Question

A wheel having moment of inertia $ 2kg{m^{ - 2}} $ about its vertical axis, rotates at the rate of $ 60rpm $ about its axis. The torque which can stop the wheel’s rotation in one minute would be,
(A) $ \dfrac{{2\pi }}{{13}}Nm $
(B) $ \dfrac{\pi }{{14}}Nm $
(C) $ \dfrac{\pi }{{15}}Nm $
(D) $ \dfrac{\pi }{{20}}Nm $

Answer 1

Hint :Use the formula of angular acceleration and using the equation of motion for a particle in rotational motion to find the torque. Torque on a body rotating with angular acceleration $ \alpha $ is $ \tau = I\alpha $ , where $ I $ is the moment of inertia about the axis of rotation. Angular acceleration of a body is the change in its angular velocity $ \omega $ over a time period $ t $ , $ \alpha = \dfrac{{{\omega _2} - {\omega _1}}}{t} $ , where, $ {\omega _2} $ is the final angular velocity, $ {\omega _1} $ is the initial angular velocity and $ t $ is the time separation between the change.

Complete Step By Step Answer:
We know that the torque on a body rotating with angular acceleration $ \alpha $ is given by, $ \tau = I\alpha $ where $ I $ is the moment of inertia about the axis of rotation
Now we know that the angular acceleration of a body is the change in its angular velocity $ \omega $ over a time period $ t $ . That can be written as, $ \alpha = \dfrac{{{\omega _2} - {\omega _1}}}{t} $ where, $ {\omega _2} $ is the final angular velocity, $ {\omega _1} $ is the initial angular velocity and $ t $ is the time separation between the change.
Here, we have, moment of inertia of the given wheel about its vertical axis, $ I = 2kg{m^{ - 2}} $ .
angular velocity is $ 60rpm $ which is equal to $ 60rpm = 60\dfrac{{2\pi }}{{60}}rad.{s^{ - 1}} $ [since, $ 1rpm = \dfrac{{2\pi }}{{60}}rad.{s^{ - 1}} $ ]
That is, $ {\omega _1} = 2\pi rad.{s^{ - 1}} $
Now , we know that to stop the wheel we have to provide a torque in the opposite direction of motion .
Hence, the axis of rotation of the wheel and the axis of torque will be opposite to each other.
Now, when the wheel stops rotating the final angular velocity must be zero, so, we can write, $ {\omega _2} = 0rad.{s^{ - 1}} $ . The wheel stops in one minute that meant, $ t = 60s $
Putting these values we get the angular acceleration as, $ \alpha = \dfrac{{0 - 2\pi }}{{60}}rad.{s^{ - 2}} $
On simplifying, $ \alpha = - \dfrac{\pi }{{30}}rad.{s^{ - 2}} $ .
Putting the value of $ \alpha $ and moment of inertia we get the torque $ \tau $ as,
 $ \tau = I\alpha $
 $ \therefore $ $ \tau = - 2\dfrac{\pi }{{30}} $
It becomes,
 $ \tau = - \dfrac{\pi }{{15}} $
The negative sign here implies that the torque is applied to the body and direction of it is opposite to the direction of angular velocity.
 $ \therefore $ To stop the wheel in one minute a torque of $ \dfrac{\pi }{{15}}Nm $ must be applied.
Hence, option ( C) is correct.

Note :
The direction of torque can be parallel or antiparallel to the axis of rotation. For, anti parallel condition, the object must be decelerating and for parallel condition the object must be accelerating. If the rotational velocity is constant then there is no torque applicable to the object.