o.k., I will try to take another shot at an explanation in keeping up with your idea of a stirrer bar and a stationary beaker...but let's change the stirrer bar for something more steady and symmetric...just a simple, smooth rod being used as a drill bit in a drill...as if you were making a milkshake.
In as much as I am going to try to stay in your frame of reference, I do want you to visualize and keep up with a few of things though...
First,
cylindrical coordinate system with origin at the bottom and center point of the cylindrical beaker. r radially out, of course, phi around and z up along the axis of the beaker.
Then, I am sure you agree, Earth gravitational force is acting in the -z direction; then, I want you to think of the centrifugal force as another gravitational force acting (perpendicular to the Earth one) in the r direction, radially out all the way around. For Earth gravity, the bottom of the beaker is the floor, ground and the surface where things tend to
fall down to and stop there; then, rotate your head 90 degrees and think for the centrifugal-force (CF) "gravity" that the walls of the beaker are the floor and ground and the surface things tend to
fall down to due to this CF "gravity"...getting this?
Hopefully, just thinking of these two forces acting on the same mass, you can start visualizing a net force resulting from these two components that starts pointing towards the bottom corner of the beaker...and that alone should start giving you a sense of why you get a "ramp" made of water (look only at left or right half of the cylinder) from the top of the beaker to the bottom center.
O.k., boundary condition: As you know, because of friction, there is no slip between beaker and water and so, the velocity of water along the
walls of the beaker is zero. But the velocity is also zero along the
bottom of the beaker...it is very important not to forget about this because it tends to eliminate the ability to produce angular momentum on the water as you approach the bottom of the beaker...so, between this condition and the column of water above this water...the radius of the funnel is necessarily smaller down here.
Getting back to the source of angular momentum...let's put the smooth rod in a drill and place down the center of the beaker, half way down for example, and rotate at, say, 500 rpm.
Boundary condition: Again, the slip between rod and water is zero and so, the water in contact with the rod moves just as fast and because of friction within water, the angular momentum starts spreading radially out (but it is being held down at r = wall of beaker). Velocity of water would be continuous, though, highest at the rod, zero along beaker surfaces.
Once water starts also experiencing CF gravity in the radial direction, it will start separating from the rod and start
falling down in this direction, too, and there is no wall (lid) there along the top of the beaker to stop it...so, it starts
rolling down a ramp "down" in the r direction and also "sideways" (up) in the z direction (you need to tilt your head here!).
For a given finite volume of water, its final position would be a balancing act between Earth gravity (and column of water above this point) and CF gravity due to the angular velocity at the same point.
gsal