JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

C.U.Q

 CENTRE OF MASS
1. When a force is applied on a body, Newton’s
 second law is applicable to
Mg 1) centre of mass 2) any part of the body
3) upper most part of body
O 4) lower most part of body
L 2. Centre of mass of the earth–moon system
sin  lies
2 1) on the surface of the earth
Net torque about the point O is 2) on the surface of the moon
L  3) with in the earth
0  Mg  sin  4) at the midpoint of the line joining their centres
2  3. A square plate and a circular plate made up
Using the second law of motion 0  I0 of same material are placed touching each
MgL ML2 3 g sin other on a horizontal table. If the side length
sin    of square plate is equal to diameter of the
2 3 2L circular plate then the centre of mass of the
(b) From above, we have combination will be
3g sin d 3g sin 1) at their point of contact
   2) inside the circular plate
2L d 2L
3) inside the square plate
3 g sin 4) outside the combination
 d  d
2L 4. A uniform straight rod is placed in vertical
Integrating within appropriate limits, we get position on a smooth horizontal surface and

0 d  2L 0 sin d
3g released. As the rod is in motion, the centre
of mass moves
1) horizontally 2) vertically down
   cos 0  1  cos 
2
3g 3g 3) in a parabolic path 4) does not move.
 5. A disc and a square sheet of same mass are
2 2L 2L
cut from same metallic sheet. They are kept
1  cos 
3g side by side with contact at a single point.
 
L Then the centre of mass of combination is
The above result can also be obtained by using 1) at point of contact 2) inside the disc
the Law of Conservation of Mechanical energy, 3) inside the square 4) outside the system
where we use LINEAR MOMENTUM OF CENTRE
 Loss inGPE   Gainin RKE  OF MASS
 of CM of Rod    of Rod  6. Two balls are thrown at the same time in
    air, while they are in air, the acceleration of
11 
1  cos 
L 1 their centre of mass
 Mg I 2
  ML2  2
1) depends on masses of the balls
2 2 23  2) depends on the direction of motion of the balls
1  cos 
3g 3) depends on speeds of the balls
  4) is equal to acceleration due to gravity
L
7. Consider a two particle system with the
(c) When the rod is horizontal  , particles having masses m1 and m2 . If the first
2 particle is pushed towards the centre of mass
3g through a distance d, by what distance should
So,  , So, the tangential linear the second particle be moved, so as to keep
2L the centre of mass at the same position?
3g [MAINS 2006]
acceleration is at  L 
2 m 2d m1d m 1d
This is greater than the acceleration of an object 1) d 2) m 3) m  m 4) m
falling freely. 1 1 2 2

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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III

VECTOR PRODUCT (OR) CROSS 15. The direction of following vectors is along the
PRODUCT line of axis of rotation
         1) angular velocity, angular acceleration only
8. If P Q  R; Q R  P and R P  Q then 2) angular velocity, angular momentum only
   3) angular velocity, angular acceleration, angular
1) P , Q and R are coplanar momentum only
  4) angular velocity, angular acceleration, angular
2) angle between P and Q may be less than 900 momentum and torque
3) P  Q  R cannot be equal to zero.
   16. A particle is moving along a fixed circular orbit
with uniform speed. Then true statement from
4) P, Q and R are mutually perpendicular
  
the following is
ROTATIONAL VARIABLES, RELATION
1) angular momentum of particle is constant only
BETWEEN LINEAR AND ANGULAR in magnitude but its direction changes from point
VARIABLES,ROTATIONAL to point
KINEMATICS,TORQUE AND 2) angular momentum of particle is constant only
MECHANICAL EQUILIBRIUM in direction but its magnitude changes from point
9. Which of the following equation is wrong to point
1)
  
2)
  
a V 3) angular momentum of particle is constant both
r F r
in magnitude and direction
     
3) at r 4) V r 4) angular momentum of particle is not constant
10. The following pair of physical quantities are both in magnitude and direction
analogous to one another in translatory 17. Class I lever is that in which
motion and rotatory motion. 1) fulcrum is between the load and effort
1) Mass , moment of inertia 2) Force,Torque 2) load is between the fulcrum and effort
3) Linear momentum , Angular momentum 3) effort is between the load and fulcrum
4) All 4) fulcrum, load and effort at one point
11. The correct relation of the following is 18. If force vector is along X-axis and radius
1)  r.F 2)  r  F
     
vector is along Y-axis then the direction of
 F
 torque is
3)   4)  r  F 1) along +ve Z-axis 2) along -ve Z-axis
  
r 3) in X-Y plane making an angle 45o with X-axis
12. Two particles p and q located at distances rp 4) in X-Y plane making an angle 135o with X-axis
and ‘ rq ’ respectively from the centre of a 19. During rotation of a body, the position vector
is along X–axis and force vector is along
rotating disc such that rp  rq . Y–axis, The direction of torque vector is
1. both p and q have the same acceleration 1) in the X-Y plane 2) along –ve Z-axis
2. both p and q do not have any acceleration 3) along +ve Z-axis 4) in the X-Z plane
3. ‘p’ has greater acceleration than ‘q’ 20. If the direction of position vector r is towards
4. ‘q’ has greater acceleration than ‘p’ south and direction of force vector F is

13. When a constant torque is applied on a rigid
body, then towards east, then the direction of torque
1) the body moves with linear acceleration vector  is
2) the body rotates with constant angular velocity 1) towards north 2) towards west
3) the body rotates with constant angular 3) vertically upward 4) vertically downward
acceleration 21. Which of the following is wrong?
4) the body undergoes equal angular displacement 1) Direction of torque is parallel to axis of rotation
in equal intervals of time 2) Direction of moment of couple is perpendicular
14. Identify the increasing order of the angular to the plane of rotation of body
velocities of the following (E-2005) 3) Torque vector is perpendicular to both position
a) earth rotating about its own axis vector and force vector
b) hours hand of a clock 4) The direction of force vector is always
c) seconds hand of a clock perpendicular to both the directions of position
d) fly wheel of radius 2m making 300 rps vector and torque vector
1)a,b,c,d 2)b,c,d,a 3)c,d,a,b 4)d,a,b,c

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JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

22. A circular disc is rotated along clockwise 28. I1 , I 2 are moments of inertia of two solid
direction in horizontal plane. The direction of spheres of same mass about axes passing
torque is through their centres If first is made of wood
1) horizontally right side 2) horizontally left side and the second is made of steel, then
3) vertically upwards 4)vertically downwards 1) I1  I 2 2) I1  I 2 3) I1  I 2 4) I1  I 2
23. Magnitude of torque is maximum in the 29. A Uniform metal rod is rotated in horizontal
following case plane about a vertical axis passing through
1) radius vector is perpendicular to force vector its end at uniform rate. The tension in the
2) radius vector is parallel to force vector rod is
3) Angle between radius vector and force vector 1) same at all points
is 45o 2) different at different points and maximum at
4) Angle between radius vector and force vector centre of rod
is 60o 3) different at different points and minimum at axis
24. A constant resultant torque rotates a wheel of rotation.
about its own axis. Then true statement of 4) different at different points and maximum at axis
of rotation
the following is 30. A boiled egg and a raw egg of same mass and
1) angular velocity of wheel is constant size are made to rotate about their own axis.
2) angular acceleration of wheel is constant
If I1 and I 2 are moments of inertia of boiled
3)angular acceleration of wheel gradually increases
4) angular momentum of wheel is constant egg and raw egg, then
25. A wheel is free to rotate about its own axis 1) I1  I 2 2) I1  I 2 3) I1  I 2 4) I1  2 I 2
without friction. A rope is wound around the 31. Raw and boiled eggs are made to spin on a
wheel. If other end of rope is pulled with a smooth table by applying the same torque.
constant force, then true statement from the The egg that spin faster is
1) Raw egg 2) Boiled egg
following is 3) Both will have same spin rate
1) constant torque is produced and the wheel is 4) Difficult to predict
rotated with constant angular velocity 32. Moment of Inertia of a body depends upon
2) constant torque is produced and the wheel is 1) distribution of mass of the body
rotated with constant angular acceleration 2) position of axis of rotation
3) variable torque is produced and the wheel is 3) temperature of the body 4) all the above
rotated with variable angular velocity 33. Of the two eggs which have identical sizes ,
4) variable torque is produced and the wheel is shapes and weights, one is raw and other is
rotated with variable angular acceleration half boiled. The ratio between the moment of
26. The following pairs of physical quantities are inertia of the raw to the half boiled egg about
not analogous to each other in translatory central axis is :
motion and rotational motion 1)  1 2)  1 3)  1 4) not comparable
34. The radius of gyration of a rotating metallic
1) force, torque 2) mass, moment of inertia disc is independent of the following physical
3) couple, torque quantity.
4) linear momentum, angular momentum 1) Position of axis of rotation 2) Mass of disc
ROTATIONAL INERTIA OF SOLID 3) Radius of disc 4) temperature of disc
BODIES, ROTATIONAL DYNAMICS 35. A brass disc is rotating about its axis. If
temperature of disc is increased then its
27. The moment of inertia of a rigid body depends 1) radius of gyration increases, but moment of
on inertia remains the same
A) mass of body 2) moment of inertia increases but radius of
B) position of axis of rotation gyration remains the same
C) time period of its rotation 3) radius of gyration, moment of inertia both
D) angular velocity of the body remain the same
1) A and B are true 2) B and C ar true 4) radius of gyration, moment of inertia both
3) C and D are true 4) A and D are true increase

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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III

36. The radius of gyration of a rotating circular 41. Identify the correct order in which the ratio
ring is maximum about following axis of of radius of gyration to radius increases for
rotation the following bodies.
1) natural axis I) Rolling solid sphere II) Rolling solid cylinder
2) axis passing through diameter of ring III) Rolling hollow cylinder
3) axis passing through tangent of ring in its plane IV) Rolling hollow sphere
4)axis passing through tangent of ring perpendicular 1) I, II, IV, III 2) I, III, II, IV
to plane of ring. 3) II, I, IV, III 4) II, I, III, IV
37. Moment of inertia of a thin circular plate is 42. Identify the increasing order of radius of
minimum about the following axis gyration of following bodies of same radius
1) axis perpendicular to plane of plate passing
I) About natural axis of circular ring
through its centre
II) About diameter of circular ring
2) axis passing through any diameter of plate
III) About diameter of circular plate
3) axis passing through any tangent of plate in its
plane IV) About diameter of solid sphere
4) axis passing through any tangent perpendicular 1) II, III, IV, I 2) III, II, IV, I
to its plane 3) III, IV, II, I 4) II, IV, III, I
38. A ring of mass ‘m’ and radius ‘r’ is melted 43. Identify the decreasing order of moments of
and then moulded into a sphere . The moment inertia of the following bodies of same mass
of inertia of the sphere will be and same radius.
1) more than that of the ring I) About diameter of circular ring
2) less than that of the ring II) About diameter of circular plate
3) equal to that of the ring III) About tangent of circular ring  r to its plane
4) none of the above IV) About tangent of circular plate in its plane
39. Two copper circular discs are of the same 1) III, IV, II, I 2) IV, III, I, II
thickness. The diameter of A is twice that of B. 3) IV, III, II, I 4) III, IV, I, II
The moment of inertia of A as compared to that 44. Three dense point size bodies of same mass
of B is are attached at three vertices of a light
1) twice as large 2) four times as large equilateral triangular frame. Identify the
3) 8 times as large 4) 16 times as large increasing order of their moment of inertia
40. The moment of inertia of a thin square plate about following axis.
ABCD of uniform thickness about an axis
I) About an axis  r to plane and passing through
passing through the centre O and
perpendicular to the plane of the plate is a corner
[IIT1992] II) About an axis  r to plane and passing through
I2 centre
I3 I1 III) About an axis passing through any side
A B
IV) About  r bisector of any side
1) IV,III, II, I 2) III, II, IV, I
I4 O 3) II, IV, III, I 4) II, III, IV, I
45. Four point size dense bodies of same mass
are attached at four corners of a light square
C D frame. Identify the decreasing order of their
moments of inertia about following axes.
I) Passing through any side
a) I 1  I 3 b) I 2  I 4 II) Passing through opposite corners
c) 2 I1  I3 d) I 1  2 I 3 III)  r bisector of any side
1) a,b are true 2) b,c are true IV)  r to the plane and passing through any corner
3) c,d are true 4) b,d are true 1) III, IV, I, II 2) IV, III, I, II
3) III, II, IV, I 4) IV, III, II, I

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46. A motor car is moving in a circular path with 1) I increases and , E decrease but L is constant
uniform speed v. Suddenly the car rotates 2) I decreases, and E increase but L is constant
through an angle . Then, the magnitude of 3) I increases, decreases, L and E are constant
change in its velocity is 4) I increases, increases but L and E are constant
  52. If polar ice caps melt, then the time duration
1) 2v cos 2) 2v sin of one day
2 2

1) increases 2) decreases

3) 2v tan 4) 2v sec 3) does not change 4) zero
2 2
53. A hollow sphere partly filled with water has
47. An electric motor rotates a wheel at a
moment of inertia I when it is rotating about
constant angular velocity   while opposing its own axis at an angular velocity . If its
torque is . The power of that electric motor angular velocity is doubled then its moment
is of inertia becomes
1) Less than I 2) More than I
1) 2) 3) 2 4) 3) I 4) zero
2 54. If most of the population on earth is migrated
48. A constant power is supplied to a rotating disc. to poles of the earth then the duration of
The relationship between the angular velocity a day
  of the disc and number of rotations (n) 1) increases
3) remains same
2) decreases
made by the disc is governed by 4) first increases then decreases
55. The law of conservation of angular momentum
1) 2)
1 2
 n3  n3 is obtained from Newton's II law in rotational
motion when
3) 4)  n2
3
n 2
1) external torque is maximum
2) external torque is minimum
ANGULAR MOMENTUM & 3) external torque is zero
CONSERVATION OF ANGULAR 4) external torque is constant
56. If earth shrinks then the duration of day
MOMENTUM 1) increases 2) decreases
49. An ice block is in a trough which is rotating
about vertical axis passing through its centre. 3) remains same
When ice melts completely, the angular 4) first increases then decreases to initial value
velocity of the system 57. A circular disc is rotating in horizontal plane
about vertical axis passing through its centre
without friction with a person standing on
the disc at its edge. If the person gently walks
to centre of disc then its angular velocity
1) increases 2) decreases
3) does not change 4 )becomes zero
58. A ballet dancer is rotating about his own
1) increases 2) decreases vertical axis.Without external torque if his
3) remains same 4)becomes double angular velocity is doubled then his rotational
50. A circular disc is rotating about its own axis, kinetic energy is
the direction of its angular momentum is 1) halved 2) doubled
1) radial 2) along axis of rotation 3) quadrupled 4) unchanged
3) along tangent 59. The following motion is based on the law of
4) perpendicular to the direction of angular velocity conservation of angular momentum
51. A ballet dancer is rotating about his own A) rotation of top B) diving of diver
vertical axis on smooth horizontal floor. I , , C) rotation of ballet dancer on smooth
L, E are moment of inertia, angular velocity, horizontal surface
angular momentum, rotational kinetic energy D) a solid sphere that rolls down on an inclined
of ballet dancer respectively. If ballet dancer plane
stretches himself away from his axis of 1) A, B and C are true 2) A, B and D are true
rotation, then 3) B, C and D are true 4) A, C and D are true

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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III

60. Two bodies with moment of inertia I1 and I 2 1) solid sphere 2) hollow sphere
 I 2  I1 
3) solid cylinder 4) hollow cylinder
are rotating with same angular 67. Solid sphere, solid cylinder, hollow sphere,
momentum. If K1 and K 2 are their K.E.s, hollow cylinder of same mass and same radii
then are rolling down freely on an inclined plane.
1) K 2  K1 2) K 2  K1 3) K1  K 2 4) K 2  K1 The body with maximum acceleration is
1) solid sphere 2) solid cylinder
61. A solid sphere is rotating in free space . If 3) hollow sphere 4) hollow cylinder
the radius of the sphere is increased keeping 68. In the case of following rolling body
mass same which one of the following will not translatory and rotational kinetic energies are
be affected? equal for
1) Moment of inertia 2) Angular momentum
1) circular ring 2) circular plate
3) Angular velocity 4) Rotational kinetic energy
62. A circular wheel is rotating in horizontal plane 3) solid sphere 4) solid cylinder
without friction about its axis. If a body is 69. A disc is rolling (without slipping) on a
gently attached to the rim of the wheel then frictionless surface . C is its centre and Q and
following is false. P are two points equidistant from C. Let
1) Moment of inertia increases but angular V p ,VQ and Vc be the magnitudes of velocities
momentum remains same of points P,Q and C respectively, then
2) Angular velocity decreases but angular [IIT-2004]
momentum remains same
3) Rotational kinetic energy decreases but angular
momentum remains same
Q
4) Angular momentum increases but angular
velocity remains same C
63. A uniform metal rod of length 'L' and mass
'M' is rotating about an axis passing through P
one of the ends perpendicular to the rod with
angular speed '  ' . If the temperature 1) VQ  VC  VP 2) VQ  VC  VP
increases by "t 0 C" then the change in its
angular velocity is proportional to which of 1
3) VQ  VP , VC  VP 4) VQ  VC  VP
the following ? (Coefficient of linear expansion 2
of rod =  ) 70. A particle performs uniform circular motion
1)  2)  3) 2 4) 1/  with an angular momentum L. If the angular
64. A gymnast standing on a rotating stool with frequency f of the particle is doubled, and
his arms outstretched, suddenly lowers his kinetic energy is halved, its angular
arms momentum becomes :
1) his angular velocity decreases L L
2) his angular velocity increases 1) 4L 2) 2 L 3) 4)
2 4
3) his moment of inertia remains same 71. If V is velocity of centre of mass of a rolling
4) his moment of inertia increases body then velocity of lowest point of that body
65. Angular momentum of the particle rotating is
with a central force is constant due to 1) 2V 2) V 3) 2V 4) Zero
[AIEEE-2007]
1) constant force 72. If the velocity of centre of mass of a rolling
2) constant linear momentum body is V then velocity of highest point of that
3) zero torque 4) constant torque body is
ROLLING MOTION &ROTATIONAL V
1) 2V 2) V 3) 2V 4)
2
KINETIC ENERGY 73. If x is ratio of rotational kinetic energy and
66. Solid sphere, hollow sphere, solid cylinder and translational kinetic energy of rolling body
hollow cylinder of same mass and same radii then the following is true
are simultaneously start rolling down from the
1
top of an inclined plane. The body that takes 1) x = 1 2) x 1 3) x 1 4) x =
longest time to reach the bottom is 2
38 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

74. A body is freely rolling down on an inclined 79. When a ring is rolling V1, V2, V3 and V4 are
plane whose angle of inclination is . If ‘a’ velocities of top most point, lowest point, end
is acceleration of its centre of mass then point of horizontal diameter, centre of ring
following is correct respectively, the decreasing order of these
1) a = gsin 2) a g sin velocities is
1) V2, V1, V4, V3 2) V2, V1, V3, V4
3) a g sin 4) a = 0 3) V1, V2, V3, V4 4) V1, V3, V4, V2
75. A Child is standing with folded hands at the 80. The increasing order of fraction of total
centre of a platform rotating about its central kinetic energy associated with translatory
axes. The K.E of the system is ‘ K ’. The child motion of the following rolling bodies is
now stretches his hands so that the moment I) circular ring II) circular plate
of inertia of the system doubles. The K.E III) solid sphere IV) hollow sphere
of the system now is 1) I, II, IV, III 2) IV, I, II, III
K K 3) I, IV, II, III 4) IV, I, III, II
1) 2K 2) 3) 4) 4K 81. A and B are two solid spheres of equal
2 4
76. A yo-yo is placed on a rough horizontal masses. A rolls down an inclined plane without
surface and a constant force F , which is less slipping from a height H. B falls vertically
than its weight, pulls it vertically. Due to this from the same height. Then on reaching the
ground.
F 1) both cannot do work
2) Acan do more work than B
3) B can do more work than A
C 4) both A and B will have different linear speeds
82. A solid sphere, a hollow sphere and a ring
are released from top of an inclined plane
///////////////////////// (frictionless) so that they slide down the
O plane. Then maximum acceleration down the
1) frictional force acts towards left, so it will move plane is (no rolling):
towards left 1) solid sphere 2) hollow sphere
2) frictional force acts towards right, so it will move 3) ring 4) same for all
towards right 83. A sphere cannot roll on
3) it will move towards left, so frictional force acts 1) a smooth horizontal surface
towards left 2) a smooth inclined surface
4) it will move towards right so friction force acts 3) a rough horizontal surface
towards right 4) a rough inclined surface.
77. When the following bodies of same radius
starts rolling down on same inclined plane, C.U.Q - KEY
identify the decreasing order of their times 01) 1 02) 3 03) 3 04) 2 05) 2 06) 4
of descent 07) 4 08) 1 09) 4 10) 4 11) 2 12) 3
I) solid cylinderII) hollow cylinder 13) 3 14) 1 15) 4 16) 3 17) 1 18) 2
III) hollow sphere IV) solid sphere 19) 3 20) 3 21) 4 22) 4 23) 1 24) 2
1) IV, I, III, II 2) II, III, I, IV 25) 2 26) 3 27) 1 28) 3 29) 4 30) 3
3) I, IV, III, II 4) II, III, IV, I 31) 2 32) 4 33) 2 34) 2 35) 4 36) 4
78. When the following bodies having same radius 37) 2 38) 2 39) 4 40) 1 41) 1 42) 3
starts rolling down on same inclined plane, 43) 4 44) 1 45) 2 46) 2 47) 2 48) 1
identify the increasing order of their
49) 2 50) 2 51) 1 52) 1 53) 2 54) 2
accelerations
I) hollow cylinder II) solid cylinder 55) 3 56) 2 57) 1 58) 2 59) 1 60) 2
III)solid sphere IV) hollow sphere 61) 2 62) 4 63) 2 64) 2 65) 3 66) 4
1) I, IV, III, II 2) IV, I, II, III 67) 1 68) 1 69) 1 70) 4 71) 4 72) 3
3) I, IV, II, III 4) I, IV, III, II 73) 2 74) 2 75) 2 76) 1 77) 2 78) 3
79) 4 80) 3 81) 4 82) 4 83) 2
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III

1) 5ms–1 2) 6ms–1 3) 8ms–1 4) Zero

LEVEL - I (C.W) 8. If two particles of masses 3kg and 6kg
which are at rest are separated by a distance
CENTRE OF MASS of 15m. The two particles are moving towards
1. A system consists of two masses connected each other under a mutual force of attraction.
by a massless rod lies along x–axis. The Then the ratio of distances travelled by the
distance of centre of mass from O is particles before collision is
m1=0.4kg m2=0.6kg 1) 2 : 1 2) 1: 2 3) 1 : 3 4) 3 :1
O 9. Two bodies of 6 kg and 4 kg masses have their
x1=2m x2=7m
1) 2m 2) 3m 3) 5m 4) 7m velocity 5iˆ 2 ˆj 10kˆ and 10iˆ 2 ˆj 5kˆ resp-
2. Four particles, each of mass 1 kg are placed ectively.Then the velocity of their centre of
at the corners of a square OABC of side 1 m. mass is
‘O’ is at the origin of the coordinate system.
OA and OC are aligned along positive X-axis 1) 5iˆ 2 ˆj 8kˆ 2) 7iˆ 2 ˆj 8kˆ
and positive Y-axis respectively. The position 3) 7iˆ 2 ˆj 8kˆ 4) 5iˆ 2 ˆj 8kˆ
vector of the centre of mass is (in ‘m’)
1 1 ˆ ˆ 10. A thin uniform rod of length “L” is bent at
1) iˆ ˆj 2) iˆ ˆj 3) iˆ ˆj 4) i j its mid point as shown in the figure. The
2 2 distance of the centre of mass from the point
3. A thick straight wire of length m is fixed at “O” is
its midpoint and then bent in the form of a
circle. The shift in its centre of mass is

1) m 2) 0.5 m 3) 2 m 4) m 
2
O
 
4. A rigid body consists of a 3kg mass located
L L
at r1  2i  5 j m and a 2kg mass located at

1) sin 2) cos
 2 2 2 2
r 2  (4iˆ  2 ˆj) m. The position of centre of mass L L
is 3) sin 4) cos
4 2 4 2
14 ˆ 19 ˆ  14 19 ˆ  11. Three identical spheres each of mass ‘m’
1) j i m 2)  ˆi  jm
5 5 5 5  and radius ‘R’ are placed touching each
 19 ˆ 14 ˆ  other so that their centres A, B and C lie
3)  i  j  m 4) 0 on a straight line. The position of their
5 5  centre of mass from centre of A is
5. A boat of mass 40kg is at rest. A dog of mass
4kg moves in the boat with a velocity of 2R 5R 4R
10m/s. What is the velocity of boat(nearly)? 1) 2) 2R 3) 4)
3 3 3
1) 4m/s 2) 2m/s 3) 8m/s 4) 1 m/s 12. A boy of mass 50kg is standing at one end of
6. Two blocks of masses 10kg and 30 kg are a boat of length 9m and mass 400kg. He
placed along a vertical line if the first block is runs to the other end. The distance through
raised through a height of 7cm then the
distance through the second mass should be which the centre of mass of the boat boy
moved to raise the centre of mass of the system system moves is
by 1cm is 1) 0 2) 1m 3) 2m 4) 3m
1)1cm up 2)1cm down 3)2 cm down 4)2 cm up 13. A dog weighing 5kg is standing on a flat boat
MOTION OF CENTRE OF MASS, so that it is 10 metres from the shore. It
LINEAR MOMENTUM OF walks 4m on the boat towards the shore and
then halts. The boat weighs 20kg and one can
CENTRE OF MASS assume that there is no friction between it
7. Two bodies of different masses 2kg and 4kg and water. The dog from the shore at the end
are moving with velocities 2m/s and 10m/s of this time is
towards each other due to mutual gravitational
attraction. Then the velocity of the centre of 1) 3.4 m 2) 6.8m 3) 12.6 m 4) 10 m
mass is
40 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

VECTOR PRODUCT (or) CROSS 19. The average angular velocity of the seconds
hand of a watch if the seconds hand of the
PRODUCT watch completes one revolution in 1 minute
14. The angular velocity of a rotating body is is
 4i  j  2k . The linear velocity of the

 
1) rads 1 2) rads 1
body whose position vector 2i  3 j  3k is 15 30
1) 5i +8 j +14k 2) 3i  8 j  10k 
3) rads 1 4) rads 1
45 7
3) 8i  3 j  2k 4) -8i +3 j +2k 20. The angular displacement of a particle is

 
15. The area of the triangle whose adjacent sides
given by   t3  t2  t  1 then, its angular
are represented by the vector 4i  3j  4k velocity at t = 2 sec is ......... rads 1
and 5i in sq. units is 1) 27 2) 17 3) 15 4) 16

 
1) 25 2) 12.5 3) 50 4) 45 21. In the above problem, the angular
acceleration of the particle at t = 2 sec is
16. The angle between the vectors i  j  k and

i  j  k  is
......... rads  2
1) 14 2) 16 3) 18 4) 24
ROTATIONAL KINEMATICS,
8 1
1) sin 1
2) sin   
1
TORQUE, MECHANICAL
3 3 3
EQUILIBRIUM
8 8 22. A stationary wheel starts rotating about its
3) cos 1 4) cos
1

3 3 own axis at uniform angular acceleration

ROTATIONALVARIABLES, RELATION 8rad / s 2 . The time taken by it to complete 77
BETWEEN LINEAR & ANGULAR rotations is
VARIABLES 1) 5.5 sec 2) 7 sec 3) 11 sec 4) 14 sec
17. The linear velocity of a point on the surface 23. A stationary wheel starts rotating about its own
of earth at a latitude of 60° is axis at constant angular acceleration. If the
wheel completes 50 rotations in first 2 seconds,
800 800 then the number of rotations made by it in next
1) m/sec 2) m/sec
3 3 two seconds is
5 2000 1) 75 2) 100 3) 125 4) 150
3) 800  m/sec 4) m/sec
18 27
24. If F 2 ˆi 3 ˆj N and r 3iˆ 2 ˆj m then
 
18. A table fan, rotating at a speed of 2400 rpm
is switched off and the resulting variation of torque  is
the rpm with time is shown in the figure. The 1) 12k 2) 13k
 3) 12k  4) 13k 
total number of revolutions of the fan before 25. A crowbar of length 120 cm has its fulcrum
it comes to rest is situated at a distance of 20cm from the load.
Rev/min
The mechanical advantage of the crow bar is
2400 1) 1 2) 3 3) 5 4) 7
ROTATIONAL INERTIA OF SOLID BODIES
26. Three particles of masses 1gm, 2gm & 3gm
are at 1cm, 2cm, & 3cm from the axis of
rotation respectively then the moment of
600 inertia of the system & radius of gyration of
the system respectively are .......gm cm2 and
0 t(s) .. cm
8 16 24
1) 63, 2.449 2) 60, 4.5
1) 420 2) 280 3) 240 4) 380 3) 36, 4.449 4) 36, 2.449

NARAYANAGROUP 41
 SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III

27. A hoop of mass 500gm & radius 10cm is moment of inertia is

placed on a nail. then the moment of inertia 1) 200   T 2) 100   T
of the hoop, when it is rotated about the nail 3) 50   T 4) 150   T
will be-- kgm2 36. Three point sized bodies each of mass M are
1) 0.05 2) 0.02 3) 0.01 4) 0.03 fixed at three corners of light triangular frame
of side length L. About an axis perpendicular
28. The ratio of moments of inertia of two solid to the plane of frame and passing through
spheres of same mass but densities in the centre of frame the moment of inertia of three
ratio 1:8 is bodies is
1) 1 : 4 2) 4 :1 3) 2 : 1 4) 8 :1
3ML2
29. The radius of a solid sphere is R and its density 1) ML2 2) 3) 3ML2 4) 3ML2
D. When it is made to rotate about an axis 2
37. In above problem, about an axis perpendicular
passing through any diameter of sphere, to the plane of frame and passing through a
expression for its moment of inertia is corner of frame the moment of inertia of three
8 8 bodies is
1) DR 5 2) DR 5
7 15 3ML2
1) ML2 2) 2ML2 3) 3ML2 4)
28 28 2
3) DR 5 4) DR 5 38. In above problem about an axis passing through
15 5 any side of frame the moment of inertia of three
30. Four point size bodies each of mass M are fixed bodies is
at four corners of a light squre frame of side 3ML2 3ML2 2ML2
length L. The moment of inertia of the four 1) ML2 2) 3) 4)
2 4 3
bodies about an axis perpendicular to the plane 39. The radius of gyration of a body is 18 cm when
of frame and passing through its centre is it is rotating about an axis passing through
1) 4ML2 2) 2 2ML2 3) 2ML2 4) 2ML2 centre of mass of body. If radius of gyration of
same body is 30 cm about a parallel axis to first
31. Four particles each of mass ‘m’ are placed axis then, perpendicular distance between two
at the corners of a square of side length '  '. parallel axes is
The radius of gyration of the system about 1) 12 cm 2) 16 cm 3) 24 cm 4) 36 cm
an axis perpendicular to the plane of square 40. The position of axis of rotation of a body is
and passing through its centre is changed so that its moment of inertia
decreases by 36%. The % change in its
 radius of gyration is
1) 2) 3)  4)

2 2 2 1) decreases by 18% 2) increases by 18%
32. In the above problem the moment of inertia 3) decreases by 20% 4) increases by 20%
41. A diatomic molecule is formed by two atoms
of four bodies about an axis perpendicular to
the plane of frame and passing through a which may be treated as mass points m1 and
corner is m2 joined by a massless rod of length r. Then
1) ML2 2) 2ML2 3) 2 2ML2 4) 4ML2 the moment of inertia of molecule about an axis
passing through centre of mass and
33. In above problem the moment of inertia of perpendicular to the rod is :
four bodies about an axis passing through 1)zero 2)  m1  m2  r 2
opposite corners of frame is
 m1m2   m1  m2 
1) 2ML2 2) 2ML2 3) ML2 4) 2 2ML2 3)  m  m  r
2
4)  m m  r
2

34. In the above problem the moment of inertia  1 2   1 2 

42. I is moment of inertia of a thin square plate
of four bodies about an axis passing through about an axis passing through opposite
any side of frame is corners of plate. The moment of inertia of
1) 4ML2 2) 2 2ML2 3) 2ML2 4) 2ML2 same plate about an axis perpendicular to the
plane of plate and passing through its centre
35. The diameter of a fly wheel is R. Its coefficient is
of linear expansion is . If its temperature is
increased by T the percentage increase in its 1) I/2 2) I / 2 3) 2I 4) 2I

