| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina suspended in equilibrium |
| Difficulty | Challenging +1.2 This is a multi-part centre of mass problem requiring composite body techniques, algebraic manipulation to verify a given formula, solving a transcendental equation numerically, and applying equilibrium conditions. While it involves several steps and the composite lamina setup is moderately sophisticated, each individual technique (standard centre of mass formulas for semicircles, subtraction method, equilibrium geometry) is well-practiced in M2. The 'show that' format reduces problem-solving demand since the target expressions guide the work. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(2(1)\sin(\pi/2) \times \pi(1^2)/2 = \frac{2r\sin(\pi/2)}{3\pi/2} \times \pi(r^2)/2 + OG(\pi/2 - \pi r^2/2)\) | M1 | Uses table of values or moment equation |
| \(OG = 4\pi(1 - r^3)/3(1 - r^2)\) | A1 | Correct moment equation |
| \(OG = 4(1 + r + r^2)/3\pi(1 + r)\) | AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = 4(1 + r + r^2)/3\pi(1 + r)\) | M1* | Sets \(r\) = answer(i) |
| \((3\pi - 4)r^2 + (3\pi - 4)r - 4 = 0\) | D* M1 | Sets up and starts solving quadratic equation |
| \(r = 0.494\) | AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(OG = 0.494\) if \(G\) on arc | M1* | |
| \(OG = 4(1 + 0.494 + 0.494^2)/3\pi(1 + 0.494)\) | D* M1 | Substitutes \(AG\) in \(OG\) expression |
| \(= 0.4937\) | AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = 0.494/1\) | M1 | |
| \(\theta = 26.3°\) | A1 | [2] |
**(i)**
$2(1)\sin(\pi/2) \times \pi(1^2)/2 = \frac{2r\sin(\pi/2)}{3\pi/2} \times \pi(r^2)/2 + OG(\pi/2 - \pi r^2/2)$ | M1 | Uses table of values or moment equation
$OG = 4\pi(1 - r^3)/3(1 - r^2)$ | A1 | Correct moment equation
$OG = 4(1 + r + r^2)/3\pi(1 + r)$ | AG | A1 | [4] | Must use $1 - r$ as a factor of $1 - r^3$ and $1 - r^2$
**(ii)**
$r = 4(1 + r + r^2)/3\pi(1 + r)$ | M1* | Sets $r$ = answer(i)
$(3\pi - 4)r^2 + (3\pi - 4)r - 4 = 0$ | D* M1 | Sets up and starts solving quadratic equation
$r = 0.494$ | AG | A1 | [3]
OR
$OG = 0.494$ if $G$ on arc | M1* |
$OG = 4(1 + 0.494 + 0.494^2)/3\pi(1 + 0.494)$ | D* M1 | Substitutes $AG$ in $OG$ expression
$= 0.4937$ | AG | A1 |
**(iii)**
$\tan\theta = 0.494/1$ | M1 |
$\theta = 26.3°$ | A1 | [2] | [9]6\\
\includegraphics[max width=\textwidth, alt={}, center]{09971be0-73b6-4c73-8dfd-c89ff877950a-3_451_775_255_685}
The diagram shows a uniform lamina $A B C D E F$, formed from a semicircle with centre $O$ and radius 1 m by removing a semicircular part with centre $O$ and radius $r \mathrm {~m}$.\\
(i) Show that the distance in metres of the centre of mass of the lamina from $O$ is
$$\frac { 4 \left( 1 + r + r ^ { 2 } \right) } { 3 \pi ( 1 + r ) } .$$
The centre of mass of the lamina lies on the $\operatorname { arc } A B C$.\\
(ii) Show that $r = 0.494$, correct to 3 significant figures.
The lamina is freely suspended at $F$ and hangs in equilibrium.\\
(iii) Find the angle between the diameter of the lamina and the vertical.
\hfill \mbox{\textit{CAIE M2 2012 Q6 [9]}}