Coefficient of Static Friction Mass on Turntable

[WahooMan]

Homework Statement

A block of mass 3.5kg rests on a rough horizontal turntable at a distance of 1.2m from the axis of rotation. If the block begins to slide when the turntable is rotating at 20.0 rev/min, what is the coefficient of static friction between the block and the turntable?

Homework Equations

Fg = -mg
Fn = mg
Ff = Fn(mu)
radial acceleration (aR) = v^2/r

The Attempt at a Solution

Fg = -mg = -9.80(3.5) = -34.3N
Fn = mg = 34.3N

c = 2pi(r) = 2pi(1.2) = 7.54m
v = 7.54m*20 = 150.8m/60s = 2.51m/s

aR = 2.51^2/1.2 = 5.26m/s^2

Now I think I am supposed to use sum of the forces = ma, but I'm not sure if there is a force acting on the mass pushing it to the outside of the turntable (I'm pretty sure there is, but I don't know how to calculate it.)

 

[Delphi51]

I agree with your aR - good work navigating through all that!
I want to give you a huge tip, if you will bear with me. The trick is to write the big picture and work with it a bit before you put in the numbers. In this case, you want the centripetal force to just equal the friction force. Any faster, and it slips. So you just write
Fc = Ff as your starting point. This is the big picture.
mv^2/r = μmg filling in some details.
Cancel the m's, solve for μ and you are ready to put in the numbers. Use the number you already calculated for v.

 

[WahooMan]

mv^2/r = mu(mg)
m's cancel
v^2/r = mu(g)
2.51^2/1.2 = mu(9.80)
5.26 = mu(9.80)
mu = 0.54

Thank you so much for helping me

 

[Delphi51]

A pleasure, WahooMan. Remember that trick. There are so many problems where you start with two forces equal.