Not the right kind of engineer but I'll give 'er a swing. (Fiber Optic and Network engineer)

Let's say all external input variables are the same. As in everything is the same except one is occurring at a 152,000 x scale (2 cups of water vs standard pool at 19000 gallons or 340,000 cups)

My semi-educated guess: We assume conservation of momentum and the same angular velocity.

The surface-area-to-volume ratio decreases with increasing volume. Therefore the pool-sized mug will experience less force of friction relative to the increase in mass/volume.

An increase in volume increases the radius and mass. Therefore the linear velocity at the edge is much higher as mentioned in another comment causing an increase in centrifugal force causing an increase in friction for the pool.

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1 cup = 14.4375 cubic inches2 cups = 28.875 cubic inches Pool sized coffee mug340,000 cups = 4,908,750 cubic inchesYeti Coffee mug on my desk external dimensions for ratio = 3.5"W x 6.6"H (r=1.75")Right cylinder volume = V=πr2hGot busy with work, and might come back later to finish my math.

Figured out how to figure it out as long as the correct answer doesn't involve calculus cause I ain't got time for that. Do the math for two spinning cylinders with friction at the edge.