Vidyamandir Classes Rotational Motion

Rotational Motion

ROTATIONAL KINEMATICS Section - 1

We have already studied the rotation of a point mass. Presently, we are going to analyse the rotation of a
rigid body. Let us consider the example of a disc of radius R and mass M rotating about a fixed axis passing
through its centre as shown :
This rotating disc can be imagined as a group of infinite point masses revolving in
circles. The points which are at different distances from axis are moving with
different speeds in circles of different radii. The time period (T) and the angular
velocity ( ) of revolution (  2 / T ) are same for each point as all of them
complete one revolution in same time interval. For example, a point at a distance
r from the axis is rotating with speed r while a point on the circumference of
the disc is rotating with speed R. Points on the axis are at rest. Points on the
circumference have maximum tangential velocities.
If the disc is rotating with constant angular velocity , its angular displacement in
time interval t is simply given by   t.
If the disc is rotating with changing angular velocity, the rate of change of angular
velocity is known as angular acceleration ( ). If  is the change in angular
velocity in t , then :


 (units of  : rad/ s 2 )
t
If the angular velocity of rotation varies with time, the magnitude of tangential velocity (v  r ) for each
point in the disc also changes with time. The rate of change of magnitude of tangential velocity is known as
tangential acceleration (at ) given as :
v 
at  r  at  r 
t t
Note :  ,  are same for all particles in the body while v and at are different for different points. For a point at a
distance r from the axis, v  r and at  r

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Rotation with constant angular acceleration

If the angular velocity increases or decreases at a constant rate, the angular acceleration  is a constant.
Let i  angular velocity at some initial instant
 f  final angular velocity t seconds later
 f  i
Angular acceleration   
t
  f  i   t
If  is the angular displacement during these t seconds,

  i t  1 / 2 t 2

 2f  i2  2 

Note : These relations can be easily compared with those in uniformly accelerated linear motion. i.e.,
v  u  at ; s  ut  1/ 2at 2 ; v 2  u 2  2as
  ,  ,  can be clockwise or anti-clockwise. We can take one direction as positive and the other
direction as negative.
  and  are always in same direction.
  and  are opposite if the rotating body slows down. (compare with the case of retardation in linear
motion)

Illustration - 1 A wheel is rotating at the rate of 50 rev/min in the anti-clockwise direction. What should
be the magnitude and direction of the angular acceleration of the wheel so that it stops in 8 s ? How many
revolutions will it cover before stopping ?
SOLUTION :

Let i  initial angular velocity Hence angular acceleration of 5 / 24 rad/ s 2 must

i  2 (50 / 60)  5 / 3 rad/ s be imparted to the block in clock-wise direction.

here,  f  0 Angular displacement  is given as :

 f  i 0  5 / 3 5  2f  i2 0   5 / 3
2

   rad/ s 2
 
t 8 24
2  2  5 / 24 
20 
  radians.
3

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 10
Number of revolutions of  revolutions Hence it completes (3 + 1/3) revolutions before
2 3 stopping.
anti-clockwise.

Illustration - 2 A flywheel of radius 30 cm starts from rest and accelerates with constant angular accel-
eration of 0.5 rad/s2. Compute the tangential, radial and resultant accelerations of a point on its circumfer-
ence :
(a) Initially at  = 0° and (b) After it has made one third of a revolution.

SOLUTION :
(a) At the start : (b) After   120(2 / 3) :
  0.5 rad / s 2
 2f  i2  2  0  2(0.5) (2 / 3)
R = 0.3 m
  i  0 rad / s 2
 f  rad/ s
3
Radial acceleration  ar   2 R  0 m / s
ar   2 R  2 / 3(0.3)   / 5 m / s 2
Tangential acceleration  at  R
= (0.3) (0.5) = 0.15 m/s2 at  R  (0.3) (0.5)  0.15 m / s 2
Net acceleration = anet
2 2
anet = ar2  at2    0.15 
= ar2  at2  02  0.152 = 0.15 m/s2 25
= 0.646 m/s2

Illustration - 3 A wheel mounted on a stationary axle starts at rest and is given the following angular
acceleration :   9  1t (in SI units )
where t is the time after the wheel begins to rotate. Find the number of revolutions that the wheel turns before
it stops (and begins to turn in the opposite direction).
SOLUTION :
The kinematic equations do not apply because the We find the elapsed time t between
angular acceleration  is not constant.
0  0 and   0 by substituting these values :
We start with the basic definition :   d / dt
0 - 0 = 9t -6t2
to write
Solving for t, we obtain t = 9/6 = 1.50s
t t 2
  0   0  dt   0  9  12t  dt  9t  6t From   d / dt , we have
(in SI units)   0

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t t
0 0  
   dt   9t  6t 2 dt  4.5 t 2  2t 3
3.375
No. of revolution   0.53
Substituting 0  0 and t = 1.5s, we obtain 2

  0  4.5 (1.5) 2  2(1.5)3  3.375 rad

KINETIC ENERGY OF ROTATING BODIES & MOMENT OF INERTIA Section - 2

Consider a system of point masses rotating with angular velocity  about a

fixed axis through O. Kinetic energy of a point mass m1 located at P distant r1
from the centre is given by :
1 1
m1 v12  m1 r12  2
2 2

Kinetic energy of the system

1 2 1 2
= m1v1  m2 v2  . . . . .
2 2
1 2 2 1 2 2
= m1 r1   m2 r2   . . . . . . . .
2 2
1
=
2
 
m1 r22  m r22  m3 r32  ......  2

2
 2

The expression m1 r1  m 2 r2  ..... is known as moment of inertia of the system about the correspond-
ing axis. It is denoted by I.
I = m1r12 + m2 r22 + m3 r32 + .............

I   m r2
Hence kinetic energy of the system  1 / 2 I  2

Moments of Inertia of Continuous Bodies :

When the distribution of mass of a system of particles is continuous, the dis-
crete sum I =  mi ri2 is replaced by an integral. We have to sum the contri-
butions of infinitesimal mass elements dm shown in figure, each of which con-
tributes dI = r2 dm to the moment of inertia.The mass element should be cho-
sen such that all the particles on it are at the same perpendicular distance from
the axis. The moment of inertia of the whole body takes the form

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I   r 2 dm

Keep in mind that here the quantity r is the perpendicular distance to an axis, not the distance to an origin.
To evaluate this integral, we must express dm in terms of dr.

Note : Comparing the expression of rotational kinetic energy with 1/2 mv2, we can say that the role of moment of
inertia (I) is same in rotational motion as that of mass in linear motion. It is a measure of the resistance
offered by a body to a change in its rotational motion .

Radius of Gyration :
If M is the mass and I is the moment of inertia of a rigid body, then the radius of gyration (k) of a body is
given by :
I
k
M
I = Mk2

Moment of Inertia of some important bodies :

1. Circular Ring :
Axis passing through the centre and perpendicular to
the plane of ring.
I = MR2

2. Hollow Cylinder :
I = MR2

3. Solid Cylinder and a Disc :

About its geometrical axis :
1
I MR 2
2
About the perpendicular axis :
 2 R2 
I M   
 12 4 
 

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4. (a) Solid Sphere :

Axis passing through the centre :
I = 2/5 MR2

(b) Hollow Sphere :

Axis passing through the centre :
I = 2/3 MR2

5. Thin Rod of length l :

(a) Axis passing through mid-point and perpendicular to the length :

M 2
I
12

(b) Axis passing through an end and perpendicular to the rod :

M 2
I
3

6. Cuboid ( l × b × h) :
Axis through centre and parallel to the height (h) :

M 2 2
I 
12

 b 
Same result will apply for a thin plate also ( h  0) .
 and be are perpendicular to the axis of rotation.

