| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Variable density rod or lamina |
| Difficulty | Challenging +1.2 This is a standard Further Maths FM2 centre of mass question with variable density. While it requires integration in polar coordinates and applying the given formula for a circular arc, the setup is straightforward with clearly defined density function and geometry. The integration is routine for FM2 students, making it slightly above average difficulty but well within expected scope. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of an appropriate element (quarter of a circle) | M1 | Use of an appropriate element (may be implied) |
| \(\delta A \approx \frac{1}{2}\pi x\,\delta x\) | A1 | Correct expression for area of element (may be implied) |
| \(\delta m \approx \frac{1}{2}\pi x\,\delta x \times \frac{4\lambda}{\pi a^4}x^2\ \left(= \frac{2\lambda}{a^4}x^3\delta x\right)\) | A1 | Use of proportionality model to obtain mass of element (may be implied) |
| \(M = \int_0^a \frac{2\lambda}{a^4}x^3\,\mathrm{d}x\) | M1 | Integrating with correct limits |
| \(M = \frac{1}{2}\lambda\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(\bar{x} = \frac{1}{M}\int x\,\mathrm{d}m\) | M1 | Use the model and correct method |
| \(= \frac{1}{M}\int\left(\frac{2\sqrt{2}x}{\pi}\right)\frac{2\lambda}{a^4}x^3\,\mathrm{d}x\) | A1 | Correct integral |
| Substitute for \(M\), integrate and substitute in limits | M1 | Use the model to complete the equation |
| \(\bar{x} = \frac{8\sqrt{2}a}{5\pi}\) | A1 | cao |
## Question 7(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of an appropriate element (quarter of a circle) | M1 | Use of an appropriate element (may be implied) |
| $\delta A \approx \frac{1}{2}\pi x\,\delta x$ | A1 | Correct expression for area of element (may be implied) |
| $\delta m \approx \frac{1}{2}\pi x\,\delta x \times \frac{4\lambda}{\pi a^4}x^2\ \left(= \frac{2\lambda}{a^4}x^3\delta x\right)$ | A1 | Use of proportionality model to obtain mass of element (may be implied) |
| $M = \int_0^a \frac{2\lambda}{a^4}x^3\,\mathrm{d}x$ | M1 | Integrating with correct limits |
| $M = \frac{1}{2}\lambda$ | A1 | cao |
**Total: 5 marks**
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## Question 7(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $\bar{x} = \frac{1}{M}\int x\,\mathrm{d}m$ | M1 | Use the model and correct method |
| $= \frac{1}{M}\int\left(\frac{2\sqrt{2}x}{\pi}\right)\frac{2\lambda}{a^4}x^3\,\mathrm{d}x$ | A1 | Correct integral |
| Substitute for $M$, integrate and substitute in limits | M1 | Use the model to complete the equation |
| $\bar{x} = \frac{8\sqrt{2}a}{5\pi}$ | A1 | cao |
**Total: 4 marks**\begin{enumerate}
\item \hspace{0pt} [In this question, you may assume that the centre of mass of a circular arc, radius $r$, with angle at centre $2 \alpha$, is a distance $\frac { r \sin \alpha } { \alpha }$ from the centre.]
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d4fc2ea6-3ffc-42f2-b462-9694adfe2ec1-26_828_561_422_753}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
A thin non-uniform metal plate is in the shape of a sector $O A B$ of a circle with centre $O$ and radius $a$. The angle $A O B = \frac { \pi } { 2 }$, as shown in Figure 5.
The plate is modelled as a non-uniform lamina.\\
The mass per unit area of the lamina, at any point $P$ of the lamina, is modelled as $k ( O P ) ^ { 2 }$, where $k = \frac { 4 \lambda } { \pi a ^ { 4 } }$ and $\lambda$ is a constant.
Using the model,\\
(a) find the mass of the plate in terms of $\lambda$,\\
(b) find, in terms of $a$, the distance of the centre of mass of the plate from $O$.
\hfill \mbox{\textit{Edexcel FM2 2021 Q7 [9]}}