Rotational MotionMedium
Question
A thin uniform rod $(X)$ of mass M and length L is pivoted at a height $\left( \frac{L}{3} \right)$ as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is $\_\_\_\_$ . ( $g =$ gravitational acceleration)
Options
A.$\sqrt{\frac{3}{2}\frac{\text{ }g}{\text{ }L}}$
B.$\frac{3}{\sqrt{2}}\sqrt{\frac{\text{ }g}{\text{ }L}}$
C.$\frac{1}{\sqrt{2}}\sqrt{\frac{\text{ }g}{\text{ }L}}$
D.$\sqrt{\frac{3\text{ }g}{\text{ }L}}$
Solution
$mg\frac{\mathcal{l}}{6} = \frac{1}{2}I\omega^{2}$
Here $I = \frac{m\mathcal{l}^{2}}{12} + \frac{m\mathcal{l}^{2}}{36} = \frac{m\mathcal{l}^{2}}{9}$
$${mg\frac{\mathcal{l}}{6} = \frac{m\mathcal{l}^{2}}{18}\omega^{2} \Rightarrow \omega^{2} = \frac{3\text{ }g}{\mathcal{l}} }{\omega = \sqrt{\frac{3\text{ }g}{\mathcal{l}}}}$$
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