Masses on a turntable

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 fs that points toward the center of the turntable.

Newtons's second law then is    Σ Fr = fs = m ar.      (1)

Now use the following facts:

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

Conclusion: the slipping criterion is ω2r > μsg or r > μsg/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!