01 Unit-Dimensions & Basic Maths
02 Errors
03 Vectors & Calculus
04 Kinematics
05 NLM & Friction
06 CM & WPE
07 Centre of Mass
08 Rotational Dynamics
09 gravitation
10 Elasticity & Thermal Expansion
11 Fluid Mechanics
12 Surface Tension & Viscosity
13 SHM
14 Waves
15 Sound Waves
16 Heat1
17 Heat2
18 Electrostatics1
19 Electrostatics2
20 capacitance
21 current electricity
22 Magnetism
23 EMI
24 AC
25 Geometrical Optics
26 Wave Optics
27 Modern Physics1
28 Modern Physics2

##### Citation preview

UNIT AND DIMENSION & BASIC MATHEMATICS THEORY AND EXERCISE BOOKLET CONTENTS

S.NO.

TOPIC

PAGE NO.

THEORY WITH SOLVED EXAMPLES ........................................ 3 – 23

EXERCISE - I ......................................................................... 24 – 34

EXERCISE - II ........................................................................ 35 – 36

EXERCISE - III ........................................................................ 37 – 39

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Page # 2

UNIT AND DIMENSION

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , [email protected]

Page # 3

UNIT AND DIMENSION 1.

PHYSICAL QUANTITY The quantites which can be measured by an instrument and by means of which we can describe the laws of physics are called physical quantities.

Types of physical quantities :

Fundamental 1.1

Derived

Supplementry

Fundamental Although the number of physical quantities that we measure is very large, we need only a limited number of units for expressing all the physical quantities since they are interrelated with one another. So, certain physical quantities have been chosen arbitrarily and their units are used for expressing all the physical quantities, such quantities are known as Fundamental, Absolute or Base Quantities (such as length, time and mass in mechanics) (i) All other quantites may be expressed in terms of fundamental quantities. (ii) They are independent of each other and cannot be obtained from one another. An international body named General Conference on Weights and Measures chose seven physical quantities as fundamental : (1) length

(2) mass

(3) time

(5) thermodynamic temperature

(4) electric current, (6) amount of substance

(7) luminous intensity. Note : These are also called as absolute or base quantities. In mechanics, we treat length, mass and time as the three basic or fundamental quantities. 1.2

Derived : Physical quantities which can be expressed as combination of base quantities are called as derived quantities.

For example : Speed, velocity, acceleration, force, momentum, pressure, energy etc. dis tan ce length = time time

Ex.1

Speed =

1.3

Supplementary : Beside the seven fundamental physical quantities two supplementary quantities are also defined, they are : (1) Plane angle

(2) Solid angle.

Note : The supplementary quantities have only units but no dimensions. 2.

MAGNITUDE : Magnitude of physical quantity = (numerical value) × (unit) Magnitude of a physical quantity is always constant. It is independent of the type of unit. ⇒ numerical value ∝ or

Ex.2

1 unit

n1u1 = n2u2 = constant

Length of a metal rod bar is unchanged whether it is measured as 2 metre or 200 cm. Observe the change in the Numerical value (from 2 to 200) as unit is changed from metre to cm.

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Page # 4 3.

UNIT AND DIMENSION

UNIT : Measurement of any physical quantity is expressed in terms of an internationally accepted certain basic reference standard called unit. The units for the fundamental or base quantities are called fundamental or base unit. Other physical quantities are expressed as combination of these base units and hence, called derived units. A complete set of units, both fundamental and derived is called a system of unit.

3.1.

Principle systems of Unit There are various system in use over the world : CGS, FPS, SI (MKS) etc Table 1 : Units of some physical quantities in different systems. System Physica l Qua ntity

CGS (Ga ussia n)

MKS (SI)

FPS (British)

Length

centimeter

meter

foot

Mass

gram

kilogram

pound

Time

second

second

second

Force

dyne

newton → N

poundal

W ork or Energy

erg

joule → J

ft-poundal

Power

erg/s

watt → W

ft-poundal/s

Fundame ntal

De rived

3.2

*

The SI system is at present widely used throughout the world. In IIT JEE only SI system is followed.

3.3

Definitions of some important SI Units (i) Metre : 1 m = 1,650, 763.73 wavelengths in vaccum, of radiation corresponding to organ-red light of krypton-86. (ii) Second : 1 s = 9,192, 631,770 time periods of a particular from Ceasium - 133 atom. (iii) Kilogram : 1kg = mass of 1 litre volume of water at 4°C (iv) Ampere : It is the current which when flows through two infinitely long straight conductors of negligible cross-section placed at a distance of one metre in vacuum produces a force of 2 × 10–7 N/m between them. (v) Kelvin : 1 K = 1/273.16 part of the thermodynamic temperature of triple point of water. (vi) Mole : It is the amount of substance of a system which contains as many elementary particles (atoms, molecules, ions etc.) as there are atoms in 12g of carbon - 12. 1   2 m of a (vii) Candela : It is luminous intensity in a perpendicular direction of a surface of   600000  black body at the temperature of freezing point under a pressure of 1.013 × 105 N/m2.

(viii) Radian : It is the plane angle between two radiia of a circle which cut-off on the circumference, an arc equal in length to the radius. (ix) Steradian : The steradian is the solid angle which having its vertex at the centre of the sphere, cut-off an area of the surface of sphere equal to that of a square with sides of length equal to the radius of the sphere.

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Page # 5

UNIT AND DIMENSION Ex.3

Find the SI unit of speed, acceleration

Sol.

speed =

meter(m) dis tan ce = = m/s sec ond(s) time

acceleration = = 4.

(called as meter per second)

velocity displacement / time = time time

displacement ( time) 2

=

meter

= m/s2 (called as meter per second square)

sec ond2

S I PREFIXES The magnitudes of physical quantities vary order a wide range. The CGPM recommended standard prefixes for magnitude too large or too small to be expressed more compactly for certain power of 10. Power of 10 18

10 15 10 12 10 9 10 6 10 3 10 2 10 1 10

5.

Prefix

Symbol

exa peta tera giga mega kilo hecto deca

Power of 10

E P T G M k h da

–1

10 –2 10 –3 10 –6 10 –9 10 –12 10 –15 10 –18 10

Prefix

Symbol

deci centi milli micro nano pico femto atto

d c m µ

n p f a

GENERAL GUIDELINES FOR USING SYMBOLS FOR SI UNITS, SOME OTHER UNITS, SOME OTHER UNITS, AND SI PREFIXES (a) Symbols for units of physical quantities are printed/written in Roman (upright type), and not in italics For example : 1 N is correct but 1 N is incorrect (b)

(i) Unit is never written with capital initial letter even if it is named after a scientist.

For example : SI unit of force is newton (correct) Newton (incorrect) (ii)

For a unit named after a scientist, the symbol is a capital letter. But for other units, the symbol is NOT a capital letter.

For example : force

newton (N)

energy

joule (J)

electric current

ampere (A)

temperature

kelvin (K)

frequency

hertz (Hz)

length

meter (m)

mass

kilogram (kg)

luminous intensity

candela (cd)

time

second (s)

For example :

Note : (c)

The single exception is L, for the unit litre.

Symbols for units do not contain any final full stop all the end of recommended letter and remain unaltered in the plural, using only singular form of the unit.

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Page # 6

UNIT AND DIMENSION

For example :

(d)

Quantity

Correct

Incorrect

25 centimeters

25 cm

25 c m 25 cms.

Use of solidus ( / ) is recommended only for indicating a division of one letter unit symbol by another unit symbol. Not more than one solidus is used.

For example : Correct m/s

2

m/s/s 2

N s / m/ m

J/K mol

J / K / mol

kg/m s

kg / m / s

N s/m

(e)

Incorrect

Prefix symbols are printed in roman (upright) type without spacing between the prefix symbol and the unit symbol. Thus certain approved prefixes written very close to the unit symbol are used to indicate decimal fractions or multiples of a SI unit, when it is inconveniently small or large.

For example megawatt

1 MW = 10 6 W

centimetre

1 cm = 10 –2 m

kilometre

1 km = 10 3 m

millivolt

1 mV = 10 –3 V

kilowatt-hour

1 kW h = 10 3 W h = 3.6 M J = 3.6 × 10 6 J

microampere

1 µ A = 1 0 –6 A

angstrom

1 Å = 0 .1nm = = 1 0 –9

– 10

nanosecond

1 ns = 10

1 pF = 10 –12 F

microsecond

1 µs = 1 0 – 6 s

gigahertz

1 GHz = 10 9 Hz

micron

1 µm = 1 0 –6 m

m

s

The unit 'fermi', equal to a femtometre or 10–15 m has been used as the convenient length unit in nuclear studies. (f)

When a prefix is placed before the symbol of a unit, the combination of prefix and symbol is considered as a new symbol, for the unit, which can be raised to a positive or negative power without using brackets. These can be combined with other unit symbols to form compound unit.

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Page # 7

UNIT AND DIMENSION

For example :

Correct

Quantity

(g)

Incorrect

cm3

(cm)3 = (0.01 m)3 = (10–2m)3 = 10–6 m3

mA2

(mA)2 = (0.001 A)2 = (10–3A)2 = 10–6 A2

0.01 m3 or 10–2 m3 0.001 A2

A prefix is never used alone. It is always attached to a unit symbol and written or fixed before the unit symbol.

For example : 103/m3 = 1000/m3 or 1000 m–3, but not k/m3 or k m–3. (h)

Prefix symbol is written very close to the unit symbol without spacing between them, while unit symbols are written separately with spacing with units are multiplied together.

For example : Quantity

Correct

Incorrect

1 metre per second

1 milli per second

1 ms

1 millisecond

1 metre second

1Cm

1 coulomb metre

1 centimetre

1 cm

1 centimetre

1 coulomb metre

1 ms

(i)

–1

The use of double profixes is avoided when single prefixes are available.

For example :

Quantity

Incorrect

10–9 m

1 nm (nanometre)

1 mµ m (milli micrometre)

10–6 m

1µm (micron)

1 m m m (milli millimetre)

10–12

1 µ µ F (micro microfarad)

1 GW (giga watt)

1 kM W (kilo megawatt)

F

109 F

(j)

Correct

The use of a combination of unit and the symbols for unit is avoided when the physical quantity is expressed by combining two or more units.

Quantity

Correct

Incorrect

joule per mole Kelvin

J/mol K or J mol–1 K–1

Joule / mole K or J/mol Kelvin or J/mole K

newton metre second

Nms

newton m second or N m second or N metre s or newton metre s

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Page # 8

UNIT AND DIMENSION

5.1. Characteristics of base units or standards : (A) Well defined (B) Accessibility (C) Invariability (D) Convenience in use 5.2

Some special types of units : 1. 1 Micron (1µ) = 10–4 cm = 10–6 m (length) 2. 1 Angstrom(1 Å) = 10–8 cm = 10–10m (length) 3. 1 fermi (1f) = 10–13 cm = 10–15 m (length) 4. 1 inch = 2.54 cm (length) 5. 1 mile = 5280 feet = 1.609 km (length) 6. 1 atmosphere = 105 N/m2 = 76 torr = 76 mm of Hg pressure (pressure) –3 3 3 7. 1 litre = 10 m = 1000 cm (volume) 8. 1 carat = 0.0002 kg (weight) 9. 1 pound (Ib) = 0.4536 kg (weight)

6.

DIMENSIONS Dimensions of a physical quantity are the power to which the fundamental quantities must be raised to represent the given physical quantity. For example,

density =

mass mass = (length )3 volume

or density = (mass) (length)–3 ...(i) Thus, the dimensions of density are 1 in mass and –3 in length. The dimensions of all other fundamental quantities are zero. For convenience, the fundamental quantities are represented by one letter symbols. Generally mass is denoted by M, length by L, time by T and electric current by A. The thermodynamic temperature, the amount of substance and the luminous intensity are denoted by the symbols of their units K, mol and cd respectively. The physical quantity that is expressed in terms of the base quantities is enclosed in square brackets. [sinθ] = [cosθ] = [tanθ] = [ex] = [M0L0T0] 7.

DIMENSIONAL FORMULA It is an expression which shows how and which of the fundalmental units are required to represent the unit of physical quantity. Different quantities with units. symbol and dimensional formula. Quantity Displacement Area Volume

s A V

Symbol

Formula  ×b ×b×h

Velocity

v

v=

Momentum

p

p = mv

Acceleration

a

a=

Force Impulse Work

F W

F = ma F × t F.d

∆s ∆t ∆v ∆t

S.I. Unit Metre or m (Metre)2 or m2 (Metre)3 or m3

D.F. M0LT0 M0L2T0 M0L3T0

m/s

M0LT–1

kgm/s

MLT–1

m/s2

M0LT–2

Newton or N N.sec N.m

MLT–2 MLT–1 ML2T–2

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Page # 9

UNIT AND DIMENSION

Energy

KE or U

K.E. =

1 mv 2 2

Joule or J

ML2T–2

watt or W

ML2T–3

kg/m3 Pascal or Pa N.m.

ML–3T0 ML–1T–2 ML2T–2

P.E. = mgh W t

Power

P

P=

Density Pressure Torque

d P τ

d = mass/volume P = F/A τ=r×F

Angular displacement θ

θ=

M0L0T0

Angular velocity

ω

ω=

θ t

M0L0T–1

Angular acceleration

α

α=

∆ω ∆t

M0L0T–2

Moment of Inertia

I

I = mr2

kg-m2

ML2T0

Frequency

v or f

f=

hertz or Hz

M0L0T–1

Stress

-

F/A

N/m2

ML–1T–2

Strain

-

∆ ∆A ∆V ; ;  A V

-

M0L0T0

Youngs modulus

Y

Y=

N/m2

ML–1T–2

T

F W or  A

N J ; m m2

ML0T–2 ML0T–2

1 T

F/A ∆ / 

(Bulk modulus of rigidity)

Surface tension

Force constant (spring)

k

F = kx

N/m

Coefficient of viscosity

η

 dv  F = η A  dx 

kg/ms(poise in C.G.S.) ML–1T–1

G

F=

Gravitation constant

Gm1 m 2 r2

Vg =

Gravitational potential Vg

PE m

N − m2 kg 2

J kg

M–1L3T–2

M0L2T–2

Temperature Heat

θ Q

Q = m × S × ∆t

Kelvin or K M0L0T0θ+1 Joule or Calorie ML2T–2

Specific heat

S

Q = m × S × ∆t

Joule kg .Kelvin

M0L2T–2θ–1

Latent heat

L

Q = mL

Joule kg

M0L2T–2

Coefficient of thermal

K

Q=

Joule m sec K

MLT–3θ–1

KA( θ1 − θ2 )t d

conductivity

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Page # 10

UNIT AND DIMENSION

Universal gas constant

R

PV = nRT

Joule mol.K

ML2T–2θ–1

Mechanical equivalent

J

W = JH

-

M0L0T0

Coulomb or C

M0L0TA

Ampere or A

M0L0T0A

of heat Charge

Q or q

I=

Q t

Current

I

-

Electric permittivity

ε0

ε0 =

1 q1q2 . 4πF r 2

Electric potential

V

V=

∆W q

Joule/coul

ML2T–3A–1

Intensity of electric field

E

E=

F q

N/coul.

MLT–3A–1

Capacitance

C

Q = CV

M–1L–2T4A2

Dielectric constant

εr

εr =

ε ε0

-

Resistance

R

V = IR

Ohm

ML2T–3A–2

Conductance

S

S=

1 R

Mho

M–1L–2T–3A2

Specific resistance

ρ

ρ=

RA 

s

σ=

1 ρ

B

F = qvBsinθ

(coul.)2 N.m 2

or

C2 N − m2

M–1L–3T4A2

M0L0T0

or relative permittivity

Ohm × meter

ML3T–3A–2

or resistivity Conductivity or

Mho/meter

M–1L–3T3A2

specific conductance Magnetic induction

Tesla or weber/m2

MT–2A–1

or F = BIL dφ dt

Magnetic flux

φ

e=

Magnetic intensity

H

B=µH

Magnetic permeability

µ0

B=

L

e = L.

Electric dipole moment

p

Magnetic dipole moment

M

µ 0 Idl sin θ 4π r 2

Weber

ML2T–2A–1

A/m

M0L–1T0A

N amp 2

MLT–2A–2

Henery

ML2T–2A–2

p = q × 2

C.m.

M0LTA

M = NIA

amp.m2

M0L2AT0

of free space or medium Coefficient of self or

dI dt

Mutual inductance

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Page # 11

UNIT AND DIMENSION

8.

USE OF DIMENSIONS

8.1

Theory of dimensions have following main uses : Conversion of units : This is based on the fact that the product of the numerical value (n) and its corresponding unit (u) is a constant, i.e., n[u] = constant or n1[u1] = n2 [u2] Suppose the dimensions of a physical quantity are a in mass, b in length and c in time. If the fundamental units in one system are M1, L1 and T1 and in the other system are M2, L2 and T2 respectively. Then we can write.

n1[M1a Lb1 T1c ] = n 2 [Ma2 Lb2 T2c ]

...(i)

Here n1 and n2 are the numerical values in two system of units respectively. Using Eq. (i), we can convert the numerical value of a physical quantity from one system of units into the other system. Ex.4 The value of gravitation constant is G = 6.67 × 10–11 Nm2/kg2 in SI units. Convert it into CGS system of units. Sol. The dimensional formula of G is [M–1 L3 T–2]. Using equation number (i), i.e.,

n1[M1–1 L31 T1–2 ] = n 2 [M2–1 L32 T2–2 ] –1

3

M  L   T  n 2 = n1  1   1   1   M2   L 2   T2 

–2

Here, n1 = 6.67 × 10–11 M1 = 1 kg, M2 = 1 g = 10–3 kg L1 = 1 m, Substituting in the above equation, we get –1

n2 = 6.67 × 10

–11

3

L2 = 1cm = 10–2 m,

 1kg   1m  1s   –3   – 2     10 kg  10 m  1s 

T1 = T2 = 1s

–2

or n2 = 6.67 × 10–8 Thus, value of G in CGS system of units is 6.67 × 10–8 dyne cm2/g2. 8.2 To check the dimensional correctness of a given physical equation : Every physical equation should be dimensionally balanced. This is called the 'Principle of Homogeneity'. The dimensions of each term on both sides of an equation must be the same. On this basis we can judge whether a given equation is correct or not. But a dimensionally correct equation may or may not be physically correct. Ex.5 Show that the expression of the time period T of a simple pendulum of length l given by T =

Sol.

l is dimensionally correct. g T = 2π

Dimensionally [T ] =

l g [L] [LT – 2 ]

= [T ]

As in the above equation, the dimensions of both sides are same. The given formula is dimensionally correct.

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Page # 12 8.3

UNIT AND DIMENSION

Principle of Homogeneity of Dimensions. This principle states that the dimensions of all the terms in a physical expression should be same. For example, in the physical expression s = ut +

1 2 1 at , the dimensions of s, ut and at2 all are same. 2 2

The physical quantities separated by the symbols +, –, =, >, < etc., have the same dimensions.

Note :

Ex.6 The velocity v of a particle depends upon the time t according to the equation v = a + bt +

c . d+ t

Write the dimensions of a, b, c and d. Sol. From principle of homogeneity [a] = [v] [a] = [LT–1]

or

Ans.

[bt] = [v] [b] =

or

[b] = [LT–2]

Similarly,

[d] = [t] = [T]

Further,

8.4

[v ] [LT –1] = [ t] [T ]

or

Ans.

[c ] = [v ] [d + t ]

or

[c] = [v] [d + t]

or

[c] = [LT–1] [T]

or

[c] = [L]

Ans.

To establish the relation among various physical quantities : If we know the factors on which a given physical quantity may depend, we can find a formula relating the quantity with those factors. Let us take an example.

Ex.7 The frequency (f) of a stretched string depends upon the tension F (dimensions of force), length l of the string and the mass per unit length µ of string. Derive the formula for frequency. Sol. Suppose, that the frequency f depends on the tension raised to the power a, length raised to the power b and mass per unit length raised to the power c. Then.

f ∝ [F]a [l ]b [µ]c or

f = k [F]a [l ]b [µ]c

Here, k is a dimensionless constant. Thus, [f ] = [F]0 [l]b [µ]c or

[M0 L0 T–1] = [MLT–2]a [L]b [ML–1]c

or

[M0L0T–1] = [Ma + c La + b – c T–2a]

For dimensional balance, the dimension on both sides should be same. Thus, and

a+c=0

...(ii)

a+b–c=0

...(iii)

– 2a = – 1

...(iv)

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Page # 13

UNIT AND DIMENSION

Solving these three equations, we get

a=

1 1 , c=– and b = – 1 2 2

Substituting these values in Eq. (i), we get

f = k(F)1/ 2 (l ) –1(µ ) –1/ 2 or

f =

k F l µ

Experimentally, the value of k is found to be Hence,

8.5

f =

1 2

1 F 2l µ

Limitations of Dimensional Analysis The method of dimensions has the following limitations : (i) By this method the value of dimensionless constant can not be calculated. (ii) By this method the equation containing trigonometrical, exponential and logarithmic terms cannot be analysed. (iii) If a physical quantity depends on more than three factors, then relation among them cannot be established because we can have only three equations by equalising the powers of M, L and T.

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Page # 14

UNIT AND DIMENSION

BASIC MATHEMATICS 9.

MENSURATION FORMULAS : r : radius ; V = Volume (a) Circle

d = diameter ; S.A = surface area

Perameter : 2πr = πd,

Area

: πr2 =

1 2 πd 4

(b) Sphere Surface area = 4πr2 = πd2 , Volume =

4 3 1 πr = πd3 3 6

(c) Spherical Shell (Hollow sphere) Surface area = 4πr2 = πd2 Volume of material used = (4πr2)(dr), dr = thickness (d) Cylinder Lateral area = 2πrh V = πr2h Total area = 2πrh + 2πr2 = 2πr (h + r) (e) Cone Lateral area = πr

r 2 + h2

h = height

1 2 2  2  Total area = π r  r + h + r  V = πr h   3

(f) Ellipse Circumference ≈ 2π

a2 + b2 2

area = πab a = semi major axis

b a

b = semi minor axis

a

(g) Parallelogram A = bh = ab sin θ

h θ

b

a = side ; h = height ; b = base θ = angle between sides a and b

b

(h) Trapezoid

h

h area = (a + b) 2

a a and b parallel sides h = height

(i) Triangle

bh ab sin γ = s(s − a)(s − b)(s − c ) = 2 2 a, b, c sides are opposite to angles α, β , γ area =

b = base ; h = height s=

1 (a + b + c ) 2

α

c

b γ

β

a

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Page # 15

UNIT AND DIMENSION

h

(j) Rectangular container lateral area = 2(b + bh + h )

b  side , b, h Mathematics is the language of physics. It becomes easier to describe, understand and apply the physical principles, if one has a good knowledge of mathematics. V =  bh

10.

LOGARITHMS : (ii) If ex = y, then x = loge y = ln y (iii) If 10x = y, then x = log10y (i) e ≈ 2.7183 (iv) log10y = 0.4343 loge y = 2.303 log10 y (v) log (ab) = log (a) + log (b) a (vi) log  = log (a) – log (b) b

(vii) log an = n log (a)

11.

TRIGONOMETRIC PROPERTIES :

(i)

Measurement of angle & relationship between degrees & radian In navigation and astronomy, angles are measured in degrees, but in calculus it is best to use units called radians because of they simplify later calculations. B Let ACB be a central angle in circle of radius r, as in figure. Then the angle ACB or θ is defined in radius as Arc length AB θ= ⇒ θ= Radius r

If

θ

C

r

A

r = 1 then θ = AB

The radian measure for a circle of unit radius of angle ABC is defined to be the length of the circular arc AB. since the circumference of the cirlce is 2π and one complete revolution of a cicle is 360°, the relation between radians and degrees is given by the following equation. π radians = 180°

ANGLE CONVERSION FORMULAS 1 degree =

π 180°

Convert

π rad to degrees : 6

45 •

Degrees to radians : multiply by

π 180°

Radians to degrees : multiply by

180° π

π π = rad 180 4

π 180 • = 30° 6 π

Ex.16 Convert 30º to radians : Sol.

30 º×

π π = rad 6 180 º

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Page # 16

UNIT AND DIMENSION

Ex.17 Convert Sol.

π 180 × = 60 3 π

(1) 30° = (4) 90° =

(2) 45° =

(7) 150° =

(3) 60° =

(5) 120° =

(6) 135° =

(Check these values yourself to see that the satisfy the conversion formulaes) (ii) Measurement of positive & Negative Angles :

y y

Positive measure

Negative Measure

x

x

An angle in the xy-plane is said to be in standard position if its vertex lies at the origin and its initial ray lies along the positive x-axis (Fig). Angles measured counterclockwise from the positive x-axis are assigned positive measures; angles measured clockwise are assigned negative measures. y

y

y

y –

x

x

x

5π 2

x

3π – 4

9π 4

(iii) Six Basic Trigonometric Functions :

y

O

r

θ x

y

oppsite side

hy po te

nu se

P(x,y)

The trigonometric fucntion of a general angle θ are defined in tems of x, y and r.

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Page # 17

UNIT AND DIMENSION

Sine :

sin θ =

opp y = hyp r

hyp r Cosecant : cos ecθ = opp = y

Cosine:

cos θ =

Secant :

opp y Tangent: tan θ = adj = x

sec θ =

adj x Cotangent: cot θ = opp = y

VALUES OF TRIGONOMETRIC FUNCTIONS If the circle in (Fig. above) has radius r = 1, the equations defining sin θ and cos θ become cosθ = x,

sinθ = y

We can then calculate the values of the cosine and sine directly from the coordinates of P. Ex.18 Find the six trigonometric ratios from given fig. (see above) Sol.

opp 4 sinθ = hyp = 5

5

adj 3 cosθ = hyp = 5

4

θ

opp 4 tanθ = adj = 3 sec θ =

cot θ =

hyp 5 = opp 3

3

cosec θ =

hyp 5 = opp 4

Ex.19 Find the sine and cosine of angle θ shown in the unit circle if coordinate of point p are as shown. y    – 1, 3   2 2    3 2

1

θ

x

1 2

Sol.

cos θ = x-coordinate of P = –

12.

1 2

sin θ = y-coordinate of P =

3 2

Values of sin θ, cos θ and tan θ for some standard angles.

Degree

0

30

37

0

π/6

37π / 180

sin θ

0

1/2

3/5

1/ 2

4/5

cos θ

1

3 /2

4/5

1/ 2

3/5

tan θ

0

1/ 3

3/4

1

4/3

45

π/ 4

53

60

90

120

135

53π / 180

π/3

π/2

2π / 3

3π / 4

π

3 /2

1

3 /2

1/ 2

0

1/2

0

–1/2

– 1/ 2

–1

3

– 3

–1

0

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180

Page # 18

UNIT AND DIMENSION

A useful rule for remembering when the basic trigonometric funcions are positive and negative is the CAST rule. If you are not very enthusiastic about CAST. You can remember it as ASTC (After school to college) y nd nd II Quadrant I Quadrant

A all positive

S sin positive

x T Tan positive IIInd Quadrant

The CAST rule RULES FOR FINDING TRIGONOMETRIC RATIO OF ANGLES GREATER THAN 90°. Step 1 →

Identify the quadrant in which angle lies.

Step 2 → (a) If angle = (nπ ± θ)

where n is an integer. Then

π   (b) If angle = ( 2n + 1) + θ where n is in interger. Then 2   π   trigonometric function of ( 2n + 1) ± θ = complimentry trignometric function of θ and 2   sign will be decided by CAST Rule.

Ex.20 Evaluate sin 120° sin 120° = sin (90° + 30°) = cos 30° =

3 2

Aliter sin 120° = sin (180° – 60°) = sin 60° =

3 2

Sol.

Ex.21 Evaluate cos 210° Sol.

cos 210° = cos (180° + 30°) = – cos 30° = –

Ex.22 tan 210° = tan (180° + 30°) = tan 30° = + 13.

3 2 1 3

IMPORTANT FORMULAS (i) sin2θ + cos2θ = 1 (ii) 1 + tan2θ = sec2θ 2 2 (iii) 1 + cot θ = cosec θ (iv) sin2θ = 2 sin θ cos θ (v) cos 2θ = 2 cos2θ – 1 = 1 – 2 sin2θ = cos2θ – sin2θ (vi) sin (A ± B) = sin A cos B ± cos A sin B (vii) cos (A ± B) = cos A cos B ∓ sin A sin B C+D C–D (viii) sin C + sin D = 2 sin  cos   2   2 

(x) cos C + cos D = 2 cos (xii) tan 2θ =

2 tan θ 2

1 – tan θ

C+D C–D cos 2 2

C–D C+D  cos  (ix) sinC – sin D = 2 sin 2    2 

(xi) cos C – cos D = 2 sin (xiii)

D–C C+D sin 2 2

tan A ± tan B tan(A ± B) = 1 ∓ tan A tan B

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Page # 19

UNIT AND DIMENSION (xiv) sin(90° + θ) = cos θ (xvi) tan (90° + θ) = – cot θ (xviii) cos(90° – θ) = sinθ (xx) sin(180° – θ) = sin θ (xxii) tan (180° + θ) = tan θ (xxiv) cos (–θ) = cos θ

sin A sin B sin C = = a b c

Sine Rule

(xv) cos (90° + θ) = – sin θ (xvii) sin(90° – θ) = cos θ (xix) cos (180° – θ) = – cos θ (xxi) cos (180° + θ) = – cos θ (xxiii) sin(– θ) = – sin θ (xxv) tan (–θ) = – tan θ

Cosine rule

B c

a

a2 = b2 + c2 – 2bc cos A

C

A b

4 Ex.23

3

90°

53°

37° x Find x :

S

o

l

.

14.

sin 90° sin 53° = x 4 x= 5

SMALL ANGLE APPROXIMATION It is a useful simplification which is only approximately true for finite angles. It involves linerarization of the trigonometric functions so that, when the angle θ is measured in radians. sin θ ~ θ cosθ ~ 1 or cos θ ~ 1 –

θ2 for the second - order approximation 2

tan θ ~ θ Geometric justification

tan

Object

arc θ

θ

D

arc

tan

d

Small angle approximation. The value of the small angle X in radians is approximately equal to its tangent. When one angle of a right triangle is small, is hypotenuse in approximately equal in length to the leg adjacent to the small angle, so the cosine is approximately 1. The short leg is approximately equal to the arc from the long leg to the hypotenuse, so the sine and tangent are both approximated by the value of the angle in radians.

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Page # 20 15.

UNIT AND DIMENSION

BINOMIAL THEOREM : (1 ± x)n = 1 ± nx +

n(n – 1)x 2 ........... 2!

(1 ± x)–n = 1 ∓ nx +

n(n + 1) 2 x ......... 2!

If x 0 ⇒ The line will pass through (0, 1)

(0,1) θ

tan θ =

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3 2

x

1 th if x 4

Page # 22

UNIT AND DIMENSION

Ex.29 Draw the graph for the equation : 2y + 4x + 2 = 0 ⇒ y = – 2x – 1 Sol. 2y + 4x + 2 = 0 m = – 2 < 0 i.e., θ > 90° c = – 1 i.e., line will pass through (0, –1)

tanθ = –2

θ

(0,–1)

: (i) If c = 0 line will pass through origin. (ii) y = c will be a line parallel to x axis.

(0,c)

(0,0) (iii) x = c will be a line perpendicular to y axis

(c,0) (0,0)

(ii)

Parabola A general quadratic equation represents a parabola. y = ax2 + bx + c a≠0 if a > 0 ; It will be a opening upwards parabola. if a < 0 ; It will be a opening downwards parabola. if c = 0 ; It will pass through origin. y ∝ x2 or y = 2x2, etc. represents a parabola passing through origin as shown in figure shown. y y

x ∝ y2

y ∝ x2

x

x

2

k = 1/2mv

(i) e.g.

(ii)

y = 4 x2 + 3x

e.g.

k=

1 mv2 2

k y=4x2+3x v 2

y=–4x +3x

Note : That in the parabola y = 2x2 or y ∝ x2, if x is doubled, y will beome four times. Graph x ∝ y2 or x = 4 y2 is again a parabola passing through origin as shown in figure shown. In this cae if y is doubled, x will become four times. y = x2 + 4 or x = y2 – 6 will represent a parabola but not passing through origin. In the first equation (y = x2 + 4), if x doubled, y will not become four times.

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Page # 23

UNIT AND DIMENSION 17.

SIMILAR TRIANGLE Two given triangle are said to be similar if

(1)

All respective angle are same or

(2)

All respective side ratio are same.

P A

R C Q As example, ABC, PQR are two triangle as shown in figure.

B

If they are similar triangle then (1)

∠A=∠P ∠B=∠Q ∠C=∠R

OR AB BC AC = = PQ QR PR

(2)

A

P 5 Ex.30

3

B

Q

x

O

6 Find x : Sol.

By similar triangle concept AB OB = PQ OQ 5 6 = 3 x

x=

18 5

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Page # 24

UNIT AND DIMENSION

(ONLY ONE OPTION IS CORRECT)

Exercise - I SECTION A : UNITS

1. Which of the following sets cannot enter into the list of fundamental quantities in any system of units ? (A) length, mass and velocity (B) length, time and velocity (C) mass, time and velocity (D) length, time and mass Sol.

2. Which of the following is not the name of a physical quantity ? (A) kilogram (B) impulse (C) energy (D) density Sol.

3. Light year is the unit of (A) speed (B) mass (C) distance (D) time Sol.

4. PARSEC is a unit of (A) Time (C) Distance Sol.

(B) Angle (D) Velocity

5. Which of the following is not the unit of time (A) solar day (B) parallactic second (C) leap year (D) lunar month Sol.

6. Which of the following system of units is NOT based on the unit of mass, length and time alone (A) FPS (B) SI (C) CGS (D) MKS Sol.

7. The SI unit of the universal gravitational constant G is (B) Nm2kg–2 (A) Nm kg–2 (C) Nm2 kg–1 (D) Nmkg–1 Sol.

8. The SI unit of the universal gas constant R is : (B) watt K–1 mol–1 (A) erg K–1 mol–1 –1 –1 (C) newton K mol (D) joule K–1 mol–1 Sol.

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Page # 25

UNIT AND DIMENSION 9. The unit of impulse is the same as that of : (A) moment force (B) linear momentum (C) rate of change of linear momentum (D) force Sol.

13. One watt-hour is equivalent to (A) 6.3 × 103 Joule (B) 6.3 × 10–7 Joule (C) 3.6 × 103 Joule (D) 3.6 × 10–3 Joule Sol.

10. Which of the following is not the unit of energy? (A) watt-hour (B) electron-volt (C) N × m (D) kg × m/sec2 Sol.

SECTION : B DIMENSIONS 14. What are the dimensions of lenth in force × displacement/time (A) –2 (B) 0 (C) 2 (D) none of these Sol.

11. A physical quantity is measured and the result is expressed as nu where u is the unit used and n is the numerical value. If the result is expressed in various units then (A) n ∝ size of u (B) n ∝ u2 (C) n ∝ √u (D) n ∝ 1/u Sol.

12. If the unit of length is micrometer and the unit of time is microsecond, the unit of velcoity will be : (A) 100 m/s (B) 10 m/s (C) micrometers (D) m/s Sol.

15. The angular frequency is measured in rad s–1. Its dimension in length are : (A) – 2 (B) –1 (C) 0 (D) 2 Sol.

16. A dimensionless quantity : (A) never has a unit (B) always has a unit (C) may have a unit (D) does not exit Sol.

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Page # 26

UNIT AND DIMENSION

17. A unitless quantity : (A) never has a nonzero dimension (B) always has a nonzero dimension (C) may have a nonzero dimension (D) does not exit Sol.

18. If a and b are two physical quantities having different dimensions then which of the following can denote a new physical quantity (A) a + b (B) a – b (E) sin (a/b) (C) a/b (D) ea/b Sol.

19. Two physical quantities whose dimensions are not same, cannot be : (A) multiplied with each other (B) divided (C) added or substracted in the same expression (D) added together Sol.

Sol.

21. The dimensions of universal gravitational constant are (A) M–1 L3 T–2 (B) M–1 L3 T–1 –1 –1 –2 (C) M L T (D) M–2 L2 T–2 Sol.

22. The SI unit of Stefan's constant is : (B) J s m–1 K–1 (A) Ws–1 m–2 K–4 –1 –2 –1 (C) J s m K (D) W m–2 K–4 Sol.

23. What are the dimensions of Boltzmann's constant? (B) ML2T–2K–1 (A) MLT–2K–1 0 –2 (C) M LT (D) M0L2T–2K–1 Sol. 20. Choose the correct statement(s) : (A) All quantities may be represented dimensionally in terms of the base quantities. (B) A base q uant i t y cannot b e re pres ente d dimensionally in terms of the rest of the base quantities. (C) The dimension of a base quantity in other base quantities is always zero. (D) The dimension of a derived quantity is never zero in any base quantity.

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Page # 27

UNIT AND DIMENSION 24. Planck's constant has the dimensions of : (A) force (B) energy (C) linear momentum (D) angular momentum Sol.

25. The velocity 'v' (in cm/s) of a particle is given in terms of time 't' (in s) by the equation b t+c The dimensions of a, b and c are a b c a (A) L2 T LT2 (B) LT2 (C) LT–2 L T (D) L Sol.

v = at +

b LT LT

c L T2

26. The position of a particle at time 't' is given by the relation

27. The time dependence of a physical quantity ? P = P0exp(–α t2) where α is a constant and t is time The constant α (A) will be dimensionless (B) will have dimensions of T–2 (C) will have dimensions as that of P (D) will have dimensions equal to the dimension of P multiplied by T–2 Sol.

28. Force F is given in terms of time t and distance x by F = A sin C t + B cos D x A C and are given by Then the dimensions of B D –2 0 0 –1 –2 (A) MLT , M L T (B) MLT , M0L–1T0 0 0 0 0 1 –1 (C) M L T , M L T (D) M0L1T–1, M0L0T0 Sol.

V0 [1 – e – αt ] α where V0 is a constant and α > 0. The dimensions of V0 and α are respectively. (A) M0L1T0 and T–1 (B) M0L1T0 and T–2 (C) M0L1T–1 and T–1 (D) M0L1T–1 and T–2 Sol.

x(t) =

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Page # 28

UNIT AND DIMENSION

29. The Van der Waal equation for 1 mole of a real gas is a    P + 2  ( V – b) = RT  V  where P is the pressure, V is the volume, T is the absolute temperature, R is the molar gas constant and a, b are Van dar Waal constants. The dimensions of a are the same as those of (A) PV (B) PV2 (C) P2V (D) P/V Sol.

30. In above question 29, the dimensions of b are the same as those of (A) P (B) V (C) PV (D) nRT Sol.

31. In above question 29, the dimensions of nRT are the same as those of (A) energy (B) force (C) pressure (D) specific heat Sol.

33. Which pair of following quantities has dimensions different from each other. (A) Impulse and linear momentum (B) Plank's constant and angular momentum (C) Moment of inertia and moment of force (D) Young's modulus and pressure Sol.

34. A pair of physical quantities having the same dimensional formula is : (A) angular momentum and torque (B) torque and energy (C) force and power (D) power and angular momentum Sol.

35. If force (F) is given by F = Pt–1 + α t, where t is time. The unit of P is same as that of (A) velocity (B) displacement (C) acceleration (D) momentum Sol.

32. In above question 29, the dimensional formula for ab is (B) ML4T–2 (C) ML6T–2 (D) ML8T–2 (A) ML2T–2 Sol.

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Page # 29

UNIT AND DIMENSION 36. The product of energy and time is called action. The dimensional formula for action is same as that for (A) power (B) angular energy (C) force × velocity (D) impulse × distance Sol.

37. Dimensions of pressure are the same as that of (A) force per unit volume (B) energy per unit volume (C) force (D) energy Sol.

39. In the above question dimensions of same as those of (A) wave velocity (C) wave amplitude Sol.

b are the c

(B) wavelength (D) wave frequency

40. What is the physical quantity whose dimensions are M L2 T–2 ? (A) kinetic energy (B) pressure (C) momentum (D) power Sol.

41. Which one of the following has the dimensions of ML–1T–2 ? (A) torque (B) surface tension (C) viscosity (D) stress Sol.

38. When a wave traverses a medium, the displacement of a particle located at x at time t is given by y = a sin (bt – cx) where a, b and c are constants of the wave. The dimensions of b are the same as those of (A) wave velocity (B) amplitude (C) wavelength (D) wave frequency Sol.

42. If force, acceleration and time are taken as fundamental quantities, then the dimensions of length will be : (B) F–1 A2 T–1 (C) FA2T (D) AT2 (A) FT2 Sol.

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Page # 30

UNIT AND DIMENSION

43. The dimensions ML–1T–2 can correspond to (A) moment of a force or torque (B) surface tension (C) pressure (D) co-efficient of viscosity   (useful relation are τ = r × F , S = F/l, F = 6 π η r v,, where symbols have usual meaning) Sol.

44. Which of the following can be a set of fundamental quantities (A) length, velocity, time (B) momentum, mass, velocity (C) force, mass, velocity (D) momentum, time, frequency Sol.

47. In a certain system of units, 1 unit of time is 5 sec, 1 unit of mass is 20 kg and unit of length is 10m. In this system, one unit of power will correspond to (A) 16 watts (B) 1/16 watts (C) 25 watts (D) none of these Sol.

48. In a book, the answer for a particular question is expressed as b =

ma  2kl   1+  here m represents k  ma 

mass, a represents accelerations, l represents length. The unit of b should be (A) m/s (B) m/s2 (C) meter (D) /sec Sol.

45. If area (A) velocity (v) and density (ρ) are base units, then the dimensional formula of force can be represented as (A) Avρ (B) Av2ρ (C) Avρ2 (D) A2vρ Sol.

49. ρ = 2 g/cm3 convert it into MKS system kg kg (B) 2 × 103 3 (A) 2 × 10–3 3 m m kg kg (C) 4 × 103 3 (D) 2 × 106 3 m m Sol.

46. The pressure of 106 dyne/cm2 is equivalent to (A) 105 N/m2 (B) 106 N/m2 7 2 (C) 10 N/m (D) 108 N/m2 Sol.

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Page # 31

UNIT AND DIMENSION F sin(βt) (here V = velocity, F = force, t = V2 time) : Find the dimension of α and β (A) α = [M1L1T0], β = [T–1] (B) α = [M1L1T–1], β = [T1] (C) α = [M1L1T–1], β = [T–1] (D) α = [M1L–1T0], β = [T–1] Sol.

50. α =

51. Given that v is the speed, r is radius and g is acceleration due to gravity. Which of the following is dimension less (A)

v 2g r

(B) v2rg

(C) vr2g

(D)

v2 rg

Sol.

54. The velocity of water waves may dpend on their wavelength λ , the density of water ρ and the acceleration due to gravity g. The method of dimensions gives the relation between these quantities as (A) v2 = kλ–1 g–1 ρ–1 (B) v2 = k g λ (C) v2 = k g λ ρ (D) v2 = k λ3 g–1 ρ–1 where k i s a di mensi onl e ss constant Sol.

Sol.

52. If E, M, J and G denote energy, mass, angular momentum and gravitational constant respectively, then

EJ2

M5 G2 (A) length Sol.

has the dimensions of (B) angle

(C) mass

(D) time

55. If the unit of force is 1 kilonewton, the length is 1 km and time is 100 second, what will be the unit of mass : (A) 1000 kg (B) 10 kg (C) 10000 kg (D) 100 kg Sol.

53. The dimensions ML–1T–2 may correspond to (A) work done by a force (B) linear momentum (C) pressure (D) energy per unit volume

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Page # 32

UNIT AND DIMENSION

56. A body moving through air at a high speed 'v' experiences a retarding force 'F' given by F = K A d vx where 'A' is the surface area of the body, 'd' is the density of air and 'K' is a numerical constant. The value of 'x' is : (A) 1 (B) 2 (C) 3 (D) 4 Sol.

p

57. The velocity of a freely falling body changes as g hq where g is acceleration due to gravity and h is the height. The values of p and q are : (A) 1, (C)

1 2

1 ,1 2

(B)

1 1 , 2 2

59. The value of G = 6.67 × 10–11 N m2 (kg)–2. Its numerical value in CGS system will be : (A) 6.67 × 10–8 (B) 6.67 × 10–6 (C) 6.67 (D) 6.67 × 10–5 Sol.

60. The density of mercury is 13600 kg m–3. Its value of CGS system will be : (A) 13.6 g cm–3 (B) 1360 g cm–3 –3 (C) 136 g cm (D) 1.36 g cm–3 Sol.

(D) 1, 1

Sol.

61. If the acceleration due to gravity is 10 ms–2 and the units of length and time are changed to kilometre and hour, respectively, the numerical value of the acceleration is : (A) 360000 (B) 72000 (C) 36000 (D) 129600 Sol. 58. Choose the correct statement(s) : (A) A dimensionally correct equation must be correct. (B) A dimensionally correct equation may be correct. (C) A dimensionally incorrect equation may be correct. (D) A dimensionally incorrect equation may be incorrect. Sol.

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Page # 33

UNIT AND DIMENSION 62. If 'c' the velocity of light 'g' the acceleration due to gravity and 'P" the atmospheric pressure are fundamental units, then the dimensions of length will be (A) c/g (B) P × c × g (C) c/P (D) c2/g Sol.

63. The units of length, velocity and force are doubled. Which of the following is the correct change in the other units ? (A) unit of time is doubled (B) unit of mass is doubled (C) unit of momentum is doubled (D) unit of energy is doubled Sol.

65. If the units of M, L are doubled then the unit of kinetic energy will become (A) 2 times (B) 4 times (C) 8 times (D) 16 times Sol.

BASIC MATHEMATICS 66. The radius of two circles are r and 4r what will be the ratio of their Area and perimeter. Sol.

67. Internal radius of a ball is 3 cm and external radius is 4 cm. What will be the volume of the material used. Sol. 64. If the units of force and that of length are doubled, the unit of energy will be : (A) 1/4 times (B) 1/2 times (C) 2 times (D) 4 times Sol.

68. Binomial (a) (99)1/2 Sol.

(b) (120)1/2

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(c) (126)1/3

Page # 34

UNIT AND DIMENSION

4 x

2

71. 69. A normal human eye can see an object making an angle of 1.8° at the eye. What is the approximate height of object which can be seen by an eye placed at a distance of 1 m from the eye.

y

Find x and y :

2 3

Sol.

h 1.8°

1m Sol.

70. Draw graph for following equations : (ii) x = 4t – 3 (i) v = v0 – at (iii) x = 4 at2 (iv) v = – gt Sol.

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Page # 35

UNIT AND DIMENSION

Exercise - II 1*. Which of the following is not the unit length : (A) micron (B) light year (C) angstrom (D) radian Sol.

4. The equation of state for a real gas at high temperature is given by P =

nRT a where − 1/ 2 V − b T V( V + b)

n, P V & T are number of moles, pressure, volume & temperature respectively & R is the universal gas constant. Find the dimensions of constant ‘a’ in the above equation. Sol.

2. A particle is in a uni-directional potential field where the potential energy (U) of a particle depends on the x-coordinate given by Ux = k(1 – cos ax) & k and ‘a’ are constants. Find the physical dimensions of ‘a’ & k. Sol. 5. The distance moved by a particle in time t from centre of a ring under the influence of its gravity is given by x = a sinωt where a & ω are constants. If ω is found to depend on the radius of the ring (r), its mass (m) and universal gravitational constant (G), find using dimensional analysis an expression for ω in terms of r, m and G. Sol.

3. The time period (T) of a spring mass system depends upon mass (m) & spring constant (k) & length of the spring (l) [k =

Force ] . Find the relation among, length

(T), (m), (l) & (k) using dimensional method. Sol.

6. If the velocity of light c, Gravitational constant G & Plank’s constant h be chosen as fundamental units, find the dimension of mass, length & time in the new system. Sol.

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Page # 36

UNIT AND DIMENSION

7. A satellite is orbiting around a planet. Its orbital velocity (v0) is found to depend upon (a) Radius of orbit (R) (b) Mass of planet (M) (c) Universal gravitation constant (G) Using dimensional analysis find an expression relating orbital velocity (v0) to the above physical quantities. Sol.

10. Use the small angle approximations to find approximate values for (a) sin 8° and

(b) tan 5°

Sol.

8. The angle subtended by the moon's diameter at a point on the earth is about 0.50°. Use this and the act that the moon is about 384000 km away to find the approximate diameter of the moon.

θ

D

rm (A) 192000 km

(B) 3350 km

(C) 1600 km

(D) 1920 km

Sol.

9. Use the approximation (1 + x)n ≈ 1 + nx, |x| 0 =

Function defined for x ≥ 0

…..

3

d  1  d −1 1 ( x ) = ( −1)x − 2 = − 2  = dx  x  dx x

Function defined for x ≥ 0 (b)

3

1 −4 / 5 x 5

derivative not defined at x = 0

Rule No.3 The Constant Multiple Rule If u is a differentiable function of x, and c is a constant, then In particular, if n is a positive integer, then

d du (cu) = c dx dx

d (cx n ) = cn x n−1 dx

Ex.34 The derivative formula d (3x 2 ) = 3 (2x ) = 6x dx

says that if we rescale the graph of y = x2 by multiplying each y-coordinate by 3, then we multiply the slope at each point by 3. Ex.35 A useful special case The derivative of the negative of a differentiable function is the negative of the function’s derivative. Rule 3 with c = – 1 gives. d d d d ( −u) = ( −1.u) = −1 . (u) = − (u) dx dx dx dx

Rule No.4 The Sum Rule The derivative of the sum of two differentiable functions is the sum of their derivatives. If u and v are differentiable functions of x, then their sum u + v is differentiable at every point where u and v are both differentiable functions in their derivatives. d d du dv du dv (u − v ) = [u + ( −1) v ] = + ( −1) = − dx dx dx dx dx dx

The sum Rule also extends to sums of more than two functions, as long as there are only finite , u2, ........ un are differentiable at x, then so if u1 + u2 + ....... + un, then 1 f u

n

c t i o

n

s

i n

t h

e

s u

m

.

I f

u

du du du d (u1 + u 2 + ...... + un ) = 1 + 2 + ........ + n dx dx dx dx

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VECTOR & CALCULUS

Page # 25

Ex.36 (a) y = x4 + 12x

(b) y = x3 +

4 2 x – 5x + 1 3

dy d 3 d 4 2 d d = x + (5 x ) + (1)  x − dx dx dx  3  dx dx

dy d 4 d = (x ) + (12 x ) dx dx dx

= 4x3 + 12

= 3x2 +

4 . 2x – 5 + 0 3

8 x−5 3 Notice that we can differentiate any polynomial term by term, the way we differentiated the polynomials in above example.

= 3x2 +

Rule No. 5 The Product Rule If u and v are differentiable at x, then if their product uv is considered, then

d dv du (uv ) = u +v . dx dx dx

The derivative of the product uv is u times the derivative of v plus v times the derivative of u. In prime notation (uv)’ = uv’ + vu’. While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance, d d 2 (x . x) = ( x ) = 2x, dx dx

while d ( x ). d ( x ) = 1.1 = 1 , which is wrong dx dx 2

+ 1) (x3 + 3)

Ex.37

F

Sol.

Using the product Rule with u = x2 + 1 and v = x3 + 3, we find

i n

d

t h

e

d

e

r

i v

a

t i v

e

s

o

f

y

=

(

x

d [( x 2 + 1)( x 3 + 3)] = (x2 + 1) (3x2) + (x3 + 3) (2x) dx

= 3x4 + 3x2 + 2x4 + 6x = 5x4 + 3x2 + 6x Example can be done as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial. We now check : y = (x2 + 1) (x3 + 3) = x5 + x3 + 3x2 + 3 dy = 5x4 + 3x2 + 6x dx

This is in agreement with our first calculation. There are times, however, when the product Rule must be used. In the following examples. We have only numerical values to work with. Ex.38 Let y = uv be the product of the functions u and v. Find y’(2) if u(2) = 3, u’(2) = – 4, v(2) = 1, and v’(2) = 2. Sol.

From the Product Rule, in the form y’ = (uv)’ = uv’ + vu’, we have y’(2) = u(2) v’(2) + v(2) u’(2) = (3) (2) + (1) (–4) = 6 – 4 = 2

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VECTOR & CALCULUS

Page # 26

Rule No.6 The Quotient Rule If u and v are differentiable at x, and v(x) ≠ 0, then the quotient u/v is differentiable at x, d u and  = dx  v 

v

du dv −u dx dx v2

Just as the derivative of the product of two differentiable functions is not the product of their derivatives, the derivative of the quotient of two functions is not the quotient of their derivatives. Ex.39 Find the derivative of y = Sol.

t2 − 1 t2 + 1

We apply the Quotient Rule with u = t2 – 1 and v = t2 + 1

dy ( t 2 + 1) 2t − ( t 2 − 1). 2t = dt ( t 2 + 1)2 =

2t 3 + 2t − 2t 3 + 2t 2

( t + 1)

2

 d  u  v( du / dt ) − u(dv / dt )   As   =  dt v2 v  

=

4t 2

( t + 1)2

Rule No. 7 Derivative Of Sine Function d (sin x ) = cos x dx

Ex.40 (a) y = x2 – sin x :

(b) y = x2 sin x :

dy d = 2x − (sin x ) = 2x – cos x dx dx

dy d = x2 (sin x ) + 2x sin x dx dx

Difference Rule

Product Rule

= x2cosx + 2x sinx dy sin x = (c) y = : dx x =

x.

d (sin x ) − sin x .1 dx x2

Quotient Rule

x cos x − sin x x2

Rules No.8 Derivative Of Cosine Function d (cos x ) = − sin x dx

Ex.41 (a) y = 5x + cos x

Sum Rule

dy d d = (5 x ) + (cos x ) = 5 – sin x dx dx dx

(b) y = sin x cos x dy d d = sin x (cos x ) + cos x (sin x ) dx dx dx

Product Rule

= sin x(– sin x) + cos x (cos x) = cos2 x – sin2 x = cos 2x

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Page # 27

Rule No. 9 Derivatives Of Other Trigonometric Functions Because sin x and cos x are differentiable functions of x, the related functions sin x 1 tan x = sec x = ; cos x cos x cot x =

cos x ; sin x

cos ec x =

1 sin x

are differentiable at every value of x at which they are defined. There derivatives, Calculated from the Quotient Rule, are given by the following formulas. d (tan x ) = sec 2 x ; dx

d (sec x ) = sec x tan x dx

d (cot x ) = − cos ec 2 x ; dx

d (cos ec x ) = − cos ec x cot x dx

Ex.42 Find dy / dx if y = tan x. Sol.

d d  sin x  (tan x ) =  = dx dx  cos x 

=

Ex.43 (a)

cos x

d d (sin x ) − sin x (cos x ) dx dx cos 2 x

cos x cos x − sin x( − sin x ) 2

cos x

=

cos 2 x + sin 2 x 2

cos x

=

1 cos 2 x

= sec 2 x

d d (3x + cot x) = 3 + (cot x) = 3 – cosec2 x dx dx

d  2  d d (b) dx  sin x  = dx (2 cosec x ) = 2 dx (cosec x )  

= 2(– cosec x cot x) = – 2 cosec x cot x

Rule No. 10 Derivative Of Logrithm And Exponential Functions d 1 (loge x ) = , dx x

d x (e ) = e x dx

Ex.44 y = ex . loge (x) dy d x d = (e ). log( x ) + [log e ( x )] e x dx dx dx

dy ex = e x . loge ( x ) + dx x

Rule No. 11 Chain Rule Or ‘Outside Inside’ Rule dy dy du = . dx du dx It sometime helps to think about the Chain Rule the following way. If y = f (g(x)), dy = f’[g(x)] . g’(x) dx In words : To find dy/dx, differentiate the “outside” function f and leave the “inside” g(x) alone; then multiply by the derivative of the inside. We now know how to differntiate sin x and x2 – 4, but how do we differentiate a composite like sin(x2 – 4)? The answer is, with the Chain Rule, which says that the derivative of the composite of two differentiable functions is the product of their derivatives evaluated at appropriate points. The Chain Rule is probably the most widely used differentiation rule in mathematics. This section describes the rule and how to use it. We begin with examples.

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Ex.45 The function y = 6x – 10 = 2(3x – 5) is the composite of the functions y = 2u and u = 3x – 5. How are the derivatives of these three functions related ? Sol.

dy = 6 , dy = 2 , du = 3 dx du dx

We have

Since 6 = 2 × 3

dy dy du = . dx du dx

Is it an accident that

dy dy du = . dx du dx

?

If we think of the derivative as a rate of change, our intution allows us to see that this relationship is reasonable. For y = f(u) and u = g(x), if y changes twice as fast as u and u changes three times as fast as x, then we expect y to change six times as fast as x. Ex.46 Let us try this again on another function. y = 9x4 + 6x2 + 1 = (3x2 + 1)2 is the composite y = u2 and u = 3x2 + 1. Calculating derivatives. We see that dy du . = 2u.6 x du dx

and

= 2 (3x2 + 1). 6x = 36x3 + 12 x

dy d = (9 x 4 + 6 x 2 + 1) = 36 x3 + 12 x dx dx dy du dy . = du dx dx

Once again,

The derivative of the composite function f(g(x)) at x is the derivative of f at g(x) times the derivative of g at x. Ex.47 Find the derivation of y = x 2 + 1 Sol.

Here y = f(g(x)), where f(u) = f ′ (u) =

1 2 u

2 u and u = g(x) = x + 1. Since the derivatives of f and g are

and g′(x) = 2x,

the Chain Rule gives

1 1 dy d = f (g( x )) = f′ (g(x)).g′(x) = .g′(x) = . (2x) = 2 g( x ) dx dx 2 x2 + 1

d sin( x 2 + x ) = cos( x 2 + x ).(2x + 1) dx Inside

Ex.49 (a)

x +1

derivative of the outside

outside

Ex.48

x 2

Inside derivative left along of the inside

d 1 (1 – x 2 )1/ 4 = (1 – x 2 ) – 3 / 4 (–2x ) dx 4

u = 1 – x2 and n = 1/4

(Function defined) on [–1, 1] =

–x 2(1 – x 2 )3 / 4

(derivative defined only on (–1, 1))

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VECTOR & CALCULUS

Page # 29

(b)

d d sin 2x = cos 2x 2x = cos 2x .2 = 2 cos 2x dx dx

(c)

d d ( A sin( ωt + φ)) = A cos (ω t + φ) (ω t + φ ) = A cos (ω t + φ). ω dt dt

Rull No. 12

= A ω cos (ω t + φ)

Power Chain Rule

d n du u = nun –1 dx dx

*

If

Ex.50

d  1  d d   = (3 x – 2) –1 = – 1 (3x – 2)–2 ( 3 x – 2) dx  3 x – 2  dx dx

= – 1 (3x – 2)–2 (3) = –

3 ( 3 x – 2) 2

In part (d) we could also have found the derivation with the Quotient Rule. Ex.51 (a) Sol.

d ( Ax + B)n dx

Here u = Ax + B,

du =A dx

d ( Ax + B )n = n( Ax + B)n –1.A dx d d 1 sin( Ax + B) = cos( Ax + B).A (c) (b) log(Ax + B) = .A dx dx Ax + B ∴

(d)

d d ( Ax +B ) e = e( Ax +B ) .A tan (Ax+B) = sec2 (Ax + B).A (e) dx dx

Note : These results are important

19.

DOUBLE DIFFERENTIATION If f is differentiable function, then its derivative f' is also a function, so f' may have a derivative of its own, denoted by ( f ' )' = f ' ' . This new function f'' is called the second derivative of because it is the derivative of the derivative of f. Using Leibniz notation, we write the second derivative of y = f(x) as

d  dy  d2 y  = dx  dx  dx 2 Another notation is f''(x) = D2 f (x). Ex.52 If f(x) = x cos x, find f" (x) Sol.

Using the Product Rule, we have f '(x) = x

d d ( x ) = – x sin x + cos x (cos x) + cos x dx dx

To find f" (x) we differentiate f'(x) : d d d d (– x sin x + cos x ) = – x (sin x ) + sin x (–x) + (cos x) dx dx dx dx = – x cos x – sinx – sinx = – x cos x – 2 sin x

f"(x) =

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Page # 30

20.

APPLICATION OF DERIVATIVE DIFFERENTIATION AS A RATE OF CHANGE dy is rate of change of 'y' with respect to 'x' : dx For examples : (i) v =

dx this means velocity 'v' is rate of change of displacement 'x' with respect to time 't' dt

(ii) a =

dv this means acceleration 'a' is rate of change of velocity 'v' with respect to time 't'. dt

(iii) F =

dp this means force 'F' is rate of change of monentum 'p' with respect to time 't'. dt

(iv) τ =

dL this means torque 'τ' is rate of change of angular momentum 'L' with respect to time 't' dt

(v) Power =

dW this means power 'P' is rate of change of work 'W' with respect to time 't' dt

π 2 D . 4 How fast is the area changing with respect to the diameter when the diameter is 10 m ? The (instantaneous) rate of change of the area with respect to the diameter is

Ex.53 The area A of a circle is related to its diameter by the equation A =

Sol.

dA π πD = 2D = dD 4 2

When D =10m, the area is changing at rate (π/2) = 5π m2/m. This mean that a small change ∆D m in the diameter would result in a changed of about 5p ∆D m2 in the area of the circle. Physical Example : Ex.54 Boyle's Law state that when a sample of gas is compressed at a constant temperature, the product of the pressure and the volume remains constant : PV = C. Find the rate of change of volume with respect to pressure. Sol.

dV C =– 2 dP P

Ex.55 (a) Find the average rate of change of the area of a circle with respect to its radius r as r changed from (i) 2 to 3

(ii) 2 to 2.5

(iii) 2 to 2.1

(b) Find the instantaneous rate of change when r = 2. (c) Show that thre rate of change of the area of a circle with respect to its radius (at any r) is equal to the circumference of the circle. Try to explain geometrically when this is true by drawing a circle whose radius is increased by an amount ∆r. How can you approximate the resulting change in area ∆A if ∆r is small ? Sol.

(a) (i) 5π (ii) 4.5 π (iii) 4.1 π (b) 4 π (c) ∆A ≈ 2 π r ∆r

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VECTOR & CALCULUS

21.

Page # 31

MAXIMA & MINIMA y Suppose a quantity y depends on another quantity x in a manner shown in figure. It becomes maximum at x1 and minimum at x2. At these points the tangent to the curve is parallel to the x-axis and hence its slope is tan θ = 0. Thus, at a maxima or a minima slope ⇒

dy =0 dx

x1

x

x2

Maxima Just before the maximum the slope is positive, at the maximum it dy is zero and just after the maximum it is negative. Thus, decrease dx dy at a maximum and hence the rate of change of is negative at dx d  dy  d  dy    < 0 at maximum. The quantity   is a maximum i.e., dx  dx  dx  dx  the rate of change of the slope. It is written d2 y

dy =0 as 2 . Conditions for maxima are : (a) dx dx

(b)

d2 y dx 2

y 2

3

4

θ4 5 θ5 slope = m1 = tan θ1 m1 > m 2 >(m3 = 0) > m4 > m5 x

θ2

θ11

O For maxima, as x increases the slope decreases

0 dx  dx 

Conditions for minima are :

dy =0 (a) dx

(b)

d2 y dx 2

O

>0

slope = m1 = tan θ1 m1 < m2 90° m > 0 ⇒ θ < 90° 0° ≤ θ < 180° m

m=tanθ

m=tanθ θ

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θ

Page # 16

KINEMATICS

Ex.23 Draw the graph for the equation : 2y = 3x + 2 3 Sol. 2y = 3x + 2 ⇒ y = x + 1 2 3 m= > 0 ⇒ θ < 90° 2 c = +1 > 0 ⇒ The line will pass through (0, 1)

(0,1)

tan θ =

θ

Ex.24 Draw the graph for the equation : 2y + 4x + 2 = 0 Sol. 2y + 4x + 2 = 0 ⇒ y = – 2x – 1 m = – 2 < 0 i.e., θ > 90° c = – 1 i.e., line will pass through (0, –1)

3 2

tanθ = –2

θ

(0,–1)

: (i) If c = 0 line will pass through origin. (0,c) (ii) y = c will be a line parallel to x axis.

(ii)

(c,0)

(0,0)

(iii) x = c will be a line perpendicular to y axis

(0,0)

Parabola A general quadratic equation represents a parabola. y = ax2 + bx + c a≠ 0 if a > 0 ; It will be a opening upwards parabola. if a < 0 ; It will be a opening downwards parabola. if c = 0 ; It will pass through origin. 2

y=4x +3x e.g.

y = 4 x2 + 3x

2

y=–4x +3x

Average velocity & instantaneous velocity from Position vs time graph Average velocity from t1 to t2 =

x2 – x1 displacement = t –t time taken 2 1

B

x2

x2–x1

= tan θ = slope of the chord AB vinstantaneous

x 2 – x1 = as lim t2 → t1 t 2 – t1

x1

A t1

θ t2–t1

t2

when t2 approaches t1 point B approaches Point A and the chord AB becomes tangent to the curve. Therefore vinstantaneous = Slope of the tangent x – t curve

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Page # 17

KINEMATICS (iii)

(A)

Reading x v/s t graphs x

(1)

x0

Explanation

Body is at rest at x0.

t x

(2)

Body starts from origin and is moving with speed θ

tan θ away from origin.

t

x (3)

Body starts from rest from origin and moves away from origin with increasing speed velocity and positive acceleration.

t

x (4)

Body starts from rest from x = x0 and moves away from origin with increasing velocity or +ve acceleration.

x0 t

(5)

x0

Body starts from x = x0 and is moving toward the origin with constant velocity passes throw origin after same time and continues to move away from origin.

t x (6)

x0

Body starts from rest at x = x0 and then moves with increasing speed towards origin ∴ acceleration is –ve

t

(7)

x

Body starts moving away from origin with some initial speed. Speed of body is decreasing till t1 and it becomes 0 momentarily of t = t1 and At this instant. Its reverses its direction and move towards the origin with increasing speed.

t2

O

t1

t

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Page # 18

(8)

KINEMATICS

x

Body starts from origin moves away from origin in the –ve x-axis at t = t1 with decreasing speed and at t= t1 it comes at rest momentarily, Reverses its direction t moves towards the origin the increasing speed. Crosses the origin at t = t2.

t1 t2

x (9)

t

Body starts from origin from rest and moves away from origin with increasing speed.

(B) V-T GRAPHS v

(1)

Body is always at rest.

t

v (2)

Body is moving with constant velocity v0

v0

t

v (3)

Body is at rest initially then it starts moving with its velocity increasing at a constant rate i.e. body is moving with constant acceleration.

t

v (4)

(5)

Body starts its motion with initial velocity v0 and continues to move with its velocity increasing at a constant rate i.e. acceleration of the body is constant.

v0 t

v

v0 t0 t v (6)

t

Body starts its motion with initial velocity v0. Then it continues to move with its velocity decreasing at a constant rate i.e. acceleration of the body is negative and constant. At t = t0 the body comes to rest instantaneously and reverses its direction of motion and then continues to move with decreasing velocity or increasing speed. For 0 < t < t0 motion of the body is deaccelerated (∴ speed is decreassing) t > t0 motion of the body is accelerated (∴ speed is increasing) Body is at rest initially. Then it starts moving with increasing velocity. As time increases its velocity is increasing more rapidly. i.e. the moving with increasing acceleration.

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Page # 19

KINEMATICS

v (7)

v0

t0

t

Body starts its motion with initial velocity v0. Its velocity is decreasing with time and at t = t0 . It becomes zero after body reverse its direction of motion and continues to move with decreasing velocity or increasing speed. Since velocity of the body is decreasing for whole motion. Therefore, its acceleration is negative.For 0 < t < t0 motion of the body is deaccelerated (speed is decreassing) t > t0 motion of the body is accelerated (∵ speed is increasing)

(C) READING OF a - t GRAPHS a

(1)

acceleration of the body is zero that means the body is moving constant velocity. t

a

(2)

Acceleration of the body is constant and positive. t

a t

(3)

Acceleration of the body is constant and negative

a

Initially the acceleration of the body is zero. Then its acceleration is increasing at a constant rate.

(4) t

a

(5)

t

The body starts accelerating(initial acceleration zero) at t = 0. Its acceleration is negative for whole of its motion and is decreasing at a constant rate.

a

(6) t

Initially acceleration of the body is zero. Its acceleration is positive for whole of its motion. Its acceleration is increasing for whole of its motion.

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Page # 20

KINEMATICS

(IV)

Drawing of graphs on the basis of given information.

(a) (i)

If acceleration of the body is zero. If the velocity of the body is v0 and it starts from origin. x v x= u

0

t

v0 t

(ii)

(iii)

t

If at t = 0, x = x0 then x t 0 +v 0 x x0 x=

v

v0

t If at t = 0, x = – x0 then x 0 –x x=

t v

t t 0 +v

v0

t

–x0

t

(b)

If a body has constant acceleration : For this section (i) u0, x0 & a0 are positive constants. (ii) u ≡ initial velocity (iii) v ≡ velocity at any time t. (iv) x ≡ Position at any time t. xi ≡ initial position

(i)

if u = 0, a = a0 1 if xi = 0, x = at 2 2 x

if xi = x0, x = x0 + (1/2)at2 x

x This is wrong because it suggest the body don't have some initial velocity

x0 t

t

t v

a

a0

slope = tanθ = a0 θ

t

t

v = a0 t (ii)

If u = u0

, a = a0

x = xi + u0t +

1 a0 t 2 2

x

v = u0 + a0t

x

v

a

a0

if xi = 0

t

if xi = x0

t

t

t

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Page # 21

KINEMATICS

(iii)

if u = u0, a = – a0 1 2 x = xi + u0t – a 0 t 2

x

x

x0

if xi = 0 t

t0

v

if xi = x0

t

a

u0 t

t0 t (iv)

–a0

if u = – u0 , a = + a0 x = xi – u0t +

x

1 a0 t 2 2

x0

t if xi = x0

if xi = 0

v

a

a0

t

–u0 (v)

t

If u = u0, a = – a0 x = xi – u0t –

x

1 a0 t 2 2

x

x0

t

v

–v0

t

if xi = 0

if xi = x0

a

t

t

–a0

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Page # 22

KINEMATICS

Ex.25 Draw the (a) position vs time graph (b) velocity vs time graph (c) acceleration vs time graph for the following cases (i) If a body is projected vertically upwards with initial velocity u. Take the projection point to be origin and upward direction as positive. x = ut –

1 2 gt 2

x

v

2

u

u 2g

u g

(ii)

a 2u g

u g

t

t

t

2u g

–g

–u

If a body is dropped from a height h above the ground. Take dropping point to be origin and upward direction as +ve. x= –

1 2 gt 2

x

v 2h g

a=–g a

2h g

t

t

t

v = – gt

–h

(iii)

If a body is projected vertically upwards from a tower of height h with initial velocity u. Take the projection point to be origin and upward direction as +ve.

x u2 2g

v

u g –h

(iv)

–g

– 2gh

2u g

t v = u – gt

a u/2g

t

t

–g

A car starting from rest accelerates uniformly at 2 ms–2 for 5 seconds and then moves with constant speed acquired for the next 5 seconds and then comes to rest retarding at 2 ms–2. Draw its (a) Position vs time graph (b) Velocity vs time graph (c) acceleration vs time graph

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Page # 23

KINEMATICS acceleration vs time graph

x (in m)

x

100 a +2

v 10

15

5

t

75

–1

10ms

15

–2

5

acceleration vs time graph

(v)

10

25

t

15 5 10 Position vs time graph

velocity vs time graph

t (in sec)

A particle starts from x = 0 and initial speed 10 ms–1 and moves with constant speed 10ms–1 for 20 sec. and then retarding uniformly comes to rest in next 10 seconds. acceleration vs time graph a v –1

20

30

10ms

t (sec)

–2

–1ms

30 20 velocity vs time graph

Acceleration vs time graph

t (sec)

x

250m

200m 30sec

20

t (sec)

Position vs time graph

(V)

Conversion of velocity v/s time graph to speed v/s time graph. As we know that magnitude of velocity represent speed therefore whenever velocity goes –ve take its mirror image about time axis.

velocity

Ex-26

speed

t (sec)

velocity

Ex-27

i m

t (sec)

speed

t

e ag m i r rro

e ag m i or irr m t

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Page # 24

Conversion of displacement vs time graph to distance vs time graph For distance time graph just make the mirror image of the displacement time graph from point of zero velocity onwards.

(VII) Conversion of v - t graphs in to x-t and a-t graphs v x

A

Time

0

t

t

v

x

a

ta nθ

=

a

0

a0 ⇒

a - t graph

t

t

t

v

x

a

tan θ = – a0 (iii)

Disp-time

B

x= v

(i)

(ii)

Dist.-time

C

t

v0

D

Dist./Displacement

(VI)

KINEMATICS

t ⇒

θ

t

t0

t

t0

–a0 at t = t0 velocity reverses its direction.

v (iv)

x – t graph From t = 0 to t = t1 acceleration = 0 therefore from t = 0 to t = t1, x - t graph will be a straight line. From t = t1 to t2 acceleration is negative ∴ It will be an opening downward parabola

v0

t1

x

t1

(v)

t2

t2

t

t

v

upto t = t1 acceleration is +ve t1 < t < t2 acceleration is zero. t > t2 acceleration is –ve x

t1

t2

t

x - t graph t1

t2

t

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Page # 25

KINEMATICS Some important points : dv ⇒ dv = adt dt ⇒ ∆v = area under the a - t curve

a=

dx ⇒ dx = vdt dt ⇒ ∆x = area under the v - t curve ⇒ displacement = area under the v - t curve

v=

a

Ex-28 If at t = 0 u = 5 ms–1 then velocity at t = 10 sec = u + change in velocity = 5 + area of the shaded part = 5 + 10 × 5 = 55 ms–1

–2

5ms

a

10 sec t

Ex-29 if at t = 0, u = 2 ms–2 find out it maximum velocity 5ms–2 Since whole motion is accelerating. Therefore velocity will be max at the end of the motion which will be 1 =2+ × 5 × 10 = 27 ms–1 t 2 10sec Ex-30 if at t = 0, u = 4 ms–1 a Find out v at t = 10 sec, t = 20 sec & t = 30 sec. –2 10ms Since for whole motion acceleration of the body is positive 1 vt= 10 sec = 4 + × 10 × 10 = 54 ms–1 t 2 10sec 20sec 30sec 1 vt = 20 sec = 4 + × 10 × 10 + 10 × 10 2 –1 = 154 ms 1 vt = 30 sec = 154 + × 10 × 10 = 204 ms–1 2 (VIII) Reading of graphs if the motion of two bodies are sketched on the same axes. (a)

Reading of x - t graphs

x

x3

B

x2 x1 O

(i) (ii) (iii) (iv) (v) (vi)

A t1

t2

t3

t

Conclusions : Body A Start its motion at t = 0 from origin and is moving away from the origin with constant velocity. Finally it ends its motion at a distance of x2m from origin at t = t3. Body B starts its motion at t = t1 from origin and is moving away from origin with constant velocity. Finally it ends its motion at a distance of x3m from origin at t = t3 Since slope of B is greater than slope of A. Therefore velocity of B is greater than velocity of A. A t = t2, Both A & B are at the same distance from starting point that means B overtakes A at t = t2 ∵ velocity of both A & B are constant ∵ acceleration of both the bodies are zero. ∴ x3 > x2 ∴ At the end of the motion B is at a greater distance from the starting point.

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Page # 26

KINEMATICS x A

x2

B

x1

Ex-31

(i) (ii) (iii) (iv) (v) (vi)

x0 t1 t0 t Conclusion : Body A starts its motion at t = 0 from origin and is moving away from the origin with constant velocity. Finally its motion ends at t = t1 at x = x2 m. Body B starts its motion at t = 0 from x = x0 and then moves with constant velocity away from the origin. Finally it ends its motion at t = t1. Velocity of A is greater than that of B. At t = t0 A overtakes B acceleration of both A & B is zero. ∵ x2 > x1 ∴ At the end of the motion A is at a greater distance from the starting point then B

x

B

A

Ex-32

(i) (ii) (iii) (iv) (v) (vi)

t1 t2 t Conclusions : Both A & B starts their motion at same time t = 0 and from same point x = 0. Both are moving away from the starting point. A is moving with constant velocity while B starts its motion from rest and its velocity is increasing with time i.e. it has some positive acceleration. ∵ At t = t1 the tangent on B's graph becomes parallel to the A's graphs ∴ At t = t1 velocity of both A & B is same. For t < t1 velocity of A is greater than velocity of B. Therefore up to t = t1, separation between A & B increases with time. For t > t1 velocity of B is greater than velocity of A. Therefore after t = t1 separation between A & B starts decreasing and it becomes zero at t = t2 where B overtakes A. Now you can try Questions 14 to 38 in Exercise I and Ques. 7 to 11 in Ex.II

4.

TWO DIMENSIONAL MOTION OR MOTION IN A PLANE Motion in a plane can be described by vector sum of two independent 1D motions along two mutual perpendicular directions (as motions along two mutual directions don’t affect each other). Consider a particle moving in X-Y plane, then its equations of motions for X and Y axes are vx = ux + axt, vy = uy + ayt 1 1 x = uxt + axt2, and ; y = uyt + ayt2, and 2 2 v 2y = u 2y + 2a y y v 2v = u2x + 2a x x where symbols have their usual meanings. Thus resultant motion would be described by the equations   and v = v i + v j r = x i + y j x

y

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Page # 27

KINEMATICS 4.1

PROJECTILE MOTION It is the best example to understand motion in a plane. If we project a particle obliquely from the surface of earth, as shown in the figure below, then it can be considered as two perpendicular 1D motions - one along the horizontal and other along the vertical.

Y

O

θ

usinθ

u +

u cos θ

x

Assume that effect of air friction and wind resistance are negligible and value of ‘acceleration due to → gravity g is constant. Take point of projection as origin and horizontal and vertical direction as +ve X and Y-axes, respectively. For X-axis For Y - axis ux = u cosθ, uy = u sinθ ax = 0, ay = – g, vx = u cosθ, and vy = u sinθ – gt, and 1 x = u cosθ × t y = u sinθ t – gt2 2 It is clear from above equations that horizontal component of velocity of the particle remains constant while vertical component of velocity is first decreasing, gets zero at the highest point of trajectory and then increases in the opposite direction. At the highest point, speed of the particle is minimum. The time, which projectile takes to come back to same (initial) level is called the time of flight (T). At initial and final points, y = 0, 1 So u sinθ t – gt2 = 0 2

⇒ t = 0 and t =

2u sin θ g

So,

T=

2u sin θ g

Range (R) The horizontal distance covered by the projectile during its motion is said to be range of the projectile u 2 sin 2θ g For a given projection speed, the range would be maximum for θ = 45°. Maximum height attained by the projectile is

R = u cosθ × T =

u 2 sin 2 θ 2g at maximum height the vertical component of velocity is 0. u sin θ T Time of ascent = Time of descent = = g 2

H=

Speed, kinetic energy, momentum of the particle initialy decreases in a projectile motion and attains a minimum value (not equal to zero) and then again increases.  θ is the angle between v and horizontal which decreases to zero. (at top most point) and again

increases in the negative direction

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Page # 28

KINEMATICS

Ex.33 A body is projected with a velocity of 30 ms–1 at an angle of 30° with the vertical. Find the maximum height, time of flight and the horizontal range. Sol. Here u = 30 ms–1, Angle of projection, θ = 90 – 30 = 60° Maximum height, u 2 sin 2 θ 30 2 sin 2 60° = = 34.44 m 2g 2 × 9.8 Time fo flight, 2u sin θ 2 × 30 sin 60° T= = = 5.3 s g 9.8 Horizontal range,

H=

R=

u 2 sin 2θ 30° sin 120° 30 2 sin 60° = = = 79.53 m. g 9.8 9.8

Ex.34 Find out the relation between uA, uB, uC (where uA, uB, uC are the initial velocities of particles A, B, C, respectively)

A

Sol.

B C

∵ Hmax is same for all three particle A, B, C

⇒ Hmax =

u2y

2g ⇒ uy is same for all

∴ uyA = uyB = uyC

 2u y  ⇒ TA = TB = TC  g    from figure

RC > RB > RA

⇒ uxC > uxB > uxA

(C)

∵R =

2u xu y g

uA < uB< uC

Coordinate of a particle after a given time t : Particle reach at a point P after time t then

Y

vy

x = ucosθ .t y = usinθ.t –

1 2 gt 2

usinθ

Position vector

P(x,y) u

vx

y θ

O ucosθ

 1   r = (u cos θ.t )ˆi +  (u sin θ)t – gt 2 ˆj 2  

(D)

v α

x

X

Velocity and direction of motion after a given time : After time 't' vx = ucosθ and vy = usinθ – gt Hence resultant velocity v =

tan α =

vy vx

=

u sin θ – gt u cos θ

2

vx + vy

2

=

u2 cos 2 θ + (u sin θ – gt) 2

–1  u sin θ – gt   ⇒ α = tan  u cos θ 

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Page # 29

KINEMATICS (E)

Velocity and direction of motion at a given height : At a height 'h', vx = ucosθ

u2 sin2 θ – 2gh

And

vy =

Resultant velocity v=

v x2 + v y 2 =

v=

u2 – 2gh

(u cos θ) 2 + u2 sin2 θ – 2gh

Note that this is the velocity that a particle would have at height h if it is projected vertically from ground with u.

Ex.35 A body is projected with a velocity of 20 ms–1 in a direction making an angle of 60° with the horizontal. Calculate its (i) position after 0.5 s and (ii) velocity after 0.5 s. Sol. Here u = 20 ms–1, θ = 60° , t = 0.5 s (i) x = (u cosθ)t = (20 cos60°) × 0.5 = 5 m y = (u sin θ) t – –

1 2 gt = (20 × sin 60°) × 0.5 2

1 × 9.8 × (0.5)2 = 7.43 m 2

(ii) vx = u cos θ = 20 cos 60° = 10 ms–1 vy = u sin θ – gt = 20 sin 60° – 9.8 × 0.5 = 12.42 ms–1 ∴ v=

v 2x + v 2y =

–1 (10) 2 + (12.42) 2 =15.95 ms

tan β =

vy vx

=

12.42 = 1.242 10

∴ β = tan–1 1.242 = 51.16°.

Equation of trajectory of a projectile. Suppose the body reaches the point P(x, y) after time t. Y

vy x

usinθ

u θ

v A α v x P(x,y) Max. y height=h

Path of projectile

m

vx=u cosθ

O ucosθ

B

R uy

θ

X

v

∵ The horizontal distance covered by the body in time t, x = Horizontal velocity × time = u cos θ. t x ucos θ For vertical motion : u = u sinθ, a = –g, so the vertical distance covered in time t is given by

or t =

s = ut +

1 2 at 2

or y = x tanθ –

or

y = u sin θ.

1 x2 g 2 2 u cos 2 θ

x 1 x2 – g. 2 u cos θ 2 u cos 2 θ

...(1)

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Page # 30

KINEMATICS

or y = px – qx2, where p and q are constants. Thus y is a quadratic function of x. Hence the trajectory of a projectile is a parabola. From equation (1)

y = x tan θ 1 –

gx cos θ

 gx    ⇒ y = x tan θ 1 – 2u 2 cos θ sin θ  2u cos θ sin θ  2

2

x  y = x tan θ 1 –  R 

...(2)

Equation (2) is another form of trajectory equation of projectile

Ex.36 A ball is thrown from ground level so as to just clear a wall 4 m high at a distance of 4 m and falls at a distance of 14 m from the wall. Find the magnitude and direction of the velocity. Sol. The ball passes through the point P(4, 4). So its range = 4 + 14 = 18m. The trajectory of the ball is, Now x = 4m, y = 4m and R = 18 m y 4   7 ∴ 4 = 4 tan θ 1 –  = 4 tanθ . P(4,4)  18  9 u or tan θ =

9 , sin θ = 7

9 130

, cosθ =

7

4m

130

θ

18 × 9.8 × 130 or u2 = = 182 2×9×7

or u =

4m

x

14m

–1 182 = 13.5 ms

Also θ = tan–1(9/7) = 52.1°

Ex.37 A particle is projected over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If α and β be the base angles and θ the angle of projection, prove that tan θ = tan α + tan β. Sol. If R is the range of the particle, then from the figure we have tan α + tan β =

y(R – x) + xy y y + = x(R – x) x R– x

or tanα + tan β =

y R × x (R – x)

Y P(x,y)

...(1)

Also, the trajectrory of the particle is

θ

x  y = x tan θ 1–   R

or tanθ =

O

y β

α x

B

R–x

A

y R × x (R – x)

From equations (1) and (2), we get tan θ = tan α + tan β .

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x

Page # 31

KINEMATICS 4.2

Projectile fired parallel to horizontal. As shown in shown figure suppose a body is projected horizontally with velocity u from a point O at a certain height h above the ground level. The body is under the influence of two simultaneous independent motions: (i) Uniform horizontal velocity u. (ii) Vertically downward accelerated motion with constant acceleration g. Under the combined effect of the above two motions, the body moves along the path OPA.

u

O

x

y x

vx

P β

h

v

vy R Y

A Ground

Trajectory of the projectile. After the time t, suppose the body reaches the point P(x, y). The horizontal distance covered by the body in time t is x x = ut ∴ t= u The vertical distance travelled by the body in time t is given by 1 2 s = ut + at 2 1 2 1 2 gt = gt 2 2 [For vertical motion, u = 0]

or

y=0×1+

or

y=

or

y = kx2 [Here k =

2

1 x  g g  =  2 2 u  2u

x  ∵ t = u   

 2 x 

g

= a constant] 2u2 As y is a quadratic function of x, so the trajectory of the projectile is a parabola. Time of flight. It is the total time for which the projectile remains in its flight (from 0 to A). Let T be its time of flight. For the vertical downward motion of the body, we use 1 2 s = ut + at 2 or h = 0 × T +

1 gT2 2

or

T=

2h g

Horizontal range. It is the horizontal distance covered by the projectile during its time of flight. It is equal to OA = R. Thus R = Horizontal velocity × time of flight = u × T 2h g Velocity of the projectile at any instant. At the instant t (when the body is at point P), let the velocity of the projectile be v. The velocity v has two rectangular components: Horizontal component of velocity, vx = u Vertical component of velocity, vy = 0 + gt = gt ∴ The resultant velocity at point P is

or

R=u

v = v2x + v2y =

u2 + g2 t2

If the velocity v makes an angle β with the horizontal, then vy  gt  gt tan β = = or β = tan–1   vx u u

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Page # 32

KINEMATICS

Ex.38 A body is thrown horizontally from the top of a tower and strikes the ground after three seconds at an angle of 45° with the horizontal. Find the height of the tower and the speed with which the body was projected. Take g = 9.8 ms–2. Sol. As shown in figure, suppose the body is thrown horizontally from the top O of a tower of height y with velocity u. The body hits the ground after 3s. Considering verticlly downward motion of the body, 1 2 1 y = uyt + gt = 0 × 3 + ×9.8 × (3)2 = 44.1 m [∴ Initial vertical velocity, uy = 0] 2 2 Final vertical velocity, vy = uy + gt = 0 + 9.8 × 3 = 29.4 ms–1 Final horizontal velocity, vx = u As the resultant velocity u makes an angle of 45° with the horizontal, so tan 45° =

vy vx

or 1 =

29.4 x

or u = 29.4 ms–1.

