A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2$\pi$ revolutions is

Given, mass of cylinder,  $m=2 \mathrm{kg}$

Radius of cylinder,  $r=4 \mathrm{cm}=4\times {10}^{-2} \mathrm{m}$

Rotational velocity,  $3\mathrm{rpm}=3\times \frac{2\pi }{60}=\frac{\pi }{10}{\mathrm{rads}}^{-1}\text{ and }$

$\theta =2\pi \times \text{ revolution }=2\pi \times 2\pi =4{\pi }^2\mathrm{rad}$

The work done in rotating an object by an angle  $\theta$ from rest is given by  $W=\tau \theta$

As the cylinder is brought to rest, so the work done will be negative.

According to work-energy theorem,

Work done = Change in rotational kinetic energy

$-\tau \theta =\frac{1}{2}I{\omega }_f^2-\frac{1}{2}I{\omega }_i^2=\frac{1}{2}I\left({\omega }_f^2-{\omega }_i^2\right)$

$\Rightarrow \tau =\frac{I\left({\omega }_i^2\right)}{2\theta }$   $\left(\because {\omega }_f-0\right)$

$=\frac{1}{2}\left(\frac{1}{2}m{r}^2\right)\frac{{\omega }_i^2}{\theta } \left[\because I=\frac{1}{2}m{r}^2\left(\text{ for cylinder }\right)\right]$

$=\frac{1}{4}m{r}^2\frac{{\omega }^2}{\theta } \left(\because {\omega }_i=\omega \right)$

$=\frac{1}{4}\times 2\times {\left(4\times {10}^{-2}\right)}^2\times {\left(\frac{\pi }{10}\right)}^2\times \frac{1}{4{\pi }^2}$

$=\frac{1}{4}\times 2\times 16\times {10}^{-4}\times \frac{{\pi }^2}{100}\times \frac{1}{4{\pi }^2}$

$=\frac{2}{100}\times {10}^{-4}=2\times {10}^{-6} \mathrm{N}-\mathrm{m}$