OCR MEI M4 (Mechanics 4) 2013 June

1 An empty railway truck of mass \(m _ { 0 }\) is moving along a straight horizontal track at speed \(v _ { 0 }\). The point P is at the front of the truck. The horizontal forces on the truck are negligible. As P passes a fixed point O , sand starts to fall vertically into the truck at a constant mass rate \(k\). At time \(t\) after P passes O the speed of the truck is \(v\) and \(\mathrm { OP } = x\).

1. Find an expression for \(v\) in terms of \(m _ { 0 } , v _ { 0 } , k\) and \(t\), and show that \(x = \frac { m _ { 0 } v _ { 0 } } { k } \ln \left( 1 + \frac { k t } { m _ { 0 } } \right)\).
2. Find the speed of the truck and the distance OP when the mass of sand in the truck is \(2 m _ { 0 }\).

2 A uniform rod AB of length 0.5 m and mass 0.5 kg is freely hinged at A so that it can rotate in a vertical plane. Attached at B are two identical light elastic strings BC and BD each of natural length 0.5 m and stiffness \(2 \mathrm {~N} \mathrm {~m} ^ { - 1 }\). The ends C and D are fixed at the same horizontal level as A and with \(\mathrm { AC } = \mathrm { CD } = 0.5 \mathrm {~m}\). The system is shown in Fig. 2.1 with the angle \(\mathrm { BAC } = \theta\). You may assume that \(\frac { 1 } { 3 } \pi \leqslant \theta \leqslant \frac { 5 } { 3 } \pi\) so that both strings are taut. \begin{figure}[h] \end{figure}

1. Show that the length of BC in metres is \(\sin \frac { 1 } { 2 } \theta\).
2. Find the potential energy, \(V \mathrm {~J}\), of the system relative to AD in terms of \(\theta\). Hence show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 1.5 \sin \theta - 1.225 \cos \theta - \frac { 0.5 \sin \theta } { \sqrt { 1.25 - \cos \theta } } - 0.5 \cos \frac { 1 } { 2 } \theta .$$
3. Fig. 2.2 shows a graph of the function \(\mathrm { f } ( \theta ) = 1.5 \sin \theta - 1.225 \cos \theta - \frac { 0.5 \sin \theta } { \sqrt { 1.25 - \cos \theta } } - 0.5 \cos \frac { 1 } { 2 } \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-2_453_1264_2021_397} \captionsetup{labelformat=empty} \caption{Fig. 2.2}
   \end{figure} Use the graph both to estimate, correct to 1 decimal place, the values of \(\theta\) for which the system is in equilibrium and also to determine their stability.

3 A model car of mass 2 kg moves from rest along a horizontal straight path. After time \(t \mathrm {~s}\), the velocity of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The power, \(P \mathrm {~W}\), developed by the engine is initially modelled by \(P = 2 v ^ { 3 } + 4 v\). The car is subject to a resistance force of magnitude \(6 v \mathrm {~N}\).

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = ( 1 - v ) ( 2 - v )\) and hence show that \(t = \ln \frac { 2 - v } { 2 ( 1 - v ) }\).
2. Hence express \(v\) in terms of \(t\). Once the power reaches 4.224 W it remains at this constant value with the resistance force still acting.
3. Verify that the power of 4.224 W is reached when \(v = 0.8\) and calculate the value of \(t\) at this instant.
4. Find \(v\) in terms of \(t\) for the motion at constant power. Deduce the limiting value of \(v\) as \(t \rightarrow \infty\).

4 A uniform lamina of mass \(m\) is in the shape of a sector of a circle of radius \(a\) and angle \(\frac { 1 } { 3 } \pi\). It can rotate freely in a vertical plane about a horizontal axis perpendicular to the lamina through its vertex O .

1. Show by integration that the moment of inertia of the lamina about the axis is \(\frac { 1 } { 2 } m a ^ { 2 }\).
2. State the distance of the centre of mass of the lamina from the axis. The lamina is released from rest when one of the straight edges is horizontal as shown in Fig. 4.1. After time \(t\), the line of symmetry of the lamina makes an angle \(\theta\) with the downward vertical. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-3_257_441_1475_322} \captionsetup{labelformat=empty} \caption{Fig. 4.1}
   \end{figure} \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-3_380_732_1635_1014} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
   \end{figure}
3. Show that \(\dot { \theta } ^ { 2 } = \frac { 4 g } { \pi a } ( 2 \cos \theta + 1 )\).
4. Find the greatest speed attained by any point on the lamina.
5. Find an expression for \(\ddot { \theta }\) in terms of \(\theta , a\) and \(g\). The lamina strikes a fixed peg at A where \(\mathrm { AO } = \frac { 3 } { 4 } a\) and is horizontal, as shown in Fig. 4.2. The collision reverses the direction of motion of the lamina and halves its angular speed.
6. Find the magnitude of the impulse that the peg gives to the lamina.
7. Determine the maximum value of \(\theta\) in the subsequent motion.