A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is 7 × 10⁻³ J. The speed of the centre of mass of the sphere is __________ cm s⁻¹

Solution

Physics sir · 3 steps

Rolling means the sphere moves forward and spins at once, so its kinetic energy has two parts.

1. Two kinds of energy. A rolling sphere has translational energy 1/2mv² and rotational energy 1/2Iω². For a solid sphere I = 2/5mR². Rolling without slipping gives ω = v/R.

   K = 1/2mv² + 1/2(2/5mR²)v²/R² = 7/10mv²

2. Put in numbers. 7 × 10⁻³ = 7/10 × 1 × v², so v² = 10⁻² m² s⁻² and v = 0.1 m s⁻¹.
3. So. 0.1 m s⁻¹ = 10 cm s⁻¹, so the speed of the centre of mass is 10.

The common slip: 0.1. Stopping at v = 0.1 leaves the answer in m s⁻¹. The blank asks for cm s⁻¹, so multiply by 100 to get 10.

The idea to keep. For rolling without slipping, write K = 1/2mv²(1 + k²/R²). For a solid sphere k²/R² = 2/5, so K = 7/10mv². Never forget the spin part.

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