CAIE M2 2004 June — Question 2 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2004
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeCentre of mass of composite lamina
DifficultyStandard +0.3 This is a standard composite lamina centre of mass problem requiring knowledge of the centroid formula for a circular sector (20/3π from the straight edge) and combining two shapes using the standard formula. The first part is essentially given (show that), and the second part is straightforward application of the composite body formula with no geometric complications. Slightly easier than average due to the guided structure.
Spec6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass

2 \includegraphics[max width=\textwidth, alt={}, center]{835616aa-0b2b-4e8c-bbbf-60b72dc5ea3e-2_291_732_822_708} A uniform lamina \(A B C D E\) consists of a rectangular part with sides 5 cm and 10 cm , and a part in the form of a quarter of a circle of radius 5 cm , as shown in the diagram.
  1. Show that the distance of the centre of mass of the part \(C D E\) of the lamina is \(\frac { 20 } { 3 \pi } \mathrm {~cm}\) from \(C E\).
  2. Find the distance of the centre of mass of the lamina \(A B C D E\) from the edge \(A B\).

Question 2:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Evaluates \(\dfrac{2r\sin\alpha}{3\alpha} \times \cos\dfrac{\pi}{4}\)M1
Obtains given answer correctlyA1 2 marks total
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
For taking moments about \(AB\)M1
\(\left(5 \times 10 + \dfrac{1}{4}\pi 5^2\right)\bar{x} = (5 \times 10) \times 5 + \dfrac{1}{4}\pi 5^2\left(10 + \dfrac{20}{3\pi}\right)\)
For the total area correct and the moment of the rectangle correct (unsimplified)A1
For the moment of \(CDE\) correct (unsimplified)A1
Distance is \(7.01\) cmA1 4 marks total
## Question 2:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Evaluates $\dfrac{2r\sin\alpha}{3\alpha} \times \cos\dfrac{\pi}{4}$ | M1 | |
| Obtains given answer correctly | A1 | **2 marks total** |

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| For taking moments about $AB$ | M1 | |
| $\left(5 \times 10 + \dfrac{1}{4}\pi 5^2\right)\bar{x} = (5 \times 10) \times 5 + \dfrac{1}{4}\pi 5^2\left(10 + \dfrac{20}{3\pi}\right)$ | | |
| For the total area correct and the moment of the rectangle correct (unsimplified) | A1 | |
| For the moment of $CDE$ correct (unsimplified) | A1 | |
| Distance is $7.01$ cm | A1 | **4 marks total** |

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2\\
\includegraphics[max width=\textwidth, alt={}, center]{835616aa-0b2b-4e8c-bbbf-60b72dc5ea3e-2_291_732_822_708}

A uniform lamina $A B C D E$ consists of a rectangular part with sides 5 cm and 10 cm , and a part in the form of a quarter of a circle of radius 5 cm , as shown in the diagram.\\
(i) Show that the distance of the centre of mass of the part $C D E$ of the lamina is $\frac { 20 } { 3 \pi } \mathrm {~cm}$ from $C E$.\\
(ii) Find the distance of the centre of mass of the lamina $A B C D E$ from the edge $A B$.

\hfill \mbox{\textit{CAIE M2 2004 Q2 [6]}}