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CHAPTER - 1
ROTATORY MOTION OF A RIGID BODY
1. Rigid Body
If an external force applied to a body does not
produce any displacement of the particles of the
body relative to each other, then the body is called
a rigid body.
No real body is perfectly rigid, However, in solid
bodies (leaving rubber, etc.) the relative
displacement by the external force is so small that
it can be neglected. Hence ordinarily, when we
speak of a body, we mean a rigid body.
2. Moment of a Force or Torque
When an external force acting on a body has a
tendency to rotate the body about an axis then the
force is said to exert a ‘torque’ upon the body about
the axis.
The moment of a force, or the torque, about an
axis of rotation is equal to the product of the
magnitude of the force and the perpendicular
distance of the line of action of the force from the
axis of rotation.
When a force F is applied on the body in the plane
of the paper, the body rotates about this axis. If r
be the perpendincular distance of the line of action
of the force from the point O, then the moment of
the force F, that is, the torque about the axis of
rotation is
(1)
τ = F x r.
If the torque tends to rotate the body anticlockwise
then it is taken as positive; if clockwise then
negative. The unit of torque is ‘newton-meter’
(N-m) and the dimensional formula is [ML 2 T -2 ].
It is clear from the above formula that is r zero i.e.
if the line of action of the force passes through O,
then the torque will be zero. In this situation the
body will not rotate how-so-ever large the force
may be. On the contrary, greater is the distance of
the line of action of the force from O, larger is
themoment of the force, or torque, about O; or
smaller the force needed to rotate the body.
Fig. 1
3. Rotatory Motion : Acceleration
When a body rotates about a fixed axis, the rotation
is called a ‘rotatory motion’ or ‘ angular motion’

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and the axis is called the ‘axis of rotation’. In
rotatory motion, every particle of the body moves
in a circle and the centres of all these circles lie at
the axis of rotation. The rotating blades of an
electric fan and the motion of a top are examples
of rotatory motion.
If the angular velocity of a rotating body about an
axis is changing with time then its motion is
‘accelerated rotatory motion’.
The rate of change of angular velocity of a body
about an axis is called the ‘angular acceleration’
of the body about that axis.
It is represented by α.
If the angular velocity of a body about an axis
changes from ω 1
to ω 2
in to second, then the angular
acceleration of the body about that axis is
α =
change in angular velocity
time−interval
= ω 2 − ω 1
t
The unit of angular acceleration is ‘radian/sec 2 , and
its dimensional formula is [T -2 ].
3.1 Relation between Angular Acceleration and
Linear Acceleration :
v = PP′
t
= r × θ
t
But θ / t = ω .
[. .
. arc PP‘ = radius r × angle θ ]
∴ v = r x ω
or ω = v . ... (i)
r
Suppose, at any instant the agnular velocity of the
body is ω 1
and after t second it becomes ω 2
. Then
the angular acceleration of the body is
α = ω 2 − ω 1
t
If at the above instants the linear velocities of the
particle P be v 1
and v 2
respectively, then according
to eq. (1), ω 1
= v 1
/ r and ω 2
= v 2
/ r. Therefore,
α = (v 2 ⁄r) − (v 1⁄ r)
t
= v 2 − v 1
rt
..(ii)
(v 2
- v 1
) is the change in the linear velocity of the
particle P in t sec. Therefore (v 2
- v 1
) / t is the rate
of change of linear velocity of the particle, that is,
it is the linear acceleration a of the particle P.
Substituting v 2 − v 1
t
= a in eq. (ii), we get
α = a r
Fig.2
Suppose a particle P of the body is at a distance r
from the point O. As the body rotates, the particle
P rotates through angle θ in t second and reaches
P ′. Thus the particle moves from P to p ′ in t second.
Therefore, its linear velocity is
or a = r x α.
Thus, the linear acceleration of a particle of a
body is equal to the product of the angular
acceleration of the body and the distance of the
particle from the axis of rotation.
4. Moment of Inertia
According to Newton’s first law of motion, a body
continues in its state of rest or uniform translatory
motion unless it is acted upon by some external
force to change its present state. The property of
bodies by virtue of which they oppose any change
in their present state is called ‘inertia’.
(2)

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In the same way, when a body rotates about an axis,
then it has a tendency to opposes any change in its
state of rotation about an axis is called the ‘moment
of inertia’ of the body about that axis. It is
represented by I.
The moment of inertia of a particle about an
axis is given by the product of the mass of the
particle and the square of the distance of the
particle from the axis of rotation.
Let there be a rigid body of mass M. We have to
find out its moment of inertia about a vertical axis
passing through O (Fig. 3). For this, let us assume
that the body is made up of a large number of
minute particles. If m 1
, m 2
, m 3
,.... be the masses of
these particles and r 1
,r 2
,r 3
, ... be their respective
distances from the axis of rotation, then their
moments of inertia about the axis of rotation will
be m 1
r 1
2
, m 2
r 2
2,
m 3
r 3
2
, ... respectively. The moment
of inertia (I) of the whole body about the axis of
rotation will be equal to the sum of the moments of
inertia of all the particles. :
I = m 1
r 1
2
+ m 2
r 2
2
+ m 3
r 3
2
+ .....
or I = Σ m r 2 .
the moment of inertia of a rigid body about a
given axis is the sum of the products of the
masses of its paticles by the square of their
respective disances from the axis of rotation.
The unit of moment of inertia is ‘kg-meter 2 ’ and its
dimensional formula is [ML 2 ].
Physical Significance of Moment of Inertia :
In order to rotate a body (initially at rest) about an
axis or to change the angular velocity of a rotating
body (i.e. to produce an angular acceleration in it),
a torque has to be applied on the body. This is
described by saying that the body has a ‘moment
of inertia’ about the axis of rotation. The greater
the moment of inertia of a body about an axis, the
greater is the torque required to rotate, or to stop,
the body about that axis.
Thus, the moment of inertia plays the same role
in the rotational motion as mass plays in
translational motion.
5.Radius of Gyration
Rms value of distance of different particles from
axis of rotation is called radius of gyration. It is
denoted by K.
2 2 2
r
K = √⎺⎺⎺⎺⎺⎺
1
+ r 2
.....+ r n
n
At distance equal to radius of gyration the entire
mass of body is supposed to be concentrated at a
point
Fig. 3
Here Σ (sigma) means the sum of all terms. Thus,
Fig. 4
In terms of radius of gyration,
(3)

