As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be:

Solution

Physics sir · 3 steps

The bob's lost potential energy is shared by its own KE and the disc's rotational KE, so balance all three.

1. Energy bookkeeping. The string does not slip, so no energy is lost. The drop in the bob's potential energy, mgh, goes into the bob's translational KE and the disc's rotational KE, with I = 1/2mr² for a disc.

   mgh = 1/2mv² + 1/2(1/2mr²)ω²

2. Link v and ω. The string unwinds off the rim, so the bob's speed equals the rim speed: v = ω r. Then mgh = 1/2mω²r² + 1/4mω²r² = 3/4mω²r².
3. So. ω² = 4gh/3r², giving ω = 1/r√(4gh/3), which is option (c).

The idea to keep. For a mass hanging from a string wound on a wheel, write mgh = translational KE + rotational KE and use v = ω r. The I of the wheel decides how the energy is split.

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