| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2006 |
| Session | June |
| Marks | 17 |
| 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 rigid body dynamics problem requiring moment of inertia about a non-center point, energy methods, and force resolution. While it involves multiple parts and careful bookkeeping of the pivot location, the techniques are all standard textbook methods for this module with no novel insights required. The algebra is moderately involved but straightforward for students who have practiced similar problems. |
| Spec | 3.03c Newton's second law: F=ma one dimension6.04e Rigid body equilibrium: coplanar forces6.05a Angular velocity: definitions6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = \frac{1}{3}m(3a)^2 + m(2a)^2\) | M1 | Using parallel axes rule |
| \(= 7ma^2\) | A1 | |
| \(mg(2a \sin \theta) = I\alpha\) | M1 | |
| \(\alpha = \frac{2g \sin \theta}{7a}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| By conservation of energy | M1 | Equation involving KE and PE |
| \(\frac{1}{2}I\omega^2 = mg(2a \cos \frac{1}{3}\pi - 2a \cos \theta)\) | A1 | Need to see how \(\frac{1}{3}\pi\) is used |
| \(\frac{7}{2}ma^2\omega^2 = mga(1 - 2\cos \theta)\) | ||
| \(\omega = \sqrt{\frac{2g(1 - 2\cos \theta)}{7a}}\) | A1 (ag) | Correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| \(mg \cos \theta - R = m(2a\omega^2)\) | M1 | For radial acceleration \(r\omega^2\) |
| \(R = mg \cos \theta - \frac{4}{7}mg(1 - 2\cos \theta)\) | A1 | |
| \(= \frac{1}{7}mg(15 \cos \theta - 4)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(mg \sin \theta - S = m(2a\alpha)\) | M1 | For transverse acceleration \(r\alpha\) |
| \(S = mg \sin \theta - \frac{4}{7}mg \sin \theta\) | A1 | |
| \(= \frac{3}{7}mg \sin \theta\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| OR \(S(2a) = I_G\alpha\) | M1A1 | Must use \(I_G\) |
| \(S = \frac{3}{7}mg \sin \theta\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| When \(\cos \theta = \frac{1}{3}\), \(\sin \theta = \frac{\sqrt{8}}{3}\), \(\tan \theta = \sqrt{8}\) | [Given values] | |
| \(R = \frac{1}{7}mg\), \(S = \frac{\sqrt{8}}{7}mg\) | M1 | |
| Angle with \(R\) is \(\tan^{-1}\frac{S}{R} = \tan^{-1}\sqrt{8} = \theta\) | A1 | |
| so the resultant force is vertical | A1 | |
| Magnitude is \(\sqrt{R^2 + S^2}\) | M1 | |
| \(= \frac{1}{7}mg\sqrt{1 + 8} = \frac{3}{7}mg\) | A1 |
| Answer | Marks |
|---|---|
| OR When resultant force is \(F\) vertically upwards | |
| \(S = F \sin \theta\), hence \(F = \frac{3}{7}mg\) | M1A1 |
| \(R = F \cos \theta\), so | |
| \(\frac{1}{7}mg(15 \cos \theta - 4) = \frac{3}{7}mg \cos \theta\) | M1 |
| \(\cos \theta = \frac{1}{3}\) | A1 |
| Answer | Marks |
|---|---|
| OR Horizontal force is \(R \sin \theta - S \cos \theta\) | |
| \(= \frac{1}{7}mg(15 \cos \theta - 4) \sin \theta - \frac{3}{7}mg \sin \theta \cos \theta\) | M1 |
| \(= \frac{1}{7}mg \sin \theta(12 \cos \theta - 4)\) | |
| \(= 0\) when \(\cos \theta = \frac{1}{3}\) | A1 |
| Vertical force is \(R \cos \theta + S \sin \theta\) | |
| \(= \frac{1}{7}mg \times \frac{1}{3} + \frac{3}{7}mg \times \frac{\sqrt{8}}{3} = \frac{3}{7}mg\) | M1A1 |