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I = M K 2
Where M = Σ m, total mass of the body.
Radius of gyration depends upon distribution of
mass of the body about the axis of rotation i.e. on
the location of axis of rotation. Greater the part of
the mass situated away from the axis of rotation,
more will be K. and hence I.
6. Equation of motion for uniformaly
accelerated rotatory motion.
Education in linear
motion
(i) v = u + at
v = linear velocity after
t second
u = initial linear velocity
a = linear acceleration
(ii) v 2 = u 2 + 2 a S, S =
linear displacement
(iii) S = u t + 1 2 a t2
Equation in rotatory
motion
(i) ω = ω o
+ α t
ω = angular velocity
after t second
ω o
= initial angaular
velocity
a = angular
acceleration
(ii) ω 2 = ω ο
2
+ 2 α θ
θ = angular diplacement
(iii) θ = ω o
t + 1 2 α t2
7. Relation between linear and angular
quantities
Quantity Linear Angular
Displacement S θ
Initial velocity u ω o
Final velocity V ω
Acceleration a α
If the distance of a particles from axis of rotation
is r then
S = r θ
V = r ω
a = r α
8. Angular Momentum
When a body moves a straight line, then the
product of the mass m and the linear velocity v of
the body is called the ‘linear momentum’ p of the
body (p = m x v).
If a body is rotating about an axis, then the sum of
the moments of the linear momenta of all the
particles about the given axis is called the ‘angular
momentum’ of the body about that axis. It is
represented by ‘J’.
Let a body be rotating about an axis with an angular
velocity ω. All the particles of the body will have
the same angular velocity, but their linear
velocities will be different. Let a particle be at a
distance r 1
from the axis of rotation. The linear
velocity of this particles given by
V 1
= r 1
ω.
If the mass of the particle be m 1
, then its linear
momentum
= m 1
v 1.
The moment of this momentum about the axis of
rotation
= momentum x distance
= m 1
v 1
× r 1
= m 1
(r 1
ω) × r 1
[ . . . v 1
= r 1
ω ]
= m 1
r 1
2
ω .
Similarly, if the masses of other particles be m 2
, m 3
... and their respective distances from the axis of
rotation be r 2
, r 3
....., then the moments of their
linear moment about the axis of rotation will be m 2
r 2
2
ω, m 3
r 3
2
ω ,…respectively. The sum of the
moments of linear momenta of all the particles, that
is, the angular momentum of the body is given by
J = m 1
r 1
2
ω + m 2
r 2
2
ω + m 3
r 3
2
ω + ....
(4)

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= (m 1
r 1
2
+ m 2
r 2 2 + m 3
r 3
2
+ ....) ω
= (Σ mr 2 ) ω.
But Σ mr 2 is the moment of inerti a I of the body
about the axis of rotation. Hence the angular
momentum of the body about the axis rotation is
J = I x w.
The unit of angular momentum is ‘kg-meter 2 /sec’
or ‘joule-sec’. Its dimensional formula is [ML 2 T -1 ].
9. Kinetic energy of rotation
Suppose a body is rotating about an axis with a
uniform angular speed ω. Angular velocity of all
particles is same but they have different linear
velocities. Suppose a particle of the body of mass
m 1
is at a distance r 1
from axis of rotation . Let v 1
be its linear velocity. As linear velocity of a particle
is equal to product of its angular velocity and its
distance from axis of rotation, we have
v 1
= r 1
ω
Kinetic energy of particle is
1
2 m v 2
= 1 1 1
2 (r 1 ω)2 = 1 2 m v 2
ω 2
1 1
If masses of other particles be m 2
, m 3
, ... and their
distance from axis of rotation be r 2
, r 3
, ... then their
kinetic energies are
1
2 m r 2
ω 2 2
, m
2 2 3
r 3
ω 2
Now kinetic energy of body is equal to sum of K.E.
of all particles
K = 1 2 m r 2
ω 2 + 1 1 1
2 m r 2
ω 2 + 1 2 2
2 m r 2
ω 2
3 3
or, K = 1 2 (m 1 r 2 2 + m 2 r 2 2 + m 3 r 3 2 ....)
or, K = 1 2 (Σ m r2 )ω 2
Here, Σ m r 2 = Moment of inertia of body about
axis of rotation
(5)
∴
K = 1 2 I ω2
If ω = 1 then I = 2 K. Thus moment of inertia of a
body rotating about an axis with unit angular
velocity equals twice the kinetic energy of rotation
about that axis.
In case of linear motion work to be done to move
a body is given by
W = Force x displacement
perpendicular to direction of force
i.e.,
W = F. x
Similarly, in case of rotational motion work done
to rotate the body is given by
W = Torque x displacement
Here, Torque = F r
= τ θ
Where, r = perpendicular distance of line of action
of force from axis of rotation.
∴
W = F r θ
Here, this F is tangential force.
If this force only produces rotational motion then
1
2 I ω2 = F r θ
but if this force results in both linear and rotational
motion then
1
2 mV2 + 1 2 I ω2 = F r θ
9.1 Relation between Angular momentum and
Rotatory kinetic energy
In linear motion, p = √⎺⎺⎺⎺⎺ 2 m E
Similarly in rotatory motion , J = √⎺⎺⎺⎺ 2 I E
9.2 Work - Energy Theorem
In linear motion : 1 2 m (v2 − u 2 ) = F × S
Similarly, in rotatory motion,

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1
2 I (ω2 − ω ο2
) = τ × θ
9.3 Expression for power
In linear motion : P = F x v
Similarly, in rotatory motion, P = τ ω
10. Law of conservation of angular
momentum
This law states that if the sum of external
torques acting on the system is zero, then the
total angular momentum of the system remain
constant.
For a system of N particles, the total external torque
is due to the sum of external torques acting due to
the external forces, since the internal force do not
contribute to the torque. Thus
(b)
and ω increases.
An ice-skater or a ballet-dancer can increaes
her angular velocity by folding her arms or by
folding her body, as this decreases moment of
inertia and increases angular velocity.
11. Theorem of Parallel axis
This theorem states that the moment of inertia I of
body about any axis is equal to its moment of
inertia I cm
about a parallel axis through its centre of
gravity plus the product of the mass M of the body
and square of the perpendicular distance between
the two axis.
τ
→ d L → (tot)
(tot) =
dt
If the sum of external torques acting on the system
is zero, then
L→ (tot) = constant vector.
or,
0 = d L→ (tot)
dt
L →+ L→ + ....L→ = constant vector.
1 2 N
We know that the angular moment can also be
written as
L = Iω ∴ Iω = constant
i.e. if I increases, ω decreases and vice-versa.
10.1 Applications of law of conservation of
angular moemtum :
Following phenomenon obey the law of
conservation of angular momentum -
(a)
The angular velocity of a planet in its orbit
round the sun increases when it gets nearer to
sun, as m.I. of the planet about sun decreases
Fig.5
In the figure shown M.I about axis AB is given by,
I = ICG + Mh2
M.I. is minimum about an axis passing through
CG
12. Theorem of perpendicular axis
This theorem states that the moment of inertia of
uniform plane lamina about an axis perpendicular
to it plane is equal to the sum of its moments of
inertia about any two mutually perpendicular axis
in its plane inter secting on the first axis.
In the figure shown, l z
= l x
+ l y
- In general, M.l about an axis perpendicular to
plane is larger as compared to M.l about an axis
lying in the plane.
13. Moment of Inertia of Some Objects
13.1 Ring
(6)

