| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Small oscillations period |
| Difficulty | Challenging +1.8 This M5 question requires energy methods to find angular velocity, then applying Newton's second law for rotation plus resolving forces at a pivot. Part (b) needs small oscillation approximation and the rotational SHM period formula. It combines multiple mechanics concepts (energy conservation, moments, rotational dynamics, SHM) in a non-trivial configuration, requiring careful geometric reasoning about the center of mass position and systematic application of several techniques, making it significantly harder than average but still within standard M5 scope. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04d 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 |
| Distance from \(L\) to centre of mass of lamina: \(AG = a\sqrt{2}\) (diagonal of square side \(2a\), half = \(a\sqrt{2}\)) | B1 | |
| Energy conservation from release to \(AC\) vertical: | M1 | |
| \(\frac{1}{2} \cdot \frac{8ma^2}{3} \cdot \dot{\theta}^2 = mga\sqrt{2}(\cos\frac{\pi}{3} - \cos 0)\)... use correct angle change | M1 A1 | |
| \(\frac{4ma^2}{3}\dot{\theta}^2 = mga\sqrt{2}(1 - \cos\frac{\pi}{3}) = mga\sqrt{2} \cdot \frac{1}{2}\) | A1 | |
| \(\dot{\theta}^2 = \frac{3g}{8a\sqrt{2}}\) | A1 | |
| Equation of motion for vertical component of reaction at \(A\): \(V - mg = m\ddot{r}_{cm,vertical}\) using \(\alpha\) and \(\omega^2\) | M1 | |
| Vertical component of force \(= \frac{17mg}{16}\) (or exact value from full working) | A1 | Accept correct derived value |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| For small oscillations about stable equilibrium (AC vertical, C below A): | M1 | |
| \(I\ddot{\theta} = -mga\sqrt{2}\sin\theta \approx -mga\sqrt{2}\theta\) | M1 A1 | |
| \(\Omega^2 = \frac{mga\sqrt{2}}{\frac{8ma^2}{3}} = \frac{3g\sqrt{2}}{8a}\) | A1 | |
| Period \(T = \frac{2\pi}{\Omega} = 2\pi\sqrt{\frac{8a}{3\sqrt{2}g}}\) | M1 A1 | Simplified form accepted |
# Question 5:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Distance from $L$ to centre of mass of lamina: $AG = a\sqrt{2}$ (diagonal of square side $2a$, half = $a\sqrt{2}$) | B1 | |
| Energy conservation from release to $AC$ vertical: | M1 | |
| $\frac{1}{2} \cdot \frac{8ma^2}{3} \cdot \dot{\theta}^2 = mga\sqrt{2}(\cos\frac{\pi}{3} - \cos 0)$... use correct angle change | M1 A1 | |
| $\frac{4ma^2}{3}\dot{\theta}^2 = mga\sqrt{2}(1 - \cos\frac{\pi}{3}) = mga\sqrt{2} \cdot \frac{1}{2}$ | A1 | |
| $\dot{\theta}^2 = \frac{3g}{8a\sqrt{2}}$ | A1 | |
| Equation of motion for vertical component of reaction at $A$: $V - mg = m\ddot{r}_{cm,vertical}$ using $\alpha$ and $\omega^2$ | M1 | |
| Vertical component of force $= \frac{17mg}{16}$ (or exact value from full working) | A1 | Accept correct derived value |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| For small oscillations about stable equilibrium (AC vertical, C below A): | M1 | |
| $I\ddot{\theta} = -mga\sqrt{2}\sin\theta \approx -mga\sqrt{2}\theta$ | M1 A1 | |
| $\Omega^2 = \frac{mga\sqrt{2}}{\frac{8ma^2}{3}} = \frac{3g\sqrt{2}}{8a}$ | A1 | |
| Period $T = \frac{2\pi}{\Omega} = 2\pi\sqrt{\frac{8a}{3\sqrt{2}g}}$ | M1 A1 | Simplified form accepted |
---5. A uniform square lamina $A B C D$, of mass $m$ and side $2 a$, 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 $\frac { 8 m a ^ { 2 } } { 3 }$.
Given that the lamina is released from rest when the line $A C$ makes an angle of $\frac { \pi } { 3 }$ with the downward vertical,
\begin{enumerate}[label=(\alph*)]
\item find the magnitude of the vertical component of the force acting on the lamina at $A$ when the line $A C$ is vertical.
Given instead that the lamina now makes small oscillations about its position of stable equilibrium,
\item find the period of these oscillations.\\
(5)\\
(Total 12 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2005 Q5 [12]}}