| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2009 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Force at pivot/axis |
| Difficulty | Challenging +1.8 This M5 question requires calculating moment of inertia for a composite body using parallel axis theorem, applying energy conservation and Newton's second law for rotation, then using small angle approximation for SHM analysis. The multi-step nature, 3D geometry visualization, and integration of multiple mechanics concepts (moments of inertia, rotational dynamics, energy methods, SHM) make this significantly harder than average, though the techniques are standard for Further Maths M5. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| MI of rod \(AB\) about \(L\): using parallel axis or direct, end at \(B\): \(\frac{1}{3}(2m)(4a)^2 = \frac{32ma^2}{3} \cdot ... \) rod length \(4a\), axis through end \(B\): \(\frac{1}{3}(2m)(4a)^2 = \frac{32ma^2}{3}\) | M1 | Correct MI for rod about \(L\) |
| MI of square lamina about axis through centre parallel to \(PQ\): \(\frac{1}{12}(4m)(a)^2 = \frac{ma^2}{3}\) | M1 | MI of square about central axis |
| Using parallel axis theorem, distance from centre of square to \(L\) is \(4a\): \(\frac{ma^2}{3} + 4m(4a)^2 = \frac{ma^2}{3}+64ma^2\) | A1 | Correct parallel axis application |
| Total \(= \frac{32ma^2}{3} + \frac{ma^2}{3} + 64ma^2 = \frac{33ma^2}{3}+64ma^2 = 11ma^2+64ma^2 = 75ma^2\) | A1 | Correct total, printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Distance of CoM from \(B\): rod CoM at \(2a\), square CoM at \(4a\) from \(B\) | B1 | Distances identified |
| Total CoM distance: \(\frac{2m(2a)+4m(4a)}{6m} = \frac{4a+16a}{6} = \frac{20a}{6} = \frac{10a}{3}\) | B1 | Correct CoM position |
| Energy equation: \(\frac{1}{2}(75ma^2)\dot\theta^2 = 6mg\cdot\frac{10a}{3}(\cos\theta - \cos\alpha)\) | M1A1 | Correct energy equation |
| \(75ma^2\dot\theta^2 = 40mga(\cos\theta-\cos\alpha)\) | A1 | Simplified |
| Equation of motion: \(75ma^2\ddot\theta = -6mg\cdot\frac{10a}{3}\sin\theta\) | M1 | Newton's 2nd law for rotation |
| \(X - 6mg\cos\theta = 6m\frac{10a}{3}\dot\theta^2 - ...\); radial equation along \(AB\): \(X - 6mg\cos\theta = 6m \cdot\frac{10a}{3}\dot\theta^2\) | M1 | Radial force equation |
| Substituting \(\dot\theta^2\): \(X = 6mg\cos\theta + \frac{8mg}{15}(\cos\theta-\cos\alpha)\cdot\frac{20}{1}\)... | M1 | Substituting energy result |
| \(X = 6mg\cos\theta + \frac{8mg}{15}\cdot 40 \cdot (\cos\theta - \cos\alpha) \cdot \frac{1}{...}\) | A1 | |
| \(X = \frac{2mg}{15}(28\cos\theta - \cos\alpha \cdot 8)\) or correct simplified form | A1 | Correct expression for \(X\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\tan\alpha = \frac{7}{24}\), so \(\sin\alpha = \frac{7}{25}\), \(\cos\alpha = \frac{24}{25}\) | B1 | Correct trig values |
| Equation of motion: \(75ma^2\ddot\theta = -20mga\sin\theta \approx -20mga\theta\) | M1 | Small angle approximation |
| \(\ddot\theta = -\frac{20g}{75a}\theta = -\frac{4g}{15a}\theta\) | A1 | Correct SHM equation |
| Period \(T = 2\pi\sqrt{\frac{15a}{4g}}\), time for angle \(\alpha\) from rest | M1 | SHM period identified |
| \(\theta = \alpha\cos(\omega t)\) where \(\omega = \sqrt{\frac{4g}{15a}}\); reaches \(\theta=0\) when \(\omega t = \frac{\pi}{2}\) | M1 | Using SHM solution |
| Time \(= \frac{\pi}{2}\sqrt{\frac{15a}{4g}} = \frac{\pi}{4}\sqrt{\frac{15a}{g}}\) | A1 | Correct time |
# Question 6:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| MI of rod $AB$ about $L$: using parallel axis or direct, end at $B$: $\frac{1}{3}(2m)(4a)^2 = \frac{32ma^2}{3} \cdot ... $ rod length $4a$, axis through end $B$: $\frac{1}{3}(2m)(4a)^2 = \frac{32ma^2}{3}$ | M1 | Correct MI for rod about $L$ |
| MI of square lamina about axis through centre parallel to $PQ$: $\frac{1}{12}(4m)(a)^2 = \frac{ma^2}{3}$ | M1 | MI of square about central axis |
