Question

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:

• A. $12\times10^{-4}Nm$
• B. $2\times10^{6}Nm$
• C. $2\times10^{-6}Nm$
• D. $2\times10^{-3}Nm$

Answer 1

The Correct Option is C

Work energy theorem .
$W = \frac{1}{2}I\left(\omega^{2}_{f} - \omega^{2}_{i}\right) \theta = 2\pi $
$ = 2 \pi \times2\pi = 4\pi^{2} $
$ W_{i} = 3\times\frac{2\pi}{60} \text{rad} / s$
$ \Rightarrow -\tau \theta = \frac{1}{2} \times\frac{1}{2} mr^{2} \left(0^{2} -\omega^{2}_{i}\right) $
$ \Rightarrow -\tau = \frac{\frac{1}{2} \times\frac{1}{2} \times2 \times\left(4 \times10^{-2}\right) \left( -3 \times\frac{2\pi}{60}\right)^{2}}{4\pi^{2}} $
$ \Rightarrow \tau = 2\times10^{-6} Nm $