# Angular velocity of system with articulated massless bars and pontual masses

I'm trying to solve a problem about rotation dynamics, but can't arrive at an answer. Any help would be much appreciated.

Consider the following system:

the two rigid bars with masses at their tips are perpendicular and connected with the fixed horizontally bar by an articulation (the three bars are in the same plane). Consider all the bars to be massless, $$g = 10 \ m/s^2$$ and $$L = 10\ cm$$. The problem asks then to find for which value of $$\omega$$ the bar with length $$2L$$ remains in the vertical position. The possible answers given are 1, 2, 3, 4 or 5 rad/s.

So, I understand that when the system rotates, the pontual masses will experience a centrifugal force (on their frame of reference), which is the responsible force for rotating the masses around the articulation axis. So, when the $$2L$$ bar is at the vertical position, the masses $$m$$ and $$2m$$ will have centrifugal forces of $$m\omega^2(2L)$$ and $$(2m)\omega^2 L$$, respectively. The masses will also have tension forces (with direction alongside the bars) and normal forces (perpendicular to the bars). So, when I use Newton's laws for the final system, considering that the sum of forces in the y-direction (vertical) for each mass has to be zero and in the horizontal it has to be equal the centripetal force (for an inertial frame of reference), I obtain some equations for the normal forces and tensions on each mass. Basically, considering the upper mass and the lower labeled as 1 and 2, respectively:$$T_1 = m\omega^2(2L)$$ $$N_1 = mg$$ $$N_2 = (2m)\omega^2L$$ $$T_2 = 2mg$$

After that, I don't know how to proceed. Tried to solve by torque too but couldn't do it.

$$\tau = -(L)\,N_1 + (2L) N_2 = 0$$
Solve the above for $$\omega$$.