6.
\begin{figure}[h]
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\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{39c2d25a-a39b-4eb9-a17b-6e741ab5ae98-09_357_606_315_717}
\end{figure}
A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). At time \(t = 0 , P\) is projected vertically downwards with speed \(\sqrt { } \left( \frac { 5 } { 2 } g a \right)\) from a point \(A\) which is at the same level as \(O\) and a distance \(a\) from \(O\). When the string has turned through an angle \(\theta\) and the string is still taut, the speed of \(P\) is \(v\) and the tension in the string is \(T\), as shown in Figure 2.
- Show that \(v ^ { 2 } = \frac { g a } { 2 } ( 5 + 4 \sin \theta )\).
- Find \(T\) in terms of \(m , g\) and \(\theta\).
The string becomes slack when \(\theta = \alpha\).
- Find the value of \(\alpha\).
The particle is projected again from \(A\) with the same velocity as before. When \(P\) is at the same level as \(O\) for the first time after leaving \(A\), the string meets a small smooth peg \(B\) which has been fixed at a distance \(\frac { 1 } { 2 } a\) from \(O\). The particle now moves on an arc of a circle centre \(B\). Given that the particle reaches the point \(C\), which is \(\frac { 1 } { 2 } a\) vertically above the point \(B\), without the string going slack,
- find the tension in the string when \(P\) is at the point \(C\).