(a) 0.394
The static frictional force provides the centripetal force that keeps the button stuck on the rotating platform. So we can write
![m\omega^2 r = \mu mg](https://tex.z-dn.net/?f=m%5Comega%5E2%20r%20%3D%20%5Cmu%20mg)
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
is the angular velocity of the platform
r is the distance of the button from the axis
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
is the angular velocity
Re-arranging the equation for
, we find the coefficient of static friction:
![\mu = \frac{\omega^2 r}{g}=\frac{(4.19)^2(0.220)}{9.8}=0.394](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B%5Comega%5E2%20r%7D%7Bg%7D%3D%5Cfrac%7B%284.19%29%5E2%280.220%29%7D%7B9.8%7D%3D0.394)
(b) 0.098 m
In this case, we know that the angular velocity is
![\omega=60.0 rev/min \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s](https://tex.z-dn.net/?f=%5Comega%3D60.0%20rev%2Fmin%20%5Ccdot%20%5Cfrac%7B2%5Cpi%20rad%2Frev%7D%7B60%20s%2Fmin%7D%3D6.28%20rad%2Fs)
And we also know now the coefficient of static friction,
![\mu=0.394](https://tex.z-dn.net/?f=%5Cmu%3D0.394)
So we can now re-arrange the equation for r:
![r=\frac{\mu g}{\omega^2}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B%5Cmu%20g%7D%7B%5Comega%5E2%7D)
And substituting, we find the maximum distance at which the button can be placed without slipping:
![r=\frac{(0.394)(9.8)}{(6.28)^2}=0.098 m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B%280.394%29%289.8%29%7D%7B%286.28%29%5E2%7D%3D0.098%20m)