Stability of square of masses on strings under rotation Imagine we have a square of masses, $m$, connected by light inextensible strings, length $l$, rotating around it's centre at angular speed, $\omega$. It's easy enough to show that there must be a tension in these strings, $T=\frac{ml\omega^2}{2}$.
However if we then increase the mass of any two opposite corners resolving forces seems to show that there is no solution for continued circular motion except when the two heavier masses are a distance $l$ from the c.o.m and the lighter ones, a distance 0. I.e. the problem reduces to a line.
Is this true, or has my logic just broken down at some point? Also, the same seems to happen if you keep all the masses equal and change any angle just a little.
Does this mean that a rotating square (or any other regular shape) of equal masses such as this is the only equilibrium position for the family of systems of shapes and masses such as this?
Here's a diagram if anyone's having trouble picturing it, just a random one from the net that had most of the right features (ignore the labels), it is colourful though...

 A: 
Here, in the above picture $M \ge m$. Note that this is the most general case. We can have $M = m$ and the angle $\theta$ can vary anywhere between $[0;\cfrac{\pi}{2}]$
(Actually, the most general case would have been to take 4 different masses but we will be going out of the bandwidth of your problem, and it would be a pointless discussion.)  
Now, suppose that the above system is in equilibrium, rotating about the COM with an angular speed $\omega$.
Equating the Tension forces to mass times centripetal acceleration:
$$2T \cos\theta = M \omega^2l \cos\theta$$ and
$$2T \sin\theta = m \omega^2l \sin\theta$$
Let me call these as eqns $(1)$ and $(2)$ respectively.
The solutions to the above equations will give the possible equilibrium states.  
Now, there are only 3 possible equilibrium states:
Case $I$ :
$$\cos\theta \ne 0,\,\sin\theta \ne 0$$
Hence, we can cancel these terms from their respective eqns.
Thus, equating Tensions, we get: $M = m$
Pay attention to this case. It says that as long as the masses are equal, the system will be in equilibrium for any value of $\theta$. (This goes against our intuitive feeling. But Einstein taught us long ago that not to value our intuition much, didn't he?)  
Case $II$ :
$$\cos\theta = 0, \ \sin\theta \ne 0$$
$$\theta = \cfrac{\pi}{2}$$
Eqn $(1)$ is satisfied. Eqn $(2)$ gives us the value of $T$.
$$T = \cfrac{m \omega^2 l}{2}$$
This is the case where the heavier masses are at the center and the lighter masses revolve around the COM of the system.  
Case $III$ :
$$\cos\theta \ne 0, \ \sin\theta = 0$$
$$\theta = 0$$
Eqn $(2)$ is satisfied. Eqn $(1)$ gives us the value of $T$.
$$T = \cfrac{M \omega^2 l}{2}$$
This is the case where the lighter masses are at the center and the heavier masses revolve around the COM of the system.  
(Just a question: How did manage to make such a beautiful diagram?)
