law, to find the currents through each resistor.

4.00 Ω 6.00 Ω I1

I4

a

I2

I3

4.00 Ω

I5

I6 (a) Find the equivalent resistance of the resistors in parallel in the upper branch:

2.00 Ω 4.00 Ω b

8.00 Ω

8.00 Ω

(R2 + R4 )R4 (R2 + R4 ) + R4 (6.00 Ω )(4.00 Ω ) =

Req,1 =

2.00 Ω + 4.00 Ω + 4.00 Ω = 2.40 Ω

Find the equivalent resistance of the 6.00-Ω resistor is in series with Req,1:

Req,2 = R6 + Req,1

Express and evaluate the equivalent resistance of the resistors in parallel in the lower branch and solve for Req,2:

Req,2 =

Find the equivalent resistance of the 4.00-Ω resistor is in series with Req,2:

Req,3 = R4 + Req,2

= 6.00 Ω + 2.40 Ω = 8.40 Ω R8 R8 = 12 R8 = 4.00Ω R8 + R8

= 4.00 Ω + 4.00 Ω = 8.00 Ω

438

Chapter 25

Finally, find the equivalent resistance of Req,2 in parallel with Req,3:

Req =

Req,2 Req,3 Req,2 + Req,3

=

(8.40 Ω )(8.00 Ω ) 8.40 Ω + 8.00 Ω

= 4.10 Ω

Vab 12.0 V = = 1.43 A Req,2 8.40 Ω

(b) Apply Ohm’s law to the upper branch to find the current I1 through the 6.00-Ω resistor:

I1 =

Find the potential difference across the 4.00-Ω and 6.00-Ω parallel combination in the upper branch:

V4 and 6 = 12.0 V − V6 = 12.0 V − I 1 R6

Apply Ohm’s law to find I4:

I4 =

V6 3.43 V = = 0.57 A R6 6.00 Ω

Apply Ohm’s law to find I3:

I3 =

V4 3.43 V = = 0.86 A R4 4.00 Ω

Apply Ohm’s law to the lower branch to find I2:

I2 =

Vab 12.0 V = = 1.50 A Req,2 8.00 Ω

Find the potential difference across the 8.00-Ω and 8.00-Ω parallel combination in the lower branch:

V8 and 8 = 12.0 V − I 2 R4

Apply Ohm’s law to find I5 = I6:

I5 = I6 =

= 12.0 V − (1.43 A )(6.00 Ω ) = 3.43 V

= 12.0 V − (1.50 A )(4.00 Ω ) = 6.0 V

V8 and 8 6.0 V = = 0.75 A 8.00 Ω 8.00 Ω

In summary, the current through the 6.00-Ω resistor is 1.43 A, the current through the lower 4.00-Ω resistor is 1.50 A, the current through the 4.00-Ω resistor that is in parallel with the 2.00-Ω and 4.00-Ω resistors is 0.86 A, the current through the 2.00-Ω and 4.00-Ω resistors that are in series is 0.57 A, and the currents through the two 8.00-Ω resistors are 0.75 A. 77 •• [SSM] A length of wire has a resistance of 120 Ω. The wire is cut into pieces that have the same length, and then the wires are connected in parallel. The resistance of the parallel arrangement is 1.88 Ω. Find the number of pieces into which the wire was cut.

`Electric Current and Direct-Current Circuits

439

Picture the Problem We can use the equation for N identical resistors connected in parallel to relate N to the resistance R of each piece of wire and the equivalent resistance Req.

Express the resistance of the N pieces connected in parallel:

1 N = Req R

where R is the resistance of one of the N pieces. Rwire N

Relate the resistance of one of the N pieces to the resistance of the wire:

R=

Substitute for R to obtain:

N2 1 = ⇒N = Req Rwire

Substitute numerical values and evaluate N:

N=

Rwire Req

120 Ω = 8 pieces 1.88 Ω

78 •• A parallel combination of an 8.00-Ω resistor and a resistor of unknown resistance is connected in series with a 16.0-Ω resistor and an ideal battery. This circuit is then disassembled and the three resistors are then connected in series with each other and the same battery. In both arrangements, the current through the 8.00-Ω resistor is the same. What is the resistance of the unknown resistor? Picture the Problem The same current I R flows in R when the resistors are connected as shown in the following diagrams. Applying Kirchhoff’s loop rule to these circuits will yield expressions for I R that we can solve simultaneously for R. 8.00 Ω 16.0 Ω

R IR

R I

24.0 Ω

IR

ε Applying Kirchhoff’s loop rule to the circuit to the left gives:

ε

ε − I (16.0Ω ) − (8.00Ω )R

8.00Ω + R

=0

440

Chapter 25

Solve for I to obtain:

I

=

ε (8.00Ω )R 16.0Ω + 8.00Ω + R

Applying Kirchhoff’s loop rule to the loop that includes the source, the 16.0-Ω resistor and R yields: Solve for I R to obtain:

ε − I (16.0Ω ) − I R R = 0

IR =

Substituting for I and simplifying gives:

ε − I (16.0Ω) R

ε−

ε (16.0Ω ) 8.00Ω )R ( 16.0Ω + 8.00Ω + R R

IR =

⎛

⎞ ⎟ ⎟ R R ⎜1+ ⎟ ⎜ ⎟ ⎝ 16.0Ω + 2 R ⎠

ε ε⎜ = − ⎜ R

Apply Kirchhoff’s loop rule to the circuit to the right to obtain: Solving for I R gives:

Equating the two expressions for I R yields:

ε − I R (24.0Ω ) − I R R = 0 IR =

ε 24.0Ω + R ⎛

⎞ ⎟ ε ⎟= R R ⎜1+ ⎟ 24.0Ω + R ⎜ ⎟ ⎝ 16.0Ω + 2 R ⎠

ε − ε ⎜⎜ R

Simplify to obtain:

1

1

ε ⎞ ⎟ R ⎟= R 24 . 0 Ω R ⎜1+ ⎟ +1 ⎜ ⎟ R ⎝ 16.0Ω + 2 R ⎠ ⎛

ε − ε ⎜⎜ R

1

or 1

1− 1+ Solving for R yields:

R 16.0Ω + 2 R

R = 4.00 Ω

=

1 24.0Ω +1 R

`Electric Current and Direct-Current Circuits 441

79 •• For the network shown in Figure 25-58, let Rab denote the equivalent resistance between terminals a and b. Find (a) R3, so that Rab = R1, (b) R2, so that Rab = R3; and (c) R1, so that Rab = R1. Picture the Problem We can find the equivalent resistance Rab between points a and b and then set this resistance equal, in turn, to R1, R3, and R1 and solve for R3, R2, and R1, respectively.

(a) Express the equivalent resistance between points a and b: Equate this expression to Rl and solve for R3 to obtain:

Rab =

R1 R2 + R3 R1 + R2

R1 =

R12 R1 R2 + R3 ⇒ R3 = R1 + R2 R1 + R2

(b) Set R3 equal to Rab:

R3 =

R1 R2 + R3 R1 + R2

Solve for R2 to obtain:

R2 = 0

(c) Set R1 equal to Rab and rewrite the result as a quadratic equation in R1:

R1 =

Solving the quadratic equation for R1 yields:

R1 R2 + R3 R1 + R2

or R12 − R3 R1 − R2 R3 = 0 R3 + R32 + 4 R2 R3 2 where we’ve used the positive sign because resistance is a non-negative quantity. R1 =

Check your results for Problem 79 using the following specific values: 80 •• (a) R1 = 4.00 Ω, R2 = 6.00 Ω; (b) R1 = 4.00 Ω, R3 = 3.00 Ω; and (c) R2 = 6.00 Ω, R3 = 3.00 Ω. Picture the Problem We can substitute the given resistances in the equations derived in Problem 79.

442

Chapter 25

(a) For R1 = 4.00 Ω and R2 = 6.00 Ω:

R12 (4.00 Ω ) = R3 = R1 + R2 4.00 Ω + 6.00 Ω 2

= 1.60 Ω and Rab = =

R1 R2 + R3 R1 + R2

(4.00 Ω )(6.00 Ω ) + 1.60 Ω 4.00 Ω + 6.00 Ω

= 4.00 Ω (b) For R1 = 4.00 Ω and R3 = 3.00 Ω:

R2 = 0

and Rab =

R1 (0 ) + R3 = 0 + 3.00 Ω R1 + 0

= 3.00 Ω (c) For R2 = 6.00 Ω and R3 = 3.00 Ω: R1 =

3.00 Ω +

(3.00 Ω )2 + 4(6.00 Ω )(3.00 Ω ) 2

and R ab =

=

3.00 Ω + 9.00 Ω = 6.00 Ω 2

R1 R 2 (6.00 Ω )(6.00 Ω ) + R3 = + 3.00 Ω = 6.00 Ω R1 + R 2 6.00 Ω + 6.00 Ω

Kirchhoff’s Rules 81 • [SSM] In Figure 25-59, the battery’s emf is 6.00 V and R is 0.500 Ω. The rate of Joule heating in R is 8.00 W. (a) What is the current in the circuit? (b) What is the potential difference across R? (c) What is the resistance r? Picture the Problem We can relate the current provided by the source to the rate of Joule heating using P = I 2 R and use Ohm’s law and Kirchhoff’s rules to find the potential difference across R and the value of r.

(a) Relate the current I in the circuit to rate at which energy is being dissipated in the form of Joule heat:

P = I 2R ⇒ I =

P R

`Electric Current and Direct-Current Circuits 443

Substitute numerical values and evaluate I:

I=

(b) Apply Ohm’s law to find VR:

VR = IR = (4.00 A )(0.500 Ω ) = 2.00 V

(c) Apply Kirchhoff’s loop rule to obtain: Substitute numerical values and evaluate r:

8.00 W = 4.00 A 0.500 Ω

ε − Ir − IR = 0 ⇒ r = ε − IR = ε I

r=

I

−R

6.00 V − 0.500 Ω = 1.00 Ω 4.00 A

82 • The batteries in the circuit in Figure 25-60 have negligible internal resistance. (a) Find the current using Kirchhoff’s loop rule. (b) Find the power delivered to or supplied by each battery. (c) Find the rate of Joule heating in each resistor. Picture the Problem Assume that the current flows clockwise in the circuit and

let ε1 represent the 12.0-V source and ε2 the 6.00-V source. We can apply Kirchhoff’s loop rule (conservation of energy) to this series circuit to relate the current to the emfs of the sources and the resistance of the circuit. In Part (b) we

can find the power delivered or absorbed by each source using P = εI and in Part (c) the rate of Joule heating using P = I 2R. (a) Apply Kirchhoff’s loop rule to the circuit to obtain: Solving for I yields:

ε 1 − IR2 − ε 2 − IR4 = 0 I=

ε1 − ε 2 R2 + R4 12.0 V − 6.00 V = 1.0 A 2.00 Ω + 4.00 Ω

Substitute numerical values and evaluate I:

I=

(b) Express the power delivered /absorbed by each source in terms of its emf and the current drawn from or forced through it:

P12 = ε 12 I = (12.0 V )(1.0 A ) = 12 W

and P6 = ε 6 I = (− 6.00 V )(1.0 A ) = − 6.0 W where the minus sign means that this source is absorbing power.

(c) Express the rate of Joule heating

P2 = I 2 R2 = (1.0 A ) (2.00 Ω ) = 2.0 W 2

444

Chapter 25

in terms of the current through and the resistance of each resistor:

and 2 P4 = I 2 R4 = (1.0 A ) (4.00 Ω ) = 4.0 W

83 •• An old car battery that has an emf of ε = 11.4 V and an internal resistance of 50.0 mΩ is connected to a 2.00-Ω resistor. In an attempt to recharge

the battery, you connect a second battery that has an emf of ε = 12.6 V and an internal resistance of 10.0 mΩ in parallel with the first battery and the resistor with a pair of jumper cables. (a) Draw a diagram of your circuit. (b) Find the current in each branch of this circuit. (c) Find the power supplied by the second battery and discuss where this power is delivered. Assume that the emfs and internal resistances of both batteries remain constant.

Picture the Problem The circuit is shown in the diagram for Part (a). (b) Let

ε1

and r1 denote the emf of the old battery and its internal resistance, ε2 and r2 the emf of the second battery and its internal resistance, and R the load resistance. Let I1, I2, and IR be the currents. We can apply Kirchhoff’s rules to determine the unknown currents. (c) The power supplied by the second battery is given by P2 = (ε 2 − I 2 r2 )I 2 . (a) The circuit diagram is shown to the right:

(b) Apply Kirchhoff’s junction rule to junction 1 to obtain: Apply Kirchhoff’s loop rule to loop 1 to obtain:

Apply Kirchhoff’s loop rule to loop 2 to obtain:

I1 + I 2 = I R

ε 1 − r1 I1 − RI R = 0 or 11.4 V − (0.0500 Ω )I1 − (2.00 Ω )I R = 0

ε 2 − r2 I 2 − RI R = 0 or 12.6 V − (0.0100 Ω )I 2 − (2.00 Ω )I R = 0

`Electric Current and Direct-Current Circuits 445

Solve the equations in I1, I2, and IR simultaneously to obtain:

I1 = −18.971 A = − 19.0 A , I 2 = 25.145 A = 25.1 A , and I R = 6.174 A = 6.17 A

where the minus sign for I1 means that the current flows in the direction opposite to the direction we arbitrarily chose, i.e., the battery is being charged. (c) The power supplied by the second battery is given by: Substitute numerical values and evaluate P2:

P2 = (ε 2 − I 2 r2 )I 2

P2 = [12.6 V − (25.145 A )(10.0 mΩ )] × (25.145 A )

= 311 W

The power delivered to the first battery is given by: Substitute numerical values and evaluate P1:

Find the power dissipated in the load resistance R:

P1 = ε 1 I 1 + I 12 r1 P1 = (11.4 V )(18.971 A ) + (18.971 A ) (0.0500 Ω ) = 216 W + 18.0 W = 234 W 2

PR = I R2 R = (6.174 A ) (2.00 Ω ) 2

= 76.2 W

In summary, battery 2 supplies 311 W. 234 W is delivered to Battery 1, and of that 234 W, 216 W goes into recharging battery 1 and 18.0 W is dissipated by the internal resistance. In addition, 76.2 W is delivered to the 2.00-Ω resistor. Remarks: Note that the sum of the power dissipated in the internal and load resistances and that absorbed by the second battery is the same (within round off errors) as that delivered by the first battery … just as we would expect from conservation of energy. 84 •• In the circuit in Figure 25-61, the reading of the ammeter is the same when both switches are open and when both switches are closed. What is the unknown resistance R?

446

Chapter 25

Picture the Problem Note that when both switches are closed the 50.0-Ω resistor is shorted. With both switches open, we can apply Kirchhoff’s loop rule to find the current I in the 100-Ω resistor. With the switches closed, the 100-Ω resistor and R are in parallel. Hence, the potential difference across them is the same and we can express the current I100 in terms of the current Itot flowing into the parallel branch whose resistance is R, and the resistance of the 100-Ω resistor. Itot, in turn, depends on the equivalent resistance of the closed-switch circuit, so we can express I100 = I in terms of R and solve for R.

Apply Kirchhoff’s loop rule to a loop around the outside of the circuit with both switches open: Solving for I yields:

Relate the potential difference across the 100-Ω resistor to the potential difference across R when both switches are closed: Apply Kirchhoff’s junction rule at the junction to the left of the 100-Ω resistor and R:

ε − (300 Ω )I − (100 Ω )I − (50.0 Ω )I = 0

I=

ε 450 Ω

=

1.50 V = 3.33 mA 450 Ω

(100 Ω )I100 = RI R

(1)

I tot = I100 + I R or I R = I tot − I100 where Itot is the current drawn from the source when both switches are closed.

Substituting for IR in equation (1) yields:

(100 Ω )I100 = R(I tot − I100 ) or I100 =

Express the current Itot drawn from the source with both switches closed:

I tot =

Express the equivalent resistance when both switches are closed:

Req =

Substitute for Req to obtain:

I tot =

RI tot R + 100 Ω

(2)

ε Req

(100 Ω )R R + 100 Ω

+ 300 Ω

1.5 V (100 Ω )R + 300 Ω R + 100 Ω

`Electric Current and Direct-Current Circuits 447

Substituting for I tot in equation (2) yields:

Solving for R yields:

I 100

⎞ ⎛ ⎟ ⎜ 1.50 V R ⎟ ⎜ = ⎟ R + 100 Ω ⎜ (100 Ω )R + 300 Ω ⎟ ⎜ ⎠ ⎝ R + 100 Ω (1.50 V )R = = 3.33 mA (400 Ω )R + 30,000 Ω 2

R = 600 Ω

Remarks: Note that we can also obtain the result in the third step by applying Kirchhoff’s loop rule to the parallel branch of the circuit. 85 •• [SSM] In the circuit shown in Figure 25-62, the batteries have negligible internal resistance. Find (a) the current in each branch of the circuit, (b) the potential difference between point a and point b, and (c) the power supplied by each battery. Picture the Problem Let I1 be the current delivered by the left battery, I2 the current delivered by the right battery, and I3 the current through the 6.00-Ω resistor, directed down. We can apply Kirchhoff’s rules to obtain three equations that we can solve simultaneously for I1, I2, and I3. Knowing the currents in each branch, we can use Ohm’s law to find the potential difference between points a and b and the power delivered by both the sources.

(a) Apply Kirchhoff’s junction rule at junction a: Apply Kirchhoff’s loop rule to a loop around the outside of the circuit to obtain:

Apply Kirchhoff’s loop rule to a loop around the left-hand branch of the circuit to obtain: Solving equations (1), (2), and (3) simultaneously yields:

I 4 Ω + I 3Ω = I 6 Ω

(1)

12.0 V − (4.00 Ω )I 4 Ω + (3.00 Ω )I 3 Ω − 12.0 V = 0 or − (4.00 Ω )I 4 Ω + (3.00 Ω )I 3 Ω = 0 12.0 V − (4.00 Ω )I 4 Ω

− (6.00 Ω )I 6 Ω = 0

(2)

(3)

I 4 Ω = 0.667 A , I 3 Ω = 0.889 A ,

and I 6 Ω = 1.56 A

448

Chapter 25

(b) Apply Ohm’s law to find the potential difference between points a and b: (c) Express the power delivered by the 12.0-V battery in the left-hand branch of the circuit: Express the power delivered by the 12.0-V battery in the right-hand branch of the circuit:

Vab = (6.00 Ω )I 6 Ω = (6.00 Ω )(1.56 A ) = 9.36 V Pleft = ε I 4 Ω = (12.0 V )(0.667 A ) = 8.00 W Pright = ε I 3 Ω = (12.0 V )(0.889 A ) = 10.7 W

86 •• In the circuit shown in Figure 25-63, the batteries have negligible internal resistance. Find (a) the current in each branch of the circuit, (b) the potential difference between point a and point b, and (c) the power supplied by each battery. Picture the Problem Let I2Ω be the current delivered by the 7.00-V battery, I3Ω the current delivered by the 5-V battery, and I1Ω, directed up, the current through the 1.00-Ω resistor. We can apply Kirchhoff’s rules to obtain three equations that we can solve simultaneously for I1, I2, and I3. Knowing the currents in each branch, we can use Ohm’s law to find the potential difference between points a and b and the power delivered by both the sources.

(a) Apply Kirchhoff’s junction rule at junction a: Apply Kirchhoff’s loop rule to a loop around the outside of the circuit to obtain: Apply Kirchhoff’s loop rule to a loop around the left-hand branch of the circuit to obtain:

Solve equations (1), (2), and (3) simultaneously to obtain:

I 2 Ω = I 3 Ω + I1Ω 7.00 V − (2.00 Ω )I 2 Ω

− (1.00 Ω )I1 Ω = 0

(1)

(2)

7.00 V − (2.00 Ω )I 2 Ω − (3.00 Ω )I 3 Ω + 5.00 V = 0 or (2.00 Ω )I 2 Ω + (3.00 Ω )I 3 Ω = 12.0 V (3) I 2 Ω = 3.00 A , I 3 Ω = 2.00 A ,

and I 1 Ω = 1.00 A

`Electric Current and Direct-Current Circuits 449

(b) Apply Ohm’s law to find the potential difference between points a and b:

(c) Express the power delivered by the 7.00-V battery: (c) Express the power delivered by the 5.00-V battery:

Vab = −5.00 V + (3.00 Ω )I 3 Ω

= −5.00 V + (3.00 Ω )(2.00 A ) = 1.00 V

P7 V = ε I 2 Ω = (7.00 V )(3.00 A ) = 21.0 W P5 V = ε I 3 Ω = (5.00 V )(2.00 A ) = 10.0 W

87 ••• Two identical batteries, each having an emf ε and an internal resistance r, can be connected across a resistance R with the batteries connected either in series or in parallel. In each situation, determine explicitly whether the power supplied to R is greater when R is less than r or when R is greater than r. Picture the Problem The series and parallel connections are shown below. The power supplied to a resistor whose resistance is R when the current through it is I is given by P = I 2 R . We can use Kirchhoff’s rules to find the currents Ip and Is and, hence, Ps and Pp, and then use the inequalities R > r and R < r to show that if R > r , then Pp < Ps and if R < r , then Pp > Ps . The steps in the proofs that

follow were motivated by having started with the desired outcome and then working backward; for example, assuming that Pp < Ps and showing that R > r .

The power Ps supplied to R in the series circuit is given by: Apply Kirchhoff’s loop rule to obtain:

Ps = I s2 R − rI s + ε − rI s + ε − RI s = 0

450

Chapter 25 2ε 2r + R

Solving for Is yields:

Is =

Substitute for Is and simplify to obtain:

4ε 2 R ⎛ 2ε ⎞ (1) Ps = ⎜ ⎟ R= (2r + R )2 ⎝ 2r + R ⎠

The power Pp supplied to R in the series circuit is given by:

Pp = I p2 R

Apply Kirchhoff’s junction rule to point a to obtain:

I p = I1 + I 2

Apply Kirchhoff’s loop rule to loop 1 to obtain:

Apply Kirchhoff’s loop rule to the outer loop to obtain:

2

− rI1 + ε − ε + rI 2 = 0 or I1 = I 2 = 12 I p

ε − RI p − rI1 = 0 or

ε − RI p − 12 rI p = 0 ⇒ I p = 1 ε 2

for Pp yields:

⎛ ε ⎞ ⎟⎟ R = Pp = ⎜⎜ 1 r + R ⎝2 ⎠

To prove that if R > r , then Pp < Ps ,

3R 2 > 3r 2

Substituting for I p in the expression

ε 2R

2

r+R

( 12 r + R )2

(2)

suppose that R > r . Then: Adding r 2 + 4rR + R 2 to both sides of the inequality yields:

r 2 + 4rR + 4 R 2 > 4r 2 + 4rR + R 2

Factor 4 from the left-hand side and recognize that the right-hand side is the square of a binomial to obtain:

4( 12 r + R ) > (2r + R ) 2

2

or

( 12 r + R )2 > (2r + R )

2

4

Taking the reciprocal of both sides of the inequality reverses the sense of the inequality:

(

1 2

1 4 < 2 (2r + R )2 r + R)

`Electric Current and Direct-Current Circuits 451

Multiplying both sides of the inequality by ε 2 R yields:

ε 2R

4ε 2 R < (2r + R )2

( 12 r + R )2

For the series combination, the power delivered to the load is greater if R > r and is greatest when R = 2r. If R = r, both arrangements provide the same power to the load. To prove that if R < r , then Pp > Ps ,

3R 2 < 3r 2

suppose that R < r . Then: Adding r 2 + 4rR + R 2 to both sides of the inequality yields:

r 2 + 4rR + 4 R 2 < 4r 2 + 4rR + R 2

Factor 4 from the left-hand side and recognize that the right-hand side is the square of a binomial to obtain:

4( 12 r + R ) < (2r + R ) 2

2

or

( 12 r + R )2 < (2r + R )

2

4

Taking the reciprocal of both sides of the inequality reverses the sense of the inequality: Multiplying both sides of the inequality by ε 2 R yields:

(

1 2

1 4 > 2 (2r + R )2 r + R)

ε 2R

( 12 r + R )2

>

4ε 2 R (2r + R )2

For the parallel combination, the power delivered to the load is greater if R < r and is a maximum when R = 12 r . 88 •• The circuit fragment shown in Figure 25-64 is called a voltage divider. (a) If Rload is not attached, show that Vout = VR2/(R1 + R2). (b) If R1 = R2 = 10 kΩ, what is the smallest value of Rload that can be used so that Vout drops by less than 10 percent from its unloaded value? (Vout is measured with respect to ground.) Picture the Problem Let the current drawn from the source be I. We can use Ohm’s law in conjunction with Kirchhoff’s loop rule to express the output voltage as a function of V, R1, and R2. In (b) we can use the result of (a) to express the condition on the output voltages in terms of the effective resistance of the loaded output and the resistances R1 and R2.

(a) Use Ohm’s law to express Vout in terms of R2 and I:

Vout = IR2

452

Chapter 25

Apply Kirchhoff’s loop rule to the circuit to obtain:

V − IR1 − IR2 = 0 ⇒ I =

V R1 + R2

Substitute for I in the expression for Vout to obtain:

⎛ V ⎞ ⎛ R2 ⎞ ⎟⎟ R2 = V ⎜⎜ ⎟⎟ Vout = ⎜⎜ ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠

(b) Relate the effective resistance of the loaded circuit Reff to R2 and Rload : Solving for Rload yields:

1 1 1 = + Reff R2 Rload Rload =

R2 Reff R2 − Reff

(1)

Letting V 'out represent the output voltage under load, express the condition that Vout drops by less than 10 percent of its unloaded value:

Vout − V' out V' = 1 − out < 0.1 Vout Vout

Using the result from (a), express V'out in terms of the effective output load Reff :

⎛ Reff V' out = V ⎜⎜ ⎝ R1 + Reff

Substitute for Vout and V'out in equation (2) and simplify to obtain:

Reff R + Reff 1− 1 < 0.1 R2 R1 + R2 or R (R + R2 ) 1 − eff 1 < 0.1 R2 (R1 + Reff )

Solving for Reff yields:

Reff >

0.9 R1R2 R1 + 0.1R2

Substitute numerical values and evaluate Reff :

Reff >

(0.90)(10 kΩ )(10 kΩ ) = 8.18 kΩ 10 kΩ + (0.10)(10 kΩ )

Finally, substitute numerical values in equation (1) and evaluate Rload :

Rload
R0 to obtain:

Br ≤ R0 (2πr ) = μ 0 I C

IC I r2 = ⇒ = I I C π r 2 π R02 R02

Br ≤ R0 (2πr ) =

Br ≤ R0 =

μ0 r 2 I R02

μ 0 rI μ 0 2 I = r 2πR02 4π R02

Br > R0 (2πr ) = μ 0 I C = μ 0 I

(1)

654

Chapter 27

Solving for Br > R0 yields:

Br >r0 =

μ0 2I 4π r

(2)

The spreadsheet program to calculate B as a function of r in the interval 0 ≤ r ≤ 10R0 follows. The formulas used to calculate the quantities in the columns are as follows: Cell B1

Formula/Content 1.00E−07

Algebraic Form

B2 B3 C6

5.00 2.55E−03 10^4*$B$1*2*$B$2*A6/$B$3^2

C17

10^4*$B$1*2*$B$2*A6/A17

I R0 μ0 2I r 4π R02 μ0 2 I 4π r

A B μ/4π= 1.00E−07 I= 5 R0= 2.55E−03

μ0 4π

C

2

1 2 3 4 5 6 7 8 9 10

R (m) 0.00E+00 2.55E−04 5.10E−04 7.65E−04 1.02E−03

R (mm) 0.00E+00 2.55E−01 5.10E−01 7.65E−01 1.02E+00

B (T) 0.00E+00 3.92E−01 7.84E−01 1.18E+00 1.57E+00

102 103 104 105 106

2.45E−02 2.47E−02 2.50E−02 2.52E−02 2.55E−02

2.45E+01 2.47E+01 2.50E+01 2.52E+01 2.55E+01

4.08E−01 4.04E−01 4.00E−01 3.96E−01 3.92E−01

N/A A m

Sources of the Magnetic Field

655

A graph of B as a function of r follows. 4

B (G)

3

2

1

0 0

2

4

6

8

10

12

14

16

18

20

22

24

26

R (mm)

86 •• A 50-turn coil of radius 10.0 cm carries a current of 4.00 A and a concentric 20-turn coil of radius 0.500 cm carries a current of 1.00 A. The planes of the two coils are perpendicular. Find the magnitude of the torque exerted by the large coil on the small coil. (Neglect any variation in magnetic field due to the current in the large coil over the region occupied by the small coil.)

r r r Picture the Problem We can use τ = μ × B to find the torque exerted on the r r small coil (magnetic moment = μ ) by the magnetic field B due to the current in the large coil. Relate the torque exerted by the large coil on the small coil to the magnetic r moment μ of the small coil and the r magnetic field B due to the current in the large coil: Express the magnetic moment of the small coil:

Express the magnetic field at the center of the large coil:

r r r τ = μ× B or, because the planes of the two coils are perpendicular, τ = μB

μ = NIA where I is the current in the coil, N is the number of turns in the coil, and A is the cross-sectional area of the coil. N'μ 0 I' where I′is the current in the 2R large coil, N′ is the number of turns in the coil, and R is its radius. B=

656

Chapter 27

Substitute for B and μ in the expression for τ to obtain:

τ=

NN'II'Aμ 0 2R

Substitute numerical values and evaluate τ :

τ=

(50)(20)(4.00 A )(1.00 A )π (0.500 cm )2 (4π × 10 −7 N/A 2 ) = 2(10.0 cm )

1.97 μN ⋅ m

87 •• The magnetic needle of a compass is a uniform rod with a length of 3.00 cm, a radius of 0.850 mm, and a density of 7.96 × 103 kg/m3. The needle is free to rotate in a horizontal plane, where the horizontal component of Earth’s magnetic field is 0.600 G. When disturbed slightly, the compass executes simple harmonic motion about its midpoint with a frequency of 1.40 Hz. (a) What is the magnetic dipole moment of the needle? (b) What is the magnetization of the needle? (c) What is the amperian current on the surface of the needle? Picture the Problem (a) We can solve the equation for the frequency f of the compass needle for the magnetic dipole moment of the needle. In Parts (b) and (c) we can use their definitions to find the magnetization M and the amperian current Iamperian.

(a) The frequency of the compass needle is given by:

4π 2 f 2 I 1 μB ⇒μ = f = 2π I B where I is the moment of inertia of the needle.

The moment of inertia of the needle is:

I = 121 mL2 = 121 ρVL2 = 121 ρπr 2 L3

Substitute for I to obtain:

μ=

π 3 f 2 ρr 2 L3 3B

Substitute numerical values and evaluate μ:

π 3 (1.40 s -1 ) (7.96 × 10 3 kg/m 3 )(0.850 × 10 −3 m ) (0.0300 m )3 2

μ=

2

(

3 0.600 × 10 − 4 T

= 5.24 × 10 − 2 A ⋅ m 2

(b) Use its definition to express the magnetization M:

M =

μ V

)

Sources of the Magnetic Field Substitute to obtain:

M =

μ V

=

π 3 f 2 ρr 2 L3 3BV

=

657

π 2 f 2 ρL2 3B

Substitute numerical values and evaluate M:

π 2 (1.40 s -1 ) (7.96 × 10 3 kg/m 3 )(0.0300 m )2 2

M=

3(0.600 × 10 T ) −4

= 7.70 × 10 5 A/m

(c) The amperian current on the surface of the needle is:

(

)

I amperian = ML = 7.70 ×10 5 A/m (0.0300 m ) = 23.1kA

88 •• A relatively inexpensive ammeter, called a tangent galvanometer, can be made using Earth’s magnetic field. A plane circular coil that has N turns and a radius R is oriented so the magnetic field Bc it produces in the center of the coil is either east or west. A compass is placed at the center of the coil. When there is no current in the coil, assume the compass needle points due north. When there is a current in the coil (I), the compass needle points in the direction of the resultant magnetic field at an angle θ to the north. Show that the current I is related to θ and to the horizontal component of Earth’s magnetic field Be 2RBe tanθ . by I = μ0N Picture the Problem Note that Be and Bc are perpendicular to each other and that the resultant magnetic field is at an angle θ with north. We can use trigonometry to relate Bc and Be and express Bc in terms of the geometry of the coil and the current flowing in it.

Express Bc in terms of Be:

Bc = Be tan θ where θ is the angle of the resultant field from north.

Express the field Bc due to the current in the coil:

Nμ 0 I 2R where N is the number of turns.

Substitute for Bc to obtain:

Nμ 0 I 2 RBe = Be tan θ ⇒ I = tan θ μ0 N 2R

Bc =

658

Chapter 27

89 •• Earth’s magnetic field is about 0.600 G at the magnetic poles, and is pointed vertically downward at the magnetic pole in the northern hemisphere. If the magnetic field were due to an electric current circulating in a loop at the radius of the inner iron core of Earth (approximately 1300 km), (a) what would be the magnitude of the current required? (b) What direction would this current have— the same as Earth’s spin, or opposite? Explain your answer. Picture the Problem The current required can be found by solving the equation for the magnetic field on the axis of a current loop for the current in the loop. We can use the right-hand rule to determine the direction of this current.

(a) The magnetic field on the axis of a current loop is given by:

μ 0 2πR 2 I Bx = 4π (x 2 + R 2 )3 2

Solving for I yields:

4π x 2 + R 2 Bx I= 2πμ 0 R 2

(

)

32

Substitute numerical values and evaluate I:

(

I=

2

⎛ 2π ⎜ 4π × 10 −7 ⎝

)

1T ⎞ ⎛ ⎜ 0.600 G × 4 ⎟ 10 G ⎠ ⎝ = 15.5 GA N ⎞ 2 ⎟(1300 km ) A2 ⎠

4π (6370 km ) + (1300 km )

2 32

(b) Because Earth’s magnetic field points down at the north pole, application of the right-hand rule indicates that the current is opposite the spin direction of Earth. r 90 •• A long, narrow bar magnet has its magnetic moment μ parallel to its long axis and is suspended at its center—in essence becoming a frictionless r compass needle. When the magnet is placed in a magnetic field B , it lines up with the field. If it is displaced by a small angle and released, show that the magnet 1 μB , will oscillate about its equilibrium position with frequency given by 2π I where I is the moment of inertia about the point of suspension. Picture the Problem We can apply Newton’s second law for rotational motion to obtain the differential equation of motion of the bar magnet. While this equation is not linear, we can use a small-angle approximation to render it linear and obtain an expression for the square of the angular frequency that we can solve for the frequency f of the motion.

Sources of the Magnetic Field Apply Newton’s second law to the bar magnet to obtain the differential equation of motion for the magnet:

For small displacements from equilibrium (θ R that we can evaluate for the given distances from the center of the cylindrical conductor. r

r

Apply Ampère’s law to a closed circular path a distance r ≤ R from the center of the cylindrical conductor to obtain:

∫ B ⋅ d l = B(r )(2πr ) = μ I

Solve for B(r) to obtain:

B(r ) =

μ0 I (r ) 2πr

Substitute for I(r):

B(r ) =

μ 0 (50 A/m) r μ 0 (50 A/m) = 2π r 2π

(a) and (b) Noting that B is independent of r, substitute numerical values and evaluate B(5.0 cm) and B(10 cm):

0 C

C

B(5.0 cm ) = B(10 cm ) =

(4π ×10

= 10 μT

−7

= μ0 I (r )

)

N/A 2 (50 A/m ) 2π

662

Chapter 27 r r B ∫ ⋅ d l = B(r )(2πr ) = μ0 I C = μ0 I (R )

(c) Apply Ampère’s law to a closed circular path a distance r > R from the center of the cylindrical conductor to obtain:

C

Solving for B(r) yields:

B(r ) =

μ 0 I (R ) 2πr

Substitute numerical values and evaluate B(20 cm): B(20 cm ) =

(4π ×10

−7

)

N/A 2 (50 A/m )(0.10 m ) = 5.0 μT 2π (0.20 m )

94 •• Figure 27- 68 shows a square loop that has 20-cm long sides and is in the z = 0 plane with its center at the origin. The loop carries a current of 5.0 A. An infinitely long wire that is parallel to the x axis and carries a current of 10 A intersects the z axis at z = 10 cm. The directions of the currents are shown in the figure. (a) Find the net torque on the loop. (b) Find the net force on the loop.

r Picture the Problem The field B due to the 10-A current is in the yz plane. The net force on the wires of the square in the y direction cancel and do not contribute r r r r r r to a net torque or force. We can use τ = l × F , F = I l × B , and the expression for the magnetic field due to a long straight wire to express the torque acting on each of the wires and hence, the net torque acting on the loop. z × 10 A

r F−10 10 cm

r B−10 5.0 A

× −10 cm

r r −10

(a) The net torque about the x axis is the sum of the torques due to the r r forces F10 and F−10 :

r r 10 0

10 cm y

5.0 A

r B10

r r r τ net = τ10 + τ -10

r F10

Sources of the Magnetic Field

r

r

Substituting for τ10 and τ -10 yields:

663

r r r r r τ net = l10 × F10 + l−10 × F−10 where the subscripts refer to the positions of the current-carrying wires.

( )

r r r F10 = I l 10× B10

The forces acting on the wires are given by:

and r r F−10 = I l

( )

−10

r × B−10

r r Substitute for F10 and F−10 to obtain: r r r r r r r τ net = l10 × [(I l )10× B10 ]+ l −10 × [(I l )−10 × B−10 ]

The lever arms for the forces acting on the wires at y = 10 cm and y = −10 cm are:

(1)

r r l10 = (0.10 m ) ˆj and l −10 = −(0.10 m ) ˆj

(

r μ 2I 1 B10 = 0 − ˆj − kˆ 4π R 2 where

The magnetic field at the wire at y = 10 cm is given by:

R=

)

(0.10 m )2 + (0.10 m )2

= 0.141 m .

r Substitute numerical values and evaluate B10 :

r 4π × 10 −7 N/A 2 2(10 A ) ˆ ˆ B10 = − j − k = (10.0 μT ) − ˆj − kˆ 0.141 m 4π 2

(

)

(

(

r B−10 = (10.0 μT ) − ˆj + kˆ

Proceed similarly to obtain:

)

)

Substitute in equation (1) and simplify to obtain:

[

]

r τ net = (0.10 m ) ˆj × (5.0 A )(0.20 m )iˆ × (10.0 μT ) (− ˆj − kˆ )

[

( )

(

− (0.10 m ) ˆj × (5.0 A )(0.20 m ) − iˆ × (10.0 μT ) − ˆj + kˆ

)]

= − (2.0 μN ⋅ m )iˆ (b) The net force acting on the loop is the sum of the forces acting on its four sides:

r r r Fnet = F10 + F−10

(2)

664

Chapter 27

r Evaluate F10 to obtain:

(

( )

r r r F10 = I l 10 × B10 = (5.0 A )(0.20 m ) iˆ × (10.0 μT ) − ˆj − kˆ = (10 μN ) iˆ × − ˆj − kˆ

[ ( )] = (10 μN )(− kˆ + ˆj )

)

r Evaluating F−10 yields:

( )

r r F−10 = I l

(

r ˆ ˆ ˆ × B −10 = (5.0 A )(− 0.20 m ) i × (10.0 μT ) − j + k −10 = (− 10 μN ) iˆ × − ˆj + kˆ

[ ( = (10 μN ) (kˆ + ˆj )

)]

)

r r Substitute for F10 and F−10 in equation (2) and simplify to obtain:

(

)

(

)

r Fnet = (10 μN ) − kˆ + ˆj + (10 μN ) kˆ + ˆj =

(20 μN ) ˆj

95 •• [SSM] A current balance is constructed in the following way: A straight 10.0-cm-long section of wire is placed on top of the pan of an electronic balance (Figure 27-69). This section of wire is connected in series with a power supply and a long straight horizontal section of wire that is parallel to it and positioned directly above it. The distance between the central axes of the two wires is 2.00 cm. The power supply provides a current in the wires. When the power supply is switched on, the reading on the balance increases by 5.00 mg. What is the current in the wire? Picture the Problem The force acting on the lower wire is given by Flower wire = IlB , where I is the current in the lower wire, l is the length of the wire on the balance, and B is the magnetic field strength at the location of the lower wire due to the current in the upper wire. We can apply Ampere’s law to find B at the location of the wire on the pan of the balance.

The force experienced by the lower wire is given by:

Flower wire = IlB

Apply Ampere’s law to a closed circular path of radius r centered on the upper wire to obtain:

B(2πr ) = μ 0 I C = μ 0 I ⇒ B =

Substituting for B in the expression for the force on the lower wire and simplifying yields:

Flower wire

2 ⎛ μ 0 I ⎞ μ 0 lI = Il⎜ ⎟= 2πr ⎝ 2πr ⎠

μ0 I 2πr

Sources of the Magnetic Field Solve for I to obtain:

Note that the force on the lower wire is the increase in the reading of the balance. Substitute numerical values and evaluate I:

I=

2πrFlower wire μ0l

I=

2π (2.00 cm ) 5.00 × 10 −6 kg 4π × 10 −7 N/A 2 (10.0 cm )

(

(

)

665

)

= 2.236 A = 2.24 A

96 •• Consider the current balance of Problem 95. If the sensitivity of the balance is 0.100 mg, what is the minimum current detectable using this current balance? Picture the Problem We can use a proportion to relate minimum current detectible using this balance to its sensitivity and to the current and change in balance reading from Problem 95.

The minimum current Imin detectible is to the sensitivity of the balance as the current in Problem 95 is to the change in the balance reading in Problem 95:

I min 2.236 A = 0.100 mg 5.00 mg

Solving for Imin yields:

⎛ 2.236 A ⎞ ⎟⎟ I min = (0.100 mg )⎜⎜ ⎝ 5.00 mg ⎠ = 44.7 mA

The ″standard″ current balance can be made very sensitive by increasing the length (i.e., moment arm) of the wire balance, which one cannot do with this kind; however, this is compensated somewhat by the high sensitivity of the electronic balance. 97 ••• [SSM] A non-conducting disk that has radius R, carries a uniform surface charge density σ, and rotates with angular speed ω. (a) Consider an annular strip that has a radius r, a width dr, and a charge dq. Show that the current (dI) produced by this rotating strip is given by ωσrdr . (b) Use your result from Part (a) to show that the magnetic field strength at the center of the disk is given by the expression 12 μ 0σωR . (c) Use your result from Part (a) to find an

expression for the magnetic field strength at a point on the central axis of the disk a distance z from its center.

666

Chapter 27

Picture the Problem The diagram shows the rotating disk and the circular strip of radius r and width dr with charge dq. We can use the definition of surface charge density to express dq in terms of r and dr and the definition of current to show that dI = ωσr dr. We can then use this current and expression for the magnetic field on the axis of a current loop to obtain the results called for in Parts (b) and (c).

(a) Express the total charge dq that passes a given point on the circular strip once each period:

dq = σdA = 2πσrdr

Letting q be the total charge that passes along a radial section of the disk in a period of time T, express the current in the element of width dr:

dI =

(c) Express the magnetic field dBx at a distance z along the axis of the disk due to the current loop of radius r and width dr:

Integrate from r = 0 to r = R to obtain:

dq 2πσrdr = = ωσrdr 2π dt

ω

dBx =

=

Bx = =

(b) Evaluate Bx for x = 0:

μ 0 2πr 2 dI 4π (z 2 + r 2 )3 2

(

μ 0ωσr 3

2 z2 + r2

μ 0ωσ 2

)

32

r3

R

∫ (z 0

dr

2

+ r2

)

32

dr

⎞ ⎜ − 2 x ⎟⎟ ⎜ 2 2 ⎠ ⎝ R +z

μ 0ωσ ⎛ R 2 + 2 z 2

Bx (0 ) =

2

μ 0ωσ ⎛ R 2 ⎞ 2

⎜ ⎜ 2 ⎝ R

⎟= ⎟ ⎠

1 2

μ 0σωR

98 ••• A square loop that has sides of length l lies in the z = 0 plane with its center at the origin. The loop carries a current I. (a) Derive an expression for the magnetic field strength at any point on the z axis. (b) Show that for z much larger

Sources of the Magnetic Field

(

667

)

than l, your result from Part (a) becomes B ≈ μμ 0 2πz 3 , where μ is the magnitude of the magnetic moment of the loop. Picture the Problem From the symmetry of the system it is evident that the magnetic fields due to each segment have the same magnitude. We can express the magnetic field at (x,0,0) due to one side (segment) of the square, find its component in the x direction, and then multiply by four to find the resultant field. z

(0, 12 l, 12 l )

(0, 12 l,0)

I

y

d R

(x,0,0) x

(0, 12 l,− 12 l )

θ1 θ2

B1

(a) B due to a straight wire segment is given by:

B=

μ0 I (sin θ1 + sin θ 2 ) 4π R

where R is the perpendicular distance from the wire segment to the field point. Use θ1 = θ 2 and R = x 2 + l 2 4 to

B1 (x,0,0) =

μ0 4π

=

μ0 2π

express B due to one side at (x,0,0):

Referring to the diagram, express sin θ1 :

l sin θ1 = 2 = d

I l2 x2 + 4 I l2 x + 4 2

l 2 l2 x + 2 2

(2 sin θ1 ) (sin θ1 )

668

Chapter 27

Substituting for sin θ1 and simplifying yields:

B1 ( x,0,0) =

μ0 2π

I

l 2

l2 x + 4 μ0 I

l2 x + 2 l

2

=

4π x 2 +

l2 4

(0,0,0)

By symmetry, the sum of the y and z components of the fields due to the four wire segments must vanish, whereas the x components will add. The diagram to the right is a view of the xy plane showing the relationship r between B1 and the angle β it makes with the x axis.

2

β

x2 +

l2 2

(12 r,0,0) x

R

(0,0, z )

r B1

β z

Express B1x: Substituting for cosβ and simplifying yields:

B1x = B1 cos β

B1x =

μ0 I

The resultant magnetic field is the sum of the fields due to the four wire segments (sides of the square):

l

l l2 2 x + 4π x + 4 2 2 μ 0 Il 2

=

l 2

2

l2 x + 4 2

⎛ l2 ⎞ l2 8π ⎜⎜ x 2 + ⎟⎟ x 2 + 4⎠ 2 ⎝

r B = 4 B1x iˆ =

μ 0 Il 2 ⎛ l2 ⎞ l2 2π ⎜⎜ x 2 + ⎟⎟ x 2 + 4⎠ 2 ⎝

iˆ

Sources of the Magnetic Field (b) Factor x2 from the two factors in the denominator to obtain:

r B=

=

For x >> l:

μ 0 Il 2 ⎛ ⎛ l ⎞ l ⎞ 2πx 2 ⎜⎜1 + 2 ⎟⎟ x 2 ⎜⎜1 + 2 ⎟⎟ ⎝ 4x ⎠ ⎝ 2x ⎠ 2

2

μ 0 Il 2 ⎛ l ⎞ l 2πx 3 ⎜⎜1 + 2 ⎟⎟ 1 + 2 2x ⎝ 4x ⎠ 2

r μ Il 2 μ0 μ ˆ B ≈ 0 3 iˆ = i 2πx 2πx 3 where μ = Il 2 .

2

iˆ

669 iˆ

670

Chapter 27

Chapter 28 Magnetic Induction Conceptual Problems 1 • [SSM] (a) The magnetic equator is a line on the surface of Earth on which Earth’s magnetic field is horizontal. At the magnetic equator, how would you orient a flat sheet of paper so as to create the maximum magnitude of magnetic flux through it? (b) How about the minimum magnitude of magnetic flux? Determine the Concept (a) Orient the sheet so the normal to the sheet is both horizontal and perpendicular to the local tangent to the magnetic equator. (b) Orient the sheet of paper so the normal to the sheet is perpendicular to the direction of the normal described in the answer to Part (a). 2 • At one of Earth’s magnetic poles, how would you orient a flat sheet of paper so as to create the maximum magnitude of magnetic flux through it? Determine the Concept (a) Orient the sheet so the normal to the sheet is vertical. (b) Any orientation as long as the paper’s plane is perpendicular to Earth’s surface at that location. 3 • [SSM] Show that the following combination of SI units is equivalent to the volt: T ⋅ m 2 s . Determine the Concept Because a volt is a joule per coulomb, we can show that T ⋅ m2 are equivalent to a volt by making a series of substitutions and the SI units s simplifications that reduces these units to a joule per coulomb.

The units of a tesla are

N : A⋅m

Substitute the units of an ampere (C/s), replace N ⋅ m with J, and simplify to obtain:

N N⋅m ⋅ m2 T⋅m = A = A⋅m s s s 2

J C 2 J T⋅m = s = s C s

671

672

Chapter 28

Finally, because a joule per coulomb is a volt:

T ⋅ m2 = V s

Show that the following combination of SI units is equivalent to the • Wb . ohm: A ⋅s

4

Determine the Concept Because a weber is a newton⋅meter per ampere, we can Wb are equivalent to an ohm by making a series of show that the SI units A ⋅s substitutions and simplifications that reduces these units to a volt per ampere.

Because a weber is a

N⋅m : A

N⋅m Wb J = A = 2 A ⋅s A ⋅s A ⋅s

Substitute the units of an ampere and simplify to obtain:

J Wb J = = C A ⋅s A ⋅ ⋅s A ⋅C s

Finally, because a joule per coulomb is a volt:

Wb V = = Ω A ⋅s A

5 • [SSM] A current is induced in a conducting loop that lies in a horizontal plane and the induced current is clockwise when viewed from above. Which of the following statements could be true? (a) A constant magnetic field is directed vertically downward. (b) A constant magnetic field is directed vertically upward. (c) A magnetic field whose magnitude is increasing is directed vertically downward. (d) A magnetic field whose magnitude is decreasing is directed vertically downward. (e) A magnetic field whose magnitude is decreasing is directed vertically upward. Determine the Concept We know that the magnetic flux (in this case the magnetic field because the area of the conducting loop is constant and its orientation is fixed) must be changing so the only issues are whether the field is increasing or decreasing and in which direction. Because the direction of the magnetic field associated with the clockwise current is vertically downward, the changing field that is responsible for it must be either increasing vertically upward (not included in the list of possible answers) or a decreasing field directed into the page. (d ) is correct.

Magnetic Induction

673

6 • Give the direction of the induced current in the circuit, shown on the right in Figure 28- 37, when the resistance in the circuit on the left is suddenly (a) increased and (b) decreased. Explain your answer. Determine the Concept The induced emf and induced current in the circuit on the right are in such a direction as to oppose the change that produces them (Lenz’s Law). We can determine the direction of the induced current in the r circuit. Note that when R is constant, B in the circuit to the right points out of the paper.

(a) If R increases, I decreases and B in the circuit to the right decreases. Lenz’s law tells us that the induced current is counterclockwise. (b) If R decreases, I increases and B in the circuit to the right increases. Lenz’s law tells us that the induced current is clockwise. 7 • [SSM] The planes of the two circular loops in Figure 28-38, are parallel. As viewed from the left, a counterclockwise current exists in loop A. If the magnitude of the current in loop A is increasing, what is the direction of the current induced in loop B? Do the loops attract or repel each other? Explain your answer. Determine the Concept Clockwise as viewed from the left. The loops repel each other. 8 • A bar magnet moves with constant velocity along the axis of a loop, as shown in Figure 28-39, (a) Make a graph of the magnetic flux through the loop as a function of time. Indicate on the graph when the magnet is halfway through the loop by designating this time t1. Choose the direction of the normal to the flat surface bounded by the flat surface to be to the right. (b) Make a graph of the induced current in the loop as a function of time. Choose the positive direction for the current to be clockwise as viewed from the left. Determine the Concept We know that, as the magnet moves to the right, the flux through the loop first increases until the magnet is half way through the loop and then decreases. Because the flux first increases and then decreases, the current will change directions, having its maximum values when the flux is changing most rapidly.

(a) and (b) The following graph shows the flux and the induced current as a function of time as the bar magnet passes through the coil. When the center of the magnet passes through the plane of the coil dφm /dt = 0 and the current is zero.

674

Chapter 28

flux current

t1

time

A bar magnet is mounted on the end of a coiled spring and is 9 • oscillating in simple harmonic motion along the axis of a loop, as shown in Figure 28-40. The magnet is in its equilibrium position when its midpoint is in the plane of the loop. (a) Make a graph of the magnetic flux through the loop as a function of time. Indicate when the magnet is halfway through the loop by designating these times t1 and t2. (b) Make a graph of the induced current in the loop as a function of time, choosing the current to be positive when it is clockwise as viewed from above. Determine the Concept Because the magnet moves with simple harmonic motion, the flux and the induced current will vary sinusoidally. The current will be a maximum wherever the flux is changing most rapidly and will be zero wherever the flux is momentarily constant. (a), (b) The following graph shows the flux, φm , and the induced current (proportional to −dφm/dt) in the loop as a function of time.

Magnetic Induction

675

flux current

0

t1

t2

Time

A pendulum is fabricated from a thin, flat piece of aluminum. At the 10 • bottom of its arc, it passes between the poles of a strong permanent magnet. In Figure 28-41a, the metal sheet is continuous, whereas in Figure 28-41b, there are slots in the sheet. When released from the same angle, the pendulum that has slots swings back and forth many times, but the pendulum that does not have slots stops swinging after no more than one complete oscillation. Explain why. Determine the Concept In the configuration shown in (a), energy is dissipated by eddy currents from the emf induced by the pendulum movement. In the configuration shown in (b), the slits inhibit the eddy currents and the braking effect is greatly reduced. 11 • A bar magnet is dropped inside a long vertical tube. If the tube is made of metal, the magnet quickly approaches a terminal speed, but if the tube is made of cardboard, the magnet falls with constant acceleration. Explain why the magnet falls differently in the metal tube than it does in the cardboard tube. Determine the Concept The magnetic field of the falling magnet sets up eddy currents in the metal tube. The eddy currents establish a magnetic field that exerts a force on the magnet opposing its motion; thus the magnet is slowed down. If the tube is made of a nonconducting material, there are no eddy currents.

A small square wire loop lies in the plane of this page, and a constant 12 • magnetic field is directed into the page. The loop is moving to the right, which is the +x direction. Find the direction of the induced current, if any, in the loop if (a) the magnetic field is uniform, (b) the magnetic field strength increases as x increases, and (c) the magnetic field strength decreases as x increases.

676

Chapter 28

Determine the Concept The direction of the induced current is in such a direction as to oppose, or tend to oppose, the change that produces it (Lenz’s Law).

(a) Because the applied field is constant and uniform, there is no change in flux through the loop and, in accord with Faraday’s law, no induced current in the loop. (b) Let the positive normal direction on the flat surface bounded by the loop be into the page. Because the strength of the applied field increases to the right, the flux through the loop increases as it moves to the right. In accord with Lenz’s law, the direction of the induced current will be such that the flux through the loop due to its magnetic field will be opposite in sign the change in flux of the applied field. Thus, on the flat surface bounded by the loop the magnetic field due to the induced current is out of the page. Using the right-hand rule, the induced current must be counterclockwise. (c) Let the positive normal direction on the flat surface bounded by the loop be into the page. Because the strength of the applied field increases to the right, the flux through the loop decreases as it moves to the right. In accord with Lenz’s law, the direction of the induced current will be such that the flux through the loop due to its magnetic field will be opposite in sign the change in flux of the applied field. Thus, on the flat surface bounded by the loop the magnetic field due to the induced current is into the page. Using the right-hand rule, the induced current must be clockwise. If the current in an inductor doubles, the energy stored in the inductor 13 • will (a) remain the same, (b) double, (c) quadruple, (d) halve. Determine the Concept The magnetic energy stored in an inductor is given by U m = 12 LI 2 . Doubling I quadruples Um. (c) is correct. 14 • Two solenoids are equal in length and radius, and the cores of both are identical cylinders of iron. However, solenoid A has three times the number of turns per unit length as solenoid B. (a) Which solenoid has the larger selfinductance? (b) What is the ratio of the self-inductance of solenoid A to the selfinductance of solenoid B? Determine the Concept The self-inductance of a coil is given by L = μ 0 n 2 Al ,

where n is the number of turns per unit length and l is the length of the coil.

Magnetic Induction

677

(a) Because the two solenoids are equal in length and radius and have identical cores, their self-inductances are proportional to the square of their number of turns per unit length. Hence A has the larger self-inductance. (b) The self-inductances of the two coils are given by:

LA = μ 0 nA2 AA l A and LB = μ 0 nB2 AB l B

Divide the self-inductance of coil A by the self-inductance of coil B and simplify to obtain:

LA μ 0 nA2 AA l A nA2 AA l A = = LB μ 0 nB2 AB l B nB2 AB l B or, because the coils have the same lengths and radii (hence, the same cross-sectional areas), LA nA2 ⎛ nA ⎞ = =⎜ ⎟ LB nB2 ⎜⎝ nB ⎟⎠

If n increases by a factor of 3, l will decrease by the same factor, because the inductors are made from the same length of wire. Hence: 15

•

2

2

LA ⎛ 3nB ⎞ ⎟ = 9 =⎜ LB ⎜⎝ nB ⎟⎠

[SSM] True or false:

(a) The induced emf in a circuit is equal to the negative of the magnetic flux through the circuit. (b) There can be a non-zero induced emf at an instant when the flux through the circuit is equal to zero. (c) The self inductance of a solenoid is proportional to the rate of change of the current in the solenoid. (d) The magnetic energy density at some point in space is proportional to the square of the magnitude of the magnetic field at that point. (e) The inductance of a solenoid is proportional to the current in it. (a) False. The induced emf in a circuit is equal to the rate of change of the magnetic flux through the circuit. (b) True. The rate of change of the magnetic flux can be non-zero when the flux through the circuit is momentarily zero (c) False. The self inductance of a solenoid is determined by its length, crosssectional area, number of turns per unit length, and the permeability of the matter in its core.

678

Chapter 28

(d) True. The magnetic energy density at some point in space is given by B2 Equation 28-20: u m = . 2μ 0 (e) False. The inductance of a solenoid is determined by its length, cross-sectional area, number of turns per unit length, and the permeability of the matter in its core.

Estimation and Approximation 16 • Your baseball teammates, having just studied this chapter, are concerned about generating enough voltage to shock them while swinging aluminum bats at fast balls. Estimate the maximum possible motional emf measured between the ends of an aluminum baseball bat during a swing. Do you think your team should switch to wooden bats to avoid electrocution? Picture the Problem The bat is swung in Earth’s magnetic field. We’ll assume that the batter swings such that the maximum linear velocity of the bat occurs at an angle such that it is moving perpendicular to Earth’s field (i.e. when the bat is aligned north-south and moving east-west). The induced emf in the bat is given by ε = vBl . A bat is roughly 1 m long, and at most its center is probably moving at 75 mph, or about 33 m/s. Earth’s magnetic field is about 0.3 G.

The emf induced in the bat is given by: Under the conditions resulting in a maximum induced emf outlined above:

ε = vBl

ε = (33 m/s)⎛⎜ 0.3 G × ⎝

1T ⎞ ⎟ (1 m ) 10 4 G ⎠

≈ 1 mV

Because 1 mV is so low, there is no danger of being shocked and no reason to switch to wooden bats. 17 • Compare the energy density stored in Earth’s electric field near its surface to that stored in Earth’s magnetic field near its surface. Picture the Problem We can compare the energy density stored in Earth’s electric field to that of Earth's magnetic field by finding their ratio. We’ll take Earth’s magnetic field to be 0.3 G and its electric field to be 100 V/m.

The energy density in an electric field E is given by:

u e = 12 ∈ 0 E 2

Magnetic Induction The energy density in a magnetic field B is given by:

B2 um = 2μ0

Express the ratio of um to ue to obtain:

B2 2μ 0 um B2 =1 = ue 2 ∈0 E 2 μ 0 ∈0 E 2

679

Substitute numerical values and evaluate um/ ue: 2

um = ue 4π × 10 −7 N/A 2

(

1T ⎞ ⎛ ⎜ 0.3 G × 4 ⎟ 10 G ⎠ ⎝ = 8.09 × 10 3 2 −12 2 2 8.854 × 10 C / N ⋅ m (100 V/m )

)(

)

or

(

)

u m ≈ 8 ×10 3 u e

18 •• A physics teacher does the following emf demonstration. She has two students hold a long wire connected to a voltmeter. The wire is held slack, so that it sags with a large arc in it. When she says ″start, ″the students begin rotating the wire as if they were playing jump rope. The students stand 3.0 m apart, and the sag in the wire is about 1.5 m. The motional emf from the ″jump rope″ is then measured on the voltmeter. (a) Estimate a reasonable value for the maximum angular speed that the students can rotate the wire. (b) From this, estimate the maximum motional emf in the wire. HINT: What field is involved in creating the induced emf? Picture the Problem We can use Faraday’s law to relate the motional emf in the wire to the angular speed with which the students turn the jump rope. Assume that Earth’s magnetic field is 0.3 G.

(a) It seems unlikely that the students could turn the ″jump rope″ wire faster than 5.0 rev/s. (b) The magnetic flux φm through the rotating circular loop of wire varies sinusoidally with time according to: Because the average value of the cosine function, over one revolution, is ½, the average rate at which the flux changes through the circular loop is:

φm = BA sin ωt ⇒

dφ m dt

dφm = BAω cos ωt dt

= 12 BAω = 12 π r 2 Bω av

680

Chapter 28

From Faraday’s law, the magnitude of the average motional emf in the loop is:

ε = dφm dt

= 12 π r 2 Bω

Substitute numerical values and evaluate ε: 2

⎛ 1T ⎞ ⎜⎜ 0.3 G × 4 ⎟⎟ (31.4 rad/s ) ≈ 0.8 mV 10 G ⎠ ⎝ 2 ⎠ ⎝

ε = 12 π ⎛⎜ 1.5 m ⎞⎟

19 •• (a) Estimate the maximum possible motional emf between the wingtips of a typical commercial airliner in flight. (b) Estimate the magnitude of the electric field between the wingtips. Picture the Problem The motional emf between the wingtips of an airliner is given by ε = vBl . Assume a speed, relative to Earth’s magnetic field, of 500 mi/h or about 220 m/s and a wingspan of 70 m. Assume that Earth’s magnetic field is 0.3 G.

(a) The motional emf between the wingtips is given by: Substitute numerical values and evaluate ε

:

ε = vBl

ε = (220 m/s)⎛⎜ 0.3 G × ⎝

1T ⎞ ⎟ (70 m ) 10 4 G ⎠

≈ 0.5 V (b) The magnitude of the electric field between the wingtips is the ratio of the potential difference between them and their separation:

E=

V 0.5 V = ≈ 7 mV/m d 70 m

Magnetic Flux A uniform magnetic field of magnitude 0.200 T is in the +x direction. 20 • A square coil that has 5.00-cm long sides has a single turn and makes an angle θ with the z axis, as shown in Figure 28-42. Find the magnetic flux through the coil when θ is (a) 0º, (b) 30º, (c) 60º, and (d) 90º.

r Picture the Problem Because the surface is a plane with area A and B is constant in magnitude and direction over the surface and makes an angle θ with the unit normal vector, we can use φm = BA cos θ to find the magnetic flux through the coil.

Magnetic Induction The magnetic flux through the coil is given by:

681

φm = BA cos θ

Substitute for B and A to obtain: ⎛

1T ⎞

⎝

⎠

φm = ⎜⎜ 2000 G ⋅ 4 ⎟⎟(5.00 × 10 −2 m ) cosθ = (5.00 × 10 −4 Wb )cosθ 10 G 2

φm = (5.00 × 10 −4 Wb )cos 0°

(a) For θ = 0°:

= 5.00 × 10 −4 Wb = 0.50 mWb

φm = (5.00 × 10 −4 Wb )cos30°

(b) For θ = 30°:

= 4.33 × 10 −4 Wb = 0.43 mWb

φ m = (5.00 × 10 −4 Wb )cos60°

(c) For θ = 60°:

= 2.50 × 10 −4 Wb = 0.25 mWb

φm = (5.00 ×10 −4 Wb )cos90° = 0

(d) For θ = 90°:

21 • [SSM] A circular coil has 25 turns and a radius of 5.0 cm. It is at the equator, where Earth’s magnetic field is 0.70 G, north. The axis of the coil is the line that passes through the center of the coil and is perpendicular to the plane of the coil. Find the magnetic flux through the coil when the axis of the coil is (a) vertical, (b) horizontal with the axis pointing north, (c) horizontal with the axis pointing east, and (d) horizontal with the axis making an angle of 30º with north.

r Picture the Problem Because the coil defines a plane with area A and B is constant in magnitude and direction over the surface and makes an angle θ with the unit normal vector, we can use φm = NBA cos θ to find the magnetic flux through the coil.

