4.
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]
\includegraphics[alt={},max width=\textwidth]{5f9a87c6-2255-4178-ab04-441bb0cc4ce0-08_558_594_299_699}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{figure}
Relative to the axes shown in Fig. 3, the centre of mass of the lamina is at \(( \bar { x } , \bar { y } )\).
- Show that \(\bar { x } = \bar { y } = \frac { a ^ { 2 } + a + 1 } { 2 ( a + 1 ) }\).
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[You may need to use the result \(\frac { a ^ { 3 } - 1 } { 2 \left( a ^ { 2 } - 1 \right) } = \frac { a ^ { 2 } + a + 1 } { 2 ( a + 1 ) }\).] - Show that the centre of mass of the lamina lies on its perimeter if \(a = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )\).
- With the value of \(a = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )\) the lamina is suspended from A and hangs in equilibrium. Find the angle that the line OA makes with the vertical.
[0pt]
[Question 4 Continued]
\section*{5.}