| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.3 This is a standard Further Maths centre of mass problem requiring composite shapes (rectangle + semicircle), standard formulas for centroids, and basic equilibrium of a suspended lamina. All techniques are routine for FM2 students with no novel insight required, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratios: \(4a^2\), \(\frac{1}{2}\pi a^2\), \((4a^2 + \frac{1}{2}\pi a^2)\) | B1 | All correct |
| \(x\): \(\frac{1}{2}a\), \(a + \frac{4a}{3\pi}\), \(\bar{x}\) and \(y\): \(2a\), \(3a\), \(\bar{y}\) | B1 | Distances could be measured from a parallel axis |
| Moments about \(OE\) | M1 | All terms needed and must be dimensionally correct |
| \(\bar{x} = \frac{(16+3\pi)a}{3(8+\pi)}\) | A1 | cao (must be in terms of \(\pi\) and \(a\)) |
| Moments about \(OA\) | M1 | All terms needed and must be dimensionally correct |
| \(\bar{y} = \frac{(16+3\pi)a}{(8+\pi)}\) | A1 | cao (must be in terms of \(\pi\) and \(a\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\alpha = \frac{\bar{x}}{\bar{y}}\) and substitute their \(\bar{x}\) and \(\bar{y}\) | M1 | Do not allow the reciprocal |
| \(\tan\alpha = \frac{1}{3}\) | A1 | cao |
## Question 1:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios: $4a^2$, $\frac{1}{2}\pi a^2$, $(4a^2 + \frac{1}{2}\pi a^2)$ | B1 | All correct |
| $x$: $\frac{1}{2}a$, $a + \frac{4a}{3\pi}$, $\bar{x}$ and $y$: $2a$, $3a$, $\bar{y}$ | B1 | Distances could be measured from a parallel axis |
| Moments about $OE$ | M1 | All terms needed and must be dimensionally correct |
| $\bar{x} = \frac{(16+3\pi)a}{3(8+\pi)}$ | A1 | cao (must be in terms of $\pi$ and $a$) |
| Moments about $OA$ | M1 | All terms needed and must be dimensionally correct |
| $\bar{y} = \frac{(16+3\pi)a}{(8+\pi)}$ | A1 | cao (must be in terms of $\pi$ and $a$) |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\alpha = \frac{\bar{x}}{\bar{y}}$ and substitute their $\bar{x}$ and $\bar{y}$ | M1 | Do not allow the reciprocal |
| $\tan\alpha = \frac{1}{3}$ | A1 | cao |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d4fc2ea6-3ffc-42f2-b462-9694adfe2ec1-02_826_649_244_708}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A letter P from a shop sign is modelled as a uniform plane lamina which consists of a rectangular lamina, $O A B D E$, joined to a semicircular lamina, $B C D$, along its diameter $B D$.
$$O A = E D = a , A B = 2 a , O E = 4 a \text {, and the diameter } B D = 2 a \text {, as shown in Figure } 1 .$$
Using the model,
\begin{enumerate}[label=(\alph*)]
\item find, in terms of $\pi$ and $a$, the distance of the centre of mass of the letter P ,\\
from (i) $O E$\\
(ii) $O A$
The letter P is freely suspended from $O$ and hangs in equilibrium. The angle between $O E$ and the downward vertical is $\alpha$.
Using the model,
\item find the exact value of $\tan \alpha$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2021 Q1 [8]}}