Chapter 31 : Kinetics of Motion of Rotation 635
and kinetic energy of the wheel,
E=(^22) (6 )
177.7 N-m
22
II
I
ωπ
== ...(ii)
Now equating the work done and kinetic energy,
177.7 I= 529.2
∴ I=
529.2=2.98kg-m 2
177.7
Ans.
Example 31.9. A solid cylindrical pulley of mass 800 kg, having 0.8 m , radius of gyration
and 2 m diameter, is rotated by an electric motor, which exerts a uniform torque of 60 kN-m. A body
of mass 3 t is to lifted by a wire wrapped round the pulley.
Find (i) acceleration of the body; and (ii) tension in the rope.
Solution. Given: Mass of pulley (M) = 800 kg; Radius of gyration
(k) = 0.8 m; Diameter of pulley (d) = 2 m or radius (r) = 1 m; Torque (T) = 60 kN-
m = 60000 N-m and mass of the body to be lifted (m) = 3 t = 3000 kg.
(i) Acceleration of the body
Let a= Acceleration of the body,
α= Angular acceleration of the pulley,
P= Tension in rope.
First of all, consider the motion of the hanging body, which is going up-
wards due to the torque. We know that forces acting on it are mg = 3000 × 9.8
= 29 400 N (downwards) and P (upwards). As the body is going upwards, there-
fore the resultant force
=P – 29 400 ...(i)
Since the body is moving with an acceleration (a), therefore force acting on the body,
=ma = 3000 a ...(ii)
Equating equations (i) and (ii),
P – 29 400 = 3000 a ...(iii)
Now consider the motion of the pulley, which is rotating about its axis due to the torque. We
know that moment of inertia of the pulley
I=Mk^2 = 800 (0.8)^2 = 512 kg-m^2
∴ Accelerating torque
T 1 =Iα = 512 α = 512 a ...(Q a = α)
and torque due to tension in the rope
T 2 =P × r = P × 1 = P
Total torque =T 1 + T 2 = 512 a + P
This total torque is equal to the torque exerted by the electric motor
∴ 60 000 = 512 a + P
or P= 60 000 – 512 a
Substituting the value of P in equation (iii),
60 000 – 512 a – 29 400 = 3000 a
3512 a= 60 000 – 29 400 = 30 600
∴ a=
30 600 8.71 m/s 2
3512
= Ans.
Fig. 31.9.