Theorems on Moment of Inertia :

(i) Parallel Axis Theorem :
Let Icm be the moment of inertia of a body about an axis through
its centre of mass and let Ip be the moment of inertia of the same
body about another axis which is parallel to the original one.
If d is the between these two parallel axes and M is the mass of
the body then according to the parallel axis theorem :

Ip = Icm + Md2

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(ii) Perpendicular Axis Theorem :

Consider a plane body (i.e., a plate of zero thickness) of mass
M. Let X and Y axes be two mutually perpendicular lines in the
plane of the body. The axes intersect at origin O.
Let Ix = moment of inertia of the body about X-axis
Let Iy = moment of inertia of the body about Y-axis
The moment of inertia of the body about Z-axis (passing through
O and perpendicular to the plane of the body) is given by :
I z = Ix + Iy
The above result is known as the perpendicular axis theorem.
Illustration - 4 Calculate the moment of inertia of a :
(a) Disc about an axis passing through its edge and perpendicular to the circular base of the disc.
(b) Solid sphere about an axis touching the sphere at its surface.
SOLUTION :
(a) I = Icm + MR2 (b) I = Icm + MR2
1 2
I = MR2 + MR2 I = MR2 + MR2
2 5
3 7
I = MR2 I  MR 2
2 5
Try Yourself
Calculate the moment of intertia of a cuboid about an axis coinciding with one edge of the cuboid.

Illustration - 5 Calculate the moment of inertia of :

(a) A ring of mass M and radius R about an axis coinciding with a diameter of the ring.
(b) A thin disc about an axis coinciding with a diameter.
SOLUTION :
Similarly for a thin disc (i.e., a circular plate)
Let X and Y axes be along two perpendicular
diameters of the ring. Moment of inertia about a diameter is
By symmetry Ix = Iy 1 1 2 1 2
= I  MR   MR
and Iz = Ix + Iy 2 2  4
But we know that Iz = MR2
MR2 = Ix + Iy

MR2 = 2 Ix
MR 2
Ix = Iy =
2

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Illustration - 6 Find the moment of inertia of a circular disk or solid cylinder of radius R about the axis
through the center and perpendicular to the flat surface.
SOLUTION :

The moment of inertia of this element is

dI  dm r 2  2   r 3dr

For the whole body,

R 1
I  2  r 3dr   R 4
0 2
The figure shows that the appropriate mass el- The mass of the whole disk or cylinder is M
ement is a circular ring of radius r and width
dr.   A   R 2 , and so
Its area is dA  2 rdr and it s mass is I = 1/2 MR2
dm   dA; where   M / A is the areal
mass density.

Illustration - 7 Find the moment of inertia of a uniform solid sphere of mass M and radius R about a
diameter

SOLUTION :
The sphere may be divided into disks perpen-
dicular to the given axis, as shown in fig. The
disk at a distance x from the center of the sphere
has radius r = (R2 - x2 )1/2 and a thickness dx.

If   M / V is the volume mass density (mass

per unit volume), the mass of this elemental disk The total moment of inertia is
is
1 R
2 I    R 4  2 R 2 x 2  x 4  dx
dm   dV   r dx, or 2  R

R
dm    ( R 2  x 2 ) dx 1  2 1  8
   R 4 x  R2 x3  x5     R5
2  3 5   R 15
From the last example, the moment of inertia
of this elemental disk is The to tal mass of the sphere is
M = (4/3 R3), so the moment of inertia may
dI = 1/2 dm r2  1 / 2  (R2 - x2)2 dx
be written as :
I = 2/5 MR2.

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Illustration - 8 A disc of mass m and radius R has a concentric hole of radius r. Its moment of inertia
about an axis through its centre and perpendicular to its plane is :
1 1 1 1
(A)
2
2
m  R  r  (B)
2

m R 2  r 2 (C)  2
2
m  R  r  (D)
2
m R2  r 2  
SOLUTION :
M   4 4 
Let   mass per unit area =  R r
 R  r 
2 2 2
If we fill the hole with mass densities
 and   then the system can be treated as
a combination of two discs of radii r and R and
densities  and  
I = I1 + I2
1 1
 M 1R12  M 2 R22
2 2
M  R4  r 4  M  2 2
1 1   R r
     r 2  r 2       R 2  R 2  2 R  r
2 2  2
2 2

Illustration - 9 The radius of gyration of a solid sphere of radius R about a certain axis is R. The distance
of this axis from the centre of the sphere is :
(A) R (B) 0.5 R (C) 0.6 R (D) 0.4 R

SOLUTION :
I  I CM  Md 2 (parallel axis theorem)

2
MK 2  MR 2  Md 2 (But K = R)
5
2
 MR 2  MR 2  Md 2
5
 d  0.6 R

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ANGULAR MOMENTUM (L) Section - 3

(a) For a particle
Angular momentum about origin (O) is given as :

L  r   mv 
where r = position vector of the particle ; v = velocity

 L  mv r sin   mv(OA) sin   mvr

where r  perpendicular distance of velocity vector from O.

(b) For a particle moving in a circle of radius r with a speed v, its linear momentum is mv, its angular
momentum (L) is given as :

L  m v r  mvr
(c) For a rigid body (about a fixed axis)
L = sum of angular momenta of all particles
= m1v1r1 + m2 v2 r2 + m3 v3 r3 + . . . . .

 m1r12  m2 r22  m3 r32  .... (v  r )

 ( m1r12  m2 r22  m3 r33 ......)

L  I
(compare with linear momentum p = mv in linear motion)
L is also a vector and its direction is same as that of  (i.e., clockwise or anticlockwise)

Illustration - 10 A body of mass m is moving with a constant velocity along a line parallel to the x-axis,
away from the origin. Its angular momentum with respect to the origin
(A) is zero (B) remains constant (C) goes on increasing (D) goes on decreasing

SOLUTION :
L  m v r
L = mv (OA)
As OA and v are both contant, the angular momentum remains constant.

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Illustration - 11 A particle of mass 1 kg is moving along the line y = x + 2 (here, x and y are in meters) with
speed 2 m/s. The magnitude of angular momentum of particle about origin is :
(A) 4 kg  m 2 / s (B) 2 2 kg  m 2 / s (C) 4 2 kg  m 2 / s (D) 2 kg  m 2 / s
SOLUTION :
L  m v r L  1 kg   2m / s   2 m 
|0  0  2|
where r   2 L  2 2 kg m 2 / s
2 2
1 1

Illustration - 12 A particle of mass m is projected with a velocity v0 making an angle of 45° with the
horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the
particle is at its maximum height h is :

(A) zero (B) 

mv3 / 4 2 g  (C) mv3 /  2g  (D) m 2 gh3

SOLUTION :

L0  m v r

At the highest point, V = speed = V0 cos 45

V 2 sin 2 45
r  h  0
2g
V   V 2  mV 3
 L0  m  0   0   0  B is correct .
 2  4g  4 2 g

TORQUE () : Section - 4

Torque is a quantity which measures the capability of a force to
rotate a body. Torque due to a force is also known as the moment
of a force. It is defined as the product of the force and the per-
pendicular distance between the line of action of the force and the
axis of rotation. This perpendicular distance is known as the force
arm.
Torque    (force)  (force arm)
  F r
Units of torque are N-m.

