A uniformly thick wheel with moment of inertia I and radius R is free to rotate about its centre of mass (see fig). A massless string is wrapped over its rim and two blocks of masses m₁ and m₂ (m₁ > m₂) are attached to the ends of the string. The system is released from rest. The angular speed of the wheel when m₁ descents by a distance h is :

Solution

Physics sir · 3 steps

Use energy: what the system loses in potential energy becomes kinetic energy of both blocks and the wheel.

1. Potential energy lost. When m₁ falls by h, the string makes m₂ rise by the same h. The net loss of potential energy is m₁ g h - m₂ g h = (m₁ - m₂)gh.
2. Kinetic energy gained. Both blocks move with the same speed v, and the wheel spins with ω. The string does not slip, so v = ω R. The total kinetic energy is 1/2(m₁ + m₂)ω² R² + 1/2Iω².

   (m₁ - m₂)gh = 1/2[(m₁ + m₂)R² + I]ω²

3. So. ω = [2(m₁ - m₂)gh/((m₁ + m₂)R² + I)]^(1/2), which is option (d).

The idea to keep. For blocks on a pulley with inertia, write one energy equation: net PE drop equals translational KE of every block plus 1/2Iω², with v = ω R.

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