(^628) A Textbook of Engineering Mechanics
31.14.KINETIC ENERGY OF ROTATION
We have already discussed in Art. 30.15 that energy is the capacity to do some work. Though
the energy exists in many forms, yet the kinetic energy is important from the subject point of view.
The kinetic energy of rotation, possesed by a body for doing work by virture of its motion of rota-
tion. Now consider a rotating body, which is brought to rest by a uniform angular retardation due to
some torque.
Let I= Mass moment of inertia of the body, and
ω= Angular velocity of the body.
The *kinetic energy of rotation,
E=
2
2
Iω
Example 31.1. A circular wheel of mass 50 kg and radius 200 mm is rotating at 300 r.p.m.
Find its kinetic energy.
Solution. Given: Mass of the wheel (M) = 50 kg; Radius of the wheel (r) = 200 mm = 0.2 m
and angular velocity (ω) = 300 r.p.m.
We know that mass moment of inertia of the circular wheel,
I= 0.5 Mr^2 = 0.5 × 50 (0.2)^2 = 1 kg-m^2
and angular velocity of the wheel,
ω= 300 r.p.m. = 5 r.p.s. = 10 π rad/s
∴ Kinetic energy of the rotating wheel,
E=
(^22) 1(10 )
= 493.5 N-m
22
Iωπ
= = 493·5 J Ans.
31.15.TORQUE AND ANGULAR ACCELERATION
Consider a body rotating about its axis.
Let M= Mass of the body
ω= Angular velocity of the body
α= Angular acceleration of the body, and
T= Torque acting on the body
Now split up the body into a number of small particles of mass m 1 , m 2 , m 3 , ...at distance of
r 1 , r 2 , r 3 , ...from the axis about which the body is rotating. We know that the linear acceleration of
the particle of mass m 1
=r 1 α
∴ Force acting on this particle
= Mass × Acceleration = m 1 (r 1 α)
and moment of this force about the axis
=m 1 (r 1 α) r 1 = m 1 r 12 α
- This equation is analogous to the equation for the kinetic energy of translation i.e.
E=
2
N-m
2
mv