Ex.39 A particle is projected horizontally with a speed u from the top of plane inclined at an angle θ with the horizontal. How far from the point of projection will the particle strike the plane? Sol. The horizontal distance covered in time t, x x = ut or t = u u The vertical distance covered in time t, θ y=0+ Also

1 x2 1 2 gt = g × 2 2 2 u

[using (1)]

y

y gx 2 = tan θ or y = x tan θ ∴ = x tan θ x 2u 2

D θ x=ut

 gx  or x  2 – tan θ  = 0  2u  2u 2 tan θ g The distance of the point of strike from the point of projection is

As x = 0 is not possible, so x =

D=

x 2 + y2 =

=x

x2 + (x tan θ)2

1 + tan2 θ = x sec θ or D =

2u2 tan θ sec θ g

Ex.40 A ball rolls off the top of a stairway with a constant horizontal velocity u. If the steps are h metre high and w meter wide, show that the ball will just hit the edge of nth step if n = Sol.

Refer to figure. For n th step, net vertical displacement = nh net horizontal displacement = nω Let t be the time taken by the ball to reach the nth step. Then R = ut nω or nω = ut or t= u 1 2 Also, y = uy t + gt 2 2

or nh = 0 +

1  nω  1 2 gt = g  2  u  2

or n =

2hu 2 gw 2

u

1st 2nd h w

nth R

2hu2

gω2

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Page # 33

KINEMATICS 4.3

Projectile at an angle θ from height h Consider the projectile as shown in the adjacent figure. Take the point of projection as the origin the X and Y-axes as shown in figure.

u

Y θ

For X-axis, = u cosθ x ax = 0 vx = u cosθ, and x = u cos θ × t u

h

For Y-axis, x

uy = u sin θ, ay = –g,

gt 2 2 Ex.41 From the top of a tower 156.8 m high a projectile is projected with a velocity of 39.2 ms–1 in a direction making an angle 30° with horizontal. Find the distance from the foot of tower where it strikes the ground and time taken to do so. Sol. The situation is shown –1 u=39.2 ms Here height of tower OA = 156.8 m H u = 39.2 ms–1 uy = usinθ θ = 30° time for which projectile remain is air = t = ? θ=30° Horizontal distance covered R = OD = ? A Now ux = u cos θ and ux = u cosθ B  uy = u sin θ be the components of velocity u . Motion of projectile from O to H to D

vy = u sin θ – gt, and y = u sin θ t –

Using equation y = uyt +

1 ay t2 2

Here : y = 156.8 m ; uy = – u sinθ = 39.2 sin 30° ay = 9.8 m/s2 ; t = ? 156.8 = – 39.2 × 0.5 t + 4.9 t2 156.8 = – 19.6 t + 4.9 t2 or 4.9 t2 – 19.6 t – 156.8 = 0 (t – 8) (t + 4) = 0 or t2 – 4t – 32 = 0 ⇒ We get t = 8 s; t = – 4s t = – 4 s is not possible, thus we take t = 8s. Now horizontal distance covered in this time R = ux × t = u cos θ × t = 39.2 × cos 30° × t R = 271.57 m

4.4

156.8 m

O

C

D

Projectile Motion in Inclined Plane Here, two cases arise. One is up the plane and the other is down the plane. Let us discuss both the cases separately. (i) Up the Plane : In this case direction x is chosen up the plane and direction y is chosen perpendicular to the plane. Hence, ax = – g sin β ux = u cos α , uy = u sin α and ay = – g cos β

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Page # 34

KINEMATICS

y

x

B

u

gsinβ β

α

β

β

O

gcosβ

g

O

C

Now, let us derive the expressions for time of flight (T) and range (R) along the plane.

Time of flight At point B displacement along y-direction is zero. So, substituting the proper values in sy = uyt +

1 ay t2 , 2

we get 0

=

u

t

s i n

1 (– g cos β ) t2 2

α+

t = 0, corresponds to point O and t =

T=

t = 0 and

2u sin α g cos β

2u sin α corresponds to point B. Thus, g cos β

2u sin α g cos β

Range Range (R) or the distance OB is also equal to be displacement of projectile along x-direction in the t = T. Therefore. R = sa = uxT +

(ii)

1 axT2 2

R = u cos α T –

Down the inclined plane : along x - axis y-axis (1) ux = ucos α (1) uy = usinα (2) ax = g sin β (2) ay = g cos β velocity at P vy = uy + ay T vx = ux + axT Time of flight T =

1 sin β T2 2

y

u sin α (0,0)

2u y ay

2u sin α = g cos β

1 2 Range Sx = ux T + a x T 2 1 2 = u cos α T + g sin β. T 2

u α

uc os α

β os c g

β

gs P in β

g

β

x

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Page # 35

KINEMATICS

Ex.42 A particle is projected at an angle α with horizontal from the foot of a plane whose inclination to horizontal is β . Show that it will strike the plane at right angles if cotβ = 2 tan (α – β) Let u be the velocity of projection so that u cos (α – β ) and u sin (α – β ) are the initial velocities Sol. respectively parallel and perpendicular to the inclined plane. The acceleration in these two directions are (–g sin β ) and (–g cos β ). The initial component of velocity perpendicular to PQ is u sin (α – β ) and the acceleration in this direction is (–g cosβ ). If T is the time the particle takes to go from P to Q then in time T the space described in a direction perpendicular to PQ is zero. 0 = u sin (α – β ).T – T=

u

1 g cos β .T2 2

Q

2u sin(α – β) g cos β

If the direction of motion at the instant when the particle hits the plane be perpendicular to the plane, then the velocity at that instant parallel to the plane must be zero. ∴ u cos (α – β ) – g sin β T = 0

α β N

P

u cos(α – β) 2u sin(α – β) =T= g sin β g cos β

∴ cosβ = 2 tan (α – β )

Ex.43 Two inclined planes OA and OB having inclinations 30° and 60° with the horizontal respectively intersect each other at O, as shwon in u

figure. a particle is projected from point P with velocity u = 10 3 m / s along a direction perpendicular to plane OA. If the particle strikes A plane OB perpendicular of flight, then calculate.

h

(a) time of flight

(d) distance PQ. (Take g = 10 m/s2) Let us choose the x and y directions along OB and OA respectively. Then, ux = u = 10 3 m/s, uy = 0 ax = – g sin 60° = – 5 3 m/s2 and ay = – g cos 60° = – 5 m/s2 (a) At point Q, x-component of velocity is zero. Hence, substituting in vx= ux + axt 10 3

= 2s Ans. 5 3 (b) At point Q, v = vy = uy + ayt ∴ v = 0 – (5) (2) = –10 m/s Ans. Here, negative sign implies that velocity of particle at Q is along negative y direction. (c) Distance PO = |displacement of particle along y-direction| = |sy| Here, sy = uyt +

t=

1 1 ay t2 = 0 – (5)(2)2 = – 10 m 2 2

PO = 10 m

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B Q 60°

30°

(c) height h of point P from point O

0 = 10 3 – 5 3t ⇒

v

P

(b) velocity with which the particle strikes the plane OB,

Sol.

x

y

O

Page # 36

KINEMATICS

1 Therefore, h = PO sin 30° = (10)   2

or

h = 5m Ans.

(d) Distance OQ = displaement of particle along x-direction = sx Here, sx = uxt + or

1 1 ax t2 = (10 3)(2) – (5 3)(2)2 = 10 3 m 2 2

OQ = 10 3 m PQ =

(PO)2 + (OQ)2

PQ = 20 m

=

(10)2 + (10 3)2 =

100 + 300 = 400

Ans.

Now you can try Questions 45 to 68 in Exercise I and Ques. 12 to 20 in Ex.II

5.

RELATIVE MOTION The word 'relative' is a very general term, which can be applied to physical, nonphysical, scalar or vector quantities. For example, my height is five feet and six inches while my wife's height is five feet and four inches. If I ask you how high I am relative to my wife, your answer will be two inches. What you did? You simply subtracted my wife's height from my height. The same concept is applied everywhere, whether it is a relative velocity, relative acceleration or anything else. So, from the above discussion → we may now conclude that relative velocity of A with respect of B (written as v AB ) is → → → v AB = v A – v B

Similarly, relative acceleration of A with respect of B is → → → a AB = a A – a B

If it is a one dimensional motion we can treat the vectors as scalars just by assigning the positive sign to one direction and negative to the other. So, in case of a one dimensional motion the above equations can be written as vAB = vA – vB and aAB = aA – aB Further, we can see that → → → → v AB = – v BA or a BA = – a AB

Ex.44 Seeta is moving due east with a velocity of 1 m/s and Geeta is moving the due west with a velocity of 2 m/s. What is the velocity of Seeta with respect to Geeta? Sol. It is a one dimensional motion. So, let us choose the east direction as positive and the west as negative. Now, given that vs = velocity of Seeta = 1 m/s and vG = velocity of Geeta = – 2m/s Thus, vSG = velocity of Seeta with respect to Geeta = vS – vG = 1 – (–2) = 3 m/s Hence, velocity of Seeta with respect to Geeta is 3 m/s due east.

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Page # 37

KINEMATICS

IMPORTANT NOTE : PROCEDURE TO SOLVE THE VECTOR EQUATION.    ...(1) A =B+C (a)

Their are 6 variables in this equation which are following :  (1) Magnitude of A and its direction  (2) Magnitude of B and its direction  (3) Magnitude of C and its direction.

(b)

We can solve this equation if we know the value of 4 varibales [Note : two of them must be directions]

(c)

If we know the two direction of any two vectors then we will put them on the same side and other on the different side.

For example    If we know the directions of A and B and C' s direction is unknown then we make equation as follows :    C = A –B

(d)

Then we make vector diagram according to the equation and resolve the vectors to know the unknown values.

Ex.45 Car A has an acceleration of 2 m/s2 due east and car B, 4 m/s2 due north. What is the acceleration of car B with respect to car A? N Sol. It is a two dimensional motion. Therefore, → a BA = acceleration of car B with respect to car A E W → → = aB = – a A → Here, a B = acceleration of car S B = 4 m/s2 (due north) → and a A = acceleration of car A = 2 m/s2 (due east) → → → a B = 4m / s 2 a BA | a BA |= (4)2 + (2)2 = 2 5m / s2 4 α = tan–1   = tan–1(2) 2 → Thus, a BA is 2 5 m/s2 at an angle of α = tan–1(2) from west towards north.

and

α → – a A = 2m / s 2

Ex.46 Three particle A, B and C situated at the vertices of an equilateral triangle starts moving simultaneously at a constant speed "v" in the direction of adjacent particle, which falls ahead in the anti-clockwise direction. If "a" be the side of the triangle, then find the time when they meet. Sol.

Here, particle "A" follows "B", "B" follows "C" and "C" follows "A". The direction of motion of each particle keeps changing as motion of each particle is always directed towards other particle. The situation after a time "t" is shown in the figure with a possible outline of path followed by the particles before they meet.

A

O

B

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C

Page # 38

KINEMATICS

This problem appears to be complex as the path of motion is difficult to be defined. But, it has a simple solution in component analysis. Let us consider the pair "A" and "B". The initial component of velocities in the direction of line v joining the initial position of the two particles is "v" and "vcosθ" as shown in the figure here : The component velocities are directed towards eachother. v cos θ Now, considering the linear (one dimensional) motion in the 60° B direction of AB, the relative velocity of "A" with respect to v "B" is : vAB = vA – vB vAB = v – (– v cos θ) = v + vcosθ In equilateral triangle, θ = 60° v 3v vAB = v + vcos60° = v + = 2 2 The time taken to cover the displacement "a" i.e. the side of the triangle 2a t= 3v

A

O

v C

QUESTIONS BASED ON RELATIVE MOTION ARE USUALLY OF FOLLOWING FOUR TYPES : (a) Minimum distance between two bodies in motion (b) River-boat problems (c) Aircraft-wind problems (d) Rain problems

(a)

Minimum distance between two bodies in motion When two bodies are in motion, the questions like, the minimum distance between them or the time when one body overtakes the other can be solved easily by the principle of relative motion. In these type of problems one body is assumed to be at rest and the relative motion of the other body is considered. By assuming so two body problem is converted into one body problem and the solution becomes easy. Following example will illustrate the statement.

Ex.47 Car A and car B start moving simultaneously in the same direction along the line joining them. Car A with a constant acceleration a = 4 m/s2, while car B moves with a constant velocity v = 1 m/s. At time t = 0, car A is 10 m behind car B. Find the time when car A overtakes car B. Sol. Given : uA = 0, uB = 1 m/s, aA = 4m/s2 and aB = 0 Assuming car B to be at rest, we have uAB = uA – uB = 0 – 1 = – 1 m/s aAB = aA – aB = 4 – 0 = 4 m/s2 Now, the problem can be assumed in simplified form as follow : 2 2 a=4m/s v=1m/s A

10m

B

+ve

Substituting the proper values in equation 2

uAB= –1m/s, aAB= 4m/s A

10m

B

At rest

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Page # 39

KINEMATICS

s = ut + we get 10 = – t +

1 2 at 2

1 (4)(t2 ) 2

or

2t2 – t – 10 = 0

1±9 1 ± 1 + 80 1 ± 81 = = or t = 2.5 s 4 4 4 Ignoring the negative value, the desired time is 2.5s. Ans.

or

t=

and – 2 s

Note : The above problem can also be solved without using the concept of relative motion as under. At the time when A overtakes B, sA = sB + 10 1 × 4 × t 2 = 1 × t + 10 ∴ 2 or 2t2 – t – 10 = 0 Which on solving gives t = 2.5 s and – 2 s, the same as we found above. As per my opinion, this approach (by taking absolute values) is more suitable in case of two body problem in one dimensional motion. Let us see one more example in support of it. Ex.48 An open lift is moving upwards with velocity 10m/s. It has an upward acceleration of 2m/s2. A ball is projected upwards with velocity 20 m/s relative to ground. Find : (a) time when ball again meets the lift. (b) displacement of lift and ball at that instant. (c) distance travelled by the ball upto that instant. Take g = 10 m/s2 Sol. (a) At the time when ball again meets the lift, sL = sB 1 1 × 2 × t2 = 20 t – × 10t2 2 2 Solving this equation, we get

10t +

t=0

t=

and

2m/s2

20m/s

Ball

+ve

2

10m/s

5 s 3

Ball will again meet the lift after

10m/s

L

Lift

B

Ball

5 s. 3

(b) At this instant 2

sL = sB = 10 ×

5 1 175 5 + ×2×  = m = 19.4 m 3 2 3 9  

(c) For the ball u ↑ ↓a . Therefore, we will first find t0, the time when its velocity becomes zero. t0 =

u 20 = = 2s a 10

 5  As t  = s  < t0 , distance and displacement are equal  3 

or d = 19.4 m Ans. Concept of relative motion is more useful in two body problem in two (or three) dimensional motion. This can be understood by the following example.

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Page # 40

KINEMATICS

Ex.49 Two ships A and B are 10 km apart on a line running south to north. Ship A farther north is streaming west at 20 km/h and ship B is streaming north at 20km/h. What is their distance of closest approach and how long do they take to reach it ? Sol. Ships A and B are moving with same speed 20 km/h in N the directions shown in figure. It is a two dimensional, vA A two body problem with zero acceleration. Let us find E  vBA vB    vBA = vB − v A B  AB=10km Here, | vBA |= (20)2 + (20)2 = 20 2 km / h  i.e., vBA is 20 2 km / h at an angle of 45º from east towards north. Thus, the given problem can be simplified as :

45º

 A is at rest and B is moving with vBA in the direction shown in figure. Therefore, the minimum distance between the two is

C

vBA

45º

smin = AC = AB sin 45º

 1  = 10   km  2

A

B = 5 2 km

Ans.

and the desired time is BC 5 2 t=  = | vBA | 20 2

(BC = AC = 5 2 km )

=

(B)

1 h = 15 min 4

Ans.

River - Boat Problems In river-boat problems we come across the following three terms : B B

→ v br

W

θ

θ

A → vr

and •

v br cos θ

y x

vbr sinθ A

vr → v r = absolute velocity of river → v br = velocity of boatman with respect to river or velocity of boatman is still water → = absolute velocity of boatman. vb → → Here, it is important to note that v br is the velocity of boatman with which he steers and v b is the actual velocity of boatman relative to ground. → → → Further, v b = v br + v r Now, let us derive some standard results and their special cases.

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Page # 41

KINEMATICS

→ A boatman starts from point A on one bank of a river with velocity v br in the direction shown in fig. → River is flowing along positive x-direction with velocity v r . Width of the river is w, then → → → v b = v br + v r Therefore, vbx = vrx + vbrx = vr – vbr sinθ and vby = vry + vbry = 0 + vbr cosθ = vbr cosθ Now, time taken by the boatman to cross the river is : w w = v by v br cos θ w or t = v cos θ ...(i) br Further, displacement along x-axis when he reaches on the other bank (also called drift) is : w x = vbx t = (vr – vbr sin θ) v cos θ br w or x = (vr – vbr sinθ) v cos θ ...(ii) br Three special are :

t=

(i)

Condition when the boatman crosses the river in shortest interval of time B From Eq.(i) we can see that time (t) will be minimum when θ = 0°, i.e., the boatman should steer his boat perpendicular to the river → vbr current. w Also, tmin = v as cos θ = 1 A → br vr

(ii)

Condition when the boatman wants to reach point B, i.e., at a point just opposite from where he started In this case, the drift (x) should be zero. B ∴ x=0 or or or

w (vr – vbr sinθ) v cos θ = 0 br

vr = vbr sin θ vr sinθ = v br

–1  v  or θ = sin  r   v br 

→ v br

θ A

→ vr

–1  v  Hence, to reach point B the boatman should row at an angle θ = sin  r  upstream from AB.  v br 

Further, since sinθ not greater than 1. So, if vr ≥ vbr, the boatman can never reach at point B. Because if vr = vbr, sin θ = 1 or θ = 90° and it is just impossible to reach at B if θ = 90°. Moreover it can be seen that vb = 0 if vr = vbr and θ = 90°. Similarly, if vr > vbr, sinθ > 1, i.e., no such angle exists. Practically it can be realized in this manner that it is not possible to reach at B if river velocity (vr) is too high.

(iii)

Shortest path Path length travelled by the boatman when he reaches the opposite shore is s=

w 2 + x2 Here, w = width of river is constant. So for s to be minimum modulus of x (drift) should be minimum. Now two cases are possible.

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Page # 42

KINEMATICS

When vr < vbr : In this case x = 0, –1  v r  when θ = sin  v   br 

or

–1  v r   at θ = sin   v br 

smin = w

When vr > vbr : In this case x is minimum, where or

d dθ

dx =0 dθ

  w (vr – vbr sin θ) = 0   vbr cos θ 

or or

–vbr cos2θ – (vr – vbr sinθ) (– sinθ) = 0 – vbr + vr sinθ = 0  vbr   or θ = sin–1   vr  Now, at this angle we can find xmin and then smin which comes out to be

 vr smin = w   vbr

  at 

–1  vbr   θ = sin   vr 

Ex.50 A man can row a boat with 4 km/h in still water. If he is crossing a river where the current is 2 km/h. (a) In what direction will his boat be headed, if he wants to reach a point on the other bank, directly opposite to starting point? (b) If width of the river is 4 km, how long will the man take to cross the river, with the condition in part (a)? (c) In what direction should he head the boat if he wants to cross the river in shortest time and what is this minimum time? (d) How long will it take him to row 2 km up the stream and then back to his starting point ? Sol. (a) Given, that vbr = 4 km/h and vr = 2 km/h  vr ∴ θ = sin–1  v  br

 2 1  = sin–1   = sin–1   = 30° 4   2 

Hence, to reach the point directly opposite to starting point he should head the boat at an angle of 30° with AB or 90° + 30° = 120° with the river flow. (b) Time taken by the boatman to cross the river w = width of river = 4 km vbr = 4 km/h and θ = 30°

t=

2 4 h = 4 cos 30° 3

Ans.

(c) For shortest time θ = 0° w 4 tmin = v cos 0° = = 1h 4 br Hence, he should head his boat perpendicular to the river current for crossing the river in shortest time and this shortest time is 1 h.

and

vbr–vr D (d)

t = tCD + tDC

or

t=

vbr+vr C

CD DC + v db – v r v br + v r

D

=

C

2 2 1 4 + = 1+ = h 4–2 4+2 3 3

Ans.

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Page # 43

KINEMATICS

30 °

(a) We have    vm,g = v m,r + v r,g

vm

,g

Ex.51 A man can swim at a speed of 3 km/h in still water. He wants to cross a 500 m wide river flowing at 2 kh/h. He keeps himself always at an angle of 120° with the river flow while swimming. (a) Find the time he takes to cross the river. (b) At what point on the opposite bank will he arrive ? Sol. The situation is shown in figure  vr,g = velocity of the river with respect to the ground Here Y B C  vm,r = velocity of the man with respect to the river  vm,g = velocity of the man with respect to the ground.

vm,r = 3km/h

...(i)

θ

Hence, the velocity with respect to the ground is along AC. Taking y-components in equation (i),

A

vr,g = 2km/h

 3 3 vm,g sinθ = 3 km/h cos 30° + 2 km/h cos 90° = km/h 2 Time taken to cross the river =

1/ 2km 1 displacement along the Y - axis = h = velocity along the Y - axis 3 3 / 2 km / h 3 3

(b) Taking x-components in equation (i),

 1 vm,g cos θ = –3km/h sin 30° + 2 km/h = km / h 2 Displacement along the X-axis as the man crosses the river = (velocity along the X-axis) (time)  1  1  1km  h = km  ×  =  2h  3 3  6 3

Ex.52 A boat moves relative to water with a velocity v and river is flowing with 2v. At what angle the boat shall move with the stream to have minimum drift? (A) 30° (B) 60° (C) 90° (D) 120° Sol. (D) Let boat move at angle θ to the normal as shown in figure then time to cross the river =

1 v cos θ

1 drift x = (2v – v sin θ) for x to be minimum v cos θ dx = 0 = 1 (2 sec θ tan θ – sec2θ) or sin θ = 1/2 dθ or θ = 30° and φ = 90 + 30 = 120°

(C)

ub = u

u sinθ

ucosθ I = width of river

ur=2v

Aircraft Wind Problems → → This is similar to river boat problem. The only difference is that v br is replaced by v aw (velocity of → → aircraft with respect to wind or velocity of aircraft in still air), v r is replaced by v w (velocity of wind) → → → → → and v b is replaced by v a (absolute velocity of aircraft). Further,, v a = v aw + v . The following w example will illustrate the theory.

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Page # 44

KINEMATICS  A

NOTE : SHORT - TRICK

  If their are two vectors A and B and their resultent   α make an anlge α with A and β with B . β A sinα then A sin α = β sin β  B sinβ Means component of A perpendicular to resultant is equal in  magnitude to the compopent of B also perpendicular to resultant.

  Ex.53 If two vectors A and B make angle 30° and 60°  with their resultent and B has magnitude equal to  10, then find magnitude of A . So B sin 60° = A sin 30° ⇒ 10 sin 60° = A sin 30°

   C = A +B

 B

 B 60°

Bsin60°

30°

 A

A sin 30°

A = 10 3

Ex.54 An aircraft flies at 400 km/h in still air. A wind of 200 2 km/h is blowing from the south. The pilot wishes to travel from A to a point B north east of A. Find the direction he must steer and time of his journey if AB = 1000 km. Sol.

Given that vw = 200 2 km/h → → vaw = 400 km/h and v a should be along AB or in north-east direction. Thus, the direction of v aw → → should be such as the resultant of v w and v aw is along AB or in north - east direction. → N Let v aw makes an angle α with AB as shown in figure. B Applying sine law in triangle ABC, we get → AC BC v a 45° → = v w = 200 2km / h sin 45° sin α 45°  200 2  1 → C 1  BC  α v aw = 400 km / h   A =  sin 45° =  or sin α =   AC   400  2 2 E

∴ α = 30° Therefore, the pilot should steer in a direction at an angle of (45° + α) or 75° from north towards east. → 400 | v a| Further, = or sin 45° sin(180°–45°–30° )

sin 105° km → | v a | = sin 45° × (400) h

 cos 15°   0.9659  km km  (400)  (400) =  =   sin 45°  0 . 707 h h

= 546.47 km/h ∴ The time of journey from A to B is

(D)

AB 1000 h ⇒ t = 1.83 h t= → = 546.47 | v a| Rain Problems → → → In these type of problems we again come across three terms v r , vm and vrm , Here, → v r = velocity of rain

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Page # 45

KINEMATICS → vm = velocity of man (it may be velocity of cyclist or velocity of motorist also) and

→ vrm = velocity of rain with respect to man.

→ Here, v is the velocity of rain which appears to the man. Now, let us take one example of this. rm

Ex.55 A man standing on a road has to hold his umbrella at 30° with the vertical to keep the rain away. He throws the umbrella and starts running at 10 km/h. He finds that raindrops are hitting his head vertically. Find the speed of raindrops with respect to (a) the road, (b) the moving man. When the man is at rest with respect to the ground, the rain comes to him at an angle 30° with the Sol. vertical. This is the direction of the velocity of raindrops with respect to the ground. The situation when the man runs is shown in the figure

30°

vm,g

30°

v r,m

(b)

(a)

vr,g

 Here vr,g = velocity of the rain with respect to the ground   vm,g = velocity of the man with respect to the ground and vr,m = velocity of the rain with respect to the man. We have,

   vr,g = vr,m + vm,g

...(i)

Taking horizontal components, equation (i) gives vr,g sin30° = um,g = 10 km/h or, v,g =

10 km / h = 20km / h sin 30°

Taking vertical components, equation (i) gives vr,g cos30° = vr,m or, vr,m = (20 km/h)

3 = 10 √ 3 km/h. 2

Ex.56 To a man walking at the rate of 3 km/h the rain appears to fall vertically. When the increases his speed to 6 km/h it appears to meet him at an angle of 45° with vertical. Find the speed of rain. Sol.

Let i and j be the unit vectors in horizontal and vertical directions respectively.. Let velocity of rain Vertical ( j )

→ ˆ v r = aiˆ + bj Then speed of rain will be → | v r |=

a2 + b2

...(i)

Horizontal ( i )

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Page # 46

KINEMATICS

→ In the first case v m = velocity of man = 3 i → → → ∴ ˆ v rm = v r – v m = (a – 3)iˆ + bj It seems to be in vertical direction. Hence, a – 3 = 0 or a = 3 → In the second case v m = 6 i

→  ˆ = – 3 i + b j v rm = (a – 6)iˆ + bj

This seems to be at 45° with vertical. Hence, |b| = 3 Therefore, from Eq. (ii) speed of rain is

→ | v r |= (3)2 + (3)2 = 3 2 km / h Ans.

Relative Motion between Two Projectiles Let us now discuss the relative motion between two projectiles or the path observed by one projectile of the other. Suppose that two particles are projected from the ground with speeds u1 and u2 at angles α1 and α2 as shown in Fig.A and B. Acceleration of both the particles is g downwards. So, relative acceleration between them is zero because a12 = a1 – a2 = g – g = 0

Y

Y

u1

u2

α1

X

(A)

α2

X

(B)

i.e., the relative motion between the two particles is uniform. Now u2x = u2 cos α2 u1x = u1 cos α1, u1y = u1 sin α1 and u2y = u2 sin α2 Therefore, u12x = u1x – u2x = u1 cos α1– u2cos α2 and u12y = u1y – u2y = u1 sin α1– u2 sin α2 u12x and u12y are the x and y components of relative velocity of 1 with respect to 2. Hence, relative motion of 1 with respect to 2 is a straight

u  line at an angle θ = tan −1 12 y  with positive x-axis.  u12 x 

y

u12y u12 θ

u12x

a12=0

x

Now, if u12x = 0 or u1 cos α1 = u2 cos α2, the relative motion is along y-axis or in vertical direction (as θ = 90º). Similarly, if u12y = 0 or u1 sin α1 = u2 sin α2, the relative motion is along x-axis or in horizontal direction (as θ = 0º).

Note : Relative acceleration between two projectiles is zero. Relative motion between them is uniform. Therefore, condition of collision of two particles in air is that relative velocity of one with respect to  the other should be along line joining them, i.e., if two projecticles A and B collide in mid air, then VAB  should be along AB or VBA along BA.

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Page # 47

KINEMATICS

Condition for collision of two projectiles : Consider the situation shown in the figure. For projectiles to collide, direction of velocity of A with respect to B has to be along line AB. Here, vABx = u1 cos α1 + u2 cos α2

u2

vABy = u1 sin α1 – u2 sin α2

Y

Let, direction of velocity vector of A(wrt B) is making an angle β with +ve X-axis, which is given by tan β =

v ABy v ABx

=

h1

u1 sin α1 − u2 sin α2 u1 cos α1 + u2 cos α2

B

u1

h2

A

X

x

For collision to take place, h2 − h1 x Ex.57 A particle A is projected with an initial velocity of 60 m/s. at an angle 30º to the horizontal. At the same time a second particle B is projected in opposite direction with initial speed of 50 m/s from a point at a distance of 100 m from A. If the particles collide in air, find (a) the angle of projection α of particle B, (b) time when the collision takes place and (c) the distance of P from A, where collision occurs. (g = 10 m/s2) 60m/s 50m/s

tan β = tan θ =

A

30º 100m

Sol. (a) Taking x and y directions as shown in figure.   Here, a = −gˆj , a = −gˆj A

B

Y

B

uAx = 60 cos 30º

= 30 3 m / s

X uAy = 60 sin 30º = 30 m/s u AB uBx = – 50 cos α and uBy = 50 sin α   Relative acceleration between the two is zero as a A = aB . Hence, the relative motion between the two  is uniform. It can be assumed that B is at rest and A is moving with u AB . Hence, the two particles will  collide, if u AB is along AB. This is possible only when

uAy = uBy i.e., component of relative velocity along y-axis should be zero. or 30 = 50 sin α ∴ = sin–1 (3/5) Ans. a

(b) Now,

 | u AB |= u Ax – uBx = (30 3 + 50 cosα)m/s =

4   30 3 + 50 ×  m / s = (30 3 + 40) m/s  5

Therefore, time of collision is

100 AB t=  = | u AB | 30 3 + 40

or

t = 1.09 s Ans.

(c)

Distance of point P from A where collision takes place is

s=

1   (u Ax t) 2 +  u Ay t – gt 2    2

2

=

1   ( 30 3 × 109 . ) 2 +  30 × 109 . – × 10 × 109 . × 109 .    2

2

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or s = 62.64 m Ans.

Page # 48

KINEMATICS y

Ex.58 Two projectile are projected simultaneously from a point on the A ground "O" and an elevated position "A" respectively as shonw in the figure. If collision occurs at the point of return of two projectiles on H the horizontal surface, then find the height of "A" above the ground and the angle at which the projectile "O" at the ground should be projected.

5m/s

10m/s

θ

O

C

x

Sol. There is no initial separation between two projectile is x-direction. For collision to occur, the relative motion in x-direction should be zero. In other words, the component velocities in x-direction should be equal to that two projetiles cover equal horizontal distance at any given time. Hence, uOx = uAx ⇒

u0cosθ = uA ⇒

cosθ =

uA 5 1 = = = cos60° uO 10 2

θ = 60°

We should ensure that collision does occur at the point of return. It means that by the time projectiles travel horizontal distances required, they should also cover vertical distances so that both projectile are at "C" at the same time. In the nutshell, their times of flight should be equal. For projectile from "O". 2uO sin θ T= g For projectile from "A",

 2H     g 

T= For projectile from "A",

2uo sin θ  2H  =   g  g  Squaring both sides and putting values, T=

⇒ H=

4u2O sin2 θ 4 × 102 sin2 60° ⇒ H= 2g 2 × 10 2

 3 = 15m H = 20   2    We have deliberately worked out this problem taking advantage of the fact that projectiles are colliding at the end of their flights and hence their times of flight should be equal. We can, however, proceed to analyze in typical manner, using concept of relative velocity. The initial separation between two projectiles in the vertical direction is "H". This separation is covered with the component of relative in vertical direction.

⇒ vOAy = uOy – uAy = u0 sin60° – 0 = 10 ×

3 = 5 3m/s 2

Now, time of flight of projectile from ground is : 2uO sin θ 2x10x sin 60° = = 3 T= g 10 Hence, the vertical displacement of projectile from "A" before collision is :

H = vOAy X T = 5 3 x 3 = 15 m/s

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Page # 49

KINEMATICS

Ex.59 Two projectiles are projected simultaneously from two towers as shwon in figure. If the projectiles collide in the air, then find the distance "s" between the towers. 10 m/s

B

10 2 m / s A

30m

45°

10m

Sol. We see here that projectiles are approaching both horizontally and vertically. Their movement in two component directions should be synchronized so that they are at the same position at a particular given time. For collision, the necessary requirement is that relative velocity and displacement should be in the same direction. It is given that collision does occur. It means that two projectiles should cover the displacement with relative velocity in each of the component directions. 10 m/s Y B In x-direction, vABx = uAx – uBx = 10 2 cos 45° – (–10) = 10 2

1 2

If "t" is time after which collision occurs, then ⇒ s = vAy – uBy

vABy = ucos45° – 0 = 10 2 ×

1 2

10 2 m / s

+ 10 = 20 m/s

= 10m / s

A

30m

45°

10m O

S

x

The initial vertical distance between points of projection is 30 – 10 = 20 m. This vertical distance is covered with component of relative velocity in vertical direction. Hence, time taken to collide, "t", is :

⇒t=

20 =2 10

Putting this value in the earlier equation for "s", we have : ⇒ s = 20t = 20x2 = 40 m

Now you can try all the questions related to relative motion.

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Page # 50

KINEMATICS

QUESTIONS FOR SHORT ANSWER  1. A vector a is turned without a change in its length  through a small angle dθ. What are | a| and ∆a?

Sol.

2. Does the speedometer of a car measure speed or velocity ? Explain Sol.

3. When a particle moves with constant velocity, its average velocity and its instantaneous velocity & speed are equal. Comment on this statement. Sol.

4. In a given time interval, is the total displacement of a particle equal to the product of the average velocity and the time interval, even when the velocity is not constant? Explain. Sol.

5. Can you have zero displacement and a non zero average velocity? Can you have a zero displacement and a non zero velocity? Illustrate your answer on a x-t graph. Sol.

6. At which point on its path a projectile has the smallest speed? Sol.

7. A person standing on the edge of a cliff at some height above the ground below throws one ball straight up with initial speed u and then throws another ball straight down with the same initial speed. Which ball, if either, has the larger speed when it hits the ground? Neglect air resistance. Sol.

8. An airplane on floor relief mission has to drop a sack of rice exactly in the center of a circle on the ground while flying at a predetermined height and speed. What is so difficult about that? Why doesn’t it just drop the sack when it is directly above the circle. Sol.

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Page # 51

KINEMATICS

9. Which of the following graphs cannot possibily represent one dimensional motion of a particle? x  l |v| t

12. Give an example from your own experience in which the velocity of an object is zero for just an instant of time, but its acceleration is not zero. Sol.

t

t

l - length of path

x - displacement Sol.

10. Can you suggest a suitable situation from observation around you for each of the following ? x x

t

t

x - displacement x

13. A ball is dropped from rest from the top of a building and strikes the ground with a speed vf. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is v0 = vf, the same speed with which the first ball will eventually strike the ground. Ignoring air resistance, decide whether the balls cross paths at half the height of the building above the halfway point, or below the halfway point. Give your reasoning. Sol.

t

Sol.

11. One of the following statements is incorrect. (a) The car traveled around the track at a constant velocity (b) The car traveled around the track at a constant speed. Which statement is incorrect and why ? Sol.

14. The muzzle velocity of a gun is the velocity of the bullet when it leaves the barrel. The muzzle velocity of one rifle with a short barrel is greater than the muzzle velocity of another rifle that has a longer barrel. In which rifle is the acceleration of the bullet larger? Explain your reasoning. Sol.

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Page # 52

KINEMATICS

15. On a riverboat cruise, a plastic bottle is accidentally dropped overboard. A passenger on the boat estimate that the boat pulls ahead of the bottle by 5 meters each second. Is it possible to conclude that the boat is moving at 5 m/s with respect to the shore? Account for your answer. Sol.

18. A child is playing on the floor of a recreational vehicle (RV) as it moves along the highway at a constant velocity. He has a toy cannon, which shoots a marble at a fixed angle and speed with respect to the floor. The connon can be aimed toward the front or the rear of the RV. Is the range towards the front the same as, less than, or greater than the range towards the rear? Answer this question (a) from the child’s point of view and (b) from the point of view of an observer standing still on the ground. Justify your answers. Sol.

16. A wrench is accidentally dropped from the top of the mast on a sailboat. Will the wrench hit at the same place on the deck whether the sailboat is at rest or moving with a constant velocity? Justify your answer. Sol. 19. Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Swimmer A swims perpendicular to the current and lands on the far shore downstream, because the current has swept him in that direction. Swimmer B swims upstream at an angle to the current and lands on the far shore directly opposite the starting point. Swimmer C swims downstream at an angle to the current in an attempt to take advantage of the current. Who crosses the river in the least time? Account for your answer. Sol. 17. Is the acceleration of a projectile equal to zero when it reaches the top of its trajectory? If not, why not? Sol.

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Page # 53

KINEMATICS

(Objective Problems)

Exercise - I 1. A hall has the dimensions 10m × 10m × 10 m. A fly starting at one corner ends up at a diagonally opposite corner. The magnitude of its displacement is nearly (A) 5 3 m

(B) 10 3 m

(C) 20 3 m

(D) 30 3 m

Sol.

3. A body covers first 1/3 part of its journey with a velocity of 2 m/s, next 1/3 part with a velocity of 3 m/s and rest of the journey with a velocity 6m/s. The average velocity of the body will be (A) 3 m/s

(B)

11 m/s 3

(C)

8 m/s 3

(D)

4 m/s 3

Sol.

2. A car travels from A to B at a speed of 20 km h–1m and returns at a speed of 30 km h–1. The average speed of the car for the whole journey is (A) 5 km h–1 (B) 24 km h–1 (C) 25 km h–1(D) 50 km h–1 Sol.

4. A car runs at constant speed on a circular track of radius 100 m taking 62.8 s on each lap. What is the average speed and average velocity on each complete lap ? (A) velocity 10 m/s speed 10 m/s (B) velocity zero, speed 10 m/s (C) velocity zero, speed zero (D) velocity 10 m/s, speed zero Sol.

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Page # 54

KINEMATICS

5. The displacement of a body is given by 2s = gt2 where g is a constant. The velocity of the body at any time t is (D) gt3/3 (A) gt (B) gt/2 (C) gt2/2 Sol.

Sol.

6. A particle has a velocity u towards east at t = 0. Its acceleration is towards west and is constant, Let xA and xB be the magnitude of displacements in the first 10 seconds and the next 10 seconds. (A) xA < xB (B) xA = xB (C) xA > xB (D) the information is insufficient to decide the relation of xA with xB. Sol.

8. A stone is dropped into a well in which the level of water is h below the top of the well. If v is velocity of sound, the time T after which the splash is heard is given by (A) T = 2h/v (C) T =

2h h + g 2v

(B) T =

2h h + g v

(D) T =

h 2h + 2g v

Sol.

7. A body starts from rest and is uniformly accelerated for 30 s. The distance travelled in the first 10s is x1, next 10 s is x2 and the last 10 s is x3. Then x1 : x2 : x3 is the same as (A) 1 : 2 : 4 (B) 1 : 2 : 5 (C) 1 : 3 : 5 (D) 1 : 3 : 9

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Page # 55

KINEMATICS

9. The co-ordinates of a moving particle at a time t, are given by, x = 5 sin 10 t, y = 5 cos 10 t. The speed of the particle is (A) 25 (B) 50 (C) 10 (D) None Sol.

11. A body of mass 1 kg is acted upon by a force  F = 2 sin 3 πt i + 3 cos 3 πt j find its position at t = 1 sec if at t = 0 it is at rest at origin. 3   3 (A)  2 ,   3 π 9π 2 

2   2 (B)  2 ,   3π 3π 2 

2   2 (C)  ,   3π 3π 2 

(D) none of these

Sol.

10. A body moves with velocity v = lnx m/s where x is its position. The net force acting on body is zero at . (C) x = em (D) x = 1 m (A) 0 m (B) x = e2m Sol.

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Page # 56

KINEMATICS

12. A force F = Be–Ct acts on a particle whose mass is m and whose velocity is 0 at t = 0. It’s terminal velocity (velocity after a long time) is : (A)

C mB

(B)

B mC

(C)

BC m

(D) –

Sol.

B mC

Sol.

13. A particle starts moving rectilinearly at time t = 0 such that its velocity ‘v’ changes with time ‘t’ according to the equation v = t2 – t where t is in seconds and v is in m/s. The time interval for which the particle retards is (A) t < 1/2 (B) 1/2 < t < 1 (C) t > 1 (D) t < 1/2 and t > 1 Sol. 15. A particle is projected vertically upwards from a point A on the ground. It takes t1 time to reach a point B but it still continues to move up. If it takes further t2 time to reach the ground from point B then height of point B from the ground is (A)

1 g( t1 + t 2 ) 2 2

(B) g t1 t2

(C)

1 g( t1 + t 2 ) 2 8

(D)

1 gt 1t 2 2

Sol.

14. A ball is thrown vertically down with velocity of 5m/s. With what velocity should another ball be thrown down after 2 seconds so that it can hit the 1st ball in 2 seconds (A) 40 m/s (B) 55 m/s (C) 15 m/s (D) 25 m/s

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Page # 57

KINEMATICS 16. Balls are thrown vertically upward in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is 5m, the number of balls thrown per minute will be (A) 40 (B) 50 (C) 60 (D) 120 Sol.

18. The displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point x D E

F

C t

(A) C Sol.

17. A disc arranged in a vertical plane has two groves of same length directed along the vertical chord AB and CD as shown in the fig. The same particles slide down along AB and CD. The ratio of the time tAB/tCD is A

Sol.

(D) F

v(m/s)

C

60º

20 10

B

(B) 1: 2

(C) E

19. The variation of velocity of a particle moving along straight line is shown in the figure. The distance travelled by the particle in 4 s is

D

(A) 1 : 2

(B) D

(C) 2 : 1

(D)

2 :1

1 2 3 4

(A) 25m Sol.

(B) 30m

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t(s)

(C) 55m

(D) 60m

Page # 58

KINEMATICS

20. The displacement time graphs of two particles A and B are straight lines making angles of respectively 30º and 60º with the time axis. If the velocity of A is vA vA and that of B is vB then the value of v is B (A) 1/2

(B) 1 / 3

3

(C)

22. Acceleration versus velocity graph of a particle moving in a straight line starting form rest is as shown in figure. The corresponding velocity-time graph would be a

(D) 1/3

Sol. v v

v

(A)

(B) t

t

v

21. If position time graph of a particle is sine curve as shown, what will be its velocity-time graph

v

(C)

(D) t

t

Sol.

x t

v

v (A)

(B)

t

t v

v (C)

(D)

t

t

Sol.

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Page # 59

KINEMATICS 23. A man moves in x - y plane along the path shown. At what point is his average velocity vector in the same direction as his instantaneous velocity vector. The man starts from point P. y

PB

C

Question No. 25 to 27 (3 questions) The x-t graph of a particle moving along a straight line is shown in figure x

D

parabola

A x

(A) A Sol.

(B) B

(C) C

0

(D) D

T

2T

25. The v-t graph of the particle is correctly shown by

v (A)

v

0

T

2T t

v

24. The acceleration of a particle which moves along the positive x-axis varies with its position as shown. If the velocity of the particle is 0.8 m/s at x = 0, the

(C)

0

T

0

(B)

2T t

v T

2T

t

(D)

0

T

2T

t

Sol.

velocity of the particle at x = 1.4 is (in m/s) 2

a (in m/s ) 0.4 0.2

O

(A) 1.6 Sol.

(B) 1.2

0.4 0.8

1.4 x (in m)

(C) 1.4

(D) none

26. The a-t graph of the particle is correctly shown by

a (A)

0

a 2T T

t

(B)

0

t

v

a (C)

0

t

(D)

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0

t

Page # 60

KINEMATICS

28. Choose the incorrect statement. The particle comes to rest at (A) t = 0 s (B) t = 5 s (C) t = 8 s (D) none of these Sol.

Sol.

27. The speed-time graph of the particle is correctly shown by speed

(A)

0

speed 2T t

T

speed

(C)

0

0

(B)

T

2T t

29. Identify the region in which the rate of change of  ∆v of the particle is maximum velocity ∆t (A) 0 to 2s Sol.

(B) 2 to 4s

(C) 4 to 6s (D) 6 to 8 s

speed 0

2T t (D)

T

T

2T t

Sol.

30. If the particle starts from the position x0 = –15 m, then its position at t = 2s will be (A) – 5m (B) 5m (C) 10 m (D) 15 m Sol.

Question No. 28 to 33 (6 questions) The figure shows a velocity-time graph of a particle moving along a straight line v(ms–1) 10 0

2

4

6

8 t(s)

–20

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KINEMATICS

31. The maximum of displacement of the particle is (A) 33.3 m (B) 23.3 m (C) 18.3 (D) zero Sol.

33. The correct displacement-time graph of the particle is shown as x (m)

x (m)

(A)

(B) 0 2 x (m)

4

6

0 2 x (m)

8 t(s)

(C)

4

6

8 t(s)

(D) 0 2 4

6

0 2 4 6 8 t(s)

8 t(s)

Sol.

34. The velocity-time graph of a body falling from rest under gravity and rebounding from a solid surface is represented by which of the following graphs ? V

(A)

V

t

(B)

V

(C) 32. The total distance travelled by the particle is (A) 66.6 m (B) 51.6 m (C) zero (D) 36.6 m Sol.

t V

t

(D)

Sol.

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t

Page # 62

KINEMATICS

35. Shown in the figure are the displacement time graph for two children going home from the school. Which of the following statements about their relative motion is true after both of them started moving ? Their relative velocity: X

37. A body A is thrown vertically upwards with such a velocity that it reaches a maximum height of h. Simultaneously another body B is dropped from height h. It strikes the ground and does not rebound. The velocity of A relative to B v/s time graph is best represented by : (upward direction is positive)

C1

VAB

VAB

(A)

C2

(B) t

O

t

VAB

VAB

T

(A) first increases and then decreases (B) first decreases and then increases (C) is zero (D) is non zero constant Sol.

36. Shown in the figure are the velocity time graphs of the two particles P1 and P2. Which of the following statements about their relative motion is true ? Theire relative velocity (A) is zero (B) is non-zero but constant (C) continuously decreases (D) continuously increases Sol.

t

v

(C)

t

(D)

t Sol.

P1 P2

O

t

T

38. An object A is moving with 10 m/s and B is moving with 5 m/s in the same direction of positive x-axis. A is 100 m behind B as shown. Find time taken by A to Meet B

10m/s

A

(A) 18 sec. Sol.

5m/s

B

100m (B) 16 sec. (C) 20 sec. (D) 17 sec.

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Page # 63

KINEMATICS 39. It takes one minute for a passenger standing on an escalator to reach the top. If the escalator does not move it takes him 3 minute to walk up. How long will it take for the passenger to arrive at the top if he walks up the moving escalator? (A) 30 sec (B) 45 sec (C) 40 sec (D) 35 sec Sol.

40. A body is thrown up in a lift with a velocity u relative to the lift and the time of flight is found to be t. The acceleration with which the lift is moving up is u – gt (A) t

2u – gt (B) t

u + gt (C) t

2u + gt (D) t

42. A point mass is projected, making an acute angle  with the horizontal. If angle between velocity v and  acceleration g is θ, then θ is given by (A) 0º < θ < 90º (C) θ = 90º Sol.

(B) θ = 90º (D) 0º < θ < 180º

43. The velocity at the maximum height of a projectile is half of its initial velocity u. Its range on the horizontal plane is 2u 2 (A) 3g

(B)

3 u2 2g

u2 (C) 3g

u2 (D) 2g

Sol.

Sol.

Question No. 44 to 46 A projectile is thrown with a velocity of 50 ms–1 at an angle of 53º with the horizontal 41. A ball is thrown upwards. It returns to ground describing a parabolic path. Which of the following remains constant ? (A) speed of the ball (B) kinetic energy of the ball (C) vertical component of velocity (D) horizontal component of velocity. Sol.

44. Choose the incorrect statement (A) It travels vertically with a velocity of 40 ms–1 (B) It travels horizontally with a velocity of 30 ms–1 (C) The minimum velocity of the projectile is 30 ms–1 (D) None of these Sol.

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KINEMATICS

45. Determine the instants at which the projectile is at the same height (A) t = 1s and t = 7s (B) t = 3s and t = 5s (C) t = 2s and t = 6s (D) all the above Sol.

48. A ball is thrown from a point on ground at some angle of projection. At the same time a bird starts from a point directly above this point of projection at a height h horizontally with speed u. Given that in its flight ball just touches the bird at one point. Find the distance on ground where ball strikes (A) 2u

h g

(B) u

2h g

(C) 2u

2h g

(D) u

h g

Sol.

46. The equation of the trajectory is given by (B) 180 y = x2 – 240x (A) 180y = 240 x – x2 2 (C) 180y = 135x – x (D) 180y = x2 – 135x Sol.

49.A ball is hit by a batsman at an angle of 37º as shown in figure. The man standing at P should run at what minimum velocity so that he catches the ball before it strikes the ground. Assume that height of man is negligible in comparison to maximum height of projectile.

47. A particle is projected from a horizontal plane (xz plane) such that its velocity vector at time t is  given by V = aˆi + (b – ct )ˆj . Its range on the horizontal plane is given by ba 2ba (A) (B) c c Sol.

3 ba (C) c

(D) None

(A) 3 ms–1 Sol.

(B) 5 ms–1

(C) 9 ms–1

(D) 12 ms–1

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KINEMATICS

Question No. 53 & 54 (2 questions) At t = 0 a projectile is fired from a point O (taken as origin) on the ground with a speed of 50 m/s at an angle of 53° with the horizontal. It just passes two points A & B each at height 75 m above horizontal as shown.

50m/s

50. A projectile is fired with a speed u at an angle θ with the horizontal. Its speed when its direction of motion makes an angle ‘α’ with the horizontal is (A) u secθ cosα (B) u secθ sinα (C) u cosθ secα (D) u sinθ secα Sol.

53°

51. Two projectiles A and B are thrown with the same speed such that A makes angle θ with the horizontal and B makes angle θ with the vertical, then (A) Both must have same time of flight (B) Both must achieve same maximum height (C) A must have more horizontal range than B (D) Both may have same time of flight Sol.

52. Suppose a player hits several baseballs. Which baseball will be in the air for the longest time? (A) The one with the farthest range. (B) The one which reaches maximum height (C) The one with the greatest initial velocity (D) The one leaving the bat at 45° with respect to the ground. Sol.

B

A 75m

O 53. The horizontal separation between the points A and B is (A) 30 m (B) 60 m (C) 90 m (D) None Sol.

54 The distance (in metres) of the particle from origin at t = 2 sec. (A) 60 2 Sol.

(B) 100

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(C) 60

(D) 120

Page # 66

KINEMATICS

55. Particle is dropped from the height of 20 m from horizontal ground. There is wind blowing due to which horizontal acceleration of the particle becomes 6 ms–2. Find the horizontal displacement of the particle till it reaches ground. (A) 6 m (B) 10 m (C) 12 m (D) 24 m Sol.

56. A ball is projected from top of a tower with a velocity of 5 m/s at an angle of 53º to horizontal. Its speed when it is at a height of 0.45 m from the point of projection is (A) 2 m/s (B) 3 m/s (C) 4 m/s (D) data insufficient Sol.

58. One stone is projected horizontally from a 20 m high cliff with an initial speed of 10 ms–1. A second stone is simultaneously dropped from that cliff. Which of the following is true ? (A) Both strike the ground with the same velocity (B) The ball with initial speed 10ms–1 reaches the ground first (C) Both the balls hit the ground at the same time (D) One cannot say without knowing the height of the building Sol.

59. An aeroplane flying at a constant velocity releases a bomb. As the bomb drops down from the aeroplane. (A) it will always be vertically below the aeroplane (B) it will always be vertically below the aeroplane only if the aeroplane is flying horizontally (C) it will always be vertically below the aeroplane only if the aeroplane is flying at an angle of 45° to the horizontal. (D) it will gradually fall behind the aeroplane if the aeroplane is flying horizontally Sol.

57. Find time of flight of projectile thrown horizontally with speed 10 ms–1 from a long inclined plane which makes an angle of θ = 45º from horizontal. (A)

2 sec

(B) 2 2 sec

(C) 2 sec

(D) none

Sol.

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Page # 67

KINEMATICS 60. A particle is projected at angle 37º with the incline plane in upward direction with speed 10 m/s. The angle of incline plane is given 53º. Then the maximum height above the incline plane attained by the particle will be (A) 3m (B) 4m (C) 5m (D) zero Sol.

Sol.

63. If time taken by the projectile to reach Q is T, than PQ = v 90° P

θ

61. On an inclined plane of inclination 30°, a ball is thrown at an angle of 60° with the horizontal from the foot of the incline with velocity of 10 3 ms–1. If g = 10 ms–2, then the time in which ball with hit the inclined plane is (A) 1.15 sec. (B) 6 sec (C) 2 sec (D) 0.92 sec Sol.

62. A projectile is fired with a velocity at right angle to the slope which is inclined at an angle θ with the horizontal. The expression for the range R along the incline is (A)

2v 2 sec θ g

(B)

2v 2 tan θ g

(C)

2v 2 tan θ sec θ g

(D)

v2 tan2 θ g

Q

(A) Tvsinθ Sol.

(B) Tvcosθ

(C) Tv secθ (D) Tv tanθ

Question No. 64 to 67 (4 questions) Two projectiles are thrown simultaneously in the same plane from the same point. If their velocities are v1 and v2 at angles θ1 and θ2 respectively from the horizontal, then answer the following questions. 64. The trajectory of particle 1 with respect to particle 2 will be (A) a parabola (B) a straight line (C) a vertical straight line (D) a horizontal straight line

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Page # 68

KINEMATICS Sol.

Sol.

65. If v1 cosθ1 = v2 cosθ2, then choose the incorrect statement (A) one particle will remain exactly below or above the other particle (B) the trajectory of one with respect to other will be a vertical straight line (C) both will have the same range (D) none of these Sol. 68. A helicopter is flying south with a speed of 50 kmh–1. A train is moving with the same speed towards east. The relative velocity of the helicopter as seen by the passengers in the train will be towards. (A) north east (B) south east (C) north west (D) south west Sol.

66. If v1sinθ1 = v2sinθ2, then choose the incorrect statement (A) the time of flight of both the particles will be same (B) the maximum height attained by the particles will be same (C) the trajectory of one with respect to another will be a horizontal straight line (D) none of these Sol.

69. Two particles are moving with velocities v1 and v2. Their relative velocity is the maximum, when the angle between their velocities is (A) zero (B) π/4 (C) π/2 (D) π Sol.

67. If v1 = v2 and θ1 > θ2, then choose the incorrect statement (A) Particle 2 moves under the particle 1 (B) The slope of the trajectory of particle 2 with respect to 1 is always positive (C) Both the particle will have the same range if θ1 > 45° and θ2 < 45° and θ1 + θ2 = 90° (D) none of these

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Page # 69

KINEMATICS 70. A ship is travelling due east at 10 km/h. A ship heading 30º east of north is always due north from the first ship. The speed of the second ship in km/h is (A) 20 2 (B) 20 3 / 2 (C) 20 (D) 20 / 2 Sol.

73. A swimmer’s speed in the direction of flow of river is 16 km h–1. Against the direction of flow of river, the swimmer’s speed is 8 km h–1. Calculate the swimmer’s speed in still water and the velocity of flow of the river. (A) 12 km/h, 4 km/h (B) 10 km/h, 3 km/h (C) 10 km/h, 4 km/h (D) 12 km/h, 2 km/h Sol.

71. A particle is kept at rest at origin. Another particle starts from (5, 0) with a velocity of – 4 ˆi + 3 ˆj . Find their closest distance of approach. (A) 3 m (B) 4 m (C) 5 m (D) 2 m Sol.

74. A pipe which can rotate in a vertical plane is mounted on a cart. The cart moves uniformly along a horizontal path with a speed v1 = 2 m/s. At what angle α to the horizontal should the pipe be placed so that drops of rain falling with a velocity v2 6 m/s move parallel to the walls of the pipe without touching them ? consider the velocity of the drops as constant due to the resistance of the air.

v1

72. Four particles situated at the corners of a square of side ‘a’ move at a constant speed v. Each particle maintains a direction towards the next particle in succession. Calculate the time particles will take to meet each other. (A) Sol.

a v

(B)

a 2v

(C)

a 3v

(D)

2a 3v

(A) tan –1( 3)

–1  1  (B) tan   3

–1  1  (C) tan   2

(D) None of these

Sol.

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Page # 70

KINEMATICS 77. A flag is mounted on a car moving due North with velocity of 20 km/hr. Strong winds are blowing due East with velocity of 20 km/hr. The flag will point it direction (A) East (B) North-East (C) South-East (D) South-West Sol.

75. A swimmer swims in still water at a speed = 5 km/ hr. He enters a 200 m wide river, having river flow speed = 4 km/hr at point A and proceeds to swim at an angle of 127° (sin37° = 0.6) with the river flow direction. Another point B is located directly across A on the other side. The swimmer lands on the other bank at a point C, from which he walks the distance CB with a speed = 3 km/hr. The total time in which he reachrs from A to B is (A) 5 minutes (B) 4 minutes (C) 3 minutes (D) None Sol.

76. A boat having a speed of 5 km/hr. in still water, crosses a river of width 1 km along the shortest possible path in 15 minutes. The speed of the river in Km/hr. (A) 1

(B) 3

(C) 4

(D)

41

78. A man is crossing a river flowing with velocity of 5 m/s. He reaches a point directly across at a distance of 60 m in 5 sec. His velocity in still water should be (A) 12 m/s (B) 13 m/s (C) 5 m/s (D) 10 m/s Sol.

79. Wind is blowing in the north direction at speed of 2 m/s which causes the rain to fall at some angle with the vertical. With what velocity should a cyclist drive so that the rain appears vertical to him (A) 2 m/s south (B) 2 m/s north (C) 4 m/s west (D) 4 m/s south

Sol.

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Page # 71

KINEMATICS

Exercise - II

(One or more than one option is correct)

1. The displacement x of a particle depend on time t as x = αt2 – β t3 (A) particle will return to its starting point after time α/β. 2α (B) the particle will come to rest after time 3β (C) the initial velocity of the particle was zero but its initial acceleration was not zero. α (D) no net force act on the particle at time 3β Sol.

Sol.

3. Mark the correct statements for a particle going on a straight line (A) if the veloci ty is zero at any instant, the acceleration should also be zero at that instant (B) if the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval (C) if the velocity and acceleration have opposite sign, the object is slowing down (D) if the position and velocity have opposite sign, the particle is moving towards the origin. Sol.

2. A particle has intial velocity 10 m/s. It moves due to constant retarding force along the line of velocity which produces a retardation of 5 m/s2. Then (A) the maximum displacement in the direction of initial velocity is 10 m (B) the distance travelled in first 3 seconds is 7.5 m (C) the distance travelled in first 3 seconds is 12.5 m (D) the distance travelled in first 3 seconds is 17.5 m

4. A particle initially at rest is subjected to two forces. One is constant, the other is a retarding force proportion at to the particle velocity. In the subsequent motion of the particle. (A) the acceleration will increase from zero to a constant value (B) the acceleration will decrease from its initial value to zero (C) the velocity will increase from zero to maximum & then decrease (D) the velocity will increase from zero to a constant value.

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Page # 72

KINEMATICS

Sol.

Sol.

  5. Let v and a denote the velocity and acceleration respectively of a body in one-dimensional motion   (A) | v| must decrease when a < 0  (B) Speed must increase when a > 0   (C) Speed will increase when both v and a are < 0   (D) Speed will decrease when v < 0 and a > 0 Sol.

6. Which of the following statements are true for a moving body? (A) If its speed changes, its velocity must change and it must have some acceleration (B) If its velocity changes, its speed must change and it must have some acceleration (C) If its velocity changes, its speed may or may not change, and it must have some acceleration (D) If its speed changes but direction of motion does not changes, its velocity may remain constant

7. Let v and a denote the velocity and acceleration respectively of a body (A) a can be non zero when v = 0 (B) a must be zero when v = 0 (C) a may be zero when v ≠ 0 (D) The direction of a must have some correlation with the direction of v Sol.

8. A bead is free to slide down a A sm ooth wi re ti g ht l y st ret ched θ between points A and B on a vertical R B circle. If the bead starts from rest at A, the highest point on the circle (A) its velocity v on arriving at B is proportional to cosθ (B) its velocity v on arriving B is proportional to tanθ (C) time to arrive at B is proportional to cosθ (D) time to arrive at B is independent of θ

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Page # 73

KINEMATICS

10. The figure shows the velocity (v) of a particle plotted against time (t)

Sol.

+v0

O –v0

v T

t

2T

(A) The particle changes its direction of motion at some point (B) The acceleration of the particle remains constant (C) The displacement of the particle is zero (D) The initial and final speeds of the particle are the same Sol.

9. Velocity-time graph for a car is semicircle as shown here. Which of the following is correct : v 1m/s 2 sec (A) Car must move in circular path (B) Acceleration of car is never zero (C) Mean speed of the particle is π/4 m/s. (D) The car makes a turn once during its motion Sol.

11. A block is thrown with a velocity of 2 ms–1 (relative to ground) on a belt, which is moving with velocity 4 ms–1 in opposite direction of the initial velocity of block. If the block stops slipping on the belt after 4 sec of the throwing then choose the correct statements(s) (A) Displacement with respect to ground is zero after 2.66 sec and magnitude of displacement with respect to ground is 12 m after 4 sec. (B) Magnitude of displacement with respect to ground in 4 sec is 4 m. (C) Magnitude of displacement with respect to belt in 4 sec is 12 m. (D) Displacement with respect to ground is zero in 8/ 3 sec. Sol.

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Page # 74

KINEMATICS 13. An observer moves with a constant speed along the line joining two stationary objects. He will observe that the two objects (A) have the same speed (B) have the same velocity (C) move in the same direction (D) move in opposite directions Sol.

12. A particle moves with constant speed v along a regular hexagon ABCDEF in the same order. Then the magnitude of the average velocity for its motion from A to (A) F is v/5 (B) D is v/3 (D) B is v (C) C is v √3/2 Sol.

14. A man on a rectilinearly moving cart, facing the direction of motion, throws a ball straight up with respect to himself (A) The ball will always return to him (B) The ball will never return to him (C) The ball will return to him if the cart moves with constant velocity (D) The ball will fall behind him if the cart moves with some acceleration Sol.

15. A projectile of mass 1 kg is projected with a velocity of

20 m/s such that it strikes on the same level as

the point of projection at a distance of 3 m. Which of the following options are incorrect. (A) the maximum height reached by the projectile can be 0.25 m (B) the minimum velocity during its motion can be 15 m/s

3 sec. 5 (D) maximum potential energy during its motion can be 6J.

(C) the time taken for the flight can be

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Page # 75

KINEMATICS

Sol.

17. If T is the total time of flight, h is the maximum height & R is the range for horizontal motion, the x & y co-ordinates of projectile motion and time t are related as : t  t (A) y = 4h   1 −   T  T

X  X  (B) y = 4h   1 −   R  R

T  T  (C) y = 4h   1 −   t t

R  R  (D) y = 4h   1 −   X  X

Sol.

16. Choose the correct alternative (s) (A) If the greatest height to which a man can throw a stone is h, then the greatest horizontal distance upto which he can throw the stone is 2h. (B) The angle of projection for a projectile motion whose range R is n times the maximum height is tan–1(4/n) (C) The time of flight T and the horizontal range R of a projectile are connected by the equation gT2 = 2Rtanθ where θ is the angle of projection. (D) A ball is thrown vertically up. Another ball is thrown at an angle θ with the vertical. Both of them remain in air for the same period of time. Then the ratio of heights attained by the two ball 1 : 1. Sol.

18. A particle moves in the xy plane with a constant acceleration ‘g’ in the negative y-direction. Its equation of motion is y = ax – bx2, where a and b constants. Which of the following are correct? (A) The x-component of its velocity is constant. (B) At the origin, the y-component of its velocity is g 2b (C) At the origin, its velocity makes an angle tan–1(a) with the x-axis (D) The particle moves exactly like a projectile. Sol. a

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Page # 76

KINEMATICS

20. A particle is projected from the ground with velocity u at angle θ with horizontal. The horizontal range, maximum height and time of flight are R, H and T respectively. They are given by, R=

2u sin θ u 2 sin 2θ u 2 sin2 θ ,H= and T = g g 2g

Now keeping u as fixed, θ is varied from 30° to 60°. Then, (A) R will first increase then decrease, H will increase and T will decrease (B) R will first increase then decrease while H and T both will increase (C) R will decrease while H and T will increase (D) R will increase while H and T will increase Sol. 19. A ball is rolled off along the edge of a horizontal table with velocity 4 m/s. It hits the ground after time 0.4s. Which of the following are correct? (A) The height of the table is 0.8 m (B) It hits the ground at an angle of 60° with the vertical (C) It covers a horizontal distance 1.6 m from the table (D) It hits the ground with vertical velocity 4 m/s Sol.

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Page # 77

KINEMATICS

Exercise - III

(Subjective Problems)

1. The position vector of a particle moving in x-y  plane is given by r = ( t 2 − 4)i + ( t − 4)j . Find (a) Equation of trajectory of the particle Sol.

3. At time t the position vector of a particle of mass  m = 3kg is given by r = 6 t i − t 3 j + cos tk . Find the re sultant force F ( t) , magnitude of its acceleration when t=

(b) Time when it crosses x-axis and y-axis Sol.

π , & speed when t = π. 2

Sol.

2. A p arti cl e move s al ong the sp ac e curv e  r = ( t 2 + t) i + ( 3 t − 2) j + (2t 3 − 4 t 2 ) k . (t in sec, r in m) Find at time t = 2 the (a) velocity, (b) acceleration, (c) speed or magnitude of velocity and (d) magnitude of acceleration. Sol.

4. The velocity time graph of a body moving in a straight line is shown. Find its

velocity in m/sec

y

60°

30° time in sec 2

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x 2.5 sec

Page # 78

KINEMATICS

(a) instantaneous velocity at t = 1.5 sec Sol.

(b) average acceleration from t = 1.5 sec. to t = 2.5 sec Sol.

6. Velocity of car v is given by v = at – bt2, where a and b are positive constants & t is time elapsed. Find value of time for which velocity is maximum & also corresponding value of velocity. Sol.

(c) draw its acceleration time graph from t = 0 to t = 2.5 sec Sol.

5. The curvilinear motion of a particle is defined by vx = 50 – 16t and y = 100 – 4t2 , where vx is in metres per second, y is in metres and t is in seconds. It is also known that x = 0 at t = 0. Determine the velocity (v) and acceleration (a) when the position y = 0 is reached. Sol.

7. The force acting on a body moving in a straight line is given by F = (3t2 – 4t + 1) Newton where t is in sec. If mass of the body is 1kg and initially it was at rest at origin. Find (a) displacement between time t = 0 and t = 2 sec Sol.

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Page # 79

KINEMATICS

9. A particle is moving along x-axis. Initially it is located 5 m left of origin and it is moving away from the origin and slowing down. In this coordinate system, the signs of the initial velocity and acceleration, are + y

+ x

– (0, 0)

(b) distance travelled between time t = 0 and t = 2 sec Sol.

v0

a

Sol.

8. A particle goes from A to B with a speed of 40 km/ h and B to C with a speed of 60 km/h. If AB = 6BC the ave rage speed i n k m/h betwe en A and C i s ____________ [Hint : Average speed =

10. Find the change in velocity of the tip of the minute hand (radius = 10 cm) of a clock in 45 minutes. Sol.

total dis tan ce travelled ] time taken

Sol.

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Page # 80

KINEMATICS

11. At a distance L = 400 m from the traffic light brakes are applied to a locomotive moving at a velocity v= 54 km/hr. Determine the position of the locomotive relative to the traffic light 1 min after the application of the breaks if its acceleration is –0.3 m/sec2. Sol.

12. A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position (s) of the train at half the maximum speed. Sol.

13. A car is moving along a straight line. It is taken from rest to a velocity of 20 ms–1 by a constant acceleration of 5ms–2. It maintains a constant velocity of 20 ms–1 for 5 seconds and then is brought to rest again by a constant acceleration of –2 ms–2. Draw a velocity-time graph and find the distance covered by the car. Sol.

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Page # 81

KINEMATICS

14. A stone is dropped from a height h. Simultaneously another stone is thrown up from the ground with such a velocity that it can reach a height of 4 h. Find the time when two stones cross each other. Sol.

15. A bal loon is ascending vertical ly with an acceleration of 0.2 m/s2 Two stones are dropped from it at an interval of 2 sec. Find the distance between them 1.5 sec after the second stone is released (use g = 9.8 m/s2). Sol.

16. From the velocity-time plot shown in figure, find the distance travelled by the particle during the first 40 seconds. Also find the average velocity during this period. V 5m/s t(s) 0 20 40 –5m/s Sol.

17. The velocity-time graph of the particle moving along a straight line is shown. The rate of acceleration and deceleration is constant and it is equal to 5 ms–2. If the average velocity during the motion is 20ms–1, then find the value of t.

o

t

Sol.

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25 sec

Page # 82

KINEMATICS

18. The fig. shows the v-t graph of a particle moving in straight line. Find the time when particle returns to the starting point. v

20. A particle is projected upwards with a velocity of 100 m/sec at an angle of 60º with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g = 10 m/sec2. Sol.

20 10

10

20

25 t

Sol.

gx2 . The 2 angle of projectile is ________ and initial velocity is _______. Sol.

21. The equation of a projectile is y = 3 x −

19. A particle is projected in the X-Y plane. 2 sec after projection the velocity of the particle makes an angle 45º with the X-axis. 4 sec after projection, it moves horizontally. Find the velocity of projection (use g = 10 ms–2). Sol.

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Page # 83

KINEMATICS

22. A ball is projected at an angle of 30º above with the horizontal from the top of a tower and strikes the ground in 5 sec at an angle of 45º with the horizontal. Find the height of the tower and the speed with which it was projected. [g =10 m/s2] Sol.

23. A rocket is launched at an angle 53º to the horizontal with an initial speed of 100 ms–1. It moves along its initial line of motion with an acceleration of 30 ms–2 for 3 seconds. At this time its engine falls & the rocket proceeds like a free body. Find : (i) the maximum altitude reached by the rocket (ii) total time of flight (iii) the horizontal range. [sin 53º = 4/5] Sol.

24. A ball is thrown horizontally from a cliff such that it strikes ground after 5 sec. The line of sight from the point of projection to the point of hitting makes an angle of 37º with the horizontal. What is the initial velocity of projection. 37º

Sol.

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Page # 84

KINEMATICS

25. A ball is projected on smooth inclined plane in direction perpendicular to line of greatest slope with velocity of 8m/s. Find it’s speed after 1 sec. 8 m/s 37º

27. The horizontal range of a projectiles is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, find the horizontal range of the projectile. Sol.

Sol.

26. Find range of projectile on the inclined plane which is projected perpendicular to the incline plane with velocity 20m/s as shown in figure. -1

u = 20 ms

28. A butterfly is flying with velocity 10 i + 12j m / s and wind is blowing along x axis with velocity u. If butterfly starts motion from A and after some time reaches point B, find the value of u. y

37º

B

Sol. A

37° x

Sol.

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Page # 85

KINEMATICS

29. In the figure shown, the two projectiles are fired simultaneously. What should be the initial speed of the left side projectile for the two projectile to hit in mid-air ? u

Sol.

20m/s 60º 45º \\\\\\\\\\\\\\\\\\\\\\\\\\ 10m

30. In the figure shown, the two projectiles are fired simultaneously. Find the minimum distance between them during their flight?

31. Two particles are moving along two long straight lines, in the same plane, with the same speed = 20 cm/s. The angle between the two lines is 60°, and their intersection point is O. At a certain moment, the two particles are located at distance 3m and 4m from O, and are moving towards O. Find the shortest distance between them subsequently? Sol.

20 3 m / s 20 m/s

60°

30° 20 m

Sol.

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Page # 86

KINEMATICS

32. Rain is falling vertically with a speed of 20 ms–1 relative to air. A person is running in the rain with a velocity of 5 ms–1 and a wind is also blowing with a speed of 15 ms–1 (both towards east). Find the angle with the vertical at which the person should hold his umbrella so that he may not get drenched. Sol.

33. A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The car moves horizontally with a speed of 2 m/s. At what angle α with the vertical should the wind screen be placed so that the rain drops falling vertically downwards with velocity 6 m/s strike the wind screen perpendicularly? Sol.

34. A man with some passengers in his boat, starts perpendicular to flow of river 200m wide and flowing with 2m/s. Boat speed in still water is 4m/s. When he reaches half the width of river the passengers asked him they want to reach the just opposite end from where they have started. (a) Find the direction due to which he must row to reach the required end. (b) How many times more total time, it would take to that if he would have denied the passengers. Sol.

35. A man crosses a river in a boat. If he crosses the river in minimum time he takes 10 minutes with a drift 120 m. If he crosses the river taking shortest path, he takes 12.5 minute, find (i) width of the river (ii) velocity of the boat with respect to water (iii) speed of the current. Assume vb/r > vr Sol.

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Page # 87

KINEMATICS

Exercise - IV

(Tough Subjective Problems)

1. A speeder in an automobile passes a stationary policeman who is hiding behind a bill board with a motorcycle. After a 2.0 sec delay (reaction time) the policeman acceleraties to his maximum speed of 150 km/hr in 12 sec and catches the speeder 1.5 km beyond the billboard. Find the speed of speeder in km/hr. Sol.

2. A large number of bullets are fired in all direction with the same speed v. What is the maximum area on ground on which these bullets can spread? Sol.

3. The speed of a particle when it is at its greatest height is

2 / 5 times of its speed when it is at its half the maximum height. The angle of projection is _________ and the velocity vector angle at half the maximum height is _________. Sol.

4. A projectile is to be thrown horizontally from the top of a wall of height 1.7m. Calculate the initial velocity of projection if it hits perpendicularly an incline of angle 37° which starts from the ground at the bottom of the wall. The line of greatest slope of incline lies in the plane of motion of projectile. Sol.

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Page # 88

KINEMATICS

5. Two inclined planes OA and OB having inclination (with horizontal) 30° and 60° respectively, intersect each other at O as shown in figure. A particle is projected from point P with velocity u = 10 3 ms –1 along

6. A particle is thrown horizontally with relative velocity 10 m/s from an inclined plane, which is also moving with acceleration 10 m/s2 vertically upward. Find the time after which it lands on the plane (g = 10 m/s2)

a direction perpendicular to plane OA. If the particle strikes plane OB perpendicularly at Q, calculate A

h

u

2

10 m/s

B Q

P

30°

Sol.

60°

30° O

(a) velocity with which particle strikes the plane OB, (b) time of flight, (c) vertical height h of P from O, (d) maximum height from O attained by the particle and (e) distance PQ Sol.

7. A, B & C are three objects each moving with constant → velocity. A’s speed is 10 m/sec in a direction PQ . The velocity of B relative to A is 6 m/sec at an angle of, cos–1(15/24) to PQ. The velocity of C relative to B is → 12 m/sec in a direction QP , then find the magnitude of the velocity of C. Sol.

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Page # 89

KINEMATICS 8. A particle is projected from point P with velocity 5 2 m/s perpendicular to the surface of a hollow right angle cone whose axis is vertical. It collides at Q normally. Find the time of the flight of the particle. y

P

Q 45°

x

Sol.

10. A hunter is riding an elephant of height 4m moving in straight line with uniform speed of 2m/sec. A deer running with a speed V in front at a distance of 4 5 m moving perpendicular to the direction of motion of the elephant. If hunter can throw his spear with a speed of 10 m/sec. relative to the elephant, then at what angle θ to it’s direction of motion must he throw his spear horizontally for a successful hit. Find also the speed ‘V’ of the deer. Sol.

9. A glass wind screen whose inclination with the vertical can be changed, is mounted on a cart as shown in figure. The cart moves uniformly along the horizontal path with a speed of 6 m/s. At what maximum angle α to the vertical can the wind screen be placed so that the rain drops falling vertically downwards with velocity 2 m/s, do not enter the cart? α v=6m/s

Sol.

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Page # 90

KINEMATICS

Exercise - V

JEE-Problems

1. Two guns, situated at the top of a hill of height 10 m, fire one shot each with the same speed 5 3 m/s at some interval of time. One gun fires horizontally and other fires upwards at an angle of 60° with the horizontal. The shots collide in air at a point P. Find (a) the time interval between the firings, and (b) the coordinates of the point P. Take origin of the coordinates system at the foot of the hill right below the muzzle and trajectories in X-Y plane.[JEE’ 1996] Sol.

3. A large heavy box is sliding without friction down a smooth plane of inclination θ. From a point P on the bottom of a box, a particle is projected inside the box. The initial speed of the particle with respect to box is u and the direction of projection makes an angle α with the bottom as shown in figure.

P

α

Q

θ

2. The trajectory of a projectile in a vertical plane is y = ax – bx2, where a, b are constants & x and y are respectively the horizontal & vertical distances of the projectile from the point of projection. The maximum height attained is ___________ & the angle of projection from the horizontal is ______, [JEE’ 1997] Sol.

(a) Find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands. (Assume that the particle does not hit any other surface of the box. Neglect air resistance). (b) If the horizontal displacement of the particle as seen by an observer on the ground is zero, find the speed of the box with respect to the ground at the instant when the particle was projected. [JEE’ 1998] Sol.

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Page # 91

KINEMATICS

5. A ball is dropped vertically from a height d above the ground it hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistances, its velocity v varies with the height h above the ground as [JEE’ 2000 (Scr)] v

v

d

(A)

h

v

v d

(C)

h

d

(B)

d

h

(D)

h

Sol.

4. In 1.0 sec. a particle goes from point A to point B moving in a semicircle of radius 1.0 m. The magnitude of average velocity is [JEE ‘99] A 1m

(A) 3.14 m/sec (C) 1.0 m/sec Sol.

B

(B) 2.0 m/sec (D) zero

6. An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0 the plank starts moving along the +x direction with an acceleration 1.5 m/s2. At the same instant a stone is projected from the origin with a velocity u as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45° to the horizontal. All the motions are in x-y plane. Find u and the time after which the stone hits the object. Take g = 10 m/s2 [JEE 2000] y A 1.25m P

u O Sol.

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3.0 m x

Page # 92

KINEMATICS

7. On a frictionless horizontal surface, assumed to be the x-y plane, a small trolley A is moving along a straight line parallel to the y-axis (see figure) with a constant velocity of ( 3 – 1) m/s. At a particular instant, when the line OA makes an angle of 45° with the x-axis, a ball is thrown along the surface from the origin O. Its velocity makes an angle φ with the x-axis and it hits the trolley. y A

45° x O (a) The motion of the ball is observed from the frame of trolley. Calculate the angle θ made by the velocity vector of the ball with the x-axis in this frame. (b) Find the speed of the ball with respect to the 4θ surface, if φ = . [JEE 2002] 3 Sol.

8. A particle starts from rest. Its acceleration (a) varsus time (t) is as shown in the figure. The maximum speed of the particle will be - [JEE’ 2004 (Scr)] a 2

10m/s

11 t(s)

(A) 110 m/s Sol.

(B) 55 m/s (C) 550 m/s (D) 660 m/s

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Page # 93

KINEMATICS

9. A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance Sn travelled from time t = n – 1 to t = n. The S is n+1 [JEE’ 2004 (Scr)] (A)

2n – 1 2n

(B)

2n + 1 2n – 1

2n – 1 2n (D) 2n + 1 2n + 1

(C)

Sol.

10. The velocity displacement graph of a particle moving along a straight line is shown. The most suitable acceleration-displacement graph will be [JEE’ 2005 (Scr)]

11. STATEMENT-1 For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary. STATEMENT-2 If the observer and the object are moving at velocities   V1 and V2 respectively with reference to a laboratory frame, the velocity of the object with respect to the   observer is V2 – V1 (A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1 (B) STATEMENT-1 is True, STATEMENT-2 is True’ STATEMENT-2 is NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is True [JEE’ 2008] Sol.

v

v0

x0

a

a

(A)

x

x

(B) x

a (C)

Sol.

a

x (D)

x

12. A train is moving along a straight line with a constant acceleration 'a'. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60° to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train [JEE’ 2011] in m/s2 is Sol.

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Page # 94

KINEMATICS

  Magnitude | a| will remain uncharged. B = a

  ∆a = B − A

= a 2 + a 2 + 2a 2 cos( π − dθ)

= 2a2 (1 − cos θ)

⇒ 2a2 (1 − 1 + 2 sin 2dθ / 2) = 2a sin d θ/2

–a

2

Speedometer measure speed of car as it only gives the magnitude.

3

When particle is moving with constant velocity its average velocity and instantaneous velocity will be

4 5

same and magnitude of instantaneous velocity will also be same.    ∆S VAvg = , ∆S = VAvg × ∆t ∆t    ∆S  VAvg = with zero displacement non zero VAvg is not ∆ S = 0 ∆t

x

 possible zero displacement and non zero V is possible if particle

is reversing and coming to starting point. Show on x-t graph by an example.

t

6

Speed of projectile is smallest at the highest point.

7

Both the ball will hit the ground with same speed.

8

If sack of rice is dropped when it is just above the centre it will fall ahead of circle because sack will have velocity same as plane in horizontal direction.

9

Ist Curve : at particular time x has more than one value hence not a 1-D motion.  IInd Curve : | V| cannot be negative IIIrd Curve : Length of a moving body can not decrease with time

10

Ist Curve : A ball moving forward collides with surface rebounds and stops after IInd collision IInd Curve : A ball repeatedly making inelastic collisions with floor. IIIrd Curve : Collision of a ball with surface. {Surface has large velocity for short time}

11

(a) is incorrect car can not travel around track with constant velocity as direction is continuously changing. (b) correct

12 13

Ball at maximum height V = 0 for just an instant but acceleration due to gravity. 1 . Let balls meet after t sec. h1 Vf = 2gH 1 2 1 2 H X h1 = gt and h2 = Vf t = gt 2 2 h2 H V0=Vf H = 2 gH t t = 2 h1 + h2 = H = Vf t 2g 1 H H ∴ h1 = g = hence they will meet above half height of building. 2 2g 4

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Page # 95

KINEMATICS

14

Initially bullet is at rest u = 0 V2 = 0 + 2as ∴ a=

V2 muzzle velocity is more for short barrl and S is also less hence acceleration will be more in that case. 2S

Hence we can not conclude that velocity of boat is 5 m/sec w.r.t. shore

15

VBottle = Vriver 16

;

VB – VR = 5

Yes wrench will hit at the same place on the deck irrespective of that boat is at rest or moving because when boat is at rest wrench will have zero horizontally velocity and when boat is moving both will have same horizontal velocity.

17

Acceleration of the projectile remains constant throughout the journey = g

18

(a) In child point of view range will be same in both the cases. (b) In ground frame of reference VCT = VC – VT VC = VCT + VT

19

For front range Vcannon = VC cos θ + VT

Range will be more

For Rear range Vcannon = VC cos θ – VT

Range will be less

d t=

Vbr

d for tmin cos θ = 1 maximum Hence A will reach opposite end in least time Vbr cos θ

EXERCISE - I

ANSWER KEY 1. B 2. B 3. A 4. B 5. A 6. D 7. C 8. B 9. B 10.

D 11.

C 12.

B 13.

B 14.

A

15.

D 16. C 17.

B 18.

C 19.

C 20

D 21.

C 22.

D 23.

C 24.

B 25.

B

26.

D 27. C 28.

B 29.

C 30.

A 31.

A 32.

A 33.

C 34.

A 35.

D 36.

D

37.

C 38. C 39.

B 40.

B 41.

D 42.

D 43.

B 44.

A 45.

D 46.

A 47.

B

48.

C 49. B 50.

C 51.

D 52.

B 53.

B 54.

A 55.

C 56.

C 57.

C 58.

C

59.

A 60. A 61.

C 62.

C 63.

D 64.

B 65.

C 66.

D 67.

B 68.

D 69.

D

70.

C 71. A 72.

A 73.

A 74.

A 75.

B 76.

B 77.

C 78.

B 79.

B

EXERCISE - II

ANSWER KEY 1. A,B,C,D 2. A,C 10. A,B,C,D

3. B,C,D 4. B,D

11. B,C,D

5. C,D

6. A,C

7. A,C

8. A,D

12. A,C,D 13. A,B,C 14. C,D 15. D,C 16. A,B,C,D

19. A,C,D 20. B

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9. C

17. A,B 18. A,B,C,D

Page # 96

KINEMATICS

EXERCISE - III

1. (a) y2 + 8y + 12 = x ; (b) crosses x axis when t = 4 sec, crosses y axis when t = ± 2 sec. 3. –18 tj – 3 cos t k ; 3π ; 3 4 + π 4

2. (a) 5i + 3j + 8k, (b) 2i + 16k, (c) 7 2 , (d) 2 65

4. (a)

7. (a)

1 3

m / s , (b)

3 m / s 2 , (c) 2

  5. v = –30 i – 40 j, a = –16 i – 8 j 6. a/2b, a2/4b

38 2 m , (b) m 8. 42 km/hr 3 3

9.

v0

a

+

π 2 10.  3  cm/min 11. 25 m  

vel

12. (a) 2.7 km; (b) 60 m/s; (c) 225 m and 2.25 km 13. 240 m 14.

16. 100 m, zero 17. 5 s 18. 36.2 sec. 19. 20 5 22. u = 50 ( 3 – 1) m/sec., H = 125 (– 24. 100/3 m/s

30. 10 m

25. 10 m/s

31. 50 3 cm

 h    8g 

20. 20 sec

15. 50 m

21. 60, 2 m/sec.

3 + 2)m 23. (i) 1503.2 m (ii) 35.54 sec (iii) 3970.56 m

26. 75 m

27. R + 2H

32. tan–1 (1/2)

28. 6 m/s 29. 20 ×

33. tan–1(3)

2/3

–1  1  4 34. θ = tan   , 2 3

35. 200 m, 20 m/min, 12 m/min

EXERCISE - IV

ANSWER KEY 1. 122.7 km/hr 2.

πv 4

3. 60°, tan

g2

16.25 m, (e) 20 m 6.

