Solution

By the work–energy theorem: $W=\frac{1}{2}I(\omega_f^2-\omega_i^2)$ $\theta=2$ revolutions $=2\times2\pi=4\pi\ rad$; $\omega_i=3\ rpm=\frac{2\pi\times3}{60}\ rad/s$ $-\tau\theta=\frac{1}{2}\cdot\frac{1}{2}mr^2(0-\omega_i^2)$ $\tau=\frac{\frac{1}{4}\times2\times(4\times10^{-2})^2\times\left(\frac{2\pi\times3}{60}\right)^2}{4\pi}$ $\tau=\frac{0.0008\times(0.1\pi)^2}{4\pi}=2\pi\times10^{-6}\ N\,m$ (the original paper lists this option as $2\pi\times10^{-6}\ N\,m$)

Concept torque to stop a rotating cylinder

System of Particles and Rotational Motion in past papers

100 questions from this chapter have appeared across 27 exam years.