$A$ 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 $\,m$)

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.

$A$ solid sphere and a solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping throughout the motion. The two climb maximum heights $h_{sph}$ and $h_{cyl}$ on the incline. The ratio $\frac{h_{sph}}{h_{cyl}}$ is given by

$A$ ring,a solid sphere,and a disc are rolled down an inclined plane from the same height. The order in which they reach the bottom is:

$A$ solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be $t_1$ and $t_2$,respectively,then

$A$ spherical object of mass $M$ and radius $R$ has a moment of inertia $I$ about its axis. If this object rolls down an inclined plane of angle $\theta$ without slipping,its acceleration is given by: