Question

A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of ${30}^o$ 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

The correct option is A $\sqrt{30}$

Work done by gravity from initial to final
position is,

$W=mg\frac{\ell }{2}sin{30}^o$

$=\frac{mg\ell }{4}$

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

$\therefore$ correct answer is (1)