φm = NBA cos θ = NBπ r 2 cosθ

The magnetic flux through the coil is given by: Substitute for numerical values to obtain: ⎛

φm = 25⎜⎜ 0.70 G ⋅ ⎝

2 1T ⎞ ⎟⎟π 5.0 × 10 −2 m cosθ = (13.7 μWb )cosθ 4 10 G ⎠

(

)

682

Chapter 28

(a) When the plane of the coil is horizontal, θ = 90°:

φm = (13.7 μWb )cos 90° = 0

(b) When the plane of the coil is vertical with its axis pointing north, θ = 0°:

φm = (13.7 μWb )cos 0° = 14 μWb

(c) When the plane of the coil is vertical with its axis pointing east, θ = 90°:

φ m = (13.7 μWb )cos 90° = 0

(d) When the plane of the coil is vertical with its axis making an angle of 30° with north, θ = 30°:

φ m = (13.7 μWb )cos 30° = 12 μWb

22 • A magnetic field of 1.2 T is perpendicular to the plane of a14 turn square coil with sides 5.0-cm long. (a) Find the magnetic flux through the coil. (b) Find the magnetic flux through the coil if the magnetic field makes an angle of 60º with the normal to the plane of the coil. Picture the Problem Because the square coil defines a plane with area A and r B is constant in magnitude and direction over the surface and makes an angle θ with the unit normal vector, we can use φm = NBA cos θ to find the magnetic flux through the coil.

The magnetic flux through the coil is given by:

φ m = NBA cosθ

Substitute numerical values for N, B, and A to obtain:

φm = 14(1.2 T ) (5.0 × 10 −2 m ) cosθ = (42.0 mWb)cosθ

(a) For θ = 0°:

φ m = (42.0 mWb )cos 0° = 42 mWb

(b) For θ = 60°:

φ m = (42.0 mWb )cos 60° = 21 mWb

2

r A uniform magnetic field B is perpendicular to the base of a 23 • hemisphere of radius R. Calculate the magnetic flux (in terms of B and R) through the spherical surface of the hemisphere.

Magnetic Induction

683

Picture the Problem Noting that the flux through the base must also penetrate the spherical surface (the ± in the answer below), we can apply its definition to express φm.

Apply the definition of magnetic flux to obtain:

φ m = ± AB = ± πR 2 B

24 • Find the magnetic flux through a 400-turn solenoid that has a length equal to 25.0 cm, has a radius equal to 1.00 cm, and carries a current of 3.00 A. Picture the Problem We can use φm = NBA cos θ to express the magnetic flux through the solenoid and B = μ0 nI to relate the magnetic field in the solenoid to

the current in its coils. Assume that the magnetic field in the solenoid is constant. Express the magnetic flux through a coil with N turns:

φm = NBA cos θ

Express the magnetic field inside a long solenoid:

B = μ0 nI where n is the number of turns per unit length.

φm = Nμ0 nIA cosθ

Substitute to obtain:

or, because n = N/L and θ = 0°, N 2 μ0 IA N 2 μ0 Iπ r 2 = φm = L L Substitute numerical values and evaluate φm:

φm =

(400)2 (4π ×10 −7 N/A 2 )(3.00 A )π(0.0100 m )2 0.250 m

= 758 μWb

25 • Find the magnetic flux through a 800-turn solenoid that has a length equal to 30.0 cm, has a radius equal to 2.00 cm, and carries a current of 2.00 A. Picture the Problem We can use φm = NBA cos θ to express the magnetic flux through the solenoid and B = μ0 nI to relate the magnetic field in the solenoid to

the current in its coils. Assume that the magnetic field in the solenoid is constant. Express the magnetic flux through a coil with N turns:

φm = NBA cos θ

684

Chapter 28

Express the magnetic field inside a long solenoid:

B = μ0 nI

where n is the number of turns per unit length.

φm = Nμ0 nIA cosθ

Substitute for B to obtain:

or, because n = N/L and θ = 0°, N 2 μ0 IA N 2 μ0 Iπ r 2 = φm = L L Substitute numerical values and evaluate φm:

φm =

(800)2 (4π ×10 −7 N/A 2 )(2.00 A )π (0.0200 m )2 0.300 m

= 6.74 mWb

26 •• A circular coil has 15.0 turns, has a radius 4.00 cm, and is in a uniform magnetic field of 4.00 kG in the +x direction. Find the flux through the coil when the unit normal to the plane of the coil is (a) iˆ , (b) jˆ , (c) iˆ + ˆj 2 , (d) kˆ , and (e) 0.60iˆ + 0.80 jˆ .

( )

Picture the Problem We can apply the definitions of magnet flux and of the dot product to find the flux for the given unit vectors.

Apply the definition of magnetic flux to the coil to obtain:

r Because B is constant:

r

φm = N ∫ B ⋅ nˆ dA S

(

r

r

)

φm = NB ⋅ nˆ ∫ dA = N B ⋅ nˆ A

(

S

)

r = N B ⋅ nˆ π r 2

r Evaluate B :

r B = (0.400 T ) iˆ

Substitute numerical values and simplify to obtain:

φ m = (15.0)[(0.400 T )]π (0.0400 m )2

(a) Evaluate φm for nˆ = iˆ :

φ m = (0.03016 T ⋅ m 2 )iˆ ⋅ iˆ = 30.2 mWb

(b) Evaluate φm for nˆ = ˆj :

φ m = (0.03016 T ⋅ m 2 )iˆ ⋅ ˆj = 0

(

)

= 0.03016 T ⋅ m 2 iˆ ⋅ nˆ

Magnetic Induction

( )

(c) Evaluate φm for nˆ = iˆ + ˆj

φm = (0.03016 T ⋅ m 2 )iˆ ⋅

2:

=

685

(iˆ + ˆj ) 2

0.03016 T ⋅ m = 21.3 mWb 2 2

φm = (0.03016 T ⋅ m 2 )iˆ ⋅ kˆ = 0

(d) Evaluate φm for nˆ = kˆ : (e) Evaluate φm for nˆ = 0.60iˆ + 0.80 ˆj :

(

)

φm = (0.03016 T ⋅ m 2 ) iˆ ⋅ 0.60iˆ + 0.80 ˆj = 18 mWb 27 •• [SSM] A long solenoid has n turns per unit length, has a radius R1, and carries a current I. A circular coil with radius R2 and with N total turns is coaxial with the solenoid and equidistant from its ends. (a) Find the magnetic flux through the coil if R2 > R1. (b) Find the magnetic flux through the coil if R2 < R1. Picture the Problem The magnetic field outside the solenoid is, to a good approximation, zero. Hence, the flux through the coil is the flux in the core of the solenoid. The magnetic field inside the solenoid is uniform. Hence, the flux through the circular coil is given by the same expression with R2 replacing R1:

(a) The flux through the large circular loop outside the solenoid is given by: Substituting for B and A and simplifying yields: (b) The flux through the coil when R2 < R1 is given by:

φm = NBA

φm = N (μ 0 nI )(πR12 ) = μ 0 nINπR12 φ m = N (μ 0 nI ) (πR22 ) = μ 0 nINπR22

28 ••• (a) Compute the magnetic flux through the rectangular loop shown in Figure 28-45. (b) Evaluate your answer for a = 5.0 cm, b = 10 cm, d = 2.0 cm, and I = 20 A. Picture the Problem We can use the hint to set up the element of area dA and express the flux dφm through it and then carry out the details of the integration to express φm.

(a) The flux through the strip of area dA is given by:

dφm = BdA where dA = bdx.

686

Chapter 28

μ0 2 I μ0 I = 4π x 2π x

Express B at a distance x from a long, straight wire:

B=

Substitute to obtain:

dφm =

Integrate from x = d to x = d + a:

μ0 I μ Ib dx bdx = 0 2π x 2π x

μ0 Ib d +a dx μ Ib ⎛ d + a ⎞ = 0 ln⎜ ⎟ ∫ 2π d x 2π ⎝ d ⎠

φm =

(b) Substitute numerical values and evaluate φm:

φm =

(4π ×10

−7

)

N/A 2 (20 A )(0.10 m ) ⎛ 7.0 cm ⎞ ⎟⎟ = 0.50 μWb ln⎜⎜ 2π ⎝ 2.0 cm ⎠

29 ••• A long cylindrical conductor with a radius R and a length L carries a current I. Find the magnetic flux per unit length through the area indicated in Figure 28-44. Picture the Problem Consider an element of area dA = Ldr where r ≤ R. We can use its definition to express dφm through this area in terms of B and Ampere’s law to express B as a function of I. The fact that the current is uniformly distributed over the cross-sectional area of the conductor allows us to set up a proportion from which we can obtain I as a function of r. With these substitutions in place we can integrate dφm to obtain φm/L.

r Noting that B is parallel to nˆ over the entire area, express the flux dφm through an area Ldr:

Apply Ampere’s law to the current contained inside a cylindrical region of radius r < R:

dφm = BdA = BLdr

(1)

r r B ∫ ⋅ d l = 2πrB = μ0 IC C

μ0 I C 2πr

Solving for B yields:

B=

Using the fact that the current I is uniformly distributed over the crosssectional area of the conductor, express its variation with distance r from the center of the conductor:

r2 I (r ) πr 2 = 2 ⇒ I (r ) = I C = I 2 I πR R

(2)

Magnetic Induction Substitute for IC in equation (2) and simplify to obtain:

μ0 I r 2 μI B= = 0 2r 2 2πr R 2πR

Substituting for B in equation (1) yields:

dφ m =

Integrate dφm from r = 0 to r = R to obtain:

μ0 LI R μ LI rdr = 0 φm = 2 ∫ 2πR 0 4π

Divide both sides of this equation by L to express the magnetic flux per unit length:

φm L

=

687

μ 0 LI rdr 2πR 2

μ0 I 4π

Induced EMF and Faraday’s Law 30 • The flux through a loop is given by φm = 0.10t2 – 0.40t, where φm is in webers and t is in seconds. (a) Find the induced emf as a function of time. (b) Find both φm and ε at t = 0, t = 2.0 s, t = 4.0 s, and t = 6.0 s. Picture the Problem Given φm as a function of time, we can use Faraday’s law to express ε as a function of time.

(a) Apply Faraday’s law to express the induced emf in the loop in terms of the rate of change of the magnetic flux:

ε (t ) = − dφm

[ (

d 0.10 t 2 − 0.40t dt dt Wb = −(0.20t − 0.40 ) s =−

)]

= − (0.20t − 0.40 ) V where (b) Evaluate φm at t = 0: Evaluate

ε

at t = 0:

Proceed as above to complete the table to the right:

ε

is in volts and t is in seconds.

φm (0 s ) = 0.10(0)2 − (0.40 )(0) = 0

ε (0 s ) = −[0.20(0) − 0.40]V =

ε φm t (s) (Wb) (V) 0 0 0.40 2.0 −0.40 0.00 4.0 0.0 −0.40 6.0 1.2 −0.80

0.40 V

688

Chapter 28

31 • The flux through a loop is given by φm = 0.10t2 – 0.40t, where φm is in webers and t is in seconds. (a) Sketch graphs of magnetic flux and induced emf as a function of time. (b) At what time(s) is the flux minimum? What is the induced emf at that (those) time(s)? (c) At what time(s) is the flux zero? What is (are) the induced emf(s) at those time(s)? Picture the Problem We can find the time at which the flux is a minimum by looking for the lowest point on the graph of ε versus t and the emf at this time by determining the value of V at this time from the graph. We can interpret the graphs to find the times at which the flux is zero and the corresponding values of the emf.

(a) The flux, φm, and the induced emf, ε , are shown as functions of t in the following graph. The solid curve represents φm, the dashed curve represents ε . 1.00

1.50

flux emf

1.00

0.50

0.00 0.0

1.0

2.0

3.0

4.0

5.0

0.00 6.0

emf (V)

flux (Wb)

0.50

-0.50

-0.50 -1.00

-1.50

t (s)

-1.00

(b) Referring to the graph, we see that the flux is a minimum when t = 2.0 s and that V(2.0 s) = 0. (c) The flux is zero when t = 0 and t = 4.0 s. ε(0) = 0.40 V and

ε(4.0 s) =

−0.40 V.

32 • A solenoid that has a length equal to 25.0 cm, a radius equal to 0.800 cm, and 400 turns is in a region where a magnetic field of 600 G exists and makes an angle of 50º with the axis of the solenoid. (a) Find the magnetic flux

Magnetic Induction

689

through the solenoid. (b) Find the magnitude of the average emf induced in the solenoid if the magnetic field is reduced to zero in 1.40 s. Picture the Problem We can use its definition to find the magnetic flux through the solenoid and Faraday’s law to find the emf induced in the solenoid when the external field is reduced to zero in 1.4 s.

(a) Express the magnetic flux through the solenoid in terms of N, B, A, and θ :

φm = NBA cosθ

Substitute numerical values and evaluate φm:

φm = (400 )(60.0 mT )π (0.00800 m )2 cos 50°

(b) Apply Faraday’s law to obtain:

ε = − Δφm

= NBπR 2 cosθ

= 3.10 mWb = 3.1 mWb

Δt

=−

0 − 3.10 mWb 1.40 s

= 2.2 mV 33 •• [SSM] A 100-turn circular coil has a diameter of 2.00 cm, a resistance of 50.0 Ω, and the two ends of the coil are connected together. The plane of the coil is perpendicular to a uniform magnetic field of magnitude 1.00 T. The direction of the field is reversed. (a) Find the total charge that passes through a cross section of the wire. If the reversal takes 0.100 s, find (b) the average current and (c) the average emf during the reversal. Picture the Problem We can use the definition of average current to express the total charge passing through the coil as a function of Iav. Because the induced current is proportional to the induced emf and the induced emf, in turn, is given by Faraday’s law, we can express ΔQ as a function of the number of turns of the coil, the magnetic field, the resistance of the coil, and the area of the coil. Knowing the reversal time, we can find the average current from its definition and the average emf from Ohm’s law.

(a) Express the total charge that passes through the coil in terms of the induced current:

ΔQ = I av Δt

Relate the induced current to the induced emf:

I = I av =

ε R

(1)

690

Chapter 28

Using Faraday’s law, express the induced emf in terms of φm:

ε = − Δφm

Substitute in equation (1) and simplify to obtain:

Δφ m 2φ ΔQ = Δt = Δt Δt = m R R R ⎛π ⎞ 2 NB⎜ d 2 ⎟ 2 NBA ⎝4 ⎠ = = R R 2 NBπd = 2R where d is the diameter of the coil.

Substitute numerical values and evaluate ΔQ:

Δt

ε

ΔQ =

(100)(1.00 T )π (0.0200 m )2 2(50.0 Ω )

= 1.257 mC = 1.26 mC (b) Apply the definition of average current to obtain:

I av =

ΔQ 1.257 mC = = 12.57 mA Δt 0.100 s

= 12.6 mA (c) Using Ohm’s law, relate the average emf in the coil to the average current:

ε av = I av R = (12.57 mA )(50.0 Ω ) = 628 mV

34 •• At the equator, a 1000-turn coil that has a cross-sectional area of 2 300 cm and a resistance of 15.0 Ω is aligned so that its plane is perpendicular to Earth’s magnetic field of 0.700 G. (a) If the coil is flipped over in 0.350 s, what is the average induced current in it during the 0.350 s? (b) How much charge flows through a cross section of the coil wire during the 0.350 s? Picture the Problem (a) Because the average induced current is proportional to the induced emf and the induced emf, in turn, is given by Faraday’s law, we can find Iav from the change in the magnetic flux through the coil, the resistance of the coil, and the time required for the flipping of the coil. (b) Knowing the average current, we can use its definition to find the charge flowing in the coil.

(a) The average induced current is given by:

I av =

ε R

Magnetic Induction The induced emf in the coil is the rate at which the magnetic flux is changing: Substituting for ε yields:

ε = Δφm Δt

I av =

=

691

2φm 2 NBA = Δt Δt

2 NBA RΔt

Substitute numerical values and evaluate I av :

I av =

2 1T ⎞⎛ ⎛ ⎛ 1 m ⎞ ⎞⎟ 2(1000 )⎜ 0.700 G × 4 ⎟ ⎜ 300 cm 2 × ⎜ 2 ⎟ 10 G ⎠ ⎜⎝ ⎝ ⎝ 10 cm ⎠ ⎟⎠

(15.0 Ω )(0.350 s )

= 800 μA

ΔQ ⇒ ΔQ = I av Δt Δt

(b) The average current is also given by:

I av =

Substitute numerical values and evaluate ΔQ:

ΔQ = (0.800 mA )(0.350 s ) = 280 μC

35 •• A current integrator measures the current as a function of time and integrates (adds) the current to find the total charge passing through it. (Because I = dq/dt, the integrator calculates the integral of the current or Q = ∫ Idt .) A

circular coil that has 300 turns and a radius equal to 5.00 cm is connected to such an instrument. The total resistance of the circuit is 20.0 Ω. The plane of the coil is originally aligned perpendicular to Earth’s magnetic field at some point. When the coil is rotated through 90º about an axis that is in the plane of the coil, a charge of 9.40 μC passes through the current integrator is measured to be 9.40 μC. Calculate the magnitude of Earth’s magnetic field at that point. Picture the Problem We can use Faraday’s law to express Earth’s magnetic field at this location in terms of the induced emf and Ohm’s law to relate the induced emf to the charge that passes through the current integrator. Δφm NBA NBπ r 2 =− = = Δt Δt Δt

Using Faraday’s law, express the induced emf in terms of the change in the magnetic flux as the coil is rotated through 90°:

ε

Solving for B yields:

B=

εΔt Nπ r 2

692

Chapter 28

Using Ohm’s law, relate the induced emf to the induced current:

ε = IR = ΔQ R

Δt where ΔQ is the charge that passes through the current integrator.

Substitute for ε and simplify to obtain:

ΔQ RΔt ΔQR B = Δt 2 = Nπ r Nπ r 2

Substitute numerical values and evaluate B:

B=

(9.40 μC)(20.0 Ω ) (300)π (0.0500 m )2

= 79.8 μT

Motional EMF 36 •• A 30.0-cm long rod moves steadily at 8.00 m/s in a plane that is perpendicular to a magnetic field of 500 G. The velocity of the rod is perpendicular to its length. Find (a) the magnetic force on an electron in the rod, (b) the electrostatic field in the rod, and (c) the potential difference between the ends of the rod. Picture the Problem We can apply the equation for the force on a charged particle moving in a magnetic field to find the magnetic force acting on an r r r electron in the rod. We can use E = v × B to find E and V = El , where l is the length of the rod, to find the potential difference between its ends.

(a) Relate the magnetic force on an electron in the rod to the speed of the rod, the electronic charge, and the magnetic field in which the rod is moving: Substitute numerical values and evaluate F:

r r r F = qv × B and F = qvB sin θ

(

)

F = 1.602 × 10 −19 C (8.00 m/s ) × (0.0500 T )sin90° = 6.4 × 10 − 20 N

(b) Express the electrostatic field r E in the rod in terms of the magnetic r field B :

r r r E = v × B and E = vB sin θ where θ is r r the angle between v and B .

Magnetic Induction Substitute numerical values and evaluate B:

693

E = (8.00 m/s )(0.0500 T )sin90° = 0.400 V/m = 0.40 V/m

(c) Relate the potential difference between the ends of the rod to its length l and the electric field E:

V = El

Substitute numerical values and evaluate V:

V = (0.400 V/m )(0.300 m ) = 0.12 V

37 •• A 30.0-cm long rod moves in a plane that is perpendicular to a magnetic field of 500 G. The velocity of the rod is perpendicular to its length. Find the speed of the rod if the potential difference between the ends is 6.00 V. r r r Picture the Problem We can use E = v × B to relate the speed of the rod to the electric field in the rod and magnetic field in which it is moving and V = El to relate the electric field in the rod to the potential difference between its ends.

r Express the electrostatic field E in the rod in terms of the magnetic field r B and solve for v:

r r r E = v × B and v =

E where θ is B sin θ r r the angle between v and B . V l

Relate the potential difference between the ends of the rod to its length l and the electric field E and solve for E:

V = El ⇒ E =

Substitute for E to obtain:

v=

V Bl sin θ

Substitute numerical values and evaluate v:

v=

6.00 V (0.0500 T )(0.300 m )sin 90°

= 400 m/s 38 •• In Figure 28-45, let the magnetic field strength be 0.80 T, the rod speed be 10 m/s, the rod length be 20 cm, and the resistance of the resistor be 2.0 Ω. (The resistance of the rod and rails are negligible.) Find (a) the induced emf in the circuit, (b) the induced current in the circuit (including direction), and (c) the force needed to move the rod with constant speed (assuming negligible friction). Find (d) the power delivered by the force found in Part (c) and (e) the rate of Joule heating in the resistor.

694

Chapter 28

Picture the Problem Because the speed of the rod is constant, an external force must act on the rod to counter the magnetic force acting on the induced current. We can use the motional-emf equation ε = vBl to evaluate the induced emf, Ohm’s law to find the current in the circuit, Newton’s second law to find the force needed to move the rod with constant speed, and P = Fv to find the power input by the force.

(a) Relate the induced emf in the circuit to the speed of the rod, the magnetic field, and the length of the rod: (b) Using Ohm’s law, relate the current in the circuit to the induced emf and the resistance of the circuit:

ε = vBl = (10 m/s)(0.80 T )(0.20 m ) = 1.60 V = 1.6 V

I=

ε R

=

1.60 V = 0.80 A 2.0 Ω

Note that, because the rod is moving to the right, the flux in the region defined by the rod, the rails, and the resistor is increasing. Hence, in accord with Lenz’s law, the current must be counterclockwise if its magnetic field is to counter this increase in flux.

(c) Because the rod is moving with constant speed in a straight line, the net force acting on it must be zero. Apply Newton’s second law to relate F to the magnetic force Fm:

∑F

Solving for F and substituting for Fm yields:

F = Fm = BIl

Substitute numerical values and evaluate F:

x

= F − Fm = 0

F = (0.80 T )(0.80 A )(0.20 m ) = 0.128 N = 0.13 N

(d) Express the power input by the force in terms of the force and the velocity of the rod:

P = Fv = (0.128 N )(10 m/s ) = 1.3 W

(e) The rate of Joule heat production is given by:

P = I 2 R = (0.80 A ) (2.0 Ω ) = 1.3 W 2

Magnetic Induction

695

39 •• A 10-cm by 5.0-cm rectangular loop (Figure 28-46) that has a resistance equal to 2.5 Ω moves at a constant speed of 2.4 cm/s through a region that has a uniform 1.7-T magnetic field directed out of the page as shown. The front of the loop enters the field region at time t = 0. (a) Graph the flux through the loop as a function of time. (b) Graph the induced emf and the current in the loop as functions of time. Neglect any self-inductance of the loop and construct your graphs to include the interval 0 ≤ t ≤ 16 s. Picture the Problem We’ll need to determine how long it takes for the loop to completely enter the region in which there is a magnetic field, how long it is in the region, and how long it takes to leave the region. Once we know these times, we can use its definition to express the magnetic flux as a function of time. We can use Faraday’s law to find the induced emf as a function of time.

(a) Find the time required for the loop to enter the region where there is a uniform magnetic field: Letting w represent the width of the loop, express and evaluate φm for 0 < t < 4.17 s : Find the time during which the loop is fully in the region where there is a uniform magnetic field:

Express φm for 4.17 s < t < 8.33 s :

t=

l side of loop

v

=

10 cm = 4.17 s 2.4 cm/s

φm = NBA = NBwvt

= (1.7 T )(0.050 m )(0.024 m/s )t = (2.04 mWb/s )t

t=

l side of loop

v

=

10 cm = 4.17 s 2.4 cm/s

that is, the loop will begin to exit the region when t = 8.33 s.

φm = NBA = NBlw

= (1.7 T )(0.10 m )(0.050 m ) = 8.50 mWb

φm = mt + b

The left-end of the loop will exit the field when t = 12.5 s. Express φm for 8.33 s < t < 12.5 s :

where m is the slope of the line and b is the φm-intercept.

For t = 8.33 s and φm = 8.50 mWb:

8.50 mWb = m(8.33 s ) + b

(1)

For t = 12.5 s and φm = 0:

0 = m(12.5 s ) + b

(2)

Solve equations (1) and (2) simultaneously to obtain:

φm = −(2.04 mWb/s) t + 25.5 mWb

696

Chapter 28

φm = 0

The loop will be completely out of the magnetic field when t > 12.5 s and:

The following graph of φm (t ) was plotted using a spreadsheet program. 9 8

Magnetic flux, mWb

7 6 5 4 3 2 1 0 -1 0

2

4

6

8

10

12

14

16

t, s

(b) Using Faraday’s law, relate the induced emf to the magnetic flux:

ε = − dφm

During the interval 0 < t < 4.17 s :

ε = − d [(2.04 mWb/s)t ] = −2.04 mV

dt

dt

During the interval 4.17 s < t < 8.33 s :

ε = − d [8.50 mWb] = 0 dt

During the interval 8.33 s < t < 12.5 s :

ε = − d [(− 2.04 mWb/s)t + 25.5 mWb] dt = 2.04 mV

For t > 12.5 s:

ε =0

The current in each of these intervals is given by Ohm’s law:

I=

ε R

Magnetic Induction The following graph of ε(t) was plotted using a spreadsheet program. 2.5 2 1.5

emf, mV

1 0.5 0 -0.5 -1 -1.5 -2 -2.5 0

2

4

6

8

10

12

14

16

t, s

The following graph of I(t) was plotted using a spreadsheet program. 1 0.8 0.6 0.4

I , mA

0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0

2

4

6

8 t, s

10

12

14

16

697

698

Chapter 28

40 •• A uniform 1.2-T magnetic field is in the +z direction. A conducting rod of length 15 cm lies parallel to the y axis and oscillates in the x direction with displacement given by x = (2.0 cm) cos (120πt), where 120π has units of rad/s . (a) Find an expression for the potential difference between the ends the rod as a function of time? (b) What is the maximum potential difference between the ends the rod? Picture the Problem The rod is executing simple harmonic motion in the xy plane, i.e., in a plane perpendicular to the magnetic field. The emf induced in the rod is a consequence of its motion in this magnetic field and is given by ε = vBl .

Because we’re given the position of the oscillator as a function of time, we can differentiate this expression to obtain v. (a) The potential difference between the ends of the rod is given by:

ε = vBl = Bl dx

Evaluate dx/dt:

dx d = [(2.0 cm )cos120π t ] dt dt = −(2.0 cm ) 120 s −1 π sin 120π t

dt

(

)

= −(7.54 m/s )sin 120π t

Substitute numerical values and evaluate

ε:

ε = −(1.2 T )(0.15 m )(7.54 m/s)sin 120π t = (b) The maximum potential difference between the ends the rod is the amplitude of the expression for ε derived in Part (a):

ε max =

− (1.4 V )sin 120π t

1.4 V

41 •• [SSM] In Figure 28-47, the rod has a mass m and a resistance R. The rails are horizontal, frictionless and have negligible resistances. The distance

between the rails is l. An ideal battery that has an emf ε is connected between points a and b so that the current in the rod is downward. The rod released from rest at t = 0. (a) Derive an expression for the force on the rod as a function of the speed. (b) Show that the speed of the rod approaches a terminal speed and find an expression for the terminal speed. (c) What is the current when the rod is moving at its terminal speed?

Picture the Problem (a) The net force acting on the rod is the magnetic force it experiences as a consequence of carrying a current and being in a magnetic field. The net emf that drives I in this circuit is the difference between the emf of the

Magnetic Induction

699

battery and the emf induced in the rod as a result of its motion. Applying a righthand rule to the rod reveals that the direction of this magnetic force is to the right. Hence the rod will accelerate to the right when it is released. (b) We can obtain the equation of motion of the rod by applying Newton’s second law to relate its acceleration to ε, B, I, R and l . (c) Letting v = v t in the equation for the current in

the circuit will yield current when the rod is at its terminal speed. (a) Express the magnetic force on the current-carrying rod:

Fm = IlB

The current in the rod is given by:

I=

ε − Blv R

(1)

Substituting for I yields:

Bl ⎛ ε − Blv ⎞ (ε − Blv ) Fm = ⎜ ⎟lB = R ⎝ R ⎠

(b) Letting the direction of motion of the rod be the positive x direction, apply ∑ Fx = ma x to the rod:

Bl (ε − Blv ) = m dv dt R

Solving for dv dt yields:

dv Bl (ε − Blv ) = dt mR

Note that as v increases, ε − Blv → 0 , dv dt → 0 and the rod approaches its terminal speed vt . Set dv dt = 0 to obtain:

ε Bl (ε − Blvt ) = 0 ⇒ vt = Bl mR

(c) Substitute vt for v in equation (1) to obtain:

I=

ε − Bl R

ε Bl = 0

42 • A uniform magnetic field is established perpendicular to the plane of a loop that has a radius equal to 5.00 cm and a resistance equal to 0.400 Ω. The magnitude of the field is increasing at a rate of 40.0 mT/s. Find (a) the magnitude of the induced emf in the loop, (b) the induced current in the loop, and (c) the rate of Joule heating in the loop.

700

Chapter 28

Picture the Problem (a) We can find the magnitude of the induced emf by applying Faraday’s law to the loop. (b) and (c) The application of Ohm’s law will yield the induced current in the loop and we can find the rate of Joule heating using P = I 2 R .

dφ m d dB dB = ( AB ) = A = πR 2 dt dt dt dt

(a) Apply Faraday’s law to express the induced emf in the loop in terms of the rate of change of the magnetic field:

ε

=

Substitute numerical values and evaluate ε :

ε

= π (0.0500 m ) (40.0 mT/s )

(b) Using Ohm’s law, relate the induced current to the induced voltage and the resistance of the loop and evaluate I:

I=

(c) Express the rate at which power is dissipated in a conductor in terms of the induced current and the resistance of the loop and evaluate P:

P = I 2 R = (0.7854 mA ) (0.400 Ω )

2

= 0.3142 mV = 0.314 mV

ε R

=

0.3142 mV = 0.7854 mA 0.400 Ω

= 0.785 mA 2

= 0.247 μW

43 •• In Figure 28-48, a conducting rod that has a mass m and a negligible resistance is free to slide without friction along two parallel frictionless rails that have negligible resistances separated by a distance l and connected by a resistance R. The rails are attached to a long inclined plane that makes an angle θ with the horizontal. There is a magnetic field directed upward as shown. (a) Show that there is a retarding force directed up the incline given by F = (B 2 l 2 v cos 2 θ ) R . (b) Show that the terminal speed of the rod is vt = (mgR sin θ ) B 2 l 2 cos 2 θ .

(

)

Picture the Problem The free-body diagram shows the forces acting on the rod as it slides down the inclined plane. The retarding force is the component of Fm acting up the incline, i.e., in the −x direction. We can express Fm using the expression for the force acting on a conductor moving in a magnetic field. r Recognizing that only the horizontal component of the rod’s velocity v produces an induced emf, we can apply the expression for a motional emf in conjunction with Ohm’s law to find the induced current in the rod. In Part (b) we can apply Newton’s second law to obtain an expression for dv/dt and set this expression equal to zero to obtain vt.

Magnetic Induction

701

y r Fn r Fm

θ θ

x

r mg

(a) Express the retarding force acting on the rod:

Express the induced emf due to the motion of the rod in the magnetic field:

F = Fm cosθ (1) where Fm = IlB and I is the current induced in the rod as a consequence of its motion in the magnetic field.

ε = Blv cosθ

ε

Blv cos θ R

Using Ohm’s law, relate the current I in the circuit to the induced emf:

I=

Substitute in equation (1) to obtain:

⎛ Blv cosθ ⎞ F =⎜ ⎟lB cosθ R ⎝ ⎠ =

(b) Apply

∑F

x

= max to the rod:

R

=

(B l v cos θ ) R 2 2

mg sin θ −

2

B 2l 2v dv cos 2 θ = m R dt

and dv B 2l 2v = g sin θ − cos 2 θ dt mR When the rod reaches its terminal speed vt , dv dt = 0 :

B 2 l 2 vt 0 = g sin θ − cos 2 θ mR

Solve for vt to obtain:

vt =

(mgR sin θ )

(B l

2 2

cos 2 θ

)

702

Chapter 28

44 ••• A conducting rod of length l rotates at constant angular speed ω about one end, in a plane perpendicular to a uniform magnetic field B (Figure 28-49). (a) Show that the potential difference between the ends of the rod is 12 l 2θ . (b) Let the angle θ between the rotating rod and the dashed line be given byθ = ωt. Show that the area of the pie-shaped region swept out by the rod during time t is 12 l 2θ .

(c) Compute the flux φm through this area, and apply ε = –dφm/dt (Faraday’s law) to show that the motional emf is given by 12 Bωl 2 . r r r Picture the Problem We can use F = qv × B to express the magnetic force acting on the moving charged body. Expressing the emf induced in a segment of the rod of length dr and integrating this expression over the length of the rod will lead us to an expression for the induced emf.

(a) Use the motional emf equation to express the emf induced in a segment of the rod of length dr and at a distance r from the pivot:

dε = Brdv = Brωdr

Integrate this expression from r = 0 to r = l to obtain:

ε

(b) Using Faraday’s law, relate the induced emf to the rate at which the flux changes:

ε

Express the area dA, for any value of θ, between r and r + dr:

dA = rθdr

l

∫ dε = Bω ∫ rdr ⇒ 0

ε=

1 2

Bωl 2

0

=

dφm dt

Integrate from r = 0 to r = l to obtain:

A = θ ∫ rdr =

(c) Using its definition, express the magnetic flux through this area:

φm = BA =

Differentiate φm with respect to time to obtain:

ε

l

1 2

θ l2

0

=

d dt

[

1 2

1 2

Bl 2θ

]

Bl 2θ = 12 Bl 2

dθ = dt

1 2

Bl 2ω

Magnetic Induction

703

45 • [SSM] A 2.00-cm by 1.50-cm rectangular coil has 300 turns and rotates in a region that has a magnetic field of 0.400 T. (a) What is the maximum emf generated when the coil rotates at 60 rev/s? (b) What must its angular speed be to generate a maximum emf of 110 V? Picture the Problem We can use the relationship

ε max = 2πNBAf to relate the

maximum emf generated to the area of the coil, the number of turns of the coil, the magnetic field in which the coil is rotating, and the angular speed at which it rotates. (a) Relate the induced emf to the magnetic field in which the coil is rotating:

ε max = NBAω = 2πNBAf

(1)

Substitute numerical values and evaluate εmax:

ε max = 2π (300)(0.400 T )(2.00 × 10 −2 m )(1.50 ×10 −2 m )(60 s −1 ) = (b) Solve equation (1) for f:

f =

14 V

ε max

2πNBA

Substitute numerical values and evaluate f: f =

110 V = 486 rev/s 2π (300)(0.400 T )(2.00 × 10 − 2 m )(1.50 × 10 − 2 m )

46 • The coil of Problem 45 rotates at 60 rev/s in a magnetic field. If the maximum emf generated by the coil is 24 V, what is the magnitude of the magnetic field? Picture the Problem We can use the relationship

ε max = NBAω to relate the

maximum emf generated to the area of the coil, the number of turns of the coil, the magnitude of the magnetic field in which the coil is rotating, and the angular speed at which it rotates. Relate the induced emf to the magnetic field in which the coil is rotating:

ε max = NBAω ⇒ B = ε max

NAω

704

Chapter 28

Substitute numerical values and evaluate B: B=

24 V = 0.71T 2π (300) 2.00 × 10 m 1.50 × 10 − 2 m (60 rev/s)

(

−2

)(

)

Inductance 47 • When the current in an 8.00-H coil is equal to 3.00 A and is increasing at 200 A/s, find (a) the magnetic flux through the coil and (b) the induced emf in the coil. Picture the Problem We can use φm = LI to find the magnetic flux through the coil. We can apply Faraday’s law to find the induced emf in the coil.

(a) The magnetic flux through the coil is the product of the selfinductance of the coil and the current it is carrying:

φm = LI

When the current is 3.00 A:

φm = (8.00 H )(3.00 A ) = 24.0 Wb

(b) Use Faraday’s law to relate ε, L, and dI dt :

ε = − L dI

Substitute numerical values and

ε = −(8.00 H )(200 A/s) =

evaluate ε :

dt − 1.60 kV

48 •• A 300-turn solenoid has a radius equal to 2.00 cm; a length equal to 25.0 cm, and a 1000-turn solenoid has a radius equal to 5.00 cm and is also 25.0-cm long. The two solenoids are coaxial, with one nested completely inside the other. What is their mutual inductance? Picture the Problem We can find the mutual inductance of the two coaxial

solenoids using M 2,1 =

φm2 I1

= μ 0 n2 n1lπr12 .

Substitute numerical values and evaluate M2,1: ⎛ 300 ⎞ ⎛ 1000 ⎞ ⎟⎟ ⎜⎜ ⎟⎟(0.250 m )π (0.0200 m )2 = 1.89 mH M 2,1 = 4π × 10 −7 N/A 2 ⎜⎜ ⎝ 0.250 m ⎠ ⎝ 0.250 m ⎠

(

)

Magnetic Induction

705

49 •• [SSM] An insulated wire that has a resistance of 18.0 Ω/m and a length of 9.00 m will be used to construct a resistor. First, the wire is bent in half and then the doubled wire is wound on a cylindrical form ( Figure 28-50) to create a 25.0-cm-long helix that has a diameter equal to 2.00 cm. Find both the resistance and the inductance of this wire-wound resistor. Picture the Problem Note that the current in the two parts of the wire is in opposite directions. Consequently, the total flux in the coil is zero. We can find the resistance of the wire-wound resistor from the length of wire used and the resistance per unit length.

Because the total flux in the coil is zero:

L= 0

Express the total resistance of the wire:

Ω⎞ ⎛ R = ⎜18.0 ⎟ L m⎠ ⎝

Substitute numerical values and evaluate R:

Ω⎞ ⎛ R = ⎜18.0 ⎟(9.00 m ) = 162 Ω m⎠ ⎝

50 •• You are given a length l of wire that has radius a and are told to wind it into an inductor in the shape of a helix that has a circular cross section of radius r. The windings are to be as close together as possible without overlapping. Show that the self-inductance of this inductor is L = 14 μ 0 rl a . Picture the Problem The wire of length l and radius a is shown in the diagram, as is the inductor constructed with this wire and whose inductance L is to be found. We can use the equation for the self-inductance of a cylindrical inductor to derive an expression for L. l 2a

...

2a

2r

... d

706

Chapter 28

The self-inductance of an inductor with length d, cross-sectional area A, and number of turns per unit length n is: The number of turns N is given by:

Assuming that a R. Picture the Problem We can apply Faraday’s law to relate the induced electric field E to the rates at which the magnetic flux is changing at distances r < R and r > R from the axis of the solenoid.

736

Chapter 28

(a) Apply Faraday’s law to relate the induced electric field to the magnetic flux in the solenoid within a cylindrical region of radius r < R:

r

r

∫ E ⋅ dl = − C

or E (2πr ) = −

dφm dt dφm dt

(1)

Express the field within the solenoid:

B = μ0 nI

Express the magnetic flux through an area for which r < R:

φm = BA = π r 2 μ0 nI

Substitute in equation (1) to obtain:

E (2πr ) = −

Because I = I 0 sin ωt :

[

d π r 2 μ0 nI dt dI = −π r 2 μ0 n dt

E r < R = − 12 rμ 0 n

]

d [I 0 sin ωt ] dt

= − 12 rμ 0 nI 0ω cos ωt

(b) Proceed as in (a) with r > R to obtain:

Solving for Er > R yields:

[

]

d π R 2 μ0 nI dt dI = −π R 2 μ0 n dt 2 = −π R μ0 nI 0ω cos ωt

E (2πr ) = −

Er >R = −

μ 0 nR 2 I 0ω 2r

cos ωt

80 ••• A coaxial cable consists of two very thin-walled conducting cylinders of radii r1 and r2 (Figure 28-62). The currents in the inner and outer cylinders are equal in magnitude but opposite in direction. (a) Use Ampère’s law to find the magnetic field as a function of the perpendicular distance r from the central axis of the cable for (1) 0 < r < r1 (2) r1 < r < r2 and (3) r > r2. (b) Show that the magnetic energy density in the region between the cylinders is given by μ m = 12 (μ 0 4π )I 2 πr 2 . (c) Show that the total magnetic energy in a cable

( )

volume of length l is given by U m = (μ 0 4π )I 2 l ln (r2 r1 ) . (d) Use the result in Part (c) and the relationship between magnetic energy, current and inductance to show that the self-inductance per unit length of this cable arrangement is given by L l = (μ 0 2π )ln (r2 r1 ) .

Magnetic Induction

737

Picture the Problem The system exhibits cylindrical symmetry, so one can use Ampère’s law to determine B inside the inner cylinder, between the cylinders, and outside the outer cylinder. We can use um = B 2 2μ0 and the expression for B from

Part (a) to express the magnetic energy density in the region between cylinders. We can integrate this expression for um over the volume between cylinders to find the total magnetic energy in a volume of length l . Finally, can use our result in Part (c) and U m = 12 LI 2 to find the self-inductance of

the the we the

cylinders per unit length. (a) For r < r1 and for r > r2 the net enclosed current is zero; consequently, in these regions:

For r1 < r < r2:

(b) Express the magnetic energy density in the region between the cylinders:

B(r < r1 ) = 0

and B(r > r2 ) = 0 2π rB = μ 0 I C ⇒ B(r1 < r < r2 ) =

um =

μ0 I 2π r

B2 2μ0 2

Substitute for B and simplify to obtain:

⎛ μ0 I ⎞ ⎟ ⎜⎜ 2π r ⎟⎠ μ0 I 2 ⎝ = um = 2μ 0 8π 2 r 2

(c) Express the magnetic energy dUm in the cylindrical element of volume dV:

μ0 I 2 dU m = um dV = 2 2 (l 2πrdr ) 8π r 2 μ I l dr = 0 ⋅ 4π r

Integrate this expression from r = r1 to r = r2 to obtain:

μ I 2 l dr μ 0 2 ⎛ r2 ⎞ = I l ln⎜⎜ ⎟⎟ Um = 0 ∫ 4π r r 4π ⎝ r1 ⎠

r2

1

(d) Express the energy in the magnetic field in terms of L and I:

U m = 12 LI 2 ⇒ L =

From our result in (c):

⎛r ⎞ U m μ0 l ln⎜⎜ 2 ⎟⎟ = 2 I 4π ⎝ r1 ⎠

2U m I2

738

Chapter 28

Substitute to obtain:

⎡μ ⎛ r ⎞⎤ μ ⎛r ⎞ L = 2 ⎢ 0 l ln⎜⎜ 2 ⎟⎟⎥ = 0 l ln⎜⎜ 2 ⎟⎟ ⎝ r1 ⎠⎦ 2π ⎝ r1 ⎠ ⎣ 4π

Express the ratio L/ l :

L μ0 ⎛ r2 ⎞ = ln⎜ ⎟ 2π ⎜⎝ r1 ⎟⎠ l

81 ••• A coaxial cable consists of two very thin-walled conducting cylinders of radii r1 and r2 (Figure 28-63). The currents in the inner and outer cylinders are equal in magnitude but opposite in direction. Compute the flux through a rectangular area of sides l and r2 – r1 between the conductors shown in Figure 28-P95. Use the relationship between flux and current (φm = LI) to show the selfinductance per unit length of the cable by is given by L l = (μ 0 2π )ln (r2 r1 ) . Picture the Problem We can use its definition to express the magnetic flux through a rectangular element of area dA and then integrate from r = r1 to r = r2 to express the total flux through the region. Substituting in L = φm/I will yield the same result found in Part (d) of Problem 80.

φm

Use the definition of self-inductance to relate the magnetic flux through the region of interest to the current I:

L=

Consider a strip of unit length l and width dr at a distance r from the axis. The flux through this area is given by:

dφ m = BdA = Bldr

Apply Ampere’s law to express the magnetic field at a distance r from the axis:

2πrB = μ0 I ⇒ B =

Substituting for B yields:

dφm =

Integrate from r = r1 to r = r2 to obtain:

μ Il dr μ Il ⎛ r ⎞ φ m = 0 ∫ ⇒ φm = 0 ln⎜⎜ 2 ⎟⎟ 2π r r 2π ⎝ r1 ⎠

(1)

I

μ 0 Il dr 2π r r2

1

Substituting for φm in equation (1) and dividing both sides by l yields:

μ0 I 2πr

μ 0 ⎛ r2 ⎞ L = ln⎜ ⎟ 2π ⎜⎝ r1 ⎟⎠ l

Magnetic Induction

739

82 ••• Figure 28-64 shows a rectangular loop of wire that is 0.300 m wide, is 1.50 m long, and lies in the vertical plane which is perpendicular to a region that has a uniform magnetic field. The magnitude of the uniform magnetic field is 0.400 T and the direction of the magnetic field is into the page. The portion of the loop not in the magnetic field is 0.100 m long. The resistance of the loop is 0.200 Ω and its mass is 50.0 g. The loop is released from rest at t = 0. (a) What is the magnitude and direction of the induced current when the loop has a downward speed v? (b) What is the force that acts on the loop as a result of this current? (c) What is the net force acting on the loop? (d) Write out Newton’s second law for the loop. (e) Obtain an expression for the speed of the loop as a function of time. (f) Integrate the expression obtained in Part (e) to find the distance the loop falls as a function of time. (g) Using a spreadsheet program, make a graph of the position of the loop as a function of time (letting t = 0 at the start) for values of y between 0 and 1.40 m (i.e., when the loop leaves the magnetic field). (h) At what time does the loop completely leave the field region? Compare this to the time it would have taken if there were no field. Picture the Problem We can use I = ε/R and ε = Bv l to find the current induced in the loop and Lenz’s law to determine its direction. We can apply the equation for the force on a current-carrying wire to find the net magnetic force acting on the loop and then sum the forces to find the net force on the loop. Separating the variables in the differential equation and integrating will lead us to an expression for v(t) and a second integration to an expression for y(t). We can solve the latter equation for y = 1.40 m to find the time it takes the loop to exit the magnetic field and our expression for v(t) to find its exit speed. Finally, we can use a constantacceleration equation to find its exit speed in the absence of the magnetic field.

ε

(a) Relate the magnitude of the induced current to the induced emf and the resistance of the loop:

I=

Relate the induced emf ε to the motion of the loop:

ε = Bvl

Substitute for ε to obtain:

R

where l is the length of the horizontal portion of the loop. I=

Bl v R

As the loop falls, the flux into it (the loop) decreases. The direction of the induced current is such that its magnetic field opposes this decrease; i.e., clockwise.

740

Chapter 28

(b) Express the velocity-dependent force that acts on the loop in terms of the current in the loop:

Fv = BIl

Substitute for I to obtain:

B 2l 2 ⎛ Bl ⎞ v Fv = B⎜ ⎟vl = R ⎝ R⎠

r r r Apply dF = Id l × B to the horizontal portion of the loop that is in the magnetic field to conclude that the net magnetic force is upward. Note that the magnetic force on the left side of the loop is to the left and the magnetic force on the right side of the loop is to the right.

(c) The net force acting on the loop is the difference between the downward gravitational force and the upward magnetic force: (d) Apply Newton’s second law of motion to the loop to obtain its equation of motion:

Fnet = mg − Fv

= mg −

mg −

B 2l 2 v R

B 2l 2 dv dv B 2l 2 v v=m ⇒ =g− dt R mR dt

Factor g to obtain an alternate form of the equation of motion:

⎛ ⎛ B 2l 2 ⎞ v dv = g ⎜⎜1 − v ⎟⎟ = g ⎜⎜1 − dt ⎝ mgR ⎠ ⎝ vt mgR where vt = 2 2 B l

(e) Separating the variables yields:

dv dv = dt or = dt 2 2 Bl a − bv v g− mR B 2l 2 where a = g and b = mR v

t

⎞ ⎟⎟ ⎠

Integrate v from 0 to v and t from 0 to t:

1 ⎛ a − bv ⎞ dv ∫0 a − bv = ∫0 dt ⇒ − b ln⎜⎝ a ⎟⎠ = t

Transforming from logarithmic to exponential form and solving for v yields:

v(t ) =

(

a 1 − e − bt b

)

Magnetic Induction Noting that vt =

(

v(t ) = vt 1 − e −t τ

a , we have: b

where vt =

741

)

v v mgR and τ = t = t . 2 2 B l a g

(

)

(f) Write v as dy/dt and separate variables to obtain:

dy = vt 1 − e −t τ dt

Integrate y from 0 to y and t from 0 to t:

y

t

0

0

(

)

−t τ ∫ dy = vt ∫ 1 − e dt

and y (t ) = vt [t − τ (1 − e −t τ )

]

(g) A spreadsheet program to generate the data for graphs of position y as a function of time t is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 B4 B5

Formula/Content 0.0500 0.200 0.400 0.300 $B$1*$B$7*$B$2/($B$3^2*$B$4^2)

Algebraic Form m R B L vt

τ

B6 $B$5/$B$7 B7 9.81 A10 0.00 B10 $B$5*(A10−$B$6*(1−EXP(−A10/$B$6))) C10 0.5*$B$7*A10^2 A 1 2 3 4 5 6 7 8 9 10 11 12

m= R= B= L= v t= τ= g= t 0.00 0.01 0.02

B 0.0500 0.200 0.400 0.300 6.813 0.694 9.81

y 0.000 0.000 0.002

g t y 2 1 2 gt C

kg ohms T m m/s s m/s2

y (no B) 0.000 0.000 0.002

742

Chapter 28 13 14 15

0.03 0.04 0.05

0.004 0.008 0.012

0.004 0.008 0.012

62 63 64 65 66 67 68 69 70 71 72 73 74

0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.60 0.61 0.62 0.63 0.64

1.049 1.085 1.122 1.159 1.196 1.234 1.273 1.311 1.351 1.390 1.430 1.471 1.511

1.326 1.378 1.430 1.484 1.538 1.594 1.650 1.707 1.766 1.825 1.885 1.947 2.009

Examination of either the table or the following graph shows that, when the loop falls in the magnetic field, y = 1.4 m when t ≈ 0.61s. In the absence of the magnetic field, y = 1.4 m when t ≈ 0.53 s. The following graph shows y as a function of t for B ≠ 0 (solid curve) and B = 0 (dashed curve). 5

4

3 y (m)

B>0 B=0 2

1

0 0.0

0.1

0.2

0.3

0.4

0.5 t (s)

0.6

0.7

0.8

0.9

1.0

Magnetic Induction

743

In the absence of the magnetic field, the loop fall a distance of 1.40 m in about 0.08 s less time than it takes to fall the same distance in the presence of the magnetic field. 83 ••• A coil of N turns and area A suspended from the ceiling by a wire that provides a linear restoring torque with torsion constant κ. The two ends of the coil are connected to each other, the coil has resistance R, and the moment of inertia of the coil is I. The plane of the coil is vertical, and parallel to a uniform horizontal r magnetic field B when the wire is not twisted (i.e., θ = 0). The coil is rotated about a vertical axis through its center by a small angle θ0 and released. The coil then undergoes damped harmonic oscillation. Show that its angle with its equilibrium position will vary with time according to θ (t ) = θ 0 e −t 2τ cos ω ' t , where

τ = RI ( NBA)2 , ω0 = κ I and ω ' = ω0 1 − (2ω 0τ )−2 . Picture the Problem Picture the Problem If the coil is rotated through an angle θ, the wire exerts a restoring torque equal to –κθ acts on it returning it to its equilibrium position. However, when it rotates with angular speed dθ dt, there

will be an emf induced in the coil. The direction of the current resulting from this induced emf will be such that its magnetic field will oppose the change in flux resulting from the rotation of the coil. The net effect is that the magnetic field exerts a torque on the coil in a direction opposite to the direction of the angular velocity of the coil. We can show that θ will vary with time according to

θ (t ) = θ 0 e −t 2τ cos ω ' t ,where τ = RI (NBA)2 , ω0 = κ I and ω ' = ω0 1 − (2ω 0τ )−2

by demonstrating that θ (t ) = θ 0 e −t 2τ cos ω ' t satisfies the differential equation obtained in our solution for Part (a). Apply

∑τ = Iα to the rotating coil

to obtain: The magnitude of the retarding (damping) torque is given by:

τ restoring − τ retarding = I

d 2θ dt 2

τ retarding = NiBA cosθ where i is the current induced in the coil whose cross-sectional area is A.

Substitute for τrestoring and τretarding to obtain:

− κθ − NiBA cosθ = I

Apply Faraday’s law to express the emf induced in the coil:

ε = − d (NBA sin θ ) = −(NBA cosθ ) dθ dt

d 2θ dt 2

(1)

dt

744

Chapter 28

From Ohm’s law, the magnitude of the induced current i in the coil is: Substitute for the induced current i in equation (1) to obtain: For small displacements from equilibrium, cosθ ≈ 1 and:

i=

ε R

=

2 ( NBA) cos 2 θ − κθ −

R

− κθ −

Rearrange to express the differential equation in standard form, then substitute using ω0 = κ I and

τ = RI ( NBA)2 :

NBA cos θ dθ R dt

(NBA)2 dθ R

dt

dθ d 2θ =I 2 dt dt

≈I

d 2θ dt 2

d 2θ (NBA) dθ κ + + θ ≈0 RI dt I dt 2 or d 2θ 1 dθ + + ω 02θ = 0 2 τ dt dt 2

(2)

Assume that the solution to equation (2) is given by θ (t ) = θ 0e − t 2τ cos ω ' t and evaluate its first and second derivatives with respect to time: dθ d − t 2τ d d − t 2τ ⎤ ⎡ e cos ω ' t = θ 0 ⎢e − t 2τ (cos ω ' t ) + cos ω ' t e = θ0 ⎥⎦ dt dt dt dt ⎣ 1 ⎛ ⎞ = θ 0 ⎜ − ω ' e − t 2τ sin ω ' t − e − t 2τ θ 0 cos ω ' t ⎟ 2τ ⎝ ⎠

[

]

(

1 ⎛ ⎞ = −θ 0e − t 2τ ⎜ ω ' sin ω ' t + cos ω ' t ⎟ 2τ ⎝ ⎠

and 1 d 2θ d ⎡ ⎤ = ⎢− θ 0ω ' e − t 2τ sin ω ' t − θ 0 e − t 2τ cos ω ' t ⎥ 2 2τ dt dt ⎣ ⎦ 1 d⎛ ⎞ = −θ 0e − t 2τ ⎜ ω ' sin ω ' t + cos ω ' t ⎟ 2τ dt ⎝ ⎠ 1 ⎛ ⎞d − θ 0 ⎜ ω ' sin ω ' t + cos ω ' t ⎟ e − t 2τ 2τ ⎝ ⎠ dt ω' ⎛ ⎞ = −θ 0e − t 2τ ⎜ ω '2 cos ω ' t − sin ω ' t ⎟ 2τ ⎝ ⎠ 1 ⎛ 1 ⎞ + θ 0 ⎜ ω ' sin ω ' t + cos ω ' t ⎟e − t 2τ 2τ ⎝ 2τ ⎠

ω' 1 ⎛ ⎞ = −θ 0e − t 2τ ⎜ ω '2 cos ω ' t − sin ω ' t − 2 cos ω ' t ⎟ τ 4τ ⎝ ⎠

)

Magnetic Induction

745

Substitute these derivatives in equation (2) to obtain:

ω' 1 ⎛ ⎞ − θ 0e − t 2τ ⎜ ω '2 cos ω ' t − sin ω ' t − 2 cos ω ' t ⎟ τ 4τ ⎝ ⎠ 1 1 ⎛ ⎞ − θ 0e − t 2τ ⎜ ω ' sin ω ' t + cos ω ' t ⎟ + ω02θ 0e − t 2τ cos ω ' t = 0 τ 2τ ⎝ ⎠ Because θ0 and e − t τ are never zero, we can divide them out of the equation and simplify to obtain:

− ω '2 cos ω ' t −

1 cos ω ' t + ω02 cos ω ' t = 0 4τ 2

or 1 ⎛ 2 2⎞ ⎜ − ω ' − 2 + ω0 ⎟ cos ω ' t = 0 4τ ⎝ ⎠ This equation is satisfied provided:

− ω '2 −

Solving for ω ' yields:

ω ' = ω0 1 − (2ω 0τ )−2

1 + ω02 = 0 4τ 2

Thus we’ve shown that the angular position of the oscillating coil, relative to its equilibrium position, varies with time according to θ (t ) = θ 0 e −t 2τ cos ω ' t , where

τ = RI ( NBA)2 , ω0 = κ I and ω ' = ω0 1 − (2ω 0τ )−2 .

746

Chapter 28

Chapter 29 Alternating-Current Circuits Conceptual Problems 1 • A coil in an ac generator rotates at 60 Hz. How much time elapses between successive peak emf values of the coil? Determine the Concept Successive peaks are one-half period apart. Hence the 1 1 1 = = 8.33 ms . elapsed time between the peaks is T = 2 2 f 2 60 s −1

(

)

2 • If the rms voltage in an ac circuit is doubled, the peak voltage is (a) doubled, (b) halved, (c) increased by a factor of 2 , (d) not changed. Picture the Problem We can use the relationship between V and Vpeak to decide the effect of doubling the rms voltage on the peak voltage. Express the initial rms voltage in terms of the peak voltage:

Vrms =

Express the doubled rms voltage in terms of the new peak voltage V' peak :

2Vrms =

Divide the second of these equations by the first and simplify to obtain:

Vpeak 2

V' peak 2 V' peak

V' peak 2Vrms = 2 ⇒ 2= Vpeak Vpeak Vrms 2

Solving for V' peak yields:

V' peak = 2Vpeak ⇒ (a ) is correct.

3 • [SSM] If the frequency in the circuit shown in Figure 29-27 is doubled, the inductance of the inductor will (a) double, (b) not change, (c) halve, (d) quadruple. Determine the Concept The inductance of an inductor is determined by the details of its construction and is independent of the frequency of the circuit. The inductive reactance, on the other hand, is frequency dependent. (b) is correct. 4 • If the frequency in the circuit shown in Figure 29-27 is doubled, the inductive reactance of the inductor will (a) double, (b) not change, (c) halve, (d) quadruple.

747

748

Chapter 29

Determine the Concept The inductive reactance of an inductor varies with the frequency according to X L = ωL. Hence, doubling ω will double XL. (a ) is

correct. 5 • If the frequency in the circuit in Figure 29-28 is doubled, the capacitive reactance of the circuit will (a) double, (b) not change, (c) halve, (d) quadruple. Determine the Concept The capacitive reactance of an capacitor varies with the frequency according to X C = 1 ωC . Hence, doubling ω will halve XC. (c) is

correct. 6 • (a) In a circuit consisting solely of a ac generator and an ideal inductor, are there any time intervals when the inductor receives energy from the generator? If so, when? Explain your answer. (b) Are there any time intervals when the inductor supplies energy back to the generator? If so when? Explain your answer. Determine the Concept Yes to both questions. (a) While the magnitude of the current in the inductor is increasing, the inductor absorbs power from the generator. (b) When the magnitude of the current in the inductor decreases, the inductor supplies power to the generator. 7 • [SSM] (a) In a circuit consisting of a generator and a capacitor, are there any time intervals when the capacitor receives energy from the generator? If so, when? Explain your answer. (b) Are there any time intervals when the capacitor supplies power to the generator? If so, when? Explain your answer. Determine the Concept Yes to both questions. (a) While the magnitude of the charge is accumulating on either plate of the capacitor, the capacitor absorbs power from the generator. (b) When the magnitude of the charge is on either plate of the capacitor is decreasing, it supplies power to the generator. 8 • (a) Show that the SI unit of inductance multiplied by the SI unit of capacitance is equivalent to seconds squared. (b) Show that the SI unit of inductance divided by the SI unit of resistance is equivalent to seconds. Determine the Concept (a) Substitute the SI units of inductance and capacitance and simplify to obtain:

V ⋅s C s ⋅ = ⋅ C = s2 A V C s

Alternating-Current Circuits (b) Substitute the SI units of inductance divided by resistance and simplify to obtain:

749

V ⋅s V ⋅s A = A = s V Ω A

9 • [SSM] Suppose you increase the rotation rate of the coil in the generator shown in the simple ac circuit in Figure 29-29. Then the rms current (a) increases, (b) does not change, (c) may increase or decrease depending on the magnitude of the original frequency, (d) may increase or decrease depending on the magnitude of the resistance, (e) decreases. Determine the Concept Because the rms current through the resistor is given by ε peak NBA ε I rms = rms = = ω , I rms is directly proportional to ω. (a ) is correct. R 2 2 10 • If the inductance value is tripled in a circuit consisting solely of a variable inductor and a variable capacitor, how would you have to change the capacitance so that the natural frequency of the circuit is unchanged? (a) triple the capacitance, (b) decrease the capacitance to one-third of its original value, (c) You should not change the capacitance.(d) You cannot determine how to change the capacitance from the data given. Determine the Concept The natural frequency of an LC circuit is given by f 0 = 1 2π LC .

Express the natural frequencies of the circuit before and after the inductance is tripled: Divide the second of the these equations by the first and simplify to obtain:

Because the natural frequency is unchanged: When the inductance is tripled:

f0 =

1 2π LC

and f 0' =

1 2π L'C'

1 f = 2π L'C' = 1 f0 2π LC ' 0

LC L'C'

1=

LC L LC ⇒ = 1 ⇒ C' = C L'C' L' L'C'

C' =

L 1 C = C⇒ 3L 3

(b )

is correct.

11 • [SSM] Consider a circuit consisting solely of an ideal inductor and an ideal capacitor. How does the maximum energy stored in the capacitor compare to the maximum value stored in the inductor? (a) They are the same and each equal to the total energy stored in the circuit. (b) They are the same and each

Chapter 29

750

equal to half of the total energy stored in the circuit. (c) The maximum energy stored in the capacitor is larger than the maximum energy stored in the inductor. (d) The maximum energy stored in the inductor is larger than the maximum energy stored in the capacitor. (e) You cannot compare the maximum energies based on the data given because the ratio of the maximum energies depends on the actual capacitance and inductance values. Determine the Concept The maximum energy stored in the electric field of the 1 Q2 capacitor is given by U e = and the maximum energy stored in the magnetic 2 C 1 field of the inductor is given by U m = LI 2 . Because energy is conserved in an 2 LC circuit and oscillates between the inductor and the capacitor, Ue = Um = Utotal. (a ) is correct.

True or false:

12

•

(a)

A driven series RLC circuit that has a high Q factor has a narrow resonance curve. A circuit consists solely of a resistor, an inductor and a capacitor, all connected in series. If the resistance of the resistor is doubled, the natural frequency of the circuit remains the same. At resonance, the impedance of a driven series RLC combination equals the resistance R. At resonance, the current in a driven series RLC circuit is in phase with the voltage applied to the combination.

(b) (c) (d)

(a) True. The Q factor and the width of the resonance curve at half power are related according to Q = ω0 Δω ; i.e., they are inversely proportional to each other. (b) True. The natural frequency of the circuit depends only on the inductance L of the inductor and the capacitance C of the capacitor and is given by ω = 1 LC . (c) True. The impedance of an RLC circuit is given by Z = R 2 + ( X L − X C ) . At 2

resonance XL = XC and so Z = R. (d) True. The phase angle δ is related to XL and XC according to ⎛ X − XC ⎞ δ = tan −1 ⎜ L ⎟ . At resonance XL = XC and so δ = 0. R ⎝ ⎠

Alternating-Current Circuits

751

True or false (all questions related to a driven series RLC circuit):

13

•

(a)

Near resonance, the power factor of a driven series RLC circuit is close to zero. The power factor of a driven series RLC circuit does not depend on the value of the resistance. The resonance frequency of a driven series RLC circuit does not depend on the value of the resistance. At resonance, the peak current of a driven series RLC circuit does not depend on the capacitance or the inductance. For frequencies below the resonant frequency, the capacitive reactance of a driven series RLC circuit is larger than the inductive reactance. For frequencies below the resonant frequency of a driven series RLC circuit, the phase of the current leads (is ahead of) the phase of the applied voltage.

(b) (c) (d) (e) (f)

(a) False. Near resonance, the power factor, given by cos δ =

R

(X L − X C )

2

+R

2

,

is close to 1. (b) False. The power factor is given by cos δ =

R

( X L − X C )2 + R 2

.

(c) True. The resonance frequency for a driven series RLC circuit depends only on L and C and is given by ω res = 1 LC (d) True. At resonance X L − X C = 0 and so Z = R and the peak current is given by I peak = Vapp, peak R . (e) True. Because the capacitive reactance varies inversely with the driving frequency and the inductive reactance varies directly with the driving frequency, at frequencies well below the resonance frequency the capacitive reactance is larger than the inductive reactance. (f) True. For frequencies below the resonant frequency, the circuit is more capacitive than inductive and the phase constant φ is negative. This means that the current leads the applied voltage. 14 • You may have noticed that sometimes two radio stations can be heard when your receiver is tuned to a specific frequency. This situation often occurs when you are driving and are between two cities. Explain how this situation can occur.

Chapter 29

752

Determine the Concept Because the power curves received by your radio from two stations have width, you could have two frequencies overlapping as a result of receiving signals from both stations.

True or false (all questions related to a driven series RLC circuit):

15

•

(a)

At frequencies much higher than or much lower than the resonant frequency of a driven series RLC circuit, the power factor is close to zero. The larger the resonance width of a driven series RLC circuit is, the larger the Q factor for the circuit is. The larger the resistance of a driven series RLC circuit is, the larger the resonance width for the circuit is.

