A thin, uniform metal rod of mass 'M' and length 'L' is swinging about a horizontal axis passing through its…

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)
  1. $\frac{\mathrm{L}^{2} \omega^{2}}{6 \mathrm{~g}}$
  2. $\frac{\mathrm{L}^{2} \omega^{2}}{\mathrm{~g}}$
  3. $\frac{\mathrm{L}^{2} \omega^{2}}{2 \mathrm{~g}}$
  4. $\frac{\mathrm{L}^{2} \omega^{2}}{3 \mathrm{~g}}$

Solution

$\begin{array}{l} \mathrm{mgh}=\frac{1}{2} \mathrm{I} \omega^{2} \\ \mathrm{I}=\frac{\mathrm{mL}^{2}}{3} \\ \therefore \quad \mathrm{mgh}=\frac{1}{2}\left(\frac{\mathrm{mL}^{2}}{3}\right) \omega^{2} \\ \quad \mathrm{~h}=\frac{1}{6 \mathrm{~g}} \mathrm{~L}^{2} \omega^{2} \end{array}$

Asked in: MHT CET 2020 (15 Oct Shift 1)

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