Solution

The ring will roll without slipping when the speed of its center of mass $v$ is equal to $\omega R$, where $\omega$ is the angular speed of rotation of the ring around its axis. This is achieved by friction $f=\mu N=\mu mg$, which increases the ring's linear momentum and at the same time decreases its angular momentum. After time interval $t$, linear impulse from friction is $\mu mgt$, and $v=\mu gt$. Similarly, angular impulse from friction is $-\mu mgRt$, and

$$\omega=\omega_0-\frac{\mu mgRt}/{mR^2}=\omega_0-\frac{\mu gt}{R}.$$

Setting $v=\omega R$ at time $t_*$, we have

$$\mu gt_*=\left(\omega_0-\frac{\mu gt_*}{R}\right)R\qquad\Rightarrow\qquad t_*=\frac{\omega_0R}{2\mu g}.$$

We also obtain the final speed of the center of mass $v_*=\omega_0R/2$ and the final angular speed of rotation of the ring around its axis $\omega_*=\omega_0/2$. Heat generated from lost kinetic energy is

$$\frac{1}{2}mR^2\omega_0^2-\frac{1}{2}mR^2\omega_*^2-\frac{1}{2}m\v_*^2=\frac{1}{2}\cdot\frac{1}{2}mR^2\omega_0^2.$$