(b) (c)

2

1 ⎞ ⎛ 2 (a) True. Because the power factor is given by cos δ = R ⎜ ωL − ⎟ + R , for ωC ⎠ ⎝ values of ω that are much higher or much lower than the resonant frequency, the term in parentheses becomes very large and cosδ approaches zero. (b) False. When the resonance curve is reasonably narrow, the Q factor can be approximated by Q = ω 0 Δω . Hence a large value for Q corresponds to a narrow resonance curve. (c) True. See Figure 29-21. 16 • An ideal transformer has N1 turns on its primary and N2 turns on its secondary. The average power delivered to a load resistance R connected across the secondary is P2 when the primary rms voltage is V1. The rms current in the primary windings can then be expressed as (a) P2/V1, (b) (N1/N2)(P2/V1), (c) (N2/N1)(P2/V1), (d) (N2/N1)2(P2/V1). Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. Assuming no loss of power in the transformer, we can equate the power in the primary circuit to the power in the secondary circuit and solve for the rms current in the primary windings.

Assuming no loss of power in the transformer:

P1 = P2

Substitute for P1 and P2 to obtain:

I1, rmsV1, rms = I 2, rmsV2, rms

Alternating-Current Circuits Solving for I1,rms and simplifying yields:

I1, rms =

(a ) [SSM]

I 2, rmsV2, rms V1, rms

=

753

P2 V1, rms

is correct.

True or false:

17

•

(a) (b) (c) (d) (e)

A transformer is used to change frequency. A transformer is used to change voltage. If a transformer steps up the current, it must step down the voltage. A step-up transformer, steps down the current. The standard household wall-outlet voltage in Europe is 220 V, about twice that used in the United States. If a European traveler wants her hair dryer to work properly in the United States, she should use a transformer that has more windings in its secondary coil than in its primary coil. The standard household wall-outlet voltage in Europe is 220 V, about twice that used in the United States. If an American traveler wants his electric razor to work properly in Europe, he should use a transformer that steps up the current.

(f)

(a) False. A transformer is a device used to raise or lower the voltage in a circuit. (b) True. A transformer is a device used to raise or lower the voltage in a circuit. (c) True. If energy is to be conserved, the product of the current and voltage must be constant. (d) True. Because the product of current and voltage in the primary and secondary circuits is the same, increasing the current in the secondary results in a lowering (or stepping down) of the voltage. (e) True. Because electrical energy is provided at a higher voltage in Europe, the visitor would want to step-up the voltage in order to make her hair dryer work properly. (f) True. Because electrical energy is provided at a higher voltage in Europe, the visitor would want to step-up the current (and decrease the voltage) in order to make his razor work properly.

Estimation and Approximation 18 •• The impedances of motors, transformers, and electromagnets include both resistance and inductive reactance. Suppose that phase of the current to a

754

Chapter 29

large industrial plant lags the phase of the applied voltage by 25° when the plant is under full operation and using 2.3 MW of power. The power is supplied to the plant from a substation 4.5 km from the plant; the 60 Hz rms line voltage at the plant is 40 kV. The resistance of the transmission line from the substation to the plant is 5.2 Ω. The cost per kilowatt-hour to the company that owns the plant is $0.14, and the plant pays only for the actual energy used. (a) Estimate the resistance and inductive reactance of the plant’s total load. (b) Estimate the rms current in the power lines and the rms voltage at the substation. (c) How much power is lost in transmission? (d) Suppose that the phase that the current lags the phase of the applied voltage is reduced to 18º by adding a bank of capacitors in series with the load. How much money would be saved by the electric utility during one month of operation, assuming the plant operates at full capacity for 16 h each day? (e) What must be the capacitance of this bank of capacitors to achieve this change in phase angle? Picture the Problem We can find the resistance and inductive reactance of the plant’s total load from the impedance of the load and the phase constant. The current in the power lines can be found from the total impedance of the load the potential difference across it and the rms voltage at the substation by applying Kirchhoff’s loop rule to the substation-transmission wires-load circuit. The power 2 lost in transmission can be found from Ptrans = I rms Rtrans . We can find the cost

savings by finding the difference in the power lost in transmission when the phase angle is reduced to 18°. Finally, we can find the capacitance that is required to reduce the phase angle to 18° by first finding the capacitive reactance using the definition of tanδ and then applying the definition of capacitive reactance to find C. Rtrans = 5.2 Ω

εsubstation

∼ f = 60 Hz

Z

(a) Relate the resistance and inductive reactance of the plant’s total load to Z and δ:

R = Z cos δ and X L = Z sin δ

Express Z in terms of the rms current I rms in the power lines and the rms voltage ε rms at the plant:

Z=

ε rms I rms

ε rms= 40 kV δ = 25°

Alternating-Current Circuits Express the power delivered to the

plant in terms of εrms, I rms , and δ and solve for I rms :

Substitute to obtain:

Substitute numerical values and evaluate Z: Substitute numerical values and evaluate R and XL:

Pav = ε rms I rms cos δ and

I rms =

Z=

Z=

Pav ε rms cos δ

(1)

2 ε rms cos δ

Pav

(40 kV )2 cos 25° = 630 Ω 2.3 MW

R = (630 Ω ) cos 25° = 571Ω = 0.57 kΩ and X L = (630 Ω )sin 25° = 266 Ω = 0.27 kΩ

(b) Use equation (1) to find the current in the power lines:

I rms =

2.3 MW = 63.4 A (40 kV )cos 25°

= 63 A Apply Kirchhoff’s loop rule to the circuit:

ε sub − I rms Rtrans − I rms Z tot = 0

Solve for εsub:

ε sub = I rms (Rtrans + Z tot )

Substitute numerical values and

ε sub = (63.4 A )(5.2 Ω + 630 Ω )

evaluate εsub:

(c) The power lost in transmission is:

= 40.3 kV 2 Ptrans = I rms Rtrans = (63.4 A ) (5.2 Ω ) 2

= 20.9 kW = 21kW

(d) Express the cost savings ΔC in terms of the difference in energy consumption (P25° − P18°)Δt and the per-unit cost u of the energy:

ΔC = (P25° − P18° )Δtu

755

756

Chapter 29

Express the power lost in transmission when δ = 18°:

P18° = I182 ° Rtrans

Find the current in the transmission lines when δ = 18°:

I18° =

Evaluate P18° :

P18° = (60.5 A ) (5.2 Ω ) = 19.0 kW

2.3 MW = 60.5 A (40 kV ) cos18° 2

Substitute numerical values and evaluate ΔC: d ⎞ ⎛ $0.14 ⎞ ⎛ h ⎞⎛ ΔC = (20.9 kW − 19.0 kW )⎜16 ⎟ ⎜ 30 ⎟⎜ ⎟ = $128 ⎝ d ⎠ ⎝ month ⎠ ⎝ kW ⋅ h ⎠ (e) The required capacitance is given by: Relate the new phase angle δ to the inductive reactance XL, the reactance due to the added capacitance XC, and the resistance of the load R:

Substituting for XC yields:

C=

1 2πfX C

tan δ =

C=

XL − XC ⇒ X C = X L − R tan δ R

1 2πf ( X L − R tan δ )

Substitute numerical values and evaluate C: C=

1 = 33 μF 2π (60 s )(266 Ω − (571Ω ) tan 18°) -1

Alternating Current in Resistors, Inductors, and Capacitors 19 • [SSM] A 100-W light bulb is screwed into a standard 120-V-rms socket. Find (a) the rms current, (b) the peak current, and (c) the peak power. Picture the Problem We can use Pav = ε rms I rms to find I rms , I peak = 2I rms to find

I peak , and Ppeak = I peakε peak to find Ppeak .

(a) Relate the average power delivered by the source to the rms voltage across the bulb and the rms current through it:

Pav = ε rms I rms ⇒ I rms =

Pav

ε rms

Alternating-Current Circuits

757

100 W = 0.8333 A = 0.833 A 120 V

Substitute numerical values and evaluate I rms :

I rms =

(b) Express I peak in terms of I rms :

I peak = 2 I rms

Substitute for I rms and evaluate I peak :

I peak = 2 (0.8333 A ) = 1.1785 A = 1.18 A

(c) Express the maximum power in terms of the maximum voltage and maximum current: Substitute numerical values and evaluate Ppeak :

Ppeak = I peakε peak

Ppeak = (1.1785 A ) 2 (120 V ) = 200 W

20 • A circuit breaker is rated for a current of 15 A rms at a voltage of 120 V rms. (a) What is the largest value of the peak current that the breaker can carry? (b) What is the maximum average power that can be supplied by this circuit? Picture the Problem We can I peak = 2I rms to find the largest peak current the

breaker can carry and Pav = I rmsVrms to find the average power supplied by this circuit. (a) Express I peak in terms of I rms :

I peak = 2 I rms = 2 (15 A ) = 21 A

(b) Relate the average power to the rms current and voltage:

Pav = I rmsVrms = (15 A )(120 V ) = 1.8 kW

21 • [SSM] What is the reactance of a 1.00-μH inductor at (a) 60 Hz, (b) 600 Hz, and (c) 6.00 kHz? Picture the Problem We can use X L = ωL to find the reactance of the inductor at any frequency.

Express the inductive reactance as a function of f:

X L = ωL = 2πfL

(a) At f = 60 Hz:

X L = 2π 60 s −1 (1.00 mH ) = 0.38 Ω

(

)

758

Chapter 29

(

)

(b) At f = 600 Hz:

X L = 2π 600 s −1 (1.00 mH ) = 3.77 Ω

(c) At f = 6.00 kHz:

X L = 2π (6.00 kHz )(1.00 mH ) = 37.7 Ω

22 • An inductor has a reactance of 100 Ω at 80 Hz. (a) What is its inductance? (b) What is its reactance at 160 Hz? Picture the Problem We can use X L = ωL to find the inductance of the inductor at any frequency.

(a) Relate the reactance of the inductor to its inductance:

X L = ωL = 2πfL ⇒ L =

Solve for and evaluate L:

L=

(b) At 160 Hz:

X L = 2π 160 s −1 (0.199 H ) = 0.20 kΩ

XL 2πf

100 Ω = 0.199 H = 0.20 H 2π 80 s −1

(

)

(

)

23 • At what frequency would the reactance of a 10-μF capacitor equal the reactance of a 1.0-μH inductor? Picture the Problem We can equate the reactances of the capacitor and the inductor and then solve for the frequency.

Express the reactance of the inductor:

X L = ωL = 2πfL

Express the reactance of the capacitor:

XC =

Equate these reactances to obtain:

Substitute numerical values and evaluate f:

1 1 = ωC 2πfC

2πfL =

f =

1 2π

1 1 ⇒ f = 2πfC 2π

1 LC

1 = 50 kHz (10 μF)(1.0 μH )

24 • What is the reactance of a 1.00-nF capacitor at (a) 60.0 Hz, (b) 6.00 kHz, and (c) 6.00 MHz?

Alternating-Current Circuits

759

Picture the Problem We can use X C = 1 ωC to find the reactance of the

capacitor at any frequency. Express the capacitive reactance as a function of f:

XC =

(a) At f = 60.0 Hz:

XC =

1 1 = ωC 2πfC 1

(

2π 60.0 s

−1

)(1.00 nF) =

2.65 MΩ

(b) At f = 6.00 kHz:

XC =

1 = 26.5 kΩ 2π (6.00 kHz )(1.00 nF)

(c) At f = 6.00 MHz:

XC =

1 = 26.5 Ω 2π (6.00 MHz )(1.00 nF)

25 • [SSM] A 20-Hz ac generator that produces a peak emf of 10 V is connected to a 20-μF capacitor. Find (a) the peak current and (b) the rms current. Picture the Problem We can use Ipeak = εpeak/XC and XC = 1/ωC to express Ipeak as

a function of εpeak, f, and C. Once we’ve evaluate Ipeak, we can use Irms = Ipeak/ 2 to find I rms . Express I peak in terms of εpeak and XC:

Express the capacitive reactance:

Substitute for XC and simplify to obtain:

(a) Substitute numerical values and evaluate I peak : (b) Express I rms in terms of I peak :

I peak =

XC =

ε peak XC

1 1 = ωC 2πfC

I peak = 2πfCε peak I peak = 2π (20 s −1 )(20 μF)(10 V ) = 25.1 mA = 25 mA

I rms =

I peak 2

=

25.1 mA = 18 mA 2

760

Chapter 29

26 • At what frequency is the reactance of a 10-μF capacitor (a) 1.00 Ω, (b) 100 Ω, and (c) 10.0 mΩ? Picture the Problem We can use X C = 1 ωC = 1 2πfC to relate the reactance of

the capacitor to the frequency. 1 1 1 = ⇒f = ωC 2πfC 2πCX C

The reactance of the capacitor is given by:

XC =

(a) Find f when XC = 1.00 Ω:

f =

1 = 16 kHz 2π (10 μF)(1.00 Ω )

(b) Find f when XC = 100 Ω:

f =

1 = 0.16 kHz 2π (10 μF)(100 Ω )

(c) Find f when XC = 10.0 mΩ:

f =

1

2π (10 μF)(10.0 mΩ )

= 1.6 MHz

27 •• A circuit consists of two ideal ac generators and a 25-Ω resistor, all connected in series. The potential difference across the terminals of one of the generators is given by V1 = (5.0 V) cos(ωt – α), and the potential difference across the terminals of the other generator is given by V2 = (5.0 V) cos(ωt + α), where α = π/6. (a) Use Kirchhoff’s loop rule and a trigonometric identity to find the peak current in the circuit. (b) Use a phasor diagram to find the peak current in the circuit. (c) Find the current in the resistor if α = π/4 and the amplitude of V2 is increased from 5.0 V to 7.0 V. Picture the Problem We can use the trigonometric identity cosθ + cos φ = 2 cos 12 (θ + φ ) cos 12 (θ − φ )

to find the sum of the phasors V1 and V2 and then use this sum to express I as a function of time. In (b) we’ll use a phasor diagram to obtain the same result and in (c) we’ll use the phasor diagram appropriate to the given voltages to express the current as a function of time. (a) Applying Kirchhoff’s loop rule to the circuit yields:

V1 + V2 − IR = 0

Solve for I to obtain:

I=

V1 + V2 R

Alternating-Current Circuits

761

Use the trigonometric identity cos θ + cos φ = 2 cos 12 (θ + φ ) cos 12 (θ − φ ) to find V1 + V2: V1 + V2 = (5.0 V )[cos(ωt − α ) + cos(ωt + α )] = (5 V )[2 cos 12 (2ωt )cos 12 (− 2α )] = (10 V ) cos

π 6

cos ωt = (8.66 V ) cos ωt

Substitute for V1 + V2 and R to obtain:

I=

(8.66 V ) cos ωt = (0.346 A ) cos ωt 25 Ω

= (0.35 A ) cos ωt where I peak = 0.35 A (b) Express the magnitude of the current in R:

I =

r V R r V2

The phasor diagram for the voltages is shown to the right.

r V

30° 30° r V1

r Use vector addition to find V :

r r V = 2 V1 cos 30° = 2(5.0 V ) cos 30°

= 8.66 V r Substitute for V and R to obtain:

I =

8.66 V

= 0.346 A 25 Ω and I = (0.35 A ) cos ωt where I peak = 0.35 A

762

Chapter 29 r V2

(c) The phasor diagram is shown to the right. Note that the phase angle r r between V1 and V2 is now 90°.

α

r V

o

90 − α

δ

r V1

Use the Pythagorean theorem to r find V :

Express I as a function of t:

r r2 r 2 V = V1 + V2 =

(5.0 V )2 + (7.0 V )2

= 8.60 V r V

cos(ωt + δ ) R where δ = 45° − (90° − α ) = α − 45° I=

⎛ 7.0 V ⎞ ⎟⎟ − 45° = tan −1 ⎜⎜ ⎝ 5.0 V ⎠ = 9.462° = 0.165 rad

Substitute numerical values and evaluate I:

I= =

8.60 V cos(ωt + 0.165 rad ) 25 Ω

(0.34 A )cos(ωt + 0.17 rad )

Undriven Circuits Containing Capacitors, Resistors and Inductors 28 • (a) Show that 1 LC has units of inverse seconds by substituting SI units for inductance and capacitance into the expression. (b) Show that ω0L/R (the expression for the Q-factor) is dimensionless by substituting SI units for angular frequency, inductance, and resistance into the expression. Picture the Problem We can substitute the units of the various physical quantitities in 1 / LC and Q = ω 0 L R to establish their units.

Alternating-Current Circuits (a) Substitute the units for L and C in the expression 1 LC and simplify to obtain: (b) Substitute the units for ω0, L, and R in the expression Q = ω 0 L R and simplify to obtain:

1 = H⋅F

1

(Ω ⋅ s )⎛⎜ s ⎞⎟

=

1 s

2

763

= s −1

⎝Ω⎠

1 V ⋅s 1 V ⋅s ⋅ ⋅ s A = s A = 1 ⇒ no units V V A A

29 • [SSM] (a) What is the period of oscillation of an LC circuit consisting of an ideal 2.0-mH inductor and a 20-μF capacitor? (b) A circuit that oscillates consists solely of an 80-μF capacitor and a variable ideal inductor. What inductance is needed in order to tune this circuit to oscillate at 60 Hz? Picture the Problem We can use T = 2π/ω and ω = 1 f) to L and C.

2π

(a) Express the period of oscillation of the LC circuit:

T=

For an LC circuit:

ω=

Substitute for ω to obtain:

T = 2π LC

Substitute numerical values and evaluate T:

T = 2π

(b) Solve equation (1) for L to obtain:

L=

Substitute numerical values and evaluate L:

L=

LC to relate T (and hence

ω 1 LC (1)

(2.0 mH )(20 μF) =

1.3 ms

T2 1 = 2 2 2 4π C 4π f C

(

1

4π 2 60 s −1

) (80 μF) 2

= 88 mH

30 • An LC circuit has capacitance C0 and inductance L. A second LC circuit has capacitance 12 C0 and inductance 2L, and a third LC circuit has capacitance 2C0 and inductance 12 L. (a) Show that each circuit oscillates with the same frequency. (b) In which circuit would the peak current be greatest if the peak voltage across the capacitor in each circuit was the same?

764

Chapter 29

Picture the Problem We can use the expression f 0 = 1 2π LC for the resonance

frequency of an LC circuit to show that each circuit oscillates with the same frequency. In (b) we can use I peak = ωQ0 , where Q0 is the charge of the capacitor at time zero, and the definition of capacitance Q0 = CV to express Ipeak in terms of

ω, C and V. Express the resonance frequency for an LC circuit: (a) Express the product of L and C0 for each circuit:

f0 =

1 2π LC

Circuit 1: L1C1 = L1C0 , Circuit 2: L2C2 = (2 L1 )( 12 C0 ) = L1C1 , and Circuit 3: L3C3 = ( 12 L1 )(2C0 ) = L1C1

Because L1C1 = L2 C 2 = L3C 3 , the resonance frequencies of the three circuits are the same. (b) Express I peak in terms of the

I peak = ωQ0

charge stored in the capacitor: Express Q0 in terms of the capacitance of the capacitor and the potential difference across the capacitor:

Q0 = CV

Substituting for Q0 yields:

I peak = ωCV or, for ω and V constant, I peak ∝ C . Hence, the circuit with capacitance 2C0 has the greatest peak current.

31 •• A 5.0-μF capacitor is charged to 30 V and is then connected across an ideal 10-mH inductor. (a) How much energy is stored in the system? (b) What is the frequency of oscillation of the circuit? (c) What is the peak current in the circuit? Picture the Problem We can use U = 12 CV 2 to find the energy stored in the

electric field of the capacitor, ω0 = 2πf 0 = 1 Q0 = CV to find I peak .

LC to find f0, and I peak = ωQ0 and

Alternating-Current Circuits

(a) Express the energy stored in the system as a function of C and V:

U = 12 CV 2

Substitute numerical values and evaluate U:

U=

(b) Express the resonance frequency of the circuit in terms of L and C:

ω0 = 2πf 0 =

Substitute numerical values and evaluate f0:

f0 =

1 2

(5.0 μF)(30 V )2 =

2.3 mJ

1 1 ⇒ f0 = LC 2π LC 1

2π

765

(10 mH )(5.0 μF)

= 712 Hz

= 0.71 kHz (c) Express I peak in terms of the

I peak = ωQ0

charge stored in the capacitor: Express Q0 in terms of the capacitance of the capacitor and the potential difference across the capacitor:

Q0 = CV

Substituting for Q0 yields:

I peak = ωCV

Substitute numerical values and evaluate I peak :

I peak = 2π 712 s −1 (5.0 μF)(30 V )

(

)

= 0.67 A

A coil with internal resistance can be modeled as a resistor and an 32 •• ideal inductor in series. Assume that the coil has an internal resistance of 1.00 Ω and an inductance of 400 mH. A 2.00-μF capacitor is charged to 24.0 V and is then connected across coil. (a) What is the initial voltage across the coil? (b) How much energy is dissipated in the circuit before the oscillations die out? (c) What is the frequency of oscillation the circuit? (Assume the internal resistance is sufficiently small that has no impact on the frequency of the circuit.) (d) What is the quality factor of the circuit?

766

Chapter 29

Picture the Problem In Part (a) we can apply Kirchhoff’s loop rule to find the initial voltage across the coil. (b) The total energy lost via joule heating is the total energy initially stored in the capacitor. (c) The natural frequency of the circuit is given by f 0 = 1 2π LC . In Part (d) we can use its definition to find the quality

factor of the circuit. (a) Application of Kirchhoff’s loop rule leads us to conclude that the initial voltage across the coil is 24.0 V . (b) Because the ideal inductor can not dissipate energy as heat, all of the energy initially stored in the capacitor will be dissipated as joule heat in the resistor:

1 1 2 U = CV 2 = (2.00 μF)(24.0 V ) 2 2

(c) The natural frequency of the circuit is:

f0 =

= 0.576 mJ

1 2π LC

=

1 2π (400 mH )(2.00 μF)

= 178 Hz (d) The quality factor of the circuit is given by: Substituting for ω0 and simplifying yields:

Substitute numerical values and evaluate Q:

Q=

Q=

Q=

ω0 L R 1 L LC = 1 L R R C

1 400 mH = 447 1.00 Ω 2.00 μF

33 ••• [SSM] An inductor and a capacitor are connected, as shown in Figure 29-30. Initially, the switch is open, the left plate of the capacitor has charge Q0. The switch is then closed. (a) Plot both Q versus t and I versus t on the same graph, and explain how it can be seen from these two plots that the current leads the charge by 90º. (b) The expressions for the charge and for the current are given by Equations 29-38 and 29-39, respectively. Use trigonometry and algebra to show that the current leads the charge by 90º. Picture the Problem Let Q represent the instantaneous charge on the capacitor and apply Kirchhoff’s loop rule to obtain the differential equation for the circuit. We can then solve this equation to obtain an expression for the charge on the capacitor as a function of time and, by differentiating this expression with respect

Alternating-Current Circuits

767

to time, an expression for the current as a function of time. We’ll use a spreadsheet program to plot the graphs. Apply Kirchhoff’s loop rule to a clockwise loop just after the switch is closed:

Q dI +L =0 C dt

Because I = dQ dt :

d 2Q Q d 2Q 1 L 2 + = 0 or + Q=0 dt 2 LC C dt

The solution to this equation is:

Q(t ) = Q0 cos(ωt − δ ) where ω =

1 LC

Because Q(0) = Q0, δ = 0 and:

Q(t ) = Q0 cos ωt

The current in the circuit is the derivative of Q with respect to t:

I=

dQ d = [Q0 cos ωt ] = −ωQ0 sin ωt dt dt

(a) A spreadsheet program was used to plot the following graph showing both the charge on the capacitor and the current in the circuit as functions of time. L, C, and Q0 were all arbitrarily set equal to one to obtain these graphs. Note that the current leads the charge by one-fourth of a cycle or 90°. 1.2

1.2

Charge 0.6

0.0

0.0

-0.6

-0.6

I (mA)

Q (mC)

Current 0.6

-1.2

-1.2 0

2

4

6

8

10

t (s)

(b) The equation for the current is:

I = −ωQ0 sin ωt

(1)

768

Chapter 29

The sine and cosine functions are related through the identity:

π⎞ ⎛ − sin θ = cos⎜θ + ⎟ 2⎠ ⎝

Use this identity to rewrite equation (1):

π⎞ ⎛ I = −ωQ0 sin ωt = ωQ0 cos⎜ ωt + ⎟ 2⎠ ⎝ Thus, the current leads the charge by 90°.

Driven RL Circuits 34 •• A circuit consists of a resistor, an ideal 1.4-H inductor and an ideal 60-Hz generator, all connected in series. The rms voltage across the resistor is 30 V and the rms voltage across the inductor is 40 V. (a) What is the resistance of the resistor? (b) What is the peak emf of the generator? Picture the Problem We can express the ratio of VR to VL and solve this expression for the resistance R of the circuit. In (b) we can use the fact that, in an LR circuit, VL leads VR by 90° to find the ac input voltage.

(a) Express the potential differences across R and L in terms of the common current through these components:

VL = IX L = IωL and VR = IR

Divide the second of these equations by the first to obtain:

⎛V ⎞ VR IR R = = ⇒ R = ⎜⎜ R ⎟⎟ωL VL IωL ωL ⎝ VL ⎠

Substitute numerical values and evaluate R:

⎛ 30 V ⎞ ⎟⎟2π 60 s −1 (1.4 H ) = 0.40 kΩ R = ⎜⎜ ⎝ 40 V ⎠

(b) Because VR leads VL by 90° in an LR circuit:

Vpeak = 2Vrms = 2 VR2 + VL2

Substitute numerical values and evaluate Vpeak :

Vpeak = 2

(

)

(30 V )2 + (40 V )2 =

71 V

35 •• [SSM] A coil that has a resistance of 80.0 Ω has an impedance of 200 Ω when driven at a frequency of 1.00 kHz. What is the inductance of the coil? Picture the Problem We can solve the expression for the impedance in an LR circuit for the inductive reactance and then use the definition of XL to find L.

Alternating-Current Circuits Express the impedance of the coil in terms of its resistance and inductive reactance:

Z = R 2 + X L2

Solve for XL to obtain:

X L = Z 2 − R2

Express XL in terms of L:

X L = 2πfL

Equate these two expressions to obtain:

2πfL = Z 2 − R 2 ⇒ L =

Substitute numerical values and evaluate L:

L=

769

Z 2 − R2 2πf

(200 Ω )2 − (80.0 Ω )2 2π (1.00 kHz )

= 29.2 mH 36 •• A two conductor transmission line simultaneously carries a superposition of two voltage signals, so the potential difference between the two conductors is given by V = V1 + V2, where V1 = (10.0 V) cos(ω1t) and V2 = (10.0 V) cos(ω2t), where ω1 = 100 rad/s and ω2 = 10 000 rad/s. A 1.00 H inductor and a 1.00 kΩ shunt resistor are inserted into the transmission line as shown in Figure 29-31. (Assume that the output is connected to a load that draws only an insignificant amount of current.) (a) What is the voltage (Vout) at the output of the transmission line? (b) What is the ratio of the low-frequency amplitude to the high-frequency amplitude at the output? Picture the Problem We can express the two output voltage signals as the product of the current from each source and R = 1.00 kΩ. We can find the currents due to each source using the given voltage signals and the definition of the impedance for each of them.

(a) Express the voltage signals observed at the output side of the transmission line in terms of the potential difference across the resistor:

V1, out = I1 R and V2, out = I 2 R

770

Chapter 29

Evaluate I1 and I2: I1 =

V1 = Z1

(10.0 V )cos100t = (9.95 mA )cos100t 2 2 -1 [ ] (1.00 kΩ ) + (100 s )(1.00 H )

I2 =

V2 = Z2

(10.0 V )cos10 4 t = (0.995 mA )cos10 4 t 2 2 4 −1 (1.00 kΩ ) + [(10 s )(1.00 H )]

and

Substitute for I1 and I2 to obtain:

V1, out = (1.00 kΩ )(9.95 mA )cos100t = (9.95 V )cos100t where ω1 = 100 rad/s and ω2 = 10 000 rad/s. and V2, out = (1.00 kΩ )(0.995 mA )cos10 4 t

(b) Express the ratio of V1,out to V2,out:

V1, out V2, out

=

(0.995 V )cos10 4 t

=

9.95 V = 10 : 1 0.995 V

37 •• A coil is connected to a 120-V rms, 60-Hz line. The average power supplied to the coil is 60 W, and the rms current is 1.5 A. Find (a) the power factor, (b) the resistance of the coil, and (c) the inductance of the coil. (d) Does the current lag or lead the voltage? Explain your answer. (e) Support your answer to Part (d) by determining the phase angle. Picture the Problem The average power supplied to coil is related to the power 2 R to find R. Because the factor by Pav = ε rms I rms cos δ . In (b) we can use Pav = I rms

inductance L is related to the resistance R and the phase angle δ according to X L = ωL = R tan δ , we can use this relationship to find the resistance of the coil. Finally, we can decide whether the current leads or lags the voltage by noting that the circuit is inductive. (a) Express the average power supplied to the coil in terms of the power factor of the circuit:

Pav = ε rms I rms cos δ ⇒ cos δ =

Substitute numerical values and evaluate cosδ:

cos δ =

Pav ε rms I rms

60 W = 0.333 = 0.33 (120 V )(1.5 A )

Alternating-Current Circuits (b) Express the power supplied by the source in terms of the resistance of the coil:

Pav 2 I rms

2 Pav = I rms R⇒R =

60 W = 26.7 Ω = 27 Ω (1.5 A )2

Substitute numerical values and evaluate R:

R=

(c) Relate the inductive reactance to the resistance and phase angle:

X L = ωL = R tan δ

Solving for L yields:

L=

771

R tan δ

ω

=

[

]

R tan cos −1 (0.333) 2πf

(26.7 Ω ) tan(70.5°) =

Substitute numerical values and evaluate L:

L=

(d) Evaluate XL:

X L = (26.7 Ω ) tan (70.5°) = 75.4 Ω

(

2π 60 s −1

)

0.20 H

Because the circuit is inductive, the current lags the voltage.

δ = cos −1 (0.333) = 71°

(e) From Part (a): 38

••

A 36-mH inductor that has a resistance of 40 Ω is connected to an

ideal ac voltage source whose output is given by ε = (345 V) cos(150πt), where t is in seconds. Determine (a) the peak current in the circuit, (b) the peak and rms voltages across the inductor, (c) the average power dissipation, and (d) the peak and average magnetic energy stored in the inductor. Picture

the

Problem

(a)

We

can

use

I peak = ε peak

R 2 + (ωL ) and 2

VL , peak = I peak X L = ωLI peak to find the peak current in the circuit and the peak voltage across the inductor. (b) Once we’ve found VL , peak we can find VL , rms using VL , rms = VL , peak

2 2 . (c) We can use Pav = 12 I rms R to find the average power

2 to find the peak and average magnetic energy dissipation, and (d) U L , peak = 12 LI peak

stored in the inductor. The average energy stored in the magnetic field of the inductor can be found using U L ,av = ∫ Pav dt .

772

Chapter 29

(a) Apply Kirchhoff’s loop rule to the circuit to obtain: Substitute numerical values and evaluate I:

ε − IZ = 0 ⇒ I = ε

=

Z

ε R 2 + (ωL )

2

(345 V )cos(150πt ) (40 Ω )2 + [(150π s −1 )(36 mH )]2 = (7.94 A ) cos(150πt )

I=

and I peak = 7.9 A . (b) Because ε = (345 V) cos(150πt) : Find VL ,rms from VL , peak :

V L , peak = 345 V VL , rms =

VL , peak 2

=

345 V = 244 V 2 2

(c) Relate the average power dissipation to Ipeak and R:

⎛I ⎞ 2 Pav = I R = ⎜⎜ peak ⎟⎟ R = 12 I peak R ⎝ 2 ⎠

Substitute numerical values and evaluate Pav :

Pav =

(d) The maximum energy stored in the magnetic field of the inductor is:

2 U L , peak = 12 LI peak =

The definition of U L ,av is:

U(t) is given by:

Substitute for U(t) to obtain:

2 rms

1 2

(7.94 A )2 (40 Ω ) = 1 2

1.3 kW

(36 mH )(7.94 A )2

= 1.1 J T

U L ,av

1 = ∫ U (t )dt T0

U (t ) =

1 2 L[I (t )] 2 L 2T

T

U L ,av =

1 2 ⎡1 2 ⎤ ⎢⎣ 2 I peak ⎥⎦T = 4 LI peak

∫ [I (t )] dt 2

0

Evaluating the integral yields:

U L ,av =

L 2T

Substitute numerical values and evaluate U L ,av :

U L ,av =

1 (36 mH )(7.94 A )2 = 0.57 J 4

Alternating-Current Circuits

773

39 •• [SSM] A coil that has a resistance R and an inductance L has a power factor equal to 0.866 when driven at a frequency of 60 Hz. What is the coil’s power factor it is driven at 240 Hz? Picture the Problem We can use the definition of the power factor to find the relationship between XL and R when the coil is driven at a frequency of 60 Hz and then use the definition of XL to relate the inductive reactance at 240 Hz to the inductive reactance at 60 Hz. We can then use the definition of the power factor to determine its value at 240 Hz.

Using the definition of the power factor, relate R and XL:

cos δ =

Square both sides of the equation to obtain:

cos 2 δ =

R = Z

R

(1)

R + X L2 2

R2 R 2 + X L2

Solve for X L2 (60 Hz ) :

⎞ ⎛ 1 X L2 (60 Hz ) = R 2 ⎜ − 1⎟ 2 ⎠ ⎝ cos δ

Substitute for cosδ and simplify to obtain:

⎛ ⎞ 1 − 1⎟⎟ = 13 R 2 X L2 (60 Hz ) = R 2 ⎜⎜ 2 ⎝ (0.866 ) ⎠

Use the definition of XL to obtain:

X L2 ( f ) = 4πf 2 L2 and X L2 ( f' ) = 4πf' 2 L2

Dividing the second of these equations by the first and simplifying yields:

X L2 ( f' ) 4πf' 2 L2 f' 2 = = 2 X L2 ( f ) 4πf 2 L2 f or 2

⎛ f' ⎞ X ( f' ) = ⎜⎜ ⎟⎟ X L2 ( f ) ⎝ f ⎠ 2 L

Substitute numerical values to obtain:

2

⎛ 240 s −1 ⎞ 2 ⎟ X L (60 Hz ) X (240 Hz ) = ⎜⎜ −1 ⎟ ⎝ 60 s ⎠ ⎛ 1 ⎞ 16 = 16⎜ R 2 ⎟ = R 2 ⎝3 ⎠ 3 2 L

Chapter 29

774

Substitute in equation (1) to obtain:

(cos δ )240 Hz

R

=

R2 +

16 2 R 3

=

3 19

= 0.397 40

••

A resistor and an inductor are connected in parallel across an ideal ac

voltage source whose output is given by ε = εpeakcosωt as shown in Figure 29-32. Show that (a) the current in the resistor is given by IR = εpeak/R cos ωt, (b) the

current in the inductor is given by IL = εpeak/XL cos(ωt – 90º), and (c) the current in the voltage source is given by I = IR + IL = Ipeak cos(ωt – δ), where

Ipeak = εmax/Z.

Picture the Problem Because the resistor and the inductor are connected in parallel, the voltage drops across them are equal. Also, the total current is the sum of the current through the resistor and the current through the inductor. Because these two currents are not in phase, we’ll need to use phasors to calculate their sum. The amplitudes of the applied voltage and the currents are equal to the magnitude r r r r of the phasors. That is ε = ε peak , I = I peak , I R = I R , peak , and I L = I L , peak .

(a) The ac source applies a voltage given by ε = ε peak cos ωt . Thus, the

ε peak cos ωt = I R R

voltage drop across both the load resistor and the inductor is: The current in the resistor is in phase with the applied voltage: Because I R , peak =

ε peak R

:

(b) The current in the inductor lags the applied voltage by 90°: Because I L , peak =

ε peak XL

:

(c) The net current I is the sum of the currents through the parallel branches:

I R = I R , peak cos ωt

IR =

ε peak R

cos ωt

I L = I L , peak cos(ωt − 90°)

IL =

ε peak XL

I = IR + IL

cos(ωt − 90°)

Alternating-Current Circuits

r

Draw the phasor diagram for the circuit. The projections of the phasors onto the horizontal axis are the instantaneous values. The current in the resistor is in phase with the applied voltage, and the current in the inductor lags the applied voltage by 90°. The net current phasor is the sum of the r r r branch current phasors I = I L + I R .

(

Z is the impedance of the combination: From the phasor diagram we have:

ε

r IR

r I

δ

ωt −δ

ωt

90°− ω t r IL

)

The peak current through the parallel combination is equal to ε peak Z , where

775

I = I peak cos(ωt − δ ) ,

where I peak =

ε peak Z

2 = I R2 , peak + I L2, peak I peak

⎛ε ⎞ ⎛ε ⎞ = ⎜⎜ peak ⎟⎟ + ⎜⎜ peak ⎟⎟ ⎝ R ⎠ ⎝ XL ⎠ 2

2

2 ⎛ 1 1 ⎞ ε peak = ε ⎜⎜ 2 + 2 ⎟⎟ = 2 XL ⎠ Z ⎝R 1 1 1 where 2 = 2 + 2 Z R XL . 2 peak

Solving for Ipeak yields:

From the phasor diagram:

I peak =

ε peak Z

where Z −2 = R −2 + X L−2

I = I peak cos(ωt − δ where

tan δ =

)

ε peak I L , peak I R , peak

=

XL

ε peak

=

R XL

R

41 •• [SSM] Figure 29-33 shows a load resistor that has a resistance of RL = 20.0 Ω connected to a high-pass filter consisting of an inductor that has inductance L = 3.20-mH and a resistor that has resistance R = 4.00-Ω. The output

of the ideal ac generator is given by ε = (100 V) cos(2πft). Find the rms currents in all three branches of the circuit if the driving frequency is (a) 500 Hz and (b) 2000 Hz. Find the fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is (c) 500 Hz and (d) 2000 Hz.

776

Chapter 29

Picture the Problem ε = V1 + V2 , where V1 is the voltage drop across R and V2 is r r r the voltage drop across the parallel combination of L and RL. ε = V1 + V2 is the r r r relation for the phasors. For the parallel combination I = I RL + I L . Also, V1 is in

phase with I and V2 is in phase with I RL . First draw the phasor diagram for the currents in the parallel combination, then add the phasors for the voltages to the diagram. r I RL

The phasor diagram for the currents in the circuit is:

r I

δ

r IL r

Adding the voltage phasors to the diagram gives:

ε r V2

r I RL

δ r I

ωt

δ

r V1

r IL

The maximum current in the inductor, I2, peak, is given by:

I 2, peak =

V2, peak

(1)

Z2

where Z 2−2 = RL−2 + X L−2 tan δ is given by:

tan δ = =

Solve for δ to obtain:

I L , peak I R , peak

=

(2)

V2, peak X L V2, peak RL

RL R R = L = L X L ωL 2πfL ⎛ RL ⎞ ⎟⎟ ⎝ 2πfL ⎠

δ = tan −1 ⎜⎜

(3)

Alternating-Current Circuits Apply the law of cosines to the triangle formed by the voltage phasors to obtain: 2 ε peak = V1,2peak + V22, peak + 2V1, peakV2, peak cos δ

or 2 2 2 I peak Z 2 = I peak R 2 + I peak Z 22 + 2 I peak RI peak Z 2 cos δ

Dividing out the current squared yields:

Z 2 = R 2 + Z 22 + 2 RZ 2 cos δ

Solving for Z yields:

Z = R 2 + Z 22 + 2 RZ 2 cos δ

The maximum current I peak in the circuit is given by: Irms is related to I peak according to: (a) Substitute numerical values in equation (3) and evaluate δ :

I peak =

I rms =

ε peak

(4)

(5)

Z 1 I peak 2

(6)

⎛

⎞ 20.0 Ω ⎟⎟ ⎝ 2π (500 Hz )(3.20 mH ) ⎠

δ = tan −1 ⎜⎜

⎛ 20.0 Ω ⎞ ⎟⎟ = 63.31° = tan −1 ⎜⎜ ⎝ 10.053 Ω ⎠

Solving equation (2) for Z2 yields:

Substitute numerical values and evaluate Z2:

Z2 =

Z2 =

1 −2 L

R + X L−2 1

(20.0 Ω )−2 + (10.053 Ω )−2

= 8.982 Ω

Substitute numerical values and evaluate Z: Z=

(4.00 Ω )2 + (8.982 Ω )2 + 2(4.00 Ω )(8.982 Ω )cos 63.31° = 11.36 Ω

Substitute numerical values in equation (5) and evaluate I peak :

I peak =

100 V = 8.806 A 11.36 Ω

777

778

Chapter 29

Substitute for I peak in equation (6) and evaluate I rms : The maximum and rms values of V2 are given by:

1 (8.806 A ) = 6.23 A 2

I rms =

V2, peak = I peak Z 2

= (8.806 A )(8.982 Ω ) = 79.095 V

and 1 V2, peak 2 1 (79.095 V ) = 55.929 V = 2

V2,rms =

The rms values of I RL ,rms and

I RL ,rms =

I L ,rms are:

V2,rms RL

=

55.929 V = 2.80 A 20.0 Ω

and I L ,rms =

V2,rms 55.929 V = = 5.56 A XL 10.053 Ω

X L = 40.2 Ω , δ = 26.4° , Z 2 = 17.9 Ω ,

(b) Proceed as in (a) with f = 2000 Hz to obtain:

Z = 21.6 Ω , I peak = 4.64 A , and I rms = 3.28 A ,

V2,max = 83.0V , V2, rms = 58.7 V , I RL ,rms = 2.94 A , and I L ,rms = 1.46 A

(c) The power delivered by the ac source equals the sum of the power dissipated in the two resistors. The fraction of the total power delivered by the source that is dissipated in load resistor is given by: −1

⎛ ⎛ P ⎞ I2 R ⎞ = ⎜1 + R ⎟ = ⎜1 + 2 rms ⎟ ⎜ I R ,rms RL ⎟ PRL + PR ⎜⎝ PRL ⎟⎠ L ⎠ ⎝ PRL

−1

Substitute numerical values for f = 500 Hz to obtain: −1

PRL PRL + PR

f =500 Hz

⎛ (6.23 A )2 (4.00 Ω ) ⎞ ⎟ = 0.502 = 50.2% = ⎜⎜1 + 2 ⎟ ( ) ( ) Ω 2 . 80 A 20 . 0 ⎝ ⎠

Alternating-Current Circuits

779

(d) Substitute numerical values for f = 2000 Hz to obtain: −1

PRL PRL + PR 42

••

f = 2000 Hz

⎛ (3.28 A )2 (4.00 Ω ) ⎞ ⎟ = 0.800 = 80.0% = ⎜⎜1 + 2 ⎟ ( ) ( ) Ω 2 . 94 A 20 . 0 ⎝ ⎠

An ideal ac voltage source whose emf ε1 is given by (20 V) cos(2πft)

and an ideal battery whose emf ε2 is 16 V are connected to a combination of two resistors and an inductor (Figure 29-34), where R1 = 10 Ω, R2 = 8.0 Ω, and L = 6.0 mH. Find the average power delivered to each resistor if (a) the driving frequency is 100 Hz, (b) the driving frequency is 200 Hz, and (c) the driving frequency is 800 Hz. Picture the Problem We can treat the ac and dc components separately. For the

dc component, L acts like a short circuit. Let ε1, peak denote the peak value of the voltage supplied by the ac voltage source. We can use P = ε 22 R to find the power dissipated in the resistors by the current from the ideal battery. We’ll apply Kirchhoff’s loop rule to the loop including L, R1, and R2 to derive an expression for the average power delivered to each resistor by the ac voltage source. (a) The total power delivered to R1 and R2 is:

The dc power delivered to the resistors whose resistances are R1 and R2 is: Express the average ac power delivered to R1: Apply Kirchhoff’s loop rule to a clockwise loop that includes R1, L, and R2: Solving for I2 yields:

P1 = P1, dc + P1, ac

(1)

and P2 = P2, dc + P2, ac

(2)

P1,dc =

P1, ac =

ε 22 and P

2 ,dc

R1

=

ε 22 R2

ε12, rms ε12, peak R1

=

2R1

R1 I1 − Z 2 I 2 = 0

I2 =

R1 R I1 = 1 Z2 Z2

ε1, peak ε1, peak R1

=

Z2

780

Chapter 29 P2, ac

Substituting in equations (1) and (2) yields:

P1 = and P2 =

Substitute numerical values and evaluate P1:

⎞ ⎛ε R2 = ⎜⎜ 1, peak ⎟⎟ R2 = I ⎝ Z2 ⎠ ε2 R = 1, peak2 2 2Z 2 2

Express the average ac power delivered to R2:

1 2

2 2 , rms

1 2

ε 22 + ε12, peak R1

2R1

ε 22 + ε12, peak R2 R2

2Z 22

2 2 ( ( 16 V ) 20 V ) P1 = + = 10 Ω 2(10 Ω )

46 W

Substitute numerical values and evaluate P2: P2 =

(16 V )2 8.0 Ω

+

(20 V )2 (8.0 Ω ) 2 2 2 (8.0 Ω ) + (2π {100 s −1 }{6.0 mH})

[

(b) Proceed as in (a) to evaluate P1 and P2 with f = 200 Hz:

(c) Proceed as in (a) to evaluate P1 and P2 with f = 800 Hz:

]=

52 W

P1 = 25.6 W + 20.0 W = 46 W P2 = 32.0 W + 13.2 W = 45 W P1 = 25.6 W + 20.0 W = 46 W P2 = 32.0 W + 1.64 W = 34 W

43 •• An ac circuit contains a resistor and an ideal inductor connected in series. The voltage rms drop across this series combination is 100-V and the rms voltage drop across the inductor alone is 80 V. What is the rms voltage drop across the resistor? Picture the Problem We can use the phasor diagram for an RL circuit to find the voltage across the resistor.

Alternating-Current Circuits

781

r VL

The phasor diagram for the voltages in the circuit is shown to the right:

ε

rms

r VR

Use the Pythagorean theorem to express VR:

VR =

2 ε rms − VL2

Substitute numerical values and evaluate VR:

VR =

(100 V ) 2 − (80 V ) 2

= 60 V

Filters and Rectifiers 44 •• The circuit shown in Figure 29-35 is called an RC high-pass filter because it transmits input voltage signals that have high frequencies with greater amplitude than it transmits input voltage signals that have low frequencies. If the input voltage is given by Vin = Vin peak cos ωt, show that the output voltage is

Vout = VH cos(ωt – δ) where VH = Vin peak

1 + (ωRC ) . (Assume that the output is −2

connected to a load that draws only an insignificant amount of current.) Show that this result justifies calling this circuit a high-pass filter. Picture the Problem The phasor diagram for the RC high-pass filter is r r shown to the right. Vapp and VR are the

phasors for Vin and Vout, respectively. Note that tan δ = − X C R. That δ is r negative follows from the fact that Vapp r lags VR by δ . The projection of r Vapp onto the horizontal axis is r Vapp = Vin, and the projection of VR onto the horizontal axis is VR = Vout. Express Vapp :

r VR

ωt − δ δ

r Vapp

ωt

r VC

Vapp = Vapp,peak cos ωt where Vapp,peak = Vpeak = I peak Z and Z 2 = R 2 + X C2

Because δ < 0:

ωt + δ = ωt − δ

(1)

782

Chapter 29

VR is given by:

VR = VR , peak cos(ωt − δ ) where VR , peak = VH = I peak R

Solving equation (1) for Z and substituting for XC yields:

⎛ 1 ⎞ Z = R +⎜ ⎟ ⎝ ωC ⎠

Because Vout = VR :

Vout = VR , peak cos(ωt − δ )

Simplify further to obtain:

(2)

= I in peak R cos(ωt − δ ) =

Using equation (2) to substitute for Z yields:

2

2

Vout =

Vout =

Vin peak Z

R cos(ωt − δ )

Vin peak ⎛ 1 ⎞ R2 + ⎜ ⎟ ⎝ ωC ⎠

2

Vin peak 1 + (ωRC )

−2

R cos(ωt − δ )

cos(ωt − δ )

or Vout = VH cos(ωt − δ ) where VH = As ω → ∞:

VH →

Vin peak 1 + (ωRC )

−2

Vin peak

= Vin peak showing that

1 + (0) the result is consistent with the highpass name for this circuit. 2

45 •• (a) Find an expression for the phase constant δ in Problem 44 in terms of ω, R and C. (b) What is the value of δ in the limit that ω → 0? (c) What is the value of δ in the limit that ω → ∞? (d) Explain your answers to Parts (b) and (c).

Alternating-Current Circuits r VR

Picture the Problem The phasor diagram for the RC high-pass filter is r r shown below. Vapp and VR are the

r Vapp

phasors for Vin and Vout, respectively. r The projection of Vapp onto the

δ

horizontal axis is Vapp = Vin, and the r projection of VR onto the horizontal axis is VR = Vout. r r (a) Because Vapp lags VR by δ .

Use the definition of XC to obtain:

Solving for δ yields:

783

ωt

r VC

tan δ = −

VC IX X =− C =− C VR IR R

1 1 tan δ = − ωC = − R ωRC ⎡

1 ⎤

δ = tan −1 ⎢− ⎣ ωRC ⎥⎦

(b) As ω → 0:

δ → − 90°

(c) As ω → ∞:

δ→ 0

r r (d) For very low driving frequencies, X C >> R and so VC effectively lags Vin by r 90°. For very high driving frequencies, X C ω0.

ω 2 ≈ ω0 +

Evaluating Δω = ω2 − ω1 yields:

Δω ≈ ω 0 +

R ω0 = L Q

From the definition of Q:

Substitute in the expression for Δω to obtain:

77

•••

R ⎛ R ⎞ R − ⎜ ω0 − ⎟= 2L ⎝ 2L ⎠ L

Δω ≈

ω0 Q

⇒Q ≈

ω0 Δω

d2 Q dQ 1 + Q =0 2 +R dt C dt cos ω ' t , where τ = 2 L R ,

Show by direct substitution that L

(Equation 29-43b) is satisfied by Q = Q0e − t τ

ω ' = 1 (LC ) − 1 τ 2 , and Q0 is the charge on the capacitor at t = 0. Picture the Problem We’ll differentiate Q = Q0e − t τ cos ω ' t twice and substitute

this function and both its derivatives in the differential equation of the circuit. Rewriting the resulting equation in the form Acosω′t + Bsinω′t = 0 will reveal that B vanishes. Requiring that Acosω′t = 0 hold for all values of t will lead to the

result that ω ' = 1 (LC ) − 1 τ 2 . Equation 29-43b is:

L

d 2Q dQ 1 +R + Q=0 2 dt C dt

Q = Q0 e −t τ cos ω ' t

Assume a solution of the form: Differentiate Q(t) twice to obtain:

[

]

dQ d −t τ e cos ω ' t = Q0 dt dt d d ⎞ ⎛ = Q0 ⎜ e −t τ cos ω ' t + cos ω ' t e −t τ ⎟ dt dt ⎠ ⎝ 1 ⎞ ⎛ = Q0e −t τ ⎜ − ω ' sin ω ' t − cos ω ' t ⎟ τ ⎠ ⎝

Alternating-Current Circuits

823

and d 2Q d ⎡ 1 ⎞⎤ ⎛ = Q0 ⎢e − t τ ⎜ − ω ' sin ω ' t − cos ω ' t ⎟⎥ 2 dt dt ⎣ τ ⎠⎦ ⎝ ⎡⎛ 1 2ω' ⎤ ⎞ sin ω't ⎥ = Q0e − t τ ⎢⎜ 2 − ω' 2 ⎟ cos ω't + τ ⎦ ⎠ ⎣⎝ τ

Substitute these derivatives in the differential equation and simplify to obtain: ⎡⎛ 1 2ω' 1 ⎤ ⎛ ⎞ ⎞ LQ0e −t τ ⎢⎜ 2 − ω' 2 ⎟ cos ω't + sin ω't ⎥ + RQ0e −t τ ⎜ − ω ' sin ω ' t − cos ω ' t ⎟ τ τ ⎦ ⎝ ⎠ ⎠ ⎣⎝ τ 1 + Q0e − t τ cos ω ' t = 0 C Because Q0 and e −t τ are never zero, divide them out of the equation and simplify to obtain: 2 Lω' 1 R ⎛1 ⎞ L⎜ 2 − ω' 2 ⎟ cos ω't + sin ω't − ω ' R sin ω ' t − cos ω ' t + cos ω ' t = 0 τ τ C ⎝τ ⎠ Rewriting this equation in the form Acosω′t + Bsinω′t = 0 yields:

(Rω' − Rω' )sin ω't + ⎡⎢ L⎛⎜ 12 − ω' 2 ⎞⎟ + 1 − R ⎤⎥ cos ω't = 0 ⎣ ⎝τ

⎠ C

τ⎦

or ⎡ ⎛1 1 R⎤ 2⎞ ⎢ L⎜ τ 2 − ω' ⎟ + C − τ ⎥ cos ω't = 0 ⎠ ⎦ ⎣ ⎝

If this equation is to hold for all values of t, its coefficient must vanish: Solving for ω′ yields:

⎛1 ⎞ 1 R L⎜ 2 − ω' 2 ⎟ + − = 0 ⎝τ ⎠ C τ

ω' =

1 ⎛ 1 ⎞ −⎜ ⎟ LC ⎝ 2L ⎠

2

,

the condition that must be satisfied if Q = Q0e − t τ cos ω't is the solution to Equation 29-43b. 78 ••• One method for measuring the magnetic susceptibility of a sample uses an LC circuit consisting of an air-core solenoid and a capacitor. The resonant frequency of the circuit without the sample is determined and then measured

Chapter 29

824

again with the sample inserted in the solenoid. Suppose you have a solenoid that is 4.00 cm long, 3.00 mm in diameter, and has 400 turns of fine wire. You have a sample that is inserted in the solenoid and completely fills the air space. Neglect end effects. (a) Calculate the inductance of your empty solenoid. (b) What value for the capacitance of the capacitor should you choose that the resonance frequency of the circuit without a sample is exactly 6.0000 MHz? (c) When a sample is inserted in the solenoid, you determine that the resonance frequency drops to 5.9989 MHz. Use your data to determine the sample’s susceptibility. Picture the Problem We can use L = μ 0 n 2 Al to determine the inductance of the

empty solenoid and the resonance condition to find the capacitance of the samplefree circuit when the resonance frequency of the circuit is 6.0000 MHz. By expressing L as a function of f0 and then evaluating df0/dL and approximating the derivative with Δf0/ΔL , we can evaluate χ from its definition. (a) Express the inductance of an aircore solenoid:

L = μ 0 n 2 Al

Substitute numerical values and evaluate L: 2

⎛ 400 ⎞ π ⎟⎟ (3.00 cm )2 (4.00 cm ) = 3.553 mH = 3.55 mH L = 4π × 10 N/A ⎜⎜ ⎝ 4.00 cm ⎠ 4

(

−7

2

)

1 2πf 0C

(b) Express the condition for resonance in the LC circuit:

X L = X C ⇒ 2πf 0 L =

Solving for C yields:

C=

1 4π f 02 L

C=

1 2 4π (6.0000 MHz ) (3.553 mH )

Substitute numerical values and evaluate C:

(1)

2

2

= 1.9803 ×10 −13 F = 0.198 pF

(c) Express the sample’s susceptibility in terms of L and ΔL:

χ=

Solve equation (1) for f0:

f0 =

ΔL L

(2)

1 2π LC

Alternating-Current Circuits Differentiate f0 with respect to L:

df 0 d −1 2 1 1 L =− L−3 2 = dL 2π C dL 4π C f 1 =− =− 0 2L 4πL LC

Approximate df0/dL by Δf0/ΔL:

f Δf 0 Δf ΔL = − 0 or 0 = − 2L f0 2L ΔL

Substitute in equation (2) to obtain:

χ = −2

Substitute numerical values and evaluate χ:

χ = −2⎜⎜

825

Δf 0 f0

⎛ 5.9989 MHz − 6.0000 MHz ⎞ ⎟⎟ 6.0000 MHz 0 ⎠ ⎝

= 3.7 × 10− 4

The Transformer 79 • [SSM] A rms voltage of 24 V is required for a device whose impedance is 12 Ω. (a) What should the turns ratio of a transformer be, so that the device can be operated from a 120-V line? (b) Suppose the transformer is accidentally connected in reverse with the secondary winding across the 120-V-rms line and the 12-Ω load across the primary. How much rms current will then be in the primary winding? Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. We can use V2 N1 = V1 N 2 and N1 I1 = N 2 I 2 to find the turns ratio and the primary current when the transformer connections are reversed.

(a) Relate the number of primary and secondary turns to the primary and secondary voltages:

V2, rms N1 = V1, rms N 2

Solve for and evaluate the ratio N2/N1:

N 2 V2, rms 24 V 1 = = = 5 N1 V1, rms 120 V

(b) Relate the current in the primary to the current in the secondary and to the turns ratio:

I1, rms =

N2 I 2, rms N1

(1)

826

Chapter 29

Express the current in the primary winding in terms of the voltage across it and its impedance: Substitute for I2, rms to obtain:

Substitute numerical values and evaluate I1, rms:

I 2, rms =

V2, rms

I1, rms =

N 2 V2, rms N1 Z 2

Z2

⎛ 5 ⎞⎛ 120 V ⎞ ⎟⎟ = 50 A I1 = ⎜ ⎟⎜⎜ ⎝ 1 ⎠⎝ 12 Ω ⎠

80 • A transformer has 400 turns in the primary and 8 turns in the secondary. (a) Is this a step-up or a step-down transformer? (b) If the primary is connected to a 120 V rms voltage source, what is the open-circuit rms voltage across the secondary? (c) If the primary rms current is 0.100 A, what is the secondary rms current, assuming negligible magnetization current and no power loss? Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. We can decide whether the transformer is a step-up or step-down transformer by examining the ratio of the number of turns in the secondary to the number of terms in the primary. We can relate the open-circuit rms voltage in the secondary to the primary rms voltage and the turns ratio.

(a) Because there are fewer turns in the secondary than in the primary it is a stepdown transformer. (b) Relate the open-circuit rms voltage V2, rms in the secondary to the rms voltage V1, rms in the primary:

V2, rms =

N2 V1, rms N1

Substitute numerical values and evaluate V2, rms :

V2, rms =

8 (120 V ) = 2.40 V 400

(c) Because there are no power losses:

V1, rms I1, rms = V2, rms I 2, rms

and I 2, rms =

Substitute numerical values and evaluate I2, rms:

I 2, rms =

V1, rms V2, rms

I1, rms

120 V (0.100 A ) = 5.00 A 2.40 V

Alternating-Current Circuits

827

81 • The primary of a step-down transformer has 250 turns and is connected to a 120-V rms line. The secondary is to supply 20 A rms at 9.0 V rms. Find (a) the rms current in the primary and (b) the number of turns in the secondary, assuming 100 percent efficiency. Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. We can use I1, rmsV1, rms = I 2, rmsV2, rms to find the current in the primary

and V2, rms N1 = V1, rms N 2 to find the number of turns in the secondary. (a) Because we have 100 percent efficiency:

I1, rmsV1, rms = I 2, rmsV2, rms

and I1, rms = I 2, rms

V2, rms V1, rms

9.0 V = 1.5 A 120 V

Substitute numerical values and evaluate I1, rms:

I1, rms = (20 A )

(b) Relate the number of primary and secondary turns to the primary and secondary voltages:

V2, rms N1 = V1, rms N 2 ⇒ N 2 =

Substitute numerical values and evaluate N2/N1:

N2 =

V2, rms V1, rms

N1

9.0 V (250) ≈ 19 120 V

82 •• An audio oscillator (ac source) that has an internal resistance of 2000 Ω and an open-circuit rms output voltage of 12.0 V is to be used to drive a loudspeaker coil that has a resistance of 8.00 Ω. (a) What should be the ratio of primary to secondary turns of a transformer, so that maximum average power is transferred to the speaker? (b) Suppose a second identical speaker is connected in parallel with the first speaker. How much average power is then supplied to the two speakers combined? Picture the Problem Note: In a simple circuit maximum power transfer from source to load requires that the load resistance equals the internal resistance of the source. We can use Ohm’s law and the relationship between the primary and secondary currents and the primary and secondary voltages and the turns ratio of the transformer to derive an expression for the turns ratio as a function of the effective resistance of the circuit and the resistance of the speaker(s).

(a) Express the effective loudspeaker resistance at the primary of the transformer:

Reff =

V1, rms I1, rms

828

Chapter 29

Relate V1, rms to V2, rms, N1, and N2:

V1, rms = V2, rms

N1 N2

Express I1, rms in terms of I2, rms, N1, and N2:

I1, rms = I 2, rms

N2 N1

Substitute for V1, rms and I1, rms and simplify to obtain:

V2, rms Reff = I 2, rms

Solve for N1/N2:

N1 = N2

N1 2 N 2 ⎛⎜ V2, rms ⎞⎟ ⎛ N1 ⎞ ⎟ ⎜ = N 2 ⎜⎝ I 2, rms ⎟⎠ ⎜⎝ N 2 ⎟⎠ N1

I 2, rms Reff V2, rms

=

Reff R2

(1)

Evaluate N1/N2 for Reff = Rcoil:

N1 2000 Ω = = 15.811 = 15.8 8.00 Ω N2

(b) Express the power delivered to the two speakers connected in parallel:

Psp = I12, rms Reff

Find the equivalent resistance Rsp of

1 1 1 ⇒ Rsp = 4.00 Ω = + Rsp 8.00 Ω 8.00 Ω

the two 8.00-Ω speakers in parallel: Solve equation (1) for Reff to obtain:

Reff

⎛N ⎞ = R2 ⎜⎜ 1 ⎟⎟ ⎝ N2 ⎠

(2)

2

Substitute numerical values and evaluate Reff :

Reff = (4.00 Ω )(15.811) = 1000 Ω

Find the current supplied by the source:

I1, rms =

2

Vrms 12.0 V = Rtot 2000 Ω + 1000 Ω

= 4.00 mA

Substitute numerical values in equation (2) and evaluate the power delivered to the parallel speakers:

Psp = (4.00 mA ) (1000 Ω ) = 16.0 mW 2

Alternating-Current Circuits

829

General Problems 83 • The distribution circuit of a residential power line is operated at 2000 V rms. This voltage must be reduced to 240 V rms for use within residences. If the secondary side of the transformer has 400 turns, how many turns are in the primary? Picture the Problem We can relate the input and output voltages to the number of turns in the primary and secondary using V2, rms N1 = V1, rms N 2 .

Relate the output voltages V2, rms to the input voltage V1, rms and the number of turns in the primary N1 and secondary N2:

V2, rms =

Substitute numerical values and evaluate N1:

⎛ 2000 V ⎞ ⎟⎟ = 3.33 × 10 3 N1 = (400 )⎜⎜ ⎝ 240 V ⎠

V N2 V1, rms ⇒ N1 = N 2 1, rms V2, rms N1

84 •• A resistor that has a resistance R carries a current given by (5.0 A) sin 2πft + (7.0 A) sin 4πft, where f = 60 Hz. (a) What is the rms current in the resistor? (b) If R =12 Ω, what is the average power delivered to the resistor? (c) What is the rms voltage across the resistor?

(I ) to relate the rms current to the current carried by the resistor and find (I ) by integrating I2.

Picture the Problem We can use its definition, I rms =

2

av

2

av

(a) Express the rms current in terms of the (I 2 )av :

I rms =

(I ) 2

av

Evaluate I2: I 2 = [(5.0 A ) sin 2πft + (7.0 A ) sin 4πft ]

(

)

(

2

)

(

)

= 25 A 2 sin 2 2πft + 70 A 2 sin 2πft sin 4πft + 49 A 2 sin 2 4πft

Find (I 2 )av by integrating I2 from t = 0 to t = T = 2π/ω and dividing by T:

(I )av = 2ωπ 2

2π ω

∫ {(25 A )sin 2

0

(

)

2

2πft + (70 A 2 )sin 2πft sin 4πft

}

+ 49 A 2 sin 2 4πft dt

830

Chapter 29

Use the trigonometric identity sin 2 x = 12 (1 − cos 2 x ) to simplify and

(I ) 2

av

= 12.5 A 2 + 0 + 24.5 A 2 = 37.0 A 2

evaluate the 1st and 3rd integrals and recognize that the middle term is of the form sinxsin2x to obtain:

( )

Substitute for I 2

av

and evaluate Irms:

I rms = 37.0 A 2 = 6.1 A

(b) Relate the power dissipated in the resistor to its resistance and the rms current in it:

2 P = I rms R

Substitute numerical values and evaluate P:

P = (6.08 A ) (12 Ω ) = 0.44 kW

(c) Express the rms voltage across the resistor in terms of R and I rms :

Vrms = I rms R

Substitute numerical values and evaluate Vrms :

Vrms = (6.08 A )(12 Ω ) = 73 V

2

85 •• [SSM] Figure 29-45 shows the voltage versus time for a squarewave voltage source. If V0 = 12 V, (a) what is the rms voltage of this source? (b) If this alternating waveform is rectified by eliminating the negative voltages, so that only the positive voltages remain, what is the new rms voltage? Picture the Problem The average of any quantity over a time interval ΔT is the integral of the quantity over the interval divided by ΔT. We can use this definition to find both the average of the voltage squared, V 2 av and then use the definition

( )

of the rms voltage. (a) From the definition of Vrms we have: Noting that − V02 = V02 , evaluate Vrms :

Vrms =

(V )

2 0 av

Vrms = V02 = V0 = 12 V

Alternating-Current Circuits (b) Noting that the voltage during the second half of each cycle is now zero, express the voltage during the first half cycle of the time interval 1 2 ΔT :

V = V0

Express the square of the voltage during this half cycle:

V 2 = V02

Calculate (V 2 )av by integrating V2 from t = 0 to t = 12 ΔT and dividing

(V ) 2

av

V2 = 0 ΔT

1 ΔT 2

by ΔT: Substitute to obtain:

Vrms =

1 2

∫ dt = 0

V02 =

V0

2

831

V02 12 ΔT 1 2 [t ] 0 = 2 V0 ΔT

=

12 V 2

= 8.5 V

86 •• What are the average values and rms values of current for the two current waveforms shown in Figure 29-46? Picture the Problem The average of any quantity over a time interval ΔT is the integral of the quantity over the interval divided by ΔT. We can use this definition to find both the average current I av , and the average of the current squared,

(I ) 2

av

1 ΔT

(I )

From the definition of I av and I rms we have:

I av =

Waveform (a) Express the current during the first half cycle of time interval ΔT:

4A t ΔT where I is in A when t and T are in seconds.

