| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Prove standard centre of mass formula |
| Difficulty | Standard +0.8 This is a two-part M3 question requiring integration to derive a standard result (3h/4 for a cone), then applying it to a composite body problem. Part (a) requires setting up and evaluating a volume integral with appropriate limits. Part (b) involves finding centres of mass of two components (hemisphere and cone) and combining them, requiring recall of the hemisphere result and careful calculation. The multi-step nature, integration setup, and composite body analysis place this above average difficulty but within reach of well-prepared M3 students. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| let \(y = \frac{r}{h}x, \rho =\) mass per unit volume | M1 | |
| \(\therefore \frac{1}{3}\rho r^2 h\bar{x} = \int_0^h \rho \rho y^2 x\,dx\) | A1 | |
| \(\frac{1}{3}r^2h\bar{x} = \int_0^h \frac{r^2}{h^2}x^4\,dx = \frac{r^2}{h^2}\left[\frac{x^4}{4}\right]_0^h\) | M1 A1 | |
| \(\frac{1}{3}r^2h\bar{x} = \frac{1}{4}r^2h \therefore \bar{x} = \frac{3}{4}h\) | M1 A1 | |
| \(\rho = \) mass per unit volume, \(y\) coords. taken vert. from vertex | M2 A3 | |
| \(\rho r^2 \times \bar{y} = \frac{2}{6}\rho r^3 \therefore \bar{y} = \frac{2}{6}r\) | M1 A1 | (13) |
| let $y = \frac{r}{h}x, \rho =$ mass per unit volume | M1 | |
| $\therefore \frac{1}{3}\rho r^2 h\bar{x} = \int_0^h \rho \rho y^2 x\,dx$ | A1 | |
| $\frac{1}{3}r^2h\bar{x} = \int_0^h \frac{r^2}{h^2}x^4\,dx = \frac{r^2}{h^2}\left[\frac{x^4}{4}\right]_0^h$ | M1 A1 | |
| $\frac{1}{3}r^2h\bar{x} = \frac{1}{4}r^2h \therefore \bar{x} = \frac{3}{4}h$ | M1 A1 | |
| $\rho = $ mass per unit volume, $y$ coords. taken vert. from vertex | M2 A3 | |
| $\rho r^2 \times \bar{y} = \frac{2}{6}\rho r^3 \therefore \bar{y} = \frac{2}{6}r$ | M1 A1 | (13) |5. (a) Use integration to show that the centre of mass of a uniform solid right circular cone of height $h$ is $\frac { 3 } { 4 } h$ from the vertex of the cone.\\
(6 marks)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ad523c3f-9109-45a8-8399-80a4c2edeff7-4_419_424_372_721}
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\caption{Fig. 3}
\end{center}
\end{figure}
A paperweight is made by removing material from the top half of a solid sphere of radius $r$ so that the remaining solid consists of a hemisphere of radius $r$ and a cone of height $r$ and base radius $r$ as shown in Figure 3.\\
(b) Find the distance of the centre of mass of the paperweight from its vertex.\\
(7 marks)\\
\hfill \mbox{\textit{Edexcel M3 Q5 [13]}}