For the pivoted slender rod of length l as sh | vedclass

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(C) The rod is pivoted at a distance $l/4$ from one end. The distance of the pivot from the center of mass $(CM)$ is $d = |l/2 - l/4| = l/4$.By the pa

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$\sqrt {\frac{24g}{7l}} $

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(C) The rod is pivoted at a distance $l/4$ from one end. The distance of the pivot from the center of mass $(CM)$ is $d = |l/2 - l/4| = l/4$.By the parallel axis theorem,the moment of inertia $I$ of the rod about the pivot is:$I = I_{CM} + md^2 = \frac{ml^2}{12} + m\left(\frac{l}{4}\right)^2 = \frac{ml^2}{12} + \frac{ml^2}{16} = \frac{4ml^2 + 3ml^2}{48} = \frac{7ml^2}{48}$.When the rod is released from the horizontal position and reaches the vertical position,the center of mass drops by a vertical distance $h = l/4$.By the law of conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy:$mgh = \frac{1}{2} I \omega^2$$mg\left(\frac{l}{4}\right) = \frac{1}{2} \left(\frac{7ml^2}{48}\right) \omega^2$$\frac{mgl}{4} = \frac{7ml^2}{96} \omega^2$$\omega^2 = \frac{mgl}{4} \cdot \frac{96}{7ml^2} = \frac{24g}{7l}$$\omega = \sqrt{\frac{24g}{7l}}$.

For the pivoted slender rod of length $l$ as shown in the figure,the angular velocity as the bar reaches the vertical position after being released from the horizontal position is:

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