CAIE M2 2012 June — Question 2 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2012
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeLamina suspended in equilibrium
DifficultyStandard +0.3 This is a standard two-part centre of mass problem requiring (i) calculation using known formulae for semicircular lamina (4r/3π) and arc (2r/π) centres of mass, then combining using weighted averages, and (ii) applying equilibrium condition tan(θ) = x̄/ȳ. While it involves multiple steps and careful coordinate work, it follows a well-established template with no novel insight required, making it slightly easier than average.
Spec6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

2 \includegraphics[max width=\textwidth, alt={}, center]{6d3892e0-8c88-44ec-940f-c526d71a7fc6-2_481_412_440_865} The diagram shows a circular object formed from a uniform semicircular lamina of weight 11 N and a uniform semicircular arc of weight 9 N . The lamina and the arc both have centre \(O\) and radius 0.7 m and are joined at the ends of their common diameter \(A B\).
  1. Show that the distance of the centre of mass of the object from \(O\) is 0.0371 m , correct to 3 significant figures. The object hangs in equilibrium, freely suspended at \(A\).
  2. Find the angle between \(A B\) and the vertical and state whether the lowest point of the object is on the lamina or on the arc.

Question 2:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((9+11)OG = \pm[9 \times 0.7/(\pi/2) - 11 \times (2 \times 0.7)/3\pi/2)]\)M1, A1 Table of value idea with signs either way round
\(OG = 0.0371\text{ m}\) AGA1 [3] Accept –ve answer
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\tan\theta = 0.0371(36..)/0.7\)M1
\(\theta = 3.0°\)A1
LaminaB1 [3]
## Question 2:

### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(9+11)OG = \pm[9 \times 0.7/(\pi/2) - 11 \times (2 \times 0.7)/3\pi/2)]$ | M1, A1 | Table of value idea with signs either way round |
| $OG = 0.0371\text{ m}$ AG | A1 [3] | Accept –ve answer |

### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\theta = 0.0371(36..)/0.7$ | M1 | |
| $\theta = 3.0°$ | A1 | |
| Lamina | B1 [3] | |

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2\\
\includegraphics[max width=\textwidth, alt={}, center]{6d3892e0-8c88-44ec-940f-c526d71a7fc6-2_481_412_440_865}

The diagram shows a circular object formed from a uniform semicircular lamina of weight 11 N and a uniform semicircular arc of weight 9 N . The lamina and the arc both have centre $O$ and radius 0.7 m and are joined at the ends of their common diameter $A B$.\\
(i) Show that the distance of the centre of mass of the object from $O$ is 0.0371 m , correct to 3 significant figures.

The object hangs in equilibrium, freely suspended at $A$.\\
(ii) Find the angle between $A B$ and the vertical and state whether the lowest point of the object is on the lamina or on the arc.

\hfill \mbox{\textit{CAIE M2 2012 Q2 [6]}}