| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Conical or hemispherical shell composite |
| Difficulty | Standard +0.8 This is a multi-step centre of mass problem requiring knowledge of standard results for hollow cylinders and hemispherical shells, followed by a toppling condition application. Part (a) involves algebraic manipulation of composite centre of mass formulas (show-that style), while part (b) requires setting up and solving a toppling equilibrium condition with trigonometry. The combination of 3D geometry, standard results recall, and applied mechanics reasoning places this above average difficulty but within reach of well-prepared M3 students. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Ratio of masses: \(4\pi a^2 \quad 4\pi a \times ka \quad 8\pi a^2 \quad 12\pi a^2 + 4k\pi a^2\) | B1 | Correct ratio of masses (any equivalent form). |
| Distances: \((0) \quad \frac{k}{2}a \quad (1+k)a \quad \bar{y}\) | B1 | Correct distances from \(O\) or a parallel axis. |
| \((0+)k \times \frac{k}{2}a + 2(1+k)a = (k+3)\bar{y}\) | M1A1ft | M1: Attempt moments equation, dimensionally correct, no extra terms. A1ft: Correct equation following through ratio of masses and distances. |
| \(\left(\frac{k^2}{2} + 2 + 2k\right)a = (k+3)\bar{y}\) | ||
| \(\bar{y} = \dfrac{(k^2+4k+4)}{2(k+3)}a\) * | A1* (5) | Correct given expression with sufficient working. Note: distance from \(O\) for combined cylinder and base is \(\frac{ak^2}{2(1+k)}\). |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\tan 60° = \dfrac{(k^2+4k+4)}{2(k+3)}a \div 2a\) | M1 | Use \(\tan 60° = \dfrac{\bar{y}}{2a}\) or \(\dfrac{2a}{\bar{y}}\). May also use \(\tan 30°\). |
| \(k^2 + 4k(1-\sqrt{3}) + (4-12\sqrt{3}) = 0\) | A1 | Obtain correct 3TQ. |
| \(k>0 \Rightarrow k = 5.8147\ldots = 5.8\) or \(5.81\) or better | A1 (3) | Correct value of \(k\). |
# Question 4(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Ratio of masses: $4\pi a^2 \quad 4\pi a \times ka \quad 8\pi a^2 \quad 12\pi a^2 + 4k\pi a^2$ | B1 | Correct ratio of masses (any equivalent form). |
| Distances: $(0) \quad \frac{k}{2}a \quad (1+k)a \quad \bar{y}$ | B1 | Correct distances from $O$ or a parallel axis. |
| $(0+)k \times \frac{k}{2}a + 2(1+k)a = (k+3)\bar{y}$ | M1A1ft | M1: Attempt moments equation, dimensionally correct, no extra terms. A1ft: Correct equation following through ratio of masses and distances. |
| $\left(\frac{k^2}{2} + 2 + 2k\right)a = (k+3)\bar{y}$ | | |
| $\bar{y} = \dfrac{(k^2+4k+4)}{2(k+3)}a$ * | A1* (5) | Correct given expression with sufficient working. Note: distance from $O$ for combined cylinder and base is $\frac{ak^2}{2(1+k)}$. |
# Question 4(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\tan 60° = \dfrac{(k^2+4k+4)}{2(k+3)}a \div 2a$ | M1 | Use $\tan 60° = \dfrac{\bar{y}}{2a}$ or $\dfrac{2a}{\bar{y}}$. May also use $\tan 30°$. |
| $k^2 + 4k(1-\sqrt{3}) + (4-12\sqrt{3}) = 0$ | A1 | Obtain correct 3TQ. |
| $k>0 \Rightarrow k = 5.8147\ldots = 5.8$ or $5.81$ or better | A1 (3) | Correct value of $k$. |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bd1e2b07-4a87-49d6-addd-c9f67467ef2f-12_659_513_246_774}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A thin uniform right hollow cylinder, of radius $2 a$ and height $k a$, has a base but no top. A thin uniform hemispherical shell, also of radius $2 a$, is made of the same material as the cylinder. The hemispherical shell is attached to the end of the cylinder forming a container $C$. The open circular rim of the cylinder coincides with the rim of the hemispherical shell. The centre of the base of $C$ is $O$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance from $O$ to the centre of mass of $C$ is
$$\frac { \left( k ^ { 2 } + 4 k + 4 \right) } { 2 ( k + 3 ) } a$$
The container is placed with its circular base on a plane which is inclined at $30 ^ { \circ }$ to the horizontal. The plane is sufficiently rough to prevent $C$ from sliding. The container is on the point of toppling.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2022 Q4 [8]}}