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...
4 bar linkage
which can collapse. $\endgroup$