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AlienTD5 t1_j5q6oc5 wrote

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welshmanec2 t1_j5qbckw wrote

Because that determines how much energy the water has (volume is equivalent of mass). That energy has to be lost to friction, which doesn't increase as a cube.

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Ragidandy t1_j5qu4v3 wrote

It depends on the un-stated specifics of OP's question. But generally, the more volume the fluid in the cylinder has (ie: the larger the cylinder), the smaller the percentage of the fluid that is in contact with the edges of the cylinder. So a smaller volume would have a greater proportion of its fluid dragging against the edges slowing the liquid.

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Lankpants t1_j5t28t9 wrote

In addition to what the other commenters have said there's also another effect at play. To get the pool spinning at the same RPM as the cup the outside needs to be moving several times faster. If both swirl once per second then the water at the edge of the pool needs to be covering the entire radius of the whirlpool every second.

What this means in practice is that if the two are spinning at a similar speed the energy even of one water molecule in the pool on average is higher. There are also more molecules moving. So the pool has far more energy. And remember, water is a liquid. The partials behave relatively independently and experience friction with the wall independently. So any of these particles at the edge of the pool have more energy to lose to friction. There's also a lower percentage of particles experiencing direct friction with the wall at any given time due to the square cube rule.

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