Answer
$\mu_s = 0.21$
Work Step by Step
The coin will stay on the turntable until the speed is so high that the force of static friction is not able to provide the centripetal force. At this point the coin will slide off the turntable:
$v = (2\pi)(0.130 ~m)(38.0 ~rpm)(\frac{1 ~min}{60 ~s})$
$v = 0.517 ~m/s$
We can use the velocity to find the coefficient of static friction:
$F_f = m\frac{v^2}{r}$
$mg ~\mu_s = m\frac{v^2}{r}$
$\mu_s = \frac{v^2}{gr}$
$\mu_s = \frac{(0.517 ~m/s)^2}{(9.8 ~m/s^2)(0.130 ~m)}$
$\mu_s = 0.21$
Therefore, the coefficient of static friction $\mu_s$ is 0.21.
