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A uniform rod of length l, hinged at the lower end is free to rotate in the vertical plane. If the rod is held vertically in the beginning and then released, the angular acceleration of the rod when it makes an angle of45° with the horizontal is $α = \frac{3 g}{2 \sqrt{n} l}$ where n =

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Detailed Solution

$mg \times \frac{l}{2} \sin 45 = Iα$

$mg \times \frac{l}{2 \sqrt{2}} = \frac{{ml}^{2}}{3} \times α$

$\Rightarrow α = \frac{3 g}{2 \sqrt{2} l}$

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