Two identical masses on a horizontal turntable are located at different distances from the center of the turntable. As the turntable's rotation rate increases, which mass slides off the turntable first?

1. The mass closest to the center.
2. The mass furthest from the center.
3. Both slide at the same time.

Slipping Criterion: If the mass doesn't slip, the only horizontal force is static friction f_s that points toward the center of the turntable.

Newtons's second law then is    Σ F_r = f_s = m a_r.      (1)

Now use the following facts:

1. For the mass to not slip when the turntable rotates with angular frequency ω, the inward acceleration a_r must equal v²/r = ω²r.
2. The maximum inward force provided by static friction is (f_s)_max = μ_smg. Thus the maximum inward acceleration provided by friction is μ_sg.
3. Use a_r = ω²r and (f_s)_max = μ_smg in Eq. (1). When the rotational acceleration, ω²r, exceeds the maximum acceleration due to friction, μ_sg, the mass slips.

Conclusion: the slipping criterion is ω²r > μ_sg or r > μ_sg/ω².

For a given ω, a mass at a larger radius slips, while a mass at a smaller radius does not. The outer masses slip first!