MHT CET · Physics · Rotational Motion

A thin circular ring of mass ' \(\mathrm{M}^{\prime}\) and radius ' \(\mathrm{R}^{\prime}\) is rotating about a transverse axis
passing through its centre with constant angular velocity ' \(\omega\) '. Two objects each of
mass 'm' are attached gently to the opposite ends of a diameter of the ring. What is the new angular velocity?

A \(\frac{M \omega}{M+2 m}\)
B \(\frac{M(\omega)}{M+m}\)
C \(\frac{(M+2 m) \omega}{M}\)
D \(\frac{(\mathrm{M}-2 \mathrm{~m}) \omega}{\mathrm{M}+2 \mathrm{~m}}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{M \omega}{M+2 m}\)

Step-by-step Solution

Detailed explanation

Initial angular momentum of the ring \(\quad=I \omega=M R^{2} \omega\)
If the new angular velocity is \(\omega^{\prime}\) then the final angular momentum \(=I^{\prime} \omega^{\prime}\) where \(I^{\prime}=M R^{2}+2 m R^{2}=(M+2 m) R^{2}\).
By law of conservation of momentum
\(\begin{array}{c}
\mathrm{I}^{\prime} \omega^{\prime}=\mathrm{I} \omega \\
(\mathrm{M}+2 \mathrm{~m}) \mathrm{R}^{2} \omega^{\prime} \\
\therefore \omega^{\prime}=\frac{\mathrm{M} \omega}{\mathrm{M}+2 \mathrm{~m}}
\end{array}=\mathrm{MR}^{2} \omega\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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