4. [The centre of mass of a uniform semicircular lamina of radius \(r\) is \(\frac { 4 r } { 3 \pi }\) from the centre.] \begin{figure}[h] \end{figure} The uniform rectangular lamina \(A B C D E F\) has sides \(A C = F D = 6 a\) and \(A F = C D = 3 a\). The point \(B\) lies on \(A C\) with \(A B = 2 a\) and the point \(E\) lies on \(F D\) with \(F E = 2 a\). The template, \(T\), shown shaded in Figure 3, is formed by removing the semicircular lamina with diameter \(B C\) from the rectangular lamina and then fixing this semicircular lamina to the opposite side, \(F D\), of the rectangular lamina. The diameter of the semicircular lamina coincides with \(E D\) and the semicircular arc \(E D\) is outside the rectangle \(A B C D E F\). All points of \(T\) lie in the same plane.

1. Show that the centre of mass of \(T\) is a distance \(\left( \frac { 9 + 2 \pi } { 6 } \right)\) a from \(A C\). The mass of \(T\) is \(M\). A particle of mass \(k M\) is attached to \(T\) at \(C\). The loaded template is freely suspended from \(A\) and hangs in equilibrium with \(A F\) at angle \(\phi\) to the downward vertical through \(A\). Given that \(\tan \phi = \frac { 3 } { 2 }\)
2. find the value of \(k\).
   \section*{\textbackslash section*\{Question 4 continued\}} \includegraphics[max width=\textwidth, alt={}, center]{c16c17b6-2c24-4939-b3b5-63cd63646b76-11_149_142_2604_1816}