| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Angular speed and period |
| Difficulty | Challenging +1.3 This M5 question requires applying the parallel axis theorem, rotational dynamics, and small angle approximations—all standard Further Maths mechanics techniques. While it involves multiple steps and careful geometric reasoning about forces at the pivot, the methods are well-practiced in M5 syllabi. The question is more demanding than typical A-level due to the Further Maths content, but follows predictable patterns for this module without requiring novel insights. |
| Spec | 3.03a Force: vector nature and diagrams6.04e Rigid body equilibrium: coplanar forces6.05f Vertical circle: motion including free fall |
| Answer | Marks |
|---|---|
| \(T = 2\pi\sqrt{\frac{3a}{2g}}\) | M1 A1 M1 A1(2) |
| Answer | Marks |
|---|---|
| \(\ddot{\theta} = -\frac{2a}{3a}\sin\theta = \ddot{z}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(= \frac{2mg\sin\theta}{3}\) | M1 A2 M1(5)(12) |
## Part (a)
$T_k = \frac{1}{2}m\dot{r}^2 + m(\frac{1}{a})^2\dot{\theta}^2 = \frac{3}{2}m\dot{z}^2$
At $A(1)$, $-mg a \leq 0 = \frac{3}{2}m\ddot{z}^2$
$-\frac{3a}{3a}\sin\theta = \ddot{z}$
For small $\theta$, $-\frac{2a}{3a}\theta = \ddot{z}$
$T = 2\pi\sqrt{\frac{3a}{2g}}$ | M1 A1 M1 A1(2) |
## Part (b)
$\ddot{\theta} = -\frac{2a}{3a}\sin\theta = \ddot{z}$ | M1 A1 |
## Part (c)
At $P(k)$: $Y - mg\sin\theta = m\ddot{z}^2$
$\Rightarrow Y = mg\sin\theta + m\ddot{z}\left(-\frac{2a}{3a}\sin\theta\right)$
$= \frac{2mg\sin\theta}{3}$ | M1 A2 M1(5)(12) |
---A uniform circular disc, of mass $m$, radius $a$ and centre $O$, is free to rotate in a vertical plane about a fixed smooth horizontal axis. The axis passes through the mid-point $A$ of a radius of the disc.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of motion for the disc when the line $AO$ makes an angle $\theta$ with the downward vertical through $A$. [5]
\item Hence find the period of small oscillations of the disc about its position of stable equilibrium. [2]
\end{enumerate}
When the line $AO$ makes an angle $\theta$ with the downward vertical through $A$, the force acting on the disc at $A$ is $\mathbf{F}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the magnitude of the component of $\mathbf{F}$ perpendicular to $AO$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2006 Q6 [12]}}