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MHT CET202313 May 2023Evening ShiftPhysicsMotion in Two DimensionsActual

A thin uniform rod A B of mass ' m ' and length ' l ' is hinged at one end A to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its end B strikes the ground is ( g = acceleration due to gravity)

Options

  1. Ag l
  2. Bmg l
  3. C3 ~g l
  4. Dmg 3 l

Correct answer

C. 3 ~g l

Step-by-step solution

We know, Loss in P.E = Gain in rotational K.E. When the centre of mass of the rod falls through a distance L 2 , Loss in P.E =m g L 2 ...(i) and, Gain in Rotational K.E = 1 2 I ^2= 1 2 [ ML ^2 3 ] ^2 ...(ii) Equating (i) and (ii) aligned & MgL 2 = 1 2 ML ^2 3 ^2 & ^2= 3 ~g ~L & = 3 ~g ~L = 3 ~g l ( here L =l) aligned ( here L =l)

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