MHT CET · Physics · Rotational Motion

A thin uniform metal rod of mass ' \(M\) ' and length ' \(L\) ' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' \(\omega\) '. Its centre of mass rises to a maximum height of ( \(\mathrm{g}=\) Acceleration due to gravity)

A \(\frac{\mathrm{L}^2 \omega^2}{3 \mathrm{~g}}\)
B \(\frac{L^2 \omega^2}{2 g}\)
C \(\frac{\mathrm{L}^2 \omega^2}{6 \mathrm{~g}}\)
D \(\frac{\mathrm{L}^2 \omega^2}{4 \mathrm{~g}}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{\mathrm{L}^2 \omega^2}{6 \mathrm{~g}}\)

Step-by-step Solution

Detailed explanation

By conservation of energy, \(\frac{1}{2} \mathrm{I} \omega^2=\mathrm{Mgh}\)
\(\therefore \quad \mathrm{h}=\frac{\mathrm{I} \omega^2}{2 \mathrm{Mg}}=\mathrm{h}...(i)\)
The M.I of a uniform rod about an axis passing through its centre is \(\frac{\mathrm{ML}^2}{12}\).
As the axis passing through the end, using parallel axis theorem,
\(\mathrm{I}=\frac{\mathrm{ML}^2}{12}+\mathrm{M}\left(\frac{\mathrm{~L}}{2}\right)^2=\frac{\mathrm{ML}^2}{3}...(ii)\)
Putting (ii) into (i),
\(h=\frac{L^2 \omega^2}{6 g}\)

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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