2 A uniform rod AB of length 0.5 m and mass 0.5 kg is freely hinged at A so that it can rotate in a vertical plane. Attached at B are two identical light elastic strings BC and BD each of natural length 0.5 m and stiffness \(2 \mathrm {~N} \mathrm {~m} ^ { - 1 }\). The ends C and D are fixed at the same horizontal level as A and with \(\mathrm { AC } = \mathrm { CD } = 0.5 \mathrm {~m}\). The system is shown in Fig. 2.1 with the angle \(\mathrm { BAC } = \theta\). You may assume that \(\frac { 1 } { 3 } \pi \leqslant \theta \leqslant \frac { 5 } { 3 } \pi\) so that both strings are taut. \begin{figure}[h] \end{figure}

1. Show that the length of BC in metres is \(\sin \frac { 1 } { 2 } \theta\).
2. Find the potential energy, \(V \mathrm {~J}\), of the system relative to AD in terms of \(\theta\). Hence show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 1.5 \sin \theta - 1.225 \cos \theta - \frac { 0.5 \sin \theta } { \sqrt { 1.25 - \cos \theta } } - 0.5 \cos \frac { 1 } { 2 } \theta .$$
3. Fig. 2.2 shows a graph of the function \(\mathrm { f } ( \theta ) = 1.5 \sin \theta - 1.225 \cos \theta - \frac { 0.5 \sin \theta } { \sqrt { 1.25 - \cos \theta } } - 0.5 \cos \frac { 1 } { 2 } \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-2_453_1264_2021_397} \captionsetup{labelformat=empty} \caption{Fig. 2.2}
   \end{figure} Use the graph both to estimate, correct to 1 decimal place, the values of \(\theta\) for which the system is in equilibrium and also to determine their stability.