A small bead \(P\), of mass \(m \mathrm {~kg}\), can slide on a smooth circular ring, with centre \(O\) and radius \(r \mathrm {~m}\), which is fixed in a vertical plane. \(P\) is projected from the lowest point \(L\) of the ring with speed \(\sqrt { } ( 3 g r ) \mathrm { ms } ^ { - 1 }\). When \(P\) has reached a position such that \(O P\) makes an angle \(\theta\) with the downward vertical, as shown, its speed is \(v \mathrm {~ms} ^ { - 1 }\).
\includegraphics[max width=\textwidth, alt={}, center]{e756c147-2711-4ee4-9ec6-7680fe84a0c5-2_355_337_262_1590}
Show that \(v ^ { 2 } = g r ( 1 + 2 \cos \theta )\).
Show that the magnitude of the reaction \(R N\) of the ring on the bead is given by
$$R = m g ( 1 + 3 \cos \theta ) .$$
Find the values of \(\cos \theta\) when
\(P\) is instantaneously at rest, (ii) the reaction \(R\) is instantaneously zero.
Hence show that the ratio of the heights of \(P\) above \(L\) in cases (i) and (ii) is \(9 : 8\).