Question

One end of a straight, uniform $1\text{m}$ long bar is pivoted on a horizontal table. It is released from rest, when it makes an angle 30° from the horizontal (see figure). Its angular speed , when it hits the table, is given as $\sqrt{n}{\text{s}}^{-1}$, where $n$ is an integer. The value of $n$ is .

Answer 1

Here, length of bar, $1=1\text{m}$

and, angle, $\theta ={30}^{\circ }$

Now, $\Delta PE=\Delta KE$

$mgh=\frac{1}{2}I{\omega }^2$

$\Rightarrow \left(mg\right)\frac{l}{2}\sin {30}^{\circ }=\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2$

$\Rightarrow g\times \frac{1}{4}=\frac{l}{6}\times {\omega }^2$

$\Rightarrow \omega =\sqrt{\frac{3g}{2l}}=\sqrt{\frac{3\times 10}{2\times 1}}=\sqrt{15}\text{rad/s}$

$\therefore n=15$