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A uniform ring of mass M and radius R starts from rest at the top of a rough inclined plane and rolls down without slipping through a vertical height h. What is the magnitude of the angular momentum of the ring about its own center of mass when it reaches the bottom of the incline?

1. MR*sqrt(gh)
2. MR*sqrt(gh/2)
3. MR*sqrt(2gh)
4. None of these

Correct answer: MR*sqrt(gh)

Solution

Energy conservation gives omega = sqrt(gh)/R, and since L = I*omega = MR² * sqrt(gh)/R = MR*sqrt(gh).

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