$A$ disc is rolling without slipping on a straight surface. The ratio of its translational kinetic energy to its total kinetic energy is

Obtain the necessary condition $v_{cm} = R\omega$ for a body rolling without slipping.

Two spheres are rolling with the same velocity (for their $C.M.$). Their ratio of kinetic energy is $2:1$ and the ratio of their radii is $2:1$. Their mass ratio will be:

$A$ solid sphere and a solid cylinder,each of mass $M$ and radius $R$,are rolling with a linear speed $v$ on a flat surface without slipping. Let $L_1$ be the magnitude of the angular momentum of the sphere with respect to a fixed point $O$ on the surface along the path of the sphere. Likewise,let $L_2$ be the magnitude of the angular momentum of the cylinder with respect to the same fixed point $O$ along its path. The ratio $\frac{L_1}{L_2}$ is

$A$ solid sphere of mass $2\,kg$ is making pure rolling on a horizontal surface with kinetic energy $2240\,J$. The velocity of the centre of mass of the sphere will be $..........\,m/s$.

If a solid sphere is rolling without slipping on a horizontal plane,then the ratio of its rotational and total kinetic energies is