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\includegraphics[max width=\textwidth, alt={}, center]{5bb3bd29-a2eb-4124-802c-fb17b68c50e4-3_598_839_1480_706} One end of a light inextensible string of length 0.5 m is attached to a fixed point \(O\). A particle \(P\) of mass 0.3 kg is attached to the other end of the string. With the string taut and at an angle of \(60 ^ { \circ }\) to the upward vertical, \(P\) is projected with speed \(2 \mathrm {~ms} ^ { - 1 }\) (see diagram). \(P\) begins to move without air resistance in a vertical circle with centre \(O\). When the string makes an angle \(\theta\) with the upward vertical, the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that \(v ^ { 2 } = 8.9 - 9.8 \cos \theta\).
2. Find the tension in the string in terms of \(\theta\).
3. \(P\) does not move in a complete circle. Calculate the angle through which \(O P\) turns before \(P\) leaves the circular path.