| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Energy considerations in circular motion |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring multiple advanced techniques: setting up and evaluating a non-uniform centre of mass integral with variable density (part a), combining centres of mass of composite bodies (part b), and applying equilibrium conditions with suspended bodies (part c). The variable density λ/x² and the geometric setup require careful coordinate work, but the question provides significant scaffolding through 'show that' parts and given information. While demanding, it follows standard FM2 patterns for centre of mass problems. |
| Spec | 4.08d Volumes of revolution: about x and y axes6.04c 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 of dome \(= \int_a^{2a} \pi y^2 \frac{\lambda}{x^2}dx = \pi\lambda\int_a^{2a}\frac{4a^2-x^2}{x^2}dx = \pi\lambda\int_a^{2a}\left(\frac{4a^2}{x^2}-1\right)dx\) | M1 | Use the model to find the mass of the dome. Allow without limits. |
| \(= \pi\lambda\left[\frac{-4a^2}{x}-x\right]_a^{2a} = \pi\lambda\left[\frac{-4a^2}{2a}-2a+\frac{4a^2}{a}+a\right] = \pi\lambda a\) (kg) | A1 | Correct integration. Correct limits seen or implied. |
| Moments: \(\int_a^{2a}\pi y^2 \frac{\lambda}{x^2}\times x\,dx = \pi\lambda\int_a^{2a}\left(\frac{4a^2}{x}-x\right)dx\) | M1 | Use the model to take moments. Allow without limits. |
| \(= \pi\lambda\left[4a^2\ln x - \frac{1}{2}x^2\right]_a^{2a} = \pi\lambda\left(4a^2\ln 2 - \frac{3a^2}{2}\right)\) | A1 | Correct integration. Correct limits seen or implied. |
| \(\Rightarrow \text{distance} = \dfrac{\pi\lambda\left(4a^2\ln 2 - \frac{3a^2}{2}\right)}{\pi\lambda a}\) | DM1 | Complete method to find the distance of the centre of mass from the centre of the plane face. |
| \(\Rightarrow \text{distance} = 4a\ln 2 - \frac{3a}{2} - a = \left(4\ln 2 - \frac{5}{2}\right)a\) (m) * | A1* | Obtain the given answer from full and correct working. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Centre of mass of cone lies \(a\) m from the plane surface | B1 | Seen or implied |
| Moments about a diameter of the plane face | M1 | Moments equation. Need all terms. Dimensionally correct. Allow use of a parallel axis. |
| \(a\times kW - a\left(4\ln 2 - \frac{5}{2}\right)\times 2W = (2+k)Wd\) | A1 | Correct unsimplified equation in \(d\) |
| \(d = \dfrac{ | k+5-8\ln 2 | }{2+k}\,a\) * |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of trigonometry | M1 | Complete method to find the required angle, e.g. by use of tangent. Condone if they ignore the modulus signs. |
| \(\tan\alpha = \dfrac{\left(\dfrac{k+5-8\ln 2}{2+k}\right)a}{\sqrt{3}a} = \dfrac{1}{2\sqrt{3}}\) | A1 | Correct unsimplified equation. |
| \(\Rightarrow k = 16\ln 2 - 8\) | A1 | Any equivalent exact form. |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass of dome $= \int_a^{2a} \pi y^2 \frac{\lambda}{x^2}dx = \pi\lambda\int_a^{2a}\frac{4a^2-x^2}{x^2}dx = \pi\lambda\int_a^{2a}\left(\frac{4a^2}{x^2}-1\right)dx$ | M1 | Use the model to find the mass of the dome. Allow without limits. |
| $= \pi\lambda\left[\frac{-4a^2}{x}-x\right]_a^{2a} = \pi\lambda\left[\frac{-4a^2}{2a}-2a+\frac{4a^2}{a}+a\right] = \pi\lambda a$ (kg) | A1 | Correct integration. Correct limits seen or implied. |
| Moments: $\int_a^{2a}\pi y^2 \frac{\lambda}{x^2}\times x\,dx = \pi\lambda\int_a^{2a}\left(\frac{4a^2}{x}-x\right)dx$ | M1 | Use the model to take moments. Allow without limits. |
| $= \pi\lambda\left[4a^2\ln x - \frac{1}{2}x^2\right]_a^{2a} = \pi\lambda\left(4a^2\ln 2 - \frac{3a^2}{2}\right)$ | A1 | Correct integration. Correct limits seen or implied. |
| $\Rightarrow \text{distance} = \dfrac{\pi\lambda\left(4a^2\ln 2 - \frac{3a^2}{2}\right)}{\pi\lambda a}$ | DM1 | Complete method to find the distance of the centre of mass from the centre of the plane face. |
| $\Rightarrow \text{distance} = 4a\ln 2 - \frac{3a}{2} - a = \left(4\ln 2 - \frac{5}{2}\right)a$ (m) * | A1* | Obtain the given answer from full and correct working. |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Centre of mass of cone lies $a$ m from the plane surface | B1 | Seen or implied |
| Moments about a diameter of the plane face | M1 | Moments equation. Need all terms. Dimensionally correct. Allow use of a parallel axis. |
| $a\times kW - a\left(4\ln 2 - \frac{5}{2}\right)\times 2W = (2+k)Wd$ | A1 | Correct unsimplified equation in $d$ |
| $d = \dfrac{|k+5-8\ln 2|}{2+k}\,a$ * | A1* | Obtain given answer from correct working. Condone if modulus signs not used. |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of trigonometry | M1 | Complete method to find the required angle, e.g. by use of tangent. Condone if they ignore the modulus signs. |
| $\tan\alpha = \dfrac{\left(\dfrac{k+5-8\ln 2}{2+k}\right)a}{\sqrt{3}a} = \dfrac{1}{2\sqrt{3}}$ | A1 | Correct unsimplified equation. |
| $\Rightarrow k = 16\ln 2 - 8$ | A1 | Any equivalent exact form. |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b070338-1de4-4c33-be29-d37ac06c9fed-24_590_469_292_484}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b070338-1de4-4c33-be29-d37ac06c9fed-24_415_554_383_1025}
\captionsetup{labelformat=empty}
\caption{Figure 6}
\end{center}
\end{figure}
The shaded region shown in Figure 5 is bounded by the line with equation $x = a$ and the curve with equation $x ^ { 2 } + y ^ { 2 } = 4 a ^ { 2 }$
This shaded region is rotated through $180 ^ { \circ }$ about the $x$-axis to form a solid of revolution.
This solid is used to model a dome with height $a$ metres and base radius $\sqrt { 3 } a$ metres.\\
The dome is modelled as being non-uniform with the mass per unit volume of the dome at the point $( x , y , z )$ equal to $\frac { \lambda } { x ^ { 2 } } \mathrm {~kg} \mathrm {~m} ^ { - 3 }$, where $a \leqslant x \leqslant 2 a$ and $\lambda$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the dome from the centre of its plane face is $\left( 4 \ln 2 - \frac { 5 } { 2 } \right) a$ metres.
A solid uniform right circular cone has base radius $\sqrt { 3 } a$ metres and perpendicular height $4 a$ metres. A toy is formed by attaching the plane surface of the dome to the plane surface of the cone, as shown in Figure 6.
The weight of the cone is $k W$ and the weight of the dome is $2 W$\\
The centre of mass of the toy is a distance $d$ metres from the plane face of the dome.
\item Show that $d = \frac { | k + 5 - 8 \ln 2 | } { 2 + k } a$
The toy is suspended from a point on the circumference of the plane face of the dome and hangs freely in equilibrium with the plane face of the dome at an angle $\alpha$ to the downward vertical.\\
Given that $\tan \alpha = \frac { 1 } { 2 \sqrt { 3 } }$
\item find the exact value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2023 Q7 [13]}}