| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Cone stability and toppling conditions |
| Difficulty | Challenging +1.8 This M3 question requires integration to find the centre of mass of a spherical cap (part a, 9 marks suggests multi-step calculus setup), then applies equilibrium conditions for toppling on an inclined plane (part b). The spherical cap geometry and coordinate setup require careful visualization and the calculus is non-trivial, making this significantly harder than routine mechanics problems but still within standard M3 scope. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| (a) \(\bar{x}\pi\int_0^{a/2}(a^2-x^2)dx = \pi\int_0^{a/2}(a^2x - x^3)dx\) | M1 A1 M1 A1 |
| \(\frac{5a^3x}{24} - \frac{9a^5}{64} = \frac{24a}{64} - \frac{27a}{40}\) | M1 A1 M1 A1 |
| \(\bar{x} = \frac{24}{64} - \frac{27a}{40}\) | |
| From O: \(\frac{27a}{40} - \frac{2}{2} = \frac{7a}{40}\) | |
| (b) Reaction acts through centre O; centre of mass G on vertical through point of contact S; let angle \(OGS = \beta\) | B1 B1 B1 |
| Sine rule in \(\triangle OGS\): \(\frac{\sin\beta}{40} = \frac{\sin 30°}{27}\) | M1 A1 |
| \(\sin\beta = \frac{20}{27}\) | |
| \(\beta = 132.2°\) | A1 |
| \(\alpha = 180° - (30° + \beta) = 17.8°\) | 15 marks total |
**(a)** $\bar{x}\pi\int_0^{a/2}(a^2-x^2)dx = \pi\int_0^{a/2}(a^2x - x^3)dx$ | M1 A1 M1 A1
$\frac{5a^3x}{24} - \frac{9a^5}{64} = \frac{24a}{64} - \frac{27a}{40}$ | M1 A1 M1 A1
$\bar{x} = \frac{24}{64} - \frac{27a}{40}$ |
From O: $\frac{27a}{40} - \frac{2}{2} = \frac{7a}{40}$ |
**(b)** Reaction acts through centre O; centre of mass G on vertical through point of contact S; let angle $OGS = \beta$ | B1 B1 B1
Sine rule in $\triangle OGS$: $\frac{\sin\beta}{40} = \frac{\sin 30°}{27}$ | M1 A1
$\sin\beta = \frac{20}{27}$ |
$\beta = 132.2°$ | A1
$\alpha = 180° - (30° + \beta) = 17.8°$ | 15 marks totalA uniform solid sphere, of radius $a$, is divided into two sections by a plane at a distance $\frac{a}{2}$ from the centre and parallel to a diameter.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of gravity of the smaller cap from its plane face is $\frac{7a}{40}$. [9 marks]
\end{enumerate}
This smaller cap is now placed on an inclined plane whose angle of inclination to the horizontal is $\theta$. The plane is rough enough to prevent slipping and the cap rests with its curved surface in contact with the plane.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item If the maximum value of $\theta$ for which this is possible without the cap turning over is 30°, find the corresponding maximum inclination of the axis of symmetry of the cap to the vertical. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [15]}}