A small sphere \(S\), of mass \(m \mathrm {~kg}\) is released from rest at the surface of a liquid in a right circular cylinder whose axis is vertical. When \(S\) is moving downwards with speed \(v \mathrm {~ms} ^ { - 1 }\), the viscous resistive force acting upwards on it has magnitude \(v ^ { 2 } \mathrm {~N}\).
Write down a differential equation for the motion of \(S\), clearly defining any symbol(s) that you introduce.
Find, in terms of \(m\), the distance \(S\) has fallen when its speed is \(\sqrt { \frac { m g } { 2 } } \mathrm {~ms} ^ { - 1 }\).
The diagram shows two identical particles, each of mass \(m \mathrm {~kg}\), connected by a thin, light inextensible string. \(P\) slides on the surface of a smooth right circular cylinder fixed with its axis, through \(O\), horizontal. \(Q\) moves vertically. \(O P\) makes an angle \(\theta\) radians with the horizontal.
The system is released from rest in the position where \(\theta = 0\).
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Show that the vertical distance moved by \(Q\) is \(\frac { \theta } { \sin \theta }\) times the vertical distance moved by \(P\).
In the position where \(\theta = \frac { \pi } { 6 }\), prove that the reaction of the cylinder on \(P\) has magnitude \(\left( 1 - \frac { \pi } { 6 } \right) m g \mathrm {~N}\).