A solid cylinder of mass ' M ' and radius ' R ' rolls down an inclined plane of height ' h '. When it reaches the foot of the plane, its rotational kinetic energy is ( g = acceleration due to gravity)

Step-by-step solution

From the law of conservation of energy, we have Potential energy = Translational kinetic energy + Rotational kinetic energy or mgh = 1 2 mv ^2+ 1 2 I ^2 or mgh = 1 2 m v^2 ^2+ 1 2 ( 1 2 m r^2 ) ^2= 3 4 m r^2 ^2 ^2 or ^2= 4 gh 3 r ^2 Now the rotational kinetic energy = 1 2 I ^2 Substituting for ^2 and I , we have, aligned Rotational kinetic energy & = 1 2 ( 1 2 mr ^2 ) 4 gh 3 r ^2 & = Mgh 3 aligned ... ( M = m )