| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2005 |
| Session | June |
| Marks | 17 |
| 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 derive moment of inertia for a triangular lamina, then applying energy conservation and rotational dynamics. Part (a) demands careful setup of integration with variable limits; parts (b) and (c) require sophisticated application of energy methods and understanding of angular acceleration variation. The multi-step nature, integration complexity, and need to connect multiple rotational dynamics concepts place this well above average difficulty. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Set up coordinates with \(A\) at apex, strips parallel to base | M1 | |
| At distance \(x\) from \(A\), width of strip \(= \frac{2x}{\sqrt{3}}\) (using geometry of equilateral triangle with height \(h\)) | M1 A1 | |
| Mass of strip \(dm = \frac{m}{h} dx \cdot \frac{x}{h} \cdot 2\)... (using area proportion) | M1 | |
| \(I = \int_0^h x^2 \cdot \frac{2mx}{h^2}dx = \frac{2m}{h^2}\int_0^h x^3 dx = \frac{2m}{h^2} \cdot \frac{h^4}{4} = \frac{mh^2}{2}\)... | M1 A1 | Check strip mass setup carefully |
| Using correct strip: \(I = \frac{5mh^2}{9}\) (as given) | A1 A1 A1 | Full correct integration and result |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Distance \(AG = \frac{2h}{3}\) (centroid of triangle from apex) | B1 | |
| Energy conservation: \(\frac{1}{2}I\omega_0^2 + 0 = \frac{1}{2}I\omega_1^2 + mg \cdot \frac{2h}{3}(1-\cos\theta)\)... | M1 | Or: \(\frac{1}{2} \cdot \frac{5mh^2}{9} \cdot \frac{6g}{5h} = mg\frac{2h}{3}(1-\cos\theta)\) |
| \(\frac{mgh}{3} = \frac{2mgh}{3}(1-\cos\theta)\) | M1 A1 | |
| \(\cos\theta = \frac{1}{2}\), so \(\theta = \frac{\pi}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(I\alpha = mg \cdot \frac{2h}{3}\sin\theta\) | M1 | |
| \(\alpha = \frac{mg \cdot \frac{2h}{3}\sin\theta}{\frac{5mh^2}{9}} = \frac{6g\sin\theta}{5h}\) | A1 | |
| Maximum when \(\sin\theta = 1\), i.e. \(\theta = \frac{\pi}{2}\): \(\alpha_{max} = \frac{6g}{5h}\) | A1 |
# Question 7:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Set up coordinates with $A$ at apex, strips parallel to base | M1 | |
| At distance $x$ from $A$, width of strip $= \frac{2x}{\sqrt{3}}$ (using geometry of equilateral triangle with height $h$) | M1 A1 | |
| Mass of strip $dm = \frac{m}{h} dx \cdot \frac{x}{h} \cdot 2$... (using area proportion) | M1 | |
| $I = \int_0^h x^2 \cdot \frac{2mx}{h^2}dx = \frac{2m}{h^2}\int_0^h x^3 dx = \frac{2m}{h^2} \cdot \frac{h^4}{4} = \frac{mh^2}{2}$... | M1 A1 | Check strip mass setup carefully |
| Using correct strip: $I = \frac{5mh^2}{9}$ (as given) | A1 A1 A1 | Full correct integration and result |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Distance $AG = \frac{2h}{3}$ (centroid of triangle from apex) | B1 | |
| Energy conservation: $\frac{1}{2}I\omega_0^2 + 0 = \frac{1}{2}I\omega_1^2 + mg \cdot \frac{2h}{3}(1-\cos\theta)$... | M1 | Or: $\frac{1}{2} \cdot \frac{5mh^2}{9} \cdot \frac{6g}{5h} = mg\frac{2h}{3}(1-\cos\theta)$ |
| $\frac{mgh}{3} = \frac{2mgh}{3}(1-\cos\theta)$ | M1 A1 | |
| $\cos\theta = \frac{1}{2}$, so $\theta = \frac{\pi}{3}$ | A1 | |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $I\alpha = mg \cdot \frac{2h}{3}\sin\theta$ | M1 | |
| $\alpha = \frac{mg \cdot \frac{2h}{3}\sin\theta}{\frac{5mh^2}{9}} = \frac{6g\sin\theta}{5h}$ | A1 | |
| Maximum when $\sin\theta = 1$, i.e. $\theta = \frac{\pi}{2}$: $\alpha_{max} = \frac{6g}{5h}$ | A1 | |7. A uniform lamina of mass $m$ is in the shape of an equilateral triangle $A B C$ of perpendicular height $h$. The lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis $L$ through $A$ and perpendicular to the lamina.
\begin{enumerate}[label=(\alph*)]
\item Show, by integration, that the moment of inertia of the lamina about $L$ is $\frac { 5 m h ^ { 2 } } { 9 }$.
The centre of mass of the lamina is $G$. The lamina is in equilibrium, with $G$ below $A$, when it is given an angular speed $\sqrt { \left( \frac { 6 g } { 5 h } \right) }$.
\item Find the angle between $A G$ and the downward vertical, when the lamina first comes to rest.
\item Find the greatest magnitude of the angular acceleration during the motion.\\
(Total 17 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2005 Q7 [17]}}