1. **Problem statement:** A coin is placed 7.0 cm (0.07 m) from the axis of a rotating circular table. The coin begins to slide outward when the rotation rate reaches 60 revolutions per minute (rpm). We want to analyze the forces involved and find the maximum static friction force or the maximum angular velocity before slipping. 2. **Relevant formulas:** - Convert angular velocity from rpm to radians per second: $$\omega = \frac{2\pi \times \text{rpm}}{60}$$ - Centripetal acceleration: $$a_c = \omega^2 r$$ - Maximum static friction force: $$f_{max} = m a_c$$ (assuming friction provides centripetal force) 3. **Convert given values:** - Radius: $$r = 7.0\ \text{cm} = 0.07\ \text{m}$$ - Angular velocity at slipping: $$\omega = \frac{2\pi \times 60}{60} = 2\pi\ \text{rad/s} \approx 6.283\ \text{rad/s}$$ 4. **Calculate centripetal acceleration at slipping:** $$a_c = \omega^2 r = (6.283)^2 \times 0.07 = 39.48 \times 0.07 = 2.7636\ \text{m/s}^2$$ 5. **Interpretation:** The coin begins to slip when the centripetal acceleration exceeds the maximum static friction acceleration. If the mass and friction coefficient were given, we could find the friction force or coefficient. Here, the key is understanding the relationship between rotation rate, radius, and slipping. 6. **Summary:** The coin at radius 0.07 m starts to slip at angular velocity approximately $$6.283\ \text{rad/s}$$, corresponding to 60 rpm, with centripetal acceleration about $$2.76\ \text{m/s}^2$$. This explains why the coin slips outward as rotation rate increases beyond this point.