A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is √(x/10) where x= ________.

Solution

Physics sir · 4 steps

Same height, same drop in potential energy: the body that spends less on spinning rolls down faster.

1. Energy at the bottom. Rolling without slipping loses no energy to friction, so mgh turns into translational plus rotational kinetic energy. With v = ω R and I = mk², the rotational part is 1/2mv² k²/R².

   mgh = 1/2mv²(1 + k²/R²) ⇒ v = √(2gh/(1 + k²/R²))

2. Each body. For the ring I = mR², so k²/R² = 1 and v_(ring) = √gh. For the solid sphere I = 2/5mR², so k²/R² = 2/5 and vₛₚₕₑᵣₑ = √(2gh/(7/5)) = √(10gh/7).
3. Take the ratio. v_(ring)/vₛₚₕₑᵣₑ = √(gh/(10gh/7)) = √(7/10). Mass, radius and height all cancel.
4. So. Comparing with √(x/10) gives x = 7.

The idea to keep. For anything rolling from rest, v = √(2gh/(1 + k²/R²)). Only k²/R² matters: ring 1, disc 1/2, hollow sphere 2/3, solid sphere 2/5.

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