Solution

The center of the ring will rotate about point $P$ at the edge of the table until it leaves the table. Suppose the moment of inertia of the ring about its center is $I$. Then the moment of inertia of the ring about $P$ is $I+mr^2$, where $m$ and $r$ are the mass and the radius of the ring, respectively. Assume that the ring leaves the table after its center has rotated through an angle $\alpha$. Then, its height drops by $r-r\cos\alpha$, and gravitational potential energy becomes rotational kinetic energy; that is,

$$mgr(1-\cos\alpha)=\frac{1}{2}(I+mr^2)\omega^2\qquad\Rightarrow\qquad\omega^2=\frac{2mgr(1-\cos\alpha)}{I+mr^2}.$$

When the ring is about to lose contact with the table, the centripetal force that keeps the center of the ring rotating about $P$ is a component of gravity. At this point,

$$mg\cos\alpha=m\omega^2r\qquad\Rightarrow\qquad\cos\alpha=\frac{2mr^2(1-\cos\alpha)}{I+mr^2}.$$

Using $I=mr^2$ for the ring, we have $\cos\alpha=1/2$, and so $\alpha=60^\circ.$ For a ball with the same mass and radius, $I<mr^2$, and hence

$$\cos\alpha>1-\cos\alpha\qquad\Rightarrow\qquad\cos\alpha>\frac{1}{2}\qquad\Rightarrow\qquad\alpha<60^\circ.$$