Question

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 is reduced to half is :

• A. $\frac{{\omega }_0\mu R}{2g}$
• B. $\frac{{\omega }_{0}g}{2\mu R}$
• C. $\frac{2{\omega }_{0}R}{\mu g}$
• D. $\frac{{\omega }_{0}R}{2\mu g}$

Answer 1

The correct option is D $\frac{{\omega }_{0}R}{2\mu g}$

Taking $\tau$ about CM

$\mu mgR=m{R}^2\alpha$

$\mu g=R\alpha$

$\alpha =\frac{\mu g}{R}$

$\omega ={\omega }_0-\frac{\mu g}{R}t=\frac{{\omega }_0}{2}$

$\frac{{\omega }_0}{2}=\frac{\mu g}{R}t\Rightarrow t=\frac{{\omega }_{0}R}{2\mu g}$