What is the force of friction on a 50 gram coin on a turntable?

[mybrohshi5]

Homework Statement

A 50 gram coin is on a horz. turntable. it rides without slipping making 1 revolution every 1.8 seconds. it is 10 cm from the center. the coefficient of static are 1 and the coefficient of kinetic is 0.5.

the size of the force of friction on the coin is

Homework Equations

F = ma

mu_s = F/N

The Attempt at a Solution

I don't think i have to use either static or kinetic coefficients for this problem but i may be wrong.

anyways i did

F = ma

F = m(v²/R)

v = 1rev/1.8s (2pi rad / 1rev)
v = 3.49 rad/s

F = .05kg(3.49^2)/(.1m)

F = 6.09 N

The answer is 60.9 mN so it just seems like i am off a few decimal places but i can't seem to find out where i went wrong or if this is even the right way to approach the problem.

Thanks for any help :)

 

[Doc Al]

F = m(v²/R)

v = 1rev/1.8s (2pi rad / 1rev)
v = 3.49 rad/s

That 'v' should be the tangential velocity--you found the angular velocity, ω.

How would you find the tangential velocity of the coin?

 

[mybrohshi5]

Thank you :)

tangential velocity is v = 2piR / T

v = .349

F = .05kg (.349^2 / .1)

F = .0609 N or 60.9 mN