| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Difficulty | Challenging +1.8 This is a challenging M3 centre of mass problem requiring multiple composite body calculations, geometric reasoning about stability conditions, and algebraic manipulation. Part (a) demands careful application of the composite body formula with three components (cylinder, removed cone, hemisphere), while part (b) involves non-trivial equilibrium analysis with suspended bodies and toppling conditions. The modulus sign explanation and the 'any position' stability criterion require deeper conceptual understanding beyond routine application of formulas. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| (a) \(M(CD): \pi r^2 h \cdot \frac{\pi}{4} - \frac{1}{3}\pi r^2 \cdot \frac{2h}{3} - \frac{2}{3}\pi r^3 \cdot \frac{3k}{8} = (\frac{2}{3}\pi r^2 h + \frac{2}{3}\pi r^3)\bar{y}\) | M1 A1 A1 A1 |
| \(\frac{h^2}{2} - \frac{h^2}{9} - \frac{\ell^2}{4} = \frac{2}{3}(h+r)\bar{y}\) | M1 A1 A1 |
| \(\bar{y} = \frac{3}{8}(h-r)\) | A1 B1 |
| Modulus needed, since \(h\) may be \(< r\) or \(> r\) | A1 B1 |
| (b) (i) \(\tan\theta = \frac{3(h-r)}{8r}\) | M1 A1 A1 |
| Hemisphere alone: \(\tan\theta = \frac{3}{8}\) | M1 A1 A1 |
| Thus \(h - r = r\), so \(h = 2r\) | A1 |
| \(h : r = 2 : 1\) | A1 |
| (ii) C. of mass must be at \(O\), so \(\bar{y} = 0\), i.e. \(h = r\) | M1 A1 A1 |
| \(h : r = 1 : 1\) | A1 |
| Total: 16 marks |
(a) $M(CD): \pi r^2 h \cdot \frac{\pi}{4} - \frac{1}{3}\pi r^2 \cdot \frac{2h}{3} - \frac{2}{3}\pi r^3 \cdot \frac{3k}{8} = (\frac{2}{3}\pi r^2 h + \frac{2}{3}\pi r^3)\bar{y}$ | M1 A1 A1 A1 |
$\frac{h^2}{2} - \frac{h^2}{9} - \frac{\ell^2}{4} = \frac{2}{3}(h+r)\bar{y}$ | M1 A1 A1 |
$\bar{y} = \frac{3}{8}(h-r)$ | A1 B1 |
Modulus needed, since $h$ may be $< r$ or $> r$ | A1 B1 |
(b) (i) $\tan\theta = \frac{3(h-r)}{8r}$ | M1 A1 A1 |
Hemisphere alone: $\tan\theta = \frac{3}{8}$ | M1 A1 A1 |
Thus $h - r = r$, so $h = 2r$ | A1 |
$h : r = 2 : 1$ | A1 |
(ii) C. of mass must be at $O$, so $\bar{y} = 0$, i.e. $h = r$ | M1 A1 A1 |
$h : r = 1 : 1$ | A1 |
| **Total: 16 marks** |A container consists of two sections made from the same material: a hollow portion formed by removing a cone (shaded in the figure) from a solid cylinder of radius $r$ and height $h$, and a solid hemisphere of radius $r$. The vertex of the removed cone coincides with the centre $O$ of the horizontal plane face of the hemisphere. $CD$ is a diameter of this plane face.
\includegraphics{figure_7}
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the container from the plane face of the hemisphere is $\left|\frac{3}{8}(h-r)\right|$. Explain why the modulus sign is necessary. [9 marks]
\item Find the ratio $h : r$ in each of the following cases:
\begin{enumerate}[label=(\roman*)]
\item When the container is suspended from the point $C$, the angle made by $CD$ with the vertical is equal to the angle which $CD$ would make with the vertical if the hemisphere alone were suspended from $C$. [4 marks]
\item The container is able to stand without toppling in any position when it is placed with the surface of the hemispherical part in contact with a smooth horizontal table. [3 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [16]}}