# Mass on a turntable - Force

1. May 8, 2010

### mybrohshi5

1. The problem statement, all variables and given/known data

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

2. Relevant equations

F = ma

mu_s = F/N

3. The attempt at a solution

I dont 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)

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 cant seem to find out where i went wrong or if this is even the right way to approach the problem.

Thanks for any help :)

2. May 8, 2010

### Staff: Mentor

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

How would you find the tangential velocity of the coin?

3. May 8, 2010

### mybrohshi5

Thank you :)

tangential velocity is v = 2piR / T

v = .349

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

F = .0609 N or 60.9 mN