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(i)
About an axis passing through centre of
gravity and perpendicular to plane
I = I d
+ mr 2
Fig.8
Fig. 6
l z
= mr 2 , m = mass of ring, R = radius of ring here,
k = r as all particles are situated at same distance
l = m r2
2 + m r2 = 3 2 m r2
(b) Perpendicualr to plane
For ring k r = l
l = k2
r 2
(ii) About an axis passing through centre of
gravity but in plane of ring (diametrical axis).
Fig.7
d = diametrical moment of inertia then by
perpendicular axis theorem
Fig.9
by parallel axis theorem
13.2 Disc
l = l z
+ m r 2
l = m r 2 + m r 2 = 2 m r 2
In case of a disc mass is uniformaly distributed e.g.
a coin.
(i)
Moment of inertia about an axis passing
through centre of gravity and perpendicular to
plane
l d
+ l d
= l z
= m r 2
l d = 1 2 m r2
About a tangential axis
(a) Parallel to plane
parallel axis theorem
m = mass of disk
Fig. 10
l z = m r2
2
(7)

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r = radius of disk
(ii)
About an axis passing through centre of
gravity and in plane i.e., about diametrical
axis
I d
= diametrical axis
Fig.11
By perpendicular axis theorem
l d
+ l d
= 2l z
2 l d
= m r2
2
By parallel axis theorem
l = l z
+ m r 2
l = l + m r2
2 + m r2
l = 3 m r2
2
Fig.13
13. 3 Rectangular Lamina
(i)
About an axis passing through centre of in
plane of lamina parallel to breadth.
or,
l d
= m r2
4
(iii) About tangential axis in plane of disc
Fig.14
By parallel axis theorem
l = l d
+ m r 2
l = m r2
4 + m r2
l = 5 4 m r2
Fig.12
(iv) About tangential axis perpendicualr to plane
of disc.
Fig.15
Its elementray strip is a rectangle of negligible
breadth and length l.
So, Ml of the strip = m l2
12
and Ml of the entire Lamina = M l2
12
where, m is the mass of elementary strip and l is
the mass of Lamina.
(8)

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So, l B
= M L2
12
- Since axis is parallel to breadth, the term of
breadth will not appear in the formula.
(ii)
About an axis parallel to length in plane of
lamina and passing through centre of gravity
l L
= M B2
12
(The term of L does not appear)
Fig.16
(iii) About an axis passing through centre of
gravity and perpendicular to plane of lamina
This cuboid can be considered to be made up of a
large number of laminas placed on one another.
Elementry section is a lamina containing length
and bredth.
∴ Moment of inertia about this axis is similar to
that about an axis passing through centre of gravity
and perpendicular to plane of lamina.
i.e., I = M 12 [l2 + b 2 ]
put here, M = mass of entire cuboid
(ii)
About an axis passing through centre of
graivity and perpendicular to plane
containing breadth and height
Fig.17
By perpendicualr axis theorem
I = I B
+ I L
I = ⎡ ML 2
⎢ ⎣
12 + MB2 ⎤
12 ⎥
⎦
I = M 12 [L2 + B 2 ]
13.4 Moment of Inertia of a cuboid
(i)
About an axis passing from centre of gravity
and perpendicular to plane containing length
and breadth
Fig.19
Moment of inertia about this axis is similar to that
about an axis passing through centre of gravity and
perpendicular to plane of lamina, this plane
contains breadth and height of cuboid.
i.e., I = M 12 [l2 + b 2 ]
13.5 Moment of inertia of solid cylinder
(i)
About an axis passing through centre of
gravity and parallel to length
Fig.18
Fig.20
Since, the elementary section is disc and given axis
(9)

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is perpendicular to plane of disc, therefore MI of
the cylinder about the given axis will be same is
that of a disc, about an axis passing thourgh C.G.,
perpendicular to plane.
∴
I = M r2
2
Where, M = mass of whole cylinder
(ii) M I of a solid cylinder about axis passing
through surface and parallel to lenght
perpendicular to length
I = Ml2
12
(iv) MI of a thick cylinder about an axis passing
through C.G and perpendicular to lengh.
Here, elementry section is a solid disc. It may
be supposed to be subjected to following two
motions symultaneously :
Fig.21
Applying theorem of parallel axis,
I = 1 2 M R2 + M R 2
(a)
Fig.23
Each disc is changing position from l on one
2
side to l on the other side, i.e., each disc is
2
rotated through 180º in space. For this MI to
I = 3 2 M R2
(iii) Moment of inertia of a thin cylindrical rod
about an axis passing through centre of
gravity perpendicular to lengh :
As it is thin cylindrical rod its radius is
negligible therefore its surface is almost
similar to a lamina
(b)
be overcome = Ml2
12
Each disc is also rotating about its diametrical
axis. For this MI = MR2
4
Total MI = = Ml2
12 + MR2
4
13.6 Moment of Inertia of Hollow Cylinder
(i)
About an axis passing through centre of
gravity and parallel to length.
Fig.22
This axis is parallel to breadth and
Fig.24
(10)

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(ii)
Here, elementary section is a ring and given
axis parallel to length.
∴ I = mR 2
About an axis parallel to length and passing
through surface.
Applying parallel axis theorem
I = 2 5 mr2
Where, m = mass of sphere
r = radius of sphere
So, K2
r 2 = 2 5
(ii) About any tangetial axis :
I = MR 2 + MR 2
I = 2 M R 2
Fig.25
(iii) About an axis passing through C.G. and
perpendicular to length.
I = Ml2
12 + MR2
2
Fig.26
13.7. Moment of inertia of a solid sphere.
(i)
About any axis passing through C.G.
In this case moment of inertia is same about
any axis passing through centre of gravity.
0r,
Fig.28
I = 2 5 M R2 + M R 2
= 7 5 M R2
K 2
R 2 = 7 5
13.8 Hollow sphere :
(i) About Any diametrical axis :
(ii)
⇒
I = 2 3 M R2
About any tangential axis
I = 5 3 M R2
If M and R are same, MI of hollow sphere is
greater.
13.9. Moment of inertia for a hollow sphere
about any diametrical axis
I = 2 3 m r2
Fig. 27
(11)
∴
Moment of inertia of hollow sphere > moment
of inertia of a solid sphere
14. Concept of Sliding and Rolling Motion
14.1. Pure Sliding :

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Pure rolling is a combination of spinning and
translatory motion.
When resultant velocity at point of contact
becomes zero then motion becomes pure rolling.
Fig.29
In case of pure sliding translatory velocity of each
particle is same but angular velocity of each
particle is zero.
e.g. A ball thrown on a frictionless floor without
exerting any torque on it.
If an object slides down smooth inclined plane of
vertical height h without rolling then its velocity at
bottom due to sliding motion is given by :
v s
= √⎺⎺ 2gh ⎺
A particle executing pure sliding motion has only
translatory kinetic energy.
14.2. Partly rolling and partly sliding motion :
When a spinning ball is thrown with a certain
velocity, it has both translatory motion as well as
angular velocity.
fig. 30
Let translatory velocity of the ball be v. The particle
also have linear counterpart r ω of the angular
velocity. So, resultant velocity of the highest point
in v + r ω and that of the lowest point (point of
contact) is v - r ω .
Such a body pusses both rotatory as well as
translatory kinetic energy.
14.3 Pure rolling :
Fig.31
Velocity of the point of contact :
Since, v = 0
v L
= v - r ω
v L
= r ω
Velocity of the highest point :
= v + v = 2 v
v H
= v + rω
Such a motion has both translatory as well as
rotatory kinetic energy.
If the axis of rotation has translatory velocity
then its translatory kinetic energy is equal to
1
2 mv2 while its rotatory kinetic energy is
1
2 I ω2
When velocity of a ball thrown on a rough
floor reduces to 5/7th of its initial velocity, its
motion become pure rolling.
14.4 Pure Rolling Motion Down An Inclined
Plane :
When an object comes down an inclined
plane of vertical height h, potential energy
lost by it is equal to m g h. This is also equal
to total kinetic energy gained.
m g h = 1 2 m v2 + 1 2 I ω2
(12)

