| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Rotation dynamics with moment of inertia |
| Difficulty | Challenging +1.8 This is a challenging Further Maths M5 question requiring moment of inertia calculations, angular dynamics with energy conservation, and force resolution at a specific angle. It demands multiple sophisticated techniques (parallel axis theorem, rotational kinematics, Newton's second law for rotation) across three interconnected parts, but follows a standard framework for this advanced topic without requiring novel geometric insight. |
| Spec | 6.03a Linear momentum: p = mv6.03f Impulse-momentum: relation6.05b Circular motion: v=r*omega and a=v^2/r6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Moment of inertia of square lamina about axis through P and Q: \(I = \frac{1}{3}m(2a)^2 = \frac{4ma^2}{3}\) | M1 | Using standard result or deriving MI about PQ |
| Impulse-momentum: \(J \times a = I\omega\) (impulse applied at midpoint of SR, distance \(a\) from axis) | M1 | Angular impulse = \(J \times a\) |
| \(Ja = \frac{4ma^2}{3}\omega \Rightarrow \omega = \frac{3J}{4ma}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| At angle \(\theta = \frac{\pi}{6}\), use energy or equation of motion: \(I\alpha = -mg\bar{d}\cos\theta\) where \(\bar{d}\) is distance of centre of mass from axis | M1 | Taking moments about axis, restoring torque due to gravity |
| Centre of mass is distance \(a\) from PQ (midpoint of square). Torque \(= mga\cos\frac{\pi}{6} = mga \cdot \frac{\sqrt{3}}{2}\) | A1 | |
| \(\frac{4ma^2}{3}\alpha = mga\cdot\frac{\sqrt{3}}{2} \Rightarrow \alpha = \frac{3g\sqrt{3}}{8a}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| First find \(\omega^2\) at \(\theta = \frac{\pi}{6}\) using energy: \(\frac{1}{2}I\omega^2 = \frac{1}{2}I\omega_0^2 - mga(1-\cos\frac{\pi}{6})\) | M1 | Energy equation from initial position |
| \(\frac{1}{2}\cdot\frac{4ma^2}{3}\omega^2 = \frac{1}{2}\cdot\frac{4ma^2}{3}\cdot\frac{9J^2}{16m^2a^2} - mga(1-\frac{\sqrt{3}}{2})\) | A1 | Substituting values |
| Perpendicular to lamina: equation of motion for centre of mass gives force component \(F_\perp\): \(F_\perp - mg\sin\frac{\pi}{6} = -ma\alpha\) (tangential equation) | M1 | Resolving perpendicular to lamina for centre of mass |
| \(F_\perp = mg\sin\frac{\pi}{6} - ma\alpha = \frac{mg}{2} - ma\cdot\frac{3\sqrt{3}g}{8a} = \frac{mg}{2} - \frac{3\sqrt{3}mg}{8}\) | A1 | |
| \(\left | F_\perp\right | = mg\left |
## Question 7:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Moment of inertia of square lamina about axis through P and Q: $I = \frac{1}{3}m(2a)^2 = \frac{4ma^2}{3}$ | M1 | Using standard result or deriving MI about PQ |
| Impulse-momentum: $J \times a = I\omega$ (impulse applied at midpoint of SR, distance $a$ from axis) | M1 | Angular impulse = $J \times a$ |
| $Ja = \frac{4ma^2}{3}\omega \Rightarrow \omega = \frac{3J}{4ma}$ | A1 | |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| At angle $\theta = \frac{\pi}{6}$, use energy or equation of motion: $I\alpha = -mg\bar{d}\cos\theta$ where $\bar{d}$ is distance of centre of mass from axis | M1 | Taking moments about axis, restoring torque due to gravity |
| Centre of mass is distance $a$ from PQ (midpoint of square). Torque $= mga\cos\frac{\pi}{6} = mga \cdot \frac{\sqrt{3}}{2}$ | A1 | |
| $\frac{4ma^2}{3}\alpha = mga\cdot\frac{\sqrt{3}}{2} \Rightarrow \alpha = \frac{3g\sqrt{3}}{8a}$ | A1 | |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| First find $\omega^2$ at $\theta = \frac{\pi}{6}$ using energy: $\frac{1}{2}I\omega^2 = \frac{1}{2}I\omega_0^2 - mga(1-\cos\frac{\pi}{6})$ | M1 | Energy equation from initial position |
| $\frac{1}{2}\cdot\frac{4ma^2}{3}\omega^2 = \frac{1}{2}\cdot\frac{4ma^2}{3}\cdot\frac{9J^2}{16m^2a^2} - mga(1-\frac{\sqrt{3}}{2})$ | A1 | Substituting values |
| Perpendicular to lamina: equation of motion for centre of mass gives force component $F_\perp$: $F_\perp - mg\sin\frac{\pi}{6} = -ma\alpha$ (tangential equation) | M1 | Resolving perpendicular to lamina for centre of mass |
| $F_\perp = mg\sin\frac{\pi}{6} - ma\alpha = \frac{mg}{2} - ma\cdot\frac{3\sqrt{3}g}{8a} = \frac{mg}{2} - \frac{3\sqrt{3}mg}{8}$ | A1 | |
| $\left|F_\perp\right| = mg\left|\frac{4-3\sqrt{3}}{8}\right|$ | A1 | Accept equivalent simplified form |7. A uniform square lamina $P Q R S$, of mass $m$ and side $2 a$, is free to rotate about a fixed smooth horizontal axis which passes through $P$ and $Q$. The lamina hangs at rest in a vertical plane with $S R$ below $P Q$ and is given a horizontal impulse of magnitude $J$ at the midpoint of $S R$. The impulse is perpendicular to $S R$.
\begin{enumerate}[label=(\alph*)]
\item Find the initial angular speed of the lamina.
\item Find the magnitude of the angular deceleration of the lamina at the instant when the lamina has turned through $\frac { \pi } { 6 }$ radians.
\item Find the magnitude of the component of the force exerted on the lamina by the axis, in a direction perpendicular to the lamina, at the instant when the lamina has turned through $\frac { \pi } { 6 }$ radians.\\
\includegraphics[max width=\textwidth, alt={}, center]{f932d7cb-1299-41d1-8248-cfbf639795ed-12_2255_50_315_1978}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2016 Q7 [11]}}