Question

It has been given a solid cylinder of mass $2kg$ and the radius $4cm$ which is rotating about its axis at the rate of $3rpm$. What will be the torque required to stop after $2\pi$ revolution?
$\begin{aligned} & A.2\times {{10}^{-6}}Nm \\\ & B.2\times {{10}^{-3}}Nm \\\ & C.12\times {{10}^{-4}}Nm \\\ & D.2\times {{10}^{6}}Nm \\\ \end{aligned}$

Answer 1

In accordance to the work energy theorem, the variation in the kinetic energy will be equivalent to the work done by the system. The work done will be equivalent to the negative of the product of the torque and the angular displacement. Substitute the values in the equation and rearrange this equation. This will help you in answering this question.

Complete step by step answer:
In accordance with the work energy theorem, the variation in the kinetic energy will be equivalent to the work done by the system. This can be written as an equation given as,
$W=\dfrac{1}{2}I\left( {{\omega }_{f}}^{2}-{{\omega }_{i}}^{2} \right)$
Here the angular displacement will be equivalent to,
$\theta =2\pi \text{revolution}$
That means we can write that,
$\theta =2\pi \times 2\pi =4{{\pi }^{2}}rad$
The angular velocity will be equivalent to,
${{\omega }_{i}}=3\times \dfrac{2\pi }{60}=\dfrac{\pi }{10}rad{{s}^{-1}}$
The work done will be equivalent to the negative of the product of the torque and the angular displacement. This can be written as an equation given as,
$W=-\tau \theta$
The moment of inertia of the body can be shown as,
$I=\dfrac{1}{2}m{{r}^{2}}$
Substituting the values in this equation can be shown as,
$W=-\tau \theta =\dfrac{1}{2}\times \dfrac{1}{2}m{{r}^{2}}\left( 0-{{\omega }_{i}}^{2} \right)$
From this the value of torque can be derived. It can be achieved as,
$W=-\tau \theta =\dfrac{1}{2}\times \dfrac{1}{2}\times 2\times {{\left( 4\times {{10}^{-2}} \right)}^{2}}\left( 0-{{\left( 3\times \dfrac{2\pi }{60} \right)}^{2}} \right)$
Rearranging this equation can be shown as,
$\tau =\dfrac{\dfrac{1}{2}\times \dfrac{1}{2}\times 2\times {{\left( 4\times {{10}^{-2}} \right)}^{2}}\left( 0-{{\left( 3\times \dfrac{2\pi }{60} \right)}^{2}} \right)}{4{{\pi }^{2}}}$
Simplifying this equation can be shown as,
$\tau =2\times {{10}^{-6}}Nm$

So, the correct answer is “Option A”.

Note: Angular displacement of a body can be defined as the angle in radians or degrees or in revolutions with which a point is revolving around a centre or a line which has been rotated in a particular way about a particular axis. Angular displacement can be considered as a vector quantity.