| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | June |
| Marks | 11 |
| 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 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 (square minus semicircle), and part (b) uses basic equilibrium geometry. The question is slightly easier than average because the semicircle formula is provided and the setup is routine for M2 students. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Shape | Square | Semi-circle |
| Relative masses | 100 | \(12\frac{1}{2}\pi (39.3)\) |
| Centre of mass from \(AB\) | 5 | \(\frac{20}{3\pi} (2.12)\) |
| Moments about \(AB\): \(100 \times 5 - 12\frac{1}{2}\pi \times \frac{20}{3\pi} = (100 - 12\frac{1}{2}\pi)\bar{x}\) | ||
| Answer: 6.86 cm |
| Answer | Marks | Guidance |
|---|---|---|
| Correct angle, diagram sufficient | ||
| Method to find \(\theta\) [or \((90 - \theta)\)] | ||
| \(\tan \theta = \frac{10 - \bar{x}}{5}\) | ||
| Answer: \(32.1°\) |
## Part (a)
| Shape | Square | Semi-circle | Lamina $L$ |
| Relative masses | 100 | $12\frac{1}{2}\pi (39.3)$ | $100 - 12\frac{1}{2}\pi (60.7)$ | M1 A1 |
| Centre of mass from $AB$ | 5 | $\frac{20}{3\pi} (2.12)$ | $\bar{x}$ | B1 B1 |
| Moments about $AB$: $100 \times 5 - 12\frac{1}{2}\pi \times \frac{20}{3\pi} = (100 - 12\frac{1}{2}\pi)\bar{x}$ | | | | M1 A1 |
| Answer: 6.86 cm | | | | A1 (cao) (7) |
## Part (b)
| Correct angle, diagram sufficient | | | | M1 |
| Method to find $\theta$ [or $(90 - \theta)$] | | | | M1 |
| $\tan \theta = \frac{10 - \bar{x}}{5}$ | | | | A1 ft |
| Answer: $32.1°$ | | | | A1 (cao) (4) |
**Total: 11 marks**
---\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 $\frac{4a}{3\pi}$ 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 2002 Q4 [11]}}