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JEE Main 2019 · Physics · Rotational Motion · medium

A uniform horizontal disc of mass M and radius R rotates freely about its vertical symmetry axis with angular velocity ω. A beetle of mass M/4 sits…

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A uniform horizontal disc of mass MM and radius RR rotates freely about its vertical symmetry axis with angular velocity ω\omega. A beetle of mass M/4M/4 sits at the rim. The beetle crawls slowly to the centre of the disc. The ratio of the final rotational kinetic energy of the system to its initial rotational kinetic energy is
  1. A. 43\dfrac43
  2. B. 22
  3. C. 32\dfrac32
  4. D. 34\dfrac34
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Answer: (C) 32\dfrac32

There is no external torque about the axis, so angular momentum is conserved. Initially I1=12MR2+M4R2=34MR2I_1=\tfrac12MR^2+\tfrac M4R^2=\tfrac34MR^2. When the beetle reaches the centre it contributes nothing, so I2=12MR2I_2=\tfrac12MR^2. Then ω2=ω I1/I2=32ω\omega_2=\omega\,I_1/I_2=\tfrac32\omega. The kinetic energy is K=L22IK=\dfrac{L^2}{2I} with constant LL, so K2K1=I1I2=3/41/2=32\dfrac{K_2}{K_1}=\dfrac{I_1}{I_2}=\dfrac{3/4}{1/2}=\dfrac32. The energy increases because the beetle does positive work against the centrifugal tendency as it crawls inward.
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