| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina suspended in equilibrium |
| Difficulty | Challenging +1.2 This is a standard Further Maths M3 centre of mass question requiring integration to find centroids of a lamina and solid of revolution, plus a straightforward equilibrium calculation. While it involves multiple parts and some algebraic manipulation, the techniques are routine for this module with no novel problem-solving required. The 'show that' parts guide students through the working, making it slightly above average difficulty but well within expected M3 standard. |
| Spec | 4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area \(= \int_1^a x^{-2}\,dx = \left[-x^{-1}\right]_1^a = 1 - \frac{1}{a}\) | M1 A1 | |
| \(\bar{x} = \frac{1}{A}\int_1^a x \cdot x^{-2}\,dx = \frac{1}{1-1/a}\int_1^a x^{-1}\,dx\) | M1 | |
| \(= \frac{\ln a}{1 - 1/a} = \frac{a\ln a}{a-1}\) | A1 | |
| \(\bar{y} = \frac{1}{A} \cdot \frac{1}{2}\int_1^a (x^{-2})^2\,dx = \frac{1}{2(1-1/a)}\int_1^a x^{-4}\,dx\) | M1 | Using \(\frac{1}{2}\int y^2\,dx\) |
| \(= \frac{a}{2(a-1)} \cdot \left[\frac{-1}{3x^3}\right]_1^a\) | M1 | |
| \(= \frac{a}{2(a-1)} \cdot \frac{1}{3}\left(1 - \frac{1}{a^3}\right)\) | A1 | |
| \(= \frac{a}{6(a-1)} \cdot \frac{a^3-1}{a^3} = \frac{a^3-1}{6a^2(a-1)}\) | M1 | |
| \(= \frac{(a-1)(a^2+a+1)}{6a^2(a-1)} = \frac{a^2+a+1}{6a^2} = \frac{a^3-1}{6(a^3-a^2)}\) ✓ | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(a=2\): \(\bar{x} = \frac{2\ln 2}{1} = 2\ln 2\), \(\bar{y} = \frac{7}{6 \times 4} \cdot ... = \frac{7}{24}\) | M1 | Substituting \(a=2\) |
| A is at \((1,1)\); displacement from A to \((\bar{x},\bar{y})\) is \((\bar{x}-1, \bar{y}-1)\) | M1 | Finding vector from A to CoM |
| \(\tan\angle = \frac{\bar{x}-1}{1-\bar{y}}\) (horizontal/vertical from A) | M1 | Using geometry |
| Angle \(\approx 76°\) (or equivalent correct answer) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{x}_{solid} = \frac{\int_1^a \pi x^{-4} \cdot x\,dx}{\int_1^a \pi x^{-4}\,dx}= \frac{\int_1^a x^{-3}\,dx}{\int_1^a x^{-4}\,dx}\) | M1 | Using \(\pi\int y^2 x\,dx / \pi\int y^2\,dx\) |
| Numerator: \(\left[\frac{-1}{2x^2}\right]_1^a = \frac{1}{2}\left(1-\frac{1}{a^2}\right) = \frac{a^2-1}{2a^2}\) | M1 A1 | |
| Denominator: \(\left[\frac{-1}{3x^3}\right]_1^a = \frac{1}{3}\left(1-\frac{1}{a^3}\right) = \frac{a^3-1}{3a^3}\) | A1 | |
| \(\bar{x} = \frac{(a^2-1)}{2a^2} \cdot \frac{3a^3}{a^3-1} = \frac{3a(a^2-1)}{2(a^3-1)}\) | M1 | |
| \(= \frac{3a(a+1)(a-1)}{2(a-1)(a^2+a+1)} = \frac{3a(a+1)}{2(a^2+a+1)}\) | A1 | |
| As \(a\to\infty\): \(\bar{x} \to \frac{3a^2}{2a^2} = \frac{3}{2} = 1.5\), and for finite \(a>1\) it is less than \(1.5\) | M1 A1 | Must show \(< 1.5\) with argument |
# Question 4:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Area $= \int_1^a x^{-2}\,dx = \left[-x^{-1}\right]_1^a = 1 - \frac{1}{a}$ | M1 A1 | |
| $\bar{x} = \frac{1}{A}\int_1^a x \cdot x^{-2}\,dx = \frac{1}{1-1/a}\int_1^a x^{-1}\,dx$ | M1 | |
| $= \frac{\ln a}{1 - 1/a} = \frac{a\ln a}{a-1}$ | A1 | |
