$A$ solid sphere rolls without slipping,first horizontally and then up to a point $X$ at height $h$ on an inclined plane before rolling down,as shown in the figure below. The initial horizontal speed of the sphere is

$A$ body rolls down an inclined plane without slipping. The kinetic energy of rotation is $50 \%$ of its translational kinetic energy. The body is :

The linear speed of a uniform spherical shell after rolling down an inclined plane of vertical height $h$ from rest is:

$A$ cylinder of mass $M_c$ and a sphere of mass $M_s$ are placed at points $A$ and $B$ of two inclines,respectively (See Figure). If they roll on the incline without slipping such that their accelerations are the same,then the ratio $\frac{\sin \theta_c}{\sin \theta_s}$ is

$A$ solid cylinder rolls down an inclined plane from a height $h$. At any moment,the ratio of rotational kinetic energy to the total kinetic energy 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.