A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω . Two objects each of mass M are attached ge - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω . Two objects 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^{'} :

Solution & Step-by-step Explanation

From the conservation of angular momentum (L_{i} = L_{f}): I_{i} ω = I_{f} ω^{'} Moment of inertia of ring I_{r} = m R^{2} .Initial I_{i} = m R^{2} .After attaching two masses M at radius R, final I_{f} = m R^{2} + 2 M R^{2} = (m + 2 M) R^{2} . m R^{2} ω = (m + 2 M) R^{2} ω^{'} ω^{'} = \frac{mω}{m + 2 M}

Try it yourself before checking the explanation above.

A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω . Two objects 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^{'} :

\frac{mω}{m + 2 M}

\frac{mω}{m + M}

\frac{( m + 2 M ) ω}{m}

\frac{mω}{m + 4 M}

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