TSny said:
That's interesting. I didn't even think to check if CD would go slack!
Right now, when I do the calculations I'm getting that CD doesn't go slack. But my attempt at the calculation is messy and tedious. Every time I recheck it, I find mistakes. Just a factor of 2 somewhere can make all the difference. So, I'm not confident. But at this point I'm getting tension in CD all the way through the motion.
OK, here are my calcs.
Let ##K=\dfrac{q^2}{4 \pi \epsilon_0 d^2}##.
Initial PE = 6Kd. Assume CD stays taut. When planar, PE = 5Kd+Kd/√3, so PE loss = Kd(1-1/√3).
If each particle now has speed v, KE = 2mv
2.
Here's a tricky bit: the instantaneous axis of rotation of the triangle ACD is the line joining the midpoints of AC, AD. So the radius of rotation of A is d(√3)/4. the centripetal force = ##\frac{mv^2}{d\frac{\sqrt 3}{4}} = 2K\frac{\sqrt 3 - 1}{3} ##
Let the tension in AC be T. The electrostatic repulsion A feels from B, C and D in the BA direction is K(√3+1/√3).
Hence T√3 = K(√3+1/√3)+mv
2/(d(√3)/4) = ##K\left(\sqrt 3+\frac {1}{\sqrt 3}+\frac{2\sqrt 3 - 2}{3}\right) = 2K(\sqrt 3 - \frac 13)##
The tensions in CA, CB etc. also act to pull C and D together. The force that way on C due to strings AC and BC = T = 2K(1-1/√3).
The electrostatic force acting oppositely on C is K(1+1/2+1/2) = 2K. There is no centripetal force required of CD.
√3 - 1/3 > 1, yes?
Edit: I think I've underestimated the centripetal acceleration of A by a factor of 2. The triangle ACD as a whole is accelerating towards the line where CD is at that instant. This will serve to make CD go slack even more surely.
