Question

A thin rod $\text{MN}$, free to rotate in the vertical plane about the fixed end $\text{N}$, is held in a horizontal position. If the rod is released from this position, speed of end $\text{M}$ when the rod makes an angle $\alpha$ with the horizontal will be proportional to

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

Answer 1

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

Let the length of the rod $=l$, mass of the rod $=m$
When the rod makes an angle $\alpha$ with horizontal, the downward displacement of center of mass $h=\frac{l}{2}\sin \alpha$


Applying mechanical energy conservation from initial to final position,
$\text{Loss in PE}=\text{Gain in}K{E}_{\text{Rot}}...\left(ii\right)$

Since rod has started from $\omega =0$, hence gain in rotational kinetic enrgy is given by,
$K{E}_{\text{Rot}}=\frac{1}{2}I{\omega }^2...\left(iii\right)$
where $I=\frac{m{l}^2}{3}$, is the $\text{MOI}$ of rod about end $\text{N}$

From Eq $\left(i\right),\left(ii\right),\left(iii\right)$
$mg\left(\frac{l\sin \alpha }{2}\right)=\frac{1}{2}I{\omega }^2$
$\Rightarrow \frac{mgl\sin \alpha }{2}=\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2$
$\therefore \omega =\sqrt{\frac{3g\sin \alpha }{l}}$

For the tangential speed $\left(v\right)$ of end $\text{M}$, applying fundamental equation,
$v=r\omega$ where $r=l$
$\Rightarrow v=l\times \omega =\sqrt{\frac{3g\sin \alpha }{l}\times l^2}$
$\Rightarrow v=\sqrt{3gl\sin \alpha }$
$\therefore v\propto \sqrt{\sin \alpha }$