42 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

43. Mass of thin long metal rod is 2 kg and its 2 mr 4 mr 2 mr 2 4 mr 2

moment of inertia about an axis perpendicular 1) 2) 3) 4)
to the length of rod and passing through its T T T T
51. If the radius of earth shrinks by 0.2% without
one end is 0.5kg m 2 . Its radius of gyration is change in its mass, the % change in its
1) 20 cm 2) 40 cm 3) 50 cm 4) 1m angular velocity is
ANGULAR MOMENTUM AND 1) increase by 0.4% 2) increase by 0.1%
CONSERVATION OF ANGULAR 3) decrease by 0.4% 4) decrease by 0.1%
MOMENTUM 52. A metallic circular plate is rotating about its
axis without friction. If the radius of plate
44. The diameter of a disc is 1m. It has a mass of expands by 0.1% then the % change in its
20kg. It is rotating about its axis with a speed moment of inertia is
of 120rotations in one minute. Its angular 1) increase by 0.1% 2) decrease by 0.1%
momentum in kg m 2/s is 3) increase by 0.2% 4) decrease by 0.2%
1)13.4 2) 31.4 3) 41.4 4) 43.4 53. A constant torque acting on a uniform circular
45. If the earth were to suddenly contract to 1/nth wheel changes its angular momentum from A
of its present radius without any change in to 4A in 4sec. The torque acted on it is
its mass, the duration of the new day will be
nearly 3A A 2A 3A
1) 2) 3) 4)
1) 24/n hours 2) 24n hours 4 4 4 2
3) 24/n2 hours 4) 24n2 hours 54. Density remaining constant, if earth
46. A particle performs uniform circular motion contracts to half of its present radius,
with an angular momentum L. If the angular duration of the day would be (in minutes)
frequency f of the particle is doubled, and 1) 45 2) 80 3) 100 4) 120
kinetic energy is halved, its angular 55. A mass is whirled in a circular path with an
momentum becomes angular momentum L. If the length of string
1) 4L 2) 2L 3) L/2 4) L/4 and angular velocity, both are doubled, the
47. A ballet dancer is rotating about his own new angular momentum is
vertical axis at an angular velocity 100 rpm 1) L 2) 4L 3) 8L 4) 16L
on smooth horizontal floor. The ballet dancer ROTATIONAL DYNAMICS
folds himself close to his axis of rotation by
which is moment of inertia decreases to 56. An automobile engine develops 100 KW
half of initial moment of inertia then his final when rotating at a speed of 1800 rev/min. The
angular velocity is torque it delivers ( in N-m )
1) 50rpm 2) 100rpm 1) 350 2) 440 3) 531 4) 628
3) 150rpm 4) 200rpm 57. An electric motor exerts a constant torque
48. A circular ring of mass M is rotating about 5Nm on a fly wheel by which it is rotated at
its own axis in horizontal plane at an angular the rate of 420rpm The power of motor is
velocity . If two point size bodies each of 1)110watt 2)150watt
mass m, are gently attached to the rim of ring
at two ends of its diameter, then the angular 3)220watt 4)300watt
velocity of ring is ROLLING MOTION
M 2m m 2M 58. A shaft rotating at 3000rpm is transmitting a
1) 2) 3) 4) power of 3.14KW. The magnitude of the
M  2m M  2m M  2m M  2m
49. A ballet dancer is rotating at angular velocity driving torque is
on smooth horizontal floor. The ballet 1) 6Nm 2) 10Nm 3) 15Nm 4) 22Nm
dancer folds his body close to his axis of 59. A solid sphere rolls down without slipping
rotation by which his radius of gyration from rest on a 30 0 incline. Its linear
decreases by 1/4th of his initial radius of acceleration is
gyration, his final angular velocity is 1) 5g/7 2) 5g/14 3) 2g/3 4) g/3
3 9 9 16 60. A hollow sphere rolls down a 30o incline of
1) 2) 3) 4)
4 4 16 9 length 6m without slipping. The speed of cen
50. A particle of mass m is moving along a circle tre of mass at the bottom of plane is
of radius r with a time period T. Its angular
momentum is 1) 6ms 1 2) 3ms 1 3) 6 2ms 1 4) 3 2ms 1

NARAYANAGROUP 43
 SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III

 L / 4
61. For a body rolling along a level surface, the m
translational and rotational K.E. are sin
ycm  2 ; rcm  x 2 cm  y 2 cm
equal.The body is m
1) Solid cylinder 2) disc
m1 x1 m2 x2 m3 x3
3) ring 4) hollow sphere 11. xcm
62. A ring and a disc of same mass roll without m1 m2 m3
slipping along a horizontal surface with same 12. Center of mass does not change
velocity. If the K.E. of ring is 8J, then that of 13. Distance from shore= 10  l  d  14. v   r
  
disc is
A B
 
1) 2J 2) 4J 3) 6J 4) 16J 1  
63. When a hollow sphere is rolling without 15. Area of triangle = A  B 16. sin 
2 AB
slipping on a rough horizontal surface then 2
the percentage of its total K.E. which is 17. v  r ; r  R cos ; 
translational is T
18. Number of revolutions = area under the curve
1) 72% 2) 28% 3) 60% 4) 40%
64. If a sphere of mass 2kg and diameter 10cm 2
19.   rads 1
is rolling at speed of 5ms 1 . Its rotational 60 30
kinetic energy is d
 t 3  t 2  t  1 ;  dt  3t  2t  1
2
20.
1)10J 2) 30J 3)50J 4) 70J
LEVEL-I - (C.W) - KEY 
d
 6t  2  12  2  14rads 2
01) 3 02) 2 03) 2 04) 2 05) 4 06) 2 21.
dt
07) 4 08) 1 09) 3 10) 4 11) 2 12) 1 1 2
13) 2 14) 2 15) 2 16) 1 17) 4 18) 2 22. Given i  0 and   t
2
19) 2 20) 2 21) 1 22) 3 23) 4 24) 4
25) 3 26) 4 27) 3 28) 2 29) 2 30) 3 23. i  0 ; t  2s ;
31) 1 32) 4 33) 3 34) 3 35) 1 36) 1  50  2   100 rad
37) 2 38) 3 39) 3 40) 3 41) 3 42) 4 2 200
43) 3 44) 2 45) 3 46) 4 47) 4 48) 1     50 rad s 2
49) 4 50) 3 51) 1 52) 3 53) 1 54) 1 t 2
4
t   50 16 
55) 3 56) 3 57) 3 58) 2 59) 2 60) 1 1 2 1
61) 3 62) 3 63) 3 64) 1 ‘ ’ in 4 sec ; 
2 2
LEVEL-I - (C.W) - HINTS = 400 rad
in the last 2 sec,  400  100  300 rad
m1 x1  m2 x2
1. xcm  m1  m2 2. rˆcm  xcmiˆ  ycm ˆj 
300
 no.of rotations = = 150
2 2
m1 r1  m2 r2
 
l 24.   r  F

4. r cm 

3. 2 r  l; r 
2 m1  m2 effort arm
mv m1y1  m2 y2 25. M A  l o a d a r m
5. vb  6. ycm 
26. I  mr 2  11  2  2   3  3
mM m1  m2 2 2 2

m1v1  m2 v2
7. vcm  m1  m2 ; Internal force does not change
I
and K  Ans: 36, 2.449
the position of centre of mass m
2
m1 v1  m2 v 2
 
27. I  mr 2  mr 2  2mr 2 28. I sphere  MR
2
9. v cm 

8. m1r1  m2 r2 5
m1  m2
2 2  4  8
29. I  MR   R  DR 
  L / 4    cos  L / 4
DR5
2 3 2
m m
5 5 3  15
10. xcm  2 2
m

44 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

I   mr 2
M L M I1 I 2 R
30.  , R2  1
1  I2
45. I1 ;
2 T T2 n
L 1
L L Here r  1
2 46. KE  L
  L  2
L  I  4 M     2 ML
2
47. I1 1  I 2 2 ; I1 n1  I 2 n2  n2  200rpm
M M   2 
48. I1 1  I 2 2

 l  4ml 2
2

31. I  mr 2
 4 m     2ml 2
3 
2
 2 2
49. I1  I2 ; mk2
 mk 2
;k
2
  k1 
4 
1 2 1 1 2 2 1 1 2
I 2ml
2
l
Radius of gyration k    2r
M 4m 2 50. L  mvr and V 
T
32. I  2  ML2   M  L 2  ;  2ML2  2ML2  4 ML2
2
2
33. 51. I  MR  constant ;  R 2  constant
2

5
I R L2  L1
  L 2  52. I  R 2 and 2 53. 
I  2 M  I R t
   ML
2

  2   54. I1 1  I 2 2 and R1 T1  R2 T2
5 5

L1 r1  1 
2

34. I  ML2  ML2  2ML2 55. L  mr 2 ; L  r 2 ;    

L2 r2  2 
I 2l 56. p  57. p  58. p 
35. I  l and   2 T
2

I l g sin 2 gl sin
36. a v
  L 2  59. k 2
60. k2
L L I  3 M 
1 2 1 2
  R R
O   3  
1 2 1 2 1 2  k2 
L  ML2 61. mv  I 62. KE  mv  1  2 
2 2 2  R 
37. I  2  ML 
2
 
 1 
 3L  KET
2
3ML 2
 100     100 1
38. I  M    63. KETOTAL 64. KErot  I 2
1 K 
2
2
 2  4 
 R2


39. K  k cm
2
d2
LEVEL - I (H.W)
 K2   I2 
40. I  MK 2  I K 2 ; K  1  100   I  1 
 1   1  CENTRE OF MASS
41. With respect to centre of mass, effective mass 1. The distance of centre of mass from ‘O’ is
mm  mm  5kg 4kg 6kg
 1 2 ; I   1 2  r 2
m1  m2  m1  m2 
0 0.3m 1m
1)0.21m 2) 0.35m 3) 0.42m 4) 0.48m
ML2 ML2 2. Four bodies of masses 1,2,3,4 kg respectively
42. I  I
, z  I  I y ; I 1
  2I
12
x
6 are placed at the comers of a square of side
L ‘a’. Coordinates of centre of mass are (take
mL2
43. I  ;K  1kg at the origion, 2kg on X-axis and 4kg on
3 3 Y-axis)
mr
 7 a a   a 7 a   a 3a   7a 3a 
2
44. L  I ;where I  ; 2 n
2 1)  ,  2)  ,  3)  ,  4)  , 
 10 2   2 10   2 10   10 2 

NARAYANAGROUP 45
SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III

3. A uniform rod of length one meter is bent at 10. Two objects of masses 200g and 500g have
its midpoint to make 900. The distance of velocities of 10i m/s and  3i  5 j  m / s
centre of mass from the centre of rod is (in
cm) respectively. The velocity of their centre of mass
1) 20.2 2) 13.4 3) 15 4) 35.36 is
4. Particles of masses 1kg and 3kg are at 5
1) 5i  25 j 2) i  25 j
 2i  5 j  13k  m and  6i  4 j  2k  m then 7
instantaneous position of their centre of mass 25 5
is 3) 5i  j 4) 25i  j
7 7
1)  16i  17 j  7k  m
1 VECTOR PRODUCT OR CROSS
4 PRODUCT
2)  8i  17 j  7k  m
1 11. The position of a particle is given by
4
r  i  2 j  k and its momentum is


3)  6i  17 j  7k  m
1
p  3i  4 j  2k . The angular momentum is

4
4)  6i  17 j  5k  m
1 perpendicular to
4 1) x-axis 2) y-axis 3) z-axis
5. A boat of mass 50kg is at rest. A dog of mass 4) line at equal angles to all the axes
5kg moves in the boat with a velocity of 20m/ 12. A uniform sphere has radius R. A sphere of
s. What is the velocity of boat? diameter R is cut from its edge as shown.
1) 4m/s 2) 2m/s 3) 8m/s 4) 1 m/s Then the distance of centre of mass of
MOTION OF CENTRE OF MASS, remaining portion from the centre of mass of
the original sphere is
LINEAR MOMENTUM OF
CENTRE OF MASS R
6. Two bodies of masses 5kg and 3kg are moving
towards each other with 2ms 1 and 4ms 1 1)R/7 2) R/14 3)2R/7 4) R/18
13. The area of the parallelogram whose adjacent
respectively. Then velocity of centre of mass is
1) 0.25ms 1 towards 3kg 2) 0.5ms 1 towards 5kg sides are P  3i  4 j; Q  5i  7 j is
3) 0.25ms 1 towards 5kg 4) 0.5ms 1 towards 3kg (in sq.units)
1)20.5 2) 82 3) 41 4) 46
7. A circular disc of radius 20cm is cut from one
14. If A  3i  j  2k and B  2i  2 j  4k and
 
edge of a larger circular disc of radius 50cm.
The shift of centre of mass is is the angle between the two vectors, then
1) 5.7cm 2) -5.7cm 3) 3.2cm 4) -3.2cm sin is equal to
8. Two particles of masses 4kg and 6kg are
separated by a distance of 20m and are 2 2 2 2
1) 2) 3) 4)
moving towards each other under mutual 3 3 7 13
force of attraction, the position of the point ROTATIONAL VARIABLES,
where they meet is RELATION BETWEEN LINEAR
1) 12m from 4kg body 2) 12m from 6kg body
3) 8m from 4kg body 4) 10m from 4kg body AND ANGULAR VARIABLES
9. A uniform metre rod is bent into L shape with 15. A particle is moving with uniform speed
the bent arms at 900 to each other. The 0.5m/s along a circle of radius 1m then the
distance of the center of mass from the bent angular velocity of particle is ( in rads-1 )
point is 1)2 2)1.5 3)1 4) 0.5
L L L L 16. The angular velocity of the seconds hand in a
1) m 2) m 3) m 4) m watch is
4 2 2 2 2 8 2 1) 0.053 rad/s 2) 0.210 rad/s
3) 0.105 rad/s 4) 0.42 rad/s
46 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

17. The angular displacement of a particle is given 24. If I is moment of inertia of a thin circular plate
by  t 3  2t  1 , where t is time in seconds. about an axis passing through tangent of plate
Its angular acceleration at t = 2s is in its plane. The moment of inertia of same
circular plate about an axis perpendicular to its
1) 14 rad s 2 2) 17 rad s 2 plane and passing through its centre is
3) 12 rad s 2 4) 9 rad s 2 4I 2I 4I 2I
ROTATIONAL KINEMATICS, 1) 2) 3) 4)
5 5 3 3
TORQUE, MECHANICAL 25. The moment of inertia of a solid sphere about
EQUILIBRIUM an axis passing through its centre is 0.8kgm 2 .
18. A circular disc is rotating about its own axis The moment of inertia of another solid sphere
at a uniform angular velocity . The disc is whose mass is same as mass of first sphere,
subjected to uniform angular retardation by but the density is 8 times density of first
sphere, about an axis passing through its
which its angular velocity is decreased to centre is
2
during 120 rotations. The number of rotations 1) 0.1kgm2 2) 0.2 kgm 2
further made by it before coming to rest is 3) 0.4 kgm 2 4) 0.5 kgm 2
1)120 2) 60 3) 40 4) 20
26. Moment of inertia of a hoop suspended from
19. The handle of a door is at a distance 40cm
a peg about the peg is
from axis of rotation. If a force 5N is applied
on the handle in a direction 300 with plane of MR 2 3MR 2
door, then the torque is 1) MR 2)2 3) 2MR 2 4)
2 2
1) 0.8 Nm 2) 1 Nm 3) 1.6 Nm 4) 2 Nm 27. Four particles each of mass 1kg are at the
20. A door can just be opened with 10N force on four corners of square of side 1m. The M.I.of
the handle of the door. The handle is at a the system about a normal axis through centre
distance of 50cm from the hinges. Then, the of square is
torque applied on the door (in Nm) is
1) 5 2) 10 3) 15 4) 20 1) 6 kgm2 2) 2 kgm 2 3)1.25 kgm2 4) 2.5 kgm 2
21. A particle of mass m is projected with an initial 28. Three identical masses, each of mass 1kg,
velocity u at an angle to horizontal.The are placed at the corners of an equilateral
torque of gravity on projectile at maximum triangle of side l. Then the moment of inertia
height about the point of projection is of this system about an axis along one side of
the triangle is
mgu 2 sin 2 3 2 3 2
1) 2) mgu 2 sin 2 l
2 1) 3l 2 2) l 2 3) 4) l
4 2
mgu 2 sin 1 29. A wire of mass m and length l is bent in the
3) 4) mu sin 2
2

2 2 form of circular ring. The moment of inertia

22. A uniform rod is 4m long and weights 10kg. of the ring about its axis is
If it is supported on a knife edge at one meter ml 2 ml 2 ml 2
from the end, what weight placed at that end 1) ml 2 2) 3) 4)
4 2 2 2 8 2
keeps the rod horizontal. 30. The moment of inertia of a thin uniform rod of
1) 8kg 2) 10kg 3) 12kg 4) 15kg mass M and length L about an axis
ROTATIONAL INERTIA OF SOLID BODIES perpendicular to the rod, through its centre is
23. The ratio of moments of inertia of a solid I.The moment of inertia of the rod about an axis
sphere about axes passing through its centre perpendicular to rod through its end point is
and tangent respectively is I I
1) 2:5 2) 2:7 3) 5:2 4) 7:2 1) 2) 3) 2I 4) 4I
4 2

NARAYANAGROUP 47
 SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III

31. Four point size bodies each of mass m are now stretches his arms so that the M.I. of
fixed at four corners of light square frame of the system doubles. The K.E. of the system
side length 1m. The radius of gyration of these now is
four bodies about an axis perpendicular to the 1) 2K 2) K/2 3) 4K 4) K/4
plane of frame passing through its centre is 40. If radius of earth shrinks by 0.1% without
1 1 change in its mass, the percentage change in
1) 2 2) 2 3) 4) the duration of one day
2 2 1) decrease by 0.1% 2) increase by 0.1%
32. Uniform square plate of mass 240 gram is 3) decrease by 0.2% 4) increase by 0.2%
made to rotate about an axis passing through 41. A ballet dancer spins about a vertical axis at
any diagonal of plate. If its moment of inertia 60rpm with his arms closed. Now he stretches
is 2  104 kgm2 then its side length is his arms such that M.I. increases by 50%.
1) 10cm 2) 12cm 3) 15cm 4) 20cm The new speed of revolution is
33. Two objects of masses 1kg and 2kg separated 1) 80rpm 2) 40rpm 3) 90rpm 4) 30rpm
by a distance of 1.2m are rotating about their 42. A metallic circular wheel is rotating about its
centre of mass. Find the moment of inertia of own axis without friction. If the radius of wheel
the system expands by 0.2%, percentage change in its
angular velocity
1) 0.96kgm2 2) 0.48kgm 2 1) increase by 0.1% 2) decrease by 0.1%
3) 0.83kgm2 4) 0.72kgm2 3) increase by 0.4% 4) decrease by 0.4%
34 The radius of gyration of a body about an axis 43. A uniform circular disc of radius R is rotating
at a distance of 4cm from its centre of mass about its own axis with moment of inertia I at
is 5cm. The radius of gyration about a parallel an angular velocity If a denser particle of
axis through centre of mass is mass m is gently attached to the rim of disc
1) 2cm 2) 5cm 3) 4cm 4) 3cm than its angular velocity is
35. The M.I. of a thin rod about a normal axis I  mR 2
1) 2) I  I  mR  3)
I
through its centre is I. It is bent at the centre 4)
I I  mR 2
such that the two parts are perpendicular to
44. A particle of mass m is rotating along a circular
each other and perpendicular to the axis. The
path of radius r. Its angular momentum is L.
M.I. of the system about the same axis will be
The centripetal force acting on the particle is
1) 2I 2) I 3) I/2 4) 4I
36. The moment of inertia of two spheres of equal L2 L2 m L2 L2
masses about their diameters are the same. 1) 2) 3) 4)
mr r mr 2 mr 3
One is hollow, then ratio of their diameters
45. F  ai  3 j  6k and r  2i  6 j  12k . The
 
1) 1:5 2) 1: 5 3) :1 4) 5 : 3
value of ‘a’ for which the angular momentum
ANGULAR MOMENTUM AND is conserved is
CONSERVATION OF ANGULAR 1) -1 2) 0 3) 1 4) 2
MOMENTUM 46. If earth shrinks to 1/64 of its volume with mass
37. A circular disc of mass 4kg and of radius 10cm remaining same, duration of the day will be
is rotating about its natural axis at the rate 1) 1.5h 2) 3h 3) 4.5h 4) 6h
of 5 rad/sec. its angular momentum is 47. A mass is whirled in a circular path with a
constant angular velocity and its angular
1) 0.25 kgm 2 s 1 2) 0.1kgm 2 s 1 momentum is L. If the length of string is now
3) 2.5kgm 2 s 1 4) 0.2 kgm 2 s 1 halved keeping the angular velocity same, the
38. If the mass of earth and radius suddenly new angular momentum is
become 2 times and 1/4th of the present value, 1) L/4 2) L/2 3) L 4) 2L
the length of the day becomes 48. A disc rotates with angular velocity and
1) 24h 2) 6h 3) 3/2h 4) 3h kinetic energy E. Then its angular momentum
39. A child is standing with folded hands at the E 2E
centre of a platform rotating about its central 1) I 2) L  3) L  4) L 
E
axis. The K.E. of the system is K. The child
48 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

ROTATIONAL DYNAMICS LEVEL-I (H.W) - KEY

2 01) 4 02) 2 03) 4 04) 1 05) 2 06) 3
49. A wheel at rest has M.I. 2 kgm . It is 07) 2 08) 1 09) 1 10) 3 11) 1 12) 2
2

rotated by a 60W motor for one minute. The 13) 3 14) 3 15) 4 16) 3 17) 3 18) 3
number of rotations made by the wheel in one 19) 2 20) 1 21) 4 22) 2 23) 2 24) 2
minute is 25) 2 26) 3 27) 2 28) 3 29) 2 30) 4
1) 90 2) 450 3) 1800 4) 1200 31) 3 32) 1 33) 1 34) 4 35) 2 36) 4
50. The shaft of a motor is making 1260rpm. The 37) 2 38) 4 39) 2 40) 3 41) 2 42) 4
torque supplied by the motor is 100Nm. the 43) 4 44) 4 45) 1 46) 1 47) 1 48) 3
power of motor is ( in KW) 49) 3 50) 3 51) 2 52) 1 53) 4 54) 1
1) 100 2) 21 3) 13.2 4) 4.8 55) 2 56) 1 57) 2 58) 2
51. An electric motor rotates a wheel at a LEVEL-I (H.W) - HINTS
constant angular velocity 10rps while m1 x1  m2 x2
opposing torque is 10Nm . The power of that 1 . xcm  m1  m2
electric motor is
1) 120W 2) 628W 3) 314W 4) 3.14W m x m x m y m y
2. xcm  1 1 2 2 ; ycm  1 1 2 2
52. The work done in increasing the angular m1  m2 m1  m2
frequency of a circular ring of mass 2kg and
radius 25cm from 10 rpm to 20rpm about is L 
3. d Sin  
axis 4 2
1)0.2058J 2)0.2040J 3)0.2085J 4)0.2004J
m1 r 1  m2 r 2
 
ROLLING MOTION 4. r cm 


53. A ring is allowed to roll down on an incline of m1  m2

1 in 10 without slipping. The acceleration of mv
its center of mass is 5. vb 
mM
1) 9.8ms 2 2) 4.9ms 2 3) 0.98ms 2 4) 0.49ms 2 m1v1  m2 v2
54. A cylinder is released from rest from the top 6. vcm  m1  m2
of an incline of inclination and length ‘l’. If
the cylinder roles without slipping, its speed r 2d
at the bottom 7. shift   2 2
R r
4 gl sin 3gl sin 8. m1r1  m2 r2
1) 2)
3 2 L
9. rcm  xcm2
 ycm
2
(or) rcm  cos
4 gl 4 g sin 4 2
3) 4)
m1 v1  m2 v 2
 
3sin 3l
10. v cm 

55. For a body rolling along a level surface, without m1  m2
slipping the translational and rotational kinetic
11. r  F  ;  x  axis
  
energies are in the ratio 2:1.The body is 
1) Hollow sphere 2) solid cylinder
r 3d
3) Ring 4) Solid sphere 12. shift  3 3
56. A solid sphere and a spherical shell roll down R r
an incline from rest from same height. The 13. Area of parallelogram  P  Q
 
ratio of times taken by them is
A B
 
21 21 25 25
1) 2) 3) 4) 14. sin 
25 25 21 21 AB
57. When a solid sphere is rolling along level v
surface the percentage of its total kinetic 15. 
energy that is translational is r
1) 29%2) 71% 3) 60% 4) 40% 2
58. A thin ring of mass 1kg and radius 1m is rolling 16. 
60
at a speed of 1ms 1 . Its kinetic energy is d
1) 2J 2) 1J 3) 0.5J 4) zero 17. 
dt
NARAYANAGROUP 49
SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III

2 2
2 2 
18. is constant,  1 2
;  1
40. I  MR 2     constant
2 3 5  T 
2

19.  rF sin  rF
20. T R
 R T  R 2 and 2
 T R
21.    i  H j   mg j
 
 2  41. I1 1  I 2 2 ; I1 n1  I 2 n2
22. clockwise torque = anticlockwise torque 42. I  mr 2  constant
2
MR 2  r
I centre 5 2  r 2 and  2
23.   r
I tan get 7 MR 2 7
43. I1 1  I 2 2 ; I1  I ; I 2  I  mR 2
5
L
5MR MR 2 44. L  mvr  v  mr
2

24. I  &I 
|

4 2
mv 2 L2
centripetal force F  
2
1 I  R   D 3
2

25.Mass is same and D  3 ; 1   1    2  r mr 3

R I 2  R2   D1  dL
 r  F and  dt  0
  
26. It is equivalent to ring rotating about an axis passing 45.
through tangent. 2
I  MR 2  constant and V  R 3
27. I   mr ; r 
l 46.
2 5
2 L1  r1  L
2

47. L  mr 2 ; L  r2 ;     L2 
 3l 
2
L2  r2  4
28. I  M  
 2  1
48. KE  L 49. p ; 2 N
l 2 t
29. I  mr 2 ; r  50. p  51. p 
2
52. W  I  2  
1
ML2 L
2 2 2

30. I  ; I  ICM  M   2
1

12 2 g sin
I a
31. I  2ml ; k  53. k2
1 2
2

4m R
Ml 2 2 gl sin
32. I z  I x  I y ; II v
12 54. 1 2
k2
 m1m2  2 R
33. I   m  m  r 1
 1 2  mV 2
2
2 
34. I  mk ; I  I 0  mr 2
2
55. 1 2 1
I
l   l  2
2 2

35. ml 2 m  2  m  2 
I  ; I   I  k2 
|

12 2 3 2 3 2l 1  2 
2 2 56.  R 
36. M.I. of solid sphere about diameter  mr t
5 g sin
2 2  
M.I. of hollow sphere about diameter  mr KE T  1 
3  100     100
mr 2 57. KE trans  1  k
2

37. L  I  38. I1 1  I 2 2  R2 
2
1
L2 58. KErot  I 2

39. KE  2
2I
50 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

6. A bomb of mass ‘m’ at rest at the coordinate

LEVEL- II (C.W) origin explodes into three equal pieces. At
a certain instant one piece is on the x–axis
CENTRE OF MASS at x=40cm and another is at x=20cm,
1. A uniform wire is bent into the form of a
rectangle of length L and width W. The y = –60cm. The position of the third piece is
coordinates of its centre of mass from a corner 1) x = 60cm, y=60cm 2) x = –60cm, y= –60cm
are 3) x = –60cm, y=60cm 4) x = 60cm, y= –60cm
L  7. Particles of masses m,2m, 3m ........... nm gram
1) (0, 0) 2)  , W  are placed on the same line at distances, l,
2 
2l, 3l, ...... nl cm from a fixed point. The
 W L W
3)  L,  4)  ,  distance of centre of mass of the particles
 2 2 2  from the fixed point in cm in
2. A uniform disc of radius R is put over another
uniform disc of radius 2R of same thickness ( 2n  1)l l
1) 2)
and density. The peripheries of the two discs 3 n 1
touch each other. The position of their centre
of mass is n(n 2  l )l 2l
1) at R/3 from the centre of the bigger disc towards 3) 4) n(n 2  l )l
2
the centre of the smaller disc
2) at R/5 from the centre of the bigger disc towards 8. Three particles each of mass 2kg are at the
the centre of the smaller disc corners of an equilateral triangle of side 3 m.
3) at 2R/5from the centre of the bigger disc towards
the centre of the smaller disc If one of the particles is removed, the shift in
4) at 2R/5from the centre of the smaller disc the centre of mass is
3. Three particles each 1kg mass are placed at 1) 0.2m 2) 0.5m 3) 0.4m 4) 0.3m
the corners of a right angled triangle AOB, 9. The mass of a uniform ladder of length 5m is
O being the origin of the co–ordinate system 20 kg. A person of mass 60kg stand on the
OA and OB along +ve x-direction and +ve y –
direction. The position vector of the centre of ladder at a height of 2m from the bottom. The
mass is (OA = OB = 1m) (in meters) position of centre of mass of the ladder and
i j i j 2 (i  j ) man from the bottom is
1) 2) 3) 4) (i–j) 1)1.256m 2) 2.532m 3) 3.513m 4)2.125m
3 3 3
4. If three particles of masses 2kg, 1kg and 10. A uniform thin rod of length 1m and mass 3kg
3kg are placed at corners of an equilateral is attached to a uniform thin circular disc of
triangle of perimeter 6m then the distance of radius 30cm and mass 1kg at its centre
centre of mass which is at origin of particles perpendicular to its plane. The centre of mass
from 1kg mass is (approximately) ( Assume
of the combination from the centre of disc is
2kg on x-axis
1) 0.375m 2) 0.25m 3) 0.125m 4)0.475m
1
1) 6 m 2) 2m 3) m 4) 2m 11. Four identical particles each of mass “m” are
2 arranged at the corners of a square of side
5. Six identical particles each of mass ‘m’ are
length “L”. If one of the masses is doubled,
arranged at the corners of a regular hexagon
of side length “L”. If the mass of one of the the shift in the centre of mass of the system.
particle is doubled, the shift in the centre of w.r.t. diagonally opposite mass
mass is
L 3 2L L L
L 1) 2) 3) 4)
1) L 2) 6L / 7 3) L / 7 4) 2 5 4 2 5 2
3

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51
SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III
12. A circular hole of radius ‘r’ is made in a MOTION OF CENTRE OF MASS
disk of radius ‘R’ and of uniform thickness
at a distance ‘a’ from the centre of the disk. AND LINEAR MOMENTUM
The distance of the new centre of mass from 16. Two particles of equal masses have velocities
the original centre of mass is
v1 = 4iˆand v 2  4jˆ . First particle has an
 

acceleration a1 =(5i  5 j ) ms–2 while the


R r
a acceleration of the other particle is zero.
The centre of mass of the two particles
aR 2 ar 2 moves in a path of
1) 2 2) 2 1) Straight line 2) Parabola
R  r2 R  r2
a( R 2  r 2 ) a( R 2  r 2 ) 3) Circle 4) Ellipse
3) 4) 17. Two particles of masses “p” and “q” (p>q)
r2 R2
13. The centre of mass of the letter F which is cut are separated by a distance “d”. The shift
from a uniform metal sheet from point A is in the centre of mass when the two particles
6 are interchanged is
A 1) d(p+q) / (p–q) 2) d(p–q) / (p+q)
2
4 3) d p/(p–q) 4) d q/ (p–q)
2 VECTOR PRODUCT OR CROSS PRODUCT
8 18. The unit vector perpendicular to
2
A  2iˆ  3 ˆj  kˆ and B  iˆ  ˆj  kˆ is
 
2 2
4 iˆ  ˆj  5 kˆ 4 iˆ  ˆj  5 kˆ
1) 2)
2 42 42

1) 15/7, 33/7 2) 15/7, 23/7 4 iˆ  ˆj  5 kˆ 4iˆ  ˆj  5kˆ

3) 22/7, 33/7 4) 33/7, 22/7 3) 4)
14. Two identical thin uniform rods of length L 42 42
each are joined to form T shape as shown 19. An electron is moving with speed 2  10 5 m / s
in the figure. The distance of centre of mass
along the positive x-direction in the presence

 
from D is
A C B
of magnetic induction B  i  4 j  3k T . The


magnitude of the force experienced by the

electron in N  e  1.6  10  1 9 C   F  q  v  B 
  

1) 18  1013 2) 28  10  13
D 3) 1.6 1013 4) 73  10 13
1) 0 2) L/4 3) 3L/4 4)L
15. Figure shows a square plate of uniform 20. A particle of mass 80 units is moving with a
thickness and side length 2 m. One fourth uniform speed v  4 2 units in XY plane,
of the plate is removed as indicated. The along a line y  x  5 . The magnitude of the
distance of centre of mass of the remaining
portion from the centre of the original square angular momentum of the particle about the
plate is origin is
1) 1600units 2) 160 2 units
O
3) 152 2 units 4) 16 2 units
1) 1/3m 2) 1/2 m 3)1/6 m 4)1/8m

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JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

ROTATIONALVARIABLES, kg is suspended to the rod at its 10 cm division.