Evaluate I av, a :

∫

ΔT

0

Idt and I rms =

av

Ia =

I av, a =

1 ΔT

ΔT

4.0 A 4.0 A ∫0 ΔT tdt = (ΔT )2 ΔT

=

Express the square of the current during this half cycle:

2

I

2 a

4.0 A ⎡ t 2 ⎤ = 2.0 A (ΔT )2 ⎢⎣ 2 ⎥⎦ 0

2 ( 4.0 A ) 2 = t (ΔT )2

ΔT

∫ tdt 0

Chapter 29

832

Noting that the average value of the squared current is the same for each time interval ΔT, calculate (I a2 )av by

(I )

2 a av

integrating I a2 from t = 0 to t = ΔT and dividing by ΔT: Substitute in the expression for I rms, a to obtain: Waveform (b) Noting that the current during the second half of each cycle is zero, express the current during the first half cycle of the time interval 12 ΔT :

(4.0 A )2 t 2 dt ∫0 (ΔT )2 ΔT 2 ( 4.0 A ) ⎡ t 3 ⎤ 16 = A2 = ⎥ 3 ⎢ (ΔT ) ⎣ 3 ⎦ 0 3 ΔT

1 = ΔT

I rms, a =

16 2 A = 2.3 A 3

I b = 4.0 A

Evaluate I av, b :

1

I av, b

ΔT

4.0 A 2 4.0 A 12 ΔT [t ] 0 dt = = ΔT ∫0 ΔT = 2.0 A

Express the square of the current during this half cycle: Calculate (I b2 )av by integrating I b2 from t = 0 to t = 12 ΔT and dividing by ΔT:

I b2 = (4.0 A )

2

(I )

2 b av

(4.0 A )2 = ΔT

Substitute in the expression for I rms, b

∫ dt 0

(4.0 A ) [t ] ΔT 0 2

=

1 Δ 2

ΔT

1 2

= 8.0 A 2

I rms, b = 8.0 A 2 = 2.8 A

to obtain: 87

••

In the circuit shown in Figure 29-47, ε1 = (20 V) cos 2πft, where

f = 180 Hz; ε2 = 18 V, and R = 36 Ω. Find the maximum, minimum, average, and rms values of the current in the resistor. Picture the Problem We can apply Kirchhoff’s loop rule to express the current in the circuit in terms of the emfs of the sources and the resistance of the resistor. We can then find Imax and I min by considering the conditions under which the time-dependent factor in I will be a maximum or a minimum. Finally, we can use

Alternating-Current Circuits I rms =

(I ) 2

av

to derive an expression for Irms that we can use to determine its

value.

ε 1, peak cos ωt + ε 2 − IR = 0

Apply Kirchhoff’s loop rule to obtain: Solving for I yields:

I=

ε 1, peak R

cos ωt +

ε2 R

or I = A1 cos ωt + A2 where A1 =

ε 1, peak R

and A2 =

ε2 R

⎛ 20 V ⎞ 18 V ⎟⎟ cos 2π 180 s −1 t + I = ⎜⎜ 36 Ω ⎝ 36 Ω ⎠ = (0.556 A ) cos 1131s −1 t + 0.50 A

( (

Substitute numerical values to obtain:

(

))

)

The current is a maximum when cos 1131s −1 t = 1 . Hence :

I max = 0.50 A + 0.556 A = 1.06 A

Evaluate I min :

I min = 0.50 A − 0.556 A = − 0.06 A

Because the average value of cosωt = 0:

I av = 0.50 A

The rms current is the square root of the average of the squared current:

I rms =

(

[I ] 2

av

)

[I ] 2

(1)

av

is given by:

[I ] = [( A cos ωt + A ) ] = [A cos ωt + 2 A A cos ωt + A ] = [A cos ωt ] + [2 A A cos ωt ] + [A ] = A [cos ωt ] + 2 A A [cos ωt ] + A [I ] = A + A ωt ] = and 2

2

av

1

Because cos 2 [cos ωt ]av = 0 :

av

2

2 1

2

2 1

2

2 1

[

833

av

1

2 2 av

2

1

av

2

2

1 2

Substituting in equation (1) yields:

av

1

av

2

2

av

I rms =

2 2 av

av

1 2

2 1

1 2

2 2

A12 + A22

2 2

Chapter 29

834

Substitute for A1 and A2 to obtain:

Substitute numerical values and evaluate Irms:

I rms =

1 2

⎛ ε1 ⎞ ⎛ ε 2 ⎞ ⎜ ⎟ +⎜ ⎟ ⎝R⎠ ⎝ R⎠

I rms =

1 2

⎛ 20 V ⎞ ⎛ 18 V ⎞ ⎜ ⎟ +⎜ ⎟ ⎝ 36 Ω ⎠ ⎝ 36 Ω ⎠

2

2

2

2

= 0.64 A 88

••

Repeat Problem 87 if the resistor is replaced by a 2.0-μF capacitor.

Picture the Problem We can apply Kirchhoff’s loop rule to obtain an expression for charge on the capacitor as a function of time. Differentiating this expression with respect to time will give us the current in the circuit. We can then find Imax and Imin by considering the conditions under which the time-dependent factor in I

(I )

will be a maximum or a minimum. Finally, we can use I rms =

2

av

expression for Irms that we can use to determine its value. Apply Kirchhoff’s loop rule to obtain: Solving for q(t) yields:

ε1 peak cos ωt + ε 2 − q(t ) = 0 C

q(t ) = C (ε 1 peak cos ωt + ε 2 ) = A1 cos ωt + A2 where A1 = Cε 1 peak and A2 = Cε 2

Differentiate this expression with respect to t to obtain the current as a function of time:

dq d = ( A1 cos ωt + A2 ) dt dt = −ωA1 sin ωt

I=

Substituting numerical values yields:

(

)

I = −2π (180 Hz )(2.0 μF)sin (2π (180 Hz )t ) = (− 2.26 mA )sin 1131 s −1 t

The current is a minimum when sin (1131s −1 )t = 1 . Hence:

I min = − 2.3 mA

The current is a maximum when sin 1131s −1 t = −1 . Hence:

I max = 2.3 mA

(

)

to derive an

Alternating-Current Circuits Because the dc source sees the capacitor as an open circuit and the average value of the sine function over a period is zero:

I av = 0

The rms current is the square root of the average of the squared current:

I rms =

[I ] 2

av

[I ] 2

835

(1)

av

is given by:

[I ] = [( A cos ωt + A ) ] = [A cos ωt + 2 A A cos ωt + A ] = [A cos ωt ] + [2 A A cos ωt ] + [A ] = A [cos ωt ] + 2 A A [cos ωt ] + A [I ] = A + A ωt ] = and 2

2

av

1

2 1

2

2 1

2

2 1

[

Because cos 2 [cos ωt ]av = 0 :

av

2

av

1

2 2 av

2

1

av

2

2 2 av

av

2 2

2

1 2

av

1

av

2

2

1 2

av

2 1

2 2

Substituting in equation (1) yields:

I rms =

1 2

A12 + A22

Substitute for A1 and A2 to obtain:

I rms =

1 2

(Cε1 )2 + (Cε 2 )2

=C

Substitute numerical values and evaluate Irms:

1 2

(ε1 )2 + (ε 2 )2

I rms = (2.0 μF)

1 2

(20 V )2 + (18 V )2

= 46 μA

89 ••• [SSM] A circuit consists of an ac generator, a capacitor and an ideal inductor⎯all connected in series. The emf of the generator is given by ε peak cos ωt . (a) Show that the charge on the capacitor obeys the equation

L

d 2Q Q + = ε peak cos ωt . (b) Show by direct substitution that this equation is dt 2 C

satisfied by Q = Qpeak cos ωt where Qpeak = −

(

ε peak

L ω 2 − ω02

can be written as I = I peak cos(ωt − δ ) , where I peak =

) . (c) Show that the current

ωε peak

L ω 2 − ω 02

=

ε peak

XL − XC

and

δ = –90º for ω < ω0 and δ = 90º for ω > ω0, where ω0 is the resonance frequency.

836

Chapter 29

Picture the Problem In Part (a) we can apply Kirchhoff’s loop rule to obtain the 2nd order differential equation relating the charge on the capacitor to the time. In Part (b) we’ll assume a solution of the form Q = Qpeak cos ωt , differentiate it twice,

and substitute for d2Q/dt2 and Q to show that the assumed solution satisfies the differential equation provided Qpeak = −

(

ε peak

L ω 2 − ω02

) . In Part (c) we’ll use our

results from (a) and (b) to establish the result for Ipeak given in the problem statement. (a) Apply Kirchhoff’s loop rule to obtain:

ε − Q − L dI

Substitute for ε and rearrange the differential equation to obtain:

L

Because I = dQ dt :

C

dt

=0

dI Q + = ε max cos ωt dt C

d 2Q Q L 2 + = ε max cos ωt C dt

(b) Assume that the solution is:

Q = Qpeak cos ωt

Differentiate the assumed solution twice to obtain:

dQ = −ωQpeak sin ωt dt and d 2Q = −ω 2 Qpeak cos ωt 2 dt

dQ d 2Q and in the Substitute for dt dt 2 differential equation to obtain:

− ω 2 LQpeak cos ωt +

Factor cosωt from the left-hand side of the equation:

Q ⎞ ⎛ ⎜⎜ − ω 2 LQpeak + peak ⎟⎟ cos ωt C ⎠ ⎝ = ε peak cos ωt

If this equation is to hold for all values of t it must be true that:

Qpeak

cos ωt C = ε peak cos ωt

− ω 2 LQpeak +

Qpeak C

= ε peak

Alternating-Current Circuits Solving for Qpeak yields:

Factor L from the denominator and substitute for 1/LC to obtain:

Qpeak =

Qpeak =

ε peak − ω 2L +

ε peak 1 ⎞ ⎛ L⎜ − ω 2 + ⎟ LC ⎠ ⎝

= −

(c) From (a) and (b) we have:

I= =

1 C

(

ε peak

)

L ω 2 − ω02

dQ = −ωQpeak sin ωt dt

ωε peak sin ωt = I peak sin ωt L(ω 2 − ω02 )

= I peak cos(ωt − δ )

where I peak =

=

ωε peak Lω −ω 2

ε peak ωL −

1 ωC

2 0

=

=

ε peak L

ω

ω 2 − ω02

ε peak

XL − XC

If ω > ω0, XL > XC and the current lags the voltage by 90° (δ = 90°). If ω < ω0, XL < XC and the current leads the voltage by 90°(δ = −90°).

837

838

Chapter 29

Chapter 30 Maxwell’s Equations and Electromagnetic Waves Conceptual Problems •

(a) (b)

The displacement current has different units than the conduction current. Displacement current only exists if the electric field in the region is changing with time. In an oscillating LC circuit, no displacement current exists between the capacitor plates when the capacitor is momentarily fully charged. In an oscillating LC circuit, no displacement current exists between the capacitor plates when the capacitor is momentarily uncharged.

(c) (d)

[SSM]

True or false:

1

(a) False. Like those of conduction current, the units of displacement current are C/s. (b) True. Because displacement current is given by I d = ∈0 dφe dt , Id is zero if dφe dt = 0 . (c) True. When the capacitor is fully charged, the electric flux is momentarily a maximum (its rate of change is zero) and, consequently, the displacement current between the plates of the capacitor is zero. (d) False. Id is zero if dφe dt = 0 . At the moment when the capacitor is momentarily uncharged, dE/dt ≠ 0 and so dφe dt ≠ 0 . 2

•

Using SI units, show that I d = ∈0 dφe dt has units of current.

Determine the Concept We need to show that ∈0 dφe dt has units of amperes. We can accomplish this by substituting the SI units of ∈0 and dφe dt and simplifying the resulting expression. N 2 ⋅m C2 C C ⋅ = = A N⋅m s s •

(a)

Maxwell’s equations apply only to electric and magnetic fields that are constant over time. The electromagnetic wave equation can be derived from Maxwell’s equations.

(b)

[SSM]

True or false:

3

839

840 (c) (d)

Chapter 30 Electromagnetic waves are transverse waves. The electric and magnetic fields of an electromagnetic wave in free space are in phase.

(a) False. Maxwell’s equations apply to both time-independent and timedependent fields. (b) True. One can use Faraday’s law and the modified version of Ampere’s law to derive the wave equation. (c) True. Both the electric and magnetic fields of an electromagnetic wave oscillate at right angles to the direction of propagation of the wave. (d) True. 4 • Theorists have speculated about the existence of magnetic monopoles, and several experimental searches for such monopoles have occurred. Suppose magnetic monopoles were found and that the magnetic field at a distance r from a monopole of strength qm is given by B = (μ0/4π)qm/r2. Modify the Gauss’s law for magnetism equation to be consistent with such a discovery. Determine the Concept Gauss’s law for magnetism would become

∫

S

Bn dA = μ0 qm, inside

where qm, inside is the total magnetic charge inside the

Gaussian surface. Note that Gauss’s law for electricity follows from the existence of electric monopoles (charges), and the electric field due to a point charge follows from the inverse-square nature of Coulomb’s law. 5 • (a) For each of the following pairs of electromagnetic waves, which has the higher frequency: (1) visible light or X rays, (2) green light or red light, (3) infrared waves or red light. (b) For each of the following pairs of electromagnetic waves, which has the longer wavelength: (1) visible light or microwaves, (2) green light or ultraviolet light, (3) gamma rays or ultraviolet light. Determine the Concept Refer to Table 30-1 to rank order the frequencies and wavelengths of the given electromagnetic radiation. (a) (1) X rays (2) green light (3) red light (b) (1) microwaves (2) green light (3) ultraviolet light

Maxwell’s Equations and Electromagnetic Waves

841

6 • The detection of radio waves can be accomplished with either an electric dipole antenna or a loop antenna. True or false: (a) (b) (c)

The electric dipole antenna works according to Faraday’s law. If a linearly polarized radio wave is approaching you head on such that its electric field oscillates vertically, to best detect this wave the normal to a loop antenna’s plane should be oriented so that it points either right or left. If a linearly polarized radio wave is approaching you such that its electric field oscillates in a horizontal plane, to best detect this wave using an dipole antenna the antenna should be oriented vertically.

(a) False. A dipole antenna is oriented parallel to the electric field of an incoming wave so that the wave can induce an alternating current in the antenna. (b) True. A loop antenna is oriented perpendicular to the magnetic field of an incoming wave so that the changing magnetic flux through the loop can induce a current in the loop. Orienting the loop antenna’s plane so that it points either right or left satisfies this condition. (c) False. The dipole antenna needs to be oriented parallel to the electric field of an incoming wave so that the wave can induce an alternating current in the antenna. 7 • A transmitter emits electromagnetic waves using an electric dipole antenna oriented vertically. (a) A receiver to detect these waves also uses an electric dipole antenna that is one mile from the transmitting antenna and at the same altitude. How should the receiver’s electric dipole antenna be oriented for optimum signal reception? (b) A receiver to detect these waves uses a loop antenna that is one mile from the transmitting antenna and at the same altitude. How should the loop antenna be oriented for optimum signal reception? Determine the Concept (a) The electric dipole antenna should be oriented vertically. (b) The loop antenna and the electric dipole transmitting antenna should be in the same vertical plane. r r r 8 • Show that the expression E × B μ0 for the Poynting vector S (Equation 30-21) has units of watts per square meter (the SI units for electromagnetic wave intensity).

r r Determine the Concept We can that E × B μ0 has units of W/m2 by substituting r r the SI units of E , B and μ0 and simplifying the resulting expression.

842

Chapter 30 N N C N⋅m J ⋅T N A N⋅m s W C = C = ⋅ = ⋅ 2 = s2 = s2 = T⋅m m C m C m m m m2 A A

9 • [SSM] If a red light beam, a green light beam, and a violet light beam, all traveling in empty space, have the same intensity, which light beam carries more momentum? (a) the red light beam, (b) the green light beam, (c) the violet light beam, (d) They all have the same momentum. (e) You cannot determine which beam carries the most momentum from the data given. Determine the Concept The momentum of an electromagnetic wave is directly proportional to its energy ( p = U c ). Because the intensity of a wave is its energy per unit area and per unit time (the average value of its Poynting vector), waves with equal intensity have equal energy and equal momentum. (d ) is correct. 10 • If a red light plane wave, a green light plane wave, and a violet light plane wave, all traveling in empty space, have the same intensity, which wave has the largest peak electric field? (a) the red light wave, (b) the green light wave, (c) the violet light wave, (d) They all have the same peak electric field. (e) You cannot determine the largest peak electric field from the data given. Determine the Concept The intensity of an electromagnetic wave is given by r EB I= S = 0 0. av 2 μ0

The intensity of an electromagnetic wave is given by:

r I= S

Because E0 = cB0:

r S

r This result tells us that S

av

av

=

av

=

E0 B0 2 μ0

E02 2cμ0

∝ E02 independently of the wavelength of the

electromagnetic radiation. Thus

(d )

is correct.

11 • Two sinusoidal plane electromagnetic waves are identical except that wave A has a peak electric field that is three times the peak electric field of wave B. How do their intensities compare? (a) I A = 13 I B (b) I A = 91 I B (c) I A = 3I B (d) I A = 9I B (e) You cannot determine how their intensities compare from the data given.

Maxwell’s Equations and Electromagnetic Waves

843

Determine the Concept The intensity of an electromagnetic wave is given by r EB I= S = 0 0. av 2 μ0

Express the intensities of the two waves: Dividing the first of these equations by the second and simplifying yields:

Because wave A has a peak electric field that is three times that of wave B, the peak magnetic field of A is also three times that of wave B. Hence:

IA =

E 0 , A B0 , A 2μ 0

and I B =

E 0 , B B0 , B 2μ 0

E 0 , A B0 , A E 0 , A B0 , A 2μ 0 IA = = E 0 , B B0 , B IB E 0 , B B0 , B 2μ 0

I A (3E0, B )(3B0, B ) = = 9 ⇒ I A = 9I B IB E0, B B0, B

(d )

is correct.

Estimation and Approximation 12 •• In laser cooling and trapping, the forces associated with radiation pressure are used to slow down atoms from thermal speeds of hundreds of meters per second at room temperature to speeds of a few meters per second or slower. An isolated atom will absorb only radiation of specific frequencies. If the frequency of the laser-beam radiation is tuned so that the target atoms will absorb the radiation, then the radiation is absorbed during a process called resonant absorption. The cross-sectional area of the atom for resonant absorption is approximately equal to λ2, where λ is the wavelength of the laser light. (a) Estimate the acceleration of a rubidium atom (molar mass 85 g/mol) in a laser beam whose wavelength is 780 nm and intensity is 10 W/m2. (b) About how long would it take such a light beam to slow a rubidium atom in a gas at room temperature (300 K) to near-zero speed? Picture the Problem We can use Newton’s second law to express the acceleration of an atom in terms of the net force acting on the atom and the relationship between radiation pressure and the intensity of the beam to find the net force. Once we know the acceleration of an atom, we can use the definition of acceleration to find the stopping time for a rubidium atom at room temperature.

(a) Apply Newton’s second law to the atom to obtain:

Fr = ma (1) where Fr is the radiation force exerted by the laser beam.

844

Chapter 30

The radiation pressure Pr and intensity of the beam I are related according to: Solve for Fr to obtain:

Substitute for Fr in equation (1) to obtain:

Pr =

Fr I = A c

Fr =

IA Iλ2 = c c

Iλ2 Iλ2 = ma ⇒ a = mc c

Substitute numerical values and evaluate a: a=

(10 W/m )(780 nm ) ⎛ ⎞ g 1 mol ⎜⎜ 85 ⎟ (2.998 × 10 × mol 6.022 × 10 particles ⎟ 2

2

⎝

23

8

⎠

m/s

)

= 1.44 × 10 5 m/s 2

= 1.4 × 10 5 m/s 2

(b) Using the definition of acceleration, express the stopping time Δt of the atom: Because vfinal ≈ 0:

Using the rms speed as the initial speed of an atom, relate vinitial to the

Δt =

v final − vinitial a

Δt ≈

− v initial a

vinitial = v rms =

3kT m

temperature of the gas: Substitute in the expression for the stopping time to obtain:

Δt = −

1 3kT a m

Substitute numerical values and evaluate Δt: 1 Δt = − − 1.44 × 10 5 m/s 2

(

)

3 1.38 × 10 − 23 J/K (300 K ) = 2.1ms ⎞ ⎛ g 1 mol ⎟⎟ ⎜⎜ 85 × 23 ⎝ mol 6.022 × 10 particles ⎠

13 •• [SSM] One of the first successful satellites launched by the United States in the 1950s was essentially a large spherical (aluminized) Mylar balloon from which radio signals were reflected. After several orbits around Earth,

Maxwell’s Equations and Electromagnetic Waves

845

scientists noticed that the orbit itself was changing with time. They eventually determined that radiation pressure from the sunlight was causing the orbit of this object to change—a phenomenon not taken into account in planning the mission. Estimate the ratio of the radiation-pressure force by the sunlight on the satellite to the gravitational force by Earth’s gravity on the satellite. Picture the Problem We can use the definition of pressure to express the radiation force on the balloon. We’ll assume that the gravitational force on the balloon is approximately its weight at the surface of Earth, that the density of Mylar is approximately that of water and that the area receiving the radiation from the sunlight is the cross-sectional area of the balloon.

The radiation force acting on the balloon is given by:

Because the radiation from the Sun is reflected, the radiation pressure is twice what it would be if it were absorbed:

Substituting for Pr and A yields:

The gravitational force acting on the balloon when it is in a near-Earth orbit is approximately its weight at the surface of Earth: Because the surface area of the balloon is 4π r 2 = π d 2 : Express the ratio of the radiationpressure force to the gravitational force and simplify to obtain:

Fr = Pr A where A is the cross-sectional area of the balloon.

Pr =

2I c

Fr =

2 I 14 πd 2 πd 2 I = c 2c

(

)

Fg = wballoon = mballoon g = ρ MylarVMylar g = ρ Mylar Asurface, ballon t g where t is the thickness of the Mylar skin of the balloon. Fg = πρ Mylar d 2 t g

πd 2 I Fr I 2c = = 2 Fg πρ Mylar d t g 2 ρ Mylart gc

846

Chapter 30

Assuming the thickness of the Mylar skin of the balloon to be 1 mm, substitute numerical values and evaluate Fr/Fg: kW 1.35 2 Fr m ≈ 2 × 10 −7 = m m kg Fg ⎞ ⎛ ⎞ ⎞⎛ ⎛ 2 ⎜1.00 × 10 3 3 ⎟ ⎜ 9.81 2 ⎟ (1 mm )⎜ 2.998 × 108 ⎟ s⎠ m ⎠⎝ s ⎠ ⎝ ⎝ 14 •• Some science fiction writers have described solar sails that could propel interstellar spaceships. Imagine a giant sail on a spacecraft subjected to radiation pressure from our Sun. (a) Explain why this arrangement works better if the sail is highly reflective rather than highly absorptive. (b) If the sail is assumed highly reflective, show that the force exerted by the sunlight on the spacecraft is given by PS A 2π r 2 c where PS is the power output of the Sun (3.8 × 1026 W), A is the surface area of the sail, m is the total mass of the spacecraft, r is the distance from the Sun, and c is the speed of light. (Assume the area of the sail is much larger than the area of the spacecraft so that all the force is due to radiation pressure on the sail only.) (c) Using a reasonable value for A, compare the force on the spacecraft due to the radiation pressure and the force on the spacecraft due to the gravitational pull of the Sun. Does the result imply that such a system will work? Explain your answer.

(

)

Picture the Problem (b) We can use the definition of radiation pressure to show that the force exerted by the sunlight on the spacecraft is given by PS A 2π r 2 c

(

26

)

where PS is the power output of the Sun (3.8 × 10 W), A is the surface area of the sail, m is the total mass of the spacecraft, r is the distance from the Sun, and c is the speed of light. (a) If the sail is highly reflective rather than highly absorptive, the radiation force is doubled. (b) Because the sail is highly reflective:

The intensity of the solar radiation on P the sail is given by I = s 2 . 4πr Substituting for I yields:

2 IA c where A is the area of the sail. Fr = Pr A =

Fr =

2 Ps A Ps A = 2 4π r c 2π r 2 c

Maxwell’s Equations and Electromagnetic Waves (c) Express the ratio of the force on the spacecraft due to the radiation pressure and the force on the spacecraft due the gravitational force of the Sun on the spacecraft:

847

PS A PS A Fr 2π r 2 c = = Fg GmM S 2π cGmM S r2

Assuming a 15-m diameter circular sail and a 500-kg spacecraft (values found using the internet), substitute numerical values and evaluate the ratio of the accelerations: ⎛π 2⎞ W )⎜ (15 m ) ⎟ Fr ⎝4 ⎠ = 2 Fg m ⎞⎛ N⋅m ⎞ ⎛ ⎟ (500 kg ) (1.99 × 10 30 kg ) 2π ⎜ 2.998 × 10 8 ⎟⎜⎜ 6.673 × 10 −11 2 ⎟ s kg ⎠ ⎝ ⎠⎝

(3.8 × 10

26

= 5.4 × 10 − 4 for A = 177 m2 and m = 500 kg. This scheme is not likely to work effectively. For any reasonable spacecraft mass, the surface mass density of the sail would to be extremely small (experimental sails have area densities of approximately 3 g/m2) and the sail would have to be huge. Additionally, unless struts are built into the sail, it would collapse during use.

Maxwell’s Displacement Current 15 • [SSM] A parallel-plate capacitor has circular plates and no dielectric between the plates. Each plate has a radius equal to 2.3 cm and the plates are separated by 1.1 mm. Charge is flowing onto the upper plate (and off of the lower plate) at a rate of 5.0 A. (a) Find the rate of change of the electric field strength in the region between the plates. (b) Compute the displacement current in the region between the plates and show that it equals 5.0 A. Picture the Problem We can differentiate the expression for the electric field between the plates of a parallel-plate capacitor to find the rate of change of the electric field strength and the definitions of the conduction current and electric flux to compute Id.

(a) Express the electric field strength between the plates of the parallelplate capacitor:

E=

Q

∈0 A

848

Chapter 30

Differentiate this expression with respect to time to obtain an expression for the rate of change of the electric field strength:

dE d ⎡ Q ⎤ I 1 dQ = ⎢ = ⎥= dt dt ⎣∈0 A ⎦ ∈0 A dt ∈0 A

Substitute numerical values and evaluate dE/dt: dE 5.0 A = = 3.40 × 1014 V/m ⋅ s 2 2 2 −12 dt (8.854 × 10 C / N ⋅ m )π (0.023 m ) = 3.4 ×1014 V/m ⋅ s (b) Express the displacement current Id :

I d =∈0

dφe dt

Substitute for the electric flux to obtain:

I d =∈0

d [EA] =∈0 A dE dt dt

Substitute numerical values and evaluate Id:

(

)

(

)

I d = 8.854 × 10 −12 C 2 / N ⋅ m 2 π (0.023 m ) 3.40 × 1014 V/m ⋅ s = 5.0 A 2

In a region of space, the electric field varies with time as 16 • E = (0.050 N/C) sin (ωt), where ω = 2000 rad/s. Find the peak displacement current through a surface that is perpendicular to the electric field and has an area equal to 1.00 m2. Picture the Problem We can express the displacement current in terms of the electric flux and differentiate the resulting expression to obtain Id in terms of dE/dt.

The displacement current Id is given by:

I d =∈0

dφe dt

Substitute for the electric flux to obtain:

I d =∈0

d [EA] =∈0 A dE dt dt

Because E = (0.050 N/C )sin 2000t :

d [(0.050 N/C)sin 2000t ] dt = 2000 s -1 ∈ 0 A(0.050 N/C ) cos 2000t

I d =∈ 0 A

(

)

Maxwell’s Equations and Electromagnetic Waves Id will have its maximum value when cos 2000t = 1. Hence:

(

849

)

I d,max = 2000 s -1 ∈ 0 A(0.050 N/C )

Substitute numerical values and evaluate Id,max: ⎛ C2 I d, max = 2000 s −1 ⎜⎜ 8.854 × 10 −12 N ⋅ m2 ⎝

(

)

⎞ ⎟⎟ 1.00 m 2 ⎠

(

)⎛⎜ 0.050 NC ⎞⎟ =

0.89 nA

⎠

⎝

17 • For Problem 15, show that the magnetic field strength between the plates a distance r from the axis through the centers of both plates is given by B = (1.9 × 10–3 T/m)r. Picture the Problem We can use Ampere’s law to a circular path of radius r between the plates and parallel to their surfaces to obtain an expression relating B to the current enclosed by the amperian loop. Assuming that the displacement current is uniformly distributed between the plates, we can relate the displacement current enclosed by the circular loop to the conduction current I.

Apply Ampere’s law to a circular path of radius r between the plates and parallel to their surfaces to obtain: Assuming that the displacement current is uniformly distributed:

r r B ∫ ⋅ d l = 2πrB = μ0 I enclosed = μ0 I C

Id I r2 = ⇒ I = Id π r 2 π R2 R2 where R is the radius of the circular plates.

Substituting for I yields:

Substitute numerical values and evaluate B:

2π rB =

B(r ) =

μ0 r 2 R

2

Id ⇒ B =

(4π × 10

−7

μ0 r Id 2π R 2

)

N / A 2 (5.0 A )

2π (0.023 m )

2

r

T⎞ ⎛ = ⎜1.9 × 10 −3 ⎟ r m⎠ ⎝ 18 •• The capacitors referred to in this problem have only empty space between the plates. (a) Show that a parallel-plate capacitor has a displacement current in the region between its plates that is given by Id = C dV/dt, where C is the capacitance and V is the potential difference between the plates. (b) A 5.00-nF

850

Chapter 30

parallel-plate capacitor is connected to an ideal ac generator so the potential difference between the plates is given by V = V0 cos ωt, where V0 = 3.00 V and ω = 500π rad/s. Find the displacement current in the region between the plates as a function of time. Picture the Problem We can use the definitions of the displacement current and electric flux, together with the expression for the capacitance of an air-coreparallel-plate capacitor to show that Id = C dV/dt.

(a) Use its definition to express the displacement current Id:

I d =∈0

dφe dt

Substitute for the electric flux to obtain:

I d =∈0

d [EA] =∈0 A dE dt dt

Because E = V/d:

I d =∈0 A

The capacitance of an air-coreparallel-plate capacitor whose plates have area A and that are separated by a distance d is given by:

C=

Substituting yields:

d ⎡V ⎤ ∈0 A dV = dt ⎢⎣ d ⎥⎦ d dt

∈0 A d

Id = C

dV dt

(b) Substitute in the expression derived in (a) to obtain: I d = (5.00 nF)

(

)

d [(3.00 V ) cos 500π t ] = −(5.00 nF)(3.00 V ) 500π s −1 sin 500π t dt

= − (23.6 μA )sin 500π t

19 •• [SSM] There is a current of 10 A in a resistor that is connected in series with a parallel plate capacitor. The plates of the capacitor have an area of 0.50 m2, and no dielectric exists between the plates. (a) What is the displacement current between the plates? (b) What is the rate of change of the electric field r r strength between the plates? (c) Find the value of the line integral ∫ B ⋅ d l , where C

the integration path C is a 10-cm-radius circle that lies in a plane that is parallel with the plates and is completely within the region between them.

Maxwell’s Equations and Electromagnetic Waves

851

Picture the Problem We can use the conservation of charge to find Id, the definitions of the displacement current and electric flux to find dE/dt, and r r Ampere’s law to evaluate B ⋅ d l around the given path.

(a) From conservation of charge we know that:

I d = I = 10 A

(b) Express the displacement current Id :

I d =∈0

Substituting for dE/dt yields:

I dE = d dt ∈0 A

Substitute numerical values and evaluate dE/dt:

10 A dE = dt ⎛ C2 ⎜⎜ 8.85 × 10 −12 N ⋅ m2 ⎝ V = 2.3 × 1012 m ⋅s

dφe d dE =∈0 [EA] =∈0 A dt dt dt

⎞ ⎟⎟ (0.50 m 2 ) ⎠

r r B ∫ ⋅ d l = μ0 I enclosed

(c) Apply Ampere’s law to a circular path of radius r between the plates and parallel to their surfaces to obtain:

C

Assuming that the displacement current is uniformly distributed and letting A represent the area of the circular plates yields:

I enclosed I d π r2 = ⇒ I = Id enclosed A A π r2

Substitute for I enclosed to obtain:

r r μ 0π r 2 ∫CB ⋅ d l = A I d

Substitute numerical values and evaluate

(

r r B ∫ ⋅ dl : C

)

r r 4π × 10 −7 N / A 2 π (0.10 m )2 (10 A ) B = 0.79 μT ⋅ m ∫C ⋅ d l = 0.50 m 2 20 ••• Demonstrate the validity of the generalized form of Ampère’s law (Equation 30-4) by showing that it gives the same result as the Biot–Savart law (Equation 27-3) in a specified situation. Figure 30-13 shows two momentarily

852

Chapter 30

equal but opposite point charges (+Q and –Q) on the x axis at x = –a and x = +a, respectively. At the same instant there is a current I in the wire connecting them, as shown. Point P is on the y axis at y = R. (a) Use the Biot–Savart law to show μ Ia 1 that the magnitude of the magnetic field at point P is given by B = 0 . 2 2π R R + a2 (b) Now consider a circular strip of radius r and width dr in the x = 0 plane that has its center at the origin. Show that the flux of the electric field through this 32 Qπr 2 strip is given by E x dA = r + a 2 dr . (c) Use the result from Part (b) to

∈0

(

)

show that the total electric flux φe through a circular surface S of radius R. is ⎞ Q⎛ a given by φe = ⎜ 1− ⎟ . (d) Find the displacement current Id through S, ∈o ⎝ a2 + R 2 ⎠ a and show that I + Id = I (e) Finally, show that the generalized form of a2 + R 2 Ampere’s law (Equation 30-4) gives the same result for the magnitude of the magnetic field as found in Part (a). Picture the Problem We can follow the step-by-step instructions in the problem statement to show that Equation 30-4 gives the same result for B as that given in Part (a).

(a) Express the magnetic field strength at P using the expression for B due to a straight wire segment:

BP =

μ0 I (sin θ1 + sin θ 2 ) 4π R

where

sin θ1 = sin θ 2 = Substitute for sinθ1 and sinθ2 to obtain:

(b) Express the electric flux through the circular strip of radius r and width dr in the yz plane: The electric field due to the dipole is:

a R + a2 2

2a μ0 I 4π R R 2 + a 2 1 μ Ia = 0 2 2π R R + a 2

BP =

dφe = E x dA = E x (2π rdr )

Ex =

2kQ 2kQa cos θ1 = 32 2 r +a r 2 + a2 2

(

)

Maxwell’s Equations and Electromagnetic Waves Substitute for Ex to obtain:

dφe = Ex dA =

(c) Multiply both sides of the expression for dφe by ∈0:

(r

2kQa 2

+ a2

)

32

2Qa

=

4π ∈0 (r 2 + a 2 )

=

Qa

32

∈0 (r 2 + a 2 )

32

∈ 0 dφ e =

(r

Qa 2

+ a2

)

32

853

(2π rdr )

(2π rdr )

rdr

rdr

Integrate r from 0 to R to obtain: R

∈0 φe = Qa ∫ 0

(r

rdr 2

+a

)

2 32

⎛ ⎛ −1 1⎞ a + ⎟⎟ = Q⎜⎜1 − = Qa⎜⎜ 2 2 a⎠ R2 + a2 ⎝ R +a ⎝

⎞ ⎟ ⎟ ⎠

a dφ e d ⎡ ⎛ = ⎢Q⎜⎜1 − 2 dt dt ⎢⎣ ⎝ R + a2 ⎛ ⎞ dQ a ⎟ = ⎜⎜1 − 2 2 ⎟ dt R +a ⎠ ⎝ ⎛ ⎞ a ⎟ = − I ⎜⎜1 − 2 2 ⎟ + R a ⎠ ⎝

(d) The displacement current is defined to be:

I d =∈0

The total current is the sum of I and Id :

⎛ a I + I d = I − I ⎜⎜1 − R2 + a2 ⎝ = I

a R + a2 2

r r B ⋅ d l = 2πRB = μ 0 (I + I d )

(e) Apply Equation 30-4 (the generalized form of Ampere’s law) to obtain:

∫

Solving for B yields:

B=

C

⎞ ⎟ ⎟ ⎠

μ0 (I + I d ) 2π R

⎞⎤ ⎟⎥ ⎟ ⎠⎥⎦

854

Chapter 30

Substitute for I + Id from (d) to obtain:

B= =

a μ0 ⎛ ⎜I ⎜ 2π R ⎝ R2 + a2

⎞ ⎟ ⎟ ⎠

μ 0 Ia 1 2 2π R R + a 2

Maxwell’s Equations and the Electromagnetic Spectrum 21 • The color of the dominant light from the Sun is in the yellow-green region of the visible spectrum. Estimate the wavelength and frequency of the dominant light emitted by our Sun. HINT: See Table 30-1. Picture the Problem We can find both the wavelength and frequency of the dominant light emitted by our Sun in Table 30-1.

Because the radiation from the Sun is yellow-green dominant, the dominant wavelength is approximately:

λ yellow-green = 580 nm

The corresponding frequency is:

f yellow-green =

c

λ yellow-green

2.998 × 10 8 m/s = 580 nm = 5.17 × 1014 Hz 22 • (a) What is the frequency of microwave radiation that has a 3.00-cmlong wavelength? (b) Using Table 30-1, estimate the ratio of the shortest wavelength of green light to the shortest wavelength of red light. Picture the Problem We can use c = fλ to find the frequency corresponding to the given wavelength.

(a) The frequency of an electromagnetic wave is the ratio of the speed of light in a vacuum to the wavelength of the wave: Substitute numerical values and evaluate f:

f =

f =

c

λ

2.998 ×108 m/s = 9.993 × 1010 Hz −2 3.00 × 10 m

= 9.99 GHz

Maxwell’s Equations and Electromagnetic Waves (b) The ratio of the shortest wavelength green light to the shortest wavelength red light is:

855

λshortest green 520 nm ≈ = 0.84 620 nm λshortest red

23 • (a) What is the frequency of an X ray that has a 0.100-nm-long wavelength? (b) The human eye is sensitive to light that has a wavelength equal to 550 nm. What is the color and frequency of this light? Comment on how this answer compares to your answer for Problem 21. Picture the Problem We can use c = fλ to find the frequency corresponding to the given wavelengths and consult Table 30-1 to determine the color of light with a wavelength of 550 nm.

(a) The frequency of an X ray with a wavelength of 0.100 nm is:

f =

c

λ

=

2.998 × 10 8 m/s 0.100 × 10 −9 m

= 3.00 × 1018 Hz (b) The frequency of light with a wavelength of 550 nm is:

f =

2.998 × 10 8 m/s = 5.45 × 1014 Hz 550 nm

Consulting Table 30-1, we see that the color of light that has a wavelength of 550 nm is yellow-green. This result is consistent with those of Problem 21 and is close to the wavelength of the peak output of the Sun. Because we see naturally by reflected sunlight, this result is not surprising.

Electric Dipole Radiation 24 •• Suppose a radiating electric dipole lies along the z axis. Let I1 be the intensity of the radiation at a distance of 10 m and at angle of 90º. Find the intensity (in terms of I1) at (a) a distance of 30 m and an angle of 90º, (b) a distance of 10 m and an angle of 45º, and (c) a distance of 20 m and an angle of 30º. Picture the Problem We can use the intensity I1 at a distance r = 10 m and at an angle θ = 90° to find the constant in the expression for the intensity of radiation from an electric dipole and then use the resulting equation to find the intensity at the given distances and angles.

Express the intensity of radiation as a function of r and θ :

I (θ , r ) =

C sin 2 θ 2 r where C is a constant.

(1)

856

Chapter 30

Express I(90°,10 m):

I (90°,10 m ) = I1 = =

(

C 100 m 2

)

Solving for C yields:

C = 100 m 2 I1

Substitute in equation (1) to obtain:

I (θ , r ) =

(a) Evaluate equation (2) for r = 30 m and θ = 90°:

(100 m )I 2

r

I (90°,30 m ) = =

(b) Evaluate equation (2) for r = 10 m and θ = 45°:

I (45°,10 m ) = =

(c) Evaluate equation (2) for r = 20 m and θ = 30°:

C sin 2 90° 2 (10 m )

I (30°,20 m ) = =

1

2

sin 2 θ

(100 m ) I 2

(30 m )

1

sin 2 90°

1

sin 2 45°

2

1 9

I1

(100 m )I 2

(10 m )

1 2

(2)

2

I1

(100 m ) I 2

(20 m )

1 16

2

1

sin 2 30°

I1

25 •• (a) For the situation described in Problem 24, at what angle is the intensity at a distance of 5.0 m equal to I1? (b) At what distance is the intensity equal to I1 when θ = 45º? Picture the Problem We can use the intensity I1 at a distance r = 10 m and at an angle θ = 90° to find the constant in the expression for the intensity of radiation from an electric dipole and then use the resulting equation to find the angle for a given intensity and distance and the distance corresponding to a given intensity and angle.

Express the intensity of radiation as a function of r and θ :

I (θ , r ) =

C sin 2 θ r2 where C is a constant.

(1)

Maxwell’s Equations and Electromagnetic Waves Express I(90°,10 m):

I (90°,10 m ) = I1 =

)

Solving for C yields:

C = 100 m 2 I1

Substitute in equation (1) to obtain:

I (θ , r ) =

(a) For r = 5 m and I(θ,r) = I1:

Solve for θ to obtain: (b) For θ = 45° and I(θ,r) = I1:

(100 m )I 2

r

(100 m )I = (5.0 m )

1

2

sin 2 θ

(2)

sin 2 θ ⇒ sin 2 θ =

1 4

θ = sin −1 ( 12 ) = 30° I1 =

(100 m ) I 2

r

(

r=

1 2

1

2

or r 2 = 12 100 m 2 Solve for r to obtain:

1

2

2

I1

C sin 2 90° 2 (10 m )

C 100 m 2

=

(

857

sin 2 45°

)

(100 m ) = 2

7.1m

You and your engineering crew are in charge of setting up a wireless 26 •• telephone network for a village in a mountainous region. The transmitting antenna of one station is an electric dipole antenna located atop a mountain 2.00 km above sea level. There is a nearby mountain that is 4.00 km from the antenna and is also 2.00 km above sea level. At that location, one member of the crew measures the intensity of the signal to be 4.00 × 10–12 W/m2. What should be the intensity of the signal at the village that is located at sea level and 1.50 km from the transmitter? Picture the Problem We can use the intensity I at a distance r = 4.00 km and at an angle θ = 90° to find the constant in the expression for the intensity of radiation from an electric dipole and then use the resulting equation to find the intensity at sea level and 1.50 km from the transmitter.

Express the intensity of radiation as a function of r and θ :

I (θ , r ) =

C sin 2 θ 2 r where C is a constant.

(1)

858

Chapter 30

Use the given data to obtain:

4 × 10 −12 W/m 2 =

Solving for C yields:

C = (4.00 km ) 4.00 × 10 −12 W/m 2

C sin 2 90° 2 (4.00 km ) C = (4.00 km )2

2

(

)

= 6.40 × 10 −5 W

6.40 ×10 −5 W 2 sin θ r2

Substitute in equation (1) to obtain:

I (θ , r ) =

For a point at sea level and 1.50 km from the transmitter:

θ = tan −1 ⎜⎜

(2)

⎛ 2.00 km ⎞ ⎟⎟ = 53.1° ⎝ 1.50 km ⎠

Evaluate I(53.1°,1.50 km): I (53.1°,1.5 km ) =

6.40 × 10 −5 W 2 sin 53.1° = 18.2 pW/m 2 2 (1.50 km )

27 ••• [SSM] A radio station that uses a vertical electric dipole antenna broadcasts at a frequency of 1.20 MHz and has a total power output of 500 kW. Calculate the intensity of the signal at a horizontal distance of 120 km from the station. Picture the Problem The intensity of radiation from an electric dipole is given by C(sin2θ)/r2, where C is a constant whose units are those of power, r is the distance from the dipole to the point of interest, and θ is the angle between the antenna r and the position vector r . We can integrate the intensity to express the total power radiated by the antenna and use this result to evaluate C. Knowing C we can find the intensity at a horizontal distance of 120 km.

Express the intensity of the signal as a function of r and θ :

sin 2 θ I (r ,θ ) = C 2 r

At a horizontal distance of 120 km from the station:

I (120 km,90°) = C

sin 2 90° (120 km )2 C = (120 km )2

(1)

Maxwell’s Equations and Electromagnetic Waves From the definition of intensity we have:

859

dP = IdA and Ptot = ∫∫ I (r , θ ) dA

where, in polar coordinates, dA = r 2 sin θ dθ dφ Substitute for dA to obtain:

2π π

Ptot =

∫ ∫ I (r ,θ ) r

2

sin θ dθ dφ

0 0

Substitute for I(r,θ):

2π π

Ptot = C ∫ ∫ sin 3 θ dθ dφ 0 0

From integral tables we find that:

π

∫ sin

3

θ dθ = − 13 cosθ (sin 2 θ + 2)] 0 = π

0

4 3

2π

Substitute and integrate with respect to φ to obtain:

4 4 8π 2π Ptot = C ∫ dφ = C [φ ]0 = C 3 0 3 3

Solving for C yields:

C=

3 Ptot 8π

Substitute for Ptot and evaluate C to obtain:

C=

3 (500 kW ) = 59.68 kW 8π

Substituting for C in equation (1) and evaluating I(120 km, 90°):

I (120 km, 90°) =

59.68 kW (120 km )2

= 4.14 μW/m 2 28 ••• Regulations require that licensed radio stations have limits on their broadcast power so as to avoid interference with signals from distant stations. You are in charge of checking compliance with the law. At a distance of 30.0 km from a radio station that broadcasts from a single vertical electric dipole antenna at a frequency of 800 kHz, the intensity of the electromagnetic wave is 2.00 × 10–13 W/m2. What is the total power radiated by the station? Picture the Problem The intensity of radiation from an electric dipole is given by C(sin2θ)/r2, where C is a constant whose units are those of power, r is the distance from the dipole to the point of interest, and θ is the angle between the electric r dipole moment and the position vector r . We can integrate the intensity to express

860

Chapter 30

the total power radiated by the antenna and use this result to evaluate C. Knowing C we can find the total power radiated by the station. From the definition of intensity we have:

dP = IdA and Ptot = ∫∫ I (r , θ ) dA

where, in polar coordinates, dA = r 2 sin θ dθ dφ Substitute for dA to obtain:

2π π

Ptot =

∫ ∫ I (r ,θ ) r

2

sin θ dθ dφ (1)

0 0

sin 2 θ r2

Express the intensity of the signal as a function of r and θ :

I (r ,θ ) = C

Substitute for I(r,θ) in equation (1) to obtain:

Ptot = C ∫ ∫ sin 3 θ dθ dφ

From integral tables we find that:

(2)

2π π

0 0

π

∫ sin

3

θ dθ = − 13 cos θ (sin 2 θ + 2 )]0 = π

0

Substitute and integrate with respect to φ to obtain: From equation (2) we have:

4 3

2π

Ptot =

C=

4 4 8π 2π C ∫ dφ = C [φ ] 0 = C 3 0 3 3

I (r ,θ )r 2 sin 2 θ

Substitute for C in the expression for Ptot to obtain:

8π I (r ,θ )r 2 3 sin 2 θ or, because θ = 90°, 8π Ptot = I (r )r 2 3

Substitute numerical values and evaluate Ptot:

Ptot =

Ptot =

8π 2 2.00 × 10−13 W/m 2 (30.0 km ) 3

(

)

= 1.51 mW

29 ••• A small private plane approaching an airport is flying at an altitude of 2.50 km above sea level. As a flight controller at the airport, you know your

Maxwell’s Equations and Electromagnetic Waves

861

system uses a vertical electric dipole antenna to transmit 100 W at 24.0 MHz. What is the intensity of the signal at the plane’s receiving antenna when the plane is 4.00 km from the airport? Assume the airport is at sea level. Picture the Problem The intensity of radiation from the airport’s vertical dipole antenna is given by C(sin2θ)/r2, where C is a constant whose units are those of power, r is the distance from the dipole to the point of interest, and θ is the angle r between the electric dipole moment and the position vector r . We can integrate the intensity to express the total power radiated by the antenna and use this result to evaluate C. Knowing C we can find the intensity of the signal at the plane’s elevation and distance from the airport.

Express the intensity of the signal as a function of r and θ :

sin 2 θ I (r ,θ ) = C 2 r

From the definition of intensity we have:

dP = IdA and Ptot = ∫∫ I (r , θ ) dA

(1)

where, in polar coordinates, dA = r 2 sin θ dθ dφ Substitute for dA to obtain:

2π π

Ptot =

∫ ∫ I (r ,θ ) r

2

sin θ dθ dφ

0 0

Substituting for I(r,θ) yields:

2π π

Ptot = C ∫ ∫ sin 3 θ dθ dφ 0 0

From integral tables we find that:

π

∫ sin

3

θ dθ = − 13 cos θ (sin 2 θ + 2)]0 = π

0

2π

Substitute and integrate with respect to φ to obtain:

4 4 8π 2π Ptot = C ∫ dφ = C [φ ] 0 = C 3 0 3 3

Solving for C yields:

C=

Substitute for C in equation (1) to obtain:

I (r ,θ ) =

3 Ptot 8π 3Ptot sin 2 θ 8π r 2

4 3

862

Chapter 30

At the elevation of the plane:

⎛ 4000 m ⎞ ⎟⎟ = 58.0° ⎝ 2500 m ⎠

θ = tan −1 ⎜⎜ and r=

Substitute numerical values and evaluate I(4717 m, 58°):

(2500 m )2 + (4000 m )2

= 4717 m

3(100 W ) sin 2 58.0° I (4717 m, 58.0°) = 8π (4717 m )2 = 386 nW/m 2

Energy and Momentum in an Electromagnetic Wave 30 • An electromagnetic wave has an intensity of 100 W/m2. Find its (a) rms electric field strength, and (b) rms magnetic field strength. Picture the Problem We can use Pr = I/c to find the radiation pressure. The intensity of the electromagnetic wave is related to the rms values of its electric and magnetic field strengths according to I = ErmsBrms/μ0, where Brms = Erms/c.

(a) Relate the intensity of the electromagnetic wave to Erms and Brms:

I=

Erms Brms

μ0

or, because Brms = Erms/c, E E c E2 I = rms rms = rms μ0 μ0c Erms = μ0cI

Solving for Erms yields:

Substitute numerical values and evaluate Erms: E rms =

(4π ×10

−7

N/A 2 )(2.998 × 108 m/s)(100 W/m 2 ) = 194 V/m

(b) Express Brms in terms of Erms:

Brms =

Erms c

Substitute numerical values and evaluate Brms:

Brms =

194 V/m = 647 nT 2.998 × 108 m/s

31 • [SSM] The amplitude of an electromagnetic wave’s electric field is 400 V/m. Find the wave’s (a) rms electric field strength, (b) rms magnetic field strength, (c) intensity and (d) radiation pressure (Pr).

Maxwell’s Equations and Electromagnetic Waves

863

Picture the Problem The rms values of the electric and magnetic fields are found from their amplitudes by dividing by the square root of two. The rms values of the electric and magnetic field strengths are related according to Brms = Erms/c. We can find the intensity of the radiation using I = ErmsBrms/μ0 and the radiation pressure using Pr = I/c.

(a) Relate Erms to E0:

E rms =

E0 2

=

400 V/m = 282.8 V/m 2

= 283 V/m (b) Find Brms from Erms:

Brms =

E rms 282.8 V/m = c 2.998 × 108 m/s

= 0.9434 μT = 943 nT (c) The intensity of an electromagnetic wave is given by:

I=

Substitute numerical values and evaluate I:

I=

Erms Brms

μ0

(282.8 V/m )(0.9434 μT ) 4π ×10 −7 N/A 2

= 212.3 W/m 2 = 212 W/m 2 (d) Express the radiation pressure in terms of the intensity of the wave:

Pr =

I c

Substitute numerical values and evaluate Pr:

Pr =

212.3 W/m 2 = 708 nPa 2.998 × 108 m/s

32 • The rms value of an electromagnetic wave’s electric field strength is 400 V/m. Find the wave’s (a) rms magnetic field strength, (b) average energy density, and (c) intensity. Picture the Problem Given Erms, we can find Brms using Brms = Erms/c. The average energy density of the wave is given by uav = ErmsBrms/μ0c and the intensity of the wave by I = uavc .

(a) Express Brms in terms of Erms:

Brms =

Erms c

864

Chapter 30

Substitute numerical values and evaluate Brms:

Brms =

400 V/m = 1.334 μT 2.998 ×108 m/s

= 1.33 μT (b) The average energy density uav is given by:

uav =

Substitute numerical values and evaluate uav:

u av =

Erms Brms μ0c

(400 V/m)(1.334 μT )

(4π ×10

−7

N/A 2 )(2.998 × 108 m/s)

= 1.417 μJ/m 3 = 1.42 μJ/m 3 (c) Express the intensity as the product of the average energy density and the speed of light in a vacuum: Substitute numerical values and evaluate I:

I = uav c

I = (1.417 μJ/m 3 )(2.998 × 108 m/s ) = 425 W/m 2

33 •• (a) An electromagnetic wave that has an intensity equal to 200 W/m2 is normal to a black 20 cm by 30 cm rectangular card absorbs 100 percent of the wave. Find the force exerted on the card by the radiation. (b) Find the force exerted by the same wave if the card reflects 100 percent of the wave. Picture the Problem We can find the force exerted on the card using the definition of pressure and the relationship between radiation pressure and the intensity of the electromagnetic wave. Note that, when the card reflects all the radiation incident on it, conservation of momentum requires that the force is doubled.

(a) Using the definition of pressure, express the force exerted on the card by the radiation:

Fr = Pr A

Relate the radiation pressure to the intensity of the wave:

Pr =

I c

Substitute for Pr to obtain:

Fr =

IA c

Maxwell’s Equations and Electromagnetic Waves Substitute numerical values and evaluate Fr:

865

(200 W/m )(0.20 m)(0.30 m ) = 2

Fr

2.998 × 108 m/s

= 40 nN (b) If the card reflects all of the radiation incident on it, the force exerted on the card is doubled:

Fr = 80 nN

34 •• Find the force exerted by the electromagnetic wave on the card in Part (b) of Problem 33 if both the incident and reflected rays are at angles of 30º to the normal. Picture the Problem Only the normal component of the radiation pressure exerts a force on the card.

Using the definition of pressure, express the force exerted on the card by the radiation:

Fr = 2 Pr A cos θ where the factor of 2 is a consequence of the fact that the card reflects the radiation incident on it.

Relate the radiation pressure to the intensity of the wave:

Pr =

I c

Substitute for Pr to obtain:

Fr =

2 IA cosθ c

Substitute numerical values and evaluate Fr: Fr =

2(200 W/m 2 )(0.20 m )(0.30 m )cos 30° = 69 nN 2.998 × 108 m/s

35 • [SSM] (a) For a given distance from a radiating electric dipole, at what angle (expressed as θ and measured from the dipole axis) is the intensity equal to 50 percent of the maximum intensity? (b) At what angle θ is the intensity equal to 1 percent of the maximum intensity? Picture the Problem At a fixed distance from the electric dipole, the intensity of radiation is a function θ alone.

(a) The intensity of the radiation from the dipole is proportional to sin2θ:

I (θ ) = I 0 sin 2 θ

(1)

where I0 is the maximum intensity.

866

Chapter 30

For I = 12 I 0 :

1 2

I 0 = I 0 sin 2 θ ⇒ sin 2 θ = 12

( )=

Solving for θ yields:

θ = sin −1

(b) For I = 0.01I 0 :

0.01I 0 = I 0 sin 2 θ ⇒ sin 2 θ = 0.01

Solving for θ yields:

θ = sin −1 0.01 = 5.7°

45°

1 2

(

)

36 •• A laser pulse has an energy of 20.0 J and a beam radius of 2.00 mm. The pulse duration is 10.0 ns and the energy density is uniformly distributed within the pulse. (a) What is the spatial length of the pulse? (b) What is the energy density within the pulse? (c) Find the rms values of the electric and magnetic fields in the pulse. Picture the Problem The spatial length L of the pulse is the product of its speed c and duration Δt. We can find the energy density within the pulse using its definition (u = U/V). The electric amplitude of the pulse is related to the energy density in the beam according to u =∈0 E 2 and we can find B from E using

B = E/c. (a) The spatial length L of the pulse is the product of its speed c and duration Δt: Substitute numerical values and evaluate L:

L = cΔt

L = (2.998 × 108 m/s )(10.0 ns ) = 2.998 m = 3.00 m

(b) The energy density within the pulse is the energy of the beam per unit volume:

u=

U U = V π r 2L

Substitute numerical values and evaluate u:

u=

20.0 J π (2.00 mm )2 (2.998 m )

= 530.9 kJ/m 3 = 531kJ/m 3 (c) E is related to u according to:

2 ⇒ E rms = u =∈ 0 E rms

u

∈0

Maxwell’s Equations and Electromagnetic Waves Substitute numerical values and evaluate Erms:

E rms =

867

530.9 kJ/m 3 8.854 ×10 −12 C 2 / N ⋅ m 2

= 244.9 MV/m = 245 MV/m Use Brms = Erms/c to find Brms:

Brms =

244.9 MV/m = 0.817 T 2.998 × 10 8 m/s

37 •• [SSM] An electromagnetic plane wave has an electric field that is parallel to the y axis, and has a Poynting vector that is given by r S ( x, t ) = 100 W/m 2 cos2 [kx − ωt ] iˆ , where x is in meters, k = 10.0 rad/m, ω = 3.00 × 109 rad/s, and t is in seconds. (a) What is the direction of propagation of the wave? (b) Find the wavelength and frequency of the wave. (c) Find the electric and magnetic fields of the wave as functions of x and t.

(

)

Picture the Problem We can determine the direction of propagation of the wave, its wavelength, and its frequency by examining the argument of the cosine r function. We can find E from S = E 2 μ 0 c and B from B = E/c. Finally, we can r use the definition of the Poynting vector and the given expression for S to find r r E and B .

(a) Because the argument of the cosine function is of the form kx − ωt , the wave propagates in the +x direction. 2π

(b) Examining the argument of the cosine function, we note that the wave number k of the wave is:

k=

Examining the argument of the cosine function, we note that the angular frequency ω of the wave is:

ω = 2πf = 3.00 ×10 9 s −1

Solving for f yields: r (c) Express the magnitude of S in terms of E:

f =

λ

= 10.0 m −1 ⇒ λ = 0.628 m

3.00 × 10 9 s −1 = 477 MHz 2π

r r E2 S = ⇒ E = μ0c S μ0c

868

Chapter 30

Substitute numerical values and evaluate E:

E=

(4π ×10

−7

)(

Because r S ( x, t ) = 100 W/m 2 cos 2 [kx − ω t ] iˆ r 1 r r E×B: and S =

(

)(

)

N/A 2 2.998 × 108 m/s 100 W/m 2 = 194.1V/m

)

μ0

r E ( x, t ) = (194 V/m ) cos[kx − ωt ] ˆj where k = 10.0 rad/m and ω = 3.00 × 109 rad/s. 194.1V/m = 647.4 nT 2.998 × 10 8 m/s

Use B = E/c to evaluate B:

B=

r 1 r r E × B , the direction Because S =

r B ( x, t ) =

r of B must be such that the cross r r product of E with B is in the +x direction:

where k = 10.0 rad/m and ω = 3.00 × 109 rad/s.

μ0

(647 nT )cos[kx − ωt ] kˆ

A parallel-plate capacitor is being charged. The capacitor consists of a 38 •• pair of identical circular parallel plates that have radius b and a separation distance d. (a) Show that the displacement current in the capacitor gap has the same value as the conduction current in the capacitor leads. (b) What is the direction of the Poynting vector in the region between the capacitor plates? (c) Find an expression for the Poynting vector in this region and show that its flux into the region between the plates is equal to the rate of change of the energy stored in the capacitor. Picture the Problem We can use the expression for the electric field strength between the plates of the parallel-plate capacitor and the definition of the displacement current to show that the displacement current in the capacitor is equal to the conduction current in the capacitor leads. In (b) we can use the definition of the Poynting vector and the directions of the electric and magnetic fields to determine the direction of the Poynting vector between the capacitor r plates. In (c), we’ll demonstrate that the flux of S into the region between the plates is equal to the rate of change of the energy stored in the capacitor by evaluating these quantities separately and showing that they are equal.

(a) The displacement current is proportional to the rate at which the flux is changing between the plates:

I d =∈0

dφe d dE =∈0 ( AE ) =∈0 A dt dt dt

Maxwell’s Equations and Electromagnetic Waves The electric field strength between the plates of the capacitor is given by:

Substituting for E yields:

E=

869

Q ∈0 A

where Q is the instantaneous charge on the capacitor plates.

I d =∈0 A

d Q dQ = = I dt ∈0 A dt

r r (b) Because E is perpendicular to the plates of the capacitor and B is tangent to circles that are concentric and whose center is through the middle of the capacitor r plates, S points radially inward toward the center of the capacitor. (c) The Poynting vector is:

r 1 r r S= E×B

μ0

r Letting the direction of E be the +x direction:

Apply Ampere’s law to a closed circular path of radius R ≤ b to obtain:

(1)

r E = Eiˆ where E is the electric field strength between the plates of the capacitor.

B(2π R ) = μ0 I d

dφe d = μ0 ∈0 EA dt dt dE = μ 0 ∈0 π R 2 dt

Substituting for Id and simplifying yields:

B(2πR ) = μ0 ∈0

Solve for B to obtain:

B=

μ 0 ∈0 2

R

dE dt

and r μ ∈ dE ˆ B=− 0 0 R j dt 2 where ˆj is a unit vector that is tangent to the concentric circles.

870

Chapter 30

r r Substitute for B and E in equation (1) and simplify to obtain: r E

x

(

Rˆ

r B

y r S

R

The rate at which energy is stored in the capacitor is:

Because E =

r ⎛ 1 ⎞ ⎛ μ ∈ dE ⎞ S = ⎜⎜ E ⎟⎟iˆ × ⎜ − 0 0 R ⎟ ˆj dt μ 2 ⎠ ⎝ 0 ⎠ ⎝ ∈ E dE ˆ ˆ = 0 R i ×−j 2 dt ∈ E dE ˆ = − 0 RR 2 dt where R ≤ b, E is the electric field strength between the plates, R is the radial distance from the line joining the centers of the plates, Rˆ is a unit vector pointing radially outward from the line joining the centers of the plates, and b is the radius of the plates.

Q : ∈0 A

(

dU d d = (uV ) = V ∈0 E 2 dt dt dt dE = Ad ∈0 E dt

Substituting for E and simplifying yields:

Because

)

⎛ Q ⎞d ⎛ Q ⎞ dU ⎟⎟ ⎟⎟ ⎜⎜ = Ad ∈0 ⎜⎜ dt ⎝∈0 A ⎠ dt ⎝∈0 A ⎠ =

Consider a cylindrical surface of length d and radius b. Because r S points inward, the energy flowing into the solenoid per unit time is:

)

Qd dQ Qd = I ∈0 A dt ∈0 A

∫ S dA = S (2π bd ) n

dE ⎞ ⎛ = ⎜ 12 ∈0 Eb ⎟ (2π bd ) dt ⎠ ⎝ dE = π ∈0 Eb 2 d dt ⎛ Q ⎞ 2 d ⎛ Q ⎞ ⎜⎜ ⎟⎟b d ⎜⎜ ⎟ dt ⎝∈0 A ⎟⎠ ⎝ ∈0 A ⎠ 2 ⎛ Q ⎞ b d dQ Qd =π⎜ 2 ⎟ = I ⎝ πb ⎠ ∈0 A dt ∈0 A

∫ S dA = π ∈ n

0

r S dA = dU dt , we’ve proved that the flux of S into the region ∫ n

between the capacitor is equal to the rate of change of the energy stored in the capacitor.

