A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be _______ ms⁻¹.

Solution

Physics sir · 3 steps

A rolling sphere carries two kinds of kinetic energy, so add the spinning part before solving for v.

1. Two energies. In pure rolling the sphere moves forward and spins at the same time. Its total kinetic energy is 1/2mv² + 1/2Iω², and pure rolling links the two motions through ω = v/R.

   K = 1/2mv² + 1/2(2/5mR²)(v/R)² = 7/10mv²

2. Put in numbers. With m = 2 kg, 7/10 × 2 × v² = 2240, so 1.4 v² = 2240 and v² = 1600 m² s⁻².
3. So. v = √1600 = 40 m s⁻¹.

The common slip: 47. Leaving out the rotational energy and using 1/2mv² = 2240 gives v = √2240 ≈ 47, not 40.

The idea to keep. For pure rolling, write K = 1/2mv²(1 + k²/R²). For a solid sphere k²/R² = 2/5, which gives K = 7/10mv².

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