Q. 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) $2 \times 10^{-6}$ N m
B) $2 \times 10^{-3}$ N m
C) $12 \times 10^{-4}$ N m
D) $2 \times 10^6$ N m
Explanation
[Solution] $I = \frac{1}{2}mr^2 = 1.6 \times 10^{-3}$; $\omega = 0.1\pi$ rad/s; $\theta = 4\pi^2$ rad; $\alpha = \omega^2/2\theta = 1.25 \times 10^{-3}$; $\tau = I\alpha = 2 \times 10^{-6}$ N m. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.