Answer to Question #88380 in Mechanics | Relativity for Mohanrao

Question

A Ring of radius R is first rotated with an angular velocity ∆ and then carefully placed on a horizontal rough surface. The coefficient of friction between the surface and the ring is ¶. Time after which its angular speed is reduced to half is

Answer 1

Let $\mu$ be the coefficient of friction, $\omega$ be the angular speed and $\alpha$ the angular acceleration. Consider a torque of force of friction:

$$\tau_\text{f}=\mu mg\cdot R,$$

and according to Newton's second law for rotational motion, the net torque is

$$\tau_\text{net}=mR^2\alpha=mR^2\cdot \frac{\omega-\omega/2}{t},$$

equate them:

$$\mu mg\cdot R=mR^2\cdot \frac{\omega}{2t},$$

$$t=\frac{\omega R}{2\mu g}.$$