Question

A coin is placed 29 cm from the center of a
horizontal turntable, initially at rest. The
turntable then begins to rotate. When the
speed of the coin is 120 cm/s (rotating at a
constant rate), the coin just begins to slip.
The acceleration of gravity is 980 cm/s
2. What is the coefficient of static friction between the coin and the turntable?

Answer 1

$$\mu mg=m\frac{v^2}{r}$$

$$\mu =\frac{v^2}{gr}$$

$$\mu =\frac{120^2}{980\cdot 29}=0.51$$