Question

A rod of length $50cm$ is pivoted at one end. It is raised such that if makes an angle of $30^{\circ }$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad{s}^{-1}$) will be $(g=10m{s}^{-2})$

• A. $\sqrt{30}$
• B. $\sqrt{\frac{30}{2}}$
• C. $\frac{\sqrt{30}}{2}$
• D. $\frac{\sqrt{20}}{3}$

Answer 1

Answer: (A)

Work done by gravity from initial to final position is,
$W=mg\frac{\ell }{2}\sin 30^{\circ }$
$=\frac{mg\ell }{4}$
According to work energy theorem
$W=\frac{1}{2}I{\omega }^2$
$\Rightarrow \frac{1}{2}\frac{m{\ell }^2}{3}{\omega }^2=\frac{mg\ell }{4}$
$\omega =\sqrt{\frac{3g}{2\ell }}=\sqrt{\frac{3\times 10}{2\times 0.5}}$
$\omega =\sqrt{30}$ rad / sec