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 35.0 rev/min , provided the button is no more than 0.240 m from the axis. what is the coefficient of static friction between the button and the platform?

Answer 1

For the position of button we can use force balance as

Friction Force = Centripetal force

so here we will have

$\mu_s mg = m\omega^2 R$

here we know that

R = radius of circle where button is placed

$\omega = 2\pi f$

f = 35 rev/min

$f = \frac{35}{60} rev/s = 0.58 rev/s$

$\omega = 2\pi (0.58) = 3.66 rad/s$

now from above equation

$\mu_s(m)(9.81) = (3.66)^2(0.240)$

$\mu_s = 0.33$

so friction coefficient will be 0.33