A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed v. The sphere and the cylinder reaches upto maximum heights h₁ and h₂ respectively, above the initial level. The ratio h₁: h₂ is n/10. The value of n is __________.

Solution

Physics sir · 4 steps

All the kinetic energy, rotation included, turns into height, so the body that spins more climbs higher.

1. Total kinetic energy. In rolling without slipping, ω = v/R. For a body with moment of inertia I = mk², the total kinetic energy is 1/2mv²(1 + k²/R²).

   1/2mv²(1+k²/R²) = mgh ⇒ h = v²/2g(1+k²/R²)

2. Solid sphere. k²/R² = 2/5, so h₁ = v²/2g × 7/5 = 7v²/10g.
3. Hollow cylinder. All its mass sits at radius R, so k²/R² = 1. That gives h₂ = v²/2g × 2 = v²/g.
4. So. h₁ : h₂ = 7/10. Matching this with n/10 gives n = 7.

The common slip: 10. Counting only the translational energy 1/2mv² makes both heights equal to v²/2g. That gives a ratio of 1 and n = 10.

The idea to keep. For rolling up an incline, h = v²/2g(1+k²/R²). Mass and radius cancel. Only k²/R² decides the height, and a larger value means a higher climb.

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