| Using parallel axis theorem, distance from centre of square to $L$ is $4a$: $\frac{ma^2}{3} + 4m(4a)^2 = \frac{ma^2}{3}+64ma^2$ | A1 | Correct parallel axis application |
| Total $= \frac{32ma^2}{3} + \frac{ma^2}{3} + 64ma^2 = \frac{33ma^2}{3}+64ma^2 = 11ma^2+64ma^2 = 75ma^2$ | A1 | Correct total, printed answer |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Distance of CoM from $B$: rod CoM at $2a$, square CoM at $4a$ from $B$ | B1 | Distances identified |
| Total CoM distance: $\frac{2m(2a)+4m(4a)}{6m} = \frac{4a+16a}{6} = \frac{20a}{6} = \frac{10a}{3}$ | B1 | Correct CoM position |
| Energy equation: $\frac{1}{2}(75ma^2)\dot\theta^2 = 6mg\cdot\frac{10a}{3}(\cos\theta - \cos\alpha)$ | M1A1 | Correct energy equation |
| $75ma^2\dot\theta^2 = 40mga(\cos\theta-\cos\alpha)$ | A1 | Simplified |
| Equation of motion: $75ma^2\ddot\theta = -6mg\cdot\frac{10a}{3}\sin\theta$ | M1 | Newton's 2nd law for rotation |
| $X - 6mg\cos\theta = 6m\frac{10a}{3}\dot\theta^2 - ...$; radial equation along $AB$: $X - 6mg\cos\theta = 6m \cdot\frac{10a}{3}\dot\theta^2$ | M1 | Radial force equation |
| Substituting $\dot\theta^2$: $X = 6mg\cos\theta + \frac{8mg}{15}(\cos\theta-\cos\alpha)\cdot\frac{20}{1}$... | M1 | Substituting energy result |
| $X = 6mg\cos\theta + \frac{8mg}{15}\cdot 40 \cdot (\cos\theta - \cos\alpha) \cdot \frac{1}{...}$ | A1 | |
| $X = \frac{2mg}{15}(28\cos\theta - \cos\alpha \cdot 8)$ or correct simplified form | A1 | Correct expression for $X$ |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\tan\alpha = \frac{7}{24}$, so $\sin\alpha = \frac{7}{25}$, $\cos\alpha = \frac{24}{25}$ | B1 | Correct trig values |
| Equation of motion: $75ma^2\ddot\theta = -20mga\sin\theta \approx -20mga\theta$ | M1 | Small angle approximation |
| $\ddot\theta = -\frac{20g}{75a}\theta = -\frac{4g}{15a}\theta$ | A1 | Correct SHM equation |
| Period $T = 2\pi\sqrt{\frac{15a}{4g}}$, time for angle $\alpha$ from rest | M1 | SHM period identified |
| $\theta = \alpha\cos(\omega t)$ where $\omega = \sqrt{\frac{4g}{15a}}$; reaches $\theta=0$ when $\omega t = \frac{\pi}{2}$ | M1 | Using SHM solution |
| Time $= \frac{\pi}{2}\sqrt{\frac{15a}{4g}} = \frac{\pi}{4}\sqrt{\frac{15a}{g}}$ | A1 | Correct time |
This image shows a blank answer/working page from a June 2009 exam paper (page 21, "Question 6 continued"). There is no mark scheme content visible on this page — it is simply a lined continuation sheet for student responses, with a barcode at the bottom (N34277RA0124).
There is no mark scheme content to extract from this image.\begin{enumerate}
\item A pendulum consists of a uniform rod $A B$, of length $4 a$ and mass $2 m$, whose end $A$ is rigidly attached to the centre $O$ of a uniform square lamina $P Q R S$, of mass $4 m$ and side $a$. The $\operatorname { rod } A B$ is perpendicular to the plane of the lamina. The pendulum is free to rotate about a fixed smooth horizontal axis $L$ which passes through $B$. The axis $L$ is perpendicular to $A B$ and parallel to the edge $P Q$ of the square.\\
(a) Show that the moment of inertia of the pendulum about $L$ is $75 m a ^ { 2 }$.
\end{enumerate}
The pendulum is released from rest when $B A$ makes an angle $\alpha$ with the downward vertical through $B$, where $\tan \alpha = \frac { 7 } { 24 }$. When $B A$ makes an angle $\theta$ with the downward vertical through $B$, the magnitude of the component, in the direction $A B$, of the force exerted by the axis $L$ on the pendulum is $X$.\\
(b) Find an expression for $X$ in terms of $m , g$ and $\theta$.
Using the approximation $\theta \approx \sin \theta$,\\
(c) find an estimate of the time for the pendulum to rotate through an angle $\alpha$ from its initial rest position.
\hfill \mbox{\textit{Edexcel M5 2009 Q6 [19]}}