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= 1 2 m v2 + 1 v2
m k2
2 r 2
= 1 2 m v2 ⎛ ⎜ ⎝
1 + k2
r 2 ⎞ ⎟⎠
v rolling
= √⎺ ⎺ 2 g h
⎛
⎜1 + k2 ⎞
⎝ r 2 ⎟⎠
substituting k 2 = I/m,
v rolling
= √⎺⎺⎺ 2 g h
⎛
⎜1 + I ⎞
⎝
mr 2 ⎟⎠
hence, v sliding
= √⎺⎺⎺ 2 g ⎺h
v rolling
= v sliding
/ √⎺⎺⎺⎺ 1 + k2
r 2
similarly it can be prove that arolling = g sin θ
1 + k2
r 2
14.5 Time taken to reach the bottom of the
inclined plane :
S = 1 2 a t2
t r
= √⎺⎺ 2 S
a r
2h
= √⎺⎺⎺⎺⎺
sin θ ⎛ g sin θ⎞
⎜ ⎟
⎜ 1 + k2
⎝
r 2 ⎟
⎠
=
or,
1
sin θ √⎺⎺2h g × √⎺⎺⎺⎺⎺⎺⎺⎺ 1 + k2 ⁄ r 2
t r
= t s
√⎺⎺⎺⎺ 1 + k2
r 2
14.6 Energies of a body in pure rolling :
- Translatory K.E. = ⎛ ⎜ ⎝
1
2 m v2 ⎞ ⎟⎠ × 1 .....(i)
- Rotatory K.E. = 1 2 I ω2
= 1 2 m k2 × v2
r 2
= ⎛ 1
⎜ ⎝
2 m ⎞ v2 ⎟⎠ × k2
r 2
- Total K.E. = Tran. K.E. + Rot. K.E.
= 1 2 I ω2 + 1 2 m v2
= 1 2 m v2 ⎛ ⎜ ⎝
1 + k2
r 2 ⎞ ⎟⎠
... (iii)
(ii)
From equations (i), (ii) and (iii) following
important results can be obtained :
⇒
⇒
⇒
Rotatory kinetic energy
Translatory energy
= k2
r 2
rotatory kinetic energy k 2 ⁄ r 2
=
Total energy 1 + k 2 ⁄ r 2
Translatory kinetic energy 1
=
Total kinetic energy 1 + k 2 ⁄ r 2
(13)

AISECT TUTORIALS : PHYSICS : SET-4
Objective Questions
1. The rotational analogue of force in linear
motion is
(a) Torque
(b) Weight
(b) Moment of inertia
(d) Angular momentum
2. The moment of momentum is called
(a) Couple
(c) Impulse
(b) Torque
(d) Angular momentum
[CPMT]
3. When a steady torque is acting on a body, the
body
[NCERT]
(a) Continues in its state of rest or uniform
(b)
(c)
(d)
motion along a straight line
Gets linear acceleration
Gets angular acceleration
Rotated at a constant speed
4. Angular momentum of a body is defined as
the product of
(a)
Mass and angular velocity
(b) Centripetal force and radius
[CPMT]
(c) Linear velocity and agnular velocity
(d) Moment of inertia and angular velocity
5. The moment of inertia of a body comes into
play
[AFMC]
(a) In motion along a curved path
(b) In linear motion
(c)
(d)
In rotational motion
None of the above
6. Moment of inertia of a ring of mass M and
radius R about an axis passing through the
centre and perpendicular to the plane is
(a) 1 2 MR2 (b) MR 2
(c) 1 4 MR2
(d) 3 4 MR2
[CPMT]
7. A hollow cylinder and a solid cylinder having
the same mass and same diameter are released
from rest simultanceously from the top of an
inclined plane. Which will reach the bottom
first
(a) The solid cylinder
(b) The hollow cyliner
(c) Both will reach the bottom together
(d) The one having greater density
[CPMT]
8. Moment of inertia of body in the case of
rotational motion plays the same role as in the
case of translatory motion is played by
(a) Velocity
(c) Mass
(b) Acceleration
(d) Force
9. A small object of mass m is attached to a light
string which passes through a hollow tube.
(14)

AISECT TUTORIALS : PHYSICS : SET-4
The tube is hold by one hand and the string by
the other. The object is set into rotation in a
circle of radius R and velocity v. The string is
then pulled down, shortening the radius of
path of r. What is conserved?
(a) Angular momentum
(b) Linear momentum
(c) Kinetic energy
(d) None of the above
10. A homegeneous disc of mass 2 kg and radius
15 cm is rotating about its axis (which is fixed)
with an angular velocity of 4 radian / s. The
linear momentum of the disc is
(a) 1.2 kg-m/s
(c) 0.67 kg-m / s
(b) 1.0 kg-m/s.
[CPMT]
(d) None of the above
11. A spherical solid ball of 1 kg mass and radius
3 cm is rotating about an axis passing through
its centre with an angular velocity of 50
radian/ s. The kinetic energy of rotation is
(a) 4500 J
(b) 90 J
(c) 910 J (d) 9
20 J
[CPMT]
12. One solid sphere and disc of same radius are
falling along an inclined plane without slip.
One reaches earlier than the other due to
(a) Different radius of gyration
(b) Different size
(c) Different friction
(d) Different moment of inertia
13. A sphere of mass 50 gm and diameter 20 cm
rolls without slipping with a velocity of 5 cm
/ sec. Its total kinetic energy is
(a) 625 ergs
(c) 875 ergs
(b) 250 ergs
(d) 875 joules
14. Moment of inertia x angular acceleration =
(a) Torque
(b) Force
(c) Angular momentum
(d) Work done
15. From an inclined plane a sphere, a dise, a righ
and a shell are rolled without slipping. The
order of their reaching at the base will be
(a) Ring, shell, disc, sphere
(b) Shell, sphere, disc ring
(c) Sphere, disc, shell, ring
(d) Ring, sphere, disc, shell
16. The flywheel is so constructed that the entire
mass of it is concentrated at its rim, because
(a) It increases the power
(b) It increases the speed
(c) It increases the moment of inertia
(d) It saves the flywheel fan breakage
17. In heat engines, flywheel plays an important
role because
(a) It accelerates the speed of the engine
(b) It supplies power to the engine
(c) It saves the engine from damage
(d) It maintains the constant speed of the
engine
18. If all of a sudden the radius of the earth
decreases, then
(a)
(b)
(c)
(d)
The angular momentum of the earth will
become greater than that of the sun
The orbital speed of the earth will
increase
The periodic time of the earth will
increase
The energy and angular momentum will
remain constant
19. The moment of inertia about an axis normal
to the plane and passing through the centre of
(15)

