$A$ sphere of mass $m$ and radius $r$ rolls on a horizontal plane without slipping with the speed $u$. Now if it rolls up an incline, the maximum height it would attain will be

$A$ solid cylinder is rolling down on an inclined plane of angle $\theta$. The coefficient of static friction between the plane and the cylinder is $\mu_s$. The condition for the cylinder not to slip is

$A$ solid sphere is rolling on a frictionless surface with a linear velocity $v \; m/s$,as shown in the figure. If the sphere climbs up to a height $h$,then the value of $v$ will be

$A$ ring is rolling on an inclined plane. The ratio of the linear and rotational kinetic energies will be

$A$ sphere is rolling down an incline without slipping. If the height of the incline is $14 \ m$,find its linear velocity at the bottom.

An inclined plane makes an angle $30^{\circ}$ with the horizontal. $A$ solid sphere rolling down an inclined plane from rest without slipping has a linear acceleration (where $g$ is the acceleration due to gravity and $\sin 30^{\circ} = 0.5$).