M4 January 2006 Q6
6.
A smooth wire with ends \(A\) and \(B\) is in the shape of a semi-circle of radius \(a\). The mid-point of \(AB\) is \(O\). The wire is fixed in a vertical plane and hangs below \(AB\) which is horizontal. A small ring \(R\), of mass \(m\sqrt{2}\), is threaded on the wire and is attached to two light inextensible strings. The other end of each string is attached to a particle of mass \(\dfrac{3m}{2}\). The particles hang vertically under gravity, as shown in Figure 1.
(a) Show that, when the radius \(OR\) makes an angle \(2\theta\) with the vertical, the potential energy, \(V\), of the system is given by \[V = \sqrt{2}mga(3\cos\theta - \cos 2\theta) + \text{constant}.\] (7)
(b) Find the values of \(\theta\) for which the system is in equilibrium. (6)
(c) Determine the stability of the position of equilibrium for which \(\theta \gt 0\). (4)
| Scheme | Marks |
|---|---|
| PE of R \(= -\sqrt{2}mga\cos 2\theta\ \ (+c)\qquad (1)\) | B1 |
| PE of LH mass \(= -\dfrac{3}{2}mg(2a - 2a\sin(45 + \theta))\ \ (+c)\qquad (2)\) | M1 A1 |
| PE of RH mass \(= -\dfrac{3}{2}mg(2a - 2a\sin(45 - \theta))\ \ (+c)\qquad (3)\) | A1 |
| \(V = (1) + (2) + (3)\qquad\) (in terms of \(\theta\) etc.) | M1 |
| \(= -\sqrt{2}mga\cos 2\theta - \dfrac{3}{2}mg\left[4a - a\sqrt{2}(\cos\theta + \sin\theta + \cos\theta - \sin\theta)\right]\) | M1 |
| \(= -\sqrt{2}mga\cos 2\theta - \dfrac{3}{2}mga\left(-2\sqrt{2}\cos\theta + 4\right)\) | |
| \(= \sqrt{2}mga(3\cos\theta - \cos 2\theta) + \text{constant}\qquad (*)\) | A1 |
| (7) |
| Scheme | Marks |
|---|---|
| \(\dfrac{dV}{d\theta} = \sqrt{2}mga(-3\sin\theta + 2\sin 2\theta)\) | M1 A1 |
| \(\dfrac{dV}{d\theta} = 0 \Rightarrow 2\sin 2\theta - 3\sin\theta = 0\) | M1 |
| \(\Rightarrow \sin\theta(4\cos\theta - 3) = 0\) | M1 |
| \(\Rightarrow \theta = 0,\ \text{or}\ \theta = \pm\arccos\dfrac{3}{4}\ (= \pm 0.723)\) | A1, A1 |
| (6) |
| Scheme | Marks |
|---|---|
| \(\dfrac{d^2V}{d\theta^2} = \sqrt{2}mga(-3\cos\theta + 4\cos 2\theta)\) | M1 A1 |
| \(\cos\theta = \dfrac{3}{4}\colon\ \dfrac{d^2V}{d\theta^2} = \sqrt{2}mga\left(-3.\dfrac{3}{4} + 4\left(2.\dfrac{9}{16} - 1\right)\right)\) | M1 |
| \(= \sqrt{2}mga\left(-\dfrac{9}{4} + \dfrac{1}{2}\right)\) | |
| \(\lt 0\ \therefore\) Unstable | A1 |
| (4) | |
| (17 marks) |