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A simple pendulum of length $\ell$ is displaced so that its taught string is horizontal and then released. A uniform bar pivoted at one end is simultaneously released from its horizontal position. If their motions are synchronous, what is the length of the bar?
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Verified Answer
The correct answers are:
$\frac{3 \ell}{2}$
Hint:
$v=\sqrt{2 g h}$
$=\sqrt{2 g \mid \sin \theta}$
$\omega_{1}=\frac{v}{l}=\sqrt{\frac{2 g \sin \theta}{1}}$
$m g h=\frac{1}{2} l \omega^{2}$
$\Rightarrow \mathrm{mg} \cdot \frac{\mathrm{l}^{\prime} \sin \theta}{2}=\frac{1}{2} \times \frac{\mathrm{ml}^{2}}{3} \times \omega_{2}^{2}$
$\omega_{2}=\sqrt{\frac{3 \mathrm{~g} \sin \theta}{\mathrm{l}}}=\omega_{1} \Rightarrow \frac{2}{\mathrm{l}}=\frac{3}{\mathrm{l}}$
$\Rightarrow \mathrm{I}^{\prime}=\frac{3}{2}I$
$v=\sqrt{2 g h}$
$=\sqrt{2 g \mid \sin \theta}$
$\omega_{1}=\frac{v}{l}=\sqrt{\frac{2 g \sin \theta}{1}}$
$m g h=\frac{1}{2} l \omega^{2}$
$\Rightarrow \mathrm{mg} \cdot \frac{\mathrm{l}^{\prime} \sin \theta}{2}=\frac{1}{2} \times \frac{\mathrm{ml}^{2}}{3} \times \omega_{2}^{2}$
$\omega_{2}=\sqrt{\frac{3 \mathrm{~g} \sin \theta}{\mathrm{l}}}=\omega_{1} \Rightarrow \frac{2}{\mathrm{l}}=\frac{3}{\mathrm{l}}$
$\Rightarrow \mathrm{I}^{\prime}=\frac{3}{2}I$
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