M4 June 2014 (R) Q5
5.
A uniform rod \(AB\), of length \(2l\) and mass \(12m\), has its end \(A\) smoothly hinged to a fixed point. One end of a light inextensible string is attached to the other end \(B\) of the rod. The string passes over a small smooth pulley which is fixed at the point \(C\), where \(AC\) is horizontal and \(AC = 2l\). A particle of mass \(m\) is attached to the other end of the string and the particle hangs vertically below \(C\).
The angle \(BAC\) is \(\theta\), where \(0 < \theta < \dfrac{\pi}{2}\), as shown in Figure 1.
(a) Show that the potential energy of the system is \[4mgl\left(\sin\frac{\theta}{2} - 3\sin\theta\right) + \text{constant}\] (4)
(b) Find the value of \(\theta\) when the system is in equilibrium and determine the stability of this equilibrium position. (10)
| Scheme | Marks |
|---|---|
| \(-12mgl\sin\theta\) rod | B1 |
| \(-mg(L - 4l\sin\frac{1}{2}\theta)\) particle | M1 A1 |
| \(4mgl(\sin\frac{1}{2}\theta - 3\sin\theta) +\) constant *given answer* | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = 4mgl(\frac{1}{2}\cos\frac{1}{2}\theta - 3\cos\theta)\) Differentiate | M1 A1 |
| \(4mgl(\frac{1}{2}\cos\frac{1}{2}\theta - 3\cos\theta) = 0\) Derivative \(= 0\) | M1 |
| \(\frac{1}{2}\cos\frac{1}{2}\theta - 3(2\cos^2\frac{1}{2}\theta - 1) = 0\) In terms of \(\cos\dfrac{1}{2}\theta\) | M1 |
| \(\cos\frac{1}{2}\theta = \frac{3}{4}\) or \(-\frac{2}{3}\) | A1 |
| \(\cos\theta = \frac{1}{8}\) or \(-\frac{1}{9}\) Solve for \(\theta\) | M1 A1 |
| \(\theta = 1.45\) as \(\theta < \frac{1}{2}\pi \qquad (82.8^\circ)\) | A1 |
| \(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = 4mgl(-\frac{1}{4}\sin\frac{1}{2}\theta + 3\sin\theta)\) Second derivative | M1 |
| When \(\theta = 1.45,\ \dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = 11.25mgl > 0\), hence stable | A1 |
| (10) | |
| (14 marks) |
Notes
(Corrected from the printed mark scheme: the angle in degrees is printed as 83.0\(^\circ\); \(\cos^{-1}\frac{1}{8} = 82.8^\circ\).)