M4 June 2008 Q7
7.
A uniform rod \(AB\), of length \(2a\) and mass \(kM\) where \(k\) is a constant, is free to rotate in a vertical plane about the fixed point \(A\). One end of a light inextensible string of length \(6a\) is attached to the end \(B\) of the rod and passes over a small smooth pulley which is fixed at the point \(P\). The line \(AP\) is horizontal and of length \(2a\). The other end of the string is attached to a particle of mass \(M\) which hangs vertically below the point \(P\), as shown in Figure 3. The angle \(PAB\) is \(2\theta\), where \(0^\circ \leqslant \theta \leqslant 180^\circ\).
(a) Show that the potential energy of the system is\[Mga(4\sin\theta - k\sin 2\theta) + \text{constant}.\] (5)
The system has a position of equilibrium when \(\cos\theta = \frac{3}{4}\).
(b) Find the value of \(k\). (5)
(c) Hence find the value of \(\cos\theta\) at the other position of equilibrium. (3)
(d) Determine the stability of each of the two positions of equilibrium. (5)
| Scheme | Marks |
|---|---|
| PE of rod \(= -kMga\sin 2\theta\) | B1 |
| \(BP = 2 \times 2a\sin\theta = 4a\sin\theta\) | M1 |
| PE of mass \(= -Mg(6a - 4a\sin\theta)\) | A1 |
| \(V = -Mg(6a - 4a\sin\theta) - kMga\sin 2\theta\) | M1 |
| \(= Mga(4\sin\theta - k\sin 2\theta) + \text{constant}\quad *\) | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = Mga(4\cos\theta - 2k\cos 2\theta)\) | M1 A1 |
| so, \(4 \times \tfrac{3}{4} - 2k\left(2\left(\tfrac{3}{4}\right)^2 - 1\right) = 0\) | M1 M1 |
| \(\Rightarrow k = 12\) | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(4\cos\theta - 24(2\cos^2\theta - 1) = 0\) | M1 |
| \(12\cos^2\theta - \cos\theta - 6 = 0\) | DM1 |
| \((4\cos\theta - 3)(3\cos\theta + 2) = 0\) | |
| \(\cos\theta = -\tfrac{2}{3}\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\dfrac{d^2V}{d\theta^2} = (Mga)(-4\sin\theta + 4k\sin 2\theta)\) | M1 A1 |
| when \(\cos\theta = \dfrac{3}{4},\ \dfrac{d^2V}{d\theta^2} = (Mga) \times 44.97\ldots \Rightarrow\) stable | M1 A1 |
| when \(\cos\theta = \dfrac{-2}{3},\ \dfrac{d^2V}{d\theta^2} = (Mga) \times -50.68\ldots \Rightarrow\) unstable | A1 |
| (5) | |
| (18 marks) |