| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2004 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Rotating disc with friction |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring application of F=mrω² with friction providing centripetal force in part (a), then adding elastic string tension in part (b). The setup is typical textbook material with straightforward inequality manipulation, though part (b) requires careful consideration of friction direction in two cases (string slack vs taut, friction inward vs outward). |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02j Conservation with elastics: springs and strings6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \((\uparrow),\ R = mg\) | B1 | |
| \(m\dfrac{4a}{3}\omega^2\) | B1 | seen and used |
| \(m\dfrac{4a}{3}\omega^2 \leq \dfrac{3}{5}mg\) | M1 | |
| \(\omega^2 \leq \dfrac{9g}{20a}\) | A1 c.s.o | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(T = \dfrac{2mg}{a}\cdot\dfrac{a}{3} = \dfrac{2mg}{3}\) | B1 | |
| \((\rightarrow),\ \dfrac{3}{5}mg + \dfrac{2mg}{3} \geq m\dfrac{4a}{3}\omega_{\max}^2\) | M1 A1 f.t | |
| \(\dfrac{19g}{20a} = \omega_{\max}^2\) | A1 | |
| \((\rightarrow),\ -\dfrac{3}{5}mg + \dfrac{2mg}{3} \leq m\dfrac{4a}{3}\omega_{\min}^2\) | M1 A1 f.t | |
| \(\dfrac{g}{20a} = \omega_{\min}^2\) | A1 | (7 marks) |
| If only one answer, must be clear whether max or min for final A1 | (11 marks) |
## Question 4:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(\uparrow),\ R = mg$ | B1 | |
| $m\dfrac{4a}{3}\omega^2$ | B1 | seen and used |
| $m\dfrac{4a}{3}\omega^2 \leq \dfrac{3}{5}mg$ | M1 | |
| $\omega^2 \leq \dfrac{9g}{20a}$ | A1 c.s.o | (4 marks) |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $T = \dfrac{2mg}{a}\cdot\dfrac{a}{3} = \dfrac{2mg}{3}$ | B1 | |
| $(\rightarrow),\ \dfrac{3}{5}mg + \dfrac{2mg}{3} \geq m\dfrac{4a}{3}\omega_{\max}^2$ | M1 A1 f.t | |
| $\dfrac{19g}{20a} = \omega_{\max}^2$ | A1 | |
| $(\rightarrow),\ -\dfrac{3}{5}mg + \dfrac{2mg}{3} \leq m\dfrac{4a}{3}\omega_{\min}^2$ | M1 A1 f.t | |
| $\dfrac{g}{20a} = \omega_{\min}^2$ | A1 | (7 marks) |
| If only one answer, must be clear whether max or min for final A1 | | **(11 marks)** |
---4. A rough disc rotates in a horizontal plane with constant angular velocity $\omega$ about a fixed vertical axis. A particle $P$ of mass $m$ lies on the disc at a distance $\frac { 4 } { 3 } a$ from the axis. The coefficient of friction between $P$ and the disc is $\frac { 3 } { 5 }$. Given that $P$ remains at rest relative to the disc,
\begin{enumerate}[label=(\alph*)]
\item prove that $\omega ^ { 2 } \leqslant \frac { 9 g } { 20 a }$.
The particle is now connected to the axis by a horizontal light elastic string of natural length $a$ and modulus of elasticity 2 mg . The disc again rotates with constant angular velocity $\omega$ about the axis and $P$ remains at rest relative to the disc at a distance $\frac { 4 } { 3 } a$ from the axis.
\item Find the greatest and least possible values of $\omega ^ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2004 Q4 [11]}}