OCR MEI M3 2010 June — Question 3 18 marks

Exam BoardOCR MEI
ModuleM3 (Mechanics 3)
Year2010
SessionJune
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeCentre of mass of lamina by integration
DifficultyChallenging +1.2 This is a standard M3/Further Mechanics question requiring systematic application of centre of mass formulas for solids of revolution and laminas. Part (i) uses standard volume integration, part (ii) applies standard lamina formulas with 1/x, part (iii) exploits symmetry, and part (iv) combines regions using the composite body formula. While lengthy with multiple parts, each step follows routine procedures without requiring novel insight—slightly above average difficulty due to the algebraic manipulation and multi-part nature.
Spec6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids
3 In this question, give your answers in an exact form.
The region \(R _ { 1 }\) (shown in Fig. 3) is bounded by the \(x\)-axis, the lines \(x = 1\) and \(x = 5\), and the curve \(y = \frac { 1 } { x }\) for \(1 \leqslant x \leqslant 5\).
  1. A uniform solid of revolution is formed by rotating the region \(R _ { 1 }\) through \(2 \pi\) radians about the \(x\)-axis. Find the \(x\)-coordinate of the centre of mass of this solid.
  2. Find the coordinates of the centre of mass of a uniform lamina occupying the region \(R _ { 1 }\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{c93aed95-f655-45cb-805f-7114a15acccf-4_849_841_735_651} \captionsetup{labelformat=empty} \caption{Fig. 3}
    \end{figure} The region \(R _ { 2 }\) is bounded by the \(y\)-axis, the lines \(y = 1\) and \(y = 5\), and the curve \(y = \frac { 1 } { x }\) for \(\frac { 1 } { 5 } \leqslant x \leqslant 1\). The region \(R _ { 3 }\) is the square with vertices \(( 0,0 ) , ( 1,0 ) , ( 1,1 )\) and \(( 0,1 )\).
  3. Write down the coordinates of the centre of mass of a uniform lamina occupying the region \(R _ { 2 }\).
  4. Find the coordinates of the centre of mass of a uniform lamina occupying the region consisting of \(R _ { 1 } , R _ { 2 }\) and \(R _ { 3 }\) (shown shaded in Fig. 3).
Question 3:
Part (i)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Volume \(= \int_1^5 \pi\left(\frac{1}{x}\right)^2 dx\)M1 \(\pi\) may be omitted throughout; *Limits not required*
\(= \pi\left[-\frac{1}{x}\right]_1^5 \left(= \frac{4}{5}\pi\right)\)A1 For \(-\frac{1}{x}\)
\(\int \pi x y^2\, dx = \int_1^5 \pi x\left(\frac{1}{x}\right)^2 dx\)M1 *Limits not required*
\(= \pi\left[\ln x\right]_1^5 \left(= \pi\ln 5\right)\)A1 For \(\ln x\)
\(\bar{x} = \frac{\pi\ln 5}{\frac{4}{5}\pi} = \frac{5\ln 5}{4} \quad (2.012)\)A1 *SR* If exact answers are not seen, deduct only the first A1 affected
Part (ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Area \(= \int_1^5 \frac{1}{x}\,dx\)M1 *Limits not required*
\(= \left[\ln x\right]_1^5 \left(= \ln 5\right)\)A1 For \(\ln x\)
\(\int xy\,dx = \int_1^5 x\left(\frac{1}{x}\right)dx \left(= \left[x\right]_1^5 = 4\right)\)M1 *Limits not required*
\(\bar{x} = \frac{4}{\ln 5} \quad (2.485)\)A1
\(\int \frac{1}{2}y^2\,dx = \int_1^5 \frac{1}{2}\left(\frac{1}{x}\right)^2 dx\)M1 For \(\int\left(\frac{1}{x}\right)^2 dx\)
\(= \left[-\frac{1}{2x}\right]_1^5 \left(= \frac{2}{5}\right)\)A1 For \(-\frac{1}{2x}\)
\(\bar{y} = \frac{\frac{2}{5}}{\ln 5} = \frac{2}{5\ln 5} \quad (0.2485)\)A1
Part (iii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
CM of \(R_2\) is \(\left(\frac{2}{5\ln 5},\ \frac{4}{\ln 5}\right)\)B1B1 ft *Do not penalise inexact answers in this part*
Part (iv)
AnswerMarks Guidance
Answer/WorkingMark Guidance
B1For CM of \(R_3\) is \(\left(\frac{1}{2}, \frac{1}{2}\right)\) (one coordinate is sufficient)
\(\bar{x} = \frac{(\ln 5)\!\left(\frac{4}{\ln 5}\right) + (\ln 5)\!\left(\frac{2}{5\ln 5}\right) + (1)\!\left(\frac{1}{2}\right)}{\ln 5 + \ln 5 + 1}\)M1 Using \(\sum mx\) with three terms
M1Using \(\frac{\sum mx}{\sum m}\) with at least two terms in each sum
CM is \(\left(\frac{4.9}{2\ln 5+1},\ \frac{4.9}{2\ln 5+1}\right) \quad (1.161,\ 1.161)\)A1 cao 
# Question 3:

## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Volume $= \int_1^5 \pi\left(\frac{1}{x}\right)^2 dx$ | M1 | $\pi$ may be omitted throughout; *Limits not required* |
| $= \pi\left[-\frac{1}{x}\right]_1^5 \left(= \frac{4}{5}\pi\right)$ | A1 | For $-\frac{1}{x}$ |
| $\int \pi x y^2\, dx = \int_1^5 \pi x\left(\frac{1}{x}\right)^2 dx$ | M1 | *Limits not required* |
| $= \pi\left[\ln x\right]_1^5 \left(= \pi\ln 5\right)$ | A1 | For $\ln x$ |
| $\bar{x} = \frac{\pi\ln 5}{\frac{4}{5}\pi} = \frac{5\ln 5}{4} \quad (2.012)$ | A1 | *SR* If exact answers are not seen, deduct only the first A1 affected |

## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $= \int_1^5 \frac{1}{x}\,dx$ | M1 | *Limits not required* |
| $= \left[\ln x\right]_1^5 \left(= \ln 5\right)$ | A1 | For $\ln x$ |
| $\int xy\,dx = \int_1^5 x\left(\frac{1}{x}\right)dx \left(= \left[x\right]_1^5 = 4\right)$ | M1 | *Limits not required* |
| $\bar{x} = \frac{4}{\ln 5} \quad (2.485)$ | A1 | |
| $\int \frac{1}{2}y^2\,dx = \int_1^5 \frac{1}{2}\left(\frac{1}{x}\right)^2 dx$ | M1 | For $\int\left(\frac{1}{x}\right)^2 dx$ |
| $= \left[-\frac{1}{2x}\right]_1^5 \left(= \frac{2}{5}\right)$ | A1 | For $-\frac{1}{2x}$ |
| $\bar{y} = \frac{\frac{2}{5}}{\ln 5} = \frac{2}{5\ln 5} \quad (0.2485)$ | A1 | |

## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| CM of $R_2$ is $\left(\frac{2}{5\ln 5},\ \frac{4}{\ln 5}\right)$ | B1B1 ft | *Do not penalise inexact answers in this part* |

## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | B1 | For CM of $R_3$ is $\left(\frac{1}{2}, \frac{1}{2}\right)$ (one coordinate is sufficient) |
| $\bar{x} = \frac{(\ln 5)\!\left(\frac{4}{\ln 5}\right) + (\ln 5)\!\left(\frac{2}{5\ln 5}\right) + (1)\!\left(\frac{1}{2}\right)}{\ln 5 + \ln 5 + 1}$ | M1 | Using $\sum mx$ with three terms |
| | M1 | Using $\frac{\sum mx}{\sum m}$ with at least two terms in each sum |
| CM is $\left(\frac{4.9}{2\ln 5+1},\ \frac{4.9}{2\ln 5+1}\right) \quad (1.161,\ 1.161)$ | A1 cao | |

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3 In this question, give your answers in an exact form.\\
The region $R _ { 1 }$ (shown in Fig. 3) is bounded by the $x$-axis, the lines $x = 1$ and $x = 5$, and the curve $y = \frac { 1 } { x }$ for $1 \leqslant x \leqslant 5$.\\
(i) A uniform solid of revolution is formed by rotating the region $R _ { 1 }$ through $2 \pi$ radians about the $x$-axis. Find the $x$-coordinate of the centre of mass of this solid.\\
(ii) Find the coordinates of the centre of mass of a uniform lamina occupying the region $R _ { 1 }$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{c93aed95-f655-45cb-805f-7114a15acccf-4_849_841_735_651}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}

The region $R _ { 2 }$ is bounded by the $y$-axis, the lines $y = 1$ and $y = 5$, and the curve $y = \frac { 1 } { x }$ for $\frac { 1 } { 5 } \leqslant x \leqslant 1$. The region $R _ { 3 }$ is the square with vertices $( 0,0 ) , ( 1,0 ) , ( 1,1 )$ and $( 0,1 )$.\\
(iii) Write down the coordinates of the centre of mass of a uniform lamina occupying the region $R _ { 2 }$.\\
(iv) Find the coordinates of the centre of mass of a uniform lamina occupying the region consisting of $R _ { 1 } , R _ { 2 }$ and $R _ { 3 }$ (shown shaded in Fig. 3).

\hfill \mbox{\textit{OCR MEI M3 2010 Q3 [18]}}
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