$A$ solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What is the speed of its centre of mass when the cylinder reaches its bottom?

An object of uniform density rolls up a curved path with an initial velocity $v_0$ as shown in the figure. If the maximum height attained by the object is $\frac{7v_0^2}{10g}$ ($g =$ acceleration due to gravity), the object is a . . . . . . .

The acceleration of a body rolling down on an inclined plane does not depend upon

$A$ circular ring and a solid sphere having the same radius roll down an inclined plane from rest without slipping. The ratio of their velocities when they reach the bottom of the plane is $\sqrt{\frac{x}{5}}$,where $x=$ . . . . . . .

$A$ solid cylinder and a solid sphere,having same mass $M$ and radius $R$,roll down the same inclined plane from the top without slipping. They start from rest. The ratio of the velocity of the solid cylinder to that of the solid sphere,with which they reach the ground,will be:

Prove that the velocity $v$ of translation of a rolling body (like a ring,disc,cylinder,or sphere) at the bottom of an inclined plane of height $h$ is given by $v^{2} = \frac{2gh}{1 + k^{2}/R^{2}}$ using dynamical considerations (i.e.,by considering forces and torques). Note: $k$ is the radius of gyration of the body about its symmetry axis,and $R$ is the radius of the body. The body starts from rest at the top of the plane.