A thin circular ring of mass m and radius R is rotating about its axis perpendicular to the plane of the ring with a constant angular velocity ω . Two point particles each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates, with an angular velocity ω 2 . Then, the ratio m M is

Question

A thin circular ring of mass m and radius R is rotating about its axis perpendicular to the plane of the ring with a constant angular velocity ω . Two point particles each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates, with an angular velocity ω 2 . Then, the ratio m M is, 1, 2, 1 2, 2

MARKS App · Accepted Answer

Angular Momentum before particle is attached to the ring, = I ω = m R 2 ω Angular Momentum after two particle is attached to the ring, = m R 2 ω 2 + 2 M R 2 ω 2 According law of conservation of momentum, (momentum before) = (momentum after) ⇒ m R 2 ω = m R 2 + 2 M R 2 ω 2 ⇒ 2 m R 2 = m R 2 + 2 M R 2 So m M = 2

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