Rotational Motion
Rigid body:
A rigid body is a body that can rotate with all its parts locked together and without any change in its shape. A
body with a perfectly definite and unchanging shape is called rigid body. Rigid bodies can undergo both
translational and rotational motion. Motion of a rigid body can always be represented as a combination of
translational motion of some point in the body and rotational motion about an axis through that point.
A rigid body of arbitrary shape in pure rotation about the z axis of a coordinate system. The position of the
reference line with respect to the rigid body is arbitrary, but it is perpendicular to the rotation axis. It is fixed
in the body and rotates with the body
Angular displacement:
If a rigid body of arbitrary shape rotating about a fixed axis and if  represents any arbitrary angle subtended
by an arc of length s on the circumference of a circle of radius r, then  (in radians) is the angular displacement
of that body and is defined as

s
r
Average and instantaneous angular velocity:
Prof. Dr. Suranjan Kumar Das
PHY109/Mech-3
1/9
The average velocity of the body is defined as the ratio of the angular displacement to the elapsed time. If
12is the angular displacement between time interval t1 and t2, then the average velocity is

 2  1
t2  t1


t
The instantaneous velocity  is defined as the limit approached by this ratio as t approaches zero, that is
  lim
t  0
 d

t dt
Average and instantaneous angular acceleration:
If the angular velocity of a body changes, it is said to have an angular acceleration. If 1 and 2 are the
instantaneous angular velocities at times t1 and t2, then the average angular acceleration is defined as

2  1
t2  t1


t
The instantaneous acceleration  is defined as the limit of this ratio when tapproaches to zero.
 d

t 0 t
dt
  lim
Relation between angular and linear velocity and acceleration:
Consider a body is moving in a circular path of radius r and at a certain time the body is at point P. If the arc
x is small then the angular displacement is related with r and as
x  r
 x  r
x


r
t
t
 v  r …………. (4)
P

x
r
This is the relation between linear and angular velocity.
Again, from (4) we can write
 v  r
v


r
t
t
 a  r ………. (5)
. Equation (5) is the relation between the linear acceleration and angular acceleration.
Kinetic energy of rotation:
We know the kinetic energy in linear motion is defined as
Prof. Dr. Suranjan Kumar Das
PHY109/Mech-3
2/9
1
k  mv 2 …………… (6)
2
In rotational motion v  r
1
1
1
 k  mv 2  m 2 r 2  mr 2  2
2
2
2
 
For a system of particles
k
1
2
k 
 m r 
2
2
i i
1 2
I ………… (7)
2
This is the expression for kinetic energy in rotational motion.
Moment of inertia:
From the definition of kinetic energy in angular motion, we get
I   mr 2
The term I is known as moment of inertia. It is seen that I plays a role of mass in rotational motion.
Radius of gyration
I   mr 2
or, I = MK2
Where, M = mass of the whole body, K= distance of the point from the axis of rotation where total mass of
the body is concentrated.
K is called the radius of gyration.
For different size and shape of the body I can have different expressions; e.g.
i)
A slender rod; axis through center
I
ii)
Rectangular plate; axis through center
I
iii)
1
Ml 2
12

1
M a 2  b2
12

Hollow cylinder

1
2
2
I  M R1  R2
2
iv)

Solid cylinder
1
I  MR 2
2
v)
Thin walled cylinder
I  MR2
Prof. Dr. Suranjan Kumar Das
PHY109/Mech-3
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vi)
Solid sphere
2
I  MR 2
5
Angular momentum:
The angular momentum of a moving body about an axis is defined as the product of its linear momentum and
the perpendicular distance from the axis to its linear motion. It is denoted by L.
𝐿⃗ = r × 𝑃⃗ => L = rP sin =r (mv)sin
 L  mvr (=90)
𝑃⃗
𝑟

Relation between angular momentum and angular velocity
Let an object rotate about an axis with angular velocity . If the object is composed of many small particles,
then we can write,
L = l1 +l2+l3 + --------------------------------- ln
(l1 l2, , l3 ----- angular momentum of the individual particle)
L= r1p1 +r2p2+ r3p3 + ----------------------- + rnpn
= r1m1 v1 + r2m2 v2 + r3m3 v3 + --------- + rnmn vn
= r1m1 (𝜔 𝑟1 )+ r2m2 (𝜔 𝑟2 )+ + r3m3 (𝜔 𝑟3 )+ + --------- + rnmn (𝜔 𝑟𝑛 )
= 𝜔{𝑚1 𝑟12 +
𝑚2 𝑟22 + 𝑚3 𝑟32 − − − − − − − − − 𝑚𝑛 𝑟𝑛2 }
= 𝜔 ∑ 𝑚𝑖 𝑟𝑖2
= 𝜔I
[ 𝑠𝑖𝑛𝑐𝑒 I 
m r
i i
2
]
∴ L= I 𝜔
Torque:
In case of translational motion the quantity related with the linear acceleration is force
similarly, in case of rotational motion the quantity related with the angular acceleration is torque or moment
of force.
Definition: The vector product of the radius vector or position vector of a rotating particle and force applied
on it is the torque.
𝜏 = 𝑟×𝐹
Prof. Dr. Suranjan Kumar Das
PHY109/Mech-3
4/9
Relation between torque and angular acceleration
 = 1 + 2 +3 + ----------------- ---------- n
= r1F1 + r2F2 + r3F3 + -------------------- rnFn
= r1m1 a1 + r2m2 a2 + r3m3 a 3
------------------ rnmn an
= r1m1 ( r1) + r2m2 ( r2) + ----------------- rnmn ( rn)
=  {𝑚1 𝑟12 +
𝑚2 𝑟22 + 𝑚3 𝑟32 − − − − − − − − − 𝑚𝑛 𝑟𝑛2 }
2
= 𝒂 ∑ 𝑚𝑟
=I
∴𝝉=I
Two Theorems relating moment of inertia:
1. Perpendicular axes theorem
2. Parallel axes theorem
Iz= Ix + Iy
Perpendicular axes theorem:
1.
Parallel axes theorem
Comparison of linear and angular motion:
Concept
Displacement
Velocity
Acceleration
Translational motion
Rotational motion
Relationship
x

