| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Solid of revolution MI |
| Difficulty | Challenging +1.8 This is a Further Mechanics question requiring integration to derive a moment of inertia formula for a 3D solid of revolution. While the setup is given (parabola rotated about x-axis) and students can use the disc MOI formula, they must correctly set up the integral with mass elements dm = ρπy²dx, apply the perpendicular axis theorem or disc formula, and manipulate the algebra including expressing mass M in terms of volume. This requires multiple sophisticated steps beyond standard A-level, though it follows a well-established method for M5 students. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(V = \pi\int_0^a 4ax\, dx\) | M1 | |
| \(= 2\pi a^3\) | A1 | |
| \(\delta m = \frac{M}{2\pi a^3}\cdot\pi\, 4ax\,\delta x \quad \left(= \frac{2M}{a^2}x\,\delta x\right)\) | M1 | |
| \(\delta I = \frac{1}{2}\frac{2M}{a^2}x\,\delta x \cdot y^2 = \frac{4M}{a}x^2\,\delta x\) | M1 A1 | |
| \(I = \frac{4M}{a}\int_0^a x^2\,dx = \frac{4}{3}Ma^2\) | DM1 A1 | (7) |
## Question 4:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $V = \pi\int_0^a 4ax\, dx$ | M1 | |
| $= 2\pi a^3$ | A1 | |
| $\delta m = \frac{M}{2\pi a^3}\cdot\pi\, 4ax\,\delta x \quad \left(= \frac{2M}{a^2}x\,\delta x\right)$ | M1 | |
| $\delta I = \frac{1}{2}\frac{2M}{a^2}x\,\delta x \cdot y^2 = \frac{4M}{a}x^2\,\delta x$ | M1 A1 | |
| $I = \frac{4M}{a}\int_0^a x^2\,dx = \frac{4}{3}Ma^2$ | DM1 A1 | **(7)** |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5180a4e0-dafe-4595-a517-e3501f7aed40-3_780_1175_242_420}
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\caption{Figure 1}
\end{center}
\end{figure}
A region $R$ is bounded by the curve $y ^ { 2 } = 4 a x ( y > 0 )$, the $x$-axis and the line $x = a ( a > 0 )$, as shown in Figure 1. A uniform solid $S$ of mass $M$ is formed by rotating $R$ about the $x$-axis through $360 ^ { \circ }$. Using integration, prove that the moment of inertia of $S$ about the $x$-axis is $\frac { 4 } { 3 } M a ^ { 2 }$.\\
(You may assume without proof that the moment of inertia of a uniform disc, of mass $m$ and radius $r$, about an axis through its centre perpendicular to its plane is $\frac { 1 } { 2 } m r ^ { 2 }$.)\\
\hfill \mbox{\textit{Edexcel M5 2007 Q4 [7]}}