Maximum Tension

If $\theta$ is the angle the trailing string makes with the horizontal, then the front string makes the angle $\theta + \tfrac13\pi$ with the horizontal, so the system has kinetic energy $T = m\ell^2\dot{\theta}^2$ and potential energy $V = -mg\ell\sin\theta - mg\ell\sin(\theta+\tfrac13\pi) = -\sqrt{3}mg\ell\cos(\theta - \tfrac13\pi)$ . Thus conservation of energy tells us that $T + V = -\tfrac12\sqrt{3}mg\ell$ , and hence that $\dot{\theta}^2 \; = \; \frac{\sqrt{3}g}{\ell}\Big[\cos(\theta - \tfrac13\pi) - \tfrac12\Big] \hspace{2cm} \ddot{\theta} \; = \; -\frac{\sqrt{3}g}{2\ell}\sin(\theta - \tfrac13\pi)$ Thus the system oscillates between $\theta = 0$ and $\theta = \tfrac23\pi$ . If the tension in the trailing string is $T$ and the thrust in the rod is $R$ , we obtain the equation of motion of the trailing particle: $m\ell\binom{\sin\theta}{-\cos\theta}\ddot{\theta} + m\ell\binom{\cos\theta}{\sin\theta}\dot{\theta}^2 \; = \; T\binom{\cos\theta}{\sin\theta} + R\binom{-\cos(\frac13\pi-\theta)}{\sin(\frac13\pi - \theta)} + mg\binom{0}{-1}$ Taking a suitable scalar product to eliminate $R$ , we deduce that $\begin{aligned} -\tfrac12m\ell^2 \ddot{\theta} + \tfrac12\sqrt{3}m\ell\dot{\theta}^2 & = \; \tfrac12\sqrt{3}T - mg\cos(\tfrac13\pi - \theta) \\ \tfrac12\sqrt{3}T & = \; \tfrac14mg\Big[ 10\cos(\theta - \tfrac13\pi) + \sqrt{3}\sin(\theta - \tfrac13\pi) - 3\Big] \\ T & = \; \tfrac{1}{2\sqrt{3}}mg\Big[\sqrt{103}\cos(\theta -\tfrac13\pi - \theta_0) - 3\Big] \end{aligned}$ where $\theta_0 = \tan^{-1}\tfrac{\sqrt{3}}{10}$ , which is smaller than $\tfrac13\pi$ . Thus the maximum value of $T$ is $\alpha mg$ , where $\alpha = \tfrac{1}{2\sqrt{3}}\big(\sqrt{103} - 3\big) \; = \; \boxed{2.063707235}$ and this maximum value is achieved when $\theta = \tfrac13\pi + \theta_0$ . Note that the tension in the leading string is just the same as the tension in the trailing string, but half a cycle of oscillation out of phase.

[Sketch of a solution]

First thing one finds is that the each particle is constrained to move along the circle of radius $l$ centered at point $O$ . Hence, the whole system has one degree of freedom represented by one independent coordinate - for example, angle between vertical and center of mass of two particles. Accordingly, reaction forces must invalidate potential motion in all other directions for each particle. These conditions generate two equations of motion per particle: one for tangential and one for normal component of acceleration. The latter must include centrifugal force due to circular trajectory and thus requires knowledge of speed which can be obtained using conservation of energy principle. In the end, one gets cord tension as a function of angle and by means of differentiation finds its maximum.

This problem is from the Chinese Physics Olympiad (Semi Finals).