| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2008 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Impulse and angular speed |
| Difficulty | Challenging +1.8 This M5 rotation question requires multiple sophisticated techniques: deriving angular velocity using energy methods, finding reaction forces perpendicular to a diagonal (requiring careful resolution and application of angular equation of motion), and impulse-momentum with moment of momentum. The geometry of forces perpendicular to AC diagonal adds complexity beyond standard rotation problems. However, it's a structured multi-part question with clear guidance ('show that') rather than requiring novel insight. |
| Spec | 6.03f Impulse-momentum: relation6.04e Rigid body equilibrium: coplanar forces6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{1}{2}(24ma^2)\dot{\theta}^2 = 2mg \cdot 3a(1-\cos\theta)\) | M1 A1 A1 | |
| \(2a\dot{\theta}^2 = g(1-\cos\theta)\) | A1 | *(4)* |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \((\searrow)\): \(2mg\sin\theta - X = 2m \cdot 3a\ddot{\theta}\) | M1 A2 | |
| \(M(L)\): \(2mg \cdot 3a\sin\theta = 24ma^2\ddot{\theta}\) | M1 A1 | |
| \(\Rightarrow X = 2mg\sin\theta - 6ma\!\left(\dfrac{g\sin\theta}{4a}\right)\) | DM1 | |
| \(= \dfrac{mg\sin\theta}{2}\) | A1 | *(7)* |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\theta = \pi\): \(2a\dot{\theta}^2 = g(1 - \cos\pi)\) | M1 | |
| \(\dot{\theta} = \sqrt{\dfrac{g}{a}}\) | A1 | |
| \(6a \cdot I = 24ma^2\sqrt{\dfrac{g}{a}}\) | M1 A1 | |
| \(\Rightarrow I = 4m\sqrt{ag}\) | A1 | Total: 16 |
## Question 7:
### Part (a):
| Working | Marks | Notes |
|---|---|---|
| $\frac{1}{2}(24ma^2)\dot{\theta}^2 = 2mg \cdot 3a(1-\cos\theta)$ | M1 A1 A1 | |
| $2a\dot{\theta}^2 = g(1-\cos\theta)$ | A1 | ***(4)*** |
### Part (b):
| Working | Marks | Notes |
|---|---|---|
| $(\searrow)$: $2mg\sin\theta - X = 2m \cdot 3a\ddot{\theta}$ | M1 A2 | |
| $M(L)$: $2mg \cdot 3a\sin\theta = 24ma^2\ddot{\theta}$ | M1 A1 | |
| $\Rightarrow X = 2mg\sin\theta - 6ma\!\left(\dfrac{g\sin\theta}{4a}\right)$ | DM1 | |
| $= \dfrac{mg\sin\theta}{2}$ | A1 | ***(7)*** |
### Part (c):
| Working | Marks | Notes |
|---|---|---|
| $\theta = \pi$: $2a\dot{\theta}^2 = g(1 - \cos\pi)$ | M1 | |
| $\dot{\theta} = \sqrt{\dfrac{g}{a}}$ | A1 | |
| $6a \cdot I = 24ma^2\sqrt{\dfrac{g}{a}}$ | M1 A1 | |
| $\Rightarrow I = 4m\sqrt{ag}$ | A1 | **Total: 16** |7. A uniform square lamina $A B C D$, of mass $2 m$ and side $3 a \sqrt { 2 }$, is free to rotate in a vertical plane about a fixed smooth horizontal axis $L$ which passes through $A$ and is perpendicular to the plane of the lamina. The moment of inertia of the lamina about $L$ is $24 m a ^ { 2 }$.
The lamina is at rest with $C$ vertically above $A$. At time $t = 0$ the lamina is slightly displaced. At time $t$ the lamina has rotated through an angle $\theta$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$2 a \left( \frac { d \theta } { d t } \right) ^ { 2 } = g ( 1 - \cos \theta )$$
\item Show that, at time $t$, the magnitude of the component of the force acting on the lamina at $A$, in a direction perpendicular to $A C$, is $\frac { 1 } { 2 } m g \sin \theta$.
When the lamina reaches the position with $C$ vertically below $A$, it receives an impulse which acts at $C$, in the plane of the lamina and in a direction which is perpendicular to the line $A C$. As a result of this impulse the lamina is brought immediately to rest.
\item Find the magnitude of the impulse.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2008 Q7 [16]}}