| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2004 |
| Session | June |
| 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 compound pendulum problem requiring parallel axis theorem, small angle approximations, and energy conservation. While it involves multiple steps and Further Maths content (M5), the techniques are routine for this specification level—showing SHM and finding period follow a template approach, and part (c) is straightforward energy conservation with small angle approximation. More challenging than typical A-level but standard for M5. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_C = \frac{1}{2}mr^2 + m(\frac{1}{2}r)^2 = \frac{3}{4}mr^2\) | M1 A1 | |
| \(M(C): \frac{3}{4}mr^2\ddot{\theta} = -mg\frac{r}{2}\sin\theta\) | M1 A1 ft | |
| \(\sin\theta \approx \theta \Rightarrow \ddot{\theta} \approx -\frac{2g}{3r}\theta\) approx. SHM | M1 A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Period \(= 2\pi\sqrt{\frac{3r}{2g}} = \pi\sqrt{\frac{6r}{g}}\) (*) | A1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dot{\theta}_{\max} = \omega\alpha \Rightarrow \frac{2}{3r}\sqrt{\frac{gr}{54}} = \sqrt{\frac{2g}{3r}}\,\alpha\) | M1 A1 | |
| \(\Rightarrow \alpha = \frac{1}{9}^c\) | A1 | (3 marks) |
| Answer | Marks |
|---|---|
| \(mg \times \frac{1}{2}r(1-\cos\alpha) = \frac{1}{2}(\frac{3}{4}mr^2)\left(\frac{gr}{54}\right)\left(\frac{2}{3r}\right)^2\) | M1 A1 |
| \(\cos\alpha = \frac{161}{162},\; \alpha = 6.4°\) (AWRT) or \(0.11^c\) (AWRT) | A1 |
## Question 4:
### Part (a)
$I_C = \frac{1}{2}mr^2 + m(\frac{1}{2}r)^2 = \frac{3}{4}mr^2$ | M1 A1 |
$M(C): \frac{3}{4}mr^2\ddot{\theta} = -mg\frac{r}{2}\sin\theta$ | M1 A1 ft |
$\sin\theta \approx \theta \Rightarrow \ddot{\theta} \approx -\frac{2g}{3r}\theta$ approx. SHM | M1 A1 | (6 marks)
### Part (b)
Period $= 2\pi\sqrt{\frac{3r}{2g}} = \pi\sqrt{\frac{6r}{g}}$ (*) | A1 | (1 mark)
### Part (c)
$\dot{\theta}_{\max} = \omega\alpha \Rightarrow \frac{2}{3r}\sqrt{\frac{gr}{54}} = \sqrt{\frac{2g}{3r}}\,\alpha$ | M1 A1 |
$\Rightarrow \alpha = \frac{1}{9}^c$ | A1 | (3 marks)
**Alternative:**
$mg \times \frac{1}{2}r(1-\cos\alpha) = \frac{1}{2}(\frac{3}{4}mr^2)\left(\frac{gr}{54}\right)\left(\frac{2}{3r}\right)^2$ | M1 A1 |
$\cos\alpha = \frac{161}{162},\; \alpha = 6.4°$ (AWRT) or $0.11^c$ (AWRT) | A1 |
---4. A uniform circular disc, of mass $m$ and radius $r$, has a diameter $A B$. The point $C$ on $A B$ is such that $A C = \frac { 1 } { 2 } r$. The disc can rotate freely in a vertical plane about a horizontal axis through $C$, perpendicular to the plane of the disc. The disc makes small oscillations in a vertical plane about the position of equilibrium in which $B$ is below $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that the motion is approximately simple harmonic.
\item Show that the period of this approximate simple harmonic motion is $\pi \sqrt { \left( \frac { 6 r } { g } \right) }$.
The speed of $B$ when it is vertically below $A$ is $\sqrt { \left( \frac { g r } { 54 } \right) }$. The disc comes to rest when $C B$ makes an angle $\alpha$ with the downward vertical.
\item Find an approximate value of $\alpha$.\\
(3)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2004 Q4 [10]}}