RELATION BETWEEN LINEAR AND The mass of rod is
ANGULAR VARIABLES 1) 0.4 kg 2) 0.8 kg 3) 1.2 kg 4) 1.6 kg
21. The linear and angular velocities of a body in 27. A metallic rod of mass 20 kg and of uniform
rotatory motion are 3 ms–1 and 6 rad/s thickness rests against a wall while the lower
respectively. If the linear acceleration is 6 end of rod is in contact with rough floor. The
m/s2 then its angular acceleration in rads–2 is rod makes an angle 60° with floor. If the
1) 6 2) 10 3) 12 4) 2 weight of rod produces a torque 150 N m about
ROTATIONAL KINEMATICS, TORQUE its lower end,the length of rod (g = 10 ms–2)
1) 1.5 m 2) 2 m 3) 3 m 4) 4 m
AND 28. A roller of mass 300 kg and of radius 50 cm
MECHANICAL EQUILIBRIUM lying on horizontal floor is resting against a
22. A stationary wheel starts rotating about its step of height 20 cm. The minimum horizontal
own axis at an angular acceleration force to be applied on the roller passing
5.5rad / s 2 . To acquire an angular velocity through its centre to turn the roller on to the
step is
420 revolutions per minute, the number of
1) 980N 2)1960N 3)2940N 4) 3920N
rotations made by the wheel is
1) 14 2) 21 3) 28 4) 35
ROTATIONAL INERTIA OF SOLID
23. A circular disc is rotating about its own axis BODIES
at constant angular acceleration. If its 29. A thin rod of mass M and length L is bent into a
angular velocity increases from 210 rpm to circular ring. The expression for moment of
inertia of ring about an axis passing through its
420 rpm during 21 rotations then the angular
diameter is
acceleration of disc is
ML2 ML2 ML2 ML2
1) 5.5rad / s 2 2) 11rad / s2 1) 2) 3) 4)
2 2 4 2 8 2 2
3) 16.5rad / s2 4) 22rad / s2 30. Two identical circular plates each of mass 0.1
24. A circular disc is rotating about its own axis kg and radius 10 cm are joined side by side as
at uniform angular velocity . The disc is shown in the figure. Their moment of inertia
subjected to uniform angular retardation by about an axis passing through their common
which its angular velocity is decreased to tangent is
/ 2 during 120 rotations. The number of
rotations further made by it before coming to
rest is
1) 120 2) 60 3) 40 4) 20S
25. Average torque on a projectile of mass m ,
initial speed u and angle of projection
between initial and final positions P and Q , 1) 1.25x10 3 kgm 2 2) 2.5x10 3 kgm 2
about the point of projection is :
3) 1.25x10 2 kgm 2 4) 2.5x10 2 kgm 2
mu 2 sin 2
1) 2) mu 2 cos 31. A wheel starting from rest is uniformly
2
mu 2 cos accelerated with  4 rad / s 2 for 10 seconds.
3) mu 2 sin 4) It is then allowed to rotate uniformly for the
2
26. A metal rod of uniform thickness and of next two seconds and is finally brought to rest
length 1 m is suspended at its 25 cm division in the next 10 seconds. Find the total angle
with help of a string. The rod remains rotated by the wheel.
horizontally straight when a block of mass 2 1) 200 rad 2) 400 rad 3) 300 rad 4) 480rad

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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III
32. Two spheres each of mass M and radius R/2 37. I is moment of inertia of a thin circular plate
are connected with a massless rod of length about its natural axis. The moment of inertia
2R as shown in the figure. The moment of of a circular ring whose mass is half of mass
inertia of the system about an axis passing of plate but radius is twice the radius of plate
through the centre of one of the spheres and about an axis passing through any tangent of
perpendicular to the rod is ring in its plane is
Y 1) 3 I 2) 4 I 3) 6 I 4) 1.5 I
Y
1

38. The moment of inertia of a uniform rod of

P M Q M length 2l and mass m about an axis xy passing
2R through its centre and inclined at an enable
R R  is
2 2

Y
1
Y
1

21 2 5 5
1) MR 2 2) MR 2 3) MR 2 4) MR 2
5 5 2 21
ml 2 ml 2
33. Moment of inertia of a thin circular plate of 1) sin 2  2) sin 2 
mass M, radius R about an axis passing 3 12
through its diameter is I . The moment of ml 2 ml 2
inertia of a circular ring of mass M, radius R 3) cos 
2
4) cos2 
6 2
about an axis perpendicular to its plane and
39. The ratio of radii of two solid spheres of same
passing through its centre is
material is 1 : 2. The ratio of moments of
I I inertia of smaller and larger spheres about
1) 2I 2) 3) 4I 4)
2 4 axes passing through their centres is
34. The mass of a thin circular plate is M and its 1) 1 : 4 2) 1 : 8 3) 1 : 16 4) 1: 32
radius is R. About an axis in the plane of 40. I is moment of inertia of a thin circular ring
plate at a perpendicular distance R/2 from about an axis perpendicular to the plane of
centre of plate, its moment of inertia is ring and passing through its centre. The same
ring is folded into 2 turns coil. The moment
MR 2 MR 2 3MR 2 3MR 2 of inertia of circular coil about an axis
1) 2) 3) 4)
4 2 4 2 perpendicular to the plane of coil and passing
35. In a rectangle ABCD (BC = 2 AB). The through its centre is
moment of inertia is maximum along axis I I
through 1) 2I 2) 4I 3) 4)
2 4
A E D
41. A metallic thin wire has uniform thickness.
From this wire, two circular loops of radii r,
F H 2r are made. If moment of inertia of 2 nd loop
B G C about its natural axis is n times moment of
inertia of 1st loop about its natural axis. The
1) BC 2) AB 3) HF 4) EG value of n is
36. M is mass and R is radius of a circular ring. 1) 2 2)4 3) 2 2 4) 8
The moment of inertia of same ring about an 42. The moment of inertia of a solid cylinder about
axis in the plane of ring at a perpendicular an axis parallel to its length and passing
2R through its centre is equal to its moment of
distance from centre of ring is inertia about an axis perpendicular to the
3
length of cylinder and passing through its
2MR 2 4 MR 2 3 MR 2 17 MR 2 centre. The ratio of radius of cylinder and its
1) 2) 3) 4) length is
3 9 8 18
1) 1: 2 2) 1 : 2 3) 1: 3 4) 1 : 3
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JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

43. The moment of inertia of a solid cylinder ANGULAR MOMENTUM &

about its natural axis is I. If its moment of CONSERVATION OFANGULAR
inertia about an axis r to natural axis of MOMENTUM
49. A thin uniform circular disc of mass M and
cylinder and passing through one end of
radius R is rotating in a horizontal plane about
cylinder is 19I/6 then the ratio of radius of an axis perpendicular to the plane at an
cylinder and its length is angular velocity . Another disc of mass M/
1) 1 : 2 2) 1 : 3 3) 1 : 4 4) 2 : 3 3 but same radius is placed gently on the
44. Two identical circular plates each of mass M first disc coaxially. The angular velocity of
and radius R are attached to each other with the system now is
their planes r to each other .The moment of 4 3 3
1) 2) 3) 4)
inertia of system about an axis passing through 3 4 8
their centres and the point of contact is 50. A turn table is rotating in horizontal plane
about its own axis at an angular velocity
MR 2 5MR 2 3 90rpm while a person is on the turn table at
1) 2) 3) MR 4) MR2
2

4 4 4 its edge. If he gently walks to the centre of

45. The radius of gyration of rod of length ‘L’and table by which moment of inertia of system
decreases by 25%, then the time period of
mass ‘M’ about an axis perpendicular to its
rotation of turn table is
length and passing through a point at a 1) 0.5sec 2) 1sec 3) 1.5sec 4) 2sec
distance L/3 from one of its ends is 51. A uniform cylindrical rod of mass m and length
7 L2 L 5 L is rotating with an angular velocity . The
1) L 2) 3) 4) L axis of rotation is perpendicular to its axis of
6 9 3 2 symmetry and passes through one of its edge
46. Two point size bodies of masses 2 kg, 3 kg are faces. If the room temperature increases by
fixed at two ends of a light rod of length 1 m. ‘t’ and the coefficient of linear expansion is
The moment of inertia of two bodies about an , the change in its angular velocity is
axis perpendicular to the length of rod and 3 t
1) 2 t 2) t 3) t 4)
passing through centre of mass of two bodies is 2 2
1)0.6 kgm2 2)0.8 kgm2 3)1 kgm2 4)1.2 kgm2 ROTATIONAL DYNAMICS
47. Three rings each of mass M and radius R 52. A constant torque of 1000Nm turns a wheel
are arranged as shown in the figure. The of M.I. 200kg m 2 about an axis through
moment of inertia of the system about AB is centre. The angular velocity after 3s is
A
1) 15 rad s 1 2) 22 rad s 1
3) 28 rad s 1 4) 60 rad s 1
53. If 484J of energy is spent in increasing the
speed of a wheel from 60rpm to 360rpm, the
M.I. of the wheel is
B
1) 1.6 kg m 2 2) 0.3kg m2
3 7
1) 3MR 2 2) MR 3) 5MR 2 4) MR
2 2

2 2 3) 0.7 kg m 2 4) 1.2 kg m2
48. Three identicalHitachi thin rods each of mass 54. The angular frequency of a fan of moment of
m and length L are joined together to form inertia 0.1kgm2 is increased from 30rpm to
an equilateral triangular frame. The moment 60rpm when a torque of 0.03Nm acts on it.
of inertia of frame about an axis perpendicular The number of revolutions made by the fan
to the plane of frame and passing through a while the angular frequency is increased from
corner is 30rpm to 60rpm
1) 7.855rev 2) 6.855rev
2mL2 3mL2 4mL2 3mL2 3) 5.855rev 4) 8.855rev
1) 2) 3) 4)
3 2 3 4
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III

55. A wheel rotating at an angular speed of angle with the horizontal. The frictional
20 rad s 1 is brought to rest by a constant force between the cylinder and the incline is
mg sin
torque in 4s. If the M.I. is 0.2 kg m 2 the work 1) mg sin 2)
3
done in first 2s is 2mg sin
1) 50J 2) 30J 3) 20J 4) 10J 3) mg cos 4)
3
ROLLING MOTION 61. A thin metal disc of radius 0.25m and mass
56. A sphere of mass m and radius r rolls on a 2kg starts from rest and rolls down an inclined
horizontal plane without slipping with a speed plane. If its rotational kinetic energy is 4J at
u. Now it rolls up vertically, then maximum the foot of the inclined plane, then its linear
height it would be attain will be velocity at the same point is
3u 2 5u 2 7u 2 u2 1) 1.2ms 1 2) 2.8ms 1 3) 20ms 1 4) 2ms 1
1) 2) 3) 4) 62. A small sphere of radius R rolls without
4g 2g 10 g 2g slipping inside a large hemispherical bowl of
57. A circular ring starts rolling down on an radius R . The sphere starts from rest at the
inclined plane from its top. Let v be velocity top point of the hemisphere. What fraction of
of its centre of mass on reaching the bottom the total energy is rotational when the small
sphere is at the bottom of the hemisphere
of inclined plane. If a block starts sliding down
on an identical inclined plane but smooth, from 7 2 5 7
1) 2) 3) 4)
its top, then the velocity of block on reaching 5 7 7 10
63. A metal disc of radius R and mass M freely
the bottom of inclined plane is rolls down from the top of an inclined plane
v v of height h without slipping. The speed of its
1) 2) 2v 3) 4) 2v centre of mass on reaching the bottom of the
2 2 inclined plane is
58. A thin rod of length L is vertically straight on
4 gh 3gh gh
horizontal floor. This rod falls freely to one 1) 2) 3) gh 4)
side without slipping of its bottom. The linear 3 4 2
velocity of centre of rod when its top end 64. A thin rod of length L is vertically straight on
horizontal floor. This rod falls freely to one
touches floor is
side without slipping at its bottom. The linear
3gL 3gL velocity of the top end of the rod with which it
1) 2gL 2) 3) 3gL 4) strikes the floor is
2 4
59. A wheel of radius ‘r’ rolls without slipping with 3gL 3gL
1) 2gL 2) 3) 3gL 4)
a speed v on a horizontal road. When it is at a 2 4
point A on the road, a small lump of mud LEVEL-II - (C.W) - KEY
separates from the wheel at its highest point 01) 4 02) 2 03) 1 04) 2 05) 3 06) 3
B and drops at point C on the ground. The 07) 1 08) 2 09) 4 10) 1 11) 4 12) 2
distance AC is 13) 1 14) 3 15) 3 16) 1 17) 2 18) 1
B 2V 19) 3 20) 1 21) 3 22) 3 23) 1 24) 3
25) 1 26) 3 27) 3 28) 4 29) 3 30) 2
31) 4 32) 1 33) 3 34) 2 35) 2 36) 4
37) 3 38) 1 39) 4 40) 4 41) 4 42) 3
A C 43) 1 44) 3 45) 3 46) 4 47) 4 48) 2
49) 3 50) 1 51) 1 52) 1 53) 3 54) 1
r r r 3r 55) 2 56) 3 57) 4 58) 4 59) 3 60) 2
1) v 2) 2v 3) 4v 4) v 61) 2 62) 2 63) 1 64) 3
g g g g
60. A solid cylinder of mass m rolls without
slipping down an inclined plane making an
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JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

LEVEL-II - (C.W) - HINTS

 
A B
 
18. n  A  B
ˆ
19. F  e v  B
  
1. CM coincides with point of intersection of
diagonals
2. Distance of C.M. from centre of big disc x =
r 2a 20. L  r  p  m  r  v  21. 
a v
r- radius of small disc
R2  r2 22. 2 N ; 2
2
R- radius of big disc a- distance between the
 1
2 2
centres of discs 23.  2
3. rcm  xcmiˆ  ycm ˆj 2
4. Coordinates of 1 kg, 2 kg, 3 kg are (0, 0), 
2 2

24.  is constant;  1 2

(2,0), (1, 3 ) respectively 2

   
2 2
mi xi mi yi  
Xcm= m ; Ycm = m ; rcm X cm Ycm2   0
2 2
2
; 2
1
3
i i

md 2 1 2 2
5. shift  25.   Force  perpendicular distance about the
M m
6. Coordinates of masses m1 = m, m2 = m, m3 = m point of projection
are (40, 0), (20, -60) & (x3, y3) respectively, 26. clockwise torque = anti clockwise torque
 xcm , ycm    0, 0 mg  25  2 g  15

 mx  1  2  3  ....  n  l
///////////////
2 2 2 2

 m 1  2  3  .....  n
7. xcm  10 50
25
md
8. shift  mg
M m
2g
9. Mass of ladder acts at 2.5m Mass of man acts at 2m L
27.  mg  cos 60
m1 y1  m2 y2 2
y
m1  m2
m2 d L
md
10. Xcm = m m 11. shift  2
1 2 M m
r 2a
x
L
12.
R2  r 2 2
mi xi mi yi 60
13. m = A ; Xcm = mi ;Ycm =
0

mi
L mg

 
m m( L)
14. X 2 28. Clock wise torque = anti clockwise torque
2m
F (50  20)  mg 502  302
cm

15. ‘m’ be the mass of each part 3m  cc2  m  cc1

 ad MR 2
L
(or) X  a - Area of removed plate A - 29. I  but 2 RL R
Aa 2 2
area of original plate d - distance between centers ML2
  I  2
16. Vcm parallel to acm 8
m2 d pd 5 1 2
17. xcm  m  m  p  q 30. I  2I 1 & I1  MR 2 31.  t t
4 2
0

 2  R 2 
32. I  2  5 M  2    M  2 R 
1 2
m1d qd
xcm  
2

m1  m2 p  q ;shift = x cm  xcm
1

   
1

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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III

MR 2 1  I
33. I  ; I '  MR 2 51. I    2 t
4 I

34. I  I c  Md  2 MR 2
M 
R
2
52. I ;  t
1
53. W  I 2
2
n
2
2  n12 
4  2
35. perpendicular distance is maximum when the axis 54. I , 2
 02  2 ,  2 N
of rotation passes through AB, hence M.I about I 1 2  k2 
AB is maximum. 55.  ;W 56. mgh  2
mv 1  2 
2R
t  R 
36. I  I c  Md 2 ; I C  MR and d  2 gh
2

3 57. for ring, v1   v for block v 2  2gh

k2
MR I  M R 2    2 R  6 I
  1 2
2 3 3 M
I
2
2  2 
37. ; 1 R
2
1 1
2
38. Take small element and use integration 3g L
58.  ; v  r and r 
2 L 2
39. I  MR 2 , M R 3 ; I R 5  k2 
5 59. R  2v  T  2v
2h
60. f  mg sin  2 2 
1 g k R 
40. Mass is same. l  2 rn  r where n is 1 1  MR2  mv 2
n 61. KErot  2 I    
2 2

number of turns 2 2  4
1 2
I1  r1   n2 
2 2
I

I  r 2 and I   r    n 
2 1
2 gh
62. 1 2  k2  63. v  64. mgh  I 

 2  1 mv 1  2  k2 2
1 2
2
2  R 
41. I  MR 2 and M  L  R  I  R3 and r
I 2  R2 
3

  LEVEL- II (H.W)
I1  R1 

42.
MR 2 ML2 MR 2
  CENTRE OF MASS
2 12 4 1. Four particles, each of mass 1kg, are placed
MR 2
19I ML2 MR 2 at the comers of square of side one meter in
43. I  ; I1    the XY plane. If the point of intersection of
2 6 3 4 the diagonals of the square is taken as the
MR 2
MR 2
3 origin,the co-ordinates of the center of mass
44. I1    MR 2 are
2 4 4
45. According to parallel axes theorem 1) (1,1)2) (-1, 1) 3) (1,-1) 4) (0,0)
2. Three identical particles each of mass 0.1kg
ML2 L 4 ML2 ML2 L
2

I M    K  are arranged at three corners of a square of

12 6 36 9 3 side 2m . The distance of the center of mass
 m1m2  2 from the fourth corner is
46. I   m  m  L 1) 2/3m 2) 4/3m 3) 1m 4) 8/3m
 1 2 
3. A bomb of mass ‘m’ at rest at the coordinate
MR 2 3 2 MR 2 7 MR 2 origin explodes into three equal pieces. At a
47. I   2  2 MR   2  3MR  2
2
2 certain instant one piece is on the x-axis at
mL2 x=60cm and another is at x=40cm, y=60cm.
48. I  2I 1  I 2 ; I1  ; I 2  I c  md 2 The position of the third piece is
3 1) x  100cm , y  60cm
mL2 3L
Ic  and d  2) x  60cm , y  60cm
12 2
3) x  60cm , y  60cm
MR 2  M
49. I1 1  I 2 2 ; I1  , I2   M   R2 4) x  60cm , y  60cm
2  3 
50. I1 1  I 2 2 ; I1  100, I 2  100  25  75

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JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

4. Masses 1kg, 1.5kg , 2kg and M kg are VECTOR PRODUCT OR CROSS PRODUCT
situated at (2,1,1), (1,2,1), (2,-2,1) and 10. The magnitude of two vectors which can be
represented in the form i  j   2 x  k is
(-1,4,3).If their center of mass is situated at
(1,1,3/2),the value of M is
1) 1kg 2) 2kg 3) 1.5kg 4) 3kg 18 .Then the unit vector that is
5. Six identical particles each of mass ‘m’ are perpendicular to these two vectors is
arranged at the corners of a regular hexagon
of side length ‘L’. If the masses of any two i  j i j i  j i  j
1) 2) 3) 4)

 
adjacent particles are doubled. The shift in the 2 8 2 8 2 2
centre of mass is
11. A proton of velocity 3i  2 j ms enters a
1

 
L 3L 3L 3L
1) 2) 3) 4)
8 8 16 4 field of magnetic induction 2i  3k T . The
6. Three particles each of mass ‘m’ are arranged accel eration produced in the proton in
at the corners of an equilateral triangle of side
‘L’. If one of masses is doubled. The shift in (specific charge of proton  0.96  108 Ckg 1 )
the centre of mass of the system

F  q v  B  
 
 
 
L L 3L L
1) 2) 4 3 3) 4) 2 3
1) 0.96  10 6i  9 j  4k
3 4 8

2) 0.96  10  6i  9 j  4k 

7. Four identical particles each of mass ‘m’ are
arranged at the corners of a square of side length 8
‘l’. If the masses of the particles at the end of a
3) 0.96  10  i  j  k 
side are doubled, the shift in the centre of mass
of the system.
8

4) 0.96  10  5i  9 j  4k 

l l l l
1) 2) 3) 4)
8

6 6 2 2 5 2
8. The co-ordinates of centre of mass of letter E ROTATIONALVARIABLES, RELATION
which is cut from a uniform metal sheet are BETWEEN LINEAR & ANGULAR
(take origin at bottom left corner and width of VARIABLES
letter 2cm every where) 12. A vehicle starts from rest and moves at uniform
6 cm acceleration such that its velocity increases
by 3ms 1 per every second. If diameter of
wheel of that vehicle is 60cm, the angular
8 cm 2 cm acceleration ofw heelis(in rad s-1)
2cm 1)5 2)10 3)15 4)20
13. Starting from rest the fly wheel of a motor
6 cm attains an angular velocity of 60 rad/sec in 5
1) (2cm, 4cm) 2) (2.4cm, 5cm) seconds. . The angular acceleration obtained is
3) (3cm, 5cm) 4) (3.3cm, 5cm) 1) 6 rad / s 2 2) 12 rad / s 2
MOTION OF CENTRE OF MASS,
LINEAR MOMENTUM OF 3) 300 rad / s 2 4) 150 rad / s 2
CENTRE OF MASS ROTATIONAL KINEMATICS, TORQUE,
9. Two particles of equal mass have velocities MECHANICAL EQUILIBRIUM
14. A ceiling fan is rotating about its own axis
V 1  8i and V 2  8 j . First particle has an
 
with uniform angular velocity . The electric

 
current is switched off then due to constant
acceleration a1  5i  5 j ms while the
 2
opposing torque its angular velocity is
acceleration of the other particle is zero.The 2
centre of mass of the two particles moves is a reduced to as it completes 30 rotations.
3
path of The number of rotations further it makes
1) straight line 2) parabola before coming to rest is
3) circle 4) ellipse 1) 18 2) 12 3) 9 4) 24

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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III
15. A wheel has a speed of 1200 revolutions per rings and passing through the point of contact
minute and is made to slow down at a rate of is
4 rad/s2. The number of revolutions it makes
before coming to rest is
1) 143 2) 272 3) 314 4) 722
16. A particle of mass 1kg is projected with an
initial velocity 10ms 1 at an angle of G
projection 450 with the horizontal. The 3 3 5 5
average torque acting on the projectile 1) MR 2 2) MR 2 3) MR 2 4) MR 2
between the time at which it is projected and 2 4 2 4
the time at which it strikes the ground about 22. The moment of inertia of a thin square plate
the point of projection in newton meter is of mass 1.2 kg is 0.2 kgm 2 when it is made to
1) 25 2) 50 3) 75 4) 100 rotate about an axis perpendicular to plane
17. A uniform metre scale of mass 1kg is placed of plate and passing through a corner of plate.
on table such that a part of the scale is beyond The side length of plate is
the edge. If a body of mass 0.25kg is hung at 1) 0.2m 2) 0.4m 3) 0.5m 4) 0.8m
the end of the scale then the minimum length 23. Three point masses m1,m2,m3 are placed at
of scale that should lie on the table so that it three corners of an equilateral triangle of side
does not tilt is a. The moment of inertia of the system about
1) 30cm 2) 80cm 3) 70cm 4) 60cm an axis coinciding with the altitude of triangle
18. A heavy wheel of radius 20cm and weight passing through m 1 is
10kg is to be dragged over a step of height
10cm, by a horizontal force F applied at the
centre of the wheel. The minimum value of F m1
is
1)20kgwt 2)1kgwt
3) 10 3 kgwt 4) 10 2 kgwt
m2 a a m3
ROTATIONAL INERTIA OF SOLID BODIES
19. Two discs one of density 7.2 g/cm3 and the 2 2
other of density 8.9 g/cm3 are of same mass  m2  m3  a 2
and thickness. Their moments of inertia are 1)  m1  m2  m3  a 2
2)
6
in the ratio
 m2  m3  a
2 2
 m2  m3  a 2
8.9 7.2 3) 4)
1) 2) 2 4
7.2 8.9 24. From a uniform wire two circular loops are
3)  8.9  7.2  :1 4) 1: 8.9  7.2 made (i) P of radius r and (ii) Q of radius nr. If
20. The mass of a circular ring is M and its radius the moment of inertia of Q about an axis
is R. Its moment of inertia about an axis in passing through its centre and perpendicular
the plane of ring at a perpendicular distance to its plane is 8 times that of P about a similar
R/2 from centre of ring is axis. The value of n is (diameter of the wire
is very much smaller than r or nr)
MR 2 MR 2 3MR 2 3MR 2 1) 8 2) 6 3) 4 4) 2
1) 2) 3) 4)
4 2 2 4 25. The moment of inertia of a uniform thin rod of
21. Two circular rings each of mass M and radius length L and mass M about an axis passing
R are attached to each other at their rims through a point at a distance of L/3 from one of
and their planes perpendicular to each other its ends and perpendicular to the rod is
as shown in the figure. The moment of inertia
of the system about a diameter of one of the 7 ML2 ML2 ML2 ML2
1) 2) 3) 4)
48 6 9 3
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26. Two small spheres of mass 5kg and 15kg are 33. The moment of inertia of thin rod of linear density
joined by a rod of length 0.5m and of and length l about an axis passing through
negligible mass. The M.I. of the system about one end and perpendicular to its length is
an axis passing through centre of rod and
normal to it is l2 l2 l3 l3
1) 2) 3) 4)
1) 10 kgm 2 2) 1.25 kgm2 3) 20 kgm 2 4) 8 kgm 2 12 3 12 3
27. Ratio of densities of materials of two 34. Moment of inertia of a bar magnet of mass M ,
circular discs of same mass and thickness is length L and breadth B is I. Then moment of
5:6.The ratio of their M.I. about natural axes inertia of another bar magnet with all these values
is doubled would be
1) 5:6 2) 6:5 3) 25:36 4) 1:1 1) 8I 2) 4I 3) 2I 4) I
28. M.I. of a uniform horizontal solid cylinder of ANGULAR MOMENTUM & CONSERVATION OF
mass M about an axis passing through its edge ANGULARMOMENTUM
and perpendicular to the axis of cylinder when 35. A circular disc is rotating without friction
its length is 6 times of its radius R is about its natural axis with an angular velocity
39 39 49 49 . Another circular disc of same material and
1) MR 2 2) MR 2 3) MR 2 4) MR 2 thickness but half the radius is gently placed
4 2 4 2
29. A circular disc of radius R and thickness R/6 over it coaxially. The angular velocity of
has moment of inertia I about an axis passing composite disc will be
through its centre and perpendicular to its 4 8 7 16
plane. It is melted and recast into a solid 1) 2) 3) 4)
3 9 8 17
sphere. The M.I. of the sphere about its
36. A ballet dancer is rotating about his own
diameter as axis of rotation is
vertical axis on smooth horizontal floor with
1) I 2) 2I/3 3) I/5 4) I/10
30. The moment of inertia of ring about an axis a time period 0.5sec. The dancer folds himself
passing through its diameter is I. Then close to his axis of rotation due to which his
moment of inertia of that ring about an axis radius of gyration decreases by 20%, then
passing through its centre and perpendicular his time period is
to its plane is 1) 0.1sec 2) 0.25sec 3) 0.32sec 4) 0.4sec
1) 2I 2) I 3) I/2 4) I/4 37. A particle of mass 1kg is moving along the line
31. A thin rod of mass 6m and length 6L is bent y  x  2 with speed 2m/sec. The magnitude
into regular hexagon. The M.I. of the of angular momentum of the particle about the
hexagon about a normal axis to its plane and origin is
through centre of system is 1) 4 kg  m 2 / sec 2) 2 2 kg  m 2 / sec
1) mL2 2) 3mL2 3) 5mL2 4) 11mL2
3) 4 2 kg  m 2 / sec 4) 2 kg  m 2 / sec
32. If I1 is moment of inertia of a thin rod about
38. A uniform metal rod of length L and mass M
an axis perpendicular to its length and passing
is rotating about an axis passing through one
through its centre and I2 is its moment of
of the ends perpendicular to the rod with
inertia when it is bent into a shape of a ring
angular speed . If the temperature
then (Axis passing through its centre and
increases by toC then the change in its
perpendicular to its plane)
angular velocity is proportional to which of
I1 I1 the following?(Coefficient of linear expansion
1) I 2  2) I 2 
4 2 2
of rod= ).
I2 2
I2 3 1
3)  4) I  2 1) 2) 3) 4)
2

I1 3 1

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ROTATIONAL DYNAMICS 10 20 30 40
39. A fly wheel of M.I. 6  102 kgm2 is rotating 1) 7 2) 3) 4)
7 7 7
with an angular velocity of 20 rad s 1 . The 45. A thin metal rod of length 0.5m is vertically
straight on horizontal floor. This rod is falling
torque required to bring it to rest in 4s is
freely to a side without slipping. The angular
1)1.6Nm 2) 0.6Nm 3) 0.8Nm 4)0.3Nm
40. When 200J of work is done on a fly wheel its velocity of rod when its top end touches the
frequency of rotation increases from 4Hz to floor is (nearly)
9Hz. The M.I. of the wheel about the axis of 1) 7rad s 1 2) 4.2rad s 1
rotation is (nearly)
1) 0.12kg m 2 2) 0.2 kg m 2 3) 3.5rad s 1 4) 2.1rad s 1
3) 0.22 kg m 2 4) 0.3kg m2 46. What should be the minimum coefficient of
41. The moment of inertia of a wheel of radius static friction between the plane and the
cylinder, for the cylinder not to slip on an
20cm is 40 kgm 2 if a tangential force of 80N inclined plane
applied on the wheel, its rotational K.E. after
1 1 2 2
4s is 1) tan 2) sin 3) tan 4) sin
1) 16.2J 2) 51.2J 3) 25.6J 4) 24.8J 3 3 3 3
ROLLING MOTION 47. A thin metal disc of radius 25cm and mass
2kg starts from rest and rolls down on an
42. An initial momentum is imparted to a
inclined plane. If its rotational kinetic energy
homogeneous cylinder, as a result of which it
is 8J at the foot of this inclined plane, then
begins to roll without slipping up an inclined
linear velocity of centre of mass of disc is
plane at a speed of v0  4m / sec The plane 1) 2 m/s 2) 4m/s 3) 6m/s 4) 8m/s
makes an angle  300 with the horizontal. LEVEL-II(H.W) - KEY
What height h will the cylinder rise to? 01) 4 02) 2 03) 1 04) 3 5) 2 06) 2
 g  10m / s 
2 07) 1
13) 2
08) 2
14) 4
09) 1
15) 3
10) 1
16) 2
11) 2
17) 4
12) 2
18) 3
1) 0.8m 2) 1.2m 3) 1.0m 4) 1.6m 19) 1 20) 4 21) 3 22) 3 23) 4 24) 4
43. A solid cylinder starts rolling down on an 25) 3 26) 2 27) 2 28) 3 29) 3 30) 1
inclined plane from its top and V is velocity of
31) 3 32) 4 33) 4 34) 1 35) 4 36) 3
its centre of mass on reaching the bottom of
37) 2 38) 2 39) 4 40) 1 41) 2 42) 2
inclined plane. If a block starts sliding down
43) 3 44) 4 45) 1 46) 1 47) 2
on an identical inclined plane but smooth, from
its top, then the velocity of block on reaching LEVEL-II(H.W) - HINTS
the bottom of inclined plane is m1 x1  m2 x2 m y  m2 y2
1. xcm  ; ycm  1 1
v 3 2 m1  m2 m1  m2
1) 2) 2v 3) v 4) v
2 2 3 2. rcm  xcm
2
 ycm
2