1 3

sec

–1

(

3/2

)

7. 5 m/sec

4. u = 3m/s 5. (a) 10 ms–1,(b) 2 sec, (c)5 m, (d) 8. 1 sec

9. 2 tan–1 (1/3) 10. θ = 37°, v = 6 m/s

EXERCISE - V

1. (a) 1 sec, (b) ( (5 3 m, 5 m)

2.

a2 , tan –1 a 4b

3. (a)

u 2 sin 2α u cos(α + θ) , (b) v = g cos θ cos θ

4. B

5. A

6. u = 7.29 m/s, t = 1 sec 7. (a) 45°, (b) 2m/sec

8. B

9. C

10. B

11. B

12. 5 m/s2

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CONSTRAINED MOTION MOTION, TION, N.L.M., FRICTION THEORY AND EXERCISE BOOKLET CONTENTS

S.NO.

TOPIC

PAGE NO.

CONSTRAINED MOTION 1. String Constraint ............................................................................. 3 – 9 2. Wedge Constraint ........................................................................... 9 – 10

NEWTON'S LAW OF MOTION 1.

Force ....................................................................................... 11– 15

2.

Newton's Ist Law of Motion ................................................................ 16

3.

Newton's 2nd Law of Motion ........................................................ 16 – 25

4.

Newton's 3rd Law of Motion ......................................................... 25 – 27

5.

Spring Force ............................................................................. 27 – 32

6.

Pseudo Force ........................................................................... 32 – 36

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Page # 2

CONSTRAINED MOTION

FRICTION S.NO.

TOPIC

PAGE NO.

1.

Friction ................................................................................... 37 – 40

2.

Minimum Force Required to move the particle .................................. 40 – 41

3.

Friction as the component of contact force .................................... 41 – 42

4.

Motion on a Rough Inclined Plane .................................................. 42 – 43

5.

Angle of Repose ............................................................................... 43

6.

Two blocks on an inclined Plane .................................................... 44 – 45

7.

Range of force for which Acceleration of body is zero ....................... 45 – 48

8.

Pulley block system involve friction ............................................... 48 – 49

9.

Two block system ...................................................................... 49 – 53

10. Friction involve pseudo concept ................................................... 54 – 55 11. Exercise - I .............................................................................. 56 – 87 12. Exercise - II ............................................................................. 88 – 96 13. Exercise - III .......................................................................... 97 – 106 14. Exercise - IV ......................................................................... 107 – 111 15. Exercise - V .......................................................................... 112 – 117 16. Answer key ........................................................................... 118 – 120

IIT-JEE Syllabus : Newton's law of motion; inertial frame of reference; Uniformly accelerated frames of reference, Static and dynamic friction.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 3

CONSTRAINED MOTION

1.

CONSTRAINED MOTION :

1.1

String constraint : When the two object are connected through a string and if the string have the following properties : The length of the string remains constant i.e., it is inextensible string Always remains taut i.e., does not slacks. Then the parameters of the motion of the objects along the length of the string have a definite relation between them.

• •

Ist format : - (when string is fixed) A

s

B

v

The block B moves with velocity v. i.e. each particle of block B moves with velocity v. If string remain attached to block B it is necessary that velocity of each particle of string is same = v (vs = v) Now we can say that Block A also moves with velocity v. v v

A

B vA = vB = v

: If pulley is fixed then the velocity of all the particles of string is same along the string. v

B

Ex.1

Sol.

A vA =?

In the above situation block B is moving with velocity v. Then speed of each point of the string is v along the string. ∴ speed of the block A is also v v B v A vA=v

Ex.2

VA = 8 m/s

A 37°

B

vB=? Sol.

∵ Block A is moving with velocity 8 ms–1. 8 m/s ∴ velocity of every point on the string must be 8m/s along the string. The real velocity of B is vB. Then the string will not break only when the compoent of vB along string is 8 m/s. ⇒ vB cos 37° = 8 ⇒

vB =

8 = 10 m/sec cos 37°

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

A 8 m/s 8 m/s 37° vB

B

Page # 4 Ex.3

CONSTRAINED MOTION

Find out the velocity of block B in a pulley block system as shown in figure.

37°

53° 10 m/s Sol.

B

A

In a given pulley block system the velocity of all the particle of string is let us assume v then.

v 53°

10m/s

A

v 37°

B

53° 10cos53° 10 m/s is the real velocity of block A then its component along string is v. ⇒ 10 cos 53° = v ...(1) If vB is the real velocity of block B then it component along string is v then vBcos37° = v ...(2)

v

37° vB

B

from (1) & (2) vB cos37° = 10 cos53° ⇒

vB =

10 × 3 / 5 30 15 = m / sec = 4/5 4 2

50/3 m/s 53°

Ex.4

A

What is the velocity of block A in the figure as shown above. Sol.

The component of velocity of ring along string = velocity of A =

50 cos 53° = vA ⇒ vA = 10 m/s 3

: In the first format only two points of string are attached or touched to moving bodies.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 5

CONSTRAINED MOTION IInd format (when pulley is also moving) To understand this format we consider the following example in which pulley is moving with velocity vp and both block have velocity vA & vB respectively as shwon in figure. If we observe the motion of A and B with respect to pulley. Then the pulley is at rest. Then from first format. vAP = – vBP

vP

vB

vA A

(–ve sign indicate the direction of each block is opposite with respect to Pulley) v A – v p = – vB + v P ⇒ vP = :-

v A + vB 2

To solve the problem put the values of vA, vB, & vP with sign.

vP

10 m/s

Ex.5 A v=? A

B

Sol.

v A + vB 2 Putting vp = 10 ms–1, vB = 0, we get vA = 20 ms–1 (upward direction)

vP =

vP= 10m/s Ex.6

5m/s

Sol.

A

B v=? B

If we take upward direction as +ve then –5 + vB 2 vB = 25 m/sec (in upward direction)

10 =

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

B

Page # 6

CONSTRAINED MOTION

7 m/s

E 8m/s

F

Ex.7

Sol.

A

D

B C 5m/s

2m/s

Find out the velocity of Block D From 2nd format of constrained motion vE =

v A + vB 2

2–5 = –3/2 (If upward direction is taken to be +ve) 2 vE = –3/2 m/s

vE =

Now

vE + vF –3 / 2 + vF = 7 m/s ⇒ 7 = 2 2

14 + 3/2 = vF ⇒ vF =

31 2

v C + vD 8 + vD 31 = vF ⇒ = ⇒ vD = 31 – 8 2 2 2 vD = 23 m/s (upward direction)

Now

C

B

D E

Ex.8 A H

F

G m

10 m/sec

Sol.

Find the velocity of point G. In string ABCD from first format of constrain VD = 10 m/s↑ Now

+ 10 =

vH + vE 2 vH = 10 m/s ↓ if upward direction is taken to be positive

vD =

−10 + v E ⇒ vE = 30 m/s ↑ 2

vF + v G –10 + v G = vE ⇒ 30 = 2 2 60 + 10 = vG vG = 70 m/s↑

Now

: In IInd format three or four Points of the string is attached to the moving bodies.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 7

CONSTRAINED MOTION III format :

1.

SOLVING STRATEGY : First choose the longest string in the given problem which contains the point of which velocity/ acceleration to be find out.

2.

Now mark a point on the string wherever it comes in contact or leaves the contact of real bodies.

3.

If due to motion of a point, length of the part of a string with point is related, increases then its speed will be taken +ve otherwise –ve.

A D

I

E H

J C Ex.9

B 5m/s

C F A

vC=?

G B

2m/s

Sol. Step 1.

We choose a longest string ABCDEFGHIJ in which we have to find out velocity of point J (vc)

Step 2.

Mark all the point A, B ................

Step 3.

Write equation vA + vB + vC + vD + vE + vF + vG + vH + vI + vJ = 0 vA = vD = vE = vH = vI = 0

(No movement of that point because attached to fixed objects) ⇒ vB + vc + vF + vG + vJ = 0

...(1)

vB = vC = 5 m/s

(increases the length)

vF = vG = 2m/s

(It also increases the length)

Let us assume C is moving upward with velocity vc so vc negative because it decreasing the length ⇒ 5 + 5 + 2 + 2 – vc = 0 vC = 14 m/sec (upward) Ex.10

4m/s F

c ↑ 2m / sec

8m / s↑ A

B ↓ 2m / s

E

1 m / s↓ D Find out the velocity of block E as shown in figure.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 8 Sol. Step-1

CONSTRAINED MOTION We first choose the longest string in which point j (block E) lie. (abcdefghij) 4m/s c

F

d

b

e x

a k 8m / s ↑ A

h

y

c ↑ 2m / s

i j

f

B ↓ 2m / s

g

E

z 1m / s ↓ D Step 2 : Now write equation according to the velocity of each point (either increase or decrease the length) va + vb + vc + vd + ve + vf + vg + vh + vi + vj= 0 ...(1) Now find value of va, vb ..... in a following way v A + vB (from second format) 2 8–2 = = 3 m/sec. (upward) 2

vk =

va =

vK + v C (from 2nd format) 2

3+2 = 5/2 m/sec. (upward) 2 vx = 4m/s (from first format of constrain) vy + vz from 2nd format of constrain vx = ∴ vz = 0 (fixed) 2 ⇒ vy = 2 vx = 8 m/s (upward) ⇒ Now va = – 5/2 m/s (decreases the length) vb = vc = vd = ve = 0 (attached to fixed object) vf = vg = 1m/s (increases the length) vh = vi = vy = 8 m/s (increase the length) Let us assume block E move upward then vj = – vE (decrease the length) Puting the above values in eq. (1) ⇒ –5/2 + 1 + 1 + 8 + 8 – vE = 0 vE = 31/2 m/s (upward)

=

: In the following figure pulley is moving with velocity v at an angle θ with the horizontal.

v sin θ

v

A

B

θ C

D

*

v

A

B

θ C v cos θ

D

Only v cos θ is responsible to increase or decrease the length AB and v sin θ is responsible to either decrease or increase the length CD. Further solving strategy is same as 3rd format

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 9

CONSTRAINED MOTION Ex.11 Find out the relation between acceleration a and b as shown in following figure.

b B Sol. Step 1.

A

a

θ

Mark the points on the string which is attached to the real object (e.f,g,h)

f

e

b

g

f

b cos θ

b θ

a

θ

A

B

os bc

h

b

g

b

θ

h

θ

a Step 2. Acceleration of each point which are responsible to effect the length of string ae = 0 (because it is attached to fixed object) af = –b (attach to pulley which is moving with wedge's acceleration & –ve because it decreases the length) ag = b cos θ (only this component is responsible to effect the length of string) ah = (a – b cos θ) (resultant velocity at point h along the string) So now from 3rd format ae + af + ag + ah = 0 ⇒ 0 + (–b) + b cos θ + (a – b cos θ) = 0 a–b=0 ⇒ a=b

2.

(i) (ii)

WEDGE CONSTRAINT : Conditions : Contact must not be lost between two bodies. Bodies are rigid. The relative velocity / acceleration perpendicular to the contact surface of the two rigid object is always zero. Wedge constraint is applicable for each contact.

v3

v2

v3

v 3 = v1 sin θ

v1

v 1 sin θ Contact Plane

θ

In other words, Components of velocity and acceleration perpendicular to the contact surface of the two objects is always equal if there is no deformation and they remain in contact.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 10

CONSTRAINED MOTION

Ex.12 Find the relation between velocity of rod and that of the wedge at any instant in the figure shown.

v θ Sol.

u

Using wedge constraint. Component of velocity of rod along perpendicular to inclined surface is equal to velocity of wedge along that direction. u cos θ = v sin θ u = tan θ v u = v tan θ

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 11

NEWTON’S LAW OF MOTION

1.

FORCE A pull or push which changes or tends to change the state of rest or of uniform motion or direction of motion of any object is called force. Force is the interaction between the object and the source (providing the pull or push). It is a vector quantity. Effect of resultant force : may change only speed may change only direction of motion. may change both the speed and direction of motion. may change size and shape of a body

• • • •

unit of force : newton and

kg.m (MKS System) s2

g.cm (CGS System) s2 1 newton = 105 dyne Kilogram force (kgf) The force with which earth attracts a 1 kg body towards its centre is called kilogram force, thus

dyne and

Force in newton g Dimensional Formula of force : [MLT–2] For full information of force we require → Magnitude of force → direction of force → point of application of the force

kgf =

Force

Electromagnetic force

Gravitational force

Contact force

Nuclear force

Normal reaction

Tension

friction

1.1

Electromagnetic Force

• • • •

Force exerted by one particle on the other because of the electric charge on the particles is called electromagnetic force. Following are the main characteristics of electromagnetic force These can be attractive or repulsive These are long range forces These depend on the nature of medium between the charged particles. All macroscopic force (except gravitational) which we experience as push or pull or by contact are electromagnetic, i.e., tension in a rope, the force of friction, normal reaction, muscular force, and force experienced by a deformed spring are electromagnetic forces. These are manifestations of the electromagnetic attractions are repulsions between atoms/molecules.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 12

1.2

NEWTON’S LAW OF MOTION

Gravitational force : It acts between any two masses kept anywhere in the universe. It follows inverse square rule (F ∝ 1 ) and is attractive in nature. dis tan ce 2 GM1M2 F= R2 The force mg, which Earth applies on the bodies, is gravitational force.

1.3

Nuclear force : It is the strongest force. It keeps nucleons (neutrons and protons) together inside the nucleus inspite of large electric repulsion between protons. Radioactivity, fission, and fusion, etc. result because of unbalancing of nuclear forces. It acts within the nucleus that too upto a very small distance.

1.4

Contact force : Forces which are transmitted between bodies by short range atomic molecular interactions are called contact forces. When two objects come in contact they exert contact forces on each other.

1.4.1 Normal force (N) : It is the component of contact force perpendicular to the surface. It measures how strongly the surfaces in contact are pressed against each other. It is the electromagnetic force. A table is placed on Earth as shown in figure. •

3

1

4

Here table presses the earth so normal force exerted by four legs of table on earth are as shown in figure. N2

N1 N3

2

N4

ground

Now a boy pushes a block kept on a frictionless surface. Block

Here, force exerted by boy on block is electromagnetic interaction which arises due to similar charges appearing on finger and contact surface of block, it is normal force. (by boy) N

Block

A block is kept on inclined surface. Component of its weight presses the surface perpendicularly due to which contact force acts between surface and block.

θ

Normal force exerted by block on the surface of inclined plane is shown in figure.

N θ

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 13

NEWTON’S LAW OF MOTION Force acts perpendicular to the surface : • •

Normal force acts in such a fashion that it tries to compress the body Normal is a dependent force, it comes in role when one surface presses the other.

Ex.1

Two blocks are kept in contact on a smooth surface as shown in figure. Draw normal force exerted by A on B. A

B

Sol.

In above problem, block A does not push block B, so there is no molecular interaction between A and B. Hence normal force exerted by A on B is zero.

Ex.2

Draw normal forces on the massive rod at point 1 and 2 as shown in figure. 2

1

Sol.

Normal force acts perpendicular to extended surface at point of contact. N2

N1

50N 30°

Ex.3

Two blocks are kept in contact as shown in figure. Find (a) forces exerted by surfaces (floor and wall) on blocks (b) contact force between two blcoks. N1

Sol.

F.B.D. of 10 kg block N1 = 10 g = 100 N N2 = 100 N

...(1) ...(2)

100 N

N2 10 g

F.B.D. of 20 kg block N2 = 50 sin 30° + N3 ∴ N3 = 100 – 25 = 75 N and N4 = 50 cos30° + 20 g N4 = 243.30 N

N4

50 N

30°

...(3)

N2

N3 20 g

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

100N

10kg 20kg

Page # 14

NEWTON’S LAW OF MOTION

R=5m B

A

3m 1m

Ex.4

Find out the normal reaction at point A and B if the mass of sphere is 10 kg. N1

N2

y O

Sol.

A

5m3m 53°

N1

N2

37° 1m 4m

53°

37° O x'

B

Now F.B.D.

x

10 g y'

Now resolve the forces along x & y direction 3N2 5 y N1sin53° = 4N1/5 N1 N2sin37°=

N2

37° 4N2 O N2cos37°= 5

53° N1cos53°=

3N1 5

100

The body is in equilibrium so equate the force in x & y direction In x-direction

3N1 4N2 = 5 5

...(1)

3N2 4N1 + = 100 5 5 after solving above equation N1 = 80 N, N2 = 60 N

In y-direction

...(2)

1.4.2 Tension : Tension in a string is an electromagnetic force. It arises when a string is pulled. If a massless string is not pulled, tension in it is zero. A string suspended by rigid support is pulled by a force ‘F’ as shown in figure, for calculating the tension at point ‘A’ we draw F.B.D. of marked portion of the string; Here string is massless. F.B.D of marked portion T A

A

F

F

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Page # 15

NEWTON’S LAW OF MOTION

⇒ T=F String is considered to be made of a number of small segments which attracts each other due to electromagnetic nature. The attraction force between two segments is equal and opposite due to newton’s third law. Conclusion : T = mg (i) Tension always acts along the string and in such a direction that it tries to reduce the length of string (ii) If the string is massless then the tension will be same along the string but if the string have some mass then the tension will continuously change along the string.

Ex.5

m mg

The system shown in figure is in equilibrium. Find the magnitude of tension in each string ; T1, T2, T3 and T4. (g = 10 m/s–2)

60°

T3 T1

Sol.

T

30°

B

T4

T A 2 10 kg

F.B.D. of block 10 kg

F.B.D. of point ‘A’

y

T0

30° T0=10 g T0=100N

T1

10g

A

T2 x

T0

∑ Fy = 0 T2 cos30° = T0 = 100 N T2 =

200 3

N

∑ Fx = 0 T1 = T2 sin 30° =

200 1 100 . = N 3 2 3

F.B.D of point of ‘B’ ∑ Fy = 0 ⇒ T4 cos60° = T2 cos 30° and ∑ Fx = 0 ⇒ T3 + T2 sin 30° = T4 sin 60° ∴

T3 =

200 3

N , T = 200 N 4

y 60°

T3

B T2

T4 x

° 30

1.4.3 Frictional force : It is the component of contact force tangential to the surface. It opposes the relative motion (or attempted relative motion) of the two surfaces in contact. (which is explained later) 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 16

2.

NEWTON’S LAW OF MOTION

NEWTON’S FIRST LAW OF MOTION : According to this law “A system will remain in its state of rest or of uniform motion unless a net external force act on it. 1st law can also be stated as “If the net external force acting on a body is zero, only then the body remains at rest.” The word external means external to the system (object under observation), interactions within the system has not to be considered. The word net means the resultant of all the forces acting on the system. Newton’s first law is nothing but Galileo’s law of inertia. Inertia means inability of a body to change its state of motion or rest by itself. The property of a body that determines its resistance to a change in its motion is its mass (inertia). Greater the mass, greater the inertia. An external force is needed to set the system into motion, but no external force is needed to keep a body moving with constant velocity in its uniform motion. Newton’s laws of motion are valid only in a set of frame of references, these frames of reference are known as inertial frames of reference. Generally, we take earth as an inertial frame of reference, but strictly speaking it is not an inertial frame. All frames moving uniformly with respect to an inertial frame are themselves inertial. We take all frames at rest or moving uniformly with respect to earth, as inertial frames.

3.

NEWTON’S SECOND LAW OF MOTION : Newton’s second law states, “The rate of change of a momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts”    dp dp or F = k i.e., F ∝ dt dt where k is a constant of proportionality.     p = mv, So F = k (dmv) dt For a body having constant mass,    dv ⇒ F = km = k ma dt From experiments, the value of k is found to be 1.   So, Fnet = ma Force can’t change the momentum along a direction normal to it, i.e., the component of velocity normal to the force doesn’t change. Newton’s 2nd law is strictly applicable to a single point particle. In case of rigid bodies or system of   particles or system of rigid bodies, F refers to total external force acting on system and a refers to acceleration of centre of mass of the system. The internal forces, if any, in the system are not to be  included in F . Acceleration of a particle at any instant and at a particular location is determined by the force (net) acting on the particle at the same instant and at same location and is not in any way depending on the history of the motion of the particle.

PROBLEM SOLVING STRATEGY : Newton’s laws refer to a particle and relate the forces acting on the particle to its mass and to its acceleration. But before writing any equation from Newton’s law, you should be careful about which particle you are considering. The laws are applicable to an extended body too which is nothing but collection of a large number of particles. Follow the steps given below in writing the equations : 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 17

NEWTON’S LAW OF MOTION

Step 1 : Select the body The first step is to decide the body on which the laws of motion are to be applied. The body may be a single particle, an extended body like a block, a combination of two blocks-one kept over another or connected by a string. The only condition is that all the parts of the body or system must have the same acceleration. Step 2 : Identify the forces Once the system is decided, list down all the force acting on the system due to all the objects in the environment such as inclined planes, strings, springs etc. However, any force applied by the system shouldn’t be included in the list. You should also be clear about the nature and direction of these forces. Step 3 : Make a Free-body diagram (FBD) Make a separate diagram representing the body by a point and draw vectors representing the forces acting on the body with this point as the common origin. This is called a free-body diagram of the body. A

Ts

Tb R B m/sec2

5

R

C

Wp Wm platform man F.B.D of Diagram

100kg

50 kg

Look at the adjoining free-body diagrams for the platform and the man. Note that the force applied by the man on the rope hasn’t been included in the FBD. Once you get enough practice, you’d be able to identify and draw forces in the main diagram itself instead of making a separate one Step 4 : Select axes and Write equations When the body is in equillibrium then choose the axis in such a fashion that maximum number of force lie along the axis. If the body is moving with some acceleration then first find out the direction of real acceleration and choose the axis one is along the real acceleration direction and other perpendicular to it. Write the equations according to the newton’s second law (Fnet = ma) in the corresponding axis.

4.

APPLICATIONS :

4.1

Motion of a Block on a Horizontal Smooth Surface. Case (i) : When subjected to a horizontal pull : The distribution of forces on the body are shown. As there is no motion along vertical direction, hence, R = mg For horizontal motion F = ma or a = F/m R a

m

F

mg

Case (ii) : When subjected to a pull acting at an angle (θ) to the horizontal : Now F has to be resolved into two components, F cosθ along the horizontal and F sin θ along the vertical direction. Fsinθ R F θ m Fcosθ mg

For no motion along the vertical direction. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 18

NEWTON’S LAW OF MOTION

we have R + F sin θ = mg or R = mg – F sin θ :

Hence R ≠ mg. R < mg For horizontal motion F cos θ m

F cos θ = ma, a =

Case (iii) : When the block is subjected to a push acting at an angle θ to the horizontal : (down ward) The force equation in this case θ F R = mg + F sin θ

R

F cos θ

θ

:

R ≠ mg, R > mg

mg

For horizontal motion F cosθ = ma, a =

4.2

F cos θ m

F F sin θ

Motion of bodies in contact. Case (i) : Two body system : Let a force F be applied on mass m1 B

F

A f m1

f m2

Free body diagrams : (vertical force do not cause motion, hence they have not been shown in diagram) F ⇒ a = m +m 1 2

m 2F and f = m + m 1 2

(i) Here f is known as force of contact. (ii) Acceleration of system can be found simply by a=

F

m1

f

f

m2

force total mass

: If force F be applied on m2, the acceleration will remain the same, but the force of contact will be different m1F i.e., f’ = m + m 1 2

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Page # 19

NEWTON’S LAW OF MOTION Ex.6

Find the contact force between the 3 kg and 2kg block as shown in figure. F1 = 100N

Sol.

B

A 3kg

F2 = 25N

2kg

Considering both blocks as a system to find the common acceleration Fnet = F1 – F2 = 100 – 25 = 75 N common acceleration Fnet = 75 N 5kg Fnet 75 2 a= = = 15 m / s 5kg 5 R To find the contact force between A & B we draw a N F.B.D of 2 kg block 25 N 2kg from (∑Fnet)x = max ⇒ N – 25 = (2) (15) ⇒ N = 55 N 2g

a

Case (ii) : Three body system : C B A m2 m3 m F 1

Free body diagrams : F ⇒ a = m +m +m 1 2 3

For A

(m 2 + m 3 ) F and f1 = (m + m + m ) 1 2 3

For B

m1

F

m2

f1 f1

For C

f2 f2

m3

F – f1 = m1a f1 – f2 = m2a f2 = m3a m 3F f2 = (m + m + m ) 1 2 3 f1 = contact force between masses m1 and m2 f2 = contact force between masses m2 and m3 Remember : Contact forces will be different if force F will be applied on mass C

Ex.7

Find the contact force between the block and acceleration of the blocks as shown in figure.

F1 = 50N

Sol.

C

B

A

5kg

2kg

3kg

F2 = 30N

Considering all the three block as a system to find the common acceleration Fnet = 50 – 30 = 20 N 20 = 2m / s2 10 To find the contact force between B & C we draw F.B.D. of 3 kg block. a=

(∑ F ) net

x

= ma

Fnet=20N

10kg

a R

N1

⇒ N1 – 30 = 3(2) ⇒ N1 = 36 N To find contact force between A & B we draw F.B.D. of 5 kg block ⇒ N2 – N1 = 5a N2 = 5 × 2 + 36 ⇒ N2 = 46 N

3kg

30N a

mg

a

N2

5kg

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N1

Page # 20

4.3

NEWTON’S LAW OF MOTION

Motion of connected Bodies Case (i) For Two Bodies : F is the pull on body A of mass m1. The pull of A on B is exercised as tension through the string connecting A and B. The value of tension throughout the string is T only. B

T

m2

A

T

m1

F

Free body diagrams : For body A

For body B

R1

R2

a

A

T

a

B

F

m1g

m2g

R1 = m1g F – T = m1a

R2 = m2g T = m2a

T

F a = m +m 1 2

Case (ii) : For Three bodies : A

T1

m1

B m2

a

C

T2

m3

F

Free body diagrams :

For A

For B

R1

For C R3

R2

A

m1g R1 = m1g T1 = m1a

T1

T1

B

T2

m2g

T2

C

F

m3g

R3 = m3g R2 = m2g F – T2 = m3a T2 – T1 = m2a ⇒ F = m3a + T2 ⇒ T2 = m2a + T1 =m a+(m 3 1+m2 )a T2 = (m2 + m1)a F=(m1+m2+m3)a F ⇒ a = m +m +m 1 2 3

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Page # 21

NEWTON’S LAW OF MOTION Ex.8

A 5 kg block has a rope of mass 2 kg attached to its underside and a 3 kg block is suspended from the other end of the rope. The whole system is accelerated upward is 2 m/s2 by an external force F0. F0 5 kg

2 kg 3 kg

Sol.

(a) What is F0? (b) What is the force on rope? (c) What is the tension at middle point of the rope? (g = 10 m/s2) For calculating the value of F0, consider two blocks with the rope as a system. F.B.D. of whole system

F0 (a)

2m/s2

10 g = 100N

F0 – 100 = 10 × 2 F = 120 N ...(1) (b) According to Newton’s second law, net force on rope. F = ma = (2) (2) = 4 N ...(2) (c) For calculating tension at the middle point we draw F.B.D. of 3 kg block with half of the rope (mass 1 kg) as shown. T – 4g = 4.(2) = 48 N

4.4

T 4g

Motion of a body on a smooth inclined plane : Natural acceleration down the plane = g sin θ Driving force for acceleration a up the plane, F=m(a+ gsinθ) and for an acceleration a down the plane, F=m(a – gsinθ) N mgcos mg

mg sin

Find out the contact force between the 2kg & 4kg block as shown in figure.

2k g

4k g

Ex.9

37º

Sol.

On an incline plane acceleration of the block is independent of mass. So both the blocks will move with the same acceleration (gsin 37º) so the contact force between them is zero. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564

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Page # 22

NEWTON’S LAW OF MOTION

20 N

2k g

3k g

Ex.10 Find out the contact force between 2kg & 3kg block placed on the incline plane as shown in figure.

37º

Considering both the block as a 5kg system because both will move the same acceleration.

20 N

5k g

Sol.

37º

20 N

5k g

N

Now show forces on the 5 kg block

4.5

c t

f o

r c e

g 5k

2 3 7 0N º

s in

( N

1

t a

N

n

5gcos 37º 37º 5g

3k g

c o

2

r

2m /s 3g si n3 7º

F o

a

5g

∵ Acceleration of 5kg block is down the incline. So choose one axis down the incline and other perpendicular to it From Newton’s second Law N = 5g cos 37º ...(i) 5gsin 37º – 20 = 5a ..(ii) 30 – 20 = 5a a = 2m/s2 (down the incline) ) between 2kg & 3kg block 1 we draw F.B.D. of 3kg block From Fnet = ma ⇒ 3gsin 37º – N1 = 3 × 2 18 – N1 = 6 N1= 12 N

N

37º 5g

Pulley block system :

Ex.11 One end of string which passes through pulley and connected to 10 kg mass at other end is pulled by 100 N force. Find out the acceleration of 10 kg mass. (g = 9.8 m/s2)

Sol.

Since string is pulled by 100N force.

100 N

So tension in the string is 100 N. F.B.D. of 10 kg block 100 – 10 g = 10 a

100 N

100 – 10 × 9.8 = 10 a a = 0.2 m/s2

10 kg

10 g

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Page # 23

NEWTON’S LAW OF MOTION Ex.12 In the figure shown, find out acceleration of each block.

10kg

2kg 4kg

Sol.

Now F.B.D. of each block and apply Newton’s second law on each F.B.D

(1)

10 kg

10kg

a1

2T

a1

10g –2 T = 10a1

...(i)

a2

2kg 4kg

a3

10g T

(2)

2kg

a2

T – 2g = 2a2

...(2)

a3

T – 4g = 4a3

...(3)

2g T

(3)

4kg

4g

...(4) from constrain relation 2a1 = a2 + a3 Solving equations (1), (2), (3) and (4) we get 800 N 23 a1 = 70/23 m/s2 (downward), a2 = 170/23 m/s2 (upward),

T=

a3 = 30/23 m/s2 (downward)

Ex.13 Find the acceleration of each block in the figure shown below; in terms of their masses m1, m2 and g. Neglect any friction. m1

m2

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Page # 24 Sol.

NEWTON’S LAW OF MOTION

Let T be the tension in the string that is assumed to be massless. For mass m1, the FBD shows that N1 = m1g Where N1 is the force applied upward by plane on the mass m1. If acceleration of m1 along horizontal is a1. then T = m1a1 ...(i) For mass m2, the FBD shows that m2g – 2T = m2a2 ...(ii) Where a2 is vertical acceleration of mass m2. Note that upward tension on m2 is 2T applied by both sides of the string. from constrain relation a a2 = 1 2 Thus, the acceleration of m1 its twice that of m2. with this input, solving (i) and (ii) we find 2m 2 g a1 = 4m + m 1 2 m2g a2 = 4m + m 1 2

N1 T

m1g

T

T 2T

m2g

a2 m2g

Ex.14 Two blocks A and B each having a mass of 20 kg, rest on frictionless surfaces as shown in the figure below. Assuming the pulleys to be light and frictionless, compute : (a) the time required for block A, to move down by 2m on the plane, starting from rest, (b) tension in the string, connecting the blocks. A B

37º

Sol. Step 1.

Draw the FBDs for both the blocks. If tension in the string is T, then we have T

NA

mAg

NB

and

T mB g

Note that mAg, should better be resolved along and perpendicular to the plane, as the block A is moving along the plane. T NA mAg sin

mAg cos

Step 2. From FBDs, we write the force equations ‘ for block A where NA = mA g cos θ = 20 × 10 × 0.8 = 160 N and mAg sin θ – T = mA a ... (i) Where ‘a’ is acceleration of masses of blocks A and B. Similarly, force equations for block B are NB = mBg = 20 × 10 = 200 N and T = mBa ...(ii) From (i) and (ii), we obtain

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Page # 25

NEWTON’S LAW OF MOTION m A g sin θ 20 × 10 × 0.6 a = m +m = = 3 ms–2 40 A B

T = mBa = 20 × 3 = 60 N Step 3. With constant acceleration a = 3 ms–2, the block A moves down the inclined plane a distance S = 2 m in time t given by S=

1 2 at or t = 2

2S 2 = sec onds. a 3

Ex.15 Two blocks m1 and m2 are placed on a smooth inclined plane as shown in figure. If they are released from rest. Find : (i) acceleration of mass m1 and m2 (ii) tension in the string (iii) net force on pulley exerted by string N1 T Sol. F.B.D of m1 : m1g sin θ – T = m1a a m1 3 g – T = 3a ...(i) θ=30° 2

m1

m2 1kg

√3kg 30°

60°

m1g F.B.D. of m2 :

T – 1.

3 g = 1.a 2

a

m2

...(ii) θ

Adding eq. (i) and (ii) we get a = 0 Putting this value in eq. (i) we get T=

N2

T

T – m2g sin θ = m2a

m2g

3g , 2

F.B.D. of pulley

5.

FR =

2T

FR =

3 g 2

T

T

FR

NEWTONS’ 3RD LAW OF MOTION : Statement : “To every action there is equal and opposite reaction”. But what is the meaning of action and reaction and which force is action and which force is reaction? Every force that acts on body is due to the other bodies in environment. Suppose that a body A   experiences a force FAB due to other body B. Also body B will experience a force FBA due to A. According to Newton third law two forces are equal in magnitude and opposite in direction Mathematically we write it as   FAB = –FBA   Here we can take either FAB or FBA as action force and other will be the reaction force.

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Page # 26

NEWTON’S LAW OF MOTION

: (i) Action-Reaction pair acts on two different bodies. (ii) Magnitude of force is same. (iii) Direction of forces are in opposite direction. (iv) For action-reaction pair there is no need of contact

Ex.16 A block of mass ‘m’ is kept on the ground as shown in figure. (i) Draw F.B.D. of block (ii) Are forces acting on block action - reaction pair (iii) If answer is no, draw action reaction pair. Sol. (i) F.B.D. of block

m

N (Normal)

m

mg (field force)

(ii) ‘N’ and Mg are not action - reaction pair. Since pair act on different bodies, and they are of same nature. (iii) Pair of ‘mg’ of block acts on earth in opposite direction.

m

mg

earth

and pair of ‘N’ acts on surface as shown in figure.

N

5.1

Climbing on the Rope : Rope

F.B.D of man T

a

a mg

Now three condition arises. if T > mg ⇒ man accelerates in upward direction T < mg ⇒ man accelerates in downward direction T = mg ⇒ man’s acceleration is zero

*

Either climbing or decending on the rope man exerts force downward

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Page # 27

NEWTON’S LAW OF MOTION Ex.17 If the breaking strength of string is 600N then find out the maximum acceleration of the man with which he can climb up the road

600N (50 kg)

Sol.

Maximum force that can be exerted on the man by the rope is 600 N. F.B.D of man ⇒ 600 – 50 g = 50 a amax = 2 m/s2

a

50 g

Ex.18 A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform.

400 N

Sol.

(i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter. (ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s ? (iii) What force should apply now to maintain the constant speed of 1 m/s? The free body diagram of the painter and the platform as a system can be drawn as shown in the figure. Note that the tension in the string is equal to the force by which he pulles the rope. (i) Applying Newton’s Second Law TT 2T – (M + m)g = (M + m)a 2T – (M + m)g M+m Here M = 60 kg; m = 15 kg ; T = 400 N g = 10 m/s2

or

a=

a

2( 400) – ( 60 + 15)(10 ) (M+m) g = 0.67 m/s2 60 + 15 (ii) To attain a speed of 1 m/s in one second the acceleration a must be 1 m/s2 Thus, the applied force is

a=

1 (M + m) (g + a) = (60 + 15) (10 + 1) = 412.5 N 2 (iii) When the painter and the platform move (upward) together with a constant speed, it is in a state of dynamic equilibrium Thus, 2F – (M + m) g = 0

F=

or F =

6.

(M + m)g (60 + 15)(10) = = 375 N 2 2

SPRING FORCE : Every spring resists any attempt to change its length; when it is compressed or extended, it exerts force at its ends. The force exerted by a spring is given by F = –kx, where x is the change in length and k is the stiffness constant or spring constant (unit Nm–1) When spring is in its natural length, spring force is zero.

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Page # 28

NEWTON’S LAW OF MOTION

 0+x

 0 F=0

Fext F = –kx

F F

x Graph between spring force v/s x Ex.19 Two blocks are connected by a spring of natural length 2 m. The force constant of spring is 200 N/m. Find spring force in following situations. 2m B

A

Sol.

(a) If block ‘A’ and ‘B’ both are displaced by 0.5 m in same direction. (b) If block ‘A’ and ‘B’ both are displaced by 0.5 m in opposite direction. (a) Since both blocks are displaced by 0.5 m in same direcetion, so change in length of spring is zero. Hence, spring force is zero. (b) In this case, change in length of spring is 1 m. So spring force is F = –Kx = – (200). (1) F = –200 N 2m Natural length

B

A 3m

1m B

A F

F

When spring is extended

B

A F

F

When spring is compressed

Ex.20 Force constant of a spring is 100 N/m. If a 10 kg block attached with the spring is at rest, then find extension in the spring (g = 10 m/s2) Sol.

6.1

In this situation, spring is in extended state so spring force acts in upward direction. Let x be the extension in the spring. F.B.D. of 10 kg block : Fs Fs = 10 g

Kx = 100

(100)x = (100)

x=1m

10 kg

10g

SPRING FORCE SYSTEM : Initially the spring is in natural length at A with block m. But when the block displaced towards right then the spring is elongated and now block is released at B then the block move towards left due to spring force (kx).

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Page # 29

NEWTON’S LAW OF MOTION Analysis of motion of block : Natural Length

v a v=0

v a m

m

C v a Av a (i)

Initial position

B

From B to A speed of block increase and acceleration decreases. (due to decrease in spring force kx)

kx (ii)

a v

m

Due to inertia block crosses natural length at A. From A to C speed of the block decreases and acceleration increases.(due to increase in spring force kx)

m

(iii)

kx

a v

At C the block stops momentarily at this instant and since the spring is compressed spring force is towards right and the block starts to move towards right. From C to A speed of block increases and acceleration decreases.(due to decrease in spring force kx) kx a v

m

(iv)

Again block crosses point A due to inertia then from A to B speed decreases and acceleration increases. kx

m

a v

In this way block does SHM (to be expalined later) if no resistive force is acting on the block. Note : N.L.

A

(i)

Release

B when the block A is released then it take some finite time to reach at B. i.e., spring force doesn’t change instantaneously. N.L.

A

Release

m

(2) B

When point A of the spring is released in the above situation then the spring forces changes instantaneously and becomes zero because one end of the spring is free. (3)

In string tension may change instantaneously. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564

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Page # 30

NEWTON’S LAW OF MOTION

Ex.21 Find out the acceleration of 2 kg block in the figures shown at the instant 1 kg block falls from 2 kg block. (at t = 0)

1kg

A Sol.

1kg

2kg

2kg B

F.B.D.s before fall of 1kg block

30N (kx) 2kg

1kg

1kg

30N (T)

2kg 30(mg)

30(mg)

after the fall of the 1 kg block tension will change instantaneously but spring force (kx) doesn’t change instantaneously. F.B.D.s just after the fall of 1 kg block

(A)

30N (kx) 2kg

20 B 2kg

20 aA =

20

30 – 20 = 5 m/s2 (upward) 2

aB = 0 m/s2

Ex.22 Two blocks ‘A’ and ‘B’ of same mass ‘m’ attached with a light spring are suspended by a string as shown in figure. Find the acceleration of block ‘A’ and ‘B’ just after the string is cut.

Sol.

When block A and B are in equilibrium position F.B.D of ‘B’ T0

A m

B m

T0=mg

...(i) mg

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Page # 31

NEWTON’S LAW OF MOTION F.B.D of ‘A’ T T = mg + T0 T = 2 mg

.....(ii)

mg T0

when string is cut, tension T becomes zero. But spring does not change its shape just after cutting. So spring force acts on mass B, again draw F.B.D. of block A and B as shown in figure F.B.D of ‘B’

T0=mg T0 – mg = m.aB aB = 0

mg F.B.D. of ‘A’ mg + T0 = m. aA 2 mg = m. aA aA = 2g (downwards)

mg T0=mg Ex.23 Find out the acceleration of 1kg, 2kg and 3kg block and tension in the string between 1 kg & 2 kg block just after cutting the string as shown in figure. Sol.

A 1kg B 2kg

F.B.D before cutting of string

6gN

A 1kg

C 3kg

1gN (mg) 5gN

B 2kg 3gN

2gN

3gN(spring force)

C 3kg 3gN(mg)

Let us assume the Tension in the string connecting blocks A & B becomes zero just after cutting the string then.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 32

NEWTON’S LAW OF MOTION 2 kg

1 kg a1 =

1g

1g = g ms–2 1

a2 =

2g + 3g (weight)

5g = 2.5 g ms–2 2

(spring force)

∵ a2 > a1 i.e., ∴ T≠0 If T ≠ 0 that means string is tight and Both block A & B will have same acceleration. So it will take as a system of 3 kg mass.

1 kg 10N

T

3kg

2 kg

a

60N

30N 20N System

T 2

20m/s

Total force down ward = 10 + 30 + 20 = 60 N 60 Total mass = 3 kg ⇒ a = = 20 m/s2 3 Now apply Fnet = ma at block B. ∵

2kg 50

50 – T = 2 × 20 T = 10 N

the spring force does not change instantaneously the F.B.D of ‘C’ 3g 3kg

2

ac= 0 m/s

3g Reference Frame : A frame of reference is basically a coordinate system in which motion of object is analyzed. There are two types of reference frames. (a) Inertial reference frame : Frame of reference moving with constant velocity or stationary (b) Non-inertial reference frame : A frame of reference moving with non-zero acceleration

: (i) Although earth is a non inertial frame (due to rotation) but we always consider it as an inertial frame. (ii) A body moving in circular path with constant speed is a non intertial frame (direction change cause acceleration)

7.

PSEUDO FORCE : Consider the following example to understand the pseudo force concept support

B m

a A

The block m in the bus is moving with constant acceleration a with respect to man A at ground. Force 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 33

NEWTON’S LAW OF MOTION required for this acceleration is the normal reaction exerted by the support So,

N = ma

..(i)

This block m is at rest with respect to man B who is in the bus (a non intertial frame). So the acceleration of the block with respect to man B is zero. N = m(0) = 0 ..(ii) But the normal force is exerted in a non-inertial frame also. So the equation (ii) is wrong therefore we conclude that Newton’s law is not valid in non-inertial frame. If we want to apply Newton’s law in non-inertial frame, then we can do so by using of the cencept pseudo force. Pseudo force is an imaginary force, which in actual is not acting on the body. But after applying it on the body we can use Newton’s laws in non-inertial frames. This imaginary force is acting on the body only when we are solving the problem in a non-inertial frame of reference. In the above example. The net force on the block m is zero with respect to man B after applying the pesudo force. ma

m

N

N = ma : 1.

Direction of pseudo force is opposite to the acceleration of frame

2.

Magnitude of pseudo force is equal to mass of the body which we are analyzing multiplied by acceleration of frame

3.

Point of application of pseudo force is the centre of mass of the body which we are analysing

Ex.24 A box is moving upward with retardation ‘a’ < g, find the direction and magnitude of “pseudo force” acting on block of mass ‘m’ placed inside the box. Also calculate normal force exerted by surface on block

'm' 'Ma'

Sol.

N

Pseudo force acts opposite to the direction of acceleration of reference frame.

N + ma = mg N = mg – ma

(Pseudo force)

pseudo force = ma in upward direction F.B.D of ‘m’ w.r.t box (non-inertial)

mg Ex.25 Figure shows a pendulum suspended from the roof of a car that has a constant acceleration a relative to the ground. Find the deflection of the pendulum from the vertical as observed from the ground frame and from the frame attached with the car. a

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Page # 34

NEWTON’S LAW OF MOTION T

θ

Sol.

ax=a

a

θ mg

ay= 0

mg

Figure represents free Body diagram of the bob w.r.t ground. In an inertial frame the suspended bob has an acceleration a caused by the horizontal component of tension T. ...(i) T sin θ = ma ...(ii) T cosθ = mg From equation (i) and (ii) tan θ =

a g

 a θ = tan–1    g

In a non-inertial frame

T

θ ma

θ

a

a=0 x

ma

mg

a= y 0 mg Figure represents free Body diagram of bob w.r.t car. In the non-intertial frame of the car, the bob is in static equilibrium under the action of three froces, T, mg and ma (pseudo force) T sin θ = ma ...(iii) ....(iv) T cos θ = mg From equation (iii) and (iv) tan θ =

 a a ⇒ θ = tan–1    g g

Ex.26 A pulley with two blocks system is attached to the ceiling of a lift moving upward with an acceleration a0. Find the deformation in the spring. K m2

Sol.

a0

m1

Non-Inertial Frame T

K

2T m2T

T m1

a0

a

T a

m2g m2a0 m g m a (pseudo) 1 1 0 (pseudo)

Let relative to the centre of pulley, m1 accelerates downward with a and m2 accelerates upwards with a. Applying Newton’s 2nd law. m1a + m1a0 – T = m1a ...(i) ...(ii) T – m2g – m2a0 = m2a

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Page # 35

NEWTON’S LAW OF MOTION On adding (iv) and (v) we get  m1 – m 2  a =  m + m  (g + a0)  1 2

...(iii)

Substituting a in equation (i) We get T =

2m1m 2 (g + a 0 ) m1 + m 2

∴ x=

4m1m 2 ( g + a 0 ) F 2T = = (m + m ) k k k 1 2

Ex.27 All the surfaces shown in figure are assumed to be frictionless. The block of mass m slides on the prism which in turn slides backward on the horizontal surface. Find the acceleration of the smaller block with respect to the prism. A

m a0

Sol.

α B C Let the acceleration of the prism be a0 in the backward direction. Consider the motion of the smaller block from the frame of the prism The forces on the block are (figure) N'

N

a0 α N Mg

α

a

α

ma0

α mg

α

(i) N normal force (ii) mg downward (gravity), (iii) ma0 forward (Psuedo Force) The block slides down the plane. Components of the forces parallel to the incline give ma0 cosα + mg sin α = ma or, a = a0 cos α + g sin α ...(i) Components of the forces perpendicular to the incline give ...(ii) N + ma0 sin α = mg cos α Now consider the motion of the prism from the ground frame. No pseudo force is needed as the frame used is inertial. The forces are (figure) (i) Mg downward (ii) N normal to the incline (by the block) (iii) N’ upward (by the horizontal surface) Horizontal components give, N sin α = Ma0 or N = Ma0 / sin α, ...(iii) Putting in (ii) Ma 0 + ma0 sin α = mg cos α sin α mg sin α cos α or, a0 = M + m sin 2 α (M + m) g sin α mg sin α cos 2 α From (i) a = + g sin α = 2 M + m sin 2 α M + m sin α

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Page # 36

8.

NEWTON’S LAW OF MOTION

WEIGHING MACHING : A weighing machine does not measure the weight but measures the force. exerted by object on its upper surface or we can say weighing machine measure normal force on the man.

8.1

Motion in a lift :

(A)

If the lift is unaccelerated (v = 0 or constant) In this case no pseudo force act on the man N In this case the F.B.D. of the man N = mg In this case machine read the actual weight

(B)

If the lift is accelerated upward. (where a = constant)

weighing machine

mg

a

N

weighing machine F.B.D of man with respect to lift So weighing machine read N = m(g + a) Apparent weight N > Actual weight (mg) (c)

mg ma(pseudo)

If the lift is accelerated down ward.

a N ma (pseudo)

weighing machine F.B.D of man with respect to lift So weighing machine read

mg

N = m(g - a) Apparent weight N < Actual weight (mg) Note : (i) If a = g ⇒ N = 0 Thus in a freely falling lift, the man will experience a state of weightlessness (ii) If the lift is accelerated downwards such that a > g : So the man will be accelerated upward and will stay at the ceiling of the lift. (iii) Apparent weight is greater than or less than actual weight only depends on the direction and magnitude of acceleration. Magnitude and direction of velocity doesn’t play any roll in apparent weight. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 37

FRICTION

1.

FRICTION : Friction is a contact force that opposes the relative motion or tendency of relative motion of two bodies. Mg

F

F f

N Consider a block on a horizontal table as shown in the figure. If we apply a force, acting to the right, the block remains stationary if F is not too large. The force that counteracts F and keeps the block in rest from moving is called frictional force. If we keep on increasing the force, the block will remain at rest and for a particular value of applied force, the body comes to state of about to move. Now if we slightly increase the force from this value, block starts its motion with a jerk and we observe that to keep the block moving we need less effort than to start its motion. So from this observation, we see that we have three states of block, first, block does not move, second, block is about to move and third, block starts moving. The friction force acting in three states are called static frictional force, limiting frictional force and kinetic frictional force respectively. If we draw the graph between applied force and frictional force for this observation its nature is as shown in figure.

1.1

Static frictional force f

b

flim fkin

a

Static region

c

Kinetic region

d

F

When there is no relative motion between the contact surfaces, frictional force is called static frictional force. It is a self-adjusting force, it adjusts its value according to requirement (of no relative motion). In the taken example static frictional force is equal to applied force. Hence one can say that the portion of graph ab will have a slope of 45º. The Direction of Static Friction The direction of static friction on a body is such that the total force acting on it keeps it at rest with respect to the body in contact. The direction of static friction is as follows. For a moment consider the surfaces to be frictionless. In absence of friction the bodies will start slipping against each other. One should then find the direction of friction as opposite to the velocity with respect to the body applying the friction.

1.2

Limiting Frictional Force This frictional force acts when body is about to move. This is the maximum frictional force that can exist at the contact surface. (i) The magnitude of limiting frictional force is proportional to the normal force at the contact surface. flim ∝ N ⇒ flim = µsN Here µs is a constant the value of which depends on nature of surfaces in contact and is called as ‘coefficient of static firction’.

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Page # 38

1.3

FRICTION

Kinetic Frictional Force Once relative motion starts between the surface in contact, the frictional force is called as kinetic frictional force. The magnitude of kinetic frictional force is also proportional to normal force. fk = µkN From the previous observation we can say that µk < µs Although the coefficient of kinetic friction varies with speed, we shall neglect any variation i.e., when relative motion starts a constant frictional force starts opposing its motion. Direction of Kinetic Friction The kinetic friction on a body A slipping against another body B is opposite to the velocity of A with respect to B. It should be carefully noted that the velocity coming into picture is with respect to the body applying the force of friction. f

f

v

Suppose we have a long truck moving on a horizontal road. A small block is placed on the truck which slips on the truck to fall from the rear end. As seen from the road, both the truck and the block are moving towards right, of course the velocity of the block is smaller than that of the truck. What is the direction of the kinetic friction acting on the block due to the truck ? The velocity of the block as seen from the truck is towards left. Thus, the friction on the block is towards right. The friction acting on the truck due to the block is towards left. Ex.1

Find the direction of kinetic friction force (a) on the block, exerted by the ground. (b) on the ground, exerted by the block. F=1N

Sol.

(a)

f1

1 kg 5 m/s w.r.t to ground

F=1N

1 kg V=5m/s ///////////////////////////////////

5m/s w.r.t to block

(b)

f2

where f1 and f2 are the friction forces on the block and ground respectively. Ex.2

Sol.

The correct relation between magnitude of f1 and f2 in above problem is : (A) f1 > f2 (B) f2 > f1 (C) f1 = f2 (D) not possible to decide due to insufficient data. By Newton’s third law the above friction forces are action-reaction pair and equal but opposite to each other in direction. Hence (C) Also note that the direction of kinetic friction has nothing to do with applied force F. 10 m/s

A

Ex.3

B

20m/s

All surfaces are rough. Draw the friction force on A & B fkBA Sol.

A

fkAB

B

fkBG Kinetic friction acts to reduce relative motion. Summary We can summarise the laws of friction between two bodies in contact as follows: 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 39

FRICTION (i)

(ii) (iii)

(iv) Ex.4

If the bodies slip over each other, the force of friction is given by fk = µk N where N is the normal contact force and µk is the coefficient of kinetic friction between the surfaces. The direction of kinetic friction on a body is opposite to the velocity of this body with respect to the body applying the force of friction. If the bodies do not slip over each other, the force of friction is given by fs ≤ µs N where µs is the coefficient of static friction between the bodies and N is the normal force between them. The direction and magnitude of static friction are such that the condition of no slipping between the bodies is ensured. The frictional force fk or fs does not depend on the area of contact as long as the normal force N is same. A block of mass 5 kg is resting on a rough surface as shown in the figure. It is acted upon by a force of F towards right. Find frictional force acting on block when (a) F = 5N (b) 25 N (c) 50 N (µs = 0.6, µk = 0.5) [g = 10 ms–2]

F

Sol.

Ex.5

Maximum value of frictional force that the surface can offer is Mg fmax = flim = µsN F = 0.6 × 5 × 10 = 30 newton Therefore, it F ≤ fmax body will be at rest and f = F or F > fmax body will more and f = fk N (a) F = 5N < Fmax So body will not move hence static frictional force will act and , fs = f = 5N (b) F = 25 N < Fmax ∴ fs = 25 N (c) F = 50 N > Fmax So body will move and kinetic frictional force will act, its value will be fk = µk N = 0.5 × 5 × 10 = 25 newton

f

A block having a mass 3 kg is initially at rest on a horizontal surface. The coefficient of static friction µs = 0.3 between the block and the surface and µk is 0.25. A constant force F of 50 N, acts on the body at the angle θ = 37º. What is the acceleration of the block ? θ

F

x

Sol.

We have two possibilities here, the block may remain at rest, or it may accelerate towards the right. The decision hinges on whether or not the x-component of the force F has magnitude, less than or greaer than the maximum static friction force. The x-component of F is Fx = Fcos θ = (50 N ) (0.8) = 40 N To find fs, max, we first calculate the normal force N, whether or not the block accelerates horizontally, the sum of the y-component of all the forces on the block is zero. N N - F sin θ – mg = 0 x f θ or N = F sin θ + mg=(50 N) (0.6) + (3 kg)(9.8ms–2) = 59.4 N The maximum static frictional force mg F fs,max = µsN = (0.3) (59.4 N) = 17.8 N This value is smaller than the x-component of F, hence the block moves. We now interpret the force f in the figure as a kinetic frictional force. This value is obtained as fK = µk N = (0.25) (59.4 N) = 14.8 N 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564

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Page # 40

FRICTION

Therefore resultant force in the x-direction is

∑F

x

= F cos θ − f = 40 N –14.8 N = 25.2 N

Then the acceleration ‘a’ of the block is a=

25.2 N = 8.4 ms −2 3 kg

Think : What would happen if the magnitude of Fx happened to be less than fs.max but larger than fk ? Ex.6

In the previous example, suppose we move the block by pulling it with the help of a massless string tied to the block as shown here. What is the force F required to produce the same acceleration in the block as obtained in the last example ? F θ m

Sol.

a

We are given that, m = 3kg, µs = 0.3, µk = 0.25, θ = 37º, and a = 8.4 ms–2 In order to determine the force F, we first draw the FBD as shown below f The equations of motion therefore, are N + Fsin θ = mg N = mg – Fsin θ F cos θ – f = ma and where f = µs N before the start of the motion, once motion is set, f = µkN. Hence, force F which produces a = 8.4 m/s2 is given by Fcosθ – µk (mg – F sin θ) = ma or F =

N

F sin F cos

mg

3(0.4 + 0.25 × 9.8) ma + µkmg = = 34.26N 0.8 + 0.25 × 0.6 cos θ + µk sin θ

: Fsin θ works out to be less than mg. Otherwise we would lift the block up in the above analysis COMMENT It is easier to pull then to push. Only about 34 N force is required to pull than 50 N required during pushing why ? Because, when we pull at an angle, the effective normal force N by which block is pressing down on surface is reduced and consequently friction is reduced. Just the contrary happens when you are pushing.

2.

MINIMUM FORCE REQUIRED TO MOVE THE PARTICLE : A body of mass m rests on a horizontal floor with which it has a coefficient of static friction µ. It is desired to make the body slide by applying the minimum possible force F. F φ m Fig. A Let the applied force F be at angle φ with the horizontal

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Page # 41

FRICTION F

R

φ

R m

R = Normal force mg

Fig. B For vertical equilibrium, R = (mg – F sinφ) ...(i) R + F sin φ = mg or, For horizontal equilibrium i.e. when the block is just about to slide, F cos φ = µR ...(ii) Substituting for R, or F = µmg / (cos φ + µ sinφ) F cosφ = µ (mg – F sinφ) for minimum F (cosφ + µsinφ) is maximum, ⇒ Let x = cos φ + µ sinφ dx = − sin φ + µ cos φ dφ for maximum of x, dx = 0 dφ tan φ = µ and at this value of φ Fmin =

3.

µmg 1 + µ2

FRICTION AS THE COMPONENT OF CONTACT FORCE : When two bodies are kept in contact, electromagnetic forces act between the charged particles at the surfaces of the bodies. As a result, each body exerts a contact force on other The magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite and hence the contact forces obey Newton’s third law. N=normal force Fc=contact force

f=friction

The direction of the contact force acting on a particular body is not necessarily perpendicular to the contact surface. We can resolve this contact force into two components, one perpendicular to the contact surface and the other parallel to it. The perpendicular component is called the normal contact force or normal force and parallel component is called friction.

f 2 + N2 = N {when fmin = 0}

Contact force = Fc min

Fc max =

µ 2N2 + N2

N ≤ Fc ≤

(µ 2 + 1) N

{when fmax = µN}

0 ≤ λ ≤ tan–1µ

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Page # 42 Ex.7

Sol.

FRICTION

A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find (a) the angle made by the contact force on the body with the vertical and (b) the magnitude of the contact force. Take g = 10 m/s2. Let the contact force on the block by the surface be Fc which makes an angle λ with the vertical (shown figure) Fc N f The component of Fc perpendicular to the contact surface is the normal force N and the component of F parallel to the surface is the firction f. As the surface is horizontal, N is vertically upward. For vertical equilirbrium, N = Mg = (0.400 kg) (10 m/s2) = 4.0 N The frictional force is f = 3.0 N f 3 = or, λ = tan–1 (3/4) = 37º N 4 (b) The magnituded of the contact force is

(a)

tan λ =

F = N2 + f 2 =

4.

(4.0 N) 2 + (3.0N) 2 = 5.0 N

MOTION ON A ROUGH INCLINED PLANE Suppose a motion up the plane takes place under the action of pull P acting parallel to the plane N = mg cos α P N Frictional force acting down the plane F = µN = µ mg cos α Appling Newton’s second law for motion up the plane mg sin α mg cos α P – (mg sin α + f) = ma f α mg P – mg sin α – µ mg cos α = ma If P = 0 the block may slide downwards with an acceleration a. The frictional force would then act up the plane mg sin α – F = ma or, mg sin α – µ mg cos α = ma

Ex.8

A 20 kg box is gently placed on a rough inclined plane of inclination 30° with horizontal. The coefficient of sliding friction between the box and the plane is 0.4. Find the acceleration of the box down the incline. N Y X F = µN mgsinα

Sol.

mg

O

mgcosα

Y' X' In solving inclined plane problems, the X and Y directions along which the forces are to be considered, may be taken as shown. The components of weight of the box are (i) mg sin α acting down the plane and (ii) mg cos α acting perpendicular to the plane. N = mg cos α mg sin α – µ N = ma ⇒ mg sin α – µ mg cos α = ma a = g sin α – µg cos α = g (sin α – µ cos α)

1 3 = 9.8 2 – 0.4 × 2  = 4.9 × 0.3072 = 1.505 m/s2   The box accelerates down the plane at 1.505 m/s2. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 43

FRICTION Ex.9

A force of 400 N acting horizontal pushes up a 20 kg block placed on a rough inclined plane which makes an anlge of 45° with the horizontal. The acceleration experienced by the block is 0.6 m/ s2. Find the coefficient of sliding friction between the box and incline.

Sol.

The horizontally directed force 400 N and weight 20 kg of the block are resolved into two mutually perpendicular components, parallel and perpendicular to the plane as shown. N = 20 g cos 45° + 400 sin 45° = 421.4 N The frictional force experienced by the block

a=0.6 m/s

F = µN = µ × 421.4 = 421.4 µN.

2

400 cos45°

R

As the accelerated motion is taking placed up the plane.

400 N

400 cos 45° – 20 g sin 45° – f = 20a 400 2

20 × 9.8 2

400 sin45°

20g sin45°

– 421.4 µ = 20a = 20 × 0.6 = 12

45°20 g 20 cos 45°

 400  196 1 282.8 – 138.6 – 12 µ= =– – 12 × = 0.3137 . =  2  4214 2 4214 . The coefficient of sliding friction between the block and the incline = 0.3137

5.

ANGLE OF REPOSE : Consider a rough inclined plane whose angle of inclination θ with ground can be changed. A block of mass m is resting on the plane. Coefficient of (static) friction between the block and plane is µ. For a given angle θ, the FBD (Free body diagram) of the block is

N

f

mg sin

mg cos

Where f is force of static friction on the block. For normal direction to the plane, we have N=mg cosθ As θ increases, the force of gravity down the plane, mg sin θ, increases. Friction force resists the slide till it attains its maximum value. fmax = µN = µ mg cos θ Which decreases with θ (because cos θ decreases as θ increases) Hence, beyond a critical value θ = θc, the blocks starts to slide down the plane. The critical angle is the one when mg sin θ is just equal of fmax, i.e., when mg sin θc = µ mg cosθc or tan θc = µ where θC is called angle of repose If θ > θc, block will slide down.For θ < θc the block stays at rest on the incline.

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Page # 44

6.

FRICTION

TWO BLOCKS ON AN INCLINED PLANE :

2

N

m

1

N

m

Consider two blocks having masses m1 & m2 placed on a rough inclined plane. µ1 & µ2 are the friction coefficient for m1 & m2 respectively. If N is the normal force between the contact surface of m1 & m2

θ

Now three condition arises. (i) If µ1 = µ2 = µ then N = 0 because, Both the blocks are in contact but does not press each other. a1 = a2 = g sin θ – µ mg cos θ (a1,a2 are acceleration of block µ1 & µ2 respectively) (ii) If µ1 < µ2 then N = 0 because, there is no contact between the blocks. a1 = g sin θ – µ1 g cos θ a2 = g sin θ – µ2 g cos θ ⇒ a1 > a2 (iii) If µ1 > µ2 then N≠0 a1 = a2

Ex.10 Mass m1 & m2 are placed on a rough inclined plane as shown in figure. Find out the acceleration of the blocks and contact force in between these surface. m2 2kg

m1 1kg

37°

As we know if µ1 > µ2 both will travel together so a1 = a2 = a F.B.D =3 .2

os gc

m2 ° 37 in s g

m1

f2

° 37

m2

2

f1 = µ1m1gcos37°=4

° 37 n i gs 37° 1

m2

m

3k g

Sol.

µ2=0.2 µ1=0.5

which is equivalent to a=

3g sin 37°–(f1 + f2 ) 18 – 7.2 a= = 3.6 m/sec2 3 2

Now F.B.D of 1 kg block is

3

f1 + f2

7° n3 37° i gs

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Page # 45

FRICTION

N 3. a=

7° in3 s g

4N 2 c /se m 6

gsin37° + N – 4 = (1) a N = 3.6 + 4 – 6 = 1.6 Newton

RANGE OF FORCE F FOR WHICH ACCELERATION OF BODY IS ZERO.

2k g

F

7.

Ex.11 37º

=8N x

2kg

12 N

m

37º

Now take different value of F Force (F) F.B.D.

fma

F =( 0. 5) m gc o ax

fm

gs in 37 º

2k g

s3 7º

F

Sol.

Find out the range of force in the above situation for which 2kg block does not move on the incline. F.B.D of 2 kg block

Acceleration

Friction Type

Fnet m

8N

a=

Direction & magnitude

N 12

2m/s2

8N

Kinetic

0m/s2

8N

Static

8N

4N

(i) F = 0N

(ii) F = 4N

4N

8N

N 12

0m/s2

(iii) F = 8N

4N

N 12

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Static

Page # 46

f= 0

12 N

FRICTION

0m/s2

(iv) F = 12N

0

Static

16 N

N 12

0m/s2

(v) F = 16N

4N

Static

8N

Static

8N

Kinetic

f= 4N

N 12

20

N

(in this condition friction change its direction to stop relative slipping)

0m/s2

(vi) F = 20N

24

N

8N

N 12

2m/s2

(vii) F = 24N 8N

N 12

From the above table block doesn't move from F = 4N (mgsinθ – µmgcosθ) to F = 20N (mgsinθ + µmgcosθ). So friction develope a range of force for which block doesn't move

: If Friction is not present then only for F = 12N the block will not move but friction develop a range of force 4N to 20N to prevent slipping. So we can write the range of force F for which acceleration of the body is zero. mg sin θ – µmg cos θ ≤ F ≤ mg sin θ + µ mg cosθ .

Ex.12 In the following figure force F is gradually increased from zero. Draw the graph between applied force F and tension T in the string. The coefficient of static friction between the block and the ground is µs.

F

Sol.

M

µs As the external force F is gradually increased from zero it is compensated by the friction and the string beares no tension. When limiting friction is achieved by increasing force F to a value till µs mg, the further increase in F is transferred to the string.

T 45° F µsmg

Ex.13 Fig. shows two blocks tied by a string. A variable force F = 5t is applied on the block. The coefficient of friction for the blocks are 0.6 and 0.5 respectively. Find the frictional force between blocks and ground as well as tension in the string at

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Page # 47

FRICTION (A) t = 1 s

(B) t = 2s B

(C) t = 3s A

1kg

2kg

µ=0.6

µ=0.5

10N

Sol.

F=5t

20N

T fA

T

fB 10N

5t

20N

(a) At t = 1s, F = 5 × 1 = 5 N Maximum value of friction force fA = µN = 0.5 × 20 = 10 N To keep the block stationary the magnitude of frictional force should be 5N. So fA = 5 N Now from the figure it becomes clear that if fA = 5N & F = 5 N, Tension T = 0 Since tension is not in application so frictional force on block B is 0 i.e., fB = 0 (b) At t = 2s, F = 5 × 2 = 10 N Maximum value of friction force f = µN = 0.5 × 20 = 10 N To keep the block stationary the magnitude of friction force should be 10 N. So fA = 10 N From the figure it is clear that if fA = 10 N and F = 10 N Tension T = 0 Hence friction force on block B is fB = 0 (c) At t = 3s, F = 5 × 3 = 15 N Maximum value of friction force f = µN = 0.5 × 20 = 10 Newton Again applying the same analogy fA = 10 N From the figure it is clear that if fA = 10 N and F = 15 N Tension T = 5 N So frictional force on block B is fB = 5 Newton

Ex.14 Find the tension in the string in situation as shown in the figure below. Forces 120 N and 100 N start acting when the system is at rest.

120 N

10 fsmax=90 N

Sol.

20

100N

fsmax=60 N

(i) Let us assume that system moves towards left then as it is clear from FBD, net force in horizontal direction is towards right. Therefore the assumption is not valid.

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Page # 48

FRICTION

120 N

10

100N

20 60 N

90 N

Above assumption is not possible as net force on system comes towards right. Hence system is not moving towards left. (ii) Similarly let us assume that system moves towards right.

120 N

10

100N

20 60 N

90 N

Above assumption is also not possible as net force on the system is towards left in this situation. Hence assumption is again not valid. Therefore it can be concluded that the system is stationary T

10

120 N

20

100N

fmax=60 N

Fmax=90 N

Assuming that the 10 kg block reaches limiting friction first then using FBD’s

10

120 N 120 = T + 90 ⇒ Also

T + f = 100

30 + f = 100 ⇒

T T 90N f

20

100N

T = 30 N f = 70 N

which is not possible as the limiting value is 60 N for this surface of block.

∴ Our assumption is wrong and now taking the 20 kg surface to be limiting we have 10

120 N

Also

T T f 60 N

T + 60 = 100 N ⇒

T = 40 N

f + T = 120 N ⇒

f = 80 N

20

100N

This is acceptable as static friction at this surface should be less than 90 N. Hence the tension in the string is

T = 40 N

8.

PULLEY BLOCK SYSTEM INVOLVING FRICTION :

If friction force is acting and value of acceleration of a particle is negative, then it means direction of friction force is opposite to that what we assumed and acceleration would be having a different numerical value.

Ex.15 Two blocks of masses 5 kg and 10 kg are attached with the help of light string and placed on a rough incline as shown in the figure. Coefficients of friction are as marked in the figure. The system is released from rest. Determine the acceleration of the two blocks.

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Page # 49

FRICTION

10kg

5kg

50°

37°

Fixed

Sol.

Let 10 kg block is sliding down, then acceleration of both the blocks are given by,

10g sin 37°– µ1 × 10g cos 37°–5g sin 53° – µ 2 × 5g cos 53°

a=

15

= – ve

It means our assumed direction of motion is wrong and 5 kg block is going to slide down, if this would be the case, the direction of friction force will reverse and acceleration of blocks would be given by :

5g sin 53°– µ 2 × 5g cos 53°– µ1 × 10g cos 37° – 10g sin 37°

a1 =

= –ve

15

It means in this direction also there is no motion. So we can conclude that the system remains at rest and friction force is static in nature.

9.

TWO BLOCK SYSTEM :

B 2kg

Ex.16

A

F

4kg frictionless

Find out the maximum value of F for which both the blocks will move together Sol.

In the given situation 2kg block will move only due to friction force exerted by the 4 kg block

F.B.D. B 2kg f

A

4kg

f F

The maximum friction force exerted on the block B is fmax = µN fmax = (0.5) (20) = 10 N So the maximum acceleration of 2 kg block is a 2 max

10 = = 5m / s 2 2

2kg

fmax = 10N

amax is the maximum acceleration for which both the block will move together. i.e., for a ≤ 5 ms–2 acceleration of both blocks will be same and we can take both the blocks as a system.

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Page # 50

FRICTION

F.B.D a

5m/s

6kg

2

Fmax

Fmax = 6 × 5 = 30 N for 0 < F < 30 Both the block move together.

Ex.17 In the above question find the acceleration of both the block when (i) F = 18 N (ii) F = 36 N (i) Since F < 30 both the blocks will move together Sol. F.B.D

6kg a=

F = 18N

18 = 3 m / s2 6

(ii) When F = 36 N When F > 30 both the blocks will move separately so we treat each block independently F.B.D of 2 kg block

B 2kg

f = 10N (Friction force)

aB = 5 m/s2 F.B.D of 4 kg block

f = 10N

aA =

A

4kg

F = 36N

36 − 10 26 = m / s2 4 4

B 2kg Ex.18

Sol.

A

4kg

F

Find out the range of force in which both the blocks move together If f1 is friction force between block A & lower surface and f2 is friction force between both the block’s surface. F.B.D f2=10N B 2kg f2=10N F 6N = f1 A 4kg f1 max = µ1N1 = (0.1) (60) = 6 N f2 max = µ2N2 = (0.5) (20) = 10 N Upper 2kg block is move only due to friction force so maximum acceleration of that block is 2kg

f2 =10N

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Page # 51

FRICTION amax =

10 = 5 m / s2 2

This is the maximum acceleration for which both the blocks will move together. Therefore for a ≤ 5ms–2 we can take both the blocks as one system. F.B.D. 2 5m/s 6kg

f1=6N

F

For F < 6 N. Blocks will not move at all. Now the value of Fmax for which both the blocks will move together. Fmax – 6 = 6 × 5 Fmax = 36 N Conclusion if 0N < F < 6N No blocks will move 6N < F < 36 N Both blocks will move together F > 36 N Both move separately.

Ex.19

B 2kg

F

A 4kg frictionless

The lower block A will move only due to friction force F.B.D.

2kg

f

4kg

F f(frictional force)

fmax = µN = (0.5) (20) = 10 N

F.B.D. of 4 kg blocks

amax

4kg

f = 10N

The maximum acceleration of 4 kg block is ⇒

amax =

10 = 2.5 m / s 2 4

This is the maximum acceleration for which both the blocks move together 2.5 m/s2

6kg

F

Fmax for which both the blocks will move together Fmax = 2.5 × 6 = 15 N

B 2kg Ex.20

F

A 4kg

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Page # 52

FRICTION

If f2 is the friction force between A & B and f1 is the friction force between A & floor f1 max = 6 N f2 max = 10 N Lower block A will move only due to friction force So amax for 4 kg block

f1 =6N

4kg

f2 = 10N

10 − 6 = 1 m / s2 4 This is the maximum acceleration for which both the blocks will move together amax =

1m/s f1 =6N

2

6kg

F

F–6=6×1 F = 12 N If F is less than 6N both the blocks will be stationary Conclusion : 0 < F < 6 N = Both blocks are stationary 6 N < F < 12 N = Both move together F > 12 N = Both move separately

Ex.21 Find the accelerations of blocks A and B for the following cases. (A) µ1 = 0 and µ2 = 0.1 (P) aA = aB = 9.5 m/s2 (B) µ2 = 0 and µ1 = 0.1 (Q) aA = 9 m/s2, aB = 10 m/s2 (C) µ1 = 0.1 and (R) aA = aB = g = 10 m/s2 µ2 = 1.0 (D) µ1 = 1.0 and (S) aA = 1, aB = 9 m/s2 µ2 = 0.1 Sol. (a) R, (b) Q, (c) P, (d) S (i) FBD in (case (i)) {µ1 = 0, µ2 = 0.1} µ2N O

1kg

N=10

A

1kg B

µ1

1 kg A

µ2

1kg

10 N

B

N=10

µ2N mg

mg

While friction’s work is to oppose the relative motion and here if relative motion will start then friction comes and without relative motion there is no friction so both the block move together with same acceleration and friction will not come.

A

B

mg

mg

2

⇒ aA = aB = 10 m/s

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Page # 53

FRICTION

0

1 (ii)

1

B 10 1kg

A 10 1kg 10 0

10

Friction between wall and block A oppose relative motion since wall is stationary so friction wants to slop block A also and maximum friction will act between wall and block while there is no friction between block.

: Friction between wall and block will oppose relative motion between wall and block only it will not do anything for two block motion.

1 A

B

10

10 2

aA = 9 m/s ; aB = 10 m/s2

f

1 A

(iii)

10

B f

10

10

Friction between wall and block will be applied maximum equal to 1N but maximum friction available between block A and B is 10 N but if this will be there then relative motion will increase while friction is to oppose relative motion. So friction will come less than 10 so friction will be f that will be static. 1 f

A

B

f 10

10

by system (20–1) = 2 × a ⇒ a = (iv)

19 =9.5 m/s2 2

aB =

10 – 1 = 9 m / s2 1

1

10

11 – 10 = 1m / s 2 aA = 1

A 10

10

B 1 10

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Page # 54

10.

FRICTION

FRICTION INVOLVING PSEUDO CONCEPT :

Ex.22 What is the minimum acceleration with which bar A should be shifted horizontally to keep the bodies 1 and 2 stationary relative to the bar ? The masses of the bodies are equal and the coefficient of friction between the bar and the bodies equal to µ. The masses of the pulley and the threads are negligible

1

A

2

while the friction in the pulley is absent. see in fig. Sol.

Let us place the observer on A. Since we have non-inertial frame we have pseudo forces. For body ‘1’ we have, T = ma + µmg

ma

....(1)

T

1 mg

For body ‘2’ we have,

µN

N = ma

From (1) and (2) amin

ma

A

mg – T – µma = 0 ∴ mg = T + µmg

....(2)

T

2

N

mg

 1− µ  = g   1+ µ 

a

µ=0.5

F

Ex.23

M=4kg m= 1kg

Find out the range of force for which smaller block is at rest with respect to bigger block. Sol.

Smaller block is at rest w.r.t. the bigger block. Let both the block travel together with acceleration a F.B.D of smaller block w.r. to the bigger block.

f

fmax = µ × N N = ma f = µ ma ⇒ f = mg

ma (Pseudo)

...(1) ...(2)

N

from (1) & (2)

mg

a = g/µ = 20 m/s2 So F = 20 (M + m) = 20 (5) = 100 N If F ≥ 100 N Both will travel together

Ex.24 The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end as shown. The coefficient of friction between the box & the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 ms–2 . At what distance from the starting point does the box fall off the truck (i.e. distance travelled by the truck) ? [Ignore the size of the box]

/////////////////////////////////////////

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Page # 55

FRICTION

Sol.

In the reference frame of the truck FBD of 40 kg block

µN

40 kg

ma (psuedo force)

Net force ⇒ ma – µN ⇒ 40 × 2 – mablock ⇒ 80 – 60 ⇒ ablock =

15 × 40 × 10 100

20 1 = m/s2 40 2

This acceleration of the block in reference frame of truck so time taken by box to fall down from truck Srel = urelt +

1 1 1 a t2 ⇒ 5 = 0 + × × t2 ⇒ t2 = 20 2 rel 2 2

So distance moved by the truck ⇒

1 × atruck × t2 2

1 × 2 × (20) = 20 meter.. 2

Ex.25 Mass m2 placed on a plank of mass m1 lying on a smooth horizontal plane. A horizontal force F = α0t (α0 is a constant) is applied to a bar. If acceleration of the plank and bar are a1 and a2 respectively and the coefficient of friction between m1 and m2 is µ. Then find acceleration a with time t.

m2

F

m1 Sol.

If F < µm2g then both blocks move with common acceleration, i.e., a1 = a2 When F > µm2g, then Equation for block of mass m

and

F – µm2g = m2a2

...(1)

µm2g = m1a1

...(2)

a2

From equation (1)

its slope positive and intercept negative. From equation (2) a1 is independent of time.

=a

2

a1

1

i.e., acceleration a2 varies with time linearly,

a

a

α0t – µm2g = m2a2

0

t0

t

So, the graph between a & t is as follow.

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Page # 56

N.L.M.

(Objective Problems)

Exercise - I (A) NEWTON'S LAW OF MOTION 1. At a given instant, A is moving with velocity of 5 m/s upwards. What is velocity of B at the time

3. The pulleys in the diagram are all smooth and light. The acceleration of A is a upwards and the acceleration of C is f downwards. The acceleration of B is

A

C

A B

(A) 15 m/s↓ Sol.

(B) 15 m/s↑

(C) 5 m/s ↓ (D) 5 m/s ↑

(A) 1/2 (f – a) up (C) 1/2 (a + f) up Sol.

B (B) 1/2 (a + f) down (D) 1/2 (a – f) up

4. If acceleration of A is 2 m/s2 to left and acceleration of B is 1 m/s2 to left, then acceleration of C is A

2. Find the velocity of the hanging block if the velocities of the free ends of the rope are as indicated in the figure. 2m/s

C 1m/s

(A) 3/2 m/s ↑ (C) 1/2 m/s ↑ Sol.

B

2

(A) 1 m/s upwards (C) 2 m/s2 downwards Sol.

(B) 1 m/s2 downwards (D) 2 m/s2 upwards

(B) 3/2 m/s ↓ (D) 1/2 m/s ↓

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Page # 57

N.L.M.

4 m/s

6 m/s 6m/s

5. In the figure shown the velocity of different blocks is shown. The velocity of C is

7. Find velocity of block 'B' at the instant shown in figure.

A B C D

(A) 6 m/s (C) 0 m/s Sol.

(B) 4 m/s (D) none of these

37°

A

(A) 25 m/s (C) 22 m/s Ans.

10 m/s

(B) 20 m/s (D) 30 m/s

6. If block A has a velocity of 0.6 m/s to the right, determine the velocity of block B.

A B (A) 1.8 m/s in downward direction (B) 1.8 m/s in upward direction (C) 0.6 m/s in downward direction (D) 0.6 m/s in upward direction Sol.

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B

Page # 58

N.L.M.

8.In the arrangement shown in fig. the ends P and Q of an unstretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed. B

A

θ

θ

Q

P

M

(A) 2 U cos θ (C)

(B) U cos θ

2U cos θ

(D)

U cos θ

Sol.

10. The velocity of end ‘A’ of rigid rod placed between two smooth vertical walls moves with velocity ‘u’ along vertical direction. Find out the velocity of end ‘B’ of that rod, rod always remains in constant with the vertical walls.

A 'u' 9. Block B moves to the right with a constant velocity v0. The velocity of body A relative to B is :

v0

B (A) u tan 2θ (C)u tan θ Sol.

(B) u cot θ (D) 2u tan θ

B

A

(A)

v0 , towards left 2

(B)

v0 , towards right 2

(C)

3v0 , towards left 2

(D)

3v0 , towards right 2

Sol.

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Page # 59

N.L.M. 11. Find the acceleration of C w.r.t. ground.

A

C

a

Sol.

B b

(A) ai – (2a + 2b)j

(B) a ˆi – (2a + b)ˆj

(C) a ˆi – (a + 2b)ˆj Sol.

(D) b ˆi – (2a + 2b)ˆj

13. The 50 kg homogeneous smooth sphere rests on the 30° incline A and bears against the smooth vertical wall B. Calculate the contact forces at A and B. A

B

30°

(A) NB = (B) NA = (C) NA = (D) NA =

N, N = A

3 1000

N, N = B

3 100 3

N, N = B

1000 3

N, N = B

Sol.

12. Find the acceleration of B.

A

1000

B

a

acos α 1 (A) cos α 2

(B)

a sin α1 a cos α 2 (C) cos α 2 cos α1

(D)

cos α1 cos α 2

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500 3 500 3 500 3 50 3

N N N N

Page # 60

N.L.M. 15. A sperical ball of mass m = 5 kg rests between two planes which make angles of 30° and 45° respectively with the horizontal. The system is in equilibrium. Find the normal forces exerted on the ball by each of the planes. The planes are smooth.

30 °

45°

14. Find out the reading of the weighing machine in the following cases.

g 2k

M

W 30º

(A) 10 3 Sol.

(B) 10 2

W

2k g

(A) N45 = 96.59 N, N30 = 136.6 N (B) N30 = 96.59 N, N45 = 136.6 N (C) N45 = 136.6 N, N30 = 96.56 N (D) none of these Sol.

M

30º

(C) 20 3

(D) 30 3

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Page # 61

N.L.M. Question No. 16 to 17 (2 questions) In the figure the tension in the diagonal string is 60 N.

18. A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force θ with the vertical in equilibrium, then the tension in the string AB is : F .

F1

45°

I f

t h

e

r o

p

e

A

B

m

a

k

e

 16. Find the magnitude of the horizontal force F1 and  F2 that must be applied to hold the system in the

position shown. 3

N

(B)

20 2

N

(C)

n

θ

W

60

a

40 2

a

n

g

l e

A

F3

(A)

s

N

(D)

60 2

B

F

M

(A) F sin θ Sol.

(B) F/sin θ

(C) F cos θ (D) F/cos θ

N

Sol.

17. In the above questions what is the weight of the suspended block ? (A) Sol.

60 2

N

(B)

40 2

N

(C)

60 3

N

(D)

50 2

N 19. Objects A and B each of mass m are connected by light inextensible cord. They are constrained to move on a frictionless ring in a vertical plane as shown in figure. The objects are released from rest at the positions shown. The tension in the cord just after release will be A T T

B

mg C

mg

(A) mg 2

(B)

mg 2

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(C)

mg 2

(D)

mg 4

Page # 62

N.L.M. 21. Two masses m and M are attached to the strings as shown in the figure. If the system is in equilibrium, then

Sol.

45°

45°

M

m

2M m 2M (C) cot θ = 1 + m Sol. (A) tan θ = 1 +

2m M 2m (D) cot θ = 1 + M (B) tan θ = 1 +

20. Three blocks A, B and C are suspended as shown in the figure. Mass of each blocks A and C is m. If system is in equilibrium and mass of B is M, then :

A

(A) M = 2m Sol.

B

(B) M < 2 m

C

(C) M > 2m (D) M = m

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Page # 63

N.L.M. 22. A flexible chain of weight W hangs between two fixed points A & B which are at the same horizontal level. The inclination of the chain with the horizontal at both the points of support is θ. What is the tension of the chain at the mid point ? A

Sol.

B θ

θ

W

W .cos ec θ 2 W .cot θ (C) 2 Sol. (A)

(B)

W .tan θ 2

(D) none

24.A stunt man jumps his car over a crater as shown (neglect air resistance) (A) during the whole flight the driver experiences weightlessness (B) during the whole flight the driver never experiences weightlessness (C) during the whole flight the driver experiences weightlessness only at the highest point

(D) the apparent weight increases during upward journey Sol. 23.A weight can be hung in any of the following four ways by string of same type. In which case is the string most likely to break ?

(A)

W

(B)

(C)

W

W (D)

W (A) A

(B) B

(C) C

(D) D

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Page # 64 Question No. 25 to 27 (3 questions) A particle of mass m is constrained to move on x-axis. A force F acts on the particle. F always points toward the position labeled E. For example, when the particle is to the left of E, F points to the right. The magnitude of F is constant except at point E where it is zero.

N.L.M. Sol.

A

+ve x E m The system is horizontal. F is the net force acting on the particle. The particle is displaced a distance A towards left from the equilibrium position E and released from rest at t = 0

25. What is the period of the motion ?  2Am   (A) 4  F  

 2Am   (B) 2  F   27. Find minimum time it will take to reach from x = −

 2 Am   (C)  F  

(D) None

A 2

to 0. mA ( 2 − 1) F mA ( 2 − 1) (C) 2 F Sol.

Sol.

(A)

3 2

(B)

mA ( 2 − 1) F

(D) none

26. Velocity-time graph of the particle is v

v t

(A)

v

v

(C)

t

(B)

t

(D)

t

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Page # 65

N.L.M. 28. A particle of mass 50 gram moves on a straight line. The variation of speed with time is shown in figure. find the force acting on the particle at t = 2, 4 and 6 seconds.

Sol.

v(m/s)

15 10 5 0

2

4

6

8 t(s)

(A) 0.25 N along motion, zero, 0.25 opposite to motion (B) 0.25 N along motion, zero, 0.25 along to motion (C) 0.25 N opposite motion, zero, 0.25 along to motion (D) 0.25 N opposite motion, zero, 0.25 opposite to motion Sol.

30. A constant force F is applied in horizontal direction as shown. Contact force between M and m is N and between m and M’ is N’ then

F

M'>M M m M'

(A) N or N’ equal (C) N’ > N Sol.

smooth (B) N > N’ (D) cannot be determined

29. Two blocks are in contact on a frictionless table. One has mass m and the other 2m. A force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio force of contact between the two block will be :

2m m (A) same

(B) 1 : 2

F

F

2m m

(C) 2 : 1

(D) 1 : 3

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Page # 66

N.L.M. 33. Force exerted by support on string. (A) 10 N (B) 15 N (C) 20 N Sol.

Question No. 31 to 33 (3 questions) A block of mass 1kg is suspended by a string of mass 1 kg, length 1m as shown in figure. (g = 10 m/ s2) Calculate :

(D) 25 N

34. A body of mass 8 kg is hanging another body of mass 12 kg. The combination is being pulled by a string –2 . The tension T1 and T2 will be respectively : (use g = 9.8 m/s2) w

i t h

a

n

a

c c e

l e

r a

t i o

n

o

f

2

. 2

m

s

T1

12kg 1m

T2 1 kg

a

8kg

31. The tension in string at its lowest point. (A) 10 N (B) 15 N (C) 20 N (D) 25 N Sol.

32. The tension in string at its mid-point (A) 10 N (B) 15 N (C) 20 N Sol.

(A) 200 N, 80 N (C) 240 N, 96 N Sol.

