Homework Help: Physics-circular force

Posted by Anonymous on Wednesday, October 28, 2009 at 5:02pm.

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?

Physics-circular force - bobpursley, Wednesday, October 28, 2009 at 5:10pm
forcefriction=mg*mu
centripetalforce= mw^2r
w= 31*2PIr/60 in rad/sec
set them equal, solve for mu.
Physics-circular force - Anonymous, Wednesday, October 28, 2009 at 5:18pm
i got 3.64 but that is wrong because it is too large
Physics-circular force - bobpursley, Wednesday, October 28, 2009 at 5:23pm
I didn't get that, I get on the order of .1

Check your work, or post it here.
Physics-circular force - Anonymous, Wednesday, October 28, 2009 at 5:34pm
mu*9.8=(((30/60)(2pi11))^2)/11
mu=11.8

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