Question

A uniform rod of mass $m$ and length $l$ is fixed at one end released from horizontal position as shown in the figure. The angular velocity of the rod when the rod makes an angle ${60}^o$ with vertical is:

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

Answer 1

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

Since total mechanical energy is conserved

$P.E.=mgl\cdot \left({\sin }^2\left(\frac{\theta }{2}\right)\right)$

$K.E.=\frac{1}{2}{I}_0{\omega }^2$

Moment of inertia of this road is

$I_0=\frac{m{l}^2}{3}$

Substituting this in the above equation we get

$K.E.=\frac{m{l}^2{\omega }^2}{6}$

$\frac{m{l}^2{\omega }^2}{6}=$$mgl\cdot \left({\sin }^2\left(\frac{\theta }{2}\right)\right)$

$\omega =\sqrt{\frac{6g}{l}{\sin }^2\left({30}^{\circ }\right)}$

$\omega =\sqrt{\frac{6g}{l}}\sin \left({30}^{\circ }\right)$

$\omega =\frac{1}{2}\sqrt{\frac{6g}{l}}$

$\omega =\sqrt{\frac{3g}{2l}}$