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One end of a straight uniform $1\; \mathrm{m}$ long bar is pivoted on horizontal table. It is released from rest when it makes an angle $30^{\circ}$ from the horizontal (see figure). Its angular speed when it hits the table is given as $\sqrt{\mathrm{n}}\; \mathrm{s}^{-1},$ where $\mathrm{n}$ is an integer. The value of $n$ is
$10$
$13$
$15$
$18$
Solution
From mechanical energy conservation,
$\mathrm{U}_{\mathrm{i}}+\mathrm{K}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{K}_{\mathrm{r}}$
$\Rightarrow \mathrm{mg} \frac{\ell}{2} \sin 30^{\circ}+0=0+\frac{1}{2} \mathrm{I} \omega^{2}$
$\Rightarrow \operatorname{mg} \times \frac{1}{2} \times \frac{1}{2}+0=0+\frac{1}{2} \times \frac{\mathrm{m}(1)^{2}}{3} \omega^{2}$
$\Rightarrow \omega^{2}=\frac{3 \mathrm{g}}{2} \Rightarrow \omega=\sqrt{15}$
$\therefore n=15$
