Q. A thin uniform heavy rod of length $l$ hangs from a horizontal axis passing through one end. The initial angular velocity $\omega$ that must be imparted to it to rotate it through $90^{\circ}$ is :

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A

$\sqrt{g / l}$

23%

B

$\sqrt{3 g / l}$

30%

C

$\sqrt{2 g / l}$

38%

D

$\sqrt{6 g / l}$

9%

Solution:

By conservation of energy we use
$\frac{1}{2} I \omega^{2}=m g \frac{l}{2}$

$\frac{m l^{2}}{3} \omega^{2}=m g l$
$\Rightarrow \omega^{2}=\frac{3 g}{l}$
$\Rightarrow \omega=\sqrt{\frac{3 g}{l}}$

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