(B) 220 N, 90 N (D) 260 N, 96 N

(D) 25 N

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Page # 67

N.L.M. 35. A rope of mass 5 kg is moving vertically in vertical position with an upwards force of 100 N acting at the upper end and a downwards force of 70 N acting at the lower end. The tension at midpoint of the rope is (A) 100 N (B) 85 N (C) 75 N (D) 105 N Sol.

37. Two blocks, each having mass M , rest on frictionless surfaces as shown in the figure. If the pulleys are light and frictionless, and M on the incline is allowed to move down, then the tension in the string will be

M

fixed

M

(A)

2 Mgsin θ 3

(B)

(C)

Mgsin θ 2

(D) 2 Mg sin θ

Sol.

36. A particle of small mass m is joined to a very heavy body by a light string passing over a light pulley. Both bodies are free to move. The total downward force in the pulley is (A) mg (B) 2 mg (C) 4 mg (D) can not be determined Sol.

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3 Mgsin θ 2

Page # 68 38. The pulley arrangements shown in figure are identical the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with cosntant downward force F = 2mg, where g is acceleration due to gravity. The acceleration of mass in case I is

N.L.M. 39. Two masses M1 and M2 are attached to the ends of a light string which passes over a massless pulley attached to the top of a double inclined smooth plane of angles of inclination α and β . The tension in the string is :

F=2mg

m 2m

m

(I)

(II)

(A) zero (B) more than that in case II (C) less than that in case II (D) equal to that in case II Sol.

α

fixed

M2

M1

β

(A)

M2 (sin β) g M1 + M2

(B)

M1(sin α ) g M1 + M2

(C)

M1 M2 (sin β + sin α ) g M1 + M2

(D) zero

Sol.

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N.L.M. 40. Calculate the acceleration of the block B in the above figure, assuming the surfaces and the pulleys P1 and P2 are all smooth and pulleys and string and light 2m

F

A

Page # 69 41. In previous Question surface is replaced by block C of mass m as shown in figure. Find the acceleration of block B. 2m

4m

4m P1

F

B

P1

A

c

P2

B P2

(A) a =

3F m/s2 17m

(B) a =

2F m/s2 17m

3F (A) a = 20m m/s2

(B) a =

(C) a =

3F m/s2 15m

(D) a =

3F m/s2 12m

2F (C) a = 21m m/s2

3F (D) a = 18m m/s2

Sol.

Sol.

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3F m/s2 21m

Page # 70 42. In the arrangement shown in the fig, the block of mass m = 2 kg lies on the wedge on mass M = 8 kg. Find the initial acceleration of the wedge if the surfaces are smooth and pulley & strings are massless.

N.L.M. 43. In the arrangement shown in figure, pulleys are massless and frictionless and threads are inextensible. The Block of mass m1 will remain at rest, if

P1

m

60°

M

P2

60°

(A) a =

30 3 m/s2 23

30 2 (C) a = m/s2 23 Sol.

(B) a =

20 3 m/s2 23

(D) none of these

m1 m2

1 1 1 (A) m = m + m 1 2 3 4 1 1 (C) m = m + m 1 2 3

m3

(B) m1 = m2 + m3 (D)

1 2 3 = + m3 m2 m1

Sol.

44. Both the blocks shown here are of mass m and are moving with constant velocity in direction shown in a resistive medium which exerts equal constant force on both blocks in direction opposite to the velocity. The tension in the string connecting both of them will be (Neglect friction) 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 71

N.L.M. Sol.

A B (A) mg Sol.

(B) mg/2

(C) mg/3

(D) mg/4

47. Block of 3 kg is initially in equilibrium and is hanging by two identical springs A and B as shown in figures. If spring A is cut from lower point at t = 0 then, find acceleration of block in ms–2 at t = 0. 45. A monkey of mass 20 kg is holding a vertical rope. The rope can break when a mass of 25 kg is suspended from it. What is the maximum acceleration with which the monkey can climb up along the rope? (A) 7 ms–2 (B) 10 ms–2 (C) 5 ms–2 (D) 2.5 ms–2 Sol.

46. Fi nd the accel eration of 3 kg mass when acceleration of 2 kg mass is 2 ms–2 as shown in figure. 3 kg

2 kg

10N

–2

(A) 3 ms–2

(B) 2 ms–2

2ms (C) 0.5 ms–2

(D) zero

A

B 3 kg

(A) 5 Sol.

(B) 10

(C) 15

(D) 0

48. Two masses of 10 kg and 20 kg respectively are connected by a massless spring as shown in figure. A force of 200 N acts on the 20 kg mass at the instant when the 10 kg mass has an acceleration of 12 ms–2 towards right, the aceleration of the 20 kg mass is : 10kg

20kg 200 N

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Page # 72 (A) 2 ms Sol.

–2

N.L.M. (B) 4 ms

–2

(C) 10 ms

–2

–2

(D) 20 ms

50. A man of mass 60 kg is standing on a weighing machine placed in a lift moving with velocity 'v' and acceleration 'a' as shown in figure. Calculate the reading of weighing machine in following situation: (g = 10 m/s2) (i) a = 0, v=0 (A) 600 N (B) 500 N (C) 450 N (D) 700 N (ii) a = 0, v = 2m/s (A) 600 N (B) 500 N (C) 450 N (D) 700 N (iii) a = 0, v = –2m/s (A) 450 N (B) 500 N (C) 600 N (D) 700 N 49. Two blocks are connected by a spring. The combination is suspended, at rest, from a string attatched to the ceiling, as shown in the figure. The string breaks suddenly. Immediately after the string breaks, what is the initial downward acceleration of the upper block of mass 2m ?

2m

m

(A) 0 Sol.

(B) 3g/2

(C) g

(D) 2g

a

v

W.M.

(iv) a = 2m/s2, v=0 (A) 600 N (B) 500 N (C) v=0 (v) a = –2m/s2 (A) 600 N (B) 480 N (C) (vi) a = 2m/s2, v = 2m/s (A) 600 N (B) 480 N (C) (vii) a = 2 m/s2, v = –2m/s (A) 600 N (B) 720 N (C) (viii) a = –2m/s2 v = –2 m/s (A) 600 N (B) 480 N (C) Sol.

450 N

(D) 720 N

450 N

(D) 700 N

450 N

(D) 720 N

450 N

(D) 700 N

450 N

(D) 700 N

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Page # 73

N.L.M.

Question No. 52 to 54 (3 questions) An object of mass 2 kg is placed at rest in a frame (S1) moving with velocity 10 i + 5 j m/s and having 51. What will be the reading of spring balance in the figure shown in following situations. (g = 10 m/s2) (i) a = 0, v = 0 (A) 100 N (B) 80 N (C) 120 N (D) 150 N (ii) a = 0, v = 2 m/s (A) 100 N (B) 80 N (C) 120 N (D) 150 N (iii) a = 0, v = – 2m/s (A) 100 N (B) 80 N (C) 120 N (D) 150 N a

acceleration 5 i + 10 j m / s 2 . The object is also seen by an observer standing in a frame (S2) moving with velocity 5 i + 10 j m / s 52. Calculate 'Pseudo force' acting on object. Which frame is responsible for this force. (A) F = – 10 i – 20 j due to acceleration of frame S1 (B) F = – 20 i – 20 j due to acceleration of frame S1 (C) F = – 10 i – 30 j due to acceleration of frame S1 (D) none of these Sol.

v

M = 10 kg

(iv) a = 2 m/s2, v =0 (A) 100 N (B) 80 N (C) 120 N (v) a = – 2m/s2, v = 0 (A) 100 N (B) 80 N (C) 120 N (vi) a = 2 m/s2, v = 2 m/s (A) 100 N (B) 80 N (C) 120 N (vii) a = 2 m/s2, v = –2m/s (A) 100 N (B) 80 N (C) 120 N (viii) a = – 2 m/s2, v = – 2 m/s (A) 100 N (B) 80 N (C) 120 N Sol.

(D) 150 N (D) 150 N (D) 150 N (D) 150 N (D) 150 N

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Page # 74 53. Calculate net force acting on object with respect to S2 frame. (B) F = 10 i + 20 j (A) F = 20ˆi + 20ˆj (C) F = 5ˆi + 20 ˆj Sol.

N.L.M.

(D) F = 10 ˆi + 5 ˆj

54. Calculate net force acting on object with respect of S1 frame. (A) 0 (B) 1 (C) 2 (D) none of these Sol.

56. A block of mass m resting on a wedge of angle θ as shown in the figure. The wedge is given an acceleration a. What is the minimum value of a so that the mass m falls freely ?

A m

a θ

B (A) g Sol.

(B) g cos θ

(C) g cot θ

C (D) g tan θ

gs in θ

55. A trolley is accelerating down an incline of angle θ with acceleration gsinθ. Which of the following is correct. (α is the constant angle made by the string with vertical)

α

m

θ

(A) α = θ (B) α = 0º (C) Tension in the string, T = mg (D) Tension in the string, T = mg sec θ Sol.

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Page # 75

N.L.M. 57. In the figure the reading of the spring balanced will be : [g = 10 m/s2] 2m/s

10kg

Sol.

2

5kg

30°

(A) 50 N Sol.

(B) 40 N

(C) 60 N

(D) 70 N

(B) FRICTION 59. Find the direction of friction forces on each block and the ground (Assume all surfaces are rough and all velocities are with respect to ground). E D C

58. A pendulum of mass m hangs from a support fixed to a trolley. The direction of the string when the trolley rolls up of plane of inclination α with acceleration a0 is (String and bob remain fixed with respect to trolley)

a0

B

5 m/s

Sol.

θ

α

(A) θ = tan–1α

 a0  (B) θ = tan–1    g

 g (C) θ = tan–1  a   0

–1  a 0 + g sin α   (D) θ = tan   g cos α 

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A

2 m/s 1 m/s 3 m/s 5 m/s

Page # 76

N.L.M. 61. A monkey of mass m is climbing a rope hanging from the roof with acceleration a. The coefficient of static friction between the body of the monkey and the rope is µ. Find the direction and value of friction force on the monkey.

(A) Upward, F = m(g + a) (B) downward, F = m(g + a) (C) Upward, F = mg (D) downward, F = mg Sol.

60. In the following figure, find the direction of friction on the blocks and ground

VA=3m/s 7N VB=6m/s

A

F = 5N

B

Sol.

62. A body is placed on a rough inclined plane of inclination θ. As the angle θ is increased from 0º to 90º the contact force between the block and the plane (A) remains constant (B) first remains constant then decreases (C) first decreases then increases (D) first increases then decreases

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Page # 77 64. If force F is increasing with time and at t=0, F=0 where will slipping first start ?

N.L.M. Sol.

F

3 2 1

(A) between 3 kg and 2 kg (B) between 2 kg and 1 kg (C) between 1 kg and ground (D) both (A) and (B) Sol.

 63. A force F = i + 4 j acts on block shown. The force of friction acting on the block is F

y 1 Kg x (A) − i Sol.

(B) − 18 . i

(C) − 2.4 i

(D) − 3 i

65. A wooden block of mass m resting on a rough µ) is pulled by a force F as shown in figure. The acceleration of the block moving horizontally is : h

o

r i z o

n

t a

l

t a

b

l e

( c o

e

f f i c i e

n

t

o

f

f r i c t i o

n

=

F θ

m

(A)

F cos θ m

(B)

(C)

F (cos θ + µ sin θ) – µg m

(D) none

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µF sin θ M

Page # 78

N.L.M.

Sol.

Sol.

66. In the arrangement shown in the figure, mass of the block B and A is 2m and m respectively. Surface between B and floor is smooth. The block B is connected to the block C by means of a string pulley system. If the whole system is released, then find the minimum value of mass of block C so that block A remains stationary w.r.t. B. Coefficient of friction between A and B is µ.

67. Two masses A and B of 10 kg and 5 kg respectively are connected with a string passing over a fricionless pulley fixed at the corner of a table as shown in figure. The coefficient of friction of A with the table is 0.2. The minimum mass of C that may be placed on A to prevent if from moving is equal to :

A

C 10kg A

B 5 kg B

C (A)

m µ

(B)

2m + 1 µ +1

(C)

3m µ −1

(D)

6m µ +1

(A) 15 kg

(B) 10 kg

(C) 5 kg

(D) zero

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Page # 79

N.L.M.

69. A body of mass m moves with a velocity v on a surface whose friction coefficient is µ. If the body covers a distance s then v will be :

Sol.

(A)

2µgs (B) µgs

(C)

µgs / 2

(D)

3µgs

Sol.

68. If the coefficient of friction between an insect and bowl is µ and the radius of the bowl, is r, the maximum height to which the insect can crawl in the bowl is : 1+ µ2

 1  r (B) 1 – 1+ µ2 

(C) r 1 + µ 2

(D) r 1 + µ 2 – 1

r

(A)

Sol.

70. With what minimum velocity should block be projected from left end A towards end B such that it reaches the other end B of conveyer belt moving with constant velocity v. Friction coefficient between block and belt is µ.

A

    (A)

µgL (B)

m

v0

B

v

2µgL (C)

Sol.

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L 3µgL

(D) 2 µgL

Page # 80

N.L.M.

71. A box 'A' is lying on the horizontal floor of the compartment of a train running along horizontal rails from left to right. At time 't', it decelerates. Then the reaction R by the floor on the box is given best by A

(A)

R

R

(B)

floor

A floor

R

R

(C)

A

A

(D)

floor

floor

Sol.

73. A small mass slides down an inclined plane of θ with the horizontal. The co-efficient of friction is µ = µ0x where x is the distance through which the mass slides down and µ0, a constant. Then the distance covered by the mass before it stops is : i n

c l i n

a

t i o

n

2 4 1 1 (A) µ tan θ (B) µ tan θ (C) 2µ tan θ (D) µ tan θ 0 0 0 0

Sol. 72. A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined an at an angle θ to the vertical. The minimum value of coefficient of friction between the block and the surface for which the block will remain in equilibrium is :

Q θ

P /////////////////////////////////////// (A)

P + Q sin θ mg + Q cos θ

(B)

P cos θ + Q mg – Q sin θ

(C)

P + Q cos θ mg + Q sin θ

(D)

P sin θ – Q mg – Q cos θ

Sol.

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Page # 81

N.L.M. 74. In the above question the speed of the mass when travelled half the maximum distance is (A)

g tan θ sin θ µ0

(B)

g tan θ sin θ 2µ 0

(C)

g tan θ sin θ 8µ 0

(D) none of these

Sol.

76. A body is moving down a long inclined plane of slope 37º. The coefficient of friction between the body and plane varies as µ = 0.3 x, where x is distance travelled down the plane. The body will have maximum 3 and g = 10 m/s2) speed. (sin 37º = 5 (A) at x = 1.16 m (B) at x = 2m (C) at bottom of plane (D) at x = 2.5 m Sol.

75. For the equilibrium of a body on an inclined plane of inclination 45º. The coefficient of static friction will be (A) greater than one (B) less than one (C) zero (D) less than zero Sol.

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Page # 82 77. Block B of mass 100 kg rests on a rough surface of friction coefficient µ = 1/3. A rope is tied to block B as shown in figure. The maximum acceleration with which boy A of 25 kg can climbs on rope without making block move is

(A) Sol.

4g 3

(B)

g 3

(C)

g 2

(D)

N.L.M. 78. Starting from rest a body slides down a 45° inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is : (A) 0.75 (B) 0.33 (C) 0.25 (D) 0.80 Sol.

3g 4

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N.L.M. 79. A block of mass 5 kg and surface area 2 m2 just begins to slide down an inclined plane when the angle of inclination is 30°. Keeping mass same, the surface area of the block is doubled. The angle at which this starts sliding down is : (A) 30° (B) 60° (C) 15° (D) none Sol.

Page # 83 81. A block placed on a rough inclined plane of inclination (θ = 30º) can just be pushed upwards by applying a force “F” as shown. If the angle of inclination of the inclined plane is increased to (θ = 60º), the same block can just be prevented from sliding down by application of a force of same magnitude. The coefficient of friction between the block and the inclined plane is

F

(A) (C)

3 +1 3 −1 3 −1 3 +1

Sol. 80. Two blocks of masses m1 and m2 are connected with a massless unstretched spring and placed over a plank moving with an acceleration 'a' as shown in figure. the coefficient of friction between the blocks and platform is µ. m1

m2 a

(A) spring will be stretched if a > µg (B) spring will be compressed if a ≤ µg (C) spring will neither be compressed nor be stretched for a ≤ µg (D) spring will be in its natural length under all conditions. Sol.

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(B)

2 3 −1 3 +1

(D) none of these

oo

Q 3m

h

Sm

N.L.M.

ug

P

Ro

m

th

Page # 84 82. A fixed wedge with both surface inclined at 45° to the horizontal as shown in the figure. A particle P of mass m is held on the smooth plane by a light string which passes over a smooth pulley A and attached to a particle Q of mass 3m which rests on the rough plane. The system is released from rest. Given that g the acceleration of each particle is of magnitude 5 2 then A

45° fixed 45°

(a) the tension in the string is : 6mg mg (A) mg (B) (C) 5 2 2 Sol.

(D)

mg 4

(c) In the above question the magnitude and direction of the force exerted by the string on the pulley is : (A)

6mg downward 5

(B)

6mg upward 5

(C)

mg downward 5

(D)

mg downward 4

Sol.

(b) In the above question the coefficient of friction between Q and the rough plane is : (A)

4 5

(B)

1 5

(C)

3 5

(D)

2 5

Sol.

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N.L.M. 83. A force F = t is applied to block A as shown in figure. The force is applied at t = 0 seconds when the system was at rest and string is just straight without tension. Which of the following graphs gives the friction force between B and horizontal surface as a function a time 't'. B A m

m

f

For Q.84. to Q.88 refer given figure (5 questions) 84. When F = 2N, the frictional force between 5 kg block and ground is 10kg

F

5kg

F

µs = µk

µ s > µk

Page # 85

(A) 2N Sol.

(B) 0

(C) 8 N

(D) 10 N

f

(A)

(B) t

t

f

f

(C)

(D) t

t

Sol.

85. When F = 2N, the frictional force between 10 kg block and 5 kg block is (A) 2N (B) 15N (C) 10N (D) None Sol.

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Page # 86 86. The maximum “F” which will not cause motion of any of the blocks (A) 10N (B) 15N (C) data insufficient (D) None Sol.

N.L.M. 89. A truck starting from rest moves with an acceleration of 5 m/s2 for 1 sec and then moves with constant velocity. The velocity w.r.t. ground v/s time graph for block in truck is (Assume that block does not fall off the truck) µ = 0.2

5 m/s

3 m/s

(A)

(B)

1 sec

87. The maximum acceleration of 5 kg block (B) 3 m/s2 (C) 0 (D) None (A) 1 m/s2 Sol.

1 sec

5 m/s (C)

(D) None of these

2.5 sec Sol.

88. The acceleration of 10 kg block when F = 30 N (B) 3 m/s2 (C) 1 m/s2 (D) None (A) 2 m/s2 Sol.

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N.L.M. 90. A board is balanced on a rough horizontal semicircular log. Equilibrium is obtained with the help of addition of a weight to one of the ends of the board when the board makes an angle θ with the horizontal. Coefficient of friction between the log and the board is

Page # 87 91. A stationary body of mass m is slowly lowered onto a massive plateform of mass M (M >> m) moving at a speed V0 = 4 m/s as shown in fig. How far will the body slide along the platform (µ = 0.2 and g = 10 m/ s2 ) ?

(A) 4 m Sol. (A) tan θ Sol.

(B) cos θ

(C) cot θ

(B) 6 m

(D) sin θ

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(C) 12 m

(D) 8 m

Page # 88

N.L.M.

(One or more than one option is correct)

Exercise - II (A) N. L. M

1. A student calculates the acceleration of m1 in figure (m1 – m 2 )g shown as a1 = m + m . Which assumption is not 1 2 required to do this calculation.

3. Two men of unequal masses hold on to the two sections of a light rope passing over a smooth light pulley. Which of the following are possible?

m1 m2 (A) pulley is frictionless (C) pulley is massless Sol.

(B) string is massless (D) string is inextensible

2. Which graph shows best the velocity-time graph for an object launched vertically into the air when air resistance is given by |D| = bv? The dashed line shows the velocity graph if there were no air resistance.

v

v t

(A)

t

(B)

v (C)

(A) The lighter man is stationary while the heavier man slides with some acceleration (B) The heavier man is stationary while the lighter man climbs with some acceleration (C) The two men slide with the same acceleration in the same direction (D) The two men move with accelerations of the same magnitude in opposite directions Sol.

v t

(D)

t

Sol.

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Page # 89

N.L.M. 4. Adjoining figure shows a force of 40 N acting at 30° to the horizontal on a body of mass 5 kg resting on a smooth horizontal surface. Assuming that the acceleration of free-fall is 10 ms–2, which of the following statements A, B, C, D, E is (are) correct?

5 kg

6. In the system shown in the figure m1 > m2. System is held at rest by thread BC. Just after the thread BC is burnt :

40 N 30°

[1] The horizontal force acting on the body is 20 N [2] The weight of the 5 kg mass acts vertically downwards [3] The net vertical force acting on the body is 30 N (A) 1, 2, 3 (B) 1, 2 (C) 2 only (D) 1 only Sol.

spring k

B m2

m1 A C

(A) acceleration of m2 will be upwards (B) magnitude of acceleration of both blocks will be  m1 – m 2  equal to  m + m  g  1 2 (C) acceleration of m1 will be equal to zero (D) magnitude of acceleration of two blocks will be non-zero and unequal. Sol.

5. For ordinary terrestrial experiments, which of the following observers below are inertial. (A) a child revolving in a “giant wheel”. (B) a driver in a sports car moving with a constant high speed of 200 km/h on a straight road. (C) the pilot of an aeroplane which is taking off. (D) a cyclist negotiating a sharp turn. Sol. 7. A particle is resting on a smooth horizontal floor. At t = 0, a horizontal force starts acting on it. Magnitude of the force increases with time according to law F = α.t, where α is a constant. For the figure shown which of the following statements is/are correct ? y

O

2 1

x

(A) Curve 1 shows acceleration against time (B) Curve 2 shows velocity against time (C) Curve 2 shows velocity against acceleration (D) none of these

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Page # 90

N.L.M.

Sol.

9. Figure shows the displacement of a particle going along the x-axis as a funtion of time :

x

E

A

8. Two blocks A and B of equal mass m are connected through a massless string and arranged as shown in figure. Friction is absent everywhere. When the system is released from rest.

B

C

D t

(A) the force acting on the particle is zero in the region AB (B) the force acting on the particle is zero in the region BC (C) the force acting o the particle is zero in the region CD (D) the force is zero no where. Sol.

A fixed

30°

mg 2 mg (B) tension in string is 4 (C) acceleation of A is g/2 3 (D) acceleration of A is g 4 Sol.

B

(A) tension in string is

10. A force of magnitude F1 acts on a particle so as to accelerate it from rest to a velocity v. The force F1 is then replaced by another force of magnitude F2 which decelerates it to rest. (A) F1 must be the equal to F2 (B) F1 may be equal to F2 (C) F1 must be unequal to F2 (D) None of these Sol.

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Page # 91

N.L.M. 11. In the figure, the blocks A, B and c of mass m each have acceleration a1. a2 and a3 respectively. F1 and F2 are external forces of magnitudes 2 mg and mg respectively.

m F1=2mg A

m B

2m

m C

Sol.

F2=mg

(A) a1 = a2 = a3 (B) a1 > a2 > a3 (C) a1 = a2, a2 > a3 (D) a1 > a2 , a2 = a3 Sol.

12. A rope is stretched between two boats at rest. A sailor in the first boat pulls the rope with a constant force of 100 N. First boat with the sailor has a mas of 250 kg whereas mass of second boat is double of that mass. If the initial distance between the boats was 100 m, the time taken for two boats to meet each other is -

(A) 13.8 s

(B) 18.3 s

(C) 3.18 s

(D) 31.8 s

13. A chain of length l is placed on a smooth spherical surface of radius r with one of its ends fixed at the top of the surface. Length of chain is assumed to be πr l< . Acceleration of each element of chain when 2 upper end is released is -

(A)

g  r  1– cos  r  

(B)

rg    1– cos  r 

(C)

g    1– sin  r  r

(D)

rg  r  1– sin   

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Page # 92

N.L.M.

Sol.

(B) FRICTION 15. A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, then (A) µs = 0.60 (B) µk = 0.52 (C) µk = 0.60 (D) µs = 0.52 Sol.

14. Five persons A, B, C, D & E are pulling a cart of mass 100 kg on a smooth surface and cart is moving with acceleration 3 m/s2 in east direction. When person ‘A’ stops pulling, it moves with acceleration 1 m/s2 in the west direction. When person ‘B’ stops pulling, it moves with acceleration 24 m/s2 in the north direction. The magnitude of acceleration of the cart when only A & B pull the cart keeping their directions same as the old directions, is : (A) 26 m/s2 (C) 25 m/s2 Sol.

(B) 3 71 m / s 2 (D) 30 m/s2

16. The coefficient of friction between 4 kg and 5 kg blocks is 0.2 and between 5 kg block and ground is 0.1 respectively. Choose the correct statements P

4 kg

Q

5 kg

F

(A) Minimum force needed to cause system to move is 17N (B) When force is 4N static friction at all surfaces is 4 N to keep system at rest. (C) Maximum acceleration of 4 kg block is 2 m/s2 (D) Slipping between 4 kg and 5 kg blocks start when F is 17 N 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 93

N.L.M. Sol.

Sol.

17. In a tug-of-war contest, two men pull on a horizontal rope from opposite sides. The winner will be the man who (A) exerts greater force on the rope (B) exerts greater force on the ground (C) exerts a force on the rope which is greater than the tension in the rope (D) makes a smaller angle with the vertical Sol.

Q

u

e

s

t i o

n

N

o

.

1

9

t o

2

0

(

2

q

u

e

s

t i o

n

s

)

In figure, two blocks M and m are tied together with an inextensible and light string. The mass M is placed on a rough horizontal surface with coefficient of friction µ and the mass m is hanging vertically against a smooth vertical wall. The pulley is frictionless. M Rough Smooth

18. A man pulls a block heavier than himself with a light horizontal rope. The coefficient of friction is the same between the man and the ground, and between the block and the ground.

m

19. Choose the correct statement(s) (A) The system will accelerate for any value of m (B) The system will accelerate only when m > M (C) The system will accelerate only when m > µM (D) Nothing can be said Sol.

(A) The block will not move unless the man also moves (B) The man can move even when the block is stationary (C) If both move, the acceleration of the man is greater than the acceleration of the block (D) None of the above assertions is correct

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Page # 94 20. Choose the correct statement(s) related to the tension T in the string (A) When m < µM, T = mg (B) When m < µM, T = Mg (C) When m > µM, µMg < T < mg (D) When m > µM, mg < T < µMg Sol.

Question No. 21 to 23 (3 questions) Imagine a situation in which the horizontal surface of block M0 is smooth and its vertical surface is rough with a coefficient of friction µ

N.L.M. Sol.

22. In above problem, choose the correct value(s) of F which the blocks M and m remain stationary with respect to M0 g µ

(B)

(C) (M0 + M + m)

mg M

(D) none of these

Smooth M F

Mo

m

m(M0 + M + m)g M – µm

(A) (M0 + M + m)

Sol.

Rough

21. Identify the correct statement(s) (A) If F=0, the blocks cannot remain stationary (B) For one unique value of F, the blocks M and m remain stationary with respect to M0 (C) The limiting friction between m and M 0 is independent of F (D) There exist a value of F at which friction force is equal to zero.

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Page # 95

N.L.M.

24. The value(s) of mass m for which the 100 kg block remains is static equilibrium is

100

23. Consider a special situation in which both the faces of the block M0 are smooth, as shown in adjoining figure. Mark out the correct statement(s) (A) 35 kg Sol.

Smooth M

F

M0

µ = 0.3

37°

(B) 37 kg

m Smooth

(A) If F = 0, the blocks cannot remain stationary (B) For one unique value of F, the blocks M and m remain stationary with respect to block M0 (C) There exists a range of F for which blocks M and m remain stationary with respect to block M0 (D) Since there is no friction, therefore, blocks M and m cannot be in equilibrium with respect to M0 Sol.

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m

(C) 83 kg

(D) 85 kg

Page # 96

N.L.M. 26. Car is accelerating with acceleration = 20 m/s2. A box that is placed inside the car, of mass m = 10 kg is put in contact with the vertical wall as shown. The friction coefficient between the box and the wall is µ = 0.6. µ = 0.6 10kg

20m/s2

(A) The acceleration of the box will be 20 m/sec2 (B) The friction force acting on the box will be 100 N (C) The contact force between the vertical wall and

25. The contact force exerted by one body on another body is equal to the normal force between the bodies. It can be said that : (A) the surface must be frictionless (B) the force of friction between the bodies is zero (C) the magnitude of normal force equals that of friction (D) It is possible that the bodies are rough and they do not slip on each other. Sol.

the box will be 100 5 N (D) The net contact force between the vertical wall and the box is only of electromagnetic in nature. Sol.

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Page # 97

N.L.M.

(SUBJECTIVE

Exercise - III

PROBLEMS)

1. Two masses A and B, lie on a frictionless table. They are attached to either end of a light rope which passes around a horizontal movable pulley of negligible mass. Find the acceleration of each mass MA = 1 kg, MB = 2 kg, MC = 4 kg. The pulley P2 is vertical.

P1

B

P2

A

C

Sol. 3. To Paint the side of a building, painter normally hoists himself up by pulling on the rope A as in figure. The painter and platform together weigh 200 N . The rope B can withstand 300 N. Find

2. Block A of mass m/2 is connected to one end of light rope which passes over a pulley as shown in the fig. Man of mass m climbs the other end of rope with a relative acceleration of g/6 with respect to rope find acceleration of block A and tension in the rope.

(a) the maximum acceleration of the painter. (b) tension in rope A (i) when painter is at rest (ii) when painter moves up with an acceleration 2 m/s2. Sol.

g/6 m

m/2 Sol.

A

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Page # 98 4. An inclined plane makes an angle 30º with the horizontal. A groove OA = 5 m cut in the plane makes an angle 30º with OX. A short smooth cylinder is free to slide down the influence of gravity. Find the time taken by the cylinder to due to reach from A to O. (g = 10 m/s2)

Sol.

37º

30°

30°

Sol.

x

5. Same spring is attached with 2 kg, 3 kg and 1 kg blocks in three different cases as shown in figure. If x1, x2 and x3 be the extensions in the spring in these three cases then find the ratio of their extensions.

2 kg

2 kg

(a)

2.5 kg

Spring balance M

er ind A cyl

O

N.L.M. 6. Find the reading of spring balance as shown in figure. Assume that mass M is in equilibrium. (All surfaces are smooth)

3 kg

2 kg

(b)

7. At what value of m1 will 8 kg mass be at rest.

8kg

1 kg

2 kg

(c) 5kg

Sol.

m1

Sol.

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Page # 99

N.L.M. 8. What force must man exert on rope to keep platform in equilibrium ?

10. Find force in newton which mass A exerts on mass B if B is moving towards right with 3 ms–2. Also find mass of A. (All surfaces are smooth) A

2

3m/s

1kg B 37º

Sol.

Sol.

9. Inclined plane is moved towards right with an acceleration of 5ms–2 as shown in figure. Find force in newton which block of mass 5 kg exerts on the incline plane. (All surfaces are smooth) 5kg

37º 5 m/s2

Sol.

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Page # 100 11. Force F is applied on upper pulley. If F = 30t where t is time in seconds. Find the time when m1 loses contact with floor.

N.L.M.

F 30t N

m1 m2 m1 = 4kg m2 = 1kg

13. The vertical displacement of block A in meter is given by y = t2/4 where t is in second. Calculate the downward acceleration aB of block B.

Sol.

A

y

B

Sol.

12. The 40 kg block is moving to the right with a speed of 1.5 m/s when it is acted upon by forces F1 & F2. These forces vary in the manner shown in the graph. Find the velocity of the block after t = 12 s Neglect friction and masses of the pulleys and cords. F(N)

F2

40

F2

F1

30

F1

20 10

t(s) 0

2

4

6

12

Sol.

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Page # 101

N.L.M. 14. An object of mass m is suspended in equilibrium using a string of length l and a spring having spring constant K (< 2 mg/l) and unstreched length l/2.

Sol.



m

(a) Find the tension in the string (b) What happens if K > 2 mg / l ?

16. A person of mass m is standing on a platform of mass M and wants to raise this platform. Massless pulleys are configured in two different ways as shown. We would like to know which configuration makes it easier to raise the platform. Answer the following questions in terms of m, M, a and constant as appropriate. [Note : Assume the rope is also massless and does not stretch.]

m

M

M

Fig(1)

15. Three monkeys A, B, and C with masses of 10, 15 & 8 kg respectively are climbing up & down the rope suspended from D. at the instant represented, A is descending the rope with an acceleration of 2 m/s2 & C is pulling himself up with an acceleration of 1.5 m/s2. Monkeys B is climbing up with a constant speed of 0.8 m/s. Treat the rope and monkeys as a complete system & calculate the tension T in the rope at D. (g = 10 m/s–2)

m Fig(2)

(a) For configuration (1) find the force, F, the person must exert straight up in order to accelerate the platform + person system with an acceleration a. Include a freebody diagram in your solution. (b) What force does the platform exert on the person when the acceleration of the system is a? Include a freebody diagram in your solution. (c) If platform is massless, M = 0, and he wants to raise it with a constant velocity find F. Does this configuration offer a mechanical advantage ? (That is, is F < mg ?) (d) Now repeat the above for configuration (2). First, find the force, F, the person must exert straight down in order to accelerate the platform+ person system with an upward acceleration a. Include a freebody diagram in your solution. (e) Now, what force does the platform exert on the

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Page # 102

N.L.M.

person when the acceleration of the system is a? Include a freebody diagram in your solution. (f) Again, if the platform is massless, M = 0, and he wants to raise it with a constant velocity find F. Does this configuration offer a mechanical advantage ? (That is, is F < mg?) Sol.

(B) FRICTION 17. Give the acceleration of blocks : 50N

µ

= 0.5

(A) µ ks = 0.4 5kg

40N

(B)

µ s = 0.5 µ k = 0.4 10kg

37°

µ s = µ k = 0.6 10kg

(C) 5kg

Sol.

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N.L.M. 18. Determine the coefficient of friction (µ), so that rope of mass m and length l does not slide down.

l/3

Page # 103 20. A rope so lies on a table that part of it lays over. The rope begins to slide when the length of hanging part is 25 % of entire length. The co-efficient of friction between rope and table is : Sol.

Sol.

21. A worker wishes to pile a cone of sand into a circular area in his yard. The radius of the circle is r, and no sand is to spill onto the surrounding area. If µ is the static coefficient of friction between each layer of sand along the slope and the sand, the greatest volume of sand that can be stored in this manner is : Sol.

19. A wooden block A of mass M is placed on a frictionless horizontal surface. On top of A, another lead block B also of mass M is placed. A horizontal force of magnitude F is applied to B. Force F is increased continuously from zero. Then draw the graph between A and F. [µk < µs] B A

F

Sol.

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Page # 104

N.L.M.

22. A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of 50 N. The coefficient of friction between the surfaces of contact is (g = 10 m/s2) T

24. A block of mass 1 kg is horizontally thrown with a velocity of 10 m/s on a stationary long plank of mass 2 kg whose surface has µ = 0.5. Plank rests on frictionless surface. Find the time when m1 comes to rest w.r.t. plank. Sol.

m horizontal

45°

Sol.

25. Block M slides down on frictionless incline as shown. Find the minimum friction coefficient so that m does not slide with respect to M.

m M

23. In the figure, what should be mass m so that block A slide up with a constant velocity.

37º Sol.

A g 1k 37º

m

=0.5

Sol.

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Page # 105

N.L.M. 26. The coefficient of static and kinetic friction between the two blocks and also between the lower block and the ground are µs = 0.6 and µk = 0.4. Find the value of tension T applied on the lower block at which the upper block begins to slip relative to lower block.

M = 2kg M = 2kg

T

Sol.

28. A body of mass 2kg rests on a horizontal plane having coefficient of friction µ = 0.5. At t = 0 a horizontal  force F is applied that varies with time F = 2t. The time constant t0 at which motion starts and distance moved in t = 2t0 second will be _______ and ________ respectively. Sol.

27. A thin rod of length 1 m is fixed in a vertical position inside a train, which is moving horizontally with constant acceleration 4 m/s2. A bead can slide on the rod, and friction coefficient between them is 1/2. If the bead is released from rest at the top of the rod, find the time when it will reach at the bottom. Sol.

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Page # 106

N.L.M.

29. Find the acceleration of the blocks and magnitude & direction of frictional force between block A and table, if block A is pulled towards left with a force of 50N.

F = 50N

A

x

5Kg

g = 10m/s2

B 4Kg

Sol.

31. Coefficient of friction between 5 kg and 10 kg block is 0.5. If friction between them is 20N. What is the value of force being applied on 5 kg. The floor is frictionless.

5kg

F

10kg Sol.

30. A block A of mass 2kg rests on another block B of mass 8kg which rests on a horizontal floor. The coefficient of friction between A and B is 0.2 while that between B and floor is 0.5. When a horizontal force F of 25N is applied on the block B, the force of friction between A and B is _________.

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Page # 107

N.L.M.

(TOUGH

Exercise - IV

1. The diagram shows particles A and B, of masses 0.2 kg and m kg respectively, connected by a light inextensible string which passes over a fixed smooth peg. The system is released from rest, with B at a height of 0.25m above the floor. B descends, hitting the floor 0.5s later. All resistances to motion may be ignored.

SUBJECTIVE PROBLEMS)

2. An ornament for a courtyard at a word’s fair is to be made up of four identical, frictionless metal sphere, each weighing 2 6 Newton. The spheres are to be arranged as shown, with three resting on a horizontal surface and touching each other; the fourth is to rest freely on the other three. The bottom three are kept from separating by spot welds at the points of contact with each other. Allowing for a factor of satety of 3N, how much tension must the spot welds with stand.

A 0.2 kg

B

m kg 0.25m

(a) Find the acceleration of B as it descends. (b) Find the tension in the string while B is descending and find also the value of m. (c) When B hits the floor it comes to rest immediately, and the string becomes slack. Find the length of time for which B remains at rest on the ground before being jerked into motion again. Sol.

Top View

Sol.

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Horizontal View

Page # 108

N.L.M.

3. A 1kg block ‘B’ rests as shown on a bracket ‘A’ of same mass. Constant foroes F1 = 20 N and F2 = 8N start to act at time t = 0 when the distance of block B from pulley is 50 cm. Time when block B reaches the pulley is _______.

Sol.

50cm F2

F1

B

A

Sol.

5. In figure shown, pulleys are ideal m1 > 2 m2. Initially the system is in equilibrium and string connecting m2 to rigid support below is cut. Find the initial acceleration of m2?

m2 m1

4. Two men of masses m1 and m2 hold on the opposite ends of a rope passing over a frictionless pulley. The mass m1 climbs up the rope with an acceleration of 1.2 m/s2 relative to the rope. The man m2 climbs up the rope with an acceleration of 2.0 m/s2 relative to the rope. Find the tension in the rope if m1 = 40 kg and m2 = 60 kg. Also find the time after which they will be at same horizontal level if they start from rest and are initially separated by 5m.

Sol.

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Page # 109

N.L.M. 6. The system shown adjacent is in equilibrium. Find the acceleration of the blocks A, B & C all of equal masses m at the instant when (Assume springs to be ideal) (a) The spring between ceiling & A is cut.

7. In the system shown. Find the initial acceleration of the wedge of mass 5M. The pulleys are ideal and the cords are inextensible. (there is no friction anywhere). M 5M

K A

2M

Sol.

B K C

(b) The string (inextensible) between A & B is cut. (c) The spring between B & C is cut. Also find the tension in the string when the system is at rest and in the above 3 cases. Sol.

8. A smooth right circular cone of semi vertical angle α = tan–1(5/12) is at rest on a horizontal plane. A rubber ring of mass 2.5 kg which requires a force of 15N for an extension of 10cm is placed on the cone. Find the increase in the radius of the ring in equilibrium. Sol.

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Page # 110

N.L.M. 10. A car begins to move at time t = 0 and then accelerates along a straight track with a speed given 2 ms–1 for 0 ≤ t ≤ 2 After the end of acceleration, the car continues to move at a constant speed. A small block initially at rest on the floor of the car begins to slip at t = 1sec. and stops slipping at t = 3sec. Find the coefficient of static and kinetic friction between the block and the floor. Sol. b

y

V

( t )

=

2

t

9. A block of mass m lies on wedge of mass M as shown in figure. Answer following parts separately. (a) With what minimum acceleration must the wedge be moved towards right horizontally so that block m falls freely. m

M

(b) Find the minimum friction coefficient required between wedge M and ground so that it does not move while block m slips down on it. Sol.

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Page # 111

N.L.M. 11. In the figure shown, (i) For what maximum value of force F can all these blocks move together. (ii) Find the value of force F at which sliding starts at other rough surfaces F

1kg 2kg 3kg

12. A particle having a mass m and velocity Vm in the y-direction is projected on to a horizontal belt that is moving with uniform velocity Vb in the x-direction as shown in figure. µ is the coefficient of friction between particle and belt. Assuming that the particle first touches the belt at the origin of the fixed xy coordinate system and remains on the belt, find the coordinates (x, y) of the point where the sliding stops. y

(iii) Find acceleration of all blocks, nature and value of friction force of force F = 18N. Sol.

vb

belt

x m

Sol.

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Page # 112

N.L.M.

Exercise - V

(JEE-PROBLEMS)

1. A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of (A) (2/3) k (B) (3/2) k (C) 3k (D) 6K [JEE 1999] Sol.

2. In the figure masses m1, m2 and M are 20 kg, 5 kg and 50 kg respectively. The co-efficient of friction between M and ground is zero. The co-efficient of friction between m1 and M and that between m2 and ground is 0.3. The pulleys and the string are massless. The string is perfectly horizontal between P1 and m1 and also between P2 and m2. The string is perfectly vertical between P1 and P2. An external horizontal force F is applied to the mass M. Take g = 10 m/s2. P1 m2

P2

m1 M

F

(a) Draw a free - body diagram for mass M, clearly showing all the forces. (b) Let the magnitude of the force of friction between m1 and M be f1 and that between m2 and ground be f2. For a particular F it is found that f1 =2f2. Find f1 and f2. Write down equations of motion of all the masses. Find F, tension in the string and accelerations of the masses. [JEE 2000] Sol.

3. The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be [JEE (Scr) 2001]

B

A

P m (A) 0°

(B) 30°

1/2

2 m (C) 45°

Q m (D) 60°

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Page # 113

N.L.M. 5. A block of mass

Sol.

3 kg is placed on a rough horizontal

surface whose coefficient of friction is 1 / 2 3 minimum value of force F (shown in figure) for which the block starts to slide on the surface. (g = 10m/s2)

60°

(A) 20 N

(B) 20 3 N

(C) 10 3 N Sol.

(D) None of these

4. A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given [JEE (Scr) 2001]

M

(A)

2 Mg

(B)

2 mg

(C)

(M + m)2 + m2 g

(D)

(M + m) 2 + M2 g

Sol.

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Page # 114

N.L.M.

6. Two blocks A and B of equal masses are released from an inclined plane of inclination 45° at t = 0. Both the blocks are initially at rest. The coefficient of kientic friction between the block A and the inclined plane is 0.2 while it is 0.3 for block B. Initially, the block A is

Sol.

2 m behind the block B. When and where their front faces will come in line. [Take g = 10m/s2].

2m

A

B

45°

A

B

Sol.

7. Two blocks A and B masses 2m and m, respectively, are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration of A and B, immediately after the string is cut, are respectively. [JEE 2006]

8. A circular disc with a groove along its diameter is placed horizontally. A block of mass 1 kg is placed as shown. The co-efficient of friction between the block and all surfaces of groove in contact is µ = 2/5. The disc has an acceleration of 25 m/s 2 . Find the acceleration of the block with respect to disc. [JEE 2006]

a=25m/s2 2m A m B

(A) g, g

(B) g, g/2

(C) g/2, g

(D) g/2, g/2

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Page # 115

N.L.M. Sol.

Sol.

9. Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance ‘a’ from the center P (as shown in the figure). Now, the midpoint of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surfaces. The magnitude of acceleration, when the separation between them becomes 2x, is [JEE 2007] F

m

a F a (A) 2m 2 a − x2

(C)

F x 2m a

m

P

10. STATEMENT-1 A cloth Covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table because STATEMENT-2 For every action there is an equal and opposite reaction (A) Statement-1 is True, Statement-2 is True; State me nt -2 i s a correc t ex pl anat i on for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement - 1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True [JEE 2007] Sol.

a F x (B) 2m 2 a − x2

(D)

F a2 − x2 2m x

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Page # 116

N.L.M.

11. STATEMENT-1 It is easier to pull a heavy object than to push it on a level ground. and STATEMENT-2 The magnitude of frictional force depends on the nature of the two surfaces in contact. (A) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 i s true, statement-2 is true’ statement-2 is NOT a correct explanati on for statement-1 (C) Statement-1 is true, statement-2 is false (D) Statement-1 is false, statement-2 is true [ JEE 2008] Sol.

13. A block of mass m is on an inclined plane of angle θ. The coefficient of friction betwen the block and the plane is µ and tan θ > µ. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to the positive. As P is varied from P = mg (sin θ – µ cos θ ) to Pz = mg (sin θ + µ cos θ), the frictional force f versus P graph will look like

P

θ

f

f

P2

(A)

P1

P

(B)

f

12. A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is 3 . The inclination θ of this inclined plane from the horizontal plane is gradually increased from 0º. Then (A) at θ = 30º, the block will start sliding down the plane (B) the block will remain at rest on the plane up to certain θ and then it will topple (C) at θ = 60º, the block will start sliding down the plane and continue to do so at higher angles (D) at θ = 60º, the block will start sliding down the plane and on further increasing θ, it will topple at certain θ [JEE 2009] Sol.

P1

P2

P1

P2

f

P1

(C)

P

P

P2

P

(D)

[JEE 2010] Sol.

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N.L.M. 14. A block is moving on an inclined plane making an angle 45º with horizontal and the coefficient of friciton is µ. the force required to just push it up the inclined plane is 3 times the force requried to just prevent it from sliding down. If we define N = 10µ, then N is [JEE 2011]

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

Page # 117

Page # 118

N.L.M.

EXERCISE - I

(A) NEWTONS'S LAW OF MOTION 1.

A

2.

A

3.

A

4.

A

5.

B

6.

A

7.

A

8.

D

9.

B

10.

C

11.

A

12.

A

13.

B

14.

A

15.

A

16.

D

17.

A

18.

B

19.

B

20.

B

21.

A

22.

C

23.

C

24.

A

25.

A

26.

A

27.

B

28.

A

29.

B

30.

B

31.

A

32.

B

33.

C

34.

C

35.

B

36.

C

37.

C

38.

C

39.

C

40.

A

41.

B

42.

A

43.

C

44.

B

45.

D

46.

B

47.

A

48.

B

49.

B

50.

(i)

A

(ii)

A

(iii)

C

(iv)

D

(v)

B

(vi)

D

(vii)

B

(viii)

B

(i)

A

(ii)

A

(iii)

A

(iv)

C

(v)

B

(vi)

C

(vii)

C

(viii)

B

A

53.

B

54.

A

55.

A

56.

C

57.

C

58.

D

51.

52.

(B) FRICTION

fDC

fED 59.

E

fED

2 m/s

1 m/s

D

3 m/s

C

fDC

fCS fBA

fCB A

5 m/s

5 m/s

B

fAg

fAg

fBA fkAB

fkAB

60.

fkAB

61.

A

62.

B

63.

A

64.

C

65.

C

66.

C

67.

A

68.

B

69.

A

70.

B

71.

C

72.

A

73.

A

74.

A

75.

A

76.

D

77.

B

78.

A

79.

A

80.

D

81.

C

82. (a) B (b) D (c) A

83.

A

84.

A

85.

A

86.

A

87.

C

88.

90.

A

91.

A

A

A

89.

C

B

fkBG

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Page # 119

N.L.M.

EXERCISE - II

(A) NEWTONS'S LAW OF MOTION

1.

C

7. 13.

2.

B

3.

A,B,D

4.

C

5.

B

6.

A,C

A,B,C 8.

B,D

9.

A,B,C

10.

B

11.

B

12.

B

B

C

14.

(B) FRICTION

15.

A,B

16.

C

17.

B

18.

A,B,C

19.

C

20.

A,C

21.

A,D

22.

B,C

23.

A,B

24.

B,C

25.

B,D

26.

A,B,C,D

EXERCISE - III

(A) NEWTON'S LAW OF MOTION 1.

4 g 2g 3 g , , 5 5 5

2. a =

4g 13mg ,T= 9 18

3. (a) 5m/s2, (b) (i) 100N, (ii) 120N

4. 2 sec

5. x2 > x1 > x3 x1 : x2 : x3 : 15 : 18 : 10

6. 12 N

7. 10/3 kg

8. 300 N

9. 55

11. 2sec

12. 12 m/s

13. aB = 4m/s2 ( ↑ )

10. 5N, 16/31 kg 14. (a) T = mg –

kl , (b) length of spring will less than ‘l’ and T = 0 in the string. 2

15. 322 N 16. (a) T =

(m + M)( a + g) (m + M)(a + g) mg , (b) N = m(a + g) + T, (c) T = , (d) T = 2 3 2

(e) N = m (a + g) – T,

(f) T =

mg 3

(B) FRICTION

17. (A)4m/s2, (B) 1.2 m/s2, (C) 0 a

18. µ = 2

aB aA

1 µ π r3 3

20. 0.33

21.

24. 4/3 sec

25. 3/4

26. 40 N

29. 10 i

30. 0

31. 30 N

19.

22. 1/2

23. 1kg

27. 1/2 sec

28. 5 sec, 125/6 m

F

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Page # 120

N.L.M.

EXERCISE - IV

ANSWER KEY 1. (a) 2 ms–2, (b) 2.4 N 0.3 (c) 0.2 s  m1 − 2m 2  g 5.  2m   2

6. (a) a A =

(c) a A = aB = g / 2 ↑ ; T =

7. 2g/23

4. 556.8 N , 1.47 sec

8. ∆ r =

mg cot α 4 π 2k

, 1cm

m sin θ cos θ

10. µs = 0.4 , µk = 0.3

m cos 2 θ + M

12. x =

11. 12 N, 21 N, 4 m/s2, 2 m/s2, 4 N, 6 N

Vb V Vm2 + Vb2 y = m Vm2 + Vb2 2µg 2µg

EXERCISE - V

1. B

3. 0.5 sec

3g ↓ = aB ; a C = 0 ; T = mg / 2 (b) a A = 2g ↑ ; aB = 2g ↓ ; aC = 0, T = 0 2

3mg ; T = 2mg 2

9. (a) a = g cotθ, (b) µ min =

2. 2 N

2. (b) a = 3/5 m/s2, T = 18 N, F = 60 N

6. 11.313 m

7. B

8. 10 m/s

12. B

13. A

14. 5N

2

3. C

4. D

9. B

10. B

5. A 11. B

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CIRCULAR MOTION & W.P.E THEORY AND EXERCISE BOOKLET CONTENTS

TOPIC

S. NO.

PAGE NO.

1.

Circular Motion ........................................................................ 3

2.

Kinematics of Circular Motion .............................................. 3 – 10

3.

Dynamics of circular Motion ................................................ 11 – 12

4.

Simple Pendulam ................................................................... 13

5.

Circular Motion in Horizontal Plane ......................................... 13

6.

Motion of motorcyclist on a curved ...................................... 14 – 16 path

7.

Circular Turning on roads .................................................... 16 – 17

8.

Death Well .............................................................................. 18

9.

Motion of a cyclist on a circular path .................................. 18 – 19

10.

Effect of earth rotation on Apparent .................................... 19 – 20 weight

11.

Some Solved Examples ..................................................... 20 – 23

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CIRCULAR MOTION

Page # 2

WORK POWER ENERGY 1. Work ................................................................................ 24–25 2. Units of Work ...................................................................... 25 3 Work Done by multiple forces ......................................... 25–27 4. Work Done by a variable force ....................................... 27 – 28 5. Area under force Displacement curve ................................. 28 6. Internal work .................................................................... 28 – 29 7. Conservative force .......................................................... 29 – 31 8. Non-conservative forces .................................................. 31–32 9. Energy ............................................................................ 32 – 33 10. Conservative force and Potential .................................. 34 – 35 energy 11. Work energy Theorem ................................................... 35–41 12. Power ........................................................................... 41 – 43 13. Vertical Circular Motion ................................................ 43 – 48 14. Exercise - I.................................................................... 49 – 68 15. Exercise - II ................................................................... 69 – 73 16. Exercise - III .................................................................. 74 – 84 17. Exercise - IV ................................................................. 85 – 89 18. Exercise- V ................................................................... 90 – 97 19. Answer key .................................................................. 98 – 100

IIT - JEE Syllbabus : Circular Motion (uniform and non-uniform), Work, Power, Kinetic Energy, Potential Energy, Conservation of Mechanical Energy.

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CIRCULAR MOTION

1.

Page # 3

CIRCULAR MOTION When a particle moves in a plane such that its distance from a fixed (or moving) point remains constant then its motion is called as the circular motion with respect to that fixed (or moving) point. That fixed point is called centre and the distance between fixed point and particle is called radius. v v v

A θ1 θ2

B The car is moving in a straight line with respect to the man A. But the man B continuously rotate dθ =0 his face to see the car. So with respect to man A dt dθ ≠0 But with respect to man B dt Therefore we conclude that with respect to A the motion of car is straight line but for man B it has some angular velocity

2.

KINEMATICS OF CIRCULAR MOTION :

2.1

Variables of Motion :

(a)

Angular Position : The angle made by the position vector with given line (reference line) is called angular position Circular motion is a two dimensional motion or motion in a plane. Suppose a particle P is moving in a circle of radius r and centre O. The position of the particle P at a given instant may be described by the angle θ between OP and OX. This angle θ is called the angular position of the particle. As the particle moves on the circle its angular position θ change. Suppose the point rotates an angle ∆θ in

Y P' ∆θ P θ r O

x

time ∆t. (b)

Angular Displacement : Definition : Angle rotated by a position vector of the moving particle in a given time interval with some reference line is called its angular displacement.

______________________________________________________________________________________________ Important point : •

It is dimensionless and has proper unit SI unit radian while other units are degree or revolution 2π rad = 360° = 1 rev

Infinitely small angular displacement is a vector quantity but finite angular displacement is not because the addition of the small angular displacement is cummutative while for large is not.     dθ1 + dθ 2 = dθ 2 + dθ1 but θ1 + θ2 ≠ θ2 + θ1

Direction of small angular displacement is decided by right hand thumb rule. When the fingers are directed along the motion of the point then thumb will represents the direction of angular displacement.

Angular displacement can be different for different observers

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CIRCULAR MOTION

Page # 4

______________________________________________________________________________________________ (c)

Angular Velocity ω (i) Average Angular Velocity ωav =

Total Angle of Rotation ; Total time taken

θ 2 – θ1 ∆θ ωav = t – t = ∆t 2 1

where θ1 and θ2 are angular position of the particle at time t1 and t2 respectively. (ii) Instantaneous Angular Velocity The rate at which the position vector of a particle with respect to the centre rotates, is called as instantaneous angular velocity with respect to the centre. dθ ∆θ lim ω = ∆t→0 = dt ∆t

_______________________________________________________________________________________________ Important points : •

It is an axial vector with dimensions [T–1] and SI unit rad/s.

For a rigid body as all points will rotate through same angle in same time, angular velocity is a characteristic of the body as a whole, e.g., angular velocity of all points of earth about its own axis is (2π/24) rad/hr.

If a body makes ‘n’ rotations in ‘t’ seconds then angular velocity in radian per second will be 2πn t If T is the period and ‘f’ the frequency of uniform circular motion ωav =

ωav =

2π × 1 = 2πf T

If θ = a – bt + ct2 then ω =

dθ = – b + 2ct dt

Relation between speed and angular velocity : ω=

dθ ∆θ lim = dt ∆t→θ ∆t

The rate of change of angular velocity is called the angular acceleration (α). Thus, α=

dω d2 θ = 2 dt dt

Y

P'

The linear distance PP’ travelled by the particle in time ∆t is ∆s = r∆θ

∆S or ∆lim t→0 ∆t

∆θ = r ∆lim t→ 0 ∆ t

or

P

∆θ O

r

X

∆s dθ =r or v = rω ∆t dt

Here, v is the linear speed of the particle It is only valid for circular motion   v v = rω is a scalar quantity ( ω ≠ ) r 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR MOTION

Page # 5

Ex.1

If θ depends on time t in following way θ = 2t2 + 3 then (b) ω at 3 sec (a) Find out ω average upto 3 sec.

Sol.

ωavg =

Total angular displacement θ f – θi = total time t 2 – t1

θf = 2 (3)2 + 3 = 21 rad θi = 2 (0) + 3 = 3 rad.

So,

ωavg =

21 – 3 = 6 rad/sec 3

dθ = 4t dt = 4 × 3 = 12 rad/sec

ωinstantaneous = ωat t = 3 sec

(d)

Relative Angular Velocity Angular velocity is defined with respect to the point from which the position vector of the moving particle is drawn Here angular velocity of the particle w.r.t. ‘O’ and ‘A’ will be different P'

A

ω PO =

O

P Ref lin

dα dβ ; ω PA = dt dt

Definition : Relative angular velocity of a particle ‘A’ with respect to the other moving particle ‘B’ is the angular velocity of the position vector of ‘A’ with respect to ‘B’. That means it is the rate at which position vector of ‘A’ with respect to ‘B’ rotates at that instant

A

VA

VB r

B

ω AB =

=

( VAB ) ⊥ rAB

here VAB ⊥ = Relative velocity ⊥ to position vector AB

Re lative velocity of A w.r.t. B perpendicu lar to line AB Seperation between A and B ( VAB ) ⊥ = VA sin θ1 + VB sin θ 2 rAB = r ω AB =

VA sin θ1 + VB sin θ 2 r

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CIRCULAR MOTION

Page # 6

_______________________________________________________________________________________ Important points : • If two particles are moving on the same circle or different coplanar concentric circles in same direction with different uniform angular speed ωA and ωB respectively, the rate of change of →

angle between OA and OB is B

B

A

dθ = ωB − ω A dt

O

A Initial line

O

Initial line

So the time taken by one to complete one revolution around O w.r.t. the other T=

If two particles are moving on two different concentric circles with different velocities then angular velocity of B relative to A as observed by A will depend on their positions and velocities. consider the case when A and B are closest to each other moving in same direction as shown in figure. In this situation   B vB v rel = | vB − v A | = vB − v A rrel

so,

TT 2π 2π = = 1 2 ω rel ω 2 − ω 1 T1 − T2

  = | rB − rA | = rB − rA

ω BA =

( v rel ) ⊥ vB − v A = rrel rB − rA

vA

A rA

r

rB

O

( v rel ) ⊥ = Relative velocity perpendicular to position vector

_______________________________________________________________________________________ Ex.2

Sol.

Two particles move on a circular path (one just inside and the other just outside) with angular velocities ω and 5 ω starting from the same point. Then, which is incorrect. 2π when their angular velocities are (a) they cross each other at regular intervals of time 4ω oppositely directed (b) they cross each other at points on the path subtending an angle of 60° at the centre if their angular velocities are oppositely directed π (c) they cross at intervals of time if their angular velocities are oppositely directed 3ω (d) they cross each other at points on the path subtending 90° at the centre if their angular velocities are in the same sense If the angular velocities are oppositely directed, they meet at intervals of 2π 2π π time t = ω = = 6ω 3ω rel Angle subtended at the centre by the crossing points π θ = ωt = = 60° 3 When their angular velocities are in the same direction,

Ans. Ex.3

2π 2π π π π ×ω = t’ = ω = = and θ’ = 4ω 2ω 2ω 2 rel (a)

Two moving particles P and Q are 10 m apart at a certain instant. The velocity of P is 8 m/s making 30° with the line joining P and Q and that of Q is 6 m/s making 30° with PQ in the figure. Then the angular velocity of Q with respect to P in rad/s at that instant is 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR MOTION

Page # 7 6 m/s 30°

P 30°

10 m

Q

8 m/s

(A) 0

(B) 0.1

(C) 0.4

(D) 0.7

6 m/s 30°

P

Sol.

30°

10 m

Q

8 m/s Angular velocity of Q relative to P =

Projection of VQP perpendicular to the line PQ Separation between P and Q

VQ sin θ 2 – VP sin θ1 6 sin 30°–(–8 sin 30° ) = = 0.7 rad/s PQ 10 ∴ (D)

(e)

Angular Acceleration α : (i) Average Angular Acceleration : Let ω1 and ω2 be the instantaneous angular speeds at times t1 and t2 respectively, then the average angular acceleration αav is defined as ω 2 – ω1 ∆ω αav = t – t = ∆t 2 1

(ii) Instantaneous Angular Acceleration : It is the limit of average angular acceleration as ∆t approaches zero, i.e., lim ∆ω = dω = ω dω α = ∆t→0 dt dθ ∆t

_______________________________________________________________________________________________ Important points : •

It is also an axial vector with dimension [T–2] and unit rad/s2

If α = 0, circular motion is said to be uniform.

As ω =

dθ dω d2θ , α= = , dt dt dt 2

i.e., second derivative of angular displacement w.r.t time gives angular acceleration. •

α is a axial vector and direction of α is along ω

if ω increases and opposite to ω if ω decreases

_______________________________________________________________________________________ (f)

Radial and tangential acceleration Acceleration of a particle moving in a circle has two components one is along e t (along tangent) and the other along − eˆ r (or towards centre). Of these the first one is the called the tangential acceleration. (at) and the other is called radial or centripetal acceleration (ar). Thus.

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CIRCULAR MOTION

Page # 8

at =

dv = rate of change of speed dt 2

2 v a r = ω2 r = r   = v and r r Here, at is the component which is responsible for changing the magnitude of speed of the particle in circular motion. ar is the component which is responsible for changing the direction of particle in circular motion. the two component are mutually perpendicular. Therefore, net acceleration of the particle will be :

 dv  a = ar2 + a 2t = (rω 2 ) 2 +    dt 

2

2

2  v2   dv  =   +    dt   r 

Following three points are important regarding the above discussion : 1. In uniform circular motion, speed (v) of the particle is constant, i.e.,

dv = 0 . Thus, dt

at = 0 and a = ar = rω2 dv = positive, i.e., at is along e t or tangential acceleration of dt  dv    et particle is parallel to velocity v because v = rω e t and ar = dt dv 3. In decelerated circular motion, = negative and hence, tangential acceleration is anti-parallel dt  to velocity v .

2. In accelerated circular motion,

(g)

Relation between angular acceleration and tangential acceleration we know that v = rω Here, v is the linear speed of the particle Differentiating again with respect to time, we have dv dω =r or at = rα dt dt dv Here, at = is the rate of change of speed (not the rate of change of velocity). dt

at =

Ex.4 Sol.

A particle travels in a circle of radius 20 cm at a speed that uniformly increases. If the speed changes from 5.0 m/s to 6.0 m/s in 2.0s, find the angular acceleration. The tangential acceleration is given by at =

v 2 – v1 dv = t –t dt 2 1

( ∵ Here speed increases uniformly at =

∆v dv = ) ∆t dt

6.0 – 5.0 m/s2 = 0.5 m/s2 2.0 The angular acceleration is α = at/r

=

= Ex-5

Sol.

0.5 m / s 2 = 2.5 rad/s2 20 cm

A particle moves in a circle of radius 20 cm. Its linear speed at any time is given by v = 2t where v is in m/s and t is in seconds. Find the radial and tangential acceleration at t = 3 seconds and hence calculate the total acceleration at this time. The linear speed at 3 seconds is 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR MOTION

Page # 9

v = 2 × 3 = 6 m/s The radial acceleration at 3 seconds 6×6 v2 = = 180 m/s2 0.2 r The tangential acceleration is given by dv = 2 , because v = 2t. dt ∴ tangential acceleration is 2 m/s2.

=

Net Acceleration =

ar 2 + a t 2 =

(180)2 + (2)2 = 180.01 m/s2

T-1

Is it possible for a car to move in a circular path in such a way that it has a tangential acceleration but no centripetal acceleration ?

Ex.6

A particle moves in a circle of radius 2.0 cm at a speed given by v = 4t, where v is in cm/s and t in seconds. (a) Find the tangential acceleration at t = 1 s. (b) Find total accleration at t = 1 s. (a) Tangential acceleration

Sol.

at =

dv dt

or

at =

ac =

v2 (4) 2 = =8 R 2

a=

d (4 t) = 4 cm/s2 dt a 2t + a2c =

2 (4) 2 + (8) 2 = 4 5 m / s

Ex.7

A boy whirls a stone in a horizontal circle of radius 1.5 m and at height 2.0 m above level ground. The string breaks, and the stone files off horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the cetripetal acceleration of the stone while in circular motion ?

Sol.

t=

Ex.8 Sol.

2h = g

2×2 = 0.64 s 9.8

v=

10 = 15.63 m/s t

a=

vB2 = 0.45 m/s2 R

Find the magnitude of the acceleration of a particle moving in a circle of radius 10 cm with uniform speed completing the circle in 4 s. The distance covered in completing the circle is 2 π r = 2π × 10 cm. The linear speed is v = 2 π r/t =

2π × 10cm = 5 π cm/s. 4s 2

The acceleration is ar = Ex.9 Sol.

v 2 ( 5 πcm / s) = =2.5 π2 cm/s2 10 cm r

A particle moves in a circle of radius 20 cm. Its linear speed is given by v = 2t where t is in second and v in meter/second. Find the radial and tangential acceleration at t = 3s. The linear speed at t = 3s is v = 2t = 6 m/s The radial acceleration at t = 3s is

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CIRCULAR MOTION

Page # 10

ar = v2/r =

36m 2 / s 2 = 180 m/s2 0.20 m

The tangential acceleration is at =

dv d( 2t) = = 2 m/s2 dt dt

Y

Ex.10 Two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds vA = 0.7 m/s and vB = 1.5 m/s, respectively. Determine the time when they collide and the magnitude of the acceleration of B just before this happens. Sol.

1.5 t + 0.7 t = 2πR = 10 π a=

∴ t=

10 π = 14.3 s 2.2

5. 0

cm

A

B

vB =1.5 m/s O x vA=0.7m/s

vB2 = 0.45 m/s2 R Non-uniform circular motion

Uniform circular motion

speed of the particle is not constant i.e. ωis not constant  d| v| ≠0 at = dt

(1) Speed of the particle is constant i.e., ω is constant  d| v| (ii) a t = =0 dt

v2 ≠0 r

ar =

ar ≠ 0    anet = ar + a t

∴ anet = ar

anet

at

ar= anet ar

(h)

Relations among Angular Variables These relations are also referred as equations of rotational motion and are ω = ω0 + αt ...(1) 1 2 θ = ω0t + αt ...(2) 2 ω2 = ω02 + 2αθ ...(3) These are valid only if angular acceleration is constant and are analogous to equations of translatory motion, i.e.,

v = u + at ; s = ut +

1 2 at and 2

v2 = u2 + 2as

dθ, ω or α

ar vo

ra

O

ds r

t

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CIRCULAR MOTION

3.

Page # 11

DYNAMICS OF CIRCULAR MOTION : In circular motion or motion along any curved path Newton’s law is applied in two perpendicular directions one along the tangent and other perpendicular to it. i.e., towards centre. The compnent of net force towards the centre is called centripetal force. The component of net force along the tangent is called tangential force. tangential force (Ft) = Mat = M

dv = M α r ; where α is the angular acceleration dt

mv 2 r Ex.11 A small block of mass 100 g moves with uniform speed in a horizontal circular groove, with vertical side walls, of radius 25 cm. If the block takes 2.0s to complete one round, find the normal contact force by the slide wall of the groove. Sol. The speed of the block is

centripetal force (Fc) = m ω2 r =

v=

2π × ( 25 cm) = 0.785 m/s 2.0 s

The acceleration of the block is (0.785m / s) 2 v2 = = 2.46 m/s2 0.25 r towards the center. The only force in this direction is the normal contact force due to the side walls. Thus from Newton’s second law, this force is N = ma = (0.100 kg) (2.46 m/s2) = 0.246 N

a=

3.1

Centripetal Force : Concepts : This is necessary resultant force towards the centre called the centripetal force. mv 2 = mω2r r A body moving with constant speed in a circle is not in equilibrium. It should be remembered that in the absence of the centripetal force the body will move in a straight line with constant speed. It is not a new kind of force which acts on bodies. In fact, any force which is directed towards the centre may provide the necessary centripetal force.

F=

(i) (ii) (iii)

Ex.12 A small block of mass 100 g moves with uniform speed in a horizontal circular groove, with vertical side walls, of radius 25 cm. If the block takes 2.0s to complete one round, find the normal contact force by the slide wall of the groove. Sol. The speed of the block is v=

2π × ( 25 cm) = 0.785 m/s 2.0 s

The acceleration of the block is (0.785m / s) 2 v2 = = 2.5 m/s2 0.25 r towards the center. The only force in this direction is the normal contact force due to the slide walls. Thus from Newton’s second law, this force is N = ma = (0.100 kg) (2.5 m/s2) = 0.25 N

a=

3.2

Centrifugal Force : When a body is rotating in a circular path and the centripetal force vanishes, the body would leave the circular path. To an observer A who is not sharing the motion along the circular path, the body appears to fly off tangentially at the point of release. To another observer B, who is sharing the motion along the circular path (i.e., the observer B is also rotating with the body which is released, 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564

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CIRCULAR MOTION

Page # 12

it appears to B, as if it has been thrown off along the radius away from the centre by some force. This inertial force is called centrifugal force.) mv 2 . Centrifugal force is a fictitious force r which has to be applied as a concept only in a rotating frame of reference to apply Newton’s law of motion in that frame) FBD of ball w.r.t non inertial frame rotating with the ball.

Its magnitude is equal to that of the centripetal force =

ω T

mω 2r mg

Suppose we are working from a frame of reference that is rotating at a constant, angular velocity ω with respect to an inertial frame. If we analyse the dynamics of a particle of mass m kept at a distance r from the axis of rotation, we have to assume that a force mrω2 act radially outward on the particle. Only then we can apply Newton’s laws of motion in the rotating frame. This radially outward pseudo force is called the centrifugal force. T-2

A particle of mass m rotates in a circle of radius r with a uniform angular speed ω. It is viewed from a frame rotating about same axis with a uniform angular speed ω0. The centrifugal force on the particle is 2

2

 ω + ω0  (C) m  r  2 

2 0

(A) mω r

(B) mω r

(D) mω0ωr

B

: A rod move with ω angular velocity then we conclude following for point A & B in a rod. αA = αB s B > sA θA = θB vB > vA ωA = ωB atB > atA

A

ω

Ex.13 Find out the tension T1, T2 is the string as shown in figure 2 rad/sec. T1

1kg

1m

T2 1m

2kg

T1

m1

T2

m2

We know that ωm1 = ωm 2 ⇒

So

T1 = m1ω2R1 + T2 T2 = m ω2 R2 T2 = 2 × 4 × 2 = 16 N T1 = (1) (2)2 (1) + 16 N = 4 + 16 N T1 = 20 N 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR MOTION

4.

Page # 13

SIMPLE PENDULUM

Ex.14 A simple pendulum is constructed by attaching a bob of mass m to a string of length L fixed at its upper end. The bob oscillates in a vertical circle. It is found that the speed of the bob is v when the string makes an angle θ with the vertical. Find the tension in the string at this instant. Sol. The force acting on the bob are (figure) (a) the tension T (b) the weight mg. As the bob moves in a vertical circle with centre at O, the radial acceleration is v2/L towards O. Taking the components along this radius and applying Newton’s second law, we get mgcos T – mgcos θ = mv2/L or, T = m(gcos θ + v2/L) mg mgsin  |Fnet | =

5.

 mv 2   (mg sin θ) +   L   

2

2

2 2 = m g sin θ +

v4 L2

CIRCULAR MOTION IN HORIZONTAL PLANE ω A ball of mass m attached to a light and inextensible string rotates in a horizontal circle of radius r with an angular speed

θ

ω about the vertical. If we draw the force diagram of the ball. We can easily see that the component of tension force along

the centre gives the centripetal force and component of tension along vertical balances the gravitation force. Such a system is called a conical pendulum.

T T cos θ θ

T sinθ

mω 2r

mg FBD of ball w.r.t ground

Ex. 15 A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r. Find (a) the speed of the particle and (b) the tension in the string. Sol.

The situation is shown in figure. The angle θ made by the string with the vertical is given by sinθ = r/L

... (i)

T

θ

The forces on the particle are (a) the tension T along the string and

r

(b) the weight mg vertically downward. The particle is moving in a circle with a constant speed v. Thus, the radial acceleration towards the centre has magnitude v2/r. Resolving the forces along the radial direction and applying. Newton’s second law, Tsin θ = m(v2/r)

mg

...(ii)

As there is no acceleration in vertical directions, we have from Newton’s law, Tcosθ = mg

...(iii)

Dividing (ii) by (iii),

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L

CIRCULAR MOTION

Page # 14

tanθ =

v2 rg

or v =

And from (iii), T = Using (i), v =

6.

rgtan θ mg cos θ

r g 2

and

2 1/ 4

(L – r )

T=

mgL 2

(L – r 2 )1/ 2

MOTION OF A MOTORCYCLIST ON A CURVED PATH. A cylist having mass m move with constant speed v on a curved path as shown in figure. B D

A

C

E

We divide the motion of cyclist in four parts : (1) from A to B (2) from B to C (3) from C to D (4) from D to E (1 and 3 are same type of motion)

(A)

Motion of cyclist from A to B N+

mv 2 = mg cos θ R

mv N = mg cosθ – R f = mg sin θ

N+

mv 2 R

f

B

2

...(1) ..(2)

mgsinθ

θ

θ os gc m θ

(1)

As cyclist move upward A mg ∵ θ decreases & cos θ increases ∴ N increases and ∵ θ decreases sin θ decreases ∴ friction force required to balance mg sin θ (As cyclist is moving with constant speed) also decreases

(B)

Motion of cyclist from B to C B

2

N+

mv = mg cos θ R

⇒ N = mg cos θ –

mv 2 R

...(1)

f = mg sinθ ...(2) Therefore from B to C Normal force decrease but friction force increase becuse θ increases. (C)

N+

f mgcos θ

mv 2 R

mg sin θ

mg

C

Motion of cyclist from D to E N=

mv 2 + mg cos θ R

f = mg sin θ

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CIRCULAR MOTION

Page # 15

from D to E θ decreases therefore mg cos θ increase So N increase but f decreases

D

f mv 2 R

E

mg Ex.16 A hemispherical bowl of radius R is rotating about its axis of symmetry which is kept vertical. A small ball kept in the bowl rotates with the bowl without slipping on its surface. If the surface of the bowl is smooth and the angle made by the radius through the ball with the vertical is α. Find the angular speed at which the bowl is rotating. Sol. Let ω be the angular speed of rotation of the bowl. ω Two force are acting on the ball. 1. normal reaction N 2. weight mg The ball is rotating in a circle of radius r (= R sin α) with R α centre at A at an angular speed ω. Thus, N N sin α = mrω2 = mRω2 sin α A r N = mRω2 ...(i) ...(ii) and N cos α = mg mg Dividing Eqs. (i) by (ii),

we get

1 ω 2R = cos α g

ω=

g Rcos α

Ex.17 If friction is present between the surface of ball and bowl then find out the range of ω for which ball does not slip (µ is the friction coefficient) Friction develop a range of ω for which the particle will be at rest. Sol. (a) When ω > ω0 In this situation ball has a tendency to slip upwards α so the friction force will act downwards. So F.B.D of ball N = mω2r sin α + mg cos α. ... f + mg sin α = mω2r cos α ...(2) ∴ fmax = µN = µ(mω2r sin α + mg cosα) 2 r = R sin α mω r Substituting the values of fmax & r in eq. (2) we get

N α

m α

⇒ µ (mω2r sin α + mg cosα) ≥ mω2r cosα –mg sin α ∴ µ(mω2R sin 2α + mg cos α) ≥ mω2 R sinα cosα – mg sin α ω≤

(b)

f

mg

µg cos α + g sin α R sin α(cos α – µ sin α )

N 2

mω r

f + mω2r cos α = mg sin α f = m (g sinα – ω2r cosα ) ...(1) N = mg cos α + mω2r sin α ...(2) fmax = µN = µ(mg cos α + mω2r sinα) for equillibrium

α

f

when ω < ω0 In this situation ball has a tendency to slip downwards so the friction force will act upwards. So F.B.D of ball ⇒

r

α

m r

α

mg

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CIRCULAR MOTION

Page # 16 ⇒ µ(g cos α + ω2r sinα ) ≥ (gsinα – ω2 r cos α) ⇒ Substituting r = Rsinα then ω≥

7.

g(sin α – µ cos α ) R sin α(µ sin α + cos α )

CIRCULAR TURNING ON ROADS : When vehicles go through turnings, they travel along a nearly circular arc. There must be some force which will produce the required centripetal acceleration. If the vehicles travel in a horizontal circular path, this resultant force is also horizontal. The necessary centripetal force is being provided to the vehicles by following three ways. 1. By Friction only 2. By Banking of Roads only 3. By Friction and Banking of Roads both. In real life the necessary centripetal force is provided by friction and banking of roads both. Now let us write equations of motion in each of the three cases separately and see what are the constant in each case.

7.1

By Friction Only Suppose a car of mass m is moving at a speed v in a horizontal circular arc of radius r. In this case, the necessary centripetal force to the car will be provided by force of friction f acting towards center mv 2 r Further, limiting value of f is µN or fL = µN = µmg (N = mg) Thus, f =

Therefore, for a safe turn without sliding

mv 2 ≤ fL r

v2 mv 2 µmg or µ ≥ or v ≤ µrg ≤ rg r Here, two situations may arise. If µ and r are known to us, the speed of the vehicle should not

or

exceed

7.2

µrg and if v and r are known to us, the coefficient of friction should be greater than

v2 . rg

By Banking of Roads Only Friction is not always reliable at circular turns if high speeds and sharp turns are involved to avoid dependence on friction, the roads are banked at the turn so that the outer part of the road is some what lifted compared to the inner part. Applying Newton’s second law along the

N

radius and the first law in the vertical direction. Nsinθ =

mv 2 r

or

N cosθ = mg θ

from these two equations, we get v2 tanθ = rg

or

v=

W

rgtanθ

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CIRCULAR MOTION 7.3

Page # 17

By Friction and Banking of Road Both If a vehicle is moving on a circular road which is rough and banked also, then three forces may act on the vehicle, of these force, the first force, i.e., weight (mg) is fixed both in magnitude and direction. N θ θ f

y mg

x

θ

Figure (ii)

The direction of second force i.e., normal reaction N is also fixed (perpendicular or road) while the direction of the third i.e., friction f can be either inwards or outwards while its magnitude can be vari ed upt o a maxi mum l i mi t (fL = µN). So the magnitude of normal reaction N and directions plus magnitude of friction f are so mv 2 towards the center. Of r these m and r are also constant. Therefore, magnitude of N and directions plus magnitude of friction mainly depends on the speed of the vehicle v. Thus, situation varies from problem to problem. Even though we can see that : (i) Friction f will be outwards if the vehicle is at rest v = 0. Because in that case the component weight mg sinθ is balanced by f. (ii) Friction f will be inwards if adjusted that the resultant of the three forces mentioned above is

v > rgtanθ (iii) Friction f will be outwards if v
90°), we say that the work done by the force is negative.

Example : When a body is lifted, the work done by the gravitational force is negative. This is because the gravitational force acts vertically downwards while the displacement is in the vertically upwards direction.

__________________________________________________________________________________ Important points about work : 1. Work is said to be done by a force when its point of application moves by some distance.Force does no work if point of application of force does not move (S = 0) Example : A person carrying a load on his head and standing at a given place does no work. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

WORK, POWER & ENERGY 2.

Page # 25

Work is defined for an interval or displacement. There is no term like instantaneous work similar to instantaneous velocity. µ=0.2

2kg frictionless

3. 4. 5. 6.

2kg

10 N 2m

10 N 2m

Work done by 10 N force in both the cases are same = 20 N For a particular displacement, work done by a force is independent of type of motion i.e. whether it moves with constant velocity, constant acceleration or retardation etc. If a body is in dynamic equilibrium under the action of certain forces, then total work done on the body is zero but work done by individual forces may not be zero. When several forces act, work done by a force for a particular displacement is independent of other forces. A force is independent of reference frame. Its displacement depends on frame so work done by a force is frame dependent therefore work done by a force can be different in different reference frame.

________________________________________________________________________________________ 2.

UNITS OF WORK : In cgs system, the unit of work is erg. One erg of work is said to be done when a force of one dyne displaces a body through one centimetre in its own direction. ∴ 1 erg = 1 dyne × 1 cm = 1 g cm s–2 × 1 cm = 1 g cm2 s–2 Note : Another name for joule is newton metre. Relation between joule and erg 1 joule = 1 newton × 1 metre 1 joule = 105 dyne × 102 cm = 107 dyne cm 1 joule = 107 erg 1 erg = 10–7 joule Dimensions of Work : [Work] = [Force] [Distance] = [MLT–2] [L] = [ML2T–2] Work has one dimension in mass, two dimensions in length and ‘–2’ dimensions in time, On the basis of dimensional formula, the unit of work is kg m2 s–2. Note that 1 kg m2 s–2 = (1 kg m s–2) m = 1 N m = 1 J.

3.

WORK DONE BY MULTIPLE FORCES :    If several forces act on a particle, then we can replace F in equation W = F . S by the net force  F where     F = F1 + F2 + F3 +......   F . S ∴ W= ...(i)  This gives the work done by the net force during a displacement S of the particle. We can rewrite equation (i) as :       W = F1. S + F2 . S + F3 . S+..... or W = W1 + W2 + W3 + ......... So, the work done on the particle is the sum of the individual work done by all the forces acting on the particle.

∑ ∑

[∑ ]

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WORK, POWER & ENERGY

Page # 26

Ex.1

Sol.

A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with horizontal. Coefficient of friction between block and surface is µ. If the block travels with uniform velocity, find the work done by this applied force during a displacement d of the block. The forces acting on the block are shown in Figure. As the block moves with uniform velocity the resultant force on it is zero. ∴ F cos θ = µN ...(i) F sin θ + N = Mg ...(ii) Eliminating N from equations (i) and (ii), N F cos θ = µ(Mg – Fsin θ) F F=

µMg cos θ + µ sin θ

M

Work done by this force during a displacement d W = F . d cos θ =

Ex.2

Sol.

µMgdcos θ cos θ + µ sin θ

Mg

 A particle moving in the xy plane undergoes a displacement S = (2.0 ˆi + 3.0ˆj)m while a constant  force F = (5.0 ˆi + 2.0ˆj)N acts on the particle.

(a) Calculate the magnitude of the displacement and that of the force. (b) Calculate the work done by the force.   (a) s = (2.0 i + 3.0 j ) F = (5.0 i + 2.0 j )  | s | = x2 + y 2 = (2.0)2 + (3.0)2 = 13 m  |F|=

Fx2 + Fy2 =

(5.0 )2 + ( 2.0) 2 = 5.4N

  (b) Work done by force, W = F . s

= (5.0 ˆi + 2.0 ˆj) . (2.0ˆi + 3.0ˆj) N. m Ex.3

= 10 + 0 + 0 + 6 = 16 N.m = 16 J

A block of mass m is placed on an inclined plane which is moving with constant velocity v in horizontal direction as shown in figure. Then find out work done by the friction in time t if the block is at rest with respect to the incline plane. v=const.

m µ θ

Sol.

F.B.D of block with respect to ground.

f f

m

v

π−θ

v θ θ m mg g sin θ B

A

θ

mg

mgsin θ

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WORK, POWER & ENERGY

Page # 27

Block is at rest with respect to wedge ⇒ f = mg sin θ In time t the displacement of block with respect to ground d = vt Work done by friction for man A Wf = (component of friction force along displacement) × displacement Wf = mgsinθ.vt cos(180°–θ) Wf = – mg vt cosθ sin θ Wf for man B = 0 (displacement is zero with respect to man B)

4.

WORK DONE BY A VARIABLE FORCE :

(A)

When F as a function of x, y, z When the magnitude and direction of a force vary in three dimensions, it can be expressed as a function of the position. For a variable force work is calculated for infinitely small displacement and for this displacement force is assumed to be constant.   dW = F. ds The total work done will be sum of infinitely small work B

WA → B =

  F. ds =

A

B





∫ (F cos θ)ds A

It terms of rectangular components,  F = Fx i + Fy j + Fzk  ds = dx i + dy j + dzk xB

WA → B =

yB

Fx dx +

xA

zB

∫ F dz

Fy dy +

z

yA

zA

Ex.4

A force F = (4.0 x i + 3.0 y j ) N acts on a particle which moves in the x-direction from the

Sol.

or ig in to x = 5.0 m. Find the work done on the object by the force. Here the work done is only due to x component of force because displacement is along x-axis. x2

i.e., W =

5

Fx dx =

x1

Ex.5

Sol.

∫ 4x dx = [2x ] 2

0

5 0

= 50 J

A force F = 0.5x + 10 acts on a particle. Here F is in newton and x is in metre. Calculate the work done by the force during the displacement of the particle from x = 0 to x = 2 metre. → Small amount of work done dW in giving a small displacement dx is given by →→ dW = F . dx or dW = Fdx cos 0° or dW = Fdx [∴ cos 0° = 1] x =2

Total work done, W =

x =2

Fdx =

x =0 x=2

=

x =2

∫ 0.5xdx + ∫ 10dx

x =0

x =0

F

∫ (0.5x + 10)dx

x = 0.5 2

+ve

0

x =0

2

10

2

x= 2 x= 2

+ 10 x x= 0 x= 0

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x

WORK, POWER & ENERGY

Page # 28 0.5 2 [2 – 02] + 10[2 – 0] = (1 + 20) = 21 J 2

= (B)

When F is given as a function of Time(t) :

Ex.6

The force F = 2t2 is applied on the 2 kg block. Then find out the work done by this force in 2sec. Initially at time t = 0, block is at rest.

at t = 0, v = 0 F=2t2

2kg

Sol.

