## University Physics with Modern Physics (14th Edition)

(a) $\mu_s = 0.394$ (b) The button can be placed a maximum distance of 0.0978 meters from the axis without slipping.
(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.