Question Detail
Question
A solid cylinder of mass M and radius R has a moment of inertia I = ½ M R². If it rolls without slipping, what fraction of its total kinetic energy is rotational kinetic energy?
Correct Answer
Option B is the correct answer.
Detailed Explanation
The total kinetic energy is KE_total = ½ M v² + ½ I ω². For rolling without slipping, ω = v/R and I = ½ M R². Substituting gives KE_rot = ½ (½ M R²) (v/R)² = ¼ M v². KE_total = ½ M v² + ¼ M v² = ¾ M v². The fraction that is rotational is (¼ M v²) / (¾ M v²) = 1/3.
Hint
Express both the translational and rotational kinetic energies in terms of M and v using the rolling condition and the given moment of inertia.