| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with removed circle/semicircle |
| Difficulty | Standard +0.8 This is a multi-part centre of mass problem requiring decomposition of a composite shape (rectangle with two semicircles), application of the given formula, algebraic manipulation to reach a specific form, and then equilibrium analysis with suspended lamina. The 7-mark part (a) demands careful coordinate setup and systematic calculation, while part (b) requires geometric reasoning about equilibrium. More demanding than standard M2 questions but follows established techniques without requiring novel insight. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Rectangle | Semicircles | Template, T |
| \(24x\) | \(4.5\pi\) | \(4.5\pi\) |
| \(x\) | \(\frac{4 \times 3}{3\pi}\) | \(\frac{4 \times 3}{3\pi}\) |
| \(24x^2 - 4.5\pi \times \left(\frac{4 \times 3}{3\pi}\right) - 4.5\pi \times \left(\frac{4 \times 3}{3\pi}\right) = (24x + 9\pi)\bar{x}\) | M1 A1 | |
| \(\text{distance} = | \bar{x} | = \frac{4 |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 2\), \( | \bar{x} | = \frac{20}{16 + 3\pi}\) |
| \(\tan\theta = \frac{6}{4 - | \bar{x} | } = \frac{6}{4 - \frac{20}{16 + 3\pi}}\) |
| \(= \frac{48 + 9\pi}{22 + 6\pi}\) | A1 (4) |
## (a)
| Rectangle | Semicircles | Template, T |
|-----------|-------------|------------|
| $24x$ | $4.5\pi$ | $4.5\pi$ | $24x + 9\pi$ | B2 |
| $x$ | $\frac{4 \times 3}{3\pi}$ | $\frac{4 \times 3}{3\pi}$ | $\bar{x}$ | B2 |
$24x^2 - 4.5\pi \times \left(\frac{4 \times 3}{3\pi}\right) - 4.5\pi \times \left(\frac{4 \times 3}{3\pi}\right) = (24x + 9\pi)\bar{x}$ | M1 A1 |
$\text{distance} = |\bar{x}| = \frac{4|2x^2 - 3|}{(8x + 3\pi)}$ **\*\*** | A1 (7) |
## (b)
When $x = 2$, $|\bar{x}| = \frac{20}{16 + 3\pi}$ | B1 |
$\tan\theta = \frac{6}{4 - |\bar{x}|} = \frac{6}{4 - \frac{20}{16 + 3\pi}}$ | M1 A1 |
$= \frac{48 + 9\pi}{22 + 6\pi}$ | A1 (4) |
**Total: [11]**
---[The centre of mass of a semi-circular lamina of radius $r$ is $\frac{4r}{3\pi}$ from the centre]
\includegraphics{figure_3}
A template $T$ consists of a uniform plane lamina $PQRQS$, as shown in Figure 3. The lamina is bounded by two semicircles, with diameters $SQ$ and $QR$, and by the sides $SP$, $PQ$ and $QR$ of the rectangle $PQRS$. The point $O$ is the mid-point of $SR$, $PQ = 12$ cm and $QR = 2$ cm.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of $T$ is a distance $\frac{4|2x^2 - 3|}{8x + 3\pi}$ cm from $SR$.
[7]
\end{enumerate}
The template $T$ is freely suspended from the point $P$ and hangs in equilibrium.
Given that $x = 2$ and that $\theta$ is the angle that $PQ$ makes with the horizontal,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item show that $\tan \theta = \frac{48 + 9\pi}{22 + 6\pi}$.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2010 Q7 [11]}}