| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Solid on inclined plane - toppling |
| Difficulty | Challenging +1.2 This is a standard M3 centre of mass question requiring volume of revolution integration, then applying toppling and sliding conditions. The integration is routine (polynomial under rotation), and both equilibrium conditions are textbook applications. More involved than basic C1/M1 questions due to the integration setup and multiple parts, but follows standard M3 patterns without requiring novel insight. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\int \pi y^2 \, dx = \frac{\pi}{4}\int x^2(6-x)^2 \, dx = \frac{\pi}{4}\int 36x^2 - 12x^3 + x^4 \, dx\) | M1 A1 | Setting up volume integral correctly |
| \(= \frac{\pi}{4}\left[12x^3 - 3x^4 + \frac{x^5}{5}\right]_2^6 = \frac{\pi}{4} \times \frac{1024}{5}\) | M1 | Integrating and substituting limits |
| \((160.8\ldots)\) | ||
| \(\int \pi y^2 x \, dx = \frac{\pi}{4}\int x^3(6-x)^2 \, dx = \frac{\pi}{4}\int 36x^3 - 12x^4 + x^5 \, dx\) | M1 A1 | Setting up moment integral |
| \(= \frac{\pi}{4}\left[9x^4 - \frac{12}{5}x^5 + \frac{1}{6}x^6\right]_2^6 = \frac{\pi}{4} \times \frac{10496}{15}\) | M1 | Integrating and substituting limits |
| \((549.5\ldots)\) | ||
| \(\Rightarrow \bar{x} = \frac{10496}{15} \times \frac{5}{1024} = 3.416\ldots\) | M1 A1 | Dividing moments by volume |
| Required distance \(\approx 3.42 - 2 = 1.42\) (cm) | A1 | Subtracting 2 to get distance from base |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| Base has radius \(\frac{1}{2} \times 2 \times 4 = 4\) cm | B1 | |
| About to topple \(\Rightarrow \tan\alpha = \frac{4}{1.42}\) | M1 A1 | Using base radius and \(\bar{x}\) from base |
| \(\alpha \approx 70.5°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| Parallel to slope: \(F = mg\sin\beta\) | ||
| Perpendicular to slope: \(R = mg\cos\beta\) | M1 A1 | Resolving forces correctly |
| About to slip: \(F = \mu R\) | ||
| \(\tan\beta = \mu = 0.3, \quad \beta \approx 16.7°\) | A1 |
# Question 7:
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\int \pi y^2 \, dx = \frac{\pi}{4}\int x^2(6-x)^2 \, dx = \frac{\pi}{4}\int 36x^2 - 12x^3 + x^4 \, dx$ | M1 A1 | Setting up volume integral correctly |
| $= \frac{\pi}{4}\left[12x^3 - 3x^4 + \frac{x^5}{5}\right]_2^6 = \frac{\pi}{4} \times \frac{1024}{5}$ | M1 | Integrating and substituting limits |
| $(160.8\ldots)$ | | |
| $\int \pi y^2 x \, dx = \frac{\pi}{4}\int x^3(6-x)^2 \, dx = \frac{\pi}{4}\int 36x^3 - 12x^4 + x^5 \, dx$ | M1 A1 | Setting up moment integral |
| $= \frac{\pi}{4}\left[9x^4 - \frac{12}{5}x^5 + \frac{1}{6}x^6\right]_2^6 = \frac{\pi}{4} \times \frac{10496}{15}$ | M1 | Integrating and substituting limits |
| $(549.5\ldots)$ | | |
| $\Rightarrow \bar{x} = \frac{10496}{15} \times \frac{5}{1024} = 3.416\ldots$ | M1 A1 | Dividing moments by volume |
| Required distance $\approx 3.42 - 2 = 1.42$ (cm) | A1 | Subtracting 2 to get distance from base |
**(9 marks)**
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| Base has radius $\frac{1}{2} \times 2 \times 4 = 4$ cm | B1 | |
| About to topple $\Rightarrow \tan\alpha = \frac{4}{1.42}$ | M1 A1 | Using base radius and $\bar{x}$ from base |
| $\alpha \approx 70.5°$ | A1 | |
**(4 marks)**
## Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| Parallel to slope: $F = mg\sin\beta$ | | |
| Perpendicular to slope: $R = mg\cos\beta$ | M1 A1 | Resolving forces correctly |
| About to slip: $F = \mu R$ | | |
| $\tan\beta = \mu = 0.3, \quad \beta \approx 16.7°$ | A1 | |
**(3 marks)**
**Total: 16 marks**7.
Diagram NOT accurately drawn
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bbd531ab-05f8-48ff-8a68-ec6f33ac0a2f-12_444_768_253_603}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The shaded region $R$ is bounded by the curve with equation $y = \frac { 1 } { 2 } x ( 6 - x )$, the $x$-axis and the line $x = 2$, as shown in Figure 1. The unit of length on both axes is 1 cm . A uniform solid $P$ is formed by rotating $R$ through $360 ^ { \circ }$ about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of $P$ is, to 3 significant figures, 1.42 cm from its plane face.
The uniform solid $P$ is placed with its plane face on an inclined plane which makes an angle $\theta$ with the horizontal. Given that the plane is sufficiently rough to prevent $P$ from sliding and that $P$ is on the point of toppling when $\theta = \alpha$,
\item find the angle $\alpha$.
Given instead that $P$ is on the point of sliding down the plane when $\theta = \beta$ and that the coefficient of friction between $P$ and the plane is 0.3 ,
\item find the angle $\beta$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2012 Q7 [16]}}