| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with attached triangle |
| Difficulty | Standard +0.3 This is a standard centre of mass problem requiring students to find the centroid of a composite shape using symmetry arguments and the formula for combining centres of mass. Part (i) uses symmetry (both components have centres of mass on OB), and part (ii) applies the standard weighted average formula with known results for semicircle and triangle centroids. It's slightly easier than average as it's a textbook application with no novel insight required. |
| Spec | 6.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Height of triangle \(= 0.36/0.3(= 1.2 \, \text{m})\) | B1 | |
| Semi-circle C of M \(= 2 \times 0.6/(3\pi/2)\) | B1 | Centre of mass lamina from BOD |
| \(0.36 \times (1.2/3) = \pi \times 0.6^2/2 \times 2 \times 0.6/(3\pi/2)\) | M1 | Equating moments idea |
| \(0.144 = 0.144\) | A1 [4] | Evidence of checking equality |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.36 \times (1.2/3) - \pi \times 0.6^2/2 \times 2 \times 0.6/(3\pi/2)\) | M1 | Table of moments idea |
| \(= \text{distance} \times \text{total area}\) | ||
| Distance \(= 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.36 \times 0.3\) | A1 | Correct sum of parts |
| \(= (0.36 + \pi \times 0.6^2/2) \times OG\) | A1 | Correct moment of whole |
| \(OG = 0.117 \, \text{m}\) | A1 [4] |
## (i)
Height of triangle $= 0.36/0.3(= 1.2 \, \text{m})$ | B1 |
Semi-circle C of M $= 2 \times 0.6/(3\pi/2)$ | B1 | Centre of mass lamina from BOD
$0.36 \times (1.2/3) = \pi \times 0.6^2/2 \times 2 \times 0.6/(3\pi/2)$ | M1 | Equating moments idea
$0.144 = 0.144$ | A1 [4] | Evidence of checking equality
**OR**
$0.36 \times (1.2/3) - \pi \times 0.6^2/2 \times 2 \times 0.6/(3\pi/2)$ | M1 | Table of moments idea
$= \text{distance} \times \text{total area}$ | |
Distance $= 0$ | A1 |
## (ii)
$0.36 \times 0.3$ | A1 | Correct sum of parts
$= (0.36 + \pi \times 0.6^2/2) \times OG$ | A1 | Correct moment of whole
$OG = 0.117 \, \text{m}$ | A1 [4] |\includegraphics{figure_6}
A uniform lamina $OABCD$ consists of a semicircle $BCD$ with centre $O$ and radius $0.6$ m and an isosceles triangle $OAB$, joined along $OB$ (see diagram). The triangle has area $0.36$ m$^2$ and $AB = AO$.
\begin{enumerate}[label=(\roman*)]
\item Show that the centre of mass of the lamina lies on $OB$. [4]
\item Calculate the distance of the centre of mass of the lamina from $O$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2012 Q6 [8]}}