A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

Solution

Physics sir · 3 steps

Write both kinetic energies in terms of mv² using v = Rω, then take the ratio.

1. Rolling condition. A ball rolling without slipping has v = Rω. At constant speed no friction is needed to keep it rolling, so the frictionless surface changes nothing here. For a solid sphere, I = 2/5mR².
2. Rotational part. Kᵣₒₜ = 1/2Iω² = 1/2 × 2/5mR² × v²/R² = 1/5mv². The translational part is Kₜᵣₐₙₛ = 1/2mv².

   Kᵣₒₜ/Kₜₒₜₐₗ = (1/5mv²)/(1/2mv² + 1/5mv²) = (1/5)/(7/10) = 2/7

3. So. The rotational kinetic energy is 2/7 of the total, which is option (b).

The idea to keep. For any rolling body with I = kmR², the rotational share of total KE is k/(1+k). Solid sphere: k = 2/5, so the share is 2/7.

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