Rod sliding and falling

Let the coordinates of the lowermost point on the rod be $(s,0)$ . Then, the coordinates of the COM are:

$x_c = s + \frac{L}{2}\sin{\theta}$ $y_c = \frac{L}{2}\cos{\theta}$

$\dot{x}_c = \dot{s} + \frac{L}{2}\cos{\theta} \dot{\theta}$ $\dot{y}_c = -\frac{L}{2}\sin{\theta} \dot{\theta}$

The total kinetic energy of the rod is the sum of kinetic and rotational energy which is:

$\mathcal{T} = \frac{m}{2} \left(\dot{x}_c^2 + \dot{y}_c^2\right) + \frac{1}{2}\left(\frac{mL^2}{12}\right)\dot{\theta}^2$

Potential energy:

$\mathcal{V} = \frac{mgL\cos{\theta}}{2}$

Applying momentum conservation along X gives:

$\dot{x}_c = 0$ $\implies \dot{s} = - \frac{L}{2}\cos{\theta} \dot{\theta}$

Eliminating $\dot{s}$ in the kinetic energy expression gives:

$\mathcal{T} = \frac{mL^2\dot{\theta}^2}{24}\left(4 - 3\cos^2{\theta}\right)$

Applying conservation of energy gives and expression for $\dot{\theta}^2$ which is:

$\dot{\theta}^2 = \frac{12g}{L}\left(\frac{1-\cos{\theta}}{4-3\cos^2{\theta}}\right)$

Now, consider the upper tip of the rod. Its coordinates are:

$x_u = s + L\sin{\theta}$ $y_u = L\cos{\theta}$

$V^2 = \dot{x}_u^2 + \dot{y}_u^2$ $V^2 = 3gL(1-\cos{\theta})$ $V \approx 4.44$

An interesting observation. If the lowermost point was hinged and the rigid rod rotates about the hinge from its initial vertical orientation, one sees that the expression for the speed of the uppermost tip is exactly the same as in this non hinged case. $V^2 = 3gL(1-\cos{\theta})$ I wonder why this is so. Any insights would be appreciated.

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