CAIE M2 2012 November — Question 1 3 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2012
SessionNovember
Marks3
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TopicCircular Motion 1
TypeCentre of mass of rotating body
DifficultyModerate -0.8 This is a straightforward application of v = rω requiring only recall of the center of mass position for a semicircular arc (2r/π from the diameter) and the basic circular motion formula. The calculation involves minimal steps and no problem-solving insight beyond standard bookwork.
Spec6.04a Centre of mass: gravitational effect6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r
\(ABC\) is a uniform semicircular arc with diameter \(AC = 0.5\) m. The arc rotates about a fixed axis through \(A\) and \(C\) with angular speed \(2.4\) rad s\(^{-1}\). Calculate the speed of the centre of mass of the arc. [3]
AnswerMarks Guidance
\(OG = 0.25 \sin(\pi/2)/(\pi/2)\)B1 0.159 (15..)
\(v = 0.159 \times 2.4\)M1
\(v = 0.382 \, \text{ms}^{-1}\)A1* [3] \(2.4 \times cv\) (OG) 
$OG = 0.25 \sin(\pi/2)/(\pi/2)$ | B1 | 0.159 (15..)
$v = 0.159 \times 2.4$ | M1 |
$v = 0.382 \, \text{ms}^{-1}$ | A1* [3] | $2.4 \times cv$ (OG)
$ABC$ is a uniform semicircular arc with diameter $AC = 0.5$ m. The arc rotates about a fixed axis through $A$ and $C$ with angular speed $2.4$ rad s$^{-1}$. Calculate the speed of the centre of mass of the arc.
[3]

\hfill \mbox{\textit{CAIE M2 2012 Q1 [3]}}
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