| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.3 This is a standard two-part centre of mass question requiring composite shapes (square + triangle), basic geometry to find individual centroids, and a straightforward equilibrium angle calculation using tan θ = horizontal/vertical distance. All techniques are routine for M2 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| i | \(-(8\cos 30/3)(8^2 \sin 60/2)\) | M1 |
| \(+ (4)(8^2)\) | A1 | \(-2.309 \times 27.7\) |
| \(= (8^2 + 8^2 \sin 60/2)(x_G)\) | A1 | |
| \(x_G = 2.09 \text{ cm}\) | A1 | |
| A1 | ||
| [5] | ||
| ii | \(\tan \theta = (2.09/4)\) | M1 |
| \(\theta = 27.6°\) | A1ft | \(ft cv(x_G)\) |
| [2] |
| **Part** | **Answer/Working** | **Marks** | **Guidance** |
|----------|-------------------|----------|-------------|
| i | $-(8\cos 30/3)(8^2 \sin 60/2)$ | M1 | Table of moments idea, may include $g$ and/or density. |
| | $+ (4)(8^2)$ | A1 | $-2.309 \times 27.7$ |
| | $= (8^2 + 8^2 \sin 60/2)(x_G)$ | A1 | |
| | $x_G = 2.09 \text{ cm}$ | A1 | |
| | | A1 | |
| | | [5] | |
| ii | $\tan \theta = (2.09/4)$ | M1 | |
| | $\theta = 27.6°$ | A1ft | $ft cv(x_G)$ |
| | | [2] | |3 A uniform lamina $A B C D E$ consists of a square $A C D E$ and an equilateral triangle $A B C$ which are joined along their common edge $A C$ to form a pentagon whose sides are each 8 cm in length.\\
(i) Calculate the distance of the centre of mass of the lamina from $A C$.\\
(ii) The lamina is freely suspended from $A$ and hangs in equilibrium. Calculate the angle that $A C$ makes with the vertical.
\hfill \mbox{\textit{OCR M2 2011 Q3 [7]}}