| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Equilibrium with applied force |
| Difficulty | Standard +0.8 This is a multi-part M2 question requiring: (i) composite centre of mass calculation with rectangle and semicircle, (ii) equilibrium analysis with limiting friction cases (about to topple at two different edges). The toppling equilibrium with applied horizontal force requires careful moment analysis and consideration of two critical cases, which is more sophisticated than standard centre of mass problems. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2(8)\sin(\pi/2)/(3\pi/2)\) | B1 | CoM of semi-circle \((3.395305\ldots)\); \((32/3\pi)\) |
| M1 | Table of values idea to get an equation/expression, using any fixed axis | |
| \((80 + 32\pi)\, x_G\) | A1 | |
| \(= 80(2) + 32\pi(4 + \text{cv(CoM)})\) | A1ft | Relative to the axis they are using; ft their CoM |
| \(x_G = 5(.00) \text{ cm}\) | A1 | \((5.004444492 \text{ cm})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 | Moments about E or A; allow wrong distance for either force | |
| \(4(20) = W(x_G \text{ from (i)} - 4)\) | A1ft | ft on \(\text{cv}(x_G)\); \(mg\) OK here; if \(x_G < 4\) need \((4 - x_G \text{ from (i)})\) |
| \(4(20) = W(x_G \text{ from (i)})\) | A1ft | ft on \(\text{cv}(x_G)\); \(mg\) OK here |
| (Greatest) \(W = 80 \text{ N}\) | A1 | 79.64601393 from exact \(x_G\); allow anything between 79.6 and 80 |
| (Least) \(W = 16 \text{ N}\) | A1 | 15.98579026 from exact \(x_G\) |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2(8)\sin(\pi/2)/(3\pi/2)$ | B1 | CoM of semi-circle $(3.395305\ldots)$; $(32/3\pi)$ |
| | M1 | Table of values idea to get an equation/expression, using any fixed axis |
| $(80 + 32\pi)\, x_G$ | A1 | |
| $= 80(2) + 32\pi(4 + \text{cv(CoM)})$ | A1ft | Relative to the axis they are using; ft their CoM |
| $x_G = 5(.00) \text{ cm}$ | A1 | $(5.004444492 \text{ cm})$ |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | Moments about E or A; allow wrong distance for either force |
| $4(20) = W(x_G \text{ from (i)} - 4)$ | A1ft | ft on $\text{cv}(x_G)$; $mg$ OK here; if $x_G < 4$ need $(4 - x_G \text{ from (i)})$ |
| $4(20) = W(x_G \text{ from (i)})$ | A1ft | ft on $\text{cv}(x_G)$; $mg$ OK here |
| (Greatest) $W = 80 \text{ N}$ | A1 | 79.64601393 from exact $x_G$; allow anything between 79.6 and 80 |
| (Least) $W = 16 \text{ N}$ | A1 | 15.98579026 from exact $x_G$ |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{8492ec9b-3327-4d89-aaa4-bf98cdf0ebdc-2_721_513_1260_762}
A uniform solid prism has cross-section $A B C D E$ in the shape of a rectangle measuring 20 cm by 4 cm joined to a semicircle of radius 8 cm as shown in the diagram. The centre of mass of the solid lies in this cross-section.\\
(i) Find the distance of the centre of mass of the solid from $A B$.
The solid is placed with $A E$ on rough horizontal ground (so the object does not slide) and is in equilibrium with a horizontal force of magnitude 4 N applied along $C B$.\\
(ii) Find the greatest and least possible values for the weight of the solid.
\hfill \mbox{\textit{OCR M2 2015 Q4 [10]}}