| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2003 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with elastic strings/springs |
| Difficulty | Challenging +1.8 This M4 question requires setting up potential energy for a rod-string-sphere system with geometric constraints, finding equilibrium via differentiation, and testing stability via second derivative. It demands careful geometry (relating extension to θ), combining gravitational and elastic PE, and multi-step calculus. While systematic, the geometric setup and algebraic manipulation of trigonometric expressions elevate it above standard mechanics questions, though it follows a recognizable framework for M4 equilibrium problems. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| P.E. of rod \(= mg \times 2a \sin 2\theta\) | B1 | |
| \(AC = a \cot \theta\) | B1 | |
| EPE in String \(= \frac{1}{2} \times \frac{3}{4} \times \frac{mg}{a}(a\cot\theta - a)^2\) | M1 A1 | |
| Total P.E. \(V = mg.2a\sin 2\theta + \frac{3}{8}\frac{mg}{a}(a\cot\theta - a)^2\) | M1 | |
| \(= \frac{mga}{8}[16\sin 2\theta + 3\cot^2\theta - 6\cot\theta + 3]\) | M1 | |
| \(V = \frac{mga}{8}[16\sin 2\theta + 3\cot^2\theta - 6\cot\theta] + \text{const}\) | A1 cso | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| \(\frac{dv}{d\theta} = \frac{mga}{8}[32\cos 2\theta - 6\cot\theta\operatorname{cosec}^2\theta + 6\operatorname{cosec}^2\theta]\) | M1 A2, 1, 0 | |
| \(\frac{dv}{d\theta}\bigg\ | _{\theta=0.535} = \frac{mga}{8}(-0.5^{..})\) | M1 |
| \(\frac{dv}{d\theta}\bigg\ | _{\theta=0.545} = \frac{mga}{8}(0.2^{...})\) | A1 |
| Change of sign \(\therefore \frac{dv}{d\theta} = 0\) in range, so \(\exists\) find a position of equilibrium | A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| \(\frac{dv}{d\theta}\bigg\ | _{0.535} < 0, \quad \frac{dv}{d\theta}\bigg\ | _{0.545} > 0\) |
| So turning point is minimum, \(\therefore\) equilibrium is stable | A1, A1 | (3) |
| (16 marks) |
## Part (a)
| Content | Marks | Notes |
|---------|-------|-------|
| P.E. of rod $= mg \times 2a \sin 2\theta$ | B1 | |
| $AC = a \cot \theta$ | B1 | |
| EPE in String $= \frac{1}{2} \times \frac{3}{4} \times \frac{mg}{a}(a\cot\theta - a)^2$ | M1 A1 | |
| Total P.E. $V = mg.2a\sin 2\theta + \frac{3}{8}\frac{mg}{a}(a\cot\theta - a)^2$ | M1 | |
| $= \frac{mga}{8}[16\sin 2\theta + 3\cot^2\theta - 6\cot\theta + 3]$ | M1 | |
| $V = \frac{mga}{8}[16\sin 2\theta + 3\cot^2\theta - 6\cot\theta] + \text{const}$ | A1 cso | (7) |
## Part (b)
| Content | Marks | Notes |
|---------|-------|-------|
| $\frac{dv}{d\theta} = \frac{mga}{8}[32\cos 2\theta - 6\cot\theta\operatorname{cosec}^2\theta + 6\operatorname{cosec}^2\theta]$ | M1 A2, 1, 0 | |
| $\frac{dv}{d\theta}\bigg\|_{\theta=0.535} = \frac{mga}{8}(-0.5^{..})$ | M1 | |
| $\frac{dv}{d\theta}\bigg\|_{\theta=0.545} = \frac{mga}{8}(0.2^{...})$ | A1 | |
| Change of sign $\therefore \frac{dv}{d\theta} = 0$ in range, so $\exists$ find a position of equilibrium | A1 | (6) |
## Part (c)
| Content | Marks | Notes |
|---------|-------|-------|
| $\frac{dv}{d\theta}\bigg\|_{0.535} < 0, \quad \frac{dv}{d\theta}\bigg\|_{0.545} > 0$ | M1 | |
| So turning point is minimum, $\therefore$ equilibrium is **stable** | A1, A1 | (3) |
| | **(16 marks)** | |
---\includegraphics{figure_1}
Figure 1 shows a uniform rod $AB$, of mass $m$ and length $4a$, resting on a smooth fixed sphere of radius $a$. A light elastic string, of natural length $a$ and modulus of elasticity $\frac{1}{4}mg$, has one end attached to the lowest point $C$ of the sphere and the other end attached to $A$. The points $A$, $B$ and $C$ lie in a vertical plane with $\angle BAC = 2\theta$, where $\theta < \frac{\pi}{4}$.
Given that $AC$ is always horizontal,
\begin{enumerate}[label=(\alph*)]
\item show that the potential energy of the system is
$$\frac{mga}{8}(16\sin 2\theta + 3\cot^2 \theta - 6\cot \theta) + \text{constant}.$$
[7]
\item show that there is a value of $\theta$ for which the system is in equilibrium such that $0.535 < \theta < 0.545$.
[6]
\item Determine whether this position of equilibrium is stable or unstable.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2003 Q4 [16]}}