M4 June 2018 Q1
1.
A uniform rod \(AB\) has mass \(m\) and length \(4a\). The end \(A\) of the rod is freely hinged to a fixed point. One end of a light elastic string, of natural length \(a\) and modulus \(\dfrac{1}{4}mg\), is attached to the end \(B\) of the rod. The other end of the string is attached to a small light smooth ring \(R\). The ring can move freely on a smooth horizontal wire which is fixed at a height \(a\) above \(A\), and in a vertical plane through \(A\). The angle between the rod and the horizontal is \(\theta\), where \(0 \lt \theta \lt \dfrac{\pi}{2}\), as shown in Figure 1. Given that the elastic string is vertical,
(a) show that the potential energy of the system is \[2mga(\sin^2\theta - \sin\theta) + \text{constant}\] (4)
(b) Show that when \(\theta = \dfrac{\pi}{6}\) the rod is in stable equilibrium. (7)
| Scheme | Marks |
|---|---|
| For the rod: GPE \(= -2mga\sin\theta\) Must be working from a fixed point | B1 |
| Extension in the string \(= 4a\sin\theta\) | B1 |
| GPE in the string \(= \dfrac{\frac{1}{4}mgx^2}{2a}\) | M1 |
| Total \(\dfrac{mg}{8a}\times(4a\sin\theta)^2 - 2mga\sin\theta = 2mga\left(\sin^2\theta - \sin\theta\right) +\) constant Given Answer | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Differentiate: \(\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = 2mga(2\sin\theta\cos\theta - \cos\theta)\) | M1A1 |
| Second derivative: \(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = 2amg\left(2\cos^2\theta - 2\sin^2\theta + \sin\theta\right)\) | M1A1 |
| Substitute \(\theta = \dfrac{\pi}{6}\) in both: | M1 |
| \(\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = 2mga\left(2\times\dfrac{1}{2}\cos\theta - \cos\theta\right) = 0\) hence equilibrium cso Allow from working to find solutions for \(\theta\) | A1 |
| \(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = 2mga\left(2\times\dfrac{3}{4} - 2\times\dfrac{1}{4} + \dfrac{1}{2}\right) = 3mga \gt 0\) hence equilibrium stable Given Answer | A1 |
| (7) | |
| (11 marks) |