| Exam Board | WJEC |
|---|---|
| Module | Further Unit 6 (Further Unit 6) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Challenging +1.2 This is a multi-part centre of mass problem requiring composite shape techniques (adding/subtracting standard shapes), followed by a suspended lamina equilibrium calculation. While it involves several steps and careful bookkeeping of multiple components, the methods are standard Further Maths mechanics procedures with no novel insight required. The equilibrium angle calculation is straightforward trigonometry once the centre of mass is found. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = 4\) (cm) | B1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| Shape GPQE: mass = 64, distance = 4 | B1 | AO1 |
| EFG: mass = \(8\pi\), distance = \(8 + \frac{16}{3\pi}\) | B1 | AO3 |
| APB: mass = \(\pi\), distance = \(\frac{8}{3\pi}\) | ||
| CQD: mass = \(\pi\), distance = \(\frac{8}{3\pi}\) | B1 | AO1, either APB or CQD |
| ABCDEFG: mass = \(64+6\pi\), distance = \(\bar{y}\) | B1 | AO1, areas |
| Moments about BC: \((64 + 6\pi)\bar{y} = 64 \times 4 + 8\pi \times (8 + \frac{16}{3\pi})\) | M1 | AO3 |
| \(- 2\pi \times \frac{8}{3\pi}\) | A1 | AO1 |
| \(\bar{y} = 5.967\) (cm) (correct to 3 d.p.) | A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| If hanging in equilibrium, vertical passes through centre of mass. | M1 | AO3, correct triangle |
| \(\theta = \tan^{-1}\left(\frac{8-5.967}{4}\right)\) | A1 | AO1 |
| \(\theta = 26.94(1954...)°\) | A1 | AO1 |
| Total | [11] |
# Question 4:
## Part (a)
$\bar{x} = 4$ (cm) | B1 | AO1
## Part (b)
| Answer | Mark | Notes |
|--------|------|-------|
| Shape GPQE: mass = 64, distance = 4 | B1 | AO1 |
| EFG: mass = $8\pi$, distance = $8 + \frac{16}{3\pi}$ | B1 | AO3 |
| APB: mass = $\pi$, distance = $\frac{8}{3\pi}$ | | |
| CQD: mass = $\pi$, distance = $\frac{8}{3\pi}$ | B1 | AO1, either APB or CQD |
| ABCDEFG: mass = $64+6\pi$, distance = $\bar{y}$ | B1 | AO1, areas |
| Moments about BC: $(64 + 6\pi)\bar{y} = 64 \times 4 + 8\pi \times (8 + \frac{16}{3\pi})$ | M1 | AO3 |
| $- 2\pi \times \frac{8}{3\pi}$ | A1 | AO1 |
| $\bar{y} = 5.967$ (cm) (correct to 3 d.p.) | A1 | AO1 |
## Part (c)
| Answer | Mark | Notes |
|--------|------|-------|
| If hanging in equilibrium, vertical passes through centre of mass. | M1 | AO3, correct triangle |
| $\theta = \tan^{-1}\left(\frac{8-5.967}{4}\right)$ | A1 | AO1 |
| $\theta = 26.94(1954...)°$ | A1 | AO1 |
| **Total** | **[11]** | |
---4. The diagram shows a uniform lamina consisting of a rectangular section $G P Q E$ with a semi-circular section EFG of radius 4 cm . Quadrants $A P B$ and $C Q D$ each with radius 2 cm are removed. Dimensions in cm are as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{3efc4ef6-8a80-4267-8e95-733200e875c5-3_758_604_497_651}
\begin{enumerate}[label=(\alph*)]
\item Write down the distance of the centre of mass of the lamina $A B C D E F G$ from $A G$.
\item Determine the distance of the centre of mass of the lamina $A B C D E F G$ from $B C$.
\item The lamina $A B C D E F G$ is suspended freely from the point $E$ and hangs in equilibrium. Calculate the angle EG makes with the vertical.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 6 Q4 [11]}}