Question

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

Answer 1

Hint: The wheel is undergoing pure angular motion; it does not show translational motion. According to the second law of motion, a force is required to change the state of rest or motion of a body. Here, torque is analogous to force, this means an external torque is required to bring the wheel at rest.

Formulas used:
$\omega ={{\omega }_{0}}+\alpha t$
$\tau =I\alpha $

Complete step-by-step solution:
A wheel is rotating about a vertical axis passing through its centre. Its angular velocity is-
$\begin{align}
  & 60rpm=\dfrac{60\times 2\pi }{1\times 60}rad\,{{s}^{-1}} \\
 & \therefore 60rpm=2\pi \,rad\,{{s}^{-1}} \\
\end{align}$
Given, its moment of inertia is,$I=2\,kg\,{{m}^{2}}$
Since no external force is acting on the system of the wheel, its angular acceleration is constant.
Using equations for angular motion along an axis,
$\omega ={{\omega }_{0}}+\alpha t$
Here,
$\omega $ is the final angular velocity
${{\omega }_{0}}$ is the initial angular velocity
$\alpha $ is the angular acceleration
$t$ is time taken
The wheel is to be stopped, therefore $\omega =0$
We substitute the given values in the above equation to get,
$\begin{align}
  & \omega ={{\omega }_{0}}+\alpha t \\
 & \Rightarrow 0=2\pi +\alpha \times 60 \\
 & \Rightarrow \alpha =-\dfrac{2\pi }{60} \\
 & \therefore \alpha =-\dfrac{\pi }{30}rad\,{{s}^{-2}} \\
\end{align}$ [1 minute = 60secs]
The wheel will have to undergo deceleration of $\dfrac{\pi }{30}rad\,{{s}^{-2}}$ to come to rest in 1 minute.
The torque required to bring the wheel at rest is given by-
$\tau =I\alpha $
Here,
$\tau $ is the torque
$I$ is the moment of inertia
Therefore, we substitute values in the above equation to get,
$\begin{align}
  & \tau =I\alpha \\
 & \Rightarrow \tau =2\times \dfrac{\pi }{30} \\
 & \therefore \tau =\dfrac{\pi }{15}\,N\,m \\
\end{align}$
The torque required to bring the wheel at rest is$\dfrac{\pi }{15}\,N\,m$.

Hence, the correct option is (C).

Note:
The torque must be applied in the direction opposite to the motion of the wheel. The moment of inertia, also called the angular mass is a quantity which is used to determine the torque needed for an angular acceleration. The linear velocity of the wheel is tangential to its motion. The equations of angular motion of a body are analogous to the equations of motion in a straight line.