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COMEDK2023Morning ShiftPhysicsRotational MotionActual

A thin circular ring of mass , M and radius R rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity . Four small spheres each of mass m (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

Options

  1. A( M+4 m M )
  2. B( M M-4 m )
  3. C( M M+4 m )
  4. DM 4 m

Correct answer

C. ( M M+4 m )

Step-by-step solution

The initial moment of inertia of the ring about the axis passing through its centre and perpendicular to its plane is I₁ = MR^2 . The initial angular momentum of the system is L = I₁ = MR^2 . When four small spheres of mass m are placed at the ends of two mutually perpendicular diameters, they are at a distance R from the axis of rotation. The new moment of inertia of the system becomes I₂ = MR^2 + 4(mR^2) = (M + 4m)R^2 . Since no external torque acts on the system, the angular momentum is conserved, so L₁ = L₂ . M

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