44. A wheel of radius 0.2m rolls without slip ping m1 x1 m2 x2 m3 x3

with a speed 10m/s on a horizontal road. When 3.  xcm , ycm    0, 0 ; 0 m1 m2 m3
it is at a point A on the road, a small lump of
m1 y1 m2 y2 m3 y3
mud separates from the wheel at its highest 0
point B and drops at point C on the ground. m1 m2 m3
The distance AC is m1 x1 m2 x2 m3 x3 m4 x4
4. xcm
B m1 m2 m3 m4
md md
5. shift  6. shift 
M m M m
A C
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md  L2 R2 
7. shift  28. I  M  3  4 
M m  
m1 x1 m2 x2 m3 x3 m4 x4 1
8. xcm
m1 m2 m3 m4 29. I  R5 ;volume of disc = volume of sphere
12
 radius of sphere  R1  
m1 y1 m2 y2 m3 y3 m4 y4
ycm R
m1 m2 m3 m4
2
A B
 
9. vcm is parallel to acm 10. nˆ   
A B MR 2
30. I  and I 1  MR 2  2 I
   
   e   2
11. F  e v  B ; a  v B

 3 
2
m ml 2
31. I rod   m  l  ; I system  6 I rod
a a 12
12.  13.   2 
r r
14. is constant ML2 L
32. I1  , I 2  MR 2 , R 
  2 / 3
12 2
 2 and 1 
2 2
2 2
 M ML L
 2 / 3  0
2 3
 ; I 
1
2
2
33.
2
L 3 3
 L  b2 
2 N  12
35. I1 1   I1  I 2 
M 2
2

15.  ;  2 34. I
t 2 12 2

 m  u 2 cos2 
dL 2 2
16.  36. I1 1  I 2 2 ; mK1  T  mK 2  T
2 2
dt
 50  x 1  x  0.25 x  40cm 37. L  mvr  1  2   2 cos 450   2 2
1 2
17.
length of the table = 100-40=60cm Y y=x+2
18. clockwise torque = anticlockwise torque
mg  20   10  F  20 10 
2 2
r
X
MR I R2 2
8.9
19. I   1   
1 2 m
2 I2 R 7.2 1 ML2
38. I 
2
2 1
-----(1) ; I   I  L2 -----(2)
MR 2 R 3
20. I  I C  md , IC  ; d 1
2

2 2 from (1) and (2)  2

MR 2 5MR 2 L
40. W  I 4  n2  n1 
21. I  2MR 2   1
2 2 39. I
2 2 2

MR 2  L 
2 2
22. I  M   0.2 1 2
6  2 41.  0  t ; KE  I
2
23. I   mr ; r1  0, r2  r3 
a
g sin h
2

2 a , v  2aS , sin 
24. Moment of inertia P  Mr 2 42. 1 2
K 2
s
Moment of inertia Q   nM   nr 2 
R
2 gh
given I Q  8 I P  n  2 v1  v
43. For cylinder, k2
1 2
ML2 L R
2

25. I M   26. I  mr 2
12 6 For block v 2  2gh
1
27. I  tR 4 2h
2 44. R  2v  T  2v and h  2R
g

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  5. A circular disc of radius R is removed from a

 k 2 / R2  bigger circular disc of radius 2R such that the
l 1
 I  tan  2  circumference of the discs coincide . The
45. Mg 46.  k  1 
2

2 2 centre of mass of the new disc is R from the

 R2  centre of the bigger disc. The value of is
1) 1/3 2) 1/2 3) 1/6 4) 1/4
1 1  mR2  6. Four identical planks each of lengths ‘L’
47. KErot  2 I   
2 2

2 2  are arranged one above the other over a

table as shown. Each projects a distance ‘a’
LEVEL- III beyond the edge of the one that is below it.
What is the maximum possible value of ‘a’
MOTION OF CENTRE OF MASS& for the system to be in equilibrium without
LINEAR MOMENTUM tripping forward?
1. Seven homogeneous bricks each of length a
L,mass M are arranged as shown. Projection a
a
L 1) L/5 2) L/ 4 3) L/3 4) L
x then x co-ordinate of C.M is
10 7. Two masses ‘m1 ’ and ‘m2’ (m1>m 2) are
-x-
-x- connected to the ends of a light inextensible
-x-
-x- string which passes over the surface of a
-x-
-x- smooth fixed pulley. If the system is
released from rest, the acceleration of the
0
22 32 42 12 centre of mass of the system will be (g =
1) L 2) L 3) L 4) L acceleration due to gravity)
35 35 35 35
2. The centre of mass of a non uniform rod of g (m1 - m 2 ) g (m1 - m 2 ) 2
length L whose mass per unit length 1) 2)
( m1  m 2 ) (m1  m2 ) 2
Kx 2
 ,Where k is a constant and x is the g (m1  m 2 ) g (m1  m 2 )
L 3) 4)
distance from one end is : (m1  m2 ) (m1  m2 )
3L L K 3K 8. Two bodies of masses m 1 and m 2 are moving
1) 2) 3) 4) with velocity v1 and v2 respectively in the
4 8 L L same direction. The total momentum of the
3. A rope of length 30 cm is on a horizontal table system in the frame of reference attached to
with maximum length hanging from edge A of
the centre of mass is (v is relative velocity
the table. The coefficient of friction between
the rope and table is 0.5. The distance of centre between the masses)
of mass of the rope from A is m1 m 2 v 2 m1 m 2 v
1) 2)
5 15 5 17 5 19 7 17 m1  m2 m1  m2
1) cm 2) cm 3) cm 4) cm
3 3 3 3 4 m1 m 2 v
4. As shown in figure from a uniform 3) zero 4)
rectangular sheet a triangular sheet is m1  m2
removed from one edge. The shift of centre 9. A shell in flight explodes into n equal
of mass is fragments k of the fragments reach the
60 cm ground earlier than the other fragments.
The acceleration of their centre of mass
O 30 cm subsequently will be
1) g 2)(n–k)g
(n  k ) g (n  k )
3) 4) g
1) 4.2 cm 2) – 4.2cm 3) 6.67 cm 4)– 6.67 cm k n

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10. A body of mass ‘m’ is dropped and another 14. Two particles A and B of mass 1 Kg and 2 Kg
body of mass M is projected vertically up with respectively are projected in the directions
speed ‘u’ simultaneously from the top of a shown in figure with speeds uA  200 m / s and
tower of height H . If the body reaches the
highest point before the dropped body reaches uB  50 m / s , Initially they were 90 m apart.
the ground, then maximum height raised by Find the maximum height attained by the
the centre of mass of the system from ground centre of mass of the particles. ( g  10 m / s 2 )
is B
u2 u2 uB
1) H  2) uA
2g 2g
A
1  Mu  1  mu 
2 2

3) H    4) H    1) 115.55m 2) 145.55 m
2g  m  M  2g  m  M  3) 4.55 m 4) 34.55 m
11. Two blocks of equal mass are tied with a light VECTOR PRODUCT OR CROSS
string, which passes over a massless pulley
as shown in figure. The magnitude of
PRODUCT
acceleration of centre of mass of both the 15. At a given instant of time the position vector
of a particle moving in a circle with a velocity
blocks is ( neglect friction everywhere )
3iˆ  4 ˆj  5kˆ is iˆ  9 ˆj  8kˆ . Its angular velocity
at that time is:

1)

13iˆ  29 ˆj  31kˆ  2)
 
13iˆ  29 ˆj  31kˆ
146
 
146
60
0
30
0

//////////////////////////////////////////////////////////////////// 3)
13iˆ  29 ˆj  31kˆ
4)
 
13iˆ  29 ˆj  31kˆ
146 146
 3 1  g  3 1
1)  4 2 g  2) ( 3  1)g 3) 4)  2  g ROTATIONAL VARIABLES,
  2   RELATION BETWEEN LINEAR
12. A rope thrown over a pulley has a ladder with & ANGULAR VARIABLES
a man of mass m on one of its ends and a 16. Two points P and Q, diametrically opposite on
counter balancing mass M on its other end. a disc of radius R have linear velocities v and
The man climbs with a velocity vr relative to 2v as shown in figure. Find the angular speed
ladder . Ignoring the masses of the pulley and of the disc.
the rope as well as the friction on the pulley v P
axis, the velocity of the centre of mass of this
system is :
2v
m m M 2M Q
1) vr 2) vr 3) vr 4) vr
M 2M m m v 2v v v
13. Two particles of masses 2 kg and 3 kg are 1) 2) 3) 4)
R R 2R 4R
projected horizontally in opposite directions 17. Point A of rod AB (l =2m) is moved upwards
from the top of a tower of height 39.2 m with against a wall with velocity v=2 m/s. Find
velocities 5 m/s and 10 m/s respectively. The angular speed of the rod at an instant when
horizontal range of the centre of mass of two
particles is  600 .
v
1) 8 2 m in the direction of 2 kg A
2) 8 2 m in the direction of 3 kg
3) 8 m in the direction of 2 kg

4) 8 m in the direction of 3 kg
B
1) 4rad/s 2)1.155rad/s 3) 2rad/s 4) 2.50rad/s
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18. A uniform circular disc of radius R lies in the 22. A wheel having radius 10 cm is coupled by a
XY plane with its centre coinciding with the belt to another wheel of radius 30cm. 1st
origin of the coordinate system. Its moment wheel increases its angular speed from rest
of inertia about an axis, lying in the XY plane, at a uniform rate of 1.57 rad s–2. The time for
parallel to the X-axis and passing through a 2nd wheel to reach a rotational speed of 100
point on the Y-axis at a distance y =2R is I1 . rev/min is...(assume that the belt does not slip)
Its moment of inertia about an axis lying in a 1) 20 sec 2) 10 sec 3) 1.5 sec 4) 15 sec
plane perpendicular to XY plane passing 23. An equilateral prism of mass m rests on a
through a point on the x-axis at a distance x rough horizontal surface with coefficient of
= d is I 2 . If I1=I2 the value ofd is friction µ . A horizontal force F is applied on
the prism as shown. If the coefficient of
19 17 15 13
1) R 2) R 3) R 4) R friction is sufficiently high so that the prism
2 2 2 2 does not slide before toppling, the minimum
ROTATIONAL KINEMATICS, TORQUE
force required to topple the prism is
MECHANICAL EQUILIBRIUM
19. A wheel rotating with uniform angular F
acceleration covers 50 revolutions in the first
five seconds after the start. Find the angular
acceleration and the angular velocity at the a a
end of five seconds.
1) 4 rad / s 2 ,80 rad / s
2) 8 rad / s 2 ,40 rad / s a
mg mg µmg µmg
3) 6 rad / s 2 , 40 rad / s 1) 2) 3) 4)
3 4 3 4
4) 6 rad / s 2 ,80 rad / s 24. The mass of a metallic beam of uniform
20. A square is made by joining four rods each of thickness and of length 6 m is 60 kg. The
mass M and length L. Its moment of inertia beam is horizontally and symmetrically lies
about an axis PQ, in its plane and passing on two vertical pillars which are separated
through one of its corner is by a distance 3 m. A person of mass 75 kg is
P walking on this beam. The closest distance
to which the person can approach one end of
the beam so that the beam does not tilt down
45 is (neglect thickness of pillars)
0

1) 30 cm 2) 20 cm 3) 15 cm 4) 10cm
25. Two persons P and Q of same height are
L
carrying a uniform beam of length 3 m. If Q
is at one end, the distance of P from the other
end so that P, Q receive loads in the ratio 5 :
Q 3 is
1) 0.5 m 2) 0.6 m 3) 0.75 m 4) 1 m
4 8 10 26. A uniform meter scale of mass 1 kg is placed
1) 6ML2 2) ML2 3) ML2 4) ML2 +
3 3 3 on table such that a part of the scale is beyond
21. A shaft is turning at 65rad /sec at time zero. the edge. If a body of mass 0.25 kg is hung
Thereafter, angular acceleration is given by at the end of the scale then the minimum
 ( 10  5t ) rad / s 2 where t is the elapsed length of scale that should lie on the table so
time. Find its angular speed at t =3sec. that it does not tilt is
1) 25 rad/sec 2) 12.5rad/sec 1) 90 cm 2) 80 cm 3) 70 cm 4) 60 cm
3) 17 rad/sec 4) 22 rad /sec

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27. A metallic cube of side length 1.5 m and of angle 15 with the horizontal is
mass 3.2 metric ton is on horizontal rough
floor. The minimum horizontal force that
should be applied on the cube at a height 1.2 15
0

m from that floor to turn the cube about its

lower edge is
1) 1.96  103 N 2) 4.9  103 N
3) 1.96  10 4 N 4) 4.9  104 N
28. A cubical block of mass m and side L rests on
a rough horizontal surface with coefficient of ML2 11ML2 7 ML2 10ML2
1) 2) 3) 4)
friction  . A horizontal force F is applied on 12 24 12 24
the block as shown. If the coefficient of 31. A thin rod of length L and mass M is bent at
friction is sufficiently high so that the block the middle point O at an angle of 600 . The
does not slide before toppling, the minimum moment of inertia of the rod about an
force required to topple the block is axis passing through O and perpendicular to
[JEE 2000] the plane of the rod will be
F
O

60
0
L/2 L/2
///////////////////
1) mg/4 2) infinitesimal
3) mg/2 4) mg (1 – u)
29. The center of an equilateral triangle is O. ML2 ML2 ML2 ML2
1) 2) 3) 4)
Three forces F1 , F2 and F3 are applied along 6 12 24 3
AB, BC and AC respectively. The magnitude 32. Four identical solid spheres each of mass M
and radius R are fixed at four corners of a light
of F3 so that the total torque about O should
square frame of side length 4R such that
be zero is centres of spheres coincide with corners of
square. The moment of inertia of 4 spheres
about an axis perpendicular to the plane of
frame and passing through its centre is
21MR 2 42MR 2 84MR 2 168MR 2
1) 2) 3) 4)
5 5 5 5
33. In the above problem moment of inertia of 4
spheres about an axis passing through any
side of square is
21MR 2 42MR 2 84MR 2 168MR 2
1) 2) 3) 4)
1)  F1  F2  2)  F1  F2  5 5 5 5
34. Thickness of a wooden circular plate is same
F1  F2
3) 4) 2  F1  F2  as the thickness of a metal circular plate but
2 density of metal plate is 8 times density of
ROTATIONAL INERTIA OF SOLID BODIES wooden plate. If moment of inertia of wooden
30. A square plate of mass M and edge L is shown plate is twice the moment of inertia of metal
in figure. The moment of inertia of the plate plate about their natural axes, then the ratio of
about the axis in the plane of plate and radii of wooden plate to metal plate is
passing through one of its vertex making an 1) 1 : 2 2) 1 : 4 3) 4 : 1 4) 2 :1
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III
35. A uniform circular disc of radius 'R' lies in Their moments of inertia about the axis
the X-Y plane with the centre coinciding with passing through the centre and perpendicular
the origin. The moment of inertia about an to their planes are in the ratio 1:m. The
axis passing through a point on the X-axis at relation between m and n is
a distance x = 2R and perpendicular to the 1) m=n 2) m = n2 3) m = n3 4) m = n4
X-Y plane is equal to its moment of inertia 40. The moment of inertia of a hollow sphere of
about an axis passing through a point on the mass M having internal and external radii R
Y-axis at a distance y = d and parallel to the and 2R about an axis passing through its
X-axis in the X-Y plane. The value of 'd' is
centre and perpendicular to its plane is
4R R R R 62
1) 2) 17   3) 15   4) 13   3 13 31
2 2 2 1) MR 2) MR 2 3) MR 2 4) MR 2
2
3
2 32 35 35
36. Two rings of the same radius R and mass M
41. Find moment of inertia of half disc of radius
are placed such that their centres coincide
and their planes are perpendicular to each R 2 and mass M about its centre. A smaller
other. The moment of inertia of the system half disc of radius R1 is cut from this disc.
about an axis passing through the common
centre and perpendicular to the plane of one
of the rings is
MR 2 3MR 2
1) 2) MR 3)
2 4) 2MR 2
2 2
37. In the above problem, the moment of inertia
of the system about an axis passing through
the diameters of both rings is 1)
M
4

R1  R 2
2 2
 2)
8

M 2
R 1  R 22 

 R 1  R 22   R 1  R 22 
MR 2 MR 2 3MR 2 M 2 M 2
1) 2) 3) 4) MR 2 3) 4)
4 2 2 16 32
38. Four thin metal rods, each of mass M and
length L, are welded to form a square ABCD
ANGULAR MOMENTUM &
as shown in figure. The moment of inertia of CONSERVATION OF ANGULAR
the composite structure about a line which MOMENTUM
bisects rods AB and CD is 42. A uniform smooth rod (mass m and length l)
placed on a smooth horizontal floor is hit by a
particle (mass m) moving on the floor, at a
distance l / 4 from one end elastically
(e =1).The distance travelled by the centre
of the rod after the collision when it has
completed three revolutions will be
1) 2 l 2) cannot be determined
3) l 4) none of these
43. A bullet of mass m is fired upward in a
direction of angle of projection 60° with an
ML2 ML2 ML2 2ML2 initial velocity u. The angular momentum of
1) 2) 3) 4) this bullet when it is crossing highest point
6 3 2 3
with respect to point of projection is
39. Two circular loops A and B are made of the
same wire and their radii are in the ratio 1:n. 2mu 3 3mu 3 2mu 3 3mu 3
1) 2) 3) 4)
5g 8g 9g 16 g
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JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

44. A particle of mass 5g is moving with a speed P

of 3 2cms 1 in X-Y plane along the line
(3/2) R
y  x  4 . The magnitude of its angular
momentum about the origin in gcm 2 s 1 is v0
C
30
1) zero 2) 60 3) 30 4) 1) increase continuously as the disc moves away
2
2) decrease continuously as the disc moves away
45. A ballot dancer is rotating about his own 3) is always equal to 2MRv0
vertical axis on smooth horizontal floor with 4) is always equal to MRv0
a time period 0.5 sec. The dancer folds 48. A disc of mass m and radius R moves in the
himself close to his axis of rotation due to X-Y plane as shown in figure. The angular
which his radius of gyration decreases by momentum of the disc about the origin O at
20%, then his new time period is the instant shown is
1) 0.1 sec 2)0.25 sec 3) 0.32 sec 4) 0.4 sec y
46. A smooth uniform rod of length L and mass v = R
M has two identical beads of negligible size, 
each of mass m, which can slide freely along 3R
the rod. Initially the two beads are at the
centre of the rod and the system is rotating x
O
with angular velocity 0 about its axis 4R
perpendicular to the rod and passing through 5 7
1)  mR k 2) mR k
2 2

its mid point (see figure). There are no 2 3

external forces. When the beads reach the 9 5
ends of the rod, the angular velocity of the 3)  mR k 4) mR k
2 2

2 2
system is [JEE - 1988] 49. A uniform sphere of mass m, radius r and
moment of inertia I about its centre of mass
axis moves along the x-axis is shown in figure.
Bead Bead Its centre of mass moves with velocity v0,
and it rotates about its centre of mass with
L L angular velocity 0 . Let L   I 0  mv0r k .
2 2 The angular momentum of the body about the
the origin O is
0 y
0

M 0 M 0 v0
1) 2) r
M  3m M  6m
M 6m  0 O
3) 4) 0
M 1) L, only if v0  0 r
47. A uniform circular disc of mass M and radius
2) greater than L, v0  0 r
R rolls without slipping on a horizontal
surface. If the velocity of its centre is v 0, then 3) less than L, if v0  0 r
the total angular momentum of the disc about 4) L, for all value of v0 and
a fixed point P at a height 3R/2 above the
0

centre C.
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III
50. In the figure shown, a ring A is initially rolling g 3g g 2g
without sliding with a velocity v on the 1) 2) 3) 4)
2 2 3 3
horizontal surface of the body of the body B
54. The arrangement shown in figure consists of
(of same mass as A). All surfaces are smooth.
two identical uniform solid cylinders each of
B has no initial velocity. What will be the
mass 5kg on which two light threads are
maximum height reached by A on B?
wound symmetrically. Find the tensions of
each thread in the process of motion. The
friction in the axle of the upper cylinder is
A v assumed to be absent.
smooth
B

3v2 v2 v2 v2
1) 2) 3) 4)
4g 4g 2g 3g
ROTATIONAL DYNAMICS
51. Calculate the linear acceleration of the blocks
in the given figure . Mass of block A = 10kg,
mass of block B = 8kg, mass of disc shaped 1) 4.9N 2) 9.8N 3) 8.8N 4) 5.8N
pulley = 2kg (take g  10m / s 2 ) 55. The top in the figure has a moment of inertia
equal to 4.0  104 kgm 2 and is initially at rest.
It is free to rotate about the stationary axis
AA1. A string wrapped around a peg along the
axis of the top is pulled in such a manner as
B to maintain a constant tension of 5.57N . If
A the string does not slip while it is unwound
from the peg. what is the angular speed of
20 19 29 20
1) m / s 2 2) m / s 2 3) m / s 2 4) m / s2 the top after 80.0cm of string has been pulled
19 20 20 29 off the peg.
52. A block of mass m is attached at the end of
an inextensible string which is wound over a A'
rough pulley of mass M and radius R as
F
shown in figure. Assume that string does not
slide over the pulley. Find the acceleration of
the block when released.
A

R
1)130rad/s 2)142rad/s 3)149rad/s 4)120rad/s
M 56. A solid cylinder of mass m=4kg and radius
m R=10cm has two ropes wrapped around it, one
near each end. The cylinder is held
2mg 2mg mg mg
1) 2) 3) 4) horizontally by fixing the two free ends of the
mM 2m  M 2m  M mM cords to the hooks on the ceiling such that
53. A uniform rod of length L and mass M is both the cords are exactly vertical. The
pivoted freely at one end (at bottom level ) cylinder is released to fall under gravity. Find
and placed in vertical position. What is the the tension along the strings.
tangential linear acceleration of the free end 1) 6.53N 2) 5.23N 3) 3.23N 4) 4.43N
when the rod is horizontal?
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JEE-ADV PHYSICS-VOL - III SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS

ROLLING MOTION 61. A tangential force F acts at the top of a thin

57. Assume the earth’s orbit around the sun as spherical shell of mass m and radius R. Find
circular and the distance between their the acceleration of the shell if it rolls without
centres as ‘D’ Mass of the earth is ‘M’ and slipping.
its radius is ‘R’ If earth has an angular velocity F
R
‘ 0 ’ with respect to its centre and ‘ ’ with
respect to the centre of the sun, the total O
kinetic energy of the earth is:
MR2    5 D  
2 2

6F 6m 3m 5m
2

1)
0
1       
5    2  R 0   1) 2) 3) 4)
5m 5F 5F 6F
0

MR 2 2  5  D 2  62. A uniform circular ring of radius R is first

2)
0
1     rotated about its horizontal axis with an
5  2  R 0  
angular velocity 0 and then carefully placed
5  5  D 2  on a rough horizontal surface as shown. The
3) 2 MR 0 1    
2 2

 2  R 0   coefficient of friction between the surface and

the rings is . Time after which its angular
2    5 D   speed is reduced to 0.5 is
2 2

4) 5 MR
2 2
1        0
   2  R 0   0
0
0

58.A solid cylinder of mass 10kg is rolling

perfectly on a plane of inclination 300 . The
force of friction between the cylinder and the
surface of the inclined plane is R g 2 0R 0R
1) 49N 2) 24.5N 3) 49/3N 4) 12.25N 1) 2) 3) 4)
0 0

2g 2 R g 2 g
59 The velocities are in ground frame and the
cylinder is performing pure rolling on the 63. A uniform circular disc of radius R rolls without
plank, velocity of point ' A ' would be slipping with its center of mass moving along
positive x=axis with a speed v. The velocity
A
of point P at the instant shown in figure is
y
VC
C r P
 v
VP
x
1) 2VC 2) 2VC  VP 3) 2VC  VP 4) 2 VC  VP 
60. A carpet of mass M made of inextensible
material is rolled along its length in the form  vr sin  vr cos
1) v p   v  i  j
of a cylinder of radius R and kept along a  R  R
rough floor. The carpet starts unrolling
 vr sin  vr cos
without sliding on the floor, when a negligibly 2) v p   v  i  j
small push is given to it. The horizontal  R  R
velocity of the axis of a cylindrical part of the vr sin vr cos
carpet, when its radius is reduced to R/2 is 3) v p  v  i j
R R
14 7 vr sin vr cos
1)
3
gR 2)
3
gR 3) gR 4) 2gR 4) v p  v  i j
R R

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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III
64. A uniform solid sphere of radius r is rolling LEVEL - III - HINTS
on a smooth horizontal surface with velocity
v and angular velocity   v  r  . The 1.
m x i i

sphere collides with a sharp edge on the wall xcm 

as shown in figure. The coefficient of friction m
between the sphere and the edge  1 / 5.
i

Just after the collision the angular velocity

 xdm 0 L dx
L L
 L4 
kx 3
of the sphere becomes equal to zero. The  
4
 3
3L
kx 2 ;x  L  L 2
linear velocity of the sphere just after the 2. dm  dx cm
0

0 dm 0 L dx  3 
kx L  4
collision is equal to L
 3. Fractional length hanging,
V edge
l l 0.5
O     l  10 cm
L 1 30 1  0.5
let ‘ ’ be the mass per unit length. The co-
v 3v v
1) v 2)
3) 4) ordinates of 20 and 10 are (10,0) and (0,5)
5 5 6 respectively from ‘A’.
65. A particle of mass ‘m’ is rigidly attached at
‘A’ to a ring of mass ‘3m’ and radius ‘r’. The Distance of C.M from A, r cm= xcm
2
 ycm
2

system is released from rest and rolls without mass of removed part  d
sliding. The angular acceleration of ring just 4. shift 
after release is Mass of remaining part Here d=20 cm
r 2a
r 5. Shift of centre of mass x = 2
A R  r2
O
m Where r = radius of removed disc
R = radius of original disc
P a = distance between the centres
g g g g Note:In this question shift must be  R for exact
1) 2) 3) 4) approach to the solution
4r 6r 8r 2r
66. A solid sphere of mass M and radius R is 6. CM of bricks, above each brick must not be
placed on a rough horizontal surface. It is mi xi
stuck by a horizontal cue stick at a height h beyond its edge. xcm = m ; xcm  L
above the surface. The value of h so that the i

sphere performs pure rolling motion L L L

x1  a  , x2  2a  , x3  3a 
immediately after it has been stuck is 2 2 2
J L
h (or) a 
n
Fext  m1  m 2  g  2T
R
7.  a cm  y    1
2R 5R 7R 9R M m1  m 2
1) 2) 3) 4)
 2
2m1 m2 g
5 2 5 5 But T 
LEVEL-III - KEY m1  m2
01) 1 02) 1 03) 2 04) 4 05) 1 06) 2
m a
9. acm 
m
07) 2 08) 3 09) 4 10) 3 11) 1 12) 2 8. Theoretical
i i

13) 2 14) 1 15) 2 16) 3 17) 3 18) 3 i

 ucm 
19) 2 20) 3 21) 2 22) 1 23) 1 24) 1 2
25) 2 26) 4 27) 3 28) 3 29) 1 30) 2
10. hmax  H 
31) 2 32) 4 33) 4 34) 4 35) 2 36) 3 2g
37) 4 38) 4 39) 3 40) 4 41) 1 42) 1 11. Acceleration of system,
43) 4 44) 2 45) 3 46) 2 47) 4 48) 1 mg sin 60  mg sin 30
49) 4 50) 2 51) 1 52) 2 53) 2 54) 1 a
55) 3 56) 1 57) 2 58) 3 59) 2 60) 1 2m
61) 4 62) 4 63) 2 64) 1 65) 2 66) 3

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 3 1 ma1  ma 2
a    g , Now a cm  t 2
0
 4 

2m 2
1

 3 1  23. About right base corner ; F  mg

here, a1 and a 2 are  4  g at right angles. 24. About a pillar clockwise torque = anticlockwise
  torque
25. Let x is distance of P from other end
2a  3  1 
Hence, acm  2    g
xFP LFQ W
L
 4 2  2
12. The masses of load, ladder and man are M, where L is the length of the rod and W is its weight
M-m and m respectively. Their velocities are
but W = FP  FQ
v(upward), -v and vr -v respectively 26. If the distance from one end is x then
 mi vi  50  x 1  x  0.25  
x
m
 vcm 
4
200  4x  x  x  40cm
i

M (v)  ( M  m)(  v)  m(v r  v) m

  vr  length on the table = 100 - 40 = 60 cm
2M 2M L
2h 27. cw  Acw ; F1.2  mg  ; F  1.96  104 N
13. Range of C.M = vcm 2
g 28. (F) (perpendicular distance) = mg(perpendicular
m1v1  m2 v 2 1  mg
But v cm  m  m distance) FL  mg    F 
2 2
29. If the perpendicular distance of any side of the
1 2
14. Maximum height attained by C.M
ucm
2 triangle from ‘O’ is ‘x’then F1 x  F2 x  F3 x  0
= Initial height of C.M +  F1  F2  F3
2g
 rv
  30. From diagram, we get x  AO sin 600
15.  2

   
r L 3 ML2 ML2
  I
; z  I
; x  I 
3iˆ  4 ˆj  5kˆ  xiˆ  yjˆ  zkˆ  iˆ  9 ˆj  8kˆ 2 2 6 12
y

v 2v v ML 2

16.   ;  x  2R ;  I AB   Mx 
x x  2R 2R 12
31. Moment of inertia of a uniform rod about one end
v
17.  mL2
L cos   moment of inertia of the system
3
 m  2R  ; I 2 
mR 2 mR 2
18. I1   md 2 ; I1  I 2 m  L / 2
2
mL2
2
4 2  2 
1 2 2 3 12
19.  2 N ,  0t  t
2 32. I  4I1 where I 1 is M.I of each sphere
20. M.I about an axis passing through the diagonal I1  I c  Md 2
2 ML2 2 L
Ig  M.I about the given axis
and I c  MR ; d  ; L  4R
2
3 5 2
 L  33. I  2 I c  I1  and I1  I c  Md 2
2

I  I g  4M  
 2 2
I c  MR 2 and d = 4R
21.   dt 5
22. At any instant linear acceleration of all points of 34. I1  2I 2 ; M1R 12 2M 2 R 22
coupled belt is same. a  r  R1  M
2

r1 1  r2 2 ;find 2 ; 2  1  2t    2 2 ;but m  D  R 2
 R2  M1
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III
Hence, distance travelled by the centre of the rod
R1 D2 R 2  R1  D
2 2 4

2 =    2 2 2 l
R2 D1 R 1  R2  D1 is s  vt  v  2 l
35. On x-axis at a distance 2R,  v 
43. Angular momentum about a point P(x,y) with
 m  4 R 2   mR 2
mR 2 9 respect to the point of projection is
I
 u 2 sin 2  3mu 2
L=-m u cos  
2 2
  (  600 )
mR 2  2 g  16 g
On y-axis at a distance ‘d’ I   md 2
4 44. From the diagram,
R 1
equating both, d  17 OC  OA sin 450  4   2 2cm
2 2

  
MR 2 Angular momentum
36. I  I1  I 2 where I1  MR 2 and I 2 
2  mv  OC   5 3 2 2 2 ;  60gm cm 2 s 1
MR 2
37. I  2I1 , where I1 
2 45. I1  I2 ; MK12 2  MK 22 2
38. I  I1  I2  I 3  I 4 T1 T2
1 2

46. Initial angular momentum = final angular

ML2 L momentum
2

I1  I 2  and I 3  I 4  M 2
12   ML2
I 0  I ; whereL
1
I
I1 M 1R12 12
39.  for a circular loop, M R
I 2 M 2 R22 ML 2
L
2

I1   2m  
I1 R13 12 2
 
 
I 2 R23 47. L  I cm  m r  vcm
 

40. If I1 and I 2 are moment of inertia of hollow

 k   M R   j   v 0i  M R v 0 k
MR 2 3
spheres of radii R and 2R respectively, then 

 
I  I 2  I1 and mass R 3 2 2
48. L  I cm  m r  vcm
 
41. I  I total  I '

42. 

