| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Challenging +1.2 This is a 3D moments problem requiring knowledge that the center of mass of a hemisphere is at 3r/8 from the base, then applying equilibrium conditions (sum of forces and moments). While it involves multiple steps and 3D visualization, the techniques are standard M2 material with straightforward geometry and moment calculations once the setup is understood. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(d = 2.25\) | B1 | 3/8x6 OG (be generous) |
| \(h = 1.125\) or \(1.12\) or \(1.13\) or \(9/8\) | B1 | 2 marks total; horizontal distance |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(T_1 + T_2 = 12\) resolving vertically | M1 | if not then next M1 ok |
| \(T_1 \times 6\cos30° = 12xh\) (their \(h\)) | M1 | or \(\text{mom}(A)T_2 \times 6\cos30° = 12(6\cos30° - h)\) |
| \(\text{mom}(O)\) their \(h\) ok for A1 | A1 | |
| \(T_1 = 2.60\) N or \(3\sqrt{3}/2\) | A1 | or \(T_2 = 9.40\) |
| \(T_2 = 9.40\) N \(\checkmark(12 - T_1)\) | A1\(\checkmark\) | 5 marks total; or \(T_1=2.60\) or \(\checkmark(12-T_2)\) |
| above \(\checkmark\) depends on at least one of the M marks \((T_s > 0)\) | 7 marks total |
# Question 3:
## Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $d = 2.25$ | B1 | 3/8x6 OG (be generous) |
| $h = 1.125$ or $1.12$ or $1.13$ or $9/8$ | B1 | 2 marks total; horizontal distance |
## Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $T_1 + T_2 = 12$ resolving vertically | M1 | if not then next M1 ok |
| $T_1 \times 6\cos30° = 12xh$ (their $h$) | M1 | or $\text{mom}(A)T_2 \times 6\cos30° = 12(6\cos30° - h)$ |
| $\text{mom}(O)$ their $h$ ok for A1 | A1 | |
| $T_1 = 2.60$ N or $3\sqrt{3}/2$ | A1 | or $T_2 = 9.40$ |
| $T_2 = 9.40$ N $\checkmark(12 - T_1)$ | A1$\checkmark$ | 5 marks total; or $T_1=2.60$ or $\checkmark(12-T_2)$ |
| above $\checkmark$ depends on at least one of the M marks $(T_s > 0)$ | | 7 marks total |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{d6d87705-be4b-407d-b699-69fb441d88a7-2_710_572_721_788}
A uniform solid hemisphere of weight 12 N and radius 6 cm is suspended by two vertical strings. One string is attached to the point $O$, the centre of the plane face, and the other string is attached to the point $A$ on the rim of the plane face. The hemisphere hangs in equilibrium and $O A$ makes an angle of $60 ^ { \circ }$ with the vertical (see diagram).\\
(i) Find the horizontal distance from the centre of mass of the hemisphere to the vertical through $O$.\\
(ii) Calculate the tensions in the strings.
\hfill \mbox{\textit{OCR M2 2006 Q3 [7]}}