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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 of $45^o$ with the horizontal ($I = ml^2/3$)
$\frac{{3g}}{{2\sqrt 2 \,{\text{l}}}}$
$\frac{{6g}}{{\sqrt 2 \,{\text{l}}}}$
$\frac{{\sqrt 2 \,g}}{{\text{l}}}$
$\frac{{2\,g}}{{\text{l}}}$
Solution
Loss in $P.E.$ $=$ gain in $K.E.$
or $\operatorname{mg} \frac{l}{2}(1-\cos \theta)=\frac{1}{2} \mathrm{I} \omega^{2}$
Differenting both sides,
$\operatorname{mg} \frac{l}{2} \sin \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{I} \omega \frac{\mathrm{d} \omega}{\mathrm{dt}}$
or $\operatorname{mg} \frac{l}{2} \sin \theta=\mathrm{I} \alpha$
$\therefore \alpha=\frac{\mathrm{mg} \frac{l}{2} \sin 45^{\circ}}{\frac{\mathrm{m} l^{2}}{3}}$
$\alpha=\frac{3 \mathrm{g}}{2 \sqrt{2} l}$