x = r
v = r
dx
dt
dv
a
dt
v
Prof. Dr. Suranjan Kumar Das
PHY109/Mech-3
d
dt
d

dt

a  r , a 
v2
r
5/9
Motion with
constant
acceleration
Mass, moment
of inertia
Kinetic energy
Momentum
v  v0  at
  0  t
1
x  x0  v0t  at 2
2
2
2
v  v0  2ax  x0 
m
   0  0 t   t 2
1
2
 2  0 2  2    0 
I
I = mr2
1
K  mv 2
2
P = mv
1 2
I
2
L = mvr
L = I
Type equation here.
K
Mathematical Problems:
1. The mass of a wheel is 6 kg and its radius of gyration is 40 cm. The wheel rotates 300 times per
minute. Determine its moment of inertia and rotational kinetic energy.
Soln. We know, Moment of inertia I = MK2
= 6 kg x (0.4m)2 = 0.96 kgm2
1
Rotational kinetic energy, K = 2 I 𝜔2
=
[angular velocity, 𝜔=
300×2𝜋
60
1
×0.96×(10𝜋)2
2
J = 473.26 J
rad/s]
2. The moment of inertia of a fly wheel rotating at 6000 rads-1 angular velocity is 80 kgm2. It is stopped
within 30 s by applying a uniform brake.
i)
what its angular acceleration if brake is applied?
ii)
how many times it will rotate during this time?
iii)
how much torque the brake will provide?
Soution:
i)
ii)
hints. 𝜔 = 𝜔0 +𝛼t
[Ans. -200 rads-1]
We know, angular displacement 

𝜔2 =𝜔02 + 2∝ 𝜃
∴ 𝜃 = (𝜔2 − 𝜔02 )/ 2∝ = (6000)2/ 2*200 = 90000 radian
𝜃
𝑜𝑟, 𝑛 =
= 14331.21 rev.
2𝜋
iii)
𝜏 = I𝛼 = (80x200) Nm = 16000 Nm
3. Diameter of a wheel is 2 m and its mass is 20 kg. What will be the angular momentum for the
wheel at an angular velocity of 1800 rpm?
Solution: hints. L= mvr= mr2 = (20x60𝜋 x 12 ) =600𝜋 kg ms-1.
1800×2𝜋
[𝜔=
= 60 rads-1]
60
4. Consider a body consisting of two spherical masses of 5.0 kg each connected by a light rigid rod
1.0 m long. Treat the spheres as point particles and neglect the mass of the rod. Determine the
rotational inertia (or moment of inertia) of the body,
a) about an axis normal to it through its center at C, and
Prof. Dr. Suranjan Kumar Das
PHY109/Mech-3
6/9
b) about an axis
Solution:
a) If the axis is normal to the center C
We have
𝐼𝐶 =∑ 𝑚𝑖 𝑟𝑖2 = mar2 + mb r2 = (5.0kg) (0.5m )2 + (5.0kg) (0.5m )2 = 2.5 kgm2
b)
If the axis is normal through A or B
Exercise 9-13𝐼𝐴 = mar2 + mb r2 =(5.0kg) (0.0m )2 + (5.0kg) (1.0 m )2
= 5.0 kgm2=======
n cccc
𝐼𝐵 = mar2 + mb r2 =(5.0kg) (1.0m )2 + (5.0kg) (0.0 m )2
= 5.0 kgm2=======
5. A particle of mass m is released from rest at point a, falling parallel to the y-axis.
a) Find the torque acting on m at any time t w.r.t origin O.
b) Find the angular momentum of m at any time t, w.r.t the same origin.
Solution:
a) the value of torque is given by 𝜏= 𝑟 × 𝐹
or, 𝜏
= rFsin

In this case rsin= b and F=mg
∴ 𝜏 = 𝑚𝑔𝑏 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
b) The angular momentum, l=rpsinbmv= bmgt
6. The angular position of a point on the rim of a rotating wheel is given by
θ = 4.0t − 3.0t2 + t3, where θ is in radians if t is given in seconds.
(a)What are the angular velocities at t = 2.0 s and t = 4.0 s?
(b) What is the average angular acceleration for the time interval that begins at t = 2.0 s
and ends at t = 4.0 s?
(c) What are the instantaneous angular accelerations at the beginning and end
of this time interval?
Solution:
a)
Prof. Dr. Suranjan Kumar Das
PHY109/Mech-3
7/9
b ) The average angular acceleration for the interval:
The average angular acceleration is 12.0 rad s -2 .
c)
c
7
777777777777.
77877
v
Calculate the rotational inertia of a meter stick with mass 0.56 kg, about an axis perpendicular to the stick and
located at the 20 cm mark.
solution: Parallel Axis Theorem tells us that the moment of inertia about our axis is given by
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PHY109/Mech-3
8/9
8.
A small 0.75 kg ball is attached to one end of a 1.25 m long massless rod, and the other end of the rod is hung
from a pivot. When the resulting pendulum is 30◦ from the vertical, what is the magnitude of the torque about
the pivot?
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PHY109/Mech-3
9/9