Hinge Velocities

It is more elegant to note that this problem can be solved using energy conservation.

Let $M$ be the mass of rod.

Now, when it is vertical, then, we know the centre of mass of the rod will lower down by $L/2$ .

Hence By Energy Conservation Law,

$\frac{MgL}{2} = \frac{1}{2} I \omega^2$

[Where $I$ = Moment of inertia of The Rod About the end that is $1/3 ML^2$ ]

hence $\omega = \sqrt {30} s^{-1} \approx 5.47722558 s^{-1}$

Relevant wiki: Rotational Kinetic Energy - Translational Kinetic Energy

Consider an infinitesimally small part of length $dx$ at a distance $x$ from the hinge. Thus it's Mass would be $(M/L)dx$ where $L$ is the length of the rod.

Let us say that the velocity of the rod varies linearly from $0$ to $V$ as the length of the rod increases from the hinge to the tip. Thus at some distance $x$ from the hinge the velocity of the rod would be $Vx/L$ . Hence;

$\begin{aligned} \textrm{KE}&=\dfrac 12\int (dm)v^2\\ &=\dfrac 12\int_0^L\dfrac{Mdx}L\times\dfrac{V^2x^2}{L^2}\\ &=\dfrac{MV^2}{2L^3}\times\left(\dfrac{x^3}3\right)_0^L\\ &=\dfrac{MV^2}6\\ \end{aligned}$

Now the change in Kinetic energy is equal to the potential energy, thus we have:

$\begin{aligned} \dfrac{MV^2}6&=\dfrac{MgL}2\\ V^2&=3gL\\ \omega^2L^2&=3gL\\ \omega&=\sqrt{\dfrac{3g}{L}}\\ &=\sqrt{30} \end{aligned}$