Unit - 5 Rotational Motion UG... · Comparison between physical quantities of linear motion and rotational motion Translational motion Rotational motion Linear displacement, d Angular

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Unit - 5Rotational Motion

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SUMMARY* Important Formula, Facts and Terms

1. Centre of mass of system of particles

n

1n n

nn

n21

nn2111cm

mrm

mmmrmrmrmR

for rigid body

nncm rmR.M

for Two body System

2211 rmrm

OR 1 2 1 2 2 1

1 12 2

r m r +r m +m+1= +1 OR =m mr r

1 2 1 22 11 1 2 1 22

r m +m rm rm = r = OR r =m m +m m +mr

2.1 2 n1 2 n

cmm v + m v + ......... + m vV =

M

1 2 .........cm nP M v p p p

similarly

Mamamama nn2211

cm

n21cm FFFaMF

3. Torque = T = Fr = I

sinrF

= product of force and perpendicular distance between point of rotation

and line of action.

Angular momentum = L r P I

| L | = rpsin = product of linear momentum and perpendicular distance between point ofrotation and line of action.

Moment of inertia = I = 2nn

222

211 rmrmrm

=

n

1n

2nnrm



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4. Low of conservation of angular momentum

As L I

pd L dI Idt dt

d Ldt

when

= then L remains constant

Its geometrical representation in planetary motion

Let dtdA

is an areal velocity

Then m dA L dA L= OR =dt 2 dt 2m

5. Radius of gyration {K}

As I = 2nn

222

211 rmrmrm

If all particles have same mass then= m 2

n2

22

1 rrr

n

rrrnmI2

n2

22

1 Here (nm = M)

2mk

n)rrr(k

2n

22

21

6. Some relations between linear and rotational motion.v = rw

v w r

Here w = ddt

2

2d w ddt dt

r Ta a a

where Twv rra a

2 2 2 2 2 2r ta a a v r

2 2 2 2 2 4 2r r r

7. Equilibrium of a rigid body.

When 0FFFF n21 it is in linear equilibrium

When 21 0nP P P P

it is in rotational equilibrium8. Two theorm for moment of inertia

YXZ III Theorm of perpendicular axis.2

cm MdII Theorm of Parallel axis.



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9. Rolling down of body on an inclined plane.

V 2 22 2

2 2 sin,1 1

gh aK K

R R

Condition for rolling without sliding

,KR1

tan

22s

For ring tan21

s (K = R)

disc

2RKtan

31

s

solid sphere

R

52Ktan

72

s

10. Rotational Kinetic Energy.

R.K.E = 2

2 2 22

1 1 &2 2

V VI MK I MKRR

I2L2

Total kinetic Energy = Rotational K.E. + Linear K.E.

2 2

2 2 22 2

1 1 1 12 2 2

V KMK MV MVR R

Now (1) 22

RK

E.LinearKE.K.R

(2) 2

2 22 2 2

2

.. 1

KR

KR

Rotational K E KTotal K E R K

Percentage rotational K.E. = %1001 2

2

22

RKR

K



111

(3)2

2

RK22

2

11

KRR

E.KTotalE.KnalTranslatio

Comparison between physical quantities of linear motion and rotational motion

Translational motion Rotational motion

Linear displacement, d Angular displacement,

Linear velocity, V Angular velocity, w

Linear acceleration, dtvda Angular acceleration,

dtwd

Mass, m Moment of inertia, I

Linear Momentum, vmP Angular momentum, wIL

Force, amF Torque,dtLd

Newton's Second Law of Motion, A result similar to newtown's Second Law,

dtPdF

dtLd

Translational kinetic energy K = 2mv21

Rotational kinetic energy K = 21 I2

m2p2

I2

L2

Work, W = dF Work, W = Power, P = Fv Power, P = wEquations of linear motion taking place Equations of rotational motion taking placewith constant linear acceleration with constant angular acceleration :

v = VO + at w = tWO

d = 2O

1v t+ at2

20 at

21tW

2ad = 220v -v 2a = 2

02 WW

Law of conservation of linear momentum Law of conservation of angular momentum

when 0F then P is constant Impulse when 0P

then L is constant Impulse

linear 12 PPtF rotational 2 1t L L



Value of V, a, t for some Rolling Bodies

Shape of Body velocity velocity acceleration Time 22

RK

Ring/Hollow cylinder gh 21

singl sing21

sing2

1

Disc/Solid cylinder gh34 2

1

singl34

sing3

2sing

32

1

Solid Sphere gh7

10 21

singl7

10

sing7

5sing

145

2

Shell/Hollow spher gh56 2

1

singl56

sing5

3sing

103

2

Moment of inertia an radius of gyration for some symmetric bodies

Body Axis Figure I K For rollingbody 2

2

RK

Thin rod of Passing through its 2ML121

32L

-

Length L centre and per pendicularto its length

Ring of Any diameter 2MR21

2R

-radius R

Ring of Passing through its 2MR R 1radius R centre and perpen-

dicular to its plane

Circular disc Passing through its 2MR21

2R

21

radius R centre and perpen-dicular to its plane of

Circular disc Any diameter 2MR41

2R -

of radius R



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Body Axis Figure I K For rollingbody 2

2

RK

Hollow Geometrical 2MR R 1cylinder of axis of theradius R cylinder

Solid cylinder Geometrical axis 2MR21

2R

21

of radius R of the cylinder

Solid sphere Any diameter 2MR52 R

52

52

of radius R

Hollow Any diameter 2MR32 R

32 2

3sphereof radius R

MCQFor the answer of the following questions choose the correct alternative from among the given ones.1. The centre of mass of a systems of two particles is

(A) on the line joining them and midway between them(B) on the line joining them at a point whose distance from each particle is proportional

to the square of the mass of that particle.(C) on the line joining them at a point whose distance from each particle inversely

propotional to the mass of that particle.(D) On the line joining them at a point whose distance from each particle is proportional

to the mass of that particle.2. Particles of 1 gm, 1 gm, 2 gm, 2 gm are placed at the corners A, B, C, D, respectively

of a square of side 6 cm as shown in figure. Find the distance of centre of mass of thesystem from geometrical centre of square.{A} 1 cm

{B} 2 cm

{C} 3 cm

{D} 4 cm3. Three particles of the same mass lie in the (X, Y) plane, The (X, Y) coordinates of their

positions are (1, 1), (2, 2) and (3, 3) respectively. The (X,Y) coordinates of the centre ofmass are{A} (1, 2) {B} (2, 2) {C} (1.5, 2) {D} (2, 1.5)



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4. Consider a two-particle system with the particles having masses 1M , and 2M . If the firstparticle is pushed towards the centre of mass through a distance d, by what distance shouldthe second particle be moved so as to keep the centre of mass at the same position?

{A} 21

1

MMdM

{B} 21

2

MMdM

{C} 2

1

MdM

{D} 1

2

MdM

5. Four particles A, B, C and D of masses m, 2m, 3m and 5mrespectively are placed at corners of a square of side x asshown in figure find the coordinate of centre of mass take Aat origine of x-y plane.

{A}

10x7,x2 {B}

7x10,x2

{C}

7x10,

2x

{D}

10x7,

2x

6. From a uniform circular disc of radius R, a circular disc of radius 6R and having centre at a

distance + 2R from the centre of the disc is removed. Determine the centre of mass of remaining

portion of the disc.

