JEE Main 2020 · 5 September · morning · Single correct · +4 / −1

A wheel is rotating freely with an angular speed ω on a shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is :

  1. (a) 0
  2. (b) 5/6
  3. (c) 1/4
  4. (d) 3/4
Answer and solution

Answer (d) 3/4

Solution

Physics sir · 3 steps

The coupling is a sudden grab, so energy is not saved, but angular momentum is. Start from that.

  1. Conserve L. No external torque acts on the shaft, so angular momentum is the same before and after coupling. Friction between the wheels is internal, so it cannot change the total.

    Iω = (I + 3I) ω' ⇒ ω' = ω/4

  2. Compare energies. Before: Kᵢ = 1/2Iω². After: K_f = 1/2(4I)(ω/4)² = 1/8Iω², which is 1/4Kᵢ.
  3. So. The fractional loss is (Kᵢ - K_f)/Kᵢ = 1 - 1/4 = 3/4, option (d).

The idea to keep. When a spinning body grabs a body at rest, conserve angular momentum, not energy. Then K_f/Kᵢ = I₁/(I₁+I₂), and the loss fraction is I₂/(I₁+I₂).

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

Study Rotational Motion

Asked about 180 times in JEE Main since 2020. This question is from 2020.