Maxwell’s Equations and Electromagnetic Waves

871

39 •• [SSM] A pulsed laser fires a 1000-MW pulse that has a 200-ns duration at a small object that has a mass equal to 10.0 mg and is suspended by a fine fiber that is 4.00 cm long. If the radiation is completely absorbed by the object, what is the maximum angle of deflection of this pendulum? (Think of the system as a ballistic pendulum and assume the small object was hanging vertically before the radiation hit it.) Picture the Problem The diagram shows the displacement of the pendulum bob, through an angle θ, as a consequence of the complete absorption of the radiation incident on it. We can use conservation of energy (mechanical energy is conserved after the collision) to relate the maximum angle of deflection of the pendulum to the initial momentum of the pendulum bob. Because the displacement of the bob during the absorption of the pulse is negligible, we can use conservation of momentum (conserved during the collision) to equate the momentum of the electromagnetic pulse to the initial momentum of the bob.

Apply conservation of energy to obtain:

θ

L

L cos θ

m h

Kf − Ki + U f − U i = 0

or, because Ui = Kf = 0 and K i = pi2 − + Uf = 0 2m

Uf is given by: Substitute for Uf:

Solve for θ to obtain:

Ug = 0

U f = mgh = mgL(1 − cos θ )

pi2 − + mgL(1 − cos θ ) = 0 2m ⎛

θ = cos −1 ⎜⎜1 − ⎝

pi2 ⎞ ⎟ 2m 2 gL ⎟⎠

pi2 , 2m

872

Chapter 30

Use conservation of momentum to relate the momentum of the electromagnetic pulse to the initial momentum pi of the pendulum bob:

U PΔt = = pi c c where Δt is the duration of the pulse. pem wave =

⎡

Substitute for pi:

θ = cos −1 ⎢1 − ⎣

2 P 2 (Δt ) ⎤ ⎥ 2m 2c 2 gL ⎦

Substitute numerical values and evaluate θ : 2 2 ⎡ ⎤ ( 1000 MW ) (200 ns ) θ = cos ⎢1 − ⎥ = 6.10° 2 2 8 2 ⎣⎢ 2(10.0 mg ) 2.998 × 10 m/s 9.81 m/s (0.0400 m ) ⎦⎥ −1

(

)(

)

Remarks: The solution presented here is valid only if the displacement of the bob during the absorption of the pulse is negligible. (Otherwise, the horizontal component of the momentum of the pulse-bob system is not conserved during the collision.) We can show that the displacement during the pulse-bob collision is small by solving for the speed of the bob after absorbing the pulse. Applying conservation of momentum (mv = P(Δt)/c) and solving for v gives v = 6.67 × 10−7 m/s. This speed is so slow compared to c, we can conclude that the duration of the collision is extremely close to 200 ns (the time for the pulse to travel its own length). Traveling at 6.67 × 10−7 m/s for 200 ns, the bob would travel 1.33 × 10−13 m—a distance 1000 times smaller that the diameter of a hydrogen atom. (Because 6.67×10−7 m/s is the maximum speed of the bob during the collision, the bob would actually travel less than 1.33 × 10−13 m during the collision.) 40 •• The mirrors used in a particular type of laser are 99.99% reflecting. (a) If the laser has an average output power of 15 W, what is the average power of the radiation incident on one of the mirrors? (b) What is the force due to radiation pressure on one of the mirrors? Picture the Problem We can use the definitions of pressure and the relationship between radiation pressure and the intensity of the radiation to find the force due to radiation pressure on one of the mirrors.

(a) Because only about 0.01 percent of the energy inside the laser "leaks out", the average power of the radiation incident on one of the mirrors is:

P=

15 W = 1.5 × 10 5 W −4 1.0 × 10

Maxwell’s Equations and Electromagnetic Waves

873

(b) Use the definition of radiation pressure to obtain:

Fr A where Fr is the force due to radiation pressure and A is the area of the mirror on which the radiation is incident.

The radiation pressure is also related to the intensity of the radiation:

2I 2P = c Ac where P is the power of the laser and the factor of 2 is due to the fact that the mirror is essentially totally reflecting.

Equate the two expression for the radiation pressure and solve for Fr:

Fr 2 P 2P ⇒ Fr = = A Ac c

Substitute numerical values and evaluate Fr:

Fr =

Pr =

Pr =

(

)

2 1.5 × 105 W = 1.0 mN 2.998 × 108 m/s

41 •• [SSM] (a) Estimate the force on Earth due to the pressure of the radiation on Earth by the Sun, and compare this force to the gravitational force of the Sun on Earth. (At Earth’s orbit, the intensity of sunlight is 1.37 kW/m2.) (b). Repeat Part (a) for Mars which is at an average distance of 2.28 × 108 km from the Sun and has a radius of 3.40 × 103 km. (c) Which planet has the larger ratio of radiation pressure to gravitational attraction. Picture the Problem We can find the radiation pressure force from the definition of pressure and the relationship between the radiation pressure and the intensity of the radiation from the Sun. We can use Newton’s law of gravitation to find the gravitational force the Sun exerts on Earth and Mars.

(a) The radiation pressure exerted on Earth is given by:

Express the radiation pressure in terms of the intensity of the radiation I from the Sun: Substituting for Pr, Earth and A yields:

Pr, Earth =

Fr, Earth

Pr, Earth =

I c

⇒ Fr, Earth = Pr, Earth A A where A is the cross-sectional area of Earth.

Fr, Earth

Iπ R 2 = c

874

Chapter 30

Substitute numerical values and evaluate Fr:

π (1.37 kW/m 2 )(6.37 × 10 6 m )

2

Fr, Earth =

2.998 × 10 8 m/s

= 5.825 × 10 8 N = 5.83 × 10 8 N The gravitational force exerted on Earth by the Sun is given by:

Gmsun mearth r2 where r is the radius of Earth’s orbit. Fg, Earth =

Substitute numerical values and evaluate Fg, Earth: Fg, Earth =

(6.673 ×10

−11

)(

)(

N ⋅ m 2 / kg 2 1.99 × 10 30 kg 5.98 × 10 24 kg

(1.50 ×10

Express the ratio of the force due to radiation pressure Fr, Earth to the gravitational force Fg, Earth:

11

m

)

2

Fr, Earth

=

Express the radiation pressure on Mars in terms of the intensity of the radiation IMars from the sun: Substituting for Pr, Mars and A yields:

Express the ratio of the solar constant at Earth to the solar constant at Mars: Substitute for I Mars to obtain:

22

N

5.825 × 10 8 N = 1.65 × 10 −14 22 3.529 × 10 N

Fg, Earth or Fr, Earth = 1.65 × 10 −14 Fg, Earth

(

(b) The radiation pressure exerted on Mars is given by:

) = 3.529 ×10

)

Pr, Mars =

Fr, Mars

Pr, Mars =

I Mars c

Fr, Mars =

2 I Marsπ RMars c

⇒ Fr, Mars = Pr, Mars A A where A is the cross-sectional area of Mars.

I Mars ⎛ rearth =⎜ I earth ⎜⎝ rMars

Fr, Mars

2

⎞ ⎛r ⎟⎟ ⇒ I Mars = I earth ⎜⎜ earth ⎠ ⎝ rMars

2 I earth π RMars = c

⎛ rearth ⎜⎜ ⎝ rMars

⎞ ⎟⎟ ⎠

2

⎞ ⎟⎟ ⎠

2

Maxwell’s Equations and Electromagnetic Waves

875

Substitute numerical values and evaluate Fr, Mars:

π (1.37 kW/m 2 )(3.40 × 10 3 km ) ⎛ 1.50 × 1011 m ⎞ 2

Fr, Mars =

⎜⎜ ⎟⎟ = 7.18 × 10 7 N 11 ⎝ 2.28 × 10 m ⎠

2.998 × 10 m/s 8

The gravitational force exerted on Mars by the Sun is given by:

2

Gmsun mMars Gmsun (0.11mEarth ) = r2 r2 where r is the radius of Mars’ orbit. Fg, Mars =

Substitute numerical values and evaluate Fg

Fg, Mars =

(6.673 ×10

−11

)(

)

(

N ⋅ m 2 / kg 2 1.99 × 10 30 kg (0.11) 5.98 × 10 24 kg

(2.28 ×10 m)

)

2

11

= 1.68 × 10 21 N Express the ratio of the force due to radiation pressure Fr, Mars to the gravitational force Fg, Mars:

Fr, Mars

=

Fg, Mars or Fr, Mars =

7.18 × 10 7 N = 4.27 × 10 −14 21 1.68 × 10 N

(4.27 × 10 )F −14

g, Mars

(c) Because the ratio of the radiation pressure force to the gravitational force is 1.65 × 10−14 for Earth and 4.27 × 10−14 for Mars, Mars has the larger ratio. The reason that the ratio is higher for Mars is that the dependence of the radiation pressure on the distance from the Sun is the same for both forces (r−2), whereas the dependence on the radii of the planets is different. Radiation pressure varies as R2, whereas the gravitational force varies as R3 (assuming that the two planets have the same density, an assumption that is nearly true). Consequently, the ratio of the forces goes as R 2 / R 3 = R −1 . Because Mars is smaller than Earth, the ratio is larger.

The Wave Equation for Electromagnetic Waves 42 • Show by direct substitution that Equation 30-8a is satisfied by the wave function Ey = E0 sin (kx − ω t ) = E0 sin k (x − ct ) where c = ω/k. Picture the Problem We can show that Equation 30-8a is satisfied by the wave function Ey by showing that the ratio of ∂2Ey/∂x2 to ∂2Ey/∂t2 is 1/c2 where c = ω/k.

Chapter 30

876

Differentiate E y = E0 sin (kx − ωt )

∂E y

∂ [E0 sin(kx − ωt )] ∂x ∂x = kE0 cos(kx − ωt )

with respect to x:

=

Evaluate the second partial derivative of Ey with respect to x:

∂2Ey

Differentiate E y = E0 sin (kx − ωt )

∂E y

∂x

2

∂ [kE0 cos(kx − ωt )] ∂x = − k 2 E0 sin(kx − ωt ) =

∂ [E0 sin(kx − ωt )] ∂t ∂t = −ωE0 cos(kx − ωt )

with respect to t:

=

Evaluate the second partial derivative of Ey with respect to t:

∂2Ey

Divide equation (1) by equation (2) to obtain:

∂2 Ey 2 2 ∂x 2 = − k E0 sin (kx − ωt ) = k ∂ 2 E y − ω 2 E0 sin (kx − ωt ) ω 2 ∂t 2 or 2 ∂2Ey k 2 ∂2Ey 1 ∂ Ey = 2 = 2 ω ∂t 2 c ∂t 2 ∂x 2 provided c = ω/k.

43

•

(1)

∂t

2

∂ [− ωE0 cos(kx − ωt )] (2) ∂t = −ω 2 E 0 sin(kx − ωt ) =

Use the values of μ0 and ∈0 in SI units to compute 1 ∈0 μ0 and

show that it is equal to 3.00 × 108 m/s. Picture the Problem Substitute numerical values and evaluate c:

c=

(4π ×10

1

−7

N/A

2

)(8.854 ×10

−12

C / N⋅m 2

2

)

= 3.00 × 10 8 m/s

r •• (a) Use Maxwell’s equations to show for a plane wave, in which E r ∂B y ∂E z ∂E z ∂B y = and . and B are independent of y and z, that = μ0 ∈0 ∂t ∂t ∂x ∂x (b) Show that Ez and By also satisfy the wave equation. 44

Maxwell’s Equations and Electromagnetic Waves

877

Picture the Problem We can use Figures 30-5 and 30-6 and a derivation similar to that in the text to obtain the given results.

∂E z Δx ∂x

In Figure 30-5, replace Bz by Ez. For Δx small:

E z ( x2 ) = E z ( x1 ) +

r Evaluate the line integral of E around the rectangular area ΔxΔz:

r r ∂E E ∫ ⋅ d l ≈ − ∂xz ΔxΔz

Express the magnetic flux through the same area:

∫ B dA = B ΔxΔz

Apply Faraday’s law to obtain:

r r ∂ E ∫ ⋅ dl ≈ − ∂ t

S

n

(1)

y

∂

∫ B dA = − ∂ t (B ΔxΔz ) n

S

y

=− Substitute in equation (1) to obtain:

∂B y ∂t

ΔxΔz

∂B ∂E z ΔxΔz = − y ΔxΔz ∂x ∂t

− or

∂E z ∂B y = ∂x ∂t In Figure 30-6, replace Ey by By and r evaluate the line integral of B around the rectangular area ΔxΔz: Evaluate these integrals to obtain:

r r B ∫ ⋅ d l = μ0 ∈0

S

En dA

provided there are no conduction currents.

∂B y ∂x

(b) Using the first result obtained in (a), find the second partial derivative of Ez with respect to x:

∫

= μ0 ∈0

∂ ⎛ ∂E z ⎜ ∂x ⎝ ∂x

∂E z ∂t

⎞ ∂ ⎛ ∂B y ⎟ = ⎜⎜ ⎠ ∂x ⎝ ∂ t

or ∂ 2 Ez ∂ ⎛ ∂B = ⎜⎜ y 2 ∂x ∂ t ⎝ ∂x

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ ⎠

878

Chapter 30

Use the second result obtained in (a) to obtain:

∂ 2 Ez ∂ ⎛ ∂E z ⎞ ∂ 2 Ez ⎜ ⎟ = = μ ε μ ∈ 0 0 0 0 ∂x 2 ∂ t ⎜⎝ ∂ t ⎟⎠ ∂t2 or, because μ0∈0 = 1/c2,

1 ∂ 2 Ez ∂ 2 Ez . = ∂x 2 c2 ∂ t 2 Using the second result obtained in (a), find the second partial derivative of By with respect to x:

∂ ⎛ ∂B y ⎜ ∂x ⎜⎝ ∂x

or ∂ 2 By

∂x Use the first result obtained in (a) to obtain:

2

∂ 2 By

⎞ ∂ ⎛ ∂E z ⎞ ⎟⎟ = μ 0 ∈0 ⎜ ⎟ ∂x ⎜⎝ ∂ t ⎟⎠ ⎠

= μ 0 ∈0

∂ ⎛ ∂E z ⎞ ⎜ ⎟ ∂ t ⎝ ∂x ⎠ ∂ ⎛ ∂B y ⎜ ∂ t ⎜⎝ ∂ t

∂ 2 By ⎞ ⎟ = μ ∈ 0 0 ⎟ ∂x 2 ∂t2 ⎠ or, because μ0∈0 = 1/c2, 2 ∂ 2 By 1 ∂ By . = ∂x 2 c2 ∂ t 2 = μ 0 ∈0

[SSM] Show that any function of the form y(x, t) = f(x – vt) or 45 •• y(x, t) = g(x + vt) satisfies the wave Equation 30-7 Picture the Problem We can show that these functions satisfy the wave equations by differentiating them twice (using the chain rule) with respect to x and t and equating the expressions for the second partial of f with respect to u.

Let u = x − vt. Then:

∂ f ∂u ∂ f ∂ f = = ∂ x ∂ x ∂u ∂u and ∂ f ∂u ∂ f ∂f = = −v ∂t ∂t ∂u ∂u

Express the second derivatives of f with respect to x and t to obtain:

2 ∂2 f ∂2 f ∂2 f 2 ∂ f and = = v ∂ x2 ∂ u 2 ∂t2 ∂ u2

Divide the first of these equations by the second to obtain:

∂2f ∂ x2 1 1 ∂2 f ∂2 f = = 2⇒ ∂ x2 v2 ∂ t 2 v ∂2f ∂t2

Maxwell’s Equations and Electromagnetic Waves Let u = x + vt. Then:

879

∂ f ∂u ∂ f ∂ f = = ∂ x ∂ x ∂u ∂u and ∂ f ∂u ∂ f ∂f = =v ∂t ∂t ∂u ∂u

Express the second derivatives of f with respect to x and t to obtain:

2 ∂2 f ∂2 f ∂2 f 2 ∂ f and = =v ∂ x2 ∂ u 2 ∂t2 ∂ u2

Divide the first of these equations by the second to obtain:

∂2f ∂ x2 1 1 ∂2 f ∂2 f = = ⇒ ∂ x2 v2 ∂ t 2 v2 ∂2f ∂t2

General Problems 46 • An electromagnetic wave has a frequency of 100 MHz and is traveling r in a vacuum. The magnetic field is given by B (z , t ) = 1.00 × 10 −8 T cos(kz − ωt )iˆ . (a) Find the wavelength and the direction of propagation of this wave. (b) Find r the electric field vector E ( z , t ) . (c) Determine the Poynting vector, and use it to find the intensity of this wave.

(

)

Picture the Problem We can use c = fλ to find the wavelength. Examination of the argument of the cosine function will reveal the direction of propagation of the wave. We can find the magnitude, wave number, and angular frequency of the electric vector from the given information and the result of (a) and use these r results to obtain E (z, t). Finally, we can use its definition to find the Poynting vector.

(a) Relate the wavelength of the wave to its frequency and the speed of light:

λ=

Substitute numerical values and evaluate λ:

2.998 × 108 m/s λ= = 3.00 m 100MHz

c f

From the sign of the argument of the cosine function and the spatial dependence on z, we can conclude that the wave propagates in the +z direction.

880

Chapter 30

(

r (b) Express the amplitude of E :

)(

E = cB = 2.998 ×108 m/s 10 −8 T = 3.00 V/m

)

ω = 2π f = 2π (100 MHz ) = 6.28 × 108 s −1

Find the angular frequency and wave number of the wave:

and

k=

2π

λ

=

2π = 2.09 m −1 3.00 m

r r Because S is in the positive z direction, E must be in the negative y direction in order to satisfy the Poynting vector expression:

[

]

r E ( z , t ) = − (3.00 V/m )cos (2.09 m −1 ) z − (6.28 × 108 s −1 )t ˆj (c) Use its definition to express and evaluate the Poynting vector:

(

)

r 1 r r − (3.00 V/m ) 10 −8 T S (z, t ) = E×B = cos 2 2.09 m −1 z − 6.28 ×108 s −1 t ˆj × iˆ μ0 4π ×10 −7 N/A 2 or

[(

) ](

) (

[

)

]

r S ( z, t ) = (23.9 mW/m 2 )cos 2 (2.09 m −1 ) z − (6.28 × 108 s −1 )t kˆ

The intensity of the wave is the average magnitude of the Poynting vector. The average value of the square of the cosine function is 1/2:

r I= S =

1 2

(23.9 mW/m ) 2

= 11.9 mW/m 2

47 •• [SSM] A circular loop of wire can be used to detect electromagnetic waves. Suppose the signal strength from a 100-MHz FM radio station 100 km distant is 4.0 μW/m2, and suppose the signal is vertically polarized. What is the maximum rms voltage induced in your antenna, assuming your antenna is a 10.0-cm-radius loop? Picture the Problem We can use Faraday’s law to show that the maximum rms voltage induced in the loop is given by ε rms = AωB0 2 , where A is the area of

the loop, B0 is the amplitude of the magnetic field, and ω is the angular frequency of the wave. Relating the intensity of the radiation to B0 will allow us to express ε rms as a function of the intensity.

Maxwell’s Equations and Electromagnetic Waves

ε = − dφ m

(

881

)

d r d B ⋅ Anˆ = − (BA) dt dt dt dB d = −A = −πR 2 (B0 sin ωt ) dt dt 2 = −πR ωB0 cos ωt = −ε peak cos ωt

The emf induced in the antenna is given by Faraday’s law:

where

=−

ε peak = πR 2ωB0 and R is the

radius of the loop antenna..

ε rms equals ε peak divided by the

ε rms =

square root of 2: The intensity of the signal is given by:

ε peak 2

=

πR 2ωB0 2

(1)

E 0 B0 2μ 0 or, because E 0 = cB0 , I=

cB0 B0 B02 c I= = 2μ 0 2μ 0 Solving for B0 yields:

2μ 0 I c

B0 =

Substituting for B0 and ω in equation (1) and simplifying yields:

ε rms =

πR 2 (2πf )

2μ 0 I c

2

= 2π 2 R 2 f

μ0 I c

Substitute numerical values and evaluate εrms: 2 2 −7 ε rms = 2π 2 (0.100 m )2 (100 MHz ) (4π × 10 N/A )(84.0 μW/m ) =

2.998 × 10 m/s

2.6 mV

48 •• The electric field strength from a radio station some distance from the electric dipole transmitting antenna is given by 1.00 × 10−4 N/C cos 1.00 × 106 rad/s t , where t is in seconds. (a) What peak voltage is picked up on a 50.0-cm long wire oriented parallel with the electric field direction? (b) What is the maximum voltage that can be induced by this electromagnetic wave in a conducting loop of radius 20.0 cm? What orientation of the loop does this require?

(

) (

)

882

Chapter 30

Picture the Problem The voltage induced in the piece of wire is the product of the electric field and the length of the wire. The maximum rms voltage induced in the loop is given by ε = AωB0 , where A is the area of the loop, B0 is the

amplitude of the magnetic field, and ω is the angular frequency of the wave. (a) Because E is independent of x:

V = El where l is the length of the wire.

Substitute numerical values and evaluate V:

V = 1.00 × 10− 4 N/C cos106 t (0.500 m )

[(

)

= (50.0 μV ) cos 106 t

]

and Vpeak = 50.0 μV (b) The maximum voltage induced in a loop is given by:

Eliminate B0 in favor of E0 and substitute for A to obtain:

ε = ωB0 A where A is the area of the loop and B0 is the amplitude of the magnetic field.

ε=

ωE0π R 2 c

Substitute numerical values and evaluate ε:

ε = (1.00 ×10

6

)(

)

s −1 1.00 × 10 −4 N/C π (0.200 m ) = 41.9 nV 2.998 × 108 m/s 2

The loop antenna should be oriented so the transmitting antenna lies in the plane of the loop. 49 ••• A parallel-plate capacitor has circular plates of radius a that are separated by a distance d. In the gap between the two plates is a thin straight wire of resistance R that connects the centers of the two plates. A time-varying voltage given by V0 sin ωt is applied across the plates. (a) What is the current drawn by this capacitor? (b) What is the magnetic field as a function of the radial distance r from the centerline within the capacitor plates? (c) What is the phase angle between the current drawn by the capacitor and the applied voltage? Picture the Problem Some of the charge entering the capacitor passes through the resistive wire while the rest of it accumulates on the upper plate. The total current is the rate at which the charge passes through the resistive wire plus the rate at which it accumulates on the upper plate. The magnetic field between the capacitor plates is due to both the current in the resistive wire and the displacement current though a surface bounded by a circle a distance r from the

Maxwell’s Equations and Electromagnetic Waves

883

resistive wire. The phase difference between the current drawn by the capacitor and the applied voltage may be calculated using a phasor diagram.

(a) The current drawn by the capacitor is the sum of the conduction current through the resistance wire and dQ/dt, where Q is the charge on the upper plate of the capacitor:

I = Ic +

Express the conduction current Ic in terms of the potential difference between the plates and the resistance of the wire:

Ic =

Because Q = CV :

dQ dV =C = ωCV0 cos ωt dt dt

Substitute in equation (1):

I=

The capacitance of a parallel-plate capacitor with plate area A and plate separation d is given by: Substituting for C in equation (2) gives:

C=

dQ dt

(1)

V V0 = sin ωt R R

V0 sin ωt + ωCV 0 cos ωt R

(2)

∈0 A ∈0 π a 2 d

=

d

⎛1 ⎞ ω ∈0 π a 2 I = V0 ⎜⎜ sin ωt + cos ωt ⎟⎟ d ⎝R ⎠

884

Chapter 30

(b) Apply the generalized form of Ampere’s law to a circular path of radius r centered within the plates of the capacitor, where I'd is the

r r B ∫ ⋅ d l = μ0 (I c + I'd ) C

displacement current through the flat surface S bounded by the path and Ic is the conduction current through the same surface: By symmetry the line integral is B times the circumference of the circle of radius r:

B(2π r ) = μ 0 (I c + I'd )

(3)

In the region between the capacitor plates there is a uniform electric field due to the surface charges +Q and – Q. The associated displacement current through S is:

dφ e d =∈ 0 ( A'E ) dt dt dE dE =∈ 0 A' =∈ 0 π r 2 dt dt provided (r ≤ a )

To evaluate the displacement current we first must evaluate E everywhere on S. Near the surface of a conductor, where σ is the surface charge density:

E = σ ∈0 , where σ = Q A = Q π a 2

Substituting for E in the equation for I'd gives:

I' d =∈ 0

(

so

E=

Q

∈0 π a 2

dE d ⎛V ⎞ =∈0 π r 2 ⎜ ⎟ dt dt ⎝ d ⎠ ∈ π r 2 dV ∈0 π r 2 d (V0 sin ωt ) = 0 = d dt d dt ∈0 π r 2 =ω V0 cos ωt d

I'd =∈0 π r 2

Solving for B in equation (3) and substituting for Ic and I'd yields: B(r ) =

)

⎞ μ0 (I c + I'd ) μ0 ⎛ V0 ∈ π r2 ⎜⎜ sin ωt + ω 0 = V0 cos ωt ⎟⎟ 2π r 2π r ⎝ R d ⎠

⎞ μ0V0 ⎛ 1 ∈0 π r 2 ⎜ ⎟⎟ sin ω ω cos ω = t + t 2π r ⎜⎝ R d ⎠

Maxwell’s Equations and Electromagnetic Waves (c) Both the charge Q and the conduction current Ic are in phase with V. However, dQ/dt, which is equal to the displacement current Id through S for r ≥ a, lags V by 90°. (Mathematically, cos ωt lags behind sin ωt by 90°.) The voltage V leads the current I = Ic + Id by phase angle δ. The current relation is expressed in terms of the current amplitudes: The values of the conduction and displacement current amplitudes are obtained by comparison with the answer to Part (a):

885

I = Ic + Id or I max sin (ωt + δ ) = I c, max sin ωt + I d, max cos ωt

I c, max =

V0 R

and

I d, max =

A phasor diagram for adding the currents Ic and Id is shown to the right. The conduction current Ic is in phase with the voltage V across the resistor and Id lags behind it by 90°:

ω ∈0 π a 2V0 d I c, max

r Ic

δ I d,max

I max

r Id

r I

From the phasor diagram we have: tan δ =

=

r V

I d,max I c,max

=

V0

ω ∈0 π a 2 d V0 R

Rω ∈0 π a 2 d

so

⎛ Rω ∈ 0 π a 2 ⎞ ⎟⎟ d ⎝ ⎠

δ = tan −1 ⎜⎜

Remarks: The capacitor and the resistive wire are connected in parallel. The potential difference across each of them is the applied voltage V0 sin ωt. 50 •• A 20-kW beam of electromagnetic radiation is normal to a surface that reflects 50 percent of the radiation. What is the force exerted by the radiation on this surface?

886

Chapter 30

Picture the Problem The total force on the surface is the sum of the force due to the reflected radiation and the force due to the absorbed radiation. From the conservation of momentum, the force due to the 10 kW that are reflected is twice the force due to the 10 kW that are absorbed.

Express the total force on the surface:

Ftot = Fr + Fa

Substitute for Fr and Fa to obtain:

Ftot =

2( 12 P ) 12 P 3P + = c c 2c

Substitute numerical values and evaluate Ftot:

Ftot =

3(20 kW ) = 0.10 mN 2 2.998 ×108 m/s

(

)

51 •• [SSM] The electric fields of two r harmonic electromagnetic waves of angular frequency ω1 and ω2 are given by E1 = E1,0 cos(k1x − ω1t )ˆj and r by E2 = E2 ,0 cos(k2 x − ω2 t + δ )ˆj . For the resultant of these two waves, find (a) the instantaneous Poynting vector and (b) the time-averaged Poynting vector. (c) Repeat Parts (ra) and (b) if the direction of propagation of the second wave is reversed so that E2 = E2 ,0 cos(k2 x + ω 2t + δ )ˆj Picture the Problem We can use the definition of the Poynting vector and the r r relationship between B and E to find the instantaneous Poynting vectors for each of the resultant wave motions and the fact that the time average of the cross product term is zero for ω1 ≠ ω2, and ½ for the square of cosine function to find the time-averaged Poynting vectors.

(a) Because both waves propagate in the x direction: Express B in terms of E1 and E2:

r r r E × B = μ 0 Siˆ ⇒ B = Bkˆ

B=

1 (E1 + E2 ) c

Substitute for E1 and E2 to obtain: r 1 B (x, t ) = [E1,0 cos(k1 x − ω1t ) + E 2,0 cos(k 2 x − ω 2 t + δ )]kˆ c

Maxwell’s Equations and Electromagnetic Waves The instantaneous Poynting vector for the resultant wave motion is given by: r 1 (E1,0 cos(k1 x − ω1t ) + E2,0 cos(k 2 x − ω 2t + δ )) ˆj S ( x, t ) =

μ0

× =

1 (E1,0 cos(k1 x − ω1t ) + E2,0 cos(k 2 x − ω 2t + δ ))kˆ c

(

(E

1

μ0c

2 cos(k1 x − ω1t ) + E 2,0 cos(k 2 x − ω 2 t + δ )) ˆj × kˆ

1, 0

1

= μ0c

[E

2 1, 0

)

cos 2 (k1 x − ω1t ) + 2 E1, 0 E 2, 0 cos(k1 x − ω1t )

]

× cos(k 2 x − ω 2 t + δ ) + E 22,0 cos 2 (k 2 x − ω 2 t + δ ) iˆ

(b) The time average of the cross product term is zero for ω1 ≠ ω2, and the time average of the square of the cosine terms is ½:

r S av =

[E 2μ c 1

2 1, 0

]

+ E22, 0 iˆ

0

r (c) In this case B2 = − Bkˆ because the wave with k = k2 propagates in the − iˆ direction. The magnetic field is then:

r 1 B (x, t ) = [E1,0 cos(k1 x − ω1t ) − E 2,0 cos(k 2 x + ω 2 t + δ )]kˆ c The instantaneous Poynting vector for the resultant wave motion is given by:

r 1 (E1,0 cos(k1 x − ω1t ) + E2,0 cos(k 2 x − ω2t + δ )) ˆj S ( x, t ) =

μ0

× =

1 (E1,0 cos(k1 x − ω1t ) − E2,0 cos(k 2 x + ω2t + δ ))kˆ c

[E μc 1

2 1, 0

]

cos 2 (k1 x − ω1t ) − E 22,0 cos 2 (k 2 x + ω 2 t + δ ) iˆ

0

The time average of the square of the cosine terms is ½:

r S av =

1 2μ0c

[E

2 1, 0

]

− E22,0 iˆ

∂Bz ∂E = μo ∈0 n (Equation 30-10) follows from ∂x ∂t r r ∂E y ∫CB ⋅ d l = −μ 0 ∈0 ∫S ∂t dA (Equation 30-6d with I = 0) by integrating along a

52

••

Show that

887

888

Chapter 30

suitable curve C and over a suitable surface S in a manner that parallels the derivation of Equation 30-9. Picture the Problem We’ll choose the curve with sides Δx and Δz in the xy plane shown in the diagram and apply Equation 30-6d to show that

∂ Ey ∂ Bz . = − μ 0 ∈0 ∂x ∂t Because Δx is very small, we can approximate the difference in Bz at the points x1 and x2 by:

Bz ( x2 ) − Bz ( x1 ) = ΔB ≈

Then:

∫ B ⋅ dl ≈ μ

r

C

r

0

∈0

∂E y ∂t

∂Bz Δx ∂x

ΔxΔz

The flux of the electric field through this curve is approximately:

∫

Apply Faraday’s law to obtain:

∂E y ∂Bz ΔxΔz = − μ 0 ∈0 ΔxΔz ∂x ∂t or ∂E y ∂Bz = − μ 0 ∈0 ∂x ∂t

S

En dA = E y ΔxΔy

53 •• For your backpacking excursions, you have purchased a radio capable of detecting a signal as weak as 1.00 × 10–14 W/m2. This radio has a 2000-turn coil antenna that has a radius of 1.00 cm wound on an iron core that increases the magnetic field by a factor of 200. The broadcast frequency of the radio station is 1400 kHz. (a) What is the peak magnetic field strength of an electromagnetic wave of this minimum intensity? (b) What is the peak emf that it is capable of inducing in the antenna? (c) What would be the peak emf induced in a straight 2.00-m long metal wire oriented parallel to the direction of the electric field? Picture the Problem We can use the relationship between the average value of the Poynting vector (the intensity), E0, and B0 to find B0. The application of Faraday’s law will allow us to find the emf induced in the antenna. The emf induced in a 2.00-m wire oriented in the direction of the electric field can be found using ε = El and the relationship between E and B.

Maxwell’s Equations and Electromagnetic Waves (a) The intensity of the signal is related the amplitude of the magnetic field in the wave:

S av = I =

E0 B0 cB02 = ⇒ B0 = 2μ0 2μ0

889

2μ0 I c

Substitute numerical values and evaluate B0:

B0 =

(

)(

)

2 4π × 10 −7 N/A 2 1.00 × 10 −14 W/m 2 = 9.16 × 10 −15 T 8 2.998 × 10 m/s

(b) Apply Faraday’s law to the antenna coil to obtain:

ε (t ) = d (BA) = A d (NK m B0 sin ωt ) dt dt = NK m AB0ω cos ωt = ε peak cos ωt where

Substitute numerical values and evaluate

ε peak = NK m AB0ω

ε peak :

ε peak = 2000(200)π (0.0100 m )2 (9.16 ×10−15 T )[2π (1400 kHz )] = (c) The voltage induced in the wire is the product of its length l and the amplitude of electric field E0: Relate E to B: Substitute for E to obtain:

10.1 μV

ε = El

E = cB = cB0 sin ωt

ε = clB0 sin ωt = ε peak sin ωt where ε peak = clB0

Substitute numerical values and evaluate ε peak :

ε peak = (2.998 × 10 8 m/s)(2.00 m ) (9.16 × 10 −15 T ) =

5.49 μV

54 •• The intensity of the sunlight striking Earth’s upper atmosphere is 1.37 kW/m2. (a) Find the rms values of the magnetic and electric fields of this light. (b) Find the average power output of the Sun. (c) Find the intensity and the radiation pressure at the surface of the Sun. Picture the Problem We can use I = ErmsBrms/μ0 and Brms = Erms/c to express Erms in terms of I. We can then use Brms = Erms/c to find Brms. The average power output

890

Chapter 30

of the Sun is given by Pav = 4πR 2 I where R is the Earth-Sun distance. The intensity and the radiation pressure at the surface of the sun can be found from the definitions of these physical quantities. (a) The intensity of the radiation is given by:

Erms Brms

I=

μ0

=

2 Erms ⇒ Erms = cμ0 I cμ0

Substitute numerical values and evaluate Erms :

E rms =

(2.998 ×10

8

)(

)(

)

m/s 4π × 10 −7 N/A 2 1.37 kW/m 2 = 718.4 V/m

= 718 V/m Use Brms = E rms c to evaluate Brms :

Brms =

718.4 V/m = 2.40 μT 2.998 × 10 8 m/s

(b) Express the average power output of the Sun in terms of the solar constant:

Pav = 4π R 2 I where R is the Earth-sun distance.

Substitute numerical values and evaluate Pav:

Pav = 4π 1.50 × 1011 m 1.37 kW/m 2

(

)( 2

= 3.874 × 10 26 W = 3.87 × 10 26 W

(c) Express the intensity at the surface of the Sun in terms of the sun’s average power output and radius r: Substitute numerical values (see Appendix B for the radius of the Sun) and evaluate I at the surface of the Sun:

Pav 4π r 2

I=

I=

3.874 × 10 26 W

4π (6.96 × 108 m )

2

= 6.363 × 10 7 W/m 2 = 6.36 × 10 7 W/m 2

Express the radiation pressure in terms of the intensity:

Pr =

I c

Substitute numerical values and evaluate Pr:

Pr =

6.363 × 10 7 W/m 2 = 0.212 Pa 2.998 ×108 m/s

)

Maxwell’s Equations and Electromagnetic Waves

891

55 ••• [SSM] A conductor in the shape of a long solid cylinder that has a length L, a radius a, and a resistivity ρ carries a steady current I that is uniformly distributed over its cross-section. (a) Use Ohm’s law to relate the electric field r r E in the conductor to I, ρ, and a. (b) Find the magnetic field B just outside the conductor. (c) Use the results from Part (a) and Part (b) to compute the Poyntingr r r r vector S = E × B μ0 at r = a (the edge of the conductor). In what direction is S ?

(

)

(d) Find the flux ∫ S n dA through the surface of the cylinder, and use this flux to

show that the rate of energy flow into the conductor equals I2R, where R is the resistance of the cylinder.

Picture the Problem A side view of the cylindrical conductor is shown in the diagram. Let the current be to the right (in the +x direction) and choose a coordinate system in which the +y direction is radially outward from the axis of the conductor. Then the +z direction is tangent to cylindrical surfaces that are concentric with the axis of the conductor (out of the plane of the diagram at the location indicated in the diagram). We can use Ohm’s law to relate the electric field strength E in the conductor to I, ρ, and a and Ampere’s law to find the r r magnetic field strength B just outside the conductor. Knowing E and B we can r find S and, using its normal component, show that the rate of energy flow into the conductor equals I2R, where R is the resistance. y r B r E

x I

Axis of the conductor

a

(a) Apply Ohm’s law to the cylindrical conductor to obtain:

r Because E is in the same direction as I:

IρL Iρ L = EL = A π a2 Iρ where E = . π a2 V = IR =

r E=

Iρ ˆ i where iˆ is a unit vector π a2

in the direction of the current. (b) Applying Ampere’s law to a circular path of radius a at the surface of the cylindrical conductor yields:

r r B ∫ ⋅ d l = B(2π a ) = μ0 I enclosed = μ0 I C

892

Chapter 30

Solve for the magnetic field strength B to obtain:

B=

Apply a right-hand rule to determine r the direction of B at the point of interest shown in the diagram:

r B=

(c) The Poynting vector is given by:

r r Substitute for E and B and simplify to obtain:

μ0 I 2π a μ0 I ˆ θ where θˆ is a unit vector 2π a

perpendicular to iˆ and tangent to the surface of the conducting cylinder.

r 1 r r S= E×B

μ0

r 1 ⎛ Iρ ⎞ ⎛ μ 0 I ⎞ ⎜ ⎟ iˆ × ⎜ ⎟ kˆ S= μ 0 ⎜⎝ π a 2 ⎟⎠ ⎜⎝ 2π a ⎟⎠ I 2ρ = − 2 3 ˆj 2π a

Letting rˆ be a unit vector directed radially outward from the axis of the cylindrical conductor yields.

r I 2ρ S = − 2 3 rˆ where rˆ is a unit 2π a vector directed radially outward away from the axis of the conducting cylinder.

(d) The flux through the surface of the conductor into the conductor is:

∫ S dA =S (2π aL )

Substitute for Sn, the inward r component of S , and simplify to obtain:

I 2ρ I 2 ρL ∫ Sn dA = 2π 2 a 3 (2π aL ) = π a 2

Because R =

ρL A

=

ρL : πa 2

n

∫ S dA = n

I 2R

Remarks: The equality of the two flow rates is a statement of the conservation of energy. 56 ••• A long solenoid that has n turns per unit length carries a current that increases linearly with time. The solenoid has radius R, length L, and the current I in the windings is given by I = at. (a) Find the induced electric field at a distance r < R from the central axis r of the solenoid. (b) Find the magnitude and direction of the Poynting vector S at r = R (just inside the solenoid windings). (c) Calculate

Maxwell’s Equations and Electromagnetic Waves

893

the flux ∫ S n dA into the region inside the solenoid, and show that this flux equals the rate of increase of the magnetic energy inside the solenoid. Picture the Problem An end view of the solenoid is shown in the diagram. Let the current be clockwise and choose a coordinate system in which the +x direction is tangent to cylindrical surfaces concentric with the axis of the solenoid. Note that the +z direction is out of the plane of the diagram. We can use Faraday’s law to express the induced electric field at a distance r < R from the solenoid axis in r terms of the rate of change of magnetic flux and B = nμ 0 at to express B in terms

of the current in the windings of the solenoid. We can use the results of (a) to find r the Poynting vector S at the cylindrical surface r = R just inside the solenoid windings. In Part (c) we’ll use the definition of flux and the expression for the r magnetic energy in a given region to show that the flux of S into the solenoid equals the rate of increase of the magnetic energy inside the solenoid. y r E

× r B

x

r I Axis of the solenoid

R

r

r

(a) Apply Faraday’s law to a circular path of radius r < R to relate the magnitude of the induced electric field to the magnitude of the rate of change of the magnetic flux:

∫ E ⋅ d l = E (2π r ) = −

Solving for E yields:

E=−

C

1 dφm 2π r dt

dφm dt

(1)

894

Chapter 30

Express the magnetic field strength inside a long solenoid:

B = nμ 0 I = nμ 0 at

The magnetic flux through a circle of radius r is:

φm = BA = nμ 0 atπ r 2

Substitute for φm in equation (1) and simplify to obtain:

r The direction of E is such that it produces an emf that opposes the increase in the current, so if the r current is clockwise, then E is in the opposite direction and:

E=−

nμ a r d nμ 0 atπ r 2 = − 0 2π r dt 2 1

[

]

r E = − 12 nμ 0 a rθˆ where θˆ is a unit vector that is tangent to the circles that are concentric with the axis of the solenoid.

r (b) Express S at r = R:

r 1 r r S= E×B

r The magnitude of B is given by:

B = nμ 0 I = nμ 0 at

Applying a right-hand rule yields:

r B = −(nμ 0 at ) kˆ

r r Substitute for E and B in equation (2) and simplify to obtain:

r 1 ⎛ nμ 0 a r ⎞ S= ⎜− ⎟ iˆ × (− nμ 0 at ) kˆ μ0 ⎝ 2 ⎠

μ0

=− r r Because E × B is a vector that points toward the axis of the r solenoid, we can also write S as:

(c) Consider a cylindrical surface of length L and radius R. Because r S points inward, the energy flowing into the solenoid per unit time is:

(2)

n 2 μ 0 a 2 Rt ˆ j 2

r S = − 12 n 2 μ0 a 2 Rtrˆ

where rˆ is a unit vector that points radially outward—away from the axis of the solenoid. ⎛ n 2 μ 0 a 2 Rt ⎞ ∫ S n dA =2π RLS = 2π RL⎜⎜⎝ 2 ⎟⎟⎠ = n 2π μ 0 R 2 La 2t

Maxwell’s Equations and Electromagnetic Waves Express the magnetic energy in the solenoid:

U B = umV =

(

B2 π R2L 2μ0

895

)

( μ 0 nat )2 (π R 2 L ) = 2μ0

=

Evaluate dUB/dt:

n 2π μ0 R 2 La 2t 2 2

dU B d ⎡ n 2π μ 0 R 2 La 2 t 2 ⎤ = ⎢ ⎥ dt dt ⎣ 2 ⎦ = n 2π μ 0 R 2 La 2 t = ∫ S n dA

Remarks: The equality of the two flow rates is a statement of the conservation of energy. 57 ••• Small particles are be blown out of the solar system by the radiation pressure of sunlight. Assume that each particle is spherical, has a radius r, has a density of 1.00 g/cm3, and absorbs all the radiation in a cross-sectional area of πr2. Assume the particles are located at some distance d from the Sun, which has a total power output of 3.83 × 1026 W. (a) What is the critical value for the radius r of the particle for which the radiation force of repulsion just balances the gravitational force of attraction to the Sun? (b) Do particles that have radii larger than the critical value get ejected from the solar system, or is it only particles that have radii smaller than the critical value that get ejected? Explain your answer. Picture the Problem We can use a condition for translational equilibrium to obtain an expression relating the forces due to gravity and radiation pressure that act on the particles. We can express the force due to radiation pressure in terms of the radiation pressure and the effective cross sectional area of the particles and the radiation pressure in terms of the intensity of the solar radiation. We can solve the resulting equation for r.

(a) Apply the condition for translational equilibrium to the particle:

The radiation pressure Pr depends on the intensity of the radiation I:

Fr − Fg = 0 or, since Fr = PrA and Fg = mg, GM s m Pr A − =0 (1) R2 Pr =

I c

Chapter 30

896

The intensity of the solar radiation at a distance R is:

I=

Substitute for I to obtain:

Pr =

Substitute for Pr, A, and m in equation (1): Solve for r to obtain:

P 4π R 2 P 4π R 2 c

3 4 P 2 3 π r ρ GM s (π r ) − R 2 = 0 4π R 2 c

r=

3P 16π ρ c GM s

Substitute numerical values and evaluate r: r=

3(3.83 × 10 26 W ) 16π (1.00 g/cm 3 )(2.998 × 108 m/s )(6.673 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg )

= 574 nm (b) Because both the gravitational and radiation pressure forces decrease as the square of the distance from the Sun, it is then a comparison of grain mass to grain area. Since mass is proportional to volume and thus varies with the cube of the radius, the larger grains have more mass and thus experience a stronger gravitational than radiation-pressure force. The critical radius is an upper limit and so particles smaller than that radius will be blown out. 58 ••• When an electromagnetic wave at normal incidence on a perfectly conducting surface is reflected, the electric field of the reflected wave at the reflecting surface is equal and opposite to the electric field of the incident wave at the reflecting surface. (a) Explain why this assertion is valid. (b) Show that the superposition of incident and reflected waves results in a standing wave. (c) Are the magnetic fields of the incident waves and reflected waves at the reflecting surface equal and opposite as well? Explain your answer. Picture the Problem r (a) At a perfectly conducting surface E = 0 . Therefore, the sum of the electric r r fields of the incident and reflected wave must add to zero, and so E i = − E r .

(b) Let the incident and reflected waves be described by:

Ei = E 0 y cos(ωt − kx ) and E r = − E 0 y cos(ωt + kx )

Maxwell’s Equations and Electromagnetic Waves

897

Use the trigonometric identity cos(α + β) = cosαcosβ − sinαsinβ to obtain: Ei + E r = E 0 y cos(ωt − kx ) − E 0 y cos(ωt + kx ) = E 0 y [cos(ωt − kx ) − cos(ωt + kx )] = E 0 y [cos ωt cos(− kx ) − sin ωt sin (− kx ) − cos ωt cos kx + sin ωt sin (kx )] = E 0 y [cos ωt cos kx + sin ωt sin kx − cos ωt cos kx + sin ωt sin kx] = 2 E 0 y sin ωt sin kx , the equation of a standing wave. r r r r (c) Because E × B = μ 0 S and S is in the direction of propagation of the wave, we r see that for the incident wave Bi = Bz cos(ωt − kx ) . Since both S and Ey are reversed for the reflected wave Br = B z cos(ωt + kx ) . So the magnetic field vectors are in the direction at the reflecting surface and add at that surface. r r Hence B = 2 Br .

59 ••• [SSM] An intense point source of light radiates 1.00 MW isotropically (uniformly in all directions). The source is located 1.00 m above an infinite, perfectly reflecting plane. Determine the force that the radiation pressure exerts on the plane. Picture the Problem Let the point source be a distance a above the plane. Consider a ring of radius r and thickness dr in the plane and centered at the point directly below the light source. Express the force on this elemental ring and integrate the resulting expression to obtain F.

The intensity anywhere along this infinitesimal ring is given by:

P 4π (r + a 2 )

The elemental force dF on the elemental ring of area 2π rdr is given by:

dF =

2

=

P rdr a 2 2 c(r + a ) r 2 + a 2 Pardr c(r 2 + a 2 )

32

where we have taken into account that only the normal component of the incident radiation contributes to the force on the plane, and that the plane is a perfectly reflecting plane. Integrate dF from r = 0 to r = ∞:

∞

F=

Pa rdr ∫ c 0 (r 2 + a 2 )3 2

898

Chapter 30

From integral tables:

∞

∫ (r 0

rdr 2

+ a2

)

32

∞

⎤ 1 = ⎥ = 2 2 r + a ⎦0 a −1

Substitute to obtain:

F=

Pa ⎛ 1 ⎞ P ⎜ ⎟= c ⎝a⎠ c

Substitute numerical values and evaluate F:

F=

1.00 MW = 3.34 mN 2.998 × 108 m/s

Chapter 31 Properties of Light Conceptual Problems 1 • [SSM] A ray of light reflects from a plane mirror. The angle between the incoming ray and the reflected ray is 70°. What is the angle of reflection? (a) 70°, (b) 140°, (c) 35°, (d) Not enough information is given to determine the reflection angle. Determine the Concept Because the angles of incidence and reflection are equal, their sum is 70° and the angle of reflection is 35°. (c ) is correct. 2 • A ray of light passes in air is incident on the surface of a piece of glass. The angle between the normal to the surface and the incident ray is 40°, and the angle between the normal and the refracted ray 28°. What is the angle between the incident ray and the refracted ray? (a) 12°, (b) 28°, (c) 40° ,(d) 68° Determine the Concept The angle between the incident ray and the refracted ray is the difference between the angle of incidence and the angle of refraction. (a ) is correct. 3 • During a physics experiment, you are measuring refractive indices of different transparent materials using a red helium-neon laser beam. For a given angle of incidence, the beam has an angle of refraction equal to 28° in material A and an angle of refraction equal to 26° in material B. Which material has the larger index of refraction? (a) A, (b) B, (c) The indices of refraction are the same. (d) You cannot determine the relative magnitudes of the indices of refraction from the data given. Determine the Concept The refractive index is a measure of the extent to which a material refracts light that passes through it. Because the angles of incidence are the same for materials A and B and the angle of refraction is smaller (more bending of the light) for material B, its index of refraction is larger than that of A. (b ) is correct. 4 • A ray of light passes from air into water, striking the surface of the water at an angle of incidence of 45º. Which, if any, of the following four quantities change as the light enters the water: (a) wavelength, (b) frequency, (c) speed of propagation, (d) direction of propagation, (e) none of the above?

899

900

Chapter 31

Determine the Concept When light passes from air into water its wavelength changes ( λ water = λair n water ), its speed changes ( v water = c n water ), and the direction of its propagation changes in accordance with Snell’s law. Its frequency does not change, so (a ) , (c ) and (d ) change. 5 • Earth’s atmosphere decreases in density as the altitude increases. As a consequence, the index of refraction of the atmosphere also decreases as altitude increases. Explain how one can see the Sun when it is below the horizon. (The horizon is the extension of a plane that is tangent to Earth’s surface.) Why does the setting Sun appear flattened? Determine the Concept The decrease in the index of refraction n of the atmosphere with altitude results in refraction of the light from the Sun, bending it toward the normal to the surface of Earth. Consequently, the Sun can be seen even after it is just below the horizon. The setting Sun appears flattened because refraction is greater for light from the lower part of the Sun than for light from the upper part. Atmosphere

Earth

6 • A physics student playing pocket billiards wants to strike her cue ball so that it hits a cushion and then hits the eight ball squarely. She chooses several points on the cushion and then measures the distances from each point to the cue ball and to the eight ball. She aims at the point for which the sum of these distances is least. (a) Will her cue ball hit the eight ball? (b) How is her method related to Fermat’s principle? Neglect any effects due to ball rotation. Determine the Concept (a) Yes. (b) Her procedure is based on Fermat’s principle of least time. The ball presumably bounces off the cushion with an angle of reflection equal to the angle of incidence, just as a light ray would do if the cushion were a mirror. The least time would also be the shortest distance of travel for the light ray. 7 • [SSM] A swimmer at point S in Figure 31-53 develops a leg cramp while swimming near the shore of a calm lake and calls for help. A lifeguard at point L hears the call. The lifeguard can run 9.0 m/s and swim 3.0 m/s. She knows physics and chooses a path that will take the least time to reach the swimmer. Which of the paths shown in the figure does the lifeguard take?

Properties of Light

901

Determine the Concept The path through point D is the path of least time. In analogy to the refraction of light, the ratio of the sine of the angle of incidence to the sine of the angle of refraction equals the ratio of the speeds of the lifeguard in each medium. Careful measurements from the figure show that path LDS is the path that best satisfies this criterion. 8 • Material A has a higher index of refraction than material B. Which material has the larger critical angle for total internal reflection when the material is in air? (a) A, (b) B, (c) The angles are the same. (d) You cannot compare the angles based on the data given. Determine the Concept Because the product of the index of refraction on the incident side of the interface and the sine of the critical angle is equal to one (the index of refraction of air), the material with the smaller index of refraction will have the larger critical angle. (b ) is correct. 9 • [SSM] A human eye perceives color using a structure which is called a cone that is is located on the retina. Three types of molecules compose these cones and each type of molecule absorbs either red, green, or blue light by resonance absorption. Use this fact to explain why the color of an object that appears blue in air appears blue underwater, in spite of the fact that the wavelength of the light is shortened in accordance with Equation 31-6. Determine the Concept In resonance absorption, the molecules respond to the frequency of the light through the Einstein photon relation E = hf. Neither the wavelength nor the frequency of the light within the eyeball depend on the index of refraction of the medium outside the eyeball. Thus, the color appears to be the same in spite of the fact that the wavelength has changed. 10 • Let θ be the angle between the transmission axes of two polarizing sheets. Unpolarized light of intensity I is incident upon the first sheet. What is the intensity of the light transmitted through both sheets? (a) I cos2 θ, (b) (I cos2 θ)/2, (c) (I cos2 θ)/4, (d) I cos θ, (e) (I cos θ)/4, (f) None of the above Picture the Problem The intensity of the light transmitted by the second polarizer is given by I trans = I 0 cos 2 θ, where I 0 = 12 I . Therefore, I trans = 12 I cos 2 θ

and (b) is correct.

902

Chapter 31

11 •• [SSM] Draw a diagram to explain how Polaroid sunglasses reduce glare from sunlight reflected from a smooth horizontal surface, such as the surface found on a pool of water. Your diagram should clearly indicate the direction of polarization of the light as it propagates from the Sun to the reflecting surface and then through the sunglasses into the eye. Determine the Concept The following diagram shows unpolarized light from the sun incident on the smooth surface at the polarizing angle for that particular surface. The reflected light is polarized perpendicular to the plane of incidence, i.e., in the horizontal direction. The sunglasses are shown in the correct orientation to pass vertically polarized light and block the reflected sunlight. Light from the sun Polaroid sunglasses

θP θP θr

Smooth surface

12 • Why is it far less dangerous to stand in front of an intense beam of red light than in front of a very low-intensity beam of gamma rays? Determine the Concept The red light photons contain considerably less energy than the gamma photons so, even though there are likely to be fewer photons in the low-intensity gamma beam, each one is potentially dangerous. 13 • Three energy states of an atom are A, B and C. State B is 2.0 eV above state A and state C is 3.00 eV above state B. Which atomic transition results in the emission of the shortest wavelength of light? (a) B → A, (b) C → B, (c) C → A, (d) A→ C Determine the Concept Because the wavelength of the light emitted in an atomic transition is inversely proportional to the energy difference between the energy levels, the highest energy difference produces the shortest wavelength light. (c ) is correct. 14 • In Problem 13, if the atom is initially in state A, which transition results in the emission of the longest wavelength light? (a) A → B, (b) B → C, (c) A → C, (d) B→ A

Properties of Light

903

Determine the Concept Because the energy required to induce an atomic transition varies inversely with the wavelength of the light that must be absorbed to induce the transition and the transition from A to B is the lowest energy transition, the transition B→ A results in the longest wavelength light. (d ) is

correct. 15

•

[SSM]

What role does the helium play in a helium–neon laser?

Determine the Concept The population inversion between the state E2,Ne and the state 1.96 eV below it (see Figure 31-51) is achieved by inelastic collisions between neon atoms and helium atoms excited to the state E2,He. 16 • When a beam of visible white light that passes through a gas of atomic hydrogen at room temperature is viewed with a spectroscope, dark lines are observed at the wavelengths of the hydrogen atom emission series. The atoms that participate in the resonance absorption then emit light of the same wavelength as they return to the ground state. Explain why the observed spectrum nevertheless exhibits pronounced dark lines. Determine the Concept Although the excited atoms emit light of the same frequency on returning to the ground state, the light is emitted in a random direction, not exclusively in the direction of the incident beam. Consequently, the beam intensity is greatly diminished at this frequency. 17 • [SSM] Which of the following types of light would have the highest energy photons? (a) red (b) infrared (c) blue (d) ultraviolet Determine the Concept The energy of a photon is directly proportional to the frequency of the light and inversely proportional to its wavelength. Of the portions of the electromagnetic spectrum include in the list of answers, ultraviolet light has the highest frequency. (d ) is correct.

Estimation and Approximation 18 • Estimate the time required for light to make the round trip during Galileo’s experiment to measure the speed of light. Compare the time of the round trip to typical human response times. How accurate do you think this experiment is? Picture the Problem We can use the distance, rate, and time relationship to estimate the time required to travel 6 km.

904

Chapter 31

Express the distance d the light in Galileo’s experiment traveled in terms of its speed c and the elapsed time Δt:

d = cΔt ⇒ Δt =

Substitute numerical values and evaluate Δt:

Δt =

Because human reaction is approximately 0.3 s:

Δt reaction 0.3 s = ≈ 2 × 10 4 Δt 2 × 10 −5 s or Δt reaction ≈ (2 × 10 4 )Δt

d c

6 km = 2 × 10 −5 s 8 2.998 × 10 m/s

Because human reaction time is so much longer than the travel time for the light, there was no way that Galileo’s experiment could demonstrate that the speed of light was not infinite. 19 • Estimate the time delay in receiving a light on your retina when you are wearing eyeglasses compared to when you are not wearing your eyeglasses. Determine the Concept We’ll assume that the source of the photon is a distance L from your retina and express the difference in the photon’s travel time when you are wearing your glasses. Let the thickness of your glasses by 2 mm and the index of refraction of the material from which they are constructed by 1.5.

When you are not wearing your glasses, the time required for a light photon, originating a distance L away, to reach your retina is given by:

Δt 0 =

L c

L−d d + c cn (n − 1)d L + (n − 1)d = = Δt 0 + c c

If glass of thickness d and index of refraction n is inserted in the path of the photon, its travel time becomes:

Δt = t in air + t in glass =

The time delay is the difference between Δt and Δt0:

t delay = Δt − Δt 0 =

Substitute numerical values and evaluate tdelay:

t delay =

(n − 1)d c

(1.5 − 1)(2 mm) ≈

2.998 ×108 m/s

3 ps

Properties of Light

905

20 •• Estimate the number of photons that enter your eye if you look for a tenth of a second at the Sun. What energy is absorbed by your eye during that time, assuming that all the photons are absorbed? The total power output of the Sun is 4.2 × 1026 W. Picture the Problem The rate at which photons enter your eye is the ratio of power incident on your pupil to the energy per photon. We’ll assume that the electromagnetic radiation from the Sun is at 550 nm and that, therefore, its photons have energy (given by E = hf = hc λ ) of 2.25 eV.

The rate at which photons enter your eye is the ratio of the rate at which energy is incident on your pupil to the energy carried by each photon: The intensity of the radiation from the Sun is given by:

Solving for Pincident yields:

Pincident dN on pupil = dt Eper photon

I Sun =

Pincident

on pupil

Apupil

Pincident =

on pupil

on pupil

(1)

=

PSun Asphere at Earth's

distance from the Sun

Apupil Asphere at Earth's

PSun (2)

distance from the Sun

Substituting in equation (1) yields:

Apupil dN = dt Asphere at Earth's

PSun

distance from the Sun

Eper photon

1 πd pupil PSun dN = 4 2 dt 4πREarth-Sun Eper photon 2

Substitute for the two areas and simplify to obtain:

⎛ d pupil = ⎜⎜ ⎝ 4 REarth-Sun

2

⎞ PSun ⎟⎟ ⎠ Eper photon

Substitute numerical values and evaluate dN/dt: dN ⎛ 1 mm = ⎜⎜ dt ⎝ 4 1.50 × 1011 m

2

⎞ 4.2 × 10 26 W ⎟⎟ = 3.237 × 1015 s −1 −19 ⎠ 2.25 eV × 1.602 × 10 J eV or, separating variables and integrating this expression,

(

)

N = (3.237 ×1015 s −1 )t

906

Chapter 31 N (0.1 s ) = (3.237 ×1015 s −1 )(0.1 s )

Evaluating N for t = 0.1 s yields:

≈ 3 ×1014 photons The energy deposited, assuming all the photons are absorbed, is the product of the rate at which energy is incident on the pupil and the time during which it is delivered: Substituting for Pincident from

E = Pincident t on pupil

E=

on pupil

equation (2) yields:

Apupil

PSun t

Asphere at Earth's

distance from the Sun

Substitute for the two areas and simplify to obtain:

E=

1 4

2 πd pupil

2 4πREarth -Sun

⎛ d pupil = ⎜⎜ ⎝ 4 REarth-Sun

PSun t 2

⎞ ⎟⎟ PSun t ⎠

Substitute numerical values and evaluate E for t = 0.1 s: 2

⎛ ⎞ 1 mm ⎟⎟ (4.2 × 10 26 W )(0.1 s ) = 0.1167 mJ ≈ 0.1 mJ E (0.1 s ) = ⎜⎜ 11 × ( ) 4 1 . 50 10 m ⎝ ⎠ 21 •• Römer was observing the eclipses of Jupiter’s moon Io with the hope that they would serve as a highly accurate clock that would be independent of longitude. (Prior to GPS, such a clock was needed for accurate navigation.) Io eclipses (enters the umbra of Jupiter’s shadow) every 42.5 h. Assuming an eclipse of Io is observed on Earth on June 1 at midnight when Earth is at location A (as shown in Figure 31-54), predict the expected time of observation of an eclipse one-quarter of a year later when Earth is at location B, assuming (a) the speed of light is infinite and (b) the speed of light is 2.998 × 108 m/s. Picture the Problem We can use the period of Io’s motion and the position of Earth at B to find the number of eclipses of Io during Earth’s movement and then use this information to find the number of days before a night-time eclipse. During the 42.5 h between eclipses of Jupiter’s moon, Earth moves from A to B, increasing the distance from Jupiter by approximately the distance from Earth to the Sun, making the path for the light longer and introducing a delay in the onset of the eclipse.

Properties of Light

907

Tearth 365.24 d 24 h = × 4 4 d = 2191.4 h

(a) Find the time it takes Earth to travel from point A to point B:

t A→ B =

Because there are 42.5 h between eclipses of Io, the number of eclipses N occurring in the time it takes for Earth to move from A to B is:

N=

t A→ B 2191.4 h = = 51.56 42.5 h TIo

Hence, in one-fourth of a year, there will be 51.56 eclipses. Because we want to find the next occurrence that happens in the evening hours, we’ll use 52 as the number of eclipses. We’ll also assume that Jupiter is visible so that the eclipse of Io can be observed at the time we determine. Relate the time t(N) at which the Nth eclipse occurs to N and the period TIo of Io: Evaluate t(52) to obtain:

Subtract the number of whole days to find the clock time t:

t ( N ) = NTIo

⎛ 1d ⎞ ⎟ t (52 ) = (52 )⎜⎜ 42.5 h × 24 h ⎟⎠ ⎝ = 92.083 d

t = t (52) − 92 d = 92.083 d − 92 d = 0.083 d ×

24 h = 1.992 h d

≈ 2 : 00 a.m.

Because June, July, and August have 30, 31, and 31 d, respectively, the date is: (b) Express the time delay Δt in the arrival of light from Io due to Earth’s location at B: Substitute numerical values and evaluate Δt:

September 1

Δt =

rearth-sun c

1.5 × 1011 m 1 min = 500 s × 8 60 s 2.998 × 10 m/s = 8.34 min

Δt =

Hence, the eclipse will actually occur at 2:08 a.m., September 1

908

Chapter 31

22 •• If the angle of incidence is small enough, the small angle approximation sin θ ≈ θ may be used to simplify Snell’s law of refraction. Determine the maximum value of the angle that would make the value for the angle differ by no more than one percent from the value for the sine of the angle. (This approximation will be used in connection with image formation by spherical surfaces in Chapter 32.) Picture the Problem We can express the relative error in using the small angle approximation and then either use 1) trial-and-error methods, 2) a spreadsheet program, or 3) the Solver capability of a scientific calculator to solve the transcendental equation that results from setting the error function equal to 0.01.

Express the relative error δ in using the small angle approximation:

δ (θ ) =

θ − sin θ θ = −1 sin θ sin θ

A spreadsheet program was used to plot the following graph of δ (θ ). 0.016 0.014

delta(theta)

0.012 0.010 0.008 0.006 0.004 0.002 0.000 0.00

0.05

0.10

0.15

0.20

0.25

0.30

theta (radians)

From the graph, we can see that δ (θ ) < 1% for θ ≤ 0.24 radians. In degree measure, θ ≤ 14° Remarks: Using the Solver program on a TI-85 gave θ = 0.244 radians.