F = ma ⇒ 2t2 = 2a ⇒

a =t2 v

dv = t2 ⇒ dt

t

dv =

0

∫ t dt 2

(At t = 0 it is at rest)

0

t3 3 Let the displacement of the block be dx from t = t to t = t +dt then, work done by the force F in this time interval dt is. dw = F.dx = 2t2.dx

⇒ v=

2 dw = 2t . w

2

0

5.

dw = 2t 2 . 0

dx . dt ⇒ dt

t3 dt ⇒ 3

dw = 2t 2 ( v)dt 2

2  t6  W= 36  

2 5 W = 3 t dt ⇒

∫ 0

2

= 0

64 Joule 9

AREA UNDER FORCE DISPLACEMENT CURVE : Graphically area under the force-displacement is the work done +ve work

Fx

Fy

+ve work

+ve work

Fz

y

x

Ex.7 Sol.

6.

–ve work

z

The work done can be positive or negative as per the area above the x-axis or below the x-axis respectively. Force acting on a particle varies with x as shown in figure. Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m. The work done by the force is equal to the area under the curve from x = 0 to x = 6.0 m. Fx(N) This area is equal to the area of the rectangular section from x = 0 to x = 4.0 m plus the area of the triangular section from 5 x = 4.0 m to x = 6.0 m. The area of the rectangle is (4.0) (5.0) N.m = 20 1 J, and the area of the triangle is (2.0), (5.0) N.m = 5.0 J. 0 1 2 3 4 5 6 x(m) 2 Therefore, the total work done is 25 J.

INTERNAL WORK : Suppose that a man sets himself in motion backward by pushing against a wall. The forces acting on the man are his weight 'W' the upward force N exerted by the ground and the horizontal force 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

WORK, POWER & ENERGY

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N′ exerted by the wall. The works of 'W' and of N are zero because they are perpendicular to the motion. The force N' is the unbalanced horizontal force that imparts to the system a horizontal acceleration. The work of N', however, is zero because there is no motion of its point of application. We are therefore confronted with a curious situation in which a force is responsible for acceleration, but its work, being zero, is not equal to the increase in kinetic energy of the system. N'

N

The new feature in this situation is that the man is a composite system with several parts that can move in relation to each other and thus can do work on each other, even in the absence of any interaction with externally applied forces. Such work is called internal work. Although internal forces play no role in acceleration of the composite system, their points of application can move so that work is done; thus the man's kinetic energy can change even though the external forces do no work. "Basic concept of work lies in following lines Draw the force at proper point where it acts that give proper importance of the point of application of force. Think independently for displacement of point of application of force, Instead of relation the displacement of applicant point with force relate it with the observer or reference frame in which work is calculated.  displacement vector of po int of  W = (Force vector ) ×  application of force as seen by  observer 

7.

    

CONSERVATIVE FORCE : A force is said to be conservative if work done by or against the force in moving a body depends only on the initial and final positions of the body and does not depend on the nature of path followed between the initial and final positions. m

m

m

m

m

m

(a)

(b)

(c)

Consider a body of mass m being raised to a height h vertically upwards as shown in above figure. The work done is mgh. Suppose we take the body along the path as in (b). The work done during horizontal motion is zero. Adding up the works done in the two vertical parts of the paths, we get the result mgh once again. Any arbitrary path like the one shown in (c) can be broken into elementary horizontal and vertical portions. Work done along the horizontal path is zero. The work done along the vertical parts add up to mgh. Thus we conclude that the work done in raising a body against gravity is independent of the path taken. It only depends upon the intial and final positions of the body. We conclude from this discussion that the force of gravity is a conservative force.

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WORK, POWER & ENERGY

Page # 30

(i) (ii) (iii) (iv)

• • •

Examples of Conservative forces. Gravitational force, not only due to Earth due in its general form as given by the universal law of gravitation, is a conservative force. Elastic force in a stretched or compressed spring is a conservative force. Electrostatic force between two electric charges is a conservative force. Magnetic force between two magnetic poles is a conservative force. Forces acting along the line joining the centres of two bodies are called central forces. Gravitational force and Electrosatic forces are two important examples of central forces. Central forces are conservative forces. Properties of Conservative forces Work done by or against a conservative force depends only on the initial and final position of the body. Work done by or against a conservative force does not depend upon the nature of the path between initial and final position of the body. Work done by or against a conservative force in a round trip is zero. If a body moves under the action of a force that does no total work during any round trip, then the force is conservative; otherwise it is non-conservative. The concept of potential energy exists only in the case of conservative forces. The work done by a conservative force is completely recoverable. Complete recoverability is an important aspect of the work done by a conservative force.

Work done by conservative forces Ist format : (When constant force is given) 

Ex.8 Sol.

Calculate the work done to displace the particle from (1, 2) to (4, 5). if F = 4 ˆi + 3ˆj    dw = F.d r ( dr = dxi + dyj + dzk ) dw = (4 i + 3 j ).(dxi + dyj) 4

w

dw =

5

4dx +

1

0

⇒ dw = 4dx + 3dy

∫ 3dy 2

4

5

⇒ w = [4 x]1 + [3 y]2

w = (16 – 4) + (15 – 6) ⇒ w = 12 + 9 = 21 Joule II format : (When F is given as a function of x, y, z)  If F = Fx i + Fy j + Fzk then dw = (Fx i + Fy j + Fzk ).( dx i + dyj + dzk ) ⇒ dw = Fxdx + Fydy + FZdz

Ex.9

  An object is displaced from position vector r1 = (2 ˆi + 3ˆj)m to r2 = (4 ˆi + 6ˆj)m under a force  F = (3x 2 ˆi + 2yˆj)N . Find the work done by this force.  rf

Sol.

 r2

 W = F.dr = (3 x 2 ˆi + 2yˆj ) • (dx ˆi + dyˆj + dzkˆ ) =

∫ ri

∫  r1

 r2

∫ (3x dx + 2ydy) = [x 2

 r1

3

+ y 2 ](( 42,,36)) = 83 J Ans.

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WORK, POWER & ENERGY

Page # 31

IIIrd format (perfect differential format)  Ex.10 If F = yiˆ + xjˆ then find out the work done in moving the particle from position (2, 3) to (5, 6)   Sol. dw = F. ds

Now

dw = ( y i + xj ).( dxi + dyj ) dw = ydx + xdy ydx + xdy = d(xy) (perfect differential equation) ⇒ dw = d(xy) for total work done we integrate both side

∫ dw = ∫ d( xy) Put xy = k then at (2, 3) ki = 2 × 3 = 6 at (5, 6) kf = 5 × 6 = 30 30

then

w=

∫ dk = [k]

30 6

⇒ w = (30 – 6) = 24 Joule

6

8.

NON-CONSERVATIVE FORCES :

A force is said to be non-conservative if work done by or against the force in moving a body depends upon the path between the initial and final positions. The frictional forces are non-conservative forces. This is because the work done against friction depends on the length of the path along which a body is moved. It does not depend only on the initial and final positions. Note that the work done by fricitional force in a round trip is not zero. The velocity-dependent forces such as air resistance, viscous force, magnetic force etc., are non conservative forces.  Ex.11 Calculate the work done by the force F = y i to move the particle from (0, 0) to (1, 1) in the following condition (a) y = x (b) y = x2 Sol. We know that   dw = F.ds ⇒ dw = ( y i ) .(dx i ) dw = ydx ...(1) In equation (1) we can calculate work done only when we know the path taken by the particle. either y = x or y = x2 so now (a) when y = x 1 1 dw = xdx ⇒ w = Joule 0 2 (b) when y = x2

1

dw =

∫ x dx 2

0

w=

1 Joule 3

Difference between conservative and Non-conservative forces

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WORK, POWER & ENERGY

Page # 32

S. No. 1 2 3

4

5

9.

Non-Conse rva tive force s W ork done does not W ork done depends on depend upon path path. W ork done in a round W ork done in a round trip trip is zero. is not zero. Forc es are veloc ityCentral in nature. dependent and retarding in nature. W hen only a c ons ervative forc e ac ts within a sy s tem , the W ork done agains t a nonk inetic energy and c ons ervative forc e m ay potential energy can be disipiated as heat c hange. However their energy. s um , the m ec hanical energy of the s y stem , does not c hange. Conse rva tive force s

W ork done is W ork done in not c om pletely rec overable. c om pletely recoverable.

ENERGY A body is said to possess energy if it has the capacity to do work. When a body possessing energy does some work, part of its energy is used up. Conversely if some work is done upon an object, the object will be given some energy. Energy and work are mutually convertiable. There are various forms of energy. Heat, electricity, light, sound and chemical energy are all familiar forms. In studying mechanics, we are however concerned chiefly with mechanical energy. This type of energy is a property of movement or position.

9.1

• 9.2

1. 2. 3. 4. 5.

Kinetic Energy Kinetic energy (K.E.), is the capacity of a body to do work by virtue of its motion. If a body of mass m has velocity v its kinetic energy is equivalent to the work, which an external force would have to do to bring the body from rest up to its velocity v. The numerical value of the kinetic energy can be calculated from the formula 1 2 ...(8) K.E. = mv 2 2 Since both m and v are always positive, K.E. is always positive and does not depend upon the direction of motion of the body. Potential Energy Potential energy is energy of the body by virtue of its position. A body is capable to do work by virtue of its position, configuration or state of strain. Now relation between Potential energy and work done is W.D = – ∆U where ∆U is change in potential energy There are two common forms of potential energy, gravitational and elastic. Important points related to Potential energy : Potential energy is a straight function (defined only for position) Potential energy of a point depends on a reference point Potential energy difference between two position doesn't depend on the frame of reference. Potential energy is defined only for conservative force because work done by conservative force is path independent. If we define Potential energy for non conservative force then we have to define P.E. of a single point through different path which gives different value of P.E. at single point that doesn't make any sense.

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9.2.1 (a) Gravitational Potential Energy : It is possessed by virtue of height. When an object is allowed to fall from one level to a lower level it gains speed due to gravitational pull, i.e., it gains kinetic energy. Therefore, in possessing height, a body has the ability to convert its gravitational potential energy into kinetic energy. The gravitational potential energy is equivalent to the negative of the amount of work done by the weight of the body in causing the descent. If a mass m is at a height h above a lower level the P.E. possessed by the mass is (mg) (h). Since h is the height of an object above a specified level, an object below the specified level has negative potential energy. Therefore GPE = ± mgh ...(9) mg

mg

h1 Specific level where P.E. is zero

h P.E.=mgh h2

fig(a)

fig(b)

P.E. of m1 is m1gh1 P.E of m2 is –m2gh2 m2g

The chosen level from which height is measured has no absolute position. It is important therefore to indicate clearly the zero P.E. level in any problem in which P.E. is to be calculated. • GPE = ± mgh is applicable only when h is very small in comparison to the radius of the earth. We have discussed GPE in detail in 'GRAVITATION'. 9.2.2 (b) Elastic Potential Energy : It is a property of stretched or compressed springs. The end of a stretched elastic spring will begin to move if it is released. The spring. therefore possesses potential energy due to its elasticity. (i.e., due to change in its configuration) The amount of elastic potential energy stored in a spring of natural length a and spring constant k when it is extended by a length x (from the natural length) is equivalent to the amount of work necessary to produce the extension. 1 2 kx ...(10) 2 It is never negative whether the spring is extended or compressed.

Elastic Potential Energy =

Proof :

N.L.

K

N.L.

K

M

M

x0 Consider a spring block system as shown in the figure and let us calculate work done by spring when the block is displaceed by x0 from the natural length. At any moment if the elongation in spring is x, then the force on the block by the spring is kx towards left. Therefore, the work done by the spring when block further displaces by dx dW = – kx dx x0

∴ Total work done by the spring, W = –

∫ kxdx = – 0

1 2 kx 0 2

Similarly, work done by the spring when it is given a compression x0 is – : We assume zero potential energy at natural length of the spring : 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

1 2 kx 0 . 2

WORK, POWER & ENERGY

Page # 34

10.

CONSERVATIVE FORCE AND POTENTIAL ENERGY : ∂U ∂s i.e. the projection of the force field , the vector F, at a given point in the direction of the displacement r equals the derivative of the potential energy U with respect to a given direction, taken with the opposite sign. The designation of a partial derivative ∂/∂s emphasizes the fact of deriving with respect to a definite direction.

Fs = –

So, having reversed the sign of the partial derivatives of the function U with respect to x, y, z, we obtain the projection Fx, Fy and Fz of the vector F on the unit vectors i, j and k. Hence, one can readily find the vector itself :  ∂U ∂U ∂U  F = Fxi + Fy j + Fzk, or F = –  ∂ x i + ∂ y j + ∂z k    The quantity in parentheses is referred to as the scalar gradient of the function U and is denoted

by grad U or ∇ U. We shall use the second, more convenient, designation where ∇ (“nabla”) signifies the symbolic vector or operator ∂ ∂ ∂ ∇ = i ∂x + j ∂y + k ∂z

Potential Energy curve : •

A graph plotted between the PE of a particle and its displacement from the centre of force field is called PE curve.

Using graph, we can predict the rate of motion of a particle at various positions.

Force on the particle is F(x) = –

dU dx

Q

S

U

B P1

A

P2 R

P O

D

C

x

Case : I

On increasing x, if U increases, force is in (–) ve x direction i.e. attraction force.

Case : II

On increasing x, if U decreases, force is in (+) ve x-direction i.e. repulsion force.

Different positions of a particle : Position of equilibrium If net force acting on a body is zero, it is said to be in equilibrium. For equilibrium

dU = 0. Points P,, dx

Q, R and S are the states of equilbrium positions. Types of equilirbium : •

Stable equilibrium : When a particle is displaced slightly from a position and a force acting on it brings it back to the initial position, it is said to be in stable equilibrium position. dU d2U = 0,and 2 =+ ve dx dx In figure P and R point shows stable equilibrium point.

Necessary conditions:–

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Unstable Equilibrium : When a particle is displaced slightly from a position and force acting on it tries to displace the particle further away from the equilibrium position, it is said to be in unstable equilibrium. dU d2U = 0 potential energy is maximum i.e. = = – ve dx dx2 Q point in figure shows unstable equilibrium point Neutral equilibrium : In the neutral equilibrium potential energy is constant. When a particle is displaced from its position it does not experience any force acting on it and continues to be in equilibrium in the displaced position. This is said to be neutral equilibrium. In figure S is the neutral point

Condition :

Condition :

dU d2U =0 , =0 dx dx2

a b – , where a x 12 x 6 and b are positive constants and x is the distance between the atoms. The system is in stable equilibrium when -

Ex.12 The potential energy between two atoms in a molecule is given by, U(x) =

(A) x = 0 Sol.

 2a  (C) x =    b 

a (B) x = 2b

1/6

 11a   (D) x =  5b 

(C) Given that, U(x) =

6b x7

12

x

b x6

du dx = (–12) a x–13 – (–6 b) x–7= 0

We, know

or

a

F=–

=

12a 4 x13

or

x6 = 12a/6b = 2a/b or

 2a  x=    b

1/ 6

Ex.13 The potential energy of a conservative system is given by U = ax2 – bx where a and b are positive constants. Find the equilibrium position and discuss whether the equilibrium is stable, unstable or neutral. Sol.

In a conservative field F = –

dU dx

∴F=–

For equilibrium F = 0 or b – 2ax = 0 ∴ x =

From the given equation we can see that

Therefore, x =

11.

d ( ax2 – bx) = b – 2ax dx

b 2a

d2U dx2

= 2a (positive), i.e., U is minimum.

b is the stable equilibrium positon. 2a

WORK ENERGY THEOREM : If the resultant or net force acting on a body is Fnet then Newton's second law states that ...(1) Fnet = ma If the resultant force varies with x, the acceleration and speed also depend on x. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564

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WORK, POWER & ENERGY

Page # 36

then

a= v

dv dx

...(2)

from eq. (1) Fnet

dv = mv dx

m

⇒ Fnet.dx = m v dv

vi

Fnet

m vf

vf

∫F

net . dx

=

∫ mvdv vi

1 1 mv 2f – mv i2 2 2 Wnet = kf – ki Wnet = ∆K ...(3)

Wnet =

Work done by net force Fnet in displacing a particle equals to the change in kinetic energy of the particle i.e. we can write eq. (3) in following way (W.D)c + (W.D)N.C + (W.D)ext. + (W.D)pseudo = ∆K

...(4)

where (W.D)c = work done by conservative force (W.D)N.C = work done by non conservative force. (W.D)ext = work done by external force (W.D)pseudo = work done by pseudo force. we know that (W.D)c = – ∆U ⇒ – ∆U + (W.D)N.C + (W.D)ext + (W.D)pseudo = ∆K ⇒ (W.D)N.C + (W.D)ext. + (W.D)pseudo = (kf + uf) – (ki + ui) ∵ k + u = Mechanical energy. ⇒ work done by forces (except conservative forces) = change is mechanical energy. If (W.D)N.C = (W.D)ext = (W.D)pseudo = 0 Kf + Uf = Ki + Ui Initial mechanical energy = final mechanical energy This is called mechanical energy conservation law. Questions Based on work Energy Theorem : (A)

When only one conservative force is acting

Ex.14 The block shown in figure is released from rest. Find out the speed of the block when the spring is compressed by 1 m. N.L.

2kg 2m

A

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WORK, POWER & ENERGY

Sol.

Page # 37

In the above problem only one conservative force (spring force) is working on the block so from mechanical energy conservation kf + uf = ki + ui

...(i)

at A block is at rest so 1 1 ui = kx12 = k( 2) 2 2 2

ki = 0

N.L.

= 2k Joule

At position B if speed of the block is v then kf =

1 1 mv 2 = × 2 × v 2 = v 2 2 2

uf =

1 2 1 k kx2 = × k × 1 = 2 2 2

B

A

2kg

2kg

x2=1m

x1=2m

Putting the above values in equation (i), we get ⇒

v2 +

k = 2k 2

2 ⇒ v =

3k ⇒ v= 2

3k m / sec 2

Ex.15 A block of mass m is dropped from height h above the ground. Find out the speed of the block when it reaches the ground. Sol. Initial situation

Ugi = mgh , ki = 0

h

Final situation v

Ugf = 0 , K f =

1 mv 2 2

Figure shows the complete description of the problem only one conservative force is working on the block. So from mechanical energy conservation kf + uf = ki + ui

1 mv 2 + 0 = 0 + mgh 2

v = 2gh m / sec

(B)

When two conservative force are acting in problem.

Ex.16 One end of a light spring of natural length d and spring constant k is fixed on a rigid wall and the other is attached to a smooth ring of mass m which can slide without friction on a vertical rod fixed at a distance d from the wall. Initially the spring makes an angle of 37º with the horizontal as shown in fig. When the system is released from rest, find the speed of the ring when the spring becomes horizontal. [sin 37º = 3/5]

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WORK, POWER & ENERGY

Page # 38

A

Ring v=0

h

I

37°

d

B

v Rod

Sol.

If l is the stretched length of the spring, then from figure d 4 = cos 37º = , i.e., 5 l

So, the stretch

l=

y=l −d=

5 d 4

5 d d− d = 4 4

5 3 3 d× = d 4 5 4 Now, taking point B as reference level and applying law of conservation of mechanical energy between A and B, EA = EB

and

h = l sin 37º =

1 2 1 ky = mv 2 2 2 [as for, B, h = 0 and y = 0]

or

mgh +

2

3 1  d 1 mgd + k   = mv 2 4 2  4 2

or

v=d

or

3g k + 2d 16m

[as for A, h =

3 1 d and y = d ] 4 4

Ans.

Ex.17 The block shown in figure is released from rest and initially the spring is at its natural length. Write down the energy conservation equation. When the spring is compressed b

y

l1 ? m

Here two conservative forces are included in the problem. (i) Gravitational force (ii) spring force N.L.

at A as shown in figure. from mechanical energy conservation kf + uf = ki + ui

...(i)

1 1 mv 2 + k 21 = mg(  1 +  ) sin θ 2 2

final position

m

m

( + 1 )sin θ

initial position

We assume zero gravitational potential energy

1

Sol.

B

Ug = mg( +  1) sin θ

Us=0, K = 0 1 Ug = 0, Us = k 12 2 1 K = mv 2 2

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WORK, POWER & ENERGY

(C)

Page # 39

When only one non conservative force is included in problem.

Ex.18 Find out the distance travelled by the block as shown in figure. If the initial speed of the block is v and µ is the friction coefficient between the surface of block and ground. m

Sol.

(D)

v

Applying work energy theorem, we get ⇒

1  2 (–µ mg ) = (0 + 0) –  mv + 0 2

1 2 v µg  = 2

v=0

v m

m

Initial

final

mg

v2 = 2µ g

When both conservative and non-conservative force in the problem

Ex.19 A particle slides along a track with elevated ends and a flat central part as shown in figure. The flat portion BC has a length l = 3.0 m. The curved portions of the track are frictionless. For the flat part the coefficient of kinetic friction is µk = 0.20, the particle is released at point A which is at height h = 1.5 m above the flat part of the track. Where does the particle finally comes to rest? Sol.

As initial mechanical energy of the particle is mgh and final is zero, so loss in mechanical energy = mgh. This mechanical energy is lost in doing work against friction in the flat part, So, loss in mechanical energy = work done against friction or mgh = µ mgs i.e.,

s=

h 15 . = = 7.5 m µ 0.2

After starting from B the particle will reach C and then will rise up till the remaining KE at C is converted into potential energy. It will then again descend and at C will have the same value as it had when ascending, but now it will move from C to B. The same will be repeated and finally the particle will come to rest at E such that

h

BC + CB + BE = 7.5 or i.e.,

D

A

B

E

C

3 + 3 + BE = 7.5 BE = 1.5

So, the particle comes to rest at the centre of the flat part. Ex.20 A 0.5 kg block slides from the point A on a horizontal track with an initial speed 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 N/m. The part AB of the track is frictionless and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.20 respectively. If the distance AB and BD are 2 m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. [g = 10 m/s2] Sol. As the track AB is frictionless, the block moves this distance without loss in its initial 1 1 mv 2 = × 0.5 × 32 = 2.25 J. In the path BD as friction is present, so work done against 2 2 friction = µk mgs = 0.2 × 0.5 × 10 × 2.14 = 2.14 J So, at D the KE of the block is = 2.25 – 2.14 = 0.11 J. Now, if the spring is compressed by x

KE =

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WORK, POWER & ENERGY

Page # 40

0.11 =

1 × k × x2 + µk mgx 2

i.e.,

0.11 =

1 × 2 × x2 + 0.2 × 0.5 × 10x 2

or

x2 + x – 0.11 = 0

A

B

D

C

which on solving gives positive value of x = 0.1 m After moving the distance x = 0.1 m the block comes to rest. Now the compressed spring exerts a force : F = kx = 2 × 0.1 = 0.2 N on the block while limiting frictional force between block and track is fL = µs mg = 0.22 × 0.5 × 10 = 1.1 N. Since, F < fL. The block will not move back. So, the total distance moved by block = AB + BD + 0.1 = 2 + 2.14 + 0.1 = 4.24 m (E) Important Examples : Ex.21 A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the paritcle with respect to the sphere as a function of the angle θ it slides. Sol. We solve the above problem with respect to the sphere. So apply a pseudo force on the particle m ma R

a with respect to sphere

Now from work energy theorem. work done by ma = change in mechanical energy ma R sin θ = (kf + uf) – (ki + ui) 1 mv 2 − mgR (1 − cos θ) ⇒ 2 ⇒ v2 = 2R(a sin θ + g – g cos θ) ⇒

maR sin θ =

1 mv 2 = maR sin θ + mgR (1 – cos θ) 2 v = [2R (a sin θ + g – g cos θ)]1/2 m/sec

Ex.22 In the arrangement shown in figure mA = 4.0 kg and mB = 4.0 kg. The system is released from rest and block B is found to have a speed 0.3 m/s after it has descended through a distance of 1m. Find the coefficient of friction between the block and the table. Neglect friction elsewhere. (Take g = 10 m/s2) Sol.

A

B

From constraint relations, we can see that v A = 2 vB Therefore,

vA = 2(0.3) = 0.6 m/s

as

vB = 0.3 m/s (given)

Applying

Wnc = ∆U + ∆K

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WORK, POWER & ENERGY

Page # 41 1 1 mAvA2 + mBvB2 2 2

we get – µ mA g SA = – mB g SB +

Here, SA = 2SB = 2m as SB = 1 m (given) ∴

– µ(4.0) (10) (2) = – (1) (10) (1) +

or – 80 µ = – 10 + 0.72 + 0.045

or

1 1 (4) (0.6)2 + (1) (0.3)2 2 2

80µ = 9.235 or µ = 0.115

Ex.23 A body of mass ‘m’ was slowly hauled up the hill as shown in the figure by a force F which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is h, the length of its base is l and the coefficient of

Ans.

m

F

h

friction is µ. Sol.

l

Four forces are acting on the body : 1. weight (mg) 2. normal reaction (N) 3. friction (f) and 4. the applied force (F) Using work-energy theorem Wnet = ∆KE or Wmg + WN + Wf + WF = 0 Here, ∆KE = 0, because Ki = 0 = Kf

ds

Wmg = – mgh ⇒ WN = 0 (as normal reaction is perpendicular to displacement at all points) Wf can be calculated as under :

A

f = µ mg cos θ

F

dl

(dWAB)f = – f ds = – (µ mg cos θ) ds = – µ mg (dl)

B

f = – µ mg ∑ dl

(as ds cos θ = dl)

= – µ mgl

Substituting these values in Eq. (i), we get WF = mgh + µmgl

: Here again, if we want to solve this problem without using work-energy theorem we will first find    magnitude of applied force F at different locations and then integrate dW ( = F. dr ) with proper limits.

12.

POWER Power is defined as the time rate of doing work. When the time taken to complete a given amount of work is important, we measure the power of the agent doing work. The average power (P or Pav ) delivered by an agent is given by ∆W

P or Pav = ∆ t =

Total work done Total time

where ∆W is the amount of work done in time ∆ t. Power is the ratio of two scalars-work and time. So, power is a scalar quantity. If time taken to complete a given amount of work is more, then power is less. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

WORK, POWER & ENERGY

Page # 42

• •

dW  The instantaneous power is, P = where dW is the work done by a force F in a small time dt. dt  dr   dW  F = F. v where v is the velocity of the body.. = . P= dt dt By definition of dot product,

P = Fvcosθ   where θ is the smaller angle between F and v

This P is called as instantaneous power if dt is very small. 12.1

Unit of Power : A unit power is the power of an agent which does unit work in unit time. The power of an agent is said to be one watt if it does one joule of work in one second. 1 watt = 1 joule/second = 107 erg/second Also,

1 watt =

1 newton × 1 metre = 1 N ms–1. 1 sec ond

Dimensional formula of power [Power ] =

[ Work ] [ML2 T –2 ] = = [ML2T–3] [ Time] [ T]

Ex.24 A one kilowatt motor pumps out water from a well 10 metre deep. Calculate the quantity of water pumped out per second. Sol.

Power, P = 1 kilowatt = 103 watt S = 10 m ; Time, t = 1 second ; Mass of water, m = ? ∴

103 =

or

m=

Power =

mg × S t

m × 9.8 × 10 1

10 3 kg 9.8 × 10

= 10.204 kg

Ex.25 The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through in time t? (b) What is the kinetic energy or the air? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30m2, v = 36 km h–1 and the density of air is 1.2 kg m–3. What is the electrical power produced? Sol.

(a) Volume of wind flowing per second = Av Mass of wind flowing per second = Avρ Mass of air passing in t second = Avρt (b) Kinetic energy of air

=

1 1 1 mv 2 = ( Avρt)v 2 = Av 3 ρt 2 2 2

(c) Electrical energy produced = Electrical power

=

25 1 3 Av 3 ρt × Av ρt = 100 2 8

Av 3 ρt Av 3 ρ = 8t 8

Now, A = 30 m2, v = 36 km h–1 = 36 ×

5 m s–1 18

= 10 m s–1, ρ = 1.2 kg ms–1

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WORK, POWER & ENERGY

∴ Electrical power =

Page # 43 30 × 10 × 10 × 12 . W = 4500 W = 4.5 kW 8

Ex.26 One coolie takes one minute to raise a box through a height of 2 metre. Another one takes 30 second for the same job and does the same amount of work. Which one of the two has greater power and which one uses greater energy? Sol.

Power of first coolie =

Work M × g × S = Time t

Power of second coolie =

=

M × 9.8 × 2 –1 Js 60

 M × 9.8 × 2  M × 9.8 × 2 –1  J s–1 = 2 × Power of first coolie Js = 2    60 30

So, the power of the second coolie is double that of the first. Both the coolies spend the same amount of energy. We know that W = Pt For the same work, W = p1t1 = P2t2 or

P2 t1 = P1 t 2

=

1minute =2 30 s

or

P2 = 2P1

13.

VERTICAL CIRCULAR MOTION

A) B)

To understand this consider the motion of a small body (say stone) tied to a string and whirled in a vertical circle. Now we study the circular motion of the body in two parts. Motion of a body from A to B. Motion of a body from B to C.

A.

C

B

R T

T1

v1

2 m A u mg mv1 R

Motion of a body from A to B.

mv12 ...(1) R During the motion of the body from A to B. θ will increase so cos θ will decrease. Due to which mg cos θ will decrease. From A to B speed of the body also decreases due to which

T1 = mg cos θ +

mv 2 decreases. Therefore tenstion in the string decreases from A to B. R But due to mg cos θ tension can never be zero.

B.

Motion of a body from B to C. T2 =

mv 22

R From B → C

− mg cos θ

speed decreases due to which

C

...(2)

T2

mg mv 22

decreases.

R θ decreases due to which mg cos θ increases. Therefore from B →C. Tension in the string decreases.

String slacks at a point where

v2

mv 22 = mg cos θ i.e., T = 0 R

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mv2 R

B

2

WORK, POWER & ENERGY

Page # 44

13.1

Minimum velocity at point A for which body can complete the vertical circle The condition for the body to complete the vertical circle is that the string should be taut all the time i.e. the tension is greater than zero. So the body can complete the vertical circle if the tension is not zero in between the region B to C. Initially. mv 2 = TC + mg ...(1) R Apply energy conservation from A to C then Kf + Uf = Ki + Ui 1 1 mv 2 + 2mgR = mu 2 + 0 ...(2) 2 2 body can complete vertical circle, when TC ≥ 0

finally

from figure (b)

mv 2 – mg ≥ 0 R 2 ⇒ v ≥ gR ...(3) Put the value from (3) to (2) and u = umin 1 1 2 ⇒ m(Rg) + 2mgR = mumin 2 2

v

mv2 C R

Tc+mg

TA u

A mg fig(a)

fig(b)

2 ⇒ umin = 5gR ⇒ umin = 5gR

It the velocity is greater than equal to

5gR then the body will complete the vertical circle.

Tension at A TA = mg +

mu2 R

If u = umin = then

TA

5gR

5mgR TA = mg + R

u mu2 mg + R

⇒ TA = 6mg

Tension at B mv 2 R energy conservation from A to B

v

TB =

1 1 2 2 mumin = mgR + mv 2 2 ⇒ v2 = 3gR ⇒ TB = 3mg

13.2

A

Condition for the body to reach B : Let us calculate the umin such that the body just reaches B. Work done by tension = 0 Only gravitational force is working on the body which is a conservative force. Therefore Applying conservation of energy, we get mgR = ∴ if u ≤

1 2 mumin ⇒ umin = 2

TB

2gR

B

mv2 R

mg

O R A u min

R

v=0, Ug=mgR K= 0

Ug=0 1 2 K = mumin 2

2gR then the body will oscillate about A.

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13.3

When

Page # 45

2gR < u < 5gR

If the velocity of projection is greater than 2gR but less than

5gR , the particle rises above the

horizontal diameter and the tension vanishes before reaching the highest point. We have seen that the tension in the string at the highest point is lower than the tension at the lowest point. At the point D, the string OD makes an angle φ with the vertical. The radial component of the weight is mg cos φ towards the centre O. T + mg cos φ =

mv 2 R

 v2  – g cos φ  T = m  R 

...(i)

1 mv 2 2 Potential energy at D = mg(AN) = mg (AO + ON) ⇒ mg(R + R cosφ) = mgR(1 + cosφ) From conservation of energy

Kinetic energy at D =

1 1 mu 2 = mv 2 + mgR(1+ cosφ) 2 2 v2 = u2 – 2gR(1 + cos φ) Substituting in equation (i),

B

N

φT

D

O θ mg

A

 u2  T = m  – 2g(1 + cos φ) – g cos φ R    u2 2   T = m – 3g cos φ +  R 3    This equation shows that the tension becomes zero. if u2 2  = 3g cos φ +  R 3 

...(ii)

If the tension is not to become zero. 2  u2 > 3Rg  cos φ +   3

Equation (ii) gives the values of φ at which the string becomes slack. cosφ +

u2 2 = 3Rg 3

cosφ =

2 u2 – 3Rg 3

cos φ =

u 2 – 2gR 3gR

It is the angle from the vertical at which tension in the string vanishes to zero. And after that its motion is projectile. 13.4

Tension in the string versus θ We may find an expression for the tension in the string when it makes an angle θ with the vertical. At C, the weight of the body acts vertically downwards, and the tension in the string is towards the centre O.

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WORK, POWER & ENERGY

Page # 46

The weight mg is resolved radially and tangentially. The radial component is mg cos θ and the tangential component is mg sin θ. T – mg cos θ =

mv 2 , where v is the velocity at C. R

 v2  + g cos θ  ...(i) i.e., T = m  R  The velocity v can be expressed in terms of velocity u at A.

B

O

v θ T M C A u θ mg cos θ mg

1 2 The total energy at A = mu 2 1 mv 2 2 The potential energy at C = mg (AM) = mg (AO – MO) = mg (R – R cosθ) = mgR (1 – cos θ) The kinetic energy at C =

1 mv 2 + mgR(1 – cos θ) 2 ∴ From conservation of energy

The total energy at C =

1 1 mu 2 = mv 2 + mgR (1 – cosθ) 2 2 u2 = v2 + 2gR (1 – cos θ) or Substituting in equation (v),

2

v2 = u – 2gR(1 – cos θ)

  mu 2 2  u2 + 3mg cos θ –  – 2g(1 – cos θ) = T = mg cos θ + R 3 R    

...(ii)

This expression gives the value of the tension in the string in terms of the velocity at the lowest point and the angle θ. Equation (i) shows that tension in the string decreases as θ increases, since the term 'g cos θ' decreases as θ increases. when u =

5gR

⇒ T = 3mg (1 + cos θ) Now θ = 0 ⇒ cos θ = 1 ⇒ if, θ = 90° ⇒ TB = 3 mg θ = 180°, cosθ = – 1 Tc = 0

T 6mg

3mg –1

1

TA = 6 mg

13.5

Different situations :

(A)

A BODY MOVING INSIDE A HOLLOW TUBE OR SPHERE The previous discussion holds good for this case, but instead of tension in the string we have the normal reaction of the

v2

N' mg R

surface. If N is the normal reaction at the lowest point, then the condition u ≥

cosθ

5Rg for the body to complete the

circle holds for this case also. All other equations (can be) similarly obtained by replacing tension T by normal reaction N.

N

u

mg

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WORK, POWER & ENERGY (B)

Page # 47

WHEN BODY IS ATTACHED TO A ROD OF LENGTH R In this case since the body is attached to a rigid rod. The body can not leave the circular path. Therefore, if the speed of the body becomes zero before the highest point C. It's motion will be oscillatory about the centre of the rod. Condition for completing the circle : If the body just reaches the highest point then it will completes the vertical circle Applying energy conservation between the lowest and highest point of circle, we get v=0

Uf = mg2R Kf = 0

u

2mgR =

Ui = 0, Ki =

1 mu 2 ⇒ u = 2

1 mu2 2

4gR

So, If the velocity at point A is greater than equal to

4gR then

body will compete the vertical circle. (C)

VERTICAL MOTION IN A DUAL RING

R u

This system will behave as the preivious system. So umin to complete vertical circle umin =

4gR

Angle at which the normal reaction on the body will change its direction from inward to outward the ring is given by cosφ = (D)

u 2 – 2gR 3 gR

BODY MOVING ON A SPHERICAL SURFACE The small body of mass m is placed on the top of a smooth sphere of radius R and the body slides down the surface. At any instant, i.e., at point C the forces are the normal reaction N and the weight mg. The radial component of the weight is mgcos φ acting towards the centre. The centripetal force is

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WORK, POWER & ENERGY

Page # 48

mv 2 , R where v is the velocity of the body at O.

m

mg cos φ – N =

B C

D

 v 2  N = m  g cos φ – ...(i) R   The body flies off the surface at the point where N becomes zero. v2 v2 ; cos φ = ...(ii) Rg R To find v, we use conservation of energy

φ O

N

mg

A

i.e., g cos φ =

1 mv 2 = mg (BD) 2 = mg (OB – OD) = mgR (1 – cos φ) v2 = 2Rg (1 – cos φ)

i.e.,

2(1 – cos φ) =

v2 Rg

...(iii)

From equation (ii) and (iii) we get cos φ = 2 – 2 cos φ ; 3 cos φ = 2 2  2 ; φ = cos–1   ...(iv) 3  3 This gives the angle at which the body goes of the surface. The height from the ground of that point = AD = R(1 + cos φ) cos φ =

 =R  1 + 

5 2  = R 3  3

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CIRCULAR & W.P.E

Page # 49

Exercise - I

(Objective Problems)

(A) CIRCULAR MOTION 1. A wheel is at rest. Its angular velocity increases uniformly and becomes 80 radian per second after 5 second. The total angular displacement is : (A) 800 rad (B) 400 rad (C) 200 rad (D) 100 rad Sol.

3. A spot light S rotates in a horizontal plane with a constant angular velocity of 0.1 rad/s. The spot of light p moves along the wall at a distance 3 m. What is the velocity of the spot P when θ = 45° ? Wall

P

θ

3m

(Top view) S(Spot light)

(A) 0.6 m/s (C) 0.4 m/s Sol.

2. The second’s hand of a watch has length 6 cm. Speed of end point and magnitude of difference of velocities at two perpendicular positions will be : (A) 2π & 0 mm/s

(B) 2 2 π & 4.44 mm/s

(C) 2 2 π & 2π mm/s

(D) 2π & 2 2 π mm/s

(B) 0.5 m/s (D) 0.3 m/s

4. Two moving particle P and Q are 10 m apart at a certain instant. The velocity of P is 8m/s making an angle 30° with the line joining P and Q and that of Q is 6m/s making an angle 30° with PQ as shown in the figure. Then angular velocity of P with respect to Q is 6m/s

Sol.

P

30°

10m 30°

(A) Zero (C) 0.4 rad/sec Sol.

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8m/s

CIRCULAR & W.P.E 5. The magnitude of displacement of a particle moving in a circle of radius a with constant angular speed ω varies with time t as ωt (A) 2 a sin ωt (B) 2a sin 2 ωt (C) 2a cos ωt (D) 2a cos 2 Sol.

6. Two bodies A & B rotate about an axis, such that angle θ A (in radians) covered by first body is proportional to square of time, & θB (in radians) covered by second body varies linearly. At t = 0, θA = θB = 0. If A completes its first revolution in π sec. & B needs 4π sec. to complete half revolution then; angular velocity ωA : ωB at t = 5 sec. are in the ratio (A) 4 : 1 (B) 20 : 1 (C) 80 : 1 (D) 20 : 4 Sol.

Page # 50  20  7. A particle moves along a circle of radius   m π with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is : (B) 40 π m/s2 (A) 160 π m/s2 2 (C) 40 m/s (D) 640 π m/s2 Sol.

8. The graphs below show angular velocity as a function of time. In which one is the magnitude of the angular acceleration constantly decreasing?

(A)

t

(B)

(C)

t

(D)

Sol.

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t

t

CIRCULAR & W.P.E

Page # 51

9. A particle moves with deaceleration along the circle of radius R so that at any moment of time its tangential and normal accelerations are equal in moduli. At the initial moment t = 0 the speed of the particle equals v0, then : (i) the speed of the particle as a function of the distance covered s will be (B) v = v0es/R (A) v = v0 e–s/R –R/s (C) v = v0e (D) v = v0eR/s (ii) the total acceleration of the particle as function of velocity and distance covered (A) a = (C) a =

v2 R R 2 v 2

(B) a =

2

(D) a =

2R v

Sol.

v R

Sol.

11. A particle moves along an arc of a circle of radius R. Its velocity depends on the distance covered as v = a s , where a is a constant then the angle α between the vector of the total acceleration and the vector of velocity as a function of s will be R (A) tanα = (B) tanα = 2s / R 2s 2R s (C) tan α = (D) tanα = s 2R Sol.

10. If angular velocity of a disc depends an angle rotated θ as ω = θ2 + 2θ, then its angular acceleration α at θ = 1 rad is (B) 10 rad/sec2 (A) 8 rad/sec2 2 (C) 12 rad/sec (D) None 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 52

Sol.

13.Tangential acceleration of a particle moving in a circle of radius 1 m varies with time t as (initial velocity of particle is zero). Time after which total cceleration of particle makes and angle of 30º with radial acceleration is

60º time(sec)

(A) 4 sec 2/3

(C) 2

(B) 4/3 sec

sec

(D)

2 sec

Sol. 12. A particle A moves along a circle of radius R = 50 cm so that its radius vector r relative to the point O (Fig.) rotates with the constant angular velocity ω = 0.40 rad/s. Then modulus of the velocity of the particle, and the modulus of its total Aacceleration will be r

R

0

(A) v = 0.4 m/s, a = 0.4 m/s2 (B) v = 0.32 m/s, a = 0.32 m/s2 (C) v = 0.32 m/s, a = 0.4 m/s2 (D) v = 0.4 m/s, a = 0.32 m/s2 Sol.

14. A particle is going in a uniform helical and spiral path separately as shown in figure with constant speed.

(A)

(B)

(A) The velocity of the particle is constant in both cases (B) The acceleration of the particle is constant in both cases (C) The magnitude of acceleration is constant in (A) and decreasing in (B) (D) The magnitude of acceleration is decreasing continuously in both the cases Sol.

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CIRCULAR & W.P.E

Page # 53 17. Three identical particles are joined together by a thread as shown in figure. All the three particles are moving on a smooth horizontal plane about point O. If the speed of the outermost particle is v0, then the ratio of the tensions in the three sections of the string is : (Assume that the string remains straight) O

15. If the radii of circular paths of two particles of same masses are in the ratio of 1 : 2, then in order to have same centripetal force, their speeds should be in the ratio of : (A) 1 : 4 (B) 4 : 1 (C) 1 :

2

(D)

(A) 3 : 5 : 7 (C) 7 : 11 : 6 Sol.

B C   (B) 3 : 4 : 5 (D) 3 : 5 : 6 A



2 :1

Sol.

16. A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal smooth surface. The maximum tension the string can withstand is 16 newton. The maximum speed of revolution of the stone without breaking it, will be : (A) 20 ms–1 (B) 16 ms–1 (C) 14 ms–1 (D) 12 ms–1 Sol.

18. A particle is kept fixed on a turntable rotating uniformly. As seen from the ground, the particle goes in a circle, its speed is 20 cm/s and acceleration is 20 cm/s2. The particle is now shifted to a new position to make the radius half of the original value. The new values of the speed and acceleration will be (A) 10 cm/s, 10 cm/s2 (B) 10 cm/s, 80 cm/s2 (C) 40 cm/s, 10 cm/s2 (D) 40 cm/s, 40 cm/s2 Sol.

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CIRCULAR & W.P.E

Page # 54 Sol.

19. A particle moving along a circular path due to a centripetal force having constant magnitude is an example of motion with : (A) constant speed and velocity (B) variable speed and velocity (C) variable speed and constant velocity (D) constant speed and variable velocity. Sol.

20. A curved section of a road is banked for a speed v. If there is no friction between road and tyres of the car, then : (A) car is more likely to slip at speeds higher than v than speeds lower than v (B) car cannot remain in static equilibrium on the curved section (C) car will not slip when moving with speed v (D) none of the above Sol.

22. A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length  . The system is rotated about the other end of the spring with an angular velocity ω, in gravity free space. The increase in length of the spring will be

k

(A) (C)

mω 2  k mω 2  k + mω 2

Sol.

21. The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered s as k = as2 where a is a positive constant. The total force acting on the particle is :

s2 (A) 2a R

 s2  (B) 2as 1 + 2   R 

(C) 2 as

(D) 2a

1/ 2

R2 S

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(B)

m

mω 2  k − mω 2

(D) None of these

CIRCULAR & W.P.E

23. A unifrom circular ring of mass per unit length λ and radius R is rotating with angular velocity ω about its own axis in a gravity free space. Tension in the ring is 1 (A) zero (B) λ R2 ω2 2 (C) λ R2 ω2 (D) λ R ω2 Sol.

24. A uniform rod of mass m and length  rotates in a horizontal plane with an angular velocity ω about a vertical axis passing through one end. The tension in the rod at distance x from the axis is : 1 1 x2 2 (A) mω x (B) mω 2 2 2  2 x 1   2 1 mω [ 2 – x2 ] (C) mω  1 –  (D) 2  2 

Page # 55 25. Water in a bucket is whirled in a vertical circle with a string attached to it. The water does not fall down even when the bucket is inverted at the top of its path. We conclude that : mv 2 R mv 2 (C) mg < R Sol.

(A) mg =

(B) mg >

mv 2 R

(D) none of these

26. A man is standing on a rough (µ = 0.5) horizontal disc rotating with constant angular velocity of 5 rad/ sec. At what distance from centre should he stand so that he does not slip on the disc ? (B) R > 0.2 m (A) R ≤ 0.2 m (C) R > 0.5 m (D) R > 0.3 m Sol.

Sol.

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CIRCULAR & W.P.E

Page # 56

27. A car travelling on a smooth road passes through a curved portion of the road in form of an arc, of circle of radius 10m. If the mass of car is 500 kg, the reaction on car at lowest point P where its speed is 20 m/s is

29. A conical pendulum is moving in a circle with angular velocity ω as shown. If tension in the string is T, which of following equations are correct ? l m

P

(A) 35 kN (C) 25 kN Sol.

(B) 30 kN (D) 20 kN

28. A pendulum bob is swinging in a vertical plane such that its angular amplitude is less than 90°. At its highest point, the string is cut. Which trajectory is possible for the bob afterwards.

(A)

(B)

(C)

(D)

(A) T = mω2l (C) T = mg cosθ Sol.

(B) T sinθ = mω2l (D) T = mω2l sinθ

30. A road is banked at an angle of 30° to the horizontal for negotiating a curve of radius 10 3 m. At what velocity will a car experience no friction while negotiating the curve? (A) 54 km/hr (B) 72 km/hr (C) 36 km/hr (D) 18 km/hr Sol.

Sol.

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CIRCULAR & W.P.E

Page # 57

31. The ratio of period of oscillation of the conical pendulum to that of the simple pendulum is : (Assume the strings are of the same length in the two cases and θ is the angle made by the string with the vertical in case of conical pendulum) (A) cosθ (C) 1 Sol.

(B) cosθ (D) none of these

33. Which vector in the figures best represents the acceleration of pendulum mass of the intermediate point in its swing?

(A)

(B)

(C)

(D)

Sol.

32. A particle is moving in a circle: (A) The resultant force on the particle must be towards the centre. (B) The cross product of the tangential acceleration and the angular velocity will be zero. (C) The direction of the angular acceleration and the angular velocity must be the same. (D) The resultant force may be towards the centre. Sol.

34. The dumbell is placed on a frictionless horizontal table. Sphere A is attached to a frictionless pivot so that B can be made to rotate about A with constant angular velocity. If B makes one revolution in period P, the tension in the rod is 2M

d

B

A

(A)

4 π 2Md P2

(B)

8 π 2Md P2

Sol.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

(C)

2Md 4π 2Md (D) P P

CIRCULAR & W.P.E

Page # 58

35. Two racing cars of masses m1 and m2 are moving in circles of radii r1 and r2 respectively. Their speeds are such that each makes a complete circle in the same time t. The ratio of the angular speeds of the first to the second car is (A) 1 : 1 (B) m1 : m2 (C) r1 : r2 (D) m1m2 : r1r2 Sol.

37. A block of mass m is suspended by a light thread from an elevator. The elevator is accelerating upward with uniform acceleration a. The work done by tension on the block during t seconds is :

T a

m

(A)

m (g + a)at 2 2

(B)

m ( g – a)at 2 2

(C)

m gat 2 2

(D) 0

Sol.

(B) WORK, POWER AND ENERGY

36. A rigid body of mass m is moving in a circle of radius r with a constant speed v. The force on the mv 2 and is directed towards the centre. What r is the work done by this force in moving the body over half the cirumference of the circle.

body is

mv 2

(A)

πr 2

mv 2

(C)

r

2

(B) Zero

(D)

πr 2 mv 2

Sol.

38. Equal force F(> mg) is applied to string in all the three cases. Starting from rest, the point of application of force moves a distance of 2 m down in all cases. In which case the block has maximum kinetic energy?

F

m (1)

(A) 1 (C) 3

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F

F m

m (3)

(2) (B) 2 (D) equal in all 3 cases

CIRCULAR & W.P.E

Page # 59

Sol.

41. A body of mass m accelerates uniformly from rest to a speed v0 in time t0. The work done on the body till any time t is

39. Two springs have their force constant as k1 and k2(k1 > k2). When they are stretched by the same force (A) No work is done by this force in case of both the springs (B) Equal work is done by this force in case of both the springs (C) More work is done by this force in case of second spring (D) More work is done by this force in case of first spring Sol.

 2  1 2 t (A) 2 mv 0  2   t0 

(B)

2 t  (C) mv 0  t   0

2 t  (D) mv 0    t0 

1 t  mv 20  0   t 2 3

Sol.

40. The work done by the frictional force on a surface in drawing a circle of radius r on the surface by a pencil of negligible mass with a normal pressing force N (coefficient of friction µk) is : (A) 4 πr 2 µ kN

(B) –2πr 2 µ kN

(C) –3πr 2 µ kN Sol.

(D) –2πrµ kN

 42. A force F = k[ yi + xj] where k is a positive constant acts on a particle moving in x-y plane starting from the point (3, 5), the particle is taken along a straight line to (5, 7). The work done by the force is : (A) zero (B) 35 K (C) 20 K (D) 15 K

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CIRCULAR & W.P.E

Sol.

Page # 60 45.When a conservative force does positive work on a body (A) the potential energy increases (B) the potential energy decreases (C) total energy increases (D) total energy decreases Sol.

43. A light spring of length 20 cm and force constant 2 kg/cm is placed vertically on a table. A small block of mass 1 kg falls on it. The length h from the surface of the table at which the ball will have the maximum velocity is (A) 20 cm (B) 15 cm (C) 10 cm (D) 5 cm Sol.

44. The work done is joules in increasing the extension of a spring of stiffness 10 N/cm from 4 cm to 6 cm is : (A) 1 (B) 10 (C) 50 (D) 100 Sol.

46. The P.E. of a certain spring when stretched from natural length through a distance 0.3 m is 10 J. The amount of work in joule that must be done on this spring to stretch it through an additional distance 0.15 m will be (A) 10 J (B) 20 J (C) 7.5 J (D) 12.5 J Sol.

47. A 10 kg block is pulled in the vertical plane along a frictionless surface in the form of an arc of a circle of radius 10 m. The applied force is 200 N as shown in the figure. If the block started from rest at A, the velocity at B would be : 60° B

F

A

(A) 1.732 m/s (C) 173.2 m/s Sol.

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(B) 17.32 m/s (D) None of these

CIRCULAR & W.P.E

Page # 61

50. In the figure shown all the surfaces are frictionless, and mass of the block, m = 1kg. The block and wedge are held initially at rest. Now wedge is given a horizontal acceleration of 10 m/s2 by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in 3 seconds is

10m/s2 m

M 48. A man who is running has half the kinetic energy of the boy of half his mass. The man speeds up by 1 m/s and then has the same kinetic energy as the boy. The original speed of the man was 2m/ s

(A)

(C) 2 m/s

(A) 30 J

(B) 60 J

(C) 150 J

(D) 100

3J

Sol.

(B) ( 2 – 1)m / s (D) ( 2 + 1)m / s

Sol.

49. A particle is released from rest at origin. It moves under influence of potential field U = x2 – 3x, kinetic energy at x = 2 is (A) 2 J (B) 1 J (C) 1.5 J (D) 0 J Sol.

51. A 1.0 kg block collides with a horizontal weightless spring of force constant 2.75 Nm–1 as shown in figure. The block compresses the spring 4.0 m from the rest position. If the coefficient of kinetic friction between the block and horizontal surface is 0.25, the speed of the block at the instant of collision is

(A) 0.4 ms–1 (B) 4 ms–1 (C) 0.8 ms–1 (D) 8 ms–1 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Sol.

Page # 62 53. The correct statement is (A) The block will cross the mean position (B) The block come to rest when the forces acting on it are exactly balanced (C) The block will come to rest when the work done by friction becomes equal to the change in energy stored in spring. (D) None Sol.

Question No. 52 to 53 (2 questions) A spring block system is placed on a rough horizontal floor. The block is pulled towards right to give spring 2µmg µmg but more than an elongation less than K K and released. 52. Which of the following laws/principles of physics can be applied on the spring block system

54. A toy car of mass 5 kg moves up a ramp under the influence of force F plotted against displacement x. The maximum height attained is given by F ymax x=0

x=11m

100 80 60 40 20 0 2 4 6 8 10 12 x

(A) conservation of mechanical energy (B) conservation of momentum (C) work energy principle (D) None Sol.

(A) ymax = 20 m (C) ymax = 11 m Sol.

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(B) ymax = 15 m (D) ymax = 5 m

CIRCULAR & W.P.E

Page # 63

55. A wedge of mass M fitted with a spring of stiffness 'k' is kept on a smooth horizontal surface. A rod of mass m is kept on the wedge as shown in the figure. System is in equilibrium. Assuming that all surfaces are smooth, the potential energy stored in the spring is :

Question No. 57 to 62 (6 questions) A block of mass m moving with a velocity v0 on a smooth horizontal surface strikes and compresses a spring of stiffness k till mass comes to rest as shown in the figure. This phenomenon is observed by two observers:

V0

m

k M

m

θ

(A)

mg2 tan 2 θ 2K

(B)

m 2 g tan2 θ 2K

(C)

m 2 g2 tan 2 θ 2K

(D)

m 2 g2 tan 2 θ K

Sol.

K

A : standing on the horizontal surface B : standing on the block 57. To an observer A, the work done by spring force is (A) negative but nothing can be said about its magnitude 1 2 (B) – mv 0 2 (C) positive but nothing can be said about its magnitude 1 2 (D) + mv 0 2 Sol.

56. A block of mass m is hung vertically from an elastic thread of force constant mg/a. Initially the thread was at its natural length and the block is allowed to fall freely. The kinetic energy of the block when it passes through the equilibrium position will be : (A) mga (B) mga/2 (C) zero (D) 2mga Sol.

58. To an observer A, the work done by the normal reaction N between the block and the spring on the block is 1 2 (A) zero (B) – mv 0 2 1 2 (C) + mv 0 (D) none of these 2 Sol.

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CIRCULAR & W.P.E

Page # 64

59. To an observer A, the net work done on the block is 2 (A) –mv 0

1 2 (C) – mv 0 2 Sol.

(B) +mv 20 (D) zero

60. According to the observer A (A) the kinetic energy of the block is converted into the potential energy of the spring (B) the mechanical energy of the spring-mass system is conserved (C) the block loses its kinetic energy because of the negative work done the conservative force of spring (D) all the above Sol.

61. To an observer B, when the block is compressing the spring (A) velocity of the block is decreasing (B) retardation of the block is increasing (C) kinetic energy of the block is zero (D) all the above Sol.

62. According to observer B, the potential energy of the spring increases (A) due to the positive work done by pseudo force (B) due to the positive work done by normal reaction between spring & wall (C) due to the decrease in the kinetic energy of the block (D) all the above Sol.

63. A car of mass 'm' is driven with acceleration 'a' along a straight level road against a constant external resistive force 'R'. When the velocity of the car is 'V', the rate at which the engine of the car is doing work will be : (A) RV (B) maV (C) (R + ma) V (D) (ma – R)V Sol.

64. A truck of mass 30,000 kg moves up an inclined plane of slope 1 in 100 at a speed of 30 kmph. The power of the truck is (given g = 10 ms–2) (A) 25 kW (B) 10 kW (C) 5 kW (D) 2.5 kW Sol.

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CIRCULAR & W.P.E

Page # 65 Sol.

65. A part i c l e m ov es wi th a v el oc i t y  v = (5 i – 3 j + 6k ) m / s under the influence of a constant  force F = (10 i + 10 j + 20k )N. The instantaneous power applied to the particle is : (A) 200 J/s (B) 40 J/s (C) 140 J/s (D) 170 J/s Sol.

66. Assume the aerodynamic drag force on a car is proportional to its speed. If the power output from the engine is doubled, then the maximum speed of the car. (A) is unchanged (B) increases by a factor of 2 (C) is also doubled (D) increases by a factor of four. Sol.

68. The diagrams represent the potential energy U of a function of the inter-atomic distance r. Which diagram corresponds to stable molecules found in nature. U U

(A)

(B)

O U

r

(C)

Sol.

O U

r

(D)

O

67. A body of mass 1 kg starts moving from rest at t = 0, in a circular path of radius 8 m. Its kinetic energy varies as a function of time as : K.E. = 2t2 Joules, where t is in seconds. Then (A) tangential acceleration = 4m/s2 (B) power of all forces at t = 2 sec is 8 watt (C) first round is completed in 2 sec. (D) tangential force at t = 2 sec is 4 newton. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

r

O

r

CIRCULAR & W.P.E

Page # 66 71. A particle originally at rest the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance h below the highest point, such that (A) h = R (B) h = R/3 (C) h = R/2 (D) h = 2R Sol.

 69. The potential energy for a force field F is given by U(x, y) = sin (x + y). The force acting on the  π particle of mass m at  0,  is 4

(A) 1

(B)

2

(C)

1 2

(D) 0

Sol.

72. A ball whose size is slightly smaller than width of the tube of radius 2.5 m is projected from bottommost point of a smooth tube fixed in a vertical plane with velocity of 10 m/s. If N1 and N2 are the normal reactions exerted by inner side and outer side of the tube on the ball

D

A

C B O

10 m/s (A) N1 > 0 for motion in ABC, N2 > 0 for motion in CDA (B) N1 > 0 for motion in CDA, N2 > 0 for motion in ABC (C) N2 > 0 for motion in ABC & part of CDA (D) N1 is always zero. Sol. 70. F = 2x2 – 3x –2. Choose correct option (A) x = –1/2 is position of stable equilibrium (B) x = 2 is position of stable equilibrium (C) x = –1/2 is position of unstable equilibrium (D) x = 2 is position of neutral equilibrium Sol.

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CIRCULAR & W.P.E

Page # 67

73. A bob attached to a string is held horizontal and released. The tension and vertical distance from point of suspension can be represented by.

T

Sol.

T

(A)

(B) h

h

T

T

(C)

(D) h

h

Sol.

74. A small cube with mass M starts at rest point 1 at a height 4R, where R is the radius of the circular part of the track. The cube slides down the frictionless track and around the loop. The force that the track exerts on the cube at point 2 is nearly _________ times the cube’s weight Mg.

1

1R

2 R

(A) 1

(B) 2

(C) 3

(D) 4

75. The tube AC forms a quarter circle in a vertical plane. The ball B has an area of cross-section slightly smaller than that of the tube, and can move without friction through it. B is placed at A and displaced slightly. It will A B

C (A) always be in contact with the inner wall of the tube (B) always be in contact with the outer wall of the tube (C) initially be in contact with the inner wall and later with the outer wall (D) initially be in contact with the outer wall and later with the inner wall

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CIRCULAR & W.P.E

Page # 68

Sol.

76. A particle is rotated in a vertical circle by connecting it to a light rod of length l and keeping the other end of the rod fixed. The minimum speed of particle when the light rod is horizontal for which the particle will complete the circle is (A)

gl

(C) 3gl Sol.

(B)

2gl

(D) none

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CIRCULAR & W.P.E

Page # 69

(Multiple Correct Problems)

Exercise - II

(A) CIRCULAR MOTION  1. A person applies a constant force F on a particle of mass m and finds that the particle moves in a circle of radius r with a uniform speed v as seen (in the plane of motion) from an inertial frame of reference. (A) This is not possible. (B) There are other forces on the particle. (C) The resultant of the other forces is

Sol.

mv 2 towards r

the centre. (D) The resultant of the other forces varies in magnitude as well as in direction. Sol.

3. A simple pendulum of length l and mass (bob) M is oscillating in a plane about a vertical line between angular limits –φ and φ. For an angular displacement θ, [|θ| < φ] the tension in the string and velocity of the bob are T and v respectively. The following relations hold good under the above conditions : (A) T cos θ = Mg Mv 2 L (C) Tangential acc. = g sin θ (D) T = Mg cos θ Sol.

(B) T – Mg cos θ =

2.A machine, in an amusement park, consists of a cage at the end of one arm, hinged at O. The cage revolves along a vertical circle of radius r (ABCDEFGH) ab out i ts hi nge O, at c onst ant l i near spe ed v = gr . The cage is so attached that the man of weight ‘w’ standing on a weighing machine, inside the cage, is always vertical. Then which of the following is correct E × F D ×

×

r ×C

G× × A

B

×

×

H

(A) the reading of his weight on the machine is the same at all positions (B) the weight reading at A is greater than the weight reading at E by 2 w. (C) the weight reading at G = w (D) the ratio of the weight reading at E to that at A = 0 (E) the ratio of the weight reading at A to that at C = 2

(B) W.P.E 4. No work is done by a force on an object if (A) the force is always perpendicular to its velocity (B) the force is always perpendicular to its acceleration (C) the object is stationary but the point of application of the force moves on the object. (D) the object moves in such a way that the point of application of the force remains fixed.

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CIRCULAR & W.P.E

Page # 70 Sol.

Sol.

7. When total work done on a particle is positive (A) KE remains constant (B) Momentum increases (C) KE decreases (D) KE increases Sol.

5. One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block pl aced on a smooth hori zontal surface. In a displacement, the work done by the spring is

1 2 kx . 2

The possible cases are : (A) the spring was initially compressed by a distance x and was finally in its natural length (B) it was initially stretched by a distance x and finally was in its natural length (C) it was initially in its natural length and finally in a compressed position. (D) it was initially in its natural length and finally in a stretched position. Sol.

8. A particle with constant total energy E moves in one dimension in a region where the potential energy is U(x). The speed of the particle is zero where (A) U(x) = E (C)

dU( x) =0 dx

Sol.

6. Work done by force of friction (A) can be zero (B) can be positive (C) can be negative (D) information insufficient

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

(B) U(x) = 0 (D)

d2U( x) dx2

=0

CIRCULAR & W.P.E

Page # 71

9. A block of mass m slides down a plane inclined at an angle θ. Which of the following will NOT increase the energy lost by the block due to friction ? (A) Increasing the angle of inclination (B) Increasing the distance that the block travels (C) Increasing the acceleration due to gravity (D) Increasing the mass of the block Sol.

11. A ball of mass m is attached to the lower end of light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstreched) length, comes to rest again after descending through a distance x. (A) x = mg/k (B) x = 2 mg/k (C) The ball will have no acceleration at the position where it has descended through x/2. (D) The ball will have an upward acceleration equal to g at its lowermost position. Sol.

10. A box of mass m is released from rest at position on the frictionless curved track shown. It slides a distance d along the track in time t to reach position 2, dropping a vertical distance h. Let v and a be the instantaneous speed and instantaneous acceleration, respectively, of the box at position 2. Which of the following equations is valid for this situation? 1 m h 2 d (A) h = vt

(B) h = (1/2)gt2

(C) d = (1/2)at2 Sol.

(D) mgh = (1/2)mv2

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CIRCULAR & W.P.E

Page # 72

12. A cart moves with a constant speed along a horizontal circular path. From the cart, a particle is thrown up vertically with respect to the cart (A) The particle will land somewhere on the circular path (B) The particle will land outside the circular path (C) The particle will follow an elliptical path (D) The particle will follow a parabolic path Sol.

Question No. 14 to 16 (3 questions) A particle of mass m is released from a height H on a smooth curved surface which ends into a vertical loop of radius R, as shown m 13. The potential energy in joules of a particle of mass 1 kg moving in a plane is given by U = 3x + 4y, the position coordinates of the point being x and y, measured in meters. If the particle is initially at rest at (6, 4), then (A) its acceleration is of magnitude 5 m/s2 (B) its speed when it crosses the y-axis is 10 m/s (C) it crosses the y-axis (x = 0) at y = –4 (D) it moves in a straight line passing through the origin (0, 0) Sol.

R H

C θ

14. Choose the correct alternative(s) if H = 2R (A) The particle reaches the top of the loop with zero velocity (B) The particle cannot reach the top of the loop (C) The particle breaks off at a height H = R from the base of the loop (D) The particle break off at a height R < H < 2R

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CIRCULAR & W.P.E

Page # 73

Sol.

16. The minimum value of H required so that the particle makes a complete vertical circle is given by (A) 5 R (B) 4 R (C) 2.5 R (D) 2 R Sol.

15. If θ is instantaneous angle which the line joining the particle and the centre of the loop makes with the vertical, then identify the correct statement(s) related to the normal reaction N between the block and the surface (A) The maximum value N occurs at θ = 0 (B) The minimum value of N occurs at N = π for H > 5R/2 (C) The value of N becomes negative for π/2 < θ < 3π/2 (D) The value of N becomes zero only when θ ≥ π/2 Sol.

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CIRCULAR & W.P.E

Page # 74

(Subjective Problems)

Exercise - III (A) CIRCULAR MOTION 1. The 10 kg block is in equilibrium. (A)

10 kg

(B)

(i) Find the tension in string A. (ii) Find the tension in string A just after the string B is cut? Sol.

2. A particle moves in the x-y plane with the velocity  v = ai + bt j . At the instant t = a 3 / b the magnitude

3. A particle moves clockwise in a circle of radius 1 m with centre at (x, y) = (1m, 0). It starts at rest at the origin at time t = 0. Its speed increases at the  π constant rate of   m/s2. (a) How long does it take 2 to travel halfway around the circle ? (b) What is the speed at that time ? Sol.

4. A point moves along a circle having a radius 20 cm with a constant tangential acceleration 5 cm/s2. How much time is needed after motion begins for the normal acceleration of the point to be equal to tangential acceleration? Sol.

of tangential, normal and total acceleration are _________________, _______ & __________. Sol.

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CIRCULAR & W.P.E

Page # 75

5. A ring rotates about z axis as shown in figure. The plane of rotation is xy. At a certain instant the acceleration of a particle P (shown in figure) on the ring is  (6 i – 8 j ) m/s2. find the angular acceleration of the ring & the angular velocity at that instant. Radius of the ring is 2m. y P

O

x

Sol.

7. Figure shows the total acceleration and velocity of a particle moving clockwise in a circle of radius 2.5 m at a given instant of time. At this instant, find :

°

30

2. 5m

a=25 m/s

6. A particle is revolving in a circle of radius 1m with an angular speed of 12 rad/s. At t = 0, it was subjected to a constant angular acceleration α and its angular speed increased to (480/π) rpm in 2 sec. Particle then continues to move with attained speed. Calculate (a) angular acceleration of the particle, (b) tangential velocity of the particle as a function of time. (c) acceleration of the particle at t = 0.5 second and at t = 3 second (d) angular displacement at t = 3 second. Sol.

a

v

(a) the radial acceleration, (b) the speed of the particle and (c) its tangential acceleration Sol.

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2

CIRCULAR & W.P.E

Page # 76

8. A stone is launched upward at 45° with speed v0. A bee follows the trajectory of the stone at a constant speed equal to the initial speed of the stone. (a) Find the radius of curvature at the top point of the trajectory. (b) What is the acceleration of the bee at the top point of the trajectory? For the stone, neglect the air resistance. Sol.

Sol.

9. A particle moves in circle of radius R with a constant speed v. Then, find the magnitude of average acceleration during a time interval

πR . 2v

Sol.

11. A rod of length 1 m is being rotated about its end in a gravity free space with a constant angular acceleration of 5 rad/s2 starting from rest. A sleeve is fitted on the rod at a distance of 0.5 m from the centre. The coefficient of friction between the rod and the sleeve is 0.05. Find the time after which sleeve will start slipping on the rod. Sol.

10. A 4 kg block is attached to a vertical rod by means of two strings of equal length. When the system rotaes about the axis of the rod, the strings are extended as shown in figure. (a) How many revolutions per minute must the system make in order for the tension in the upper chord to be 20 kgf? (b) What is the tension in the lower chord? 5m 8m

θ

A

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CIRCULAR & W.P.E

Page # 77

12. A mass m rotating freely in a horizontal circle of radius 1 m on a frictionless smooth table supports a stationary mass 2m, attached to the other end of the string passing through smooth hole O in table, hanging vertically. Find the angular velocity of rotation.

O

14. The blocks are of mass 2 kg shown is in equilibrium. At t = 0 right spring in fig. (i) and right string in fig. (ii) breaks. Find the ratio of instantaneous acceleration of blocks ? 37° 37°

m

2m

2 kg

Sol.

37°

37°

2 kg figure (ii)

Sol.

13. A beam of mass m is attached to one end of a spring of natural length

3 R and spring constant

( 3 + 1)mg . The other end of the spring is fixed at R point A on a smooth fixed vertical ring of radius R as shown in the figure. What is the normal reaction at B just after the bead is released? B k=

A

Sol.

60°

(B) WORK, POWER AND ENERGY 15. A block of mass m is pulled on a rough horizontal surface which has a friction coefficient µ. A force F isapplied which is capable of moving the body uniformly with speed v. Find the work done on the block in time t by (a) weight of the block, (b) Normal reaction by surface on the block, (c) friction, (d) F. Sol.

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CIRCULAR & W.P.E

Page # 78

(Force dyne)

18. The relationship between force and position is shown in the figure given (in one dimensional case). What will be the work done by the force in displacing a body from x = 1 cm to x = 5 cm. 20 10 0

1

10

2

3

4

5 6 x(cm)

20

Sol.

16. Calculate the work done against gravity by a coolie in carrying a load of mass 10 kg on his head when he walks uniformly a distance of 5 m in the (i) horizontal direction (ii) vertical direction. (Take g = 10 m/s2) Sol.

17. A body is constrained to move in the y-direction. It is subjected to a force (–2i + 15 j + 6k ) newton. What is the work done by this force in moving the body through a distance of 10 m ? Sol.

19. It is well known that a raindrop or a small pebble falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop or small pebble of 1 g falling from a cliff of height 1.00 km. It hits –1 . What is the work done by the unknown resistive force ? Sol. t h

e

g

r o u

n

d

w

i t h

a

s p

e

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

e

d

o f

5

0

. 0

m

s

CIRCULAR & W.P.E

Page # 79

20. A rigid body of mass 2 kg initially at rest moves under the action of an applied horizontal force 7 N on a table with coefficient of kinetic friction = 0.1. Calculate the (a) work done by the applied force on the body in 10 s. (b) work done by friction on the body in 10 s. (c) work done by the net force on the body in 10 s. (d) change in kinetic energy of the body is 10 s. Sol.

21. A rigid body of mass 0.3 kg is taken slowly up an inclined plane of length 10 m and height 5 m, and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.15. Using g = 9.8 m/s2 find the

(a) work done by the gravitational force over the round trip. (b) work done by the applied force (assuming it to be parallel to the inclined plane) over the upward journey (c) work done by frictional force over the round trip. (d) kinetic energy of the body at the end of the trip? Sol.

22. A block of mass m sits at rest on a frictionless table in a rail car that is moving with speed vc along a straight horizontal track (fig.) A person riding in the car pushes on the block with a net horizontal force F for a time t in the direction of the car's motion.

Ground

Train

m

s

F

s1

(a) What is the final speed of the block according to a person in the car ? 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Page # 80

Sol. (g) How much work does each say the force did ? Sol.

(b) According to a person standing on the ground outside the train? Sol.

(h) Compare the work done to the K gain according to each person. Sol.

(c) How much did K of the block change according to the person in the car ? Sol.

(i) What can your conclude from this computation? Sol. (d) According to the person on the ground ? Sol.

(e) In terms of F, m, & t, how far did the force displace the object according to the person in car ? Sol.

23. In the figure shown, pulley and spring are ideal. Find the potential energy stored in the spring (m1 > m2) Sol.

(f) According to the person on the ground ? Sol.

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k

m1 m2

CIRCULAR & W.P.E

Page # 81

24. A block of mass m placed on a smooth horizontal surface is attached to a spring and is held at rest by a force P as shown. Suddenly the force P changes its direction opposite to the previous one. How many times is the maximum extension l2 of the spring longer compared to its initial compression l2 ?

26. A labourer lifts 100 stones to a height of 6 metre in two minute. If mass of each stone be one kilogram, calculate the average power. Given : g = 10 ms–2. Sol.

l1 P

Sol.

27. An engine develops 10 kW of power. How much time will it take to lift a mass of 200 kg through a height of 40 m? Given : g = 10 ms–2 Sol.

25. (a) Power applied to a particle varies with time as P = (3t2 – 2t + 1) watt, where t is in second. Find the change in its kinetic energy between time t = 2 s and t = 4 s. Sol.

28. Two trains of equal masses are drawn along smooth level lines by engines; one of then X exerts a constant force while the other Y works at a constant rate. Both start from rest & after a time t both again have the same velocity v. Find the ratio of travelled distance during the interval. Sol.

(b) The potential function for a conservative force is given by U = k(x + y). Find the work done by the conservative force in moving a particle from the point A(1, 1) to point B (2, 3). Sol.

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CIRCULAR & W.P.E

Page # 82 30. A particle moves along a straight line. A force acts on the particle which produces a constant power. It starts with initial velocity 3 m/s and after moving a distance 252 m its velocity is 6 m/s. Find the time taken. Sol.

29. Water is pumped from a depth of 10m and delivered through a pipe of cross section 10–2m2 upto a height of 10m. If it is needed to deliver a volume 0.2 m3 per second, find the power required. [Use g = 10 m/s2] Sol.

31. A force F = x2y2i + x2y2j (N) acts on a particle which moves in the XY plane.

Y

a

C

D a X A B Find the work done by F as it moves the particle from A to C (fig.) along each of the paths ABC, ADC, and AC. 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

CIRCULAR & W.P.E

Sol.

Page # 83 32. Calculate the forces F(y) associated with the following one-dimensional potential energies : (a) U = –ωy (b) U = ay3 – by2 (c) U = U0 sin β y Sol.

33. Consider the shown arrangement when a is bob of mass ‘m’ is suspended by means of a string connected  to peg P. If the bob is given a horizontal velocity u having magnitude 3gl , find the minimum speed of the bob in subsequent motion. P l

Sol.

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u

CIRCULAR & W.P.E

34. A person rolls a small ball with speed u along the floor from point A. If x = 3R, determine the required speed u so that the ball returns to A after rolling on the circular surface in the vertical plane from B to C and becoming a projectile at C. What is the minimum value of x for which the game could be played if contact must be maintained to point C ? Neglect friction.

Page # 84 35. A toy rocket of mass 1 kg has a small fuel of mass 0.02 kg which it burns out in 3 s. Starting from rest on a horizontal smooth track, it gets a speed of 20 ms–1 after the fuel is burnt out. What is the average thrust of the rocket? What is the energy content per unit mass of the fuel? (Ignore the small mass variation of the rocket during fuel burning). Sol.

Sol.

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CIRCULAR & W.P.E

Page # 85

(Tough Subjective Problems)

Exercise - IV Q.1 A particle which moves along the curved path shown passes point O with a speed of 12 m/s and slows down to 5m/s at point A in a distance of 18 m measured along the curve from O. The deceleration measured along the curve it proportional to distance from O. If the total acceleration of the particle is 10 m/s2 on it passes A. Find the radius of curvature of A.

A

O

Sol.

Q.3 A small is block can move in a straight horizontal line a along AB. Flash lights from one side projects its shadow on a vertical wall which has horizontal cross section as a circle. Find tangential & normal acceleration of shadow of the block on the wall as a function of time if the velocity of the block is constant (v). B

R

v=const

A Top View

Q.2 A ball of mass 1 kg is released from position A inside a wedge with a hemispherical cut of radius 0.5 m as shown in the figure. Find the force exerted by the vertical wall OM on wedge, when the ball is in position B. (neglect friction everywhere) Take (g = 10m/s2) M A C

Sol.

60° O

N

Sol.

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CIRCULAR & W.P.E

Page # 86

Q.4 A particle is confined to move along the +x axis under the ac tion of a force F(x) that is derivable from thePotential U(x) = ax3 – bx. U

Sol.

x x1 x0 (a) Find the expression for F(x) (b) When the total energy of the particle is zero, the particle can be trapped with in the interval x = 0 to x = x1. For this case find the values of x1. (c) Determine the maximum kinetic energy that the trapped particle has in its motion. Express all answers in terms a and b. Sol.

O

Q.5 A particle of mass 5 kg is free to slide on a smooth ring of radius r = 20 cm fixed in a vertical plane. The particle is attached to one end of a spring whose other end is fixed to the top point O of the ring. Initially the particle is at rest at a point A of the ring such that ∠OCA = 60º, C being the centre of the ring. The natural length of the spring is also equal to r = 20 cm. After the particle is released and slides down the ring the contact force between the particle & the ring becomes zero when it reaches the lowest position B. Determine the force constant of the spring.

Q.6 Two blocks of mass m1 = 10kg and m2 = 5kg connected to each other by a massless inextensible string of length 0.3 m are placed along a diameter of a turn table. The coefficient of friction between the table and m1 is 0.5 while there is no friction between m2 and the table. The table is rotating with an angular velocity of 10 rad/sec about a vertical axis passing through its centre. The masses are placed along the diameter of the table on either side of the centre O such that m1 is at a distance of 0.124 m from O. The masses are observed to be at rest with respect to an observer on the turn table. (i) Calculate the frictional force on m1 (ii) What should be the minimum angular speed of the turn table so that the masses will slip from this position (iii) How should the masses be placed with the string remaining taut, so that there is no frictional force acting on the mass m1. Sol.

O

A 60°

C

B

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CIRCULAR & W.P.E

Page # 87 Q.8 Two identical beads of mass 1 kg each are connected by an inextensible massless string & they can slide along the two arms AB and BC of a rigid smooth wire frame in vertical plane. If the system is released from rest, find the speeds of the particles when they have moved by a dis tance of 0.1 m. Also find tension in the string. 0.4m B A

0.3m

C Q.7 A ring of mass m can slide over a smooth vertical rod. The ring is connected to a spring of force con4mg stant K = where 2R is the natural length of the R spring. The other end of the spring is fixed to the ground at a horizontal distance 2R from the base of the rod. The mass is released at a height of 1.5 R from ground

Sol.

3R/2

A 2R (a) calculate the work done by the spring (b) calculate the velocity of the ring as it reaches the ground. Sol.

Q.9 The ends of spring are attached to blocks of mass 3 kg and 3 kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is in equilibrium producing a compression of 1cm of the spring. The 2kg mass must be compressed further by at least ________, so that when it is released, the 3 kg block may be lifted off the ground.

2kg

3 kg Sol.

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Page # 88

Q.10 A uniform rod of mass m length L is sliding along its length on a horizontal table whose top is partly smooth & rest rough with friction coefficient µ. If the rod after moving through smooth part, enters the rough with velocity v0. (a) What will be the magnitude of the friction force when its x length (< L) lies in the rough part during sliding. (b) Determine the minimum velocity v0 with which it must enter so that it lies completely in rough region before coming to rest. (c) If the velocity is double the minimum velocity as calculated in part (a) then what distance does its front end A would have travelled in rough region before rod comes to rest.

Sol.

L

v0 B

M

A

m

Sol. Q.12 A small bead of mass m is free to slide on a fixed smooth vertical wire, as indicated in the diagram. One end of a light elastic string, of unstretched length a and force constant 2 mg/a is attached to B. The string passes through a smooth fixed ring R and the other end of the string is attached to the fixed point A, AR being horizontal. The point O on the wire is at same horizontal level as R, and AR = RO = a. (i) In the equilibrium, find OB (ii) The bead B is raised to a point C of the wire above O, where OC = a, and is released from rest. Find the speed of the bead as it passes O, and find the greatest depth below O of the bead in the subsequent motion.

O B Sol.

Q.11 Find the velocity with which a block of mass 1 kg must be horizontally projected on a conveyer belt moving uniformly at a velocity of 3 m/s so that maximum heat is liberated. Take coefficient of friction of 0.1. Also find the corresponding amount of heat liberated. What happens when belt velocity is 5 m/s ? 1kg v=3m/s 8m

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a R

A

CIRCULAR & W.P.E

Page # 89

Q.13 A small block of mass m is projected horizontally from the top the smooth hemisphere of radius r with speed u as shown. For values of u ≥ u0, it does not slide on the hemisphere (i.e. leaves the surface at the top itself) (a) For u = 2u0 it lands at point P on ground Find OP. (b) For u = u0/3, Find the height from the ground at which it leaves the hemisphere. (c) Find its net acceleration at the instant it leaves the hemisphere. r

Sol.

u

o

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Page # 90

JEE-Problems

Exercise - V  Q.1 A force F = −K( y i + x j ) where K is a positive con-

stant, acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work  done by the force F on the particle is [JEE-98] (A) –2Ka2 (B) 2Ka2 (C) –Ka2 (D) Ka2 Sol.

Q.3 A particle is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is at a distance L/8 from O as shown. The object given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Find u. [JEE-99] A O

u

Sol.

Q.2 A stone is tied to a string of length l is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in its velocity at it reaches a position where the string is horizontal is [JEE-98] (A)

(u 2 − 2gl )

(B)

2gl

(C)

(u2 − gl )

(D)

2(u2 − gl )

Sol.

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B

CIRCULAR & W.P.E

Page # 91

Q.4 A long horizontal rod has bead which can slide along its length, and initially placed at a distance L from one end of A of the rod. The rod is set in angular motion about A with constant angular acceleration α. If the coefficient of friction between the rod and the bead is µ and gravity is neglected, then the time after which is bead starts slipping is [JEE-2000] µ α

(A)

(B)

µ α

1

(C)

µα

(D) infinitesimal

(A)

v

(B)

v

(C)

v

(D)

v

Sol.

Sol.

Q.6 An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle α with the vertical, the maximum possible value of α is given by [JEE(Scr.)-2001]

(A) cot α = 3 (C) sec α = 3 Sol.

Q.5 A small block is shot into each of the four tracks as shown below. Each of the tracks risks to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in [JEE-2001] 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

(B) tan α = 3 (D) cosec α = 3

CIRCULAR & W.P.E

Q.7 A small ball of mass 2 × 10–3 Kg having a charge of 1µc is suspended by a string of length 0.8m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball so that it can make complete revolution. [JEE-2001] Sol.

Page # 92

(A)

(B)

(C)

(D)

Sol.

Q.9 A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle x of the particle from the origin as F(x) = –kx + ax2. Here k and a are positive constants. For x ≥ 0, the functional form of the potential energy U(x) of the particle is [JEE(Scr.)-2002] U(x)

U(x) x

(A) Q.8 A simple pendulum is oscillating without damping. When the displacement of the bob is less that maxi mum, its acceleration vector a is correctly shown in [JEE(Scr.)-2002]

(B)

U(x)

(C)

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x

U(x)

x

(D)

x

CIRCULAR & W.P.E

Sol.

Page # 93 (a) Express the total normal reaction force exerted by the spheres on the ball as a function of angle θ. Sphere B

d

Q.10 An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is [JEE(Scr.)-2002] (A) 4 Mg/k (B) 2 Mg/k (C) Mg/k (D) Mg/2k Sol.

R

O

Sphere A (b) Let NA and NB denote the magnitudes of the normal reaction force on the ball exerted by the spheres A and B, respectively. Sketch the variations of NA and NB as functions of cosθ in the range 0 ≤ θ ≤ π by drawing two separate graphs in your answer book, taking cosθ on the horizontal axes. Sol.

Q.12 In a region of only gravitational field of mass ‘M’ a particle is shifted from A to B via three different paths in the figure. The work done in different paths are W1, W2, W3 respectively then [JEE(Scr.)-2003] Q.11 A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see figure.) The smaller sphere A has a radius R and the space between the two spheres has a width d. The ball has a diameter very slightly less than d. All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the upward vertical is denoted by θ (shown in the figure) [JEE-2002]

(3)

B

C

M (2) (1) A

(A) W1 = W2 = W3 (C) W1 > W2 > W3

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(B) W1 = W2 > W3 (D) W1 < W2 < W3

CIRCULAR & W.P.E

Page # 94 U(x)

Sol. U(x) x

(A)

U(x)

U(x)

(C)

x

(B)

x

(D)

x

Sol. Q.13 A particle of mass m, moving in a circular path of radius R with a constant speed v2 is located at point (2R, 0) at time t = 0 and a man starts moving with velocity v1 along the +ve y-axis from origin at time t = 0. Calculate the linear momentum of the particle w.r.t. the man as a function of time. [JEE-2003] y v2 v1 R (0,0)

m

x

Sol.

Q.14 A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). If U(0) = 0, the graph of U(x) versus x will be (where U is the potential energy function) [JEE(Scr.)-2004]

Q.15 STATEMENT-1 A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30º with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. because STATEMENT-2 The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination. [JEE-2007] (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True

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CIRCULAR & W.P.E

Page # 95

Sol.

[JEE 2009] Sol. 16. A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A, satisfies

[JEE 2008]

B

L A (A) θ =

π 4

π 3π (C) < θ < 2 4 Sol.

V (B)

π π v1

v 2 – v1 e = u –u 1 2 Note : Coefficient of restitution is a factor between two colliding bodies which is depends on the material of the body but independent of shape. We can say e is a factor which relates deformation and reformation of the body. 0  e1

Ex.35 If a body falls normally on a surface from height h, what will be the height regained after collision if coefficient of restitution is e?

h

Sol.

If a body falls from height h, from equations of motion we know that it will hit the ground with a velocity say u =

2gh which is also the velocity of approach here.

Now if after collision it regains a height h1 then again by equations of motion v =

2gh1 which is also

the velocity of separation. So, by definition of e, e=

2gh1 2gh

or h1 = e2h

Ex.36 A block of mass 2 kg is pushed towards a very heavy object moving with 2 m/s closer to the block (as 2m/s very shown). Assuming elastic collision and frictionless heavy 10m/s surface, find 2kg object the final velocities of the blocks. Sol. Let v1 and v2 be the final velocities of 2kg block 2m/s very and heavy object respectively then, heavy 14m/s v1 = u1 + 1 (u1 – u2) = 2u1 – u2 2kg object = – 14 m/s v2 = – 2m/s Ex.37 A ball is moving with velocity 2 m/s towards a heavy wall moving towards the ball with speed 1 m/s as shown in fig. Assuming collision to be elastic, find the velocity of the ball immediately after the collision.

2m/s

Sol.

1m/s

The speed of wall will not change after the collision. So, let v be the velocity of the ball after collision in the direction shown in figure. Since collision is elastic (e = 1).

