| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| 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 question requiring integration to find centre of mass of an elliptical sector (non-trivial setup with ellipse parametrization), then applying equilibrium conditions. The integration requires careful handling of the ellipse equation and the suspended lamina part requires geometric reasoning about the equilibrium angle. More demanding than standard A-level but routine for FM2 students who have practiced these techniques. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Moments about the \(y\)-axis: \(\int x \rho y \, dx\) | M1 | 3.1a |
| \(\int_0^4 x \left(36 - \frac{9x^2}{4}\right) dx = k\left[36 - \frac{9x^2}{4}\right] = 32\) | M1 | 2.1 |
| \(\left[\frac{4}{27}\left(36 - \frac{9x^2}{4}\right)^{3/2}\right]_0^4 = 32\) | A1 | 1.1b |
| \(\bar{x} = \frac{\int x \rho y \, dx}{\int \rho y \, dx}\) | DM1 | 3.1a |
| \(\bar{x} = \frac{32}{6} = \frac{16}{3}\) | A1* | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Moments about the \(x\)-axis: \(\frac{1}{2}\int \rho y^2 \, dx\) or \(\int \frac{1}{2}\rho(y) \left(36 - \frac{9x^2}{4}\right) dx\) | M1 | 3.1a |
| \(\frac{1}{2}\left[36x - \frac{9x^4}{4}\right]_0^4 = \frac{1}{2}(144 - 144) = 48\) | A1 | 1.1b |
| \(\bar{y} = \frac{\int \frac{1}{2}\rho y^2 \, dx}{32} = \frac{48}{6}\) | DM1 | 2.1 |
| \(\bar{y} = 8\) | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Correct use of trigonometry | M1 | 3.1a |
| \(\tan \theta = \frac{\bar{y}}{\frac{16}{3} - 4} = \frac{8}{4/3}\) | A1ft | 1.1b |
| \(\theta = 47.9°\) or \(48°\) or better | A1 | 1.1b |
## 4(a)
Moments about the $y$-axis: $\int x \rho y \, dx$ | M1 | 3.1a
$\int_0^4 x \left(36 - \frac{9x^2}{4}\right) dx = k\left[36 - \frac{9x^2}{4}\right] = 32$ | M1 | 2.1
$\left[\frac{4}{27}\left(36 - \frac{9x^2}{4}\right)^{3/2}\right]_0^4 = 32$ | A1 | 1.1b
$\bar{x} = \frac{\int x \rho y \, dx}{\int \rho y \, dx}$ | DM1 | 3.1a
$\bar{x} = \frac{32}{6} = \frac{16}{3}$ | A1* | 2.2a
**Notes:**
M1: Correct method for moments about the $y$-axis: $\int \frac{1}{2}\rho(y) x^2 \, dy$ or $\int \rho(x) x y \, dx$. Integrand should be in one variable only.
M1: Integrate to obtain $k\left[A - Bx^2\right]^{3/2}$. Ignore limits and/or constant of integration
A1: Correct integration with correct limits seen or implied.
DM1: Complete method to obtain $\bar{x}$. Dependent on first M1.
A1*: Obtain given answer from correct working
## 4(b)
Moments about the $x$-axis: $\frac{1}{2}\int \rho y^2 \, dx$ or $\int \frac{1}{2}\rho(y) \left(36 - \frac{9x^2}{4}\right) dx$ | M1 | 3.1a
$\frac{1}{2}\left[36x - \frac{9x^4}{4}\right]_0^4 = \frac{1}{2}(144 - 144) = 48$ | A1 | 1.1b
$\bar{y} = \frac{\int \frac{1}{2}\rho y^2 \, dx}{32} = \frac{48}{6}$ | DM1 | 2.1
$\bar{y} = 8$ | A1 | 2.2a
**Notes:**
M1: Correct method for moments about the $x$-axis: $\int \frac{1}{2}\rho(x) y^2 \, dx$ or $\int \rho(y) y \, dy$
A1: Correct integration with correct limits seen or implied.
DM1: Complete method to obtain $\bar{y}$. Dependent on previous M1.
A1: Correct exact equivalent
## 4(c)
Correct use of trigonometry | M1 | 3.1a
$\tan \theta = \frac{\bar{y}}{\frac{16}{3} - 4} = \frac{8}{4/3}$ | A1ft | 1.1b
$\theta = 47.9°$ or $48°$ or better | A1 | 1.1b
**Notes:**
M1: Correct use of trigonometry to find a relevant angle
A1ft: Correct unsimplified expression for $\tan \theta$ or its reciprocal. Follow through their $\bar{y}$
A1: $48°$ or better ($47.8823\ldots°$). $0.836$ radians is A0
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c14975b7-6afa-44ce-beab-1cba2e82b249-14_675_528_242_772}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A uniform lamina $O A B$ is in the shape of the region $R$.\\
Region $R$ lies in the first quadrant and is bounded by the curve with equation $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 36 } = 1$, the $x$-axis, and the $y$-axis, as shown shaded in Figure 3.
The point $A$ is the point of intersection of the curve and the $x$-axis.\\
The point $B$ is the point of intersection of the curve and the $y$-axis.\\
One unit on each axis represents 1 m .\\
The area of $R$ is $6 \pi$\\
The centre of mass of $R$ lies at the point with coordinates $( \bar { x } , \bar { y } )$
\begin{enumerate}[label=(\alph*)]
\item Use algebraic integration to show that $\bar { x } = \frac { 16 } { 3 \pi }$
\item Use algebraic integration to find the exact value of $\bar { y }$
The lamina is freely suspended from $A$ and hangs in equilibrium with $O A$ at angle $\theta ^ { \circ }$ to the downward vertical.
\item Find the value of $\theta$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2024 Q4 [12]}}