Properties of Light

909

The Speed of Light 23 • Mission Control sends a brief wake-up call to astronauts in a spaceship that is far from Earth. 5.0 s after the call is sent, Mission Control can hear the groans of the astronauts. How far from Earth is the spaceship? (a) 7.5 × 108 m, (b) 15 × 108 m, (c) 30 × 108 m, (d) 45 × 108 m, (e) The spaceship is on the moon. Picture the Problem We can use the distance, rate, and time relationship to find the distance to the spaceship.

Relate the distance d to the spaceship to the speed of electromagnetic radiation in a vacuum and to the time for the message to reach the astronauts:

d = cΔt

Noting that the time for the message to reach the astronauts is half the time for Mission Control to hear their response, substitute numerical values and evaluate d:

d = 2.998 × 10 8 m/s (2.5 s )

(

)

= 7.5 × 10 m 8

and (a) is correct.

The distance from a point on the surface of Earth to a point on the 24 • surface of the moon is measured by aiming a laser light beam at a reflector on the surface of the moon and measuring the time required for the light to make a round trip. The uncertainty in the measured distance Δx is related to the uncertainty in the measured time Δt by Δx = 12 cΔt. If the time intervals can be measured to ±1.00 ns, (a) find the uncertainty of the distance. (b) Estimate the percentage uncertainty in the distance. Picture the Problem We can use the given information that the uncertainty in the measured distance Δx is related to the uncertainty in the time Δt by Δx = cΔt to evaluate Δx.

(a) The uncertainty in the distance is:

Δx = 12 cΔt

Substitute numerical values and evaluate Δx:

Δx =

1 2

(2.998 ×10

= ± 15.0 cm

8

)

m/s (± 1.00 ns )

910

Chapter 31

(b)The percent uncertainty in the distance to the Moon is:

ΔxEarth to Moon 15.0 cm = xEarth to Moon 3.84 ×108 m ≈ 10 −8 %

25 •• [SSM] Ole Römer discovered the finiteness of the speed of light by observing Jupiter’s moons. Approximately how sensitive would the timing apparatus need to be in order to detect a shift in the predicted time of the moon’s eclipses that occur when the moon happens to be at perigee ( 3.63× 105 km ) and those that occur when the moon is at apogee ( 4.06 × 105 km )? Assume that an instrument should be able to measure to at least one-tenth the magnitude of the effect it is to measure. Picture the Problem His timing apparatus would need to be sensitive enough to measure the difference in times for light to travel to Earth when the moon is at perigee and at apogee.

The sensitivity of the timing apparatus would need to be one-tenth of the difference in time for light to reach Earth from the two positions of the moon of Jupiter: The time required for light to travel between the two positions of the moon is given by:

Sensitivity = 101 Δt where Δt is the time required for light to travel between the two positions of the moon.

Δt =

d moon

at apogee

Sensitivity =

Δt =

at perigee

c

Substituting for Δt yields:

Substitute numerical values and evaluate the required sensitivity:

− d moon

d moon

at apogee

− d moon

at perigee

10c

4.06 × 10 5 km − 3.63 × 10 5 km (10)(2.998 × 108 m/s)

= 14 ms Remarks: Instruments with this sensitivity did not exist in the 17th century.

Reflection and Refraction 26 • Calculate the fraction of light energy reflected from an air–water interface at normal incidence. Picture the Problem Use the equation relating the intensity of reflected light at normal incidence to the intensity of the incident light and the indices of refraction of the media on either side of the interface.

Properties of Light

911

2

Express the intensity I of the light reflected from an air-water interface at normal incidence in terms of the indices of refraction and the intensity I0 of the incident light:

⎛ n − nwater I = ⎜⎜ air ⎝ nair + nwater

Solve for the ratio I/I0:

I ⎛ nair − nwater =⎜ I 0 ⎜⎝ nair + nwater

Substitute numerical values and evaluate I/I0:

I ⎛ 1.00 − 1.33 ⎞ =⎜ ⎟ = 2.0% I 0 ⎝ 1.00 + 1.33 ⎠

⎞ ⎟⎟ I 0 ⎠

⎞ ⎟⎟ ⎠

2

2

A ray of light is incident on one of a pair of mirrors set at right angles 27 • to each A ray of light is incident on one of two mirrors that are set at right angles to each other. The plane of incidence is perpendicular to both mirrors. Show that after reflecting from each mirror, the ray will emerge traveling in the direction opposite to the incident direction, regardless of the angle of incidence. Picture the Problem The diagram shows ray 1 incident on the vertical surface at an angle θ1, reflected as ray 2, and incident on the horizontal surface at an angle of incidence θ3. We’ll prove that rays 1 and 3 are parallel by showing that θ1 = θ4, i.e., by showing that they make equal angles with the horizontal. Note that the law of reflection has been used in identifying equal angles of incidence and reflection.

1

θ1 θ1 3 2

θ2

θ3 θ3

We know that the angles of the right triangle formed by ray 2 and the two mirror surfaces add up to 180°:

θ 2 + 90° + (90° − θ1 ) = 180° or θ1 = θ 2

The sum of θ2 and θ3 is 90°:

θ 3 = 90° − θ 2

Because θ1 = θ 2 :

θ 3 = 90° − θ1

The sum of θ4 and θ3 is 90°:

θ 3 + θ 4 = 90°

θ4

912

Chapter 31

(90° − θ1 ) + θ 4 = 90° ⇒ θ1 =

Substitute for θ3 to obtain:

θ4

28 •• (a) A ray of light in air is incident on an air–water interface. Using a spreadsheet or graphing program, plot the angle of refraction as a function of the angle of incidence from 0º to 90º. (b) Repeat Part (a), but for a ray of light in water that is incident on a water–air interface. [For Part (b), there is no reflected ray for angles of incidence that are greater than the critical angle.] Picture the Problem Diagrams showing the light rays for the two cases are shown below. In (a) the light travels from air into water and in (b) it travels from water into air.

(a)

(b)

θ1

n1

Air Water

θ2

Air Water

θ1

n2

n2 n1

θ2

(a) Apply Snell’s law to the airwater interface to obtain:

Solving for θ2 yields:

n1 sin θ1 = n2 sin θ 2 where the angles of incidence and refraction are θ1 and θ2, respectively. ⎛ n1 ⎞ sin θ1 ⎟⎟ ⎝ n2 ⎠

θ 2 = sin −1 ⎜⎜

Properties of Light A spreadsheet program to graph θ2 as a function of θ1 is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 A6 A7 B6

Content/Formula 1 1.33333 0 A6 + 5 A6*PI()/180

Algebraic Form n1 n2 θ1 (deg) θ1 + Δθ

θ1 ×

C6

ASIN(($B$1/$B$2)*SIN(B6))

D6

C6*180/PI()

π

180

⎛n ⎞ sin −1 ⎜⎜ 1 sin θ1 ⎟⎟ ⎝ n2 ⎠ 180 θ2 ×

π

1 2 3 4 5 6 7 8 9 21 22 23 24

A B n1= 1 n2= 1.33333

C

D

θ1

θ1

θ2

θ2

(deg) 0 1 2 3

(rad) 0.00 0.02 0.03 0.05

(rad) 0.000 0.013 0.026 0.039

(deg) 0.00 0.75 1.50 2.25

87 88 89 90

1.52 1.54 1.55 1.57

0.847 0.847 0.848 0.848

48.50 48.55 48.58 48.59

913

914

Chapter 31

A graph of θ2 as a function of θ1 follows: 50 Angle of refraction, deg

45 40 35 30 25 20 15 10 5 0 0

10

20

30

40

50

60

70

80

90

Angle of incidence, deg

(b) Change the contents of cell B1 to 1.33333 and the contents of cell B2 to 1 to obtain the following graph: 90 Angle of refraction, deg

80 70 60 50 40 30 20 10 0 0

10

20

30

40

50

Angle of incidence, deg

Note that as the angle of incidence approaches the critical angle for a water-air interface (48.6°), the angle of refraction approaches 90°. 29 •• The red light from a helium-neon laser has a wavelength of 632.8 nm in air. Find the (a) speed, (b) wavelength, and (c) frequency of helium-neon laser light in air, water, and glass. (The glass has an index of refraction equal to 1.50.) Picture the Problem We can use the definition of the index of refraction to find the speed of light in the three media. The wavelength of the light in each medium is its wavelength in air divided by the index of refraction of the medium. The frequency of the helium-neon laser light is the same in all media and is equal to its value in air. The wavelength of helium-neon laser light in air is 632.8 nm.

Properties of Light

The speed of light in a medium whose index of refraction is n is given by:

v=

The wavelength of light in a medium whose index of refraction is n is given by:

λn =

The frequency of the light is equal to its frequency in air independently of the medium in which the light is propagating:

2.998 ×10 8 m/s f = = 632.8 nm λ

Substitute numerical values in equation (1) and evaluate vwater:

c n

915

(1)

λ air n

(2)

c

= 4.74 × 1014 Hz

v water =

2.998 × 10 8 m/s 1.33

= 2.25 × 10 8 m/s Substitute numerical values in equation (2) and evaluate λwater:

λ water =

632.8 nm = 476 nm 1.33

The other speeds and wavelengths are found similarly and are summarized in the following table: (a) speed (m/s)

(b) wavelength (c) frequency (nm) (Hz)

Air

3.00 × 108

633

4.74 × 1014

Water

2.25 × 108

476

4.74 × 1014

glass

2.00 × 108

422

4.74 × 1014

30 •• The index of refraction for silicate flint glass is 1.66 for violet light that has a wavelength in air equal to 400 nm and 1.61 for red light that has a wavelength in air equal to 700 nm. A ray of 700-nm–wavelength red light and a ray of 400-nm-wavelength violet light both have angles of refraction equal to 30º upon entering the glass from air. (a) Which is greater, the angle of incidence of the ray of red light or the angle of incidence of the ray of violet light? Explain your answer. (b) What is the difference between the angles of incidence of the two rays? Picture the Problem Let the subscript 1 refer to the air and the subscript 2 to the silicate glass and apply Snell’s law to the air-glass interface.

916

Chapter 31

(a) Because the index of refraction for violet light is larger than that of red light, for a given incident angle violet light would refract more than red light. Thus to exhibit the same refraction angle, violet light would require an angle of incidence larger than that of red light. (b) Express the difference in their angles of incidence:

Δθ = θ 1, violet − θ 1, red

Apply Snell’s law to the air-glass interface to obtain:

n1 sin θ1 = n 2 sin θ 2

Solving for θ1 yields:

(1)

⎛ n2 ⎞ sin θ 2 ⎟⎟ ⎝ n1 ⎠

θ 1 = sin −1 ⎜⎜

Substitute for θ 1, violet and θ 1, red in equation (1) to obtain: ⎛n ⎞ ⎛n ⎞ Δθ = sin −1 ⎜⎜ violet sin θ 2, violet ⎟⎟ − sin −1 ⎜⎜ red sin θ 2, red ⎟⎟ ⎝ nair ⎠ ⎝ nair ⎠

For θ 2, violet = θ 2, red = 30° : ⎛n ⎞ ⎛n ⎞ ⎛n Δθ = sin −1 ⎜⎜ violet sin 30° ⎟⎟ − sin −1 ⎜⎜ red sin 30° ⎟⎟ = sin −1 ⎜⎜ violet ⎝ n air ⎠ ⎝ n air ⎠ ⎝ 2n air

⎞ ⎛n ⎟⎟ − sin −1 ⎜⎜ red ⎠ ⎝ 2n air

⎞ ⎟⎟ ⎠

Substitute numerical values and evaluate Δθ: ⎛ 1.66 ⎞ ⎛ 1.61 ⎞ ⎟⎟ − sin −1 ⎜⎜ ⎟⎟ = 56.10° − 53.61° = 2.49° Δθ = sin −1 ⎜⎜ ⎝ 2(1.00) ⎠ ⎝ 2(1.00) ⎠ Remarks: Note that Δθ is positive. This means that the angle for violet light is greater than that for red light and confirms our answer in Part (a). 31 •• [SSM] A slab of glass that has an index of refraction of 1.50 is submerged in water that has an index of refraction of 1.33. Light in the water is incident on the glass. Find the angle of refraction if the angle of incidence is (a) 60º, (b) 45º, and (c) 30º. Picture the Problem Let the subscript 1 refer to the water and the subscript 2 to the glass and apply Snell’s law to the water-glass interface.

Properties of Light Apply Snell’s law to the waterglass interface to obtain: Solving for θ2 yields:

(a) Evaluate θ2 for θ1 = 60°:

(b) Evaluate θ2 for θ1 = 45°:

(c) Evaluate θ2 for θ1 = 30°:

917

n1 sin θ1 = n 2 sin θ 2

⎛ n1 ⎞ sin θ 1 ⎟⎟ ⎝ n2 ⎠

θ 2 = sin −1 ⎜⎜

⎛ 1.33 ⎞ sin 60° ⎟ = 50° ⎝ 1.50 ⎠

θ 2 = sin −1 ⎜

⎛ 1.33 ⎞ sin 45° ⎟ = 39° ⎝ 1.50 ⎠

θ 2 = sin −1 ⎜

⎛ 1.33 ⎞ sin 30° ⎟ = 26° ⎝ 1.50 ⎠

θ 2 = sin −1 ⎜

32 •• Repeat Problem 31 for a beam of light initially in the glass that is incident on the glass–water interface at the same angles. Picture the Problem Let the subscript 1 refer to the glass and the subscript 2 to the water and apply Snell’s law to the glass-water interface.

Apply Snell’s law to the waterglass interface to obtain: Solving for θ2 yields:

(a) Evaluate θ2 for θ1 = 60°:

(b) Evaluate θ2 for θ1 = 45°:

(c) Evaluate θ2 for θ1 = 30°:

n1 sin θ1 = n 2 sin θ 2

⎛ n1 ⎞ sin θ 1 ⎟⎟ ⎝ n2 ⎠

θ 2 = sin −1 ⎜⎜

⎛ 1.50 ⎞ sin 60° ⎟ = 78° ⎝ 1.33 ⎠

θ 2 = sin −1 ⎜

⎛ 1.50 ⎞ sin 45° ⎟ = 53° ⎝ 1.33 ⎠

θ 2 = sin −1 ⎜

⎛ 1.50 ⎞ sin 30° ⎟ = 34° ⎝ 1.33 ⎠

θ 2 = sin −1 ⎜

33 •• A beam of light in air strikes a glass slab at normal incidence. The glass slab has an index of refraction of 1.50. (a) Approximately what percentage of the incident light intensity is transmitted through the slab (in one side and out the other)? (b) Repeat Part (a) if the glass slab is immersed in water.

918

Chapter 31

Picture the Problem Let the subscript 1 refer to the medium to the left (air) of the first interface, the subscript 2 to glass, and the subscript 3 to the medium (air) to the right of the second interface. Apply the equation relating the intensity of reflected light at normal incidence to the intensity of the incident light and the indices of refraction of the media on either side of the interface to both interfaces. We’ll neglect multiple reflections at glass-air interfaces.

(a) Express the intensity of the transmitted light in the second medium:

Express the intensity of the transmitted light in the third medium:

n1 = 1.00

n2 = 1.50 I r, 2

I r,1

I2

I3

⎛ n − n2 = I 1 − ⎜⎜ 1 ⎝ n1 + n 2

⎞ ⎟⎟ I 1 ⎠

I1

I 2 = I 1 − I r,1

n3 = 1.00

⎡ ⎛n −n 2 = I 1 ⎢1 − ⎜⎜ 1 ⎢⎣ ⎝ n1 + n 2

⎞ ⎟⎟ ⎠

2

2

⎤ ⎥ ⎥⎦ 2

I 3 = I 2 − I r,2

⎛ n − n3 ⎞ ⎟⎟ I 2 = I 2 − ⎜⎜ 2 ⎝ n 2 + n3 ⎠

⎡ ⎛ n − n ⎞2 ⎤ 3 ⎟⎟ ⎥ = I 2 ⎢1 − ⎜⎜ 2 + n n ⎢⎣ ⎝ 2 3 ⎠ ⎥ ⎦

Substitute for I2 to obtain:

⎡ ⎛n −n 2 I 3 = I 1 ⎢1 − ⎜⎜ 1 + n n ⎢⎣ ⎝ 1 2

Solve for the ratio I3/I1 to obtain:

I 3 ⎡ ⎛ n1 − n2 = ⎢1 − ⎜ I 1 ⎢ ⎜⎝ n1 + n2 ⎣

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ ⎠ 2

2

⎤⎡ ⎛ n − n ⎞2 ⎤ 3 ⎟⎟ ⎥ ⎥ ⎢1 − ⎜⎜ 2 + n n ⎥⎦ ⎢⎣ ⎝ 2 3 ⎠ ⎥ ⎦

⎤⎡ ⎛ n − n ⎞2 ⎤ 3 ⎟⎟ ⎥ ⎥ ⎢1 − ⎜⎜ 2 ⎥⎦ ⎢⎣ ⎝ n 2 + n3 ⎠ ⎥⎦

Substitute numerical values and evaluate I3/I1: 2 2 I 3 ⎡ ⎛ 1.00 − 1.50 ⎞ ⎤ ⎡ ⎛ 1.50 − 1.00 ⎞ ⎤ = ⎢1 − ⎜ ⎟ ⎥ = 92% ⎟ ⎥ ⎢1 − ⎜ I 1 ⎢⎣ ⎝ 1.00 + 1.50 ⎠ ⎥⎦ ⎢⎣ ⎝ 1.50 + 1.00 ⎠ ⎥⎦

Properties of Light

919

(b) With n1 = n3 = 1.33: 2 2 I 3 ⎡ ⎛ 1.33 − 1.50 ⎞ ⎤ ⎡ ⎛ 1.50 − 1.33 ⎞ ⎤ = ⎢1 − ⎜ ⎟ ⎥ = 99% ⎟ ⎥ ⎢1 − ⎜ I 1 ⎢⎣ ⎝ 1.33 + 1.50 ⎠ ⎥⎦ ⎢⎣ ⎝ 1.50 + 1.33 ⎠ ⎥⎦

34 •• This problem is a refraction analogy. A band is marching down a football field with a constant speed v1. About midfield, the band comes to a section of muddy ground that has a sharp boundary making an angle of 30º with the 50-yd line, as shown in Figure 31-55. In the mud, each marcher moves at a speed equal to 12 v1 in a direction perpendicular to the row of markers they are in. (a) Diagram how each line of marchers is bent as it encounters the muddy section of the field so that the band is eventually marching in a different direction. Indicate the original direction by a ray and the final direction by a second ray. (b) Find the angles between these rays and the line normal to the boundary. Is their direction of motion ″bent″ toward the normal or away from it? Explain your answer in terms of refraction. Picture the Problem We can apply Snell’s law to find the angle of refraction of the line of marchers as they enter the muddy section of the field

(a)

30°

θ1 θ2 (b) Apply Snell’s law at the interface to obtain: Solving for θ2 yields:

The ratio of the indices of refraction is the reciprocal of the ratio of the speeds of the marchers in the two media:

n1 sin θ1 = n2 sin θ 2

⎡ n1 ⎤ sin θ1 ⎥ ⎣ n2 ⎦

θ 2 = sin −1 ⎢

v n1 v1 v2 12 v1 = = = = v n2 v1 v1 v2

1 2

920

Chapter 31

Because the left and right sides of the 30° angle and θ1 are mutually perpendicular, θ1 = 30°. Substitute numerical values and evaluate θ2:

θ 2 = sin −1 [12 sin 30°] = 14°

As the line enters the muddy field, its speed is reduced by half and the direction of the forward motion of the line is changed. In this case, the forward motion in the muddy field makes an angle θ2 with respect to the normal to the boundary line. Note that the separation between successive lines in the muddy field is half that in the dry field. 35 •• [SSM] In Figure 31-56, light is initially in a medium that has an index of refraction n1. It is incident at angle θ1 on the surface of a liquid that has an index of refraction n2. The light passes through the layer of liquid and enters glass that has an index of refraction n3. If θ3 is the angle of refraction in the glass, show that n1 sin θ1 = n3 sin θ3. That is, show that the second medium can be neglected when finding the angle of refraction in the third medium. Picture the Problem We can apply Snell’s law consecutively, first to the n1-n2 interface and then to the n2-n3 interface.

Apply Snell’s law to the n1-n2 interface:

n1 sin θ1 = n 2 sin θ 2

Apply Snell’s law to the n2-n3 interface:

n2 sin θ 2 = n3 sin θ 3

Equate the two expressions for n 2 sin θ 2 to obtain:

n1 sin θ1 = n3 sin θ 3

36 •• On a safari, you are spear fishing while wading in a river. You observe a fish gliding by you. If your line of sight to the fish is 64.0o degrees below the horizontal in air, and assuming the spear follows a straight-line path through the air and water after it is released, determine the angle below the horizontal that you should aim your spear gun in order to catch dinner. Assume the spear gun barrel is 1.50 m above the water surface, the fish is 1.20 m below the surface, and the spear travels in a straight line all the way to the fish. Picture the Problem The following pictorial representation summarizes the information given in the problem statement. We can use the geometry of the diagram and apply Snell’s law at the air-water interface to find the aiming angle α.

Properties of Light

α θ1 n1

Air Water

n2

d = 1.20 m

h = 1.50 m

64.0°

θ2

L

l ⎡h + d ⎤ ⎥ ⎣L+l⎦

Use the pictorial representation to express the aiming angle α:

α = tan −1 ⎢

Referring to the diagram, note that:

L = h tan θ 1 and l = d tan θ 2

Substituting for L and l yields:

Apply Snell’s law at the air-water interface to obtain: Solving for θ2 yields:

⎡

⎤ h+d ⎥ ⎣ h tan θ 1 + d tan θ 2 ⎦

α = tan −1 ⎢

n1 sin θ 1 = n 2 sin θ 2

⎡ n1 ⎤ sin θ 1 ⎥ ⎣ n2 ⎦

θ 2 = sin −1 ⎢

Substitute for θ2 to obtain: ⎡ ⎤ ⎢ ⎥ h+d ⎥ −1 ⎢ α = tan ⎢ ⎛ ⎡ ⎤⎞⎥ ⎢ h tan θ1 + d tan⎜ sin −1 ⎢ n1 sin θ1 ⎥ ⎟ ⎥ ⎜ ⎟ ⎢⎣ ⎣ n2 ⎦ ⎠ ⎥⎦ ⎝

921

Chapter 31

922

Noting that θ1 is the complement of 64.0°, substitute numerical values and evaluate α: ⎤ ⎡ ⎥ ⎢ 1.50 m + 1.20 m −1 ⎢ ⎥ = 66.9° α = tan ⎢ ⎞ ⎛ −1 ⎡1.00 ⎤ ⎥ sin 26.0°⎥ ⎟⎟ ⎥ ⎢ (1.50 m ) tan 26.0° + (1.20 m ) tan⎜⎜ sin ⎢ ⎣1.33 ⎦⎠⎦ ⎝ ⎣ That is, you should aim 66.9° below the horizontal.

37 ••• You are standing on the edge of a swimming pool and looking directly across at the opposite side. You notice that the bottom edge of the opposite side of the pool appears to be at an angle of 28° degrees below the horizontal. However, when you sit on the pool edge, the bottom edge of the opposite side of the pool appears to be at an angle of only 14o below the horizontal. Use these observations to determine the width and depth of the pool. Hint: You will need to estimate the height of your eyes above the surface of the water when standing and sitting. Picture the Problem The following diagrams represent the situations when you are standing on the edge of the pool (the diagram to the left) and when you are sitting on the edge of the pool (the diagram to the right). We can use Snell’s law and the geometry of the pool to determine the width and depth of the pool. Sitting

Standing hstanding

α θ1

Air Water

n2

θ2 L

hsitting

n1

α'

θ1'

Air Water

n1 n2

θ2'

h

h l

L' d

d

Use the fact that angles α and θ1 are complementary, as are α′ and θ1′ to determine θ1 and θ1′: Express the distances L and L′ in terms of θ1 and θ1′:

' l

θ 1 = 90° − α = 90° − 28° = 62° and θ 1 ' = 90° − α ' = 90° − 14° = 76° L = hstanding tan θ 1 and L' = hsitting tan θ 1 '

Properties of Light Assuming that your eyes are 1.7 m above the level of the water when you are standing and 0.7 m above the water when you are sitting, evaluate L and L′: Referring to the pictorial representations, note that:

923

L = (1.7 m ) tan 62° = 3.197 m and L' = (0.7 m ) tan 76° = 2.808 m

tan θ 2 = and tan θ 2 ' =

l d −L = h h

(1)

l' d − L ' = h h

Divide the first of these equations by the second to obtain:

tan θ 2 d − L = tan θ 2 ' d − L'

Solving for d yields:

d=

Apply Snell’s law to the air-water interface when you are standing:

θ 2 = sin −1 ⎢

Substitute numerical values and evaluate θ 2 :

θ 2 = sin −1 ⎢

Apply Snell’s law to the air-water interface when you are sitting:

θ 2 ' = sin −1 ⎢

Substitute numerical values and evaluate θ 2 ' :

θ 2 ' = sin −1 ⎢

L' tan θ 2 − L tan θ 2 ' tan θ 2 − tan θ 2 '

(2)

⎡ n1 ⎤ sin θ1 ⎥ ⎣ n2 ⎦

⎡1.00 ⎤ sin 62°⎥ = 41.60° ⎦ ⎣1.33

⎡ n1 ⎤ sin θ1 '⎥ ⎣ n2 ⎦

⎡1.00 ⎤ sin 76°⎥ = 46.85° ⎣1.33 ⎦

Substitute numerical values in equation (2) and evaluate d: d=

(2.808 m ) tan 41.60° − (3.197 m ) tan 46.85° = 5.130 m = tan 41.60° − tan 46.85°

5.1 m wide

Solving equation (1) for h yields:

h=

d −L tan θ 2

Substitute numerical values and evaluate h:

h=

5.130 m − 3.197 m = 2.2 m deep tan 41.60°

924

Chapter 31

38 ••• Figure 31-57 shows a beam of light incident on a glass plate of thickness d and index of refraction n. (a) Find the angle of incidence so that the separation b between the ray reflected from the top surface and the ray reflected from the bottom surface and exiting the top surface is a maximum. (b) What is this angle of incidence if the index of refraction of the glass is 1.60? (c) What is the separation of the two beams if the thickness of the glass plate is 4.0 cm? Picture the Problem Let x be the perpendicular separation between the two rays and let l be the separation between the points of emergence of the two rays on the glass surface. We can use the geometry of the refracted and reflected rays to express x as a function of l, d, θr, and θi. Setting the derivative of the resulting equation equal to zero will yield the value of θi that maximizes x.

l θi θi

x θi

Air

θr θr

d

Glass

Air

(a) Express l in terms of d and the angle of refraction θr:

l = 2d tan θ r

Express x as a function of l, d, θr, and θi:

x = 2d tan θ r cos θ i

Differentiate x with respect toθi: ⎛ ⎞ dx d (tan θ r cos θ i ) = 2d ⎜⎜ − tan θ r sin θ i + sec 2 θ r cos θ i dθ r ⎟⎟ (1) = 2d dθ i dθ i dθ i ⎠ ⎝

Apply Snell’s law to the air-glass interface:

n1 sin θ i = n 2 sin θ r or, because n1 = 1 and n2 = n, sin θ i = n sin θ r

Differentiate implicitly with respect toθI to obtain:

cos θ i dθ i = n cos θ r dθ r or dθ r 1 cos θ i = dθ i n cos θ r

(2)

Properties of Light

925

Substitute in equation (1) to obtain: ⎛ 1 cos 2 θ i sin θ r sin θ i ⎛ sin θ r 1 cos θ i cos θ i ⎞ dx ⎟ ⎜ sin θ i + = 2d ⎜ − 2 ⎟ = 2d ⎜⎜ n cos 3 θ − cos θ cos θ cos dθ i n θ cos θ r r r ⎠ r r ⎝ ⎝

Substitute 1 − sin 2 θ i for cos 2 θ i 1 and sin θ i for sin θ r to obtain: n

⎛ 1 − sin 2 θ i sin 2 θ i dx = 2d ⎜⎜ − 3 dθ i ⎝ n cos θ r n cos θ r

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ ⎠

Multiply the second term in parentheses by cos 2 θ r cos 2 θ r and simplify to obtain: ⎛ 1 − sin 2 θ i sin 2 θ i cos 2 θ r dx = 2d ⎜⎜ − 3 dθ i cos θ n n cos 3 θ r r ⎝

⎞ 2d ⎟⎟ = 1 − sin 2 θ i − sin 2 θ i cos 2 θ r 3 ⎠ n cos θ r

(

Substitute 1 − sin 2 θ r for cos 2 θ r :

[

(

dx 2d = 1 − sin 2 θ i − sin 2 θ i 1 − sin 2 θ r 3 dθ i n cos θ r Substitute

)]

1 sin θ i for sin θ r to obtain: n dx 2d = dθ i n cos 3 θ r

⎡ 1 ⎛ ⎞⎤ 2 2 2 ⎢1 − sin θ i − sin θ i ⎜1 − n 2 sin θ i ⎟⎥ ⎝ ⎠⎦ ⎣

Factor out 1/n2, simplify, and set equal to zero to obtain:

[

]

dx 2d = 3 sin 4 θ i − 2n 2 sin 2 θ i + n 2 = 0 for extrema 3 dθ i n cos θ r If dx/dθ1 = 0, then it must be true that: Solve this quartic equation for θi to obtain:

sin 4 θ i − 2n 2 sin 2 θ i + n 2 = 0

⎛

θ i = sin − 1 ⎜ n 1 − 1 − ⎜ ⎝

1 ⎞⎟ n 2 ⎟⎠

)

926

Chapter 31

(b) Evaluate θI for n = 1.60:

⎛

θ i = sin −1 ⎜1.6 1 − 1 − ⎜ ⎝

1

(1.60)2

⎞ ⎟ ⎟ ⎠

= 48.5°

(c) In (a) we showed that: Solve equation (2) for θr:

Substitute numerical values and evaluate θr: Substitute numerical values and evaluate x:

x = 2d tan θ r cos θ i ⎞ ⎛ n1 sin θ i ⎟⎟ ⎠ ⎝ n2

θ r = sin −1 ⎜⎜

⎛ 1 ⎞ sin 48.5° ⎟ = 27.9° ⎝ 1.60 ⎠

θ r = sin −1 ⎜

x = 2(4.0 cm ) tan 27.9° cos 48.5° = 2.8 cm

Total Internal Reflection 39 • [SSM] What is the critical angle for light traveling in water that is incident on a water–air interface? Picture the Problem Let the subscript 1 refer to the water and the subscript 2 to the air and use Snell’s law under total internal reflection conditions.

Use Snell’s law to obtain:

n1 sin θ 1 = n 2 sin θ 2

When there is total internal reflection:

θ1 = θ c and θ 2 = 90°

Substitute to obtain:

n1 sin θ c = n2 sin 90° = n2

Solving for θc yields:

Substitute numerical values and evaluate θc:

⎛ n2 ⎞ ⎟⎟ n ⎝ 1⎠

θ c = sin −1 ⎜⎜

⎛ 1.00 ⎞ ⎟ = 48.8° ⎝ 1.33 ⎠

θ c = sin −1 ⎜

40 • A glass surface (n = 1.50) has a layer of water (n = 1.33) on it. Light in the glass is incident on the glass–water interface. Find the critical angle for total internal reflection.

Properties of Light Picture the Problem Let the index of refraction of glass be represented by n1 and the index of refraction of water by n2 and apply Snell’s law to the glasswater interface under total internal reflection conditions.

n2 = 1.33

θ2

Water Glass

n1 = 1.50

θ1

Apply Snell’s law to the glass-water interface:

n1 sin θ1 = n2 sin θ 2

At the critical angle, θ1 = θc and θ2 = 90°:

n1 sin θ c = n2 sin 90°

Solve for θc:

Substitute numerical values and evaluate θc:

927

⎡ n2 ⎤ sin 90°⎥ ⎣ n1 ⎦

θ c = sin −1 ⎢

⎡1.33 ⎤ sin 90°⎥ = 62.5° ⎣1.50 ⎦

θ c = sin −1 ⎢

41 • A point source of light is located 5.0 m below the surface of a large pool of water. Find the area of the largest circle on the pool’s surface through which light coming directly from the source can emerge. Air

Picture the Problem We can apply Snell’s law to the water-air interface to express the critical angle θc in terms of the indices of refraction of water (n1) and air (n2) and then relate the radius of the circle to the depth d of the point source and θc.

r

n2 = 1.00

Water

n1 = 1.33 d = 5.0 m

θc

Express the area of the circle whose radius is r:

A = π r2

Relate the radius of the circle to the depth d of the point source and the critical angle θc:

r = d tan θ c

Apply Snell’s law to the water-air interface to obtain:

n1 sin θ c = n2 sin 90° = n2

90º

θc

928

Chapter 31

Solving for θc yields:

Substitute for r and θc to obtain:

⎛ n2 ⎞ ⎟⎟ ⎝ n1 ⎠

θ c = sin −1 ⎜⎜

A = π [d tan θ c ]

2

⎡ ⎛ ⎧ n ⎫ ⎞⎤ = π ⎢d tan⎜⎜ sin −1 ⎨ 2 ⎬ ⎟⎟⎥ ⎢⎣ ⎩ n1 ⎭ ⎠⎥⎦ ⎝ Substitute numerical values and evaluate A:

2

⎡ ⎛ ⎧ 1 ⎫ ⎞⎤ A = π ⎢(5.0 m ) tan⎜⎜ sin −1 ⎨ ⎬ ⎟⎟⎥ ⎩1.33 ⎭ ⎠⎦ ⎝ ⎣

2

= 1.0 × 10 2 m 2 42 •• Light traveling in air strikes the largest face of an isosceles-righttriangle prism at normal incidence. What is the speed of light in this prism if the prism is just barely able to produce total internal reflection? Picture the Problem We can use the definition of the index of refraction to express the speed of light in the prism in terms of the index of refraction n1 of the prism. The application of Snell’s law at the prism-air interface will allow us to relate the index of refraction of the prism to the critical angle for total internal reflection. Finally, we can use the geometry of the isosceles-righttriangle prism to conclude that θc = 45°.

45º

n1 θc 45º

n2 = 1.00

Express the speed of light v in the prism in terms of its index of refraction n1:

v=

Apply Snell’s law to the prism-air interface to obtain:

n1 sin θ c = n2 sin 90° = 1

Solving for n1 yields:

n1 =

Substitute for n1 and simplify to obtain:

v = c sin θ c

c n1

1 sin θ c

Properties of Light

(

929

)

v = 2.998 × 10 8 m/s sin 45°

Substitute numerical values and evaluate v:

= 2.1×10 8 m/s

43 •• A point source of light is located at the bottom of a steel tank, and an opaque circular card of radius 6.00 cm is placed horizontally over it. A transparent fluid is gently added to the tank so that the card floats on the fluid surface with its center directly above the light source. No light is seen by an observer above the surface until the fluid is 5.00 cm deep. What is the index of refraction of the fluid? Picture the Problem The observer above the surface of the fluid will not see any light until the angle of incidence of the light at the fluid-air interface is less than or equal to the critical angle for the two media. We can use Snell’s law to express the index of refraction of the fluid in terms of the critical angle and use the geometry of card and light source to express the critical angle. r

n2 n1

θ2 θc

d

θc

Apply Snell’s law to the fluid-air interface to obtain:

n1 sin θ1 = n2 sin θ 2

Light is seen by the observer when θ1 = θc and θ2 = 90°:

n1 sin θ c = n2 sin 90° = n2

Because the medium above the interface is air, n2 = 1. Solve for n1 to obtain:

n1 =

From the geometry of the diagram:

tan θ c =

Substitute for θ c to obtain:

n1 =

1 sin θ c

r ⎛r⎞ ⇒ θ c = tan −1 ⎜ ⎟ d ⎝d ⎠ 1

⎡ ⎛ r ⎞⎤ sin ⎢ tan −1 ⎜ ⎟⎥ ⎝ d ⎠⎦ ⎣

930

Chapter 31

Substitute numerical values and evaluate n1:

n1 =

1 ⎡ ⎛ 6.00 cm ⎞⎤ ⎟⎟⎥ sin ⎢ tan −1 ⎜⎜ ⎝ 5.00 cm ⎠⎦ ⎣

= 1.30

44 •• An optical fiber allows rays of light to propagate long distances by using total internal reflection. Optical fibers are used extensively in medicine and in digital communications. As shown in Figure 31-58 the fiber consists of a core material that has an index of refraction n2 and radius b surrounded by a cladding material that has an index of refraction n3 < n2. The numerical aperture of the fiber is defined as sinθ1, where θ1 is the angle of incidence of a ray of light that impinges on the center of the end of the fiber and then reflects off the corecladding interface just at the critical angle. Using the figure as a guide, show that

the numerical aperture is given by sin θ1 = n 22 − n 32 assuming the ray is initially in air. Hint: Use of the Pythagorean theorem may be required. Picture the Problem We can use the geometry of the figure, the law of refraction at the air-n1 interface, and the condition for total internal reflection at the n1-n2

interface to show that the numerical aperture is given by sin θ1 = n22 − n32 . Referring to the figure, note that:

Apply the Pythagorean theorem to the right triangle to obtain: Solving for

b yields: c

Substitute for

b a and to obtain: c c

Use the law of refraction to relate θ1 and θ2: Substitute for sinθ2, let n1 = 1 (air), and simplify to obtain:

sin θ c =

n3 a b = and sin θ 2 = n2 c c a 2 b2 + =1 c2 c2

a 2 + b 2 = c 2 or b a2 = 1− 2 c c n32 sin θ 2 = 1 − 2 n2

n1 sin θ1 = n2 sin θ 2

sin θ1 = n2 1 −

n32 = n22

n22 − n32

45 •• [SSM] Find the maximum angle of incidence θ1 of a ray that would propagate through an optical fiber that has a core index of refraction of 1.492, a core radius of 50.00 μm, and a cladding index of 1.489. See Problem 44.

Properties of Light

931

Picture the Problem We can use the result of Problem 44 to find the maximum angle of incidence under the given conditions.

From Problem 44:

sin θ1 = n22 − n32

Solve for θ1 to obtain:

θ1 = sin −1 n22 − n32

Substitute numerical values and evaluate θ1:

θ1 = sin −1 ⎛⎜ (1.492 )2 − (1.489)2 ⎞⎟

(

)

⎝

⎠

= 5° 46 •• Calculate the difference in time needed for two pulses of light to travel down 15.0 km of the fiber that is described in Problem 44. Assume that one pulse enters the fiber at normal incidence, and the second pulse enters the fiber at the maximum angle of incidence calculated in Problem 45. In fiber optics, this effect is known as modal dispersion. Picture the Problem We can derive an expression for the difference in the travel times by expressing the travel time for a pulse that enters the fiber at the maximum angle of incidence and a pulse that enters the fiber at normal incidence. Examination of Figure 31-58 reveals that, if the length of the tube is L, the distance traveled by the pulse that enters at an angle θ1 is the ratio of c to a multiplied by L.

The difference in the travel times Δt is given by:

Δt = tθ1 − t normal

The travel time for the pulse that enters the fiber at the maximum angle of incidence is:

c distance traveled tθ1 = = a c speed in the medium n2

The travel time for the pulse that enters the fiber normally is:

Substitute for tθ1 and t normal

incidence

equation (1) to obtain:

in

(1)

incidence

L

=

t normal

incidence

L c n2

c L Δt = a − c c n2 n2 L

932

Chapter 31

n2 L ⎛ c ⎞ ⎜ − 1⎟ c ⎝a ⎠

Simplify to obtain:

Δt =

Referring to the figure, note that:

sin θ c =

a c

From Snell’s law, the sine of the critical angle is also given by:

sin θ c =

n3 a n ⇒ = 3 n2 c n2

Substitute for c/a in equation (2) to obtain:

Δt =

n2 L ⎛ n2 ⎞ ⎜ − 1⎟ c ⎜⎝ n3 ⎟⎠

Substitute numerical values and evaluate Δt:

Δt =

(1.492)(15 km ) ⎛ 1.492 − 1⎞

⎜ 2.998 × 108 m/s ⎝ 1.489

(2)

⎟ ⎠

= 150 ns 47 ••• Investigate how a thin film of water on a glass surface affects the critical angle for total reflection. Use n = 1.50 for glass and n = 1.33 for water. (a) What is the critical angle for total internal reflection at the glass–water interface? (b) Does a range of incident angles exist such that the angles are greater than θc for glass-to-air refraction and for which the light rays will leave the glass, travel through the water and then pass into the air? Picture the Problem Let the index of refraction of glass be represented by n1, the index of refraction of water by n2, and the index of refraction of air by n3. We can apply Snell’s law to the glass-water interface under total internal reflection conditions to find the critical angle for total internal reflection. The application of Snell’s law to glass-air and glass-water interfaces will allow us to decide whether there are angles of incidence greater than θc for glass-to-air refraction for which light rays will leave the glass and the water and pass into the air. Air

n3 = 1.00

θ3 θ2

Water

n2 = 1.33

Glass

n1 = 1.50

θ2 θ1

Properties of Light (a) Apply Snell’s law to the glasswater interface:

n1 sin θ1 = n2 sin θ 2

At the critical angle, θ1 = θc and θ2 = 90°:

n1 sin θ c = n2 sin 90°

Solving for θc yields:

⎤ ⎡ n2 sin 90°⎥ ⎦ ⎣ n1

θ c = sin −1 ⎢

⎡1.33 ⎤ sin 90°⎥ = 62.5° ⎣1.50 ⎦

Substitute numerical values and evaluate θc:

θ c = sin −1 ⎢

(b) Apply Snell’s law to a water-air interface at the critical angle for a water-air interface:

n2 sin θ c = n3 sin 90°

Solving for θ c yields:

⎛ n3 ⎞ ⎟⎟ ⎝ n2 ⎠

θ c = sin −1 ⎜⎜

⎛ 1.00 ⎞ ⎟ = 48.8° ⎝ 1.33 ⎠

Substitute numerical values and evaluate θ c :

θ c = sin −1 ⎜

Apply Snell’s law to a ray incident at the glass-water interface:

n1 sin θ1 = n2 sin θ 2 and ⎞ ⎛n θ1 = sin −1 ⎜⎜ 2 sin θ 2 ⎟⎟ ⎠ ⎝ n1

For θ 2 = θ c :

Substitute numerical values and evaluate θ1:

933

⎡⎛ n2 ⎞ ⎛ n3 ⎞⎤ −1 ⎛ n3 ⎞ ⎟⎟ ⎜⎜ ⎟⎟⎥ sin ⎜⎜ ⎟⎟ n ⎝ n1 ⎠ ⎣⎝ 1 ⎠ ⎝ n2 ⎠⎦

θ1 = sin −1 ⎢⎜⎜

⎛ 1.00 ⎞ ⎟ = 41.8° ⎝ 1.50 ⎠

θ1 = sin −1 ⎜

Yes, if θ ≥ 41.8°, where θ is the angle of incidence for the rays in glass that are incident on the glass-water boundary, the rays will leave the glass through the water and pass into the air. 48 ••• A laser beam is incident on a plate of glass that is 3.0-cm thick (Figure 31-57). The glass has an index of refraction of 1.5 and the angle of incidence is 40º. The top and bottom surfaces of the glass are parallel. What is the distance b

934

Chapter 31

between the beam formed by reflection off the top surface of the glass and the beam reflected off the bottom surface of the glass. Picture the Problem The situation is shown in the adjacent figure. We can use the geometry of the diagram and trigonometric relationships to derive an expression for d in terms of the angles of incidence and refraction. Applying Snell’s law will yield θr.

b θ i θi θi x

Air Glass

n = 1.5

θr t

θr

Express the distance x in terms of t and θr:

x = 2t tan θ r

The separation of the reflected rays is:

b = x cosθ i

Substitute for x to obtain:

b = 2t tan θ r cosθ i

Apply Snell’s law at the air-glass interface to obtain:

⎛ sin θ i ⎞ sin θ i = n sin θ r ⇒ θ r = sin −1 ⎜ ⎟ ⎝ n ⎠

Substitute for θ r in equation (1) to obtain:

⎡ ⎛ sin θ i b = 2t tan ⎢sin −1 ⎜ ⎝ n ⎣

Substitute numerical values and evaluate b:

⎡ ⎛ sin 40° ⎞⎤ b = 2(3.0 cm ) tan ⎢sin −1 ⎜ ⎟⎥ cos 40° ⎝ 1.5 ⎠⎦ ⎣

(1)

⎞⎤ ⎟⎥ cosθ i ⎠⎦

= 2.2 cm

Dispersion A beam of light strikes the plane surface of silicate flint glass at an 49 • angle of incidence of 45º. The index of refraction of the glass varies with wavelength (see Figure 31-59). How much smaller is the angle of refraction for violet light of wavelength 400 nm than the angle of refraction for red light of wavelength 700 nm? Picture the Problem We can apply Snell’s law of refraction to express the angles of refraction for red and violet light in silicate flint glass.

Properties of Light

935

Express the difference between the angle of refraction for violet light and for red light:

Δθ = θ r,red − θ r,violet

Apply Snell’s law of refraction to the interface to obtain:

⎛ 1 ⎞ sin 45° = n sin θ r ⇒ θ r = sin −1 ⎜ ⎟ ⎝ 2n ⎠

Substituting for θ r in equation (1) yields:

⎛ 1 ⎞ ⎛ 1 ⎞ ⎟ − sin −1 ⎜ ⎟ Δθ = sin −1 ⎜⎜ ⎟ ⎜ ⎟ n 2 n 2 red ⎠ violet ⎠ ⎝ ⎝

Substitute numerical values and evaluate Δθ :

⎛ ⎞ 1 ⎟ Δθ = sin −1 ⎜⎜ ⎟ ( ) 2 1 . 60 ⎝ ⎠ ⎛ ⎞ 1 ⎟ − sin −1 ⎜⎜ ⎟ ( ) 2 1 . 66 ⎝ ⎠

(1)

= 26.2° − 25.2° = 1.0° 50 •• In many transparent materials, dispersion causes different colors (wavelengths) of light to travel at different speeds. This can cause problems in fiber-optic communications systems where pulses of light must travel very long distances in glass. Assuming a fiber is made of silicate crown glass (see Figure 31-59), calculate the difference in travel times that two short pulses of light take to travel 15.0 km in this fiber if the first pulse has a wavelength of 700 nm and the second pulse has a wavelength of 500 nm. Picture the Problem The transit times will be different because the speed with which light of various wavelengths propagates in silicate crown glass is dependent on the index of refraction. We can use Figure 31-19 to estimate the indices of refraction for pulses of wavelengths 500 and 700 nm.

Express the difference in time needed for two short pulses of light to travel a distance L in the fiber:

Δt =

L L − v500 v700

Substitute for L, v500, and v700 and simplify to obtain:

Δt =

n500 L n700 L L − = (n500 − n700 ) c c c

Use Figure 31-59 to find the indices of refraction of silicate crown glass for the two wavelengths:

n500 ≈ 1.51 and n700 ≈ 1.50

936

Chapter 31

Substitute numerical values and evaluate Δt:

Δt =

15.0 km (1.51 − 1.50) 2.998 ×108 m/s

≈ 0.5 μs

Polarization 51 • [SSM] What is the polarizing angle for light in air that is incident on (a) water (n = 1.33), and (b) glass (n = 1.50)? Picture the Problem The polarizing angle is given by Brewster’s law: tan θ p = n2 n1 where n1 and n2 are the indices of refraction on the near and far

sides of the interface, respectively. Use Brewster’s law to obtain:

⎛ n2 ⎞ ⎟⎟ n ⎝ 1⎠

θ p = tan −1 ⎜⎜

⎛ 1.33 ⎞ ⎟ = 53.1° ⎝ 1.00 ⎠

(a) For n1 = 1 and n2 = 1.33:

θ p = tan −1 ⎜

(b) For n1 = 1 and n2 = 1.50:

θ p = tan −1 ⎜

⎛ 1.50 ⎞ ⎟ = 56.3° ⎝ 1.00 ⎠

52 • Light that is horizontally polarized is incident on a polarizing sheet. It is observed that only 15 percent of the intensity of the incident light is transmitted through the sheet. What angle does the transmission axis of the sheet make with the horizontal? Picture the Problem The intensity of the transmitted light I is related to the intensity of the incident light I0 and the angle the transmission axis makes with the horizontal θ according to I = I 0 cos 2 θ .

Express the intensity of the transmitted light in terms of the intensity of the incident light and the angle the transmission axis makes with the horizontal:

⎛ I ⎞ ⎟ I = I 0 cos 2 θ ⇒ θ = cos −1 ⎜⎜ ⎟ I ⎝ 0⎠

Substitute numerical values and evaluate θ :

θ = cos −1 0.15 = 67°

(

)

Properties of Light

937

53 • Two polarizing sheets have their transmission axes crossed so that no light gets through. A third sheet is inserted between the first two so that its transmission axis makes an angle θ with the transmission axis of the first sheet. Unpolarized light of intensity I0 is incident on the first sheet. Find the intensity of the light transmitted through all three sheets if (a) θ = 45º and (b) θ = 30º. Picture the Problem Let In be the intensity after the nth polarizing sheet and use I = I 0 cos 2 θ to find the intensity of the light transmitted through all three sheets

for θ = 45° and θ = 30°. (a) The intensity of the light between the first and second sheets is:

I1 = 12 I 0

The intensity of the light between the second and third sheets is:

I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 45° = 14 I 0

The intensity of the light that has passed through the third sheet is:

I 3 = I 2 cos 2 θ 2,3 = 14 I 0 cos 2 45° =

(b) The intensity of the light between the first and second sheets is:

I1 = 12 I 0

The intensity of the light between the second and third sheets is:

I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 30° = 83 I 0

The intensity of the light that has passed through the third sheet is:

I 3 = I 2 cos 2 θ 2,3 = 83 I 0 cos 2 60° =

1 8

I0

3 32

I0

54 • A horizontal 5.0 mW laser beam that is vertically polarized is incident on a sheet that is oriented with its transmission axis vertical. Behind the first sheet is a second sheet that is oriented so that its transmission axis makes an angle of 27º with respect to the vertical. What is the power of the beam transmitted through the second sheet? Picture the Problem Because the light is polarized in the vertical direction and the first polarizer is also vertically polarized, no loss of intensity results from the first transmission. We can use Malus’s law to find the intensity of the light after it has passed through the second polarizer.

The intensity of the beam is the ratio of its power to cross-sectional area:

I=

P A

938

Chapter 31

Express the intensity of the light between the first and second polarizers:

I1 = I 0 and P1 = P0

Express the law of Malus in terms of the power of the beam:

P P0 = cos 2 θ ⇒ P = P0 cos 2 θ A A

Express the power of the beam after the second transmission:

P2 = P1 cos 2 θ1, 2 = P0 cos 2 θ12

Substitute numerical values and evaluate P2:

P2 = (5.0 mW ) cos 2 27° = 4.0 mW

55 •• [SSM] The polarizing angle for light in air that is incident on a certain substance is 60º. (a) What is the angle of refraction of light incident at this angle? (b) What is the index of refraction of this substance? Picture the Problem Assume that light is incident in air (n1 = 1.00). We can use the relationship between the polarizing angle and the angle of refraction to determine the latter and Brewster’s law to find the index of refraction of the substance.

(a) At the polarizing angle, the sum of the angles of polarization and refraction is 90°:

θ p + θ r = 90° ⇒ θ r = 90° − θ p

Substitute for θp to obtain:

θ r = 90° − 60° = 30°

(b) From Brewster’s law we have:

tan θ p =

n2 n1

or, because n1 = 1.00, n2 = tan θ p Substitute for θp and evaluate n2:

n 2 = tan 60° = 1.7

56 •• Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that its transmission axis makes an angle θ with the transmission axis of the first sheet. (a) Derive an expression for the intensity of the transmitted light as a function of θ. (b) Show that the intensity transmitted through all three sheets is maximum when θ = 45º.

Properties of Light

939

Picture the Problem Let In be the intensity after the nth polarizing sheet and use I = I 0 cos 2 θ to find the intensity of the light transmitted through the three sheets.

(a) The intensity of the light between the first and second sheets is:

I1 = 12 I 0

The intensity of the light between the second and third sheets is:

I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 θ

Express the intensity of the light that has passed through the third sheet and simplify to obtain:

I 3 = I 2 cos 2 θ 2,3 = 12 I 0 cos 2 θ cos 2 (90° − θ ) = 12 I 0 cos 2 θ sin 2 θ

= 18 I 0 (2 cos θ sin θ )

2

=

1 8

I 0 sin 2 2θ

(b) Because the sine function is a maximum when its argument is 90°, the maximum value of I3 occurs when θ = 45°. 57 •• If the middle polarizing sheet in Problem 56 is rotating at an angular speed ω about an axis parallel with the light beam, find an expression for the intensity transmitted through all three sheets as a function of time. Assume that θ = 0 at time t = 0. Picture the Problem Let In be the intensity after the nth polarizing sheet, use I = I 0 cos 2 θ to find the intensity of the light transmitted through each sheet, and

replace θ with ωt. The intensity of the light between the first and second sheets is:

I1 = 12 I 0

The intensity of the light between the second and third sheets is:

I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 ωt

Express the intensity of the light that has passed through the third sheet and simplify to obtain:

I 3 = I 2 cos 2 θ 2,3 = 12 I 0 cos 2 ωt cos 2 (90° − ωt ) = 12 I 0 cos 2 ωt sin 2 ωt

= 18 I 0 (2 cos ωt sin ωt )

2

=

1 8

I 0 sin 2 2ωt

940

Chapter 31

58 •• A stack of N + 1 ideal polarizing sheets is arranged so that each sheet is rotated by an angle of π/(2N) rad with respect to the preceding sheet. A linearly polarized light wave of intensity I0 is incident normally on the stack. The incident light is polarized along the transmission axis of the first sheet and is therefore perpendicular to the transmission axis of the last sheet in the stack. (a) Show that the intensity of the light transmitted through the entire stack is given by I0 cos 2N ⎡⎣π (2N )⎤⎦ . (b) Using a spreadsheet or graphing program, plot

the transmitted intensity as a function of N for values of N from 2 to 100. (c) What is the direction of polarization of the transmitted beam in each case? Picture the Problem Let In be the intensity after the nth polarizing sheet and use I = I 0 cos 2 θ to find the ratio of In+1 to In.

(a) Find the ratio of In+1 to In:

I n+1 π = cos 2 In 2N

Because there are N such reductions of intensity:

I N +1 I N +1 ⎛ π ⎞ = = cos 2 N ⎜ ⎟ I1 I0 ⎝ 2N ⎠

and ⎛ π ⎞ I N +1 = I 0 cos 2 N ⎜ ⎟ ⎝ 2N ⎠

(b) A spreadsheet program to graph IN+1/I0 as a function of N is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell A2 A3 B2

Content/Formula 2 A2 + 1 (cos(PI()/(2*A2))^(2*A2)

Algebraic Form N N+1 ⎛ π ⎞ cos 2 N ⎜ ⎟ ⎝ 2N ⎠

1 2 3 4 5

A N 2 3 4 5

B I/I0 0.250 0.422 0.531 0.605

95 96 97 98 99

95 96 97 98 99

0.974 0.975 0.975 0.975 0.975

Properties of Light 100

100

941

0.976

A graph of I/I0 as a function of N follows. 1.0 0.9 0.8

I /I0

0.7 0.6 0.5 0.4 0.3 0.2 0

10

20

30

40

50

60

70

80

90

100

N

(c) The transmitted light, if any, is polarized parallel to the transmission axis of the last sheet. (For N = 2 there is no transmitted light.) Remarks: The correct values for the intensity of the transmitted light would be significantly smaller than predicted above because of reflection losses. 59 •• [SSM] The device described in Problem 58 could serve as a polarization rotator, which changes the linear plane of polarization from one direction to another. The efficiency of such a device is measured by taking the ratio of the output intensity at the desired polarization to the input intensity. The result of Problem 58 suggests that the highest efficiency is achieved by using a large value for the number N. A small amount of intensity is lost regardless of the input polarization when using a real polarizer. For each polarizer, assume the transmitted intensity is 98 percent of the amount predicted by the law of Malus and use a spreadsheet or graphing program to determine the optimum number of sheets you should use to rotate the polarization 90º. Picture the Problem Let In be the intensity after the nth polarizing sheet and use I = I 0 cos 2 θ to find the ratio of In+1 to In. Because each sheet introduces a 2%

loss of intensity, the net transmission after N sheets (0.98)N. Find the ratio of In+1 to In:

I n+1 π = (0.98) cos 2 In 2N

942

Chapter 31 I N +1 ⎛ π ⎞ N = (0.98) cos 2 N ⎜ ⎟ I0 ⎝ 2N ⎠

Because there are N such reductions of intensity:

A spreadsheet program to graph IN+1/I0 for an ideal polarizer as a function of N, the percent transmission, and IN+1/I0 for a real polarizer as a function of N is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell A3 B2

Content/Formula 1 (cos(PI()/(2*A2))^(2*A2)

C3

(0.98)^A3

D4

B3*C3

A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

B Ideal Polarizer 0.000 0.250 0.422 0.531 0.605 0.660 0.701 0.733 0.759 0.781 0.798 0.814 0.827 0.838 0.848 0.857 0.865 0.872 0.878 0.884

Algebraic Form N ⎛ π ⎞ cos 2 N ⎜ ⎟ ⎝ 2N ⎠ (0.98)N

(0.98)N cos 2 N ⎛⎜

π ⎞

⎟ ⎝ 2N ⎠

C Percent Transmission 0.980 0.960 0.941 0.922 0.904 0.886 0.868 0.851 0.834 0.817 0.801 0.785 0.769 0.754 0.739 0.724 0.709 0.695 0.681 0.668

D Real Polarizer 0.000 0.240 0.397 0.490 0.547 0.584 0.608 0.624 0.633 0.638 0.639 0.638 0.636 0.632 0.626 0.620 0.613 0.606 0.598 0.590

Properties of Light

943

A graph of I/I0 as a function of N for the quantities described above follows: 1.0

0.8

0.6 I /I 0

Ideal Polarizer

0.4

Percent Transmission 0.2

Real Polarizer

0.0 0

5

10

15

20

Number of sheets (N )

Inspection of the table, as well as of the graph, tells us that the optimum number of sheets is 11. Remarks: The correct values for the intensity of the transmitted light would be significantly smaller than predicted above because of reflection losses. 60 •• Show mathematically that a linearly polarized wave can be thought of as a superposition of a right and a left circularly polarized wave. Picture the Problem A circularly polarized wave is said to be right circularly polarized if the electric and magnetic fields rotate clockwise when viewed along the direction of propagation and left circularly polarized if the fields rotate counterclockwise.

For a circularly polarized wave, the x and y components of the electric field are given by:

E x = E0 cos ωt and E y = E0 sin ωt or E y = − E0 sin ωt for left and right circular polarization, respectively.

For a wave polarized along the x axis:

r r E right + Eleft = E0 cos ωt iˆ + E0 cos ωt iˆ = 2 E0 cos ωt iˆ

61 •• Suppose that the middle sheet in Problem 53 is replaced by two polarizing sheets. If the angle between the transmission axes in the second, third, and fourth sheets in the stack make angles of 30º, 60º and 90º, respectively, with the transmission axis of the first sheet, (a) what is the intensity of the transmitted

944

Chapter 31

light? (b) How does this intensity compare with the intensity obtained in Part (a) of Problem 53? Picture the Problem Let In be the intensity after the nth polarizing sheet and use I = I 0 cos 2 θ to find the intensity of the light transmitted by the four sheets.

(a) The intensity of the light between the first and second sheets is:

I1 = 12 I 0

The intensity of the light between the second and third sheets is:

I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 30° = 83 I 0

The intensity of the light between the third and fourth sheets is:

I 3 = I 2 cos 2 θ 2,3 = 83 I 0 cos 2 30° =

The intensity of the light to the right of the fourth sheet is:

I 4 = I 3 cos 2 θ 3, 4 =

9 32

9 32

I0

27 I 0 cos 2 30° = 128 I0

= 0.211I 0

(b) The intensity with four sheets at angles of 0°, 30°, 30° and 90° is greater the intensity of three sheets at angles of 0°, 45° and 90° by a factor of 1.69. Remarks: We could also apply the result obtained in Problem 58(a) to solve this problem. 62 •• Show that the electric field of a circularly polarized wave propagating parallel with the x axis can be expressed by r E = E0 sin (kx + ωt ) ˆj + E0 cos(kx + ωt )kˆ . Picture the Problem A circularly polarized wave may be resolved into two linearly polarized waves, of equal amplitude, 90 degrees apart and with their planes of polarization at right angles to each other. The components of the electric r field E of a circularly polarized wave propagating parallel to the x axis are given by E x = E0 sin ωt and E y = E0 cos ωt .

r The electric field E is the vector sum of its components:

For an electric field propagating parallel with the x axis and in the −x direction:

r E = E x ˆj + E y kˆ

E x = E0 sin (kx + ωt ) and E y = E0 cos(kx + ωt )

Properties of Light

945

Substituting for Ex and Ey yields: r E = E0 sin (kx + ωt ) ˆj + E0 cos(kx + ωt )kˆ 63 •• [SSM] A circularly polarized wave is said to be right circularly polarized if the electric and magnetic fields rotate clockwise when viewed along the direction of propagation and left circularly polarized if the fields rotate counterclockwise. (a) What is the sense of the circular polarization for the wave described by the expression in Problem 62? (b) What would be the expression for the electric field of a circularly polarized wave traveling in the same direction as the wave in Problem 60, but with the fields rotating in the opposite direction? Picture the Problem We can apply the given definitions of right and left circular polarization to the electric field and magnetic fields of the wave.

(a) The electric field of the wave in Problem 62 is: r E = E0 sin (kx + ωt ) ˆj + E0 cos(kx + ωt )kˆ

The corresponding magnetic field is: r B = B0 sin (kx + ωt )kˆ − B0 cos(kx + ωt ) ˆj

Because these fields rotate clockwise when viewed along the direction of propagation, the wave is right circularly polarized. (b) For a left circularly polarized wave traveling in the opposite direction: r E = E0 sin (kx + ωt ) ˆj − E0 cos(kx + ωt )kˆ

Sources of Light 64 • A helium–neon laser emits light that has a wavelength equal to 632.8 nm and has a power output of 4.00 mW. How many photons are emitted per second by this laser? Picture the Problem We can express the number of photons emitted per second as the ratio of the power output of the laser and energy of a single photon.

946

Chapter 31

Relate the number of photons per second n to the power output of the pulse and the energy of a single photon Ephoton :

n=

The energy of a photon is given by:

E photon =

Substitute for Ephoton to obtain:

n=

P Ephoton

hc

λ

λP hc

Substitute numerical values and evaluate n: n=

(632.8 nm )(4.00 mW ) 1240 eV ⋅ nm

1eV = 1.27 × 1016 photons/s −19 1.602 × 10 J

65 • The first excited state of an atom of a gas is 2.85 eV above the ground state. (a) What is the maximum wavelength of radiation for resonance absorption by atoms of the gas that are in the ground state? (b) If the gas is irradiated with monochromatic light that has a wavelength of 320 nm, what is the wavelength of the Raman scattered light? Picture the Problem We can use the Einstein equation for photon energy to find the wavelength of the radiation for resonance absorption. We can use the same relationship, with ERaman = Einc − ΔE where ΔE is the energy for resonance absorption, to find the wavelength of the Raman scattered light.

