A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is π: 22 then, the value of its angular speed will be ____________ rad / s.

Solution

Physics sir · 4 steps

Write both L and the total energy in terms of MR²ω, and M and R cancel in the ratio.

1. Angular momentum. The axis of rotation passes through the centre, and for a solid sphere I = 2/5MR². So L = Iω = 2/5MR²ω.
2. Total energy. Rolling without slipping means v = Rω. The energy is translational plus rotational: E = 1/2MR²ω² + 1/2·2/5MR²ω² = 7/10MR²ω².

   L/E = (2/5MR²ω)/(7/10MR²ω²) = 4/7ω = π/22

3. Solve for ω. ω = 88/7π. With π = 22/7, 7π = 22, so ω = 88/22.
4. So. The angular speed is 4 rad s⁻¹.

The common slip: 14. Using only the rotational energy 1/5MR²ω² as the total energy gives L/E = 2/ω and ω = 14. The translational energy must be included.

The idea to keep. For a rolling body, total KE is 1/2Mv²(1 + K²/R²). Never drop the translational part, and take L about the axis the question names.

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