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CENTRE OF MASS

2m/s

Page # 31

1m/s

Before Collision

1m/s

v

After Collision

separation speed = approach speed or v–1=2+1 or v = 4 m/s Ans. Ex.38 A ball is dropped from a height h on to a floor. If in each collision its speed becomes e times of its striking value (a) find the time taken by ball to stop rebounding (b) find the total change in momentum in this time (c) find the average force exerted by the ball on the floor using results of part (a) and (b). Sol. (a) When the ball is dropped from a height h, time taken by it to reach the ground will be

t0 =

2h and its speed v0 = g

2gh

h

v0

v1 v2

t0

t1

t0

Now after collision its speed will becomes e times, i.e., v1 = ev0 = e 2gh and so, it will take time to go up till its speed becomes zero = (v1/g). The same time it will take to come down. So total time between I and II collision will be t1 = 2v1/g. Similarly, total time between II and III collision t2 = 2v2/g. So total time of motion T = t0 + t1 + t2 +......... or

T = t0 +

2 v1 2v 2 + ....... g g

or

T = t0 +

2ev 0 2e 2 v 0 + ....... g g [as v2 = ev1 = e2v0]

or

T =

2h [1  2e(1  e  e2 ....)] g

2h   1    1  2e g  1 e 

2h  1  e  g  1 – e 

(b) Change in momentum in I collision = mv1 – (–mv0) = m (v1 + v0) Change in momentum in II collision = m(v2 + v1) Change in momentum in nth collision = m(vn + vn–1) Adding these all total change in momentum p = m[v0 + 2v1 + ....+ 2vn–1 + vn] or

p = mv0[1 + 2e + e2 + .....]

or

 1 e   1  p = mv0 1  2e 1 – e    m 2gh  1 – e     

...(2)

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CENTRE OF MASS

Page # 32   dp p (C) Now as F  so, Fav =  T dt

Substituting the value of T and p from Eqns. (1) and (2) 1 e  Fav = m 2gh   × 1 – e 

7.1

g 1 – e  = mg 2h  1  e 

...(3)

Line of Motion The line passing through the centre of the body along the direction of resultant velocity.

7.2

Line of Impact The line passing through the common normal to the surfaces in contact during impact is called line of impact. The force during collision acts along this line on both the bodies. Direction of Line of impact can be determined by : (a) Geometry of colliding objects like spheres, discs, wedge etc. (b) Direction of change of momentum. If one particle is stationary before the collision then the line of impact will be along its motion after collision. Examples of line of impact (i) Two balls A and B are approaching each other such that their centres are moving along line CD. Line of impact and line of motion

C

D

A

B

(ii) Two balls A and B are approaching each other such that their centre are moving along dotted lines as shown in figure.

B Line of motion of ball B

Line of motion of ball A D

A Line of impact (iii)

Ball is falling on a stationary wedge.

Line of motion of ball

Line of impact

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Page # 33

Note : In previous discussed examples line of motion is same as line of impact. But in problems in which line of impact and line of motion is different then e will be

velocity of seperation along line of impact velocity of approach along line of impact

e=

Ex.40 A ball of mass m hits a floor with a speed v making an angle of incident  with the normal. The coefficient of restitution is e. Find the speed of the reflected ball and the angle of reflection of the ball. Sol.

Suppose the angle of reflection is  and the speed after the collision is v  (shown figure) The floor exerts a force on the ball along the normal during the collision. There is no force parallel to the surface. Thus, the parallel component of the velocity of the ball remains unchanged. This gives v  sin   = v sin 

...(i)

For the components normal to the floor, the velocity of separation is v cos  and the velocity of approach is v cos .

v

 '

v'

Hence, v  cos   = ev cos  sin 2   e 2 cos 2 

From (i) and (ii), v  = v

tan e For elastic collision, e = 1, so that  =  and v = v.

v cos 

tan   =

Hence,

v'cos '

v sin

Initial velocity

v' sin  ' Final velocity

Ex.41 A ball is projected from the ground at some angle with horizontal. Coefficient of restitution between the ball and the ground is e. Let a, b and c be the ratio of times of flight, horizontal range and maximum height in two successive paths. Find a, b and c in terms of e?

1 2

Sol.

Let us assume that ball is projected with speed u at an angle  with the horizontal. Then Before first collision with the ground. Time fo flight T 

2u y

euy

g

Horizontal range R 

u I

u y  u sin 

2u x u y g

u x  u cos 

u2y Maximum Height Hmax =

ux

...(1)

2g

uy

After striking the ground the component uy is change into e uy, so 2eu y

Time of flight T =

'  Hmax

g

, R' 

2u x ( eu y ) g

(eu y )2 2g

...(2)

from eq (1) & (2) Now

T 1 a T' e

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II

CENTRE OF MASS

Page # 34 Hmax 1 R 1  2 =c b ; ' Hmax e R' e

Ex.42 A ball is projected from the ground with speed u at an angle  with horizontal. It collides with a wall at a distance a from the point of projection and returns to its original position. Find the coefficient of restitution between the ball and the wall. Sol.

A ball is a projected with speed u at an angle  with horizontal. It collides at a distance a with a wall parallel to y-axis as shown in figure. Let vx and vy be the components of its velocity along x and y-directions at the time of impact with wall. Coefficient of restitution between the ball and the wall is e. Component of its velocity along y-direction (common tangent) vy will remain unchanged while component of its velocity along x-direction (common normal) vx will becomes evx is opposite direction. *Further, since vy does not change due to collision, the time of flight (time taken by the ball to return to the same level) and maximum height attained by the ball remain same as it would had been in the absence of collision with the wall. Thus, vy

v B

A

u

vx

evx

C

vy

B

y

O

O

x

a

a

From O A B, R = a = u cos  . tOAB from BCO, R = a = eucos. tBCO T = tOAB + tBCO

or

2u sin a a = + g u cos  eu cos 

or

a 2u 2 sin  cos  – ag  eu cos  gu cos 

2u sin a a – = g eu cos  u cos 

or

ag 

e=

2

2u sin  cos  – ag

1 e =  u 2 sin 2  Ans.  – 1 ag   Ex.43 To test the manufactured properties of 10 N steel balls, each ball is released from rest as shown and strikes a 45° inclined surface. If the coefficient of restitution is to be e = 0.8. determine the distance s to where the ball must strike the horizontal plane at A. At what speed does the ball strike at A? (g = 9.8 m/s2) or

B 1.5m

1.0m

45° A s

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CENTRE OF MASS Sol.

v0 =

2gh =

Page # 35 2  9.8  15 . = 5.42 m/s

Component of velocity parallel and perpendicular to plane at the time of collision. v0

v1 = v2 =

= 3.83 m/sec.

2 ev2=0.8v2

C

C

v1

v2

v1

1.0 m 45°

v0 45°

D

x

A

E s x

y

Component parallel to plane (v1) remains unchanged, while component perpendicular to plane becomes ev2, where ev2 = 0.8 × 3.83 = 3.0 m/s  Component of velocity in horizontal direction after collision ( v1  ev 2 ) vx =

(3.83  3.0) =

2

2

= 4.83 m/s

While component of velocity in vertical direction after collision. v1 – ev 2 vy =

2

3.83 – 3.0 =

2

= 0.59 m/s

Let t be the time, the particle takes from point C to A, then 1.0 = 0.59 t +

1 × 9.8 × t2 ; 2

t = 0.4 sec

Solving this we get, 

DA = vxt = (4.83)(0.4) = 1.93 m

S = DA – DE = 1.93 – 1.0 S = 0.93 m vyA = vyc + gt = (0.59) + (9.8) (0.4) = 4.51 m/s vxA = vxC = 4.83 m/s

vA =

( v xA ) 2  ( v yA ) 2 = 6.6 m/s

Ex.44 A ball of mass m = 1 kg falling vertically with a velocity v0 = 2m/s strikes a wedge of mass M = 2kg kept on a smooth, horizontal surface as shown in figure. The coefficient of restitution 1 between the ball and the wedge is e = . Find the velocity of the wedge and the ball immediately 2 after collision.

m v0 M

Sol.

30°

Given M = 2kg and m = 1kg

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CENTRE OF MASS

Page # 36

v3

v1

Jcos30°

M

m

v2

m

Jsin30° J

30°

J Jsin30°

30°

Jcos30°

Let, J be the impulse between ball and wedge during collision and v1, v2 and v3 be the components of velocity of the wedge and the ball in horizontal and vertical directions respectively. Applying

impulse = change in momentum

we get

J sin 30° = Mv1 = mv2 J = 2v1 = v2 2

or

J cos 30° = m(v3 + v0)

3 J = (v3 + 2) 2

or

...(i)

...(ii)

Applying, relative speed of separation = e (relative speed of approach) in common normal direction, we get (v1 + v2) sin 30° + v3 cos30° =

1 ( v 0 cos 30 ) 2 Common normal direction

3 ...(iii) 2 Solving Eqs. (i), (ii) and (iii), we get or

v1 + v2 +

30°

1 v1 =

3v3 =

3

m/s

2

m / s and v = 0 3 3 Thus, velocities of wedge v2 =

1 and ball are v1 = 1

v1 

3

m/s

v2 

m/s

3

2

m/s

3

30°

2 and v2 =

3

m / s in horizontal direction as shown in figure.

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8.

Page # 37

COLLISION OR IMPACT Collision is an event in which an impulsive force acts between two or more bodies for a short time, which results in change of their velocities. Note :

In a collision, particles may or may not come in physical contact.

The duration of collision, t is negligible as compared to the usual time intervals of observation of motion.

In a collision the effect of external non impulsive forces such as gravity are not taken into account as due to small duration of collision (t) average impulsive force responsible for collision is much larger than external forces acting on the system.

The collision is in fact a redistribution of total momentum of the particle : Thus law of conservation of linear momentum is indepensible in dealing with the phenomenon of collision between particles. Consider a situation shown in figure. Two balls of masses m1 and m2 are moving with velocities v1 and v2 (< v1) along the same straight line in a smooth horizontal surface. Now let us see what happens during the collision between two particles. v1

v2

m1

m2

figure (a) v1 '

N

N

v2 '

N

N

figure(b)

figure(c) figure (a) : Balls of mass m1 is behind m2. Since v1 > v2, the balls will collide after some time. figure (b) : During collision both the balls are a little bit deformed. Due to deformation two equal and opposite normal forces act on both the balls. These forces decreases the velocity of m1 and increase the velocity of m2 figure (c): Now velocity of ball m1 is decrease from v1 to v1 and velocity of ball m2 is increase from v2 to v2. But still v1 > v2 so both the ball are continuously deformed. figure(d) : Contact surface of both the balls are deformed till the velocity of both the balls become equal. So at maximum deformation velocities of both the blocks are equal v1 ''

v 2 ''

figure(d)

at maximum deformation v1 ''  v 2 '' figure(e) : Normal force is still in the direction shown in figure i.e. velocity of m1 is further decreased

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Page # 38

and that of m2 increased. Now both the balls starts to regain their original shape and size. v1 ''

v 2 '' N

N

v 2 ''  v1 ''

figure(e) figure (f) : These two forces redistributes their linear momentum in such a manner that both the blocks are separated from one another, Velocity of ball m2 becomes more than the velocity of block m1 i . e . , v2 > v1 v1

m1

v2

m2

v2>v1

figure(f)

The collision is said to be elastic if both the blocks regain their original form, The collision is said to be inelastic. If the deformation is permanent, and the blocks move together with same velocity after the collision, the collision is said to be perfectly inelastic.

8.1

Classification of collisions

(a)

On the basis of line of impact (i) Head-on collision : If the velocities of the colliding particles are along the same line before and after the collision. (ii) Oblique collision : If the velocities of the colliding particles are along different lines before and after the collision.

(b)

On the basis of energy : (i) Elastic collision : (a) In an elastic collision, the colliding particles regain their shape and size completely after collision. i.e., no fraction of mechanical energy remains stored as deformation potential energy in the bodies. (b) Thus, kinetic energy of system after collision is equal to kinetic energy of system before collision. (c)

e=1

(d) Due to Fnet on the system is zero linear momentum remains conserved. (ii) Inelastic collision : (a) In an inelastic collision, the colliding particles do not regain their shape and size completely after collision. (b) Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. Thus, the kinetic energy of the particles no longer remains conserved. (c) However, in the absence of external forces, law of conservation of linear momentum still holds good. (d) (Energy loss)Perfectly Inelastic > (Energy loss)Partial Inelastic (e) 0 < e < 1

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(iii) Perfectly Inelastic collision : (i) In this the colliding bodies do not return to their original shape and size after collision i.e. both the particles stick together after collision and moving with same velocity (ii) But due to Fnet of the system is zero linear momentum remains conserved. (iii) Total energy is conserved. (iv) Initial kinetic energy > Final K.E. Energy (v) Loss in kinetic energy goes to the deformation potential energy (vi) e = 0

8.2

Value of Velocities after collision : Let us now find the velocities of two particles after collision if they collide directly and the coefficient of restitution between them is given as e. m2

m1

m2

m1

u1

u2

v1

v2

(a) Before Collision

(b) After Collision

u1 > u2

v2 > v1

v 2 – v1 e = u –u 1 2  (u1 – u2)e = (v2 – v1)

...(i)

By momentum conservation m1u1 + m2u2 = m1v1 + m2v2

...(ii)

v2 = v1 + e(u1 – u2)

...(iii)

from above equation

v1 =

m 1u1  m 2u 2  m 2 e(u 2 – u1 ) m1  m 2

...(iii)

v2 =

m 1u1  m 2 u 2  m 1e(u 1 – u 2 ) m1  m 2

...(iv)

Special cases : u

1.

If m1 >> m2 and u2 = 0 and u1 = u and

m1

e=1 m1 = m2

from eq. (iii) & (iv) v1 =

m1u – m 2u u(m1 – m 2 ) = m1  m 2 m1  m 2

v1 ~– u

v2 =

m1u  m 2u 2m1u = m1  m 2 m1  m 2

; v2 = 2u

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m2

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Page # 40 2.

u1

If m1 = m2 = m and e = 1 then

m

from eq. (iii) & (iv)

v1 =

u2 m

m(u1  u 2 )  m(u2 – u1 ) 2m

v 1 = u2 In this way v2 = u1

8.3

i.e when two particles of equal mass collide elastically and the collision is head on, they exchange their velocities. Collision in two dimension (oblique) :

1.

A pair of equal and opposite impulses act along common normal direction. Hence, linear momentum of individual particles change along common normal direction. If mass of the colliding particles remain constant during collision, then we can say that linear velocity of the individual particles change during collision in this direction.

2.

No component of impulse act along common tangent direction. Hence, linear momentum or linear velocity of individual particles (if mass is constant) remain unchanged along this direction.

3.

Net impulse on both the particles is zero during collision. Hence, net momentum of both the particles remain conserved before and after collision in any direction.

4.

Definition of coefficient of restitution can be applied along common normal direction, i.e., along common normal direction we can apply Relative speed of separation = e (relative speed of approach)

Ex.45 A ball of mass m makes an elastic collision with another identical ball at rest. Show that if the collision is oblique, the bodies go at right angles to each other after collision. Sol.

In head on elastic collision between two particles, they exchange their velocities. In this case, the component of ball 1 along common normal direction, v cos  becomes zero after collision, while v sin v sin

1 v

1 v cos

2 2 v cos After collision

Before collision

that of 2 becomes v cos . While the components along common tangent direction of both the particles remain unchanged. Thus, the components along common tangent and common normal direction of both the balls in tabular form are given a head :

Ball

Component along common tangent direction Before collision After collision

Component along common normal direction Before collision

After collision

1

v sin 

v sin 

v cos 

0

2

0

0

0

v cos 

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From the above table and figure, we see that both the balls move at right angles after collision with velocities v sin  and v cos . Note : When two identical bodies have an oblique elastic collision, with one body at rest before collision, then the two bodies will go in  directions. Ex.46 Two spheres are moving towards each other. Both have same radius but their masses are 2kg and 4kg. If the velocities are 4m/s and 2m/s respectively and coefficient of restitution is e = 1/3, find. (a) The common velocity along the line of impact. (b) Final velocities along line of impact. 2kg A R

4m/s 2m/s

R B

4kg

(c) Impulse of deformation. (d) impulse of reformation (e) Maximum potential energy of deformation (f) Loss in kinetic energy due to collision.

Sol.

In ABC sin =

BC R 1 = = AB 2R 2

or  = 30°

A 4m/s 2kg  R

C

Line of motion

R R 4kg Line of motion 2m/s B

Line of impact

(a) By conservation of momentum along line of impact. LOI 2kg

4sin30°

4m/s

30°

4sin30°

4cos30° 2cos30°

2m/s

30°

B 4kg

2sin30°

v 2cos30° Maximum Deformed State

Just Before Collision Along LOI

2(4 cos 30°) – 4(2cos30°) = (2 + 4)v or

v = 0 (common velocity along LOI)

(b) Let v1 and v2 be the final velocity of A and B respectively then, by conservation of momentum along line of impact, 2(4 cos 30°) – 4(2cos30°) = 2(v1) + 4(v2)

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Page # 42 4sin30°

A 2kg

v1

4kg

B

v2

2sin30° Just After Collision Along LOI

or

0 = v1 + 2v2 ........(1)

By coefficient of restitution, e=

velocity of separation along LOI velcoity of approach along LOI v 2 – v1 1 = 4 cos 302 cos 30 3

or

or

v2 – v1 =

3

...(2)

from the above two equations, –2 v1 =

3

m / s and v = 2

1 3

m/s

(c) J0 = m1(v – u1) = 2 (0 – 4 cos 30°) = – 4 3 N-s (d) JR = eJ0 =

1 4 N s (–4 3 ) = – 3 3

(e) Maximum potential energy of deformation is equal to loss in kinetic energy during deformation upto maximum deformed state,

U=

1 1 1 1 1 1 m1(u1 cos ) 2  m 2 (u 2 cos ) 2 – (m1  m 2 )v 2 = 2( 4 cos 30 ) 2  4(–2 cos 30 ) 2 – ( 2  4)(0) 2 2 2 2 2 2

or U = 18 Joule (f)

Loss in kinetic energy

KE =

1 1 1 1 2 2 m1(u1 cos ) 2 + m 2 (u 2 cos ) 2 –  m1v1  m 2 v 2    2 2 2 2

 1  2  2 1  1  2 1 1    4   = 2 (4 cos 30°) + 4 (–2 cos 30°) –  2 2 2  3   3 2 2 

KE = 16 Joule

9.

VARIABLE MASS In our discussion of the conservation linear momentum, we have so far dealt with systems whose system whose mass remains constant. We now consider those mass is variable, i.e., those in which mass enters or leaves the system. A typical case is that of the rocket from which hot gases keep on escaping thereby continuously decreasing its mass.  In such problem you have nothing to do but apply a thrust force (Ft ) to the main mass in addition to the all other force acting on it. This thrust force is given by,    dm  Ft  v rel    dt 

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 Here v rel is the velocity of the mass gained or mass ejected relative to the main mass. In case of

rocket this is sometimes called the exhaust velocity of the gases.

dm is the rate at which mass is dt

increasing or decreasing.

vr

v

dm

m

v + dv m–dm system

The expression for the thrust force can be derived from the conservation of linear momentum in the absence of any external forces on a system as follows :  Suppose at some moment t = t mass of a body is m and its velocity is v . After some time at t = t + dt   its mass becomes (m – dm) and velocity becomes v  dv . The mass dm is ejected with relative velocity     v r . Absolute velocity of mass ‘dm’ is therefore ( v r  v  dv) . If no external forces are acting on the system, the linear momentum of the system will remain conserved,

or

  Pi  Pf

or

        mv  mv  mdv  dmv  (dm)(dv)  dm v  v r dm  (dm)((dv)

  mdv   v r dm

or

  dv    dm  m    vr     dt   dt   

Here,

    dv  m   = thrust force (F1 )  dt   

or

      mv  (m  dm)( v  dv)  dm ( v r  v  dv)

and

dm = rate at which mass is ejecting dt

Problems related to variable mass can be solved in following three steps 1.

2.

Make a list of all the forces acting on the main mass and apply them on it.   dm   Apply an additional thrust force Ft on the mass, the magnitude of which is v r   dt  and direction is   given by the direction of v r in case the mass is increasing and otherwise the direction of  v r if it is decreasing.

3.

Find net force on the mass and apply 

 dv Fnet  m dt

(m = mass at that particular instant)

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Rocket Propulsion Let m0 be the mass of the rocket at time t = 0. m its mass at any time t and v its velocity at that moment. Initially let us suppose that the velocity of the rocket is u. u

u At

t=0 v=u m = m0

At

t=t m=m v=v

Exhaust velocity = vr   dm   be the mass of the gas ejected per unit time and vr the exhaust velocity of the Further, let  dt    dm   and vr are kept constant throughout the journey of the rocket. Now, let us gases. Usually  dt  write few equations which can be used in the problems of rocket propulsion. At time t = t

1. Thrust force on the rocket  dm  Ft  v r     dt 

(upwards)

2. Weight of the rocket W = mg

(downwards)

3. Net force on the rocket Fnet = Ft – W or

(upwards)

 – dm  Fnet  v r   – mg  dt 

4. Net acceleration of the rocket a 

or

dv v r   dm     g dt m  dt 

or

  dm  dv  v r    g dt  m 

or

 dv  v 

or

m  v – u = vr In  0   gt  m

Thus,

 m0   v = u – gt + vr In  m

v u

m r

m0

F m

t dm  g dt 0 m

...(i)

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dm  dm  Note : 1. Ft  v r   is negative.  is upwards, as vr is downwards and dt  dt   m0  . 2. If gravity is ignored and initial velocity of the rocket u = 0, Eq. (i) reduces to v = vr In  m

Ex.47 A uniform chain of mass per unit length  begins to fall with a velocity v on the table. Find the thrust force exerted by the chain on the table. Sol.

Let us assume that the mass of the chain is m and length . We assume that after time t, x length of the chain has fallen on the table. Then the speed of the upper part of the chain is

2gx as shown in figure.

m

x

2gx  v  v r

at t =0

at time t = t

Now its time t + dt, length of chain has fallen on the table is v dt. Then the mass of chain has fallen on the table is dm 

m .vdt 

Now the rate of increase of mass

x

t t + dt

vdt

dm m m  v 2gx   dt

Here v is downward and mass is increasing so thrust force act in down ward direction and is given by ft  v r

=

dm dt

2gx

at time t + dt

m ( 2gx ) 

ft =  v2

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Objective Problems

Exercise - I 1. A semicircular portion of radius 'r' is cut from a uniform rectangular plate as shown in figure. The distance of centre of mass 'C' of remaining plate, from point 'O' is r

2r

C

O

(A)

2r (3 – )

(B)

3r 2(4 – )

(C)

2r (4   )

(D)

2r 3(4 – )

2. From a circle of radius a, an isosceles right angled triangle with the hypotenuse as the diameter of the circle is removed. The distance of the centre of gravity of the remaining position from the centre of the circle is (  – 1)a (A) 3( – 1)a (B) 6 a a (C) (D) 3(  – 1) 3(   1) 3. In the figure shown a hole of radius 2 cm is made in semicircular disc of radius 6 at a distance 8 cm from the centre C of the disc. The distance of the centre of mass of this system from point C is 2cm

8cm (A) 4 cm (C) 6 cm

(B) 8 cm (D) 12 cm

4. Centre of mass of two thin uniform rods of same length but made up of different materials & kept as shown, can be, if the meeting point is the origin of co-ordinates

5. A man of mass M stands at one end of a plank of length L which lies at rest on a frictionless surface. The man walks to other end of the plank. If the mass M of the plank is , then the distance that the man 3 moves relative to ground is : 3L L (A) (B) 4 4 4L L (C) (D) 5 3 6. A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal to (A)

3R 4

(B)

3R 2

(C)

5R 4

(D) 3R

7. A man weighing 80 kg is standing at the centre of a flat boat and he is 20 m from the shore. He walks 8 m on the boat towards the shore and then halts. The boat weight 200 kg. How far is he from the shore at the end of this time? (A) 11.2 m (B) 13.8 m (C) 14.3 m (D) 15.4 m 8. Two particles having mass ratio n : 1 are interconnected by a light inextensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is : 2

2

(A) (n –1) g 2

 n – 1  g (C)   n  1

 n  1  g (B)   n – 1  n  1 g (D)  n – 1

Question No. 9 to 10 (2 questions) A uniform chain of length 2L is hanging in equilibrium position, if end B is given a slightly downward displacement the imbalance causes an acceleration. Here pulley is small and smooth & string is inextensible

y

L

x L

(A) (L/2, L/2)

(B) (2L/3, L/2)

(C) (L/3, L/3)

(D) (L/3, L/6) A

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B

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Page # 47

9. The acceleration of end B when it has been displaced by distance x, is x 2x x g (A) g (B) (C) g (D) g L L 2 10. The velocity v of the string when it slips out of the pulley (height of pulley from floor > 2L) (A)

gL 2

(B)

(C)

gL

(D) none of these

2gL

11. Internal forces can change (A) the linear momentum but not the kinetic energy of the system. (B) the kinetic energy but not the linear momentum of the system. (C) linear momentum as well as kinetic energy of the system. (D) neither the linear momentum nor the kinetic energy of the system. 12. A small sphere is moving at a constant speed in a vertical circle. Below is a list of quantities that could be used to describe some aspect of the motion of the sphere I - kinetic energy II - gravitational potential energy III - momentum Which of these quantities will change as this sphere moves around the circle ? (A) I and II only (B) I and III only (C) III only (D) II and III only 13. Which of the following graphs represents the graphical relation between momentum (p) and kinetic energy (K) for a body in motion ?

ln p

16. A system of N particles is free from any external forces (a) Which of the following is true for the magnitude of the total momentum of the system ? (A) It must be zero (B) It could be non-zero, but it must be constant (C) It could be non-zero, and it might not be constant (D) It could be zero, even if the magnitude of the total momentum is not zero. (b) Which of the following must be true for the sum of the magnitudes of the momenta of the individual particles in the system ?

ln p

(D) none

17. An isolated rail car of mass M is moving along a straight, frictionless track at an initial speed v0. The car is passing under a bridge when a crate filled with N bowling balls, each of mass m, is dropped from the bridge into the bed of the rail car. The crate splits open and the bowling balls bounce around inside the rail car, but none of them fall out. (a) Is the momentum of the rail car + bowling balls system conserved in this collision ?

ln K

ln K

ln p (C)

15. There are some passengers inside a stationary railway compartment. The track is frictionless. The centre of mass of the compartment itself (without the passengers) is C1, while the centre of mass of the compartment plus passengers system is C2. If the passengers move about inside the compartment along the track. (A) both C1 and C2 will move with respect to the ground (B) neither C1 nor C2 will move with respect to the ground (C) C1 will move but C2 will be stationary with respect to the ground (D) C2 will move but C1 will be stationary with respect to the ground

(A) It must be zero (B) It could be non-zero, but it must be constant (C) It could be non-zero, and it might not be constant (D) The answer depends on the nature of the internal forces in the system

(B)

(A)

(A) depends on the direction of the motion of the balls (B) depends on the masses of the two balls (C) depends on the speeds of the two balls (D) is equal to g

ln K 14. Two balls are thrown in air. The acceleration of the centre of mass of the two balls while in air (neglect air resistance)

(A) Yes, the momentum is completely conserved (B) Only the momentum component in the vertical direction is conserved (C) Only the momentum component parallel to the track is conserved (D) No components are conserved

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(b) What is the average speed of the rail car + bowling balls system some time after the collision ? (A) (M + Nm)v0/M (B) Mv0/(Nm + M) (C) Nmv0/M (D) The speed cannot be determined because there is not enough information Question No. 18 to 21 A small ball B of mass m is suspended with light inelastic string of length L from a block A of same mass m which can move on smooth horizontal surface as shown in the figure. The ball is displaced by angle  from equilibrium position & then released. A L

L u=0

B

18. The displacement of block when ball reaches the equilibrium position is L sin (B) L sin  (A) 2 (C) L (D) none of these 19. Tension in string when it is vertical, is (A) mg (B) mg(2 – cos ) (C) mg (3 – 2 cos) (D) none of these 20. Maximum velocity of block during subsequent motion of the system after release of ball is (A) [gl(1 – cos )]1/2 (B) [2gl(1 – cos )]1/2 (C) [glcos]1/2 (D) informations are insufficient to decide 21. The displacement of centre of mass of A + B system till the string becomes vertical is L (A) zero (B) (1 – cos ) 2 L (C) (1 – sin ) (D) none of these 2 Question No. 22 to 26 (5 questions) Two persons of mass m1 and m2 are standing at the two ends A and B respectively, of a trolley of mass M as shown. m2

m1

22. When the person standing at A jumps from the trolley towards left with urel with respect to the trolley, then (A) the trolley moves towards right m1urel (B) the trolley rebounds with velocity m  m  M 1 2 (C) the centre of mass of the system remains stationary (D) all the above 23. When only the person standing at B jumps from the trolley towards right while the person at A keeps standing, then (A) the trolley moves towards left m 2urel (B) the trolley mones with velocity m  m  M 1 2 (C) the centre of mass of the system remains stationary (D) all the above 24. When both the persons jump simultaneously with same speed then (A) the centre of fmass of the systyem remains stationary (B) the trolley remains stationary (C) the trolley moves toward the end where the person with heavier mass is standing (D) None of these 25. When both the persons jump simultaneously with urel with respect to the trolley, then the velocity of the trolley is | m1  m 2 |urel | m1  m 2 |urel (B) (A) m  m  M M 1 2 m1urel m 2urel (C) m  M  m  M 2 1

(D) none of these

26. Choose the incorrect statement, if m1 = m2 = m and both the persons jump one by one, then (A) the centre of mass of the system remains stationary (B) the final velocity of the trolley is in the direction of the person who jumps first murel   murel   (C) the final velocity of the trolley is  M  m M  2m  (D) none of these 27. In the diagram shown, no friction at any contact surface. Initially, the spring has no deformation. What will be the maximum deformation in the spring ? Consider all the strings to be sufficiency large. Consider the spring constant to be K

M A

B

L

F

2M

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M

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Page # 49

(A) 4F / 3K (C) F / 3K

(B) 8F / 3K (D) none

28. A super-ball is to bounce elastically back and forth between two rigid walls at a distance d from each other. Neglecting gravity and assuming the velocity of super-ball to be v0 horizontally, the average force being exerted by the super-ball on each wall is :

1 mv 20 (A) 2 d

mv 20 (B) d

2mv 20 (C) d

4mv 20 (D) d

is masselss and frictionless. Initially the system is at rest when, a bullet of mass 'm' moving with a velocity 'u' as shown hits the block 'B' and gets embedded into it. The impulse imparted by tension force to the block of mass 3m is :

m u m B

A 3m 5mu 4 2mu (C) 5

4mu 5 3mu (D) 5

(A) 29. In the figure (i), (ii) & (iii) shown the objects A, B & C are of same mass. String, spring & pulley are massless. C strikes B with velocity ‘u’ in each case and sticks to it. The ratio of velocity of B in case (i) to (ii) to (iii) is

(i)

(iii)

(ii) C B

A

A

B

C A

B

(B)

34. A ball strikes a smooth horizontal ground at an angle of 45° with the vertical. What cannot be the possible angle of its velocity with the vertical after the collision. (Assume e  1). (A) 45° (B) 30° (C) 53° (D) 60°

C

(A) 1 : 1 : 1 (C) 3 : 2 : 2

(B) 3 : 3 : 2 (D) none of these

30. A force exerts an impulse I on a particle changing its speed from u to 2u. The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is 3 1 (A) Iu (B) Iu 2 2 (C) Iu (D) 2 Iu

35. As shown in the figure a body of mass m moving vertically with speed 3 m/s hits a smooth fixed inclined plane and rebounds with a velocity vf in the horizontal direction. If  of inclined is 30°, the velocity vf will be vf

(A) 3 m/s 31. A parallel beam of particles of mass m moving with velocity v impinges on a wall at an angle  to its normal. The number of particles per unit volume in the beam is n. If the collision of particles with the wall is elastic, then the pressure exerted by this beam on the wall is (A) 2 mn v2 cos  (B) 2 mn v2 cos2  (C) 2 mn v cos  (D) 2 mn v cos2  32. A boy hits a baseball with a bat and imparts an impulse J to the ball. The boy hits the ball again with the same force, except that the ball and the bat are in contact for twice the amount of time as in the first hit. The new impulse equals. (A) half the original impulse (B) the original impulse (C) twice the original impulse (D) four times the original impulse

(B)

m

3 m/s

(C) 1 / 3 m/s (D) this is not possible 36. Two massless string of length 5 m hang from the ceiling very near to each other as shown in the figure. Two balls A and B of masses 0.25 kg and 0.5 kg are attached to the string. The ball A is released from rest at a height 0.45 m as shown in the figure. The collision between two balls is completely elastic Immediately after the collision. the kinetic energy of ball B is 1 J. The velocity of ball A just after the collision is ,

A

33. A system of two blocks A and B are connected by an inextensible massless strings as shown. The pulley

0.45m

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B

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(A) 5 ms–1 to the right (C) 1 ms–1 to the right

(B) 5 ms–1 to the left (D) 1 ms–1 to the left

37. Two balls A and B having masses 1 kg and 2 kg, moving with speeds 21 m/s and 4 m/s respectively in opposite direction, collide head on. After collision A moves with a speed of 1 m/s in the same direction, then the coefficient of restitution is (A) 0.1 (B) 0.2 (C) 0.4 (D) None 38. A truck moving on horizontal road east with velocity 20ms–1 collides elastically with a light ball moving with velocity 25 ms–1 along west. The velocity of the ball just after collision (A) 65 ms–1 towards east (B) 25 ms–1 towards west (C) 65 ms–1 towards west (D) 20 ms–1 towards east 39. Two perfectly elastic balls of same mass m are moving with velocities u1 and u2. They collide elastically n times. The kinetic energy of the system finally is : (A)

1m 2 u1 2 u

(B)

1m 2 (u1  u 22 ) 2 u

(C)

1 m(u12  u22 ) 2

(D)

1 mn(u12  u 22 ) 2

40. In the figure shown, the two identical balls of mass M and radius R each, are placed in contact with each other on the frictionless horizontal surface. The third ball of mass M and radius R/2, is coming down vertically and has a velocity = v0 when it simultaneously hits the two balls and itself comes to rest. The each of the two bigger balls will move after collision with a speed equal to

(A) 4 v 0 / 5

(B) 2 v 0 / 5

(C) v 0 / 5

(D) none

41. In the above, suppose that the smaller ball does not stop after collision, but continues to move downwards with a speed = v0/2, after the collision. Then, the speed of each bigger ball after collision is (A) 4 v 0 / 5

(B) 2 v 0 / 5

(C) v 0 / 2 5

(D) none

42. A body of mass ‘m’ is dropped from a height of ‘h’. Simultaneously another body of mass 2m is thrown up vertically with such a velocity v that they collide at the height h/2. If the collision is perfectly inelastic, the velocity at the time of collision with the ground will be (A)

5gh 4

(B)

(C)

gh 4

(D)

gh 10 gh 3

43. A sphere of mass m moving with a constant velocity hits another stationary sphere of the same mass, if e is the coefficient of restitution, then ratio of speed of the first sphere to the speed of the second sphere after collision will be  1– e  (A)  1 e

 1 e  (B)  1– e

 e  1  (C)  e – 1

 e – 1  (D)  e  1

44. In a smooth stationary cart of length d, a small block is projected along it’s length with velocity v towards front. Coefficient of restitution for each collision is e. The cart rests on a smooth ground and can move freely. The time taken by block to come to rest w.r.t. cart is d

v

(A)

ed (1 e) v

(B)

ed (1 e) v

(C)

d e

(D) inifinite

45. Three blocks are initially placed as shown in the figure. Block A has mass m and initial velocity v to the right. Block B with mass m and block C with mass 4 m are both initially at rest. Neglect friction. All collisions are elastic. The final velocity of block A is V B A C m m 4m (A) 0.60 v to the left (C) v to the left

(B) 1.4 v to the left (D) 0.4 v to the left

46. A block of mass m starts from rest and slides down a frictionless semi-circular track from a height h as shown. When it reaches the lowest point of the

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track, it collides with a stationary piece of putty also having mass m. If the block and the putty stick together and continue to slide, the maximum height that the block-putty system could reach is

h (A) h/4 (C) h

(B) h/2 (D) independent of h

47. Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball 2. Initially, ball 1 moves with a speed v towards ball 2 which is at rest. Immediately after the collision, ball 1 travels at a speed of v/3 in the same direction. What type of collision has occured ? (A) inelastic (B) elastic (C) completely inelastic (D) Cannot be determined from the information given 48. The diagram shows the velocity - time graph for two masses R and S that collided elastically. Which of the following statements is true ?

V(ms–1) 1.2 0.8 0.4

R

S

1 2 3 4 t(s) I. R and S moved in the same direction after the collision. II. The velocities of R and S were equal at the mid time of the collision. III. The mass of R was greater than mass of S. (A) I only (B) II only (C) I and II only (D) I, II and III 49. A ball is dropped from a height h. As is bounces off the floor, its speed is 80 percent of what it was just before it hit the floor. The ball will then rise to a height of most nearly (A) 0.80 h (B) 0.75 h (C) 0.64 h (D) 0.50 h 50. A ball is thrown vertically downwards with velocity 2gh from a height h. After colliding with the ground it just reaches the starting point. Coefficient of restitution is (A) 1 / 2

(B) 1/2

(C) 1

(D)

2

51. A ball is dropped from height 5m. The time after which ball stops rebounding if coefficient of restitution between ball and ground e = 1/2, is

(A) 1 sec (C) 3 sec

(B) 2 sec (D) infinite

52. A ball is projected from ground with a velocity V at an angle  to the vertical. On its path it makes an elastic collision with a vertical wall and returns to ground. The total time of flight of the ball is (A)

2v sin g

(B)

2v cos  g

(C)

v sin2 g

(D)

v cos  g

53. The Gardener water the plants by a pipe of diameter 1 mm. The water comes out at the rate of 10 cm3/ sec. The reactionary force exerted on the hand of the Gardener is : (density of water is 103 kg/m3) (A) zero (B) 1.27 × 10–2 N –4 (C) 1.27 × 10 N (D) 0.127 N 54. An open water tight railway wagon of mass 5 × 103 kg coasts at an initial velocity 1.2 m/s without friction on a railway track. Rain drops fall vertically downwards into the wagon. The velocity of the wagon after it has collected 103 kg of water will be (A) 0.5 m/s (B) 2 m/s (C) 1 m/s (D) 1.5 m/s 55. If the force on a rocket which is ejecting gases with a relative velocity of 300 m/s, is 210 N. Then the rate of combustion of the fuel will be (A) 10.7 kg/sec (B) 0.07 kg/sec (C) 1.4 kg/sec (D) 0.7 kg/sec 56. A rocket of mass 4000 kg is set for vertical firing. How much gas must be ejected per second so that the rocket may have initial upwards acceleration of magnitude 19.6 m/s2. [Exhaust speed of fuel = 980 m/ s] (A) 240 kg s–1 (B) 60 kg s–1 –1 (C) 120 kg s (D) none 57. A wagon filled with sand has a hole so that sand leaks through the bottom at a constant rate . An  external force F acts on the wagon in the direction of motion. Assuming instantaneous velocity of the wagon to be v and initial mass of system to be m0, the force equation governing the motion of the wagon is :       dv dv  v – v (B) F  m 0 (A) F  m 0 dt dt      dv dv F  ( m –  t ) F  ( m –  t )  v (C) (D) 0 0 dt dt

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Multiple Choice Problem

Exercise - II

1. A body has its centre of mass at the origin. The xcoordinates of the particles (A) may be all positive (B) may be all negative (C) may be all non-negative (D) may be positive for some cases and negative in other cases 2. An object comprises of a uniform ring of radius R and its uniform chord AB (not necessarily made of the same material) as shown. Which of the following can not be the centre of mass of the object. y

(A) must not move (C) may move

(B) must not accelerate (D) may accelerate

Question No. 6 to 12 (7 Questions) A particle of mass m moving horizontal with v0 strikes a smooth wedge of mass M, as shown in figure. After collision, the ball starts moving up the inclined face of the wedge and rises to a height h.

h

B

m v0 A

x

6. The final velocity of the wedge v2 is (A)

(A) (R/3, R/3)

(B) (R3, R/2)

(C) (R/4, R/4)

(D) (R / 2, R / 2 )

3. In which of the following cases the centre of mass of a an rod is certainly not at its centre ? (A) the density continuously increases from left to right (B) the density continuously decreases from left to right (C) the density decreases from left to right upto the centre and then increase (D) the density increases from left to right upto the centre and then decreases 4. Consider following statements [1] CM of a uniform semicircular disc of radius R = 2R/ from the centre [2] CM of a uniform semicircular ring of radius R = 4R/3 from the centre [3] CM of a solid hemisphere of radius R = 4R/3 from the centre [4] CM of a hemisphere shell of radius R = R/2 from the centre Which statements are correct? (A) 1, 2, 4 (B) 1, 3, 4 (C) 4 only (D) 1, 2 only 5. If the external forces acting on a system have zero resultant, the centre of mass

M

mv 0 M

(C) v0

(B)

mv 0 Mm

(D) insufficient data

7. When the particle has risen to a height h on the wedge, then choose the correct alternative(s) (A) The particle is stationary with respect to ground (B) Both are stationary with respect to the centre of mass (C) The kinetic energy of the centre of mass remains constant (D) The kinetic energy with respect to centre of mass is converted into potential energy 8. The maximum height h attained by the particle is 2  m  v0  (A)  m  M 2g 2  M  v0  (C)  m  M 2g

2  m v0 (B)   M 2g

(D) none of these.

9. Identify the correct statement(s) related to the situation when the particle starts moving downward. (A) The centre of mass of the system remains stationary (B) Both the particle and the wedge remain stationary with respect to centre of mass (C) When the particle reaches the horizontal surface its velocity relative to the wedge is v0 (D) None of these

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10. Suppose the particle when reaches the horizontal surfaces, its velocity with respect to ground is v1 and that of wedge is v2. Choose the correct statement (s)

v2

v1

m

(A) mv1 = Mv2 (C) v1 + v2 = v0

M (B) Mv2 – mv1 = mv0 (D) v1 + v2 < v0

11. Choose the correct statement(s) related to particle m  mM  (A) Its kinetic energy is K f    gh  m  M  M  m  (B) v1  v 0  M  m

(C) The ratio of its final kinetic energy to its initial kinetic energy is

Kf  M    K i  m  M

2

(D) It moves opposite to its initial direction of motion 12. Choose the correct statement related to the wedge M

 4 m2   gh (A) Its kinetic energy is K f    m  M  2m   v0 (B) v 2   m  M  4 mM   1  mv 20   (C) Its gain in kinetic energy is K   2    (m  M)   2 (D) Its velocity is more that the velocity of centre of mass 13. Two blocks A (5kg) and B(2kg) attached to the ends of a spring constant 1120 N/m are placed on a smooth horizontal plane with the spring undeformed. Simultaneously velocities of 3m/s and 10m/s along the line of the spring in the same direction are imparted to A and B then

A

3m/s

10m/s

5

2

B

(A) when the extension of the spring is maximum the velocities of A and B are zero. (B) the maximum extension of the spring is 25 cm

(C) maximum extension and maximum compression occur alternately. (D) the maximum compression occur for the first time  after sec. 56 14. Two identical balls are interconnected with a massless and inextensible thread. The system is in gravity free space with the thread just taut. Each ball is imparted a velocity v, one towards the other ball and the other perpendicular to the first, at t = 0. Then, (A) the thread will become taut at t = (L/v) (B) the thread will become taut at some time t < (L/ v). (C) the thread will always remain taut for t > (L/v) (D) the kinetic energy of the system will always remain mv2. 15. A ball moving with a velocity v hits a massive wall moving towards the ball with a velocity u. An elastic impact lasts for a time t. (A) The average elastic force acting on the ball is m(u  v) t (B) The average elastic force acting on the ball is 2m(u  v) t (C) The kinetic energy of the ball increases by 2mu (u + v) (D) The kinetic energy of the ball remains the same after the collision. 16. A particle moving with kinetic energy = 3 joule makes an elastic head on collision with a stationary particle when has twice its mass during the impact. (A) The minimum kinetic energy of the system is 1 joule (B) The maximum elastic potential energy of the system is 2 joule. (C) Momentum and total kinetic energy of the system are conserved at every instant. (D) The ratio of kinetic energy to potential energy of the system first decreases and then increases. 17. Two balls A and B having masses 1 kg and 2 kg, moving with speeds 21 m/s and 4 m/s respectively in opposite direction, collide head on. After collision A moves with a speed of 1 m/s in the same direction, then correct statements is :

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(A) The velocity of B after collision is 6 m/s opposite to its direction of motion before collision. (B) The coefficient of restitution is 0.2. (C) The loss of kinetic energy due to collision is 200 J. (D) The impulse of the force between the two balls is 40 Ns. 18. The diagram to the right shown the velocity-time graph for two masses R and S that collided elastically. Which of the following statements is true? –1 v(ms )

1.2 0.8

S R

0.4 1

2 3 4 t (milli sec)

(I) R and S moved in the same direction after the collision. (II) Kinetic energy of the system (R & S) is minimum at t = 2 milli sec. (III) The mass of R was greater than mass of S. (A) I only (B) II only (C) I and II only (D) I, II and III 19. In an inelastic collision, (A) the velocity of both the particles may be same after collision. (B) kinetic energy is not conserved (C) linear momentum of the system is conserved. (D) velocity of separation will be less than velocity of approach.

L 2V L (B) The last block starts moving at t = (n – 1) V (C) The centre of mass of the system will have a final speed v/n (D) The centre of mass of the system will have a final speed v (A) The last block starts moving at t = n(n–1)

22. An isolated rail car original moving with speed v0 on a straight, frictionless, level track contains a large amount of sand. A release value on the bottom of the car malfunctions, and sand begins to pour out straight down relative to the rail car. (a) Is momentum conserved in this process? (A) The momentum of the rail car along is conserved (B) The momentum of the rail car + sand remaining within the car is conserved (C) The momentum of the rail car + all of the sand, both inside and outside the rail car, is conserved (D) None of the three previous systems have momentum conservation (b) What happens to the speed of the rail car as the sand pours out? (A) The car begins to roll faster (B) The car maintains the same speed (C) The car begins to slow down (D) The problem cannot be solved since momentum is not conserved

20. In a one-dimensional collision between two par ticles, their relative velocity is v1 before the collision  and v 2 after the collision   (A) v1  v 2 if the collision is elastic   (B) v1  – v 2 if the collision is elastic   (C) | v 2 | | v1| in all cases   (D) v1  – kv 2 in all cases, where k  1 21. A set of n-identical cubical blocks lie at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surfaces of any two adjacent blocks is L. The block at one end is given a speed V towards the next one at time t = 0. All collision are completely inelastic, then

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Subjective Problem

Exercise - III 1. The mass of an uniform ladder of length l increases uniformly from one end A to the other end B, (a) Form an expression for linear mass density as a function of distance x from end A where linear mass density 0. The density at one end being twice that of the other end. (b) find the position of the centre of mass from end A. 2. Find the distance of centre of mass of a uniform plate having semicircular inner and other boundaries of radii a and b from the centre O.

6. In the arrangement, mA = 2 kg and mB = 1 kg. String is light and inextensible. Find the acceleration of centre of mass of both the blocks. Neglect friction everywhere. 7. A small cube of mass m slides down a circular path of radius R cut into a large block of mass M. M rests on a table and both blocks move without friction. The blocks initially are at rest and m starts from the top of the path. Find the velocity v of the cube as it leaves the block. m R

M

a b O 3. A thin sheet of metal of uniform thickness is cut into the shape bounded by the line x = a and y = ± k x2, as shown. Find the coordinates y of the centre of mass. y=kx2 x

a y=–kx2

4. Two balls of equal masses are projected upwards simultaneously, one from the ground with speed 50 m/s and other from a 40m high tower with initial speed 30 m/s. Find the maximum height attained by their centre of mass. 5. In the figure shown, when the persons A and B exchange their positions, then

B

A m1

M

m2

2m m1=50kg, m2 = 70kg, M = 80 kg

8. A (trolley + child) of total mass 200 kg is moving with a uniform speed of 36 km/h on a frictionless track. The child of mass 20 kg starts running on the trolley from one end to the other (10 m away) with a speed of 10 ms–1 relative to the trolley in the direction of the trolley's motion and jumps out of the trolley with the same relative velocity. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run? 9. In the figure shown the spring is compressed by ‘x0’ and released. Two blocks ‘A’ and ‘B’ of masses ‘m’ and ‘2m’ respectively are attached at the ends of the spring. Blocks are kept on a smooth horizontal surface and released. k A

B

(a) Find the speed of block A by the time compression of the spring is reduced to x0/2. (b) Find the work done by the spring on ‘A’ by the time compression of the spring reduced to x0/2. 10. The figure showns the force versus time graph for a particle. (i) Find the change in momentum p of the particle (ii) Find the average force acting on the particle

100 N (i) the distance moved by the centre of mass of the system is__________. (ii) the plank moves towards_________ (iii) the distance moved by the plank is _________. (iv) the distance moved by A with respect to ground is ____________ (v) the distance moved by B with respect to ground is ____________.

0

0.2

0.4

t(a)

11. A force F acts on an object (mass = 1kg) which is initially at rest as shown in the figure. Draw the graph showing the momentum of the object varying during the time for which the force acts.

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F(N) 5

50

100 t(ms)

12. A man hosing down his driveway hits the wall by mistake. Knowing that the velocity of the stream is 25 m/s and the crosssectional area of the stream is 300 mm2, determine the force exerted on the wall. Assume that streamstrikes wall horizontally and after striking the wall, stream comes to rest. Also find the pressure exerted on the wall by stream.

13. A bullet of mass m strikes an obstruction and deviates off at 60° to its original direction. If its speed is also changed from u to v, find the magnitude of the impulse acting on the bullet. 14. A neutro intially at rest, decays into a proton, an electron and an antineutrio. The ejected electron has a momentum of p1 = 1.4 × 10–28 kg-m/s and the antineutrino p2 = 6.5 × 10–27 kg-m/ s. Find the recoil speed of the proton (a) if the electron and the antineutrino are ejected along the the same direection and (b) if they are ejected along perpendicular direction. Mass of the proton mp = 1.67 × 10–27 kg. 15. A steel ball of mass 0.5 kg is dropped from a height of 4 m on to a horizontal heavy steel slab. The collision is elastic and the ball rebounds to its original height. (a) Calculate the impulse delivered to the ball during impact. (b) If the ball is in contact with the slab for 0.002 s, find the average reaction force on the ball during impact. 16. A particle A of mass 2 kg lies on the edge of a table of height 1m. It is connected by a light inelastic string of length 0.7 m to a second particle B of mass 3 kg which is lying on the table 0.25 m from the edge (line joining A & B is perpendicular to the edge). If A is pushed gently so that it start falling from table then, find the speed of B when it starts to move. Also find the imulsive tension in the string at that moment. 17. Two particles, each of mass m, are connected by a light inextensible string of length 2l. Initially they lie on a smooth horizontal table at points A and B distant l apart. The particle at A is projected across the table with velocity u. Find the speed

Page # 56 with which the second particle begins to move if the direction of u is, (a) along BA, (b) at an angle of 120° with AB, (c) perpendicular to AB. In each case calculate (in terms of m and u) the impulsive tension in the string. 18. Two particle P of mass 2m and Q of mass m are subjected to mutual force of attraction and no other act on them. At time t = 0, P is at rest at a fixed O and Q is directly moving away from O with a speed 5 u. At a later instant when t = T before any collision has taken place Q is moving towards O with speed u. (a) Find in terms of m and u the total work done by the forces of attraction during the time interval 0  t  T. (b) At the instantt = T, impulses of magnitude J and K and applied to P and Q bringing them to rest. Find the values of J and K 19. A block of mass m moving with a velocity v, enters a region where it starts colliding with the stationary dust particles. If the desnsity of dust particles is , & all colliding particle stick to its front surface of cross-sectional area A. The velocity of block after it has covered a distance x in this region is __________________. 20. A football approaches a player at v = 12 m/s. At what speed u and in which direction should the player’s foot move in order to stop the ball upon contact? Assume that the mass of the foot is much greater than that of the ball and that the collision is elastic. 21. Three carts move on a frictioless track with inertias and velocities as shown. The carts collide and stick together after successive collisions.

m1 = 2 kg v1 = 1 m/s

m2 = 1 kg v2 = 1 m/s

A

B

m3 = 2 kg v3 = 2 m/s C

(a) Find loss of mechanical energy when B & C stick together. (b) Find magnitude of impulse experienced by A when it sticks to combined mass (B & C). 22. A sphere of mass m1 in motion hits directly another sphere of mass m2 at rest and sticks to it, the total kinetic energy after collision is 2/3 of their total K.E. before collision. Find the ratio of m1 : m2. 23. A body is thrown vertically upwards from ground with a speed of 10 m/s. If coefficient of restitution of ground, e = 1/2. Find (a) the total distance travelled by the time it almost stops. (b) time elapsed (after the ball has been thrown) when it is at its subsequent maximum height for the third time.

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24. A ball of mass 'm' is suspended by a massless string of length '' from a fixed point. A ball of mass 2m strikes in the direction of  = 45° from horizontal and sticks to it. (a) What should be the initial velocity of 2m so that system  deflects by  = 2 (b) If at  = 60° the stirng is cut then what will be the velocity at highest point of trajectory.

2m 45°

m

A

2kg

90° 1.5 m 4 kg B

(b) the maximum displacement of B after the impact. 28. A small block of mass 2m initially rests at the bottom of a fixed circular, vertical track, which has a radius of R. The contact surface between the mass and the loop is frictionless. A bullet of mass m strikes the block horizontally with initial speed v0 and remain embedded in the block as the block and the bullet circle the loop. Determine each of the following in terms m, v0, R and g.

25. A ball is thrown horizontally from a cliff 10 m high with a velocity of 10 m/s. It strikes the smooth ground and rebounds as shown. The coefficient of restitution e for collision with the ground e = 1/ 2 . Find

R

10 m/s

m, v0 (a) The speed of the masses immediately after the impact.

(a) velocity of ball just before striking ground. (b) angle of velocity vector with horizontal before striking. (c) angle of velocity vector with horizotal after striking. (d) range of ball after first collision. 26. A wedge free to move of mass 'M' has one face making an angle  with horizotnal and is resting on a smooth rigid floor. A particle of mass 'm' hits the inclined face of the wedge with a horizontal velocity v0. It is observed that the particle rebounds in vertical direction after impact. Neglect friction between particle and the wedge & take M = 2m, v0 = 10 m/s, tan  = 2, g = 10 m/s2

(b) The minimum initial speed of the bullet if the block and the bullet are to successfully execute a complete ride on the loop. 29. A Cart of total mass M0 is at rest on a rough horizontal road. It ejects bullets at rate of  kg/s at an angle  with the horizontal and at velocity 'u' (constant) relative to the cart. The coefficient of friction between the cart and the ground is . Find the velocity of the cart in terms of time 't'. The cart moves with sliding.

M

(a) Determine the coefficient of restitution for the impact. (b) Assume that the inclined face of the wedge is sufficiently long so that the particle hits the same face once more during its downward motion. Calculate the time elapsed between the two impacts. 27. A sphere A is released from rest in the position shown and strikes the block B which is at rest. If e = 0.75 between A and B and k = 0.5 between B and the support, determine (a) the velocity of A just after the impact

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CENTRE OF MASS

Page # 58

Tough Subjective Problem

Exercise - IV

1. In a game of Carom Board, the Queen (a wooden disc of radius 2 cm and mass 50 gm) is placed at the exact center of the horizontal board. The striker is a smooth plastic disc of radius 3 cm and mass 100 gm. The board is frictionless. Th striker is given an initial velocity ‘u’ parallel to the sides BC or AD so that is hits the Queen inelastically with same coefficient of restitution = 2/3. The impact parameter for the collision is ‘d’ (shown in the figure). The Queen rebounds from the edge AB of the board inelastically with same coefficient of restitution = 2/3. and enters the hole D following the dotted path shown. The side of the board is L. Find the value of impact parameter ‘d’ and the time which the Queen takes to enter hole D after collision with the striker. A

L

L

B

5. A massive vertical wall is approaching a man at a speed u. When it is at a distance of 10m, the man throws a ball with speed 10 m/s an at angle of 37° which after completely elastic rebound reaches back directly into his bands. Find the velocity u of the wall. 6. Mass m1 hits & sticks with m2 while sliding horizontally with velocity v along the common line of centres of the three equal masses (m1 = m2 = m3 =m). Initially masses m2 and m3 are stationary and the spring is unstretched. Find

m1

k

v

m2

m3

Frictionless (a) the velocities of m1, m2 and m3 immediately after impact.

(b) the maximum kinetic energy of m3.

u d

(c) the minimum kinetic energy of m2. (d) the maximum compression of the spring.

D

C

2. A flexible chain has a length l and mass m. It is lowered on the table top with constant velocity v. Find the force that the chain exerts on the table as a function of time.

7. Two masses A and B connected with an inextensible string of length l lie on a smooth horizontal plane. A is giv en a velo city of v m/s alon g the gro und perpendicular to line AB as shown in figure. Find the tension in string during their sub sequent motion. B m

3. A 24-kg projectile is fired at an angle of 53° above the horizontal with an initial speed of 50 m/s. At the highest point in its trajectory, the projectile explodes into two fragments of equal mass, the first of which falls vertically with zero initial speed. (a) How far from the point of firing does the second fragment strike the ground? (Assume the ground is level.) (b) How much energy was released during the explosion?

l

A 2m

v

8. The simple pendulum A of mass mA and length l is suspended from the trolley B of mass mB. If the system is released from rest at  = 0, determine the velocity vB of the trolley and tension in the string when  = 90°. Friction is negligible.

4. A particle is projected from point O on level ground towards a smooth vertical wall 50m from O and hits the wall. The initial velocity of the particle is 30m/s at 45° to the horizontal and the coefficient of restitution between the particle and the wall is e. Find the distance from O of the point at which the particle hits the ground again if (a) e = 0, (b) e = 1, (c) e = 1/2

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B l

A

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Page # 59

9. A ball with initial speed 10m/s collides elastically with two other identical ball whose centres are on a line perpendicular to the initial velocity and which are initially in contact with each other. All the three ball are lying on a smooth horizontal table. The first ball is aimed directly at the contact point of the other two balls All the balls are smooth. Find the velocities of the three balls after the collision.

11. A cart is moving along +x direction with a velocity of 4m/s. A person in the cart throws a stone with a velocity of 6m/s relative to himself. In the frame of reference of the cart the stone is thrown in y-z plane making an angle of 30° with the vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from branch of a tree by means of a string of length L. A completely inelastic collosion occurs, in which the stone gets embedded in the object. Determine (a) the speed of the combined mass immediately after collision with respect to an observer on the ground.

10 m/s . 10. A mass m1 with initial speed v0 in the positive xdirection collides with a mass m2 = 2m1 which is initially at rest at the origin, as shown in figure. After the collision m1 moves off with speed v1 = v0/2 in the negative y-direction, and m2 moves off with speed v2 at angle . (A) Find the velocity (magnitude and direction) of the center of mass before the collision, as well as its velocity after the collision. (B) Write down the x and y-components of the equation of conservation of momentum for the collision. (C) Determine tan, and find v2 in terms of v0.

(b) the length L of the string such that tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass.

12. Twp equal sphere of mass ‘m’ are suspended by vertical strings so that they are in contact with their centres at same level. A third equal spheres of mass m falls vertically and strikes the other two simultaneously so that their centres at the instant of impact form an equilateral triangle in a vertical plane. If u is the velocity of m just before impact, find the velocities just after impact and the impulsive tension of the strings.

(D) Determine how much (if any) energy was gained or lost in the collision, and state whether the collision was elastic or inelastic. y

u

v2

y

m

m2

v 

x v0 m2 = 2m1

v

m

m1 After

Before

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m

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Page # 60

JEE-Problems

Exercise - V 1. Two trolleys A and B of equal masses M are moving   in opposite directions with velocities v and  v respectively on separate horizontal frictionless parallel tracks. When they start crossing each other, a ball of mass m is thrown from B to A and another of same is  thrown from A to B with velocities normal to v . The balls may be thrown in following two ways (i) balls from A to B to A are thrown simultaneously. (ii) ball is thrown from A to B after the ball thrown from B reaches A. Which procedure would lead to a larger change in the velocities of the trolleys ? [REE-2000] 2. A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to [IIT(Scr.)-2000] (A) v (B) v2 3 (C) v (D) V4 3. Two particles of masses m1 and m2 in projectile   motion have velocities v 1 and v 2 respectively at time t = 0. They collide at time t0. Their velocities become   v 1 and v 2 at time 2t0 while still moving in air. The     value of [(m1v 1  m 2 v 2 )  (m1u1  m 2u 2 )] is [IIT(Scr.)-2001)] (A) zero (B) (m1 + m2)gt0 (C) 2(m1 + m2)gt0 (D) ½(m1 + m2)gt0 4. A car P is moving with a uniform speed of 5(31/2)m/ s towards a carriage of mass, 9 Kg at rest kept on the rails at a point B as shown in fig. The height AC is 120 m. Cannon balls of 1 Kg are fired from the car with an initial velocity 100 m/s at an angle 30º with the horizontal. The first canon ball hits the stationary carriage after a time t0 ans stricks to it. Determine t0. At t0, the second cannon ball is fired. Assume that the resistive force between the rails and the carriage is constant and ignore the vertical motion of the carriage throughout. If the second ball also hits and sticks to the carriage. What will be the horizontal velocity of the carriage just after the second impact ? [IIT-2001] C P

5. Two block of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is [IIT(Scr.)-2002] (A) 30 m/s (B) 20 m/s (C) 10 m/s (D) 5 m/s 6. STATEMENT-1 In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. because STATEMENT-2 In an elastic collision, the linear momentum of the system is conserved (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True  7. The balls, having linear momenta p1  p i and  p 2  –p i , undergo a collision in free space. There is   no external force acting on the balls. Let p1 and p 2 be their final momenta. The following options(s) is (are) NOT ALLOWED for any non-zero value of p, a1, a2, b1, b2, c1 and c2. [JEE 2008]  (A) p'1  a1ˆi  b1ˆj  c 1kˆ    (B) p'1  c1kˆ ; p' 2  a 2 ˆi  b 2 ˆj ; p'2  c 2kˆ  (C) p'1  a1ˆi  b1ˆj  c 1kˆ    (D) p'1  a1ˆi  b1ˆj ; p' 2  a 2 ˆi  b 2 ˆj – c1k ; p' 2  a 2 ˆi  b1ˆj

Paragraph for Question No. 8 to 10 A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from 60° to 30° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g = 10 m/s2). Figure : [JEE 2008] M A v 60° B

30° A

B

3m

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3 3m

C

CENTRE OF MASS

Page # 61

8. The speed of the block at point B immediately after it strikes the second incline is (A)

60 m / s

(B)

45 m / s

(C)

30 m / s

(D)

15 m / s

9. The speed of the block at point C, immediately before it leaves the second incline is (A)

120 m / s

(B)

105 m / s

(C)

90 m / s

(D)

75 m / s

10. If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is (A)

30 m / s

(B)

15 m / s

(D) – 15 m / s

(C) 0

11. Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink unsed to draw the outer circle is 6m. The coordinates of the centres of the different parts are outer circle (0, 0), left inner circle (–a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, – a). The y-coordinate of the centre of mass of the ink in this drawing is [JEE 2009] m (-a, a)

m (a, a) (0, 0) 7m

13. Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These have masses m, 2m and m, respectively. The object A moves towards B with a speed 9 ms–1 and makess an elastic collision with it. There after, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in ms–1 ) of the object C. [JEE 2009]

2m

m A

B

m C

14. A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverse its direction and moves with a speed of 2 ms–1. Which of the following statement(s) is (are ) correct for the system of these two masses ? (A) Total momentum of the system is 3 kg ms–1 (B) Momentum of 5 kg mass after collision is 4 kg ms–1 (C) Kinetic energy of the centre of mass is 0.75 J (D) Total kinetic energy of the system is 4 J [JEE 2010] 15. A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg traveling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity V of the bullet is v m/s

m (0, -a)

12. Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circle orbit. Their tangential velocities are v and 2v, respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particulars will again reach the point A ?

V

(A) 4 (C) 2

A

0

(A) 250 m/s (C) 400 m/s

20

(B) 250 2 m/s (D) 500 m/s

2V

(B) 3 (D) 1

100

[JEE 2009]

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[JEE 2011]

Page # 62

CENTRE OF MASS

Exercise-I 1.

D

2.

C

8. 15. 22. 29. 36. 43.

C C D B D A

9. 16. (a) 23. 30. 37. 44.

50. 57.

A D

51.

3.

4.

D

5.

B

6.

C

7.

C

A 10. C B, (b) C 17. (a) C, (b) B D 24. A B 31. B B 38. A D 45. A

11. 18. 25. 32. 39. 46.

B A A C C A

12. 19. 26. 33. 40. 47.

D D D D C B

13. 20. 27. 34. 41. 48.

D A B B C D

14. 21. 28. 35. 42. 49.

D B B B D C

C

53.

D

54.

C

55.

D

56.

C

B,C

6.

B

7.

B,D

52.

B

B

Exercise-II 1.

C,D

2.

B,D

3.

A,B

4.

C

5.

8.

C

9.

C

10.

B,C

11.

B

12. A,B,C,D

13.

B,C

14.

A,C

15.

B,C

16.

17.

A,B,C

18.

D

19.

20.

B,D

21.

A,C

22.

(a) A,C (b) B

4.

100 m

A,B,D

A,B,C,D

Exercise-III 4  b3  a3  2. y  3  2 2  b – a 

5 x , (b) L L 9

1.

(a) ( x)   

5.

(i) zero; (ii) right ; (iii) 20 cm ; (iv) 2.2 m ; (v) 1.8 m

8.

9 m/s, 9 m

9. (a)

Kx20 kx 20 , (b) 4 2m

3.

3 a 4

6. g/9 downwards

7. v 

2gR m 1 M

10. (i) 20Ns, (ii) 50 N

P(N-sec) 0.25

11.

0.125

12. 187.5 N, 625 kPa 50

13. m  u 2  uv  v 2

100 t(ms)

p1  p 2  12.3 m / sec , (b) mp

14.

(a)

16.

1.5 m/s, 3.6 Ns

18.

W = –3mu2; J = 6 mu, K = mu

p12  p 2 2 mp

= 9.4 m/s

15. (a) 4 5 N , (b) 2000 5 N

17. (a) u/2, mu/2; (b) u 13 / 8 , m u 13 / 8 (c) u 3 / 4 , mu 3 / 4 mv 19. (m  Ax)

20. 6 m/s in the direction of football’s velocity

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Page # 63

CENTRE OF MASS 12 Ns 5

21.

(a) 3J, (b)

22. 2 : 1

24.

(a) v  3 gl , (b) v =

26.

(a) e = cot2 +

28.

(a) v0/3, (b) 3 5gR

gl 2

23. (a)

40 m , (b) te = 3.25 s 3

25. (a) 10 3 , (b) tan–1 2 , (c) 45º, (d) 20 m

m 3 = , (g) t = 3 sec. M 4

27. v A  g / 12 m / s , Smax = 49/48 m  M0  29. v = (ucos –  u sin  ) ln  M – t  – gt  0 

Exercise-IV m v( v  gt) 

3. (a) 360 m, (b) 10800 J

4. (a) 50 m, (b) 10 m, (c) 30 m

1. ( 5 / 17 cm , 153L / 80u

2.

5. 13/3 m/s

6. (a) v/2, v/2, 0; (b) 2mv2/9, (c) mv2/72, (d) x  m / 6k v

7. 2mv2/3l

mA 8. vB  m B

2gl 2m2A g ; T = 3m g + A 1+ m A / mB mB

10. (a) v0/3, (b) mv0 = 2mv2 cos, 0 = 2mv2sin–

11. 2.5 m/sec, 0.312 m

12. v 

9. -2m/s, 6.93 m/s 30º

mv 0 mv 20 1 5 v 0 , (d) , (c) , 2 2 4 16

2 3u 5u 6 , u  , T  mv 7 7 7

Exercise-V 1. 2 in case I

2. C

3. C

4. t0 = 12 sec, v = 100 3 /11 1

5. C

6. B

7. A, D

8. B

9. B

10. C

11. a/10

12. C

13. 4

14. A,C

15. D

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Nurturing potential through education

ROTATIONAL MOTION THEORY AND EXERCISE BOOKLET CONTENTS S.NO.

TOPIC

PAGE NO.

1.

Rigid Body .............................................................................................. 3 – 5

2.

Moment of Inertia ................................................................................... 5 – 10

3.

Theorems of Moment of Inertia ............................................................. 11 – 14

4.

Cavity Problems ................................................................................... 14 – 15

5.

Torque ................................................................................................... 15 – 18

6.

Body In Equilibrium ............................................................................... 18 – 22

7.

Relation between Torque & ................................................................... 23 – 25 Angular acceleration

8.

Angular Momentum ............................................................................... 25 – 27

9.

Conservation of Angular Momentum ..................................................... 27 – 30

10.

Angular Impulse .................................................................................... 30 – 31

11.

Combined Translational & ..................................................................... 31 – 39 Rotational Motion

12.

Uniform Pure Rolling ............................................................................. 39 – 47

13.

Toppling ................................................................................................ 47 – 49

14.

Instantaneous Axis of Rotation ............................................................. 49 – 51

15.

Exercise - I ........................................................................................... 52 – 59

16.

Exercise - II .......................................................................................... 60 – 62

17.

Exercise - III ......................................................................................... 63 – 66

18

Exercise - IV ........................................................................................ 67 – 69

19.

Exercise - V .......................................................................................... 70 – 76

20.

Answer key ........................................................................................... 77 – 78

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Page # 2

ROTATIONAL DYNAMICS

Syllabus Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment of inertia of uniform bodies with simple geometrical shapes; Angular momentum; Torque; Conservation of angular momentum; Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres; Equilibrium of rigid bodies; Collision of point masses with rigid bodies.

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Page # 3

ROTATIONAL DYNAMICS 1.

RIGID BODY : Rigid body is defined as a system of particles in which distance between each pair of particles remains constant (with respect to time) that means the shape and size do not change, during the motion. Eg. Fan, Pen, Table, stone and so on. Our body is not a rigid body, two blocks with a spring attached between them is also not a rigid body. For every pair of particles in a rigid body, there is no velocity of seperation or approach between the particles. In the figure shown velocities of A and B with respect to   ground are VA and VB respectively

A VA sin1

A

A

VA cos 1

 1 VA B

B

VBA

B

VB  2 VB sin 2

VB cos  2

If the above body is rigid VA cos 1 = VB cos 2 Note : With respect to any particle of rigid body the motion of any other particle of that rigid body is circular. VBA = relative velocity of B with respect to A. Types of Motion of rigid body

Pure Translational Motion

1.1.

Pure Rotational Motion

Combined Translational and Rotational Motion

Pure Translational Motion :

A body is said to be in pure translational motion if the displacement of each particle is same   during any time interval however small or large. In this motion all the particles have same s, v  & a at an instant. example. A box is being pushed on a horizontal surface.

10

6

6

10 16   Vcm  V of any particle,   Scm  S of any particle

  a cm  a of any particle

For pure translational motion :v m2

v m1 v m4

v

m7

v m5

v m3

v vm6

m8

m2

m1

m3

m4

m5

m7

m6 m8

    Fext  m1a1  m 2 a 2  m 3 a 3 ............. IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 4

ROTATIONAL DYNAMICS

Where m1, m2, m3, ......... are the masses of different particles of the body having accelerations    a1, a 2 , a 3 ,............... respectively..     But acceleration of all the particles are same So, a1  a 2  a 3  .........  a   Fext  Ma Where M = Total mass of the body  a = acceleration of any particle or of centre of mass of body     P  m1v1  m 2 v 2  m 3 v 3 ............. Where m1, m2, m3 ...... are the masses of different particles of the body having velocities    v 1, v 2 , v 3 ............. respectively     But velocities of all the particles are same so v1  v 2  v 3 ..........  v   P  Mv  Where v = velocity of any particle or of centre of mass of the body.. Total Kinetic Energy of body =

1 1 1 m1v 12  m 2 v 22  .......... .  Mv 2 2 2 2

1.2. Pure Rotational Motion : A body is said to be in pure rotational motion if the perpendicular distance of each particle remains constant from a fixed line or point and do not move parallel to the line, and that line    is known as axis of rotation. In this motion all the particles have same ,  and  at an instant. Eg. : - a rotating ceiling fan, arms of a clock. For pure rotation motion :m3 m5

m2

m2

s Where  = angle rotated by the particle r s = length of arc traced by the particle. r = distance of particle from the axis of rotation. 

m1

m3 m6

m1

m5

m6

m4

m4

d  Where  = angular speed of the body.. dt

d Where  = angular acceleration of the body.. dt All the parameters ,  and  are same for all the particles. Axis of rotation is perpendicular to the plane of rotation of particles. Special case : If  = constant,  = 0 +  t Where 0 = initial angular speed 

  0t  2 = 02

1 2 t t = time interval 2 + 2

Total Kinetic Energy 

1 1 m1v12  m 2 v 22 ................. 2 2

1 [m1r12  m 2r22 ................]  2 2

1 2 I Where I = Moment of Inertia = m1r12  m 2r22 ....... 2  = angular speed of body. 

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Page # 5

ROTATIONAL DYNAMICS

1.3

Combined translation and rotational Motion A body is said to be in translation and rotational motion if all the particles rotates about an axis of rotation and the axis of rotation moves with respect to the ground.

2.

MOMENT OF INERTIA

Like the centre of mass, the moment of inertia is a property of an object that is related to its mass distribution. The moment of inertia (denoted by I) is an important quantity in the study of system of particles that are rotating. The role of the moment of inertia in the study of rotational motion is analogous to that of mass in the study of linear motion. Moment of inertia gives a measurement of the resistance of a body to a change in its rotaional motion. If a body is at rest, the larger the moment of inertia of a body the more difficuilt it is to put that body into rotational motion. Similarly, the larger the moment of inertia of a body, the more difficult to stop its rotational motion. The moment of inertia is calculated about some axis (usually the rotational axis). Moment of inertia depends on : (i) density of the material of body (ii) shape & size of body (iii) axis of rotation In totality we can say that it depends upon distribution of mass relative to axis of rotation. Note : Moment of inertia does not change if the mass : (i) is shifted parallel to the axis of the rotation (ii) is rotated with constant radius about axis of rotation 2.1

2.2

Moment of Inertia of a Single Particle For a very simple case the moment of inertia of a

r

single particle about an axis is given by, I = mr2 ...(i) Here, m is the mass of the particle and r its distance from the axis under consideration. Moment of Inertia of a System of Particles The moment of inertia of a system of particles about an axis is given by, I=

2 i i

m r

...(ii)

i

r1

m1

r2

m2

r3

m3

where ri is the perpendicular distance from the axis to the ith particle, which has a mass mi. Ex.1

Two heavy particles having masses m1 & m2 are situated in a plane perpendicular to line AB at a distance of r1 and r2 respectively. C A

E

(i) (ii)

r1

m1

r2 m2

F

D B What is the moment of inertia of the system about axis AB? What is the moment of inertia of the system about an axis passing through m1 and perpendicular to the line joining m1 and m2 ? IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 6 (iii) Sol.

(i)

(ii)

(iii)

ROTATIONAL DYNAMICS What is the moment of inertia of the system about an axis passing through m1 and m2? Moment of inertia of particle on left is I1 = m1r12. Moment of Inertia of particle on right is I2 = m2r22. Moment of Inertia of the system about AB is I = I1+ I2 = m1r12 + m2r22 Moment of inertia of particle on left is I1 = 0 Moment of Inertia of the system about CD is I = I1 + I2 = 0 + m2(r1 + r2)2 Moment of inertia of particle on left is I1 = 0 Moment of inertia of particle on right is I2 = 0 Moment of Inertia of the system about EF is I = I1 + I2 = 0 + 0

Ex.2

Sol.

Three light rods, each of length 2, are joined together to form a triangle. Three particles A, B, C of masses m, 2m, 3m are fixed to the vertices of the triangle. Find the moment of inertia of the resulting body about (a) an axis through A perpendicular to the plane ABC, (b) an axis passing through A and the midpoint of BC. (a) B is at a distant 2 from the axis XY so the moment of inertia of B (IB) about XY is 2 m (2)2 Similarly Ic about XY is 3m (2)2 and IA about XY is m(0)2

X

A m Y

2l Therefore the moment of inertia of the body about XY is 2m (2)2 + 3 m(2)2 + m(0)2 = 20 m2 B (b) IA about X' Y' = m(0)2 2 IB about X' Y' = 2m () 2m 2 IC about X' Y' = 3m () Therefore the moment of inertia of the body about X' Y' is m(0)2 + 2m()2 + 3m()2 = 5 m2

2l

C 3m

X' A

m

B

C 3m

2m Y' Ex.3

Four particles each of mass m are kept at the four corners of a square of edge a. Find the moment of inertia of the system about a line perpendicular to the plane of the square and passing through the centre of the square.

Sol.

The perpendicular distance of every particle from the given line is a / 2 . The moment of inertia of

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Page # 7

ROTATIONAL DYNAMICS one particle is, therefore, m(a / 2 ) 2 =

1 ma 2 . The 2

m

m

moment of inertia of the system is, /

2

1 ma2 = 2 ma2. 2

a

therefore, 4 

m

2.3

m

Moment of Inertia of Rigid Bodies For a continuous mass distribution such as found in a rigid body, we replace the summation of I

2 i i

m r

by an integral. If the system is divided

i

r

into infinitesimal element of mass dm and if r is the distance from a mass element to the axis of rotation, the moment of inertia is, I=

r

2

dm

where the integral is taken over the system. (A)

Uniform rod about a perpendicular bisector Consider a uniform rod of mass M and length l figure and suppose the moment of inertia is to be calculated about the bisector AB. Take the origin at the middle point O of the rod. Consider the element of the rod between a distance x and x + dx from the origin. As the rod is uniform, Mass per unit length of the rod = M / l so that the mass of the element = (M/l)dx. The perpendicular distance of the element from the line AB is x. The moment of inertia of this element about AB is dI 

A

B

x

dx

0

M dx x 2 . l

When x = – l/2, the element is at the left end of the rod. As x is changed from – l/2 to l/2, the elements cover the whole rod. Thus, the moment of inertia of the entire rod about AB is l/2

I

M x3  M 2 x dx    l  l 3 

l / 2

(B)

l/2

 –l / 2

Ml 2 12

Moment of inertia of a rectangular plate about a line parallel to an edge and passing through the centre The situation is shown in figure. Draw a line parallel to AB at a distance x from it and another at a distance x + dx. We can take the strip enclosed between the two lines as the small element. x A b dx l It is “small” because the perpendiculars from different points of the strip to AB differ by not B

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Page # 8

ROTATIONAL DYNAMICS

more than dx. As the plate is uniform, its mass per unit area =

M bl

M M b dx  dx . l bl The perpendicular distance of the strip from AB = x. Mass of the strip =

The moment of inertia of the strip about AB = dI =

M dx x 2 . The moment of inertia of the given l

plate is, therefore, l/2

I

M 2 Ml 2 x dx  l 12

l / 2

The moment of inertia of the plate about the line parallel to the other edge and passing through the centre may be obtained from the above formula by replacing l by b and thus,

Mb 2 . 12 Moment of inertia of a circular ring about its axis (the line perpendicular to the plane of the ring through its centre) Suppose the radius of the ring is R and its mass is M. As all the elements of the ring are at the same perpendicular distance R from the axis, the moment of inertia of the ring is I

(C)

I  r 2 dm  R2 dm  R2 dm  MR2 .

(D)

Moment of inertia of a uniform circular plate about its axis Let the mass of the plate be M and its radius R. The centre is at O and the axis OX is perpendicular to the plane of the plate. X dx

0 x R

Draw two concentric circles of radii x and x + dx, both centred at O and consider the area of the plate in between the two circles. This part of the plate may be considered to be a circular ring of radius x. As the periphery of the ring is 2 x and its width is dx, the area of this elementary ring is 2xdx. The area of the plate is  R2. As the plate is uniform, M

Its mass per unit area = M

Mass of the ring 

2

 R2

2  x dx 

2 M x dx

R R2 Using the result obtained above for a circular ring, the moment of inertia of the elementary ring about OX is  2 Mx dx  2 dI   x .  R2  The moment of inertia of the plate about OX is R

I

2M

R 0

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2

x 3 dx 

MR 2 . 2

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Page # 9

ROTATIONAL DYNAMICS (E)

Moment of inertia of a hollow cylinder about its axis Suppose the radius of the cylinder is R and its mass is M. As every element of this cylinder is at the same perpendicular distance R from the axis, the moment of inertia of the hollow cylinder about its axis is

I  r 2 dm  R 2 dm  MR 2

(F)

Moment of inertia of a uniform solid cylinder about its axis Let the mass of the cylinder be M and its radius R. Draw two cylindrical surface of radii x and x + dx coaxial with the given cylinder. Consider the part of the cylinder in between the two surface. This part of the cylinder may be considered to be a hollow cylinder of radius x. The area of cross-section of the wall of this hollow cylinder is 2 x dx. If the length of the cylinder is l, the volume of the material of this elementary hollow cylinder is 2 x dxl. The volume of the solid cylinder is  R2 l and it is uniform, hence its mass per unit volume is M 

 R2 l

The mass of the hollow cylinder considered is M  R2 l

2 x dx l 

2M R2

x dx .

dx

As its radius is x, its moment of inertia about the given axis is

x  2M  dI   2 x dx x2 . R 

The moment of inertia of the solid cylinder is, therefore, R

I

2M

R 0

(G)

2

x 3 dx 

MR 2 2 .

Note that the formula does not depend on the length of the cylinder. Moment of inertia of a uniform hollow sphere about a diameter Let M and R be the mass and the radius of the sphere, O its centre and OX the given axis (figure). The mass is spread over the surface of the sphere and the inside is hollow. Let us consider a radius OA of the sphere at an angle  with the axis OX and rotate this radius about OX. The point A traces a circle on the sphere. Now change  to  + d and get another circle of somewhat larger radius on the sphere. The part of the sphere between these two circles, shown in the figure, forms a ring of radius R sin. The width of this ring is Rd and its periphery is 2R sin. Hence, the area of the ring = (2R sin) (Rd). x

M Mass per unit area of the sphere 

M The mass of the ring 

4 R

4 R2

R sin A Rd

.

(2R sin )(Rd)  2

M sin  d. 2

R 0

d

The moment of inertia of this elemental ring about OX is M  d I   sin  d. (R sin ) 2  M R2 sin 3  d 2  2

As  increases from 0 to , the elemental rings cover the whole spherical surface. The

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Page # 10

ROTATIONAL DYNAMICS

moment of inertia of the hollow sphere is, therefore, 

I

    M 2 MR2  MR 2   (1  cos 2 )sin  d   (1  cos 2 ) d (cos ) R sin3  d  2 2  2    0 0   0  

MR2  2

(H)

 cos 3   2 2 cos     MR 3  3  0

Moment of inertia of a uniform solid sphere about a diameter Let M and R be the mass and radius of the given solid sphere. Let O be centre and OX the given axis. Draw two spheres of radii x and x + dx concentric with the given solid sphere. The thin spherical shell trapped between these spheres may be treated as a hollow sphere of radius x.

x dx

0 x

The mass per unit volume of the solid sphere

=

M 3M  3 4 R 3 4 R 3

The thin hollow sphere considered above has a surface area 4x2 and thickness dx. Its volume is 4  x2 dx and hence its mass is  3M  3M  (4  x 2 dx) = 3 x 2 dx =  3 R  4 R  Its moment of inertia about the diameter OX is, therefore, 2  3M 2  x dx x2 3  R 3 

2M

x 4 dx R3 If x = 0, the shell is formed at the centre of the solid sphere. As x increases from 0 to R, the shells cover the whole solid sphere.

dl =

=

The moment of inertia of the solid sphere about OX is, therefore, R

I=

2M

R

3

x 4 dx =

0

Ex.4

2 MR 2 . 5

Find the moment of Inertia of a cuboid along the axis as shown in the figure.

I

b a c

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Page # 11

ROTATIONAL DYNAMICS

M(a 2  b 2 ) 12

Sol.

After compressing the cuboid parallel to the axis I =

3.

THEOREMS OF MOMENT OF INERTIA There are two important theorems on moment of inertia, which, in some cases enable the moment of inertia of a body to be determined about an axis, if its moment of inertia about some other axis is known. Let us now discuss both of them. Theorem of parallel axes A very useful theorem, called the parallel axes theorem relates the moment of inertia of a rigid body about two parallel axes, one of which passes COM through the centre of mass.

3.1

Two such axes are shown in figure for a body of mass M. If r is r the distance between the axes and ICOM and I are the respective moments of inertia about them, these moments are related by, I = ICOM + Mr2 Theorem of parallel axis is applicable for any type of rigid body whether it is a two dimensional or three dimensional

*

A

Ex 5. Three rods each of mass m and length l are joined together to form an equilateral triangle as shown in figure. Find the moment of inertia of the system about an axis passing through its centre of mass and perpendicular to the plane of triangle. Sol.

Moment of inertia of rod BC about an axis perpendicular to plane of triangle ABC and passing through the midpoint of rod BC (i.e., D) is

COM B

C

ml 2 12 From theorem of parallel axes, moment of inertia of this rod about the asked axis is I1 =

A

2

 l  ml 2 ml 2   m   I2 = I1 + mr = 12 6 2 3

COM

2

r  Moment of inertia of all the three rod is

B  ml 2  ml 2   I  3I2  3 2  6 

30° D

C

Ex.6. Find the moment of inertia of a solid sphere of mass M and radius R about an axis XX shown in figure.

x

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Page # 12

ROTATIONAL DYNAMICS

x Sol.

From theorem of parallel axis, IXX = ICOM + Mr2 =

2 MR2  MR2 5

=

7 MR 2 5

COM

x

r=R Ex.7. Consider a uniform rod of mass m and length 2l with two particles of mass m each at its ends. Let AB be a line perpendicular to the length of the rod passing through its centre. Find the moment of inertia of the system about AB. A Sol. IAB = Irod + Iboth particles

3.2

m(2l )2  2(ml 2 ) 12 7 2 ml 3

I

I

m

m Ans.

B

Theorem of perpendicular axes The theorem states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about two axes perpendicular to each other, in its own plane and intersecting each other, at the point where the perpendicular axis passes through it. Let x and y axes be chosen in the plane of the body and z-axis perpendicular, to this plane, three axes being mutually perpendicular, then the theorem states that.

z y

xi ri

P yi x

O

(i) (ii) (iii) Ex.8

Iz = Ix + Iy Important point in perpendicular axis theorem This theorem is applicable only for the plane bodies (two dimensional). In theorem of perpendicular axes, all the three axes (x, y and z) intersect each other and this point may be any point on the plane of the body (it may even lie outside the body). Intersection point may or may not be the centre of mass of the body. Find the moment of inertia of uniform ring of mass M and radius R about a diameter. B

Z

C

Sol.

