A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is x/5. The value of x is _________.

Solution

Physics sir · 3 steps

Write both energies in terms of v using v = ωR, and the ratio comes out without M or R.

1. Moment of inertia. For a hollow sphere (a thin spherical shell), I = 2/3MR². Rolling without slipping means v = ω R.
2. Each energy. Rotational: 1/2Iω² = 1/2 × 2/3MR² × v²/R² = 1/3Mv². Translational: 1/2Mv². Total: 1/3Mv² + 1/2Mv² = 5/6Mv².

   Kᵣₒₜ/Kₜₒₜₐₗ = (1/3Mv²)/(5/6Mv²) = 2/5

3. So. The ratio is 2/5, so x = 2.

The common slip: 3. Taking translational KE over total KE gives 3/5, so x = 3. The question asks for the rotational part.

The idea to keep. For any rolling body with I = kMR², rotational KE over total KE is k/(1+k). Hollow sphere k = 2/3, solid sphere 2/5, ring 1, disc 1/2.

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