Stir plate magnetic field RPM relative to stir bar Observing the use of a variable speed stir plate prompted some discussion about the stir plate motor RPM vs the stir bar RPM and we did not have any physicists handy.
The stir plate has a beaker filled with water on it and a stir bar inside the beaker. The stir plate measures RPM based on the RPM of the motor with attached magnet.
We noticed that if the stir plate RPM setting is rapidly increased or set to a high setting the stir bar seems to "decouple" from the magnetic field and flop around in the beaker like a fish out of water. Here we assumed the polarity of the stir bar reversed relative to the stir plate magnet due to friction of the liquid causing the stir bar to lag behind the stir plate.
1) Is our assumption about the decoupling correct?
2) Is there a technical term for this decoupling?
The main question is:
3) Due to friction can the RPM of the stir plate exceed the RPM of the stir bar without the magnetic field decoupling? We are wondering if there could be a substantial difference in revolutions between the stir plate and stir bar over a couple of minutes with a low viscosity liquid. "Substantial" being the stir plate running at 100 RPM and the stir bar running at 80, all without decoupling.
 A: The magnetic bar experiences two different moments while rotating: one from the magnet attached to the motor, the other from the fluid. 
At speed zero the magnet bar moves until it gets in a position of minimum moment experienced (in this case zero). When the motor starts rotating the bar has to follow this minimum. Since drag increases with speed, in order to reach equilibrium and rotate with the same speed as the motor, the bar has to place itself in a position relative to the motor where the experienced moment is exactly equal to the friction. This position is a little behind the minimum position at speed zero.
As long as the rotation speed is low enough, bar's speed will be equal to the motor speed. With increasing speed the bar will take and equilibrium position further and further from the minimum (the point at speed zero). Eventually the bar will be so far from the equilibrium point that the moment applied by the magnet attached to the motor is no longer sufficient to keep the bar in place. That's when the decoupling takes place.
In practice, the motor magnet acts as a spring: it exerts a recalling force on the bar. This force doesn't extend indefinitely and for sufficient distances it drops, as if the spring breaks. 

Due to friction can the RPM of the stir plate exceed the RPM of the stir bar without the magnetic field decoupling? We are wondering if there could be a substantial difference in revolutions between the stir plate and stir bar over a couple of minutes with a low viscosity liquid. "Substantial" being the stir plate running at 100 RPM and the stir bar running at 80, all without decoupling.

I don't think so. It might happen that the bar would reach a frequency that is an integer fraction of the motor speed, but it is very unlikely and also very unstable.