{A} 70R

{B} 70R

{C} 7R

{D} 7R

7. A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter42 cm. is removed from +ve x edge of the plate. Find the position of centre of mass of theremaining portion with respect to centre of mass of whole plate.{A} - 7 cm {B} + 9 cm {C} - 9 cm {D} + 7 cm

8. Two blocks of masses 10 kg an 4 kg are connected by a spring of negligible mass and placedon a frictionless horizontal surface. An impulse gives velocity of 14 m/s to the heavier block inthe direction of the lighter block. The velocity of the centre of mass is :{A} 30 m/s {B} 20 m/s {C} 10 m./s {D} 5 m/s

9. A particle performing uniform circular motion has angular momentum L., its angular frequency isdoubled and its K.E. halved, then the new angular momentum is :{A} ½ {B} ¼ {C} 2L {D} 4L

10. A circular disc of radius R is removed from a bigger disc of radius 2R. such that the circumferencesof the disc coincide. The centre of mass of the remaining portion is R from the centre of massof the bigger disc. The value of is.{A} ½ {B} 1/6 {C} ¼ {D} 1/3

11. Three point masses M1, M2 and M3 are located at the vertices of an equilateral triangle ofside 'a'. what is the moment of inertia of the system about an axis along the attitude of the trianglepassing through M1, ?

{A} 4aMM

221 {B} 4

aMM2

32 {C} 4aMM

231 {D} 4

aMMM2

321



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12. A body of mass m is tied to one end of spring and whirled round in a horizontal planewith a anstant angular velocity. The elongation in the spring is one centimeter. If the angularvelocity is doubted, the elongation in the spring is 5 cm. The original length of spring is…{A} 16 cm {B} 15 cm {C} 14 cm {D} 13 cm

13. A cylinder of mass 5 kg and radius 30 cm, and free to rotate about its axis, receives an angularimpulse of 3 kg M2S-1 initially followed by a similar impulse after every 4 sec. what is the angularspeed of the cylinder 30 sec after initial imulse ? The cylinder is at rest initially.

{A} 1Srad7.106 {B} 1Srad7.206 {C} 107.6 rad S-1 {D} 207.6 rad S-1

14. Two circular loop A & B of radi ra and rb respectively are made from a uniform wire. The ratioof their moment of inertia about axes passing through their centres and perpendicular to their planes

is 8II

A

B then rarb

Ra is equal to…

{A} 2 {B} 4 {C} 6 {D} 815. If the earth were to suddenly contract so that its radius become half of it present radius, without

any change in its mass, the duration of the new day will be…{A} 6 hr {B} 12 hr {C} 18 hr {D} 30 hr

16. In HC1 molecule the separation between the nuclei of the two atoms is about m10A1A27.1 10 .

The approximate location of the centre of mass of the molecule is iA with respect of Hydrogenatom ( mass of CL is 35.5 times of mass of Hydrogen){A} 1 {B} 2.5 {C} 1.24 {D} 1.5

17. Two bodies of mass 1kg and 3 kg have position vector kj2i and (-3i-2j+k) respectivelythe center of mass of this system has a position vector……

{A} k2i2 {B} kji2 {C} kji2 {D} kji 18. Identify the correct statement for the rotational motion of a rigid body

{A} Individual particles of the body do not undergo accelerated motion{B} The center of mass of the body remains unchanged.{C} The center of mass of the body moves uniformly in a circular path{D} Individual particle and centre of mass of the body undergo an accelerated motion.

19. A car is moving at a speed of 72 km/hr the radius of its wheel is 0.25m. If the wheels arestopped in 20 rotations after applying breaks then angular retardation produced by the breaksis ……

{A} -25.5 2srad {B} -29.5 2s

rad {C} -33.5 2srad {D} -45.5 2s

rad

20. A wheel rotates with a constant acceleration of 2.0 2secrad If the wheel start from rest. Thenumber of revolution it makes in the first ten seconds will be approximately.{A} 8 {B} 16 {C} 24 {D} 32

21. Two discs of the same material and thickness have radii 0.2 m and 0.6 m their moment of inertiaabout their axes will be in the ratio{A} 1 : 81 {B} 1 : 27 {C} 1 : 9 {D} 1 : 3



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22. A wheel of mass 10 kg has a moment of inertia of 160 kg 2m about its own axis. Theradius of gyration will be _______ m.{A} 10 {B} 8 {C} 6 {D} 4

23. One circular rig and one circular disc both are having the same mass and radius. The ratio oftheir moment of inertia about the axes passing through their centres and perpendicular to theirplanes, will be……{A} 1 : 1 {B} 2 : 1 {C} 1 : 2 {D} 4 : 1

24. One solid sphere A and another hollow sphere B are of the same mass and same outer radii.The moment of inertia about their diameters are respectively AI and BI such that…

{A} BA II {B} BA II {C} BA II {D} dBdA

II

B

A (radio of their densities)

25. A ring of mass M and radius r is melted and then molded in to a sphere then the moment ofinertia of the sphere will be…..{A} more than that of the ring {B} Less than that of the ring{C} Equal to that of the ring {D} None of these

26. A circular disc of radius R and thickness R/6 has moment of inertia I about an axis passing throughits centre and perpendicular to its plane. It is melted and recasted in to a solid sphere. The momentof inertia of the sphere about its diameter as axis of rotation is …

{A} I {B} 8I2

{C} 5I

{D} 10I

27. One quater sector is cut from a uniform circular disc of radius R. Thissector has mass M. It is made to rotate about a line perpendicular toits plane and passing through the centre of the original disc. Its momentof inertia about the axis of rotation is…

{A} 2MR21 {B} 2MR4

1

{C} 2MR81 {D} 2MR2

28. A thin wire of length L and uniform linear mass density is bentin to a circular loop with centre at O as shown in figure. The momentof inertia of the loop about the axis xx' is ….

{A} 2

2L

8

{B}

3

2L

16

{C}3

25 L

16

{D}

3

23 L

8

29. Two disc of same thickness but of different radii are made of two different materials such that

their masses are same. The densities of the materials are in the ratio 1:3. The moment of inertiaof these disc about the respective axes passing through their centres and perpendicular to theirplanes will be in the ratio.{A} 1 : 3 {B} 3 : 1 {C} 1 : 9 {D} 9 : 1



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30. Let I be the moment of inertia of a uniform square plate about an axis AB that passesthrough its centre and is parallel to two of its sides CD is a line in the plane of the platethat passes through the centre of the plate and makes an angle of Q with AB. The momentof inertia of the plate about the axis CD is then equal to….

{A} I {B} 2sinI {C} 2cosI {D} 2cosI 2

31. A small disc of radius 2 cm is cut from a disc of radius 6 cm. If the distance between theircentres is 3.2 cm, what is the shift in the centre of mass of the disc…{A} -0.4 cm {B} -2.4 cm {C} -1.8 cm {D} 1.2 cm

32. A straight rod of length L has one of its ends at the origin and the other end at x=L If the massper unit length of rod is given by Ax where A is constant where is its center of mass.{A} L/3 {B} L/2 {C} 2L / 3 {D} 3L / 4

33. A uniform rod of length 2L is placed with one end in contact with horizontal and is then inclinedat an angle to the horizontal and allowed to fall without slipping at contact point. When it becomeshorizontal, its angular velocity will be…..

{A} w =L2

sing3 {B} w = sing3

L2{C} w =

Lsing6

{D} w = singL

34. A cubical block of side a is moving with velocity V on ahorizontal smooth plane as shown in figure. It hits a ridge atpoint O. The angular speed of the block after it hits O is ….

{A} 34

va {B} 3

2v

a {C} 32

va {D} zero

35. Consider a body as shown in figure, consisting of two identicalbulls, each of mass M connected by a light rigid rod. If animpulse J = MV is imparted to the body at one of its ends,what would be its angular velocity. What is V ?{A} V / L {B} 2V / L {C} V / 3L {D} V / 4L

36. A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocityw. Two objects each of mass m are attached gently to the opposite ends of a diameter of thering. The ring will now rotate with an angular velocity….