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Torque is a vector quantity. Torque is clock-wise if the tendency of force

is to produce clock-wise rotation and vice-versa. In vector form, torque
due to a force F acting at a point A (whose position vector is r) is :
  r F
Taking magnitude, we can see that :   r F sin 
   F (r sin  )  F r

Couple
A pair of two equal and opposite forces acting along parallel lines but
having different lines of action constitutes a couple.
Moment of couple or torque  F 
= (force) × (perpendicular distance between forces)

Work done by a Torque

Consider a rigid body acted upon by a force F at perpendicular distance r from
the axis of rotation. Suppose that under this force, the body rotates through an
angle  .

Work done = force × displacement

W  F r. 

W   

Work done = (torque) × (angular displacement)

dW d
Power =   
dt dt
Newton’s IInd Law for Rotation
According to Newton’s IInd Law of rotation :
The rate of change angular momentum of a body is equal to the net torque acting on it.
L

t

   
I
t
    t 
 
  I

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Note : This important equation  I  can be compared with Newton’s IInd Law of motion (F = ma). Hence in
rotation, torque ( ) plays the role corresponding to force in linear motion.

Illustration - 13 A grind-stone is in the form of a solid cylinder has a radius of 0.5 m and a mass 50 kg.
(a) What torque will bring it from rest to an angular velocity of 300 rev/min in 10 s ?
(b) What is the kinetic energy when it is rotating at 300 rev/min ?

SOLUTION :
Let i  0 rad / s , (b) Kinetic energy of a rotating body
(RKE) :
 f  2 (300 / 60)  10  rad / s 2
1 2 1 1 
RK E I   MR 2   2
 f  i 10  0 2 2 2 
(a)     rad / s 2
t 10 1  1   2   2
  50 0.5  10 
1 2
2 2 
Torque required =   I   MR  
2  R K E = 3084 J

   1 / 2 (50) (0.5) 2   19.6 Nm

Illustration - 14 Calculate the torque developed by an airplane engine whose output is 2000 hp at an
angular velocity of 2400 rev/min.
SOLUTION :

  2 (2400 / 60)  80  rad / s. Power = work done per sec  
t
Power   
Work done by torque
2000  746
= (torque) × (angular displacement)   5937 Nm
80

Illustration - 15 In the given figure, calculate the linear acceleration of the blocks.
Mass of block A = 10 kg
Mass of block B = 8 kg
Mass of disc shaped pulley = 2 kg
(take g = 10 m/s2)
SOLUTION :
Let R be the radius of the pulley and T1 and T2 Let m1 = 10 kg ; m2 = 8 kg ; M = 2 kg.
be the tensions in the left and right portions of Let a be the acceleration of blocks.
the string.

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1
(iii) T2 R  T1R = MR 2 
2
The linear acceleration of blocks is same as
the tangential acceleration of any point on the
circumference of the pulley which is R  .
(iv) a  R
Dividing (iii) by R and adding to (i) and (ii),
For the blocks (linear motion) M
m2g - m1 g = m2a + m1a + R
2
(i) T1 - m1g = m1a
 M
(ii) m2g - T2 = m2a  m2 g  m1g   m2  m1  a
 2 
m2  m1
a g
For the pulley (rotation) M
m2  m1 
2
Net torque  I 
10  8 g 
20
m / s2
=
2 19
10  8 
2

Illustration - 16 A uniform rod of length L and mass M is pivoted freely at one end.
(a) What is the angular acceleration of the rod when it is at angle  to the vertical ?
(b) What is the tangential linear acceleration of the free end when the rod is horizontal ? The moment of
inertia of a rod about one end is 1/3 ML2.

SOLUTION :
The Figure shows the rod at an angle  to the
vertical.
If we take torques about the pivot we need not
be concerned with the force due to the pivot.
The torque due to the weight is mgL/2 sin,
so the second law for the rotational motion is When the rod is horizontal   / 2 and
  3g/ 2 L.
mgL ml 2
sin    The tangential linear acceleration of the free end
2 3
is
3g sin 
Thus   3g
2L at   L 
2

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Vidyamandir Classes Rotational Motion

GENERAL RIGID BODY MOTION Section - 5

We now analyse the motion of objects that move through space while also rotating : for example, a ball
rolling down an incline, or an object thrown into the air tumbling end-over-end as it moves. This complex
motion can be described simply as a combination of translational motion of the CM and a pure rotational
motion about the C.M.
Kinetic Energy
The total kinetic energy of a body which is moving through space as well as rotating is given by :
K = Ktranslational + Krotation

1 2 1
K MVCM  I CM  2
2 2
where VCM = velocity of the centre of mass
ICM = moment of inertia about CM
 = angular velocity of rotation
Instantaneous Velocity of a point
All points on a rigid body have instantaneous velocity v that is the vector sum of two velocities : the velocity

Vcm of the CM plus the tangential velocity Vt of the point relative to CM (at right angles to the radius vector
r of the point drawn from the CM). For a point A,
V A  V CM  V t

Where V t  r (r = distance of A from CM)

And Vt is directed along the tangent to the circular trajectory about CM.

Roll ing without Slipping

The figure shows a wheel rolling without slipping along the horizontal stationary surface. As the wheel turns
through an angle  = s/r, the arc length s equals the horizontal distance x that the CM of the wheel moves.
In a sense, the arc length s is “laid down” along the horizontal distance x. We can think of the wheel’s motion
as made up of two separate motions : the linear translational motion of the CM combined with pure
rotation about the CM (ignoring the linear motion of CM).
The relations between corresponding variables are :
x  r
(As P is at rest, VCM and r balance).

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(As P is at rest, VCM and r balance).

Vcm  r 

Acm  r 
Another feature of rolling without slipping is that the instantaneous velocity of the point of contact with the
surface is always zero. The highest point Q has a velocity  Vcm  r  2r

Illustration - 17 A rigid body of radius of gyration k and radius R rolls (without slipping) down a plane
inclined at an angle  with horizontal. Calculate its acceleration and the frictional force acting on it.

SOLUTION :
When the body is placed on the inclined plane,  v  R and ACM  R 
it tries to slip down and hence a static friction f
acts upwards. This friction provides a torque Solve the following three equations for a and
which causes the body to rotate. Let ACM be f:
the linear acceleration of centre of mass and  mg sin   f  ma
be the angular acceleration of the body.
fR  mk 2
From force diagram :
For linear motion parallel to the plane ACM  R 
mg sin   f  ma mg sin 
g sin  f 
ACM  and
k2 R2
For rotation around the axis through centre of 1 1
mass R2 k2

Net torque  I   f R  (mk 2 ) We can also derive the condition for pure
rolling (rolling without slipping) :

To avoid slipping , f   s N
g sin 
 s mg cos 
1  R2 / k 2
tan 
 s 
R2
1
k2
As there is no slipping, the point of contact of
the body with plane is instantaneously at rest. This is the condition on µs so that the body rolls
without slipping.

