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JEE Main 2013 (25 Apr, Shift 1) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2013 (25 Apr, Shift 1)

Q79JEE Main 2013 · 25 APR · SHIFT 1Medium

Angular Momentum & its Conservation

A ring of mass M and radius R is rotating about its axis with angular velocity ω\omega. Two identical bodies each of mass m are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be:

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Answer: (c)

  1. Loss when two bodies are coupled: ΔK=12I1I2I1+I2ω2\Delta K = \frac{1}{2}\frac{I_1I_2}{I_1+I_2}\omega^2 (second body initially at rest).
  2. I1=MR2I_1 = MR^2 (ring), I2=2mR2I_2 = 2mR^2 (two masses).
  3. ΔK=12⋅MR2⋅2mR2(M+2m)R2ω2=MmM+2mω2R2\Delta K = \frac{1}{2}\cdot\frac{MR^2\cdot 2mR^2}{(M+2m)R^2}\omega^2 = \frac{Mm}{M+2m}\omega^2R^2
  4. So the answer is (c).
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