3. [The centre of mass of a semicircular lamina of radius \(r\) is \(\frac { 4 r } { 3 \pi }\) from the centre.]
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{88731f1c-5177-4096-841b-cd9c3f87782b-08_581_460_374_740}
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\caption{Figure 2}
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Figure 2 shows the uniform lamina \(A B C D E\), such that \(A B D E\) is a square with sides of length \(2 a\) and \(B C D\) is a semicircle with diameter \(B D\).
- Show that the distance of the centre of mass of the lamina from \(B D\) is \(\frac { 20 a } { 3 ( 8 + \pi ) }\).
The lamina is freely suspended from \(D\) and hangs in equilibrium.
- Find, to the nearest degree, the angle that \(D E\) makes with the downward vertical.