| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Energy methods for rotation |
| Difficulty | Challenging +1.8 This M5 question requires applying the parallel axis theorem, rotational dynamics with variable angle, and resolving forces in a rotating reference frame. While the setup is standard for Further Maths mechanics, it demands careful application of I = I_center + md² and Newton's second law for rotation at a non-equilibrium position, plus decomposing the reaction force—more sophisticated than typical A-level but follows established M5 patterns. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moment of inertia about axis \(L\) (tangent at \(A\)): \(I = \frac{1}{2}ma^2 + ma^2 = \frac{3}{2}ma^2\) | M1 A1 | Using parallel axis theorem; correct \(I\) |
| When turned through \(\frac{\pi}{3}\), perpendicular distance of \(O\) from vertical through \(A\): torque \(= mga\sin\frac{\pi}{3} = \frac{mga\sqrt{3}}{2}\) | M1 A1 | Correct torque (moment of weight about \(L\)) |
| \(I\ddot{\theta} = mga\sin\frac{\pi}{3}\): \(\frac{3}{2}ma^2\ddot{\theta} = \frac{mga\sqrt{3}}{2}\) | ||
| \(\ddot{\theta} = \frac{g\sqrt{3}}{3a} = \frac{g}{\sqrt{3}a}\) | A1 | Correct angular acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using energy: \(\frac{1}{2}I\omega^2 = mga(1-\cos\frac{\pi}{3})= mga \cdot \frac{1}{2}\) | M1 | Energy equation from rest |
| \(\frac{3}{2}ma^2\omega^2 = \frac{mga}{2} \Rightarrow \omega^2 = \frac{g}{3a}\) | A1 | Correct \(\omega^2\) |
| Component of force perpendicular to disc \(= m\omega^2 a - mg\cos\frac{\pi}{3}\) (centripetal equation for \(O\) relative to \(A\), perpendicular component) | M1 | Equation of motion for centre perpendicular to disc |
| \(F_\perp = m\omega^2 a - mg\cos\frac{\pi}{3} = m\cdot\frac{g}{3a}\cdot a - \frac{mg}{2} = \frac{mg}{3} - \frac{mg}{2} = -\frac{mg}{6}\) | ||
| Magnitude \(= \frac{mg}{6}\) | A1 | Correct magnitude |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moment of inertia about axis $L$ (tangent at $A$): $I = \frac{1}{2}ma^2 + ma^2 = \frac{3}{2}ma^2$ | M1 A1 | Using parallel axis theorem; correct $I$ |
| When turned through $\frac{\pi}{3}$, perpendicular distance of $O$ from vertical through $A$: torque $= mga\sin\frac{\pi}{3} = \frac{mga\sqrt{3}}{2}$ | M1 A1 | Correct torque (moment of weight about $L$) |
| $I\ddot{\theta} = mga\sin\frac{\pi}{3}$: $\frac{3}{2}ma^2\ddot{\theta} = \frac{mga\sqrt{3}}{2}$ | | |
| $\ddot{\theta} = \frac{g\sqrt{3}}{3a} = \frac{g}{\sqrt{3}a}$ | A1 | Correct angular acceleration |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using energy: $\frac{1}{2}I\omega^2 = mga(1-\cos\frac{\pi}{3})= mga \cdot \frac{1}{2}$ | M1 | Energy equation from rest |
| $\frac{3}{2}ma^2\omega^2 = \frac{mga}{2} \Rightarrow \omega^2 = \frac{g}{3a}$ | A1 | Correct $\omega^2$ |
| Component of force perpendicular to disc $= m\omega^2 a - mg\cos\frac{\pi}{3}$ (centripetal equation for $O$ relative to $A$, perpendicular component) | M1 | Equation of motion for centre perpendicular to disc |
| $F_\perp = m\omega^2 a - mg\cos\frac{\pi}{3} = m\cdot\frac{g}{3a}\cdot a - \frac{mg}{2} = \frac{mg}{3} - \frac{mg}{2} = -\frac{mg}{6}$ | | |
| Magnitude $= \frac{mg}{6}$ | A1 | Correct magnitude |\begin{enumerate}
\item A uniform circular disc, of mass $m$ and radius $a$, is free to rotate about a fixed smooth horizontal axis $L$. The axis $L$ is a tangent to the disc at the point $A$. The centre $O$ of the disc moves in a vertical plane that is perpendicular to $L$.
\end{enumerate}
The disc is held at rest with its plane horizontal and released.\\
(a) Find the angular acceleration of the disc when it has turned through an angle of $\frac { \pi } { 3 }$\\
(b) Find the magnitude of the component, in a direction perpendicular to the disc, of the force of the axis $L$ acting on the disc at $A$, when the disc has turned through an angle of $\frac { \pi } { 3 }$\\
\hfill \mbox{\textit{Edexcel M5 2015 Q5 [9]}}