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 ' \(\omega^{\prime}\). Its centre of mass rises to a maximum height of \((\mathrm{g}=\) acceleration due to gravity)
- A \(\frac{\mathrm{L}^{2} \omega^{2}}{6 \mathrm{~g}}\)
- B \(\frac{\mathrm{L}^{2} \omega^{2}}{\mathrm{~g}}\)
- C \(\frac{\mathrm{L}^{2} \omega^{2}}{2 \mathrm{~g}}\)
- D \(\frac{\mathrm{L}^{2} \omega^{2}}{3 \mathrm{~g}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{L}^{2} \omega^{2}}{6 \mathrm{~g}}\)
Step-by-step Solution
Detailed explanation
\( \begin{array}{l} \mathrm{mgh}=\frac{1}{2} \mathrm{I} \omega^{2} \\ \mathrm{I}=\frac{\mathrm{mL}^{2}}{3} \\ \therefore \mathrm{mgh}=\frac{1}{2}\left(\frac{\mathrm{mL}^{2}}{3}\right) \omega^{2} \\ \mathrm{~h}=\frac{1}{6 \mathrm{~g}} \mathrm{~L}^{2} \omega^{2} \end{array} \)
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