| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with removed circle/semicircle |
| Difficulty | Standard +0.3 This is a standard centre of mass question with circular holes requiring systematic application of the formula for composite bodies. Part (i) and (ii) involve routine calculations using mass ratios and coordinates (negative mass method). Part (iii) applies the equilibrium condition that the centre of mass lies vertically below the pivot. While it requires careful bookkeeping with multiple holes and coordinates, it follows a well-established procedure with no novel insight needed, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
\includegraphics{figure_6}
A uniform circular disc has centre $O$ and radius $1.2\,\text{m}$. The centre of the disc is at the origin of $x$- and $y$-axes. Two circular holes with centres at $A$ and $B$ are made in the disc (see diagram). The point $A$ is on the negative $x$-axis with $OA = 0.5\,\text{m}$. The point $B$ is on the negative $y$-axis with $OB = 0.7\,\text{m}$. The hole with centre $A$ has radius $0.3\,\text{m}$ and the hole with centre $B$ has radius $0.4\,\text{m}$. Find the distance of the centre of mass of the object from
\begin{enumerate}[label=(\roman*)]
\item the $x$-axis, [4]
\item the $y$-axis. [3]
\end{enumerate}
The object can rotate freely in a vertical plane about a horizontal axis through $O$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Calculate the angle which $OA$ makes with the vertical when the object rests in equilibrium. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2015 Q6 [9]}}