Question

A body starts rotating from rest due to a torque of 10Nm. If it completes 300 rotations in one minute, then find the moment of inertia of the body.

Answer 1

Hint :First find the angular acceleration of the body with the help of a suitable rotational kinematic equation. Then use the relation between torque, moment of inertia and angular acceleration of a body.
 $ \theta ={{\omega }_{0}}t+\dfrac{1}{2}\alpha {{t}^{2}} $ , where $ \theta $ is the angle rotated by a body in time t, moving with a constant angular acceleration $ \alpha $ and $ {{\omega }_{0}} $ is the initial angular velocity of the body.
  $ \tau =I\alpha $ , where $ \tau $ is the torque on the body and I is the moment of inertia of the body about the axis of rotation.

Complete Step By Step Answer:
It is given that the body starts from rest and rotates due to a torque of 10 Nm. It is said that it completes 300 rotations in one minute. To find the moment of inertia of the body we shall first find its angular acceleration.
We know that one rotation is equal to an angle of $ 2\pi $ radians. This means that 300 rotations are equal to $ 300\times 2\pi =600\pi $ radians.
Therefore, in this case $ \theta =600\pi $ .
The body completes this angle in a time interval of one minute. This means that $ t=1\min =60s $ .
Now, we shall use the kinematic equation $ \theta ={{\omega }_{0}}t+\dfrac{1}{2}\alpha {{t}^{2}} $ …. (i)
Since the body started from rest, $ {{\omega }_{0}}=0 $
Substitute the known values in (i).
 $ \Rightarrow 600\pi =(0)(60)+\dfrac{1}{2}\alpha {{(60)}^{2}} $
 $ \Rightarrow \alpha =\dfrac{2\times 600\pi }{{{(60)}^{2}}}\approx 1.05{{s}^{-2}} $
In this case, the torque on the body is $ \tau =10Nm $
We know that $ \tau =I\alpha $ .
   $ \Rightarrow 10=I(1.05) $
 $ \Rightarrow I=\dfrac{10}{1.05}=9.8kg{{m}^{2}} $
Therefore, the moment of inertia of the body in this case is equal to $ 9.8kg{{m}^{2}} $ .

Note :
If you do not know the formulae of the rotational kinematic, then you can make out the formulae involved in the motion of a body (moving constant acceleration) in one dimension. This is because the rotational motion is analogous to the translation motion.