Posted by **Physics** on Saturday, September 5, 2009 at 5:18pm.

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

- Physics -
**drwls**, Saturday, September 5, 2009 at 7:02pm
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

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