Physicscircular force
posted by Anonymous on .
A coin is placed 11.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 31 rpm is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?

forcefriction=mg*mu
centripetalforce= mw^2r
w= 31*2PIr/60 in rad/sec
set them equal, solve for mu. 
i got 3.64 but that is wrong because it is too large

I didn't get that, I get on the order of .1
Check your work, or post it here. 
mu*9.8=(((30/60)(2pi11))^2)/11
mu=11.8