Rotation Angle vs. Reaction Force

The above shows the external forces acting on the system (system consists of the two rods).

Since the direction of the net reaction force is not known, I've assumed it in terms of its x and y components, denoted by $F_x$ and $F_y$ respectively.

(Note that the reaction force at the hinge point is an internal force.)

The above shows the net acceleration of the system. (Since $\omega$ is constant, $\alpha$ is zero.)

Writing Newton's Law of Motion for the system:

$\vec{F_{ext}} = \sum m_i\vec{a_i}$

$\implies 2mg(-\hat{j}) + F_y(\hat{j}) + F_x(-\hat{i}) = \frac{ml\omega^2}{2} (-\cos(\theta)\hat{i} - \sin(\theta)\hat{j})$

Equating the components:

$F_x = \frac{ml\omega^2}{2} \cos(\theta)$

$F_y = 2mg -\frac{ml\omega^2}{2}\sin(\theta)$

Magnitude of net reaction force: $\sqrt{F_x^2 + F_y^2}$

$\therefore \sqrt{F_x^2 + F_y^2} = 2mg$

$\implies (\frac{ml\omega^2}{2} \cos(\theta))^2 + (2mg -\frac{ml\omega^2}{2}\sin(\theta))^2 = 4m^2g^2$

$\implies \frac{m^2l^2\omega^4}{4} \cos^2(\theta) + 4m^2g^2 - 2m^2gl\omega^2\sin(\theta) +\frac{m^2l^2\omega^4}{4}\sin^2(\theta) = 4m^2g^2$

$\implies \frac{m^2l^2\omega^4}{4} = 2m^2gl\omega^2\sin(\theta)$

$\implies \sin(\theta) = \frac{l\omega^2}{8g}$

$\therefore \theta = \boxed{\arcsin\bigg(\dfrac{\pi^2}{80}\bigg)}$