| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Challenging +1.2 This is a standard M3/Further Mechanics question requiring systematic application of centre of mass formulas for solids of revolution and laminae. Part (i) uses the standard formula for solids of revolution, part (ii) applies lamina integration formulas, and part (iii) uses composite body techniques. While it involves multiple parts and careful integration of x^{-1/3}, these are routine procedures for Further Maths students with no novel problem-solving required. The calculations are somewhat lengthy but conceptually straightforward, placing it slightly above average difficulty. |
| 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/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V = \int_1^8 \pi(x^{-\frac{1}{3}})^2\, dx\) | M1 | \(\pi\) may be omitted throughout |
| \(= \pi\left[3x^{\frac{1}{3}}\right]_1^8 = 3\pi\) | A1 | |
| \(V\bar{x} = \int_1^8 \pi x(x^{-\frac{1}{3}})^2\, dx\) | M1 | |
| \(= \pi\left[\frac{3}{4}x^{\frac{4}{3}}\right]_1^8 = \frac{45}{4}\pi\) | A1 | |
| \(\bar{x} = \frac{\frac{45}{4}\pi}{3\pi} = \frac{15}{4} = 3.75\) | M1 A1 | [6] Dependent on previous M1M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = \int_1^8 x^{-\frac{1}{3}}\, dx\) | M1 | |
| \(= \left[\frac{3}{2}x^{\frac{2}{3}}\right]_1^8 = \frac{9}{2} = 4.5\) | A1 | |
| \(A\bar{x} = \int_1^8 x(x^{-\frac{1}{3}})\, dx\) | M1 | |
| \(= \left[\frac{3}{5}x^{\frac{5}{3}}\right]_1^8 = \frac{93}{5} = 18.6\) | A1 | |
| \(\bar{x} = \frac{18.6}{4.5} = \frac{62}{15}\) \((\approx 4.13)\) | A1 | |
| \(A\bar{y} = \int_1^8 \frac{1}{2}(x^{-\frac{1}{3}})^2\, dx\) | M1 | If \(\frac{1}{2}\) omitted, award M1A0A0 |
| \(= \left[\frac{3}{2}x^{\frac{1}{3}}\right]_1^8 = \frac{3}{2} = 1.5\) | A1 | |
| \(\bar{y} = \frac{1.5}{4.5} = \frac{1}{3}\) | A1 | [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((1)\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} + (3.5)\begin{pmatrix}4.5\\0.25\end{pmatrix} = (4.5)\begin{pmatrix}62/15\\1/3\end{pmatrix} = \begin{pmatrix}18.6\\1.5\end{pmatrix}\) | M1 | Attempt formula for CM of composite body (one coordinate sufficient) |
| M1 | Formulae for both coordinates; signs must be correct, but areas (1 and 3.5) may be wrong | |
| \(\bar{x} = 2.85\) | A1 | ft only if \(1 < \bar{x} < 8\) |
| \(\bar{y} = 0.625\) | A1 | [4] ft only if \(0.5 < \bar{y} < 1\) |
# Question 4:
## Part 4(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V = \int_1^8 \pi(x^{-\frac{1}{3}})^2\, dx$ | M1 | $\pi$ may be omitted throughout |
| $= \pi\left[3x^{\frac{1}{3}}\right]_1^8 = 3\pi$ | A1 | |
| $V\bar{x} = \int_1^8 \pi x(x^{-\frac{1}{3}})^2\, dx$ | M1 | |
| $= \pi\left[\frac{3}{4}x^{\frac{4}{3}}\right]_1^8 = \frac{45}{4}\pi$ | A1 | |
| $\bar{x} = \frac{\frac{45}{4}\pi}{3\pi} = \frac{15}{4} = 3.75$ | M1 A1 | **[6]** Dependent on previous M1M1 |
## Part 4(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = \int_1^8 x^{-\frac{1}{3}}\, dx$ | M1 | |
| $= \left[\frac{3}{2}x^{\frac{2}{3}}\right]_1^8 = \frac{9}{2} = 4.5$ | A1 | |
| $A\bar{x} = \int_1^8 x(x^{-\frac{1}{3}})\, dx$ | M1 | |
| $= \left[\frac{3}{5}x^{\frac{5}{3}}\right]_1^8 = \frac{93}{5} = 18.6$ | A1 | |
| $\bar{x} = \frac{18.6}{4.5} = \frac{62}{15}$ $(\approx 4.13)$ | A1 | |
| $A\bar{y} = \int_1^8 \frac{1}{2}(x^{-\frac{1}{3}})^2\, dx$ | M1 | If $\frac{1}{2}$ omitted, award M1A0A0 |
| $= \left[\frac{3}{2}x^{\frac{1}{3}}\right]_1^8 = \frac{3}{2} = 1.5$ | A1 | |
| $\bar{y} = \frac{1.5}{4.5} = \frac{1}{3}$ | A1 | **[8]** |
## Part 4(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(1)\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} + (3.5)\begin{pmatrix}4.5\\0.25\end{pmatrix} = (4.5)\begin{pmatrix}62/15\\1/3\end{pmatrix} = \begin{pmatrix}18.6\\1.5\end{pmatrix}$ | M1 | Attempt formula for CM of composite body (one coordinate sufficient) |
| | M1 | Formulae for both coordinates; signs must be correct, but areas (1 and 3.5) may be wrong |
| $\bar{x} = 2.85$ | A1 | ft only if $1 < \bar{x} < 8$ |
| $\bar{y} = 0.625$ | A1 | **[4]** ft only if $0.5 < \bar{y} < 1$ |
*Other methods: M1A1 for $\bar{x}$, M1A1 for $\bar{y}$. In each case M1 requires a complete and correct method leading to a numerical value.*4 Fig. 4.1 shows the region $R$ bounded by the curve $y = x ^ { - \frac { 1 } { 3 } }$ for $1 \leqslant x \leqslant 8$, the $x$-axis, and the lines $x = 1$ and $x = 8$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c470e80e-b346-4335-9c08-beb5a46cc506-4_597_1018_411_566}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}
(i) Find the $x$-coordinate of the centre of mass of a uniform solid of revolution obtained by rotating $R$ through $2 \pi$ radians about the $x$-axis.\\
(ii) Find the coordinates of the centre of mass of a uniform lamina in the shape of the region $R$.\\
(iii) Using your answer to part (ii), or otherwise, find the coordinates of the centre of mass of a uniform lamina in the shape of the region (shown shaded in Fig. 4.2) bounded by the curve $y = x ^ { - \frac { 1 } { 3 } }$ for $1 \leqslant x \leqslant 8$, the line $y = \frac { 1 } { 2 }$ and the line $x = 1$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c470e80e-b346-4335-9c08-beb5a46cc506-4_595_1015_1610_607}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}
\hfill \mbox{\textit{OCR MEI M3 2008 Q4 [18]}}