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JEE Main201912 Apr 2019Morning ShiftPhysicsRotational MotionActual

A person of mass M is sitting on a swing of length L and swinging with and an angular amplitude θ 0 . If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance l l < < L , is close to:

Options

  1. AM g l   ( 1 - θ 0 2 )
  2. BM g l ( 1 + θ 0 2 2 )
  3. CM g l
  4. DM g l ( 1 + θ 0 2 )

Correct answer

D. M g l ( 1 + θ 0 2 )

Step-by-step solution

From angular momentum conservation M V 0 L = M V ( L - l ) V = V 0 L L - l Now W = M g l + 1 2 M V 0 2 L L - l 2 - V 0 2 Since l ≪ L W = M g l + 1 2 M V 0 2 1 - l L - 2 - V 0 2 = M g l + 1 2 M V 0 2 1 + 2 l L - V 0 2 = M g l + M V 0 2 l L From SHM V 0 = ω A = g L × L θ 0 = g L θ 0 Using this W = M g l + M g L × θ 0 2 × l L W = M g l 1 + θ 0 2

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