$A$ body is rolling down an inclined plane. If the kinetic energy of rotation is $40\%$ of the translational kinetic energy,then the body is a

$A$ uniform disk of mass $m$ and radius $R$ rolls without slipping down an inclined plane of length $l$ and inclination $\theta$. Initially,the disk was at rest at the top of the inclined plane. Its angular momentum about the point of contact with the inclined plane when it reaches the bottom will be equal to:

$A$ solid cylinder is suspended symmetrically by two massless strings,as shown in the figure. The distance the cylinder should fall by unbinding the strings to achieve a speed of $4\,ms^{-1}$ is $........cm$. (Take $g=10\,ms^{-2}$)

$A$ solid cylinder is rolling down on an inclined plane of angle $\theta$. The coefficient of static friction between the plane and the cylinder is $\mu_s$. The condition for the cylinder not to slip is

$A$ disc rolls without slipping on an inclined plane. What fraction of its total energy is in the form of rotational kinetic energy?

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.