| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.8 This is a multi-part Further Maths mechanics question requiring: (a) standard sector centre of mass formula application, (b) composite lamina calculation with three components of different types (two squares and a sector), and (c) suspended equilibrium analysis using moments. While the individual techniques are standard for FM2, the combination of multiple shapes and the final equilibrium calculation with angle-finding requires careful coordinate work and trigonometry, placing it moderately above average difficulty. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using sector: distance \(OG = \dfrac{2 \times 3a\sin\frac{\pi}{4}}{3 \times \frac{\pi}{4}}\) | B1 | Correct application of standard result from formula booklet. Must substitute for \(a\) but need not simplify. Implied if you see \(\left(=\dfrac{4\sqrt{2}a}{\pi}\right)\) |
| Using Pythagoras: \(2d^2 = \dfrac{32a^2}{\pi^2}\) where \(d^2 + d^2 = OG^2\), or using trigonometry: Distance from \(OC = OG\cos 45° = OG\sin 45°\) | M1 | Correct strategy to find the distance for the quadrant. Need to see use of \(\dfrac{1}{\sqrt{2}}\) or \(\dfrac{\sqrt{2}}{2}\) somewhere in the solution |
| \(d = \sqrt{\dfrac{16a^2}{\pi^2}} = \dfrac{4a}{\pi}\) * | A1* | Obtain the given result from correct working |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using semicircle of radius \(3a\): \(\bar{y} = \dfrac{4 \times 3a}{3\pi}\left(=\dfrac{4a}{\pi}\right)\) | B1 | 1.1b |
| Moments about diameter: \(\dfrac{9\pi a^2}{2} \times \dfrac{4a}{\pi} = 2 \times \dfrac{9\pi a^2}{4} \times d\) | M1 | 2.1 |
| \(\Rightarrow d = \dfrac{4a}{\pi}\) * | A1* | 2.2a |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratios and distances from \(FC\): ABCO: mass \(9\), distance \(-\dfrac{3a}{2}\); ODEF: mass \(9\), distance \(\dfrac{3a}{2}\); ODC: mass \(\dfrac{9\pi}{4}\), distance \(\dfrac{4a}{\pi}\) | B1 | Correct masses and distances from \(FC\) or a parallel axis or \(BOE\). Seen or implied (a bright candidate might realise that if taking moments about \(FC\) then the two squares cancel each other) |
| Moments about \(FC\): | M1 | Moments about \(FC\) or a parallel axis or \(BOE\). All terms required, and dimensionally correct. Condone sign errors. Accept as part of a vector equation |
| \(-9\times\dfrac{3a}{2} + 9\times\dfrac{3a}{2} + \dfrac{9\pi}{4}\times\dfrac{4a}{\pi} = \left(18 + \dfrac{9\pi}{4}\right)\bar{x}\ (= 9a)\) | A1 | Correct unsimplified equation for their axis |
| \(\bar{x} = \dfrac{4a}{8+\pi}\) | A1 | Or equivalent with no errors seen. Accept \(0.36a\) or better \((0.3590\ldots a)\) |
| (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratios from \(BOE\): ABCO: mass \(9\), distance \(0\); ODEF: mass \(9\), distance \(0\); ODC: mass \(\dfrac{9\pi}{4}\), distance \(\dfrac{4\sqrt{2}a}{\pi}\) | B1 | 1.2 |
| Moments about \(BOE\): \(\left(18 + \dfrac{9\pi}{4}\right)d = \dfrac{9\pi}{4}\times\dfrac{4\sqrt{2}a}{\pi}\) | M1 | 3.1a |
| \(\bar{x} = d\cos 45° = \dfrac{4a}{8+\pi}\) | A1 | 1.1b |
| (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{y} = \dfrac{4a}{8+\pi}\) from \(OD\) or \(\bar{y} = 3a + \dfrac{4a}{8+\pi}\) from \(FE\) | B1ft | Allow use of symmetry seen or implied. Accept \(\bar{y} = \bar{x}\). From FE, \(\bar{y} = \dfrac{28a+3\pi a}{8+\pi}\). Accept \(+/-\) |
| Complete method to find a relevant angle | M1 | (\(\theta\) or \(90-\theta\)). Need to substitute their values of \(\bar{x}\) and distance from \(F \neq \dfrac{4a}{\pi}\) |
| \(\theta° = \tan^{-1}\left(\dfrac{\bar{x}}{3a+\bar{y}}\right) = \tan^{-1}\left(\dfrac{4a}{28a+3\pi a}\right)\) | A1ft | Correct unsimplified expression for a relevant angle. Follow their \(\bar{x}\) and \(\bar{y}\) |
| \(\theta = 6.1\) | A1 | 6.1 or better \((6.10067\ldots)\). The question defines \(\theta\) as measured in degrees. \(0.106\) can score B1M1A1ftA0. Do not ISW |
| (4 marks) |
# Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using sector: distance $OG = \dfrac{2 \times 3a\sin\frac{\pi}{4}}{3 \times \frac{\pi}{4}}$ | B1 | Correct application of standard result from formula booklet. Must substitute for $a$ but need not simplify. Implied if you see $\left(=\dfrac{4\sqrt{2}a}{\pi}\right)$ |
