5 A ship \(S\) is travelling with constant speed \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on a course with bearing \(345 ^ { \circ }\). A patrol boat \(B\) spots the ship \(S\) when \(S\) is 2400 m from \(B\) on a bearing of \(050 ^ { \circ }\). The boat \(B\) sets off in pursuit, travelling with constant speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a straight line.

1. Given that \(v = 16\), find the bearing of the course which \(B\) should take in order to intercept \(S\), and the time taken to make the interception.
2. Given instead that \(v = 10\), find the bearing of the course which \(B\) should take in order to get as close as possible to \(S\).
   \includegraphics[max width=\textwidth, alt={}, center]{181fad74-6e60-4435-a176-3edff5062c32-4_337_954_278_544} A uniform rod \(A B\) has mass \(m\) and length \(2 a\). The point \(P\) on the rod is such that \(A P = \frac { 2 } { 3 } a\). The rod is placed in a horizontal position perpendicular to the edge of a rough horizontal table, with \(A P\) in contact with the table and \(P B\) overhanging the edge. The rod is released from rest in this position. When it has rotated through an angle \(\theta\), and no slipping has occurred at \(P\), the normal reaction acting on the rod at \(P\) is \(R\) and the frictional force is \(F\) (see diagram).
3. Show that the angular acceleration of the rod is \(\frac { 3 g \cos \theta } { 4 a }\).
4. Find the angular speed of the rod, in terms of \(a , g\) and \(\theta\).
5. Find \(F\) and \(R\) in terms of \(m , g\) and \(\theta\).
6. Given that the coefficient of friction between the rod and the edge of the table is \(\mu\), show that the rod is on the point of slipping at \(P\) when \(\tan \theta = \frac { 1 } { 2 } \mu\).
   \includegraphics[max width=\textwidth, alt={}, center]{181fad74-6e60-4435-a176-3edff5062c32-5_677_624_269_753} A smooth circular wire, with centre \(O\) and radius \(a\), is fixed in a vertical plane. The highest point on the wire is \(A\) and the lowest point on the wire is \(B\). A small ring \(R\) of mass \(m\) moves freely along the wire. A light elastic string, with natural length \(a\) and modulus of elasticity \(\frac { 1 } { 2 } m g\), has one end attached to \(A\) and the other end attached to \(R\). The string \(A R\) makes an angle \(\theta\) (measured anticlockwise) with the downward vertical, so that \(O R\) makes an angle \(2 \theta\) with the downward vertical (see diagram). You may assume that the string does not become slack.
7. Taking \(A\) as the level for zero gravitational potential energy, show that the total potential energy \(V\) of the system is given by $$V = m g a \left( \frac { 1 } { 4 } - \cos \theta - \cos ^ { 2 } \theta \right) .$$
8. Show that \(\theta = 0\) is the only position of equilibrium.
9. By differentiating the energy equation with respect to time \(t\), show that $$\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } = - \frac { g } { 4 a } \sin \theta ( 1 + 2 \cos \theta ) .$$
10. Deduce the approximate period of small oscillations about the equilibrium position \(\theta = 0\).