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]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-2_328_732_1032_667}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}

(i) Show that the length of BC in metres is $\sin \frac { 1 } { 2 } \theta$.\\
(ii) Find the potential energy, $V \mathrm {~J}$, of the system relative to AD in terms of $\theta$. Hence show that

$$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 1.5 \sin \theta - 1.225 \cos \theta - \frac { 0.5 \sin \theta } { \sqrt { 1.25 - \cos \theta } } - 0.5 \cos \frac { 1 } { 2 } \theta .$$

(iii) 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]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-2_453_1264_2021_397}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\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.

\hfill \mbox{\textit{OCR MEI M4 2013 Q2 [13]}}