A body of mass 'm' and radius 'r' rolling on a horizontal floor with velocity 'v', rolls up an inclined plane up to vertical height (3V^(2))/(4g). Find the moment of inertia of body and comment on its shape.

A body of mass 'm' and radius 'r' rolling on a horizontal floor with v

1.	A body of mass 'm' and radius 'r' rolling on a horizontal floor with velocity 'v', rolls up an inclined plane up to vertical height (3V^(2))/(4g). Find the moment of inertia of body and comment on its shape.
Answer» Solution :The total KE of the body `K=K_(T)+K_(R )=(1)/(2)mv^(2)[1+(I)/(mr^(2))]` When rolls up an inclined plane of HEIGHT `H=(3V^(2))/(4g)` its KE is converted into PE So `(1)/(2)mv^(2)[1+(I)/(mr^(2))]=mg((3v^(2))/(4g))` on simplification `I=(mr^(2))/(2)` HENCE the body is either a DISC or CYLINDER.