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NEET 2012 (AIPMT, Mains) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from NEET 2012 (AIPMT, Mains)

Q82NEET 2012 · AIPMT · MAINSMedium

Angular Momentum & its Conservation

A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is 200 kg m2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 ms−1{}^{-1} relative to the ground. Time taken by the man to complete one revolution is

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Answer: (c)

  1. Total L=0L = 0: man's L=mvR=50(1)(2)=100 kg m2/sL = mvR = 50(1)(2) = 100\ \text{kg m}^2\text{/s}, so platform gets Iωp=100I\omega_p = 100.
  2. ωp=100200=0.5 rad/s\omega_p = \frac{100}{200} = 0.5\ \text{rad/s}, opposite to the man.
  3. Man's angular speed (ground) =vR=0.5 rad/s= \frac{v}{R} = 0.5\ \text{rad/s}, so relative to the platform ωrel=0.5+0.5=1 rad/s\omega_{rel} = 0.5 + 0.5 = 1\ \text{rad/s}.
  4. One revolution on the platform: t=2πωrel=2π st = \frac{2\pi}{\omega_{rel}} = 2\pi\ \text{s}
  5. So the answer is (c) 2π2\pi s.
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