1
2

MR 2 k  m 4 Ri  3 R j  Ri   
5
  MR 2 k
V 2
1/4 49. L  R  Mv 0  I 0 ;  Mv 0 r  I 0 , which is
V
m v constant
 i
50. When the ring is at the maximum height, the wedge
mv  mv'  mV  v  v'  V and the ring have the same horizontal component
Applying conservation of angular momentum about of velocity. As all the surfaces are smooth, in the
point of collision. absence of friction between the ring and wedge
 ml 2  and surface angular velocity of the ring remains
  mV    l  3V  ii 
l
0 constant.
 12  4 From conservation of mechanical energy
Applying restituting equation,
 u1  u2 n   v2  v1 n   v  0  (V  v' (iii) 1 1
Mv 2  I 2

1 1 1
Mv12  I
Mv12  mgh
2

Solving Eqs. (i), (ii), and (iii) we get V=v and 2 2 2 2 2
3v v
 where v1 = final common velocity 
l 2
Time taken to complete three revolutions v 2
h
   6  ; t    3v  v
6 6 l 2 4g

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ROTATIONAL DYNAMICS g
m2  m1 mgR  MR 2 ; g  R  
51. a  g R
M g 0R
m2  m1    t  0 ; t 
2 0
R 2 2 g
 MR 2  v
52. mg  T  ma; a  R; TR    63. Hence 
 2  R
2T 2mg
 ;  a
MR 2m  M  90-
mgl 3g sin v
53.  sin ; 
2 2l
When rod is horizontal  r
2  v  v 
3g 3g v px   v  r sin  i ; v py    r cos  j
   at  l   R  R 
2l 2  vr sin   vr cos 
mg vp  v  i   j
54. T   R   R 
10 64. Impulse provided by the edge in the horizontal
55. w   5.57 N    0.800m   4.46 J direction
  Ndt   mV |   mV  ----(1)
I 
1
K  2
 2

2
f i
Friction impulse in the vertical direction
R  Ndt  mR 2   ---(2)
mg 2 V 
56. T 
6 5 R
1 1 1 from eq(1) & (2) we get
57. Total kinetic energy   mv2  I
 Ndt  2mV and V  V
I 2 2

2 2 2
0 |

1 2  1 1 2  65. The distance of CM from the ring centre O

 MD 2   MR 2 
3m  0   m  r  r
 MR 2  2 2 2

2 5  2 2 5 
0

x 
MR 2 2
 2
5 D2 2
 3m  m 4
 0
1    we can apply torque equation about point of
5 
2
0 2 R2 2
0  contact as the ring is rolling
mg sin R2 N
58. f  where 2
R2 K2 4mg
1 2
K x
59. VC  r  Vp  1 
O A
VA  VC  r  2VC  Vp 3mg
60. Gain in KE = loss in PE CM mg
1 2  K2  f
mv 1  2   Mgh2  mgh1 P
2  R   Ip

4mg     3mr 2  mr 2   m  AP  
p
where M = mass of carpet of radius R
r
R
2

m = mass of carpet of radius 4  

 
2
 mgr   4mr 2  m 2r 
2
R R  
2

h2 R and h1 and also m  

2 2 g
61. F  f  ma & FR  fR  I  mgr  6mr 2 a
6r
62. Taking the about C.M.

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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III
66. Let v be the velocity of the centre of mass of the its plane
sphere and be the angular velocity of the body 5
about an axis passing through the centre of mass. c) about natural axis g) MR 2
4

J  Mv ; J  h  R   5 MR 
2 2 1
d) about any tangent h) MR 2
2

from the above two equations v  h  R   r

2 2  r to its plane
5 5. Match the following
from the condition of pure rolling, v  R A disc rolls on ground without slipping.
2R 7R Velocity of centre of mass is v. There is a point
hR  h P on circumference of disc at angle . Suppose
5 5
v p is the speed of this point. Then, match the
LEVEL- IV following table:

Matching Type Questions

1. Match the following :
List– I List – II
Vcm

A. Position of centre of e. is zero 

mass P

B. The algebraic sum f. in non uniform

of moments of all the gravitational field Column-I Column-II
masses about centre
of mass g.is independent a) If  600 p) v p  2v
C. Centre of mass and of frame of b) If  900 q) v p  v
centre of gravity coincide reference
D. Centre of mass and h. in uniform c) If  1200 r) v p  2v
centre of gravity do gravitational
d) If  1800 s) v p  3v
not coincide field
2. Match the following: 6. Column-I Column-II
List - 1 List - 2 2
a) Moment of inertia p) MR
2
a) torque e) mass 5
b) moment of inertia f) linear momentum of annular disc of inner
c) angular momentum g) linear acceleration
d) angular acceleration h) force radius R1 and outer
3. If R is radius and K is radius of gyration then radius R2 about
in the case of following rolling bodies match symmetric axis
the ratio K 2 : R 2 3
List - 1 List - 2 b) Moment of inertia q) MR 2
a) solid sphere e) 1 : 1 10
b) solid cylinder f) 2 : 3 of elliptical disc of two
c) hollow sphere g) 1 : 2 radii R1 and R2
d) hollow cylinder h) 2 : 5 about symmetric axis

 R1  R22 
4. Match the following moment of inertia of thin M 2
circular plate about different axes of rotation c) Moment of inertia r)
List - 1 List -2 4
of solid sphere of radius
3
a) about any diameter e) MR 2 R about symmetric axis

 R1  R22 
2
1 M 2
b) about any tangent in f) MR 2 d) Moment of inertia s)
4 2
of solid cone of base

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radius R about symmetric axis (c) If assertion is true but reason is false.
7. A solid spherical ball of mass M and radius (d) If the assertion and reason both are false.
R rolls without slipping down a surface 10. Assertion : The centre of mass of a two particle
inclined to horizontal at an angle . system lies on the line joining the two particles,
Considering that ball is uniform solid sphere being closer to the heavier particle.
and that ball and surface are perfectly rigid. Reason: Product of mass of one particle and its
Column-I Column-II distance from centre of mass is numerically equal
to product of mass of other particle and its distance
a) Friction force p) Zero from centre of mass.
involved 11. Assertion: The centre of mass of system of n
b) Minimum value of q)  2 / 7  Mg sin particles is the weighted average of the position
coefficient of friction vector of the n particles making up the system.
for pure rolling Reason: The position of the centre of mass of a
c) Work done against r) Static friction system is independent of coordinate system.
frictional force 12. Assertion: The centre of mass of an isolated
s)  2 / 7  tan
system has a constant velocity.
d) Force of kinetic Reason: If centre of mass of an isolated system
friction is already at rest, it remains at rest.
8. A solid sphere, hollow sphere, solid cylinder, 13. Assertion: The centre of mass of a body may lie
hollow cylinder and ring each of mass M and where there is no mass.
radius R are simultaneously released at rest Reason: Centre of mass of a body is a point,
from top of incline and allowed to roll down where the whole mass of the body is supposed to
the incline. be concentrated.
Column-I Column-II 14. Assertion: Aparticle is moving on a straight line
a) Time taken to reach p) Solid sphere with a uniform velocity, its angular momentum is
bottom is maximum for always zero.
b) Angular acceleration q) Hollow cylinder Reason: The momentum is zero when particle
maximum for moves with a uniform velocity.
c) Kinetic energy at r) Hollow sphere 15. Assertion: The centre of mass of a proton and
bottom is same for an electron, released from their respective
d) Rotational kinetic s) Ring positions remains at rest.
energy is maximum for Reason: The centre of mass remain at rest, if no
9. A rigid body of mass M and radius R rolling -external force is applied.
without slipping on an inclined plane. The 16. Assertion: The position of centre of mass of a
magnitude of force of friction body does not depend upon shape and size of the
Column-I Column-II body.
Mg sin Reason: Centre of mass of a body lies always at
a) For ring p) the centre of the body
2.5
17. Assertion: A shell at rest, explodes. The centre
Mg sin of mass of fragments moves along a straight path.
b) For solid sphere q)
3 Reason: In explosion the linear momentum of
Mg sin the system is not conserved.
c) For solid cylinder r) 18. Assertion: A judo fighter in order to throw his
3.5 opponent on to the mattress he initially bend his
Mg sin opponent and then rotate him around his hip.
d) For hollow sphere s) Reason: As the mass of the opponent is brought
2
Assertion & Reason closer to the fighter’s hip, the force required to
Read the assertion and reason carefully to mark throw the opponent is reduced.
the correct option out of the options given below 19. Assertion: The centre of mass of an electron and
(a) If both assertion and reason are true and the proton, when released moves faster towards
reason is the correct explanation of the assertion. proton.
(b) If both assertion and reason or true but reason Reason: Proton is lighter than electron.
is not the correct explanation of the assertion.
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III
20. Assertion: At the centre of earth, a body has Reason: In rolling down, a body acquired both,
centre of mass, but no centre of gravity. kinetic energy of translation and rotation.
Reason: Acceleration due to gravity is zero at 32. Assertion: In rolling, all points of a rigid body
the centre of earth. have he same linear speed.
21. Assertion: When a body dropped from a height Reason: The rotational motion does not affect
explodes in mid air, its centre of mass keeps moving the linear velocity of rigid body.
in vertically downward direction. 33. Assertion: A wheel moving down a perfectly
Reason: Explosion occur under internal forces frictionless inclined plane will undergo slipping (not
only. External force is zero. rolling motion).
22. Assertion: It is harder to open and shut the door Reason: For perfect rolling motion, work done
if we apply force near the hinge. against the friction is zero.
Reason: Torque is maximum for the door. 34. Assertion: The total kinetic energy of a rolling
23. Assertion: Moment of inertia of a particle solid sphere is the sum of translational and
changes, when the axis of rotation changes. rotational kinetic energies.
Reason: Moment of inertia depends on mass Reason: For all solid bodies total kinetic energy
and distance of the particles. is always twice the translational kinetic energy.
24. Assertion: Inertia and moment of inertia are
different quantities. Statement type questions
Reason: Inertia represents the capacity of a body 1) Statement A& B are true
to oppose its state of motion or rest. 2) Statement A is true, Statement B is false
25. Assertion: If earth shrink (without change in mass) 3) Statement A is false, Statement B is true
to half its present size, length of the day would 4) Statement A & B are false
become 6 hours. 35. Consider the following two statements A and B
Reason: As size of the earth changes its moment and identify the correct answer
of inertia changes. Statement A : The centre of mass of a system of
26. Assertion: Torque due to force is maximum when particles depends on forces on the particles.
angle between r and F is 900. Statement B : In the absence of external force,
Reason: The unit of torque is newton-metre. the centre of mass of system moves with uniform
27. Assertion: Radius of gyration of body is a velocity
variable quantity. 36. Consider the two statements A and B and identify
Reason: The radius of gyration of a body about the correct answer
an axis of rotation may be defined as the root Statement A : A wooden sphere and a copper
mean square distance of the particles from sphere of same radius are kept in contact with
the axis of rotation. each other the centre of mass will be with in
28. Assertion: A ladder is more apt to slip, when the wooden sphere.
you are high up on it than when you just begin to Statement B: Three identical spheres each of
climb. radius R are placed touching each other on
Reason: At the high up on a ladder, the torque is horizontal table. The centre of mass of the system
large and it is small when one just begins to climb. is located at the point of intersection of the
29. Assertion: Torque is equal to rate of change of medians of the triangle formed by the centres of
angular momentum. spheres.
Reason:Angular momentum depends on moment 37. Consider the following two statements A and B
of inertia and angular velocity. and identify the correct choice
30. Assertion:The speed of whirlwind in a tornado Statement A : The rotational kinetic energy of a
is alarmingly high. rolling body is always greater than its translatory
Reason: If no external torque acts on a body, its kinetic energy
angular velocity remains conserved. Statement B: The maximum value of radius of
31. Assertion: The velocity of a body at the bottom gyration of a rolling body can not be greater than
of an inclined plane of given height, is more when the radius of that body
it slides down the plane, compared to, when it 38. Consider the following two statements A and B
rolling down the same plane. and identify the correct choice

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Statement A : Spokes are used in a bicycle of centre of mass [EAMCET-2012]

wheel to increase the strength of wheel 1) Both A and B are correct
Statement B: Mass of fly wheel is concentrated 2) Both B and C are wrong
on its rim to increase its moment of inertia 3) Both A and C are wrong
39. Consider the following two statements A and B 4) Both A and D are wrong
and identify the correct choice 46. A shell is projected at some angle with horizontal.
Statement A : Moment of inertia of circular plate When the shell is at its highest point, it explodes
is minimum about its natural axis into two pieces.
Statement B : Inertia of rotation of a rotating Statement A : : The law of conservation of linear
body is proportional to its angular momentum momentum can be used for the small interval of
40. Consider the following two statements A and B explosion
and identify the correct choice Statement B : The net force on the shell at highest
Statement A : The torques produced by two point is zero
forces of couple are opposite to each other. 47. Statement A : Impulsive force on a particle may
Statement II : The direction of torque is always change its kinetic energy and its momentum
perpendicular to plane of rotation of body Statement B : Momentum of a particle changes
41. Consider the following two statements A and B only when kinetic energy of the particle changes
and identify the correct choice 48. A uniform rod is held vertically on a smooth
Statement A : The torque required to stop a horizontal surface. Now the rod is released, given
rotating body in a given time is directly proportional it simultaneously a gentle push
to its initial angular momentum Statement A : Centre of mass of the rod moves
Statement B: If radius of earth shrinks then its in vertical direction as the rod falls
rotational kinetic energy increases Statement B : The rod is falling freely
42. Statement A : Mechanical advantage of a lever 49. A particle is thrown vertically upward from ground,
is always < 1 while another is thrown simultaneously vertically
Statement B: Mechanical advantage of a lever downward from some height
can be increased by increasing its effort arm or by Statement A : In the reference frame of centre
decreasing its load arm. of mass of the system, the particles move uniformly
43. Consider the following two statements A and B Statement B : Acceleration of the centre of mass
and identify the correct choice of the system is zero
Statement A : When a rigid body is rotating about 50. Statement A : Momentum of a system w.r.t centre
its own axis, at a given instant all particles of body of mass of the system is zero
possess same angular velocity. Statement B : Centre of mass can acceleration
Statement B: When a rigid body is rotating about only under the action of external forces.
its own axis, the linear velocity of a particle is 51. Statement A : Internal force cannot change
directly proportional to its perpendicular distance kinetic energy of a system of particles
from axis Statement B : Internal forces cannot change
44. Consider the following two statements A and B momentum of a system of particles.
and identify the correct choice 52. Statement A : Linear momentum of a system of
Statement A : The moment of inertia of a rigid particles with respect to centre of mass must be
body is independent of its angular velocity zero
Statement B: The radius of gyration of a rotating Statement B : Linear momentum of a system of
metallic disc is dependent on its temperature particles is the vector-sum of linear momenta of
45. Choose correct statement. all particles of the system.
(A) The position of centre of mass of a system is 53. Statement A : Work done by a force on a body
dependent on the choice of coordinate system whose centre of mass does not move may be non-
(B) Newton’s second law of motion is applicable zero
to the centre of mass of the system. Statement B : Work done by a force depends
(C) Internal forces cannot change the state of on the displacement of the centre of mass
centre of mass. 54. Statement A : Net work done by internal force
(D) Internal forces can change the state in a system may be zero.
Statement B : Net force on the centre of mass of
the system by internal mechanism is zero
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SYSTEM OF PARTICLES AND RIGID BODY DYNAMICS JEE-ADV PHYSICS-VOL - III
55. Statement A : In collision between two bodies, Statement B : Both spheres take same time to
they remain in contact with each other for a very reach bottom of the incline.
short interval of time before they separate. During 65. Statement A : A particle in uniform motion may
the period of restitution, the bodies try to regain have non-zero angular momentum about a point
their original shape. in space.
Statement B : During the period of contact, Statement B : A particle may be moving on a
bodies exchange their momentum and energy curved path with uniform speed.
56. Statement A : For a system of particles, total 66. Statement A : Rolling on a stationary surface
energy of the system can change even if net force can be treated as pure rotation about the point of
acting on the system is zero contact
Statement B : If net force on a system of particles Statement B : Point of contact of the body is
is zero, total momentum can not change. instantaneous centre of rotation, as it is
57. Statement A : In pure rolling motion, net work instantaneously at rest during rolling.
done by friction is zero. Multi Option Questions
Statement B : Sum of translational work done 67. Identify the correct one from the following
by friction and rotational work done by friction is statements.
zero A. the position of centre of mass in a co–ordinate
58. Statement A : For a system of particles under system does not change if a man moves from
central force field, total angular momentum is the one end to other end on a floating wooden
conserved about the centre log in still water.
Statement B : Torque acting on such a system is B. When a man moves from one end to other
zero about the centre end on a floating wooden log in still water, it
59. Statement A : A ball is rolling on a rough moves in opposite direction
horizontal surface. It gradually slows down and C. Due to action and reaction the wooden log
stops. floating in still water moves in opposite direction
Statement B :: Force of rolling friction decreases as the man on it moves from one end to the other
linear velocity end
60. Statement A : Aring is rolling without slipping on 1) B & C are true 2) A & D are true
rough surface as shown in figure. The force of 3) A, B & C are true 4) All are correct
friction necessary for ring to purely roll is in forward 68. If external forces acting on a system have
direction. zero resultant, the centre of mass
Statement B : Force of friction is zero when A. may move B. may accelerate
external force acts at top of ring. C. must not move D.must not accelerate
61. Statement A : Velocity acquired by a rolling body 1) A & B are correct 2) B & C are correct
depends on inclination of plane on which it rolls 3) C & D are correct 4) A & D are correct
down without slipping 69. In which of the following cases, the centre of
Statement B : Velocity depends upon height of mass of a rod is certainly not at its centre?
descent of body A. The density increases from left to right upto
62. Statement A : A cylinder rolls up an incline plane, the centre and then decrease
reaches some height and then rolls down. The B. The density decreases from left to right upto
direction of friction force acting on cylinder is up the centre and then increase
the incline while ascending as well as descending. C. The density continuously increases from left
Statement B : Direction of force of friction is in to right
accordance with sense of angular acceleration. D. The density continuously decreases from left
63. Statement A : Angular momentum of a particle to right
executing uniform circular motion is constant. 1) A & B are correct 2) B & C are correct
Statement B : Momentum of a particle executing 3) C & D are correct 4) A & D are correct
uniform circular motion is constant. 70. If a circular concentric hole is made in a disc
64. Two solid spheres (of masses m and 4m and radii then about an axis passing through the centre
r and 16r) roll down without slipping on an incline. of the disc and perpendicular to its plane.
Statement A : Both reach the bottom of incline 1) moment of inertia decreases
with same kinetic energies.
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2) moment of inertia increases a smooth table. One particle describes a

3) radius of gyration increases circular path on the table with angular
4) radius of gyration decreases velocity 1 , and the other describes a conical
71. A rotor of radius r is rotating about its own
vertical axis and a person in contact with inner pendulum with angular velocity 2 below the
wall of rotor remains in equilibrium without table. If l1 and l2 are the lengths of portions
slipping down. If  is angular velocity of rotor
of the string above and below the table, then
and  is minimum coefficient of friction
between person and the wall of rotor then l1 l1 2

1) l  2) 
2 2
following is correct 2 1 l2 2
1
1 1
A)   2 B)   C)   2 D)   r 1

1
ml

1 1 l cos
 2 
r  3) 4)
1) A and B are true 2) A and D are true
2
1 g 2
2 1
2
2 g
3) B and C are true 4) C and D are true 76. A symmetrical body of mass M, radius R and
72. A particle of mass m is executing uniform radius of gyration k is rolling on a horizontal
circular motion on a path of radius r. If v is surface without slipping. If linear velocity of
speed and p the magnitude of its linear centre of mass is vc and angular velocity ;
momentum, then the radial force acting on then
the particle is
1 2  k2 
1)
mv 2
2)
pm
3)
vp
4)
p2 1) the total KE of body is 2 mvc 1  R 2 
r r r mr  
 
73. In circular motion if v is velocity vector, a is 1
2) the rotational KE is MR 2 2
acceleration vector, r is instantaneous 2


1

position vector, and p is momentum vector
3) the translational KE is Mvc
2

and is angular velocity of particle. Then 2
  4) Total energy = 0
1) v, and r are mutually perpendicular

77. A ring type flywheel of mass 100kg and
  
2) p, v and are mutually perpendicular diameter 2m is rotating at the rate of
5
3) r  v  0 and r   0
   
rev/sec. Then
11
4) r.v  0 and r.  0
  
1) the moment of inertia of the wheel is
74. The length of second hand of a watch is 1cm.
Then 100kg  m 2
1) The linear speed of tip of second hand is 2) the kinetic energy of rotation of flywheel is
5  103 J
cm / s 3) the angular momentum associated with the
30
2) The linear speed of the tip of second hand is flywheel is 103 joule-sec
uncertain 4) the flywheel, if subjected to a retarding torque
3) The change in linear velocity vector in 15 250N  m , will come to rest in 4sec.
78. In which of the following case(s), the angular
seconds is 2cm / s 2 momentum is conserved?
30
4) The change in acceleration vector in 15minutes 1) The planet Neptune moves in elliptical orbit
around the sun with sun at one focus
2
2 2) A solid sphere rolling on an inclined plane
is cm / s 2
1800 3) An electron revolving around the nucleus in
75. Two particles, each of mass m are attached elliptical orbit
to the two ends of a light string of length L 4) An  particle approaching a heavy nucleus
which passes through a hole at the centre of from sufficient distance.
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79. A solid sphere is in pure rolling motion on an 53) 2 54) 3 55) 1 56) 1 57) 1 58) 1
inclined surface having inclination 59) 1 60) 3 61) 3 62) 1 63) 2 64) 3
[IIT-2006] 65) 1 66) 2
1) frictional force acting on sphere is µmg cos MULTI OPTION QUESTIONS
67) 3 68) 4 69) 3 70) 1,3
2) f is dissipative force
71) 3 72) 1,3,4 73) 1,4 74) 1,3,4
3) friction will increase its angular velocity and
75) 2,4 76) 1,3 77) 1,2,3,4
decrease its linear velocity 78) 1,3,4 79) 3,4 80) 1,2,3 81) 1,4
4) If decreases, friction will decrease. LEVEL-IV-HINTS
Assertion and reasoning type

80. The torque on a body about a given point is
found to be equal to A  L , where A is a 10. If centre of mass of system lies at origin
  

then r cm  0

constant vector and L is angular momentum

of the body about that point. From this it y
follows that

dL
1) is perpendicular to L at all instants of time

dt
2) the component of L in the direction of A does
  x
not change with time r1 r2
3) the magnitude of L does not change with time

m1 r1  m 2 r2
 
4) L does not change with time. r cm 

81. A sphere is rolled on a rough horizontal surface. m1  m 2
It gradually slows down and stops. The force  m1 r1  m 2 r2  0 or m1r1  m 2 r2
 
of friction tends to
1) decrease linear velocity It is clear that if m1  m 2 then r 2 > r1
2) increase linear momentum 12. External force on the system

 
3) decrease angular velocity d 
4) increase angular velocity Fext  M v cm If system is isolated i.e.
dt
LEVEL - IV - KEY
Fext  0 then vcm =constant.Initially if the velocity

Matching Type
1) A g; B e; C h; D f of centre of mass is zero then it will remain zero.
2) a  h; b  e; c  f ; d  g 13. As the concept of centre of mass is only theoretical,
therefore in practice no mass may lie at the
3) a - h; b - g; c - f; d - e
centre of mass. For example, centre of mass of a
4) a-f; b-g; c-h; d-e
uniform circular ring is at the centre of the ring where
5) a-q, b-p, c-s, d-r
there is no mass.
6) a-s; b-r; c-p; d-q 
7) a-q,r; b-s; c-p; d-q 14. When particle moves with constant velocity v then
8) a-q,s; b-p; c-pqrs; d-q,s its linear momentum has some inite value
9) a-s; b-r; c-q; d-p
 
P  mv . Angular momentum (L) = Linear
 
Assertion & Reason Type Questions
10) 1 11)2 12)2 13) 1 14) 4 15) 1 momentum (P) x Perpendicular distance of line of
16) 4 17) 4 18) 1 19) 4 20) 1 21) 1 action of linear momentum form the point of
22) 3 23) 1 24) 2 25) 1 26) 2 27) 1 rotation(d). So if d  0 then L  0, but if d = 0
28) 1 29) 2 30) 3 31) 1 32) 4 33) 2 then L may be zero. So we can conclude that angular
34) 3 momentum of a particle moving with constant
Statement Type Questions velocity is not always zero.
35) 3 36) 1 37) 3 38) 3 39)3 40) 3 15. Initially the electron and proton were at rest so
41) 1 42) 3 43) 1 44) 1 45) 4 46) 2 then centre of mass will be at rest. When they move
47) 2 48) 2 49) 2 50)) 1 51) 3 52) 1 towards each other under mutual attraction then

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velocity of centre of mass remains unaffected  T IR 2 . It means if size of the earth changes
because external force on the system is zero. then its moment of inertia changes. In the problem
16. The position of centre of mass of a body depends radius becomes half so time period (Length of the
on shape, size and distribution of mass of the body. day) will becomes 1/4 of the present value i.e.24/
The centre of mass does not lie necessarily at the 4=6 hr.
centre of the body.
17. As the shell is initially at rest and after explosion, 26.   rFsin . If   900 then  max  rF Unit of
according to law of conservation of linear torque is N - m.
momentum, particles move in all direction,such that 27. Radius of gyration of body is not a constant
total momentum of all parts is equal to zero. quantity. Its value changes with the change in
18. Through bending weight of opponent is made to location of the axis of rotation. Radius of gyration
pass through the hip of judo fighter to make of a body about a given axis is given as
its torque zero.
r12  r22  .....  rn2
19. The position of centre of mass of electron and K
proton remains at rest. As their motion is due to n
internal force of electrostatic attraction, which is 28. When a person is high up on the ladder, then a
conservative force. No external force is acting on large torque is produced due to his weight about
the two particles, therefore centre of mass remain the point of contact between the ladder and the
at rest. floor. Whereas when he starts climbing up, The
20. At the centre of earth, g = 0. Therefore a body has torque is small. Due to this reason, the ladder is
no weight at the centre of earth and have no centre more apt to slip, when one is high up on it.
of gravity (centre of gravity of a body is the point  dL

where the resultant force of attraction or the weight 29. r and L  I
of the body acts). But centre of mass of a body dt
depends on mass and position of particles and is 30. In a whirlwind in a tornado, the air from nearby
independent of weight. regions gets concentrated in a small space thereby
21. Explosion is due to internal forces. As no external decreasing the value of its moment of inertia
force is involved, the vertical down ward motion considerably. Since, I  constant, so due to
of centre of mass is not affected. decrease in moment of inertia of the air, its angular
22. Torque = Force x perpendicular distance of line of speed increases to a high value. If no external torque
action of force from the axis of rotationHence for dL
a given applied force, torque or true tendency of acts, then   0   0 or L = constant
rotation will be high for large value of d. If distance dt
d is smaller, then greater force is required to cause  I =constant. As in the rotational motion, the
the same torque, hence it is harder to open or shut moment of inertia of the body can change due to
down the door by applying a force near the hinge. the change in position of the axis of rotation, the
23. The moment of inertia of a particle about an axis of angular speed may not remain conserved.
rotation is given by the product of the mass of the 31. In sliding down, the entire potential energy is
particle and the square of the perpendicular distance converted into kinetic energy. While in rolling down
of the particle from the axis of rotation. some part of potential energy is converted into
For different axis, distance would be different, K.E. of rotation. Therefore linear velocity acquired
therefore moment of inertia of a particle changes is less.
with the change in axis of rotation. 32. In rolling all points of rigid body have the same
24. There is a difference between inertia and moment angular speed but different linear speed.
of inertia of a body. The inertia of a body depends 33. Rolling occurs only on account of friction which is
only upon the mass of the body but the moment of a tangential force capable of
inertia of a body about an axis not only depends providing torque. When the inclined plane is
upon the mass of the body but also upon the perfectly smooth, body will simply slip under the
distribution of mass about the axis of rotation. effect of its own weight.
25. When earth shrinks it angular momentum remains 34. K N  K R  K T This equation is correct for any
constant. i.e. body which is rolling without slipping. For the ring
2 2 and hollow cylinder only K R  K T i.e.
L  I  mR 2   cons tan t
5 T K N  2K T

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ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

LEVEL-V
m

SINGLE ANSWER QUESTIONS A

1. A rigid massless beam is balanced by a 
particle of mass 4m in left hand side and a
pulley particle system in right hand side. The 2 2
(a) tan (b) tan
x 3 9
value of y is: 1
(c) 6 tan (d) none of these

x Y 5. A disc of mass m is connected with an ideal

spring of stiffness k . If it is released from
m rest., it rolls without sliding on an inclined
4m
plane. The maximum elongation of the spring
is :
m
2m k

7 5
(a) 6 (b) 6 (c) 1:1 (d) 11/12 m

2. A uniform box is kept on a rough inclined plane.

It begins to topple when is equal to : 

mg sin 2mg sin 3mg sin 2mg sin

(a) (b) (c) (d)
2x k 3k k k
x 6. A massless thin hollow sphere is completely
filled with water of mass m . If the hollow
sphere rolls with a velocity v . the kinetic
energy of the sphere of water is :(Asune water

is non viscous)
1
(a) 300 (b) 600 (c) tan 1 (d) 45°
2
3. A rod touches a disc kept on a smooth V
horizontal plane. If the rod moves with an
accleration a , the disc rolls on the rod without
sliding.Then, the acceleration of the disc w.r.t
the rod is
1 1 7 5
(a) mv 2 (b) mv 2 (c) mv 2 (d) mv 2
2 3 10 6
a 7. A particle P collides elastically at M with a
speed v . The change in angular momentum of
the particle about the point N during collision
60°
is :
a 3a P
(a) 3 (b) (c) zero (d) a/2
2 
4. A uniform cylinder of mass m is kept on an M
accelerating wedge. If the wedge moves with N I

an acceleration a  3g tan , the minimum V

coefficient of friction between the cylinder and
wedge to avoid relative sliding between them (a) 2mvl cos  (b) 2mvl cos 
is (c) zero (d) mvl cos 
84 NARAYANAGROUP
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8. Aball is attached to a string that is attached to 12. A hemispherical shell of mass M and radius
a thick pole. When the ball is hit, the string R is hinged at point O and palced on a
warps around the pole and the ball spirals horizontal surface. A ball of mass M moving
inwards sliding on the frictionless surface. with a velocity u inclined at an
Neglecting air resistance, what happens as the
ball swings around the pole? 1
angle  tan 1   strikes the shell at point A
(a) The mechnical energy and angular momentum 2
are conserved
(b) The angular momentum of the ball is conserved (as shown in the figure) and stops. What is the
at the mechanical energy of the ball increases minimum speed u if the given shell is to reach
(c) The angular momentum of the ball is conserved the horizontal surface OP?
and the mechanical energy of the ball decreases O P
(d) The mechanical energy of the ball is conserved
and angular momentum of the ball decreases
9. The free end of a thread wound on a bobbin is 
A
passed round a nail A hammered into the wall. m u
The thread is pulled at a constant velocity’v’
Assuming pure rolling of bobbin, find the 2 gR gR
(a) Zero (b) (c)
velocity v0 of the centre of the bobbin at the 3 5
instant when the thread forms an angle with d) it cannot come on the surface for any value of u
the vertical:(R and r are outer and inner radii 13. A hollow sphere of mass 2kg is kept on a rough
off the babbin) horizontal surface. A force of 10 N is applied
at the centre of the sphere as shown in the
figure. Find the minimum value of so that
the sphere starts pure rolling.
V (Take g=10m/s2)
F = 10N

vR vR 30°
(a) (b)
R sin  r R sin  r
2vR v
(c) (d)
R sin  r R sin  r
10. A thin wire of length L and uniform linear mass
m
density is bent into a circular loop with centre
at O as shown. The moment of inertia of the (a) 3  0.16 (b) 3  0.08
loop about the axis XX ' is :
x x
1 (c) 3  0.1 (d) Data insufficient
14. A cubical block of side L and mass m is placed
O on a horizontal floor. In the arrangement as
shown, a force F is applied at the end of the
plane sheet PQ which is firmly attached with
L3 L3 5 L3 3 L3
(a) (b) (c) (d) the block. What should be the minimum value
8 16 16 2 8 2
of PQ so that block may be tipped about an
2 2

11. A partice moves in a circular path with

decreasing speed. Choose the correct edge?
statement F 30° L (a) L / 3
(a) Angular momentum remains constant P Q
(b) Acceleration  a  is towards the centre (b) L / 2
 a

(c) Particle moves in a spiral path wth decreasing L

radius (c) L 2 / 3
(d) The direction of angular momentum remains
constant O
(d) L 3 / 2