{A} ( 2 )

2M m

M m

{B} 2

MM m {C}

MM m

{D} 2M m

M

37. A smooth sphere A is moving on a frictionless horizontal plane with angular speed and centreof mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglectfriction everywhere. After the collision, their angular speeds are A and B respectively, Then{A} A < B {B} A = B {C} A = {D} = B

38. Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m andof negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniformangular speed. The point on the rod through which the axis should pass in order that the workrequired for rotation of the rod is minimum, is located at a distance of …..{A} 0.4 m from mass of 0.3 kg {B} 0.98 m from mass of 0.3 kg{C} 0.7 m from mass of 0.7 kg {D} 0.98 m from mass of 0.7 kg



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39. In a bicycle the radius of rear wheel is twice the radius of front wheel. If Fr and rr arethe radius, vF and vr are speed of top most points of wheel respectively then...{A} vr = 2vF {B} vF = 2vr {C} vF = vr {D} vF > vr

40. From a circular disc of radius R and mass 9M, a small disc of radiusR/3 is removed from the disc. The moment of inertia of the remainingportion about an axis perpendicular to the plane of the disc and passingthrough O is….

{A} 2MR4 {B} 2MR940

{C} 2MR10 {D} 2MR9

37

41. A child is standing with folded hands at the centre of a platform rotating about its central axisthe kinetic energy of the system is K. The child now stretches his arms so that the moment ofinertia of the system doubles. The kinetic energy of the system now is…{A} 2 K {B} K/2 {C} K/4 {D} 4K

42. If the earth is treated as a sphere of radius R and mass M. Its angular momentum about theaxis of rotation with period T is…..

{A} 3MR

T

{B} T

MR 2 {C}

T5MR2 2

{D} T5

MR4 2

43. If the angular momentum of any rotating body increases by 200%, then the increase in its kineticenergy will be…..{A} 400% {B} 800% {C} 200% {D} 100%

44. The M.I. of a body about the given axis is 1.2 kgm2 initially the body is at rest. In order toproduce a rotational kinetic energy of 1500 J. an angular acceleration of 2secrad25 must beapplied about that axis for duration of ….{A} 4 sec {B} 2 sec {C} 8 sec {D} 10 sec

45. An automobile engine develops 100kw when rotating at a speed of 1800 r.p.m. what torque doesit deliver ?{A} 350 Nm {B} 440 Nm {C} 531 Nm {D} 628 Nm

46. The moment of inertia of two rotating bodies A and B are AI and BI . BA II and their angular

momentum are equal. If their K.E. be AK and BK respectively then….

{A} KA, KB {B} 1KAKB {C} 1KB

KA {D} 21

KAKB

47. The centre of mass of the disc undergoes S.H.M. with angular frequency equal to..

{A} mk

{B} mk2

{C} m3k2

{D} m3k4

48. Three rings, each of mass P and radius are arranged as shown in thefigure the moment of inertia of the arrangement about YY' axis will be.

{A} 2P27 {B} 2P7

2 {C} 2P52 {D} 2P2

5



119

49. If distance of the earth becomes three times that of the present distance from the sun thennumber of days in one year will be ….

{A} 365 3 {B} 365 27 {C} 33365 {D} 33365

50. A solid sphere and a solid cylinder having same mass and radius roll down the same incline theratio of their acceleration will be….

{A} 15 : 14 {B} 14 : 15 {C} 5 : 3 {D} 3 : 551. The ratio of angular momentum of the electron in the first allowed orbit to that in the second

allowed orbit of hydrogen atom is ……

{A} 2 {B} 21

{C} ½ {D} 2

52. A player caught a cricket ball of mass 150 gm moving at a rate of 20 m/s If the catching processis Comlitad in 0.1 sec the force of the flow exerted by the ball on the hand of theplayer ….. N{A} 3 {B} 30 {C} 150 {D} 300

53. Two disc one of the density 7.2 gm/cc and other of density 8.9 gm/cc are of the same massand thickness their moment of inertia are in the ratio of ……

{A} 9.82.7 {B} 2.79.8

1

{C} 2.79.8 {D} 2.79.8

54. Two identical hollow spheres of mass M and radius R are joinedtogether and the combination is rotated about an axis tangential toone sphere and perpendicular to the line connecting their centers.The moment of inertia of the combination is ________.

{A} 10 2MR {B} 2MR34 {C} 2MR3

32 {D} 2MR334

55. A rod of length L rotate about an axis passing through its centre and normal to its length withan angular velocity . If A is the cross-section and D is the density of material of rod. Findits rotational K.E.

{A} 312 AL D {B} 31

6 AL D {C} 3124 AL D {D} 31

12 AL D

56. Initial angular velocity of a circular disc of mass M is w1 Then two spheres of mass m are attachedgently two diametrically oppsite points on the edge of the disc what is the final angular velocityof the disc?

{A} 1wM

mM

{B} 1wM

m4M

{C} 1wm4M

M

{D} 1wm2M

M

57. A circular disc x of radius R is made from an iron plate of thickness t. and another disc Y ofradius 4R is made from an iron plate of thickness t/4 then the rotation between the moment of

inertia XI and yI is ___________

{A} xy I64I {B} xy I32I {C} xy I16I {D} xy II

58. A Pulley of radius 2 m is rotated about its axis by a force Nt5t20F 2 where t is in sec

applied tangentially. If the moment of inertia of the Pulley about its axis of rotation is 10 ,KgM 2



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the number of rotations made by the pulley before its direction of motion is reversed is :{A} more than 3 but less then 6 {B}more than 6 but less then 9{C} more than 9 {D} Less then 3

59. Two spheres each of mass M and radius R/2 are connected witha mass less rod of length 2R as shown in figure. What will bemoment of inertia of the system about an axis passing throughcentre of one of the spheres and perpendicular to the rod?

{A} 2MR521

{B} 2MR52

{C} 2MR25

{D} 2MR215

60. Four particles each of mass 'm' are lying symmetrically on therim of the disc of mass M and radius R moment of inertiaof this system about an axis passing through one of the particlesand perpendicular to plane of disc is _____

{A} 2MR16 {B} 2

RM16M32

{C} 2

RM12M32

{D} Zero

61. A mass M is supported by a mass less string wound arounda uniform cylinder of mass M and radius R as in figure. Withwhat acceleration will the mass fall on release?{A} 2/3g {B} g/2{C} g {D} 4g/3

62. Calculate rotational K.E. of earth due to its rotation about its own axis.24

eM =6×10 kg Re=6400 Km

{A} 29102.6 Joule {B} 29106.2 Joule {C} 291062 Joule {D} 291026 Joule63. A cord is wound round the circumference of wheel of radius r. the axis of the wheel is horizontal

and moment of inertia about it is I A weight mg is attached to the end of the cord and fallsfrom the rest. After falling through the distance h. the angular velocity of the wheel will be….

{A} mrI

gh2

{B} 21

2mrImgh2

{C} 2

1

2mr2Imgh2

{D} gh2

64. If rotational K.E. is 50% of translational K.E. then the body is …..{A} Ring {B} solid cylinder {C} Hollow sphere {D} Solid sphere

65. A meter stick of mass 400 gm is pivoted at one end and displaced through an angle 600 theincrease in its P.E. is ______ J.{A} 2 {B} 3 {C} Zero {D} 1

66. Tow uniform rod of equal length but different masses are rigidlyjoined to form L shaped body which is then pivoted as shown.If in equilibrium the body is in the shown configuration ratioM/m will be…….{A} 2 {B} 3

{C} 2 {D} 3



121

67. A light rod carries three equal masses A, B and C as shown inthe figure the velocity of B in vertical position of rod if it is releasedfrom horizontal position as shown in the figure is …….

