$A$ solid sphere rolling without friction on a horizontal surface with a constant speed of $2 \,m/s$, rolls up on an inclined ramp which is inclined at $30^{\circ}$. The maximum distance travelled by the sphere on the inclined ramp is (acceleration due to gravity $g=10 \,m/s^2, \sin 30^{\circ}=1/2$) (in $\,m$)

$A$ solid sphere having mass $m$ and radius $r$ rolls down an inclined plane. Then its kinetic energy is

$A$ sphere,a disc,a ring,and a hollow sphere of the same radius are rolled down from the same height on an inclined plane simultaneously. The order in which these objects reach the bottom is:

$A$ disc of radius $R$ and mass $M$ is rolling horizontally without slipping with speed $v$. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is:

$A$ 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?

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.