Mechanical Rod with archery and bow.

Key principles:

1) The angle spanned by the rod is 60 degrees (left as an exercise for the reader to derive)
2) The distance from the center of the rod to the center of the circle is $\frac{\sqrt{3}}{2} R$ (left as an exercise for the reader to derive)

Rod Moment of Inertia (from Parallel Axis Theorem):

$\large{I = \frac{1}{12} M R^2 + \frac{3}{4} M R^2 = \frac{5}{6} M R^2}$

Define $\theta$ as the angle between the line through the center of the rod and the horizontal. The angle starts at 30 degrees. We need to evaluate the motion when the angle gets to 45 degrees. The change in gravitational potential energy relative to the start point is equal to the kinetic energy at that time.

$\large{\Delta U = M g \frac{\sqrt{3}}{2} R (sin{45^\circ} - sin{30^\circ}) = M g \frac{\sqrt{3}}{2} R \Big(\frac{1}{\sqrt{2}} - \frac{1}{2} \Big) }$

Equating Change in Potential Energy to Kinetic Energy:

$\large{\Delta U = \frac{1}{2} I \omega^2 \\ M g \frac{\sqrt{3}}{2} R \Big(\frac{1}{\sqrt{2}} - \frac{1}{2} \Big) = \frac{1}{2} \, \frac{5}{6} M R^2 \, \omega^2}$

Plugging in numbers gives $\omega = \sqrt{6} \approx 2.449$