AISECT TUTORIALS : PHYSICS : SET-4
gravity will be maximum for which of the
following bodies of the same mass
(a) A disc of radius a
(b) A ring of radous a
(c) A square of side 2 a
[CPMT]
(d) A square made by four rods of length 2a
20. A mass of 10 kg connected at the end of a rod
of negligible mass is rotating in a circle of
radius 30 cm with an angular velocity of 10
rad / sec. If this mass is brought to rest in 10
sec by a brake, what is the magnitude of the
torque applied?
(a) 0.9 N-m
(c) 2.3 N-m
(b) 1.2 N-m
(d) 0.5 N-m
21. The moment of inertia of a solid cylinder of
mass M and radius R about a line parallel to
the axis of the cylinder and lying on the
surface of the cylinder is
(a) 2 5 MR2
(c) 3 2 MR2
(b) 3 5 MR2
(d) 5 2 MR2
[MP PET]
22. A wheel rotates with a constant acceleration
of 2.0 radian / sec 2 . If the wheel starts from
rest the number of revolutions it makes in the
first ten seconds will be approximately
(a) 8 (b) 16
(c) 24 (d) 32
[MP PET]
23. A disc is of mass M and radius r. The moment
of inertia of it About an axis tangential to its
edge and in plane of the disc or parallel to its
diameter is
[MP PET; CBSE]
(a) 5 4 Mr2
(c) ML2
12
(b) Mr2
4
(d) ML2
3
24. The moment of inertia of a uniform thin rod
of length L and mass M about an axis passing
through a point at a distance of L from one of
3
its ends and perpendicular to the rod is
(a) 7ML2
48
(c) ML2
12
(b) ML2
9
(d) ML2
3
[MP PMT]
25. The moments of inertia of two freely rotating
bodies A and B are I A and I B respectively. I A
> I B and their angular momenta are equal. If
K A and K B are their kinetic energies, then
(a) K A
= K B
(b) K A
> K B
[MP PET]
(c) K A
< K B
(d) K A
= 2K B
26. The radius of a rotating disc is suddenly
reduces to half without any change in its mass.
Then its angular velocity will be
(a) Four times (b) Double
(c) Half
(d) Unchanged
[MP PMT]
27. The moment of inertia of a uniform ring of
mass M and radius r about a tangent lying in
its own plane is
(a) 2Mr 2
(c) Mr 2
(b) 3 2 Mr2
(d) 1 2 Mr2
[MP PMT]
28. Two rings have their moments of inertia in the
ratio 2 : 1 and their diameters are in the ratio
2 : 1. The ratio of their masses will be
(16)

AISECT TUTORIALS : PHYSICS : SET-4
(a) 2 : 1 (b) 1 : 2
(c) 1 : 4 (d) 1 :1
[MP PMT / PET ]
29. A solid cylinder has mass M, length L and
radius R. The moment of inetia of this
cylinder about a generator is
(a) M ⎛ L 2
⎜ ⎝
12 + R2 ⎞
4 ⎟
⎠
(c) 1 2 MR2
(b) ML2
4
(d) 3 2 MR2
[MP PET]
30. A wheel is rotating at 900 r.p.m. about its axis.
When the power is cut-off, it comes to rest in
1 minute. The angular retardation in radian /
s 2 is
(a) π ⁄ 2 (b) π ⁄ 4
(c) π ⁄ 6 (d) π ⁄ 8
[MP PET]
31. The moment of inertia of thin square plate
ABCD of uniform thickness about an axis
passing through the centre O and
perpendicular to the plane of the plate is
Fig.32
(a) I 1
+ I 2
(b) I 3
+ I 4
[IIT]
(c) I 1
+ I 3
(d) I 1
+ I 2
+ I 3
+ I 4
Where I 1, I 2, I 3, and I 4 are respectively
moments of inertia about axes 1,2,3, and 3 and
4 which are in the plane of the plate
32. Two circular iron discs are of the same
thickness. The diameter of A is twice that of
B. The moment of inertia of A as compared to
that of B is
(a) Twice as large
(b) 8 times as large
[CPMT ]
(b) Four times as large
(d) 16 times as large
33. What remains constant when the earth
revolves around the sun
(a) Angular momentum
(b) Linear momentum
(c) Angular kinetic energy
(d) Linear kinetic energy
[Raj. PMT]
34. If the moment of inertia of a disc about an axis
tangential and parallel to its surface be I, then
what will be the moment of inertia about the
axis tangential but perpendicular to the
surface
(a) 6 5 I
(c) 3 2 I
(b) 3 4 I
(d) 5 4 I
[Raj. PET]
35. A body is rolling without slipping on a
horizontal surface and its rotational kinetic
energy is equal to the translational kinetic
energy. The body is
(a) Disc
(c) Cylinder
(b) Sphere
(d) Ring
[Raj. PMT]
36. A loop rolls down on inclined plane. The
fraction of its total kinetic energy that is
associated with rotational motion is
(a) 1 : 2 (b) 1 : 3
(c) 1 : 4 (d) 2 : 3
37. Two masses m 1 and m 2, m 1 > m 2 are connected
(17)

AISECT TUTORIALS : PHYSICS : SET-4
to the ends of masseless rope and allowed to
move as shown in the figure. The acceleration
of the centre of mass assuming pulley is
massless and frictionless, is
(a) m 1 − m 2
m 1
+ m 2
g (b) zero
2
(c) ⎛ m 1
− m 2
⎞
⎜ ⎟ g (d) ⎛ m 1
+ m 2
⎞
⎜ ⎟ g
⎝
m 1
+ m 2 ⎠
⎝
m 1
− m 2 ⎠
38. A ring of mass m and radius r rotates about an
axis passing through its centre and
perpendicular to its plane with angular
velocity ω. Its kinetic energy is
(a) 1 2 m r2 ω
(b) mrω
2
sphere and a disc of the same mass and the
same diameter to roll down through the same
distance from rest on a smooth inclened plane
is
(a) 15 : 14
(c) 15 2 : 14 2
(b) √⎺ ⎺15 : √⎺ ⎺14
(d) √⎺ ⎺14 : √⎺ ⎺15
[AFMC]
41. In figure (a) , a meter stick, half of which is
wood and the other half steel is pivoted at the
wooden end at O and a force F is applied to
the steel end a. In figure (b) the stick is pivoted
at the steel end at O’ and the same force F is
applied at the wooden end at a’. The angular
acceleration
(c) m r 2 ω
(d) 1 2 mrω
39. The moment of inertia of a thin square plate
ABCD (fig) of uniform thickness about an
axis passing through the centre O and
perpendicular to the plane of plate is
(a) I 1
+ I 2
(b) I 3
+ I 4
(c) I 1
+ I 3
(d) I 1
+ I 2
+I 3
+I 4
Fig.34
(a) in (a) is greater than in (b)
(b) in (b) is greater than in (a)
(c) equal both in (a) and (b)
(d) none of the above.
42. A cylinder of mass ‘M’ is suspended by two
strings wrapped around it as shown. The
acceleration ‘a’ and the tension T, when the
cylinder falls and the string unwinds itself is
Fig. 33
Where I 1 ,I 2 ,I 3 and I 4 are moments of inertia
about axis 1, 2, 3 and 4 which are in the plane
of plate.
40. The ratio of times taken by a uniform solid
(18)
Fig.35

