| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Cone stability and toppling conditions |
| Difficulty | Challenging +1.2 Part (a) requires a standard calculus-based derivation using integration (disc method or shells) to find centre of mass - a bookwork result often seen in M3. Parts (b) and (c) involve applying equilibrium conditions with careful attention to geometry and moments about a point. While multi-step and requiring spatial reasoning about forces and their lines of action, these are standard M3 techniques without requiring novel insight. The 16 total marks reflect length rather than exceptional difficulty. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 06.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| (a) \(x^2 + y^2 = r^2\) | B1 M1 A1 |
| \(\bar{x}\int_0^l \pi y^2 dx = \int_0^l \pi xy^2 dx\) | |
| \(\pi x \int_0^l r^2 - x^2 dx = \pi \int_0^l x^2 - x^3 dx\) | M1 A1 A1 A1 |
| \(\frac{2x^2}{3}\bar{x} = \frac{l^2}{3}\) | |
| \(\bar{x} = \frac{3x}{8}\) | |
| (b) Forces shown: weight, normal reaction (thro' O), friction at plane | B1 B1 B1 |
| (c) Resolve \(\parallel\) plane: \(F = W\sin\alpha\) | B1 M1 A1 |
| \(M(O)\): \(Fr = W\frac{3x}{8}\cos\alpha\) | |
| \(Wr\sin\theta = \frac{3}{8}W\cos\alpha\) | M1 A1 A1 |
| \(\cos\alpha = \frac{3}{8} \times \frac{10}{4} = \frac{3}{4}\) |
**(a)** $x^2 + y^2 = r^2$ | B1 M1 A1 |
$\bar{x}\int_0^l \pi y^2 dx = \int_0^l \pi xy^2 dx$ | |
$\pi x \int_0^l r^2 - x^2 dx = \pi \int_0^l x^2 - x^3 dx$ | M1 A1 A1 A1 |
$\frac{2x^2}{3}\bar{x} = \frac{l^2}{3}$ | |
$\bar{x} = \frac{3x}{8}$ | |
**(b)** Forces shown: weight, normal reaction (thro' O), friction at plane | B1 B1 B1 |
**(c)** Resolve $\parallel$ plane: $F = W\sin\alpha$ | B1 M1 A1 |
$M(O)$: $Fr = W\frac{3x}{8}\cos\alpha$ | |
$Wr\sin\theta = \frac{3}{8}W\cos\alpha$ | M1 A1 A1 |
$\cos\alpha = \frac{3}{8} \times \frac{10}{4} = \frac{3}{4}$ | |
**Total: 16 marks**\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of a uniform solid hemisphere of radius $r$ is at a distance $\frac{3r}{8}$ from the centre $O$ of the plane face. [7 marks]
\end{enumerate}
The figure shows the vertical cross-section of a rough solid hemisphere at rest on a rough inclined plane inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac{3}{10}$.
\includegraphics{figure_7}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Indicate on a copy of the figure the three forces acting on the hemisphere, clearly stating what they are, and paying special attention to their lines of action. [3 marks]
\item Given that the plane face containing the diameter $AB$ makes an angle $\alpha$ with the vertical, show that $\cos \alpha = \frac{4}{5}$. [6 marks]
\end{enumerate]
\hfill \mbox{\textit{Edexcel M3 Q7 [16]}}