A rigid bar of negligible mass has three particles whose masses are the same and are joined to the bar as indicated in the figure. The bar is free to rotate in a vertical plane about an axis with no friction perpendicular to the bar through point $P$ and is released from rest on the horizontal position at $t=0$. Find the maximum angular speed in radians per second attained by the bar. You may consider $g=9.8\frac{m}{s^2}$ and $d=4.2\pi$

The alternatives given are:

$\begin{array}{ll}

1.&1\,\frac{rad}{s}\\

2.&2\,\frac{rad}{s}\\

3.&3\,\frac{rad}{s}\\

4.&4\,\frac{rad}{s}\\

\end{array}$

What I've attempted to solve this problem was to equate the potential energy and the rotational energy of the masses assuming there's a superposition of the moment of inertia of the masses.

This is translated as:

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

$mgh=\left(\frac{1}{2}m\left(\frac{4d}{3}\right)^{2}+\frac{1}{2}m\left(\frac{d}{3}\right)^{2}+\frac{1}{2}m\left(\frac{2d}{3}\right)^{2}\right) \times \omega^2$

But after following the procedure I can't find a way to cancel the 4.2$\pi$. Supposedly, the answer is the first choice. How can I get to that result?