AISECT TUTORIALS : PHYSICS : SET-4
(a) a = g, T = Mg
2
(b) a = g 2 , T = Mg
2
(c) a = g 3 , T =Mg 3
(d) a = 2 3
g, T =
Mg
6
43. A loaded truck has to take a sharp turn to the
left. The centre of gravity of the truck can be
altered by shifting concentrated load. To
avoid toppling, the load must be shifted
(a) up and left
(c) down and left
(b) up and right
(d) down and right
44. If the radius of earth contracts to half of its
present day value, the mass remaining
unchanged, the duration of the day will be
(a) 48 hrs
(c) 12 hrs
(b) 24 hrs
(d) 6 hrs
45. If a running boy jumps on a rotating table.
Which of the following is conserved?
(a) linear momentum
(b) K.E.
(c) angular momentum
(d) none of the above.
46. A hemispherical bowl of radius R is set
rotating about its axis of sysmmetry. A small
body kept in the bowl rotates with the bowl
without slipping on its surface. If radius
through the body makes with the axis an angle
θ and assuming that the surface of the bowl is
smooth, the angular velocity with which bowl
is rotating is given by
(a)
R
g cos θ
(c) √⎺⎺⎺ g ⎺
R cos θ
(b) √⎺⎺⎺⎺
R
g cos θ
(d)
g
R cos θ
47. The rate of change of angular momentum is
equal to :
(a) Force
(b) Angular acceleration
(c) Torque
(d) Moment of Inertia
48. A square Lamina lies in the X-Y plane as
shown in fig. The z-axis is passing through the
centre and perpendicular to the plane of the
Lamina. I 1, I 2, I 3 and I 4 represent the moment of
Inertia of the Lamina about the axis shown.
The moment of Inertia I z of the
Lamina about z-axis is :
Fig.36
(a) I 1
+ I 4
(b) I 1
+ I 2
(c) I 2
+ I 3
(d) I 3
+I 4
49. M.I. of a circular loop of radius R about the
axis given in figure is :
(a) MR 2 (b) (3/4) MR 2
(c) MR 2 /2 (d) 2MR 2
Fig 37
(19)

AISECT TUTORIALS : PHYSICS : SET-4
50. Moment of inerta of a solid sphere of mass M
and radius R about any tangent is :
(a) 2 5 MR2
(c) 2 3 MR2
(b) 7 5 MR2
(d) 5 3 MR2
51. When a steady torque acts on the body, the
body:
(a)
(b)
(c)
(d)
continues in its state of rest or of
uniform circular motion
gets linear acceleration
gets angular acceleration
rotates at constat speed
52. When the torque acting on a system is zero,
which of the following will be constant :
(a)
force
(b) linear momentum
(c)
(d)
angular momentum
linear impulse
53. A mass is revolving in a circle which is in the
plane of paper. The direction of angular
acceleration is :
(a) upward the radius
(b) towards the radius
(c) tangential
(d) at right angle to angular velocity
54. Which of the following has largest moment of
inetia :
(a) ring about an axis perpendicular to its
(b)
(c)
plane
disc about an axis perpendicular to its
plane
solid sphere
(d) bar magnet
55. The moment of inertia of a body does not
depend upon :
(a)
(b)
(c)
(d)
the angular velocity of the body
the mass of the body
the distribution of mass in the body
the axis of rotation of the body
56. A hollow cylinder and a solid cylinder having
the same mass and same diameter are released
from rest simultaneously from the top of an
inclined plane. Which will reach the bottom
first :
(a) the solid cylinder
(b) the hollow cylinder
(c)
(d)
both will reach the bottom together
the one having greater density
57. The moment of inertia of a body comes into
play :
(a)
(b)
(c)
(d)
in motion along a curved path
in linear motion
in rotational motion
none of the above
58. A man turns on a rotating table with an
angular speed ω. He is holding two equal
masses at arm’s lenght. Without moving his
arms, he just drops the two masses. How will
his angular speed change?
(a)
(b)
(c)
(d)
it will be less than ω
it will be more than ω
it will remain equal to ω
may be less than, greater than or equal
to ω depending on the equantity of
masses.
59. If I, α and τ are the moment of inertia, angular
acceleration and torque respectively of a body
acceleration and torque respectively of a body
rotating about any axis with angular velocity
ω, then :
(a) τ = Iα
(b) τ = Iω
(20)

(c) Ι = τ
(d) α = τ ω
60. One solid sphere and disc of same radius are
falling along an inclined plane without
slipping. One reaches earlier than the other
due to :
(a) different radius of gyration
(b) different sizes
(c) different friction
(d) different moment of inertia
61. A disc rolls over a horizontal floor without
slipping with a linear speed of 5 cm/sec. Then
the linear speed of a particle on its rim, with
respect to the floor, when it is in its highest
position is :
(a) 10 cm/sec
(c) 2.5 cm / sec (d) 0
(b) 5 cm / sec
62. The M.I. of a body about the given axis is 1.2
kg x m -2 . Initially the body is at rest. In order
to produce a rotational kinetic energy of 1500
joule, an angular acceleration of 25 rad/sec 2
must be applied about that axis for a duration
of :
(a) 4 sec
(c) 8 sec
(b) 2 sec
(d) 10 sec
[CBSE]
63. A disc like reel with massless thread unrolls
itself while fallin vertically downwards. The
acceleration of its fall is :
(a) g
(b) zero
(c) (2/3) g (d) g / 2
64. A ring of radius r and mass m rotates about an
axis passing through its centre and
perpendicular to its plane with angular
velocity ω. Its kinetic energy is :
(a) mrω 2 (b) mrω 2 /2
(c) mr 2 ω 2 (d) mr 2 ω 2 / 2
65. A spherical ball rolls on a table without
AISECT TUTORIALS : PHYSICS : SET-4
(21)
slipping. Then the fraction of its total energy
associated with rotation is :
(a) 2/5 (b) 2/7
(c) 5/7 (d) 3/5
66. Four masses are held rigidly by a massless
circular frame of radius ‘R’ as shown in
figure. The moment of inertia of the system
about an axis through the centre of ring and
perpendicular to the plane of the ring is :
Fig.38
(a) mR 2 (b) 3 mR 2
(c) 5 mR 2 (d) 7 mR 2
67. The angular velocity of a body change from
ω 1 to ω 2 without applying a torque but
changing its M.I. The ratio of moment of
inertia in two cases is :
(a) ω 1
: ω 2
(b) √⎺ ⎺ω 2
: √⎺ ⎺ω 1
(c) √⎺ ⎺ω 1
: √⎺ ⎺ω 2
(d) ω 2
: ω 1
68. Rotatioanl analogue of mass in linear motion
is :
(a) Weight
(b) Moment of Inertia
(c) Torque
(d) Angular momentum
69. A boy comes running and sits on the rotating
platform. Which of the following is
conserved:
(a) Kinetic energy
(b) Angular momentum

