| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 14 |
| 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 composite shapes (rectangle + semicircle) and suspended lamina equilibrium. The calculations are straightforward: find centroids of components, use composite formula, then apply equilibrium condition (centre of mass directly below suspension point). While it requires multiple steps and careful coordinate work, it follows a well-practiced template with no novel insight required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | (i), (ii) | M4 A2 |
| portion | mass | \(x\) |
| rectangle | \(32\rho\) | 4 |
| semicircle | \(2\pi\rho\) | 6 |
| total | \((32 + 2\pi)\rho\) | \(\bar{x}\) |
| \(\bar{x} = \frac{(128+12\pi)\rho}{(32+2\pi)\rho} = 4.33\text{ cm}\) from \(OD\) (3sf) | M1 A1 | |
| \(\bar{y} = \frac{(8\pi+\frac{208}{3})\rho}{(32+2\pi)\rho} = 2.47\text{ cm}\) from \(OA\) (3sf) | M1A1 | |
| (b) | \(4 - 2.47 = 1.53\) from m'pt. of \(BC\) vertically | M1 |
| \(6 - 4.33 = 1.67\) from m'pt. of \(BC\) horizontally | M1 | |
| \(\tan\theta = \frac{1.53}{1.67}\) \(\therefore \theta = 42.5°\) (3sf) | M1 A1 | (14 marks total) |
(a) | (i), (ii) | M4 A2 |
| portion | mass | $x$ | $y$ | $mx$ | $my$ |
|---------|------|-----|-------|--------|---------|
| rectangle | $32\rho$ | 4 | 2 | $128\rho$ | $64\rho$ |
| semicircle | $2\pi\rho$ | 6 | $4 + \frac{8}{3\pi}$ | $12\pi\rho$ | $(8\pi + \frac{16}{3})\rho$ |
| total | $(32 + 2\pi)\rho$ | $\bar{x}$ | $\bar{y}$ | $(128 + 12\pi)\rho$ | $(8\pi + \frac{208}{3})\rho$ |
| $\bar{x} = \frac{(128+12\pi)\rho}{(32+2\pi)\rho} = 4.33\text{ cm}$ from $OD$ (3sf) | M1 A1 |
| $\bar{y} = \frac{(8\pi+\frac{208}{3})\rho}{(32+2\pi)\rho} = 2.47\text{ cm}$ from $OA$ (3sf) | M1A1 |
(b) | $4 - 2.47 = 1.53$ from m'pt. of $BC$ vertically | M1 |
| $6 - 4.33 = 1.67$ from m'pt. of $BC$ horizontally | M1 |
| $\tan\theta = \frac{1.53}{1.67}$ $\therefore \theta = 42.5°$ (3sf) | M1 A1 | (14 marks total)
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**Total: 75 marks**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f5449ec3-ead0-464f-9d03-f225cd21bca6-5_536_848_191_397}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
Figure 3 shows a hotel 'key' consisting of a rectangle $O A B D$, where $O A = 8 \mathrm {~cm}$ and $O D = 4 \mathrm {~cm}$, joined to a semicircle whose diameter $B C$ is 4 cm long. The thickness of the key is negligible and the same material is used throughout.
The key is modelled as a uniform lamina.\\
Using this model,
\begin{enumerate}[label=(\alph*)]
\item find, correct to 3 significant figures, the distance of the centre of mass from
\begin{enumerate}[label=(\roman*)]
\item OD ,
\item $O A$.
A small circular hole of negligible diameter is made at the mid-point of $B C$ so that the key can be hung on a smooth peg. When the key is freely suspended from the peg,
\end{enumerate}\item find, correct to 3 significant figures, the acute angle made by $O A$ with the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [14]}}