[The centre of mass of a uniform semicircular lamina of radius $$\(r\)$$ is $$\(\frac{4 r}{3 \pi}\)$$ from the centre.] The uniform rectangular lamina $$\(A B C D E F\)$$ has sides $$\(A C=F D=6 a\)$$ and $$\(A F=C D=3 a\)$$. The point $$\(B\)$$ lies on $$\(A C\)$$ with $$\(A B=2 a\)$$ and the point $$\(E\)$$ lies on $$\(F D\)$$ with $$\(F E=2 a\)$$. The template, $$\(T\)$$, shown shaded in Figure 3, is formed by removing the semicircular lamina with diameter $$\(B C\)$$ from the rectangular lamina and then fixing this semicircular lamina to the opposite side, $$\(F D\)$$, of the rectangular lamina. The diameter of the semicircular lamina coincides with $$\(E D\)$$ and the semicircular arc $$\(E D\)$$ is outside the rectangle $$\(A B C D E F\)$$. All points of $$\(T\)$$ lie in the same plane. (a) Show that the centre of mass of $$\(T\)$$ is a distance $$\(\left(\frac{9+2 \pi}{6}\right) a\)$$ from $$\(A C\)$$. (4) The mass of $$\(T\)$$ is $$\(M\)$$. A particle of mass $$\(k M\)$$ is attached to $$\(T\)$$ at $$\(C\)$$. The loaded template is freely suspended from $$\(A\)$$ and hangs in equilibrium with $$\(A F\)$$ at angle $$\(\phi\)$$ to the downward vertical through $$\(A\)$$. Given that $$\(\tan \phi=\frac{3}{2}\)$$ (b) find the value of $$\(k\)$$. (6)

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