AISECT TUTORIALS : PHYSICS : SET-4
(c) Linear momentum
(d) None of these
70. Angular momentum of the body is conserved
:
(a) always
(b) never
(c) in the presence of external torque
(d) in the absence of external torque
71. A disc, a solid sphere, a ring of same mass and
radius are made to slide down an inclined
plane. Which one reaches the bottom first :
(a) sphere
(b) ring
(c) disc
(d) all will reach the ground at the same
time
72. The product of moment of inertia and angular
velocity is called :
(a) torque
(b) work
(c) angular momentum
(d) kinetic energy
73. A disc has a mass of 16 kg and radius 25 cm
and is to be rotated about an axis through its
centre and perpendicular to its plane. What
torque will increase its angular velocity from
zero to 8π rad / sec in 8 seconds :
(a) π Nm
(c) π/4 Nm
(b) π/2Nm
(d) 2π Nm
74. If mass as well as the radius of gyration of the
body are doubled, then its moment of inertia
becomes :
(a) two times
(c) six times
(b) four times
(d) eight times
75. If the mass of a body moves towards the axis
of rotation, its moment of inertia
(a) incerases
(b) remains constant
(c) decreases
(d) none of these is true
76. The rotational analogue of mass (in linear
motion) is called
(a) weight
(b) torque
(c) moment of inertia
(d) angular momentum
77. The moment of momentum is called
(a) angular momentum(b) torque
(c) impulse
(d) couple.
78. If I is M.I. of the body about the axis of
rotation and ω is angular velocity, then
angular momentum of the body will be
(a) 1 2 I ω2 (b) I ω 2
(c) 2 I ω 2 (d) I ω.
79. If a body is moving in a circle of radius r metre
with a constant speed V m/sec, its angular
velocity will be
(a) n 2 / r
(b) vr
(c) v / r (d) r / v.
80. The unit of M.I. in M.K.S. system is
(a) kg-m 2 (b) kg / m 2
(c) kg-m (d) kg / m.
81. The M.I. of a circular ring with mass M and
radius R about an axis passing through its
centre and perpendicular to its place is
(a) MR 2 / 4 (b) MR 2
[CPMT]
(c) MR 2 / 2 (d) 3 4 MR2 .
82. The moment of inertia of a circular lamina
about an axis passing through its centre and
(22)

AISECT TUTORIALS : PHYSICS : SET-4
perpendicular to the plane of mass M and
radius R is
(a) 1 4 MR2
(c) MR 2
(b) 1 2 MR2
(d) 5 4 R2
83. The work done by a torque in rotating a body
about an axis is equal to
(a) (K.E.) rot
(b) (K.E.) trans
(c) moment of momentum of a body
(d) none of the above
84. The variation of angular momentum with the
frequency of the rotation (n) in fig. 39 is
represented by the curve
(a) P
(c) R
Fig.39
(b) Q
(d) S
85. A sphere is rolling down an inclined plane of
angle θ. Then its acceleration will be
(a) 2 5 g sin θ
(c) 5 7 g sin θ
(b) 2 3 g sin θ
(d) 7 5 g sin θ
86. A wheel is at rest. Its angular velocity
increases uniformaly and becomes 60 rad/sec
after 5 sec. The total angular displacement is
(a) 600 rad
(c) 300 rad
(b) 150 rad
(d) 75 rad
87. For a bar pendulum the distance between two
points of minimum time period is
(a) equal to the lenght of simple pendulum
(b) equal to the length of compound
(c)
pendulum
equal to the radius of gyration
(d) twice the radius of gyration
88. Energy of 484 joules is spent in increasing the
speed of a flywheel from 60 r.p.m. to 360
r.p.m. The M.I. of the flywheel in kg x m 2 is
(a) 7 (b) 0.7
(c) 6 (d) 0.6
89. A force of 5N acts on a body of weight 9.80
N. What is the acceleration produced in m/s 2
?
(a) 49.00 (b) 5.00
(c) 1.96 (d) 0.51
90. The magnitude of momentum of a particle is
increased by 100% Increase in kinetic energy
is
(a) 100% (b) 200%
(c) 300% (d) 400%
91. An athlete complete one round of a circular
track of radius ‘R’ in 40 seconds. What will
be his displacement at the end of 2 minutes 20
seconds
(a) zero
(b) 2 R
(c) 2π R (d) 7π R.
(23)

AISECT TUTORIALS : PHYSICS : SET-4
Answer Sheet
Q.N. Ans Q.N. Ans Q.N. Ans Q.N. Ans Q.N. Ans Q.N. Ans.
1. a
17. d
33. a
49. b
65. b
81. b
2. d
18. b
34. a
50. b
66. d
82. b
3. c
19. d
35. d
51. c
67. d
83. a
4. d
20. a
36. a
52. c
68. b
84. d
5. c
21. c
37. d
53. c
69. b
85. c
6. b
22. b
38. a
54. a
70. d
86. b
7. a
23. a
39. a,b,c
55. a
71. a
87. d
8. c
24. b
40. d
56. a
72. c
88. b
9. a
25. c
41. b
57. c
73. b
89. b
10. a
26. a
42. d
58. b
74. d
90. c
11. d
27. b
43. c
59. a
75. c
91. b
12. a,b
28. b
44. d
60. d
76. c
13. c
29. c
45. c
61. a
77. a
14. a
30. a
46. c
62. b
78. d
15. c
31. abc
47. c
63. c
79. c
16. c
32. d
48. d
64. d
80. a
(24)

AISECT TUTORIALS : PHYSICS : SET-4
Hints and Solutions
1. (a) Force = Mass x Acceleration
Rotational analogue of mass = Inertia
Rotational analogue of acc n = Angular
acc n
2. (d) r x mv = angular momentum
3. (c) T = Iα
4. (d) I = mk 2 , v = kω
A.M. = mvk ⇒ mk 2 ω = Iω
5. (c) It is the specific characteristic of the
6. (b)
body to oppose rotational motion.
7. (a) It will have smaller moment of inertia.
8. (c) Rotational analogue of mass is inertia.
9. (a) No external torque is there.
10. (a) Since linear momentum p = mv and v =
rω
∴ p = mrω = 2 x 0.15 x 4 = 1.2 kg m/s
11. (d) Since K.E. of rotation =
1
2 Iω2 , I = 2 5 MR2
∴ K.E. = 9 20 J
12. (a,d) Let M.I. about the point of contact be
given by I s and I d, then
Mg sin θ r = Is . θ
⇒ a s =
For disc a d
Mg sin θ . r2
I s
= g sing θ
(1+2 ⁄ 5) = 5 7 g sinθ
=
Mg sin θ . r2
I d
= g sing θ 2g sing θ
=
(1+1 ⁄2) 3
∴ a s > a d which is due to difference in I s
and I d and therefore different radii of
gyration.
13. (c) 1
2 mv cm
2 + 1 2 I ω2 = 1 2 mv cm
2 + 1 2 . 2 5 mv cm
2
= 7
10 . 50 . (5)2 = 875 erg
14. (a) τ = Iα for rigid bodies having fixed axis
of rotation.
15. (c) In increasing order of their M.I.
More M.I. ⇒ lesser acceleration. (Refer
Q.12)
16. (c) M.I. depends upon product of mass and
17. (d)
square of the distance of mass from axis
of rotation.
18. (b) Due to decrease in radius, I decreases
19. (d)
and as no external torque is effective.
∴ L = constant and angular velocity
increases.
For disc = 1 2 Ma2 , For ring = Ma 2
For square of side 2a
= M [(2a)2 + (2a) 2 ]
12
= 2 3 Ma2
(25)