0

D

A Let AB and CD be two mutually perpendicular diameters of the ring. Take them ax X and Y-

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Page # 13

ROTATIONAL DYNAMICS

axes and the line perpendicular to the plane of the ring through the centre as the Z-axis. The moment of inertia of the ring about the Z-axis is I = MR2. As the ring is uniform, all of its diameter equivalent and so Ix = Iy, From perpendicular axes theorem, Iz = Ix + Iy

Hence Ix =

Iz MR2 = 2 2

Similarly, the moment of inertia of a uniform disc about a diameter is MR2/4 Ex.9

Two uniform identical rods each of mass M and length  are joined to form a cross as shown in figure. Find the moment of inertia of the cross about a bisector as shown dotted in the figure.

Sol.

Consider the line perpendicular to the plane of the figure through the centre of the cross. The moment of inertia of each rod about this line is

M 2 and hence the moment of inertia of the 12

M 2 . The moment of inertia of the cross about the two bisector are equal by 6 symmetry and according to the theorem of perpendicular axes, the moment of inertia of the

cross is

M 2 . 12

Ex.10 In the figure shown find moment of inertia of a plate having mass M, length  and width b about axis 1,2,3 and 4. Assume that C is centre and mass is uniformly distributed 1

2

4

C 3

b 

Sol.

Moment of inertia of the plate about axis 1 (by taking rods perpendicular to axis 1) l1 = Mb2/3 Moment of inertia of the plate about axis 2 (by taking rods perpendicular to axis 2) I 2 = M2/12 Moment of inertia of the plate about axis 3 (by taking rods perpendicular to axis 3)

Mb 2 12 Moment of inertia of the plate about axis 4(by taking rods perpendicular to axis 4) I4 = M2/3 I3 

3.3

Moment of Inertia of Compound Bodies Consider two bodies A and B, rigidly joined together. The moment of inertia of this compound body, about an axis XY, is required. If IA is the moment of inertia of body A about XY. IB is the moment of inertia of body B about XY.Then, moment of Inertia of compound body I = IA + IB Extending this argument to cover any number of bodies rigidly joined together, we see that the moment of inertia of the compound body, about a specified axis, is the sum of the moments of inertia of the separate parts of the body about the same axis. IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 14

ROTATIONAL DYNAMICS

A Y

X B

Ex.11 Two rods each having length l and mass m joined together at point B as shown in figure.Then findout moment of inertia about axis passing thorugh A and perpendicular to the plane of page as shown in figure.  A B × 

Sol.

C We find the resultant moment of inertia I by dividing in two parts such as I = M.I of rod AB about A + M.I of rod BC about A I = I1 + I2 ... (1) first calculate I1 : B  A ×

m 2 ...(2) 3 Calculation of I2 : use parallel axis theorem I2 = ICM + md2 I1 =

/2

...(3)

m  2 m  2 5 2m   3 12 4

I= 4.

d

COM ×

 2  m 2 m 2 5 2   2  m   m = 12  4  = 12 4   Put value from eq. (2) & (3) into (1) I=

×

m 2 (4  1  15 ) 12

I=

5m 2 3

CAVITY PROBLEMS :

Ex.12 A uniform disc having radius 2R and mass density  as shown in figure. If a small disc of radius R is cut from the disc as shown. Then find out the moment of inertia of remaining disc around the axis that passes through O and is perpendicular to the plane of the page.

2R O

Sol.

R

We assume that in remaning part a disc of radius R and mass density ±  is placed. Then

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Page # 15

ROTATIONAL DYNAMICS M1  (2R) 2

M2  –R2 2R

2R O   R

× I1

 when  is taken

I2

R

+ × when – is takes

Total Moment of Inertia I = I1 + I2 I1 =

M1(2R)2 2

4R2 .4R2 = 8  R4 2 To calculate I2 we use parallel axis theorem. I2 = ICM + M2R2 I1 =

I2 =

M2R2 + M2R2 2

3 3 M2R2 = (– R 2 )R2 2 2 Now I = I1 + I2 I2 =

4 I = 8 R –

3 R4 2

I2 = –

I=

3 R4 2

13 R 4 2

R/ 3

Ex.13 A uniform disc of radius R has a round disc of radius R/3 cut as shown in Fig. The mass of the remaining (shaded) portion of the disc equals M. Find the moment of inertia of such a disc relative to the axis passing through geometrical centre of original disc and perpendicular to the plane of the disc.

Sol.

O

R

Let the mass per unit area of the material of disc be . Now the empty space can be considered as having density –  and . Now I0 = I + I–  (R2)R2/2 = M.I of  about O – (R / 3) 2 (R / 3) 2  [– (R / 3) 2 ]( 2R / 3) 2 2

I– =

= M.I of –  about 0 

5.

I0 =

4 R 4 Ans. 9

TORQUE : Torque represents the capability of a force to produce change in the rotational motion of the body

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Page # 16

ROTATIONAL DYNAMICS Line of action of force

F

P r

r sin Q

5.1 Torque about point :     Torque of force F about a point   r F  where F = force applied P = point of application of force Q = point about which we want to calculate the torque.  r = position vector of the point of application of force from the point about which we want to determine the torque.    rF sin  = rF = rF

where  = angle between the direction of force and the position vector of P wrt. Q. r = perpendicular distance of line of action of force from point Q. F = force arm

SI unit to torque is Nm Torque is a vector quantity and its direction is determined using right hand thumb rule. Ex.14 A particle of mass M is released in vertical plane from a point P at x = x0 on the x-axis it falls vertically along the y-axis. Find the torque  acting on the particle at a time t about origin? x0 P O x 

r 

Sol. 

mg Torque is produced by the force of gravity    r F sin  k or

  r F  x 0 mg

Ex.15 Calculate the total torque acting on the body shown in figure about the point O

10N

15N

O 4cm

90°

6c m

37°

3c 30° m

150°

20N

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5N

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Page # 17

ROTATIONAL DYNAMICS

15sin37°

15N

10N 90°

4cm

O

4cm

Sol.

6c m

37° 5N 20N

30°

150° 20sin30° 0 = 15sin37 × 6 + 20 sin 30° × 4 = 54 + 40 – 40 = 54 N-cm 0 = 0.54 N-m

– 10 × 4

Ex.16 A particle having mass m is projected with a velocity v0 from a point P on a horizontal ground making an angle  with horizontal. Find out the torque about the point of projection acting on the particle when

V0 

(a) it is at its maximum height ? P

Q

(b) It is just about to hit the ground back ? Sol. (a)

F = mg ; r 

p =

r

R 2

v 20 sin 2 R  P = mg = mg  2g 2

(b)

v0

Particle is at maximum height then  about point P is  p  r F

mg

P

mv 20 sin 2 2

when particle is at point Q then  about point P is  p'  rF r  R ; F = mg

 p'

v 02 sin2  mgR = mg g

Q

P mg

Ex.17 In the previous question, during the motion of particle from P to Q. Torque of gravitational force about P is : (A) increasing (B) decreasing (C) remains constant (D) first increasing then decreasing Sol.

Torque of gravitational force about P is increasing because r is increasing from O to R. (Range)

5.2

Torque about axis :      r F  where  = torque acting on the body about the axis of rotation  r = position vector of the point of application of force about the axis of rotation. IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 18

ROTATIONAL DYNAMICS  F = force applied on the body..      net   1   2   3 .....

To understand the concept of torque about axis we take a general example which comes out in daily life. Figure shows a door ABCD. Which can rotate about axis AB. Now if we apply force. F at point.

D

A r

×

y x

B

C F

in inward direction then AB = r F and direction of this AB is along y axis from right hand thumb rule. Which is parallel to AB so gives the resultant torque. Now we apply force at point C in the direction as shown   figure. At this time r & F are perpendicular to each other which gives  AB  rF

But door can’t move when force is applied in this direction because the direction of  AB is perpendicular to AB according to right hand thumb rule. So there is no component of  along AB which gives res  0  Now conclude Torque about axis is the component of r  F parallel to axis of rotation. Note : The direction of torque is calculated using right hand thumb rule and it is always perpendicular to the plane of rotation of the body.

F2 r2

F3

r3 × r1 F1

If F1 or F2 is applied to body, body revolves in anti-clockwise direction and F3 makes body revolve in clockwise direction. If all three are applied.   resul tan t  F1r1  F2r2 – F3 r3 (in anti-clockwise direction) 6.

BODY IS IN EQUILIBRIUM : We can say rigid body is in equillibrium when it is in (a) Translational equilibrium  i.e. Fnet  0

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Page # 19

ROTATIONAL DYNAMICS Fnet x = 0 and Fnet y = 0 and (b) Rotational equillibrium   net  0 i.e., torque about any point is zero

Note : (i) If net force on the body is zero then net torque of the forces may or may not be zero. example. A pair of forces each of same magnitude and acting in opposite direction on the rod.

B

A F (2)

F C

2

 A  2F If net force on the body is zero then torque of the forces about each and every point is same  about B

 B  F  + F

 B  2F  about C

 C  2F

Ex.18 Determine the point of application of third force for which body is in equillibrium when forces of 20 N & 30 N are acting on the rod as shown in figure 20N

A 10cm C 20cm Sol.

(i)

(ii)

B

30N Let the magnitude of third force is F, is applied in upward direction then the body is in the equilibrium when  Fnet  0 (Translational Equillibrium)  20 + F = 30  F = 10 N So the body is in translational equilibrium when 10 N force act on it in upward direction. Let us assume that this 10 N force act. 10N 20N Then keep the body in rotational equilibrium x So Torque about C = 0 i.e. c = 0 B A C 20cm  30 × 20 = 10 x 30N x = 60 cm so 10 N force is applied at 70 cm from point A to keep the body in equilibrium.

Ex.19 Determine the point of application of force, when forces are acting on the rod as shown in figure.

10N

5N 5cm

5cm 3N

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Page # 20

Sol.

ROTATIONAL DYNAMICS

 Since the body is in equillibrium so we conclude F net  0 and torque about any point is zero  i.e.,  net  0 10N

6

5N F2 A

x 

F

37° 8N

F1 3N

Let us assume that we apply F force downward at A angle  from the horizontal, at x distance from B From

 F net  0

Fnet x = 0 which gives F2 = 8 N From Fnet y = 0  5 + 6 = F1 + 3  F1 = 8 N If body is in equillibrium then torque about point B is zero.  3 × 5 + F1. x – 5 × 10 = 0  15 + 8x – 50 = 0 x=

35 9

x = 4.375 cm

Ex.20 A uniform rod length , mass m is hung from two strings of equal length from a ceiling as shown in figure. Determine the tensions in the strings ?

/4

Sol.

B A Let us assume that tension in left and right string is TA and TB respectively. Then   Rod is in equilibrium then Fnet  0 & net  0  From Fnet  0

mg = TA + TB ...(1) From net = 0 about A mg

 3  TB  0 2 4

TB =

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TA

TB  /4

 /2

B

A mg

2mg 3

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Page # 21

ROTATIONAL DYNAMICS

from eq. (1) TA 

2mg = mg  3

TA =

mg 3

Ladder Problems : Ex.21 A stationary uniform rod of mass ‘m’, length ‘’ leans against a smooth vertical wall making an angle  with rough horizontal floor. Find the normal force & frictional force that is exerted by the floor on the rod?

smooth

rough Sol.

As the rod is stationary so the linear acceleration and angular acceleration of rod is zero. i.e., acm = 0 ;  = 0. A N2 N2 = f  acm =0 N = mg Torque about any point of the rod should also be zero =0 A = 0  mg cos 

N1 cos  = sin  f +

f=

mgcos 

1

N1  mg

B f Free Body Diagram

 + f  sin  = N1 cos .  2

mgcos  2

mgcos  mgcot  = 2 sin  2

Ex.22 The ladder shown in figure has negligible mass and rests on a frictionless floor. The crossbar connects the two legs of the ladder at the middle. The angle between the two legs is 60°. The fat person sitting on the ladder has a mass of 80 kg. Find the contanct force exerted by the floor on each leg and the tension in the crossbar.

W 1m 60°

T

N 1m

Sol.

N

° 30

The forces acting on different parts are shown in figure. Consider the vertical equilibrium of “the ladder plus the person” system. The forces acting on this system are its weight (80 kg)

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Page # 22

ROTATIONAL DYNAMICS

g and the contact force N + N = 2 N due to the floor. Thus 2 N = (80 kg) g or N = (40 kg) (9.8 m/s2) = 392 N Next consider the equilibrium of the left leg of the ladder. Taking torques of the forces acting on it about the upper end, 2 N (2m) tan 30° = T (1m)

or

T=N

3

2 = (392 N) ×

3

= 450 N

Ex.23 A thin plank of mass m and length  is pivoted at one end and it is held stationary in horizontal position by means of a light thread as shown in the figure then find out the force on the pivot.

Sol.

(i)

T

N2

Free body diagram of the plank is shown in figure.  Plank is in equilibrium condition So Fnet & net on the plank is zero

N1

O

A mg

from Fnet = 0  Fnet x = 0 N1 = 0 Now Fnet  0 y 

N2 + T = mg

...(i)

from net = 0  net about point A is zero so N2 .  = mg . /2 

N2 

mg 2

Ex.24 A square plate is hinged as shown in figure and it is held stationary by means of a light thread as shown in figure. Then find out force exerted by the hinge.

square plate

T

Sol.

F.B.D.

 Body is in equilibrium and N

T and mg force passing through one line so from

net = 0,

N=0

mg

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Page # 23

ROTATIONAL DYNAMICS

7.

RELATION BETWEEN TORQUE AND ANGULAR ACCELERATION The angular acceleration of a rigid body is directly proportional to the sum of the torque components along the axis of rotation. The proportionality constant is the inverse of the moment of inertia about that axis, or 

 I

Thus, for a rigid body we have the rotational analog of Newton's second law ;

   I

...(iii)

Following two points are important regarding the above equation. (i) The above equation is valid only for rigid bodies. If the body is not rigid like a rotating tank of water, the angular acceleration  is different for different particles. (ii) The sum  in the above equation includes only the torques of the external forces, because all the internal torques add to zero. Ex.25 A uniform rod of mass m and length  can rotate in vertical plane about a smooth horizontal axis hinged at point H.  × A H (i) Find angular acceleration  of the rod just after it is released from initial horizontal position from rest? (ii) Calculate the acceleration (tangential and radial) of point A at this moment. Sol. (i) H = IH   3g m 2  = = 2 2 3

mg.

(ii) aA = a =

3g 3g . = 2 2

aCA = 2r = 0. = 0

(  = 0 just after release)

×

Ex.26 A uniform rod of mass m and length  hinged at point H can rotate in vertical plane about a smooth horizontal axis. Find force exerted by the hinge just after the rod is released from rest, from an initial position making an angle of 37° with horizontal ?

Sol.

37°

H Just After releasing at 37º from horizontal F.B.D. of plank from

net = I

 m 2 .  about point A = A = mg cos 37° = 3 2

 N1

R

37º

mg mgcos 37º

A

N2

6g = rad/sec2 5 Now Tangential acceleration of centre of mass  3g m / s2 = 2 5 just after release vcm = 0  ar = 0 Now resolving of at in horizontal and vertical direction as shwon in figure

at = .

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Page # 24

ROTATIONAL DYNAMICS

a t || 

N1

37°

R

N2

a t 

9g 25

N1

3g/5

mg

12g 25

N2

from Fnet = ma in both horizontal and vertical direction  9g  N2= m  25

Now

R=

N1 =

13mg 25

N12  N22

mg 10 5 PULLEY BLOCK SYSTEM If there is friction between pulley and string and pulley have some mass then tension is different on two sides of the pulley. R=

Reason : To understand this concept we take a pulley block system as shown in figure.

 B

a

R

C

T1

A M

D a M>m m

Let us assume that tension induced in part AB of the string is T1 and block M move downward. If friction is present between pulley and string then it opposes the relative slipping between pulley and string, take two point e and f on pulley and string respectively. If friction is there then due to this, both points wants to move together. So friction force act on e and d in the direction as shown is figure T2 This friction force f acting on point d increases the tension T1 by a small amount dT. f Then T1 = T2 + dT C d e or we can say T2 = T1 – f f In this way the tension on two side of pulley is different If there is no relative slipping between pulley and string at a then   = R R

T1

Ex.27 The pulley shown in figure has moment of inertia l about its axis and radius R. Find the acceleration of the two blocks. Assume that the string is light and does not slip on the pulley. Sol. Suppose the tension in the left string is T1 and that in the right string is T2. Suppose the block of mass M goes down with an acceleration a and the other block moves up with the same acceleration. This is also the tangential acceleration of the rim of the wheel as the string

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Page # 25

ROTATIONAL DYNAMICS does not slip over the rim. The angular acceleration of the wheel  =

a . R

The equations of motion for the mass M, the mass m and the pulley are as follows ; Mg – T1 = Ma ...(i) T2 – mg = ma ...(ii)

R

m

Ia M T1R – T2R = I = ...(iii) R Substituting for T1 and T2 from equations (i) and (ii) in equation (iii) [M(g – a) – m (g + a)]R =

Ia R

Solving, we get

(M – m)gR2 a=

I  (M  m)R2

8.

ANGULAR MOMENTUM

8.1

Angular momentum of a particle about a point.     L = r p sin  L  r P  |L|  r  P  |L|  P  r  Where P = momentum of partilcle  r = position of vector of particle with respect to point about which

Pcos    r

 P

P sin 

angular momentum is to be calculated.    = angle between vectors r & p

O r = perpendicular distance of line of motion of particle from point O.

P = perpendicular component of momentum. SI unit of angular momentum is kgm2/sec. Ex.28 A particle of mass m is moving along the line y = b, z = 0 with constant speed v. State whether the angular momentum of particle about origin is increasing, decreasing or constant. Sol.

y

 | L |  mvr sin  = mvr mvb

P  r

 v

r  b

  X | L | = constant as m, v and b all are constants. O   Direction of r  v also remains the same. Therefore, angular momentum of particle about origin remains constant with due course of time.

 Note : In this problem | r | is increasing,  is decreasing but r sin , i.e., b remains constant. Hence, the angular momentum remains constant.

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Page # 26

ROTATIONAL DYNAMICS

Ex.29 A particle of mass m is projected with velocity v at an angle  with the horizontal. Find its angular momentum about the point of projection when it is at the highest point of its trajectory. Sol. At the highest point it has only horizontal velocity y vx = v cos . Length of the perpendicular to the horizontal velocity from 'O' is the maximum height, where Hmax 

v 2 sin 2  2g

H

 Angular momentum L =

8.2

mv 3 sin 2  cos  2g

O

x

Angular Momentum of a rigid body rotating about a fixed axis Suppose a particle P of mass m is going in a circle of radius r and at some instant the speed of the particle is v. For finding the angular momentum of the particle about the axis of rotation, the origin may be chosen anywhere on the axis. We choose it at the centre of the

   circle. In this case  r and P are perpendicular to each other and r  P is along the axis.   Thus, component of r  P along the axis is mvr itself. The angular momentum of the whole rigid body about AB is the sum of components of all particles, i.e., L=

m r v

i i i

i

Here, 

vi = r i  L=

m r

i i

2

 i or L = 

i

m r

2

i i

i

or L = I Here, I is the moment of inertia of the rigid body about AB.  Note : Angular momentum about axis is the component of I along the axis. In most of the cases angular momentum about axis is I.

Ex.30 Two small balls A and B, each of mass m, are attached rigidly to the ends of a light rod of length d. The structure rotates about the perpendicular bisector of the rod at an angular speed . Calculate the angular momentum of the individual balls and of the system about the axis of rotation. d A

Sol.

O

B

Consider the situation shown in figure. The velocity of the ball A with respet to the centre O d is v = . 2 The angular momentum of the ball with respect to the axis is

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Page # 27

ROTATIONAL DYNAMICS

1  d   d  L1 = mvr = m    = md2. The same the angular momentum L2 of the second ball. The  2   2 4 angular momentum of the system is equal to sum of these two angular momenta i.e., L = 1/2 md2. 9.

CONSERVATION OF ANGULAR MOMENTUM : The time rate of change of angular momentum of a particle about some referenence point in an inertial frame of reference is equal to the net torques acting on it.   dL ....(i) or net  dt    dL Now, suppose that  net  0 , then  0 , so that L = constant. dt "When the resultant external torque acting on a system is zero, the total vector angular momentum of the system remains constant. This is the principle of the conservation of angular momentum. For a rigid body rotating about an axis (the z-axis, say) that is fixed in an inertial reference frame, we have Lz = I It is possible for the moment of inertia I of a rotating body to change by rearrangement of its parts. If no net external torque acts, then Lz must remains constant and if I does change, there must be a compensating change in . The principle of conservation of angular momentum in this case is expressed. I = constant.

Ex.31 A wheel of moment of inertia I and radius R is rotating about its axis at an angular speed 0. It picks up a stationary particle of mass m at its edge. Find the new angular speed of the wheel. Sol. Net external torque on the system is zero. Therefore, angular momentum will remain conserved. Thus, I1 1 I11 = I22 or 2 = I 2 Here, I1 = I, 1 = 0, I2 = I + mR2 I 0  2 = I  mR 2 Note : A

Hinge m

L O m

u

Case I

O m

u m

Case II

Comments on Linear Momentum : In case I : Linear momentum is not conserved just before and just after collision because during collision hinge force act as an external force. In case II : Linear momentum is conserved just before and just after collision because no external force on the string. Comments on Angular Momentum : In case I : Hinge force acts at an external force during collision but except point A all the other reference point given net  0. So angular momentum is conserved only for point A.

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Page # 28

ROTATIONAL DYNAMICS

In case II : angular momentum is conserved at all points in the world. Ex.32 A uniform rod of mass m and length  can rotate freely on a smooth horizontal plane about a vertical axis hinged at point H. A point mass having same mass m coming with an initial speed u perpendicular to the rod, strikes the rod in-elastically at its free end. Find out the angular velocity of the rod just after collision? m,  × H u

Sol.

m Angular momentum is conserved about H because no external force is present in horizontal plane which is producing torque about H.   m 2 2 3u mu =  3  m    w =   4

Ex.33 A uniform rod of mass m and length  can rotate freely on a smooth horizontal plane about a vertical axis hinged at point H. A point mass having same mass m coming with an initial speed u perpendicular to the rod, strikes the rod and sticks to it at a distance of 3/4 from hinge point. Find out the angular velocity of the rod just after collision?

m,  H× u m m,

H

Sol.

Initial position

3/4

u m

from angular momentum conservation about H initial angular momentum = final angular momentum 2

m. u

ml 3  3   m   +  4 4 3

3mu 1 9   m 2    4  3 16 

2

 m,

H m

3u  16  27    4   48 



36 u 43 

Ex.34 A uniform rod AB of mass m and length 5a is free to rotate on a smooth horizontal table about a pivot through P, a point on AB such that AP = a. A particle of mass 2m moving on the table strikes AB perpendicularly at the point 2a from P with speed v, the rod being at rest. If the coefficient of restitution between them is

Sol.

1 , find their speeds 4

immediately after impact. Let the point of impact be Q so that IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 29

ROTATIONAL DYNAMICS PQ = 2a Let P be the point of pivot that AP = a 5a Pm

A

Q

C

a

B

v 2m

Before Collision

Let the velocities of point, Q and the particle after impact be vq and vp respectively then from momentum conservation about point P. Li = L f 2a(2mv) = Ip + (2a) (2mvp) ...(i) Vq  P 2

IP 

1  5a   3a  m   m    2 3  2

2

C

use parallel    axis theorem

13 ma 2 3 use equation (ii) in equation (i) 

...(ii)

13 ma2  3 12(v – vp) = 13a

....(iii)

Q

3a/2

Vp

After Collision

4ma(v – vp) =

coefficient of restitution e =

velocity of seperation velocity of approach

1 vq  vp  4 v

vq – vp =

v 4

...(iv)

vq = 2a ...(v) Put value of  from eq (iii) to equation (v)  12  vq  2   (v  vp )  13  So now from equation (iv) v 24 83 v ( v – vp )  v p   vp  13 4 148 So in this way we get  

15 v 37 a

Ex.35 A person of mass m stands at the edge of a circular platform of radius R and moment of inertia. A platform is at rest initially. But the platform rotate when the person jumps off from the platform tangentially with velocity u with respect to platfrom. Determine the angular velocity of the platform. Sol. Let the angular velocity of platform is . Then the velocity of person with respect to ground v. vmD = vmG – VDG u = vm + R vm = u –  R Now from angular momentum conservation Li = L f 0 = mvmR – I  

I  = m (u –  R) . R

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R M R

u

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Page # 30

ROTATIONAL DYNAMICS muR

=

I  mR 2

Ex.36 Consider the situation of previous example. If the platform is rotating intially with angular velocity 0 and then person jumps off tangentially. Determine the new angular velocity of the platform. Sol. Let the angular velocity of platfrom after jumps off the mass is . Then velocity Of man. 0

R

R

u

Initially

vm = vmp + vp vm = u – R From Angular momentum conservation (I + mR2) 0 = I  – m (u –  R) R I0 + mR2 0 = I  – m u R + m  R2 

10.



(I  mR 2 )0  mu R (I  mR 2 )

ANGULAR IMPULSE t2 

The angular impulse of a torque in a given time interval is defined as

t1

 dt

Here,   is the resultant torque acting on the body. Further, since   dL       dt  d L dt t2     dt = angular impulse = L 2 – L 1 or

t1

Thus, the angular impulse of the resultant torque is equal to the change in angular momentum. Let us take few examples based on the angular impulse. Ex.37 Figure shows two cylinders of radii r1 and r2 having moments of inertia I1 and I2 about their respective axes. Initially, the cylinders rotate about their axes with angular speeds 1 and 2 as shown in the figure. The cylinders are moved closer to touch each other keeping the axes parallel. The cylinders first slip over each other at the contact but the slipping finally ceases due to the friction between them. Find the angular speeds of the cylinders after the slipping ceases. 2 1 I2 r I1 r 1

Sol.

2

When slipping ceases, the linear speeds of the points of contact of the two cylinders will be equal. If  1 ' and  2 ' be the respective angular speeds, we have  1 ' r1   2 ' r2

.....(i)

The change in the angular speed is brought about by the frictional force which acts as long as the slipping exists. If this force f acts for a time t, the torque on the first cylinder is fr1 and that on the second is f r2. Assuming 1 r1 > 2 r2, the corresponding angular impluses are – f r1 t and f r2 t. We, therefore, have

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Page # 31

ROTATIONAL DYNAMICS – f r1 t = I 1 (  1 '   1 ) and

f r2 t = I 2 (  2 '   2 )

or,

I1 I ( 1 '– 1 ) = 2 ( 2 '– 2 ) r1 r2

...(ii)

Solving (i) and (ii),  1' =

I1 1r2  I2  2r1 I2r12  I1r22

r2 and  2 ' 

I1 1r2  I2  2r1 I2r12  I1r22

r1

Kinetic Energy of a rigid body rotating about a fixed axis. Suppose a rigid body is rotating about a fixed axis with angular speed . Then, kinetic energy of the rigid body will be : K=

 i

1 2 = 2

1 mi v i2 = 2 2 i i

m r

=

i

 i

 ri

1 mi (ri ) 2 2

mi

1 2 I 2

(as

2 i i

m r

 I)

i

1 2 I 2 Sometimes it is called the rotational kinetic energy. Ex.38 A uniform rod of mass m and length  is kept vertical with the lower end clamped. It is slightly pushed to let it fall down under gravity. Find its angular speed when the rod is passing through its lowest position. Neglect any friction at the clamp. What will be the linear speed of the free end at this instant? Sol. As the rod reaches its lowest position, the centre of mass is lowered by a distance . Its gravitational potential energy is decreased by mg. As no energy is lost against friction, this should be equal to the increase in the kinetic energy. As the rotation occurs about the horizontal axis through the clamped end, the moment of inertia is I = m 2/3. Thus, Thus, KE =

1 2 I  mg 2 or

1  m 2  2   = mg   2  3 

6g 

=

The linear speed of the free end is v =  =

11.

6g

COMBINED TRANSLATIONAL AND ROTATIONAL MOTION OF A RIGID BODY : We have already learnt about translational motion caused by a force and rotational motion about a fixed axis caused by a torque. Now we are going to discuss a motion in which body undergoes translational as well as rotational motion. Rolling is an example of such motion. If the axis of rotation is moving then the motion is combined translational and rotational motion. To understand the concept of combined translational and rotational motion we consider a uniform disc rolling on a horizontal surface. Velocity of its centre of mass is Vcom and its angular speed is  as shown in figure.

v R

A

v

Let us take a point A on the disc and concentrate on its motion. IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 32

ROTATIONAL DYNAMICS

Path of point A with respect to ground will be a cycloid as shown in figure. A

 v A

A

A A Motion of point A with respect to center of mass is pure rotational while center of mass itself is moving in a straight line. So for the analysis of rolling motion we deal translational motion seperately and rotational motion seperately and then we combine the result to analyses the over all motion. The velocity of any point A on the rigid body can be obtained as    VA  VCOM  VA COM  | VCOM |  V

 | V A.COM | r in the direction  to line OA   Thus, the velocity of point A is the vector sum of VCOM and VP.COM as shown in figure  VA O

r A

VCOM

Important points in combined Rotational + translation motion : 1.

Velocity of any point of the rigid body in combined R + T motion is the vector sum of v(velocity of centre of mass) and r for example A disc of radius r has linear velocity v and angular velocity  as shown in figure then find velocity of point A. B, C, D on the disc C

r B

D

v

 A We divide our problem in two parts (1) Pure Rotational + (2) Pure Translational about centre of mass. C

r

r

v

D

B r

+

v

v

r

r

A

v

Then combine the result of above both

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Page # 33

ROTATIONAL DYNAMICS r

C

D

v

2

(

r )2

v  r

v

B 2

v  (r )2

A v – r r

In combined rotational and translational motion angular velocity of any point of a rigid body with respect to other point in the rigid body is always same. For example : C 2v v 2v

v

B

D

v

v

D 2r

2.

(v = r)

r v=0

2v v

A

2v

A

2v Now for DA

DA =

2r

=

v r

2v

C

2R

For CA :

A

CA

2v v = = 2r r

For DB :

2v B

2r

D 2v

2v

3.

DB =

2r

vDB =

2v

2v 2v

v

=

r

Distance moved by the centre of mass of the rigid body in one full rotation is 2R.

v 

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Page # 34

ROTATIONAL DYNAMICS

This can be shown as under : In one rotation angular displacement  = 2 = t  2  s = v . T = ( R)    2R  

In forward slipping

s > 2R

and in backward slipping 4.

s < 2R

(as v > R) (as v < R)

The speed of a point on the circumference of the body at any instant t is 2R sin

Proof : vxp = v – v cos  vyp = v sin 

= v[1 – cos ]

 | v p |  v 2 sin2   v 2 (1 – cos ) 2

v= =

t 2

v

v = R

2 v 2 – 2 v 2 cos 

 v

P

2v(1 – cos )1/ 2

 ωt   t  = 2v sin    = 2 v sin  = 2R  sin    2  2   2 5.

The path of a point on circumference is a cycloid and the distance moved by this point in one full rotation is 8R. A3 A2

A4

A5

A1

In the figure, the dotted line is a cycloid and the distance A1 A2 ......A5 is 8R. This can be proved as under. According to point (3), speed of point A at any moment is,  t  vA = 2R sin   2

Distance moved by A in time dt is,

c

c ds = vA

 t  dt = 2R sin   dt 2

Therefore, total distance moved in one full rotation is,

v  A t=0

A

t=t

T  2 / 

 ds

S=

0

T  2 / 

or

S=

 0

 t  2 R sin  dt  2

On integration we get,

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s = 8R

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Page # 35

ROTATIONAL DYNAMICS 6.

x and y coordinates of the bottommost point at any time t. At time t the bottommost point will rotate an angle  = t with respect to the centre of the disc C. The centre C will travel a distance s = vt. y In the figure, PQ = R sin  = R sin t CQ = R cos = R cost Coordinates of point P at time t are,

C P

x = OM – PQ = vt – R sin t and y = CM – CQ = R – R cos t  (x, y)  (vt – R sin t, R – R cos t) 11.1

R

C 

Q

M

O s=vt

Angular momentum of a rigid body in combined rotation and translation

COM  r0

 v0

O Let O be a fixed point in an inertial frame of reference. Angular momentum of the body about O is.     L  L cm  M( r 0  v 0 )

 The first term L cm represents the angular momentum of the body as seen from the centre of   mass frame. The second term M( r 0  v 0 ) equals the angular momentum of centre of mass

Ex.39 A circular disc of mass m and radius R is set into motion on a horizontal floor with a v in clockwise direction R as shown in figure. Find the magnitude of the total angular momentum of the disc about bottommost point O of the disc.     ...(i) L  L cm  m( r0  v 0 )  v Here, L cm  I (perpendicular to paper inwards)  1 2  v  O   mR    2  R  r0 1  mvR 2   90º  and m( r0  v 0 )  mRv (perpendicular to paper inwards) v0 Since, both the terms of right hand side of Eq. (i) are in the linear speed v in the forward direction and an angular speed  

Sol.

same direction.  or

 1 | L | mvR  mvR 2  3 | L | mvR 2 IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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O

Ans.

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Page # 36 11.2

ROTATIONAL DYNAMICS

Kinetic Energy of a Rolling Body If a body of mass M is rolling on a plane such that velocity of its centre of mass is V and its angular speed is , its kinetic energy is given by KE =

1 1 Mv 2  I  2 2 2

I is moment of inertia of body about axis passing through centre of mass. In case of rolling without slipping. KE =

1 1 M  2 R2 + I  2 [ v = R] 2 2

1 MR2  I  2 2

=

1 Ic  2 2

Ic is moment of inertia of the body about the axis passing through point of contact. Ex.40 A uniform rod of mass M and length a lies on a smooth horizontal plane. A particle of mass m moving at a speed v perpendicular to the length of the rod strikes it at a distance a/4 from the centre and stops after the collision. Find (a) the velocity of the centre of the rod and (b) the angular velocity of the rod about its centre just after the collision.

a/4

A

a

r0 A

v

Sol.

(a)

(b)

The situation is shown in figure. Consider the rod and the particle together as the system. As there is no external resultant force, the linear momentum of the system will remain constant. Also there is no resultant external torque on the system and so the angular momentum of the system about any line will remain constant. Suppose the velocity of the centre of the rod is V and the angular velocity about the centre is . (a) The linear momentum before the collision is mv and that after the collision is MV. Thus, mv = MV, or V =

m v M

(b) Let A be the centre of the rod when it is at rest. Let AB be the line perpendicular to the plane of the figure. Consider the angular momentum of "the rod plus the particle" system about AB. Initially the rod is at rest. The angular momentum of the particle about AB is L = mv(a/4) After the collision, the particle mass to rest. The angular momentum of the rod about A is

    L  L cm  M( r 0  V ) As

  r 0 || V ,

  r0  V  0   L  L cm

Thus,

Hence the angular momentum of the rod about AB is

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Page # 37

ROTATIONAL DYNAMICS

Ma 2  12

L  I 

Thus,

mva Ma 2   4 12

or,

=

3mv Ma

Ex.41 A uniform rod of length  lies on a smooth horizontal table A particle moving on the table has a mass m and a speed v before the collision and it sticks to the rod after the collision. The rod has a mass M then find out. (a) (b) Sol.

The moment of inertia of the system about the vertical axis passing through the centre of mass C after the collision. The velocity of the centre of mass C and the angular velocity of the system about the centre of mass after the collision. Figure shows the situation of system just before and just after collision. Initially the centre of mass of the rod is at point O. After collision when the particle sticks to the rod. Centre of mass is shifted from point O to C as shown in figure. Now the system is rotated about axis passing through C

v

A 1 

m /2

v'

C

O

M

M 2(m  M)

2 

O

m 2(M  m)

 /2

M Before collision After collision Now from linear momentum conservation mv = (M + m) v (a)

v' 

mv Mm

Let us assume that moment of inertia of the system about C is 1. Then I  I(rod)C  I(part ) C

I  I0  M 22  m 21 M 2 Mm2  2 mM2  2 M(M  4m) 2   I   2 2 12 12(m  M) 4(m  M) 4(m  M) From Angular momentum conservation about A Li = L f 0 + 0 = I  – (m + M) v 1  I = (m + M) v 1 Put the value of I, v, & 1 we get I

(b)

= 11.3 (A)

6mv (M  4m)

Acceleration of a point on the circumference of the body in R + T motion : Both  & v are constant : C  2r

r

 2r  2r 2  r IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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D

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Page # 38 (B)

ROTATIONAL DYNAMICS

When  is constant and v is variable. 

v1, a

v2= v1+at

t=0 t=t So acceleration of different point on the body is given by following figure.

a 2

a

2

 2r

+

 r  2r a 2  r

=

 2r

(C)

a

a (Combined R + T)

(Rotational)

(Translational)

 2r

 r

 2r

When  is variable and v is constant :    i  t  i ,

v

v

t=0 t=t So acceleration of different point on the body is given by following way R

 2r

R

 2R

 2r  2r

2

2

 R

 R

 2r

R

R

(D)

a=0

 2R

When both  & v are variable :  i,   f   i  t v2=v1+at

v1, a

 time t =0 time t =t Now the net acceleration of different points on the rigid body is given by following way. R a R R R  2r  2R  2r  2R = a + a 2  2R 2  r  R R (Translational)

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R

R

(Rotational)

R

a

(combined Rotational + Translational)

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Page # 39

ROTATIONAL DYNAMICS

Ex.42 A force F acts at the centre of a thin spherical shell of mass m and radius R. Find the acceleration of the shell if the surface is smooth. N Sol.  Force F, mg & N passes through centre so net = 0, i.e., body is in rotational equilibrium R  F But F net  F so body moves with constant acceleration a = m

F

mg

Ex.43 In a previous problem if force F applied at a distance x above the centre then find out linear and angular acceleation. Sol. This force F translate the body linearly as well as rotate it. So, Net toruqe about O it 0 = Fx N From rotational motion 0 = I  F  Fx x R   3Fx a I 2MR2   O 2 MR 2 mg 3 smooth From linear motion of sphere F = ma 

a=

F m

Ex.44 A rigid body of mass m and radius r starts coming down an inclined plane of inclination . Then find out the acceleration of centre of mass if friciton is absent. Sol. Friction is absent so body is moving down the incline with out rolling so acceleration of centre of mass is g sin   in gs 

12.

UNIFORM PURE ROLLING Pure rolling means no relative motion (or no slipping at point of contact between two bodies.) For example, consider a disc of radius R moving with linear velocity v and angular velocity  on a horizontal ground. The disc is said to be moving without slipping if velocities of points P and Q (shown in figure b) are equal, i.e.,

v

COM

 (a)

 R

P

v

Q (b)

vp = vQ or v – R = 0 or v = R If vp > vQ or v > R, the motion is said to be forward slipping and if vp < vQ < R, the motion is said to be backward slipping. Now, suppose an external force is applied to the rigid body, the motion will no longer remain uniform. The condition of pure rolling on a stationary ground is, a = R Thus, v = R, a = R is the condition of pure rolling on a stationary ground. Sometime it is simply said rolling. Note : We can represent the moment of inertia of a different rigid body in a following way. I = CMR2 1 value of C = 1 for circular ring (R), C = for circular disc (D) and solid cylinder (S.C.) 2 2 2 C= for Hollow sphere (H.S) , C= for solid sphere (S.S) 3 5 IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 40

ROTATIONAL DYNAMICS

Ex.45 A rigid body I = CMR2 is set into a motion on a rough horizontal surface with a linear speed v0 in the forward direction at time t = 0 as shown in figure. After what time slipping finally stop and pure rolling starts. Find the linear speed of the body after it starts pure rolling on the surface. v0

O R

at t = 0

Sol.

According to the given condition in problem the point P in the body move with speed v0 while the point Q on the ground is at rest. So the friciton acts on the body is in backward direction which gives the resultant torque on the body and increase the angualr speed  as shown in figure. 1 v0

O f

P

(kinetic)

Q at t = 0

O  1R f (kinetic)

v0

v1

1

v1   1R O v1

Q at t = t1

v1 R v1

R friction static

v1

Q at t = t

As shown in above figure initially v > R so forward slipping takes place. After introducing the friciton speed decreases and  increases and at time t = t the relation v = r  is satisfied. Therefore pure rolling starts. Initially the friciton is kinetic untill the motion is in slipping condition. Afterwards at v = r fricition is static. We divide the above problem in two parts. (1)

(2)

Translational Motion : Linear acceleration a = – g So after time t, v = v0 – gt ...(1) Rotational Motion : From net = I  Only friction force is responsible for providing torque. So torque about O is f. R = I  mgR = CmR2 ...(2)  is angular acceleration of the body μg

from eq. (2) from

=

CR

f = i +  t  = t   =

g .t CR

=

v at pure rolling condition. R

So,

v=

μgt C from eq. (1) & (3)

...(3)

v 0C

μgt

 v0 – gt =

C

 t=

μg(1  C)

...(4)

Equation (4) gives the time after the pure rolling starts.

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Page # 41

ROTATIONAL DYNAMICS Put the value from eq. (4) to eq. (1) v = v0 –

v0 v 0C  v= 1 C (1  C)

...(5)

Equation (5) gives the linear speed at pure rolling situation. Alternate solution : Net torque on the body about the bottom most point A is zero. Therefore angular momentum of the body will remain conserved about the bottom most point Net torque about A A = 0 v  from Angular momentum conservation Li = Lf  R mv0R = I + mvR v0

v

v mv0R = CmR2 + mvR R v0 = Cv + v

v=

A f

v0 1 C

A f

Ex.46 In the previous problem take rigid body a solid cylinder then find out the work done by friciton from time t = 0 to t = t (at v = r) Sol.

Let us suppose that in between time t = 0 to t = t cylinder displaced s. t=t t=0

v0

v  R

S Translational work done by friciton + Rotational work Done by friciton Now calculate each type of work done one by one (A)

Translational work done by friciton : for solid cylinder c =

from eq. (5) v 

v0 1

1 2

1 2

2 v0 3 2

2  2   v 0   ( v 0 ) – 2gs 3 

from eq. v 2f  ui2  2as

s=

5 v 20 18 g

Translation W.D by friciton = – f.s ( w.D) f T  – mg.

(B)

5 v 20 18 g

= –

5mv 20 18

Rotational W.D. by friciton : We known that =

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 I

=I ...(a)

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Page # 42

ROTATIONAL DYNAMICS

from  f 2 –  i2  2   2

2  2v 0  .     3R  I

Put I =

mR2 2

 =

v 02m 9

Rotation W.D by friciton W = . Wf R =

(C)

v 20 m 9

So total W.D. by friciton W = Wf

W= –

T

+ Wf

R

= –

v 2m 5 mv 20  0 18 9

mv 20 6

Alternative Method : from work – Energy Theorem work done by friciton = change in kinetic energy (W.D)f = K = kf – ki Now kf =

1 1 mv 2f + I 2 2 2

1  2v 0   kf = m  2  3 

2

1 mR 2  2v 0    + 2 2  3R 

kf =

mv 20 3

ki =

1 mv 20 2

2

2v 0    v f   3  

mv 20 1 mv 20 ( w.D)f  – – mv 20  6 3 2 To calculate work done mostly prefer alternative method. So., (w.D)f =

Ex.47 A solid sphere of radius r is gently placed on a rough horizontal ground with an initial angular speed 0 and no linear velocity. If the coefficient of friciton is , find the linear velocity v and angular velocity  at the end of slipping.

0

Sol.

m be the mass of the sphere. Since, it is a case of backward slipping, force of friction is in forward direction. Limiting friciton will act in this case. Net torque on the sphere about the bottommost IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 43

ROTATIONAL DYNAMICS point is zero. Therefore, angular momentum of the sphere will remain conserved about the bottommost point. Li = L f  I0 = I + mrv

12.1

or

2 2 2 mr  0 = mr 2   mr (r ) 5 5

=

v

0

fmax

0

2 2  0 and v = r = r 0 7 7

Pure rolling when force F act on a body : Suppose a force F is applied at a distance x above the centre of a rigid body of radius R, mass M and moment of inertia CMR2 about an axis passing through the centre of mass. Now, the applied force F can produces by itself (i) a linear acceleration a and (ii) an angular acceleration  If a = R, then there is no need of friction and force of friction f = 0, If a < R, then to support the linear momentum the force of friciton f will act in forward direction, Similarly, if a > R, then no support the angular motion the force of friciton will act in backward direction. So, in this case force of friction will be either backward, forward or even zero also. It all depends on M, I and R. For calculation you choose any direction of friction. Let we assume it in forward direction. Let, a = linear acceleration,   = angualr acceleration F from linear motion x a F + f = Ma ...(1) C from rotational motion. Fx – f R = I  f Fx – f R

Fx – f R = CMaR from eq. (1) and (2) F(x+r) = MaR (C + 1) a=

F(R  x ) MR(C  1)

a R ...(2)

= CMR2.

...(3)

Put the value from eq. (3) to eq. (1) f=

F(x – RC) R(C  1)

f should be   s mg for pure rolling Ex.48 Consider the arrangment shown in figure. The string is wrapped around a uniform cylinder which rolls without slipping. The other end of the string is passed over a massless, frictionless pulley to a falling weight. Determine the acceleration of the falling mass m in terms of only the mass of the cylinder M, the mass m and g. M

m

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Page # 44 Sol.

ROTATIONAL DYNAMICS

Let T be the tension the string and f the force of (static) friction, between the cylinder and the surface a1 = acceleration of centre of mas of cylinder towards right a2 = downward acceleration of block m  = angular acceleration of cylinder (clockwise) Equations of motion are : For block mg – T = ma2 ...(i) For cylinder, T + f = Ma1 ...(ii) 

( T – f )R 1 MR 2 2

...(iii)

The string attaches the mass m to the highest point of the cylinder, hence vm = vCOM + R Differentiating, we get a2 = a1 + R ...(iv) We also have (for rolling without slipping) a1 = R ...(v) Solving these equations, we get

a2 

8mg 3M  8m

Note : Work done by friction in pure rolling on a stationary ground is zero as the point of application of the force is at rest. Therefore, machanical energy can be conserved if all other dissipative forces are ignored. 12.2 Pure Rolling on an Inclined Plane: A rigid body of radius R, and mass m is released at rest from height h on the incline whose inclination with horizontal is  and assume that f  friciton is sufficient for pure rolling then. a = R and v = R a From figure mg sin  – f = ma ...(1)  in {Fnet = ma} gs m 2 a cmR . f.R = ...(2) R {Fnet = I} from eq. (1) & (2) g sin 1 c So body which have low value of C have greater acceleration. value of C = 1 for circular ring (R) 1 C= for circular disc (D) and solid cylinder (S.C.) 2 2 C= for Hollow sphere (H.S) 3 a=

2 for solid sphere (S.S) 5 So, descending order of a aS.S > aD = aS.C > aH.S. > aR and order of time of descend is ts.s < tD = ts.c < tH.S < tR

C=

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Page # 45

ROTATIONAL DYNAMICS Kinetic Eneregy : Work done by friction in pure rolling is zero. Therefore, Increase in kinetic energy = change in potential energy  K.E. = mgh i.e., kinetic energy is constant for all rigid body rolling down the incline. Requirement of Friction : From eq. f = Cma f

mg sin  1  1    C

...(2)

...(3)

from eq. (3) as the value of C increase requirement of friciton is increases. Ex.49 A cylinder of mass M is suspended through two strings wrapped around it as shown in figure. Find the tension in the string and the speed of the cylinder as it falls through a distance h. Sol. The portion of the strings between ceiling and cylinder are at rest. Hence the points of the cylinder where the strings leave it are at rest also. The cylinder is thus rolling without slipping on the strings. Suppose the centre of cylinder falls with an acceleration a. The angular acceleration of cylinder about its axis given by =

a R

...(i)

as the cylinder does not slip over the strings. The equation of motion for the centre of mass of cylinder is Mg – 2T = Ma and for the motion about the centre of mass it is

 MR2  MR2  , where I = 2T.R =  2  2  Ma MR2 a  2T= 2 2 R From (i) and (ii) on adding 2TR=

T

T

mg ...(ii)

Ma 3a  Ma ; g 2 2 2g a= 3 M 2g Mg  2T= .  T= 2 3 6 As the centre of cylinder starts moving from rest, the velocity after it has fallen a height h is given by

Mg =

 2g  v2 = 2  h or v = 3

4gh 3

Ex.50 A thin massless thread is wound on a reel of mass 3kg and moment of inertia 0.6 kgm2. The hub radius is R = 10 cm and peripheral radius is 2R = 20 cm. The reel is placed on a rough table and the friction is enough to prevent slipping. Find the acceleration of the centre of reel and of hanging mass of 1 kg.

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Page # 46

ROTATIONAL DYNAMICS

2R R

A

Sol.

Let,

a1 = acceleration of centre of mass of reel a2 = acceleration of 1 kg block  = angular acceleration of reel (clockwise) T = tension in the string and f = force of friction Free body diagram of reel is as shown below : (only horizontal forces are shown). Equations of motion are : T – f = 3a1 ...(i)

a1 T f

f T  f (2R) – T.R 0.2f – 0.1T   = – ...(ii) 3 6 I I 0.6 Free body diagram of mass is, Equation of motion is, 10 – T = a2 ...(iii) For no slipping condition, a1 = 2R or a1 = 0.2 ...(iv) and a2 = a1 – R or a2 = a1 – 0.1 ...(v) 

T a2

Solving the above five equations, we get 10N a1 = 0.27 m/s2 and a2 = 0.135 m/s2 Ex.51 Determine the maximum horizontal force F that may be applied to the plank of mass m for which the solid sphere does not slip as it begins to roll on the plank. The sphere has a mass M and radius R. The coefficient of static and kinetic friction between the sphere and the plank are s and k respectively. M R

F

m

Sol.

The free body diagrams of the sphere and the plank are as shown below : Writing equations of motion : For sphere : Linear acceleration  sMg  sg a1 = M

(  sMg)R 5  s g  2 2 R MR 2 5

a1

...(i)

 s Mg  s Mg

Angular acceleration 

a2 F

..(ii)

For plank : Linear acceleration

M R

B A

m

a2

a1  R F

F –  sMg ..(iii) m For no slipping acceleration of point B and A is same, a2 

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Page # 47

ROTATIONAL DYNAMICS so : a2 = a1 + R Solving the above four equation, we get 7   F   s g M  m  2  7   Thus, maximum value of F can be  s g M  m 2

Ex.52 Find out the maximum height attained by the solid sphere on a friciton less track as shown in figure.

0  v0

R

Sol.

v0 R

Let us assume that sphere attain a maximum height H on the track. 0

Final Position v=0

0 

v0 R

H

v0

R

Initial Position

As the sphere move upward speed is decreased due to gravity but there is no force to change the 0 (friction less track). So from energy conservation 1 1 1 mv 20  I 20 = mg Hmax + I 20 2 2 2 Hmax =

13.

v 20 2g

TOPPLING You might have seen in your practical life that if a force F is applied to a block A of smaller width it is more likely to topple down, before sliding while if the same force F is applied to an another block B of broader base, chances of its sliding are more compared to its toppling. Have you ever throught why it happens so. To understand it better let us take an example. F F B A Suppose a force F is applied at a height b above the base AE of the block. Further, suppose the friction f is sufficient to prevent sliding. In this case, if the normal reaction N also passes through C, then despite the fact that the block is in translational equilibium (F = f and N = mg), an unbalanced torque (due to the couple of forces F and f) is there. This torque has a tendency to topple the block about point E. To cancel the effect of this unbalanced torque the normal reaction N is shifted towards right a distance 'a' such that, net anticlockwise torque is equal to the net clockwise torque or N D B F C b f

E

A W=mg

Fb = (mg) a IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 48

ROTATIONAL DYNAMICS

or

a=

Fb mg N

N D

B C f

A

D

B

F

C

b

a

f

E mg

A

(a)

F b

E mg (b)

Now, as F or b (or both) are increased, distance a also increases. But it can not go beyond the right edge of the block. So, in extreme case (beyond which the block will topple down), the normal reaction passes through E as shown in figure. Now, if F or b are further increased, the block will topple down. This is why the block having the broader base has less chances of toppling in comparison to a block of smaller base. Because the block of larger base has more margin for the normal reaction to shift. Why the rolling is so easy on the ground.

N F

C f mg Because in this case the normal reaction has zero margin to shift. so even if the body is in translational equilibrium (F = f, N = mg) an unbalanced torque is left behind and the body starts rolling clockwise. As soon as the body starts rolling the force of friction is so adjusted (both in magnitude and direction) that either the pure rolling starts (if friciton is sufficient enough) or the body starts sliding. Let us take few examples related to toppling. Ex.53 A uniform block of height h and width a is placed on a rough inclined plane and the inclination of the plane to the horizontal is gradually increased. If  is the coefficient of friction then under condition the block will (A) slide before toppling : The block will slide when mg sin  > f  mg sin  >  mg cos   tan  >  i.e., block is at rest when tan . ...(1) (B)

Now suppose the friction f is sufficient to prevent sliding. Then we assume that N is shifted towards downward a distance x to prevent toppling Therefore. torque about O is zero. h =Nx 2 h mg sin . = mg cos .x 2 tan .h x= 2 Maximum value of x is a/2  f.

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N

h f xO a

 in s g m 

os gc m

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Page # 49

ROTATIONAL DYNAMICS

so to prevent toppling x  

a 2

tan .h  a/2 2

 tan  

a h

...(2)

So, the block topple before sliding from (1) & (2) s >

14.

a h

INSTANTANEOUS AXIS OF ROTATION The combined effects of translation of the centre of mass and rotation about an axis through the centre of mass are equivalent to a pure rotation with the same angular speed about an axis passing through a point of zero velocity. Such an axis is called the instantaneous axis of rotation. (IAOR). This axis is always perpendicular to the plane used to represent the motion and the intersection of the axis with this plane defines the location of instantaneous centre of zero velocity (IC).

v   IC For example consider a wheel which rolls without slipping. In this case the point of contact with the ground has zero velocity. Hence, this point represents the IC for the wheel. If it is imagined that the wheel is momentarily pinned at this point, the velocity of any point on the wheel can be found using v = r. Here r is the distance of the point from IC. Similarly, the kinetic energy of the body can be assumed to be pure rotational about IAOR or, P v P vP r  r vP  r   v r  IC

1 I IAOR 2 2 Rotation + Translation  Pure rotation about IAOR passing through IC K

KE = 14.1

1 1 mv 2COM  ICOM 2  2 2

KE 

1 I IAOR 2 2

Location of the IC If the location of the IC is unknown, it may be determined by using the fact that the relative position vector extending from the IC to a point is always perpendicular to the velocity of the point. Following three possibilities exist. (i) Given the velocity of a point (normally the centre of mass) on the body and the angular velocity of the body  If v and  are known, the IC is located along the line drawn perpendicular to v at P, such that v the distance from P to IC is, r  . Note that IC lie on that side of P which causes rotation    about the IC, which is consistent with the direction of motion caused by  and v . IIT-JEE|AIEEE CBSE|SAT|NTSE OLYMPIADS

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Page # 50

ROTATIONAL DYNAMICS

P

v

r

IC

Ex.54 A rotating disc moves in the positive direction of the x-axis. Find the equation y(x) describing the position of the instantaneous axis of rotation if at the initial moment the centre c of the disc was located at the point O after which it moved with constant velocity v while the disc started rotating counter clockwise with a constant angular acceleration . The initial angular velocity is equal to zero. y

O

Sol.

x x   t  and v v The position of IAOR will be at a distance

y

t

y

v 

or

y

x

c v

IC

y c v 

O

v x v

x

x

v2 v2  constant or xy  x  This is the desired x-y equation. This equation represents a rectangular hyperbola. or

y

(ii) Given the lines of action of two non-parallel velocities   Consider the body shown in figure where the line of action of the velocities v A and v B are known. Draw perpendiculars at A and B to these lines of action. The point of intersection of these perpendiculars as shown locates the IC at the instant considered. A

 vB

 vA IC B

(iii) Given the magnitude and direction of two parallel velocities When the velocities of points A and B are parallel and have known magnitudes vA and vB then the location of the IC is determined by proportional triangles as shown in figure.  A In both the cases, and In fig. (a)

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v rA,IC  A  v rB,IC  B  rA, I C + rB, I C = d

vA

d

IC

IC  vA

A

d  vB

B

B

(a)

(b)

 vB

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Page # 51

ROTATIONAL DYNAMICS

and in fig (b) rB, I C – rA, I C = d As a special case, if the body is translating, vA = vB and the IC would be located at infinity, in which case  = 0. Ex.55 A uniform thin rod of mass m and length l is standing on a smooth horizontal surface. A slight disturbance causes the lower end to slip on the smooth surface and the rod starts falling. Find the velocity of centre of mass of the rod at the instant when it makes an angle  with horizontal. Sol. As the floor is smooth, mechanical energy of the rod will remain conserved. Further, no horizontal force acts on the rod, hence the centre of mass moves vertically downwards in a straight line. Thus velocities of COM and the lower end B are in the direction shown in figure.   The location of IC at this instant can be found by drawing perpendiculars to v C and v B at respective points. Now, the rod may be assumed to be in pure rotational motion about IAOR passing through IC with angular speed . A

COM IC

 vC

h

l (1  sin ) 2

l sin 2 

 vB

B

Applying conservation of mechanical energy. Decrease in gravitational potential energy of the rod = increase in rotational kinetic energy about IAOR mgh 

1 I IAOR 2 2

 2 1  ml 2 ml 2 l 2 or mg 2 (1  sin )  2  12  4 cos    

Solving this equation, we get 

12g(1  sin ) l (1  3 cos 2 )

 l  | v C |   cos   2 

Now,

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3gl (1  sin ) cos 2  (1  3 cos 2 )

Ans.

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Page # 52

ROTATIONAL DYNAMICS

(Objective Problems)

Exercise - I (A) MOMENT OF INERTIA

1. The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire through the centre is (A) Mr2

(C)

1 2 Mr 4

(B)

1 2 Mr 2

(D)

2 2 Mr 5

2. Let IA and IB be moments of inertia of a body about two axes A and B respectively. The axis A passes through the centre of mass of the body but B does not. (A) IA < IB (B) If IA < IB, the axes are parallel. (C) If the axes are parallel, IA < IB (D) If the axes are not parallel, IA  IB 3. Three bodies have equal masses m. Body A is solid cylinder of radius R, body B is a square lamina of side R, and body C is a solid sphere of radius R. Which body has the smallest moment of inertia about an axis passing through their centre of mass and perpendicular to the plane (in case of lamina) (A) A (B) B (C) C (D) A and C both 4. For the same total mass which of the following will have the largest moment of inertia about an axis passing through its centre of mass and perpendicular to the plane of the body (A) a disc of radius a (B) a ring of radius a (C) a square lamina of side 2a (D) four rods forming a square of side 2a 5. Two rods of equal mass m and length l lie along the x axis and y axis with their centres origin. What is the moment of inertia of both about the line x = y : ml 2 ml 2 (B) (A) 3 4 2 ml ml 2 (C) (D) 12 6 6. Moment of inertia of a rectangular plate about an axis passing through P and perpendicular to the plate is I. Then moment of PQR about an axis perpendicular to the plane of the plate :

7. A thin uniform rod of mass M and length L has its moment of inertia I1 about its perpendicular bisector. The rod is bend in the form of a semicircular arc. Now its moment of inertia through the centre of the semi circular arc and perpendicular to its plane is I2. The ratio of I1 : I2 will be _________________ (A) < 1 (B) > 1 (C) = 1 (D) can’t be said 8. Moment of inertia of a thin semicircular disc (mass = M & radius = R) about an axis through point O and perpendicular to plane of disc, is given by : O R

1 1 2 MR 2 (B) MR 4 2 1 2 (C) MR (D) MR2 8 9. A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given by I = 2x2 – 12x + 27 The x-coordinate of centre of mass is (A) x = 2 (B) x = 0 (C) x = 1 (D) x = 3

(A)

10. A square plate of mass M and edge L is shown in figure. The moment of inertia of the plate about the axis in the plane of plate passing through one of its vertex making an angle 15° from horizontal is. axis 15° L

L

ML2 (A) 12

(C)

7 ML2 12

(B)

11ML2 24

(D) none

Q

P

R

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11. Consider the following statements Assertion (A) : The moment of inertia of a rigid body reduces to its minimum value as compared to any other parallel axis when the axis of rotation 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No:0744-2439051,0744-2439052,0744-2439053,www.motioniitjee.com, [email protected]

Page # 53

ROTATIONAL DYNAMICS passes through its centre of mass. Reason (R) : The weight of a rigid body always acts through its centre of mass in uniform gravitational field. Of these statements : (A) both A and R are true and R is the correct explanation of A (B) both A and R are true but R is not a correct explanation of A (C) A is true but R is false (D) A is false but R is true Question No. 12 to 14 (3 questions) The figure shows an isosceles triangular plate of mass M and base L. The angle at the apex is 90°. The apex lies at the origin and the base is parallel to X - axis. Y  M X 12. The moment of inertia of the plate about the z-axis is

ML2 ML2 (B) 12 24 ML2 (C) (D) none of these 6 13. The moment of inertia of the plate about the x-axis is (A)

(A)

ML2 8

ML2 (C) 24

(B)

ML2 32

ML2 (D) 6

14. The moment of inertia of the plate about its base parallel to the x-axis is ML2 ML2 (B) 18 36 ML2 (C) (D) none of these 24 15. The moment of inertia of the plate about the y-axis is

17. One end of a uniform rod of mass m and length I is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity . The force exerted by the clamp on the rod has a horizontal component (A) m2 l (B) zero 1 2 (C) mg (D) m  2 18. A rod of length 'L' is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod when it is in vertical position is (A)

2g L

(B)

3g L

(C)

g 2L

(D)

g L

(B) TORQUE AND PURE ROTATIONAL MOTION 19. A horizontal force F = mg/3 is applied on the upper surface of a uniform cube of mass ‘m’ and side ‘a’ which is resting on a rough horizontal surface having s = 1/2. The distance between lines of action of ‘mg’ and normal reaction ‘N’ is : (A) a/2 (B) a/3 (C) a/4 (D) None 20. A man can move on a horizontal plank supported symmetrically as shown. The variation of normal reaction on support A with distance x of the man from the end of the plank is best represented by : x=0 A B

(A)

ML2 6 ML2 (C) 24

(A)

(B)

ML2 8

1m

N

1m

N

(A)

(B) x

x

N

(D) none of these

SECTION (D) ; FIXED AXIS 16. A body is rotating uniformly about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is (A) vertical (B) horizontal and skew with the axis (C) horizontal and intersecting the axis (D) none of these

4m

N

(C)

(D) x

x

21. A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :

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Nurturing potential through education

Page # 54

ROTATIONAL DYNAMICS

(A) Downwards at any point between A and B. (B) Downwards at mid point of AB. (C) Downwards at a point C such that AC = 1m. (D) Downwards at a point D such that BD = 1m. 22. A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is : l

A

B

l

(A) left half (B) right half (C) both applies equal pressure (D) the answer depend upon coefficient of friction 26. Consider the following statements Assertion (A) : A cyclist always bends inwards while negotiating a curve Reason (R) : By bending he lowers his centre of gravity Of these statements, (A) both A and R are true and R is the correct explanation of A (B) both A and R are true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true 27. A solid cone hangs from a frictionless pivot

C

mg 2 mg (B) 3 3 mg (C) (D) mg 2 23. In an experiment with a beam balance on unknown mass m is balanced by two known mass m is balanced by two known masses of 16 kg and 4 kg as shown in figure.

(A)

l1

l1

l2

m

16kg

at the origin O, as shown. If i , j and k are unit vectors, and a, b, and c are positive constants, which of the following forces F applied to the rim of the cone at a point P results in a torque  on the cone with a negative component Z ? z k o

i x

l2

m

y j

4kg

b

The value of the unknown mass m is (A) 10 kg (B) 6 kg (C) 8 kg (D) 12 kg

(A) F = a k , P is (0, b, –c) (B) F = –a k , P is (0, –b, –c)

24. A uniform cube of side ‘b’ and mass M rest on a rough horizontal table. A horizontal force F is applied normal to one of the face at a point, at a height 3b/4 above the base. What should be the coefficient of friction () between cube and table so that is will tip about an edge before it starts slipping?

(C) F = a j , P is (–b, 0, –c) (D) None 28. A rod is hinged at its centre and rotated by applying a constant torque starting from rest. The power developed by the external torque as a function of time is : Pext

F b

Pext

(A)

3b/4

(B) time

2 1 (A)   (B)   3 3 3 (C)   (D) none 2 25. A homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is :

Nurturing potential through education

c

time

Pext

Pext

(C)

(D) time

time

29. A pulley is hinged at the centre and a massless thread is wrapped around it. The thread is pulled with a constant force F starting from rest. As the time increases, F

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Page # 55

ROTATIONAL DYNAMICS (A) its angular velocity increases, but force on hinge remains constant (B) its angular velocity remains same, but force on hinge increases (C) its angular velocity increases and force on hinge increases (D) its angular velocity remains same and force on hinge is constant. 30. The angular momentum of a flywheel having a moment of inertia of 0.4 kg m2 decreases from 30 to 20 kg m2/s in a period of 2 second. The average torque acting on the flywheel during this period is : (A) 10 N.m (B) 2.5 N.m (C) 5 N.m (D) 1.5 N.m 31. A rod hinged at one end is released from the horizontal position as shown in the figure. When it becomes vertical its lower half separates without exerting any reaction at the breaking point. Then the maximum angle ‘’ made by the hinged upper half with the vertical is : C B A

B

(A) 30°

(B) 45°

B

C (C) 60° (D) 90°

32. A block of mass m is attached to a pulley disc of equal mass m, radius r by means of a slack string as shown. The pulley is hinged about its centre on a horizontal table and the block is projected with an initial velocity of 5 m/s. Its velocity when the string becomes taut will be

(C) ANGULAR MOMENTUM 34. A particle moves with a constant velocity parallel to the X-axis. Its angular momentum with respect to the origin. (A) is zero (B) remains constant (C) goes on increasing (D) goes on decreasing. 35. A thin circular ring of mass 'M' and radius 'R' is rotating about its axis with a constant angular velocity . Two objects each of mass m, are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velcoity. (A)

M (M  m)

(B)

M (M  2m)

(C)

M (M – 2m)

(D)

(M  3m) M

36. A person sitting firmly over a rotating stool has his arms streatched. If he folds his arms, his angular momentum about the axis of rotation (A) increases (B) decreases (C) remains unchanged (D) doubles. 37. A small bead of mass m moving with velocity v gets threaded on a stationary semicircular ring of mass m and radius R kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system?

R

O

(A) 3 m/s (C) 5/3 m/s

(B) 2.5 m/s (D) 10/3 m/s

33. A particle starts from the point (0m, 8m) and moves with uniform velocity of 3 i m/s. After 5 seconds, the angular velocity of the particle about the origin will be : y 3m/s

(A)

x

24 rad / s (C) 289

(B)

38. A man, sitting firmly over a rotating stool has his arms streched. If he folds his arms, the work done by the man is (A) zero (B) positive (C) negative (D) may be positive or negative. 39. A particle of mass 2 kg located at the position

8m O

(A) v/R (C) v/2R

v m (B) 2v/R (D) 3v/R

8 rad / s (D) 17

( i  j ) m has a velocity 2(  i – j  k ) m/s. Its angular momentum about z-axis in kg-m2 /s is : (A) zero (B) +8 (C) 12 (D) – 8 40. A thin uniform straight rod of mass 2 kg and length 1 m is free to rotate about its upper end when at rest. It receives an impulsive blow of 10

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Page # 56

ROTATIONAL DYNAMICS

Ns at its lowest point, normal to its length as shown in figure. The kinetic energy of rod just after impact is

10 NS (A) 75 J (C) 200 J

(B) 100 J (D) none

Question No. 44 & 45 (2 questions) A uniform rod is fixed to a rotating turntable so that its lower end is on the axis of the turntable and it makes an angle of 20° to the vertical. (The rod is thus rotating with uniform angular velocity about a vertical axis passing through one end.) If the turntable is rotating clockwise as seen from above. 20°

41. A ball of mass m moving with velocity v, collide with the wall elastically as shown in the figure. After impact the change in angular momentum about P is : P

d 

(A) 2 mvd (C) 2 mvd sin

(B) 2 mvd cos (D) zero

42. A uniform rod of mass M is hinged at its upper end. A particle of mass m moving horizontally strikes the rod at its mid point elastically. If the particle comes to rest after collision find the value of M/m = ?

v m M

(A) 3/4 (C) 2/3

(B) 4/3 (D) none

43. A child with mass m is standing at the edge of a disc with moment of inertia I, radius R, and initial angular velocity . See figure given below. The child jumps off the edge of the disc with tangential velocity v with respect to the ground. The new angular velocity of the disc is

44. What is the direction of the rod’s angular momentum vector (calculated about its lower end) (A) vertically downwards (B) down at 20° to the horizontal (C) up at 20° to the horizontal (D) vertically upwards 45. Is there a torque acting on it, and if so in what direction? (A) yes, vertically (B) yes, horizontally (C) yes at 20° to the horizontal (D) no 46. One ice skater of mass m moves with speed 2v to the right, while another of the same mass m moves with speed v toward the left, as shown in figure I. Their paths are separated by a distance b. At t = 0, when they are both at x = 0, they grasp a pole of length b and negligible mass. For t > 0, consider the system as a rigid body of two masses m separated by distance b, as shown in figure II. Which of the following is the correct formula for the motion after t = 0 of the skater initially at y = b/2 ? y

y

m 2v b/2

v

b

x

x

t=0

(t FC = FD (D) none Sol.

Q.15 The maximum separation between their centres after their first collision (A) 4R (B) 6R (C) 8R (D) 12R Sol.

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GRAVITATION

Page # 32

Q.17 A satellite of the earth is revolving in circular orbit with a uniform velocity V. If the gravitational force suddenly disappears, the statellite will (A) continue to move with the same velocity in the same orbit (B) move tangentially to the original orbit with velocity V (C) fall down with increasing velocity (D) come to a stop somewhere in its original orbit Sol.

Q.20

Select the correct choice(s) :

(A) The gravitational field inside a spherical cavity, within a spherical planet must be non zero and uniform. (B) When a body is projected horizontally at an appreciable large height above the earth, with a velocity less than for a circular orbit, it will fall to the earth along a parabolic path (C) A body of zero total mechanical energy placed in a gravitational field will escape the field (D) Earth’s satellite must be in equatorial plane.

Q.18 A satellite revolves in the geostationary orbit but in a direction east to west. The time interval between its successive passing about a point on the

Sol.

equator is (A) 48 hrs

(B) 24 hrs

(C) 12 hrs

(D) never

Sol.

Q.21 A satellite of mass m, initially at rest on the earth, is launched into a circular orbit at a height

Q.19

Tw o po int ma sses o f ma ss 4m and m

respectively separated by d distance are revolving under mutual force of attraction. Ratio of their kinetic energies will be (A) 1 : 4

(B) 1 : 5

(C) 1 : 1

equal to the radius of the earth. The minimum energy required is (A)

3 mgR 4

(B)

1 mgR 2

Sol.

(D) 1 : 2

Sol.

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(C)

1 3 mgR (D) mgR 4 4

GRAVITATION

Page # 33 Sol.

Q.22 The figure shows the

A Energy

variation of energy with the orbit radius of a body in circular planetary motion. Find the

C B

curves A, B and C (A) A shows the kinetic energy, B the total energy and C the potential energy of the system (B) C shows the total energy, B the kinetic energy and A the potential energy of the system (C) C and A are kinetic and potential energies respectively and B is the total energy of the system (D) A and B are kinetic and potential energies and C is the total energy of the system Sol.

Q.23 When a satellite moves around the earth in a certain orbit, the quantity which remains constant is (A) angular velocity (B) kinetic energy (C) aerial velocity Sol.

(D) potential energy

Q.24 A satellite of mass 5M orbits the earth in a circular orbit. At one point in its orbit, the satellite explodes into two pieces, one of mass M and the other of mass 4M. After the explosion the mass M ends up travelling in the same circular orbit, but in opposite direction. After explosion the mass 4M is in (A) bound orbit (B) unbound orbit (C) partially bound orbit (D) data is insufficient to determine the nature of the orbit

Q.25 A satellite can be in a geostationary orbit around earth at a distance r from the centre. If the angular velocity of earth about its axis doubles, a satellite can now be in a geostationary orbit around earth if its distance from the center is r r r r (B) (C) (D) (A) 1/ 3 (4) (2)1/ 3 2 2 2 Sol.

Q.26 A planet of mass m is in an elliptical orbit about the sun (m t2

X

(C) t1 < t2

(D) nothing can be concluded

Y

(A) Z is total energy, Y is kinetic energy and X is potential energy (B) X is kinetic energy, Y is total energy and Z is potential energy

Sol.

(C) X is kinetic energy, Y is potential energy and Z is total energy (D) Z is kinetic energy, X is potential energy and Y is total energy Sol. Q.28 If U is the potential energy and K kinetic energy then |U| > |K| at (A) Only D (B) Only C (D) neither D nor C

(C) both D & C

Sol.

Q.31 Statement - I : Assuming zero potential at infinity, gravitational potential at a point cannot be positive.

Q.29 Satellites A and B are orbiting around the earth in orbits of ratio R and 4R respectively. The ratio of their areal velocities is (A) 1 : 2 Sol.

(B) 1 : 4

(C) 1 : 8

(D) 1 : 16

Statement - 2 : Magnitude of gravitational force between two particle has inverse square dependence on distance between two particles. (A) Statement - 1 is true, statement-2 is true and statement-2 is correct explanation for statement-1 (B) Statement -1 is true, statement-2 is true and statement - 2 is NOT the correct explanation for statement-1 (C) Statement - 1 is true, statement - 2 is false. (D) Statement - 1 is false, statement - 2 is true. Sol.

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GRAVITATION

Page # 35

Exercise - II 1. Assuming the earth to be a sphere of uniform density the acceleration due to gravity (A) at a point outside the earth is inversely proportional to the square of its distance from the center (B) at a point outside the earth is inversely proportional to its distance from the centre (C) at a point inside is zero (D) at a point inside is proportional to its distance from the centre Sol.

3. In side a hollow spherical shell (A) everywhere gravitational potential is zero (B) everywhere gravitational field is zero (C) everywhere gravitational potential is same (D) everywhere gravitational field is same Sol.

4. A geostationary satellite is at a height h above the surface of earth. If earth radius is R 2.Two masses m1 and m2 (m1 < m2) are released from rest from a finite distance. They start under their mutual gravitational attraction (A) acceleration of m1 is more than that of m2 (B) acceleration of m2 is more than that of m1 (C) centre of mass of system will remain at rest in all the references frame (D) total energy of system remains constant Sol.

R R

h

(A) The minimum colatitude on earth upto which the satellite can be used for communication is sin–1 (R/R + h) (B) The maximum colatitudes on earth upto which the satellite can be used for communication is sin–1 (R/R + h) (C) The area on earth escaped from this satellite is given as 2πR2(1 + sinθ) (D) The area on earth escaped from this satellite is given as 2πR2(1 + cosθ) Sol.

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GRAVITATION

Page # 36

7. An earth satellite is moved from one stable circular orbit to another larger and stable circular orbit. The following quantities increase for the satellite as a result of this change (A) gravitational potential energy (B) angular velocity (C) linear orbital velocity (D) centripetal acceleration Sol.

5. When a satellite in a circular orbit around the earth enters the atmospheric region, it encounters small air resistance to its motion. Then (A) its kinetic energy increases (B) its kinetic energy decreases (C) its angular momentum about the earth decreases (D) its period of revolution around the earth increases Sol.

6. A communications Earth satellite (A) goes round the earth from east to west (B) can be in the equatorial plane only (C) can be vertically above any place on the earth (D) goes round the earth from west to east Sol.

8. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth (A) the acceleration of S is always directed towards the centre of the earth (B) the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant (C) the total mechanical energy of S varies periodically with time (D) the linear momentum of S remains constant in magnitude Sol.

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GRAVITATION

9. If a satellite orbits as close to the earth’s surface as possible, (A) its speed is maximum (B) time period of its rotation is minimum (C) the total energy of the ‘earth plus satellite’ system is minimum (D) the total energy of the ‘earth plus satellite’ system is maximum Sol.

Page # 37 10. For a satellite to orbit around the earth, which of the following must be true ? (A) It must be above the equator at some time (B) It cannot pass over the poles at any time (C) Its height above the surface cannot exceed 36,000 km (D) Its period of rotation must be > 2π R / g where R is radius of earth Sol.

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GRAVITATION

Page # 38

(SUBJECTIVE PROBLEMS)

Exercise - III m

Q.1 Four masses (each of m) are placed at the vertices of a regular pyramid (triangular base) of side ‘a’. Find the work done by the system m while taking them apart so that they a form the pyramid of side ‘2a’. Sol.

m

m

Q.2 A small mass and a thin uniform rod each of mass ‘m’ are positioned along the same straight line as shown. Find the force of gravitational attraction exerted by the rod on the small mass. 2L L m m Sol.

Q.3 An object is projected vertically upward from the surface of the earth of mass M with a velocity such that the maximum height reached is eight times the radius R of the earth. Calculate : (i) the initial speed of projection (ii) the speed at half the maximum height. Sol.

Q.4 A satellite close to the earth is in orbit above the equator with a period of rotation of 1.5 hours. If it is above a point P on the equator at some time, it will be above P again after time ___________. Sol.

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GRAVITATION

Page # 39

Q.5 A satellite is moving in a circular orbit around the earth. The total energy of the satellite is E = –2 ×105 J. The amount of energy to be imparted to the satellite to transfer it to a circular orbit where its po tential energ y i s U = –2×10 5 J is equal to ___________. Sol.

Q.7 A point P lies on the axis of a fixed ring of mass M and radius a, at a distance a from its centre C. A small particle starts from P and reaches C under gravitational attraction only. Its speed a C will be ________. Sol.

Q.6 Find the gravitational field strength and potential at the centre of arc of linear mass density λ subtending an angle 2α at the centre.

2α Sol.

R

Q.8 Calculate the distance from the surface of the earth at which abo ve and below the sur face acceleration due to gravity is the same. Sol.

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GRAVITATION

Page # 40

Q.9 Consider two satellites A and B of equal mass m, moving in the same circular orbit of radius r around the earth E but in opposite sense of rotation and therefore on a collision course (see figure).

A

r

Sol.

B

Me

(a)In terms of G, Me, m and r find the total mechanical energy EA + EB of the two satellite plus earth system before collision. (b) If the collision is completely inelastic so that wreckage remains as one piece of tangle d material (mass = 2m), find the total mechanical energy immediately after collision. (c) Describe the subsequent motion of the wreckage. Sol.

Q.11 A satellite of mass m is orbiting the earth in a circular orbit of radius r. It starts losing energy due to small air resistance at the rate of C J/s. Then the time taken for the satellite to reach the earth is ________. Sol.

Q.12 Find the potential energy of a system of eight particles placed at the vertices of a cube of side L. Neglect the self energy of the particles. Sol.

Q.10 A particle is fired vertically from the surface of the earth with a velocity kυe, where υe is the escape velocity and k R

O Q.14 Two small dense stars rotate about their common centre of mass as a binary system with the period 1year for each. One star is of double the mass of the other and the mass of the lighter one is 1/3 of the mass of the sun. Find the distance between the stars if distance between the earth & the sun is R. Sol.

x

(ii) Show that the gravitational field inside the hole is uniform, find its magnitude and direction. Sol.

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GRAVITATION

Page # 42 Q.17 A thin spherical shell of total mass M and radius R is held fixed. There is a small hole in the shell. A mass m is released from rest a distance R from the hole along a line that passes through the hole and also through the centre of the shell. This mass subsequently moves under the gravitational force of the shell. How long does the mass take to travel from the hole to the point diametrically opposite. Sol.

Q.16 A body moving radially away from a planet of mass M, when at distance r from planet, explodes in such a way that two of its many fragments move in mutually perpendicular circular orbits around the planet. What will be (a) then velocity in circular orbits. (b) maximum distance between the two fragments before collision and (c) magnitude of their relative velocity just before they collide. Sol.

Q.18 A remote sensing satellite is revolving in an orbit of radius x the equator of earth. Find the area on earth surface in which satellite can not send message.

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GRAVITATION

Page # 43

(TOUGH SUBJECTIVE PROBLEMS)

Exercise - IV

Q.1 A satellite P is revolving around the earth at a height h = radius of earth (R) above equator. Another satellite Q is at a height 2h revolving in opposite direction. At an instant the two are at same vertical line passing through centre of sphere. Find the least time of after which again they are in this is situation.

Earth M

PQ

Q.2 A certain triple-star system consists of two stars, each of mass m, revolving about a central star, mass M, in the same circular orbit. The two stars stay at opposite ends of a diameter of the circular orbit, see figure. Derive an expression for the period of revolution of the stars; the radius of the orbit is r.

m

r M

Q.5 A ring of radius R is made from a thin wire of radius r. If ρ is the density of the material of wire then what will be the gravitational force exerted by the ring on the material particle of mass m placed on the axis of ring at a distance x from its centre. Show that the force will be maximum when x = R / 2 and the ma xi mum va lue of for ce w il l be g iven a s

Fmax =

4 π 2 Gr 2ρm ( 3) 3 / 2 R

Q.6 A man can jump over b = 4m wide trench on earth. If mean density of an imaginary planet is twice that of the earth, calculate its maximum possible radius so that he may escape from it by jumping. Given radius of earth = 6400 km. Q.7 A launching pad with a spaceship is moving along a circular orbit of the moon, whose radius R is triple that of moon Rm. The ship leaves the launching pad with a relative velocity equal to the launching pad’s  initial orbital velocity v 0 and the launching pad then falls to the moon. Determine the angle θ with the horizontal at which the launching pad crashes into the surface if its mass is twice that of the spaceship m.

m Q.3 Find the gravitational force of interaction between the mass m and an infinite rod of varying mass density λ such that λ(x) = λ/x, where x is the distance from mass m. Given that mass m is placed at a distance d from the end of the rod on its axis as shown in figure. x d O m λ λ ( x) = x Q.4 Inside an isolated fixed sphere of radius R and uniform density r, there is a spherical cavity of radius R/2 such that the surface of the cavity passes through the centre of the sphere as in figure. A particle of mass m is released from rest at centre B of the cavity. Calculate velocity with which particle strikes the centre A of the sphere.

A R

B R/2

Q.8 A satellite of mass m is in an elliptical orbit around the earth of mass M(M >>m). The speed of the satellite 6 GM 5R where R = its closest distance to the earth. It is desired to transfer this satellite into a circular orbit around the earth of radius equal its largest distance from the earth. Find the increase in its speed to be imparted at the apogee (farthest point on the elliptical orbit). Q.9 A body is launched from the earth’s surface a an angl e α = 30º to the hor i zo ntal a t a sp eed

at its nearest point to the earth (perigee) is

15 . GM . Neglecting air resistance and earth’s R rotation, find (a) the height to which the body will rise. (ii) The radius of curvature of trajectory at its top point. Q.10 Assume that a tunnel is dug across the earth (radius = R) passing through its centre. Find the time a particle takes to reach centre of earth if it is projected into the tunnel from surface of earth with speed needed for it to escape the gravitational field of earth. v0 =

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GRAVITATION

Page # 44

(JEE PROBLEMS)

Exercise - V Q.1 If the distance between the earth and the sum were half its present value, the number of days in a year would have been [JEE’ 96] (A) 64.5 (B) 129 (C) 182.5 (C) 730 Sol.

Q.2 Distance between the centres of two stars is 10a. The masses of these starts are M and 16M and their radii a and 2a respectively. A body of a mass m is fired at night from the surface of the larger star towards the maller star. What should be its minimum initial speed to reach the surface of the smaller start? Obtain the expression in terms of G, M and a. [JEE’ 96] Sol.

Q.3 An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.)E0. Its potential energy is [JEE’ 97] (A) –E0 (B) 1.5 E0 (C) 2E0 (D) E0 Sol.

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GRAVITATION

Page # 45

Q.4 A cord of length 64 m is used to connect a 100 kg astronaut to spaceship whose mass is much larger than that of the astronaut. Estimate the value of the tension in the cord. Assume that the spaceship is orbiting near earth surafce. Assume that the spaceship and the astronaut fall on a straight line from the earth centre. The radius of the earth is 6400 km. [REE 98] Sol.

Sol.

Q.6 A body is projected vertically upwards from the bottom of a crater of moon of depth R/100 where R is the radius of moon with a velocity equal to the escape velocity on the surface of moon. Calculate maximum height attained by the body from the surface of the moon. [JEE’ 2003] Sol.

In a region of only gravitational field of mass ‘M’ a particle is shifted from A to B via three different paths in the figure. The work done in different paths are W1, W2, W3 respectively then Q

. 5

B

(1) (A) W1 = W2 = W3 (C) W1 = W2 > W3

(3) M (2)

C

A [JEE’ (Scr.) 2003] (B) W1 > W2 > W3 (D) W1 < W2 < W3

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GRAVITATION

Page # 46

Q.7 A system of binary stars of masses mA and mB are moving in circular orbits of radii rA and rB respectively. If TA and TB are the time periods of masses mA and mB respectively, then [JEE 2006] (A) TA > TB (if rA > rB) (B) TA > TB (if mA > mB)  TA   (C)   TB 

2

 rA  =    rB 

3

(D) TA = TB

Sol.

Q.9 STATEMENT-1 An astronaut in an orbiting space station above the Earth experiences weightlessness. [JEE 2008] and STATEMENT-2 An object moving around the Earth under the influence of Earth’s gravitaitonal force is in a state of ‘free-fall’. (A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1 (B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is True Sol.

Q.8 A spherically symmetric gravitational system of particles has a mass density [JEE 2008]

ρ for r ≤ R ρ= 0  0 for r > R where ρ0 is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed V as a function of distance r (0 < r < ∞) from the centre of the system is represented by

v

Q.10 A thin uniform annular disc (see figure) of mass M has outer radius 4 R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is [JEE 2010] P 4R

v 3R

4R

(B)

(A)

R

r

v

R

r

R

r

v

(C)

(D)

R

r

2GM ( 4 2 – 5) 7R GM (C) 4 R (A)

2GM ( 4 2 – 5) 7R 2GM (D) 5 R ( 2 − 1) (B) –

Sol.

Sol.

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GRAVITATION

Page # 47

Q.11 A binary star consists of two stars A (mass 2.2 Ms) and B (mass 11 Ms), where Ms is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is. [JEE 2010] Sol.

Q.13 A satellite is moving with a constant speed 'V' in a circular orbit about the earth. An object of mass 'm' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is (A)

1 mV2 2

(B) mV2

(C)

3 mV2 2

(D) 2mV2

[JEE 2011] Sol.

