| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cone from cone or cylinder |
| Difficulty | Standard +0.8 This is a multi-step centre of mass problem requiring composite body techniques (subtraction method), algebraic manipulation to prove a given result, and then applying toppling conditions with geometric reasoning. While the individual techniques are standard M3 content, the combination of proving a specific result and applying it to a toppling scenario with the constraint h=2r requires careful systematic work and understanding of stability principles, placing it moderately above average difficulty. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Vol: Large cone \(\frac{1}{3}\pi(2r)^2(2h)\), small cone \(\frac{1}{3}\pi r^2 h\), \(S\ \frac{7}{3}\pi r^2 h\) (accept ratios 8:1:7) | B1 | |
| C of M: Large cone \(\frac{1}{2}h\), small cone \(\frac{5}{4}h\), \(S\ \bar{x}\) | B1, B1 | (or equivalent) |
| \(\frac{8}{3}\pi r^2 h\cdot\frac{1}{2}h - \frac{1}{3}\pi r^2 h\cdot\frac{5}{4}h = \frac{7}{3}\pi r^2 h\cdot\bar{x}\) | M1 | or equivalent |
| \(\Rightarrow \bar{x} = \frac{11}{28}h\) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\theta = \frac{2r}{\bar{x}} = \frac{2r}{\frac{11}{28}h} = \frac{2r}{\frac{11}{14}r} = \frac{28}{11}\) | M1, A1 | |
| \(\theta \approx 68.6°\) or \(1.20\) radians | A1 | (3) |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Vol: Large cone $\frac{1}{3}\pi(2r)^2(2h)$, small cone $\frac{1}{3}\pi r^2 h$, $S\ \frac{7}{3}\pi r^2 h$ (accept ratios 8:1:7) | B1 | |
| C of M: Large cone $\frac{1}{2}h$, small cone $\frac{5}{4}h$, $S\ \bar{x}$ | B1, B1 | (or equivalent) |
| $\frac{8}{3}\pi r^2 h\cdot\frac{1}{2}h - \frac{1}{3}\pi r^2 h\cdot\frac{5}{4}h = \frac{7}{3}\pi r^2 h\cdot\bar{x}$ | M1 | or equivalent |
| $\Rightarrow \bar{x} = \frac{11}{28}h$ | A1 | (5) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\theta = \frac{2r}{\bar{x}} = \frac{2r}{\frac{11}{28}h} = \frac{2r}{\frac{11}{14}r} = \frac{28}{11}$ | M1, A1 | |
| $\theta \approx 68.6°$ or $1.20$ radians | A1 | (3) |
Special case: obtains complement by using $\tan\theta = \frac{2r}{x}$ giving $21.4°$ or $.374$ radians — M1A0A0
**Total: 8 marks**
---3.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{39c2d25a-a39b-4eb9-a17b-6e741ab5ae98-04_519_709_315_603}
\end{center}
\end{figure}
A uniform solid $S$ is formed by taking a uniform solid right circular cone, of base radius $2 r$ and height $2 h$, and removing the cone, with base radius $r$ and height $h$, which has the same vertex as the original cone, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of $S$ from its larger plane face is $\frac { 11 } { 28 } h$.
The solid $S$ lies with its larger plane face on a rough table which is inclined at an angle $\theta ^ { \circ }$ to the horizontal. The table is sufficiently rough to prevent $S$ from slipping. Given that $h = 2 r$,
\item find the greatest value of $\theta$ for which $S$ does not topple.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2008 Q3 [8]}}