| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2014 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | MI with removed region |
| Difficulty | Challenging +1.8 This is a challenging M5 compound pendulum problem requiring multiple advanced techniques: parallel axis theorem applied to a composite body with removed sections, rotational equation of motion, and small angle approximation. The moment of inertia calculation in part (a) involves careful bookkeeping of four holes, making it algebraically demanding. However, it follows a standard M5 template with clear scaffolding through parts (b)-(d), and students who know the techniques can execute systematically. The difficulty is elevated above typical M5 questions but not exceptionally so for Further Maths Mechanics. |
| Spec | 6.04b Find centre of mass: using symmetry6.05f Vertical circle: motion including free fall |
| Answer | Marks |
|---|---|
| (a) Show \(I_O = \frac{55Ma^2}{24}\) | (8 marks) |
| - Full disc MOI about \(O\): \(\frac{1}{2}M_{disc}(2a)^2\) | M1 |
| - Mass of each small disc: use area ratio \(\frac{(a/2)^2}{(2a)^2} \times M_{disc}\) | M1 |
| - MOI of each hole about its own centre: \(\frac{1}{2}m_{hole}(a/2)^2\) | M1 A1 |
| - Parallel axis theorem for each hole | DM1 A1 |
| - Subtract 4 holes from full disc | M1 A1 |
| (b) Show \(\frac{d^2\theta}{dt^2} = -\frac{48g}{151a}\sin\theta\) | (4 marks) |
| - Torque equation about \(A\): \(-Mg(2a)\sin\theta = I_A \ddot{\theta}\) | M1 A1 |
| - \(I_A = I_O + M(2a)^2\) by parallel axis theorem | M1 A1 |
| (c) Period of small oscillations | (2 marks) |
| - \(T = 2\pi\sqrt{\frac{151a}{48g}}\) | M1 A1 |
| (d) Find \(X\) | (4 marks) |
| - Tangential equation: \(X - Mg\sin\theta = Ma\ddot{\theta} \cdot 2a / (2a)\) | M1 A1 |
| - \(X = Mg\sin\theta - \frac{96Mg}{151}\sin\theta\) | M1 |
| - \(X = \frac{55Mg\sin\theta}{151}\) | A1 |
## Question 8:
**(a) Show $I_O = \frac{55Ma^2}{24}$** | **(8 marks)**
- Full disc MOI about $O$: $\frac{1}{2}M_{disc}(2a)^2$ | M1 |
- Mass of each small disc: use area ratio $\frac{(a/2)^2}{(2a)^2} \times M_{disc}$ | M1 |
- MOI of each hole about its own centre: $\frac{1}{2}m_{hole}(a/2)^2$ | M1 A1 |
- Parallel axis theorem for each hole | DM1 A1 |
- Subtract 4 holes from full disc | M1 A1 |
**(b) Show $\frac{d^2\theta}{dt^2} = -\frac{48g}{151a}\sin\theta$** | **(4 marks)**
- Torque equation about $A$: $-Mg(2a)\sin\theta = I_A \ddot{\theta}$ | M1 A1 |
- $I_A = I_O + M(2a)^2$ by parallel axis theorem | M1 A1 |
**(c) Period of small oscillations** | **(2 marks)**
- $T = 2\pi\sqrt{\frac{151a}{48g}}$ | M1 A1 |
**(d) Find $X$** | **(4 marks)**
- Tangential equation: $X - Mg\sin\theta = Ma\ddot{\theta} \cdot 2a / (2a)$ | M1 A1 |
- $X = Mg\sin\theta - \frac{96Mg}{151}\sin\theta$ | M1 |
- $X = \frac{55Mg\sin\theta}{151}$ | A1 |
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To get the **actual mark scheme**, you would need the official Edexcel mark scheme document for this paper.8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{57b98cdd-4121-4495-b500-185cbf3ff1a8-13_739_739_276_607}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform circular disc of radius $2 a$ has centre $O$. The points $P , Q , R$ and $S$ on the disc are the vertices of a square with centre $O$ and $O P = a$. Four circular holes, each of radius $\frac { a } { 2 }$, and with centres $P , Q , R$ and $S$, are drilled in the disc to produce the lamina $L$, shown shaded in Figure 1. The mass of $L$ is $M$.
\begin{enumerate}[label=(\alph*)]
\item Show that the moment of inertia of $L$ about an axis through $O$, and perpendicular to the plane of $L$, is $\frac { 55 M a ^ { 2 } } { 24 }$
The lamina $L$ is free to rotate in a vertical plane about a fixed smooth horizontal axis which is perpendicular to $L$ and which passes through a point $A$ on the circumference of $L$. At time $t , A O$ makes an angle $\theta$ with the downward vertical through $A$.
\item Show that $\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } = - \frac { 48 g } { 151 a } \sin \theta$
\item Hence find the period of small oscillations of $L$ about its position of stable equilibrium.
The magnitude of the component, in a direction perpendicular to $A O$, of the force exerted on $L$ by the axis is $X$.
\item Find $X$ in terms of $M , g$ and $\theta$.
\includegraphics[max width=\textwidth, alt={}, center]{57b98cdd-4121-4495-b500-185cbf3ff1a8-14_159_1662_2416_173}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2014 Q8 [18]}}