| Using Pythagoras: $2d^2 = \dfrac{32a^2}{\pi^2}$ where $d^2 + d^2 = OG^2$, or using trigonometry: Distance from $OC = OG\cos 45° = OG\sin 45°$ | M1 | Correct strategy to find the distance for the quadrant. Need to see use of $\dfrac{1}{\sqrt{2}}$ or $\dfrac{\sqrt{2}}{2}$ somewhere in the solution |
| $d = \sqrt{\dfrac{16a^2}{\pi^2}} = \dfrac{4a}{\pi}$ * | A1* | Obtain the given result from correct working |
| **(3 marks)** | | |
# Question 5(a) alt:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using semicircle of radius $3a$: $\bar{y} = \dfrac{4 \times 3a}{3\pi}\left(=\dfrac{4a}{\pi}\right)$ | B1 | 1.1b |
| Moments about diameter: $\dfrac{9\pi a^2}{2} \times \dfrac{4a}{\pi} = 2 \times \dfrac{9\pi a^2}{4} \times d$ | M1 | 2.1 |
| $\Rightarrow d = \dfrac{4a}{\pi}$ * | A1* | 2.2a |
| **(3 marks)** | | |
# Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios and distances from $FC$: ABCO: mass $9$, distance $-\dfrac{3a}{2}$; ODEF: mass $9$, distance $\dfrac{3a}{2}$; ODC: mass $\dfrac{9\pi}{4}$, distance $\dfrac{4a}{\pi}$ | B1 | Correct masses and distances from $FC$ or a parallel axis or $BOE$. Seen or implied (a bright candidate might realise that if taking moments about $FC$ then the two squares cancel each other) |
| Moments about $FC$: | M1 | Moments about $FC$ or a parallel axis or $BOE$. All terms required, and dimensionally correct. Condone sign errors. Accept as part of a vector equation |
| $-9\times\dfrac{3a}{2} + 9\times\dfrac{3a}{2} + \dfrac{9\pi}{4}\times\dfrac{4a}{\pi} = \left(18 + \dfrac{9\pi}{4}\right)\bar{x}\ (= 9a)$ | A1 | Correct unsimplified equation for their axis |
| $\bar{x} = \dfrac{4a}{8+\pi}$ | A1 | Or equivalent with no errors seen. Accept $0.36a$ or better $(0.3590\ldots a)$ |
| **(4 marks)** | | |
# Question 5(b) alt:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios from $BOE$: ABCO: mass $9$, distance $0$; ODEF: mass $9$, distance $0$; ODC: mass $\dfrac{9\pi}{4}$, distance $\dfrac{4\sqrt{2}a}{\pi}$ | B1 | 1.2 |
| Moments about $BOE$: $\left(18 + \dfrac{9\pi}{4}\right)d = \dfrac{9\pi}{4}\times\dfrac{4\sqrt{2}a}{\pi}$ | M1 | 3.1a |
| $\bar{x} = d\cos 45° = \dfrac{4a}{8+\pi}$ | A1 | 1.1b |
| **(4 marks)** | | |
# Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{y} = \dfrac{4a}{8+\pi}$ from $OD$ or $\bar{y} = 3a + \dfrac{4a}{8+\pi}$ from $FE$ | B1ft | Allow use of symmetry seen or implied. Accept $\bar{y} = \bar{x}$. From FE, $\bar{y} = \dfrac{28a+3\pi a}{8+\pi}$. Accept $+/-$ |
| Complete method to find a relevant angle | M1 | ($\theta$ or $90-\theta$). Need to substitute their values of $\bar{x}$ and distance from $F \neq \dfrac{4a}{\pi}$ |
| $\theta° = \tan^{-1}\left(\dfrac{\bar{x}}{3a+\bar{y}}\right) = \tan^{-1}\left(\dfrac{4a}{28a+3\pi a}\right)$ | A1ft | Correct unsimplified expression for a relevant angle. Follow their $\bar{x}$ and $\bar{y}$ |
| $\theta = 6.1$ | A1 | 6.1 or better $(6.10067\ldots)$. The question defines $\theta$ as measured in degrees. $0.106$ can score B1M1A1ftA0. Do not ISW |
| **(4 marks)** | | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1f39620e-c10f-4344-89f1-626fff36d187-16_567_602_260_735}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The uniform plane lamina shown in Figure 3 is formed from two squares, $A B C O$ and $O D E F$, and a sector $O D C$ of a circle with centre $O$. Both squares have sides of length $3 a$ and $A O$ is perpendicular to $O F$. The radius of the sector is $3 a$\\[0pt]
[In part (a) you may use, without proof, any of the centre of mass formulae given in the formulae booklet.]
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the sector $O D C$ from $O C$ is $\frac { 4 a } { \pi }$
\item Find the distance of the centre of mass of the lamina from $F C$
The lamina is freely suspended from $F$ and hangs in equilibrium with $F C$ at an angle $\theta ^ { \circ }$ to the downward vertical.
\item Find the value of $\theta$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2022 Q5 [11]}}