Answer
(a) $\mu_s = 0.394$
(b) The button can be placed a maximum distance of 0.0978 meters from the axis without slipping.
Work Step by Step
(a) We can find the speed $v$ of the button.
$v = (40.0~rev/min)(\frac{1~min}{60~s})(\frac{(2\pi)(0.220~m)}{1~rev})$
$v = 0.922~m/s$
In this situation, the force of static friction provides the centripetal force.
$F_f = \frac{mv^2}{r}$
$mg~\mu_s = \frac{mv^2}{r}$
$\mu_s = \frac{v^2}{gr}$
$\mu_s = \frac{(0.922~m/s)^2}{(9.80~m/s^2)(0.220~m)}$
$\mu_s = 0.394$
(b) We can find the speed $v$ of the button.
$v = (60.0~rev/min)(\frac{1~min}{60~s})(\frac{(2\pi)(r)}{1~rev})$
$v = (2\pi~r)~m/s$
We can find maximum radius $r$ for the button's rotation.
$F_f = \frac{mv^2}{r}$
$mg~\mu_s = \frac{m(4\pi^2~r^2)}{r}$
$g~\mu_s = (4\pi^2)~r$
$r = \frac{g~\mu_s}{4~\pi^2}$
$r = \frac{(9.80~m/s^2)(0.394)}{4~\pi^2}$
$r = 0.0978~m$
The button can be placed a maximum distance of 0.0978 meters from the axis without slipping.