(a) Use the Einstein equation for photon energy to relate the wavelength of the radiation to energy of the first excited state:

λ=

hc E

Substitute numerical values and evaluate λ:

λ=

1240 eV ⋅ nm = 435 nm 2.85 eV

(b) The wavelength of the Raman scattered light is given by:

λRaman =

Relate the energy of the Raman scattered light ERaman to the energy of the incident light Einc:

E Raman = Einc − ΔE

1240 eV ⋅ nm E Raman

1240 eV ⋅ nm − 2.85 eV 320 nm = 1.025 eV =

Properties of Light Substitute numerical values and evaluate λRaman:

λRaman =

947

1240 eV ⋅ nm = 1210 nm 1.025 eV

66 •• A gas is irradiated with monochromatic ultraviolet light that has a wavelength of 368 nm. Scattered light that has a wavelength equal to 368 nm is observed, and scattered light that has a wavelength of 658-nm is also observed. Assuming that the gas atoms were in their ground state prior to irradiation, find the energy difference between the ground state and the excited state obtained by the irradiation. Picture the Problem The incident radiation will excite atoms of the gas to higher energy states. The scattered light that is observed is a consequence of these atoms returning to their ground state. The energy difference between the ground state and the atomic state excited by the irradiation is given hc . by ΔE = hf =

