Suppose a body of mass $M$ and radius $R$ is allowed to roll on an inclined plane without slipping from its topmost point $A$. The acceleration of the body down the plane is given by (where $\beta = 1 + \frac{I}{MR^2}$):

$A$ 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

$A$ disc of radius $R$ and mass $M$ is rolling horizontally without slipping with speed $v$. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is:

If a sphere is rolling,the ratio of its rotational energy to the total kinetic energy is given by

$A$ body slides down a smooth inclined plane having angle $\theta$ and reaches the bottom with velocity $v$. If the body is a solid sphere rolling down the same plane,then its linear velocity at the bottom of the plane is

$A$ uniform spherical object of mass $M$ and radius $R$ has a moment of inertia $I$. It rolls without slipping down an inclined plane of angle $\theta$. What is its acceleration?