Question

In the given figure, an $L$-shaped frame is free to rotate in a vertical plane about a horizontal axis passing through a smooth hinge $O$. Each side of the frame has a length $L$ and mass $M$. The frame is allowed to fall, with one side horizontal and the other vertical.


The linear speed of end $A$ just before striking the ground is

• A. $\sqrt{\frac{3gL}{10}(5-\sqrt{3})}$
• B. $\sqrt{\frac{3gL}{10}(5+\sqrt{3})}$
• C. $\sqrt{\frac{3g}{10L}(5-\sqrt{3})}$
• D. $\sqrt{\frac{3g}{10L}(5+\sqrt{3})}$
  

Answer 1

The correct option is A $\sqrt{\frac{3gL}{10}(5-\sqrt{3})}$

Initial
Final:

Let $\omega$ be the angular speed of point $A$ just before hitting the ground.
Decrease in P.E = Increase in Rotational K.E

$Mg[\frac{L}{2}-\frac{L}{4}]+Mg[L-\frac{\sqrt{3}L}{4}]=\frac{1}{2}{I}_O{\omega }^2)$

$I_O=I_{OA}+I_{AB}$
$I_{OA}=\frac{M{L}^2}{3}$

Using parallel axis theorem:

$I_{AB}\left(\text{about O}\right)=[\frac{M{L}^2}{12}+M(\frac{\sqrt{5}L}{2}{)}^2]=\frac{4M{L}^2}{3}$
$\therefore I_O=\frac{5M{L}^2}{3}$

So,$\frac{MgL}{4}(5-\sqrt{3})=\frac{1}{2}\frac{5M{L}^2}{3}{\omega }^2$

On solving the above equation we get:
$\omega =\sqrt{\frac{3g}{10L}(5-\sqrt{3})}$
$\Rightarrow V_A=\omega \left(OA\right)=\omega L=\sqrt{\frac{3g}{10L}(5-\sqrt{3})}\times L=\sqrt{\frac{3gL}{10}(5-\sqrt{3})}$