| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with removed triangle/rectangle/square |
| Difficulty | Standard +0.8 Part (a) requires setting up and evaluating a double integral for centre of mass of a triangular lamina with appropriate coordinate system and limits, which is a standard M3 calculus proof but involves careful integration. Part (b) requires composite body techniques with circular sectors removed, involving sector centre of mass formula and algebraic manipulation to reach the given answer. This is moderately challenging for M3 level, requiring multiple techniques and careful bookkeeping, but follows standard methods without requiring novel insight. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
\includegraphics{figure_3}
Figure 3 shows a uniform equilateral triangular lamina $PRT$ with sides of length $2a$.
\begin{enumerate}[label=(\alph*)]
\item Using calculus, prove that the centre of mass of $PRT$ is at a distance $\frac{2\sqrt{3}}{3}a$ from $R$. [6]
\end{enumerate}
\includegraphics{figure_4}
The circular sector $PQU$, of radius $a$ and centre $P$, and the circular sector $TUS$, of radius $a$ and centre $T$, are removed from $PRT$ to form the uniform lamina $QRSU$ shown in Figure 4.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the distance of the centre of mass of $QRSU$ from $U$ is $\frac{2a}{3\sqrt{3} - \pi}$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2012 Q6 [12]}}