| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cylinder or hemisphere from solid |
| Difficulty | Challenging +1.2 This is a standard Further Maths centre of mass problem requiring systematic application of the formula for composite bodies (cylinder minus hemisphere) and toppling condition. Part (a) involves routine calculation with given standard results for component centres of mass, while part (b) applies the standard toppling criterion (vertical through COM passes through edge). The algebra is moderately involved but follows a predictable template for FM2 questions. |
| 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratios: cylinder \(4\pi a^2 h\), hemisphere \(\frac{2}{3}\pi a^3\), solid \(V = 4\pi a^2 h - \frac{2}{3}\pi a^3\) | B1 | Correct mass ratios |
| Distances from \(O\): cylinder \(\frac{h}{2}\), hemisphere \(\frac{3}{8}a\), solid \(d\) | B1 | Correct distances |
| Moments about diameter through \(O\) | M1 | All three terms dimensionally correct; parallel axis allowed but final answer must be distance from \(O\) |
| \(4\pi a^2 h \times \frac{h}{2} - \frac{2}{3}\pi a^3 \times \frac{3}{8}a = 2\pi a^2\left(2h - \frac{1}{3}a\right) \times d\) | A1 | Correct unsimplified equation |
| \(d = \dfrac{h^2 - \dfrac{a^2}{8}}{2h - \dfrac{a}{3}} = \dfrac{3(8h^2 - a^2)}{8(6h-a)}\) | A1* | Deduce given answer; working must make derivation clear |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 5a \Rightarrow d = 2.573...a\) | B1 | Distance of com from base |
| About to topple so com above tipping point | M1 | Condone tan the wrong way up |
| \(\Rightarrow \tan\phi = \dfrac{2a}{5a - 2.573a}\) | A1ft | Correct unsimplified expression for trig ratio for \(\phi\) following their \(d\) |
| \(\phi = 39.5°\) or \(0.689\) rads | A1 |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios: cylinder $4\pi a^2 h$, hemisphere $\frac{2}{3}\pi a^3$, solid $V = 4\pi a^2 h - \frac{2}{3}\pi a^3$ | B1 | Correct mass ratios |
| Distances from $O$: cylinder $\frac{h}{2}$, hemisphere $\frac{3}{8}a$, solid $d$ | B1 | Correct distances |
| Moments about diameter through $O$ | M1 | All three terms dimensionally correct; parallel axis allowed but final answer must be distance from $O$ |
| $4\pi a^2 h \times \frac{h}{2} - \frac{2}{3}\pi a^3 \times \frac{3}{8}a = 2\pi a^2\left(2h - \frac{1}{3}a\right) \times d$ | A1 | Correct unsimplified equation |
| $d = \dfrac{h^2 - \dfrac{a^2}{8}}{2h - \dfrac{a}{3}} = \dfrac{3(8h^2 - a^2)}{8(6h-a)}$ | A1* | Deduce given answer; working must make derivation clear |
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## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 5a \Rightarrow d = 2.573...a$ | B1 | Distance of com from base |
| About to topple so com above tipping point | M1 | Condone tan the wrong way up |
| $\Rightarrow \tan\phi = \dfrac{2a}{5a - 2.573a}$ | A1ft | Correct unsimplified expression for trig ratio for $\phi$ following their $d$ |
| $\phi = 39.5°$ or $0.689$ rads | A1 | |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f06704e5-454c-41c1-9577-b1210f60480d-06_608_924_226_541}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A uniform solid cylinder has radius $2 a$ and height $h ( h > a )$.\\
A solid hemisphere of radius $a$ is removed from the cylinder to form the vessel $V$.\\
The plane face of the hemisphere coincides with the upper plane face of the cylinder.\\
The centre $O$ of the hemisphere is also the centre of the upper plane face of the cylinder, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of $V$ is $\frac { 3 \left( 8 h ^ { 2 } - a ^ { 2 } \right) } { 8 ( 6 h - a ) }$ from $O$.
The vessel $V$ is placed on a rough plane which is inclined at an angle $\phi$ to the horizontal. The lower plane circular face of $V$ is in contact with the inclined plane.
Given that $h = 5 a$, the plane is sufficiently rough to prevent $V$ from slipping and $V$ is on the point of toppling,
\item find, to three significant figures, the size of the angle $\phi$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 Q3 [9]}}