A solid sphere is rolling down an inclined plane. Then the ratio of its translational kinetic energy to its rotational kinetic energy is

One twirls a circular ring (of mass $M$ and radius $R$ ) near the tip of one's finger as shown in Figure $1$ . In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is $\mathrm{r}$. The finger rotates with an angular velocity $\omega_0$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure $2$). The coefficient of friction between the ring and the finger is $\mu$ and the acceleration due to gravity is $g$.

(IMAGE)

($1$) The total kinetic energy of the ring is

$[A]$ $\mathrm{M} \omega_0^2 \mathrm{R}^2$   $[B]$ $\frac{1}{2} \mathrm{M} \omega_0^2(\mathrm{R}-\mathrm{r})^2$   $[C]$ $\mathrm{M \omega}_0^2(\mathrm{R}-\mathrm{r})^2$   $[D]$ $\frac{3}{2} \mathrm{M} \omega_0^2(\mathrm{R}-\mathrm{r})^2$

($2$) The minimum value of $\omega_0$ below which the ring will drop down is

$[A]$ $\sqrt{\frac{g}{\mu(R-r)}}$  $[B]$ $\sqrt{\frac{2 g}{\mu(R-r)}}$  $[C]$ $\sqrt{\frac{3 g}{2 \mu(R-r)}}$    $[D]$ $\sqrt{\frac{g}{2 \mu(R-r)}}$

Givin the answer quetion ($1$) and ($2$)

A circular disc is rolling on a horizontal plane. Its total kinetic energy is  $300\,\,J.$  ……… $J$ is its translational  $K.E.$

The angular momentum of a rigid body of mass m about an axis is $n$ times the linear momentum $(P)$ of the body. Total kinetic energy of the rigid body is