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MHT CET202210 Aug 2022Morning ShiftPhysicsRotational MotionActual

A uniform rod of mass M and length L is suspended from the rigid support. A small bullet of mass m hits the rod with velocity v and gets embedded into the rod. The angular velocity of the system just after impact is

Options

  1. A3 M V (M+m) L
  2. B3 M V (M+3 m) L
  3. C3 m V (M+3 m) L
  4. D3 m V (M+m) L

Correct answer

D. 3 m V (M+m) L

Step-by-step solution

Before impact, the bullet is moving with velocity v . The initial impact, angular momentum of the system about O the hinge point is J=L m v=m v L After the bullet gets embedded in the rod, suppose the system attains angular velocity . The moment of inertia of the bullet rod system about the axis through O is, I= (M.I Of bullet + M.I of rod) =mL^2+ 1 3 M L^2I= ( M+3 m 3 ) L^2 Final angular momentum of the system is By conservation of angular momentum, J=J^ ( M+3 m 3 ) L^2 =m v L ( M+3 m 3 ) L^2 =m v L So, the angula

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