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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)
Options:
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)
Solution:
2362 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{L}^{2} \omega^{2}}{6 \mathrm{~g}}$
$$
\begin{array}{l}
\mathrm{mgh}=\frac{1}{2} \mathrm{I} \omega^{2} \\
\mathrm{I}=\frac{\mathrm{mL}^{2}}{3} \\
\therefore \quad \mathrm{mgh}=\frac{1}{2}\left(\frac{\mathrm{mL}^{2}}{3}\right) \omega^{2} \\
\quad \mathrm{~h}=\frac{1}{6 \mathrm{~g}} \mathrm{~L}^{2} \omega^{2}
\end{array}
$$
\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}
$$
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