Q.12

Gravitational acceleration on the surface of a

6 g. where g is the gravitational accelera11 tion on the surface of the earth. The average mass planet is

density of the planet is

2 times that of the earth. If 3

the escape speed on the surface of the earth is taken to be 11 kms–1, the escape speed on the surface of [JEE 2010] the planet in kms–1 will be

Q.14 Two spherical planets P and Q have the same uniform density ρ , masses Mp and MQ, and surface areas A and 4A, respectively. A spherical planet R also has uniform density ρ and its mass is (MP + MQ). The escape velocities from the planets P, Q and R, are Vp, VQ and VR, respectively. Then [JEE 2012] (A) VQ > VR > VP

(B) VR > VQ > VP

(C) VR / VP = 3

(D) VP / VQ =

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1 2

GRAVITATION

Page # 48

ANSWER EX-I B D B A C B

1. 7. 13. 19. 25. 31.

(OBJECTIVE PROBLEMS) 2. 8. 14. 20. 26.

C A B C A

A A A D B

3. 9. 15. 21. 27.

1. 7.

A B B C A

5. 11. 17. 23. 29.

B B C B C

6. 12. 18. 24. 30.

(MULTIPLE CHOICE PROBLEMS) A,D A

2. 8.

B,C,D 4. A,B,C 10.

3. 9.

1. –

B B A D C

4. 10. 16. 22. 28.

A,C A,D

A,C

5.

B,D

6.

(SUBJECIVE PROBLEMS) 2.

Gm 2

3. (i)

2

3L

2 2 Gm Gm , (ii) 3 5R R

4 3

4. 1.6 hours if is rotating from west to east, 24/17 hours if it is rotating from west to east. 5. 1 × 105 J

6.

2Gλ (sin α), (–Gλ 2α) R

9. (a) –GmMe/r, (b) –2GmMe/r 13. (i) –

 3Gm  m + M , (ii)  R  3 

10.

7.

2GM  1  1 −  a  2

1− k 2

     πGρ0R  1 8   2πGρ0R  – 2 i , g= – i 15. g = + 2   6 x 3 R   x –      2

2πR3 / 2 (6 6 )

8.

GM( 2 2 + 3 3 )

GM  2 8  –   R  3 15 

2. vmin =

6. h = 99R 13. B

7. D 14.B, D

Gm ; (b) r 2 ; (c) r

2GM r

(TOUGH SUBJECTIVE PROBLEMS) 4 πr 3 / 2

2.

G(4M + m)

3.

Gmλ 2

2d

4.

2 πGρR2 3

 7  + 1 R, (b) 1.13 R 9. (a) h =   2 

16. (a)

x2 – R2  4 πR 2  x 

 3 1  + 3 +  2 3 

14. R

3

17. 2 × R3 / GM

5 –1 R 2

–4GM2 12. L

GMm  1 1 11. t = 2C  R – r   e 

R ek 2

 G m + M  R 3 

  18.  1 – 

8. h =

6.

6.4 km

7. cos θ =

3 10

 1  Re 10. T = sin –1   3 g

(JEE PROBLEMS) 3 2

5GM a 8. C

3. C

4. T = 3 × 10–2 N

9. A

10. A

5. A 11. 6

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12. 3

Nurturing potential through education

ELASTICITY & THERMAL EXPANSION THEORY AND EXERCISE BOOKLET

CONTENTS

S.NO.

TOPIC

PAGE NO.

1. Elasticity ........................................................................................ 2 2. Stress ......................................................................................... 2 – 3 3. Strain .......................................................................................... 3 – 5 4. Young Modulus ............................................................................ 5 – 6 5. Thermal Expansion .................................................................... 6 – 12 6. Exercise - I ................................................................................ 13 – 16 7. Exercise - II ............................................................................... 17 – 18 8. Exercise - III .................................................................................. 19 9. Answer key ................................................................................... 20

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Page # 2

1.

(a) (b)

2.

ELASTICITY & THERMAL EXPANSION

DEFINATION Elasticity is that property of the material of a body by virtue of which the body opposes any change in its shape or size when deforming forces are applied to it, and recovers its original state as soon as the deforming forces are removed. On the basis of defination bodies may be classified in two types : Perfectly Elastic (P.E.) : If body regains its original shape ans size completely after removal of force. Nearest approach P.E. : quartz-fibre Perfectly Plastic (P.P.) : If body does not have tendency to recover its original shape and size. Nearest Approach P.P. : Peetty Limit of Elasticity : The maximum deforming force upto which a body retains its property of elasticity is called the limit of elasticity of the material of the body. STRESS When a deforming force is applied to a body, it reacts to the applied force by developing a reaction (or restoring force which, from Newton's third law, is equal in magnitude and opposite in direction to the applied force. Thereaction force per unit area of the body which is called into play due to the action of the applied force is called stress. Stress is measured in units of force per unit area, i.e. Nm–2. Thus. F A where F is the applied force and A is the area over which it acts. A

Stress =

10 N Stress = 10/A Unit of stress : N/m2 Dimension of stress : M1L-1T-2 2.1

Types of stress : Three Types of Stress :

(A)

Tensile Stress : Pulling force per unit area.

F

A

F

It is applied parallel to the length It causes increase in length or volume (B)

Compressive Stress : Pushing force per unit area. It is applied parallel to the length

F

A

F

It causes decrease in length or volume (C)

Tangential Stress : Tangential force per unit area. It causes shearing of bodies.

Note : 1. If the stress is normal to surface called normal stress. 2. Stress is always normal to surface in case of change in length of a wire or volume of body. 3. When external force compresses the body  Nature of atomic force will be repulsive. 4. When external forces expanses the body  Nature of atomic force will be attractive. Difference between Pressure v/s Stress : S. No.

1 2 3

Pressure Pressure is always normal to the area.

Stress Stress can be normal or tangential May be compressive or Always compressive in nature tensile in nature. Scalar Tensor

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Page # 3

ELASTICITY & THERMAL EXPANSION Ex.1

Sol.

3.

A 4.0 m long copper wire of cross sectional area 1.2 cm2 is stretched by a force of 4.8 × 103 N stress will be (A) 4.0 × 107 N/mm2 (B) 4.0 × 107 KN/m2 (C) 4.0 × 107 N/m2 (D) None [C] 4.8  10 3 N F Stress = = = 4.0 × 107 N/m2 12 .  10 4 m 2 A STRAIN When a deforming force is applied to a body, it may suffer a change is size or shape. Strain is defined as the ratio of the change in size or shape to the original size or shape of the body. Strain is a number; it has no units or dimensions. The ratio of the change in length to the original length is called longitudinal strain. The ratio of the change in volume to the original volume is called volume strain. The strain resulting from a change in shape is called shearing strain.

Strain 

L final length – original length  =  T,, L0 original length

Note : Original and final length should be at same temperature. F

3.1

Types of strain : Three Types of Strain :

(A)

Linear Strain : Change in length per unit length is called linear strain Linear Strain =

Change in length Original length

L L Volume Strain : Change in volume per unit volume is called volume strain.

=

(B)

V  V

Volume sirain

Volume Strain Change in volume V = Original volume V Shear Strain : Angle through which a line originally normal to fixed surface is turned. =

(C)

 =

x L

x

L

Note : Strain is unitless. Ex.2

A copper rod 2m long is stretched by 1mm. Strain will be - Shear strain (A) 10-4, volumetric (B) 5 × 10-4, volumetric (C) 5 × 10-4, longitudinal (D) 5 × 10-3, volumetric

Sol.

[C] Strain =

 1 10 3 = 5 × 10–4, longitudinal =  2

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Page # 4

4.

ELASTICITY & THERMAL EXPANSION

THERMAL STRESS If the ends of a rod are rigidly fixed and its temperature is changed, then compressive stresses are set up in the rod. These developed stress are called thermal stress. Thermal Stress = Y  t Y  modulus of elasticity,   Coefficient of linear expansion t  change in temperature

5.

WORK DONE IN STRETCHING A WIRE In stretching a wire work is done against internal restoring forces. This work is stored in body as elastic potential energy or strain energy. If L = length of wire & A = Cross-sectional Area. F/A YA Y =  F = x x/L L work done to increase dx length YA dW = Fdx = xdx L L YA 1 YA Total work done = W = xdx = (L)2 0 L 2 L

2

 L  W 1  Work done per unit volume = = Y  V 2  L  W 1 = Y (strain)2 V 2 W 1 = x stress x strain V 2 W 1 ( stress)2 W 1 F L =  = × V 2 AL 2 A L Y 1 1 W = F × L = load x elongation 2 2

[ V = AL]

[ Y =

Stress ] Strain

stress

D E C If we increase the load gradually on a vertical B suspended metal wire, In Region OA : A Strain is small (< 2%) Stress  Strain  Hook's law is valid. O strain Slope of line OA gives Young's modulus Y of the material. In Region AB : Stress is not proportional to strain, but wire will still regain its original length after removing of stretching force. In region BC : Wire yields  strain increases rapidly with small change in stress. This behavior is shown up to point C known as yield point. In region CD : Point D correspondes to maximum stress, which is called point of breaking or tensile strength. In region DE : The wire literally flows. The maximum stress corresponding to D after which wire begin to flow. In this region strain increase even if wire is unloaded and rupture at E.

6.

STRESS-STAIN CURVE

7.

HOOKES' LAW Hookes' law states that, within the elastic limit, the stress developed in a bodyis proportional to the strain produced in it. Thus the ratio of stress to strain is a constant. This constant is called the modulus of elasticity. Thus stress Modulus of elasticity = strain Since strain has no unit, the unit of the modulus of elasticity is the same as that of stress, namely, Nm–2

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Page # 5

ELASTICITY & THERMAL EXPANSION 8.

YOUNG'S MODULUS Suppose that a rod of length l and a uniform crossectional area a is subjected to a logitudinal pull. In other words, two equal and opposite forces are applied at its ends. F A The stress in the present case is called linear stress, tensile stress, or extensional stress. If the direction of the force is reversed so that L is negative, we speak of compressional strain and compressional stress. If the elastic limit is not exceeded, then from Hooke's law Stress  strain or Stress = Y × strain

Stress =

Y

or

stress F L  . strain A L

...(1)

where Y, the constant of proportionality, is called the Young's modulus of the material of the rod and may be defined as the ratio of the linear stess to linear strain, provided the elastic limit is not exceeded. Since strain has no unit, the unit of Y is Nm–2. Consider a rod of length  0 which is fixed between to rigid end separated at a distance  0 now if the temperature of the rod is increased by  then the strain produced in the rod will be : length of the rod at new temperatrue – natural length of the rod at new temperature natural length of the rod at new temperature

strain =

F

 0 –  0 (1  ) –  0  = =  0 (1  )  0 (1  )

F

0

 is very small so

strain = –  (negative sign in the answer represents that the length of the rod is less than the natural length that means is compressed by the ends.) We know that  

stress then F = T A strain

Note : (A) For Loaded Wire : L =

FL 2

r Y

FL  2  Y  AL & A  r   

for rigid body L = 0 so Y =  i.e. elasticity of rigid body is infinite. (B)

If same stretching force is applied to different wire of same material. L

[As F and Y are const.] r2 Greater the value L, greater will be elongation. L 

(C)

Elongation of wire by its own weight : In this case F = Mg acts at CG of the wire so length of wire which is stretched will be L/2 (Mg)  L / 2 FL MgL gL2 = = = 2 r Y AY 2AY 2Y M = AL]

    L = [

L =

gL2 2Y

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Page # 6 Ex.3

Sol.

7.2

A wire of length 1m and area of cross section 4 x 10-8 m2 increases in length by 0.2 cm when a force of 16 N is applied. Value of Y for the material of the wire will be (A) 2 × 106 N/m2 (B) 2 × 1011 kg/m2 (C) 2 × 1011 N/mm2 (D) 2 × 1011 N/m2 [D] By Hook's law F/A FL Y = =  /L A 16  1 Y = = 2 × 10111 N/m2 ( 4  10 8 ) (0.2  10 2 ) Bulk Modulus : B =

7.3

8.

Volume stress = Volume strain

P VP  B = – V V  V

Compressibility : k =

7.4

ELASTICITY & THERMAL EXPANSION

1 1 = – B V

FG V IJ H P K

Modulus of Rigidity : tan gentialstress F/A  =   =  tan gentialstrain Only solid can have shearing as these have definite shape.

D

POISSION'S RATIO L

Lateralstrain d/D dL  = =   = Linear strain L / L LD Interatomic force constant = Young Modulus x Interatomic distance. 9.

 d

THERMAL EXPANSION

Most substances expand when they are heated. Thermal expansion is a consequence of the change in average separation between the constituent atoms of an object. Atoms of an object can be imagined to be connected to one another by stiff springs as shown in figure. At ordinary temperatures, the atoms in a solid oscillate about their equilibrium positions with an amplitude of approximately 10–11 m. The average specing between the atom is about 10–10 m. As the temperature of solid increases, the atoms oscillate with greater amplitudes, as a result the average separation between them increases, consequently the object expands.

9.1

LINEAR EXPANSION When the rod is heated, its increase in length L is proportional to its original length L0 and change in temperture T where T is in °C or K.

L0

Before heating

L  L 0  L

After heating

dL =  L0dT  L =  L0T If a T B & T  then FB  and T  then FB  v(1   B T)

9.8

v 0 g  mg

Barometer Their is a capaillary tube which have coefficient of linear expansion c and a liquid of volume v of volume expansion coefficient v of volume expansion coefficient of  at

A

temperature Ti. and given 3 c    . The Area of crosssection of capillary tube is A. Now temperature increases to Tf, So volume of liquid rises in the capillary. Let it rises to height H. So volume rises in tube = V V = V[1 +  T] – V[1+ 3 c T] = V ( – 3c ) T And Area of cross section of capillary = A = A [1 + 2CT] V VT(   – 3 C ) So height in capillary tube H'  A'  A(1  2 T) C

Ti

V

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c

Page # 12

ELASTICITY & THERMAL EXPANSION

Ex.10

What will happen to the water level if the vessel is heated ?

Sol.

(i) if     c then overflow occure and overflow



= AH(1    t ) – AH (1 + c T)

H

(ii) if     c final volume Vfc = AH (1  C T) final volume Vf = AH (1   T) Now So

Note

c

AF = A[1  2 C T] H[1  y  T ] H = final height = [1  2 T ] c

If two strips of equal length but of different metals are placed on each other and riveted, the single strip so formed is called 'bimetallic strip' [see given fig.]. This strip has the characteristic property of bending on heating due to unequal linear expansion of the two metals. The strip will bend with metal of greater  on outer side, i.e., convex side. This strip finds its application in auto-cut or thermostat in electric heating circuits. It has also been used as thermometer by calibrating its bending. T2 T1 Fe

Fe

Cu

Cu

(T2 > T1) (A) Ex.11

(B)

When the two rods having expansion cofficient 1, 2 (2 > 1) and width d are heated then the radius of the rod after expansion. 2

2 d

T

dI

1

(2 > 1) 

1

R

d R = (  –  ) T 2 1

Proof :  2  (1   2 t)  (R  d)   1  (1  1t )  R R  d (1   2 T )  R (1   1T )

from binomial theorem

d R = ( –  )T 2 1

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Page # 13

ELASTICITY & THERMAL EXPANSION

(OBJECTIVE QUESTIONS)

EXERCISE-I 1.

A steel scale is to be prepared such that the millimeter intervals are to be accurate within 6 × 10–5 mm. The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is ( = 12 × 10–6 k–1) (A) 4.0 ºC

(B) 4.5 ºC

(C) 5.0 ºC

(D) 5.5 ºC

2.

A steel rod 25 cm long has a cross-sectional area of 0.8 cm2. Force that would be required to stretch this rod by the same amount as the expansion produced by heating it through 10ºC is : (Coefficient of linear expansion of steel is 10–5/ºC and Young’s modulus of steel is 2 × 1010 N/m2.) (A) 160 N (B) 360 N (C) 106 N (D) 260 N

3.

Two rods of different materials having coefficients of thermal expansion 1, 2 and Young’s moduli Y1, Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If 1 : 2 = 2 : 3, the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to (A) 2 : 3 (B) 1 : 1 (C) 3 : 2 (D) 4 : 9

4.

If I is the moment of inertia of a solid body having -coefficient of linear expansion then the change in I corresponding to a small change in temperature T is (A)  I T

5.

1  I T 2

(C) 2  I T

(D) 3  I T

A metallic wire of length L is fixed between two rigid supports. If the wire is cooled through a temperature difference T (Y = young’s modulus,  = density,  = coefficient of linear expansion) then the frequency of transverse vibration is proportional to : (A)

6.

(B)

Y 

(B)

Y

(C)

 Y

(C)

Y

A metal wire is clamped between two vertical walls. At 20°C the unstrained length of the wire is exactly equal to the separation between walls. If the temperature of the wire is decreased the graph between elastic energy density (u) and temperature (T) of the wire is u

(A)

u

u

(B) T (in °C) 20

u

(C) T (in °C)

T (in °C)

20

20

(D)

T (in °C) 20

7.

A steel tape gives correct measurement at 20°C. A piece of wood is being measured with the steel tape at 0°C. The reading is 25 cm on the tape, the real length of the given piece of wood must be : (A) 25 cm (B) < 25 cm (C) >25 cm (D) can not say

8.

A rod of length 20 cm is made of metal. It expands by 0.075 cm when its temperature is raised from 0°C to 100°C. Another rod of a different metal B having the same length expands by 0.045 cm for the same change in temperature, a third rod of the same length is composed of two parts one of metal A and the other of metal B. Thus rod expand by 0.06 cm for the same change in temperature. The portion made of metal A has the length. (A) 20 cm (B) 10 cm (C) 15 cm (D) 18 cm

9.

A sphere of diameter 7 cm and mass 266.5 gm floats in a bath of a liquid. As the temperature is raised, the sphere just begins to sink at a temperature 35°C. If the density of a liquid at 0°C is 1.527 gm/cc, then neglecting the expansion of the sphere, the coefficient of cubical expansion of the liquid is f : (A) 8.486 × 10–4 per °C

(B) 8.486 × 10–5 per °C (C) 8.486 × 10–6 per °C

(D) 8.486 × 10–3 per °C

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Page # 14 10.

ELASTICITY & THERMAL EXPANSION

The volume of the bulb of a mercury thermometer at 0°C is V0 and cross section of the capillary is A0. The coefficient of linear expansion of glass is ag per °C and the cubical expansion of mercury m per °C. If the mercury just fills the bulb at 0°C, what is the length of mercury column in capillary at T°C. V0 T(  m  3ag ) (A) A (1  2a T) 0 g

11.

V0 T(  m – 3ag ) (B) A (1  2a T) 0 g

(B) EA t/(1 +  t)

(D) E/t

(B) W 0[1 – (s – 1) t]

(C) W 0[ (s – 1) t]

(D) W 0t / (s – 1)

A thin walled cylindrical metal vessel of linear coefficient of expansion 10–3 °C–1 contains benzenr of volume expansion coefficient 10–3 °C–1. If the vessel and its contents are now heated by 10°C, the pressure due to the liquid at the bottom. (A) increases by 2%

14.

(C) EA t/(1 – t)

The loss in weight of a solid when immersed in a liquid at 0°C is W 0 and at t°C is W. If cubical coefficient of expansion of the solid and the liquid by s and 1 respectively, then W is equal to : (A) W 0[1 + (s – 1) t]

13.

V0 T(  m – 2a g ) (D) A (1  3a T ) 0 g

A metallic rod 1 cm long with a square cross-section is heated through 1°C. If Young’s modulus of elasticity of the metal is E and the mean coefficient of linear expansion is  per degree Celsius, then the compressional force required to prevent the rod from expanding along its length is : (Neglect the change of cross-sectional area) (A) EAt

12.

V0 T(  m  2ag ) (C) A (1  3a T) 0 g

(B) decreases by 1%

(C) decreases by 2%

(D) remains unchanged

A rod of length 2m at 0°C and having expansion coefficient  = (3x + 2) × 10–6 °C–1 where x is the distance (in cm) from one end of rod. The length of rod at 20 °C is : (A) 2.124 m

(B) 3.24 m

(C) 2.0120 m

(D) 3.124 m

15.

A copper ring has a diameter of exactly 25 mm at its temperature of 0°C. An aluminium sphere has a diameter of exactly 25.05 mm at its temperature of 100°C. The sphere is placed on top of the ring and two are allowed to come to thermal equilibrium, no heat being lost to the surrounding. The sphere just passes through the ring at the equilibrium temperature. The ratio of the mass of the sphere & ring is : (given : Cu = 17 × 10–6/°C, Al = 2.3 × 10–5/°C, specific heat of Cu = 0.0923 Cal/g°C and specific heat of Al = 0.215 cal/g°C) (A) 1/5 (B) 23/108 (C) 23/54 y (D) 216/23

16.

A cuboid ABCDEFGH is anisotropic with x = 1 × 10–5/°C, y = 2 × 10–5/°C, z = 3 × 10–5/°C. Coefficient of superficial expansion of faces can be (A) ABCD = 5 × 10–5 /°C

(B) BCGH = 4 × 10–5 /°C

(C) CDEH = 3 × 10–5/°C

(D) EFGH = 2 × 10–5/°C

A

B C

D F

G

x

E H z

17.

An open vessel is filled completely with oil which has same coefficient of volume expansion as that of the vessel. On heating both oil and vessel, (A) the vessel can contain more volume and more mass of oil (B) the vessel can contain same volume and same mass of oil (C) the vessel can contain same volume but more mass of oil (D) the vessel can contain more volume but same mass of oil

18.

A metal ball immersed in Alcohol weights W 1 at 0°C and W 2 at 50°C. The coefficient of cubical expansion of the metal ()m is less than that of alcohol ()Al. Assuming that density of metal is large compared to that of alcohol, it can be shown that (A) W 1 > W 2

19.

(B) W 1 = W 2

(C) W 1 < W 2

(D) any of (A), (B) or (C)

A solid ball is completely immersed in a liquid. The coefficients of volume expansion of the ball and liquid are 3 × 10–6 and 8 × 10–6 per °C respectively. The percentage change in upthrust when the temperature is increased by 100°C is (A) 0.5 %

(B) 0.11 %

(C) 1.1%

(D) 0.05 %

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Page # 15

ELASTICITY & THERMAL EXPANSION 20.

A thin copper wire of length L increase in length by 1% when heated from temperature T1 to T2. What is the percentage change in area when a thin copper plate having dimensions 2L × L is heated from T1 to T2 ? (A) 1%

21.

(B) 2%

(C) 3%

(D) 4%

If two rods of length L and 2L having coefficients of linear expansion  and 2 respectively are connected so that total length becomes 3L, the average coefficient of linear expansion of the composition rod equals : 3 5 5  (B)  (C)  (D) none of these 2 2 3 The bulk modulus of copper is 1.4 × 1011 Pa and the coefficient of linear expansion is 1.7 × 10–5 (C°)–1. What hydrostatic pressure is necessary to prevent a copper block from expanding when its temperature is increased from 20°C to 30°C ?

(A)

22.

(A) 6.0 × 105 Pa

(B) 7.1 × 107 Pa

(C) 5.2 × 106 Pa

(D) 40 atm

23.

The coefficients of thermal expansion of steel and a metal X are respectively 12 × 10–6 and 2 × 10–6 per °C, At 40°C, the side of a cube of metal X was measured using a steel vernier callipers. The reading was 100 mm. Assuming that the calibration of the vernier was done at 0°C, then the actual length of the side of the cube at 0°C will be (A) > 100 mm (B) < 100 mm (C) = 100 mm (D) data insufficient to conclude

24.

A glass flask contains some mercury at room temperature. It is found that at different temperature the volume of air inside the flask remains the same. If the volume of mercury in the flask is 300 cm3, then volume of the flask is (given that coefficient of volume expansion of mercury and coefficient of linear expansion of glass are 1.8 × 10–4(°C)–1 and 9 × 10–6(°C)–1 respectively) (A) 4500 cm3 (B) 450 cm3 (C) 2000 cm3 (D) 6000 cm3

Question No. 25 to 29 (5 question) Solids and liquids both expand on heating. The density of substance decreases on expanding according to the relation 1 2 = 1  ( T – T ) 2 1

25.

where, 1  density at T1 2  density at T2   coeff. of volume expansion of substances when a solid is submerged in a liquid, liquid exerts an upward force on solid which is equal to the weight of liquid displaced by submerged part of solid. Solid will float or sink depends on relative densities of solid and liquid. A cubical block of solid floats in a liquid with half of its volume submerged in liquid as shown in figure (at temperature T) s  coeff. of linear expansion of solid L  coeff. of volume expansion of liquid s  density of solid at temp. T L  density of liquid at temp. T The relation between densities of solid and liquid at temperature T is (A) S = 2L (B) S = (1/2) L (C) S = L (D) S = (1/4) L

26.

If temperature of system increases, then fraction of solid submerged in liquid (A) increases (B) decreases (C) remains the same (D)inadequate information

27.

Imagine fraction submerged does not change on increasing temperature the relation between L and S is (A) L = 3S (B) L = 2S (C) L = 4S (D) L = (3/2)S

28.

Imagine the depth of the block submerged in the liquid does not change on increasing temperature then (A) L = 2 (B) L = 3 (C) L = (3/2) (D) L = (4/3)

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Page # 16

ELASTICITY & THERMAL EXPANSION

29.

Assume block does not expand on heating. The temperature at which the block just begins to sink in liquid is (A) T + 1/L (B) T + 1/(2L) (C) T + 2/L (D) T + L/2

30.

The coefficient of apparent expansion of a liquid in a copper vessel is C and in a silver vessel is S. The coefficient of volume expansion of copper is C. What is the coefficient of linear expansion of silver? (A)

C   c  S) 3

(B)

C –  c  S) 3

(C)

C   c – S) 3

(D)

C –  c – S) 3

31.

An aluminium container of mass 100 gm contains 200 gm of ice at –20°C. Heat is added to the system at the rate of 100 cal/s. The temperature of the system after 4 minutes will be (specific heat of ice = 0.5 and L = 80 cal/ gm, specific heat of Al = 0.2 cal/gm/°C) (A) 40.5°C (B) 25.5°C (C) 30.3°C (D) 35.0°C

32.

Two vertical glass tubes filled with a liquid are connected by a capillary tube as shown in the figure. The tube on the left is put in an ice bath at 0°C while the tube on the right is kept at 30° C in a water bath. The differenece in the levels of the liquid in the two tubes is 4 cm while the height of the liquid column at 0° C is 120 cm. The coefficient of volume expansion of liquid is (Ignore expansion of glass tube) (A) 22 × 10–4/°C (B) 1.1 × 10–4/°C –4 (C) 11 × 10 /°C (D) 2.2 × 10–4 /°C

33. 34.

A difference of temperature of 25ºC is equivalent to a difference of : (A) 45º F (B) 72º F (C) 32º F

4 cm

Water

120cm 30°C 0°C

(D) 25º F

Two thermometers x and y have fundamental intervals of 80º and 120º. When immersed in ice, they show the reading of 20º and 30º. If y measures the temperature of a body as 120º, the reading of x is : (A) 59º (B) 65º (C) 75º (D) 80º MULTIPLE CHOICE QUESTIONS

35.

When an enclosed perfect gas is subjected to an adiabatic process : (A) Its total internal energy does not change (B) Its temperature does not change (C) Its pressure varies inversely as a certain power of its volume (D) The product of its pressure and volume is directly proportional to its absolute temperature.

36.

Four rods A, B, C, D of same length and material but of different radii r, r 2 , r 3 and 2r respectively are held between two rigid walls. The temperature of all rods is increased by same amount. If the rods donot bend, then (A) the stress in the rods are in the ratio 1 : 2 : 3 : 4 (B) the force on the rod exerted by the wall are in the ratio 1 : 2 : 3 : 4 (C) the energy stored in the rods due to elasticity are in the ratio 1 : 2 : 3 : 4 (D) the strains produced in the rods are in the ratio 1 : 2 : 3 : 4

37.

A body of mass M is attached to the lower end of a metal wire, whose upper end is fixed. The elongation of the wire is l. (A) Loss in gravitational potential energy of M is Mgl (B) The elastic potential energy stored in the wire is Mgl (C) The elastic potential energy stored in the wire is 1/2 Mgl (D) Heat produced is 1/2 Mgl

38.

When the temperature of a copper coin is raised by 80ºC, its diameter increases by 0.2%. (A) Percentage rise in the area of a face is 0.4% (B) Percentage rise in the thickness is 0.4% (C) Percentage rise in the volume is 0.6% (D) Coefficient of linear expansion of copper is 0.25 × 10–4Cº–1.

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Page # 17

ELASTICITY & THERMAL EXPANSION

EXERCISE-II

(SUBJECTIVE QUESTIONS)

1.

We have a hollow sphere and a solid sphere of equal radii and of the same material. They are heated to raise their temperature by equal amounts. How will the change in their volumes, due to volume expansions, be related ? Consider two cases (i) hollow sphere is filled with air, (ii) there is vaccum inside the hollow sphere.

2.

The time represented by the clock hands of a pendulum clock depends on the number of oscillation performed by pendulum every time it reach to its extreme position the second hand of the clock advances by one second that means second hand move by two second when one oscillation in complete (a) How many number of oscillations completed by pendulum of clock in 15 minutes at calibrated temperature 20°C (b) How many number of oscillations are completed by a pendulum of clock in 15 minute at temperature of 40°C if  = 2 × 10–5c (c) What time is represented by the pendulum clock at 40°C after 15 minutes if the initial time shown by the clock is 12: 00 pm ? (d) If the clock gains two second in 15 minutes then find - (i) Number of extra oscillation (ii) New time period (iii) change in temperature.

3.

Consider a cylindrical container of cross section area ‘A’, length ‘h’ having coefficient of linear expansion c. The container is filled by liquid of real expansion coefficient L up to height h1. When temperature of the system h is increased by  then h1 (a) Find out new height, area and volume of cyclindrical container and new volume of liquid. (b) Find the height of liquid level when expansion of container is neglected. (c) Find the relation between L and c for which volume of container above the liquid level. (i) increases (ii) decreases (iii) remains constant. (d) If y  > 3 C and h = h1 then calculate, the volume of liquid overflow (e) What is the relation between   and  c for which volume of empty space becomes independent of (f) (i) (ii) (iii) (1)

change of temp. If the surface of a cylindrical container is marked with numbers for the measurement of liquid level of liquid filled inside it. If we increase the temperature of the system be  then Find height of liquid level as shown by the scale on the vessel. Neglect expansion of liquid Find height of liquid level as shown by the scale on the vessel. Neglect expansion of container Find relation between L and c so that height of liquid level with respect to ground increases (2) decreases (3)remains constant.

4.

A loaded glass bulb weighs 156.25 g in air. When the bulb is immersed in a liquid at temperature 15ºC, it weighs 56.25 g. On heating the liquid, for a temperature upto 52ºC the apparent weight of the bulb becomes 66.25 g. Find the coefficient of real expansion of the liquid. (Given coefficient of linear expansion of glass = 9 × 10–6/ºC).

5.

A body is completely submerged inside the liquid. It is in equilibrium and in rest condition at certain temperature. It L volumetric expansion coefficient of liquid s = linear expansion coefficient by of body. It we increases temperature by  amount than find (a) New thrust force if initial volume of body is V0 and density of liquid is d0. (b) Relation between s and L so body will (i) move upward (ii) down ward (iii) remains are rest

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Page # 18

ELASTICITY & THERMAL EXPANSION

6.

A clock pendulum made of invar has a period of 0.5 sec at 20°C. If the clock is used in a climate where average temperature is 30° C, aporoximately. How much fast or slow will the clock run in 106 sec. (invar = 1 × 10–6/°C)

7.

An iron bar (Young’s modulus = 1011 N/m2,  = 10–6/°C) 1 m long and 10–3 m2 in area is heated from 0°C to 100°C without being allowed to bend or expand. Find the compressive force developed inside the bar.

8.

Three aluminium rods of equal length form an equilateral triangle ABC. Taking O (mid point of rod BC) as the origin.Find the increase in Y-coordinate of center of mass per unit change in temperature of the system. Assume the length of the each rod is 2m, and al = 4 3 × 10–6/ °C A

B

O

C

9.

If two rods of length L and 2L having coefficients of linear expansion  and 2 respectively are connected so that total length becomes 3L, determine the average coefficient of linear expansion of the composite rod.

10.

A thermostatted chamber at small height h above earth’s surface maintained at 30°C has a clock fitted in it with an uncompensated pendulum. The clock designer correctly designs it for height h, but for temperature of 20°C. If this chamber is taken to earth’s surface, the clock in it would click correct time. Find the coefficient of linear expansion of material of pendulum.(earth’s radius is R)

11.

The coefficient of volume expansion of mercury is 20 times the coefficient of linear expansion of glass Find the volume of mercury that must be poured into a glass vessel of volume V so that the volume above mercury may remain constant at all temperature.

12.

A metal rod A of 25 cm lengths expands by 0.050 cm. When its temperature is raised from 0°C to 100°C. Another rod B of a different metal of length 40cm expands by 0.040 cm for the same rise in temperature. A third rod C of 50 cm length is made up of pieces of rods A and B placed end to end expands by 0.03 cm on heating from 0° C to 50°C. Find the lengths of each portion of the composite rod.

13.

The figure shows three temperature scales with the freezing and boiling points of water indicated. 70ºx

120ºW

90ºY

–20ºx

30ºW

0ºY

Bolling Point

Freezing Point

(a) Rank the size of a degree on these scales, greatest first. (b) Rank the following temperatures, highest first 50ºX, 50ºW and 50ºY. 14.

What is the temperature at which we get the same reading on both the centigrade and Fahrenheit scales ?

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Page # 19

ELASTICITY & THERMAL EXPANSION

EXERCISE-III 1.

(JEE PROBLEMS)

The apparatus shown in the figure consists of four glass columns connected by horizontal sections. The height of two central columns B & C are 49 cm each. The two outer columns A & D are open to the atmosphere. A & C are maintained at a temperature of 95° C while the columns B & d are maintained at 5°C. The height of the liquid in A & D measured from the base line are 52.8 cm & 51 cm respectively. Determine the coefficient of thermal expansion of the liquid. [JEE ‘97]

A 95°

B 5°

C 95°

D 5°

2.

A bimetallic strip is formed out of two identical strips one of copper and the other of brass. The coefficient of linear expansion of the two metals are C and B. On heating, the temperature of the strip goes up by T and the strip bends to form an arc of radius of curvature R. Then R is : [JEE ‘99] (A) proportional at T (B) inversely proportional to T (C) proportional to |B – C| (D) inversely proportional to |B – C|

3.

Two rods one of aluminium of length l1 having coefficient of linear expansion a, and other steel of length l2 having coefficient of linear expansion S are joined end to end. The expansion in both the rods is same on l1 [JEE’ (Scr) 2003] variation of temperature. Then the value of l  l is 1 2 s s a  s (A*)    (B)  –  (C) (D) None of these s a s a s

4.

A cube of coefficient of linear expansion s is floating in a bath containing a liquid of coefficient of volume expansion l. When the temperature is raised by T, the depth upto which the cube is submerged in the liquid remains the same. Find the relation between s and l, showing all the steps. [JEE 2004]

5.

A 0.1 kg mass is suspended from a wire of negligible mass The length of the wire is 1 m and its crosssectional area is 4.9 × 10–7 m2. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 red s–1. If the Young's modulus of the material of the wire is n × 109 Nm–2, the value of n is [JEE 2010]

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Page # 20

ELASTICITY & THERMAL EXPANSION

: : ANSWER KEY : :

(OBJECTIVE PROBLEMS)

1. 7. 13. 19. 25. 31.

C B C D B B

2. 8. 14. 20. 26. 32.

37.

A,C,D 38.

A B C B A C

3. 9. 15. 21. 27. 33.

C A C C A A

4. 10. 16. 22. 28. 34.

C B C B A D

5. 11. 17. 23. 29. 35.

B B D A A C,D

6. 12. 18. 24. 30. 36.

B A C C C B,C

A,C,D

(SUBJECTIVE PROBLEMS)

1. (i) hollow sphere > solid sphere, (ii) hollow sphere = solid sphere 1 900 s s (iii) 2. (a) 450 (b) 449 (c) 12 : 14 : 59 (d) (i) 1 (ii) 451 450  10 –5 3. (a) h1 = h {1 + c },

A1 = A {1 + 2s },

v1 = Ah {1 + 3s }

volume of liquid Vw = Ah1(1 + L ) (b) h1 = h {1 + L }

(c) (i) L < 3c (ii) L > 3c (iii) L = 3c.

(d) V = Ah (L – 3c ) 

(e) 3hc = h1L

(f) (i) h1 (1 – 3c), (ii) h1(1 + L ),

 1  3 s   5. (a) V0d0g  1     (b) (i) L < 3s (ii) L > 3s (iii) L = 3s.   L

1 –6  4. YR =   27  37  10  /  C 9

6. 5 sec slow

7. 10000 N

8. 4 × 10–6 m/°C 9. 5 /3

13. (a) All tie (b) 50°X, 50°Y, 50°W.

2 × 10–4 C

10. h/5R 11. 3V/20 12.10 cm, 40 cm

14. –40°C or –40°F

(JEE PROBLEMS )

(iii) (1) L > 2c (2) L < 2c (3) L = 2c .

2.

B,D

3.

A

4.

l = 2s 5.

4

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FLUID THEORY AND EXERCISE BOOKLET CONTENTS S.NO.

TOPIC

PAGE NO.

1.

Fluid .......................................................................................................... 3

2.

Pressure in a fluid ................................................................................ 3 – 9

3.

Pascal's Principle ............................................................................... 9 – 10

4.

Archimede's Principle ....................................................................... 10 – 16

5.

Equation of Continuity ....................................................................... 16 – 17

6.

Bernoullis Equation ........................................................................... 17 – 21

14. Exercise - 1 ........................................................................................ 22 – 39 15. Exercise - 2 ........................................................................................ 40 – 42 16. Exercise - 3 ........................................................................................ 43 – 48 17. Exercise - 4 ........................................................................................ 49 – 51 18. Exercise - 5 ........................................................................................ 52 – 58 19. Answer key ......................................................................................... 59 – 60

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Page # 2

FLUID

IIT-JEE Syllabus : FLUID Pressure in a fluid; Pascal's law; Byoyancy, Streamline flow, Equation of continuity Bernoulli's theorem and its applications.

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FLUID

1.

FLUID: Fluid mechanics deals with the behaviour of fluids at rest and in motion. A fluid is a substance that deforms continuously under the application of a shear (tangential) stress no matter how small the shear stress may be. Thus, fluids comprise the liquid and gas (or vapor) phase of the physical forms in which matter exists. Density () : Mass of unit volume, Called density Density at a point of liquid described by   Lim V  0

m dm  V dV

density is a positive scalar quantity. SI unit = Kg/m3 CGS unit = gm/cm3 Dimension = [ML–3] Relative Density : It is the ratio of density of given liquid to the density of pure water at 4°C

R.D. 

Density of given liquid Density of pure water at 4C

Relative density or specific gravity is unit less, dimensionless. It is a positive scalar physical Quantity Value of R.D. is same in SI and CGS system due to dimensionless/unitless Specific Gravity : It is the ratio of weight of given liquid to the weight of pure water at 4°C Weight of given liquid

Specific Gravity =

Weight of pure water at 4C(9.81 kN / m3 )

   g =   g   = Relative density of w w

liquid i.e. than specfic gravity of a liquid is approximately equal to the relative density. For calculation they can be interchange

2.

PRESSURE IN A FLUID When a fluid (either liquid or gas) is at rest, it exerts a force pependicular to any surface in contact with it, such as a container wall or a body immersed in the fluid. While the fluid as a whole is at rest, the molecules that makes up the fluid are in motion, the force exerted by the fluid is due to molecules colliding with their surroundings. If we think of an imaginary surface within the fluid, the fluid on the two sides of the surface exerts equal and opposite forces on the surface, otherwise the surface would acceleratate and the fluid would not remain at rest. Consider a small surface of area dA centered on a point on the fluid, the normal force exerted by the fluid on each side is dF . The pressure P is defined at that point as the normal force per unit area, i.e.,, dF dA If the pressure is the same at all points of a finite plane surface with area A, then P

P

F A

where F is the normal force on one side of the surface. The SI unit of pressure is pascal, where 1 pascal = 1 Pa = 1.0 N/m2 One unit used principally in meterology is the Bar which is equal to 105 Pa. 1 Bar = 105 Pa

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FLUID

2.1

Atmospheric Pressure (P0) It is pressure of the earth's atmosphere. This changes with weather and elevation. Normal atmospheric pressure at sea level (an average value) is 1.013 × 105 Pa. Thus 1 atm = 1.013 × 105 Pa Note : Fluid pressure acts perpendicular to any surface in the fluid no matter how that surface is oriented. Hence, pressure has no intrinsic direction of its own, its a scalar. By contrast, force is a vector with a definite direction.

2.2

Variation in Pressure with depth If the weight of the fluid can be neglected, the pressure in a fluid is the same throughout its volume. But often the fluid's weight is not negligible and under such condition pressure increases with increasing depth below the surface. Let us now derive a general relation between the presure P at any point in a fluid at rest and the elevation y of that point. We will assume that the density  and the acceleration due to gravity g are the same throughout the fluid. If the fluid is in equilibrium, every volume element is in equilibrium.

dW

dy

dy y PA

Consider a thin element of fluid with height dy. The bottom and top surfaces each have area A, and they are at elevations y and y + dy above some reference level where y = 0. The weight of the fluid element is dW = (volume) (density) (g) = (A dy) () (g) or dW = g A dy What are the other forces in y-direction of this fluid element ? Call the pressure at the bottom surface P, the total y component of upward force is PA. The pressure at the top surface is P + dP and the total y-component of downward force on the top surface is (P + dP) A. The fluid element is in equilibrium, so the total y-component of force including the weight and the forces at the bottom and top surfaces must be zero. Fy = 0  PA – (P + dP) A – gAdy = 0 or

dP  – g dy

...(i)

This equation shows that when y increases, P decreases, i.e., as we move upward in the fluid, pressure decreases. If P1 and P2 be the pressures at elevations y1 and y2 and if  and g are constant, then integrating Eq.(i) , we get P2

P1

or

dP  – g

y2

P2 y2

P1

y1

dy

y1

P2 – P1 = – g (y2 – y1)

...(ii)

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FLUID

It's often convenient to express Eq. (ii) in terms of the depth below the surface of a fluid. Take point 1 at depth h below the surface of fluid and let P represents pressure at this point. Take point 2 at the surface of the fluid, where the pressure is P0 (subscript zero for zero depth). The depth of point 1 below the surface is, h = y2 – y 1 and Eq. (ii) becomes P0 – P = – g (y2 – y1) = –  gh 

P = P0 + gh

...(iii)

Thus, pressure increases linearly with dpeth, if  and g are uniform, A graph between P and h is shown below. P

P  P0  gh

P0

P0

A

B

h

P0 h

PA  PB  P0  gh Further, the pressure is the same at any two points at the same level in the fluid. The shape of the container does not matter. 2.3.

Barometer

Vacuum (P = 0)

It is a device used to measure atmospheric pressure. In principle, any liquid can be used to fill the barometer, but

h

mercury is the substance of choice because its great density

1

makes possible an instrument of reasonable size.

2

P1 = P 2 Here,

P1 = atompsheric pressure (P0)

and

P2 = 0 + gh =  gh

Here,

 = density of mercury P0 = gh

Thus, the mercury barometer reads the atmosphereic pressure (P0) directly from the height of the mercury column. For example if the height of mercury in a barometer is 760 mm, then atmospheric pressure will be, P0 = gh = (13.6 × 103) (9.8) (0.760)= 1.01 × 105 N/m2 2.4

Force on Side Wall of Vessel Force on the side wall of the vesel can not be directly determined as at different depths pressures are different. To find this we cosider a strip of width dx at a depth x from the surface of the liquid as shown in figure, and on this strip the force due to the liquid is given as : dF = xg × bdx This force is acting in the direction normal to the side wall. Net force can be evaluated by integrating equation h

F  dF 

 xgbdx 0

F

gbh2 2

x

dx dF

h b a

...(2.4)

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FLUID

Average Pressure on Side Wall The absolute pressure on the side wall cannot be evaluated because at different depths on this wall pressure is different. The average pressure on the wall can be given as :  p  av 

F 1 gbh 2 1  gh = bh 2 bh 2

...(2.5)

Equation (2.5) shows that the average pressure on side vertical wall is half of the net pressure at the bottom of the vessel. 2.6

Torque on the Side Wall due to Fluid Pressure As shown in figure, due to the force dF, the side wall experiences a torque about the bottom edge of the side which is given as

d  dF  (h – x) = xgb dx (h – x) h

  d  gb(hx – x2 )dx

This net torque is

 0

 h3 h3  1  gb –   gbh 3 2 3  6  2.7

Manometer It is a device used to measure the pressure of a gas inside a container. The U-shaped tube often contains mercury. P1 = P 2 Here, P1 = pressure of the gas in the container (P) and 

P0

P2 = atmospheric pressure (P0) + gh

h

P = P0+ hg 1

This can also be written as

2

P – P0 = gauge pressure = hg Here,  is the density of the liquid used in U - tube Thus by measuring h we can find absolute (or gauge) pressure in the vessel.

Ex.1

Two liquid which do not react chemically are placed in a bent tube as shown in figure. Find out the displacement of the liquid in equillibrium position.

x 2

Sol.

 The pressure at the interface must be same, calculated via either tube. Since both tube all open to the atmosphere, we must have.

x

x

2g( – x) = g( + x)  x = /3

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FLUID Ex.2

Three liquid which do not react chemically are placed in a bent tube as shown in figure (initially) then fluid out the displacement of the liquid in equillibrium position.

 

Sol.

3 

2

Let us assume that level of liquid having density 3 displaced below by x as shown in figure below.

x x  – x

g + 2gx = 3( – x)g 2.8

x = 2/5 Pressure Distribution in an Accelerated Frame We've already discussed that when a liquid is filled in a container, generally its free surface remains horizontal as shown in figure (a) as for its equilibrium its free surface must be normal to gravity i.e. horizontal. Due to the same reason we said that pressure at every point of a liquid layer parallel to its free surface remains constant. Similar situation exist when liquid is in an accelerated frame as shown in figure (b). Due to acceleration of container, liquid filled in it experiences a pseudo force relative to container and due to this the free surface of liquid which normal to the gravity now is filled as  a   tan –1    g

...(2.22) a

A

a 

geff

g

(b) (a) Now from equilibrium of liquid we can state that pressure at every point in a liquid layer parallel to the free surface (which is not horizontal), remains same for example if we find pressure at a point A in the acceleratd container as shown in figure (a) is given as

PA = P0 + h

a2  g2

...(2.23)

Where h is the depth of the point A below the free surface of liquid along effective gravity and P0 is the atmopheric pressure acting on free surface of the liquid.

a

l1

h 

A

(c)

a

h

l2

(d)

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FLUID

The pressure at point A can also obtained in an another way as shown in figure (b). If l1 and l2 are the vertical and horizontal distances of point A from the surface of liquid then pressure at point A can also be given as PA = P0 + l1 g = P0 + l2 a

(2.24)

Here l1 g is the pressure at A due to the vertical height of liquid above A and according to Pascal's Law pressure at A is given as PA = P0 + l1 g

...(2.25) h a 2  g2 g

Here we can write l1 as

l1  h sec  

or from equation (2.25)

PA  P0  h a 2  g2

Similarly if we consider the horizotnal distance of point A from free surface of liquid, which is l2 then due to pseudo acceleration of container the pressure at point A is given as P A = P 0 + l2  a Here l2 is given as

l2  h cos ec 

...(2.26)

h g2  a 2 a

PA  P0  h g2  a 2

From equation (2.24), we have

Here students should note that while evaluating pressure at point A from vertical direction we haven't mentioned any thing about pseudo acceleration as along vertical length l1, due to pseudo acceleration at every point pressure must be constant similarly in horizontal direction at every point due to gravity pressure reamins constant. Ex.3

Figure shows a tube in which liquid is filled at the level. It is now rotated at an angular frequency w about an axis passing through arm A find out pressure difference at the liquid interfaces.  B A A

Sol.

 To solve the problem we take a small mass dm from the

B

axis at ‘a’ distance x in displaced condition. Net inward force = (P + dP) A – PA dm  Adx

P + dP

P

x0

This force is balanced by centripetal force in equilibrium 

2

 A dP = dm 2x = Adx 2x 

 dP    xdx x0

xw

dx

x  – x0

P = 2

 xdx = x0g

x0

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FLUID 

Ex.4

A liquid of density  is in a bucket that spins with angular velocity  as shown in figure. Show that the pressure at a radial distance r from the axis is

2 r 2 2 where P0 is the atmospheric pressure. P  P0 

Sol.

Consider a fluid particle P of mass m at coordinates (x, y). From a non-inertial rotating frame of reference two forces are acting on it, (i) pseudo force (mx 2) (ii) weight (mg) in the directions shown in figure. Net force on it should be perpendicular to the free surface (in equilibrium). Hence. tan  

mx 2 x 2  mg g

y

x

dy 

y

0

0

or

P x

dy x2  dx g

P  ( x, y )

x2 .dx g

x 2 2 2g

P

mx2 

This is the equation of the free surface of the liquid, which is a parabola.

3.

y

mg

Fnet

r 22 2g

At x = r,

y

P(r) = P0 + gy

or

P(r) = P0 +

P0 yP(r) x=r

 2r 2 2

PASCAL'S PRINCIPLE Some times while dealing with the problems of fluid it is desirable to know the pressure at one point is pressure at any other point in a fluid is known. For such types of calculations Pascal's Law is used extensively in dealing of static fluids. It is stated as "The pressure applied at one point in an enclosed fluid is transmitted uniformly to every part of the fluid and to the walls of the container." One more example can be considered better to explain the concept of Pascal's Principle. Consider the situation shown in figure, a tube having two different cross section S1 and S2, with pistons of same cross sections fitted at the two ends. 2 1 F1

F2

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FLUID

If an external force F1 is applied to the piston 1, it creates a pressure p1 = F1/S1 on the liquid enclosed. As the whole liquid is at the same level, everywhere the pressure in the liquid is increased by p1. The force applied by the liquid on the piston 2 can be given as F2 =p2 × S2, and as the two pistons are at same level p2 = p1. Thus F2 = p2 × S2 F1 F2 = S  S2 1

....(2.21)

Equation (2.21) shows that by using such a system the force can be amplified by an amount equal to the ratio of the cross section of the two pistons. This is the principle of hydraulic press, we'll encounter in next few pages.

3.1

The Hydraulic Lift Figure shows how Pascal's principle can be made the basis for a hydraulic lift. In operation, let an external force of magnitude F1 be exerted downward on the left input piston, whose area is S1. It result a force F2 which will act on piston 2 by the incompressible liquid in the device. Here

F2 = p2 × S2

And

p2 = pB –  gh F1

S1

S2 F2 h

Where pB is the pressure on the bottom of the device which can be given as : pB = p1 +  gh Thus p2 = p1 and F2 = p1 S2 S2 or F2 = F1 × S 1 If S2 >> S1  F2 >> F1 4.

ARCHIMEDE'S PRINCIPLE If a heavy object is immersed in water, it seems to weight less than when it is in air. This is because the water exerts an upward force called buoyant force. It is equal to the weight of the fluid displaced by the body. A body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the displaced fluid. This result is known as Archimedes' principle. Thus, the magnitude of buoyant force (F) is given by, F  Vi L g Here, Vi = immersed volume of solid and

L = density of liquid

g = acceleration due to gravity

Note : Point of Application of buoyant force is centre of liquied displaced

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Page # 11

FLUID Proof Consider an arbitraily shaped body of volume V placed in a container filled with a fluid of density L. The body is shown completely immersed, but complete immersion is not essential to the proof. To begin with, imagine the situation before the body was immersed. The region now occupied by the body was filled with fluid, whose weight was VL g . Because the fluid as a whole was in hydrostatic

V equilibrium, the net upwards force (due to difference in pressure at different depths) on the fluid in that region was equal to the weight of the fluid occuping that region. Now, consider what happens when the body has displaced the fluid. The pressure at every point on the surface of the body is unchanged from the value at the same location when the body was not present. This is because the pressure at any point depends only on the depth of that point below the fluid surface. Hence, the net force exerted by the surrounding fluid on the body is exactly the same as the exerted on the region before the body was present. But we know the latter to be VL g , the weight of the displaced fluid. Hence, this must also be the buoyant force exerted on the body. Archimedes' principle is thus, proved. Ex.5

Beaker cicular cross-section of radius 4 cm is filled with mercury upto a height of 10 cm. Find the force exerted by the mercury on the bottom of the beaker. The atmopheric pressure = 105 N/m2. Density of mercury = 13600 kg/m3. Take g = 10 m/s2

Sol.

The pressure at the surface = atmospheric pressure = 105 N/m2. The pressure at the bottom = 105 N/m2 + hg kg   = 105 N/m2 + (0.1 m)  13600 3  m

m   10 2   s 

= 105 N/m2 + 13600 N/m2 = 1.136 × 105 N/m2 The force exerted by the mercury on the bottom = (1.136 × 105 N/m2) × (3.14 × 0.04 m × 004 m) = 571 N Ex.6

A cubical block of iron 5 cm on each side is floating on mercury in a vessel. (i) What is the height of the block above mercury level ? (ii) What is poured in the vessel until it just covers the iron block. What is the height of water column. Density of mercury = 13.6 gm/cm3 Density of iron 7.2 gm/cm3

Sol.

Case-I : Suppose h be the height of cubical block of iron above mercury. Volume of iron block = 5 ×5 × 5 = 125 cm3 Mass of iron block = 125 × 7.2 = 900 gm Volume of mercury displaced by the block = 5 × 5 × (5 – h) cm3 Mass of mercury displaced

= 5 × 5 (5 – h) × 13.6 gm

By the law floatation, weight of mercury displaced = weight of iron block 5 × 5 (5 – h) × 13.6 = 900 or

(5 – h) =

900 = 2.65  25  13.6

h = 5 – 2.65 = 2.35 cm

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FLUID

Case - II : Suppose in this case height of iron block in water be x. The height of iron block in mercury will be (5 – x) cm. Water

h

Mercury

Mercury

(a)

(b)

Mass of the water displaced = 5 × 5 × (x) × 1 Mass of mercury displaced = 5 × 5 × (5 – x) × 13.6 So, weight of water displaced + weight of mercury displaced = weight of iron block or 5 × 5 × x × 1 + 5 × 5 × (5 – x) × 13.6 = 900 or x = (5 – x) × 13.6 = 36 x = 2.54 cm Ex.7 A tank contianing water is placed on spring balanced. A stone of weight w is hung and lowered into the water without touching the sides and the bottom of the tank. Explain how the reading will change. Sol. The situation is shown in figure. Make free-body diagrams of the bodies separately and consider their equilibrium. Like all other forces, buoyancy is also exerted equally on the two bodies in contact. Hence it the water exerts a buoyant force, say, B on the stone upward, the stone exerts the same force on the water downward. The forces acting on the 'water + container' system are : W, weight of the system downward, B, buoyant force of the stone downard, and the force R of the spring in the upward direction. For equilibrium R=W+B Thus the reading of the spring scale will increase by an amount equal to the weight of the liquid displaced, that is, by an amount equal to the buoyant force. Ex.8

Sol.

A cylindrical vessel containing a liquid is closed by a smooth piston of mass m as shown in figure. The area of cross-section of the piston is A. If the atmopheric pressure is P0, find the pressure of the liquid just below the prism. Let the pressure of the liquid just below the piston be P. The forces acting on the piston are (a) its weight, mg (downward) (b) force due to the air above it, P0A(downward) (c) force due to the liquid below it, PA (upward) If the piston is in equilibrium PA = P0A + mg or P = P0 +

mg A

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FLUID

Ex.9 A rubber ball of mass m and radius r is submerged in water to a depth h released. What height will the ball jump up to above the surface of the water ? Neglect the resistance of water and air. Sol. Let the ball go up by x above the level of water. Let us now consider energy conservation between the initial and final positions. In both the positions kinetic energy of the body is zero. The potential energy in the first position with reference to the water 4 3  level is – mgh plus the work done by an external agent against the buoyant force which is  r g h, 3

where  is the density of the water or

4 3  –mgh +  r g h = mgx  3

x

( 4 / 3)r 3  – m h m

Ex.10 A cube of wood supporting a 200 g mass just floats in water. When the mass is removed, the cube rises by 2 cm. What is the size of the cube ? Sol. If, l = side of cube, h = height of cube above water and  = density of wood. Mass of the cube = l3  Volume of cube in water = l2 (l – h) Volume of the displaced water

= l 2 ( l – h)

As the tube is floating weight of cube + weight of wood = weight of liquid displaced or

l3 + 200 = l2 (l – h)

...(2.10)

After the removal of 200 gm mass, the cube rises 2 cm. = l2 × {l – (h + 2)} Volume of cube in water or

l2 × {l – (h + 2)} = l3 

...(2.11)

Substituting the value of l  from equation (2.11) in equation (2.10), we get 3

l2 × {l – (h + 2)} + 200 = l2 (l – h) or

l3 – l2h – 2l2 + 200 = l3 – l2 2l2 = 200  l = 10 cm

Ex.11 A boat floating in water tank is carrying a number of large stones. If the stones were unloaded into water, what will happen to water level ? Given the reason in brief. Sol.

Suppose W and w be the weights of the boat and stones respectively. First, we consider that the boat is floating. It will displaced (W + w) × 1 cm3 of water. Thus displaced water = (W + w) cm3

[As density of water = 1 gm/cm3]

Secondly, we consider that the stones are unloaded into water. Now the boat displaces only W × 1 cm3 of water. If  be the density of stones, the volume of water displaced by stones = w/ cm3 As  > 1, hence w/ < w, thus we have Now

(W + w/) < (W + w)

This shows that the volume of water displaced in the second case is less than the volume of water displaced in the first case. Hence the level of water will come down.

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FLUID

Ex.12 Two solid uniform spheres each of radius 5 cm are connected by a light string and totally immersed in a tank of water. If the specific gravities of the sphere are 0.5 and 2, find the tension in the string and the contact force between the bottom of tank and the heavier sphere. Sol. The situation is shown in figure Let the volume of each sphere be V m3 and density of water be  kg/m3. Upward thrust on heavier sphere = v  g Weight of the heavier sphere = V × 2 × g For heavier sphere, T T+R+Vg=V×2×g ...(2.12) where R is the reaction at the bottom. Similarly for lighter sphere T + V × 0.5 ×  g = V  g ..(2.13) R Subtracting equation (2.13) from equation (2.12), we have R + 0.5 V  g = V  g ...(2.14) or R = 0.5 V  g ...(2.15) From equation (2.13) T = 0.5 V  g 4   0.5    3.14  5 3  10 6  × 1000 × 9.8 = 2.565 N 3  R = 2.565 N Similarly A rod of length 6 m has a mass of 12 kg. If it is hinged at one end at a distance of 3 m below a water surface, (i) What weight must be attached to other end of the rod so that 5 m of the rod is submerged ? (ii) Find the magnitude and direction of the force exerted by the hinge on the rod. The specific gravity of the material of the rod is 0.5. Sol.

Let AC be the submerged part of the rod AB hinged at A

B

FB

as shown in figure. G is the centre of gravity of the rod and G is the centre of buoyancy through which force of buoyancy FB acts vertically upwards.

Water Surface

C G

R

Since the rod is uniform,

w x

G' The weight of part AC will be

5  12  10kg 6

The buoyance force on rod at G is FB 

A [Because AB = 6 m and W AC = 5 m] Hinge

10 = 20 kg weight 0.5

(i) Let x be weight attached at the end B. Balancing torques about A, we get W × AG + x × AB = FB × AG 12 + 3 + x × 6 = 20 × (5/2) [As AG = 5/2] Solving we get x = 2.33 kg (ii) Suppose R be the upward reaction on the hinge, then in equilibrium position, we have W + x = FB + R or R = W + x – FB = 12 + 2.33 – 20 = – 5.67 kg. wt. Negative sign shows that the reaction at the hinge is acting in the downward direction. The magnitude of the reaction is 5.67 kg. wt.

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Page # 15

FLUID

Ex.13 A cylinder of area 300 cm2 and length 10 cm made of material of speicifc gravity 0.8 is floated in water with its axis vertical. It is then pushed downward, so as to be just immersed. Calculate the work done by the agent who pushes the cylinder into the water. Sol. Weight of the cylinder = (300 × 10–4) × (10 × 10–2) × 800 kgf = 2.4 kgf Let x be the length of the cylinder inside the water. Then by the law of floatation 2.4 g = (300 × 10–4 x) × 1000 g or

x = 0.08 m

When completely immersed, Fb(buoyant force) = (300 × 10–4 × 0.1) × 1000 × g = 3 g N Thus to immerse the cylinder inside the water the external agent has to push it by 0.02 m against average upward thrust. Increase in upward thrust = 3g – 2.4 g = 0.6 g N Since this increase in upthrust takes place gradually from 0 to 0.6 g, we may take the average upthrust against which work is done as 0.3 g N. work done = 0.3 g × 0.02 = 0.0588 J Ex.14 A piece of an alloy of mass 96 gm is composed of two metals whose specific gravities are 11.4 and 7.4. If the weight of the alloy is 86 gm in water, find the mass of each metal in the alloy. Sol. Suppose the mass of the metal of specific gravity 11.4 be m and the mass of the second metal of specific gravity 7.4 will be (96–m) m cm 3 114 .

Volume of first metal =

96 – m cm 3 7.4

Volume of second metal =

Total volume =

Buoyancy force in water

m 96 – m  114 . 7.4

96 – m   m    gm weight  114 . 7.4 

 m  (96 – m)    Apparent wt. in water = 96 –  114 7.4   .  According to the given problem,  m  ( 96 – m)  96 –     86 .  7.4   114

or

m ( 96 – m)   10 114 . 7.4

Solving we get,

m = 62.7 gm

Thus mass of second metal is

= 96 – 62.7 = 33.3 gm

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Page # 16

FLUID

NOTE :ASSUMPTIONS OF IDEAL FLUID (1) Fluid is incompressible : density of fluid remain constant through out the fluid. (2) Fluid is non-viscous : fluid friction is absent (3) Doesn't show rotational effect : If we release any body in the flowing section there it will not rotate about its C.O.M. (4) Stream line flow : velocity of fluid at any particular point remains constant with time It may vary with position. 5.

EQUATION OF CONTINUITY

A2 This equation defines the steady flow of fluid in a tube. It states that if flow of a fluid is a steady then the mass of fluid entering per second at one end is equal to the mass of fluid leaving per second

v2

at the other end. Figure shown a section of a tube in which at the ends, the cross

A1

sectional area are A1 and A2 and the velocity of the fluid are V1 and v2 respectivley. According to the equation of continuity, if flow is steady mass of fluid entering at end A1 per second = mass of fluid leaving the end A2 per second. dV  A1v1 dt Hence mass entering per second at A1 is = A1 v1  Similarly mass leaving per second at A2 is = A2 v2  According to the definition of steady flow A1 v1  = A2 v2  or

A1v1 = A2v2

Equation above in known as equation of continuity, which gives that in steady flow the product of cross-section and the speed of fluid everywhere remains constant.

5.1

Freely Falling Liquid When liquid falls freely under gravity, the area of cross section of the stream continuously decreases, as the velocity inreases. For example, we consider water coming out from a tap, as shown in figure. Let its speed near the mouth of tap is v0 and at a depth h it is v, then we have

v 2  v 20  2gh If cross section of tap is A then according to the equation of continuity, the cross section at point M (say a) can be given as v 0 A  a v 20  2gh

or

a

v0 A v 20

 2gh

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Page # 17

FLUID 6.

BERNOULLIS EQUATION It relates the variables describing the steady laminer of liquid. It is based on energy conservation. Assumptions The fluid is incompressible, non-viscous, non rotational and streamline flow. dm

T

T A2 V2 at time t = 0

P2

at time t + dt

P2

A1

A1 S P1

A2

dx2

h2

S

P1

V1 h1

dx1

Mass of the fluid entering from side S dm1 =  A1 dx1 =  dv1 The work done in this displacement dx1 at point S is WP = F1dx1 = P1A1dx1 1

WP = P1dV1

{ A1dx1 = dV1}

1

At the same time the amount of fluid moves out of the tube at point T is dm2 = dV2 According to equation of coutniuity

dm1 dm 2  dV1 = dV2 = dv dt dt

The work done in the displacement of dm2 mass at point T WP = P2dV2 2

Now applying work energy theorem. 1  1  2 2 WP + WP = (Kf + Uf) – (Ki + Ui)  P1dV – P2dV =  dV  2  dVgh 2  –  dV 1  dVgh 1  1 2 2  2  1 1 P1 – P 2 = V22 + gh2 – V12 + gh1 2 2

P1 + gh1 +

1 1 1 V12 = P2 + gh2 + V2  P + gh + V2 = constant 2 2 2

P 1 v2 h = constant g 2 g P where = pressure head g

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Page # 18 6.1

FLUID

APPLICATION OF BERNOULLIS PRINCIPLE Magnus effect : When a spinning ball is thrown, it deviates from its usual path in flight. This effect is called Magnus effect and plays an important role in tennis, cricket and soccer, etc., as by applying appropriate spin the moving ball can be made to curve in any desired direction. If a ball is moving from left to right and also spinning about a horizontal axis perpendicular to the direction of motion as shown in figure, then relative to the ball air will be moving from right to left. V+r

V O

Vertical plane Curved path Usual path

v

V A0 A 1

(A)

(B)

(C)

The resultant velocity of air above the ball will be (V+r) while below it (V –r) (shown figure). So in accordance with Bernoulli's principle pressure above the ball will be less than below it. Due to this difference of pressure an upward force will act on the ball and hence the ball will deviate from its

usual path OA0 and will hit the ground at A1 following the path OA1 (figure shown) i.e., if a ball is thrown with back spin, the pitch will curve less sharply prolonging the flight. Similarly if the spin is clockwise, i.e., the ball is thrown with top-spin, the force due to pressure difference will act in the direction of gravity and so the pitch will curve more sharply shortening the flight. A1 A0 A0

A2

Horizontal plane

Furthermore, if the ball is spinning about a vertical axis, the curving will be sideays as shown in figure. producing the so called out swing or in swing. Action of Atomiser : The action of aspirator, carburettor, paint-gun, scent-spray or insect-sprayer is based on Bernoulli's principle. In all these by means of motion of a piston P in a cylinder C high speed air is passed over a tube T dipped in liquid L to be sprayed. High speed air creates low pressure over the tube due to which liquid (paint, scent, insecticide or petrol) rises in it and is then blown off in very small droplets with expelled air.

p

T c

L

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Page # 19

FLUID

Working of Aeroplane : This is also based on Bernouilli's principle. The wings of the aeroplane are having tapering as shown in figure. Due to this specific shape of wings when the aeroplane runs, air passes at higher speed over it as compared to its lower surface. This difference of air speeds above and below the wings, in accordance with Bernoulli's principle, creates a pressure difference, due to which an upward force called ' dynamic lift' ( = pressure difference × area of wing) acts on the plane. If this force becomes greater than the weight of the plane, the plane will rise up.

v large, p small

v small, p large

A

Ex.15 If pressure and velocity at point A is P1 and V1 respectively & at point B is P2, V2 is the figure as shown. Comment on P1 and P2. Sol.

P1

From equation of continuity A1V1 = A2V2

V1

B V2

here A1 > A2  V1 < V2

....(1)

from Bernoulli's equation. We can write 1 P1 + V12 = P2 + 1/2 V22 = 2 after using equation (1) P1 > P2 TORRICIELLI'S LAW OF EFFLUX (VELOCITY OF EFFLUX) Crossectional Area of hole at A is greater than B If water is come in tank with velocity vA and going out side with velocity vB then A1 vA = A2vB  A1 > A2  vB >> vA on applying bernoulli theorem at A and B PA  ghA 

P0

h v

H B

1 2 1 v A  PB  ghB  vB2 2 2

PA = P B = P 0 and hA – hB = h 1 2 2  gh = (vB – v A ) 2 1 gh =  v2 2 v=

A

H–h

[vB >> vA] [vB2 – vA2 = v2]

2 gh

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Page # 20

FLUID

Range (R) Let us find the range R on the ground. Considering the vertical motion of the liquid. (H – h) =

1 2 gt 2

t

or

2(H – h) g

Now, considering the horizontal motion,

 2(H – h)    R = vt or R  ( 2gh ) or R  2 h(H – h)  g   From the expression of R, following conclusions can be drawn, (i) Rh = R H – h as R h  2 h(H – h) and R H – h  2 (H – h)h This can be shown as in Figure

h

H–h H

H and Rmax = H. (ii) R is maximum at h  2 Proof : R2 = 4 (Hh – h2)

v

O h

H–h

dR 2 0 dh or H – 2h = 0 or h = H/2 That is, R is maximum at h=H/2 For R to be maximum.

H H H –   H Proved 2 2 Ex.16 A cylindrical dark 1 m in radius rests on a platform 5 m high. Initially the tank is filled with water up to a height of 5 m. A plug whose area is 10–4 m2 is removed from an orifice on the side of the tank at the bottom Calculate (a) initial speed with which the water flows from the orifice (b) initial speed with which the water strikes the ground and (c) time taken to empty the tank to half its original volume (d) Does the time to be emptied the tank depend upon the height of stand. Sol. The situation is shown in figure (a) As speed of flow is given by A vH = ( 2gh) R max  2

and

or

=

2  10  5

~ –

5m

10 m/s

A0

(b) As initial vertical velocity of water is zero, so its vertical velocity when it hits the ground v V  2gh =

2  10  5

~ –

10 m/s

5m

So the initial speed with which water strikes the ground. v  vH2  v 2V = 10 2 = 14.1 m/s (c) When the height of water level above the hole is y, velocity of flow will be v  2gy and so rate of flow dV  A 0 v  A 0 2gy dt

or

–Ady = ( 2gy ) A0 dt

[As dV = – A dy]

Which on integration improper limits gives 0

t

A 2   12 [ H – H'] So t  A0 g 2gy 0 10 –4 H (d) No, as expression of t is independent of height of stand.

A 0 dt  t 

2 [ 5 – (5 / 2)] = 9.2 × 103s 10

~ –

2.5 h

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Page # 21

FLUID 6.1

Venturimeter Figure shows a venturimeter used to measure flow speed in a pipe of non - uniform cross-section. We apply Bernoulli's equation to the wide (point 1) and narrow (point 2) parts of the pipe, with h1 = h2 1 1 P0 P0 P1  v 12  P2  v 22 2 2 h A 1v 1 From the continuity equation v2 = A 2 v2 Substituting and rearranging, we get A v1 H 2 A2 A  1 P1 – P2   v 12  12 – 1 2  A2  Because A1 is greater than A2, v2 is greater than v1 and hence the pressure P2 is less than P1. A net force to the right acceleration the fluid as it enters the narrow part of the tube (called throat) and a net force to the left slows as it leaves. The pressure difference is also equal to  gh, where h is the difference in liquid level in the two tubes. Substituting in Eq. (i), we get v1 

2gh 2

 A1    – 1  A2  6.2

Pitot Tube It is a device used to measure flow velocity of fluid. It is a U shaped tube which can be inserted in a tube or in the fluid flowing space as shown in vg figure shown. In the U tube a liquid which is immiscible with the fluid is filled upto a level C and the short opening M is placed in the fluid flowing space against the flow so that few of the fluid particles entered into the tube and B A h exert a pressure on the liquid in limb A of U tube. Due to this the liquid level changes as shown in figure shown. At end B fluid is freely flowing, which exert approximately negligible pressure on this liquid. The pressure difference at ends A and B can be given by measuring the liquid level difference h as It is a gas, then PA – PB = hg It if the a liquid of density , then PA – PB = h( – g)g Now if we apply Bernoulli's equation at ends A and B we'l have 1 2 1 2 v g  PA – PB  hg 0 + 0 + PA = v g + 0 + PB or 2 2 Now by using equations, we can evaluate the velocity v, with which the fluid is flowing. Note : Pitot tube is also used to measure velocity of aeroplanes with respect to wind. It can be mounted at the top surface of the plain and hence the velocity of wind can be measured with respect to plane.

6.3

SIPHON : It is a pipe used to drain liquid at a lower height but the pipe initially rises and then comes down let velocity of outflow is v & the pipe is of uniform cross-section A. Applying bernoulli's equation between P (top of tank) & R (opening of pipe) we get

Q h2

P h=0 h1

R V

1 1 1 2 (P + gh + v2)P = (P + gh + v2)R  P + 0 + 0 = P0 – gh1 + v 2 2 2 here velocity is considered zero at P since area of tank is very large compared to area of pipe  v  2gh Naturally for siphon to work h1 > 0 Now as area of pipe is constant so by equation of continuity as Av = constant so velcoity of flow inside siphon is also constant between Q & R (P + gh + 1/2 v2)Q = (P + gh + 1/2 v2)R  PQ + gh2 = P0 – gh1 (v is same)  PQ = P0 – g (h1 + h2) as PQ = 0  P0  g (h1 + h2) means (h1 + h2) should not be more than P/g for siphon to work

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Page # 22

FLUID

(ONLY ONE OPTION IS CORRECT)

Exercise - I

1. STATIC FLUID 1. A bucket contains water filled upto a height = 15 cm. The bucket is tied to a rope which is passed over a frictionless light pulley and the other end of the rope is tied to a weight of mass which is half of that of the (bucket + water). The water pressure above atmosphere pressure at the bottom is (A) 0.5 kPa (B) 1 kPa (C) 5 kPa (D) None 2. Some liquid is filled in a cylindrical vessel of radius R. Let F1 be the force applied by the liquid on the bottom of the cylinder. Now the same liquid is poured into a vessel of uniform square cross-section of side R. Let F2 be the force applied by the liquid on the bottom of this new vessel. Then : F2 (A) F1 = F2 (B) F1 = (C) F1 = F2 (D) F1 = F2  3. A liquid of mass 1 kg is filled in a flask as shown in figure. The force exerted by the flask on the liquid is (g = 10 m/s2) [Neglect atmospheric pressure] (A) 10 N (B) greater than 10 N (C) less than 10 N (D) zero 4. A U-tube having horizontal arm of length 20 cm, has uniform cross-sectional area = 1 cm2. It is filled with water of volume 60 cc. What volume of a liquid of density 4 g/cc should be poured from one side into the U-tube so that no water is left in the horizontal arm of the tube? (A) 60 cc (B) 45 cc (C) 50 cc (D) 35 cc 5 . In the figure shown, the heavy cylinder (radius R)

resting on a smooth surface separates two liquids of densities 2 and 3. The height ‘h’ for the equilibrium of cylinder must be 3 R h 2 R 3 (C) R 2 (D) None 2 6. A light semi cylindrical gate of radius R is piovted at its mid point O, of the diameter as shown in the figure holding liquid of density . The force F required to prevent the rotation of the gate is equal t

(A) 3R/2

7. The pressure at the bottom of a tank of water is 3P where P is the atmospheric pressure. If the water is drawn out till the level of water is lowered by one fifth., the pressure at the bottom of the tank will now be (A) 2P (B) (13/5) P (C) (8/5) P (D) (4/5) P 8. An open-ended U-tube of uniform cross-sectional area contains water (density 1.0 gram/centimeter3) standing initially 20 centimeters from the bottom in each arm. An immiscible liquid of density 4.0 grams/ centimeter 3 is added to one arm until a layer 5 centimeters high forms, as shown in the figure above. What is the ratio h2/h1 of the heights of the liquid in the two arms ? 5cm h2

h1

(A) 3/1

(B) 5/2

(C) 2/1

(D) 3/2

9. The vertical limbs of a U shaped tube are filled with a liquid of density  upto a height h on each side. The horizontal portion of the U tube having length 2h contains a liquid of density 2. The U tube is moved horizontally with an accelerator g/2 parallel to the horizontal arm. The difference in heights in liquid levels in the two vertical limbs, at steady state will be (A) 2h/7 (B) 8h/7 (C) 4h/7 (D) None 10. The area of cross-section of the wider tube shown in figure is 800 cm2. If a mass of 12 kg is placed on the massless piston, the difference in heights h in the level of water in the two tubes is : 12kg

h

(B) R

(A) 10 cm

(B) 6 cm

(C) 15 cm

(D) 2 cm

2. ACCELERATED FLUID 11. A fluid container is containing a liquid of density  is is accelerating upward with acceleration a along the inclined place of inclination  as shwon. Then the angle of inclination  of free surface is :

O

a 

R F

(A) 2R3g (C)

2R2 lg 3

(B) 2gR3l (D) none of these

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Page # 23

FLUID

 g  (A) tan    g cos  

–1  a  g sin   (B) tan    g cos  

–1  a – g sin   (C) tan    g(1  cos ) 

–1  a – g sin   (D) tan    g(1 – cos ) 

–1

12. Figure shows a three arm tube in which a liquid is filled upto levels of height l. It is now rotated at an angular frequency  about an axis passing through arm B. The angular frequency  at which level of liquid of arm B becomes zero.

B

A

 l

2g 3l

(B)

g l

(A) 2

abc g

(B) 2

g da

(C) 2

bc da (D) 2 dg g

l

l

(A)

17. A cuboidal piece of wood has dimensions a, b and c. Its relative density is d. It is floating in a larger body of water such that side a is vertical. It is pushed down a bit and released. The time period of SHM executed by it is :

18. A slender homogeneous rod of length 2L floats partly immersed in water, being supported by a string fastened to one of its ends, as shown. The specific gravity of the rod is 0.75. The length of rod that extends out of water is

C

l

are connected by weightless wire and placed in a large tank of water. Under equilibrium the lighter cube will project above the water surface to a height of (A) 50 cm (B) 25 cm (C) 10 cm (D) zero

3g l

(C)

(D)

3g 2l

13. An open cubical tank was initially fully filled with water. When the tank was accelerated on a horizontal plane along one of its side it was found that one third of volume of water spilled out. The acceleration was (A) g/3 (B) 2g/3 (C) 3g/2 (D) None 3. PASCAL'S LAW & ARCHIMEDE'S PRINCIPLE 14. A cone of radius R and height H, is hanging inside a liquid of density  by means of a string as shown in the figure. The force, due to the liquid acting on the slant surface of the cone is (neglect atmospheric pressure)

2L

(A) L

(B)

1 L 2

(C)

1 L 4

(D) 3 L

19. A dumbbell is placed in water of density . It is observed that by attaching a mass m to the rod, the dumbbell floats with the rod horizontal on the surface of water and each sphere exactly half submerged as shown in the figure. The volume of the mass m is negligible. The value of length l is l 2M,V

M,V m Water

H

d R

(A)gHR2

(B) HR2

4 2 (C) gHR2 (D) gHR2 3 3

15. A heavy hollow cone of radius R and height h is placed on a horizontal table surface, with its flat base on the table. The whole volume inside the cone is filled with water of density . The circular rim of the cone’s base has a watertight seal with the table’s surface and the top apex of the cone has a small hole. Neglecting atmospheric pressure find the total upward force exerted by water on the cone is (A) (2/3)R2 hg (B) (1/3)R2 hg (C) R2 hg (D) None 16. Two cubes of size 1.0 m sides, one of relative density 0.60 and another of relative density = 1.15

d(V  3M) (A) 2(V  2M) 

d(V  2M) (B) 2(V  3M) 

d(V  2M) (C) 2(V  3M) 

d(V  2M) (D) 2(V  3M) 

20. Two bodies having volumes V and 2V are suspended from the two arms of a common balance and they are found to balance each other. If larger body is immersed in oil (density d1 = 0.9 gm/cm3) and the smaller body is immersed in an unknown liquid, then the balance remain in equilibrium. The density of unknown liquid is given by : (A) 2.4 gm/cm3 (B) 1.8 gm/cm3 3 (C) 0.45 gm/cm (D) 2.7 gm/cm3

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Page # 24

FLUID

21. A container of large surface area is filled with liquid of density . A cubical block of side edge a and mass M is floating in it with four-fifth of its volume submerged. If a coin of mass m is placed gently on the top surface of the block is just submerged. M is (A) 4m/5 (B) m/5 (C) 4m (D) 5m 22. A boy carries a fish in one hand and a bucket (not full) of water in the other hand. If the places the fish in the bucket, the weight now carried by him (assume that water does not spill) : (A) is less than before (B) is more than before (C) is the same as before (D) depends upon his speed 23. A cork of density 0.5 gcm–3 floats on a calm swimming pool. The fraction of the cork’s volume which is under water is (A) 0% (B) 25% (C) 10% (D) 50% 24. Two cyllinders of same cross-section and length L but made of two material of densities d1 and d2 are cemented together to form a cylinder of length 2L. The combination floats in a liquid of density d with a length L/2 above the surface of the liquid. If d1 > d2 then : 3 d d  d1 (A) d1  d (B)  d1 (C) (D) d < d1 4 2 4 25. A piece of steel has a weight W in air, W1 when completely immersed in water and W2 when completely immersed in an unknown liquid. The relative density (specific gravity) of liquid is : W  W1 (A) W  W 2

W  W2 (B) W  W 1

W1  W2 W1  W2 (C) W  W (D) W  W 1 2

26. A ball of relative density 0.8 falls into water from a height of 2m. The depth to which the ball will sink is (neglect viscous forces) : (A) 8m (B) 2m (C) 6m (D) 4m 27. A small wooden ball of density  is immersed in water of density  to depth h and then released. The height H above the surface of water up to which the ball will jump out of water is (A)

h 

  (B)   1  h  

(C) h

(D) zero

28. A hollow sphere of mass M and radius r is immersed in a tank of water (density w). The sphere would float if it were set free. The sphere is tied to the bottom of the tank by two wires which makes angle 45º with the horizontal as shown in the figure. The tension T1 in the wire is :

R M

T1 45º 45º

4 R 3w g  Mg 2 3 3 (A) (B) R wg  Mg 3 2 4 3 R w g  Mg 4 3 (C) 3 (D) R w g  Mg 3 2 29. A metal ball of density 7800 kg/m3 is suspected to have a large number of cavities. It weighs 9.8 kg when weighed directly on a balance and 1.5 kg less when immersed in water. The fraction by volume of the cavities in the metal ball is approximately : (A) 20% (B) 30% (C) 16% (D) 11%

30. A sphere of radius R and made of material of relative density  has a concentric cavity of radius r. It just floats when placed in a tank full of water. The value of the ratio R/r will be 1/3

   (A)     1

   1  (B)    

1/ 3

   1  (C)    

1/ 3

   1  (D)     1

1/ 3

31. A body having volume V and density  is attached to the bottom of a container as shown. Density of the liquid is d(>). Container has a constant upward acceleration a. Tension in the string is

a

(A) V[Dg – (g + a)] (C) V (d – ) g

(B) V (g + a) (d – ) (D) none

32. A hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density  is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity  is given by (A) 1.0 (B) 1.5 (C) 2.1 (D) 1.9 33. A beaker containing water is placed on the platform of a spring balance. The balance reads 1.5 kg. A stone of mass 0.5 kg and density 500 kg/m3 is immersed in water without touching the walls of beaker. What will be the balance reading now ? (A) 2 kg (B) 2.5 kg (C) 1 kg (D) 3 kg 34. There is a metal cube inside a block of ice which is floating on the surface of water. The ice melts completely and metal falls in the water. Water level in the container (A) Rises (B) Falls (C) Remains same (D) Nothing can be concluded

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Page # 25

FLUID 35. A uniform solid cylinder of density 0.8 g/cm3 floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3 and 1.2 gm/cm3. The height of liquid A is hA = 1.2 cm and the length of the part of cylinder immersed in liquid B is hB = 0.8 cm. Then the length of the cylinder in air is (A) 0.21 m (B) 0.25 cm (C) 0.35 cm (D) 0.4 cm 36. A cylindrical block of area of cross-section A and of material of density  is placed in a liquid of density one-third of density of block. The block compresses a spring and compression in the spring is one-third of the length of the block. If acceleration due to gravity is g, the spring constant of the spring is

P

P

(C)

(D) x

x

41. A cylindrical tank of height 1 m and cross section area A = 4000 cm2 is initially empty when it is kept under a tap of cross sectional area 1 cm2. Water starts flowing from the tap at t = 0, with a speed = 2 m/s. There is a small hole in the base of the tank of crosssectional area 0.5 cm2. The variation of height of water in tank (in meters) with time t is best depicted by h

h

1

0.8

(A)

(B) t

O

O

h

(A) Ag

(B) 2Ag

(C) 2Ag/3

(D) Ag/3

37. A body of density ’ is dropped from rest at a height h into a lake of density , where  > . Neglecting all disipative froces, calculate the maximum depth to which the body sinks before returning of float on the surface. h h' h' h (A)  – ' (B)  (C)  – ' (D)  – ' 4. FLUID FLOW & BERNOULLI'S PRINCIPLE 38. A rectangular tank is placed on a horizontal ground and is filled with water to a height H above the base. A small hole is made on one vertical side at a depth D below the level of the water in the tank. The distance x from the bottom of the tank at which the water jet from the tank will hit the ground is 1 DH (A) 2 D(H  D) (B) 2 DH (C) 2 D (H  D ) (D) 2 39.A jet of water with cross section of 6 cm2 strikes a wall at an angle of 60º to the normal and rebounds elastically from the wall without losing energy. If the velocity of the water in the jet is 12 m/s, the force acting on the wall is (A) 0.864 Nt (B) 86.4 Nt (C) 72 Nt (D) 7.2 Nt 40. The cross sectional area of a horizontal tube increases along its length linearly, as we move in the direction of flow. The variation of pressure, as we move along its length in the direction of flow (xdirection), is best depicted by which of the following graphs P

P

(A)

(B) x

x

t

h

1

0.8

(C)

(D) t

O

O

t

42. A cubical box of wine has a small spout located in one of the bottom corners. When the box is full and placed on a level surface, opening the spout results in a flow of wine with a initial speed of v0 (see figure). When the box is half empty, someone tilts it at 45º so that the spout is at the lowest point (see figure). When the spout is opened the wine will flow out with a speed of

V0

(A) v0

(B) v0/2

(C) v 0 / 2

(D) v 0 / 4 2

43. Water is flowing steadily through a horizontal tube of non uniform cross-section. If the pressure of water is 4 × 104 N/m2 at a point where cross-section is 0.02 m2 and velocity of flow is 2 m/s, what is pressure at a point where cross-section reduces to 0.01 m2 (A) 1.4 × 104 N/m2 (B) 3.4 × 104 N/m2 (C) 2.4 × 10–4 N/m2 (D) none of these 44. A vertical cylindrical container of base area A and upper cross-section area A1 making an angle 30° with the horizontal is placed in an open rainy field as shown near another cylindrical container having same base area A. The ratio of rates of collection of water in the

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Page # 26

FLUID

two containers will be. p

A1 60º

60º

30º

A

(A) 2 / 3

p