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Illustration - 18 A solid cylinder rolls down an inclined plane of height h and inclination . Calculate its
speed at the bottom of the plane using acceleration-method and energy-method. Also calculate the time
taken to reach bottom.
SOLUTION :
Energy Method :
Let VCM ,  be the velocity of centre of mass
and the angular velocity of cylinder respectively
at the bottom of the plane.
As the cylinder rolls down without rubbing, NO
energy is lost due to friction as heat.
Loss in G.P.E. = gain in KE
2
mgh 
1 2
mVCM
1
 I 2
Using VCM  02  2 ACM S
2 2
as down the plane
As the cylinder is rolling without slipping, (taking downward direction positive)

h
VCM  R  v 0  2a
sin 
2 
1 2 1  mR 2   VCM
 mgh  mVCM    2  2  h 4gh
2 2 2  R   2  g sin   
3  sin  3
4 gh
 VCM  Time to reach bottom = t
3

Acceleration Method : 4gh

0
From the result of last illustration v0 3
 
a 2
(using k2 = R2/2) g sin 
3
Acceleration of the cylinder is :
g sin  2 3h 1
 g sin  =
1 3 g sin 
1
2

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Illustration - 19 A solid cylinder of mass m = 4 kg and radius R = 10 cm has two ropes wrapped around it,
one near each end. The cylinder is held horizontally by fixing the two free ends of the cords to the hooks on the
ceiling such that both the cords are exactly vertical. The cylinder is released to fall under gravity. Find the
tension in the cords when they unwind and the linear acceleration of the cylinder.

SOLUTION :
Also, the linear acceleration of cylinder is same
Let ACM = linear acceleration of CM. as the tangential acceleration of any point on its
  angular acceleration of the cylinder.. surface.

For the linear motion of the cylinder : ACM  R 

mg - 2T = ma Combining the three equations, we get :
m
For the rotational motion : mg  ma  a
2
Net torque = ICM 
2g
 a  6.53 m / s 2
1 2 3
2TR =  mR  
2  mg  ma
and T  6.53 N
2
Illustration - 20 A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at
the top of a smooth incline and released. Least time will be taken in reaching the bottom by :
(A) the solid sphere (B) the hollow sphere (C) the disc (D) all will take same time.
SOLUTION :
On a smooth incline, the only accelerating force
is mg sin  acting at the CM.
N = mg cos 

mg sin   mACM  ACM  g sin 

 CM  0    0 2 2
t = same for all = A 
CM g sin 
Hence all the bodies will slip down with same
acceleration and no spin. where   length of the incline.
 D is correct .

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Illustration - 21 A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at
the top of an incline and released. The friction coefficients between the objects and the incline are same and
not sufficient to allow pure rolling. Least time will be taken in reaching the bottom by
(A) the solid sphere (B) the hollow sphere (C) the disc (D) all will take same time

SOLUTION :
Due to slipping, kinetic friction k mg cos 
will act on all.

mg sin  – mg cos  = mACM

 ACM  g (sin    cos  ) is agan 2
Hence all will take same time t = g  sin    cos  
same for all bodies.
 D is correct .

Illustration - 22 In the previous question, the smallest kinetic energy at the bottom of the incline will be
achieved by
(A) the solid sphere (B) the hollow sphere
(C) the disc (D) all will achieve same kinetic energy
SOLUTION :

As  CM    mg cos   R  I CM 
1 2 1
 KE  MVCM  ICM  2
 mgR cos   gR cos  2 2
 
2
mk k2 1 1   gR cos  2 t 2
2
 MACM t 2  Mk 2
 will be different for each one of the bodies 2 2 k4
and hence they will gain different angular velocities
1 1 M
at the bottom of the plane given by :  2
MACM t2 
2
  gR cos  2 t 2
2 2 k
  0   t
KE will be least for the body whose k2 is greatest.
 gR cos  t Hence hollow sphere will have least KE.
0 
k2  B is correct .

Illustration - 23 A hoop rolls on a horizontal ground without slipping with linear

speed v. Speed of a particle P on the circumference of the hoop at angle  is :
   
(A) 2v sin   (B) v sin  (C) 2v cos   (D) v cos 
2 2

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SOLUTION :
For rolling without slipping, (velocity of P)2 = (VCM  r cos  ) 2  ( r sin  ) 2

VCM  r 2
 VCM (1  cos  ) 2  V 2 sin 2  (as r  VCM )

2 2
= 4VCM sin
2

 velocity of P  2VCM sin
2

The velocity of P is the resultant of CM velocity

and the tangential velocity r due to clockwise
rotation

Illustration - 24 A ring of mass m and radius R has three particles attached to

the ring as shown in figure. The centre of the ring has a speed v0. The kinetic
energy of the system is : (Slipping is absent).
(A) 6 mv02 (B) 12 mv02 (C) 4 mv02 (D) 8 mv02

SOLUTION :
Velocity of any point in the ring is the vector
sum of V CM and the tangential velocity R 
due to spin about CM.

For rolling without slipping, VCM  R 

V A  VCM  R  2VCM 1

2

m V A2  VD2  2 VB2 
2
VB  VCM  R 2 2  2 VCM  2 1 
  mVCM  ICM  2 
2 2 
2
VD  VCM  R 2 2  2 VCM 1 2
 m  4  2  4  VCM
2
1 1 1 1  2 2 
Ktotal  mVA2  mVD2   2m  VB2  K ring 2 VCM
2 2 2  m  VCM  R   6m V 2
2  R2  CM

 The correct answer is A .

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NOW ATTEMPT IN-CHAPTER EXERCISE-B BEFORE PROCEEDING AHEAD IN THIS EBOOK

Vidyamandir Classes Rotational Motion

CONSERVATION OF ANGULAR MOMENTUM Section - 6

(a) Angular impulse about a point (J)

Linear impulse, I   F dt

Similarly, angular impulse J    dt

J    dt

J    Fr  dt

   F dt r 
J  I r (Compare with   Fr to remember)

(b) Impulse-Momentum Theorems :

(i) Linear impulse = I   Fdt

dp
 dt
dt

  dp

 p
 I  p

(ii) Angular impulse = J    dt

dL
= J  dt
dt
=  dL  L
 J  L

This equation is usually applied about CM.

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(c) Conservation of Angular Momentum

If external angular impulse about a point P during the collision is zero, then the angular momentum
about that point can be conserved.
If JP = 0
 L p  0

 ( Li ) p  ( L f ) p

Illustration - 25 A billiard ball, initially at rest, is given a sharp impulse by a rod. The rod is held horizontally
at a height h above the centre of the ball. The ball immediately begins to roll without slipping after the
impact. Calculate the height h in terms of the radius of the ball.
SOLUTION :
The horizontal force creates a linear impulse
and angular impulse.

I   Fdt

J    dt  Ih
Linear impulse = change in linear momentum Dividing (ii) by (i) we get :
 I  MVCM  0 . . . .(i) 2Mr 2
h (Using VCM  r )
Angular impulse about CM = change in angular 5MVCM
momentum 2
h r
2 5
J  Ih  mr 2  0 . . . .(ii)
5

Illustration - 26 In a “zero - g” environment, a thin uniform rod of length  is initially at rest with
respect to an inertial frame of reference. The rod is tapped at one end perpendicular to its length. How far
does the center of mass translate while the rod completes one revolution about its center of mass ?
SOLUTION : Impulse on the rod :
The “impulsive tap” delivered perpendicularly dP
F 
to the rod at one end gives some linear dt
momentum to the CM of the rod and also some
angular momentum about the CM.

