- \hspace{0pt} [The centre of mass of a semi-circular lamina of radius \(r\) is \(\frac { 4 r } { 3 \pi }\) from the centre]
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8dac5891-0dfd-49b4-ada4-0ecb875cf6aa-11_656_1274_421_355}
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\caption{Figure 3}
\end{figure}
A template \(T\) consists of a uniform plane lamina \(P Q R O S\), as shown in Figure 3. The lamina is bounded by two semicircles, with diameters \(S O\) and \(O R\), and by the sides \(S P , P Q\) and \(Q R\) of the rectangle \(P Q R S\). The point \(O\) is the mid-point of \(S R , P Q = 12 \mathrm {~cm}\) and \(Q R = 2 x \mathrm {~cm}\).
- Show that the centre of mass of \(T\) is a distance \(\frac { 4 \left| 2 x ^ { 2 } - 3 \right| } { 8 x + 3 \pi } \mathrm {~cm}\) from \(S R\).
The template \(T\) is freely suspended from the point \(P\) and hangs in equilibrium.
Given that \(x = 2\) and that \(\theta\) is the angle that \(P Q\) makes with the horizontal, - show that \(\tan \theta = \frac { 48 + 9 \pi } { 22 + 6 \pi }\).