Question

A uniform rod is hinged, as shown in the figure. It is released from horizontal position. The angular velocity of the rod, as it passes through the vertical position, is :

• A. $\sqrt{\frac{4g}{l}}$
• B. $\sqrt{\frac{2g}{3l}}$
• C. $\sqrt{\frac{24g}{7l}}$
• D. $\sqrt{\frac{3g}{7l}}$

Answer 1

The correct option is C $\sqrt{\frac{24g}{7l}}$

The moment of inertia of the rod about the hinge is,

$I_o=I_{com}+m{d}^2$

$\Rightarrow I_o=\frac{m{l}^2}{12}+\frac{m{l}^2}{16}=\frac{7m{l}^2}{48}$

By work energy theorem,

$W_{\text{All forces}}=\Delta KE$

$\therefore W_g=\frac{1}{2}{I}_o{\omega }^2$

Here, the displacement of the center of mass of the rod, during the given interval, is equal to $\frac{l}{4}$ below the initial horizontal position.

$\therefore mg\frac{l}{4}=\frac{1}{2}\times \frac{7m{l}^2}{48}\times {\omega }^2$

$\Rightarrow \omega =\sqrt{\frac{24g}{7l}}$