| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rotation about fixed axis: angular acceleration and velocity |
| Difficulty | Challenging +1.2 This is a standard M5 rigid body dynamics problem requiring application of the parallel axis theorem, energy conservation, and force analysis at a pivot. While it involves multiple steps and careful bookkeeping of distances from the pivot (O is not at the center), the techniques are routine for this module: (a) uses τ = Iα with parallel axis theorem, (b) uses energy conservation, (c) combines centripetal acceleration with weight components. The non-central pivot adds mild complexity but follows standard M5 patterns. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_O = \frac{1}{12}m(AB)^2 + ma^2 = \frac{7}{3}ma^2\) | M1 A1 | |
| \(mga = \frac{7ma^2}{3}\ddot{\theta}\) | M1 | |
| \(\ddot{\theta} = \frac{3g}{7a}\) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2} \times \frac{7ma^2}{3}\dot{\theta}^2 = mga \Rightarrow a\dot{\theta}^2 = \frac{6g}{7}\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R(\downarrow): mg - Y = ma \cdot \frac{3g}{7a} \Rightarrow Y = \frac{4mg}{7}\) | M1 A1 | |
| \(R(\leftarrow): X = ma\ddot{\theta}^2 = \frac{6mg}{7}\) | M1 | |
| \(R = \frac{mg}{7}\sqrt{4^2 + 6^2} = \frac{mg}{7}\sqrt{52}\) | M1 A1 | (5 marks) |
Total: 11 marks
## Part (a)
$I_O = \frac{1}{12}m(AB)^2 + ma^2 = \frac{7}{3}ma^2$ | M1 A1 |
$mga = \frac{7ma^2}{3}\ddot{\theta}$ | M1 |
$\ddot{\theta} = \frac{3g}{7a}$ | A1 | (4 marks)
## Part (b)
$\frac{1}{2} \times \frac{7ma^2}{3}\dot{\theta}^2 = mga \Rightarrow a\dot{\theta}^2 = \frac{6g}{7}$ | M1 A1 | (2 marks)
## Part (c)
$R(\downarrow): mg - Y = ma \cdot \frac{3g}{7a} \Rightarrow Y = \frac{4mg}{7}$ | M1 A1 |
$R(\leftarrow): X = ma\ddot{\theta}^2 = \frac{6mg}{7}$ | M1 |
$R = \frac{mg}{7}\sqrt{4^2 + 6^2} = \frac{mg}{7}\sqrt{52}$ | M1 A1 | (5 marks)
Total: 11 marks
---A uniform rod $AB$ of mass $m$ and length $4a$ is free to rotate in a vertical plane about a horizontal axis through the point $O$ of the rod, where $OA = a$. The rod is slightly disturbed from rest when $B$ is vertically above $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the angular acceleration of the rod when it is horizontal. [4]
\item Find the angular speed of the rod when it is horizontal. [2]
\item Calculate the magnitude of the force acting on the rod at $O$ when the rod is horizontal. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 Q6 [11]}}