A ring of radius$R$is rotating with an angular speed ${\omega }_0$about a horizontal axis. It is placed on a rough horizontal table. The coefficient of kinetic friction is ${\mu }_k.$ The time after which it starts rolling is

Acceleration produced in the centre of mass due to friction

$a=\frac{f}{M}=\frac{{\mu }_{k}Mg}{M}={\mu }_{k}g$

where $M$ is the mass of the ring.

Angular retardation produced by the torque due to friction

$$\begin{gathered} \alpha =-\frac{\tau }{I}=-\frac{fR}{I}=-\frac{{\mu }_{k}MgR}{I} \\ \text{ As } v=u+at \\ \therefore v=0+{\mu }_{k}gt\left(Using\right(i\left)\right) \\ \text{ As } \omega ={\omega }_0+\alpha t \\ \therefore \omega ={\omega }_0-\frac{{\mu }_{k}MgR}{I}t\left(Using\right(ii\left)\right) \\ \text{ For rolling without slipping } \\ v=R\omega \\ \therefore \frac{v}{R}={\omega }_0-\frac{{\mu }_{k}MgR}{I}t \end{gathered}$$

For ring, $I=M{R}^2$

$\therefore t=\frac{R{\omega }_0}{{\mu }_{k}g\left(1+\frac{M{R}^2}{M{R}^2}\right)}=\frac{R{\omega }_0}{2{\mu }_{k}g}$