$A$ solid sphere of mass $m$ rolls down an inclined plane without slipping,starting from rest at the top of an inclined plane. The linear speed of the sphere at the bottom of the inclined plane is $v$. The kinetic energy of the sphere at the bottom is

$A$ solid cylinder $P$ rolls without slipping from rest down an inclined plane,attaining a speed $v_p$ at the bottom. Another smooth solid cylinder $Q$ of the same mass and dimensions slides without friction from rest down the inclined plane,attaining a speed $v_q$ at the bottom. The ratio of the speeds $\frac{v_q}{v_p}$ is

$A$ 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 $\,m$)

Derive expressions for the kinetic energy and velocity of a body rolling without sliding down an inclined plane of inclination $\theta$.

$A$ 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

$A$ 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.