Standard +0.3 This is a standard centre of mass problem for a composite body requiring knowledge of the standard result for a semicircular wire (distance 2r/π from diameter) and application of the composite centre of mass formula. It involves straightforward substitution into formulas with no geometric complications or novel problem-solving required, making it slightly easier than average.
1
\includegraphics[max width=\textwidth, alt={}, center]{ae809dfc-c5af-4c0a-9c88-009949d3e9f9-2_618_441_253_852}
A frame consists of a uniform semicircular wire of radius 20 cm and mass 2 kg , and a uniform straight wire of length 40 cm and mass 0.9 kg . The ends of the semicircular wire are attached to the ends of the straight wire (see diagram). Find the distance of the centre of mass of the frame from the straight wire.
$c$ of $m$ of arc $= 20\sin(\pi/2)/(\pi/2)$ | $B1, M1$ | For attempting to take moments about the diameter
$(2 + 0.9)\bar{x} = 2 \times 20\sin(\pi/2)/(\pi/2)$
Distance is 8.78cm | $A1, A1$ [4] |
---
1\\
\includegraphics[max width=\textwidth, alt={}, center]{ae809dfc-c5af-4c0a-9c88-009949d3e9f9-2_618_441_253_852}
A frame consists of a uniform semicircular wire of radius 20 cm and mass 2 kg , and a uniform straight wire of length 40 cm and mass 0.9 kg . The ends of the semicircular wire are attached to the ends of the straight wire (see diagram). Find the distance of the centre of mass of the frame from the straight wire.
\hfill \mbox{\textit{CAIE M2 2010 Q1 [4]}}