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flight_recorder t1_j5pb01t wrote

That depends on how much velocity you give the water. If it’s spiralling at a similar rate then the outside water of the pool will have a higher velocity than the outside of the cup, that higher velocity means a massively larger amount of energy which will take a LOT longer to dissipate

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AlienTD5 t1_j5pjhw3 wrote

True, but because of the larger velocity, the friction from the edges of the pool would be much higher as well, right? Hard to say which effect would dominate

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Ragidandy t1_j5podj7 wrote

It isn't harder to say. The surface area of a cylinder goes up much slower than the volume as the cylinder gets bigger. The larger volume's energy would dissipate slower.

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Idyotec t1_j5pt4c9 wrote

But it would require more energy to achieve the same result. One stir for the pool would equate to many stirs for the cup. The resistance of the larger spoon would be higher. OP's question isn't specific enough to give a proper answer.

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AliceInMyDreams t1_j5rb9cq wrote

Everyone is saying that, but assuming the pool is an isotropocally scaled version of the cup, its energy scales with r^3, but the dissipation scales with S*v ~ r^2 * rw ~ r^3

Unless I'm too tired and missing something obvious this doesn't seem that simple

Edit : thinking about it for another minute I do believe I'm too tired after all, because I didn't take into account that the energy doesn't linearly scales with the volume, but rather in r^5 assuming constant angular velocity

And so r^5 >>> r^3 and the pool indeed triumphs over the cup

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Angdrambor t1_j5ps1wn wrote

What is this type of analysis called? In Compsci it's called "Big O notation" but I'm pretty sure that only applies to computational complexity.

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AlienTD5 t1_j5q6oc5 wrote

>The larger volume's energy would dissipate slower.

Why though? Why is volume relevant here

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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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