4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{deb9e495-3bfb-4a46-9ee7-3eb421c33499-07_606_883_260_532}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

The uniform lamina $O B C$ is one quarter of a circular disc with centre $O$ and radius 4 m . The points $A$ and $D$, on $O B$ and $O C$ respectively, are 3 m from $O$. The uniform lamina $A B C D$, shown shaded in Figure 1, is formed by removing the triangle $O A D$ from $O B C$.

Given that the centre of mass of one quarter of a uniform circular disc of radius $r$ is at a distance $\frac { 4 \sqrt { 2 } } { 3 \pi } r$ from the centre of the disc,
\begin{enumerate}[label=(\alph*)]
\item find the distance of the centre of mass of the lamina $A B C D$ from $A D$.

The lamina is freely suspended from $D$ and hangs in equilibrium.
\item Find, to the nearest degree, the angle between $D C$ and the downward vertical.

\begin{center}

\end{center}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{deb9e495-3bfb-4a46-9ee7-3eb421c33499-09_915_1269_118_356}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2016 Q4 [9]}}