When a sphere with moment of inertia $I$ rolls down an inclined plane,what percentage of its total energy is rotational kinetic energy?

$A$ cylinder of mass $M$ and radius $R$ rolls on an inclined plane. The gain in kinetic energy is

$A$ solid sphere rolls down two different inclined planes of the same height,but of different inclinations. In both cases:

$A$ solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60\,cm$. If the cylinder rolls without slipping,its speed upon reaching the bottom of the inclined plane is $...........\,ms^{-1}$. (Given $g = 10\,ms^{-2}$)

$A$ small sphere rolls without slipping from the top of a vertical track. The track has an inclined part and a horizontal part. The horizontal part is $1.0 \ m$ above the ground,and the top of the track is $2.4 \ m$ above the ground. The sphere falls to the ground at point $E$. The horizontal distance from the point directly below $C$ to point $E$ is $R$. Find the value of $R$ in meters.

$A$ 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