Physics
posted by Karlee .
A coin is placed 17.0 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 33 is reached and the coin slides off.What is the coefficient of static friction between the coin and the turntable?

forcefriction=forcecentripetal
mu*mg=m*v^2/r
At this point, I don't know what your 33 is? m/s? rpm? 
oops sorry, 33 rpm and 17 cm

I have a very similar problem due on mastering physics in an hour.. I came here to get help but now 30 minutes later I'm actually going to be giving help :). Hopefully it wont be too late depending on your time zone.
u= coefficient of static friction
force of friction = u*mg
IM NOT THAT SMART SO THIS MIGHT BE WRONG BUT HOPEFULLY IT WILL HELP.. I ALSO SUCK AT SIG FIGS SO ITS PROBABLY SLIGHTLY OFF.
set force of friction = to centripetal force
so u*m*g = m*v^2/r
m cancels
so u*g = v^2/r.
r = 17cm = .17m
the circumference of the circle = 2pi*r
which roughly = 1.068m (only 2 sig figs tho)
convert 33 rpm to rps to get .55 rps.
so the object move .55 rotations in one second so it moves .55*1.068m in one second. so it moves .5874m/s reduce this to 2 sig figs to get .59m/s
use the equation we formed earlier :
u*g = v^2/r.
u= v^2/rg
u= (.59m/s)^2/((.17m)(9.80m/s^2))
u=.21 
THANK YOU!!!!!!!

TY! TY!
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