A rigid uniform bar AB of length L is slipping from its vertical...

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

$∴$Displacement = L - \dfrac{L}{2}cos\text{\theta} = \dfrac{L}{2} \left(1 - cos\text{\theta}\right)</p><p>Since, force on COM is only along the vertical direction, hence, COM is falling vertically downward.</p><p>Now,x = $\dfrac{L}{2}$sin$\text{\theta}</p><p>$y = Lcos$\text{\theta}</p><p>\dfrac{x^{2}}{{\left(\dfrac{L}{2}\right)}^{2}} + \dfrac{y^{2}}{L^{2}} =$1$</p><p><img src="/api/proxy-image?url=https$A$is an ellipse.</p><p>Torques about the point of contact,$\overset{→}{τ} = \overset{→}{r} $\times \overset{→}{F}$⇒ τ = \d$dfrac{mgL}{2}sin\text{\theta}$