Moderate -0.3 This is a straightforward centre of mass problem requiring recall of the standard result that a semicircular wire's centre of mass is at distance 2r/π from its diameter, then applying the composite centre of mass formula with two components. It's slightly easier than average as it's a direct application of known formulas with clear geometry and minimal calculation steps.
\includegraphics{figure_1}
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.
[4]
\(c \text{ of } m \text{ of arc} = 20\sin(\pi/2)/(\pi/2)\)
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]
$c \text{ of } m \text{ of arc} = 20\sin(\pi/2)/(\pi/2)$ | 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] |
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\includegraphics{figure_1}
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.
[4]
\hfill \mbox{\textit{CAIE M2 2010 Q1 [4]}}