OCR M4 2012 June — Question 4 7 marks

Exam BoardOCR
ModuleM4 (Mechanics 4)
Year2012
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeProve MI by integration
DifficultyChallenging +1.2 This is a standard M4 moment of inertia problem requiring integration with a straightforward exponential function. While it involves setting up the integral MI = ∫ky²dA and substituting y = e^(x/2), the integration itself is routine (exponential functions) and the mass per unit area calculation is direct. It's harder than average A-level due to being Further Maths content, but represents a typical textbook M4 exercise without requiring novel insight or particularly complex manipulation.
Spec6.04d Integration: for centre of mass of laminas/solids
4 A uniform lamina of mass 18 kg occupies the region bounded by the \(x\)-axis, the \(y\)-axis, the line \(x = \ln 9\) and the curve \(y = \mathrm { e } ^ { \frac { 1 } { 2 } x }\) for \(0 \leqslant x \leqslant \ln 9\). The unit of length is the metre. Find the moment of inertia of this lamina about the \(x\)-axis.
Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
Area is \(\int_0^{\ln 9} e^{\frac{1}{2}x}\,dx\)M1 For \(\int e^{\frac{1}{2}x}\,dx\)
\(= \left[2e^{\frac{1}{2}x}\right]_0^{\ln 9} = 4\)A1
Mass per unit area is \(\rho = \frac{18}{4} = 4.5\)M1
\(I = \sum \frac{1}{3}(\rho\, y\,\delta x)y^2 = \frac{1}{3}\rho\int y^3\,dx\)M1 For \(\int \ldots y^3\,dx\)
\(= \frac{1}{3}\rho\int_0^{\ln 9}\left(e^{\frac{1}{2}x}\right)^3 dx\)A1 Correct integral expression for \(I\)
\(= \frac{1}{3}\rho\left[\frac{2}{3}e^{\frac{3}{2}x}\right]_0^{\ln 9} = \frac{\rho}{3} \times \frac{52}{3}\)A1 For \(\int_0^{\ln 9}\left(e^{\frac{1}{2}x}\right)^3 dx = \frac{52}{3}\)
MI is \(26\) kg m²A1 [7] 
# Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Area is $\int_0^{\ln 9} e^{\frac{1}{2}x}\,dx$ | M1 | For $\int e^{\frac{1}{2}x}\,dx$ |
| $= \left[2e^{\frac{1}{2}x}\right]_0^{\ln 9} = 4$ | A1 | |
| Mass per unit area is $\rho = \frac{18}{4} = 4.5$ | M1 | |
| $I = \sum \frac{1}{3}(\rho\, y\,\delta x)y^2 = \frac{1}{3}\rho\int y^3\,dx$ | M1 | For $\int \ldots y^3\,dx$ |
| $= \frac{1}{3}\rho\int_0^{\ln 9}\left(e^{\frac{1}{2}x}\right)^3 dx$ | A1 | Correct integral expression for $I$ |
| $= \frac{1}{3}\rho\left[\frac{2}{3}e^{\frac{3}{2}x}\right]_0^{\ln 9} = \frac{\rho}{3} \times \frac{52}{3}$ | A1 | For $\int_0^{\ln 9}\left(e^{\frac{1}{2}x}\right)^3 dx = \frac{52}{3}$ |
| MI is $26$ kg m² | A1 [7] | |

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4 A uniform lamina of mass 18 kg occupies the region bounded by the $x$-axis, the $y$-axis, the line $x = \ln 9$ and the curve $y = \mathrm { e } ^ { \frac { 1 } { 2 } x }$ for $0 \leqslant x \leqslant \ln 9$. The unit of length is the metre. Find the moment of inertia of this lamina about the $x$-axis.

\hfill \mbox{\textit{OCR M4 2012 Q4 [7]}}
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Q1 6 Q2 7 Q3 10 Q4 7 Q5 15 Q6 12 Q7 15