Question

A thin uniform rod of mass $m$ and length $L$ is hinged vertically at the frictionless pivot point $O$ as shown in the figure. It is allowed to fall to the ground, then with what speed does the free end of the rod strike the ground?

• A. $\sqrt{6gL}$
• B. $\sqrt{3gL}$
• C. $\sqrt{5gL}$
• D. $\sqrt{2gL}$

Answer 1

The correct option is B $\sqrt{3gL}$

As the end of the rod is fixed at $O$, work done by hing force will be zero $(W_N=0)$ and dissipative force (friction) is absent, hence, we can apply mechanical energy conservation on the rod.


The center of mass of the rod is at a height $\frac{L}{2}$ above the ground.
Loss in potential energy $=$ Gain in kinetic energy
$\Rightarrow mg\left(\frac{L}{2}\right)=\frac{1}{2}I{\omega }^2$
Here, $I=\frac{m{L}^2}{3}$ (about one of the ends of the rod)
$\Rightarrow \frac{mgL}{2}=\frac{m{L}^2{\omega }^2}{2\times 3}$
$\Rightarrow \omega =\sqrt{\frac{3g}{L}}$
The speed with which the free end will strike the ground will be the tangential speed.
$\Rightarrow v=r\omega =L\sqrt{\frac{3g}{L}}$
$\Rightarrow v=\sqrt{3gL}$