$A$ solid ball of mass $m$ and radius $r$ rolls without slipping along the track shown in the figure. The radius of the circular part of the track is $R$. The ball starts rolling down the track from rest from a height of $8R$ from the ground level. When the ball reaches the point $P$,then its velocity will be

$A$ solid sphere of mass $1 kg$ and radius $1 m$ rolls without slipping on a fixed inclined plane with an angle of inclination $\theta = 30^{\circ}$ from the horizontal. Two forces of magnitude $1 N$ each,parallel to the incline,act on the sphere,both at a distance $r = 0.5 m$ from the center of the sphere,as shown in the figure. The acceleration of the sphere down the plane is . . . $m s^{-2}$. (Take $g = 10 m s^{-2}$.)

$A$ solid cylinder rolls up an inclined plane of angle of inclination $30^{\circ}$. At the bottom of the inclined plane,the centre of mass of the cylinder has a speed of $5 \; m/s$.
$(a)$ How far will the cylinder go up the plane?
$(b)$ How long will it take to return to the bottom?

The moment of inertia of a solid cylinder about its axis is $I$. It is allowed to roll down an incline without slipping. If its angular velocity at the bottom is $\omega$,then the total kinetic energy $(K.E.)$ of the cylinder will be:

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

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 $\sin 30^{\circ} = 0.5$).