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

Answer 1

The correct option is D $\sqrt{\sin \alpha }$

When the rod makes an angle of $\alpha$ displacement of centre of mass$=\frac{l}{2}sin\alpha$

$mg\frac{l}{2}sin\alpha =\frac{1}{2}I{\omega }^2$

$mg\frac{l}{2}sin\alpha =\frac{m{l}^2}{6}{\omega }^2$

$\omega =\sqrt{\frac{3gsin\alpha }{l}}$

speed of end$=\omega \times l$=$\sqrt{3gsin\alpha l}$

hence $\omega$ is proportional to $\sqrt{sin\alpha }$