Question

A thin rod $MN$, free to rotate in the vertical plane about the fixed end $N$, is held horizontal. When the end $M$ is released the speed of this end, when the rod makes an angle $\alpha$ with the horizontal, will be proportional to : (see figure)

• A. $\sqrt{\sin \, \alpha}$
• B. ${\sin \, \alpha}$
• C. $\sqrt{\cos\, \alpha}$
• D. ${\cos\, \alpha}$

Answer 1

Answer: (A)

$\sqrt{\sin \, \alpha}$

Answer 2

When the end $M$ is released and rod makes angle $\alpha$ with horizontal, the displacement of centre of mass is $\frac{L}{2} \sin \alpha .$ Now,
we know $m g \frac{L}{2} \sin \alpha=\frac{1}{2} I \omega^{2}$
Here, $I=\frac{m L^{2}}{3} ;$ therefore,
$m g \frac{L}{2} \sin \alpha=\frac{1}{2} \frac{m L^{2}}{3} \omega^{2}$
$\Rightarrow \omega^{2}=\frac{3 g \sin \alpha}{L}$
$\Rightarrow \omega=\sqrt{\frac{3 g \sin \alpha}{L}}$
$\Rightarrow \omega \propto \sqrt{\sin \alpha}$