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A solid cylinder and a solid sphere, each having the same mass M and radius R, roll down the same inclined plane from the top without slipping, starting from rest. Find the ratio of the speed of the solid cylinder to that of the solid sphere when they reach the bottom.

1. sqrt(5/3)
2. sqrt(4/5)
3. sqrt(3/5)
4. sqrt(14/15)

Correct answer: sqrt(14/15)

Solution

For rolling: v² = 2gh / (1 + I/(MR²)). Cylinder: I = MR²/2 → v_cyl² = 2gh/(3/2) = 4gh/3. Sphere: I = 2MR²/5 → v_sph² = 2gh/(7/5) = 10gh/7. Ratio: v_cyl/v_sph = sqrt((4/3)/(10/7)) = sqrt(28/30) = sqrt(14/15).

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