| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Frame with circular arc or semicircular arc components |
| Difficulty | Standard +0.8 This is a multi-part Further Maths mechanics question requiring: (a) application of a quoted formula with coordinate geometry, (b) systematic calculation of composite centre of mass for 5 components with careful bookkeeping, and (c) moments equilibrium with the calculated centre of mass. While methodical rather than requiring deep insight, the composite nature with curved arcs, the algebraic manipulation, and being FM2 content places 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\alpha = \frac{\pi}{4} \Rightarrow \frac{r\sin\alpha}{\alpha} = r \times \frac{1}{\sqrt{2}} \times \frac{4}{\pi} = \frac{2\sqrt{2}r}{\pi}\) | B1 | 1.1b - Correct use of given formula; seen or implied |
| Distance from \(O\): \(A = \frac{2\sqrt{2}r}{\pi} \times \cos\frac{\pi}{4} = \frac{2\sqrt{2}r}{\pi \times \sqrt{2}} = \frac{2r}{\pi}\) * | B1* | 2.2a - Obtain given result from correct working, e.g. by use of trigonometry or Pythagoras |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Moments about \(OA\) | M1 | 3.1a - Condone dimension error in arc length, but must be dimensionally correct equation; need all terms; allow use of parallel axis |
| \(r \times \frac{r}{2} + r \times r \times \frac{2r}{\pi} \times \frac{\pi r}{2} + \frac{2r}{\pi} \times \frac{\pi r}{2} = (3r + \pi r)d\) | A1, A1 | 1.1b - Unsimplified with at most one error; correct unsimplified equation (error in arc length affects 3 terms — count as single error) |
| \(\frac{7r^2}{2} = (3r + \pi r)d \Rightarrow d = \frac{7r}{2(3+\pi)}\) * | A1* | 2.2a - Obtain given answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Moments about \(A\) or any other complete method to obtain \(F\) in terms of \(W\) | M1 | 3.1a - Dimensionally correct equation with distances perpendicular to forces |
| \(W \times \frac{7r}{2(3+\pi)} = F \times 2r\) | A1 | 1.1b - Correct unsimplified equation |
| \(F = \frac{7}{4(3+\pi)}W\) | A1 | 1.1b - Any equivalent form; \(0.28W\) or better \((0.28494...W)\) |
# Question 3(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\alpha = \frac{\pi}{4} \Rightarrow \frac{r\sin\alpha}{\alpha} = r \times \frac{1}{\sqrt{2}} \times \frac{4}{\pi} = \frac{2\sqrt{2}r}{\pi}$ | B1 | 1.1b - Correct use of given formula; seen or implied |
| Distance from $O$: $A = \frac{2\sqrt{2}r}{\pi} \times \cos\frac{\pi}{4} = \frac{2\sqrt{2}r}{\pi \times \sqrt{2}} = \frac{2r}{\pi}$ * | B1* | 2.2a - Obtain given result from correct working, e.g. by use of trigonometry or Pythagoras |
---
# Question 3(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Moments about $OA$ | M1 | 3.1a - Condone dimension error in arc length, but must be dimensionally correct equation; need all terms; allow use of parallel axis |
| $r \times \frac{r}{2} + r \times r \times \frac{2r}{\pi} \times \frac{\pi r}{2} + \frac{2r}{\pi} \times \frac{\pi r}{2} = (3r + \pi r)d$ | A1, A1 | 1.1b - Unsimplified with at most one error; correct unsimplified equation (error in arc length affects 3 terms — count as single error) |
| $\frac{7r^2}{2} = (3r + \pi r)d \Rightarrow d = \frac{7r}{2(3+\pi)}$ * | A1* | 2.2a - Obtain given answer from correct working |
---
# Question 3(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Moments about $A$ or any other complete method to obtain $F$ in terms of $W$ | M1 | 3.1a - Dimensionally correct equation with distances perpendicular to forces |
| $W \times \frac{7r}{2(3+\pi)} = F \times 2r$ | A1 | 1.1b - Correct unsimplified equation |
| $F = \frac{7}{4(3+\pi)}W$ | A1 | 1.1b - Any equivalent form; $0.28W$ or better $(0.28494...W)$ |
---\begin{enumerate}
\item \hspace{0pt} [In this question you may quote, without proof, the formula for the distance of the centre of mass of a uniform circular arc from its centre.]
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b070338-1de4-4c33-be29-d37ac06c9fed-08_816_483_338_790}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Five pieces of a uniform wire are joined together to form the rigid framework $O A B C O$ shown in Figure 1, where
\begin{itemize}
\item $O A , O B$ and $B C$ are straight, with $O A = O B = B C = r$
\item arc $A B$ is one quarter of a circle with centre $O$ and radius $r$
\item arc $O C$ is one quarter of a circle of radius $r$
\item all five pieces of wire lie in the same plane\\
(a) Show that the centre of mass of arc $A B$ is a distance $\frac { 2 r } { \pi }$ from $O A$.
\end{itemize}
Given that the distance of the centre of mass of the framework from $O A$ is $d$,\\
(b) show that $\mathrm { d } = \frac { 7 r } { 2 ( 3 + ) }$
The framework is freely pivoted at $A$.\\
The framework is held in equilibrium, with $A O$ vertical, by a horizontal force of magnitude $F$ which is applied to the framework at $C$.
Given that the weight of the framework is $W$\\
(c) find $F$ in terms of $W$
\hfill \mbox{\textit{Edexcel FM2 2023 Q3 [9]}}