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Using Impulse-Momentum equation, Let t be the time required for one complete
I  MVCM  0 revolution of 2 rad. Multiplying both sides
by t gives
 M 2
I  0  2 
2 12 
VCM  t      t
2  12 
 M 2
 MVCM   X CM 
 
2 12
6

X CM  as   2
3

Illustration - 27 A turntable rotates about a fixed vertical axis, making one revolution in 10 s. The moment
of inertia of the turntable about the axis is 1200 kg m2. A man of mass 80 kg initially standing at the centre
of the turntable, runs out along a radius. What is the angular velocity of the turntable when the man is 2m
from the centre ?
SOLUTION :

I0 = initial moment of inertia of the system I = 80(2)2 + 1200 = 1520 kg m2

I0 = Iman + Itable By conservation of angular momentum:

I0 = 0 + 1200 = 1200 kg m2 I 00  I

(Iman = 0 as the man is at the axis)
Now 0  2 / T0  2 / 10   / 5 rad / s
I = final moment of inertia of the system
I  1200  
I = Iman + Itable  0 0 
I 1520  5
I = mr2 + 1200
 0.49 rad / s

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Illustration - 28 The cylinder shown in figure has a fixed axis and

is initially at rest. The block of mass M is initially moving to the
right without friction with speed v1. It passes over the cylinder to
the dotted position. When it first makes contact with the cylinder, it
slips on the cylinder but the friction is large enough so that slipping
ceases before M loses contact with the cylinder.
Calculate the final speed v2 in terms of v1 , M, I (moment of inertia
of cylinder) and radius (R).
SOLUTION :
Let  be the angular velocity acquired by the As the slipping ceases, v2  R
cylinder as a result of contact. The only impulse on
v
the system is acting at O. Hence angular impulse  Mv1R  M v2 R  I 2
about O is zero. R
Applying conservation of angular momentum to the Mv1R v1
 v2  
block-cylinder system about the axis through O : I I
MR  1
R MR 2
Mv1R  Mv2 R  I 

Illustration - 29 A small block of mass 4 kg is attached to a cord passing through a hole in a horizontal
frictionless surface. The block is originally revolving in a circle of radius of 0.5 m about the hole, with a
tangential velocity of 4 m/s. The cord is then pulled slowly from below, shortening the radius of the circle in
which the block revolves. The breaking strength of the cord is 600 N. What will be the radius of the circle
when the cord breaks ?
SOLUTION :
 T2  600 N
mv1r1 = mv2r2 and T2 = mv22/r2

r2 T2
 mv1 r1  m r2
m
The tension of the rope is the only net force on
1/ 3 1/ 3
the block and it does not exert any torque about  mv 2 r 2   4  16  0.25 
the axis of rotation. Hence the angular momentum r2   1 1   
of the block about the axis should remain  T2   600 
conserved. 1/ 3
 16 
 m v r  constant  r2    3. 0 m
 600 
Let r1 = 0.5 m and v1 = 4 m/s.
Note :The tension in the string is inversely proportional
Let r2, v2, T2 be the radius, velocity and tension
to the cube of the radius.
when the string breaks.

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Illustration - 30 A horizontal disc rotates freely about a vertical axis through its centre. A ring, having the
same mass and radius as the disc, is now gently placed on the disc. After some time, the two rotate with a
common angular velocity.
Which of the following are true :
(A) Some friction exists between the disc and the ring
(B) The angular momentum of the ‘disc plus ring’ is conserved
2
(C) The final common angular velocity is rd of the initial angular velocity of the disc
3
2
(D) rd of the initial kinetic energy changes to heat
3

SOLUTION : (ABD)
(A) As the ring is placed on the rotating disc, they
rub against each other exerting kinetic friction
as shown. This kinetic friction produces torque
which opposes the motion of the disc and slows
it down to angular velocity . The kinetic friction
on the ring accelerates it from rest to final
angular velocity . When  for both becomes
same, they stop rubbing.

(B) As the torques of frictional forces are internal,

the angular momentum of system remains
conserved. (D) Loss in KE = Ki – Kf

MR 2 2  MR 2  2
(C) I disc 0  0  I disc   I ring   0    MR 2  0
2  2  9
( MR 2 / 2) 0 0 1
 2
MR / 2  MR 2

3

3

MR 2 02 
20 2 1 2 2 2
    MR 0   K i
3 3 2  3

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Illustration - 31 A uniform rod AB of mass m and length l is at rest on a smooth horizontal surface. An
impulse I is applied to the end B perpendicular to the rod in horizontal direction. Speed of particle P at a
 ml
distance l/6 from the centre towards A of the rod time t  is :
12 I
I I I I
(A) 2 (B) (C) (D) 2
m 2m m m

SOLUTION :
Using impulse-momentum equations, In the final position of rod,
I = MVCM – 0

 M 2
I  0
2 12
I
VCM 
M
6I
and 
M  2 2
VP2  VCM
2

M 36
After time t 
12 I I2 36 I 2 2 2I 2
VP2   
 M   6I   M2 36 M 2  2 M2
  t   
12 I  M   2
2I
Thus the rod has rotated through a right angle. VP   D is correct.
M

NOW ATTEMPT IN-CHAPTER EXERCISE-C BEFORE PROCEEDING AHEAD IN THIS EBOOK

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SUBJECTIVE SOLVED EXAMPLES

Example - 1 The end A of the rod AB is being pulled on the floor with a
constant velocity v0 as shown. Taking the length of the rod as  , Calculate :
(a) The velocity of end B (b) The angular velocity of the rod

(c) The velocity of the CM of the rod at the instant when   37.

SOLUTION :
Let OA = x, OB = y d v0 5v
   0
dt  sin 37 3
x 2  y 2  2 and x   cos .
x
(c) xCM 
(a) Differentiate x 2  y 2  2 with respect to time. 2
1 dx v0
2x
dx
 2y
dy
0
 VCM x   
2 dt 2
dt dt
y
dy x dx 4 yCM 
VB     VA cot   V0  2
dt y dt 3
1 dy vB 2
(b) Differentiate x   cos  with respect to time.  VCMy    v0 
2 dt 2 3
dx d
   sin  v02 4v02
dt dt  VCM  
4 9
d VA
 5 4
dt  sin   v0 at tan 1 below horizontal.
6 3
(negative sign indicates that  is decreasing)

Example - 2 Consider the arrangement shown in figure. The string is

wrapped around a uniform cylinder which rolls without slipping. The other
end of the string is passed over a massless, frictionless pulley to a falling
weight. Determine the acceleration of the falling mass in terms of only the
mass of the cylinder M, the mass m and g.

SOLUTION :
Let T be the tension in the string and f be the force of (static) friction between the cylinder and the surface.

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Subjective to Remember
Solved Examplees 27
 Ratational Motion Vidyamandir Classes

a = downward acceleration of block m Vm = velocity of the highest point of cylinder

Vm  VCM  r 

Taking derivatives,
a  ACM  r  …(iv)
We also have : (for rolling without slipping)
ACM  r  …(v)
Multiply (i) by r, (ii) by 2r and add to (ii).
ACM  acceleration of the CM of cylinder
2 mgr  MACM r  2 m a r  1 / 2 Mr 2
towards right
  angular acceleration o f cylinder From (iv) and (v), we have a  2r ,
(clockwise) ACM  r 

T  f  M ACM …(i)

2mgr  Mr 2  4mr 2  1 / 2Mr 2  
Taking torque about the CM,
4mg


Tr  fr  1 / 2 Mr 2   …(ii)  3M  8m  r
The acceleration of falling mass m is :
mg  T  ma …(iii)
8mg
The string attaches the mass m to the highest a  2r 
point of the cylinder. 3M  8 m

Example - 3 A rectangular rigid fixed block has a long

horizontal edge. A solid homogenous cylinder radius R is placed
horizontally at rest with its length parallel to the edge such that
the axis of the cylinder and the edge of the block are in the same
vertical plane as shown in figure. There is sufficient friction present
at the edge so that a very small displacement causes the cylinder
to roll of the edge without sipping. Determine :

(a) The angel c through which the cylinder rotates before it

leaves contact with the edge,
(b) The speed of the centre of mass of the cylinder before leaving
contact with the edge, and
(c) The ratio of the transitional and rotational kinetic energies
of the cylinder when its centre of mass is in horizontal line
with the edge.

