. The moment of inertia of a solid cylinderabout its axis is given by 0.5MR2. If

Question

. 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)

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