JEE Main 2021 · 22 July · evening · Single correct · +4 / −1
Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter.
The correct statement for this situation is
- (a) All of them will have same velocity.
- (b) The ring has greatest and the cylinder has the least velocity of the centre of mass at the bottom of the inclined plane.
- (c) The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.
- (d) The cylinder has the greatest and the sphere has the least velocity of the centre of mass at the bottom of the inclined plane.
Answer and solution
Answer (c) The sphere has the greatest and the ring has the least velocity of the centre…
Solution
Physics sir · 3 steps
Whatever energy goes into spinning is lost to forward speed, so the smallest k²/R² wins the race.
- Energy at the bottom. Rolling without slipping means friction does no work, so the loss in potential energy mgh becomes translational plus rotational kinetic energy. With v = Rω and I = mk², the rotational part is 1/2mv² k²/R².
mgh = 1/2mv²(1 + k²/R²) ⇒ v = √(2gh/(1 + k²/R²))
- What decides speed. Mass and radius cancel out. Only the shape factor k²/R² matters, and a smaller value gives a larger v.
Body k²/R² v at the bottom Ring 1 √gh, least Solid cylinder 1/2 √(4/3gh) ≈ √(1.33 gh) Solid sphere 2/5 √(10/7gh) ≈ √(1.43 gh), greatest - So. The solid sphere reaches the bottom with the greatest speed and the ring with the least, which is option (c).
The idea to keep. For bodies rolling from rest, v = √(2gh/(1 + k²/R²)). Mass and size do not matter; the body with mass closest to the axis (smallest k²/R²) is fastest.
Answer from the NTA final answer key. Solution by PrepWiser.