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SOLUTION :
1 2 1
As the cylinder rotates about the edge, its CM mgr 1  cos    mVCM  I CM 2
moves along a circular arc of radius r centred at 2 2
the edge. Let N be the normal reaction from
where VCM  r 
edge at the instant when cylinder has rotated
through an angle . Combine all the three equations to get :

4 4
  cos 1 and VCM  rg
7 7

1 
(c) The rotational kinetic energy  I CM 2 
2 
become constant after the cylinder leaves the
mv 2 edge.
mg cos θ  N 
r
1 11 4 g M gr
At jump off point, N = 0 K rot  ICM 2   Mr 2  
2 22 7 r 7
Total kinetic energy at the instant when CM is in
mv 2
 mg cos   horizontal line with edge is K.
r
K = loss in GPE from initial position = m g r
We can also apply conservative of mechanical
energy as there is no slipping. The gravitational mgr 6
K trans  mgr  K rot  mgr   mgr
potential energy decreases and hence the kinetic 7 7
energy increases. K trans
 6
K rot

Example - 4 A solid sphere of radius r and mass m rolls without slipping

down the track shown in the figure. At the end of its run at point Q its center-
of-mass velocity is directed upward. The lower portion of track is circular of
radius R.
(a) Determine the force with which the sphere presses against the track
at B.
(b) Upto what height does the CM rise after it leaves the track ?

SOLUTION :
1 2 1
From A to B mg 10 R   mVcm  I cm 12
2 1 2
Loss in GPE = gain in KE
For rolling without slipping on a fixed surface.

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Vcm1  R1

The CM follows a circular path of radius R – r

AT B, net force towards centre
2
mVcm
 N  mg 
Rr

From Q to P,  does not change because

m 100 gR  mg 107 R  7r 
 N  mg   about C.M torque is zero in air.
7R  r 7R r
gain in GPE = loss in KE

(b) From A to Q, mg  9 R  r  1 2
 mg  gain in height  mVcm
2 2

2
1 2 1  Vcm2  V 2cm2 5
 mvcm  I cm    h  9R  r 
2 2 2  r 2g 7
 
52 R 5r
 height above the base  R  h  
7 7
Vcm2  rw2 at Q 
Example - 5 The descending pulley (disc shaped) shown in figure
have a radius 20 cm and moment of inertia 0.20 kg-m2. The fixed pulley
is light and the horizontal plane frictionless. Find the acceleration of the
block if its mass is 1.0 kg.

SOLUTION :
Let T1 and T2 be tension in the strings to the left Mg  T1  T2  MA …(ii)
and right of the pulley.
Taking torque about CM of the pulley
a = acceleration of block towards right
A = acceleration of CM of pulley downward T2 R  T1R  I  …(iii)

  angular acceleration of pulley (anticlock wise) The string attaches the mass m to the left point
of the cylinder.

 a  A  R …(iv)

1
For disc, I  MR 2 …(v)
2
We also have : (for right part of the string

A R …(vi)
T1  ma …(i)

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Solving these equation for a, we get : 4  0.2  9.8

 a 2
4 Ig 4 1   0.2   3  0.2

4m2  3I  5.96 m / s 2

Example - 6 In the arrangement shown in figure a weight A

possesses mass m, a pulley B possesses mass M. Also known are the
moment of inertia I of the pulley relative to its axis and the radii of
the pulley R and 2R. The mass of the threads is negligible. Find the
acceleration of the weight A after the system is set free.

SOLUTION :

T1  tension in string attached to ceiling

T2  tension in string attached to block mg  T2  ma …(ii)

Taking torque about CM.

2T1  R   T2  2 R   I  …(iii)
For the string winding over smaller part of the
spool, X is at rest.
Acm  R …(iv)
And for the string winding over the bigger part of
spool, acceleration of point Y = acc. of block.
Acm : acceleration of CM of spool downward Acm  2 R  a …(v)
a: acceleration of block down wards Solving these five equation for a, we get :

 : angular acceleration of spool clockwise. 3 g  M  3m 

a
Mg  T2  2T1  MAcm …(i) M  9m  I / R 2

Example - 7 A grindstone 1m in diameter, of mass 50 kg, is rotating at 900 rev/min. A tool is pressed
normally against the rim with a force of 200 N, and the grindstone comes to rest in 10 seconds. Find the
coefficient of friction between the tool and the gridstone. Neglect the friction in the bearings.

SOLUTION :
Let   coefficient friction between the tool and
the grid stone.

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Initial angular velocity i  900 rev/min Using  f  i  t, we have

900  2
 red / sec  2 F  
60  0  30   t
 MR 
 30 rad/sec (anticlock wise)
30 MR 
 
Final angular velocity   f  0 2 Ft

Taking torque about CM. 30  50  0.5  3.14

  0.5890
1 2  200 10
fR  I   fR  MR 2
2
2 f 2F
and   (clockwise)
MR MR

Example - 8 A carpet of mass M made of an inextensible material is rolled along its length in the form
of a cylinder of radius R and is kept on a rough floor. The carpet starts unrolling without sliding on the floor
when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part
of the carpet when its radius reduces to R/2.

SOLUTION :
At radius is reduced to R/2, mass of rolling
part reduced to M/4
Loss in GPE = gain in KE

M R
 MgR  g  For rolling without slipping,
4 2
R
1 M  2 1 Vcm   …(iii)
   Vcm  I 2 …(i) 2
2 4  2
I = moment of inertia of rolling part Solving these equation, we get :

2
1  M   R  MR 2 14 g R
     Vcm 
…(ii) 3
2 4   2  32

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Example - 9 A spherical ball is given a translation velocity equal to v0 and pushed along a horizontal
surface where the coefficient of friction is . Calculate the time after which it begins to roll without sipping.
Also calculate the velocity of entre of mass after it begins to roll without slipping.

SOLUTION :

The point of contact has a forward velocity when We also have v  r  for pure rolling …(iii)
the ball starts moving. This causes slipping and
hence a kinetic friction equal to mg act in  v k2  1
backward direction. This frictional force provides t  0 
 k 2  r 2  g
 
 where I  mk 2 
a clock-wise torque and hence the angular
velocity increases from zero  in time t. The
same frictional force reduces the linear velocity v0 r 2
and v 
of centre of mass from v0 to v in time t. k2  r2
If the ball begins to roll without slipping after time
t, For a solid sphere k 2  2 / 5 r 2

v  r 2v0
time (t) = and v  5 / 7 v0
7 g
 v  v0   gt …(i)
2v0
Using  f  i  t and   I for rotation : Hence seconds, the translational speed of
7 g
  mgr  ball reduces to 5 / 7 v0 and it begins to roll
  0  t …(ii)
 I  without slipping.

Example - 10 A Solid sphere is set into motion on a rough horizontal

surface with a linear speed v0 in the forward direction and an angular speed
0 the anticlockwise direction as shown in the figure.

V0
Find the value of ration so that sphere comes to rest permently after
r0
some time calculate this time also. The coefficient of friction is .