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ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III
15. A rigid body of moment of inertial I is F
projected with velocity V making an angle of
450 with horizontal. The magnitude of angular F1 Q P F2
momentum of the projectile about the point of
projection when the body is at its maximum
IV 3 A B
height is given by
2 2 gR 2
where R is the
(a)  7 F1  3F2  (b)  5 F1  3F2 
2 2
radius of the rigid body. The ridid body is:
(a) sphere (b)spherical shell (c)  3F1  F2   3F1  F2 
(d)
(c) disc (d) none of these 2
16. An equilateral triangle ABC formed a uniform 18. Consider the following statements:
wire has two small identical beads initially s1 : Zero net torque on a body means always
located at A. The triangle is set rotating about absence of rotational motion of the body.
the vertical axis AO.Then the beads are s2 : A particle may have angular momentum
relased from rest simultanously and allowed even though the particle is not moving in a
to slide down,one along Ab and other AC as circle.
shown.Meglecting frictional effects,the s3 : A ring of rolling without sliding on a fixed
quantities that are conserved as beads slides
surface. the centripetal acceleration of each
down are: particle with respect to the centre of the ring
is same.
A
State in order, whether s1 , s2 , s3 are ture or
g false.
(a) FTT (b) FFT (c) TTF (d) FTF
19. A uniform rod of length L (in between the
supports) and mass m is placed on two suports
B O C
A and B. The rod breaks suddenly at length
L/10 from the support B. Find the reaction at
(a) angular velocity and total energy(kinetic support A immediately after the rod breaks.
and potential) L/10
(b) total angular momentum and moment of
interia about the axis of rotation
A D B
(c) angular velocity and moment of interia
about the aixs of rotation
(d) total angular momentum and total energy L
17. On a smooth horizontal table, a sphere is 9 19 mg 9
A) mg B) mg C) D) mg
pressed by blocks A and B by forces F1 and 40 40 2 20
20. A uniform disc of mass m and radius R is
F2 respectively  F1  F2  exactly normal to rolling up a rough inclined plane which makes
the tangent at the point of contact of blocks an angle of 300 with the horizontal. If the
and sphere. A force F is applied on the sphere coefficients of static and kinetic friction are
along a diameter perpendicular to another each equal to and the only forces acting
diameter OP, which is the line of action of are gravitatinal and frictional, then value of
forces F1 and F2 . The sphere moves out of for maximum magnitude of the frictional
force acting on the disc is
block A and B. FInd minimum value of F. the
coefficient of friction is at all contacts: 1 1 1 1
A) B) C) D)
3 2 3 3 3 3
86 NARAYANAGROUP
 JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS

? 2  r2  r1   ? 1r2 ? 2  r2  r1   ? 1r1
21. On a particle moving on a circular path with a
constant speed v , light is thrown from a
(A) (B)
projectors placed at the centre ofthe circular r2 r2
path. The shadow of the particle is formed on
the wall. The velocity of shadow up the wall is (C) ? 1 (D) ? 2
Wall 24 . Two vertical walls are separated by a distance
v
of 2 metres. Wall ‘A’ is smooth while wall B is
rough with a coefficient of friction  0.5 A
uniform rod is probed between them. The
length of the longest rod that can be probed
 between the walls is equal to
2m
(A) v sec 2 f (B) v cos 2 f P
(C) v cos f (D) None of these
22. A rod of length l is travelling with velocity Q

v and rotating with angular velocity ? such

?l
Wall Wall

 v . The distance travelled by end

A B
that
2 (a) 2 metres (b) 2 2 metres
B of the rod when rod rotates by an angle,
17
p (c) 2 metres metres (d)
is 2
2
25. A disc is rotating at an angular velocity 0 .A
A
constant retardation torque is applied on it to
stop the disc. After a certain time at which
 V some number of rotation of the disc have been
performed so that total angle rotated is 1 and
2
B
that only th of these rotatios will further stop
3
5
(a) 2l (b) l (c) 3l (d) 4l the disc. Find the retarding force.
2
23. A large rectangular box has been rotated with 11 02 9 02 5 02 3 02
a) 2) 3) 4)
a constant angular velocity ? 1 about its axis 14 1 14 1 14 1 14 1
as shown in the figure. Another small box kept 26. A square plate hinged at A, of side a and mas
inside the bigger box is rotated in the same M is placed in  x  z  plane. The plate is
allowed to fall upto  x  y  plane. Find its
sense with angular velocity ? 2 about its axis
(which is fixed to floor of bigger box). A particle
P has been identified, its angular velocity about angular velocity.
AB would be Z

A
1
y
C
2
x
P A

 3g   24 g  12 g   12 g 
1/2 1/2 1/2 1/2
r1 D
a)   b)  c)  d)  
B r2 a 2   7 a   7a   7 2a 

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ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III
27. A disc of mass m and radius r is placed on a 31. A disc is given an initial angular velocity 0
rough horizotal surface. A cue of mass m hits
and placed on rough horizontal surface as
the disc at a height h from the axis passing
shown. The quantities which will not depend
through centre and parallel to the surface. The
on the coefficient of friction is/are
disc will start pure rolling for.
r r r r
a) h  b) h  c) h  d) h 
3 2 2 2
0
MULTIPLE ANSWER QUESTIONS
28. A solid cylinder is rolling down the inclined
plane without slipping. Which of the following
is/are correct (a) The time until rolling begins
(a) The friction force is dissipative (b) The displacement of the disc until rolling begins
(b) The friction force is necessarily changing (c) The velocity when rolling begins
(c) The friction force will aid rotation but opposes (d) The work done by the force of friction
translation 32. A thin rod AB of mass M and length L is
(d) The friction force is reduced if is reduced rotating with angular speed 0 about vertical
29. An impulse I is applied at the end of a uniform axis passing through its end B on a horizontal
rod if mass m. then : smooth table as shown. If at some instant the
I2 hinge at end B of rod is opened then which of
(a) KE of translaton of the rod is the following statement is/are correct about
2m
motion of rod ?
I2
(b) KE of rotation of the rod is A M
6m

3I 2
(c) KE of rotation of the rod is L
2m
B
I 2I 2 0
(d) KE of the rod is
m
30. A uniform rod of mass m and length l is in (a) The angular speed of rod after opening the
equilibrium under the constraints of horizontal
and vertical rough surfaces. Then : hinge will remain 0
y (b) The angular speed of rod after opening the
hinge will be less than 0
(c) In the process of opening the hinge the kinetic
energy of rod will remain conserved.
(d) Angular momentum of rod will remain
1
conserved about centre of mass of rod in the
process of opening the hinge

x 33. A cylinder rolls without slipping on a rough
O
floor, moving with a speed v. It makes an elastic
(a) the net torque of normal reaction about O is collision with smooth vertical wall. After impact
mgl (a) it will move with a speed v initially
equal to cos (b) its motion will be rolling without slipping
2
(b) the net rorque due to friction about O is zero (c) its motion will be rolling with slipping initially
(c) the net torque due to normal reactions is and its rotational motion will stop momentarily at
numerically equl to the net torque due to the some instant
frictional force abou the CM of the rod (d) its motion will be rolling without slipping only
(d) all of the above after some time

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34. A sphere of radius 0.10m and mass 10kg rests 37. A wheel is under pure rotational motion about
in the corner formed by a 30° inclined plane an axis passing through its centre. It moves
and a smooth vertical wall. Choose the correct with constant angular velocity.
options a) if angular velocity is increasing then acceleration
of particles on a spoke if moved from centre to
periphery remains constant
f mg b) acceleration of particles on a spoke if moved
from centre to periphery continuously increases
30°
c) acceleration of particles on a spoke if moved
from centre to periphery continuously increases and
on peripherial points,it remains same
N2
d) accelerations of particles in both the cases remain
(a) N1  56.5 N (b) N 2  113 N same
(c) f  0 (d) f  0 COMPREHENSION TYPE QUESTIONS
35. If the resultant of all the external forces acting Passage - I :
on a system of particles is zero, then from an A string is wrapped several times on a cylinder of
inertial frame, one can surely say that mass M and radius R. the cylinder is pivoted about
(a) linear momentum of the system does not change its axis of symmetry. A block of mass m tied to the
in time string rests on a support so that the string is slack.
(b) kinetic energy of the system changes in time the block is lifted upto a height h and the support is
(c) angular momentum of the system does not removed. (shown in figure)
change in time
(d) potential energy of the system does not change
in time R
36. A rod of length ‘l’ is pivoted smoothly at O is
resting on a block of height h. If the block m
moves with a constant velocity V, pick the m h h
current alternatives.

 V h
38. What is the angular velocity of cylinder just
before the string becomes taut
O 2gh gh 2 gh
(a) zero (b) (c) (d)
R R R
V cos ?
(a) angular velocity of rod is 39. When the string experience a jerk, a large
h impulsive force is generated for a short
(b) angular acceleration of rod is duration, so that contribution of weight mg can
2V 2 cos3 ? sin ? be neglected during this duration. Then what
will be speed of block m, just after string has
h2 become taut
V cos 2 ? 2gh 2gh gh gh
(c) angular velocity of rod is
h (a) 1  M  (b)  M (c) 1  M  (d)  M
(d) tangential velocity of free end of rod is  m  1  2m   m  1  2m 

lV cos 2 ? 40. If M = m, what fraction of KE is lost due to

the jerk developed in the string
h (a) 1/2 (b) 2/3 (c) 1/3 (d) 1/4
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ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III
Passage-II : Passage - IV :
A man of mass 100 kg stands at the rim of a turn . Three particles each of mass m can slide on fixed
table of radius 2m, moment of inertia 4000 kg. The frictinless circular tracks in the same horizontal plane
table is mounted on a vertical smooth axis, through as shown . Particle m1 (= m)moves with veocity
its center. The whole system is initially at rest. The v0 and hits patricle m2 (= m), the cofficient of
man now walks on table with a velocity 1m/s relative
restitution being e = 0.5. Assume that m2 and
to earth
41. With what angular velocity will the turn table m3 (=m)are at rest initially and lie along a radial line
rotate before impact, and the spring is initially unstretched.
(a) 0.5 rad/sec (b) 0.1 rad/sec V0
m2
(c) 0.05 rad/sec (d) 0.2 rad/sec m1
42. Through what angle will the turn table have k
rotated when the man reaches his initial m3
position on it 2R

 3
(a) rad/sec (b) rad/sec R
11 11
2 4
(c) rad/sec (c) rad/sec
11 11 46. Velocity of m2 immediately after impact is
43. Through what angle will it have rotated when v0 3v0
v0 3v0
the man reaches his initial position relative to (a) (b) (c) (d)
4 42 2
earth
47. The maximum veloctiy of m3 is
 2
(a) rad/sec (b) rad/sec 3 3 3 3
5 5 (a) 5 v0
(b) 10 v0 (c) v0 (d) v0
4 2
2  48. The maximum stretch of the spring is
(c) rad/sec (d) rad/sec
11 11 3 m 3 m
Passage-III : (a) v0 (b) v0
4 5R 2 5R
A homogeneous rod AB of length L and mass M is
hinged at the centre O in such a way that it can (c) v0
3 m
(d) 3 v m
rotate freely in the vertical plane. The rod is initially 5 5R
10
0
5R
in horizontal position. An insect S of the same mass Passage - V :
M falls vertically with speed V on point C, midway A plank of length 20 m and mass 1 kg is kept on a
between the points O and B. Immediately after horizontal smooth surface. A cylinder of mass 1kg
falling, the insect starts to move towards B such is kept near one end of the plank. The coefficient
that the rod rotates with a constant angular of friction between the two surfaces is 0.5. The
velocity . plank is suddenly given a velocity 20m/s towards
left.

m m

A B
O C l = 20m
44.Calculate angular velocity in terms of V and L 49. Which of the following statement is correct?
12V V 7V 3V (a) Intitial acceleration of cylinder is 5m / s 2 towards
(a) (b) (c) (d) left
7L L 12 L 2L
45. If insect reaches the end B when the rod has (b) Initial acceleration of cylinder is 5m / s 2 towards
turned through an angle of 900 calculate V right
(c) Initial acceleration of cylinder is 10 m / s 2 towards
interms of L
right
3 7 1 2 (d) Initial acceleration of cylinder is 10 m / s2 towards
(a) 2gL (b) 2gL (c) gL (d) 2gL
7 12 12 7 left

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50. Which of the following statement is correct? 57. Choose the correct statement
(a) Pure rolling of cylinder takes place immediately (a) Linear momentum of disc is conserved as the
(b) Intitally cylinder slips and then pure rolling begins spring force is always perpendicular to velocity of
(c) Pure rolling never begins disc.
(d) There is no lose in kinetic energy during its entire (b) Angular momentum of disc about fixed end of
motion. spring is conserved.
51. Velocity of plank when pure rolling begins is (c) Kinetic energy of disc is conserved
(a) 10m/s (b) 1.5sec (c) 20m/s (d) 25m/s (d) Angular velocity of disc remains constant
52. Time in which plank and cylinder separate 58. In the subsequent motion of disc, maximum
(a) 1 sec (b) 1.5 sec (c) 2.5 sec (d) 2 sec elongation of spring is l0 /10 . The velocity of
Passage - VI : disc at this instant is:
A ring of radius R is rolling purely on the outer (a) 11 m / s (b) 10 m / s
surface of a pipe of radius 4R. At some instant, the (c) 5 m / s (d) 7 m / s
centre of the ring has a constnat speed of v.
59. What is the force constant of spring?
53. The acceleration of the point on the ring which
is in contact with the surface of the pipe is (a) 210 N / m (b) 100 N / m
(a) 4v 2 / 5R (b) 3v 2 / 5R (c) v2 / 4 R (d) zero (c) 110 N / m (d) 200 m / s
54. The acceleration of the point on the ring which Passage - IX:
is farthest from the centre of the pipe at the A thin uniform rod of mass m and length L is hinged
given moment is : at one end and from other end a light string is
(a) 4v 2 / 5R (b) 3v 2 / 5R (c) 3v 2 / 4R (d) 6v 2 / 5R attached. The string is wound over a frictionless
pullely (having mass 2m) and a block of mass 2m
Passage - VII :
is connected to string on other side of pulley as
A uniform rod of mass ‘m’ and length L is released
shown. The system is released from rest when the
from rest, with its lower end touching a frictionless
horizontal floor. At the initial moment, the rod is rod is making an angle of 37 0 with horizontal.Based
inclined at an angle of 300 with the vertical. on above information answer the following
55. Then,the value of normal reaction from the questions:
floor just after release,will be:
a) 4mg/7 b) 5mg/9 c) 2mg/5 d) mg/5 H2
56. In the above problem, the initial acceleration
of the lower end of the rod will be:
a) g 3 / 4 b) g 3 / 5 c) 3 g 3 / 7 d) g 3 / 7
mL
Passage-VIII:
One end of an ideal spring of unstretched length 37° 2m
lO  1m , is fixed on a frictionless horizontal table.
H
The other end has a small disc of mas 0.1 kg attahed
to it. The disc is projected with a velocity 60. Just after release of the system from rest,
acceleration of block is
?0  11 m / s perpendicular to the spring:
72 g 48 g
v1 (a) , downwards (b) , downwards
121 119
11l0 90 g 90 g
(c) , downwards (d) , upwards
v0 10 121 121
61. Just after release of the system, the resultant
O force exerted by hinge on rod is
l0 (a) 0.7mg (b) 0.92mg (c) 0.53mg (d) mg

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ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III
62. Just after release of the system from rest, the 65. A rigid body of mass M and Radius R rolls
resultant force exerted by hinge H 2 on pulley without slipping on an inclined plane of
is inclination under gravity Match the type of
body with magnitude of the force of friction.
46
(a) mg in upward direction C olum n I C olum n II
121
Mg sin
46 a) For ring p)
(b) mg in downward direction 2.5
121
Mg sin
438 b) For solid sphere q)
(c) mg in upward direction 3
121
Mg sin
438 c) For solid cylinder r)
(d) mg in downward direction 3.5
121
Mg sin
MATRIX MATCHING QUESTIONS d) For hollow s)
63. For the following statements, except gravity 2
and contact force between the contact spherical shell
surfaces, no other force is acting on the body. 66. A rigid body is rolling without slipping on
Column I Column II horizontal surface.At given instant BD is
(a) When a sphere is (p) Upward direction perfectly horizontal and CD is perfectly
in pure–rolling on a vertical.
fixed horizontal surface. C
(b) When a cylinder (q) vcm > R   = v/R
is in pure rolling on
a fixed inclined plane in upward D R
B
direction then friction force acts in
(c) When a cylinder is (r) vcm < R 
in pure rolling
down a fixed incline plane, Column I Column II
friction force acts in
(d) When a sphere of (s) No frictional force a) Velocity at pointA, vA p) v 2
radius R is rolling acts b) Velocity at point B, vB q) Zero
with slipping on a fixed horizontal
surface, the relation between vcm and is c) Velocity at point C, vC r) v
(t) Work done by the d) Velocity at point D, vD s) 2v
frictional force is zero
67. A horizontal table can rotate about its axis. A
64. A uniform disc is acted upon by some forces
and it rolls on a horizontal plank without block is placed at a certain distance from
slipping from north to south. The plank, in turn center as shown in figure. The table rotates
lies on a smooth horizontal surface. Match the such that block does not slide. Select possible
following regarding this situation : direction of net acceleration of block at the
Column I Column II instant shown in figure. Then match the
(a) Frictional force on (p) May be columns.
the disc by the surface directed towards 4
north
(b) Velocity of the (q) May be directed
lowermost point of the disc towards south 3
(c) Acceleration of (r) May be zero
centre of mass of the disc
(d) Vertical component (s) Must be zero 2
of the acceleration of 1
centre of mass
92 NARAYANAGROUP
 JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS
Column I Column II Column I Column II
a) When rotation is p) 1
1. No friction between p. Angular acceleration
clockwise with constant pulley and string, and of pulley is 0
b) When rotation is q) 2 moment of inertia of
clockwise with decreasing pulley is not
negligible.
c) When rotation is r) 3
clockwise with increasing 2. Friction is there q. T1 = T2
d) Just after clockwise s) 4 between pulley and
string, and pulley is
rotation begins form rest light.
68. An uniform disc rolls without slipping on a 3. Friction is not there r. T1  T2
rough horizontal surface with uniform angular and pulley is light.
velocity. Point O is the centre of disc and P is D. Friction is there and s. Angular
a point on disc as shown. In each situation of pulley is having acceleration of
some moment of pulley  0
column I a statement is given and the inertia.
corresponding result are given in column –II.
Match the statements in coloumn-I with the 70. A smooth ball of mass m moving with a uniform
results in column-II velocity v0 strikes a smooth uniform rod AB
of equal mass m, lying on a frictionless
P horizontal table. The ball strikes the rod at
one end A, perpendicular to the rod, as shown
in the figure. The collision is perfectly elastic.
O
Some physicaL quantities pertaining to this
situation are given in COLUMN-1 while their
values are given in COLUMN-2 in a different
order . Match the values in COLUMN-II and
Column I Column II the quantities in COLUMN-I
a) The velocity of p) Change at point P in m A
disc magnitude with time v0
b) The acceleration q) Is always directed of
point P on disc from that point not m
towards centre of disc.
c) The tangential r) is always zero
acceleration of point B
P on disc Column-I Column-II
d) The acceleration s) is non-zero and
of point on remains constant in 2
A) Final kinetic energy of ball p)
disc which is in contact magnitude 5
with rough horizontal surface Initial kinetic energy of ball
69. A light string is wrapped on a pulley and two B) Impulse delivered to the rod
blocks of masses m1 and m2 are attached to 3
Initial momentum of ball q)
5
free end of string as shown in figure. T1 and C) Angular momentum of rod about its
T2 are the tension in string on two sides of centre of mass
pulley. In column I, some information is Initial angular momentum of the ball about
mentioned about friction between of inertia of 9
the centre of mass of the rod r)
pulley, while in colunm II the effect of the 25
information mention in column I on the motion Final kinetic energy of rotation of the rod
of system is given. Match the entries of Final kinetic energy of translation of the rod s) 3
column I with the entries of column II. D) Final kinetic energy of rotation of the rod
Final kinetic energy of translation of the rod
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ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

INTEGER TYPE QUESTIONS LEVEL-V - KEY

71. A rod of mass m and length  is released SINGLE ANSWER QUESTIONS
from rest from vertical position as shown in
1.D 2.C 3. A 4.D 5. D 6. A 7. C
the figure. The normal force as a function of  ,
which is exerted on the rod by the ground as it 8. D 9. A 10.D 11. D 12. D 13. B 14. A
falls downward, assuming that it does not slip 15. C 16. D 17. A 18. D 19. A 20. C 21. A
22. A 23. B 24. D 25. C 26. A 27. B
 3cos   1 
2

is mg   then n = MULTIPLE ANSWER QUESTION

 n 
28. C,D 29. A,C,D 30. A,B,C,D
31. C,D 32. A,C,D 33. A,C,D
34. A,B,C 35. A,B 36. B,C 37.B,C
 COMPREHENSION QUESTIONS
38. A 39. B 40. C 41. C 42.A 43. A 44. A
45. B 46. B 47.B 48. A 49. B 50. B 51. B
72. One end of a uniform rod of mass M and 52.B 53. A 54. D 55. A 56. C 57. B 58. B
length L is supported by a frictionless hinge 59. A 60. A 61. C 62. C
which can with stand a tension of 1.75 Mg. The MATRIX MATCHING TYPE
rod is free to rotate in a vertical plane. The
63. A  s, t; B  p; C  p; D  q, r
maximum angle should the rod be rotated from
the vertical position so that when left, the hinge 64. A  p, q, r; B  p, q, r; C  p, q, r; D  s
65. A  s; B  r; C  q; D  p
does not break is 66. A  q; B  p; C  s; D  r
n
67. A  r; B  s; C  q; D  p
73. A thin uniform bar of mass m and length 2L is 68. A  p; B  q, s; C  p; D  q
held at an angle 300 with the horizontal by 69. A  p, q; B  q, s; C  p, q; D  p, q, r, s
means of two vertical inextensible strings, at 70. A  r ; B  P ; C  P ; D  S
each end as shown in figure. If the string at INTEGER TYPE QUESTIONS
the right end breaks, leaving the bar to swing
the tension in the string at the left end of the 71. 2 72. 3 73. 4 74. 5
bar immediately after string breaks is LEVEL-V - HINTS
n DETIAL SOLUTINS
T mg
13 SINGLE ANSWER QUESTIONS

a
 2m  m  g  g
1. for T in lower string ,
3m 3
g
T  2mg  2m  4mg / 3
3
30° 8mg
&T 1  2T 
R 3
74. A uniform sphere of radius starts rolling
16 8mg 11mg
T 1  mg   mg 
down without slipping from the top of another 3 3
sphere of radius R = 1 m. The angular velocity Let has take movent about hinge
of the sphere in rad s 1 , after it leaves the 11mg x 11
surface of the larger sphere is 8 x n. Where n 4mg  x  y 
3 y 12
= --.

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 JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS

x 9. When the thread is pulled, the bobbin rolls to the

2. mg sin  x  mg cos  right. Resultant velocity of point B along the thread
2
(taking moment aobut right bottom corner) is  0 sin  r , where 0 sin is the
1
component of translation velocity along the thread
 tan  and r linear velocity due to rotation. As the
2
bobbin rolls without slipping, 0  R. Solving the
3. ma cos  f  mA  (1)
f  r  I   2  ; A  r   3
R
obtained equations, we get  .
R sin r
0

a
From about equation A  1 3 3  l  3 l3
2

3 10. I  mR  mR  mR  l    2
2 2 2

2 2 2 2  8
a 1 2 11. L  mVR, hear V dicreases so L never be
A  where  60 & I  mr 
0

3 2  constant, Also ar & at both are acting & thind point

 
4.
contradicts the given question. But direction L is

ma sin always constant.
N 12. There is neither torque nor angular momentam about
f the O. (because line of action passes through O)
O So u  0
ma p
mg cos 13. Form given figure we get
mg sin N  F sin  mg  1 ; F cos  f  ma   2 
fR  I   3
mg 

ma sin Slove above equation we can easily get

mg sin  ma cos  f  ma .....(1)  0.08  3

14. Block will be tipped about point O.
N  ma sin  mg cos .....(2)
f  N ..........(3) M O   F cos 300  a  F sin 300 L  mgL / 2

a 
3 mgL
 mg sin  mR 2  ......(4) F
 p
2 3L
a  R ........(5)
Solving above equitions we get m For F  0, a 3  L ; a  L / 3.
15. At highest point, velocity  V cos 450  V / 2
5. We use work energy principal
V V 2 sin 2 450 V 2
1 2mg sin Lm .r  ; Here r  h  
mg sin  x  kx2  0  x  2 2g 4g
2 k
 given  ; So, m  I2
6. Liquid gets only tracanslatory motion in side any mV 3 IV 3
L 
1 2 4 2 g 2 2 gR 2 R
rollny mosum. This kE  mv
2

2 1
I mR 2 . It is a disc.
 taking component  2
L  mv sin  l  
7.  durring strike 
 mv sin l  0
8. The force is conservative. so ME is conserved V cos 45°

(friction less) & Angular momentum  L  I 

h
decreases due to reverse swing compare to that of
initial swing before strike.
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ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

16. Hear force is conservative  Fg  . Thus energy is

20. It may happen if friction is upwards and disc
acceleration downwards. The equations of motion
conserved. & No external torqued. L  constant. are


 As  30 0 
17. N1  F1 and fi  N1  F1 mg sin  f g f
a   ;
f 2  F2 ; F  f1  f2  ma  1
m 2 m
fR 2f
 f  f  R  5  f1  f 2  R
 1 2 
5  f1  f 2  
I

 mR / 2  mR
2

I 2mR 2 2mR
For no slipping a  R
a  R  5  f1  f 2  / 2m
g f
From eq.(1),   2 f / m or f  mg / 6
2 m
5  f1  f 2 
F  f1  f 2  mg 3
2 f  f max ;  mg cos 300  mg
6 2
f1  f 2  f1  f 2   7 F1  3F2 
5 5
F 1
2 2 2 
3 3
F

V
Vy
f1  f2
y
21.


18. In case of pure rolling,torque due to friction about

contact is zero. y  R tan ?
 9 
; Vy  R sec 2 ?  ? 
9 dy dv
19. Torque =  mg  L   R sec2 ? 
10  20  dt dt
m 9  3g v
Vy  R sec 2 ?    V sec 2 ?
2

I   L ; a
3  10  2L R
Acceleration 22. Here the point B is following a curvilinerar path
3 g  9 L  27 g whose velocity at any instant is given by vector sum
aCM   AC    
2 L  20  40
 
 ?L
of V and . When rod rotates by an angle
9 2
NA  mg
40 ? , the situation is as shown in the figure. The speed
(9L/20) of end point at this instant is
A C D A
9 
10
 V

9 L 
mg 2
10

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25. Let n1 rotations performed upoto a cenrtain time

?L
 cos  p  ? 
?L
2

v1  v     2v 
2
n1   /2 
 2  2

 1  cos ?  
1

2 2  2  1
v1  2v n2  n1   1    .
3 32  5  2
1

? ?  2 .....(1) and
 2v  2 cos  2v cos
2 2
1 0 1

2 2 2 
0  2  1  ....(2)
2
?L ? ds d? ? 5 
v1   2 cos ;   L cos
1

2 2 dt dt 2 From eqns. (1) and (2)

 ds  L 
? 2L 4 5 02
cos d?   2L  2 
s p /2
 ;
1 2
0 0 2 2 5
0 1
14 1

23. V , CD  ? 2  r2  r1  ; V , CD, AB  ? 1r1

  26. AC  a 2
The plate falls by AO  a / 2
  
VP , AB  VP , CD  VCD , AB I I  I2
I (diagonal DB )  z  1
 ? 2  r2  r   ? 1r1
2 2
C
Angular velocity of P about AB
vPAB ? 2  r2  r1   ? 1  r1  a
?  
r2 r2
D B
24. N  mg

P I2 I1
A
N1
 com
f = N2 1  Ma 2  Ma2
= 2 3  6
 
mg Q N2
; I A  I DB  M  AO 
Ma 2
Since I1  I 2 
2

  2
& N1  N 2 Ma 2 2
 M a/ 2    Ma 2  Ma 2
2

Take the torque about the com 6 3 3

1 l l a 1 12
N1 sin  N 2 sin  N 2 cos Mg  I 2
   Ma 2 2

2 2 2 2 2 23
1 l  3g 
N1 sin  2  N 2  cos  3g 
1/2 1/2

2 2 Which gives   =  
 2a  a 2 
1
tan  ; sec 2  1  27. Linear momentum is conserved, mv  mv ' .Angular
2 2 momentum abouty centre of the disc.
l2 l 0.25 17 mr 2 2vh
sec  1  ;  1 ; L mt  mvh   I  or 
4 2 4 2 2 r2

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ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

2vh
For pure rolling v  r 
N
r
or h = r / 2 , for pure rolling. 0
m vr = 

MULITPULE ANSWER 31. r 

28. Self explanatory mg

1 2 Using conservation angular momentum and equation
29. KET  mv ..............(1) & I  mv ....(2)
2
r 0
I 1 of kinematics t  so t depends on
3 g
2
so KET  ; again KER  I .......(3)
2

2m 2
r
2 2

l 1 6I s 0
; if depends on .
we have I   ml    18 g
2

2 12 ml
Again at bigining condition v = 0 and 
1 1 2 36I 2 3I 2
0

Thus KE   ml  2 2  both indipendent to

2 12 ml 2m
R
Again work by friction
2I 2 W f    mgr    gr
and KE  KET  KER  slipping
m
30. r
2 2
r
2 2

2 g 2
f2
Thus it is independent to
32. Initialy 0 is not changed. After that 0 decreases
because I 0  I' '
initialy I 0 about B & later
N2
0

I ' is about com.

