| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2003 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with cone and cylinder |
| Difficulty | Standard +0.3 This is a standard M3 centre of mass question with routine integration (volume of revolution, centre of mass formula) followed by a straightforward equilibrium problem using moments. All techniques are textbook exercises requiring no novel insight, though the multi-part structure and 16 total marks make it slightly above average length. |
| Spec | 3.04b Equilibrium: zero resultant moment and force4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = \pi\int y^2\,dx = \frac{1}{4}\pi\int(x-2)^4\,dx\) | M1 | |
| \(\int(x-2)^4\,dx = \frac{1}{5}(x-2)^5\) | M1 A1 | |
| \(V = \frac{8\pi}{5}\) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(\pi\int xy^2\,dx = \frac{1}{4}\pi\int x(x-2)^4\,dx\) | M1 | |
| Correct strategy to integrate [e.g. substitution, expand, by parts] | M1 | |
| [e.g. \(\frac{1}{4}\pi\int(u-2)^5\,du\); \(\frac{1}{4}\pi\int(x^5 - 8x^4 + 24x^3 - 32x^2 + 16x)\,dx\)] | ||
| \(= \frac{1}{4}\pi\left[\frac{2u^5}{5} + \frac{u^6}{6}\right]\) or \(\frac{1}{4}\pi\left[\frac{x^6}{6} - \frac{8x^5}{5} + 6x^4 - \frac{32x^3}{3} + 8x^2\right]\) | M1 A1 | |
| \(= \frac{8\pi}{15}\) | (limits need to be used correctly) | A1 |
| \(V_c(\rho)\bar{x} = \pi(\rho)\int xy^2\,dx\) | (seen anywhere) | M1 |
| \(\bar{x} = \frac{4}{3}\) cm (*) | (no incorrect working seen) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Moments about \(B\): \(8A = 10W - 2W(\frac{1}{3})\) | M1 A1 A1 | |
| \(A = \frac{59W}{12}\) (4.9W) | M1 A1 | (5 marks) |
## (a)
$V = \pi\int y^2\,dx = \frac{1}{4}\pi\int(x-2)^4\,dx$ | M1 |
$\int(x-2)^4\,dx = \frac{1}{5}(x-2)^5$ | M1 A1 |
$V = \frac{8\pi}{5}$ | A1 | (4 marks)
## (b)
Using $\pi\int xy^2\,dx = \frac{1}{4}\pi\int x(x-2)^4\,dx$ | M1 |
Correct strategy to integrate [e.g. substitution, expand, by parts] | M1 |
[e.g. $\frac{1}{4}\pi\int(u-2)^5\,du$; $\frac{1}{4}\pi\int(x^5 - 8x^4 + 24x^3 - 32x^2 + 16x)\,dx$] |
$= \frac{1}{4}\pi\left[\frac{2u^5}{5} + \frac{u^6}{6}\right]$ or $\frac{1}{4}\pi\left[\frac{x^6}{6} - \frac{8x^5}{5} + 6x^4 - \frac{32x^3}{3} + 8x^2\right]$ | M1 A1 |
$= \frac{8\pi}{15}$ | (limits need to be used correctly) | A1 | (7 marks)
$V_c(\rho)\bar{x} = \pi(\rho)\int xy^2\,dx$ | (seen anywhere) | M1 |
$\bar{x} = \frac{4}{3}$ cm (*) | (no incorrect working seen) | A1 |
## (c)
Moments about $B$: $8A = 10W - 2W(\frac{1}{3})$ | M1 A1 A1 |
$A = \frac{59W}{12}$ (4.9W) | M1 A1 | (5 marks)
**(16 marks total)**\includegraphics{figure_3}
The shaded region $R$ is bounded by part of the curve with equation $y = \frac{1}{4}(x - 2)^2$, the $x$-axis and the $y$-axis, as shown in Fig. 3. The unit of length on both axes is 1 cm. A uniform solid $S$ is made by rotating $R$ through $360°$ about the $x$-axis. Using integration,
\begin{enumerate}[label=(\alph*)]
\item calculate the volume of the solid $S$, leaving your answer in terms of $\pi$, [4]
\item show that the centre of mass of $S$ is $\frac{4}{5}$ cm from its plane face. [7]
\end{enumerate}
\includegraphics{figure_4}
A tool is modelled as having two components, a solid uniform cylinder $C$ and the solid $S$. The diameter of $C$ is 4 cm and the length of $C$ is 8 cm. One end of $C$ coincides with the plane face of $S$. The components are made of different materials. The weight of $C$ is $10W$ newtons and the weight of $S$ is $2W$ newtons. The tool lies in equilibrium with its axis of symmetry horizontal on two smooth supports $A$ and $B$, which are at the ends of the cylinder, as shown in Fig. 4.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the magnitude of the force of the support $A$ on the tool. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2003 Q7 [16]}}