Understanding Centripetal Force and Friction
When an object, like a coin, moves in a circular path on a rotating disc, it requires a centripetal force to stay on that path. Without this force, the coin would move in a straight line due to inertia. In this setup, the force providing the necessary centripetal acceleration is static friction.
The Physics Principles
- Centripetal Force ($F_c$): The force directed toward the center of rotation required for circular motion. The formula is $F_c = m \cdot r \cdot \omega^2$, where $m$ is the mass, $r$ is the radius, and $\omega$ is the angular velocity in radians per second.
- Static Friction ($f_s$): The force opposing relative motion between surfaces. The maximum static friction is $f_{s,max} = \mu_s \cdot N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force ($N = mg$ on a horizontal surface).
- Equilibrium Condition: For the coin to remain on the disc without sliding, the centripetal force must be less than or equal to the maximum static friction: $m \cdot r \cdot \omega^2 \le \mu_s \cdot m \cdot g$.
Step-by-Step Solution
Given:
- Angular velocity $\omega = 33 \frac{1}{3} \text{ rev/min} = \frac{100}{3} \text{ rev/min}$
- Convert to rad/s: $\omega = \frac{100}{3} \times \frac{2\pi}{60} = \frac{200\pi}{180} = \frac{10\pi}{9} \text{ rad/s}$
- Radius $r = 10 \text{ cm} = 0.1 \text{ m}$
- Gravity $g \approx 9.8 \text{ m/s}^2$
Derivation: At the point of slipping, $m r \omega^2 = \mu_s m g$. Therefore, $\mu_s = \frac{r \omega^2}{g}$.
Calculation:
- $\omega^2 = (\frac{10\pi}{9})^2 = \frac{100\pi^2}{81} \approx \frac{100 \times 9.869}{81} \approx 12.18$
- $\mu_s = \frac{0.1 \times 12.18}{9.8} \approx 0.124$
The coefficient of static friction required is approximately 0.124.