Q. A ring of radius $R$ is first rotated with an angular velocity $\omega_{0}$ and then carefully placed on a rough horizontal surface. The coefficient of friction between the surface and the ring is $\mu$. Time after which its angular speed if reduced to half is
A) $\frac{\omega_{0}\mu R}{2g}$
B) $\frac{2\omega_{0}R}{\mu g}$
C) $\frac{\omega_{0}R}{2\mu g}$
D) $\frac{\omega_{0}g}{2\mu R}$
Explanation
(c) Angular retardation, $a = \frac{\tau}{I} = \frac{\mu\ mgR}{mR^{2}} = \frac{\mu g}{R}$
As, $\omega = \omega_{0} - \alpha t$
$\therefore\ \ t = \frac{\omega_{0} - \omega}{\alpha} = \frac{\omega_{0} - \omega_{0}/2}{\mu g/R} = \frac{\omega_{0}R}{2\mu g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.