3 A uniform heavy lamina occupies the region shaded in Fig. 3. This region is formed by removing a square of side 1 unit from a square of side $a$ units (where $a > 1$ ).

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{71d839d8-12ca-4806-8f74-c572e7e21891-4_597_624_338_731}
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\caption{Fig. 3}
\end{center}
\end{figure}

Relative to the axes shown in Fig. 3, the centre of mass of the lamina is at $( \bar { x } , \bar { y } )$.\\
(i) Show that $\bar { x } = \bar { y } = \frac { a ^ { 2 } + a + 1 } { 2 ( a + 1 ) }$.\\[0pt]
[You may need to use the result $\frac { a ^ { 3 } - 1 } { 2 \left( a ^ { 2 } - 1 \right) } = \frac { a ^ { 2 } + a + 1 } { 2 ( a + 1 ) }$.]\\
(ii) Show that the centre of mass of the lamina lies on its perimeter if $a = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )$.

In another situation, $a = 4$.\\
A particle of mass one third that of the lamina is attached to the lamina at vertex B ; the lamina with the particle is freely suspended from vertex A and hangs in equilibrium. The positions of A and B are shown in Fig. 3.\\
(iii) Calculate the angle that AB makes with the vertical.

\hfill \mbox{\textit{OCR MEI M2 2015 Q3 [18]}}