| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Prove MI by integration |
| Difficulty | Challenging +1.8 This is a challenging Further Maths M5 question requiring integration to prove moment of inertia, then applying rotational dynamics with energy conservation. Part (a) demands careful setup of integration using the parallel axis theorem for elemental strips. Part (b) combines impulse-momentum with energy methods through a 120° rotation. While systematic, it requires multiple advanced techniques and careful geometric reasoning beyond standard A-level, typical of harder M5 problems. |
| Spec | 6.03f Impulse-momentum: relation6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Surface density \(\rho = \frac{M}{\frac{1}{2}a \cdot 2a} = \frac{M}{a^2}\) | B1 | Correct surface density |
| Strip of width \(\delta x\) at distance \(x\) from \(OA\), length \(= \frac{2a \cdot x}{a} = 2x\) | M1 | Forming a strip parallel to \(OA\) with correct length |
| Mass of strip \(= \rho \cdot 2x \cdot \delta x = \frac{2Mx}{a^2}\delta x\) | A1 | Correct mass element |
| \(I = \int_0^a \frac{1}{3}(2x)^2 \cdot \frac{M}{a^2} \cdot 2x \, dx\) using rod formula \(\frac{4}{3}ml^2\) with \(l=x\) | M1 | Using given rod MI formula correctly |
| \(I = \int_0^a \frac{4}{3}x^2 \cdot \frac{2Mx}{a^2} \, dx = \frac{8M}{3a^2}\int_0^a x^3 \, dx\) | A1 | Correct integrand |
| \(= \frac{8M}{3a^2} \cdot \frac{a^4}{4} = \frac{2}{3}Ma^2\) | A1 | Correct completion, printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Centre of mass of lamina: \(\bar{x} = \frac{2a}{3}\) from \(O\) along \(OA\), \(\bar{y} = \frac{2a}{3}\) from \(OA\) | B1 | Distance of CoM from axis \(OA\) is \(\frac{2a}{3}\) |
| Initial position: \(B\) is vertically below \(A\), so CoM is at distance \(\frac{2a}{3}\) below \(OA\) | M1 | Identifying initial and final positions of CoM |
| After rotating \(120°\), CoM has moved; change in height of CoM \(= \frac{2a}{3} - \frac{2a}{3}\cos 120° \cdot ...\); using energy: \(\frac{1}{2}I\omega^2 = Mg\Delta h\) | M1 | Using work-energy with correct \(I\) |
| Height of CoM initially below axis: \(\frac{2a}{3}\); after \(120°\) rotation, height relative to axis: \(\frac{2a}{3}\cos(30°)\) above axis | M1 | Correct geometry for height change |
| \(\Delta h = \frac{2a}{3} + \frac{2a}{3}\cos 30° = \frac{2a}{3}(1 + \frac{\sqrt{3}}{2})\) | A1 | Correct height change |
| \(\frac{1}{2} \cdot \frac{2}{3}Ma^2 \cdot \omega^2 = Mg \cdot \frac{2a}{3}(1+\frac{\sqrt{3}}{2})\) | A1 | Correct energy equation |
| Angular impulse: \(J \cdot 2a = I\omega = \frac{2}{3}Ma^2\omega\) | M1 | Taking moments of impulse about \(OA\) |
| \(J = \frac{Ma\omega}{3}\), then substituting for \(\omega\) | DM1 | Correct impulse-momentum equation |
| \(J = \frac{M}{3}\sqrt{ag(3+\sqrt{3})}\) | A1 | Correct final answer |
# Question 4:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Surface density $\rho = \frac{M}{\frac{1}{2}a \cdot 2a} = \frac{M}{a^2}$ | B1 | Correct surface density |
| Strip of width $\delta x$ at distance $x$ from $OA$, length $= \frac{2a \cdot x}{a} = 2x$ | M1 | Forming a strip parallel to $OA$ with correct length |
| Mass of strip $= \rho \cdot 2x \cdot \delta x = \frac{2Mx}{a^2}\delta x$ | A1 | Correct mass element |
| $I = \int_0^a \frac{1}{3}(2x)^2 \cdot \frac{M}{a^2} \cdot 2x \, dx$ using rod formula $\frac{4}{3}ml^2$ with $l=x$ | M1 | Using given rod MI formula correctly |
| $I = \int_0^a \frac{4}{3}x^2 \cdot \frac{2Mx}{a^2} \, dx = \frac{8M}{3a^2}\int_0^a x^3 \, dx$ | A1 | Correct integrand |
| $= \frac{8M}{3a^2} \cdot \frac{a^4}{4} = \frac{2}{3}Ma^2$ | A1 | Correct completion, printed answer |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Centre of mass of lamina: $\bar{x} = \frac{2a}{3}$ from $O$ along $OA$, $\bar{y} = \frac{2a}{3}$ from $OA$ | B1 | Distance of CoM from axis $OA$ is $\frac{2a}{3}$ |
| Initial position: $B$ is vertically below $A$, so CoM is at distance $\frac{2a}{3}$ below $OA$ | M1 | Identifying initial and final positions of CoM |
| After rotating $120°$, CoM has moved; change in height of CoM $= \frac{2a}{3} - \frac{2a}{3}\cos 120° \cdot ...$; using energy: $\frac{1}{2}I\omega^2 = Mg\Delta h$ | M1 | Using work-energy with correct $I$ |
| Height of CoM initially below axis: $\frac{2a}{3}$; after $120°$ rotation, height relative to axis: $\frac{2a}{3}\cos(30°)$ above axis | M1 | Correct geometry for height change |
| $\Delta h = \frac{2a}{3} + \frac{2a}{3}\cos 30° = \frac{2a}{3}(1 + \frac{\sqrt{3}}{2})$ | A1 | Correct height change |
| $\frac{1}{2} \cdot \frac{2}{3}Ma^2 \cdot \omega^2 = Mg \cdot \frac{2a}{3}(1+\frac{\sqrt{3}}{2})$ | A1 | Correct energy equation |
| Angular impulse: $J \cdot 2a = I\omega = \frac{2}{3}Ma^2\omega$ | M1 | Taking moments of impulse about $OA$ |
| $J = \frac{Ma\omega}{3}$, then substituting for $\omega$ | DM1 | Correct impulse-momentum equation |
| $J = \frac{M}{3}\sqrt{ag(3+\sqrt{3})}$ | A1 | Correct final answer |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a11940cb-73a8-4f33-bfbc-73841320c1dc-07_515_415_210_758}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform lamina of mass $M$ is in the shape of a right-angled triangle $O A B$. The angle $O A B$ is $90 ^ { \circ } , O A = a$ and $A B = 2 a$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Prove, using integration, that the moment of inertia of the lamina $O A B$ about the edge $O A$ is $\frac { 2 } { 3 } M a ^ { 2 }$.\\
(You may assume without proof that the moment of inertia of a uniform rod of mass $m$ and length $2 l$ about an axis through one end and perpendicular to the rod is $\frac { 4 } { 3 } m l ^ { 2 }$.)
The lamina $O A B$ is free to rotate about a fixed smooth horizontal axis along the edge $O A$ and hangs at rest with $B$ vertically below $A$. The lamina is then given a horizontal impulse of magnitude $J$. The impulse is applied to the lamina at the point $B$, in a direction which is perpendicular to the plane of the lamina. Given that the lamina first comes to instantaneous rest after rotating through an angle of $120 ^ { \circ }$,
\item find an expression for $J$, in terms of $M , a$ and $g$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2009 Q4 [13]}}