OCR M4 2005 June — Question 5 8 marks

Exam BoardOCR
ModuleM4 (Mechanics 4)
Year2005
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeSolid of revolution MI
DifficultyChallenging +1.2 This is a standard M4 moment of inertia problem requiring integration by parts and exponential integration. While it involves multiple steps (finding mass, setting up MOI integral, integrating), the techniques are routine for this module and the exponential function makes the calculus straightforward. The 'show that' format provides a target to work towards, reducing problem-solving demand.
Spec1.08h Integration by substitution6.04d Integration: for centre of mass of laminas/solids
5 In this question, \(a\) and \(k\) are positive constants.
The region enclosed by the curve \(y = a \mathrm { e } ^ { - \frac { x } { a } }\) for \(0 \leqslant x \leqslant k a\), the \(x\)-axis, the \(y\)-axis and the line \(x = k a\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid of mass \(m\). Show that the moment of inertia of this solid about the \(x\)-axis is \(\frac { 1 } { 4 } m a ^ { 2 } \left( 1 + \mathrm { e } ^ { - 2 k } \right)\).
Question 5:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(m = \rho\int\pi y^2\,dx = \rho\pi\int_0^{ka} a^2 e^{-\frac{2x}{a}}\,dx\)M1 Integral of \(\left(e^{-\frac{x}{a}}\right)^2\) (when finding mass or volume)
\(= \rho\pi\left[-\frac{1}{2}a^3 e^{-\frac{2x}{a}}\right]_0^{ka}\)A1 For \(\int e^{-\frac{2x}{a}}\,dx = -\frac{1}{2}a e^{-\frac{2x}{a}}\)
\(= \frac{1}{2}\rho\pi a^3(1-e^{-2k})\)A1 For mass or volume
\(I = \int\frac{1}{2}\rho\pi y^4\,dx\) Integral of \(y^4\)
\(= \frac{1}{2}\rho\pi\int_0^{ka} a^4 e^{-\frac{4x}{a}}\,dx\)M1 Correct integral expression (in terms of \(x\))
\(= \frac{1}{2}\rho\pi\left[-\frac{1}{4}a^5 e^{-\frac{4x}{a}}\right]_0^{ka} = \frac{1}{8}\rho\pi a^5(1-e^{-4k})\)A1 ft
\(= \frac{\frac{1}{4}ma^2(1-e^{-4k})}{1-e^{-2k}}\)A1
\(= \frac{\frac{1}{4}ma^2(1-e^{-2k})(1+e^{-2k})}{1-e^{-2k}} = \frac{1}{4}ma^2(1+e^{-2k})\)M1 Dependent on previous M1M1; intermediate step not required, provided no wrong working seen
A1 (ag)
Total: 8 
# Question 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $m = \rho\int\pi y^2\,dx = \rho\pi\int_0^{ka} a^2 e^{-\frac{2x}{a}}\,dx$ | M1 | Integral of $\left(e^{-\frac{x}{a}}\right)^2$ (when finding mass or volume) |
| $= \rho\pi\left[-\frac{1}{2}a^3 e^{-\frac{2x}{a}}\right]_0^{ka}$ | A1 | For $\int e^{-\frac{2x}{a}}\,dx = -\frac{1}{2}a e^{-\frac{2x}{a}}$ |
| $= \frac{1}{2}\rho\pi a^3(1-e^{-2k})$ | A1 | For mass or volume |
| $I = \int\frac{1}{2}\rho\pi y^4\,dx$ | | Integral of $y^4$ |
| $= \frac{1}{2}\rho\pi\int_0^{ka} a^4 e^{-\frac{4x}{a}}\,dx$ | M1 | Correct integral expression (in terms of $x$) |
| $= \frac{1}{2}\rho\pi\left[-\frac{1}{4}a^5 e^{-\frac{4x}{a}}\right]_0^{ka} = \frac{1}{8}\rho\pi a^5(1-e^{-4k})$ | A1 ft | |
| $= \frac{\frac{1}{4}ma^2(1-e^{-4k})}{1-e^{-2k}}$ | A1 | |
| $= \frac{\frac{1}{4}ma^2(1-e^{-2k})(1+e^{-2k})}{1-e^{-2k}} = \frac{1}{4}ma^2(1+e^{-2k})$ | M1 | Dependent on previous M1M1; intermediate step not required, provided no wrong working seen |
| | A1 (ag) | |
| **Total: 8** | | |

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5 In this question, $a$ and $k$ are positive constants.\\
The region enclosed by the curve $y = a \mathrm { e } ^ { - \frac { x } { a } }$ for $0 \leqslant x \leqslant k a$, the $x$-axis, the $y$-axis and the line $x = k a$ is rotated through $2 \pi$ radians about the $x$-axis to form a uniform solid of mass $m$. Show that the moment of inertia of this solid about the $x$-axis is $\frac { 1 } { 4 } m a ^ { 2 } \left( 1 + \mathrm { e } ^ { - 2 k } \right)$.

\hfill \mbox{\textit{OCR M4 2005 Q5 [8]}}
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