5.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{4c8ebad3-0ebb-4dfe-8036-54b651deb9cf-08_781_541_223_687}
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\caption{Figure 2}
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A uniform lamina \(A B C D\) is made by joining a uniform triangular lamina \(A B D\) to a uniform semi-circular lamina \(D B C\), of the same material, along the edge \(B D\), as shown in Figure 2. Triangle \(A B D\) is right-angled at \(D\) and \(A D = 18 \mathrm {~cm}\). The semi-circle has diameter \(B D\) and \(B D = 12 \mathrm {~cm}\).
- Show that, to 3 significant figures, the distance of the centre of mass of the lamina \(A B C D\) from \(A D\) is 4.69 cm .
Given that the centre of mass of a uniform semicircular lamina, radius \(r\), is at a distance \(\frac { 4 r } { 3 \pi }\) from the centre of the bounding diameter,
- find, in cm to 3 significant figures, the distance of the centre of mass of the lamina \(A B C D\) from \(B D\).
The lamina is freely suspended from \(B\) and hangs in equilibrium.
- Find, to the nearest degree, the angle which \(B D\) makes with the vertical.