| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving a composite lamina with removed sections. Part (a) requires symmetry recognition and the composite body formula with straightforward calculations. Part (b) applies basic equilibrium conditions (vertical line through centre of mass and suspension point). While it involves multiple steps, the techniques are routine for M2 students and the regular hexagon geometry simplifies calculations significantly. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cf960066-46b8-42a3-8a8b-d8deb76e7c70-06_736_725_258_607}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The uniform lamina $A B C D E F$ is a regular hexagon with centre $O$ and sides of length 2 m , as shown in Figure 1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cf960066-46b8-42a3-8a8b-d8deb76e7c70-06_574_723_1288_605}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The triangles $O A F$ and $O E F$ are removed to form the uniform lamina $O A B C D E$, shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of $O A B C D E$ from $O$.
The lamina $O A B C D E$ is freely suspended from $E$ and hangs in equilibrium.
\item Find the size of the angle between $E O$ and the downward vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q4 [11]}}