| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations: rigid body compound pendulum |
| Difficulty | Challenging +1.2 This is a standard M5 rotational dynamics problem requiring perpendicular axis theorem for moment of inertia, small angle approximation for SHM, and period calculation. While it involves multiple steps and proof elements, the techniques are well-practiced in mechanics modules and follow predictable patterns. The 'show that' format provides guidance, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.04d Integration: for centre of mass of laminas/solids6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_{AB} = \frac{1}{3}ma^2\) | B1 | |
| \(I_A = 2 \times \frac{1}{3}ma^2\) (perpendicular axes) | M1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(M(A), mg\frac{a}{\sqrt{2}}\sin\theta = -\frac{2}{3}ma^2\ddot{\theta}\) | M1 A1 A1 | |
| \(\ddot{\theta} \approx -\frac{3g}{2a\sqrt{2}}\theta\) for small \(\theta\), hence SHM | M1 A1 | (5 marks) |
| Answer | Marks |
|---|---|
| \(t = \frac{1}{4} \times \text{period} = \frac{\pi}{2}\sqrt{\frac{2a\sqrt{2}}{3g}}\) | M1 A1 |
| \(= \pi\sqrt{\frac{a\sqrt{2}}{6g}}\) | (2 marks) |
Total: 10 marks
## Part (a)
$I_{AB} = \frac{1}{3}ma^2$ | B1 |
$I_A = 2 \times \frac{1}{3}ma^2$ (perpendicular axes) | M1 A1 | (3 marks)
## Part (b)
$M(A), mg\frac{a}{\sqrt{2}}\sin\theta = -\frac{2}{3}ma^2\ddot{\theta}$ | M1 A1 A1 |
$\ddot{\theta} \approx -\frac{3g}{2a\sqrt{2}}\theta$ for small $\theta$, hence SHM | M1 A1 | (5 marks)
## Part (c)
$t = \frac{1}{4} \times \text{period} = \frac{\pi}{2}\sqrt{\frac{2a\sqrt{2}}{3g}}$ | M1 A1 |
$= \pi\sqrt{\frac{a\sqrt{2}}{6g}}$ | (2 marks)
Total: 10 marks
---A uniform square lamina $ABCD$ of side $a$ and mass $m$ is free to rotate in vertical plane about a horizontal axis through $A$. The axis is perpendicular to the plane of the lamina. The lamina is released from rest when $t = 0$ and $AC$ makes a small angle with the downward vertical through $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that the moment of inertia of the lamina about the axis is $\frac{2}{3}ma^2$. [3]
\item Show that the motion of the lamina is approximately simple harmonic. [5]
\item Find the time $t$ when $AC$ is first vertical. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 Q5 [10]}}