| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| 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 compound system (gravitational + elastic) with non-uniform rod rotating about a fixed point, then finding equilibrium via calculus. The geometry is non-trivial (relating rod angle to string extension through the constraint), and part (b) requires solving a transcendental equation. While systematic, it demands careful coordinate work, energy formulation for rotation, and multi-step algebraic manipulation beyond standard textbook exercises. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| \(GPE = -mg L \cos 2\theta\) | B1 |
| \(EPE = \frac{m_g(6\sin\theta - L)^2}{6}\) | M1 |
| Answer | Marks |
|---|---|
| \(= mg(3\cos^2\theta - \cos\theta) + C\) | M1 |
| Answer | Marks |
|---|---|
| \(= mgL(\cos^2\theta - \cos\theta) + c\) | M1, M1 (guidance: "unclear2DF..."), A1 (6) |
| Answer | Marks |
|---|---|
| \(\sin\theta = 0\) or \(\cos\theta = \frac{1}{2}\) | M1 A1, M1 |
| \(\theta = 0\) or \(\theta = \pm \frac{\pi}{3}\) | A1, A1 (6), (12) |
## Part (c)
$GPE = -mg L \cos 2\theta$ | B1 |
$EPE = \frac{m_g(6\sin\theta - L)^2}{6}$ | M1 |
$= \frac{m_g}{12}(36\cos^2\theta - 12\cos\theta + L^2)$
$= mg(3\cos^2\theta - \cos\theta) + C$ | M1 |
$V = -mgL(2\cos^2\theta - 1) + mg L(3\cos^2\theta - \cos\theta) + c$
$= mgL(\cos^2\theta - \cos\theta) + c$ | M1, M1 (guidance: "unclear2DF..."), A1 (6) |
## Part (d)
$\frac{dV}{d\theta} = mgL(-2\cos\theta \sin\theta - \sin\theta) = 0$
$\sin\theta(-2\cos\theta + 1) = 0$
$\sin\theta = 0$ or $\cos\theta = \frac{1}{2}$ | M1 A1, M1 |
$\theta = 0$ or $\theta = \pm \frac{\pi}{3}$ | A1, A1 (6), (12) |
---A non-uniform rod $BC$ has mass $m$ and length $3l$. The centre of mass of the rod is at distance $l$ from $B$. The rod can turn freely about a fixed smooth horizontal axis through $B$. One end of a light elastic string, of natural length $l$ and modulus of elasticity $\frac{mg}{6}$, is attached to $C$. The other end of the string is attached to a point $P$ which is at a height $3l$ vertically above $B$.
\begin{enumerate}[label=(\alph*)]
\item Show that, while the string is stretched, the potential energy of the system is
$$mgl(\cos^2 \theta - \cos \theta) + \text{constant},$$
where $\theta$ is the angle between the string and the downward vertical and $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. [6]
\item Find the values of $\theta$ for which the system is in equilibrium with the string stretched. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2005 Q5 [12]}}