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A uniform rod of mass “M” and length “L” is free to rotate in a vertical plane about a pivot at one end as shown in the figure. Initially the rod is in the horizontal position and friction at the pivot may be neglected. When the rod is released, the angular velocity of the rod when it is in the vertical position will be

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$\sqrt{\frac{3g}{L}}$

$\sqrt{\frac{3g}{2L}}$

$\sqrt{\frac{2g}{L}}$

$\sqrt{\frac{6g}{L}}$

answer is B.

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Detailed Solution

mg, $\frac{l}{2} = \frac{1}{2} I ω^{2}$ mg $\frac{l}{2} = \frac{1}{2} \frac{m l^{2}}{3} \times ω^{2}$ $ω = \sqrt{\frac{3 g}{l}}$

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