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 plane is

Step-by-step solution

The body of mass m descends from height h on an inclined plane. For sliding without friction: Mechanical energy conservation gives mgh = 1 2 mV^2 , so V^2 = 2gh . For a ring rolling without slipping: The moment of inertia is I = mR^2 , and the velocity relation is = V_ ring /R . Energy conservation yields mgh = 1 2 mV_ ring ^2 + 1 2 I ^2 = 1 2 mV_ ring ^2 + 1 2 mR^2(V_ ring /R)^2 = mV_ ring ^2 . Thus, V_ ring ^2 = gh . Substituting gh = V^2/2 from the sliding case, V_ ring ^2 = V^2/2 , so V_ ring = V/ 2 . The linea