Question

A uniform rod of mass $M$ and length $L$ is hinged about end $A$. It is released from rest from the vertical position as shown. The velocity of centre of mass of rod when it makes an angle of $45°$ with the vertical is

• A. $\sqrt{\frac{3gL}{\sqrt{2}}}$
• B. $\sqrt{\frac{3gL}{4\sqrt{2}}}$
• C. $\sqrt{\frac{3gL(\sqrt{2}-1)}{4\sqrt{2}}}$
• D. $\sqrt{\frac{2}{3}gL(1-\frac{1}{\sqrt{2}})}$

Answer 1

The correct option is C $\sqrt{\frac{3gL(\sqrt{2}-1)}{4\sqrt{2}}}$

Let us suppose the COM falls vertically downward by height $h$, when the rod makes an angle $\theta$ with the verticle.

$h=\frac{L}{2}(1-\cos \theta )...\left(i\right)$

Now, applying energy conservation, we have:

$mgh=\frac{1}{2}\left(\frac{m{L}^2}{3}\right)\times {\omega }^2...\left(ii\right)$

Substituting $\theta ={45}^{\circ }$ in the above equation,

$\omega =\sqrt{\frac{3g}{L}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)}$

Now, the velocity of COM will be given by,

$\therefore v=\omega \left(\frac{L}{2}\right)=\sqrt{\frac{3gL(\sqrt{2}-1)}{4\sqrt{2}}}$

Hence, option $\left(D\right)$ is correct.