{A} g2 {B} 7g18

{C} 3g4

{D} 8g3

68. A gramophone record of mass M and radius R is rotating with angular speed W. If two piecesof wax each of mass M are kept on it at a distance of R/2 from the centre on opposite sidethen the new angular velocity will be…..

{A} 2

{B} m

M m

{C} M

M m

{D} M mM

69. A solid cylinder rolls down a smooth inclined plane 4.8m high without slipping what is itslinear speed at the bottom of the plane, if it starts rolling from the top of the plane?(take g = 10 m/S2){A} 4 m/S {B} 2 m/S {C} 10 m/S {D} 8 m/S

70. The M.I of a disc of mass M and radius R about an axis passing

through the centre O and perpendicular to the plane of disc is 2

MR 2

.

If one quarter of the disc is removed the new moment of inertiaof disc will be…..

{A} 3

MR 2

{B} 4

MR 2

{C} 2MR83

{D} 2MR23

71. The moment of inertia of a uniform rod about a perpendicular axis passing through one of itsends is I1. The same rod is bent in to a ring and its moment of inertia about a diameter is

I2, Then 2

1I

I is.

{A} 3

2{B}

34 2

{C} 3

8 2{D}

316 2

72. A molecule consist of two atoms each of mass 'm' and separated by a distance of 'd' If 'K' isthe average rotational K.E. of the molecule at particular temperature then its angular frequency is….

{A} mk

d2

{B} mk

2d {C} k

md2 {D} km

4d

73. A car is moving with a constant speed the wheels of the car make 120 rotations per minutethe breaks are applied and the car comes to rest in 8 sec how many rotation are completedby the wheels before the car is brought to rest.{A} 4 {B} 6 {C} 8 {D} 10

74. The angular momentum of a wheel changes from 2L to 5L in 3 seconds what is the magnitudesof torque acting on it?{A} L {B} L/2 {C} L/3 {D} L/5



122

75. A uniform disc of mass 500kg and radius 2 m is rotating at the rate of 600 r.p.m. what isthe torque required to rotate the disc in the opposite direction with the same angular speed ina time of 100 sec ?{A} Nm600 {B} Nm500 {C} Nm400 {D} Nm300

76. The moment of inertia of a meter scale of mass 0.6kg about an axis perpendicular to the scaleand passing through 30 cm position on the scale is given by (Breath of scale is negligible). ________{A} 0.104 2mkg {B} 0.208 2mkg {C} 0.074 2mkg {D} 0.148 2mkg

77. How much constant force should be applied tangential to equator of the earth to stop its rotationin one day ?

{A} N103.1 22 {B} N1026.8 28 {C} 231.3 10 N {D} None of these78. A constant torque of 1500 Nm turns a wheel of moment of inertia 300 kg 2m about an axis

passing through its centre the angular velocity of the wheel after 3 sec will be…….... rad/sec{A} 5 {B} 10 {C} 15 {D} 20

79. A disc of mass M and radius R is rolling with angularspeed w on a horizontal plane, as shown in figure. Themagnitude of angular momentum of the disc about theorigin O is ______

{A} 21 MR2 {B} 2MR {C} 23 MR2 {D} 22 MR

80. A mass m is moving with a constant velocity along the line parallel to the x-axis, away from theorigin. Its angular momentum with respect to the origin{A} Zero {B} remains constant {C} goes on increasing {D} goes on decreasing

81. A body is rolling down an incline plane. If the rotational K.E. of the body is 40% of its translationalK.E. then the body is ….{A} ring {B} Cylinder {C} solid sphere {D} hollow sphere

82. A spherical ball rolls on a table without slipping, then the fraction of its total energy associatedwith rotation is .{A} 2/5 {B} 3/5 {C} 2/7 {D} 3/7

83. A binary star consist of two stars A (2.2 Ms) and B(mass 11Ms) where Ms is the mass of 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 _____.{A} 6 {B} ¼ {C} 12 {D} ½

84. A small object of uniform density rolls up a curved surface with initial

velocity 'u'. It reaches up to maximum height of 23

4v

g with respect

to initial position then the object is ____.{A} ring {B} solid sphere {C} disc {D} hollow sphere

85. A particle of mass m slides down on inclined plane and reaches the bottom with linearvelocity V. If the same mass is in the form of ring and rolls without slipping down the same inclinedplane. Its velocity will be______.

{A} V {B} 2V {C} 2V

{D} 2V



123

GRAPHICAL QUESTIONS.86. Moment of inertia of a sphere of mass M and radius R is I. keeping mass constant if

graph is plotted between I and R then its form would be.

{A} {B} {C} {D}

87. According to the theorem of parallel axis 2cm mdII the graph between dI will be

{A} {B} {C} {D}

88. The graphs between angular momentum L and angular velocity w will be.

{A} {B} {C} {D}

89. The graphs between loge L and loge P is ____ where L is angular momentum and P islinear momentum

{A} {B} {C} {D}

90. Let Er is the rotational kinetic energy and L is angular momentum then the graph betweenLEr elogandLoge can be

{A} {B} {C} {D}



124

MACHING COLUMN TYPE91. Match list I with list II and select the correct answer.

List - I List - II System Moment of inertia

(x) A ring about it axis2

MR)1(2

(y) A uniform circular disc about it axis 2MR52)2(

(z) A solid sphere about any diameter 2MR57)3(

(w) A solid sphere about any tangent 2MR)4( s

2MR59)5(

Select correct option

Option? X Y Z W

{A} 2 1 3 4

{B} 4 3 2 5

{C} 1 5 4 3

{D} 4 1 2 3

92. Match the shape of graph with given pair of physical quantities.Physical Quantities{X} Moment of inertiadistance {Z}Angular momentum(L) letargen parallel axis angular geloaty (w){Y} re Elog Lloge {W} Lloge Ploge

Shape of graph

{P} {Q} {R} {S}

Select Correct OptionOption? X Y Z W

{A} R S P Q{B} Q P S R{C} S Q R P{D} P R Q S



125

ASSERTION - REASONING TYPE92. In the following questions statement - 1 (Assertion) is followed by statement - 2 (Reason). Each

question has the following four choices out of which only one choice is correct.{A} Statement - 1 is correct (true), Statement - 2 is true and Statement- 2 is correct explanationfor Statement - 1{B} Statement -1 is true, statement -2 is true but statement-2 is not the correct explanation fourstatement -1.{C} Statement - 1 is true, statement-2 is false{D} Statement-2 is false, statement -2 is true

93. Statement -1 — The angular momentum of a particle moving in a circular orbit with a constantspeed remains conserved about any point on the circumference of the circle.Statement -2— If no net torque outs, the angular momentum of a system is conserved.

94. Statement -1— A sphere and a cylinder slide without rolling from rest from the top of an inclinedplane. They will reach the bottom with the same speed.Statement -2 — Bodies of all shapes, masses and sides slide down a plane with the sameacceleration.

95. Statement -1— Friction is necessary for a body to roll on surfaceStatement -2— Friction provides the necessary tangential force and torque.

96. Statement -1— A body tied to a string is moved in a circle with a uniform speed. If the stringsuddenly breaks the angular momentum of the body becomes zero.Statement -2 — The torque on the body equals to the rate of change of angular momentum.

97. Statement -1 — If there is no external torque on a body about its centre of mass, then the velocityof the centre of mass remains constant.Statement -2 — The Linear momentum of an isolated system remains constant.