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SOLUTION :
The point of contact of the sphere and ground is
slipping in forward direction. A kinetic friction acts
in backward direction which causes backward
acceleration of C.M. and clockwise torque.
Therefore, velocity of centre of mass and angular
velocity both decrease with time.  mgr 5 gt
  t   0  t  0   0 …(ii)
2 2 2 r
f k  mg  mACM CM  f k r  ICM  mr
5

  mgr From (i) and (ii) we get :

 ACM g and   CM 
ICM 2 mr 2 v0 2
5 
0 r 5
v  t   v0  gt  0 …(i) v
t 0
g

Example - 11 A board of mass M, whose upper surface is rough and under surface smooth, rests on a
smooth horizontal plane. A sphere of mass m is placed on the board and the board is suddenly given a velocity
v0 in the direction of its length. Find the time after which the sphere begins pure rolling, if the coefficient of
friction between the board and the sphere is .

SOLUTION :
Let a be the acceleration of block of mass M. Let ACM be the acceleration of C.M. of sphere. Let  be the
angular acceleration of sphere.

Force diagram When pure rolling stars, velocity of points P

and Q will become equal.

VP  VQ

VCM  r  v …(i)
(i) f k  Ma (ii) f k  MACM

mg
a ACM   g
M

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m 2
v  t   v0  gt …(ii) ICM  Mr 2
M 5
Solving, (i), (ii), (iii) and (iv) we get :
VCM  t   0   gt …(iii)
v
mgr t
t   0  t …(iv) 7 m 
ICM    g
2 M 

Example - 12 A billiard ball (of radius R), initially at rest is given a

sharp impulse by a cue. The cue is held horizontally a distance h above
the central line as shown. The ball leaves the cue with a speed V0, and
after some time acquires a final of 9/7 V0. Show that h = 4/5 R.

SOLUTION :

Let v0 be the linear velocity and 0 be the angular For rotation :

velocity imparted. frictional torque (mgR) opposes rotation and
hence decreases .
 linear impulse   Fdt  mv0  0
 mgR mgR …(iii)
   0  t
2 I 2 2
angular impulse   Fhdt  mr 2  0  mR 
5 5 

 h
 2 / 5 mR 2  0
After time t : V  9 v0  R
mv0 7
Combining all the equations,
2
2 R 0
 v0  …(i) 9  9v  2R
5h V  v0  v0    0  0 
7  7R  5
As linear velocity increase to 9/7 v0, friction must
be in forward direction and hence oppose angular [Eliminating  g t from (ii) and (iii)]
motion.
Substitute for v0 from I.
For linear motion :
friction (mgR) increase velocity. 2  2 R 20   9 2 R0  2 R
     0 
7  5 h   7 5h  5
(a = g) (t = time taken to start rolling)
V = V0 + gt …(ii) 4R
 h .
5

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Example - 13 A man of mass 100 kg stands at the rim of a turntable of radius 2 m and moment of inertia
4000 kg.m2 mounted on a vertical frictionless shaft at its centre. The whole system is initially at rest. The man
now walks along the outer edge of turntable with a velocity of 1 m/s relative to the earth.
(a) With that angular velocity and in what direction does the turntable rotates ?
(b) Through what angle will it have rotated when the man reaches his initial position on the turntable ?
(c) Through what angle will it have rotated when the man reaches his initial position relative to earth ?

SOLUTION :
Let the man be moving anti-clockwise 2  mt  t t (where t is the time taken)
(a) By conservation of angular momentum on the
2 2 2
man-table system t   s
m  t 0.5   0.05  0.55
Li  L f
 angular displacement of table is :
0  0  I m m  It t
 2 
t  t t   0.05   
I   0.55 
 t   m m where m  v / r  1 / 2
It
 2 
rad/s     radians
 11 
2 1
100  2     2
 t   2 The table rotates through radians clockwise.
11
4000
(c) If the man completes one revolution relative to
1 the earth, then :
 t   rad / s
20
m  2
Thus the table rotates clock-wise (opposite to
man) with angular velocity 0.05 rad/s. time  2 / m  2 / 0.5
(b) If the man completes one revolution relative to During this time, angular displacement of the table
the table, then :
 2 
mt  2  2  m  t t  t (time)  0.05   
 0.5 

t    / 5 radians.

t  36 in clock-wise direction

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Example - 14 Two skaters, each of mass 50 kg, approach each other along parallel paths separated by 3
m. They have equal and opposite velocities of 10 m/s. The first skater carries a long light pole, 3 m long, and
the second skater grabs the end of it as he passes (assume frictionless ice).
(a) Describe quantitatively the motion of the skaters after they are connected by the pole.
(b) By pulling one the pole the skaters reduce their distance apart to 1 m. What is their motion then ?
(c) Compare the KE’s of the system in parts (a) and (b). Where does the change come from ?

SOLUTION :

(a) As the net linear momentum of the system (skater (b) As the separation reduces to 2 ' = 1 m,
+ pole) is zero, the centre of mass will be at rest
before and after the collision. I   I ' '
The skaters and the pole will rotate around the (conservation of angular momentum)
centre of mass (at the mid point of the of the
pole). I  2m2
'    9  60 rad/s
Applying the conservation of angular momentum I ' 2m 2
about an axis through C and perpendicular to
the plane of the figure,  angular velocity increases

2 1
mv  mv  I  where I  2m    I ' ' 2
K .E f 2  ' 2 ' 2
(c)   2 2 9
K .Ei 1 2  
 2mv  / I  v /  I
2
 20 / 3 rad/s.
The kinetic energy increases because the skaters
do positive work in pulling themselves towards
the centre of pole.

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Example - 15 Two uniform thin rods A and B of length 0.6 m each and
of masses 0.01 kg and 0.02 kg respectively are rigidly joined end to end. The
combination is pivoted at the lighter end, P as shown in figure such that it
can freely rotate about the point P in a vertical plane. A small object of mass
0.05 kg, moving horizontally hits the lower end of the combination and sticks
to it. What should be the velocity of the object so that the system could just
be raised to the horizontal position ?

SOLUTION :
[by parallel axis theorem
2
 
I P  ICM  M 2     ]
 2
Substituting the values in (ii),

Let velocity with which mass (m = 0.05 kg) I  9  102 Kgm 2

hits the rod. The only external impulse act at P.
Substituting for I in (i),
Apply conservation of angular momentum about
P as the angular impulse about P is zero.

 0.05 u 1.2   2u
Li  L f 9  102 3

= angular velocity of the system after impact. After impact, loss in KE of rod = gain in PE

 0.05 u  0.6  0.6   I  . . . . (i) (as the rod rotates)

where I = total moment of inertia of the system 1 2   3 

I   M1g    M 2 g    mg  2 
about P 2 2 2
2
I  I1  I 2  m  2  . . . . (ii) 1  4u 2  8
2

9 102   9  2 1




M  3M 2  4 m 
2 2
where I1  I A  M1  0.01  0.6
3 3
u 2  g / 2  30  0.27 
 M 2 2
   u  6.3 m / s
I2  I B   2  M 2     
 12  2  


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Example - 16
A target in a shooting gallery consists of a vertical square wooden board, 0.2 m on a side,
of mass 2kg, and negligible thickness pivoted on an axis along its top edge. It is struck at the centre by a bullet
of mass 5 gm, travelling 300 m/s. The bulled gets embedded in the board.
(a) What is the angular velocity of the board just after bullet’s impact ?
(b) What maximum height above the equilibrium position does the centre of the board reach before starting
to swing down again ?
(c) What bullet speed would be required for the board to swing all the way over after impact ?

