| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2012 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Prove MI by integration |
| Difficulty | Challenging +1.2 This is a structured M5 question requiring standard integration techniques for moment of inertia (part a), straightforward application of a proven result (part b), and decomposition of a composite shape using parallel axis theorem (part c). While it involves multiple steps and careful geometric reasoning, the techniques are all standard for M5 students and the question provides significant scaffolding through its parts. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(\rho = \frac{2m}{bh}\) | B1 | |
| \(\delta m = \rho \frac{b(h - x)}{h}\delta x = \frac{2m}{h^2}(h - x)\delta x\) | M1 | |
| \(\delta I = \frac{2m}{h^2}(h - x)x^2\delta x\) | A1 | |
| \(I = \int_0^h \frac{2m}{h^2}(h - x)x^2 dx = \frac{2m}{h^2}\left[\frac{hx^3}{3} - \frac{x^4}{4}\right]_0^h\) | M1 A1 | |
| \(= \frac{1}{6}mh^2\) | DM1 A1 | PRINTED ANSWER |
| Answer | Marks |
|---|---|
| \(I = 2 \times \frac{1}{6}m(a\sqrt{2})^2 = \frac{2}{3}ma^2\) | B1 |
| \(k = \sqrt{\frac{I}{M}} = \sqrt{\frac{\frac{2}{3}ma^2}{2m}} = \frac{a}{\sqrt{3}}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| MI of square about QS = \(\frac{18M}{3 \cdot 7}a^2 = \frac{8M}{21}a^2\) | M1 A1 | |
| MI of square about XY = \(\frac{8M}{21}a^2 + \frac{8M}{7}\left(\frac{a\sqrt{2}}{2}\right)^2 = \frac{20Ma^2}{21}\) | M1 A1 | |
| Hence, \(I_{ROYS} = \frac{20Ma^2}{21} - \frac{1M}{6 \cdot 7}\left(\frac{a}{\sqrt{2}}\right)^2 = \frac{79Ma^2}{84}\) | M1 A1 | PRINTED |
## (a)
$\rho = \frac{2m}{bh}$ | B1 |
$\delta m = \rho \frac{b(h - x)}{h}\delta x = \frac{2m}{h^2}(h - x)\delta x$ | M1 |
$\delta I = \frac{2m}{h^2}(h - x)x^2\delta x$ | A1 |
$I = \int_0^h \frac{2m}{h^2}(h - x)x^2 dx = \frac{2m}{h^2}\left[\frac{hx^3}{3} - \frac{x^4}{4}\right]_0^h$ | M1 A1 |
$= \frac{1}{6}mh^2$ | DM1 A1 | PRINTED ANSWER |
## (b)
$I = 2 \times \frac{1}{6}m(a\sqrt{2})^2 = \frac{2}{3}ma^2$ | B1 |
$k = \sqrt{\frac{I}{M}} = \sqrt{\frac{\frac{2}{3}ma^2}{2m}} = \frac{a}{\sqrt{3}}$ | M1 A1 |
## (c)
MI of square about QS = $\frac{18M}{3 \cdot 7}a^2 = \frac{8M}{21}a^2$ | M1 A1 |
MI of square about XY = $\frac{8M}{21}a^2 + \frac{8M}{7}\left(\frac{a\sqrt{2}}{2}\right)^2 = \frac{20Ma^2}{21}$ | M1 A1 |
Hence, $I_{ROYS} = \frac{20Ma^2}{21} - \frac{1M}{6 \cdot 7}\left(\frac{a}{\sqrt{2}}\right)^2 = \frac{79Ma^2}{84}$ | M1 A1 | PRINTED |\begin{enumerate}[label=(\alph*)]
\item A uniform lamina of mass $m$ is in the shape of a triangle $ABC$. The perpendicular distance of $C$ from the line $AB$ is $h$. Prove, using integration, that the moment of inertia of the lamina about $AB$ is $\frac{1}{6}mh^2$.
[7]
\item Deduce the radius of gyration of a uniform square lamina of side $2a$, about a diagonal.
[3]
\end{enumerate}
The points $X$ and $Y$ are the mid-points of the sides $RQ$ and $RS$ respectively of a square $PQRS$ of side $2a$. A uniform lamina of mass $M$ is in the shape of $PQXYS$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that the moment of inertia of this lamina about $XY$ is $\frac{79}{84}Ma^2$.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2012 Q7 [16]}}