$A$ solid sphere is rolling down an inclined plane without slipping. The ratio of its rotational kinetic energy to its total kinetic energy is:

Three bodies: a ring,a solid cylinder,and a solid sphere,roll down an inclined plane without slipping. They start from rest. Which of the bodies reaches the bottom of the plane with the minimum velocity?

$A$ solid spherical ball rolls on a horizontal surface at $10 \ m \ s^{-1}$ and continues to roll up on an inclined surface as shown in the figure. If the mass of the ball is $11 \ kg$ and frictional losses are negligible, the value of $h$, where the ball stops and starts rolling down the inclination is $($Assume $g = 10 \ m \ s^{-2} )$ (in $m$)

Two solid spheres of different mass,radii,and density roll down a rough inclined plane under identical situations. Their time to come down is independent of their ............

$A$ solid cylinder and a disc of same radii are allowed to roll down a rough inclined plane from the top of a plane. The ratio of their times taken to reach the bottom of the inclined plane is

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.