M4 June 2006 Q4
4.
A uniform rod \(PQ\) has mass \(m\) and length \(2l\). A small smooth light ring is fixed to the end \(P\) of the rod. This ring is threaded on to a fixed horizontal smooth straight wire. A second small smooth light ring \(R\) is threaded on to the wire and is attached by a light elastic string, of natural length \(l\) and modulus of elasticity \(kmg\), to the end \(Q\) of the rod, where \(k\) is a constant.
(a) Show that, when the rod \(PQ\) makes an angle \(\theta\) with the vertical, where \(0 \lt \theta \leqslant \dfrac{\pi}{3}\), and \(Q\) is vertically below \(R\), as shown in Figure 1, the potential energy of the system is \[mgl\left[2k\cos^2\theta - (2k + 1)\cos\theta\right] + \text{constant}.\] (7)
Given that there is a position of equilibrium with \(\theta \gt 0\),
(b) show that \(k \gt \tfrac{1}{2}\). (5)
| Scheme | Marks |
|---|---|
| PE of rod \(= -mgl\cos\theta\) | B1 |
| EPE of string \(= \dfrac{kmg}{2l}(2l\cos\theta - l)^2\) | M1 A1 |
| Total PE of system, \(\ V = -mgl\cos\theta + \dfrac{kmgl}{2}(2\cos\theta - 1)^2 + c\) | M1 |
| \(= -mgl\cos\theta + \dfrac{kmgl}{2}\left(4\cos^2\theta - 4\cos\theta + 1\right) + c\) | M1 A1 |
| \(= mgl\left(-\cos\theta + 2k\cos^2\theta - 2k\cos\theta\right) + c^{\prime}\) | |
| \(= mgl\left[2k\cos^2\theta - (2k + 1)\cos\theta\right] + c^{\prime}\ \ *\) | A1 |
| (7) |
Notes
The published mark scheme for this paper is handwritten.
| Scheme | Marks |
|---|---|
| \(\dfrac{dV}{d\theta} = mgl\left(-4k\cos\theta\sin\theta + (2k + 1)\sin\theta\right)\) | M1 A1 |
| At equil\(^{m}\), \(\ mgl\sin\theta\left(-4k\cos\theta + (2k + 1)\right) = 0\) | M1 |
| \(\Rightarrow \sin\theta = 0\quad\) or \(\quad\cos\theta = \dfrac{2k + 1}{4k}\) | |
| \(\Rightarrow \theta = 0\qquad (\theta \gt 0)\quad \dfrac{2k + 1}{4k} \lt 1\) | M1 |
| \(2k + 1 \lt 4k\) | |
| \(1 \lt 2k\) | |
| \(\dfrac{1}{2} \lt k\ \ *\) | A1 |
| (5) | |
| (12 marks) |