A solid sphere at rests rolls down an inclined plane of vertical height h without sliding. Its speed on reaching the bottom of plane is ( g= acceleration due to gravity)

Step-by-step solution

Applying conservation of mechanical energy for the solid sphere rolling without slipping down the inclined plane, the initial potential energy mgh converts to both translational and rotational kinetic energy at the bottom. For a solid sphere, I = 2 5 mr^2 and = v/r . The rotational kinetic energy becomes 1 2 I ^2 = 1 2 2 5 mr^2 v^2 r^2 = 1 5 mv^2 . The total kinetic energy is then 1 2 mv^2 + 1 5 mv^2 = 7 10 mv^2 . Equating mgh = 7 10 mv^2 and solving for velocity yields v = 10gh 7 . ( 10 g h 7 )^ 1 2