| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Cone stability and toppling conditions |
| Difficulty | Challenging +1.2 This is a multi-part M3 centre of mass question requiring integration to prove the cone COM formula (standard but requires careful setup), then applying COM principles to a composite body, and finally using toppling conditions. While it involves several steps and careful bookkeeping, all techniques are standard M3 material with no novel insights required—slightly above average due to the algebraic manipulation and multi-stage reasoning. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| (a) \(\bar{x}\int_0^h my^2 dx = \int_0^h m y^2 x dx\) | M1 A1 B1 M1 |
| \(y = \frac{r}{h}x\), so \(\bar{x}\int_0^h \frac{r^2x^2}{h^2}dx = \int_0^h \frac{r^2x^3}{h^2}dx\) | M1 A1 A1 |
| \(\bar{x}\frac{r^2}{3} = \frac{r^2}{4}\) | |
| \(\bar{x} = \frac{3h}{4}\) | |
| (b) M(vertex): \(m\frac{3h}{3} + m\frac{3h}{4} = 2mkh\) | M1 A1 A1 |
| \(k = \frac{3h}{15}\) | |
| (c) \(\tan \alpha = r + \frac{3h}{15}\) | M1 A1 M1 A1 |
| \(r = \frac{3h}{15}\tan 45° = \frac{3h}{15}\) or \(0.47h\) | M1 A1 M1 A1 |
(a) $\bar{x}\int_0^h my^2 dx = \int_0^h m y^2 x dx$ | M1 A1 B1 M1 |
$y = \frac{r}{h}x$, so $\bar{x}\int_0^h \frac{r^2x^2}{h^2}dx = \int_0^h \frac{r^2x^3}{h^2}dx$ | M1 A1 A1 |
$\bar{x}\frac{r^2}{3} = \frac{r^2}{4}$ | |
$\bar{x} = \frac{3h}{4}$ | |
(b) M(vertex): $m\frac{3h}{3} + m\frac{3h}{4} = 2mkh$ | M1 A1 A1 |
$k = \frac{3h}{15}$ | |
(c) $\tan \alpha = r + \frac{3h}{15}$ | M1 A1 M1 A1 |
$r = \frac{3h}{15}\tan 45° = \frac{3h}{15}$ or $0.47h$ | M1 A1 M1 A1 |
**Total: 14 marks**\begin{enumerate}[label=(\alph*)]
\item Prove that the centre of mass of a uniform solid right circular cone of height $h$ and base radius $r$ is at a distance $\frac{3h}{4}$ from the vertex. [7 marks]
\end{enumerate}
An item of confectionery consists of a thin wafer in the form of a hollow right circular cone of height $h$ and mass $m$, filled with solid chocolate, also of mass $m$, to a depth of $kh$ as shown. The centre of mass of the item is at $O$, the centre of the horizontal plane face of the chocolate.
\includegraphics{figure_3}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $k = \frac{8h}{15}$. [3 marks]
\end{enumerate}
In the packaging process, the cone has to move on a conveyor belt inclined at an angle $\alpha$ to the horizontal as shown. If the belt is rough enough to prevent sliding, and the maximum value of $\alpha$ for which the cone does not topple is $45°$,
\includegraphics{figure_4}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the radius of the base of the cone in terms of $h$. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [14]}}