Particle on circular wire/arc

4 \includegraphics[max width=\textwidth, alt={}, center]{ab5f2781-e5ce-4fce-bc95-9d7f55ea66d9-2_515_583_1388_781} A smooth wire is in the form of an \(\operatorname { arc } A B\) of a circle, of radius \(a\), that subtends an obtuse angle \(\pi - \theta\) at the centre \(O\) of the circle. It is given that \(\sin \theta = \frac { 1 } { 4 }\). The wire is fixed in a vertical plane, with \(A O\) horizontal and \(B\) below the level of \(O\) (see diagram). A small bead of mass \(m\) is threaded on the wire and projected vertically downwards from \(A\) with speed \(\sqrt { } \left( \frac { 3 } { 10 } g a \right)\).

1. Find the reaction between the bead and the wire when the bead is vertically below \(O\).
2. Find the speed of the bead as it leaves the wire at \(B\).
3. Show that the greatest height reached by the bead is \(\frac { 1 } { 8 } a\) above the level of \(O\).

A small bead \(P\), of mass \(m\) kg, can slide on a smooth circular ring, with centre \(O\) and radius \(r\) m, which is fixed in a vertical plane. \(P\) is projected from the lowest point \(L\) of the ring with speed \(\sqrt{(3gr)}\) ms\(^{-1}\). When \(P\) has reached a position such that \(OP\) makes an angle \(\theta\) with the downward vertical, as shown, its speed is \(v\) ms\(^{-1}\). \includegraphics{figure_5}

1. Show that \(v^2 = gr(1 + 2 \cos \theta)\). [5 marks]
2. Show that the magnitude of the reaction \(RN\) of the ring on the bead is given by $$R = mg(1 + 3 \cos \theta).$$ [4 marks]
3. Find the values of \(\cos \theta\) when
   1. \(P\) is instantaneously at rest,
   2. the reaction \(R\) is instantaneously zero. [2 marks]
4. Hence show that the ratio of the heights of \(P\) above \(L\) in cases (i) and (ii) is \(9:8\). [3 marks]

\includegraphics{figure_2} A smooth wire is shaped into a circle of centre \(O\) and radius 0.8 m. The wire is fixed in a vertical plane. A small bead \(P\) of mass 0.03 kg is threaded on the wire and is projected along the wire from the highest point with a speed of \(4.2 \, \text{m s}^{-1}\). When \(OP\) makes an angle \(\theta\) with the upward vertical the speed of \(P\) is \(v \, \text{m s}^{-1}\) (see diagram).

1. Show that \(v^2 = 33.32 - 15.68\cos\theta\). [4]
2. Prove that the bead is never at rest. [1]
3. Find the maximum value of \(v\). [2]