| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of composite lamina |
| Difficulty | Standard +0.3 This is a standard composite lamina problem requiring knowledge of standard results (semicircle COM at 4r/3π from diameter, triangle COM at centroid) and systematic application of the composite body formula. The 'show' part in (i) is straightforward by symmetry, and (ii) involves routine calculation with given numerical values. Slightly above average due to being Further Maths content and requiring careful coordinate setup, but follows a well-practiced method. |
| Spec | 6.04b 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 |
|---|---|---|
| \(0.36 \times (1.2/3) - \pi \times 0.6^2/2 \times2 \times 0.6/(3\pi/2)\) | M1 | Table of moments idea |
| \(= \text{distance} \times \text{total area}\) | M1 | |
| 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**
$0.36 \times (1.2/3) - \pi \times 0.6^2/2 \times2 \times 0.6/(3\pi/2)$ | M1 | Table of moments idea
$= \text{distance} \times \text{total area}$ | M1 |
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] |6\\
\includegraphics[max width=\textwidth, alt={}, center]{2c6b2e42-09cb-4653-9378-6c6add7771cc-3_582_862_577_644}
A uniform lamina $O A B C D$ consists of a semicircle $B C D$ with centre $O$ and radius 0.6 m and an isosceles triangle $O A B$, joined along $O B$ (see diagram). The triangle has area $0.36 \mathrm {~m} ^ { 2 }$ and $A B = A O$.\\
(i) Show that the centre of mass of the lamina lies on $O B$.\\
(ii) Calculate the distance of the centre of mass of the lamina from $O$.
\hfill \mbox{\textit{CAIE M2 2012 Q6 [8]}}