posted by Physics .
A coin is placced 12.0 cm from the axis of a roatating turntable of variable speed. When the speed of the turntable is slowly incraeased, the coin remains fixed on the turntable until a rate of 50 rpm is reached at which point the coin slides off. What is the coefficent of static friction between the turntable and the coin?
ok it was rpm not mph
so then in order to do one round it would take .83 seconds
Us = (rg)^-1 (t^-1 2 pi r)^2
i got .70
how do i do this
First of all, convert rpm to radians per second, omega (w).
50 rpm = (50 rev/min)*(2 pi rad/rev)*(1 min/60 s) = 5.236 radians/s
When sliding begins, the static friction force equals the centripetal force. That means:
(mus)*M g = M r w^2
mus is the static friction coefficient. M cancels out.
mus = r w^2/g
r = 0.12 m.
Solve for mus