| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Rotating disc with friction |
| Difficulty | Moderate -0.5 This is a straightforward circular motion problem requiring standard formulas (v = rω, F = mv²/r) with routine conversions (rpm to rad/s). Part (c) involves equating friction to centripetal force, which is a textbook application. The calculations are mechanical with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\omega = \frac{45}{60} \times 2\pi = \frac{3}{2}\pi\) | M1 | |
| \(v = \omega r = \frac{3}{2}\pi \times 0.1 = \frac{3}{20}\pi\) or 0.47 ms\(^{-1}\) (2sf) | A1 | |
| (b) Resolve \(\uparrow\): \(R - mg = 0\) \(\therefore R = 0.005 \times 9.8 = 0.049\) N | M1 A1 | |
| Resolve \(\leftarrow\): \(F = ma = m\omega^2 r = 0.005 \times 0.1 \times (\frac{3}{2}\pi)^2 = 0.011\) N (2sf) | M1 A1 | |
| \(\therefore\) horiz. and vert. components are 0.011 N and 0.049 N respectively | ||
| (c) Limiting friction \(\therefore F = \mu R\) | M1 | |
| \(0.01110 = 0.049\mu\) \(\therefore \mu = \frac{0.01110}{0.049} = 0.23\) (2sf) | M1 A1 | (9) |
(a) $\omega = \frac{45}{60} \times 2\pi = \frac{3}{2}\pi$ | M1 |
$v = \omega r = \frac{3}{2}\pi \times 0.1 = \frac{3}{20}\pi$ or 0.47 ms$^{-1}$ (2sf) | A1 |
(b) Resolve $\uparrow$: $R - mg = 0$ $\therefore R = 0.005 \times 9.8 = 0.049$ N | M1 A1 |
Resolve $\leftarrow$: $F = ma = m\omega^2 r = 0.005 \times 0.1 \times (\frac{3}{2}\pi)^2 = 0.011$ N (2sf) | M1 A1 |
$\therefore$ horiz. and vert. components are 0.011 N and 0.049 N respectively | |
(c) Limiting friction $\therefore F = \mu R$ | M1 |
$0.01110 = 0.049\mu$ $\therefore \mu = \frac{0.01110}{0.049} = 0.23$ (2sf) | M1 A1 | (9)3. A coin of mass 5 grams is placed on a vinyl disc rotating on a record player. The distance between the centre of the coin and the centre of the disc is 0.1 m and the coefficient of friction between the coin and the disc is $\mu$. The disc rotates at 45 revolutions per minute around a vertical axis at its centre and the coin moves with it and does not slide.
By modelling the coin as a particle and giving your answers correct to an appropriate degree of accuracy, find
\begin{enumerate}[label=(\alph*)]
\item the speed of the coin,
\item the horizontal and vertical components of the force exerted on the coin by the disc.
Given that the coin is on the point of moving,
\item show that, correct to 2 significant figures, $\mu = 0.23$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q3 [9]}}