OCR MEI M4 (Mechanics 4) 2012 June

1 A rocket in deep space has initial mass \(m _ { 0 }\) and is moving in a straight line at speed \(v _ { 0 }\). It fires its engine in the direction opposite to the motion in order to increase its speed. The propulsion system ejects matter at a constant mass rate \(k\) with constant speed \(u\) relative to the rocket. At time \(t\) after the engines are fired, the speed of the rocket is \(v\).

1. Show that while mass is being ejected from the rocket, \(\left( m _ { 0 } - k t \right) \frac { \mathrm { d } v } { \mathrm {~d} t } = u k\).
2. Hence find an expression for \(v\) at time \(t\).

2 A light elastic string AB has stiffness \(k\). The end A is attached to a fixed point and a particle of mass \(m\) is attached at the end B . With the string vertical, the particle is released from rest from a point at a distance \(a\) below its equilibrium position. At time \(t\), the displacement of the particle below the equilibrium position is \(x\) and the velocity of the particle is \(v\).

1. Show that $$m v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - k x$$
2. Show that, while the particle is moving upwards and the string is taut, $$v = - \sqrt { \frac { k } { m } \left( a ^ { 2 } - x ^ { 2 } \right) }$$
3. Hence use integration to find an expression for \(x\) at time \(t\) while the particle is moving upwards and the string is taut.

3 A uniform rigid rod AB of length \(2 a\) and mass \(m\) is smoothly hinged to a fixed point at A so that it can rotate freely in a vertical plane. A light elastic string of modulus \(\lambda\) and natural length \(a\) connects the midpoint of AB to a fixed point C which is vertically above A with \(\mathrm { AC } = a\). The rod makes an angle \(2 \theta\) with the upward vertical, where \(\frac { 1 } { 3 } \pi \leqslant 2 \theta \leqslant \pi\). This is shown in Fig. 3. \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system relative to A in terms of \(m , \lambda , a\) and \(\theta\). Show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 2 a \cos \theta ( 2 \lambda \sin \theta - 2 m g \sin \theta - \lambda ) .$$ Assume now that the system is set up so that the result (*) continues to hold when \(\pi < 2 \theta \leqslant \frac { 5 } { 3 } \pi\).
2. In the case \(\lambda < 2 m g\), show that there is a stable position of equilibrium at \(\theta = \frac { 1 } { 2 } \pi\). Show that there are no other positions of equilibrium in this case.
3. In the case \(\lambda > 2 m g\), find the positions of equilibrium for \(\frac { 1 } { 3 } \pi \leqslant 2 \theta \leqslant \frac { 5 } { 3 } \pi\) and determine for each whether the equilibrium is stable or unstable, justifying your conclusions.

4

1. Show by integration that the moment of inertia of a uniform circular lamina of radius \(a\) and mass \(m\) about an axis perpendicular to the plane of the lamina and through its centre is \(\frac { 1 } { 2 } m a ^ { 2 }\). A closed hollow cylinder has its curved surface and both ends made from the same uniform material. It has mass \(M\), radius \(a\) and height \(h\).
2. Show that the moment of inertia of the cylinder about its axis of symmetry is \(\frac { 1 } { 2 } M a ^ { 2 } \left( \frac { a + 2 h } { a + h } \right)\). For the rest of this question take the cylinder to have mass 8 kg , radius 0.5 m and height 0.3 m .
   The cylinder is at rest and can rotate freely about its axis of symmetry. It is given a tangential impulse of magnitude 55 Ns at a point on its curved surface. The impulse is perpendicular to the axis.
3. Find the angular speed of the cylinder after the impulse. A resistive couple is now applied to the cylinder for 5 seconds. The magnitude of the couple is \(2 \dot { \theta } ^ { 2 } \mathrm { Nm }\), where \(\dot { \theta }\) is the angular speed of the cylinder in rad s \({ } ^ { - 1 }\).
4. Formulate a differential equation for \(\dot { \theta }\) and hence find the angular speed of the cylinder at the end of the 5 seconds. The cylinder is now brought to rest by a constant couple of magnitude 0.03 Nm .
5. Calculate the time it takes from when this couple is applied for the cylinder to come to rest.