| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Angular speed and period |
| Difficulty | Challenging +1.2 This is a standard M4 rotational dynamics problem requiring moment of inertia calculation (using parallel axis theorem), energy methods, and angular motion equations. While it involves multiple steps and careful bookkeeping of energy terms including friction, the techniques are routine for M4 students and the question provides clear guidance through its structure. The calculations are straightforward once the correct approach is identified, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 6.04a Centre of mass: gravitational effect6.05a Angular velocity: definitions6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = \frac{1}{3}(20)(0.3^2 + 0.9^2) + 20 \times 0.9^2\) | M1 | MI of lamina about any axis; Use of parallel axes rule (or perp) axes rule |
| \(= 22.2 \text{ kg m}^2\) | A1 (ag) | Correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| OR \(I = \frac{1}{3} \times 20 \times 0.3^2 + \frac{4}{3} \times 20 \times 0.9^2\) | M1M1 | As above |
| \(= 22.2 \text{ kg m}^2\) | A1 |
| Answer | Marks |
|---|---|
| Total moment is \(20 \times 9.8 \times 0.9 \cos \theta - 44.1\) | M1 |
| Angular acceleration is zero when moment is zero | M1 |
| \(\cos \theta = \frac{44.1}{20 \times 9.8 \times 0.9} = 0.25\) | A1 (ag) |
| Answer | Marks | Guidance |
|---|---|---|
| Maximum angular speed when \(\cos \theta = 0.25\) | M1 | |
| \(\theta = 1.318\) | A1 | |
| Work done against couple is \(44.1 \times 1.318\) | A1 | |
| By work energy principle, | M1 | Equation involving work, KE and PE |
| \(\frac{1}{2}I\omega^2 = 20 \times 9.8 \times 0.9 \sin \theta - 44.1\theta\) | A1 ft | |
| \(\omega = 3.19 \text{ rad s}^{-1}\) | A1 |
## Part (i)
$I = \frac{1}{3}(20)(0.3^2 + 0.9^2) + 20 \times 0.9^2$ | M1 | MI of lamina about any axis; Use of parallel axes rule (or perp) axes rule
$= 22.2 \text{ kg m}^2$ | A1 (ag) | Correctly obtained
**Total: 3 marks**
**OR** $I = \frac{1}{3} \times 20 \times 0.3^2 + \frac{4}{3} \times 20 \times 0.9^2$ | M1M1 | As above
$= 22.2 \text{ kg m}^2$ | A1 |
---
## Part (ii)
Total moment is $20 \times 9.8 \times 0.9 \cos \theta - 44.1$ | M1 |
Angular acceleration is zero when moment is zero | M1 |
$\cos \theta = \frac{44.1}{20 \times 9.8 \times 0.9} = 0.25$ | A1 (ag) |
**Total: 3 marks**
---
## Part (iii)
Maximum angular speed when $\cos \theta = 0.25$ | M1 |
$\theta = 1.318$ | A1 |
Work done against couple is $44.1 \times 1.318$ | A1 |
By work energy principle, | M1 | Equation involving work, KE and PE
$\frac{1}{2}I\omega^2 = 20 \times 9.8 \times 0.9 \sin \theta - 44.1\theta$ | A1 ft |
$\omega = 3.19 \text{ rad s}^{-1}$ | A1 |
**Total: 5 marks**
---A uniform rectangular lamina $ABCD$ has mass 20 kg and sides of lengths $AB = 0.6$ m and $BC = 1.8$ m. It rotates in its own vertical plane about a fixed horizontal axis which is perpendicular to the lamina and passes through the mid-point of $AB$.
\begin{enumerate}[label=(\roman*)]
\item Show that the moment of inertia of the lamina about the axis is 22.2 kg m$^2$. [3]
\end{enumerate}
\includegraphics{figure_5}
The lamina is released from rest with $BC$ horizontal and below the level of the axis. Air resistance may be neglected, but a frictional couple opposes the motion. The couple has constant moment 44.1 N m about the axis. The angle through which the lamina has turned is denoted by $\theta$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that the angular acceleration is zero when $\cos \theta = 0.25$. [3]
\item Hence find the maximum angular speed of the lamina. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR M4 2006 Q5 [11]}}