Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is √x/2. Then, the value of x is _____________.

Solution

Physics sir · 4 steps

Energy at the bottom splits between sliding and spinning, and the ring spends more of it on spinning.

1. Energy conservation. Rolling without slipping means no energy is lost to friction, and ω = v/R. The loss in potential energy Mgh becomes 1/2Mv² + 1/2Iω². Write I = Mk², where k is the radius of gyration.

   v = √(2gh/(1 + k²/R²))

2. Put in each body. Mass and radius cancel, so only k²/R² matters. For the ring it is 1, so v_(ring) = √gh. For the solid cylinder it is 1/2, so v_(cyl) = √(4gh/3).
3. Take the ratio. v_(ring)/v_(cyl) = √(gh/(4gh/3)) = √(3/4) = √3/2.
4. So. Comparing with √x/2 gives x = 3.

The idea to keep. For rolling down an incline, the speed at the bottom depends only on k²/R², not on mass or radius. A larger k²/R² means a slower body, so the ring always loses the race.

Answer from the NTA final answer key. Solution by PrepWiser.