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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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colcob t1_j5pe6hk wrote

Yes. The rate of slowing of the spinning water must be a function of the kinetic energy of the water against the drag of the side walls.

Even if the max velocity of the water is the same in the pool (ie a very slow rotation rate) then the energy of the water will scale somewhat with the volume of the water, ie cubicly, while the drag of the walls will scale somewhat with the area of the side walls, ie. squared, therefore the larger the cup, the more the kinetic energy of the water outweighs the drag of the sides.

If you do actually stir at the same rotational rate, then as the other commenter says, the max velocity of the water will be very significantly higher than in the coffee cup, and as energy scales to the square of the velocity then you'll have vastly more energy proportional to drag than in the coffee cup.

Plus anecdotally, as a kid we used to run around those circular backyard swimming pools and get the water spinning faster and faster, and they kept going for ages!

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Starbucks__Coffey t1_j5pjrxr wrote

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.

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Ragidandy t1_j5podj7 wrote

Replying to AlienTD5 (#1,470,522)

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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Starbucks__Coffey t1_j5ppyy6 wrote

Replying to colcob (#1,469,922)

>Even if the max velocity of the water is the same in the pool (ie a very slow rotation rate) then the energy of the water will scale somewhat with the volume of the water, ie cubicly, while the drag of the walls will scale somewhat with the area of the side walls, ie. squared, therefore the larger the cup, the more the kinetic energy of the water outweighs the drag of the sides.

The centripetal force increases friction aswell angular momentum. So it increases the energy lost due to friction aswell as the increase in surface area.

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Idyotec t1_j5pt4c9 wrote

Replying to Ragidandy (#1,471,018)

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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Wadsworth_McStumpy t1_j5puj23 wrote

It depends on whether you stir with the same linear velocity or the same angular velocity.

The coffee cup is about 9 inches around, and if you stir it once per second, the coffee is moving at 9 inches per second. In a 26 foot pool, 9 inches per second will barely move the water, and it will stop almost as soon as you lift the spoon.

On the other hand, if you stir the pool at one stir (360 degrees) per second, the water will be moving at 980 inches per second (55 miles per hour!), and will keep moving for a very, very long time.

Also, that will probably generate enough outward force to destroy your pool. (I'm too busy to calculate the force right now, but if somebody else wants to, I'd be interested to see it.)

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Sprinklypoo t1_j5pzecj wrote

Yes. Mostly because of momentum and friction.

Your small cup has a much higher friction at the edges in proportion to the momentum of the circling liquid.

You could of course balance this by stirring the swimming pool at a slower rate - but that wasn't part of your question...

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bildramer t1_j5qa2qa wrote

Ignoring the effect from length scaling and drag from walls, it gets more complicated. Liquids have viscosity, and do in fact behave differently at different scales. We use the Reynolds number to describe the different behaviors - laminar or turbulent flow. At small enough scales, your vortices will dissipate very quickly, and at large enough scales, you can ignore viscosity.

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imbluedabedeedabedaa t1_j5qdycm wrote

Replying to colcob (#1,469,922)

Otherwise known as the square-cube law!

Another great (but completely unrelated) example of this law explains the design differences between large and small animals. As a creature gets larger, its mass increases at a cubic rate (volume, m^3) but muscle/bone strength is proportional to their cross sectional area (m^2). This is why an elephant has such massive muscles/bones when (proportionally) compared to a dog.

It also impacts body temperature--since heat generation is proportional to body volume, but the ability to reject heat is proportional to the skin area. This explains the elephants' massive ears, and also why most cold blooded animals are very small.

If you scaled a dog up to elephant size it would collapse under its own weight before dying of heat. If you scaled an elephant down to dog size it would probably freeze to death. All because of math!

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Ragidandy t1_j5qu4v3 wrote

Replying to AlienTD5 (#1,473,091)

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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ShelfordPrefect t1_j5qzs5e wrote

Replying to colcob (#1,469,922)

Another factor in favour of the big pool: at the same rotational rate, the big pool will have less shear because the difference in speed between two points a given distance apart is smaller than the same two points in the small cup, so I would imagine less energy would be lost to turbulence

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crackaryah t1_j5r4hxl wrote

Replying to silvercup011 (#1,476,269)

It wouldn't freeze. Endotherms regulate their body temperatures by modulating their metabolisms. Not to mention that modern elephants don't live in cold places. Ancient elephants (mammoths) were highly insulated. Metabolism can be modulated a lot. There are small "warm blooded" animals that have a body temperature above 37 C in the summer, but hibernate during the winter, dropping their body temperatures to below 0 C. They don't freeze and they certainly don't die. There is even some evidence that hibernation contributes to longevity.

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AliceInMyDreams t1_j5rb9cq wrote

Replying to Ragidandy (#1,471,018)

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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thereisafrx t1_j5rz7fz wrote

Think about it this way, if you put a quarter on a small hill and rolled it down the hill, and a large stone disk on a proportionally-larger hill, would the large stone disk roll for longer?

Same for the two cases of the fluids in large/small containers, except with the solids there are no viscous forces to consider.

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colcob t1_j5sdytd wrote

Replying to Old_comfy_shoes (#1,480,351)

So by rotational rate i mean the angular velocity, ie. degrees per second, RPM, not the tangential velocity.

I guess I would stir a cup of about 1 revolution per second, so I'm imagining that if you stir the swimming pool cup at 1 revolution per second, the water's going pretty damn fast at the edge!

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sentientskeleton t1_j5siqen wrote

Replying to bildramer (#1,473,459)

The only thing that will make the water stop spinning (reduce its total angular momentum) is drag from the walls, so you can't ignore it, no matter how large the Reynolds number. Even at high Re you can't completely ignore viscosity either, its effects only get concentrated into very thin regions (like boundary layers) but there it is extremely important.

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Lankpants t1_j5t28t9 wrote

Replying to AlienTD5 (#1,473,091)

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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GotTheNameIWanted t1_j5t29eh wrote

Yes.

Think of it this way, it would require more energy to move that larger spoon compared to the smaller spoon. This energy is stored in the waters movement once you stop stirring as kinetic energy. More input (i.e. larger spoon) would mean more kinetic energy.

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thereisafrx t1_j5v6882 wrote

Replying to Psychotic_EGG (#1,481,241)

Not necessarily.

Lots of people have rolled stones down hills, or pushed something down the stairs, etc.

As such, a lot more people have a more intuitive concept of how solid mass interacts with the surrounding environment, but fluids are a different beast entirely so it wouldn't be something a lot of folks would have common experience with.

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