## Essential University Physics: Volume 1 (3rd Edition)

We know that $\omega=2\pi(\frac{200}{60})=20.94\frac{rad}{s}$ We also know that the force of friction should be equal to the required centripetal force $\mu mg=mr\omega^2$ This can be rearranged as $r=\frac{\mu g}{\omega^2}$ We plug in the known values to obtain: $r=\frac{(1.2)(9.8)}{(20.94)^2}=0.027m$