SOLUTION :
(b) Loss in K.E. = Gain in P.E.

2
1 1   
I A 2  m     M  m  gh
2 2  2 

 h = 0.0214 m
(c) Loss in K.E. = Gain in P.E.
(a) During ccollision impulse on the system is acting
2
at point A. Angular momentum can be conserved 1 1  
I A2  m      M  m  g 
only about A. 2 2  2

 LA i   LA  f 24  M  m  g
   4m  3m  
  m 2 
mv  0  IA  
2  4 
  v  
 4M  3m 
(from (i))
6m
6m v
   5.6 rad / sec …(i) v = 918 m/sec
 4 M  3m    
Note that the moment of inertia of the square board will be same as that of a rod of length l as one of
the dimensions of the board perpendicular to the axis is negligible.

Example - 17 A particle of mass m is subject to an attractive central force of magnitude k/r2 where k is
a constant. A the instant when the particle is at an extreme position in its closed elliptical orbit, its distance

from the centre of force is ‘a’ its speed is k

. Calculate its distance from force-centre when it is at the
2ma
other extreme position.

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SOLUTION :
1 2  k  1 2  k 
mv1     mv2   
2  r1  2  r2 
From conservation of angular momentum about
C,

m v1 r1  m v2 r2
Let P be the particle and C be the force-centre. We have to find r2. Hence we eliminate v2.
P1 and P2 are its extreme positions at distances
r1 to r2 from C. 2
1 2 k 1  v1 r1  k
mv1   m   
k 2 r1 2  r2  r2
We have r1 = a and v1 
2ma
k
As the force is directed towards C, torque about Substituting v1  and r1  a
2ma
C is zero.
Hence we will apply conservation of angular 1 k k 1 ma 2 k 1
m   
momentum about C and conservation of energy. 2
2 2ma a 2 r2 2ma r2

F  k / r2 2
 3r2  4ar2  a 2  0
 potential energy (U) = -k/r
 r2  a,a / 3
(compare the expression of force with
gravitational force) The other extreme positon is at a distance of
From conservation of energy, a/3 from C.

total energy at P1 = total energy at P2 Note : If the mass m is a satellite in an elliptical orbit
around a planet of mass M, then k = GmM and
the planet is at force-centre C.

Example - 18 A meter stick lies on a frictionless horizontal table. It

has a mass M and is free to move in any way on the table. A hockey puck
m, moving as shown with speed v collide elastically with stick.
(a) What is the velocity of the puck after impact ?
(b) What is the velocity of the CM and the angular velocity of the stick
after impact ?

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SOLUTION : (iii) At colliding points

Vsep  eVapp

  
 VCM   v1   ev …(iii)
 2 
e =1 (Elastic collision)
There is no external impulse on the system.
Soving (i), (ii) and (iii) we get :
 Linear momentum is conserved
 4m  M 
and Angular momentum about any point is v1   v
conserved.  4m  M 

(i) Pi  Pf 2m
VCM  v
 4m  M 
mv  mv1  MVCM …(i)
 12m  v
(ii)  LCM i   LCM  f about CM of rod.  
  4m  M   l
 
 mv 
mv  0  1  I  …(ii)
2 2 CM

Example - 19 A rod of mass m and length  is held vertically on a

smoot horizontal floor. If the rod begins to fall from this position, find
the speed of its C.M. When it makes an angle  with vertical.

SOLUTION :
When rod is falling freely due to gravity, the only
force acting is that of gravity in vertical direction. 
YCM  cos 
2
External force in horizontal direction = 0
 d
 VCM ,x remains constant at 0 m/s.  VCM    sin  
2 dt
i.e. C.M. Moves in vertical direction.

VCM   sin  …(i)
Let VCM be the velocity of CM and  be the 2
angular velocity of rod when it make angle  with Loss in P.E. = gain in K.E.
the vertical.

Mg 1  cos  
2

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1 1
2
 MVCM  ICM 2 …(ii) g 1 2 2  3 sin 2   1 
2 2 1  cos       

2 2  12 

12 g 1  cos  


 1  3 sin 2 


VCM  sin 
2

  
 sin 2 sin  
Using (i) and (ii) VCM  6 g   
 1  3 sin2  
 
 1  2 2  2
g 1  cos     sin 2    2
2 2 4  24


NOW ATTEMPT OBJECTIVE WORKSHEET BEFORE PROCEEDING AHEAD IN THIS EBOOK

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Vidyamandir Classes Rotational Motion

THINGS TO REMEMBER - ROTATIONAL MOTION

1. Rotation about a fixed axis :

(a) ,  are same for all particles in the body while v and at are different for different points. For a point
at a distance r from the axis, v = r and at = r 
1 2
(b)  = I (c) L = I (d) K I
2
In all these results I, , L are to be taken about the fixed axis of rotation.

2. Rotation with constant angular acceleration :

If the angular velocity increases or decreases at a constant rate, the angular acceleration  is a constant.
(a) f = i +  t (b)  = i t + 1/2  t2 (c) f2 = i2 + 2  

3. Theorems on Moment of Inertia :

(i) Parallel Axis Theorem :
Let Icm be the moment of inertia of a body about an axis through its centre of mass and let Ip be the
moment of inertia of the same body about another axis which is parallel to the original one.
If d is the between these two parallel axes and M is the mass of the body then according to the parallel
axis theorem : Ip = Icm + Md2

4. Perpendicular Axis Theorem :

For a plane body lying in the x-y plane.
The moment of inertia of the body about Z-axis (passing through O and perpendicular to the plane of the
body) is given by : Iz = Ix + Iy

5. Angular Momemtum (L)

(a) For a particle :

The angular momentum about origin (O) is given as : L  r   mv  or L  mvr

where r = perpendicular distance of velocity vector from O.

(b) For a particle moving in a circle of radius r with a speed v, its linear momentum is mv, its angular
momentum (L) is given as : L  m v r  mvr

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(c) For a rigid body (about a fixed axis) :

L = I
(compare with linear momentum p = mv in linear motion)
L is also a vector and its direction is same as that of  (i.e., clockwise or anticlockwise)

6. General Rigid Body Motion :

This complex motion can be described simply as a combination of translational motion of the CM and a
pure rotational motion about the C.M.

(a) Kinetic Energy :

The total kinetic energy of a body which is moving through space as well as rotating is given by :

K  K translational  K rotation

1 2 1
K MVCM  ICM 2
2 2

(b) Instantaneous Velocity of a point :

V A  V CM  V t

Where |V t |  rw (r = distance of A from CM )

And V t is directed along the tangent to the circular trajectory about CM.

(c) Rolling without Slipping :

The relations between corresponding variables for a body rolling without slipping on a stationary
surface.

x  r

Vcm  r 

Acm  r 

Another feature of rolling without slipping is that the instantaneous velocity of the point of contact with
the surface is always zero.

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7. Coservation of angular momentum

(a) Angular impulse about a point (J) :

Linear impulse, I   F dt

Similarly, angular impulse J    dt

J  I r (Compare with   Fr to remember)

(b) Impulse-Momentum Theorems :

(i) I  p (ii) J  L
This equation is usually applied about CM.

(c) Conservation of Angular momentum :

If external angular impulse about a point P during the collision is zero, then the angular momentum
about that point can be conserved.

If Jp  0

 L p  0

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