98. Statement -1 —Two cylinder one hollow and other solid (wood) with the same mass and identicaldimensions are simultaneously allowed to roll without slipping down an inclined plane from thesame height. The hollow will reach the bottom of inclined plane first.Statement -2 — By the principle of conservation of energy, the total kinetic energies of both thecylinders are identical when they reach the bottom of the incline.

99. Statement -1 —A thin uniform rod AB of mass M and length L is hinged at one end A to thehorizontal floor initially it stands vertically. It is allowed to fall freely on the floor in the vertical

plane, The angular velocity of the rod when its ends B strikes the floor Lg3

Statement -2 — The angular momentum of the rod about the hinge remains constant through outits fall to the floor.

100 .Statement -1 —If the cylinder rolling with angular speed- w. suddenly breaks up in to two equalhalves of the same radius. The angular speed of each piece becomes 2w.Statement -2—If no external torque outs, the angular momentum of the system is conserved.



126

PASSAGE BASED QUESTIONSPassage - I

A Solid sphere of mass M and radius R is released from rest at the top of a frictionless inclined planeof length 'd' and inclination ?.

In case (a) it rolls down the plane without slipping and in case (b) it slides down the plane

101. The ratio of the acceleration of the sphere in case (a) to that in case (b) is[A] 1[B] 2/3 [C] 5/7 [D] 7/9

102. The ratio of the velocity of the spheres when it reaches the bottom of the plane in case (a) tothat in case (b) is

[A] 2 [B] 32

[C] 75

[D] 37

103. The ratio of time taken by the sphere to reach the bottom in case (a) to that in case (b) as

[A] 1 [B] 23

[C] 2 [D] 57

Passage - IIA uniform disc of mass M and radius R rolls without slipping down a plane inclined at an angle with the horizontal.

104. The acceleration of the centre of mass of the disc is

[A] sing [B] 3sing2

[C] 3sing

[D] 3cosg2

105. The frictional force on the disc is

[A] 3sinMg

[B] 3sinMg2

[C] sinMg [D] None

106. The magnitude of torque acting on the disc is

[A] MgR [B] sinMgR [C] 3sinMgR2

[D] 3sinMgR

107. If the disc is replaced by a ring of the same mass M and the same radius R, the ratio of thefrictional force on the ring to that on the disc will be[A] 3/2 [B] 2 [C] 2 [D] 1

Passage - IIIA solid cylinder of mass M and R is mounted on a frictionless horizontalaxle so that it can freely rotate about this axis. A string of negligible massis wrapped round the cylinder and a body of mass m is hung from thestring as shown in figure the mass is released from rest then___

108. The acceleration with which the mass falls is

[A] g [B] Mmg

[C]

mMmg

[D] m2MMng2



127

109. The tension in string is

[A] mg [B] mMMmg [C] m2M

Mmg2 [D] 2

MgM m

110. The angular speed of cylinder is proportional to hn, where h is the height through which massfalls, Then the value of n is___[A] zero [B] 1[C] ½ [D] 2

111. The moment of inertia of a uniform circular disc of mass M and radius R about any of its diameteris ¼ 2MR , what is the moment of inertia of the disc about an axis passing through its centreand normal to the disc?

[A] 2MR [B] 2MR21

[C] 2MR23

[D] 2MR2

112. A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angularvelocity of the cylinder when it reaches the bottom of the plane will be.

[A] ghR2

[B] 2

ghR2

[C] 3

ghR2

[D] ghR21

113. A cylinder of mass m and radius r is rotating about its axis with constant speed v Its kineticenergy is _____

[A] 2mv2 [B] mv2 [C] 12 mv2 [D] mv2

114. A circular disc of mass m and radius r is rolling on a smooth horizontal surface with a constantspeed v. Its kinetic energy is _____

[A] 14 mv2 [B] 1

2 mv2 [C] 34 mv2 [D] mv2

115. A solid sphere is rotating about a diameter at an angular velocity w. if it cools so that its radiusreduces to 1/n of its original value. Its angular velocity becomes_____

[A] n [B] 2n

[C] n [D] n2

116. In above question (115)If the original rotational K.E. of the sphere is K, Its new value will be_____

[A] 2nK [B] 4n

K [C] Kn2 [D] Kn4

117. A solid sphere is rotating about a diameter due to increase in room temperature, its volume increasesby 5%, If no external torque acts. The angular speed of the sphere will.[A] increase by nearly 1/3 % [B] decrease by nearly 1/3 %[C] increase by nearly ½ % [D] decrease nearly by ½ %

118. A cylinder of mass M has length L that is 3 times its radius what is the ratio of its momentof inertia about its own axis and that about an axis passing through its centre and perpendicularto its axis?

[A] 1 [B] 31

[C] 3 [D] 23



128

119. A uniform rod of length L is suspended from one end such that it is free to rotate about anaxis passing through that end and perpendicular to the length, what maximum speed must be impartedto the lower end so that the rod completes one full revolution?

[A] gL2 [B] gL2 [C] gL6 [D] gL22120. The height of a solid cylinder is four times that of its radius. It is kept vertically at time t=o on

a belt which is moving in the horizontal direction with a velocity v = 2t45.2 where v in m/sand t is in second. If the cylinder does not slip, it will topple over a time t = ____[A] 1 second [B] 2 sec. [C] 3 sec. [D] 4 sec.

121. The moment of inertia of a thin rod of mass M and length L about an axis passing through thepoint at a distance L/4 from one of its ends and perpendicular to the rod is _____

[A] 48ML7 2

[B] 12

ML2

[C] 9

ML2

[D] 3

ML2

122. A thin uniform rod AB of mass M and length L is hinged at one end A to the horizontal floor.Initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane. The angularvelocity of the rod when its end B strikes the floor is ____

[A] Lg

[B] Lg2

[C] Lg3

[D] Lg2

123. A circular disc of radius R is free to oscillate about an axis passing through a point on its rimand perpendicular to its plane. The disc is turned through an angle of 60? and released. Its angularvelocity when it reaches the equilibrium position will be__

[A] R3g

[B] R3g2

[C] Rg2

[D] Rg22

124. The moment of inertia of a hollow sphere of mass M and inner and outer radii R and 2R aboutthe axis passing through its centre and perpendicular to its plane is

[A] 2MR23

[B] 2MR3213

[C] 2MR3531

[D] 2MR3562

125. If a is aerial velocity of a planet of mass M its angular momentum is[A] M [B] 2 MA [C] MA2 [D] 2AM

126. A wheel having moment of inertia 2 kg 2M about its vertical axis, rotates at the rate of 60 rpmabout this axis. The torque which can stop the wheels rotation in one minute will be..