| $\bar{y} = \frac{1}{A} \cdot \frac{1}{2}\int_1^a (x^{-2})^2\,dx = \frac{1}{2(1-1/a)}\int_1^a x^{-4}\,dx$ | M1 | Using $\frac{1}{2}\int y^2\,dx$ |
| $= \frac{a}{2(a-1)} \cdot \left[\frac{-1}{3x^3}\right]_1^a$ | M1 | |
| $= \frac{a}{2(a-1)} \cdot \frac{1}{3}\left(1 - \frac{1}{a^3}\right)$ | A1 | |
| $= \frac{a}{6(a-1)} \cdot \frac{a^3-1}{a^3} = \frac{a^3-1}{6a^2(a-1)}$ | M1 | |
| $= \frac{(a-1)(a^2+a+1)}{6a^2(a-1)} = \frac{a^2+a+1}{6a^2} = \frac{a^3-1}{6(a^3-a^2)}$ ✓ | A1 | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| When $a=2$: $\bar{x} = \frac{2\ln 2}{1} = 2\ln 2$, $\bar{y} = \frac{7}{6 \times 4} \cdot ... = \frac{7}{24}$ | M1 | Substituting $a=2$ |
| A is at $(1,1)$; displacement from A to $(\bar{x},\bar{y})$ is $(\bar{x}-1, \bar{y}-1)$ | M1 | Finding vector from A to CoM |
| $\tan\angle = \frac{\bar{x}-1}{1-\bar{y}}$ (horizontal/vertical from A) | M1 | Using geometry |
| Angle $\approx 76°$ (or equivalent correct answer) | A1 | |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{x}_{solid} = \frac{\int_1^a \pi x^{-4} \cdot x\,dx}{\int_1^a \pi x^{-4}\,dx}= \frac{\int_1^a x^{-3}\,dx}{\int_1^a x^{-4}\,dx}$ | M1 | Using $\pi\int y^2 x\,dx / \pi\int y^2\,dx$ |
| Numerator: $\left[\frac{-1}{2x^2}\right]_1^a = \frac{1}{2}\left(1-\frac{1}{a^2}\right) = \frac{a^2-1}{2a^2}$ | M1 A1 | |
| Denominator: $\left[\frac{-1}{3x^3}\right]_1^a = \frac{1}{3}\left(1-\frac{1}{a^3}\right) = \frac{a^3-1}{3a^3}$ | A1 | |
| $\bar{x} = \frac{(a^2-1)}{2a^2} \cdot \frac{3a^3}{a^3-1} = \frac{3a(a^2-1)}{2(a^3-1)}$ | M1 | |
| $= \frac{3a(a+1)(a-1)}{2(a-1)(a^2+a+1)} = \frac{3a(a+1)}{2(a^2+a+1)}$ | A1 | |
| As $a\to\infty$: $\bar{x} \to \frac{3a^2}{2a^2} = \frac{3}{2} = 1.5$, and for finite $a>1$ it is less than $1.5$ | M1 A1 | Must show $< 1.5$ with argument |4 In this question, $a$ is a constant with $a > 1$.\\
Fig. 4 shows the region bounded by the curve $y = \frac { 1 } { x ^ { 2 } }$ for $1 \leqslant x \leqslant a$, the $x$-axis, and the lines $x = 1$ and $x = a$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b209dbe7-769c-4301-a2f3-108c27c8cefb-5_447_848_543_612}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
This region is occupied by a uniform lamina ABCD , where A is $( 1,1 ) , \mathrm { B }$ is $( 1,0 ) , \mathrm { C }$ is $( a , 0 )$ and D is $\left( a , \frac { 1 } { a ^ { 2 } } \right)$. The centre of mass of this lamina is $( \bar { x } , \bar { y } )$.\\
(i) Find $\bar { x }$ in terms of $a$, and show that $\bar { y } = \frac { a ^ { 3 } - 1 } { 6 \left( a ^ { 3 } - a ^ { 2 } \right) }$.\\
(ii) In the case $a = 2$, the lamina is freely suspended from the point A , and hangs in equilibrium. Find the angle which AB makes with the vertical.
The region shown in Fig. 4 is now rotated through $2 \pi$ radians about the $x$-axis to form a uniform solid of revolution.\\
(iii) Find the $x$-coordinate of the centre of mass of this solid of revolution, in terms of $a$, and show that it is less than 1.5.
\hfill \mbox{\textit{OCR MEI M3 2007 Q4 [18]}}