Question

A uniform meter stick of mass $m$ is pivoted about a horizontal axis through its lower end $O$. initially it is held vertical and is allowed to fall freely. Its angular velocity at the instant when it makes an angle ${60}^o$ with the vertical is

• A. $\sqrt{\frac{g}{3}}$
• B. $\sqrt{\frac{3g}{2}}$
• C. $\sqrt{\frac{g}{4}}$
• D. $\sqrt{\frac{2g}{3}}$

Answer 1

The correct option is B $\sqrt{\frac{3g}{2}}$

Initial Energy $=\frac{L}{2}$

$x=\frac{1}{2}\cos {60}^o=\frac{L}{2}\times \frac{1}{2}=\frac{L}{4}$

Final Energy $=mg\frac{L}{4}+\frac{1}{2}I{w}^2=mg\frac{L}{4}+\frac{1}{2}\frac{m{L}^2}{3}.w^2$

conservation of energy gives

$mg\frac{1}{2}=mg\frac{L}{4}+\frac{1}{2}\frac{m{t}^2}{3}{w}^2$

$\Rightarrow mg\frac{L}{24}=\frac{1}{2}\frac{m{L}^2}{3}{w}^2\Rightarrow \frac{39}{2L}=w^2$

Given thta $L=1$ (meter stick) $\Rightarrow w^2=\frac{3g}{2}$ or $\theta =\sqrt{\frac{3g}{2}}$