Question

A rod of length $50\text{cm}$ 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 ${\text{rad s}}^{-1}$) will be ($g=10{\text{m s}}^{-2}$)

• A. $\frac{30}{2}$
• B. $\sqrt{\frac{30}{7}}$
• C. $\sqrt{\frac{20}{3}}$
• D. $\sqrt{30}$

Answer 1

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



By the law of conservation of energy,
Loss of P.E. of rod = Gain in Rotational K.E.
$mg\frac{l}{2}\sin \theta =\frac{1}{2}I{\omega }^2$( <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\because I=\frac{m{l}^2}{3}$ and <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\theta$= <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ${30}^{\circ }$)
$\Rightarrow mg\frac{l}{2}\sin {30}^{\circ }=\frac{1}{2}\frac{m{l}^2}{3}{\omega }^2\Rightarrow mg\frac{l}{2}\times \frac{1}{2}=\frac{1}{2}\frac{m{l}^2}{3}{\omega }^2$
For complete length of rod,

$\omega =\sqrt{3g/2l}=\sqrt{\frac{3\left(10\right)}{2\left(0.5\right)}}=\sqrt{30}{\text{rad s}}^{-1}$