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JEE Main20199 Apr 2019Evening ShiftPhysicsRotational MotionActual

A thin smooth rod of length L and mass M is rotating freely with angular speed ω 0 about an axis perpendicular to the rod and passing through center. Two beads of mass m and negligible size are at the center of the rod initially. The beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the oppo

Options

  1. AM ω 0 M + 3 m
  2. BM ω 0 M + 2 m
  3. CM ω 0 M + 6 m
  4. DM ω 0 M + m

Correct answer

C. M ω 0 M + 6 m

Step-by-step solution

From conservation of angular momentum. M L 2 12 ω 0 = M L 2 12 + m L 2 2 + m L 2 2 ω ' M L 2 12 ω 0 = M L 2 12 + m L 2 2 ω ' ω ' = M ω 0 M + 6 m

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