| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina in equilibrium with applied force |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass problem requiring recall of the semicircle centroid formula (4r/3π), basic trigonometry for the hanging angle, and a straightforward moments equation about point A. All three parts follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(OG = (0.1061) = 0.106\) m | B1 [1] | \(OG = (2 \times 0.25\sin\pi/2)/(3\pi/2)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\theta = 0.1061/0.25\) | M1 | Candidate's OG |
| \(\theta = 23(.0)°\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | Takes moments about A | |
| \(0.1061W = (6\cos45) \times (2 \times 0.25)\) | A1ft | ft \(\text{cv}(OG(\text{i}))\) |
| \(W = 20(.0)\) N | A1 [3] [6] |
## Question 2:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $OG = (0.1061) = 0.106$ m | B1 [1] | $OG = (2 \times 0.25\sin\pi/2)/(3\pi/2)$ |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\theta = 0.1061/0.25$ | M1 | Candidate's OG |
| $\theta = 23(.0)°$ | A1 [2] | |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | Takes moments about A |
| $0.1061W = (6\cos45) \times (2 \times 0.25)$ | A1ft | ft $\text{cv}(OG(\text{i}))$ |
| $W = 20(.0)$ N | A1 [3] [6] | |
---2 A uniform semicircular lamina of radius 0.25 m has diameter $A B$. It is freely suspended at $A$ from a fixed point and hangs in equilibrium.\\
(i) Find the distance of the centre of mass of the lamina from the diameter $A B$.\\
(ii) Calculate the angle which the diameter $A B$ makes with the vertical.
The lamina is now held in equilibrium with the diameter $A B$ vertical by means of a force applied at $B$. This force has magnitude 6 N and acts at $45 ^ { \circ }$ to the upward vertical in the plane of the lamina.\\
(iii) Calculate the weight of the lamina.
\hfill \mbox{\textit{CAIE M2 2013 Q2 [6]}}