Identical Charges are corrected

In the subsequent motion, the triangles $ACD$ and $BCD$ remain taut, and the triangles $ACD$ and $BCD$ move symmetrically, while the centre of mass of the system remains constant. Looking at the system from the side, we deduce that $A,B,C,D$ have position vectors (we write $\ell = 2r$ here, temporarily) relative to the centre of mass of the system (note that the apparent length of $BC$ is the length $r\sqrt{3}$ of the altitude of the triangle $BCD$ :

$\mathbf{a} \; = \; \left(\begin{array}{c} -\sqrt{3}r\sin\theta \\ 0 \\ \frac12\sqrt{3}r\cos\theta \end{array}\right) \hspace{0.7cm} \mathbf{b} \; = \; \left(\begin{array}{c} \sqrt{3}r\sin\theta \\ 0 \\ \frac12\sqrt{3}r\cos\theta \end{array}\right) \hspace{0.7cm} \mathbf{c} \; = \; \left(\begin{array}{c} 0 \\r \\ -\frac12\sqrt{3}r\cos\theta \end{array}\right) \hspace{0.7cm} \mathbf{d} \; = \; \left(\begin{array}{c} 0 \\ - r \\ -\frac12\sqrt{3}r\cos\theta \end{array}\right)$ Thus $\dot{\mathbf{a}} \; = \; \left(\begin{array}{c} -\sqrt{3}r\cos\theta \\ 0 \\ -\frac12\sqrt{3}r\sin\theta \end{array}\right)\dot{\theta} \hspace{0.6cm} \dot{\mathbf{b}} \; = \; \left(\begin{array}{c} \sqrt{3}r\cos\theta \\ 0 \\ -\frac12\sqrt{3}r\sin\theta \end{array}\right)\dot{\theta} \hspace{0.6cm} \dot{\mathbf{c}} \; = \; \left(\begin{array}{c} 0 \\0 \\ \frac12\sqrt{3}r\sin\theta \end{array}\right)\dot{\theta} \hspace{0.6cm} \dot{\mathbf{d}} \; = \; \left(\begin{array}{c} 0 \\ 0 \\ \frac12\sqrt{3}r\sin\theta \end{array}\right)\dot{\theta}$ and so $|\dot{\mathbf{a}}|^2 \; = \; |\dot{\mathbf{b}}|^2 \; = \; \tfrac34r^2(\sin^2\theta + 4\cos^2\theta)\dot{\theta}^2 \hspace{2cm} |\dot{\mathbf{c}}|^2\; = \; |\dot{\mathbf{d}}|^2 \; = \; \tfrac34r^2\sin^2\theta \dot{\theta}^2$ so the kinetic energy of the system is $T \; = \; \tfrac12m\left(|\dot{\mathbf{a}}|^2 + |\dot{\mathbf{b}}|^2 + |\dot{\mathbf{c}}|^2 + |\dot{\mathbf{d}}|^2\right) \; = \; m\left(|\dot{\mathbf{a}}|^2 + |\dot{\mathbf{c}}|^2\right) \; = \; \tfrac34mr^2(2\sin^2\theta + 4\cos^2\theta)\dot{\theta}^2 \; = \; \tfrac32mr^2(1 + \cos^2\theta)\dot{\theta}^2$ Since most of the particles are held in relative place by the remaining strings, the only electrostatic bond that does work is the bond between $A$ and $B$ , we we have a potential energy $V \; = \; \frac{q^2}{4\pi \varepsilon_0} \frac{1}{AB} \; = \; \frac{q^2}{4\pi \varepsilon_0} \frac{1}{2r\sqrt{3}\sin\theta}$ so conservation of energy tells us that $\tfrac32mr^2(1 + \cos^2\theta)\dot{\theta}^2 + \frac{q^2}{4\pi \varepsilon_0} \frac{1}{2r\sqrt{3}\sin\theta} \; = \; \frac{q^2}{4\pi \varepsilon_0} \frac{1}{2r\sqrt{3}\sin\theta_0}$ where $\theta_0 = \sin^{-1}\tfrac{1}{\sqrt{3}}$ is the initial angle of the system, so $r^2\dot{\theta}^2 \; = \; \frac{q^2}{12\pi \varepsilon_0 mr\sqrt{3}} \frac{\sqrt{3}-\csc\theta}{1 + \cos^2\theta}$ Thus the system oscillates in the range $\theta_0 \le \theta \le \pi - \theta_0$ , and we have $\begin{aligned} |\dot{\mathbf{a}}|^2 \; = \; |\dot{\mathbf{b}}|^2 & = \; \frac{q^2}{16 \pi \varepsilon_0 mr\sqrt{3}} \frac{(\sqrt{3} - \csc\theta)(1 + 3\cos^2\theta)}{1 + \cos^2\theta} \; = \; \frac{q^2}{\pi \varepsilon_0 m \ell} \frac{(\sqrt{3} - \csc\theta)(1 + 3\cos^2\theta)}{8\sqrt{3}(1 + \cos^2\theta)} \\ |\dot{\mathbf{c}}|^2 \; = \; |\dot{\mathbf{d}}|^2 & = \; \frac{q^2}{16 \pi \varepsilon_0 mr\sqrt{3}} \frac{(\sqrt{3} - \csc\theta)\sin^2\theta}{1 + \cos^2\theta} \; = \; \frac{q^2}{\pi \varepsilon_0 m \ell} \frac{(\sqrt{3} - \csc\theta)\sin^2\theta}{8\sqrt{3}(1 + \cos^2\theta)} \end{aligned}$ If $\xi$ is the maximum of $\frac{(\sqrt{3} - \csc\theta)(1 + 3\cos^2\theta)}{8\sqrt{3}(1 + \cos^2\theta)}$ over the range $\theta_0 \le \theta \le \pi - \theta_0$ , then we see that the maximum speeds of $A$ and $B$ are $V_A \; = \; V_B \; = \; \sqrt{\frac{q^2}{\pi \varepsilon_0 m\ell}} \sqrt{\xi}$ while the maximum speeds of $C$ and $D$ are $V_C \; = \; V_D \; = \; \sqrt{\frac{q^2}{\pi \varepsilon_0 m\ell}} \sqrt{\frac{\sqrt{3}-1}{8\sqrt{3}}}$ , achieved when $\theta = 90^\circ$ , and hence $\alpha \; = \; 2\sqrt{\xi} + 2\sqrt{\frac{\sqrt{3}-1}{8\sqrt{3}}}$ Numerical methods tell us that $\xi = 0.0591279$ (not achieved when $\theta = 90^\circ$ , but when $\theta = 64.5615^\circ$ and $180^\circ - 64.5615^\circ$ - the plot of $\frac{(\sqrt{3} - \csc\theta)(1 + 3\cos^2\theta)}{8\sqrt{3}(1 + \cos^2\theta)}$ on the right measures the angle in radians, though), and hence $\alpha = \boxed{0.946025}$ . Note that the maximum speeds of $A$ and $B$ are not attained when $\theta = 90^\circ$ , although the book from which this problem came claimed that it did.