N1 Then kinetic energy and angular momentum both
are conserved due to no external torque.
mg  33. Let look synopsis given.
f1
v '  ev    , e  1 so v '  v hence a, c, d are
N1 = mg  f 2 .............(1); N 2  f1 ........ (2) correct
34. If we take moments about an axis through the
Now taking moment about A
center of the sphere. Only f can have a
torque.   0 ; Take  Fy  0 yields
l
N 2  l sin  f 2  l cos  mg  cos .....(3)
2
Take moment about point B N 2 cos 300  mg  10 kg   9.8m / s 2 

F
l
N1  l cos  mg  cos  f1  l sin ...(4)  0 yields N 2 sin 300  N1  0
2
x

or N1  56.5N , N2  113 N
 of normals  =
0
35. As external force is zero,P is constant but K.E
mg increases.
N1l cos  N 2  l sin 
2 Due to external torque ,angular momentum
changes. P.E also changes.
mg
l cos  cot  2mgl cos / 2
2 h V cos2 ?
36. V cos ?  ? ; ? 
B,C & D are easily explain cos ? h
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2K 0
40. For M  m,K1  , so the fraction lost is
3

V  K 0  K1  1
 K   3
Vcos

0
h
41 to 43
41. By conservation of angular momentum on the man-
table system,
d? 2V 2 cos 3 ?.sin ?
 L i  L f or 0  0  I m m  I t t
 
(B) a 
dt h2
I m m v 1
37. Acceleration is given by 2 r and it increases if t      rad / s
I1 where m r 2
moved from centre to periphery. Accelerations of
all end points will be same. If is varying
 t  100  2 
1/ 2 1
acceleration will be  r    r 2   . On
  rad / s
2

2 1/2
4000 20
Thus the table rotates clockwise (opposite to man)
moving from centre to periphery acceleration with angular velocity 0.05 rad/s.
(When is varying) will increses 42. If the man completes one revolution relative to the
COMPHRENSION TYPE QUESTIONS table then  mt  2;2   m   t
38 to40
2   m t   t t (where t is the time taken)
38. Just before the string becomes taut, the block falls
freely, so v 0  2gh. There is no tension in the t  2 /  m   t   2  0.5  0.05
string, so nothing causes the cylinder to spin, so Angular displacement of table is
 0  0. 2
 t   t t  5.05   2 / 0.55   radian
39. When the string experiences a jerk, the large 11
impulse developed is of very short duration so that 43. If the man completes one revolution relative to the
the contribution of weight mg can be neglected earth, then  m  2
during this time interval.
The angular momentum of the system is conserved, 2 2
as the tension is internal force for the system. Thus time =   0.5
m
we have Li  Lf
 
During this time, angular displacement on the table,
1 2
mv1R  MR 2 1  mv 0 R  m 2ghR
2  t   t  time   0.05  ,
0.5
The string is inextensible, so v1  R1 . On solving 
 t   radian ,  t  36° in clockwise direction
2gh 5
for 1 we get 1  R 1   M / 2m  44 to 45
 
A angular momentum of system of rod and insect,
2gh just after collision = intial angular momentum of
v1  R1 
1   M / 2m  insect about o
The final kinetic energy K1 is given by  ML2  L   L
2

 M    M
1 1 1 11 2   v1   4  
 12 4
2

K1  mv12  I12  mv1   MR   2 

2

2 2 2 2 2 R 

1 M 2 1  mv02 
  m   v1  
2 1   M / 2m 

2 2 0 
12V


K0  1  x B 7L
 K o  mvo2 
1   M / 2m 
, 
2  Mg

NARAYANAGROUP 99
ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

Let the insect fall on the rod at time t  0, then at MR 2

49-51: f R  ; f  5 N ; M  1kg ;
time t, inclination of rod with horizontal,  t 2
 Angular momentum f
R  10 m / s2 ; a1   5m / s 2 ; a2  5m / s 2 ;
 ML 2 M
LI   Mx 2  For acontact  a1 ; vcontact   a1  R  t  15t .... 1
 12 
Torque produced by weight of insect , a1 
 Mg x cos = Mgx.cos t
d a2 f
 L ; F= N   0.2 f
dt
46-48 :First we apply conservation of momentum on For plank vcontact  20  5t ......... 2
m1 and m2 : m1 v0  m1v1  m2 v2
From eqns.1 and  2 , we get
v0  v1  v2 .............. 1 As m1  m2
t = 1 sec
Where v1 and v2 are veolcities of m1 and m2 Till the cylinder sllips on plank
immediately after impact.
From definition of coefficient of restitution, 1 1
Srel  urel  are1t2 ,  20 1 10 12  15m
v v 2 2
  2 1 or v0  2  v2  v1  ........ 2 
1
e Velocity of plank when pure rolling begins
2 0  v0
v = 20 - 5t = 15m/s
On sovling eqns. 1 and  2 ,
Velocity of cylinder  a1t  5m / s
v0 3v When pure rolling begins, friction force vanishes,
we get v1  and v2  0
4 4 velocity of plank and cylinder is constant. After pure
The spring has maximum extension when angular
rolling beings.
velocity w of m2 and m3 about O is same. Now we
apply conservation of angular momentum.
 v2 
a AC   
51-54: 5R ; a BA  a BG  a AG ;
3 
m  v0  2R  mR 2  m  2 R  o r
3 v0
 ;
2

 4  2 5 R D
3
Velocity of m2  2 R  v0
5 v aAG
A
3
Velocity of m3  R  v0 B
10
m2 4R
2R
aBA
C

v2
v3 = 0
m3  v2  v 2 v 2 4v 2
R a BA  ; a BG    ; similarly
R R 5R R
O  v 2 v 2 6v 2
a DG   
In order to determine maximum extension in the R 5R 5R
spring, we apply law of conservation of energy. 55-56: From constraint condition
1  3  1 2  3 v0  1 2 3 v  1
m  v0   mR    m  2R   0   k xmax
2
L
2 4  2  2 5R  2  2 5R  2 ; 0 a  0
sin 30
2
.........1 ; mg N ma...2 ; N sin300   ........... 3
L L  mL
2
3 m a
On solving for x , we get x 2
 v0 . 4 2  12
4 5R
max

100 NARAYANAGROUP
 JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS
Let a be the angular acceleration of rod about hinge
 and a be the acceleration of block just after the
L system is released from rest.

2 30°  The free body diagram of rod, pulley and block
are shown in figure. In free body diagram of rod,
a acceleration of cm the force exerted by hinge on rod is not shown.
L
For rod, mg  cos 370  T  L cos 37 0
c 2
From eqns (1), )2) and (3)  mL2 
12 g 3g 4 mg   Ia  I  
we get  ;a  ;N   3 
7l 7 7
Acceleration of lower end of rod is For block, 2mg  T  2 ma
L
 
L 12 g 3 ˆ 3 3g ˆ
    From constraint, a  La cos37 
4 La

ax  cos 300 iˆ    i  i 0
2 2 7l 2 7
5
57-59
Solving above equations, we get
v1
72 g 90 g 98mg
a ,a  ,T 
11l0 121 121L 121
v0 10 Reaction force exerted by H 2 on pulley
438
is, N1  2T  2mg 
O
l0 mg
121
Angular momentum of disc about fixed end is Now draw the complete free body diagram of rod
conserved, as the spring force passes through O as follows
 11l  10? L T
m?0l0  m?1  0  ; ?1  10 m / s  0 T cos37°
 10 
2 T sin37°
11
From conservation of energy we get mg sin37°
1 1 1 l 
2
R1
mV02  mV12  K 2  0 
2 2 2  10  R2 mg mg cos37°

 K 2  210 N / m mL
 R1  T cos37  mg cos 37  a
0 0
59- 62 2
 R2  T sin 370  mg sin 370
N1

Net reaction force on rod due to hinge is,

F  R12  R22
T

MATRIX MATCHING TYPE

T T

 2mg 63. At the time of pure rolling,friction = 0and work by

mg
friction is zero.
Due to downward reletive motion the direction
friction is upwards.
a
Due to downward reletive motion, friction is
upwards.
2mg
There are two cases possible,q and r
NARAYANAGROUP 101
ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III
64. There are possibilities of p,q,r that dependence on
relative velocity of contact point. The velocity of    
 V  2   v 
COM is always parallel to the motion of rolling a 1     
0  v0   V  2  v  v0

body. 
mg sin  
f 
 mr 2  ; mg sin
65. 1   For  w
2  v  v0  V  .......(4)
 I  2
mg sin      m 
For hollowsphere f 
2

2.5 mv
Also 0  2   mv 
 2   12 
     
mg sin
For solid cylider f 
3  v0  v 

.......(5)
6
mg sin
For solid sphere f 
3.5 From (3) and (5) we get v 

6
66. V p  2V sin 2
So, from (4), we get v  v0 
2
67. Angular velocity is constant,tangential acceleration
 ......... (6)
3
is zero.But radial acceleration does not zero.
12v0
68. (a) at point P ,tangential velocity changes with time. Solving (5) and (6), we get 
At the point contact of pure rolling frictional is 5
zero.So acceleration is towards centre. v0 V
F F
69. A  As there is no friction between pulley and
rope and pulley so T1  T2 and torque on pulley 
V
so angular acceleration is 0.
B  Since there is friction so acceleration of pulley
is non - zero. So, by torque equation
just before During Just after
T1  T2  I impact impact impact

But as I  0  T1  T2 3v0 2v
C  Since pulley is light, no friction between rope So, we get v  ,V  0
5 5
and pulley so T1  T2 and torque on pulley is 0 1 2
mv
and  0. 
Final KE of ball 2
 
9
Initial KE of ball 1 mv 2 25
D  Friction is there between pulley and rope
2
0

and moment of inertia of pulley is not negligible.

So, pulley has angular acceleration and by equation Im pulse Delivered to rod 2
 
T1  T2  I Initial Momentum of ball ( mv0 ) 5
 T1  T2
This condition is valid only when M 1  M 2  L1 rod
 m 2 

I
L   
 L1 ball about CM
12 

2
If M 1  M 2 then T1  T2  mg and 0   L 
 1 ball
1 rod

mv0    5
70. mv  mv0   Fdt   J
mv0  
......(1) 2 2
mV  J .......(2)
 m2 
mv0  mv  mV  K rod rotation  12 
   3
 v0  v  V ........(3)  K rod translation mv2

102 NARAYANAGROUP
 JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS
71. At angle 
TL cos 300 3 3T
  
m  2L 
1 2 3g
I   mg (1  cos ) or, 2  (1  cos ) I 2ml
 2
2 2 
…(i) 12
Differentiating w.r.t. q, now just after the string breaks acceleration of point
mg

sin 
A in vertical directon should be zero solving above
 2 3 g sin  equations we get
m 2  … (ii)
2 
3 4 6 3g
T mg and 
 2 3g 3 13 13 L
an    (1  cos ) and at    g sin 

2 2 2 4 74. The equation of motion for the centre of the sphere
f  max  m (at cos   a n sin ) at the moment of breaking off, N  0 is
3 3 g sin   mv 2
 m  g sin  cos   (1  cos )   mg cos
4 2  Rr

3 3 
mg sin   cos   1
where v is the speed of the centre of the sphere at
2  2  that moment and is the coresponding angle. The
speed v can be found by using the Law of
C x
conservation of energy,
an 
at y

r
Further, m g  N  m ay or,, h
N  m ( g  at )
N  m [ g  (at sin   an cos )]
R 
 3 3 g cos  
 m  g  g sin 2  (1  cos ) 
 4 2 
mg mg
 [4  3 sin 2   6 cos   6 cos2 ]
4
mg
 (1  3cos )2 .
4
The rod does not slip until N  0
1 mv2 I 2
i.e., cos . According to which mgh = 
1

3 2 2
2M  L  2 2
2
72. 1.75Mg  Mg    (i) where I  mr , v  r and
L  2  5
h   R  r 1  cos 
1  cos   
L 1 ML2 2
Mg  (ii) From the equations we get
2 2 3
Solving (i) and (ii) we get  = 60º 10 g  R  r 

17r 2
ax  0
; a   Fy  mg  T
  40rads 1 .
73. F x 0
m m
y

NARAYANAGROUP 103
ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

LEVEL-VI k
SINGLE ANSWER QUESTIONS c
m
1. The point P of a string is pulled up with an Q
acceleration g. then the acceleration of the Q Q
hanging disc (w.r.t ground) over which the
string is wrapped, is (a) C  0 for ct  t0 (b) f  0 for t  t0
ma
P (c) x  , where x = deformation of the spring
2g g k
(a)  (b) 
(d)  KEmax  ma2t02 , where  KE max is the maximum
3 3 1
2
KE of the rolling body
m 4g g
(c)  (d) 
3 3 5. A linear impulse  Fdt acts at a point C of the
2. A sphere of mass m1 is placed on a plank of smooth rod AB . The value of x is so that the
end A remains stationary just after the impact
mass m2 . The coeffcient of friction between is :
the plank and sphere is . If the inclined plane A
is smooth, the frictional force between the
plank and sphere : (a)
l
(b)
l
4 3
O

m1  x
m2 C
 l l
Fdt (c) (d)
6 5
 B
6. Two light vertical springs with equal natural
(a) depends on m1 (b) depends on m2
lengths and spring constants k1 and k2 are
(c) 0 (d)  m1 g cos sparated by a distance l . Their upper ends
3. Four beads each of mass m are glued at the are fixed to the ceiling and their lower ends to
top, bottom and the ends of the horizontal the ends A and B of a light horizontal rod AB.
diameter of a ring of mass m . If the ring rolls A vertical downwards force F is applied at point
without sliding with the velocity v of its , the C on the rod. AB will remain horizontal in
kinetic energy of the system (beads +ring) is: equilibrium if the distance AC is :
m

m m
k1 k2

(a) 5mv 2 (b) 4mv 2 (c) 2mv 2 (d) mv 2

4. A rolling body is connected with a trolley car x
by a spring of stiffness k . It does not slide A C B
and remains in equilibrium relative to the
accelerating trolley car. If the trolly car is
l lk1 lk 2 lk2
stopped after a time t  t0 :( the rolling body (a) (b) k  k (c) k (d) k  k
2
touches the trolly)
1 2 1 1 2

104 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS
7. Let I be the moment of inertia of a uniform 11. A ring of mass M and radius R lies in x-y
square plate about an axis AB that passes plane with its centre at origin as shown. The
through its centre and is parallel to two of its mass distribution of rings is non-uniform such
sides. CD is a line in the plane of the plate that at any point P on the ring, the mass per
that passes through the centre of the plate and unit length is given by  0 cos 2 ( where
makes an angle with AB. Then the moment
of inertia of the plate about the axis CD is equal 0 is a positive constant). Then the moment
to : of inertia of the ring about z- axis is
(a) I (b) I sin 2 (c) I cos 2 (d) I cos2  / 2 y

8. Two point masses A of mass M and B of mass M (a) MR 2 (b)

1
MR 2
4M are fixed at the ends of a rod of length l R P 2

and of negligible mass. The rod is set rotation
x

about an axis perpendicular to its length with MR MR

(c) 2 (d) 5
a uniform anguular speed. The work required
for rotating the rod will be minimum when the 12. As shown in figure, the hinges A and B hold a
distance of axis of rotation from the mass A is uniform 400 N door in place. the upper hinge
at supports the entire weight of the door. find the
2 8 4 l resultant force exerted on the door at the
(a) l (b) l (c) l (d)
5 5 5 5 h
9. A spool of mass M and internal and external hinges . the width of the door is, where h is
radii R and 2R hanging from a rope touches a 2
curved surface, as shown. A block of mass m the distance between the hinges.
plaed on a rough surface inclined at an angle
with horizontal is attached with other end of y
the rope. The pulley is massless and system
is in equilibrium. Find the coefficient of friction M
3mg  2 Mg A
m (a) 3mg  2 Mg
3mg sin  2 Mg cos
 (b) 3mg cos  2Mg sin 400N
B
3mg cos  2 Mg sin
(c) 3mg sin  2Mg cos
3mg  2 Mg tan (a) 312 N (b) 280 N (c) 412 N (d) 480 N
(d) 3mg  2Mg tan
13. A thin wire of lenght L is is bent into a circular
10. A ring of mass m and radius R is rolling down wire of uniform linear density . When
on a rough inclined plane of angle with circular wire is in a vertical plane find the
horizontal. Plot the angular momentum of the moment of inertia of loop about an axis
ring about the point of contact of ring and the BC,pasing through centre of the loop and
plane as a function of time.
which makes an angle with the tangent at
L L
the topmost point of the loop
B
A
a) b)  

t t
L L

c) d) L3 L3 L3 L3
a) b) c) d)
t t 8 2
2 2 2
4 2
3 2

NARAYANAGROUP 105
ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

14. A box of mass 1 kg is mounted with two 17. A uniform rod AB of length three times the
cylinders each of mass 1kg, moment of radius of a hemisphered bowl remains in
inertia0.5kg m 2 and radius 1m as shown in equilibrium in the bowl as shown. Neglecting
figure, Cylinders are mounted on their control friction find the inclination of the rod with the
axis of rotation and this system is placed on a horizontal.
rough horizontal surface, the rear cylinder is r C
B
connected to battery operated motor which 
provides a torque of 100n-m to this vcylinder 3r
via a belt as shown. if sufficient friction is A
present between cylinder and horizontal
surface for pure rolling, find acceleration of (a) sin 1 (0.92) (b) cos1 (0.92)
m (c) cos 1 (0.49) (d) tan 1 (0.92)
the vehicle in . ( Neglect mass of motor,, 18. A particle of mass m is released from rest at
s2
belt and other accessories of vehicle). point A in the figure falling freely under gravity
parallel to the vertical Y-axis. the magnitude
Electric meter of angular momentum of particle about point
O when it reaches B is
m ( whereOA=b and AB=h)
O b
A
m m m m 
(a) 20 (b) 10 2 (c) 25 2 (d) 30 2
s 2
s s s
h
15. Two identical rings Aand Bare acted upon by
torques A and B respectively.A is rotating
about an axis passing through the centre of Y B
mass and perpendicular to the plane of the
ring. B is rotating about a chord at a mh
(a) (b) mb 2 gh (c) mb 3 gh (d) 2mb gh
1 bg
distance times the radius from the centre 19. The end B of the rod AB which makes an angle
2
with the floor is being pulled with a constant
of the ring. if the angular acceleration of the
rings is the same, then velocity V0 as shown in the figure. The length
(a) A  B (b) A  B (c) A  B of the rod is l . At the instant when  370
(d) Nothing can be said about A and B as data Y
are insufficient
16. A uniform plank of weight W and total length
2L is placed as shown in figure with its ends in A l
contact with the inclined planes. the angle.of
 V0
friction is 150 . determine the maximum value X
O B
of the angle a at which slipping impends.
W 2
(a) Velocity of end A is V0 downwards
L 3
L
 5 V0
(b) angular velocity of rod is
3 l
60°
45° (c) angular velocity of rod is constant
(a) 18.10 (b) 48.40 (c) 36.2 0 (d) 88.80 (d) velocity of end A is constant

106 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS
20. A block having equilateral triangular cross- 23. A uniform rod oflength l is released from the
section of side a and mass m is placed on a position shown in the figure. The acceleration
rough inclined surface, so that it remains in due to gravity is g . There is no friction at any
equilibrium as shown in figure. The torque of surfae. Find the intial angular acceleration of
normal force acting on the block about its the rod.
centre of mass is

60° 30°

 3 3g 5 3g 3 3g 5 3g
(a) (b) (c) (d)
10l 7l 11l 19l
3 1
(a) mga sin (b) 2 3 mga sin 24. Consider an arrangement shown in the figure.
2 The pulley P is frictionless and the threads are
massless. The mass of the spools is m and
1
(c) 2 3 mga cos (d) Zero 1
moment of inertia of the spool is mR 2 . The
2
21. A thin horizontal uniform rod AB of mass m mass of the disc of radius R is also m. The
and and length l can rotate freely about a surface below the spool is rough to ensure
vertical axis passing thorough its end A. At a pure rolling of spool. The mass of the block is
certian moment the end B starts experiencing m and the surface below the block is smooth.
a constant force F which is always Find the initial acceleration of the block when
perpendicular to the original position of the the system is released from rest.
stationary rod and directed in a horizontal
plane. The angular velocity counted relative P R Spool

to the intial position is

R/2

6F 6F
m

(a) sin (b) cos

ml ml
R
Disc Rough Smooth

8F 8F
(c) sin (d) cos
ml ml 4 2 8 10
22. Ablock of mass m moves on a horizontal circle (a) g (b) g (c) g (d) g
37 37 37 37
against the wall of a cylindrical room of radius 25. Find the moment of inertia of a hemisphere of
R. The floor of the room, on which the block mass M and radius R shown in the figure,
moves, is smooth but the friction coefficient
about an axis AA' tangential to the
between the wall and the block is . The block
hemisphere.
is given an initial speed V0 . The power A
developed by the resultant force acting on the  9 
(a) I    mR
2
block as a function of distance travelled s is R  20 
M

mV03 3R s  13 
(b) I  
m 03
3 s

(b)   mR
2
(a) e e
R R A'  20 
m V 03 mV03 3R s  7   3 
(c) (d) e (c) I    mR (d) I    mR
2 2

R R  20   20 
NARAYANAGROUP 107
ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

MULTIPLE ANSWER QUESTIONS (a) The rod is in translational and rotational

26. A wheel rolls purely between a rough horizontal equilibrium.
surface below it and a horizontal plank above (b) The rod is in rotational equilibrium only.
itunder the action ofa horizontal force F (c) The magnitude of the froce exerted by the rod
  on the pivot is 503N
applied on the plank. If at any time v p and vc
(d) The rod is in rotational equilibrium about P only
represent velocity of plank and velocity of 29. A light rod of length 4 m can be maintained in
centre of mass of wheel and equilibrium position as shown in the figure if
 
a p and ac represent acceleration of plank we apply single force on it.
and acceleration of centre of mass of wheel Y
repectively then which of the following is/are
4m
correct.
X
F Vp
37

2kg
Vc

(a) v p  2 vc (b) a p  2 ac
   
5kg

(c) v p  vb (d) a p  ac The required force

   
(a) would have magnitude of 77
27. A small block of mass m is released from rest (b) Would have a line of action making an angle of
from position A inside a smooth hemispherical
bowl of radius R as shown in figure. Choose tan 1 17 / 9  with negative x- axis
the wrong option(s) 48
(c) would be appiled at a distance of m from
17
A R the right end
(d) the rod cannot be maintained in equilibrium
under the action of a single force.
30. Two particles of masses m1 and m2 aree
B connected with a rigid rod of length l . If a force
(a) Acceleration to block is constant throughout F acts perpendicular to the rod then (a1 & a2
(b) Acceleration of block is g at A are instantaneous acceleration of m 1 & m2)
(c) Acceleration of block is 3g at B
m1 m2
(d) Acceleration of block is 2g at A
28. Consider a uniform rod of mass 40 kg and
length 8m, pivoted about a point P 3m from
one end as shown in the figure. Few external
forces are acting on the rod as shown in figure.
3m 5m
200 N 50 N F
 F
P (a) a2  0 (b) a1  m
1

F  m1  m2 
100 N 20 N
mark out the correct statement (s). F
(c) aCM  m  m (d) 
m1 m2 l
Take g  m / s 2  1 2

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31. A uniform solid sphere of mass m is placed on D P
a sheet of paper on a horizontal surface. The
coefficient of friction between paper and
sphere is . If the paper is pulled horizontally
with an acceleration
B
S
C (a) The acceleration of the block B is double the
acceleration of the centre of D
P (b) The force of friction exerted by D on S acts to
a
the left
(c) The horizontal and the vertical sections of the
(a) the tension in the string is equal to mg sin string has the same tension
mg sin (d) The sum of the kinetic energies of D and B is
(b) force acting on the cylinder is less than the loss in the potential energy of B as it
2
moves down
mg sin
(c) tension in the string is equal to 35. A triangular block ABC of mass m and side 2a
2
lies on a smooth horizontal plane is shown.
(d) frictional force acting on the cylinder is zero
32. A rigid body is in pure rotation, that is, There point masses of mass m each strikes
undergoing fixed axis rotation. Then which of the block at A, B and C with speed as shown.
the following statement(s) are true After the collision the particle come to rest.
(a) You can find two points in the body in a plane Then:
perpendicular to the axis of rotation
having same velocity Y
A
(b) You can find two points in the body in a plane m
v
perpendicular to the axis of rotation having same
60 X
acceleration
(c) Speed of all the particles lying on the curved
surface of a cylinder whose axis coincides with the m
axis of rotation is same
(d) Angular speed of the body is same as seen
v
from any point in the body 60 60
33. A rough disc of mas m rotates freely with an B v C
angular velocity . If another rough disc of
m m
mass and same radius but spinning in
2
opposite sense with angular speed is kept (a) the centre of mass of ABC remains
on the first disc. Then: stationary after collision
(a) the final angular speed of the dise is 3 (b) the centre of mass of ABC moves with a
(b) the net work done by friction is zero velocity v along x- axis after collision
(c) the friction does a positive work on the lighter (c) the triangular block rotates with an angular
disc 2 3mva
velocity  about its centriod axis
mR 2 2 I
(d) the net work done by friction is perpendicular to its plane
3
34. In the figure, the disc D does not slip on the  1 
surface S, the pulley P has mass and the string (d) a point lying at a distance of   from
 2 3ma 
does not slip on it. The string is wound around centroid G on perpendicular bisector of BC is at
the disc. rest just after collision
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36. A rod leans against a stationary cylindrical 7 11 0 Al 3
body as shwon in figure, and its right end slides (c) KE  mgl (d) I 
to the right on the floor with a constant speed 39 36
v. Choose the correct option(s) 39. The torque on a body about a given point is
 found to be equal to A x L where A is a constant
 vector, and L is the angular momentum of the
R body about that point. From this it follows that
R 
dL
(a) is perpendicular to L at all instants of time
 Rv 2  2 x 2  R 2 
X
dt
(b) the component of L in the direction of A does
x2  x2  R 2 
(a) the angular speed is 3/2 not change with time
(c) the magnitude of L does not change with time
Rv (d) L does not change with time
(b) the angular acceleration is 40. Consider a sphere of mass ‘m’ radius ‘R’ doing
x x2  R2 pure rolling motion on a rough surface having
Rv velocity v 0 as shown in the Figure. It makes

(c) the angular speed is
x x2  R2 an elastic impact with the smooth wall and
 Rv 2  2 x 2  R 2  moves back and starts pure rolling after some
(d) the angular acceleration
 
is x 2 x 2  R 2 3 / 2 time again.
37. The uniform 120 N board shown in figure is
supported by two ropes. A 400 N weight is
suspended one-fourth of the way from the left
end. Choose the correct options

 30 T V0
T2 3

0.25L 0.75L O
400N (a) Change in angular momentum about ‘O’ in the
(a) T1  185 N (b) T2  371N entirem otion equals2m v0R in magnitude.
(b) Moment of impulse provided by the wall during
(c) T2  185 N (d) tan  0.257 impact about O equals 2mv0R in magnitude.
38. The KE and moment of inertia about the given
end point of a rod of mass m and length l and 3
(c) Final velocity of ball will be v 0
cross sectional area A which is rotating with 7
g 3
 as shown in the Fig. will be [ density (d) Final velocity of ball will be – v 0
l 7
41. If a cylinder is rolling down a rough inclined
 x
of the rod varies as  0  1   , x is the with initial sliding.
 l (a) after some time it may start pure rolling
distance measured from O) ] (b) after sometime it must start pure rolling
(c) it may be possible that it will never start pure
g
 rolling
l (d) cannot conclude anything
42. Which of the following statements are correct.
O
(a) friction acting on a cylinder without sliding on
an inclined surface is always upward along the
incline irrespective of any external force acting on
7 7 Al 3 it.
(a) KE  mgl (b) I  0
(b) friction acting on a cylinder without sliding on
36 36

110 NARAYANAGROUP
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an inclined surface is may be upward may be
2gr0 gr0 gr0 4gr0
downwards depending on the external force acting (a) (b) (c) (d) .
on it. tan  2 tan  tan  tan 
(c) friction acting on a cylinder rolling without sliding Passage - II : (46-48)
may be zero depending on the external force acting Arod AB of mass 3m and length 4a is falling freely
on it. in a horizontal position and c is a point distant a
(d) nothing can be said exactly about it as it depends from A. When the speed of the rod is u, the point c
on the friction coefficient on inclined plane. collides with a particle of mass m which is moving
COMPREHENSION TYPE QUESTIONS vertically upwards with speed u. If the impact
Passage - I : (43-45) between the particle and the rod is perfectly elastic
A small particle of mass m is given an initial velocity find
v0 tangent to the horizontal rim of a smooth cone at C G
a radius r0 from the vertical centerline as shown at A B
point A. As the particle slides to point B, a vertical u l
distance h below A and a distance r from the vertical 2
centerline, its velocity v makes an angle  with the
horizontal tangent to the cone through B. 3mg
43. The value of  is 46. The velocity of the particle immediately after
the impact
r0
29 19
D A (a) u down (b) u down
19 29
h
 29 27
(c) u,up (d) u down
B 19 19
r 47. The angular velocity of the rod immediately
after the impact

19u 12u 29u 19u
(a) (b) (c) (d)
12a 19a 19a 29a
48. The speed of B immediately after the impact
v0 r0
cos1 is
(a) (r0  h tan  ) v20  2gh
19 19
(a) u down (b) u up
vr 27 27
(b) cos  r  h tan  v 2  2 gh
1 0 0

0 0

27 27
v0 r0 (c) u down (d) u up
(c) cos  r  h tan
1

 v02  2 gh 19 19
Passage - III : (49-50)
0

v0 r0 An uniform rod of mass m=30kg and length

cos1
(d) r0 v02  2gh l=0.80m is free to rotate about a horizontal axis O
passing through its centre. A particle P of mass
44. The speed of particle at point B M=11.2kg falls vertically through a height
(a) v 20  2gh (b) v 20  2gh 36
h m and collides elastically with the rod at a
245
(c) v 20  gh (d) 2v02  2gh
l
45. The minimum value of v0 for which particle will distance from O. At the instant of collision the
be moving in a horizontal circle of radius r0. 4

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rod was stationary and was at an angle  37 0 K 2K 2K 4K

with horizontal as shown in figure (a) (b) (c) (d)
M M 3M 3M
53. The maximum value of V0 for which the disc
h
will roll without slipping.
37 M M
0 (a) g (b) g
l
2
K 2K
l
3M 5M
4 (c) g (d) g
l
4
K 2K
Passage - V : (54-56)
49. Calculate angular velocity of the rod just after
collision is A wheel of radius R, mass m with an axle of radius
(a) 1 rad/s (b) 3 rad/s (c) 2 rad/s (d) 4 rad/s r is placed on a horizontal surface. Its moment of
50. Velocity of particle P after collision is inertia is I  mR 2 .Unwinding a rope from its axel
( g  10ms 2 ) a force F is applied to pull it along a horizontal
surface. The friction is sufficient enough for its pure
rolling    00 
7 1 9 1
(a) ms (b) 7ms 1 (c) ms (d) 1ms 1
9 7
Passage - IV : (51-53)
wheel F
A uniform thin cylinder M and radius R is attached
to two identical massless springs of spring constant
K, which are fixed to the wall, as shown. The spring 
are attached to the axle of the disc symmetrically
on either side at distance d from its centre. The R
r
axle is massless and both the springs and the axle Axle
are in horizontal plane. The unstretched length of
each spring is L. The disc is initially at its equilibrium
position with its centre of mass (CM) at a distance
L from the wall. The disc rolls without slipping with 54. Find the linear acceleration of the wheel
F  I / m   Rr  2 F  I / m   Rr 
velocity V 0  V0iˆ . The coefficient of friction is


    I / m   r 2 
(a)  I / m  r 2  (b)

F  2I / m   2 Rr 
 
F  I 2 / m  Rr  
 I / m   r   I / m   r 
(c) 2 (d) 2
d
Y 2d V0
R 55. Find the condition for which frictional force acts
in backward direction
X
(a)  I / m   Rr (b)  2 I / m   Rr
L
51. The net external force acting on the disc when I 2  I 
its CM is at displacement x with respect to (c)  m   Rr (d)    Rr
its equilibrium position is   m 2
2 Kx 4 Kx 56. Find the condition for which frictional force acts
(a)  Kx (b) 2Kx (c)  (d)  in forward direction
3 3
52. The centre of mass of the disc undergoes SHM (a)  I / m   Rr (b)  2 I / m   Rr
with angular velocity , equal to

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I 2  I  4 15 R 6R
(c)  m   Rr (d)    Rr (a)
3 8a
(b) 4
4a
  m 2
6R 3 15 R
Passage - VI : (57-59) (c) 3 (d)
Consider a cylinder of mass M = 1kg and radius 7a 4 8a
R=1 m lying on a rough horizontal plane. It has a 61. The angular velocity of disc is
plank lying on its stop as shown in the figure. 1 7a 8a 4 7a 16 8a
(a)  (b) (c)  (d) 
m = 1kg
3 6R 15R 9 6R 9 15R
60
62. The revolution made by disc in time interval
A computed in Q.No. (i) is

M
6 5p 2
R (a) 8 (b) (c) (d)
5p 6 3p
B Passage - VIII :(63-65)
A disc of a mass M and radius R can rotate freely
A force F = 55 N is applied on the plank such that in vertical plane about a horizontal axis at O. distant
the plank moves and causes the cylinder to roll. r from the centre of disc as shown in the figure.
The plank always remains horizontal. There is no The disc is relased from rest in the shown position.
slipping at any point of contact.
57. The acceleration of cylinder is M,R
(a) 20 m/s2 (b) 10 m/s2 (c) 5 m/s2 (d) 12 m/s²
58. The value of frictional force at A is
(a) 7.5 N (b) 5.0 N (c) 2.5 N (d) 1.5 N 0 C
59. The value of frictional force at B is
(a) 7.5 N (b) 5.0 N (c) 2.5 N (d) 1.5 N
Passage - VII :(60-62) 63. The angular acceleration of disc when OC
A cabin is falling freely and inside thecabin a disc rotates by an angle of 370 , is
of mass M and radius R is made to undergo
8rg 5rg
uniform pure rolling motion with the help of some
(a) 5  R 2  2r 2  (b) 4  R 2  2r 2 
external agent. Inside the cabin wind is blowing in    
horizontal direction which imparts an acceleration
a to all the objects present in cabin in horizontal 10rg 8rg
direction. [Disc still performs uniform pure rolling (c) 3  R 2  2r 2  (d)
motion]. A very small particle gets separated from
  5R 2
disc from point P and after some time it passes 64. The angular velocity of disc in above described
through the centre of disc O. Based on above case is
information, answer the following questions: 8 gr 6 gr
(a) 5  R 2  2r 2  (b) 5  R 2  2r 2 

12 gr 12 gr
0 g
(c) 5  R 2  2r 2  (d)
5R 2
37 65. Reaction force exerted by hinge on disc at this
P
instant is
60. The time taken by particle to reah from
(a) 5  R  2r 2   g  R 2  6r 2    4 R 2 
P to O is Mg 2 2
2

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(b) 5  R  2r 2   3  R  6r 
Mg applied on cylinder at different positions with
respect to its centre O in each of four
2 2
2
situations of column-1, due to which magnitude
4 Mg Mg of acceleration of centre of mass of cylinder
(c) 5  R  2r   R (d) 5  R  2r 2   4 R is ‘a’ Match the appropriate results in column-
2 2
2 2 2
II for conditions of columnI
MATRIX MATCHING TYPE QUESTIONS Column-I
66. A rod of length L and weight w is kept in F F

equilibrium on the two support separated by R R/2

L (a) O
(b) O

as shown in the figure. The right support is

2
taken out at time t = 0.
Match the following questions based on the
above information F O
(c) (d)
O
R/2
w,L F