3.77 eV 658 nm

~~~ 1.88 eV 368 nm

~~~ 368 nm

~~~

λ

The energy difference between the ground state and the atomic state excited by the irradiation is given by: Substitute 368 nm for λ and evaluate ΔE:

ΔE = hf =

ΔE =

hc

λ

0

=

1240 eV ⋅ nm

λ

1240 eV ⋅ nm = 3.37 eV 368 nm

67 •• [SSM] Sodium has excited states 2.11 eV, 3.20 eV, and 4.35 eV above the ground state. Assume that the atoms of the gas are all in the ground state prior to irradiation. (a) What is the maximum wavelength of radiation that will result in resonance fluorescence? What is the wavelength of the fluorescent radiation? (b) What wavelength will result in excitation of the state 4.35 eV above the ground state? If that state is excited, what are the possible wavelengths of resonance fluorescence that might be observed? Picture the Problem The ground state and the three excited energy levels are shown in the diagram to the right. Because the wavelength is related to the energy of a photon by λ = hc/ΔE, longer wavelengths correspond to smaller energy differences.

3

4.35 eV

2

3.20 eV

1

2.11 eV

0

0

948

Chapter 31

(a) The maximum wavelength of radiation that will result in resonance fluorescence corresponds to an excitation to the 3.20 eV level followed by decays to the 2.11 eV level and the ground state:

λmax =

1240 eV ⋅ nm = 388 nm 3.20 eV

The fluorescence wavelengths are:

λ2→1 =

1240 eV ⋅ nm = 1140 nm 3.20 eV − 2.11eV

and

(b) For excitation:

λ1→0 =

1240 eV ⋅ nm = 588 nm 2.11eV − 0

λ 0 →3 =

1240 eV ⋅ nm = 285 nm 4.35 eV

(not in visible the spectrum) The fluorescence wavelengths corresponding to the possible transitions are:

λ3→2 =

1240 eV ⋅ nm = 1080 nm 4.35 eV − 3.20 eV (not in visible spectrum)

λ2→1 =

1240 eV ⋅ nm = 1140 nm 3.20 eV − 2.11eV (not in visible spectrum)

1240 eV ⋅ nm = 588 nm 2.11eV − 0 1240 eV ⋅ nm λ3→1 = = 554 nm 4.35 eV − 2.11eV

λ1→0 =

and

λ 2→0 =

1240 eV ⋅ nm = 388 nm 3.20 eV − 0 (not in visible spectrum)

68 •• Singly ionized helium is a hydrogen-like atom that has a nuclear charge of +2e. Its energy levels are given by En = –4E0/n2, where n = 1, 2, … and E0 = 13.6 eV. If a beam of visible white light is sent through a gas of singly ionized helium, at what wavelengths will dark lines be found in the spectrum of the transmitted radiation? (Assume that the ions of the gas are all in the same state with energy E1 prior to irradiation.)

Properties of Light

949

Determine the Concept The energy difference between the ground state and the first excited state is 3E0 = 40.8 eV, corresponding to a wavelength of 30.4 nm. This is in the far ultraviolet, well outside the visible range of wavelengths. There will be no dark lines in the transmitted radiation. 69 • [SSM] A pulse from a ruby laser has an average power of 10 MW and lasts 1.5 ns. (a) What is the total energy of the pulse? (b) How many photons are emitted in this pulse? Picture the Problem We can use the definition of power to find the total energy of the pulse. The ratio of the total energy to the energy per photon will yield the number of photons emitted in the pulse.

(a) Use the definition of power to obtain:

E = PΔt

Substitute numerical values and evaluate E:

E = (10 MW )(1.5 ns ) = 15 mJ

(b) Relate the number of photons N to the total energy in the pulse and the energy of a single photon Ephoton :

N=

The energy of a photon is given by:

E photon =

Substitute for Ephoton to obtain:

N=

E E photon

hc

λ

λE hc

Substitute numerical values (the wavelength of light emitted by a ruby laser is 694.3 nm) and evaluate N: N=

(694.3 nm )(15 mJ ) 1240 eV ⋅ nm

1eV = 5.2 × 1016 −19 1.602 × 10 J

General Problems 70 • A beam of red light that has a wavelength of 700 nm in air travels in water. (a) What is the wavelength in water? (b) Does a swimmer underwater observe the same color or a different color for this light?

950

Chapter 31

Picture the Problem We can use v = fλ and the definition of the index of refraction to relate the wavelength of light in a medium whose index of refraction is n to the wavelength of light in air.

(a) The wavelength λn of light in a medium whose index of refraction is n is given by: Substitute numerical values and evaluate λwater:

λn =

v c λ = = 0 f nf n

λwater =

700 nm 700 nm = = 526 nm nwater 1.33

(b) Because the color observed depends on the frequency of the light, a swimmer observes the same color in air and in water. 71 •• [SSM] The critical angle for total internal reflection for a substance is 48º. What is the polarizing angle for this substance? Picture the Problem We can use Snell’s law, under critical angle and polarization conditions, to relate the polarizing angle of the substance to the critical angle for internal reflection.

Apply Snell’s law, under critical angle conditions, to the interface:

n1 sin θ c = n2

(1)

n1 sin θ p = n2 sin (90° − θ p ) = n2 cosθ p

Apply Snell’s law, under polarization conditions, to the interface:

or

Solve equation (1) for the ratio of n2 to n1:

n2 = sin θ c n1

Substitute for n2/n1 in equation (2) to obtain:

θ p = tan −1 (sin θ c )

Substitute numerical values and evaluate θp:

θ p = tan −1 (sin 48°) = 37°

tan θ p =

⎛n ⎞ n2 ⇒ θ p = tan −1 ⎜⎜ 2 ⎟⎟ (2) n1 ⎝ n1 ⎠

72 •• Show that when a flat mirror is rotated through an angle θ about an axis in the plane of the mirror, a reflected beam of light (from a fixed incident beam) that is perpendicular to the rotation axis is rotated through 2θ.

Properties of Light

951

Picture the Problem Let φ be the initial angle of incidence. Since the angle of reflection with the normal to the mirror is alsoφ, the angle between incident and reflected rays is 2φ. If the mirror is now rotated by a further angle θ, the angle of incidence is increased by θ to φ +θ, and so is the angle of reflection. Consequently, the reflected beam is rotated by 2θ relative to the incident beam.

φ +θ φ φ

φ +θ θ

73 •• [SSM] Use Figure 31-59 to calculate the critical angles for light initially in silicate flint glass that is incident on a glass–air interface if the light is (a) violet light of wavelength 400 nm and (b) red light of wavelength 700 nm. Picture the Problem We can apply Snell’s law at the glass-air interface to express θc in terms of the index of refraction of the glass and use Figure 31-59 to find the index of refraction of the glass for the given wavelengths of light.

Apply Snell’s law at the glass-air interface:

n1 sin θ1 = n2 sin θ 2

If θ1 = θc and n2 = 1:

n1 sin θ c = sin 90° = 1 and ⎛1⎞ ⎟⎟ n ⎝ 1⎠

θ c = sin −1 ⎜⎜

⎛ 1 ⎞ ⎟ = 36.8° ⎝ 1.67 ⎠

(a) For violet light of wavelength 400 nm, n1 = 1.67:

θ c = sin −1 ⎜

(b) For red light of wavelength 700 nm, n1 = 1.60:

θ c = sin −1 ⎜

⎛ 1 ⎞ ⎟ = 38.7° ⎝ 1.60 ⎠

74 •• Light is incident on a slab of transparent material at an angle θ1, as shown in Figure 31-60. The slab has a thickness t and an index of refraction n. Show that n = sin (θ1 ) tan −1 (d t ) , where d is the distance shown in the figure.

[

]

952

Chapter 31

Picture the Problem We can apply Snell’s law at the air-slab interface to express the index of refraction n in terms of θ1 and θ2 and then use the geometry of the figure to relate θ2 to t and d.

θ1 n

θ2

t

d

sin θ1 sin θ 2

Applying Snell’s law to the first interface yields:

sin θ1 = n sin θ 2 ⇒ n =

From the diagram:

⎛d ⎞ d = t tan θ 2 ⇒ θ 2 = tan −1 ⎜ ⎟ ⎝t⎠

Substitute for θ2 to obtain:

n=

sin θ1 ⎡ ⎛ d ⎞⎤ sin ⎢ tan −1 ⎜ ⎟⎥ ⎝ t ⎠⎦ ⎣

75 •• A ray of light begins at the point (–2.00 m, 2.00 m, 0.00 m), strikes a mirror in the y = 0 plane at some point (x, 0, 0), and reflects through the point (2.00 m, 6.00 m, 0.00 m). (a) Find the value of x that makes the total distance traveled by the ray a minimum. (b) What is the angle of incidence on the reflecting plane? (c) What is the angle of reflection? y, m

Picture the Problem We can write an expression for the total distance traveled by the light as a function of x and set the derivative of this expression equal to zero to find the value of x that minimizes the distance traveled by the light. The adjacent figure shows the two points and the reflecting surface. The x and y coordinates are in meters.

(a) Express the total distance D traveled by the light:

6

4 d2

2 d1

−2

x, m 0

xmin

2

D = d1 + d 2 =

(x + 2)2 + 4 + (2 − x )2 + 36

Properties of Light

953

Differentiate D with respect to x: dD d ⎡ = dx dx ⎢⎣ =

1 2

(x + 2)2 + 4 + (2 − x )2 + 36 ⎤⎥

[(x + 2)

2

+4

]

− 12

[

⎦

2( x + 2 ) + 12 (2 − x ) + 36 2

]

− 12

2(2 − x )(− 1)

Setting this expression equal to zero for extreme values yields: xmin + 2

(xmin + 2)2 + 4 Solve for xmin to obtain:

−

2 − xmin

(2 − xmin )2 + 36

=0

x min = − 1.00 m

(b) Letting the coordinates of the point at which the ray originates be (x1, y1) and the coordinates of the terminal point be (x2, y2), the tangent of the angle of incidence is given by:

tan θ i =

⎛ x − x1 x − x1 ⇒ θ i = tan −1 ⎜⎜ y − y1 ⎝ y − y1

⎞ ⎟ ⎟ ⎠

where (x, y) = (−1.00 m, 0).

Substitute numerical values and evaluate θi: ⎡ − 1.00 m − (− 2.00 m ) ⎤ ⎥ = 26.6° 0 − 2.00 m ⎣ ⎦

θ i = tan −1 ⎢

(c) Letting the coordinates of the point at which the ray originates be (x, y) and the coordinates of the terminal point be (x2, y2), the tangent of the angle of reflection is given by:

tan θ r =

⎛ x −x⎞ x2 − x ⎟⎟ ⇒ θ r = tan −1 ⎜⎜ 2 y2 − y ⎝ y2 − y ⎠

where (x, y) = (−1.00 m, 0).

Substitute numerical values and evaluate θr: ⎡ 2.00 m − (− 1.00 m ) ⎤ ⎥ = 26.6° 6.00 m − 0 ⎣ ⎦

θ r = tan −1 ⎢

76 •• To produce a polarized laser beam a plate of transparent material, (Figure 31-61) is placed in the laser cavity and oriented so the light strikes it at the polarizing angle. Such a plate is called a Brewster window. Show that if θP1 is the

954

Chapter 31

polarizing angle for the n1 to n2 interface, then θP2 is the polarizing angle for the n2 to n1 interface. Picture the Problem Let the angle of refraction at the first interface by θ1 and the angle of refraction at the second interface be θ2. We can apply Snell’s law at each interface and eliminate θ1 and n2 to show that θ2 = θP2.

Apply Brewster’s law at the n1-n2 interface:

tan θ P1 =

n2 n1

Draw a reference triangle consistent with Brewster’s law:

2

2

n1

2 +n

n2

θP1 n1 Apply Snell’s law at the n1-n2 interface: Solve for θ1 to obtain:

Referring to the reference triangle we note that:

n1 sin θ P1 = n2 sin θ1

⎛ n1 ⎞ sin θ P1 ⎟⎟ ⎝ n2 ⎠

θ1 = sin −1 ⎜⎜

⎛n

⎞ ⎟ ⎜ n2 n 2 + n 2 ⎟ 1 2 ⎠ ⎝ ⎛ ⎞ n1 ⎟ = sin −1 ⎜ ⎜ n2 + n2 ⎟ 2 ⎠ ⎝ 1

θ1 = sin −1 ⎜

1

n2

that is, θ1 is the complement of θp1. Apply Snell’s law at the n2-n1 interface: Solve for θ2 to obtain:

n2 sin θ1 = n1 sin θ 2

⎛ n2 ⎞ sin θ1 ⎟⎟ ⎝ n1 ⎠

θ 2 = sin −1 ⎜⎜

Properties of Light ⎛n

955

⎞ ⎟ ⎜ n1 n 2 + n 2 ⎟ 1 2 ⎠ ⎝ ⎛ ⎞ n2 ⎟= θ = sin −1 ⎜ P2 ⎜ n2 + n2 ⎟ 2 ⎠ ⎝ 1

Refer to the reference triangle again to obtain:

θ 2 = sin −1 ⎜

Equate these expressions for n2 sin θ1 to obtain:

n1 sin θ P = n1 sin θ 2 ⇒ θ 2 = θ P

2

n1

77 •• [SSM] From the data provided in Figure 31-59, calculate the polarization angle for an air–glass interface, using light of wavelength 550 nm in each of the four types of glass shown. Picture the Problem We can use Brewster’s law in conjunction with index of refraction data from Figure 31-59 to calculate the polarization angles for the airglass interface.

From Brewster’s law we have:

⎛ n2 ⎞ ⎟⎟ ⎝ n1 ⎠

θ p = tan −1 ⎜⎜

or, for n1 = 1, θ p = tan −1 (n2 ) For silicate flint glass, n2 ≈ 1.62 and:

θ p,silicate flint = tan −1 (1.62) = 58.3°

For borate flint glass, n2 ≈ 1.57 and:

θ p,borate flint = tan −1 (1.57 ) = 57.5°

For quartz glass, n2 ≈ 1.54 and:

θ p, quartz = tan −1 (1.54) = 57.0°

For silicate crown glass, n2 ≈ 1.51 and:

θ p,silicate crown = tan −1 (1.51) = 56.5°

78 •• A light ray passes through a prism with an apex angle of α, as shown in Figure 31-63. The ray and the bisector of the apex angle bisect at right angles. Show that the angle of deviation δ is related to the apex angle and the index of refraction of the prism material by sin ⎡⎣ 12 (α + δ )⎤⎦ = n sin (12 α ).

956

Chapter 31

Picture the Problem The diagram to the right shows the angles of incidence, refraction, and deviation at the first interface. We can use the geometry of this symmetric passage of the light to express θr in terms of α and δ1 in terms of θr and α. We can then use a symmetry argument to express the deviation at the second interface and the total deviation δ. Finally, we can apply Snell’s law at the first interface to complete the derivation of the given expression.

α

90 ° − θ r θi

n1

δ1 θr

n2

With respect to the normal to the left face of the prism, let the angle of incidence be θi and the angle of refraction be θr. From the geometry of the figure, it is evident that:

θ r = 12 α

Express the angle of deviation at the refracting surface:

δ1 = θ i − θ r = θ i − 12 α

Referring to triangle ABC, we see that:

δ = 2δ 1 = 2(θ1 − 12 α ) = 2θ i − α

α

A

n1

B δ δ1 δ 1

n2

C

δ

n1

(α + δ )

Solving for θI yields:

θi =

Apply Snell’s law, with n1 = 1 and n2 = n, to the first interface:

sin θ i = n sin 12 α

1 2

δ

n1

Properties of Light Substitute for θI to obtain:

sin

α +δ 2

= n sin

α

957

(1)

2

79 •• [SSM] (a) For light rays inside a transparent medium that is surrounded by a vacuum, show that the polarizing angle and the critical angle for total internal reflection satisfy tan θp = sin θc. (b) Which angle is larger, the polarizing angle or the critical angle for total internal reflection? Picture the Problem We can apply Snell’s law at the critical angle and the polarizing angle to show that tan θp = sin θc.

(a) Apply Snell’s law at the medium-vacuum interface:

n1 sin θ1 = n2 sin θ r

For θ1 = θc, n1 = n, and n2 = 1:

n sin θ c = sin 90° = 1

For θ1 = θp, n1 = n, and n2 = 1:

Because both expressions equal one:

tan θ p =

n2 1 = ⇒ n tan θ p = 1 n1 n

tan θ p = sin θ c

(b) For any value of θ :

tan θ sinθc = tanθp

θp θc tan θ > sin θ ⇒ θ c > θ p

sin θ

90

o

80 •• Light in air is incident on the surface of a transparent substance at an angle of 58º with the normal. The reflected and refracted rays are observed to be mutually perpendicular. (a) What is the index of refraction of the transparent substance? (b) What is the critical angle for total internal reflection in this substance? Picture the Problem Let the numeral 1 refer to the side of the interface from which the light is incident and the numeral 2 to the refraction side of the interface. We can apply Snell’s law, under the conditions described in the problem statement, at the interface to derive an expression for n as a function of the angle of incidence (also the polarizing angle).

958

Chapter 31

(a) Apply Snell’s law at the airmedium interface:

sin θ1 = n sin θ 2

Because the reflected and refracted rays are mutually perpendicular: Substitute for θ2 to obtain:

θ1 + θ 2 = 90° ⇒ θ 2 = 90° − θ1 sin θ1 = n sin (90° − θ1 ) = n cos θ1 or n = tan θ1 = tan θ p

Substitute for θp and evaluate n:

n = tan 58° = 1.6

(b) Apply Snell’s law at the interface under conditions of total internal reflection:

n2 sin θ c = n1 sin 90° = n1

Because n1 = 1:

Substitute for n and evaluate θc:

⎛1⎞ 1 ⎟⎟ = sin −1 ⎛⎜ ⎞⎟ ⎝n⎠ ⎝ n2 ⎠

θ c = sin −1 ⎜⎜

⎛ 1 ⎞ ⎟ = 39° ⎝ 1.6 ⎠

θ c = sin −1 ⎜

81 •• A light ray in dense flint glass that has an index of refraction of 1.655 is incident on the glass surface. An unknown liquid condenses on the surface of the glass. Total internal reflection on the glass–liquid interface occurs for a minimum angle of incidence on the glass–liquid interface of 53.7º. (a) What is the refractive index of the unknown liquid? (b) If the liquid is removed, what is the minimum angle of incidence for total internal reflection? (c) For the angle of incidence found in Part (b), what is the angle of refraction of the ray into the liquid film? Does a ray emerge from the liquid film into the air above? Assume the glass and liquid have parallel planar surfaces. Picture the Problem We can apply Snell’s law at the glass–liquid and liquid–air interfaces to find the refractive index of the unknown liquid, the minimum angle of incidence (glass-air interface) for total internal reflection, and the angle of refraction of a ray into the liquid film.

(a) Apply Snell’s law, under critical-angle conditions, at the glass–liquid interface:

sin θ c =

nliquid nglass

⇒ nliquid = nglass sin θ c

Properties of Light Substitute numerical values and evaluate nliquid : (b) With the liquid removed:

nliquid = (1.655) sin 53.7° = 1.33

⎛ 1 ⎞ ⎟ ⎟ n ⎝ glass ⎠

θ c = sin −1 ⎜⎜

⎛ 1 ⎞ ⎟ = 37.2° ⎝ 1.655 ⎠

Substitute numerical values and evaluate θc:

θ c = sin −1 ⎜

(c) Apply Snell’s law at the glass−liquid interface:

nglass sin θ1 = nliquid sin θ 2

Solve for θ2:

959

⎡ nglass

θ 2 = sin −1 ⎢

⎣⎢ nliquid

Letting θ1 be the angle of incidence found in Part (b) yields:

θ 2 = sin −1 ⎢⎜⎜

Substitute numerical values and evaluate θ2:

θ 2 = sin −1 ⎜

⎤ sin θ1 ⎥ ⎦⎥

⎡⎛ nglass ⎞⎛ 1 ⎞⎤ ⎟⎜ ⎟⎥ ⎟ ⎜ n n ⎢⎣⎝ liquid ⎠⎝ glass ⎟⎠⎥⎦ ⎛ 1 ⎞ ⎟ = sin −1 ⎜ ⎜n ⎟ ⎝ liquid ⎠ ⎛ 1 ⎞ ⎟ = 48.6° ⎝ 1.333 ⎠

Because 48.6° is also the angle of incidence at the liquid-air interface and because it is equal to the critical angle for total internal reflection at this interface, no light will emerge into the air. 82 ••• (a) Show that for normally incident light, the intensity transmitted through a glass slab that has an index of refraction of n and is surrounded by air is 2

2 approximately given by I T = I0 ⎡ 4n (n + 1) ⎤ . (b) Use the Part (a) result to find ⎣ ⎦ the ratio of the transmitted intensity to the incident intensity through N parallel slabs of glass for light of normal incidence. (c) How many slabs of a glass that has an index of refraction of 1.5 are required to reduce the intensity to 10 percent of the incident intensity?

960

Chapter 31

Picture the Problem We’ll neglect multiple reflections at the glass-air interfaces. We can use the expression (Equation 31-7) for the reflected intensity at an interface to express the intensity of the light in the glass slab as the difference between the intensity of the incident beam and the reflected beam. Repeating this analysis at the glass-air interface will lead to the desired result.

(a) Express the intensity of the light transmitted into the glass:

The intensity of the light reflected at the air-glass interface is given by Equation 31-7: Substitute and simplify to obtain:

Glass

Air

I glass

I0 I R,1

IT

I R,2

n

I glass = I 0 − I R,1 where IR,1 is the intensity of the light reflected at the air-glass interface. 2

I R,1

⎛1− n ⎞ =⎜ ⎟ I0 ⎝1+ n ⎠ 2

⎛1− n ⎞ I glass = I 0 − ⎜ ⎟ I0 ⎝1+ n ⎠ ⎡ ⎛ 1 − n ⎞2 ⎤ = I 0 ⎢1 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ 1 + n ⎠ ⎥⎦ ⎡ 4n ⎤ = I0 ⎢ 2⎥ ⎣ (1 + n ) ⎦

Express the intensity of the light transmitted at the glass-air interface: The intensity of the light reflected at the glass-air interface is:

I T = I glass − I R,2 where IR,2 is the intensity of the light reflected at the glass-air interface. 2

I R,2

⎛1− n ⎞ =⎜ ⎟ I glass ⎝1+ n ⎠ 2

⎛ 1 − n ⎞ ⎡ 4n ⎤ =⎜ I ⎟ ⎢ 2⎥ 0 ⎝ 1 + n ⎠ ⎣ (1 + n ) ⎦

Properties of Light Substitute and simplify to obtain:

2

961

⎡ 4n ⎤ ⎛ 1 − n ⎞ ⎡ 4n ⎤ −⎜ IT = I0 ⎢ I ⎟ ⎢ 2⎥ 2⎥ 0 ⎣ (1 + n ) ⎦ ⎝ 1 + n ⎠ ⎣ (1 + n ) ⎦ ⎡ ⎛ 1 − n ⎞ 2 ⎤ ⎡ 4n ⎤ = I 0 ⎢1 − ⎜ ⎟ ⎥⎢ 2⎥ ⎢⎣ ⎝ 1 + n ⎠ ⎥⎦ ⎣ (1 + n ) ⎦ ⎡ 4n ⎤ ⎡ 4 n ⎤ = I0 ⎢ 2 ⎥⎢ 2⎥ ⎣ (1 + n ) ⎦ ⎣ (1 + n ) ⎦ ⎡ 4n ⎤ = I0 ⎢ 2⎥ ⎣ (1 + n ) ⎦

2

2

(b) From Part (a), each slab reduces the intensity by the factor:

⎡ 4n ⎤ ⎢ 2⎥ ⎣ (n + 1) ⎦

For N slabs:

⎡ 4n ⎤ I t = I0 ⎢ 2⎥ ⎣ (n + 1) ⎦ and

2N

⎡ 4n ⎤ It = ⎢ 2⎥ I0 ⎣ (n + 1) ⎦

2N

(c) Begin the solution of equation (1) for N by taking the logarithm (arbitrarily to base 10) of both sides of the equation:

⎡ 4n ⎤ ⎛I ⎞ log⎜⎜ t ⎟⎟ = log ⎢ 2⎥ ⎝ I0 ⎠ ⎣ (n + 1) ⎦

Solve for N:

⎛I ⎞ log⎜⎜ t ⎟⎟ ⎝ I0 ⎠ N= ⎡ 4n ⎤ 2 log ⎢ 2⎥ ⎣ (n + 1) ⎦

Substitute numerical values and evaluate N:

N=

(1)

2N

⎡ 4n ⎤ = 2 N log ⎢ 2⎥ ⎣ (n + 1) ⎦

log(0.1) = 28.2 ≈ 28 ⎡ 4(1.5) ⎤ 2 log ⎢ 2⎥ ⎣ (1.5 + 1) ⎦

962

Chapter 31

83 ••• Equation 31-14 gives the relation between the angle of deviation φd of a light ray incident on a spherical drop of water in terms of the incident angle θ1 and the index of refraction of water. (a) Assume that nair = 1, and derive an expression for dφd/dθ1. Hint: If y = sin–1 x, then dy/dx = (1 – x2)–1/2. (b) Use this result to show that the angle of incidence for minimum deviation θ1m is given by cos θ1m =

1 3

(n

2

)

− 1 . (c) The index of refraction for a certain red light in water is

1.3318 and that the index of refraction for a certain blue light in water is 1.3435. Use the result of Part (a) to find the angular separation of these colors in the primary rainbow. Picture the Problem (a) We can follow the directions given in the problem statement and use the hint to establish the given result. (b) Treating the result of Part (a) as an extreme-value problem will lead to the given result. (c) We can use Equation 31-14 and the result of Part (b) to find the angular separation of these colors in the primary rainbow.

(a) Equation 31-14 is:

⎛ nair sin θ1 ⎞ ⎟⎟ ⎝ nwater ⎠

φd = π + 2θ1 − 4 sin −1 ⎜⎜

⎛ sin θ1 ⎞ ⎟ ⎝ n ⎠

For nair = 1and nwater = n:

φd = π + 2θ1 − 4 sin −1 ⎜

Use the hint to differentiate φd with respect to θ1:

dφd d = dθ1 dθ1

(b) Set dφd/dθ1 = 0:

2−

⎡ −1 ⎛ sin θ1 ⎞ ⎤ ⎢π + 2θ1 − 4 sin ⎜ n ⎟⎥ ⎝ ⎠⎦ ⎣ 4 cos θ1 = 2− n 2 − sin 2 θ1 4 cos θ1 n 2 − sin 2 θ1

= 0 for extrema

(

Simplify to obtain:

16 cos 2 θ1 = 4 n 2 − sin 2 θ1

Replace sin2θ1 with 1 − cos2θ1 and simplify to obtain:

12 cos 2 θ1 = 4n 2 − 4

Solve for cosθ1 = cosθ1m:

cos θ1m =

n2 − 1 3

)

Properties of Light (c) Express the angular separation Δφ of blue and red:

Δφ = φd,blue − φd,red

From Equation 31-18, with nair = 1 and nwater = n:

φd = π + 2θ1 − 4 sin −1 ⎜

(1)

⎛ sin θ1 ⎞ ⎟ ⎝ n ⎠

⎡ n2 − 1 ⎤ ⎥ 3 ⎥⎦ ⎢⎣

From Part (b):

θ1m = cos −1 ⎢

Substitute to obtain: 2 ⎛ ⎧⎪ ⎫⎞ ⎜ sin cos −1 ⎡⎢ n − 1 ⎤⎥ ⎪ ⎟ ⎬⎟ ⎨ ⎜ ⎪ 2 3 ⎡ ⎤ ⎢ ⎥ − n 1 ⎣ ⎦ ⎪⎭ ⎟ ⎩ −1 φd = π + 2 cos −1 ⎢ ⎥ − 4 sin ⎜ ⎜ ⎟ 3 ⎥⎦ n ⎢⎣ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ Evaluate φd for blue light in water:

⎡

φd,blue = π + 2 cos −1 ⎢ ⎢⎣

2 ⎛ ⎧ ⎫⎞ ⎜ sin ⎪cos −1 ⎡⎢ (1.3435) − 1 ⎤⎥ ⎪ ⎟ ⎬ ⎜ ⎨ 3 ⎢⎣ ⎥⎦ ⎪⎭ ⎟⎟ (1.3435)2 − 1 ⎤⎥ − 4 sin −1 ⎜ ⎪⎩ ⎜ ⎟ 3 1.3435 ⎥⎦ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠

= 139.42° Evaluate φd for red light in water:

⎡

φd,red = π + 2 cos −1 ⎢ ⎢⎣

= 137.75°

2 ⎛ ⎧ ⎫⎞ ⎜ sin ⎪cos −1 ⎡⎢ (1.3318) − 1 ⎤⎥ ⎪ ⎟ ⎬ ⎜ ⎨ 3 ⎢⎣ ⎥⎦ ⎪⎭ ⎟⎟ (1.3318)2 − 1 ⎤⎥ − 4 sin −1 ⎜ ⎪⎩ ⎜ ⎟ 3 1.3318 ⎥⎦ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠

963

964

Chapter 31

Substitute numerical values in equation (1) and evaluate Δφ:

Δφ = 139.42 − 137.75° = 1.67°

84 ••• Show that the angle of deviation δ is a minimum if the angle of incidence is such that the ray and the bisector of the apex angle α (Figure 31-62) intersect at right angles. Picture the Problem The figure below shows the prism and the path of the ray through it. The dashed lines are the normals to the prism faces. The triangle formed by the interior ray and the prism faces has interior angles of α, 90° − θ2, and 90° − θ3. Consequently, θ 2 + θ 3 = α . We can apply Snell’s law at both

interfaces to express the angle of deviation δ as a function of θ3 and then set the derivative of this function equal to zero to find the conditions on θ3 and θ2 that result in δ being a minimum. a α

δd θ1

θu22

θ3 u3

θ4

n

Express the angle of deviation:

δ = θ1 + θ 2 − α

(1)

Apply Snell’s law to relate θ1 to θ2 and θ3 to θ4:

sin θ1 = n sin θ 2 and n sin θ 3 = sin θ 4

(2)

Solve equation (2) for θ1 and equation (3) for θ4:

(3)

θ1 = sin −1 (n sin θ 2 ) and θ 4 = sin −1 (n sin θ 3 )

Substitute in equation (1) to obtain:

δ = sin −1 (n sin θ 2 ) + sin −1 (n sin θ 3 ) − α = sin −1 [n sin (θ 3 − α )] + sin −1 (n sin θ 3 ) − α

Properties of Light

965

Note that the only variable in this expression is θ3. To determine the condition that minimizesδ, take the derivative of δ with respect to θ3 and set it equal to zero. dδ d = sin −1 [n sin (θ 3 − α )] + sin −1 (n sin θ 3 ) − α dθ 3 dθ 3

{

=−

n cos(α − θ 3 )

1 − [n sin (α − θ 3 )]

2

+

n cos θ 3

1 − (n sin θ 3 )

2

}

= 0 for extrema

This equation is satisfied provided:

α − θ 3 = θ 3 ⇒ θ 3 = 12 α

Because θ 2 = α − θ 3 :

θ 2 = α − 12 α = 12 α

Because θ 2 = θ 3 , we can conclude that the deviation angle is a minimum if the ray p through the prism symmetrically. Remarks: Setting dδ/dθ3 = 0 establishes the condition on θ3 that δ is either a maximum or a minimum. To establish that δ is indeed a minimum when θ3 = θ2 = 12 α, we can either show that d 2 δ dθ32 , evaluated at θ3 = 12 α , is positive or, alternatively, plot a graph of δ (θ3) to show that it is concave upward at θ3 = 12 α .

966

Chapter 31

Chapter 32 Optical Images Conceptual Problems 1 • Can a virtual image be photographed? If so, give an example. If not, explain why. Determine the Concept Yes. Note that a virtual image is ″seen″ because the eye focuses the diverging rays to form a real image on the retina. For example, you can photograph the virtual image of yourself in a flat mirror and get a perfectly good picture. 2 • Suppose the x, y, and z axes of a 3-D coordinate system are painted a different color. One photograph is taken of the coordinate system and another is taken of its image in a plane mirror. Is it possible to tell that one of the photographs is of a mirror image? Or could both photographs be of the real coordinate system from different angles? Determine the Concept Yes; the coordinate system and its mirror image will have opposite handedness. That is, if the coordinate system is a right-handed coordinate system, then the mirror image will be a left-handed coordinate system, and vice versa. 3

•

[SSM]

True or False

(a) The virtual image formed by a concave mirror is always smaller than the object. (b) A concave mirror always forms a virtual image. (c) A convex mirror never forms a real image of a real object. (d) A concave mirror never forms an enlarged real image of an object. (a) False. The size of the virtual image formed by a concave mirror when the object is between the focal point and the vertex of the mirror depends on the distance of the object from the vertex and is always larger than the object. (b) False. When the object is outside the focal point, the image is real. (c) True. (d) False. When the object is between the center of curvature and the focal point, the image is enlarged and real.

967

Chapter 32

968

4 • An ant is crawling along the axis of a convex mirror that has a radius of curvature R. At what object distances, if any, will the mirror produce (a) an upright image, (b) a virtual image, (c) an image smaller than the object, and (d) an image larger than the object? Determine the Concept Let s be the object distance and f the focal length of the mirror. (a) If s < 12 R , the image is virtual, upright, and larger than the object. (b) If s < 12 R , the image is virtual, upright, and larger than the object. (c) If s > R , the image is real, inverted, and smaller than the object. (d) If

1 2

R < s < R , the image is real, inverted, and larger than the object.

5 • [SSM] An ant is crawling along the axis of a concave mirror that has a radius of curvature R. At what object distances, if any, will the mirror produce (a) an upright image, (b) a virtual image, (c) an image smaller than the object, and (d) an image larger than the object? Determine the Concept (a) The mirror will produce an upright image for all object distances. (b) The mirror will produce a virtual image for all object distances. (c) The mirror will produce an image that is that is smaller than the object for all object distances. (d) The mirror will never produce an enlarged image. 6 •• Convex mirrors are often used for rearview mirrors on cars and trucks to give a wide-angle view. ″Warning, objects are closer than they appear.″ is written below the mirrors. Yet, according to a ray diagram, the image distance for distant objects is much shorter than the object distance. Why then do they appear more distant? Determine the Concept They appear more distant because in a convex mirror the angular sizes of the images are smaller than they would be in a flat mirror. The y' angular size of the image is , where y′ is the height of the image, s′ is the s' + L distance between the image and the mirror, and L is the distance between the observer and the mirror. For a flat mirror, y′ = y and s′ = −s.

Optical Images We wish to prove that, for a convex mirror:

y' y < s' + L s + L

If equation (1) is valid, then:

s' + L

>

y' or s' y'

+

(1)

s+L y

L s L > + y' y y

(2)

Because the lateral magnification is y' s' m= =− : y s

y' y = s' s

(3)

Combining this with equation (2) yields:

L L > ⇒ y' < y y' y

(4)

It follows that, if y′ < y, then:

s' < s

To show that s ' < s we begin with

1 1 1 + = s' s f

the mirror equation: Solving for s′ yields:

s' = s

Because f is negative and s is positive:

s' = s

969

f s− f f s+ f

⇒ s' < s

Thus, we have proven that, if equation (1) is valid, then equation (4) is valid. To prove that equation 1 is valid, we merely follow this proof in reverse order. 7 • As an ant on the axis of a concave mirror crawls from a great distance to the focal point of a concave mirror, the image of the ant moves from (a) a great distance toward the focal point and is always real, (b) the focal point to a great distance from the mirror and is always real, (c) the focal point to the center of curvature of the mirror and is always real, (d) the focal point to a great distance from the mirror and changes from a real image to a virtual image.

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Picture the Problem The ray diagram shows three object positions 1, 2, and 3 as the moves from a great distance to the focal point F of a concave mirror. The real images corresponding to each of these object positions are labeled with the same numeral. (b) is correct.

8 • A kingfisher bird that is perched on a branch a few feet above the water is viewed by a scuba diver submerged beneath the surface of the water directly below the bird. Does the bird appear to the diver to be closer to or farther from the surface than the actual bird? Explain your answer using a ray diagram. Picture the Problem The diagram shows two rays (from the bundle of rays) of light refracted at the air-water interface. Because the index of refraction of water is greater than that of air, the rays are bent toward the normal. The diver will, therefore, think that the rays are diverging from a point above the bird and so the bird appears to be farther from the surface than it actually is.

.

.

Image of the bird

The bird

Air Water

9 • An object is placed on the axis of a diverging lens whose focal length has a magnitude of 10 cm. The distance from the object to the lens is 40 cm. The image is (a) real, inverted, and diminished, (b) real, inverted, and enlarged, (c) virtual, inverted, and diminished, (d) virtual, upright, and diminished, (e) virtual, upright, and enlarged. Picture the Problem We can use a ray diagram to determine the general features of the image. In the diagram shown, the parallel ray and central ray have been used to locate the image.

971

>

Optical Images

s = 40 cm

f = 10 cm

> From the diagram, we see that the image is virtual (only one of the rays from the head of the object actually pass through the head of the image), upright, and diminished. (d ) is correct. 10 •• If an object is placed between the focal point of a converging lens and the optical center of the lens, the image is (a) real, inverted, and enlarged, (b) virtual, upright, and diminished, (c) virtual, upright, and enlarged, (d) real, inverted, and diminished.

>

Picture the Problem We can use a ray diagram to determine the general features of the image. In the diagram shown, the ray parallel to the principle axis and the central ray have been used to locate the image. Note that the object has been placed farther to the right than suggested in the problem statement. This was done so that the image would not be so large and so far to the left as to make the diagram excessively large.

F

F'

>

From the diagram, we see that the image is virtual (neither ray from the head of the object passes through the head of the image), upright, and enlarged. (c) is correct. 11 • A converging lens is made of glass that has an index of refraction of 1.6. When the lens is in air, its focal length is 30 cm. When the lens is immersed in water, its focal length (a) is greater than 30 cm, (b) is between zero and 30 cm, (c) is equal to 30 cm, (d) has a negative value.

Chapter 32

972

Picture the Problem We can apply the lens maker’s equation to the air-glass lens and to the water-glass lens to find the ratio of their focal lengths. Apply the lens maker’s equation to the air-glass interface: Apply the lens maker’s equation to the water-glass interface: Divide the first of these equations by the second to obtain:

⎛1 1⎞ 1 = (1.6 − 1)⎜⎜ − ⎟⎟ f air ⎝ r1 r2 ⎠ 1 f water

f water f air

⎛1 1⎞ = (1.6 − 1.33)⎜⎜ − ⎟⎟ ⎝ r1 r2 ⎠

⎛1 1⎞ (1.6 − 1)⎜⎜ − ⎟⎟ ⎝ r1 r2 ⎠ = 2.2 = ⎛1 1⎞ (1.6 − 1.33)⎜⎜ − ⎟⎟ ⎝ r1 r2 ⎠

or f water = 2.2 f air and (a)

is correct.

True or false:

12

•

(a) (b) (c)

A virtual image cannot be displayed on a screen. A negative image distance implies that the image is virtual. All rays parallel to the axis of a spherical mirror are reflected through a single point. A diverging lens cannot form a real image from a real object. The image distance for a converging lens is always positive.

(d) (e)

(a) True. (b) True. (c) False. Where the rays intersect the axis of a spherical mirror depends on how far from the axis they are reflected from the mirror. (d) True. (e) False. The image distance for a virtual image is negative. 13 • Both the human eye and the digital camera work by forming real images on light-sensitive surfaces. The eye forms a real image on the retina and the camera forms a real image on a CCD array. Explain the difference between the ways in which these two systems accommodate. That is, the difference between how an eye adjusts and how a camera adjusts (or can be adjusted) to form a focused image for objects at both large and short distances from the camera.

Optical Images

973

Determine the Concept The muscles in the eye change the thickness of the lens and thereby change the focal length of the lens to accommodate objects at different distances. A camera lens, on the other hand, has a fixed focal length so that focusing is accomplished by varying the distance between the lens and the light-sensitive surface. 14 • If an object is 25 cm in front of the naked eye of a farsighted person, an image (a) would be formed behind the retina if it were not for the fact that that the light is blocked (by the back of the eyeball) and the corrective contact lens should be convex, (b) would be formed behind the retina if it were not for the fact that that the light is blocked (by the back of the eyeball) and the corrective contact lens should be concave, (c) is formed in front of the retina and the corrective contact lens should be convex, (d) is formed in front of the retina and the corrective contact lens should be concave. Determine the Concept The eye muscles of a farsighted person lack the ability to shorten the focal length of the lens in the eye sufficiently to form an image on the retina of the eye. A convex lens (a lens that is thicker in the middle than at the circumference) will bring the image forward onto the retina. (a) is correct. 15 •• Explain the following statement: A microscope is an object magnifier, but a telescope is an angle magnifier. Hint: Take a look at the ray diagram for each magnifier and use it to explain the difference in adjectives. Determine the Concept The objective lens of a microscope ordinarily produces an image that is larger than the object being viewed, and that image is angularly magnified by the eyepiece. The objective lens of a telescope, on the other hand, ordinarily produces an image that is smaller than the object being viewed (see Figure 32-52), and that image is angularly magnified by the eyepiece. The telescope never produces a real image that is larger than the object.

Estimation and Approximation 16 • Estimate the location and size of the image of your face when you hold a shiny new tablespoon a foot in front of your face and with the convex side toward you. Picture the Problem We can model the spoon as a convex spherical mirror and use the mirror equation and the lateral magnification equation to estimate the location and size of the image of your face. The mirror equation is:

1 1 1 + = s s' f

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Chapter 32

Because f = r 2 , where r is the radius of curvature of the mirror:

1 1 2 rs + = ⇒ s' = s s' r r − 2s

Let the distance from your face to the surface of the spoon be 15 cm. If the radius of curvature of the spoon is 2 cm, then:

s' =

(2.0 cm) (15 cm) ≈ −1.1 cm 2.0 cm − 2(15 cm)

or about 1.1 cm behind the spoon y' s' ⎛ s' ⎞ = − ⇒ y ' = −⎜ ⎟ y y s ⎝s⎠

The lateral magnification equation is:

m=

Assume your face is 25 cm long. Substitute numerical values and evaluate y′:

⎛ − 1.1 cm ⎞ y ' = −⎜ ⎟ (25 cm ) ⎝ 15 cm ⎠ ≈

2 cm

17 • Estimate the focal length of the ″mirror″ produced by the surface of the water in the reflection pools in front of the Lincoln Memorial on a still night. Determine the Concept The surface of the ″mirror″ produced by the surface of the water in the Lincoln Memorial reflecting pool is approximately spherical with a radius of curvature approximately equal to the radius of Earth. The focal length of a spherical mirror is half its radius of curvature. Hence f = 12 REarth 18 •• Estimate the maximum value that could be obtained for the magnifying power of a simple magnifier, using Equation 32-20. Hint: Think about the smallest focal length lens that could be made from glass and still be used as a magnifier. Picture the Problem Because the focal length of a spherical lens depends on its radii of curvature and the magnification depends on the focal length, there is a practical upper limit to the magnification.

Use Equation 32-20 to relate the magnification M of a simple magnifier to its focal length f: Use the lens-maker’s equation to relate the focal length of a lens to its radii of curvature and the index of refraction of the material from which it is constructed:

M=

xnp f

⎛1 1⎞ 1 = (n − 1) ⎜⎜ − ⎟⎟ f ⎝ r1 r2 ⎠

Optical Images For a plano-convex lens, r2 = ∞. Hence:

r 1 n −1 = ⇒ f = 1 f r1 n −1

Substitute in the expression for M and simplify to obtain:

M=

975

(n − 1)xnp r1

Note that the smallest reasonable value for r1 will maximize M. A reasonable smallest value for the radius of a magnifier is 1.0 cm. Use this value and n = 1.5 to estimate Mmax:

M max =

(1.5 − 1)(25 cm ) = 1.0 cm

13

Plane Mirrors [SSM] The image of the object point P in Figure 32-57 is viewed by 19 • an eye, as shown. Draw rays from the object point that reflect from the mirror and enter the eye. If the object point and the mirror are fixed in their locations, indicate the range of locations where the eye can be positioned and still see the image of the object point. Determine the Concept Rays from the source that are reflected by the mirror are shown in the following diagram. The reflected rays appear to diverge from the image. The eye can see the image if it is in the region between rays 1 and 2.

20 • You are 1.62 m tall and want to be able to see your full image in a vertical plane mirror. (a) What is the minimum height of the mirror that will meet your needs? (b) How far above the floor should the bottom of mirror in (a) be placed, assuming that the top of your head is 14 cm above your eye level? Use a ray diagram to explain your answer.

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Chapter 32

Determine the Concept A ray diagram showing rays from your feet and the top of your head reaching your eyes is shown to the right. (a) The mirror must be half your height, i.e., 81 cm

(b) The top of the minimum-height mirror must be 7 cm below the top of your head, or 155 cm above the floor. The bottom of the mirror should be placed 155 cm − 81 cm = 74 cm above the floor. 21 •• [SSM] (a) Two plane mirrors make an angle of 90º. The light from a point object that is arbitrarily positioned in front of the mirrors produces images at three locations. For each image location, draw two rays from the object that, after one or two reflections, appear to come from the image location. (b) Two plane mirrors make an angle of 60º with each other. Draw a sketch to show the location of all the images formed of an object on the bisector of the angle between the mirrors. (c) Repeat Part (b) for an angle of 120º. Determine the Concept (a) Draw rays of light from the object (P) that satisfy the law of reflection at the two mirror surfaces. Three virtual images are formed, as shown in the following figure. The eye should be to the right and above the mirrors in order to see these images. Virtual image

Virtual image

P

Virtual image

(b) The following diagram shows selected rays emanating from a point object (P) located on the bisector of the 60° angle between the mirrors. These rays form the two virtual images below the horizontal mirror. The construction details for the

Optical Images

977

two virtual images behind the mirror that is at an angle of 60° with the horizontal mirror have been omitted due to the confusing detail their inclusion would add to the diagram. Virtual image

Virtual image

P

Virtual image

Virtual image

(c) The following diagram shows selected rays emanating from a point object (P) on the bisector of the 120° angle between the mirrors. These rays form the two virtual images at the intersection of the dashed lines (extensions of the reflected rays): P

Virtual image

Virtual image

22 •• Show that the mirror equation (Equation 32-4 where f = r/2) yields the correct image distance and magnification for a plane mirror. Picture the Problem We can use Equation 32-4 and the definition of the lateral magnification of an image to show that the mirror equation yields the correct image distance and magnification for a plane mirror.

978

Chapter 32

The mirror equation is:

1 1 1 + = s s' f

Because f = r 2 :

1 1 2 + = s s' r

For a plane mirror, r = ∞ . Hence:

1 1 + = 0 ⇒ s' = − s s s'

The lateral magnification of the image is given by:

M =−

−s s' =− = +1 s s

23 •• When two plane mirrors are parallel, such as on opposite walls in a barber shop, multiple images arise because each image in one mirror serves as an object for the other mirror. An object is placed between parallel mirrors separated by 30 cm. The object is 10 cm in front of the left mirror and 20 cm in front of the right mirror. (a) Find the distance from the left mirror to the first four images in that mirror. (b) Find the distance from the right mirror to the first four images in that mirror. (c) Explain why each more distant image becomes fainter and fainter. Determine the Concept (a) The first image in the mirror on the left is 10 cm behind the mirror. The mirror on the right forms an image 20 cm behind that mirror or 50 cm from the left mirror. This image will result in a second image 50 cm behind the left mirror. The first image in the left mirror is 40 cm from the right mirror and forms an image 40 cm behind the right mirror or 70 cm from the left mirror. That image gives an image 70 cm behind the left mirror. The fourth image behind the left mirror is 110 cm behind that mirror.

(a) For the mirror on the left, the images are located 10 cm, 50 cm, 70 cm, and 110 cm behind the mirror on the left (b) For the mirror on the right, the images are located 20 cm, 40 cm, 80 cm, and 100 cm behind the mirror on the right. (c) The successive images are dimmer because the light travels farther to form them. The intensity falls of inversely with the square of the distance the light travels. In addition, at each reflection a small percentage of the light intensity is lost. Real mirrors are not 100% reflecting.

Spherical Mirrors 24 • A concave mirror has a radius of curvature equal to 24 cm. Use ray diagrams to locate the image, if it exists, for an object near the axis at distances of

Optical Images

979

(a) 55 cm, (b) 24 cm, (c) 12 cm, and (d) 8.0 cm from the mirror. For each case, state whether the image is real or virtual; upright or inverted; and enlarged, reduced, or the same size as the object. Picture the Problem The easiest rays to use in locating the image are (1) the ray parallel to the principal axis and passes through the focal point of the mirror, (2) the ray that passes through the center of curvature of the spherical mirror and is reflected back on itself, and (3) the ray that passes through the focal point of the spherical mirror and is reflected parallel to the principal axis. We can use any two of these rays emanating from the top of the object to locate the image of the object.

(a) The ray diagram follows. The image is real, inverted, and reduced.

F C

(b) The ray diagram is shown to the right. The image is real, inverted, and the same size as the object.

Object

C

F Image

(c) The ray diagram is shown to the right. The object is at the focal plane of the mirror. The emerging rays are parallel and do not form an image.

C

F

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Chapter 32

(d) The ray diagram is shown to the right. The image is virtual, erect, and enlarged. C

F

25 • [SSM] (a) Use the mirror equation (Equation 32-4 where f = r/2) to calculate the image distances for the object distances and mirror of Problem 24. (b) Calculate the magnification for each given object distance. Picture the Problem In describing the images, we must indicate where they are located, how large they are in relationship to the object, whether they are real or virtual, and whether they are upright or inverted. The object distance s, the image 1 1 1 distance s′, and the focal length of a mirror are related according to + = , s s' f 1 where f = 2 r and r is the radius of curvature of the mirror. In this problem,

f = 12 cm because r is positive for a concave mirror. (a) Solve the mirror equation for s′:

s' =

fs r where f = s− f 2

When s = 55 cm:

s' =

(12 cm )(55 cm ) = 15.35 cm 55 cm − 12 cm

= 15 cm When s = 24 cm:

When s = 12 cm:

When s = 8.0 cm:

s' =

s' =

s' =

(12 cm )(24 cm ) = 24 cm − 12 cm

(12 cm )(12 cm ) 12 cm − 12 cm

24 cm

is undefined

(12 cm )(8.0 cm ) = −24.0 cm 8.0 cm − 12 cm

= − 0.2 m

Optical Images (b) The lateral magnification of the image is:

m=−

s' s

When s = 55 cm, the lateral magnification of the image is:

m=−

s' 15.35 cm =− = − 0.28 s 55 cm

When s = 24 cm, the lateral magnification of the image is:

m=−

s' 24 cm =− = − 1.0 s 24 cm

When s = 12 cm: When s = 8.0 cm, the lateral magnification of the image is:

981

m is undefined.

m=−

− 24 cm s' =− = 3.0 s 8.0 cm

Remarks: These results are in excellent agreement with those obtained graphically in Problem 24. 26 • A convex mirror has a radius of curvature that has a magnitude equal to 24 cm. Use ray diagrams to locate the image, if it exists, for an object near the axis at distances of (a) 55 cm, (b) 24 cm, (c) 12 cm, (d) 8.0 cm, (e) 1.0 cm from the mirror. For each case, state whether the image is real or virtual; upright or inverted; and enlarged, reduced, or the same size as the object. Picture the Problem The easiest rays to use in locating the image are (1) the ray parallel to the principal axis and passes through the focal point of the mirror, (2) the ray that passes through the center of curvature of the spherical mirror and is reflected back on itself, and (3) the ray that passes through the focal point of the spherical mirror and is reflected parallel to the principal axis. We can use any two of these rays emanating from the top of the object to locate the image of the object. Note that, due to space limitations, the following ray diagrams are not to the same scale as those in Problem 24.

(a) The ray diagram follows. The image is virtual, upright, and reduced.

F

C

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Chapter 32

(b) The ray diagram follows. The image is virtual, upright, and reduced.

F

C

(c) The ray diagram follows. The image is virtual, upright and reduced.

F

C

(d) The ray diagram follows. The image is virtual, upright, and reduced.

F

C

(e) The ray diagram follows. The image is virtual, upright, and reduced.

F

C

Optical Images

983

27 • (a) Use the mirror equation (Equation 32-4 where f = r/2) to calculate the image distances for the object distances and mirror of Problem 26. (b) Calculate the magnification for each given object distance. Picture the Problem In describing the images, we must indicate where they are located, how large they are in relationship to the object, whether they are real or virtual, and whether they are upright or inverted. The object distance s, the image 1 1 1 distance s′, and the focal length of a mirror are related according to + = , s s' f 1 where f = 2 r and r is the radius of curvature of the mirror. In this problem,

f = −12 cm because r is negative for a convex mirror. (a) Solve the mirror equation for s′:

s' =

fs s− f

When s = 55 cm:

s' =

(− 12 cm )(55 cm ) = −9.85 cm 55 cm − (− 12 cm )

= − 9.9 cm When s = 24 cm:

s' =

(− 12 cm )(24 cm ) = 24 cm − (− 12 cm )

− 8.0 cm

When s = 12 cm:

s' =

(− 12 cm )(12 cm ) = 12 cm − (− 12 cm )

− 6.0 cm

When s = 8.0 cm:

s' =

(− 12 cm )(8.0 cm ) = 8.0 cm − (− 12 cm )

(b) The lateral magnification is given by:

m=−

s' s

When s = 55 cm, the lateral magnification of the image is:

m=−

− 9.85 cm = 0.18 55 cm

When s = 24 cm, the lateral magnification of the image is:

m=−

− 8.0 cm = 0.33 24 cm

When s = 12 cm, the lateral magnification of the image is:

m=−

− 6.0 cm = 0.50 12 cm

− 4.8 cm

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Chapter 32

When s = 8.0 cm, the lateral magnification of the image is:

m=−

− 4.80 cm = 0.60 8.0 cm

Remarks: These results are in excellent agreement with those obtained graphically in Problem 26. 28 •• Use the mirror equation (Equation 32-4 where f = r/2) to prove that a convex mirror cannot form a real image of a real object, no matter where the object is placed. Picture the Problem We can solve the mirror equation for 1/s′ and then examine the implications of f < 0 and s > 0.

Solve the mirror equation for 1/s′:

1 1 1 s− f = − = s' f s sf

For a convex mirror:

f 0, the numerator is positive and the denominator negative. Consequently:

1 < 0 ⇒ s' < 0 s'

29 • [SSM] A dentist wants a small mirror that will produce an upright image that has a magnification of 5.5 when the mirror is located 2.1 cm from a tooth. (a) Should the mirror be concave or convex? (b) What should the radius of curvature of the mirror be? Picture the Problem We can use the mirror equation and the definition of the lateral magnification to find the radius of curvature of the dentist’s mirror.

(a) The mirror must be concave. A convex mirror always produces a diminished virtual image. (b) Express the mirror equation:

1 1 1 2 2ss' + = = ⇒r = s s' f r s '+ s

The lateral magnification of the mirror is given by:

m=−

Substitute for s′ in equation (1) to obtain:

r=

s' ⇒ s' = −ms s

− 2ms 1− m

(1)

Optical Images Substitute numerical values and evaluate r:

r=

985

− 2(5.5)(2.1cm ) = 5.1cm 1 − 5.5

Convex mirrors are used in many stores to provide a wide angle of 30 •• surveillance for a reasonable mirror size. Your summer job is at a local convenience store that uses the mirror shown in Figure 32-58. This setup allows you (or the clerk) to survey the entire store when you are 5.0 m from the mirror. The mirror has a radius of curvature equal to 1.2 m. Assume all rays are paraxial. (a) If a customer is 10 m from the mirror, how far from the mirror is his image? (b) Is the image in front of or behind the mirror? (c) If the customer is 2.0 m tall, how tall is his image? Picture the Problem We can use the mirror equation and the relationship between the focal length of a mirror and its radius of curvature to find the location of the image. We can then use the definition of the lateral magnification of the mirror to find the height of the image formed in the mirror.

(a) Solve the mirror equation for s′:

s' =

Substitute for f and simplify to obtain:

s' =

rs rs = 1 s − 2 r 2s − r

Substitute numerical values and evaluate s′:

s' =

(− 1.2 m )(10 m ) = −56.6 cm 2(10 m ) − (− 1.2 m )

fs where f = 12 r s− f 1 2

and the image is 57 cm from the mirror. (b) Because the image distance is negative:

the image is behind the mirror.

(c) The lateral magnification of the mirror is given by Equation 32-5:

m=

Substitute numerical values and evaluate y′:

y' = −

y' s' s' = − ⇒ y' = − y y s s − 0.566 m (2.0 m ) = 11cm 10 m

31 •• A certain telescope uses a concave spherical mirror that has a radius equal to 8.0 m. Find the location and diameter of the image of the moon formed by this mirror. The moon has a diameter of 3.5 × 106 m and is 3.8 × 108 m from Earth.

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Chapter 32

Picture the Problem We can use the mirror equation to locate the image formed in this mirror and the expression for the lateral magnification of the mirror to find the diameter of the image.

Solve the mirror equation for the location of the image of the moon: Because f = 12 r :

s' =

s' =

fs s− f

1 2

rs rs = 1 s − 2 r 2s − r

(8.0 m )(3.8 ×108 m )

Substitute numerical values and evaluate s′:

s' =

2(3.8 × 10 8 m ) − 8.0 m

Express the lateral magnification of the mirror:

m=

s' y' s' = − ⇒ y' = − y s y s

Substitute numerical values and evaluate y′:

y' = −

= 4.0 m

(

4.0 m 3.5 × 10 6 m 3.8 × 10 8 m = −3.7 cm

)

The 3.7-cm-diameter image is 4.0 m in front of the mirror. 32 •• A piece of a thin spherical shell that has a radius of curvature of 100 cm is silvered on both sides. The concave side of the piece forms a real image 75 cm from the piece. The piece is then turned around so that its convex side faces the object. The piece is moved so that the image is now 35 cm from the piece on the concave side. (a) How far was the piece moved? (b) Was it moved toward the object or away from the object? Picture the Problem We can use the mirror equation to find the focal length of the piece of the thin spherical shell and then apply it a second time to find the object position after the piece has been moved.

(a) Solve the mirror equation for f to obtain:

f =

ss' s' + s

Substitute numerical values and evaluate f:

f =

(100 cm )(75 cm) = 42.86 cm 75 cm + 100 cm

Optical Images Solving the mirror equation for s yields:

s=

fs' s' − f

With the piece turned around, f = −42.86 cm and s′ = − 35 cm. Substitute numerical values and evaluate s:

s=

(− 42.86 cm )(− 35 cm ) ≈ 191cm − 35 cm − (− 42.86 cm )

The distance d the mirror moved is:

d = 191 cm − 100 cm = 91 cm

987

(b) The piece was moved away from the object. 33 •• Two light rays parallel to the optic axis of a concave mirror strike that mirror as shown in Figure 32-59. This mirror has a radius of curvature equal to 5.0 m. They then strike a small spherical mirror that is 2.0 m from the large mirror. The light rays finally meet at the vertex of the large mirror. Note: The small mirror is shown as planar, so as not to give away the answer, but it is not actually planar. (a) What is the radius of curvature of the small mirror? (b) Is that mirror convex or concave? Explain your answer. Picture the Problem Denote the larger mirror on the right using the numeral 1, the smaller mirror using the numeral 2, and the distance between the mirrors by d. Applying the mirror equation to the two mirrors will lead us to an expression for the radius of curvature of the small mirror.

(a) Apply the mirror equation to mirror 1:

1 1 2 + = s1 s1' r1

Because s1 = ∞ :

1 2 = ⇒ s1' = 12 r1 s1' r1

The image formed by mirror #1 serves as a virtual object for mirror 2. Apply the mirror equation to mirror 2 to obtain:

1 1 2 + = s 2 s 2' r2 or, because s 2 = d − s'1 = d − 12 r1 , 1 1 2 + = 1 d − 2 r1 s 2' r2

Because the image distance for mirror 2 is d:

2 1 2 + = 2d − r1 d r2

988

Chapter 32

Substituting numerical values yields:

2 1 2 + = 2(2.0 m ) − 5.0 m 2.0 m r2

Finally, solve for r2 to obtain:

r2 = − 1.3 m

(b) Because f 2 = 12 r2 < 0 , the small mirror is convex .

Images Formed by Refraction 34 •• A very long 1.75-cm-diameter glass rod has one end ground and polished to a convex spherical surface that has a 7.20-cm radius. The glass material has an index of refraction of 1.68. (a) A point object in air is on the axis of the rod and 30.0 cm from the spherical surface. Find the location of the image and state whether the image is real or virtual. (b) Repeat Part (a) for a point object in air, on the axis, and 5.00 cm from the spherical surface. Draw a ray diagram for each case. Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions.

Use the equation for refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

Solving for s′ yields:

s' =

(a) Substitute numerical values (s = 30.0 cm, n1 = 1.00, n2 = 1.68 and r = 7.20 cm) and evaluate s′:

1.68 = 27 cm 1.68 − 1.00 1.00 − 7.20 30.0 where the positive distance tells us that the image is 27 cm in back of the surface and is real. s' =

(1)

n2 n2 − n1 n1 − r s

Optical Images

989

In the following ray diagram, the object is at P and the image at P′.

C

P Air

P'

Glass n =1.68

1.68 = − 16 cm 1.68 − 1.00 1.00 − 7.20 5.00 where the minus sign tells us that the image is 16 cm in front of the surface and is virtual.

(b) Substitute numerical values (s = 5.00 cm, n1 = 1.00, n2 = 1.68 and r = 7.20 cm) and evaluate s′:

s' =

In the following ray diagram, the object is at P and the image at P′.

P'

C

P Air

Glass n =1.68

35 • [SSM] A fish is 10 cm from the front surface of a spherical fish bowl of radius 20 cm. (a) How far behind the surface of the bowl does the fish appear to someone viewing the fish from in front of the bowl? (b) By what distance does the fish’s apparent location change (relative to the front surface of the bowl) when it swims away to 30 cm from the front surface? Picture the Problem The diagram shows two rays (from the bundle of rays) of light refracted at the water-air interface. Because the index of refraction of air is less than that of water, the rays are bent away from the normal. The fish will, therefore, appear to be closer than it actually is. We can use the equation for refraction at a single surface to find the distance s′. We’ll assume that the glass bowl is thin enough that we can ignore the refraction of the light passing through it.

Air Water

Image of Fish

Fish

r

990

Chapter 32

(a) Use the equation for refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

Solving for s′ yields:

s' =

Substitute numerical values (s = −10 cm, n1 = 1.33, n2 = 1.00 and r = 20 cm) and evaluate s′:

1.00 = 8.6 cm 1.00 − 1.33 1.33 − − 10 cm 20 cm and the image is 8.6 cm from the front surface of the bowl.

(b) For s = 30 cm:

s' =

The change in the fish’s apparent location is:

35.9 cm − 8.6 cm ≈ 27 cm

n2 n2 − n1 n1 − r s

s' =

1.00 = 35.9 cm 1.00 − 1.33 1.33 − − 30 cm 20 cm

A very long 1.75-cm-diameter glass rod has one end ground and 36 •• polished to a concave spherical surface that has a 7.20-cm radius. The glass material has an index of refraction of 1.68. A point object in air is on the axis of the rod and 15.0 cm from the spherical surface. Find the location of the image and state whether the image is real or virtual. Draw a ray diagram. Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions.

Use the equation for refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

Solving for s′ yields:

s' =

n2 n2 − n1 n1 − r s

(1)

Optical Images Substitute numerical values (s = 15.0 cm, n1 = 1.00, n2 = 1.68, and r = −7.20 cm) and evaluate s′:

991

1.68 = − 10 1.68 − 1.00 1.00 − − 7.20 cm 15.0 cm where the minus sign tells us that the image is 10 cm in front of the surface of the rod and is virtual. s' =

In the following ray diagram, the object is at P and the virtual image is at P′.

P

P' C Air

Glass n =1.68

37 •• [SSM] Repeat Problem 34 for when the glass rod and the object are immersed in water and (a) the object is 6.00 cm from the spherical surface, and (b) the object is 12.0 cm from the spherical surface. Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions.

Use the equation for refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

Solving for s′ yields:

s' =

(a) Substitute numerical values (s = 6.00 cm, n1 = 1.33, n2 = 1.68, and r = 7.20 cm) and evaluate s′:

1.68 = − 9.7 cm 1.68 − 1.33 1.33 − 7.20 cm 6.00 cm where the negative distance tells us that the image is 9.7 cm in front of the surface and is virtual. s' =

(1)

n2 n2 − n1 n1 − r s

Chapter 32

992

In the following ray diagram, the object is at P and the virtual image is at P′.

P'

P

C

Water n1 = 1.33

(b) Substitute numerical values (s = 12.0 cm, n1 = 1.33, n2 = 1.68, and r = 7.20 cm) and evaluate s′:

Glass n2 = 1.68

1.68 = − 27 cm 1.68 − 1.33 1.33 − 7.20 cm 12.0 cm where the negative distance tells us that the image is 27 cm in front of the surface and is virtual. s' =

In the following ray diagram, the object is at P and the virtual image is at P′.

P'

P

Water n1 = 1.33

C

Glass n2 = 1.68

38 •• Repeat Problem 36 for when the glass rod and the object are immersed in water and the object is 20 cm from the spherical surface. Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions.

Use the equation for refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

Solving for s′ yields:

s' =

n2 n2 − n1 n1 − r s

(1)

Optical Images

993

1.68 = − 15 1.68 − 1.33 1.33 − − 7.20 cm 20 cm where the minus sign tells us that the image is 15 cm in front of the surface of the rod and is virtual

Substitute numerical values (s = 20 cm, n1 = 1.33, n2 = 1.68, and r = −7.20 cm) and evaluate s′:

s' =

In the following ray diagram the object is at P and the virtual image is at P′.

P

P'

C

Water

n1 = 1.33

Glass

n2 = 1.68

39 •• A rod that is 96.0-cm long is made of glass that has an index of refraction equal to 1.60. The rod has its ends ground to convex spherical surfaces that have radii equal to 8.00 cm and 16.0 cm. An object is in air on the long axis of the rod 20.0 cm from the end that has the 8.00-cm radius. (a) Find the image distance due to refraction at the 8.00-cm-radius surface. (b) Find the position of the final image due to refraction at both surfaces. (c) Is the final image real or virtual? Picture the Problem (a) We can use the equation for refraction at a single surface to find the images due to refraction at the ends of the glass rod. (b) The image formed by the refraction at the first surface will serve as the object for the second surface. (c) The sign of the final image distance will tell us whether the image is real or virtual.

(a) Use the equation for refraction at a single surface to relate the image and object distances at the surface whose radius is 8.00 cm:

n1 n2 n2 − n1 + = s s' r

Solving for s′ yields:

s' =

n2 n2 − n1 n1 − r s

994

Chapter 32

Substitute numerical values (s = 20.0 cm, n1 = 1.00, n2 = 1.60, and r = 8.00 cm) and evaluate s′:

s' =

(b) The object for the second surface is 96 cm − 64 cm = 32 cm from the surface whose radius is −16.0 cm. Substitute numerical values (note that now n1 = 1.60 and n2 = 1.00) and evaluate s′:

s' =

1.60 = 64 cm 1.60 − 1.00 1.00 − 8.00 cm 20.0 cm 1.00 = − 80 cm 1.00 − 1.60 1.60 − − 16.0 cm 32 cm

(c) The final image is inside the rod and 96 cm − 80 cm = 16 cm from the surface whose radius of curvature is 8.00 cm and is virtual. 40 •• Repeat Problem 39 for an object in air on the axis of the glass rod 20.0 cm from the end that has the 16.0-cm radius. Picture the Problem (a) We can use the equation for refraction at a single surface to find the images due to refraction at the ends of the glass rod. (b) The image formed by the refraction at the first surface will serve as the object for the second surface. (c) The sign of the final image distance will tell us whether the image is real or virtual.

(a) Use the equation for refraction at a single surface to relate the image and object distances at the surface that has a 16.0 cm radius:

n1 n2 n2 − n1 + = s s' r

Solving for s′ yields:

s' =

Substitute numerical values (s = 20.0 cm, n1 = 1.00, n2 = 1.60 and r = 16.0 cm) evaluate s′:

s' =

n2 n2 − n1 n1 − r s 1.60 1.60 − 1.00 1.00 − 16.0 cm 20.0 cm

= −128 cm = − 1.3 × 10 2 cm where the minus sign tells us that the image is 1.3 × 102 cm to the left of the surface whose radius of curvature is 16.0 cm.

Optical Images (b) The object for the second surface is 96 cm + 128 cm = 224 cm from the surface whose radius of curvature is −8.00 cm. Substitute numerical values (note that now n1 = 1.60 and n2 = 1.00) and evaluate s′:

s' =

995

1.00 = 14.7 cm 1.00 − 1.60 1.60 − − 8.00 cm 224 cm

= 15 cm

(c) The final image is outside the rod, 15 cm from the end whose radius of curvature is 8.00 cm, and is real.

Thin Lenses 41 • [SSM] A double concave lens that has an index of refraction equal to 1.45 has radii whose magnitudes are equal to 30.0 cm and 25.0 cm. An object is located 80.0 cm to the left of the lens. Find (a) the focal length of the lens, (b) the location of the image, and (c) the magnification of the image. (d) Is the image real or virtual? Is the image upright or inverted? Picture the Problem We can use the lens-maker’s equation to find the focal length of the lens and the thin-lens equation to locate the image. We can use s' m = − to find the lateral magnification of the image. s

(a) The lens-maker’s equation is:

⎞⎛ 1 1 ⎞ 1 ⎛ n = ⎜⎜ − 1⎟⎟ ⎜⎜ − ⎟⎟ f ⎝ nair ⎠ ⎝ r1 r2 ⎠ where the numerals 1 and 2 denote the first and second surfaces, respectively.

Substitute numerical values to obtain:

1 ⎛ 1.45 ⎞ ⎛ 1 1 ⎞ ⎟ =⎜ − 1⎟ ⎜⎜ − f ⎝ 1.00 ⎠ ⎝ − 30.0 cm 25.0 cm ⎟⎠

Solving for f yields:

f = −30.3 cm = − 30 cm

(b) Use the thin-lens equation to relate the image and object distances:

1 1 1 fs + = ⇒ s' = s s' f s− f

996

Chapter 32

Substitute numerical values and evaluate s′:

s' =

(− 30.3 cm )(80.0 cm ) = −21.98 cm 80.0 cm − (− 30.3 cm )

= −22 cm The image is 22 cm from the lens and on the same side of the lens as the object. (c) The lateral magnification of the image is given by:

m=−

s' s

Substitute numerical values and evaluate m:

m=−

− 21.98 cm = 0.27 80.0 cm

(d) Because s' < 0 and m > 0, the image is virtual and upright. 42 • The following thin lenses are made of glass that has an index of refraction equal to 1.60. Make a sketch of each lens and find each focal length in air: (a) r1 = 20.0 cm and r2 = 10.0 cm, (b) r1 = 10.0 cm and r2 = 20.0 cm, and (c) r1 = –10.0 cm and r2 = –20.0 cm. Picture the Problem We can use the lens-maker’s equation to find the focal length of each of the lenses described in the problem statement.

The lens-maker’s equation is:

(a) For r1 = 20.0 cm and r2 = 10.0 cm:

⎞⎛ 1 1 ⎞ 1 ⎛ n = ⎜⎜ − 1⎟⎟ ⎜⎜ − ⎟⎟ f ⎝ nair ⎠ ⎝ r1 r2 ⎠ where the numerals 1 and 2 denote the first and second surfaces, respectively. 1 ⎛ 1.60 ⎞ ⎛ 1 1 ⎞ ⎟ =⎜ − 1⎟ ⎜⎜ − f ⎝ 1.00 ⎠ ⎝ 20.0 cm 10.0 cm ⎟⎠

and f = − 33 cm A sketch of the lens is shown to the right:

Optical Images (b) For r1 = 10.0 cm and r2 = 20.0 cm:

997

1 ⎞ 1 ⎛ 1.60 ⎞ ⎛ 1 ⎟ =⎜ − 1⎟ ⎜⎜ − f ⎝ 1.00 ⎠ ⎝ 10.0 cm 20.0 cm ⎟⎠

and f = 33 cm A sketch of the lens is shown to the right:

(c) For r1 = −10.0 cm and r2 = −20.0 cm:

⎞ 1 ⎛ 1.60 ⎞ ⎛ 1 1 ⎟ =⎜ − 1⎟ ⎜⎜ − f ⎝ 1.00 ⎠ ⎝ − 10.0 cm − 20.0 cm ⎟⎠

and f = − 33 cm A sketch of the lens is shown to the right:

Remarks: Note that the lenses that are thicker on their axis than on their circumferences are positive (converging) lenses and those that are thinner on their axis are negative (diverging) lenses. 43 • The following four thin lenses are made of glass that has an index of refraction of 1.5. The radii given are magnitudes. Make a sketch of each lens and find each focal length in air: (a) double-convex that has radii of curvature equal to 15 cm and 26 cm, (b) plano-convex that has a radius of curvature equal to 15 cm, (c) double concave that has radii of curvature equal to 15 cm, and (d) planoconcave that has a radius of curvature equal to 26 cm. Picture the Problem We can use the lens-maker’s equation to find the focal length of each of the lenses.

The lens-maker’s equation is:

⎞⎛ 1 1 ⎞ 1 ⎛ n = ⎜⎜ − 1⎟⎟ ⎜⎜ − ⎟⎟ f ⎝ nair ⎠ ⎝ r1 r2 ⎠ where the numerals 1 and 2 denote the first and second surfaces, respectively.

998

Chapter 32

(a) Because the lens is double convex, the radius of curvature of the second surface is negative. Hence r1 = 15 cm and r2 = −26 cm:

1 ⎛ 1.50 ⎞ ⎛ 1 1 ⎞ ⎟ =⎜ − 1⎟ ⎜⎜ − f ⎝ 1.00 ⎠ ⎝ 15 cm − 26 cm ⎟⎠

and f = 19 cm

A double convex lens is shown to the right:

(b) Because the lens is planoconvex, the radius of curvature of the second surface is negative. Hence r1 = ∞ and r2 = −15 cm:

1 ⎛ 1.50 ⎞ ⎛ 1 1 ⎞ ⎟ =⎜ − 1⎟ ⎜⎜ − f ⎝ 1.00 ⎠ ⎝ ∞ − 15 cm ⎟⎠

and f = 30 cm

A plano-convex lens is shown to the right:

(c) Because the lens is double concave, the radius of curvature of its first surface is negative. Hence r1 = −15 cm and r2 = +15 cm:

1 ⎛ 1.50 ⎞ ⎛ 1 1 ⎞ ⎟ =⎜ − 1⎟ ⎜⎜ − f ⎝ 1.00 ⎠ ⎝ − 15 cm 15 cm ⎟⎠

and f = − 15 cm

A double concave lens is shown to the right:

(d) Because the lens is planoconcave, the radius of curvature of its second surface is positive. Hence r1 = ∞ and r2 = +26 cm: A plano-concave lens is shown to the right:

1 ⎛ 1.50 ⎞ ⎛ 1 1 ⎞ ⎟ =⎜ − 1⎟ ⎜⎜ − f ⎝ 1.00 ⎠ ⎝ ∞ 26 cm ⎟⎠

and f = − 52 cm

Optical Images

999

44 • Find the focal length of a glass lens that has an index of refraction equal to 1.62, a concave surface that has a radius of curvature of magnitude 100 cm, and a convex surface that has a radius of curvature of magnitude 40.0 cm. Picture the Problem We can use the lens-maker’s equation to find the focal length of the lens.

The lens-maker’s equation is:

⎞⎛ 1 1 ⎞ 1 ⎛ n = ⎜⎜ − 1⎟⎟ ⎜⎜ − ⎟⎟ f ⎝ nair ⎠ ⎝ r1 r2 ⎠ where the numerals 1 and 2 denote the first and second surfaces, respectively.

Substitute numerical values to obtain:

⎞ 1 ⎛ 1.62 ⎞ ⎛ 1 1 ⎟ =⎜ − 1⎟ ⎜⎜ − f ⎝ 1.00 ⎠ ⎝ − 100 cm − 40.0 cm ⎟⎠

Solving for f yields:

f = 1.1 m

45 •• [SSM] (a) An object that is 3.00 cm high is placed 25.0 cm in front of a thin lens that has a power equal to 10.0 D. Draw a ray diagram to find the position and the size of the image and check your results using the thin-lens equation. (b) Repeat Part (a) if the object is placed 20.0 cm in front of the lens. (c) Repeat Part (a) for an object placed 20.0 cm in front of a thin lens that has a power equal to –10.0 D. Picture the Problem We can find the focal length of each lens from the definition of the power P, in diopters, of a lens (P = 1/f ). The thin-lens equation can be applied to find the image distance for each lens and the size of the image y' s' can be found from the magnification equation m = = − . y s

>

(a) The parallel and central rays were used to locate the image in the diagram shown below.

F' F >

The image is real, inverted, and diminished.

1000 Chapter 32 Solving the thin-lens equation for s′ yields:

fs 1 1 1 + = ⇒ s' = s s' f s− f

Use the definition of the power of the lens to find its focal length:

f =

1 1 = = 0.100 m = 10.0 cm P 10.0 m −1

Substitute numerical values and evaluate s′:

s' =

(10.0 cm )(25.0 cm ) = 16.7 cm

Use the lateral magnification equation to relate the height of the image y′ to the height y of the object and the image and object distances:

m=

y' s' s' = − ⇒ y' = − y y s s

Substitute numerical values and evaluate y′:

y' = −

(1)

25.0 cm − 10.0 cm

(2)

16.7 cm (3.00 cm ) = −2.00 cm 25.0 cm

s' = 16.7 cm , y' = −2.00 cm . Because s′ > 0, the image is real, and because y′/y = −0.67 cm, the image is inverted and diminished. These results confirm those obtained graphically.

>

(b) The parallel and central rays were used to locate the image in the following diagram.

F' F >

The image is real and inverted and appears to be the same size as the object. Use the definition of the power of the lens to find its focal length:

f =

1 = 0.100 m = 10.0 cm 10.0 m −1

Substitute numerical values in equation (1) and evaluate s′:

s' =

(10.0 cm )(20.0 cm ) =

Substitute numerical values in equation (2) and evaluate y′:

y' = −

20.0 cm − 10.0 cm

20.0 cm

20.0 cm (3.00 cm ) = − 3.00 cm 20.0 cm

Optical Images 1001 Because s′ > 0 and y′ = −3.00 cm, the image is real, inverted, and the same size as the object. These results confirm those obtained from the ray diagram. (c) The parallel and central rays were used to locate the image in the following diagram. >

F'

>

F

The image is virtual, upright, and diminished. Use the definition of the power of the lens to find its focal length:

f =

1 = −0.100 m = − 10.0 cm − 10.0 m −1

Substitute numerical values in equation (1) and evaluate s′:

s' =

(− 10.0 cm)(20.0 cm ) = 20.0 cm − (− 10.0 cm )

Substitute numerical values in equation (2) and evaluate y′:

y' = −

− 6.67 cm

− 6.67 cm (3.00 cm ) = 1.00 cm 20.0 cm

Because s′ < 0 and y′ = 1.00 cm, the image is virtual, erect, and about one-third the size of the object. These results are consistent with those obtained graphically. 46 •• The lens-maker’s equation has three design parameters. They consist of the index of refraction of the lens and the radii of curvature for its two surfaces. Thus, there are many ways to design a lens that has a particular focal length in air. Use the lens-maker’s equation to design three different thin converging lenses, each having a focal length of 27.0 cm and each made from glass that has an index of refraction of 1.60. Sketch each of your designs. Picture the Problem We can use the lens-maker’s equation to obtain a relationship between the two radii of curvature of the lenses we are to design.

1002 Chapter 32 For a thin lens of focal length 27.0 cm and index of refraction of 1.60:

⎛1 1⎞ 1 = (1.60 − 1) ⎜⎜ − ⎟⎟ 27.0 cm ⎝ r1 r2 ⎠ or 1 1 1 − = r1 r2 16.2 cm

One solution is a plano-convex lens (one with a flat surface and a convex surface). Let r2 = ∞. Then r1 = 16.2 cm and r2 = ∞ .

Another design is a double convex lens (one with both surfaces convex and radii of curvature that are equal in magnitude) obtained by letting r2 = −r1. Then r1 = 32.4 cm and r2 = − 32.4 cm .

A third possibility is a double convex lens with unequal curvature, e.g., let r2 = −12.0 cm. Then r1 = − 6.89 cm and r2 = − 12.0 cm .

47 •• Repeat Problem 46, but for a diverging lens that has a focal length in air of the same magnitude. Picture the Problem We can use the lens-maker’s equation to obtain a relationship between the two radii of curvature of the lenses we are to design.

For a thin lens of focal length −27.0 cm and index of refraction of 1.60:

⎛1 1⎞ 1 = (1.60 − 1) ⎜⎜ − ⎟⎟ − 27.0 cm ⎝ r1 r2 ⎠

or 1 1 1 − = r1 r2 − 16.2 cm

Optical Images 1003 One solution is a plano-concave lens (one with a flat surface and a concave surface), Let r2 = ∞. Then r1 = − 16.2 cm and r2 = ∞ .

Another design is a biconcave lens (one with both surfaces concave) by letting r2 = −r1. Then r1 = − 32.4 cm and r2 = 32.4 cm .

A third possibility is a lens with r2 = −8.10 cm. Then r1 = − 5.40 cm and r2 = − 8.10 cm . 48 •• (a) What is meant by a negative object distance? Describe a specific situation in which a negative object distance can occur. (b) Find the image distance and the magnification for a thin lens in air when the object distance is –20 cm and the lens is a converging lens that has a focal length of 20 cm. Describe the image—is it virtual or real, upright or inverted? (c) Repeat Part (b) if the object distance is, instead, –10 cm, and the lens is diverging and has a focal length (magnitude) of 30 cm. Picture the Problem The parallel and central rays were used to locate the image in the diagram shown below. The power P of the lens, in diopters, can be found s' from P = 1/f and the lateral magnification of the image from m = − . s

(a) A negative object distance means that the object is virtual; i.e., that extensions of light rays, and not the light rays themselves, converge on the object rather than diverge from it. A virtual object can occur in a two-lens system when converging rays from the first lens are incident on the second lens before they converge to form an image. (b) The thin-lens equation is:

1 1 1 fs + = ⇒ s' = s s' f s− f

1004 Chapter 32

(20 cm )(− 20 cm ) = 10 cm − 20 cm − (20 cm )

Substitute numerical values and evaluate s′:

s' =

The lateral magnification is given by:

m=−

s' 10 cm =− = 0.50 s − 20 cm

>

The parallel and central rays were used to locate the image in the following ray diagram:

F'

F

>

s′ = 10 cm, m = 0.50. Because s′ > 0, the image is real, and because m > 0, the image is erect. In addition, it is one-half the size of the virtual object. (c) Proceed as in Part (b) with s = −10 cm and f = −30 cm:

s' =

(− 30 cm )(− 10 cm ) = 15 cm − 10 cm − (− 30 cm )

and m=−

15 cm s' =− = 1.5 s − 10 cm

The parallel and central rays were used to locate the image in the following ray diagram: >

Object

Image

F'

>

F

s′ > 0 and m = 1.5. Because s′ > 0, the image is real, and because m > 0, the image is erect. In addition, it is one and one-half times the size of the virtual object.

Optical Images 1005 49 •• [SSM] Two converging lenses, each having a focal length equal to 10 cm, are separated by 35 cm. An object is 20 cm to the left of the first lens. (a) Find the position of the final image using both a ray diagram and the thin-lens equation. (b) Is the final image real or virtual? Is the final image upright or inverted? (c) What is the overall lateral magnification of the final image? Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens.

(a) The parallel, central, and focal rays were used to locate the image formed by the first lens and the parallel and central rays to locate the image formed by the second lens.

Apply the thin-lens equation to express the location of the image formed by the first lens:

s1' =

Substitute numerical values and evaluate s1' :

s1' =

Find the lateral magnification of the first image:

m1 = −

Because the lenses are separated by 35 cm, the object distance for the second lens is 35 cm − 20 cm = 15 cm. Equation (1) applied to the second lens is:

s2' =

Substitute numerical values and evaluate s2' :

s2' =

f1s1 s1 − f1

(1)

(10 cm)(20 cm ) = 20 cm 20 cm − 10 cm

s1' 20 cm =− = −1.0 s 20 cm

f 2 s2 s2 − f 2

(10 cm )(15 cm ) = 30 cm 15 cm − 10 cm

and the final image is 85 cm to the right of the object.

1006 Chapter 32 Find the lateral magnification of the second image:

m2 = −

s2' 30 cm =− = −2.0 s 15 cm

(b) Because s' 2 > 0 , the image is real and because m = m1m2 = 2.0, the image is erect and twice the size of the object. (c) The overall lateral magnification of the image is the product of the magnifications of each image:

m = m1m2 = (− 1.0 )(− 2.0 ) = 2.0

50 •• Repeat Problem 49 but with the second lens replaced by a diverging lens that has a focal length equal to −15 cm. Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens.

(a) The parallel, central, and focal rays were used to locate the image formed by the first lens and the parallel and central rays to locate the image formed by the second lens.

Apply the thin-lens equation to express the location of the image formed by the first lens:

s1' =

Substitute numerical values and evaluate s1' :

s1' =

Find the lateral magnification of the first image:

m1 = −

Because the lenses are separated by 35 cm, the object distance for the second lens is 35 cm − 20 cm = 15 cm. Equation (1) applied to the second lens is:

s2' =

f1s1 s1 − f1

(10 cm)(20 cm ) = 20 cm 20 cm − 10 cm

s'1 20 cm =− = −1.0 s 20 cm

f 2 s2 s2 − f 2

(1)

Optical Images 1007 Substitute numerical values and evaluate s2' :

s2' =

(− 15 cm )(15 cm ) = −7.5 cm 15 cm − (− 15 cm )

and the final image is 48 cm to the right of the object. Find the lateral magnification of the second image:

m2 = −

− 7.5 cm s2' =− = 0.50 s 15 cm

(b) Because s'2 < 0 and m = m1m2 = −0.50, the image is virtual, inverted, and half the height of the object. (c) The overall lateral magnification of the image is the product of the magnifications of each image:

m = m1m2 = (− 1.0 )(0.50 ) = − 0.50

51 •• (a) Show that to obtain a magnification of magnitude |m| using a converging thin lens of focal length f, the object distance must be equal to −1 1 + m f . (b) You want to use a digital camera which has a lens whose focal

(

)

length is 50.0 mm to take a picture of a person 1.75 m tall. How far from the camera lens should you have that person stand so that the image size on the light receiving sensors of your camera is 24.0 mm? Picture the Problem We can use the thin-lens equation and the definition of the lateral magnification to derive the result.

(a) Start with the thin-lens equation:

1 1 1 + = s s' f

Express the magnitude of the lateral magnification of the image and solve for s′: (Note that this ensures that s′ is positive for this situation, as it must be because the image is real.)

m=−

Substitute for s′ to obtain:

1 1 1 −1 + = ⇒ s = 1+ m f s ms f

(b) The magnitude of the magnification m is:

s' ⇒ s' = m s s

(

m= −

)

y' 24.0 mm = = 0.0137 y 1.75 m

1008 Chapter 32 Substitute numerical values and evaluate s:

s=

(0.0137 + 1)(50.0 mm) = 0.0137

3.70 m

52 •• A converging lens has a focal length of 12.0 cm. (a) Using a spreadsheet program or graphing calculator, plot the image distance as a function of the object distance, for object distances ranging from 1.10f to 10.0f where f is the focal length. (b) On the same graph used in Part (a), but using a different y axis, plot the magnification of the lens as a function of the object distance. (c) What type of image is produced for this range of object distances—real or virtual, upright or inverted? (d) Discuss the significance of any asymptotic limits your graphs have. Picture the Problem We can plot the first graph by solving the thin-lens equation for the image distance s′ and the second graph by using the definition of the magnification of the image.

(a) and (b) Solve the thin-lens equation for s′ to obtain:

s' =

The magnification of the image is given by:

m=−

fs s− f s' s

A spreadsheet program to calculate s′as a function of s is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 A4 A5 B4

Content/Formula 12 13.2 A4 + 1 $B$1*A4/(A4 − $B$1)

C5

−B4/A4 A

1 2 3 4 5 6 7 8 9

B f= 12

s 13.2 14.2 15.2 16.2 17.2 18.2

s' 132.00 77.45 57.00 46.29 39.69 35.23

Algebraic Form f s s + Δs fs s− f s' − s C

cm m −10.00 −5.45 −3.75 −2.86 −2.31 −1.94

Optical Images 1009 108 109 110 111

117.2 118.2 119.2 120.2

13.37 13.36 13.34 13.33

−0.11 −0.11 −0.11 −0.11

A graph of s′ and m as functions of s follows. 140

0.0

120

-2.0

s'

100

-4.0

80 -6.0

m

s ' (cm)

m

60 -8.0 40 -10.0

20 0 0

20

40

60

80

100

-12.0 120

s (cm)

(c) The images are real and inverted for this range of object distances. (d) The equations for the horizontal and vertical asymptotes of the graph of s′ versus s are s′ = f and s = f, respectively. These indicate that as the object moves away from the lens image distance the image moves toward the first focal point, and that as the object moves toward the second focal point the image distance becomes large without limit. The equations for the horizontal and vertical asymptotes of the graph of m versus s are m = 0 and s = f, respectively. These indicate that as the object moves away from the lens, the image size of the image approaches zero, and that as the object moves toward the second focal point the image becomes large without limit. 53 •• A converging lens has a focal length of 12.0 cm. (a) Using a spreadsheet program or graphing calculator, plot the image distance as a function of the object distance, for object distances ranging from 0.010f to s = 0.90f, where f is the focal length. (b) On the same graph used in Part (a), but using a different y axis, plot the magnification of the lens as a function of the object distance. (c) What type of image is produced for this range of object distances—real or virtual, upright or inverted? (d) Discuss the significance of any asymptotic limits your graphs have.

1010 Chapter 32 Picture the Problem We can plot the first graph by solving the thin-lens equation for the image distance s′ and the second graph by using the definition of the magnification of the image.

(a) and (b) Solve the thin-lens equation for s′ to obtain:

s' =

The magnification of the image is given by:

m=−

fs s− f s' s

A spreadsheet program to calculate s′as a function of s is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 A4 A5 B4

Content/Formula 12 0.12 A4 + 0.1 $B$1*A4/(A4 − $B$1)

C5

−B4/A4 A

B f= 12

Algebraic Form f s s + Δs fs s− f s' − s C

1 2 3 4 5 6 7 8 9

cm

s 0.12 0.22 0.32 0.42 0.52 0.62

s' −0.12 −0.22 −0.33 −0.44 −0.54 −0.65

m 1.01 1.02 1.03 1.04 1.05 1.05

108 109 110 111

10.52 10.62 10.72 10.82

−85.30 −92.35 −100.50 −110.03

8.11 8.70 9.37 10.17

Optical Images 1011

10

12.0

-10

10.0

-30

8.0

s' m

-50

6.0

-70

4.0

-90

2.0

-110

m

s ' (cm)

A graph of s′ and m as functions of s follows.

0.0 0

2

4

6

8

10

12

s (cm)

(c) The images are virtual and erect for this range of object distances. (d) The equation for the vertical asymptote of the graph of s′ versus s is f = s. This indicates that, as the object moves toward the second focal point, the magnitude of the image distance becomes large without limit. The equation for the vertical asymptote of the graph of m versus s is s = f. This indicates that, as the object moves toward the second focal point, the image becomes large without limit. In addition, as s approaches zero, s′ approaches negative infinity and m approaches 1, so as the object moves toward the lens the image becomes the same size as the object and the magnitude of the image distance increases without limit 54 •• An object is 15.0 cm in front of a converging lens that has a focal length equal to 15.0 cm. A second converging lens that also has a focal length equal to15.0 cm is located 20.0 cm in back of the first. (a) Find the location of the final image and describe its properties (for example, real and inverted) and (b) draw a ray diagram to corroborate your answers to Part (a). Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens.

(a) Apply the thin-lens equation to express the location of the image formed by the first lens:

s1' =

f1s1 s1 − f1

(1)

1012 Chapter 32 Substitute numerical values and evaluate s1' :

s1' =

(15.0 cm)(15.0 cm) = ∞ 15.0 cm − 15.0 cm

i.e., no image is formed by the first lens. With s1' = ∞, the thin-lens equation applied to the second lens becomes:

1 1 = ⇒ s 2' = f 2 = 15.0 cm s2' f 2

The overall magnification of the two-lens system is the magnification of the second lens:

m = m2 = −

s2' 15.0 cm =− = − 1.00 s1 15.0 cm

The final image is 50 cm from the object, real, inverted, and the same size as the object. (b) A corroborating ray diagram is shown below: Lens 2 >

>

Lens 1

F1 '

F2

F1

F2 ' >

>

55 •• [SSM] An object is 15.0 cm in front of a converging lens that has a focal length equal to 15.0 cm. A diverging lens that has a focal length whose magnitude is equal to 15.0 cm is located 20.0 cm in back of the first. (a) Find the location of the final image and describe its properties (for example, real and inverted) and (b) draw a ray diagram to corroborate your answers to Part (a). Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens.

Apply the thin-lens equation to express the location of the image formed by the first lens:

s1' =

Substitute numerical values and evaluate s1' :

s1' =

f1s1 s1 − f1

(15.0 cm)(15.0 cm) = ∞ 15.0 cm − 15.0 cm

(1)

Optical Images 1013 With s'1 = ∞, the thin-lens equation applied to the second lens becomes:

1 1 = ⇒ s2' = f 2 = − 15.0 cm s2' f 2

The overall magnification of the two-lens system is the magnification of the second lens:

m = m2 = −

s2' − 15.0 cm =− = 1.00 s1 15.0 cm

The final image is 20 cm from the object, virtual, erect, and the same size as the object. A corroborating ray diagram follows: Lens 1

Lens 2

>

>

F1 '

F2

F1 >

>

F2 '

56 ••• In a convenient form of the thin-lens equation used by Newton, the object and image distances x and x′ are measured from the focal points F and F’, and not from the center of the lens. (a) Indicate x and x′ on a sketch of a lens and show that if x = s – f and x′ = s′ – f, the thin-lens equation (Equation 32-12) can be rewritten as xx' = f 2 . (b) Show that the lateral magnification is given by m = –x′/f = –f/x. Picture the Problem We can substitute x = s − f and x′ = s′ − f in the thin-lens equation and the equation for the lateral magnification of an image to obtain Newton’s equations.

>

(a) The variables x, f, s ,and s′ are shown in the following sketch:

s f

x Object

s' x'

f

F'

F

Image

>

1014 Chapter 32 Express the thin-lens equation:

1 1 1 + = s s' f

If x = s − f and x′ = s′ − f, then:

1 1 1 + = x + f x' + f f

Expand this expression to obtain:

f ( x' + x + 2 f ) = ( x + f )( x' + f ) = xx' + xf + x'f + f 2 or, simplifying, xx' = f

(b) The lateral magnification is:

Solve equation (1) for x:

s' s or, because x = s − f and x′ = s′ − f, x' + f m=− x+ f x=

f2 x'

m=−

The lateral magnification is also given by:

m=−

Substitute to obtain:

(1)

m=−

Substitute for x and simplify to obtain:

From equation (1) we have:

2

x' =

x' + f x' x' + f = − =− 2 ( ) f f x' + f f +f x' x' x'+ f x+ f

f2 x

f2 + f x m=− =− x+ f

⎞ ⎛f f ⎜ + 1⎟ ⎠= − f ⎝x f⎞ x ⎛ x⎜ 1 + ⎟ x⎠ ⎝

Optical Images 1015 57 ••• [SSM] In Bessel’s method for finding the focal length f of a lens, an object and a screen are separated by distance L, where L > 4f. It is then possible to place the lens at either of two locations, both between the object and the screen, so that there is an image of the object on the screen, in one case magnified and in the other case reduced. Show that if the distance between these two lens locations is D, then the focal length is given by f = 14 L2 − D 2 L . Hint: Refer to Figure 32-60.

(

)

Picture the Problem The ray diagram shows the two lens positions and the corresponding image and object distances (denoted by the numerals 1 and 2). We can use the thin-lens equation to relate the two sets of image and object distances to the focal length of the lens and then use the hint to express the relationships between these distances and the distances D and L to eliminate s1, s1′, s2, and s2′ and obtain an expression relating f, D, and L.

Relate the image and object distances for the two lens positions to the focal length of the lens:

1 1 1 1 1 1 + = and + = s1 s1' f s2 s2' f

Solve for f to obtain:

f =

The distances D and L can be expressed in terms of the image and object distances:

s1s1' s s' = 2 2 s1 + s1' s2 + s2'

L = s1 + s1' = s 2 + s 2' and D = s 2 − s1 = s1' − s 2'

Substitute for the sums of the image and object distances in equation (1) to obtain:

f =

From the hint:

s1 = s2' and s1' = s2

Hence L = s1 + s2 and:

L − D = 2 s1 and L + D = 2 s 2

Take the product of L − D and L + D to obtain:

(L − D )(L + D ) = L2 − D 2

From the thin-lens equation:

4 s1 s 2 = rs1 s1' = 4 fL

Substitute to obtain:

L2 − D 2 4 fL = L − D ⇒ f = 4L

s1 s1' s 2 s 2' = L L

= 4s1 s 2 = 4s1 s1'

2

2

(1)

1016 Chapter 32 58 •• You are working for an optician during the summer. The optician needs to measure an unknown focal length and you suggest using Bessel’s method (see Problem 56) which you used during a physics lab. You set the object-toimage distance at 1.70 m. The lens position is adjusted to get a sharp image on the screen. A second sharp image is found when the lens is moved a distance of 72 cm from its first location. (a) Sketch the ray diagram for the two locations. (b) Find the focal length of the lens using Bessel’s method. (c) What are the two locations of the lens with respect to the object? (d) What are the magnifications of the images when the lens is in the two locations? Picture the Problem We can use results obtained in Problem 57 to find the focal length of the lens and the two locations of the lens with respect to the object. Let 1 refer to the first lens position and 2 to the second lens position.

>

>

(a) The distances defined in Problem 57 have been identified in the following ray diagram. Note that, for clarity, the two images have been slightly offset from the plane of the screen. 1st lens position 2nd lens position Screen

f >

>

D

L

(b) From Problem 57 we have:

f =

L2 − D 2 4L

2 2 ( 170 cm ) − (72 cm ) = 4(170 cm )

Substitute numerical values and evaluate f:

f

(c) Solve the thin-lens equation for the image distance to obtain:

s' =

fs f −s

= 35 cm

(1)

Optical Images 1017 In Problem 57 it was established that:

L − D = 2 s1 and L + D = 2 s 2

Solve for s1 and s2:

s1 =

L−D L+D and s 2 = 2 2

Substitute numerical values and evaluate s1 and s2:

s1 =

170 cm − 72 cm = 49 cm 2

and s2 =

170 cm + 72 cm = 121cm 2

(d) The magnification when the lens was in its first position is given by:

m1 = −

s'1 D − s1 =− s1 s1

Substituting numerical values yields:

m1 = −

170 cm − 49 cm = − 2.5 49 cm

The magnification when the lens was in its second position is given by:

m2 = −

s' 2 L − s2 =− s2 s2

Substitute numerical values and evaluate m2:

m2 = −

170 cm − 121 cm = − 0.41 121 cm

59 ••• An object is 17.5 cm to the left of a lens that has a focal length of +8.50 cm. A second lens, which has a focal length of −30.0 cm, is 5.00 cm to the right of the first lens. (a) Find the distance between the object and the final image formed by the second lens. (b) What is the overall magnification? (c) Is the final image real or virtual? Is the final image upright or inverted? Picture the Problem The ray diagram shows four rays from the head of the object that locate images I1 and I2. We can use the thin-lens equation to find the location of the image formed in the positive lens and then, knowing the separation of the two lenses, determine the object distance for the second lens and apply the thin lens a second time to find the location of the final image. >

Lens 2 >

Lens 1

F1'

Image 1 Image 2 F2'

>

>

F2

F1

1018 Chapter 32 (a) Express the object-to-image distance d:

d = s1 + 5 cm + s2'

Apply the thin-lens equation to the positive lens:

1 1 1 fs + = ⇒ s1' = 1 1 s1 s1' f1 s1 − f1

Substitute numerical values and evaluate s1' :

s1' =

Find the object distance for the negative lens:

s2 = 5.00 cm − s1' = 5.00 cm − 16.53 cm

(1)

(8.50 cm)(17.5 cm) = 16.53 cm 17.5 cm − 8.50 cm

= −11.53 cm

The image distance s2′ is given by:

s2' =

f 2 s2 s2 − f 2

Substitute numerical values and evaluate s2′:

s2' =

(− 30.0 cm )(− 11.53 cm) = 18.73 cm − 11.53 cm − (− 30.0 cm )

Substitute numerical values in equation (1) and evaluate d:

d = 17.5 cm + 5.00 cm + 18.73 cm = 41cm and the final image is 18.7 cm to the right of the second lens.

(b) The overall lateral magnification is given by:

m = m1m2

Express m1 and m2:

m1 = −

Substitute for m1 and m2 to obtain:

⎛ s ' ⎞⎛ s ' ⎞ s 's ' m = ⎜⎜ − 1 ⎟⎟⎜⎜ − 2 ⎟⎟ = 1 2 ⎝ s1 ⎠⎝ s2 ⎠ s1s2

Substitute numerical values and evaluate m:

m=

s1' s' and m2 = − 2 s1 s2

(16.53 cm )(18.73 cm ) = (17.5 cm )(− 11.53 cm )

− 1.53

(c) Because m < 0, the image is inverted. Because s2' > 0, the image is real.

Optical Images 1019

Aberrations Chromatic aberration is a common defect of (a) concave and convex 60 • lenses, (b) concave lenses only, (c) concave and convex mirrors, (d) all lenses and mirrors. Determine the Concept Chromatic aberrations are a consequence of the differential refraction of light of differing wavelengths by lenses. (a ) is correct. 61 • Discuss some of the reasons why most telescopes that are used by astronomers are reflecting rather than refracting telescopes. Determine the Concept Reasons for the preference for reflectors include: 1) no chromatic aberrations, 2) cheaper to shape one side of a piece of glass than to shape both sides, 3) reflectors can be more easily supported from rear instead of edges, preventing sagging and focal length changes, and 4) support from rear makes larger sizes easier to handle. 62 • A symmetric double-convex lens has radii of curvature equal to 10.0 cm. It is made from glass that has an index of refraction equal to 1.530 for blue light and to 1.470 for red light. Find the focal length of this lens for (a) red light and (b) blue light. Picture the Problem We can use the lens-maker’s equation to find the focal length the this lens for the two colors of light.

The lens-maker’s equation relates the radii of curvature and the index of refraction to the focal length of the lens: (a) For red light:

⎞⎛ 1 1 ⎞ 1 ⎛ n = ⎜⎜ − 1⎟⎟ ⎜⎜ − ⎟⎟ f ⎝ nair ⎠ ⎝ r1 r2 ⎠

1

f red

⎞ 1 ⎛ 1.470 ⎞ ⎛ 1 ⎟⎟ =⎜ − 1⎟ ⎜⎜ − ⎝ 1.00 ⎠ ⎝ 10.0 cm − 10.0 cm ⎠

and f red = 10.6 cm (b) For blue light:

1

f blue

⎞ 1 ⎛ 1.530 ⎞ ⎛ 1 ⎟⎟ =⎜ − 1⎟ ⎜⎜ − ⎝ 1.00 ⎠ ⎝ 10.0 cm − 10.0 cm ⎠

and f blue = 9.43 cm

1020 Chapter 32

The Eye Find the change in the focal length of the eye when an object originally 63 • at 3.0 m is brought to 30 cm from the eye. Picture the Problem We can apply the thin-lens equation for the two values of s to find Δf.

Express the change Δf in the focal length:

Δf = f s =0.30 m − f s =3.0 m

Solve the thin-lens equation for s:

f =

Substitute for f to obtain:

ss' s' + s s 0.30 m s' 0.30 m

Δf =

s' 0.30 m + s 0.30 m

−

s3.0 m s' 3.0 m s' 3.0 m + s3.0 m

Because s'3.0 m = s'0.30 m : ⎡ ⎤ s 0.30 m s3.0 m − Δf = s' 3.0 m ⎢ ⎥ ⎢⎣ s' 0.30 m + s 0.30 m s' 3.0 m + s3.0 m ⎥⎦ Substitute numerical values and evaluate Δf: ⎡ ⎤ 30 cm 300 cm Δf = (2.5 cm ) ⎢ − ⎥ = −0.172 cm = − 1.7 mm ⎣ 2.5 cm + 30 cm 2.5 cm + 300 cm ⎦

64 • A farsighted person requires lenses that have powers equal to 1.75 D to read comfortably from a book that is 25.0 cm from his eyes. What is that person’s near point without the lenses? Picture the Problem We can use the thin-lens equation and the definition of the power of a lens to express the near point distance as a function of P and s.

From the thin-lens equation we have:

s 1 1 1 + = = P ⇒ s' = s s' f Ps − 1

Substitute numerical values and evaluate s′:

s' =

25.0 cm = −44.4 cm 1.75 m (0.250 m ) − 1

(

−1

)

Optical Images 1021 The person’s near point without lenses is 44.4 cm . 65 • [SSM] If two point objects close together are to be seen as two distinct objects, the images must fall on the retina on two different cones that are not adjacent. That is, there must be an unactivated cone between them. The separation of the cones is about 1.00 μm. Model the eye as a uniform 2.50-cmdiameter sphere that has a refractive index of 1.34. (a) What is the smallest angle the two points can subtend? (See Figure 32-61.) (b) How close together can two points be if they are 20.0 m from the eye? Picture the Problem We can use the relationship between a distance measured along the arc of a circle and the angle subtended at its center to approximate the smallest angle the two points can subtend and the separation of the two points 20.0 m from the eye.

(a) Relate θmin to the diameter of the eye and the distance between the activated cones:

d eyeθ min ≈ 2.00 μm ⇒ θ min =

2.00 μm d eye

2.00 μm = 80.0 μrad 2.50 cm

Substitute numerical values and evaluate θmin:

θ min =

(b) Let D represent the separation of the points R = 20.0 m from the eye to obtain:

D = Rθ min = (20.0 m )(80.0 μrad ) = 1.60 mm

Suppose the eye were designed like a camera that has a lens of fixed 66 • focal length equal to 2.50 cm that could move toward or away from the retina. Approximately how far would the lens have to move to focus the image of an object 25.0 cm from the eye onto the retina? Hint: Find the distance from the retina to the image behind it for an object at 25.0 cm. Picture the Problem We can use the thin-lens equation to find the distance from the lens to the image and then take their difference to find the distance the lens would have to be moved.

Express the distance d that the lens would have to move:

d = s' − f

Solve the thin-lens equation for s′:

s' =

fs s− f

1022 Chapter 32 Substitute for s′ to obtain:

d=

fs −f s− f

Substitute numerical values and evaluate d:

d=

(2.50 cm )(25.0 cm ) − 2.50 cm

25.0 cm − 2.50 cm = 0.28 cm

That is, the lens would have to move 0.28 cm toward the object. 67 •• [SSM] The Model Eye I: A simple model for the eye is a lens that has a variable power P located a fixed distance d in front of a screen, with the space between the lens and the screen filled by air. This ″eye″ can focus for all values of object distance s such that xnp ≤ s ≤ xfp where the subscripts on the variables refer to ″near point″ and ″far point″ respectively. This ″eye″ is said to be normal if it can focus on very distant objects. (a) Show that for a normal ″eye,″ of this type, the required minimum value of P is given by Pmin = 1 d . (b) Show that the maximum value of P is given by Pmax = 1 x np + 1 d . (c) The

difference between the maximum and minimum powers, symbolized by A, is defined as A = Pmax − Pmin and is called the accommodation. Find the minimum power and accommodation for this model eye that has a screen distance of 2.50 cm, a far point distance of infinity and a near point distance of 25.0 cm. Picture the Problem The thin-lens equation relates the image and object distances to the power of a lens.

(a) Use the thin-lens equation to relate the image and object distances to the power of the lens: Because s′ = d and, for a distance object, s = ∞: (b) If xnp is the closest distance an object could be and still remain in clear focus on the screen, equation (1) becomes: (c) Use our result in (a) to obtain:

1 1 1 + = =P s s' f

Pmin =

Pmax =

Pmin =

1 1 = s' d

1 1 + xnp d

1 = 40.0 D 2.50 cm

(1)

Optical Images 1023 Use the results of (a) and (b) to express the accommodation of the model eye:

A = Pmax − Pmin =

Substitute numerical values and evaluate A:

A=

1 1 1 1 + − = xnp d d xnp

1 = 4.00 D 25.0 cm

68 •• The Model Eye II: In an eye that exhibits nearsightedness, the eye cannot focus on distant objects. (a) Show that for a nearsighted model eye capable of focusing out to a maximum distance xfp, the minimum value of the power P is greater than that of a normal eye (that has a far point at infinity) and is given by Pmin = 1 xfp + 1 d . (b) To correct for nearsightedness, a contact lens may

be placed directly in front of the model-eye’s lens. What power contact lens would be needed to correct the vision of a nearsighted model eye that has a far point distance of 50.0 cm? Picture the Problem The thin-lens equation relates the image and object distances to the power of a lens.

(a) Use the thin-lens equation to relate the image and object distances to the power of the lens: Because s′ = d and s = xfp:

(b) To correct for the nearsightedness of this eye, we need a lens that will form an image 25.0 cm in front of the eye of an object at the eye’s far point:

1 1 1 + = =P s s' f

Pmin =

Pmin =

1 1 + xfp d 1 1 + 50.0 cm − 25.0 cm

= − 2.00 D

69 •• The Model Eye III: In an eye that exhibits farsightedness, the eye may be able to focus on distant objects but cannot focus on close objects. (a) Show that for a farsighted model eye capable of focusing only as close as a distance ' x np + 1 d . (b) Show ′ , the maximum value of the power P is given by Pmax = 1 x np

that, compared to a model eye capable of focusing as close as a distance xnp (where xnp < x np ′ ), the maximum power of the farsighted lens is too small by ' 1 x np − 1 x np . (c) What power contact lens would be needed to correct the vision

of a farsighted model eye, with x np ′ = 150 cm, so that the eye may focus on objects as close as 15 cm?

1024 Chapter 32 Picture the Problem The thin-lens equation relates the image and object distances to the power of a lens.

(a) Use the thin-lens equation to relate the image and object distances to the power of the lens: Because s′ = d and s = x′np:

(b) For a normal eye:

The amount by which the power of the lens is too small is the difference between equations (2) and (1):

1 1 1 + = =P s s' f

P'max =

1 1 + x'np d 1 1 + xnp d

(2)

Pmax − P' max =

1 1 ⎛ 1 1⎞ + −⎜ + ⎟ xnp d ⎜⎝ x' np d ⎟⎠

Pmax =

=

(c) For xnp = 15 cm and x′np = 150 cm:

(1)

Pmax − P'max =

1 1 − xnp x' np

1 1 − 15 cm 150 cm

= 6.0 D 70 •• A person who has a near point of 80 cm needs to read from a computer screen that is only 45 cm from her eye. (a) Find the focal length of the lenses in reading glasses that will produce an image of the screen at a distance of 80 cm from her eye. (b) What is the power of the lenses? Picture the Problem We can use the thin-lens equation to find f and the definition of the power of a lens to find P.

(a) Solve the thin-lens equation for f:

f =

ss' s' + s

Noting that s′ < 0, substitute numerical values and evaluate f:

f =

(45 cm)(− 80 cm) = 103 cm − 80 cm + 45 cm

= 1.0 m (b) Use the definition of the power of a lens to obtain:

P=

1 1 = = 0.97 D f 1.03 m

Optical Images 1025 71 •• A nearsighted person cannot focus clearly on objects that are more distant than 2.25 m from her eye. What power lenses are required for her to see distant objects clearly? Picture the Problem We can use the thin-lens equation to find f and the definition of the power of a lens to find P.

1 f

Express the required power of the lens:

P=

The thin-lens equation is:

1 1 1 + = s s' f

For s = ∞:

1 1 = ⇒ f = s' s' f

Substitute for f to obtain:

P=

1 s'

Substitute for s′ and evaluate P:

P=

1 = 0.444 D 2.25 m

Because the index of refraction of the lens of the eye is not very 72 •• different from that of the surrounding material, most of the refraction takes place at the cornea, where the index changes abruptly from 1.00 (air) to approximately 1.38. (a) Modeling the cornea, aqueous humor, lens and vitreous humor is a transparent homogeneous solid sphere that has an index of refraction of 1.38, calculate the sphere’s radius if it focuses parallel light on the retina a distance 2.50 cm away. (b) Do you expect your result to be larger or smaller than the actual radius of the cornea? Explain your answer. Picture the Problem We can use the equation for refraction at a single surface (Equation 32-6) with s = ∞ to find the radius of the cornea modeled as a homogeneous sphere with an index of refraction of 1.38.

(a) Use Equation 32-6 to relate the radius of the cornea to its index of refraction and that of air:

n1 n2 n2 − n1 + = s s' r

Because n2 = n, n1 = 1, and s = ∞:

n n −1 s' (n − 1) ⎛ 1 ⎞ = ⇒r = = ⎜1 − ⎟ s' n s' r ⎝ n⎠

1026 Chapter 32 Substitute numerical values and evaluate r:

1 ⎞ ⎛ r = ⎜1 − ⎟ (2.50 cm ) = 0.688 cm ⎝ 1.38 ⎠

(b) The calculated value for the distance to the retina is too large. The eye is not a homogeneous sphere, and in a real eye additional refraction occurs at the lens. By modeling the eye as a homogeneous solid sphere we are ignoring the refraction that takes place at the lens. 73 •• The near point of a certain person’s eyes is 80 cm. Reading glasses are prescribed so that he can read a book at 25 cm from his eye. The glasses are 2.0 cm from the eye. What diopter lenses should be used in the glasses? Picture the Problem We can use the definition of the power of a lens and the thin-lens equation to find the power of the lens that should be used in the glasses. 1 1 + f eye f glasses

Express the power of the lens that should be used in the glasses:

P = Peye + Plens =

Because the glasses are 2.0 cm from the eye:

s' = −80 cm + 2.0 cm = −78 cm and s = 25 cm − 2.0 cm = 23 cm

Apply the thin-lens equation to the eye with s′ = ∞:

1 1 ⇒ f eye = s = s f eye

Apply the thin-lens equation to the glasses with s = ∞:

1 1 ⇒ f glasses = s' = s' f glasses

Substitute for f eye and f glasses in

P=

1 1 + s s'

P=

1 1 + = 3.1 D 0.23 m − 0.78 m

equation (1) to obtain: Substitute numerical values and evaluate P:

(1)

74 ••• At age 45, a person is fitted for reading glasses that have a power equal to 2.10 D in order to read at 25 cm. By the time she reaches the age of 55, she discovers herself holding her newspaper at a distance of 40 cm in order to see it clearly with her glasses on. (a) Where was her near point at age 45? (b) Where is her near point at age 55? (c) What power is now required for the lenses of her reading glasses so that she can again read at 25 cm? Assume the glasses are placed 2.2 cm from her eyes.

Optical Images 1027 Picture the Problem We can use the thin-lens equation and the distance from her eyes to her glasses to derive an expression for the location of her near point.

(a) Express her near point, xnp, at age 45 in terms of the location of her glasses:

xnp = s' + 2.2 cm

Because the glasses are 2.2 cm from her eye:

s = 25 cm − 2.2 cm = 22.8 cm

Apply the thin-lens equation to the glasses:

1 1 1 + = =P s s' f glasses

Solving for s′ yields:

s' =

(1)

s 1 = Ps − 1 P − 1 s

Substitute in equation (1) to obtain: xnp =

Substitute numerical values and evaluate xnp:

x np =

1 1 P− s

+ 2.2 cm

1 1 2.10 m − 0.228 m

(2)

+ 2.2 cm

−1

= 46 cm

(b) At age 55: Substitute numerical values in equation (2) and evaluate s′:

s = 40 cm − 2.2 cm = 37.8 cm

x np =

1 1 2.10 m − 0.378 m

= 1.9 m

(c) Solve the thin-lens equation for f:

f =

ss' s' + s

−1

+ 2.2 cm

1028 Chapter 32 From the definition of P:

P=

1 s' + s = f s's

For s = 22.8 cm and s′ = 183.3 cm:

P=

183.3 cm + 22.8 cm = 4.9 D (183.3 cm)(22.8 cm)

The Simple Magnifier 75 • [SSM] What is the magnifying power of a lens that has a focal length equal to 7.0 cm when the image is viewed at infinity by a person whose near point is at 35 cm? Picture the Problem We can use the definition of the magnifying power of a lens to find the magnifying power of this lens.

The magnifying power of the lens is given by:

M =

Substitute numerical values and evaluate M:

M=

xnp f

35 cm = 5.0 7.0 cm

76 •• A lens that has a focal length equal to 6.0 cm is used as a simple magnifier by one person whose near point is 25 cm and by another person whose near point is 40 cm. (a) What is the effective magnifying power of the lens for each person? (b) Compare the sizes of the images on the retinas when each person looks at the same object with the magnifier. Picture the Problem Let the numerals 1 and 2 denote the 1st and 2nd persons, respectively. We can use the definition of magnifying power to find the effective magnifying power of the lens for each person. The relative sizes of the images on the retinas of the two persons is given by the ratio of the effective magnifying powers.

(a) The magnifying power of the lens is given by:

M =

Substitute numerical values and evaluate M1 and M2:

M1 =

xnp f

25 cm = 4.17 = 4.2 6.0 cm

and M2 =

40 cm = 6.67 = 6.7 6.0 cm

Optical Images 1029 If xnp = 25 cm, then M = 4.2, and if xnp = 40 cm, then M = 6.7. (b) From the definition of magnifying power we have:

y1 y M1 f = = 1 M 2 y2 y2 f

Substitute for M1 and M2 and evaluate the ratio of y1 to y2:

y1 4.17 = = 0.63 y2 6.67

For a person with xnp = 40 cm, the image on the retina is 1.6 times larger than it is for a person with xnp = 25 cm. 77 •• [SSM] In your botany class, you examine a leaf using a convex 12-D lens as a simple magnifier. What is the angular magnification of the leaf if image formed by the lens (a) is at infinity and (b) is at 25 cm? Picture the Problem We can use the definition of angular magnification to find the expected angular magnification if the final image is at infinity and the thinlens equation and the expression for the magnification of a thin lens to find the angular magnification when the final image is at 25 cm.

(a) Express the angular magnification when the final image is at infinity:

M =

xnp

= xnp P f where P is the power of the lens.

(

)

Substitute numerical values and evaluate M:

M = (25 cm ) 12 m −1 = 3.0

(b) Express the magnification of the lens when the final image is at 25 cm:

m=−

Solve the thin-lens equation for s:

s=

Substitute for s and simplify to obtain:

m=−

s' s

fs' s' − f

s' − f s' s' =− = − +1 fs' f f s' − f = 1 − s'P

1030 Chapter 32 Substitute numerical values and evaluate m:

(

)

m = 1 − (− 0.25 m ) 12 m −1 = 4.0

The Microscope 78 •• Your laboratory microscope objective has a focal length of 17.0 mm. It forms an image of a tiny specimen at 16.0 cm from its second focal point. (a) How far from the objective is the specimen located? (b) What is the magnifying power for you if your near point distance is 25.0 cm and the focal length of the eyepiece is 51.0 mm? Picture the Problem We can use the thin-lens equation to find the location of the object and the expression for the magnifying power of a microscope to find the magnifying power of the given microscope for a person whose near point is at 25.0 cm.

(a) Using the thin-lens equation, relate the object distance s to the focal length of the objective lens f0:

f s' 1 1 1 + = ⇒s = 0 s s' f 0 s' − f 0

The image distance for the image formed by the objective lens is:

s' = f 0 + L = 1.70 cm + 16.0 cm = 17.7 cm

Substitute numerical values and evaluate s:

s=

(b) Express the magnifying power of a microscope:

M =−

Substitute numerical values and evaluate M:

⎛ 16.0 cm ⎞ ⎛ 25.0 cm ⎞ ⎟⎟ ⎜⎜ ⎟⎟ = − 46.1 M = −⎜⎜ 1.70 cm 5 . 10 cm ⎝ ⎠⎝ ⎠

(1.70 cm )(17.7 cm ) = 17.7 cm − 1.70 cm

18.8 mm

L xnp f0 fe

79 •• [SSM] Your laboratory microscope objective has a focal length of 17.0 mm. It forms an image of a tiny specimen at 16.0 cm from its second focal point. (a) How far from the objective is the specimen located? (b) What is the magnifying power for you if your near point distance is 25.0 cm and the focal length of the eyepiece is 51.0 mm? Picture the Problem The lateral magnification of the objective is mo = − L f o and the magnifying power of the microscope is M = mo M e .

Optical Images 1031 (a) The lateral magnification of the objective is given by:

mo = −

Substitute numerical values and evaluate mo:

mo = −

(b) The magnifying power of the microscope is given by:

M = mo M e

Substitute numerical values and evaluate M:

L fo 16 cm = −18.8 = − 19 8.5 mm

where Me is the angular magnification of the lens. M = (− 18.8)(10) = − 1.9 × 10 2

80 •• A crude, symmetric handheld microscope consists of two 20-D lenses fastened at the ends of a tube 30 cm long. (a) What is the tube length of this microscope? (b) What is the lateral magnification of the objective? (c) What is the magnifying power of the microscope? (d) How far from the objective should the object be placed? Picture the Problem We can find the tube length from the length of the tube to which the lenses are fastened and the focal lengths of the objective and eyepiece. We can use their definitions to find the lateral magnification of the objective and the magnifying power of the microscope. The distance of the object from the objective can be found using the thin-lens equation.

(a) The tube length L is given by:

L = D − fo − fe = 0.30 m −

2 = 20 cm 20 m −1

L 20 cm =− = − 4.0 fo 5.0 cm

(b) The lateral magnification of the objective mo is given by:

mo = −

(c) The magnifying power of the microscope is given by:

M = mo M e = mo

Substitute numerical values and evaluate M:

M = (− 4 )

(d) From the thin-lens equation we have:

1 1 1 + = so so' f o where so' = f o + L

xnp fe

25 cm = − 20 5.0 cm

1032 Chapter 32 Substitute for so' to obtain:

f ( f + L) 1 1 1 + = ⇒ so = o o so f o + L f o L

Substitute numerical values and evaluate so:

so =

(5.0 cm )(5.0 cm + 20 cm ) 20 cm

= 6.3 cm 81 •• A compound microscope has an objective lens that has a power of 45 D and an eyepiece that has a power of 80 D. The lenses are separated by 28 cm. Assuming that the final image is formed by the microscopy is 25 cm from the eye, what is the magnifying power? Picture the Problem The magnifying power of a compound microscope is the product of the magnifying powers of the objective and the eyepiece.

Express the magnifying power of the microscope in terms of the magnifying powers of the objective and eyepiece: The magnification of the eyepiece is given by:

M = mo me

me =

xnp fe

(1)

+ 1 = Pe xnp + 1

The magnification of the objective is given by:

L fo where L = D − f o − f e

Substitute for L to obtain:

mo = −

Substitute for me and mo in equation (1) to obtain:

⎛ D − fo − fe ⎞ ⎟⎟ M = (Pe xnp + 1)⎜⎜ − fo ⎝ ⎠

mo = −

D − fo − fe fo

Substitute numerical values and evaluate M: ⎛ 28 cm − 2.22 cm − 1.25 cm ⎞ ⎟⎟ = − 230 M = [(80 D )(0.25 m ) + 1] ⎜⎜ − 2.22 cm ⎝ ⎠

Optical Images 1033 82 ••• A microscope has a magnifying power of 600. The eyepiece has an angular magnification of 15.0. The objective lens is 22.0 cm from the eyepiece. Calculate (a) the focal length of the eyepiece, (b) the location of the object so that it is in focus for a normal relaxed eye, and (c) the focal length of the objective lens. Picture the Problem We can find the focal length of the eyepiece from its angular magnification and the near point of a normal eye. The location of the object such that it is in focus for a normal relaxed eye can be found from the lateral magnification of the eyepiece and the magnifying power of the microscope. Finally, we can use the thin-lens equation to find the focal length of the objective lens.

(a) The focal length of the eyepiece is related to its angular magnifying power:

Me =

xnp fe

⇒ fe =

xnp Me

25.0 cm = 1.67 cm 15.0

Substitute numerical values and evaluate fe:

fe =

(b) Relate s to s′ through the lateral magnification of the objective:

mo = −

Relate the magnifying power of the microscope M to the lateral magnification of its objective m0 and the angular magnification of its eyepiece Me:

M = mo M e ⇒ mo =

Substitute for mo to obtain:

s=−

Evaluate s′:

s' = 22 cm − f e

s' s' ⇒ s=− s mo M Me

s'M e M

= 22 cm − 1.67 cm = 20.33 cm Substitute numerical values and evaluate s:

s=−

(20.33 cm )(15.0) =

(c) Solving the thin-lens equation for fo yields:

fo =

ss' s' + s

− 600

0.508 cm

1034 Chapter 32 Substitute numerical values and evaluate fo:

fo =

(0.508 cm)(20.33 cm) 20.33 cm + 0.508 cm

= 0.496 cm

The Telescope 83 • [SSM] You have a simple telescope that has an objective which has a focal length of 100 cm and an eyepiece which has a focal length of 5.00 cm. You are using it to look at the moon, which subtends an angle of about 9.00 mrad. (a) What is the diameter of the image formed by the objective? (b) What angle is subtended by the image formed at infinity by the eyepiece? (c) What is the magnifying power of your telescope? Picture the Problem Because of the great distance to the moon, its image formed by the objective lens is at the focal point of the objective lens and we can use D = f oθ to find the diameter D of the image of the moon. Because angle subtended by the final image at infinity is given by θ e = Mθ o = Mθ, we can solve

(b) and (c) together by first using M = −fo/fe to find the magnifying power of the telescope. (a) Relate the diameter D of the image of the moon to the image distance and the angle subtended by the moon:

D = so'θ

Because the image of the moon is at the focal point of the objective lens:

so' = f o and D = f oθ

Substitute numerical values and evaluate D:

D = (100 cm )(9.00 mrad ) = 9.00 mm

(b) and (c) Relate the angle subtended by the final image at infinity to the magnification of the telescope and the angle subtended at the objective:

θ e = Mθ o = Mθ

Express the magnifying power of the telescope:

M =−

fo fe

Optical Images 1035 Substitute numerical values and evaluate M and θe:

M =−

100 cm = − 20.0 5.00 cm

and θ e = − 20.0 (9.00 mrad ) = 180 mrad The objective lens of the refracting telescope at the Yerkes 84 • Observatory has a focal length of 19.5 m. The moon subtends an angle of about 9.00 mrad. When the telescope is used to look at the moon, what is the diameter of the image of the moon formed by the objective? Picture the Problem Because of the great distance to the moon, its image formed by the objective lens is at the focal point of the objective lens and we can use D = f oθ to find the diameter D of the image of the moon.

Relate the diameter D of the image of the moon to the image distance and the angle subtended by the moon:

D = so'θ

Because the image of the moon is at the focal point of the objective lens:

so' = f o and D = f oθ

Substitute numerical values and evaluate D:

D = (19.5 m )(9.00 mrad ) = 17.6 cm

The 200-in (5.08-m) diameter mirror of the reflecting telescope at Mt. 85 •• Palomar has a focal length of 16.8 m. (a) By what factor is the light-gathering power increased over the 40.0-in (1.02-m) diameter refracting lens of the Yerkes Observatory telescope? (b) If the focal length of the eyepiece is 1.25 cm, what is the magnifying power of the 200-in telescope? Picture the Problem Because the light-gathering power of a mirror is proportional to its area, we can compare the light-gathering powers of these mirrors by finding the ratio of their areas. We can use the ratio of the focal lengths of the objective and eyepiece lenses to find the magnifying power of the Palomar telescope.

1036 Chapter 32 (a) Express the ratio of the lightgathering powers of the Palomar and Yerkes mirrors:

π

2 d Palomar PPalomar 4 mirror mirror = = π 2 PYerkes AYerkes d Yerkes mirror 4 mirror

APalomar

=

2 d Palomar mirror 2 Yerkes mirror

d

Substitute numerical values and evaluate PPalomar/PYerkes:

PPalomar (200 in )2 = = 25.0 PYerkes (40.0 in )2

(b) Express the magnifying power of the Palomar telescope:

M =−

fo fe

Substitute numerical values and evaluate M:

M =−

16.8 m = − 1340 1.25 cm

86 •• An astronomical telescope has a magnifying power of 7.0. The two lenses are 32 cm apart. Find the focal length of each lens. Picture the Problem We can use the expression for the magnifying power of a telescope and the fact that the length of a telescope is the sum of focal lengths of its objective and eyepiece lenses to obtain simultaneous equations in fo and fe.

The magnifying power of the telescope is given by:

M =−

The length of the telescope is the sum of the focal lengths of the objective and eyepiece lenses:

L = f o + f e = 32 cm

Solve these equations simultaneously to obtain:

f o = 28 cm and f e = 4 cm

fo = −7 fe

87 •• [SSM] A disadvantage of the astronomical telescope for terrestrial use (for example, at a football game) is that the image is inverted. A Galilean telescope uses a converging lens as its objective, but a diverging lens as its eyepiece. The image formed by the objective is at the second focal point of the eyepiece (the focal point on the refracted side of the eyepiece), so that the final image is virtual, upright, and at infinity. (a) Show that the magnifying power is given by M = –fo/fe, where fo is the focal length of the objective and fe is that of the

Optical Images 1037 eyepiece (which is negative). (b) Draw a ray diagram to show that the final image is indeed virtual, upright, and at infinity. Picture the Problem The magnification of a telescope is the ratio of the angle subtended at the eyepiece lens to the angle subtended at the objective lens. We can use the geometry of the ray diagram to express both θe and θo.

Lens 1

Lens 2

>

(b) The ray diagram is shown below:

>

F1' F2

θo

h θ e >

>

F1

(a) Express the magnifying power M of the telescope:

M =

θe θo

Because the image formed by the objective lens is at the focal point, F'1 :

θo =

h fo

where we have assumed that θo