## Part (i)
$I = \frac{1}{3}m(3a)^2 + m(2a)^2$ | M1 | Using parallel axes rule
$= 7ma^2$ | A1 |
$mg(2a \sin \theta) = I\alpha$ | M1 |
$\alpha = \frac{2g \sin \theta}{7a}$ | A1 |
**Total: 4 marks**
---
## Part (ii)
By conservation of energy | M1 | Equation involving KE and PE
$\frac{1}{2}I\omega^2 = mg(2a \cos \frac{1}{3}\pi - 2a \cos \theta)$ | A1 | Need to see how $\frac{1}{3}\pi$ is used
$\frac{7}{2}ma^2\omega^2 = mga(1 - 2\cos \theta)$ | |
$\omega = \sqrt{\frac{2g(1 - 2\cos \theta)}{7a}}$ | A1 (ag) | Correctly obtained
**Total: 3 marks**
---
## Part (iii)
$mg \cos \theta - R = m(2a\omega^2)$ | M1 | For radial acceleration $r\omega^2$
$R = mg \cos \theta - \frac{4}{7}mg(1 - 2\cos \theta)$ | A1 |
$= \frac{1}{7}mg(15 \cos \theta - 4)$ | A1 |
---
$mg \sin \theta - S = m(2a\alpha)$ | M1 | For transverse acceleration $r\alpha$
$S = mg \sin \theta - \frac{4}{7}mg \sin \theta$ | A1 |
$= \frac{3}{7}mg \sin \theta$ | A1 |
**Total: 6 marks**
**OR** $S(2a) = I_G\alpha$ | M1A1 | Must use $I_G$
$S = \frac{3}{7}mg \sin \theta$ | A1 |
---
## Part (iv)
When $\cos \theta = \frac{1}{3}$, $\sin \theta = \frac{\sqrt{8}}{3}$, $\tan \theta = \sqrt{8}$ | | [Given values]
$R = \frac{1}{7}mg$, $S = \frac{\sqrt{8}}{7}mg$ | M1 |
Angle with $R$ is $\tan^{-1}\frac{S}{R} = \tan^{-1}\sqrt{8} = \theta$ | A1 |
so the resultant force is vertical | A1 |
Magnitude is $\sqrt{R^2 + S^2}$ | M1 |
$= \frac{1}{7}mg\sqrt{1 + 8} = \frac{3}{7}mg$ | A1 |
**Total: 4 marks**
**OR** When resultant force is $F$ vertically upwards | |
$S = F \sin \theta$, hence $F = \frac{3}{7}mg$ | M1A1 |
$R = F \cos \theta$, so | |
$\frac{1}{7}mg(15 \cos \theta - 4) = \frac{3}{7}mg \cos \theta$ | M1 |
$\cos \theta = \frac{1}{3}$ | A1 |
---
**OR** Horizontal force is $R \sin \theta - S \cos \theta$ | |
$= \frac{1}{7}mg(15 \cos \theta - 4) \sin \theta - \frac{3}{7}mg \sin \theta \cos \theta$ | M1 |
$= \frac{1}{7}mg \sin \theta(12 \cos \theta - 4)$ | |
$= 0$ when $\cos \theta = \frac{1}{3}$ | A1 |
Vertical force is $R \cos \theta + S \sin \theta$ | |
$= \frac{1}{7}mg \times \frac{1}{3} + \frac{3}{7}mg \times \frac{\sqrt{8}}{3} = \frac{3}{7}mg$ | M1A1 |\includegraphics{figure_7}
A uniform rod $AB$ has mass $m$ and length $6a$. It is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through the point $C$ on the rod, where $AC = a$. The angle between $AB$ and the upward vertical is $\theta$, and the force acting on the rod at $C$ has components $R$ parallel to $AB$ and $S$ perpendicular to $AB$ (see diagram). The rod is released from rest in the position where $\theta = \frac{1}{4}\pi$. Air resistance may be neglected.
\begin{enumerate}[label=(\roman*)]
\item Find the angular acceleration of the rod in terms of $a$, $g$ and $\theta$. [4]
\item Show that the angular speed of the rod is $\sqrt{\frac{2g(1 - 2\cos\theta)}{7a}}$. [3]
\item Find $R$ and $S$ in terms of $m$, $g$ and $\theta$. [6]
\item When $\cos\theta = \frac{1}{3}$, show that the force acting on the rod at $C$ is vertical, and find its magnitude. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M4 2006 Q7 [17]}}