| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 11 |
| 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-part centre of mass problem requiring: (1) application of composite body formula with similar cones (involving volume ratios and careful coordinate work to show a specific result), and (2) equilibrium analysis with hanging geometry. The algebraic manipulation is substantial and the hanging equilibrium setup requires spatial reasoning, placing it moderately above average difficulty. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | portion | mass |
| large cone | \(\rho\frac{1}{3}\pi(3r)^3 h = 9\rho\pi r^3 h\) | \(\frac{1}{4} \times 3h = \frac{3}{4}h\) |
| small cone | \(\rho\frac{1}{3}\pi(2r)^2 2h = \frac{8}{3}\rho\pi r^2 h\) | \(h + \frac{1}{4} \times 2h = \frac{3}{2}h\) |
| frustum | \(\frac{19}{3}\rho\pi r^2 h\) | \(\bar{y}\) |
| \(\rho =\) mass per unit volume, \(y\) coords. taken vert. from base | M2 A3 | |
| \(\frac{19}{3}\rho\pi r^2 h \times \bar{y} = \frac{11}{4}\rho\pi r^2 h^2\) \(\therefore \bar{y} = \frac{11}{4}h + \frac{19}{3} = \frac{33}{76}h\) | M1 A1 | |
| (b) \(\tan\alpha = \frac{\frac{33}{76} \times 2r}{3r} = \frac{11}{38}\) | M2 A1 | |
| \(\therefore \alpha = 16°\) (nearest degree) | A1 | (11) |
(a) | portion | mass | y | my |
|---------|-------|---------|----------|
| large cone | $\rho\frac{1}{3}\pi(3r)^3 h = 9\rho\pi r^3 h$ | $\frac{1}{4} \times 3h = \frac{3}{4}h$ | $\frac{27}{4}\rho\pi r^2 h^2$ |
| small cone | $\rho\frac{1}{3}\pi(2r)^2 2h = \frac{8}{3}\rho\pi r^2 h$ | $h + \frac{1}{4} \times 2h = \frac{3}{2}h$ | $4\rho\pi r^2 h^2$ |
| frustum | $\frac{19}{3}\rho\pi r^2 h$ | $\bar{y}$ | $\frac{11}{4}\rho\pi r^2 h^2$ |
$\rho =$ mass per unit volume, $y$ coords. taken vert. from base | M2 A3 |
$\frac{19}{3}\rho\pi r^2 h \times \bar{y} = \frac{11}{4}\rho\pi r^2 h^2$ $\therefore \bar{y} = \frac{11}{4}h + \frac{19}{3} = \frac{33}{76}h$ | M1 A1 |
(b) $\tan\alpha = \frac{\frac{33}{76} \times 2r}{3r} = \frac{11}{38}$ | M2 A1 |
$\therefore \alpha = 16°$ (nearest degree) | A1 | (11)4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cab238c9-f4e2-4637-a079-f74779548f49-3_447_506_205_657}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
A stand used to reach high shelves in a storeroom is in the shape of a frustum of a cone. It is modelled as a uniform solid formed by removing a right circular cone of height $2 h$ from a similar cone of height $3 h$ and base radius $3 r$ as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of the stand is a distance of $\frac { 33 } { 76 } h$ from its larger plane face.\\
(7 marks)\\
The stand is stored hanging in equilibrium from a point on the circumference of the larger plane face. Given that $h = 2 r$,
\item find, correct to the nearest degree, the acute angle which the plane faces of the stand make with the vertical.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q4 [11]}}