Question

A small button placed on a horizontal rotating platform with diameter 0.520 m will revolve with the platform when it is brought up to a speed of 40.0 rev/min, provided the button is no more than 0.220 m from the axis. (a) What is the coefficient of static friction between the button and the platform? (b) How far from the axis can the button be placed, without slipping, if the platform rotates at 60.0 rev/min?

Answer 1

(a) 0.394

The static frictional force provides the centripetal force that keeps the button stuck on the rotating platform. So we can write

$m\omega^2 r = \mu mg$

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

$\omega$ is the angular velocity of the platform

r is the distance of the button from the axis

$\mu$ 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

$\omega = 40.0 rev/min \cdot \frac{2\pi rad/rev}{60 s/min}=4.19 rad/s$ is the angular velocity

Re-arranging the equation for $\mu$, we find the coefficient of static friction:

$\mu = \frac{\omega^2 r}{g}=\frac{(4.19)^2(0.220)}{9.8}=0.394$

(b) 0.098 m

In this case, we know that the angular velocity is

$\omega=60.0 rev/min \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s$

And we also know now the coefficient of static friction,

$\mu=0.394$

So we can now re-arrange the equation for r:

$r=\frac{\mu g}{\omega^2}$

And substituting, we find the maximum distance at which the button can be placed without slipping:

$r=\frac{(0.394)(9.8)}{(6.28)^2}=0.098 m$