# 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... • Sounds about right. For even more fun, google up the devotees who proved that Ringworld is fundamentally in an unstable "orbit" . Dec 18, 2013 at 12:43
• You need a diagonal spring to keep them in place. What you have now is a 4 bar linkage which can collapse. Dec 18, 2013 at 15:59
• Well, it is in equilibrium when the masses are exactly equal and the angles are perfect right-angles. I was just wondering if there were any other equilibrium configurations for different masses or angles. Dec 18, 2013 at 16:28 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?)

• Excellent answer, thankyou. Just one quick follow up, what's the stability of each of these cases (excluding case I as clearly any mass variation will break the equilibrium) I imagine III is stable and II is not, but perhaps both are? Dec 20, 2013 at 10:27
• First, you need to understand what 'equilibrium of the system' here means: You said "any MASS VARIATION will break the equilibrium". Its like, there is a system of 4 masses rotating with $\omega$ in equilibrium and you are changing one or more of the masses then. I think that 'EQUILIBRIUM of the SYSTEM' here would be an uncomfortable thing to talk about. I would rather talk about the equilibrium in terms of angle change i.e. if the angle $\theta$ is changed slightly,(for the given system), what would happen? In Case I, we get what is called as the Neutral Equilibrium. See next comment. Dec 22, 2013 at 6:01
• Neutral equilibrium is, when we change the angle $\theta$ slightly, the system remains there itself i.e. the new angle is $\theta + d\theta$. The system does not go back to the original angle $\theta$, as is the case in Stable Equilibrium. As for case II and III, truthfully, I don't know the answer. We'll have to calculate the force and take its double derivative to test the type of equilibrium. My problem is that I think there comes a thing called Coriollis Force in the picture. And I haven't been taught Coriollis force yet. My advice: Don't go by intuition here. Trust Physics and Maths. Dec 22, 2013 at 6:07