Column-II
(p) Friction force on cylinder will not zero
L F F
2 q) a  r) a 
m m
s) friction force acting on cylinder is zero
68. Column I Column II
(Object) (Moment of inertia)
Column I Column II 8MR 2
(a) Uniform rod p)
(a) The moment (p) 3g/7 11
of inertia of the rod
M
about the support point at t = 0 is
12 g
(b) The angular (q) 30
7L
l=R
acceleration of rod about
the support point at t = 0 is
4
(c) The linear (r)
7 MR 2
acceleration of centre (b) Uniform semicircular ring. q)
12
of mass of rod at t = 0 is
Axis is perpendicular
7  L2 to plane of ring
(d) The normal (s) 48 g
M

reaction on the rod

L2
by the support at t = 0 is (t) 3 g
67. A uniform solid cylinder of mass m and radius [ = 22/7]

R is placed on a rough horizontal surface 13MR 2

where friction is sufficient to provide pure (c) Uniform triangular r)
8
rolling. A horizontal force of magnitude F is plate of mass M
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R
60 0
60
R v0
R (V0 > R0)
(C)
A
MR 2
(d) Uniform disk of initial mass s)
8
m2 
M from which circular Portion of radius
R is then removed M.I of remaining mass
about axis which is perpendicular to plane
of plate and passing through its centre (D)
m1
F

Column-II
R p) The angular momentum of disc about point A
2R (as shown in figure) remains conserved.
q) The kinetic energy of disc after it starts rolling
without slipping is less than its initial kinetic energy.
r) In the duration disc rolls with slipping, the
69. In each situation of column-I, a uniform disc friction acts on disc towardsleft
of mass m and radius R rolls on a rough fixed s) Before rolling starts acceleration of the disc
horizontal surface as shown. At t=0(initially) remain constant in magnitude and direction.
the angular velocity of disc is and velocity t) Final angular velocity is independent of friction
o
coefficient between disc and the surface.
of centre of mass of disc is V0 (in horizontal INTEGER ANSWER TYPE QUESTIONS
direction). The relation between V0 and 0 for 70. A plank of mass m1 with a uniform solid sphere
each situation and also initial sense of of mass m2 placed on it rests and a force F is
applied to the plank. The acceleration of the
rotation is given for each situation in column-
plank provided there is no sliding between
I. Then match the statements in column-I with
the corresponding results in column-II
F
the plank and the sphere is m1 
n
m2 then the
Column-I 7
value of n is
0

v0
(V0 > R0 )
(A) 71. A uniform cylinder of radius r is rotating about
A
its axis at the angular velocity 0 . It is now
placed into a corner as shown in figure. The
0 coefficient of friction between the wall and the
cylinder as well as the ground and the cylinder
v0 is  . The number of turns, the cylinder
(V0 > R0)
(B) completes before it stops, are given by
A  20 r  1  2 
   the value of n is
 ng   (1  ) 

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74. A ball of radius R=20cm has mass m=0.75kg

WALL and moment of inertia (about its diameter)
I  0.0125 kgm 2 . The ball rolls without sliding
over a rough horizontal floor with velocity
V0  10 ms 1 towards a smooth vertical wall. If
coefficient of restitution between the wall and
the ball is e=0.7, velocity V of the ball long
after the collision is ( g  10 ms 2 )
72. The pulley shown in figure, has a radius 10 cm 75. A uniform square plate of mass ‘m’ is
and moment of inertia 0.5 kg-m2 about its axis. supported as shown. If the cable suddenly
Assuming the inclined planes to be frictionless, breaks, assuming centre of mass is on
the acceleration of the 4.0 kg block is
1
that horizontal ine passing through A determine ;
n
mg
value of n is The reaction at A is that n is
n
2.0kg 4.0kg
B

45 45

73. In the arrangement shown in figure, ABC is a A

straight, light and rigid rod of length 90cm. End C
A is pivoted so that the rod can rotate freely
about it, in vertical plane. A pulley, having
b b
internal and external radii R=7.5cm and r=5cm
is fixed to a shaft of radius 5cm. The pulley -
shaft system can rotate about a fixed
horizontal axis O. B is point of contact of the 76. In the figure shown there is a fixed wedge ‘W’
pulley and the rod. From free end C of the rod of inclination . A is a block, B is a disc and
a mass m2  2kg is suspended by a thread. ‘C’ is a solid cylinder. A, B and C each has
Another thread is wound over the shaft and a mass ‘m’. Assuming there is no sliding
anywhere and string to be of negligible mass
block of mass m1  4kg is suspended from it.
find :
If coefficient of friction between the rod and The friction force acting on the cylinder due
the pulley surface is  0.4 and moment of
1  n sin  that n is
mg
inertia of pulley-shaft system about axis O is to the wedge is
15
I  0.045 kg  m2 , the acceleration of block
B
m1 , when the system is released
( g  10 ms 2 ) is
A B 30cm C
60cm C
A
W
O fixed
r
R 

77. In the figure shown a uniform ringh of mass

m1 m2
m is placed on arough horizontal fixed surface.
The coefficient of friction between left half of
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ring and table is µ1 whereas between right half m3

and table is µ2 at the moment shown. The ring
48kg , then m is
2
has angular velcoity in clockwise sense in the
figure shown. At this moment find the
magnitude of acceleration  in m / s2  of m4

centre C of ring. [Given g  10 m / s 2 ]

m3
m2 m1

80. An isosceles right triangular plate ABC of

1 2
mass m is free to rotate in vertical plane about
a fixed horizontal axis through A. It is
supported by a string such that the side AB is
A C horizontal. The reaction at the support A is
p  mg 
thus p  q   .
q
78. In the given diagram a sphere of mass m and
radius R is rolling without slipping on a rough A l

inclined surface of inclination  p / 6  . Centre

R l
of mass of sphere is at C which is
3
distiance from centre in a direction parallel to
inclined plane. Moment of a intertia of the
C
sphere about point of contact is I 0 (given). At
81. The densities of two solid spheres A and B of
the given instant sphere is rotating with
the same radii R vary with radial distance r as
constant velcity ? 0 . Calculate the angular
r
A r   k   B  r   k   ,
r
5
accel eration of sphere at this instant to near and
est integer?  
R R
0
m, R
respectively, where k is a constant. The
R 3 moments of inertial of the individual spheres
about axes passing through their centres are
R
C
IA n
I A and I B , respectively. If I  10 , the value
Centre of
No slipping mass
B

30 of the n is
82. A horizontal circular platform of radius 0.5 m
[Given that m  2kg , R  0.5m and mass 0.45 kg is free to rotate about its
g  10m / s 2 , ? 02  3 in axis. Two massless spring toy-guns, each
SI unit and
carrying a steel ball of mass 0.05 kg are
I 0  10kg  m2 ] attached to the platform at a distance 0.25 m
79. Figure shows an arrangement of masses from the centre on its either sides along its
hanging from a ceilling. In equilibrium each diameter (see figure). Each gun
rod is horizontal, has negligible mass and simultaneously fires the balls horizontally and
extends three times as far to the right of the perpendicular to the diameter in opposite
directions. After leaving the platform, the balls
wire supporting is as to the left. If mass m4 is

NARAYANAGROUP 117
ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

have horizontal speed of 9 ms 1 with respect MULTIPLE ANSWER QUESTIONS

to the ground. The rotational speed of the 26. A,B 27. A,C 28. A,C
platform in rads-1 after the balls leave the 29. A,B,C 30. A,B,C 31. B,D
platform is 32. C,D 33. A,D 34. A,B,D
35. B,C 36. C,D 37. A,B,D
38. A,B 39. A,B,C 40. A,B,D
41. A,C 42. B,C
COMPREHENSION QUESTIONS
43. A 44. A 45. C 46. A 47. B 48. C 49. D
50. C 51. D 52. D 53. C 54. A 55. A 56. A
83. A ring of mass M and radius R is rotating with
angular speed  about a fixed vertical axis 57. B 58. A 59. C 60. A 61. B 62. D
passing through its centre O with two point 63. A 64. C 65. A
MATRIX MATCHING TYPE
M
masses each of mass and rest at O. These 66. A  s ; B  q ; C  p ; D  r
8
masses can move radially outwards along two 67. A  p ; B  q , s ; C  p , r ; D  p , r
maseless rods fixed on the ring as shown in 68. A  q; B  p; C  s; D  r
the figure. At some instant the angular speed 69. A  p, q, r; B  p, q, r; C  p, q; D  p, q, r
8 INTEGER TYPE QUESTIONS
of the system is  and one of the masses is 70. 2 71. 8 72. 4 73. 1 74. 2 75. 4 76. 7
9
77. 4 78. 1 79. 4 80. 5 81. 6 82. 4 83. D
3
at a distance of
5
R from O. At this instant LEVEL-VI - HINTS
the distance of the other mass from O is SINGLE ANSWER TYPE
(JEE_ADV-15) 1. 2mg-T=ma ; TR = I
a g
 ; solving 
R 3
2. m1 g sin  Fpseudo ........(1)
m1 g sin  f  m1 g sin  f  0
1  2
m 4v   2v    2v   02  +
2 2
3.
O 2  
1 2 1 1
mv  I  m  4v 2  4v 2   mv 2 = 5mv 2
2 
2

2 2
V
2V
2v

V
V

LEVEL-VI - KEY
2v
V V
V
SINGLE ANSWER QUESTIONS
1. D 2. C 3. A 4.B 5. C 6. D 7.A 4. f  0 for t  t0 until it can stop no friction acts
8.C 9.B 10.D 11.A 12.C 13.A 14.A because it neither slides nor rotates due to action
of the rolling
15.A 16.C 17.B 18.B 19.B 20.B 21.A
5. Let J be the impluse acting on the rod
22.B 23.A 24.A 25.B

118 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS

1 11. Divide the ring into infinitely small lengths of mass

J  mvcm ; Jx  ml 2 . Since the end A is dm. Even though mass distribution is non- uniform,
12
each mass dm1 is at same distance R from origin
l J  12 Jx  l
stationary VA  Vcm    2   0  MI of ring about z  axis is
2 m  ml  2
6. Let x be the distance from the end of spring constant  dm1 R 2  dm2 R 2  ......dmn R 2  MR 2
K1. If y is the elongation of the rod, kasking 12. Only a horizontal force acts at hinge B, because
moments about the point C, hinge A is assumed to support the door,s weight.Let
us take torques about A as axis.
 K1 yx  K2 y ( L  x)
1 F 0 or F2  H  0

F
x

7. I AB  ml 2  ICB ( irrespective an angle)

6  0 or V  400 N  0
y

1 2 We find from these that

8. W I H  100 N and V  400 N
2
Let x is the distance of CM from A To find the resultant force R

on the hinge at A,
I  Mx2  4 M  2  l  x    1  2Mx  8Ml
dI 4
we have R   400 2
 100
2
  412N
 10  8Ml  0; x  l
dx 5 13. M  L ; L  2 R so R  L / 2
9. Torque about point of contact of the spool will be
1 1
zero I ' XX  MR 2 ; I 'YY  MR 2
2 2
2
T .3R  Mg .2 R or T  Mg I 0  I Z  I ' XX  I 'YY  MR 2
3
Equating the forces acting on the block along and Y
perpendicular to the incline
N  T sin  mg cos
 90
2
or N  mg cos  Mg sin X X'
3
Also mg sin  T cos  f  N
2  2 
mg sin  Mg cos   mg cos  3 Mg sin 
C
Y'
3
N Now I BC  I B ' C '  I 0  IZ  MR 2
f

  L 2  2
MR 2 1 L2 L3
so I BC 
2 2 4 8
T  14. For whole sytem f1  f 2  3 1 a.... 1
For rear cylinder 100  f1  0.5  a  ....  2 
For rear cylinder f 2  0.5  a  .... 3 
mg

3mg sin  2Mg cos


3mg cos  2Mg sin 100  4a; a  20m / s 2
L 15. I B ' ( in new given condition )
10. L   mg sin  Rt ; Since,  so, L   t 
t
1  R 
2
The curve between L and time t will be a straight  MR 2  M   MR 2  I A

line. 2  2

NARAYANAGROUP 119
ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

thus A  B  B '  I 18. L  2 gh  mb

16. Appling equilibrium equations,we get 19. This equation is on example of combined rotational
X  F cos 60  N A sin 60  FB cos 45  N B sin 45  0 and translational motion, here we are solving this
with the help of instantaneous axis of rotation is
A

Also we know htat

FA  0.268N A and FB  0.268N B lying at a distance of l sin from end B and
Solving above equations we get N A 0.158 N B perpendicular to V0 . Let is the angular velocity

Y  N
of rod. Then.
cos 60  FA sin 60  N B cos 45 
VA   l cos
A

FB sin 45  W and V  V   l sin

Solving above equations we get B 0

20. Here, torque due to normal force balances the

N B  0.966W and F  0.259W torque due to frictional force which is equal to
Taking moment about A and equating it to zero,
we get 1
mga sin .
M  W  L cos    N B cos 450  2 L cos  2 3

  N B cos 450  2 L cos 

A
F

  FB sin 450  2L sin cos 

  FB cos 450  2 L sin sin 0
21. l

By putting the values of known quantities in above A t=0 B

equation we get  36.2 0
F
17. The focrces acting on the rod are:
(i) Weight W of the rod acting veritcally r
downwards from centre of gravity.
(ii) Reaction R at A acting normally at A i.e along
AO 
(iii) Reaction R1 at C actiong at right angle to rod.
For equilibrium the three forces will be concurrent.
By geometry, 
OCA  OAC  GDA  3F
AC  CD cos  2r cos , DC  2r sin l  Fl cos ;  cos
ml
and AG  1.5r
d 3F
In triangle GDC, .  cos
GC AC  AG sin 2r cos  1.5r d ml
tan   or 

 
DC DC cos 2r cos 3F 6F
 2 sin 2  2 cos 2  1.5 cos .d  cos d ;  sin
0 ml 0 ml
or 1  cos 2  cos 2  0.75cos 22. f   ma
or 2cos 2  0.75 cos  1  0
 N   ma
0.75   0.75   4 2
2

solving it , cos  
mV 2
 m  V
dV
4 R dS
0.75  8.5625
  0.92  u sin g the  sign  
V  dV

4 R dS
  cos1  0.92   23 dV 
  dS
V R
120 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS

V V  R  dS
dV 
V S
 T2

0 0 p R/2


a

S
R
V
a1

 ln 
T1 B
m
V0 R
1
T2
2

     1
a2 f
 S mg

 V  V0 e R

 R
Now a  a 1  R 
Now power co n s umed by friction,

P   f .V  2
  NV a1 R
 a .....(i) ; a1  a1R ......(ii)
mV 2 2
a2  a1  2 R   a 2  R  .....(iii)
 V
R

        2 For the block ma  T1 ......(iv)

mV 3

R For the spool ma1  T2  T1  f .......(v)
substitutethe value of V from eq 1 , we get
 mR 2 
 m 3 R   a1  T2 R  T1R  fR
3 S
P .....(vi)
R
V0 e  2 
23. Suppose C is the point through which For the disc ma2  mg  T2 ........(vii)
theinstantaneous axis of rotation passes and G is
1 2 2
 mR  a  T2 R
the centre of mass of the rod. From the geometry
......(viii)
of the figure 2 
C
4
For these equations, we get a  g
NB 30
30 37
25. Moment of inertia about an axis through
l/2 NA the flat face of hemisphere
B
G l/2 A
mg I0 IG I
60 30
A
CG  l cos30 0

The moment of inertia about C 5

 m  l cos300   ml 2
R
8
ml2
5
I
2
G
12 6 3
If a is the angular acceleration, then 8
R

5 2 l 0 3 3g
 ml  a  mg  cos 30   a 
 6   2  10l
24. Suppose the acceleration of the block B is a, A'
acceleration of disc is a2 and the acceleration fo
2
centre of mass of spool is a1 , also suppose the I0  mR 2
angular accelerations of spool and disc are a1 and 5
a 2 respectively.. From the parallel axis theorem, the moment of
inertial through c.m. of the hemisphere

 3   83 
2

I0  I0  m  R     mR
2

 8   320 
NARAYANAGROUP 121
ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

Using parallel axis theorem, Moment of inertia about vertical components of force exetred by pivot on
the axis AA' , rod, respectively.
Ry
 13 
R
5 
2

I  I G  m  R   I    mR 2 200 N 50 N
8   20 
MUTLIPLE ANSWER QUESTIONS Rx

26. For pure rolling of wheel wrt horizontal surface 100 N 400 N 20 N
below it RX  200  50  150 N
For pure rolling of wheel wrt plank
RY  400  100  20  480 N
Then ,
105   480 
Vp
R   503 N
2 2

Vc +R
The rod will exert equal and opposite force on the
 pivot.
Vc 29. For equilibrium of the rod, let us say force R is
R Vc appilied whose X and Y components are
Vc = R
RX and RY as shown in figure.
VP  Vc  R  2Vc RX  36 N , RY  48  20  68 N
dv p 2dv c Ry
  ; a p  2ac R
dt dt X 4-X
27. Acceleration of block is not constantthrought. 36 N
Acceleration of block at B is V / R where 2
Rx 60 N
V  2 gR .
2
20 N 48N

28. The rod is in translation equilibrium in any case as For rotational equilibrium, x  20  48   4  x 
it is privoted, now let us check for its rotational 48
equilibrium.  x m ; So, R  362  682  77 N .
17
For rotational equilibrium, the net torque acting
about any point must be zero F l
30. acm  ; F  I
m1  m2 2
3m 1m 4m
l l
200 N 50 N a1  acm  ; a2  acm 
2 2
 31. ma  f  ma0
p
2mR 2
fR  I ; I  ; a0  R
100 N 400 N 20 N 5
Let us take the torque about P. 32. All points in the body, in plane perepndicualr to the


axis of rotation, revolve in concentric circles. All
 400 1  20  5 100  3  0 points lying on the circle of same radius have same
[ Taking clockwise as  ve speed (and also same magnitude of acceleration)
and anticlock wise as  ve ] but different directions of velocity (also different
direction of acceleration)
So rod is in rotational equilibrium, also. Hence there cannot be two points in teh given plane
If a body is in rotational equilibrium then

with same velocity or with same acceleration. As
 0 about any point.
ext mentioned above, points lying on circle of same
The force exterted by a pivot on rod maintains the radius have same speed.
translational equilibrium in horizontal and vertucal Angular speed of body at any instant w.r.t any point
directions. Let Rx and Ry be the horizontal and on the body is same by definition.
122 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS
Dividing throughout by L and solving, we get
1 2 1m 2 1 1 
33 mr   r    mr 2  mr 2  ' T1  185 N
2 22 2 4 
Substituting into our earlier equations, we get
 '  /3
d) use work energy principle T1  185 N
Substituting into our earlier equations, we get
13 
w f  wg    KE    mr 2  2
/9 T2  sin  92.5N and T2  cos  360 N
24  Dividing th equation yields
1 1  tan  0.257, or  14.40
  mr 2 2
 mr 2 2

4 8 Than 0.249T2  92.5 and T2  371N
r 2 2
One can always check moment problem results by
 w f  0  m taking moments about another point, such as the
3
right end of the bar for this problem
34. According to the given figure VB  2Vdisc ......(1)
38. I   3dmx
1
l

After taking derivatative aB  2adisc here b & d

2

are conceptual 0

35. By law conservation of linear movementum we  x3 

 A   x 2 dx  dx 
 x
 1    Adx  x 
1 1

 
l l
2

mviˆ  mvjˆ  mv  ˆj  0 0 3  l 3 0  l 
0 0

have  1  l 3 1 l 4  7 0 Al 3
 p t riangular wedge  0  A
0    
3 3 l 4 36
 p  mviˆ since the net linear momentum

39  A  L when A is constant vector . Accroding
  
imparted to the tangular which is along x-axis end to given condition, it is cross product so  is
in non zero, so the center of mass of the wedge
perpendiculr to L and also to A .
 
ABC will move along x-axis.
R dl
36. Form the geometry, x  Forther more  Al sin the component L in
sin dt
the direction of A ( i.e l cos ) will not change

d
Also,   . Therefore, which time. otherwise AL sin will not satishfied.
dt
The magnitude of L does not change becose AL
 
dx d  R 
v    is perpendicular to 
dt dt  sin 
 R  d / dt  cos R cos 40.
 
sin 2
sin 2
v sin 2 Rv
 
R sin x x2  R2
Rv 2  2 x 2  R 2 
V0
0
d  
 
d Rv
    (before collision)
x2 x2  R2
3/2
dt dt  x x  R 
2 2

37. Taking moments about the left edge and resolving

T1 into x and y components,
  0 yields LT1 cos 300   0.25 L  400 
  0.5L 120   0 V0
0

(just after collision) P

NARAYANAGROUP 123
ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

From equation (i)

Eliminating v and substituting r = r0 – h tan
v 0 r0  v 20  2gh (r0  h tan ) cos 

v0 r0
V cos  
 v  2gh (r0  h tan )
2
0
P f
( at the point of pure rolling ) mv02
N sin  = mg …(ii) N cos   …(iii)
Taking angular momentum about the point P r0
I  mV0 R  I  mVR 
0
gr0
Solving (ii) and (iii) tan   .
2 V 2 V v02
mR 2  0  mV0 R  mR 2   mVR

 3m   4a 2  .  4ma2
5 R 5 R
1
3V J .a  I G 
V   0 12
7 46-48
41 If friction is enough to support pure rolling then its In order to use the law of restitution. we need the
starts pure rolling . other wise doesnot do this
speed of point C, which is u1  a (downwards)
42 the direction of the friction depons on the directoin
of net extrenal force acting on the body
COMPREHENSION TYPE J impulse = F x t
43-45.From angular momentum conservation about axis
of cone. O 
T h e
r0
J a
D C v1
 v2
h
E law of restitution now gives relative velocity of
r separation at point of impact = e (relative velocity
of approch ) or u2   u1  a  e  u  u 


m v0 r0 = m v r cos hence 2u  u2  u1  a
v0 r0 = v r cos …(i) For the rod the speed of the particle is
from energy conservation ; E 1 = E 2 29
(downwards)
19
1 2 1
mv0  mgh  mv 2  0 ; v  v02  2gh 49 - 50
2 2 Velocity component of particle, normal to rod (just
r0 – r = h tan ; r = (r0 – h tan )
J  M 0 cos
n 
N sin after collision ) is M
N Since, the collision is elastic, therefore, there is no
loss of kinetic energy during collision.
Hence kinetic energy of system of rod and particle
just after collision = kinetic energy of pariticle just

Ncos before collision .

 I 2  M  t2  n2   m 02
1 1 1

2 2 2
J  24 Ns
mg

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 JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS

MV02 1  MR 2   V0  Kx02
2

    
2 2  2  R  2
J Elongation of spring in extreme position
O
3M 3M
x0  V0 f  fL V0  g
4K K
54-56 F  f  ma ; F  r   f  R   I
 From equation 1  3 rad / sec tangential
component of velocity of particle aR
36 F r  R R
t  0 sin
 
solving a  I  mR 2 and f  F  ma
 2 gh sin 370  ms 1
35
51-53
51 When CM is displaced by x , 2Kx  f  Ma F [ I / m   Rr ]

 I / m   r 2 
MR 2
fR  I  and a  R
2 55. f is positive for  I / m   Rr for frictional force
4 Kx acts in backward direction.
On solving a 
3M 56. f is negative for (I/m) <Rr or frictional force
4 Kx acts in forward direction
Net force on the disc = Ma  57-59
3

Drawing the F.B.D of the plank and the cylinder
N1
Fsin

Fcos
Kx Kx
f1
f

4 Kx ˆ
mg

52. F  

i ( as disc is displaced toward right) Equations of motion are
F cos  f1  ma ........ 1
3

F sin  N1  mg ........  2 
4K 4K
a x ; 
3M 3M
f1  f2  MA ........  3
a 4 Kx
 
........  4 
53
R 3MR f1 R  f 2 R  I
MR MR  4 Kx  2 Kx A R ........  5 
so f    
2 2  3MR  3 1
4  55 
a 4 Kx MR 4 F cos 2
  ; so, f  a   10m / s 2
3M  8m  3 1   8 1 
.
R 3MR 2
MR  4 Kx  2 Kx 1
   3  1  55 
2  3MR  3 3MF cos 2  7.5 N
f1  
f is maximum when x is maximum 3M  8m 3 1  8  1
f  f L   Mg  for pure rolling.Energy 1
1  55 
conservation at equilibrium and extreme position and f 2  MF cos  2  2.5 N
3M  8m 3  1 8 1
60-62
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ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III
Y From energy-conservation,
I? 2
 Mg  r sin 370
2
 MR 2 ?2
O
3
  Mr 2   Mgr 
 2  2 5
X
P
Let us solve this situation wrt cabin
frame of reference. Acceleration of detached 12 gr
? 
particle wrt cabin in horizontal direction is a  5  R 2  2r 2 
(towards right) and is zero in vertical direction. Let
angular velcoty of disc be ? and velocity of its From FBD of disc,
centre of mass be v , then form pure rolling motion
Rx  Mg sin 370  Mar  M? 2 r
condition, v  R?  0 i.e., v  R? Initial
velocity of particle is,
RY

u p   v  R? cos37  i   R? sin 37 j
RX
 ^ ^
0 0
O 37°

R? ^ 3 R? ^ 2r
 i j ar
5 5
For required situation, position vector of final at
location of particle is, Mg

r   R sin 37  vt  i   R cos37 j

0
^
0
^ Mg cos37 0  Ry  Mat  Mra

where t is the time taken by particle to 3Mg  R 2  6r 2 

 Rx   ,
reach from P to O. 5  R 2  2r 2 
3R? 4
R cos37 0  tt  Mg  4 R 2 
5 3? Ry   
5  R 2  2r 2 
 R?  1 2
and R sin 37  vt    t  at
0
Mg
 5  2 R  Rx2  Ry2 
5  R 2  2r 2 
8a 4 15R
,t 
    
where v  R?  ? 
15 R 3 8a  2 2
 g R 2
 6 r 2 2
 4 R
Number of revolutions made in time t is,
?t 2 MATRIX MATCHING QUESTIONS
n 
2p 3p ML2 ML2 7WL2
63-65 66. (A) I   I CM  Ma 2    (S)
12 16 48 g
O L 12 g
(B) W   I     (Q)
O C 37° 4 7L
L 3g
(C) a    (P)
C 4 7
3 Mg 4W
From t  Ia (D) N  Mg 
7

7
(R)
 MR 2  67. Assume friction to be absent and horizontal Force
 Mg  r cos37 0    Mr 2  a is applied at a distance x above centre
 2 
F mR 2 2Fx
8rg a and Fx  or R 
a  m 2 mR
5  R 2  2r 2 
126 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS

R Now we substitute the value of a p from equation

if a = R then x = (4) in equation (5), hence
2
F F  m 2 ac  m1  ac  r  … (6)
The fricton force will be zero and a 
m From equations (2) and (3) or , r  ac
5
2
R
If a  R or x  ,Friction force is Substituting this value in equation (6), we have
2
7  7 
F or, F  m 2 ac  m 1ac or, F   m 2  m1  ac or,
towards left anf a  2  2 
m F
ac 
R 7
if a  R or x  Fricton force towards right m 2  m1
2
2
F 7 F
and a  Now a p ac r
2
ac
2 .
m m1 m2
7
1 2 1 71. The free body diagram is shown in the figure.
68. (a) I  ml  sin  ml
2 2

3 12 1
The initial energy  I 20
b) I  I COM  mr ( centre of mass of the reaging
2
2
where I is the moment of inertia of the cylinder
2R
will be art r  from the centre of the ring and is given by I  Mr 2
1
2
69. Angular momentum each conserved about the point (M = Mass of the cylinder and r = Radius)
of contact with the ground.Angular mo mentum also  Initial Kinetic energy of cylinder
is conserved in all cases about any point on the line
1
passing through point of contact and parallel to the  Mr 2 02 … (1)
4
velocity of the centre of mass. Then kinetic energy
decreases in all cases due to work done by the
R
f1
friction. We have to calculate the relative velocity
of contact point and the direction of friction in A,B
and D towards the left. and in case of C, the friction N

direction towards right. Those direction never be

changed in any given cases. f2

INTEGER TYPE
Mg
Here, there is no motion of the centre of gravity of
70. The situation is as shown in the figure. the cylinder, hence,
Here we have, R  N  Mg … (2) ; N  R … (3)
F  f  m1 ap … (1) Solving for R and N,
and, f  m 2 ac … (2) Mg Mg
  2  … (4) ; 1  2 
R N  … (5)
2 1
Further, f  r  I   m 2 r 2 … (3)
5
The total initial energy is dissipated against frictional
and, a p  ac  r … (4) forces.
m2
1 r 2 o2
ac  Mr
2 2
N R .2 n ; n 
4 0
8g
m1
f where n is the number of turns made by the cylinder
F
before it stops.
ap Putting the values of N and R, and solving for n
gives the final result..
Substituting the value of f from equation (2) in
equation (1), we get 72.  T2  T1  R  I
F  m 2 a c  m 1a p … (5)
NARAYANAGROUP 127
ROTATIONAL DYNAMICS JEE-ADV PHYSICS-VOL - III

 T2  T1  R  I    i 
a
mg
R 
4 g sin 4 5 0  T2  4a   ii  w

T1  2g sin 45 0  2a   iii 
O
m a

Solving (i) (ii) and (iii) we get a = 0.25 m / s 2

73. Taking moment  about A of forces acting on the
N
N
rod. 0.60 N   0.60  0.30  T2 . or N  30 N
A 60cm 30cm
 2ms 1
mb 2 mb2 mb2  1 
75. I    1  
N f T2
6 2 2  3
T2 T1
f
N

w O
C
r Nx
R
Ny

T1 mg mg mg

T1 .05    .4  30  .075   .045  20a 

2mb 2 mgb 2mb 2
3g
I g ; Hence  
For vertical forces acting on block m1 3 2 3 2 2b
m1 g  T1  m1a ; T1  18a  18 Velocity of O is zero So N x  0

4 0  1 8 a  1 8 
b b  3g  3mg mg
4 a ; a 1 m / s 2 mg  N y  m m   4 N y 
2 2  2 2b  4
74. For vetical forces, N  mg 76. (a) mgT1 ma 1
For vertical forces, N  ma. or a  g  10 .
 T2  f  mg sin  m .....  2 
a
Taking moments (about o) of forces acting on the 2
ball, NR  I or  120
T2  f  r  mr 2  ......  3
1 a
Long after the collision, there will be no sliding or it 2 2r
will be pure rolling . Let sliding stop after a time t
after collision, the final transtional velocity,
  7  at  or  7  10 t and final

angular velocity,   0  t

or  120 t  50  rad s 1  clockwise 

But at that instant  R T2
T1
 7  10 t   0.2 120 t  50  t  0.5

128 NARAYANAGROUP
JEE-ADV PHYSICS-VOL - III ROTATIONAL DYNAMICS

T1 T2 4m1 g  3  m3 g i.e., m3  12m1

a/2 16m1 g  3  m4 g
f m1  1kg
A
 2mg 
a 80.  
mg sin  3 
The FBD of plate it as shown in fiigure,
mg
l T
 T1  T2  r  mr 2  ......  4 
1 a
2 r
By 1 ,  2  ,  3 &  4
4 g  2  sin 
a  substitute this a value we get
15 l mg

1  7  sin
mg
f 
15
77. Net force,
2(Rd.).g
 For transtional equilibrium
mg  R  T
1 Rd.g d For rotational equilibrium,
 
A l 2l
 R T 
3 3
2mg
 R
3

F  2  µ1Rd?? g cos ? 81. I   4 r r dr

p /2 2 2 2

0 3

2   µ1Rd?? g cos ? I A    r   r 2  r 2  dr
p/2

I B    r 5  r 2  r 2  dr
0

 2  µ1  µ2  .R?g 

F  2 µ1  µ2  R?g
IB 6
 
a   I A 10
m 2pR?
2  m b v.d
 µ  µ2  g  4
 1 82.
=
1
 4rads 1
m p .r 2
p 2
R 83. Using conservation of angular momentum
78. mgR  m? 0 R  I0a
2
3  8   m 9 R 2 8 
mR 2   mR 2       
a1.051 ( nearest value)  9   8 25 9 

79. m2 g  1  m1 g  3 i.e., m2  3m1  m 2 8  4R

 x   x 
8 9  5
NARAYANAGROUP 129