Moderate -0.5 This is a straightforward application of circular motion kinematics (v = rω) combined with basic center of mass location for a right-angled isosceles triangle. Students need to find the distance from the rotation axis to the center of mass (which is 4 cm from the midpoint of the hypotenuse) and multiply by the angular speed. The geometry is standard and the calculation is direct, making this slightly easier than average but still requiring proper understanding of the setup.
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\includegraphics[max width=\textwidth, alt={}, center]{fcf239a6-6558-43ec-b404-70aa349af6a9-2_373_552_260_799}
A uniform isosceles triangular lamina \(A B C\) is right-angled at \(B\). The length of \(A C\) is 24 cm . The lamina rotates in a horizontal plane, about a vertical axis through the mid-point of \(A C\), with angular speed \(5 \mathrm { rad } \mathrm { s } ^ { - 1 }\) (see diagram). Find the speed with which the centre of mass of the lamina is moving. [0pt]
[3]
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\includegraphics[max width=\textwidth, alt={}, center]{fcf239a6-6558-43ec-b404-70aa349af6a9-2_373_552_260_799}
A uniform isosceles triangular lamina $A B C$ is right-angled at $B$. The length of $A C$ is 24 cm . The lamina rotates in a horizontal plane, about a vertical axis through the mid-point of $A C$, with angular speed $5 \mathrm { rad } \mathrm { s } ^ { - 1 }$ (see diagram). Find the speed with which the centre of mass of the lamina is moving.\\[0pt]
[3]
\hfill \mbox{\textit{CAIE M2 2002 Q1 [3]}}