[A] Nm15

[B] Nm18

[C] mN152

[D] Nm12

127. A wheel is rotating at 900 rpm about its axis. When power is cut off it comes to rest in 1 minute,the angular retardation in rad / sec is ___

[A] 2

[B] 4

[C] 6

[D] 8

128. What is the moment of inertia of a solid sphere of density and radius R about its diameter?

[A] 5105176

R [B] 5176105

R [C] 2105176

R [D] 2176105

R



129

129. A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocityis zero. In the first two second it rotate through an angle 1, in the next 2 sec. it rotates through

an angle 2, find the ratio 2

1

= ____

[A] 1 [B] 2 [C] 3 [D] 4130. A gramophone record of mass M and radius R is rotating at an angular velocity w. A win of

mass M is gently placed on the record at a distance R/2. from its centre. The new angular velocityof the system is

[A] mM2wM2 [B] m2M

wM2 [C] [D] M

wm

KEYNOTEQ.No. Ans Q.No. Ans Q.No. Ans Q.No. Ans Q.No. Ans

1 C 31 A 61 A 91 D 121 A2 A 32 C 62 B 92 B 122 C3 B 33 A 63 B 93 D 123 B4 C 34 64 B 94 A 124 D5 D 35 A 65 D 95 A 125 B6 A 36 B 66 D 96 D 126 A7 C 37 C 67 D 97 D 127 A8 C 38 B 68 C 98 D 128 B9 B 39 C 69 D 99 C 129 C

10 D 40 A 70 C 100 A 130 A11 B 41 B 71 C 101 C12 B 42 D 72 A 102 C13 A 43 B 73 C 103 D14 A 44 B 74 A 104 B15 A 45 C 75 C 105 A16 C 46 B 76 C 106 D17 B 47 D 77 A 107 A18 B 48 A 78 C 108 D19 A 49 C 79 C 109 D20 B 50 A 80 110 C21 A 51 C 81 C 111 B22 D 52 B 82 C 112 C23 B 53 C 83 A 113 D24 C 54 D 84 C 114 C25 B 55 C 85 C 115 D26 C 56 D 86 D 116 C27 A 57 A 87 C 117 B28 D 58 A 88 A 118 A29 B 59 A 89 B 119 C30 A 60 B 90 B 120 A

B

A



130

Hints1. Let Rcm is at origin

2211cm rmrmRM

1 1 2 2O = m r + m r

1 1 2 2m r m r

1 2

2 1

r m = r m – ve sign ignore as distance

2. Here^ ^

Ar = 0 i + 6J

^ ^

Br = 6 i + 6J

^ ^

Cr = 6 i + 0J

^ ^

Dr = 0 i + 0J

M = 1 + 1 + 2 + 2 = 6 gm

DDCCBBAAcm rmrmrmrmRM

3. The x and y co.ordinates of centre of mass are

1 1 2 2 3 3

1 2 3

m x + m x + m xx = m + m + m as m1 = m2 = m3

1 2 31 (x + x + x) = 23

similtaraly for y = 2

(x, y) = (2, 2)

4. m1x1 = m2x2

and m1(x1 – d) = m2 (x2 – d’)

1 2 m d = m d ' 1

2

m d d ' = m

5. Here at origin A (0, 0), B (x, 0), C (x, x), D (0, x)

(m 0) + (2mx) + (3mx) + (4m 0) = m + 2m + 3m + 4mcmx

cm(m × 0) + (2m × 0) + (3mx) + (4m )y =

m + 2m + 3m + 4mx

r1 r2

m1 m2



131

6. Let mass per unit area of disc = m

Mass of disc = M = 2R m

Mass of removed disc = 2 2R R mM ' = m =

6 36

from figrue R00' = 2

RM 0 = M' + (M – M')x2

RM 'x = M' + Mx2

M' Rx = M – M' 2

7. Let mass per unit area of Plote = m

Mass of whole Plote = 256M = m

2

Mass of removed part = 2

142M = m2

Mass of remaining Portion M2 = M – M1C.M of whole disc R = O at originC.M of removed Plote = r1 = 28 – 21 = 7cmC.M of remaining Portion r2 = ?

i i 2 2M O = M r + M r

8. The Velocity of C.M. is given by

1 1 2 2

1 2

m + m = m + mcmv vV

9. 21 1 1E = = = 2 2 2

I I L

2E 2E' L = L' = '

10. Let m is the mass of unit area then mass of big disc = 2(2R) m = M

Let m is the mass of unit area then mass of small disc = R2m = 1MM = 4

Mass of remaining Portion = M2 = M – M1

23MM = 4



132

Let G be the C.M of remaining PortionM2(OG) = M1(OO’)

3M M ( R) = R4 4

1 = 3

11. The moment of inertia about AD = ?I = m1 (Perpendicular distance of m1 from AD)2

+ m2 (Perpendicular distance of m2 from AD)2

+ m3 (Perpendicular distance of m3 from AD)2

2 2

2 39 9 0 + m + m 2 2

2

2 3a (m + m ) 4

12. Let L is original length & K spring anstant thenm (L + x1) w1

2 = kx1 & M (L + x2) w22 = kx2

Taking ratio2

1 1 12

2 2

L + = L +

x xx x

given x1 = 1cm, x2 = 5cm and 2 = 1

13. Inital angular momentum = Li = Ii = I x 0 = 0angular mumentum after initial inpulse = 3kgm2s–1

angular mumentum after initial 4 sec = 3 + 3 = 6kgm2s–1

angular mumentum after initial 8 sec = 6 + 3 = 9kgm2s–1

angular mumentum after initial 28 sec = 24 kgm2s–1

angular mumentum after initial 30 sec = 24 kgm2s–1

I = 24 here 2

2 2MR 1I = = 5 (0.3) = 0.225kgm2 2

24 24 = = = 106.7I 0.225

rad s–1

14. IA = ma ra2, IB = mb rb

2

2

b bB

A a a

m rI = I m r

Let K is the mass of unit length of the wire then

a am = (2 r )k and b bm = (2 r )k

b b

a a

m r = m r

2 3

B b b b

A a a a

I m r r = 8 = = I m r r

b

a

r = 2r



133

15. Let M be the mass and R1 the initial radius of the earth 1 is the angular veloalty of the rotation

of the earth, the duration T1 of the day is 1 21 2

2 2T = and T =

According to law of conservation of angular momentumI11 = I12

16. m1 = 1 m2 = 35.5 r1 = 0^

2r = 1.27 i

1 1 2 2cm

1 2

m r + m rr = m + m

17. 1 1 2 2cm

1 2

m r + m rr = m + m

18. Theory [B] The centre of mass of lucky remains uncharged.

19. oG 72 1000/3600 = = 80rad/sec0.25o r

= O, = 2 n = 2 20 = 40 rad As 2 2o2 = w – w

20. 21 = wot + t = 100 rad2

21. 2 2 21 1I = MR = ( R t )R2 2

As t are same

4I R4

1 1

2 2

I R = I R

22. I = MK2 = 160

2 160 160 K = = = 16m 10

K = 4

23.2

2

I ring MR 2 = = 1I Disc 1MR2

2 : 1

24. 2 2A Solid

2I = I = MR = 0.4MR5

and 2 2B hollow

2I = I = MR = 0.66MR3 A B I < I

25. Iring = MR12 As Volume and Mass remain same

2Solid 2

2I = MR5 R2 <<< R1



134

26. Volume of disc = 3

2 11 1

RV = R t = 6

3312

R 4 = R6 3

Volume of Sphere = 32 2

4V = R3 3 3

1 2R = 8R 1 2 R = 2R

I1 = M.I of disc = 21

1I = MR2

I2 = M.I of sphere = 2

2 12

2 MR I MR = = 5 5 2 5

27. Mass of the ontire disc would be AM and its moment of inertia about the given axis would

be 212 (AM)R . For the given section the moment of inertia about the since axis be one qvarter

of this is 212 (AM)R .

28. Mass per unit length of the wire = Mass of L length M = LWhen it is lent in form of circularring

2 r = LL R = 2

Moment of inertia of ring about given axis = 23 MR2

29. M.I of disc = 2

21 1 M 1 M MR = M = 2 2 t 2 t

22

M M = R = R t t

As their mass & thickness are some1I

30. Let IZ be the M.I of square plote about the axis passing through the centre and perpendicularto the plane of square, hence according to Perfendicular axis theorm.IZ = IAB + IA'B' Also IZ = ICD + IC'D'

As axis are symmetric IAB = IA'B' = ZI2

And ICD = IC'D' = ZI2

So we can say that IAB = IA'B' = ICD = IC'D' = I

31. Let the radius of complete disc is a & that of small disc is b After small disc is cut from completedisc let the C.M. shift to O2 at distance x2 flem original centre O.The Position of new C.M. is givenly let 6 is mass percunitarea.

