| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Conical or hemispherical shell composite |
| Difficulty | Challenging +1.2 This is a standard M3 centre of mass question requiring composite body techniques (subtracting hemispheres) and equilibrium with moments. Part (a) involves routine application of the hemisphere COM formula with careful arithmetic. Part (b) requires taking moments about the contact point, which is a standard technique. The calculations are somewhat involved but follow predictable patterns for this topic, making it moderately above average difficulty for A-level but well within M3 expectations. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | portion | mass |
| large hemisphere | \(\frac{1}{2} \times \rho \times \frac{4}{3}\pi\left(\frac{r}{2}\right)^3 = \frac{2}{3}\rho r^3\) | \(\frac{3}{8} \times \frac{r}{8} = \frac{19}{64}r\) |
| small hemisphere | \(\frac{1}{2} \times \rho \times \frac{4}{3}\pi r^3 = \frac{2}{3}\rho r^3\) | \(\frac{3}{8}r\) |
| bowl | \(\frac{19}{12}\rho r^3\) | \(\bar{y}\) |
| \(\rho = \) mass per unit volume; y coords. taken vert. from plane face | M2 A3 | |
| \(\frac{19}{12}\rho r^3 \times \bar{y} = \frac{65}{64}\rho r^3\) \(\therefore \bar{y} = \frac{65}{64}r \times \frac{19}{12} = \frac{195}{304}r\) | M1 A1 | |
| (b) mom. about vert. through pt of contact: \(Mg \times \frac{195}{304}r \sin \alpha = \frac{1}{2}Mg \times \frac{2}{3}r \cos \alpha\) \(\therefore \tan \alpha = \frac{\frac{1}{3}}{\frac{195}{304}} = \frac{76}{65}\) | M2 A2; M1 A1 | (13) |
(a) | portion | mass | y | my |
|---|---|---|---|
| large hemisphere | $\frac{1}{2} \times \rho \times \frac{4}{3}\pi\left(\frac{r}{2}\right)^3 = \frac{2}{3}\rho r^3$ | $\frac{3}{8} \times \frac{r}{8} = \frac{19}{64}r$ | $\frac{19}{64}\rho r^3$ |
| small hemisphere | $\frac{1}{2} \times \rho \times \frac{4}{3}\pi r^3 = \frac{2}{3}\rho r^3$ | $\frac{3}{8}r$ | $\frac{1}{4}\rho r^3$ |
| bowl | $\frac{19}{12}\rho r^3$ | $\bar{y}$ | $\frac{65}{64}\rho r^3$ |
$\rho = $ mass per unit volume; y coords. taken vert. from plane face | M2 A3 |
$\frac{19}{12}\rho r^3 \times \bar{y} = \frac{65}{64}\rho r^3$ $\therefore \bar{y} = \frac{65}{64}r \times \frac{19}{12} = \frac{195}{304}r$ | M1 A1 |
(b) mom. about vert. through pt of contact: $Mg \times \frac{195}{304}r \sin \alpha = \frac{1}{2}Mg \times \frac{2}{3}r \cos \alpha$ $\therefore \tan \alpha = \frac{\frac{1}{3}}{\frac{195}{304}} = \frac{76}{65}$ | M2 A2; M1 A1 | (13)
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8b7133ed-3748-46cb-99d2-570ee33c7393-4_526_620_196_598}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Figure 1 shows a bowl formed by removing from a solid hemisphere of radius $\frac { 3 } { 2 } r$ a smaller hemisphere of radius $r$ having the same axis of symmetry and the same plane face.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of the bowl is a distance of $\frac { 195 } { 304 } r$ from its plane face.\\
(7 marks)\\
The bowl has mass $M$ and is placed with its curved surface on a smooth horizontal plane. A stud of mass $\frac { 1 } { 2 } M$ is attached to the outer rim of the bowl.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8b7133ed-3748-46cb-99d2-570ee33c7393-4_517_729_1318_539}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
When the bowl is in equilibrium its plane surface is inclined at an angle $\alpha$ to the horizontal as shown in Figure 2.
\item Find tan $\alpha$.\\
(6 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [13]}}