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 gentl

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

TestHub · 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,

$= m R^{2} \frac{ω}{2} + 2 M R^{2} \frac{ω}{2}$

According law of conservation of momentum, (momentum before) = (momentum after)

$\Rightarrow m R^{2} ω = (m R^{2} + 2 M R^{2}) \frac{ω}{2}$

$\Rightarrow 2 m R^{2} = m R^{2} + 2 M R^{2}$

So $\frac{m}{M} = 2$

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