2

cm 2 2

–6 b 1X = 6 a – 6 b



135

32. Let the mass of an element of length dx of the rod located at a distance X away from left

and is M dxL . the x cordinate of the C.M. is given by..

Total mass of rod = 1 2

0

ALAx dx = 2

1

cm 20

1 1x = xdm = x (Ax dx)M AL

2

33. By Conservation of Energy P.E. of rod = Rotational K.E. M g 12 sind =

12 I2 =

12

22mL

3

3g sin = L

As here l = 2L

3g sin = 2L

34. Angular momentum of Block w.r.t. O before collision with

O = MV a2 on collision the block will rotate about the side

passing through O. Now its angular momentum = IwAcc. to law of conservation of angular momentum

MV a2

= I 2 2M M +

6 2a a

3 = 4vv

here I is the M.I of block about the axis perpendicular to the plane passing through O.

35. Given system of two purtides will rotate about its centre of mass.

initial angular momentum = LM2

v

Final angular momentum 2L 2I = 2M

2

By law of conservation of angular momentum

LMV2

= 2L2M

2

V = L

36. Initial angular momentum of ring L = I = MR2Final angular momentum of ring and particles = (MR2 + 2mR2) 'As No external forque so According to Law of conservation of angular momentum.MR2 = (MR2 + 2mR2) '

wM ' = (M + 2m)



136

37. As it is head-on elastic collision between two idential balls there fore they will exchange theirlinear vecocity is A comes to rest and B starts moving with linear velocity V.As there is no friction any where, forque on both the spheres about their centre of mass iszero and their angular velocities remains unchanged.Therefore A = and B = 0

38. I = 0.3x2 + 0.7 (1.4 – x)2

For minimum work moment of inertia of the system should be minimum is

dI = 0 = 03 2x – 0.7 2 (1.4 – x) = 0dx

x = 0.98 m [B] 0.98 m from mass of 0.3 kg39. Angular speed for both wheels are different but lincar speed for both same so VF = Vr40. M.I of complete disc about O point

I Total = 212 (gM) R

Radius of removed disc = R3

As 2M = R i.e. 2M R

Mass of removed portion = gM Mg

M.I of removed disc about it own axis = 2 21 R MRM =

2 3 18

M.I of removed disc about O. I removed = 22 2

2cm

MR 2R MRI + Mx = + M = 18 3 2

Itotal = I removed disc + I remaining part

41.2LE = = K

2I given 1K

I If L is constant when child stretches his arms the moment of inertia

of system get doubled so kinetic energy will becomes half i.e. K2

42. Moment of inertia of sphere 22I = MR5

about its axis

L = I

43.2LE =

2I2E L

44. Rotational K.E. = 21 I = 15002

According to w = wo + t



137

45. = 1800 rpm 2 1800 = 60 rad/sec

60

P =

46. As 2IK =

2I

2

AA

JK = 2I

and2

BB

JK = 2I

47. As M is the mass of disc, the force is producing angular accoration in the disc, therefore

24 4k– = –m( ) = 3 3mkx x

48. For Ring 1 and 2 I1 and I2 about ' 23YY = MR2

and for Ring 3. I3 about ' 21YY = MR2

M.I. of system 1 2 3I = I + I + I

49.2 3

2 22 3

1 1

T r = T r

50. 2 2

g sina = (1 + k / R )

1

5g sin a = 7

and 2

2g sina = 3

51. nhL = 2

52. Impulse = Change in momentum = F t

L F = t

53.

221 1

1 12

22 22 2

1 M RI R2 = = 1I RM R2

54. Apply theorm of parallel axis

55. Rotational 21K.E. = I2

56. If = 0 I1w1 = I2w2

57. M ass = Volume = M = R2t

M.I of x is 2x 1 1

1I = m R2

58. Here direction of Motion will be reversed when force F = 0 or 20 t – 5r2 = 0 or t = 4sec.

If is angular accellaration then forque = I = F.r OR 210 = (20t – 5t ) 2 OR

2 = 4t – t and dw = dt

Also d = tdt



138

2 2 3 d = .t dt = (4t – t ) t dt = (4t t ) Taking intigration3 44t t = –

3 4 If n rotations are completed in As then Putting t = 4

4 64 64 = 2 n = – 64 = 3 3

n = 3.4 which is 3 < 3.4 < 6

59.2 2

22 R 2 RI = M + M(2R) + M 5 2 5 2

221 MR5

60. According to the Parallel axis theorm M.I of disc about an axis passing through particle (3)

and perpendicular to plane of disc is 2 2 21 3 MR + MR = MR2 2

61. In falling through a height h.P.E. = K.E

2 2 2 2 21 1 1 1 1 = + = + 2 2 2 2 2

mgh mv I mv mr

According to V2 – v2 = 2ad taking v = 0 & d = h, V2 = 2ah so 3mgh = m 2ah4

62. 2 24 6 2 37 22 2I = MR = 6 10 (6.4 10 ) = 9.83 10 kgm5 5

Angularvelocity of earth 2 T

K.E. of rotation of the earth 21 I 2

63. We know 22

kr

2gh = 1 +

v 2 2 2 2

V 2gh 2mgh = = 2 r + k mr + mk

64. kr = 50% kT

65. Centre of mass of stick is at midpoint when it is displaced by 600 Its c.m. rises up to height

h from figure = – cos = (1 – cos )2 2 2l l lh

so increase in P.E. = mgh = (1 – cos )2lmg

o10.4 10 (1 – cos60 )2

= 1 Joule



139

66. Net forque about O should be zero

Hence o o1 1Mg sin 60 = Mg sin 302 2

67. Loss in PE = gain in agular kinetic energy

22 + mg + mg = 3 3 2l l lMg l Iw

Here I = Moment of inertia about fixed point2 2

2 22 14 M + m m = m3 3 9l l l l

From fig. 2 21 14 36g 2 36 892 = = As V r = 2 9 14l 3 14 7

lg lmgl mll

68. I = I''2 2 2MR MR R = + 2m '

2 2 4

M = (M + m) '

69. 2

2

2gh = k1 +

V

r

70. New Mass 3 M4 M.I of disc =

2MR2

New MI = ' 2 21 3 3 = = 2 4 8

I M R MR

[C]

71. For a rod of mass M and length L., the MI about a perpendicular axis passing through one

and is 2

1MLI =

3

when it is bent to form a ring, then L = 2 RL R = 2

The M.I of the ring about its diameter is 2 2 2

2 2 2

MR ML MLI = = = 2 4 2 8

72. The M.I of the molecule = 2 2d dM + m

2 2

2 2d mdI = 2m = 4 2

The Rotational K.E. of the moldule 21( ) = 2

K I

2K = I



140

73.2 120 = = 4 rad/sec

60o

Now = 0 + t

Total angle descrited in 8 second is

2o

1 = w t + t2

74.dL 5L – 2L 3L = = = = Ldt 3 3

75. For disc 2MR 500 4I = =

2 2

angular speed = 2 600 =

60

so angular momentum L = I

final angular momentum in opposite direction = 2kgm–1000 20

sec

So change in angular momentum = 2L = 2 1000 20 kgm /sec

dL 2 1000 20 = = = 400 N.m.dt 100

76. By Parallel axis therom solved problem.77. 1 = 2 rad/day and 2 = 0, t = 1 day

2 1w – w = t

Forque required to stop the earth = T = I = F.R.

