A body slides down a smooth inclined plane of inclination and reaches the bottom with velocity 'V'. If the same body is a ring which rolls down the same inclined plane then linear velocity at the bottom of the plane is

Step-by-step solution

Let the height of the inclined plane be h . For the body sliding down the smooth inclined plane, the conservation of mechanical energy gives: mgh = 1 2 mV^2 V = 2gh For the ring rolling down the same inclined plane, the total kinetic energy at the bottom is the sum of translational and rotational kinetic energy: mgh = 1 2 mv^2 + 1 2 I ^2 For a ring, the moment of inertia is I = mR^2 and for pure rolling, = v R . Substituting these values: mgh = 1 2 mv^2 + 1 2 (mR^2) ( v R )^2 mgh = 1 2 mv^2 + 1 2 mv^2 = mv^2 v = gh