| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2004 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cylinder or hemisphere from solid |
| Difficulty | Standard +0.8 This is a multi-part M3 centre of mass question requiring: (a) composite body calculation with removed hemisphere using standard formulae (6 marks), (b) toppling condition analysis (3 marks), and (c) friction/sliding calculation (3 marks). While it involves multiple techniques and careful coordinate work, all methods are standard M3 procedures without requiring novel insight—moderately above average difficulty for A-level. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| Cylinder: \((\rho)\pi(2a)^2(\frac{3}{2}a)\) | M1A1 |
| \([6\pi a^3]\) | [18] |
| Hemisphere: \((\rho)\frac{2}{3}\pi a^3\) | |
| [2] | |
| \(S\): \((\rho)(\frac{16}{3}\pi a^3)\) | |
| [16] |
| Answer | Marks |
|---|---|
| \(\frac{1}{4}a\) | B1B1 |
| \(\frac{3}{8}a\) | |
| \(\bar{x}\) | |
| Moments equation: \(6\pi a^3(3a) - \frac{2}{3}\pi a^3\left(\frac{3}{8}a\right) = \frac{16}{3}\pi a^3\bar{x}\) | M1 |
| \(\bar{x} = \frac{51}{64}a\) | A1 (6) |
| (b) G above "A" seen or implied | M1 |
| or \(mg\sin\alpha(GX) = mg\cos\alpha(AX)\) | |
| \(\tan\alpha = \frac{AX}{XG} = \frac{2a}{\frac{3}{2}a - \bar{x}}\) | M1 |
| \([GX = \frac{3}{2}a - \frac{51}{64}a = \frac{45}{64}a\), \(\tan\alpha = \frac{128}{45}]\) | A1 (3) |
| \(\alpha = 70.6°\) | |
| (c) Finding \(F\) and \(R\): \(R = mg\cos\beta\), \(F = mg\sin\beta\) | M1 |
| Using \(F = \mu R\) and finding \(\tan\beta\) \([= 0.8]\) | M1 |
| \(\beta = 38.7°\) | A1 (3) |
| [12] |
(a) **Masses**
Cylinder: $(\rho)\pi(2a)^2(\frac{3}{2}a)$ | M1A1 |
$[6\pi a^3]$ | [18] |
Hemisphere: $(\rho)\frac{2}{3}\pi a^3$ | |
| [2] |
$S$: $(\rho)(\frac{16}{3}\pi a^3)$ | |
| [16] |
**Distance of CM from O**
$\frac{1}{4}a$ | B1B1 |
$\frac{3}{8}a$ | |
$\bar{x}$ | |
Moments equation: $6\pi a^3(3a) - \frac{2}{3}\pi a^3\left(\frac{3}{8}a\right) = \frac{16}{3}\pi a^3\bar{x}$ | M1 |
$\bar{x} = \frac{51}{64}a$ | A1 (6) |
(b) G above "A" seen or implied | M1 |
or $mg\sin\alpha(GX) = mg\cos\alpha(AX)$ | |
$\tan\alpha = \frac{AX}{XG} = \frac{2a}{\frac{3}{2}a - \bar{x}}$ | M1 |
$[GX = \frac{3}{2}a - \frac{51}{64}a = \frac{45}{64}a$, $\tan\alpha = \frac{128}{45}]$ | A1 (3) |
$\alpha = 70.6°$ | |
(c) Finding $F$ and $R$: $R = mg\cos\beta$, $F = mg\sin\beta$ | M1 |
Using $F = \mu R$ and finding $\tan\beta$ $[= 0.8]$ | M1 |
$\beta = 38.7°$ | A1 (3) |
| [12] |
---6.
Figure 2\\
\includegraphics[max width=\textwidth, alt={}, center]{c4b453e7-8a32-458b-8041-58c9e4ef9533-5_691_1067_241_584}
A uniform solid cylinder has radius $2 a$ and height $\frac { 3 } { 2 } a$. A hemisphere of radius $a$ is removed from the cylinder. The plane face of the hemisphere coincides with the upper plane face of the cylinder, and the centre $O$ of the hemisphere is also the centre of this plane face, as shown in Fig. 2. The remaining solid is $S$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of $S$ from $O$.\\
(6)
The lower plane face of $S$ rests in equilibrium on a desk lid which is inclined at an angle $\theta$ to the horizontal. Assuming that the lid is sufficiently rough to prevent $S$ from slipping, and that $S$ is on the point of toppling when $\theta = \alpha$,
\item find the value of $\alpha$.\\
(3)
Given instead that the coefficient of friction between $S$ and the lid is 0.8 , and that $S$ is on the point of sliding down the lid when $\theta = \beta$,
\item find the value of $\beta$.\\
(3)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2004 Q6 [12]}}