I.F = R

78.d = I = Idt

79. The disc have two types of motion translational and rotational so there will be two types ofangularmotion there fore total angular momentum should be total of both

T RL = L + L80. L = Momentum x perpenlicular distance between

point of rotation and line of action= m.V.y all remain constantL = remaing constant

81.40R.K.E. = T.K.E

100

2 2 21 4 1 1 = = 2 10 2 5

I mv mv 2

2 22

1 V 2 mk = 2 5

mvr



141

82. Total energy 2 2 2 2 2 21 1 1 2 1 = + = + 2 2 2 5 2

E I mv mr mr

83. Here C.M. wrt A.

5

5 5

11M 5d d = 11M + 2.2M 6

2 2

5 55d d = 2.2 M w = 55M 6 36AL w

2 2

5 5d d = 11 M = 11M 6 36BL

84. At the highest point the whole energy is conserted to P.E. of the object.

P.E. = mgh,23vh =

4g

2 23v 3m PE = mg = 4 4

vg

Total K.E. of body = 2

2 2 22

1 1 1 1 + = + 2 2 2 2

IVmv Iw mvr

Now, K.E. = P.E.

85. Here in first case 21 = 2

mg mvh = 2ghv

In second case

12

22

KR

2gh' = 1 +

v

As for the ring K = R ' = ghv

86. 22I = MR5

2 I R

This relation shows that the graph between I and R will be paracola symatric to I axis

87. Graph should be paracola symmetric to I - axis, but it should not pass from origin becausethere is a constant value Icm plesent for d = 0

88. L = Iw L w If I constant so graph between L and w will be straight line with constantstop and passing from origin.

89. = L r P

e e e log L = log P + log rIf graph is drawn between loge L and loge P then it will be straight line which will not passthrough origin at loge P = 0 loge L = loge r

90.2

rLE = 2I

so Log Er = 2 log L – log (2I)

So the graph rlog E log L will be straightline with constant stop and when Log L = 0.Log Er = – log (2I)



142

93. The correct choice is [D] since the centrifetal force is radial. Forque is zero so L = constant.94. The correct choice is [A] If a body slides down an indined plane its acaleration is

a = g (sin – cos ) which depends only on g, and .

98.cm

2

g sina = I1 + mR

Icm of hollow cylinder is loss, so it will have more accderation and will take less time to reachbottom.

99. Loss in P.E. is equal to gain in rotational K.E. As the centre of mass of the rod falls through

the distance L2

22 2L ML 3gMg I =

2 2 3 L

100. If I is the M.I. of the complete cylinder, The M.I of each Riece becomes I2 since L = I,

the angular speed of each Piece becomes 2.

101. In case (a) accelaration down the plane is

1

2

g sin 5 = = g sinI 71 +

MR

a

So 1

2

a 5 = a 7

In ase (b) 2a = g sin102. using V2 = 2ad we can find.

21 1 1

22 22

v v 5 = = v 7v

aa

103. From 21d = at2

, we find that 1

2

d 7 = d 5

104. Mg sin – f = Ma

= I . As 21I = MR2

, a = R

and = fR

Hence 21 a 1fR = MR = MRa2 R 2

105.1f = Ma2

Mg sin f = 3

106.MgR sin fR =

3



143

107. For a ring I = M R2 Correct choice is

108. Mg – T = Ma, T is Tension in string forque on cylinder = TR = I

22

I 1 a 1 = = MR = MaR 2 R 2

2mga = (M + 2m) [D]

109.MMgT =

M + 2M [D]

110. From conservation of energy we have

2 21 1 = + 2 2

Mgh MV I

2 2 2 2 2 21 1 1 1 + (2 + M)

2 2 2 4MR MR m R

12

24mgh =

(M + 2m) R

As 12 h

111. According to perpendicular axis theorm

2x y

1I = I = MR4 (Considering two perpendicular diameter in x & y directions)

So 2 2 2x y c

1 1 1I I = I = MR + MR = MR4 4 2

112. For solid cylinder 21I = MR2

As 2 21 1 = + 2 2

Mgh MV I

113. K.E. in Rotation 21 2

Iw Here 21I mr2

and V = r

therefore 2

2 21 1 1K.E = MR = 2 2 r 4

v mr

114. Rotational 21K.E = I2 & Translational 21K.E =

2mv

2 21 1K.E = + 2 2

Mv I



144

115. A solid sphere 22M.I = 5

mr

As 2

21 1 1 1

2 2 = I = = 5 5

rIw mr mn

2112 = = n

n

116. 21K = 2

I and 21 1 1

1 = 2

K I

22 2 2 2

1 1 2

1 2 r 1 2 1K = m = mr (n )2 5 n 2 5 n

2 2 21

1K = I kn2

n

117. 34V = r3

4log V = log + 3 log r3

differentiating we have dv dr = 3v r

dr 1 dv 1 1 = = 0.5% = %r 3 v 3 6

No external forque so I = constant 22 mr5

= constant

2 r = constant 2 log r + log = log c

dr d 2 + = 0r

d dr 1 1 = –2 = –2 % = – %

dt 6 3

-ve sign for decrensing

118.2 2

21 2

1 L = , = M + 2 4 12

RI MR I

119. In one full revolution the incrase in P.E = MgL, where M is the mass of rod.2

2 21 1 MgL = = 2 2 3

MLI

120. The cylinder will topple when the forque mgr equals the forque h ma2

so hma = mgr2

2gr g a = = h 2

Now V = 2.45t2 and dV = dt

a

2da = [2.45(t) ] = 4.gtdt



145

121. Using I = I cm + Md2 Parallel axis theorm

L L Ld = – = 2 4 4 and

2

cmMLI = 12

2 2 2ML ML 7ML I = + = 12 16 48

122. Loss in P.E = gain in Rotational K.E. centre of Mass of rod is at L2 .

MgLP.E = 2

Gain in 2

2 21 1 1R.E = = 2 2 2 3

MLI

2 2MgL ML 3g = = 2 6 L

123. P.E. at 600 = Mgh (1 – cos )

Eqvillibrium at 21K.E = 2

I

For I According to Parallel axis theorm (as d = R)

2 2 2 2cm

1 3I = I + MR = MR + MR = MR2 2

Here 2 o1 = Mgh (1 – cos 60 )2

I

124. We can obtain hollow sphere as it solid sphere of radius R is removed from a solid sphere

of 2R mass of hollow sphere M = M1 – M2 It is density 31

4M = (2 )3

R and 32

4M = 3

R

228 = 3

M R

Moment of inertia of hollow sphere

2 21 2

2 2I = M (2R) – M R5 5 By substituting the values of M1 and M2

125. Areal Velocity 2RA =

T and 2T =

2 2

2

R R 2AA = = = 2 2 R

= IL



146

126. Here I = 2kgm2 n1 = 60rpm = 1rps

T = ? n2 = 0 t = 1mn = 60 sec

2 1I ( – )T = I. = t

127. Angular retardation 2 1 – = t

128. For sphere 2 3 22 2 4 = = 5 5 3

I MR R R

129. From 21 = wt + t2

21

1 = 0 + (2) = 22

21 2

1 + = 0 + (4) = 82

130. Angular momentum of Recoul L = I Where 21I = MR2

Let w' be the angular velocity after putting coin of mass m at distance R2 from centre the angular

momentum of system L' ( I + mr2) ' since T = 0 so L' = L 2 (I + mr ) ' I

2

2 2 2 2

2

12

12

I ' = = = I + mr + mr 2mr1 +

MRMR

MR

but Rr = 2

2 M ' = 2M + m



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Unit - 5 Rotational Motion UG... · Comparison between physical quantities of linear motion and rotational motion Translational motion Rotational motion Linear displacement, d Angular