AISECT TUTORIALS : PHYSICS : SET-4
For square by four rods of length 2a
4 = ⎡ ⎢
⎣
M (2a) 2
12
+ Ma 2 ⎤
⎥⎦ = 16
3 Ma2
20. (a) Torque τ = I α and we know that
ω = ω ο + αt
⇒ 0 = 10 + α . 10 ⇒ α = − 1 rad ⁄ sec 2
Negative sign means retardation.
∴ τ = mr 2 . α
= 10 (0.30) 2 . 1 = 0.9 N−m
21. (b) Applying parallel axis theorem
22. (b)
fig. 40
I A
= I G
+ MR 2 = MR2
2 + MR2 = 3 2 MR2
θ = ω ο t + 1 2 αt2 ⇒ θ = 100 rad
Number of revolution
= 100
2π = 16(approx).
23. (a) Moment of inertia w.r.t. axis
AB (diameter) = Mr2
4
Fig.41
24. (b)
Moment of inertia w.r.t parallel to axis
AB and passing through point P
= Mr2
4 + Mr2 = 5 4 Mr2
I g = ML2
12
(about middle point)
Now I = I g + Ma 2 ⎡ ⎢a = L
⎣
2 − L 3 = L ⎤
6⎥
⎦
∴ I = ML2
12 + ML2
36 = ML2
9
25. (c) In a rotating body, kinetic energy (K) =
26. (a)
27. (b)
28. (b)
29. (c)
30. (a)
1
2 Iω2 and angular momentum (J) = I ω
Hence K = J2
2I
If K A and K B are the kinetic energies of
A and B respectively having the same
angular momentum J. Then
K A
= J2 ⁄2I A
= I B
K B J 2 ⁄ 2 I B I A
If I A > I B ,.then K B > K A
I = MR 2 ; ∴ M 1
M 2
= I 1
I 2
× R 2 2
R 1
2 = 2 1 × 1 4 = 1 : 2
W i = 900
60 = × 2π = 30π rad ⁄ s, W f = 0, t = 60 s
W f = W i + αt
∴α = W f − W i
t
= O − 30π
60
= π 2 rad ⁄ s
2
31.(a,b,c) Applying the theorem of perpendicular
axis is I = I 1 + I 2 = I 3 + I 4
Because of symmetry, we have
I 1 = I 2 and I 3 = I 4 Hence I = 2I 2 = 2I 3 i.e.I 2 = I 3
(26)

AISECT TUTORIALS : PHYSICS : SET-4
32.(d) Mass of disc is proportional to the area.
Mass of A is 4 times that of B.
A → m d→ v 1
1 + m d → v 2
dt dt
cm =
(m
33. (a) The angular momentum remains
1 + m 2 )
constant in the orbit of the earth.
A→
cm = m 1a → 1 + m 2 a → 2
34. (a)
I || = MR2
4 + MR2 = 5 m 1 + m 2
4 MR2
⎛m 1 − m 2 ⎞
m 1⎜⎝ ⎟⎠ g → + m m 1 − m 2 )
2 (−g → )
I 1 = MR2
2 + MR2 = 3 A
→ m 1 + m 2 m 1 + m 2
cm =
2 MR2
(m 1 + m 2 )
Hence I I
= 6 I II 5 ⇒ I I = 6 5 I A cm = ⎛ m 1 − m 2 ⎞ ⎛m 1 − m 2 ⎞
II
⎜ ⎟⎠ ⎜⎝ ⎟⎠ g
⎝ m 1 + m 2 m 1 + m 2
35. (d) For ring K = R, hence
E k = 1 A cm = ⎛ 2
m 1 − m 2 ⎞
⎜ ⎟⎠ g
2 Mv2
⎝ m 1 + m 2
38.(a)
E r = 1 2 Iω2 = 1 2 MK2 ω 2 = 1 2 MR2 ω 2 = 1 (KE) rot = 1 2 Iω2
2 Mv2
36.(a)
(K.E.) rot
= 1
Fraction =
2 × mr2 ω 2
(KE) rot + (KE) trans
39.(a,b,c) By the use of theorem of perpendicular
1
axis.
2 Iω2
1
2 Iω2 + 1 I = I 1 + I 2 = I 3 I 4
2 mv2
40.(b)
S = 1
1
2 × 2 at2
mr2 ω 2
1
2 × mr2 ω 2 + 1 a = g sin θ
2 mv2
1 + k2
r 2
1
2 mv2
=
1
2 mv2 + 1 = 1 For solid sphere :
2
S = 1
2 mv2 2 × g sin θ 2
37.(d)
R→ = m 1r → 1 + m 2 r → 1 + 2 t 1
5
2
For disc :
m 1 + m 2
d R → m d r→ 1
+ m d r→ 2
S = 1 g singθ 2
2
dt dt
2
=
1 + 1 t 2
dt
2
m 1 + m 2
V→
cm = m 1v → 1 + m 2 v → t 1
2
=
t 2
√⎺⎺ 14
15
m 1 + m 2
d V → cm
= d m 1 v → 1 + m 2 v → 41.(b) τ = Fr = I 1 α 1 = I 2 α 2
2
dt dt
I
m 1 + m
1 > I 2
2
α 1 < α 2
(27)

AISECT TUTORIALS : PHYSICS : SET-4
43.(c)
or α 2 > α 1
Down and left, as the inner wheel has a
tendency to leave the ground first. To
avoid toppling the centre of gravity
should be as low as possible.
44.(d) I 1 ω 1 = I 2 ω 2
46.(c)
2 2π
5 MR2 24 = 2 5 M ⎛ R⎞
⎜ ⎝
2⎟
⎠
T’ = 6 hours.
2
× 2π T
The situation is shown in the figure
88.(b)
(2) rco2
; tan θ =
(1) g
R sin θ ω 2
=
g
R sin θ ω2
g
= sin θ
cosθ
or ω = √⎺⎺⎺ g ⎺
R cos θ
Change in K.E. = Work done
= 1 2 I ω 2 2 − 1 2 Iω 1 2
∴ 1 2 I [(12π)2 −(2π) 2 ] = 484
r
R
= Sin θ
Fig.42
N cos θ = mg (1)
N sin θ = mrω 2 (2)
∴ I = 0.7 kg ⁄ ms.
89.(b) W = mg ∴ 9.80 N = m 9.8
∴ m = 1 kg
F = ma ∴ 5 N = 1 a
∴ a = 5m /sec 2
91.(d) Circumference of a circle = 2π R.
Athlete completes 2π R distance in 40
seconds and hence in 140 seconds he
will cover 7π R distance
(28)