Introduction to Circular Motion and Friction
When an object, like a coin, rests on a rotating disc, it stays in place due to the force of static friction acting as a centripetal force. The centripetal force is the net force required to keep an object moving in a circular path. If the rotation speed exceeds a certain limit, the required centripetal force becomes greater than the maximum available static friction, causing the object to slide off.
The Problem Statement
A coin is placed on a disc rotating at a speed of $33 \frac{1}{3} \text{ rev min}^{-1}$. The coin remains stationary relative to the disc as long as it is not more than $10 \text{ cm}$ from the axis of rotation. We need to find the coefficient of static friction $\mu$ between the coin and the disc.
Step-by-Step Solution
1. Identify Given Data
- Angular Frequency ($f$): $33 \frac{1}{3} \text{ rev min}^{-1} = \frac{100}{3} \text{ rev min}^{-1}$.
- Convert to seconds: $f = \frac{100}{3 \times 60} = \frac{100}{180} = \frac{5}{9} \text{ rev s}^{-1}$.
- Angular Velocity ($\omega$): $\omega = 2\pi f = 2\pi \times \frac{5}{9} = \frac{10\pi}{9} \text{ rad s}^{-1}$.
- Radius ($r$): $10 \text{ cm} = 0.1 \text{ m}$.
- Acceleration due to gravity ($g$): $\approx 9.8 \text{ m/s}^2$ (or $10 \text{ m/s}^2$ for simplicity).
2. Physical Principle
For the coin to not slide, the maximum static friction must be greater than or equal to the required centripetal force: $F_f \ge F_c$ $\mu N \ge m r \omega^2$ Since $N = mg$ (the normal force equals the weight of the coin), we have: $\mu mg = m r \omega^2$ $\mu = \frac{r \omega^2}{g}$
3. Calculation
Substitute the values into the equation: $\mu = \frac{0.1 \times (\frac{10\pi}{9})^2}{9.8}$ $\mu = \frac{0.1 \times \frac{100\pi^2}{81}}{9.8}$ $\mu = \frac{0.1 imes 98.696}{81 imes 9.8}$ $\mu \approx \frac{9.8696}{793.8} \approx 0.0124$
Summary
By equating the centripetal force to the static frictional force, we can determine the conditions under which an object will remain fixed on a rotating platform. In this case, the coefficient of static friction is approximately 0.0124, assuming standard gravity.