OCR M4 2013 June — Question 3 8 marks

Exam BoardOCR
ModuleM4 (Mechanics 4)
Year2013
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 the formula I = ∫ρπy²·y² dx for rotation about the x-axis. While it involves exponential functions and requires careful algebraic manipulation through multiple steps (finding mass, setting up the integral, integrating e^(2x/a), and expressing in terms of M), it follows a well-established template that M4 students would have practiced extensively. The 8 marks reflect the length rather than conceptual difficulty—it's procedurally demanding but not requiring novel insight.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)4.08d Volumes of revolution: about x and y axes6.04e Rigid body equilibrium: coplanar forces
3 The region \(R\) is bounded by the \(x\)-axis, the \(y\)-axis, the curve \(y = a \mathrm { e } ^ { \frac { x } { a } }\) and the line \(x = a \ln 2\) (where \(a\) is a positive constant). A uniform solid of revolution, of mass \(M\), is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis. Find, in terms of \(M\) and \(a\), the moment of inertia about the \(x\)-axis of this solid of revolution.
[0pt] [8]
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(M = \rho\int\pi y^2\,dx = \int_0^{a\ln 2}\rho\pi a^2 e^{\frac{2x}{a}}\,dx\)M1 For \(\int\!\left(e^{\frac{x}{a}}\right)^2 dx\)
\(= \left[\rho\pi\dfrac{a^3}{2}e^{\frac{2x}{a}}\right]_0^{a\ln 2} = \dfrac{3}{2}\rho\pi a^3\)A1 For equation \(M = \frac{3}{2}\rho\pi a^3\) oe
\(I = \sum\frac{1}{2}(\rho\pi y^2\delta x)y^2 = \frac{1}{2}\rho\pi\int y^4\,dx\)M1 For \(\int y^4\,dx\)
\(= \int_0^{a\ln 2}\frac{1}{2}\rho\pi\!\left(ae^{\frac{x}{a}}\right)^4 dx\)A1 Correct integral expression for \(I\)
\(= \left[\frac{1}{2}\rho\pi\dfrac{a^5}{4}e^{\frac{4x}{a}}\right]_0^{a\ln 2}\)M1 Integral is a multiple of \(e^{\frac{4x}{a}}\)
\(= \dfrac{15}{8}\rho\pi a^5\)A1
\(= \dfrac{15}{8}\!\left(\dfrac{2M}{3\pi a^3}\right)\pi a^5\)M1 Obtaining \(I\) in terms of \(M\) and \(a\) — Dependent on first two M1M1
\(= \dfrac{5}{4}Ma^2\)A1 A0 for \(\dfrac{e^{4\ln 2}-1}{4(e^{2\ln 2}-1)}Ma^2\) etc — Accept \(\dfrac{15}{12}Ma^2\) etc
[8] 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $M = \rho\int\pi y^2\,dx = \int_0^{a\ln 2}\rho\pi a^2 e^{\frac{2x}{a}}\,dx$ | M1 | For $\int\!\left(e^{\frac{x}{a}}\right)^2 dx$ |
| $= \left[\rho\pi\dfrac{a^3}{2}e^{\frac{2x}{a}}\right]_0^{a\ln 2} = \dfrac{3}{2}\rho\pi a^3$ | A1 | For equation $M = \frac{3}{2}\rho\pi a^3$ oe |
| $I = \sum\frac{1}{2}(\rho\pi y^2\delta x)y^2 = \frac{1}{2}\rho\pi\int y^4\,dx$ | M1 | For $\int y^4\,dx$ |
| $= \int_0^{a\ln 2}\frac{1}{2}\rho\pi\!\left(ae^{\frac{x}{a}}\right)^4 dx$ | A1 | Correct integral expression for $I$ |
| $= \left[\frac{1}{2}\rho\pi\dfrac{a^5}{4}e^{\frac{4x}{a}}\right]_0^{a\ln 2}$ | M1 | Integral is a multiple of $e^{\frac{4x}{a}}$ |
| $= \dfrac{15}{8}\rho\pi a^5$ | A1 | |
| $= \dfrac{15}{8}\!\left(\dfrac{2M}{3\pi a^3}\right)\pi a^5$ | M1 | Obtaining $I$ in terms of $M$ and $a$ — Dependent on first two M1M1 |
| $= \dfrac{5}{4}Ma^2$ | A1 | A0 for $\dfrac{e^{4\ln 2}-1}{4(e^{2\ln 2}-1)}Ma^2$ etc — Accept $\dfrac{15}{12}Ma^2$ etc |
| **[8]** | | |

---
3 The region $R$ is bounded by the $x$-axis, the $y$-axis, the curve $y = a \mathrm { e } ^ { \frac { x } { a } }$ and the line $x = a \ln 2$ (where $a$ is a positive constant). A uniform solid of revolution, of mass $M$, is formed by rotating $R$ through $2 \pi$ radians about the $x$-axis. Find, in terms of $M$ and $a$, the moment of inertia about the $x$-axis of this solid of revolution.\\[0pt]
[8]

\hfill \mbox{\textit{OCR M4 2013 Q3 [8]}}
This paper (7 questions)
View full paper
Q1 5 Q2 7 Q3 8 Q4 12 Q5 14 Q6 12 Q7 14