When the bar makes an angle $\text{θ}$, the height of its COM (mid point) is  $\frac{L}{2}\cos \text{θ}$.

$\therefore$Displacement $=L-\frac{L}{2}\cos \text{θ}=\frac{L}{2}\left(1-\cos \text{θ}\right)$

Since, force on COM is only along the vertical direction, hence, COM is falling vertically downward.

Now, $x=\frac{L}{2}\sin \text{θ}$

$y=L\cos \text{θ}$

$\frac{x^2}{{\left(\frac{L}{2}\right)}^2}+\frac{y^2}{L^2}=1$

Put of $A$ is an ellipse.

Torques about the point of contact, $$\begin{gathered} \overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F} \\ \Rightarrow \tau =\frac{mgL}{2}\sin \text{θ} \end{gathered}$$