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

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-
$\begin{aligned} & 60rpm=\dfrac{60\times 2\pi }{1\times 60}rad\,{{s}^{-1}} \\\ & \therefore 60rpm=2\pi \,rad\,{{s}^{-1}} \\\ \end{aligned}$
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,
$\begin{aligned} & \omega ={{\omega }_{0}}+\alpha t \\\ & \Rightarrow 0=2\pi +\alpha \times 60 \\\ & \Rightarrow \alpha =-\dfrac{2\pi }{60} \\\ & \therefore \alpha =-\dfrac{\pi }{30}rad\,{{s}^{-2}} \\\ \end{aligned}$ [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,
$\begin{aligned} & \tau =I\alpha \\\ & \Rightarrow \tau =2\times \dfrac{\pi }{30} \\\ & \therefore \tau =\dfrac{\pi }{15}\,N\,m \\\ \end{aligned}$
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.