A 'Classic' Challenge in Classical Mechanics

If the shape of one of the ropes is given by the vector $\mathbf{r} = \mathbf{r}(s)$ , where the origin is the centre of rotation, and where $s$ is the arc-length of the curve, then the equation for the shape of the rope is $\frac{d}{ds}\big[T\mathbf{t}\big] + \rho\omega^2\mathbf{r} \; = \; \mathbf{0}$ where $\rho$ is the linear density of the rope (assumed constant), $T = T(s)$ is the tension in the rope and $\mathbf{t} = \frac{d\mathbf{r}}{ds}$ is the unit tangent of the curve. Thus $\mathbf{0} \; = \; \mathbf{r} \wedge \frac{d}{ds}\big[T \mathbf{t}\big] \; = \; \frac{d}{ds}\big[T \mathbf{r} \wedge \mathbf{t} \big]$ and hence $\mathbf{h} =T\mathbf{r} \wedge \mathbf{t}$ is constant along the rope.

In the middle of the rope we have $T=1$ , $\mathbf{r} = \tfrac12\mathbf{i}$ and $\mathbf{t} = \mathbf{j}$ , and hence $\mathbf{h} = \tfrac12\mathbf{k}$ . At the end of the rope we have $T = T_1$ , $\mathbf{r}= \tfrac12L\mathbf{j}$ and $\mathbf{t} = -\sin\tfrac12\theta\mathbf{i} + \cos\tfrac12\theta\mathbf{j}$ , and hence $\mathbf{h} = \tfrac12LT_1\sin\tfrac12\theta \mathbf{k} = \tfrac14LT_1\mathbf{k}$ , and so we deduce that $T_1 = \tfrac{2}{L}$ .

Considering the circular motion of the particle at the end of the rope, we see that $\tfrac{2}{L}\sqrt{3} \;=\; 2 \times T_1 \cos\tfrac12\theta \; = \; m \times \tfrac12L\omega^2 \;=\; \tfrac12L$ and hence $L^4 = \boxed{48}$ .