(a) 0.394
The static frictional force provides the centripetal force that keeps the button stuck on the rotating platform. So we can write
[tex]m\omega^2 r = \mu mg[/tex]
where the term on the left is the centripetal force, and the term on the right is the frictional force, and
m is the mass of the button
[tex]\omega[/tex] is the angular velocity of the platform
r is the distance of the button from the axis
[tex]\mu[/tex] is the coefficient of static friction
g = 9.8 m/s^2 is the acceleration of gravity
In this situation, we have:
r = 0.220 m is the distance of the button from the axis
[tex]\omega = 40.0 rev/min \cdot \frac{2\pi rad/rev}{60 s/min}=4.19 rad/s[/tex] is the angular velocity
Re-arranging the equation for [tex]\mu[/tex], we find the coefficient of static friction:
[tex]\mu = \frac{\omega^2 r}{g}=\frac{(4.19)^2(0.220)}{9.8}=0.394[/tex]
(b) 0.098 m
In this case, we know that the angular velocity is
[tex]\omega=60.0 rev/min \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s[/tex]
And we also know now the coefficient of static friction,
[tex]\mu=0.394[/tex]
So we can now re-arrange the equation for r:
[tex]r=\frac{\mu g}{\omega^2}[/tex]
And substituting, we find the maximum distance at which the button can be placed without slipping:
[tex]r=\frac{(0.394)(9.8)}{(6.28)^2}=0.098 m[/tex]
