| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina with hole removed |
| Difficulty | Challenging +1.2 This is a standard M3 centre of mass question with three routine parts: (a) algebraic integration for a cone (show that result), (b) composite body calculation using standard formula, (c) toppling condition geometry. All techniques are textbook exercises with clear methods, though the multi-step nature and algebraic manipulation required place it slightly above average difficulty. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \dfrac{\pi\int_0^{4r} x\left(\frac{1}{4}x\right)^2 dx}{\dfrac{4\pi r^3}{3}}\) or \(\bar{x} = \dfrac{\pi\int_0^{4r} x\left(r-\frac{1}{4}x\right)^2 dx}{\dfrac{4\pi r^3}{3}}\) | M1A1 | Correct method to find distance of CoM from vertex or plane face using \(\bar{x} = \frac{\pi\int xy^2 dx}{\frac{4\pi r^3}{3}}\). Formula must be correct but allow constant multiple if it appears in both numerator and denominator or cancelled \(\pi\). \(y\) must be replaced with \(y=\frac{1}{4}x\) or \(y=r-\frac{1}{4}x\). Must attempt to integrate numerator. Denominator \(\frac{4\pi r^3}{3}\) is given. |
| \(= \dfrac{3}{256r^3}\left[x^4\right]_0^{4r}\) | A1 | Correct expression after integration and division by \(\frac{4\pi r^3}{3}\). Limits must be correct. |
| \(= 3r\) * | A1* | Given answer from complete and correct working. If distance found from plane face must subtract to find required distance. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratios: Cone \(\frac{4\pi r^3}{3}\), Cylinder \(\pi\left(\frac{1}{2}r\right)^2 \times r\), S \(\left(\frac{4\pi r^3}{3}-\pi\left(\frac{1}{2}r\right)^2 \times r\right)\); simplified: \(\frac{4}{3}\), \(\frac{1}{4}\), \(\frac{13}{12}\) | B1 | Correct mass ratios |
| Distances from vertex: Cone \(3r\), Cylinder \(\left(4r-\frac{1}{2}r\right)\), S \(\bar{y}\); Distances from plane face: Cone \(r\), Cylinder \(\frac{1}{2}r\), S \(\bar{y}\) | B1 | Correct distances for their parallel axis. Ignore signs. |
| \(\left(\frac{4\pi r^3}{3}\times 3r\right) - \left(\pi\left(\frac{1}{2}r\right)^2\times r\right)\left(4r-\frac{1}{2}r\right) = \left(\frac{4\pi r^3}{3}-\pi\left(\frac{1}{2}r\right)^2\times r\right)\bar{y}\) | M1A1 | Form moments equation with correct number of terms. Must be dimensionally correct (mass ratio \(\times\) distance). Correct unsimplified equation. |
| \(\bar{y} = \dfrac{75}{26}r\) * | A1* | Given answer from complete and correct working. Working should include a line of simplification. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\alpha = \dfrac{r}{4r - \dfrac{75}{26}r}\) | M1A1 | Use of tan to obtain equation for relevant angle; allow reciprocal. Correct equation, condone reciprocal. |
| \(\tan\alpha = \dfrac{26}{29}\) | A1 | Correct answer. Must be exact value for \(\tan\alpha\). A0 if they go straight to \(\alpha\). |
## Question 5:
### Part 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \dfrac{\pi\int_0^{4r} x\left(\frac{1}{4}x\right)^2 dx}{\dfrac{4\pi r^3}{3}}$ or $\bar{x} = \dfrac{\pi\int_0^{4r} x\left(r-\frac{1}{4}x\right)^2 dx}{\dfrac{4\pi r^3}{3}}$ | M1A1 | Correct method to find distance of CoM from vertex or plane face using $\bar{x} = \frac{\pi\int xy^2 dx}{\frac{4\pi r^3}{3}}$. Formula must be correct but allow constant multiple if it appears in both numerator and denominator or cancelled $\pi$. $y$ must be replaced with $y=\frac{1}{4}x$ or $y=r-\frac{1}{4}x$. Must attempt to integrate numerator. Denominator $\frac{4\pi r^3}{3}$ is given. |
| $= \dfrac{3}{256r^3}\left[x^4\right]_0^{4r}$ | A1 | Correct expression after integration and division by $\frac{4\pi r^3}{3}$. Limits must be correct. |
| $= 3r$ * | A1* | Given answer from complete and correct working. If distance found from plane face must subtract to find required distance. |
### Part 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios: Cone $\frac{4\pi r^3}{3}$, Cylinder $\pi\left(\frac{1}{2}r\right)^2 \times r$, S $\left(\frac{4\pi r^3}{3}-\pi\left(\frac{1}{2}r\right)^2 \times r\right)$; simplified: $\frac{4}{3}$, $\frac{1}{4}$, $\frac{13}{12}$ | B1 | Correct mass ratios |
| Distances from vertex: Cone $3r$, Cylinder $\left(4r-\frac{1}{2}r\right)$, S $\bar{y}$; Distances from plane face: Cone $r$, Cylinder $\frac{1}{2}r$, S $\bar{y}$ | B1 | Correct distances for their parallel axis. Ignore signs. |
| $\left(\frac{4\pi r^3}{3}\times 3r\right) - \left(\pi\left(\frac{1}{2}r\right)^2\times r\right)\left(4r-\frac{1}{2}r\right) = \left(\frac{4\pi r^3}{3}-\pi\left(\frac{1}{2}r\right)^2\times r\right)\bar{y}$ | M1A1 | Form moments equation with correct number of terms. Must be dimensionally correct (mass ratio $\times$ distance). Correct unsimplified equation. |
| $\bar{y} = \dfrac{75}{26}r$ * | A1* | Given answer from complete and correct working. Working should include a line of simplification. |
### Part 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\alpha = \dfrac{r}{4r - \dfrac{75}{26}r}$ | M1A1 | Use of tan to obtain equation for relevant angle; allow reciprocal. Correct equation, condone reciprocal. |
| $\tan\alpha = \dfrac{26}{29}$ | A1 | Correct answer. Must be exact value for $\tan\alpha$. A0 if they go straight to $\alpha$. |
---\begin{enumerate}
\item A uniform right solid circular cone $C$ has radius $r$ and height $4 r$.\\
(a) Show, using algebraic integration, that the distance of the centre of mass of $C$ from its vertex is $3 r$.\\[0pt]
[You may assume that the volume of $C$ is $\frac { 4 } { 3 } \pi r ^ { 3 }$ ]
\end{enumerate}
A uniform solid $S$, shown below in Figure 3, is formed by removing from $C$ a uniform solid right circular cylinder of height $r$ and radius $\frac { 1 } { 2 } r$, where the centre of one end of the cylinder coincides with the centre of the plane face of $C$ and the axis of the cylinder coincides with the axis of $C$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{176ae8f8-7de9-4993-825a-6067614526ae-12_661_1194_861_440}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
(b) Show that the distance of the centre of mass of $S$ from the vertex of $C$ is $\frac { 75 } { 26 } r$
A rough plane is inclined at an angle $\alpha$ to the horizontal.\\
The solid $S$ rests in equilibrium with its plane face in contact with the inclined plane.\\
Given that $S$ is on the point of toppling,\\
(c) find the exact value of $\tan \alpha$
\hfill \mbox{\textit{Edexcel M3 2024 Q5 [12]}}