| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with removed circle/semicircle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question using the composite body method with a given formula for the semicircle. Part (a) requires straightforward application of the formula for composite bodies (subtraction), and part (b) uses the standard suspended lamina principle with basic trigonometry. The question is slightly easier than average because the semicircle formula is provided and the geometry is simple, requiring only routine calculations rather than problem-solving insight. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
\includegraphics{figure_1}
A uniform lamina $L$ is formed by taking a uniform square sheet of material $ABCD$, of side 10 cm, and removing the semi-circle with diameter $AB$ from the square, as shown in Fig. 2.
\begin{enumerate}[label=(\alph*)]
\item Find, in cm to 2 decimal places, the distance of the centre of mass of the lamina $L$ from the mid-point of $AB$.
[7]
\end{enumerate}
[The centre of mass of a uniform semi-circular lamina, radius $a$, is at a distance $\frac{4a}{3π}$ from the centre of the bounding diameter.]
The lamina is freely suspended from $D$ and hangs at rest.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find, in degrees to one decimal place, the angle between $CD$ and the vertical.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q4 [11]}}