. The moment of inertia of a solid cylinderabout its axis is given by 0.5MR2. If this cylinder rolls withoutslipping, the ratio of its rotational kinetic energy to itstranslational kinetic energy is ? Please give step by step instructions with solutions for 14points . The moment of inertia of a solid cylinderabout its axis is given by 0.5MR2. If this cylinder rolls withoutslipping, the ratio of its rotational kinetic energy to itstranslational kinetic energy is ? Please give step by step instructions with solutions for 14points Please give step by step instructions with solutions for 14points
Explanation / Answer
The ratio of its rotational kinetic energy to itstranslational kinetic energy is = ( 1/ 2) I w ^ 2 / ( 1/ 2 ) mv ^ 2 where I = moment of inertia = ( 1/ 2) m R^ 2 w = v / R So, The ratio of its rotational kinetic energy to itstranslational kinetic energy is = [ ( 1/ 2) ( 1/ 2) m R ^ 2 ( v / R ) ^ 2 ] / [ ( 1/ 2 ) m v ^ 2] = ( 1/ 2)