OCR M4 2012 June — Question 2 7 marks

Exam BoardOCR
ModuleM4 (Mechanics 4)
Year2012
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeCentre of mass of solid of revolution
DifficultyChallenging +1.2 This is a standard M4 centre of mass problem requiring integration of πy² for volume and πy²x for moment. The algebra involves polynomial expansion and straightforward integration, though the parametric form and multiple integration steps elevate it slightly above average difficulty for A-level.
Spec6.04d Integration: for centre of mass of laminas/solids
2 A uniform solid of revolution is formed by rotating the region bounded by the \(x\)-axis and the curve \(y = x \left( 1 - \frac { x ^ { 2 } } { a ^ { 2 } } \right)\) for \(0 \leqslant x \leqslant a\), where \(a\) is a constant, about the \(x\)-axis. Find the \(x\)-coordinate of the centre of mass of this solid.
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(V = \int_0^a \pi x^2\left(1 - \frac{x^2}{a^2}\right)^2 dx\)M1 For \(\int x^2\left(1-\frac{x^2}{a^2}\right)^2 dx\)
\(= \pi\left[\frac{x^3}{3} - \frac{2x^5}{5a^2} + \frac{x^7}{7a^4}\right]_0^a\) \(\left(= \frac{8\pi a^3}{105}\right)\)A1 For \(\frac{x^3}{3} - \frac{2x^5}{5a^2} + \frac{x^7}{7a^4}\)
\(V\bar{x} = \int \pi xy^2\,dx = \int_0^a \pi x^3\left(1 - \frac{2x^2}{a^2} + \frac{x^4}{a^4}\right)dx\)M1 For \(\int xy^2\,dx\)
\(= \pi\left[\frac{x^4}{4} - \frac{x^6}{3a^2} + \frac{x^8}{8a^4}\right]_0^a\) \(\left(= \frac{\pi a^4}{24}\right)\)A2 For \(\frac{x^4}{4} - \frac{x^6}{3a^2} + \frac{x^8}{8a^4}\); Give A1 if one error
\(\bar{x} = \frac{\frac{1}{24}\pi a^4}{\frac{8}{105}\pi a^3}\)M1 Dependent on previous M1M1
\(= \frac{35a}{64}\)A1 [7] Accept \(0.547a\) 
# Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = \int_0^a \pi x^2\left(1 - \frac{x^2}{a^2}\right)^2 dx$ | M1 | For $\int x^2\left(1-\frac{x^2}{a^2}\right)^2 dx$ |
| $= \pi\left[\frac{x^3}{3} - \frac{2x^5}{5a^2} + \frac{x^7}{7a^4}\right]_0^a$ $\left(= \frac{8\pi a^3}{105}\right)$ | A1 | For $\frac{x^3}{3} - \frac{2x^5}{5a^2} + \frac{x^7}{7a^4}$ |
| $V\bar{x} = \int \pi xy^2\,dx = \int_0^a \pi x^3\left(1 - \frac{2x^2}{a^2} + \frac{x^4}{a^4}\right)dx$ | M1 | For $\int xy^2\,dx$ |
| $= \pi\left[\frac{x^4}{4} - \frac{x^6}{3a^2} + \frac{x^8}{8a^4}\right]_0^a$ $\left(= \frac{\pi a^4}{24}\right)$ | A2 | For $\frac{x^4}{4} - \frac{x^6}{3a^2} + \frac{x^8}{8a^4}$; Give A1 if one error |
| $\bar{x} = \frac{\frac{1}{24}\pi a^4}{\frac{8}{105}\pi a^3}$ | M1 | Dependent on previous M1M1 |
| $= \frac{35a}{64}$ | A1 [7] | Accept $0.547a$ |

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2 A uniform solid of revolution is formed by rotating the region bounded by the $x$-axis and the curve $y = x \left( 1 - \frac { x ^ { 2 } } { a ^ { 2 } } \right)$ for $0 \leqslant x \leqslant a$, where $a$ is a constant, about the $x$-axis. Find the $x$-coordinate of the centre of mass of this solid.

\hfill \mbox{\textit{OCR M4 2012 Q2 [7]}}
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