Two water-filled buckets. One swinging. Which empties first?

[Neil9327]

Two water-filled buckets with holes in bottom hanging from strings. One swinging one not. Which empties first?

 

Two identical buckets each filled with water. One set swinging from side to side and the other stationary. Which one will empty first, and why? Or will they empty at the same rate?

 

Assume no surface tension or evaporation effects. Swing is 45 degrees either side (not over the top)

[CaptainPanic]

Flow through an orifice is what you want to study.

 

So, the swinging will cause pressure changes. When the bucket is most left and most right, then the gravity plays a smaller role (and if it would be swinging up to 90 degrees it's basically in free fall, and the gravity plays no role at all). When the bucket passes through the lowest point, it will experience more than the standard 9.81 m/s2, due to gravity and its circular motion.

 

So, how does the force change as a function of the position?

How does the flow change as a function of the force (probably better to express the force as a pressure at the bottom of the bucket).

[J.C.MacSwell]

Add to CP's remarks, how does the water move, or slosh, relative to the bucket?

 

It is unlikely to be as regular or in sync with the pendulum swing, so it is quite a complex problem.

 

If you swung it in a circle at 45 degrees it should empty faster, but that would not be the same thing.

[Neil9327]

I don't want a complex answer involving fluid mechanics. So assume no sloshing about. Assume that the rate of flow of water out is proportional to the acceleration inline with the pendulum. Assume the bucket is small so the centre of gravity is close to the hole in the bucket.

[J.C.MacSwell]

I don't want a complex answer involving fluid mechanics. So assume no sloshing about. Assume that the rate of flow of water out is proportional to the acceleration inline with the pendulum. Assume the bucket is small so the centre of gravity is close to the hole in the bucket.

 

 

If it is proportional to the force, it would be faster for the swinging pendulum, since on average the vertical force will be the same, and there are additional components to the force in that case.

[swansont]

If it is proportional to the force, it would be faster for the swinging pendulum, since on average the vertical force will be the same, and there are additional components to the force in that case.

 

Would it? The bucket and water have the same acceleration due to gravity. If the line were not taught, no water would leak out at all, so the additional pressure is dependent on the tension, which is varying sinusoidally in time.

 

The time-average of sine oscillating between 0 and 1 is 1/2. It's possible (I haven't actually worked it out) that the effects cancel, and the buckets take the same amount of time (to within one period of oscillation) to empty.

[Bignose]

I don't want a complex answer involving fluid mechanics.

 

Not to sound too trite, but it is kind of silly to pose a reasonably complex problem but not desire a sufficiently complex analysis to try to answer the problem in a way that doesn't just make grossly inaccurate and unlikely assumptions. Like, no sloshing. Because that's not what will happen, the fluid will have an inertia either way that won't immediately dissipate and the sloshing will have a significant effect.

[swansont]

Not to sound too trite, but it is kind of silly to pose a reasonably complex problem but not desire a sufficiently complex analysis to try to answer the problem in a way that doesn't just make grossly inaccurate and unlikely assumptions. Like, no sloshing. Because that's not what will happen, the fluid will have an inertia either way that won't immediately dissipate and the sloshing will have a significant effect.

 

Not if the swinging is started incrementally. Once the bucket is swinging, the water won't slosh — the only nonradial force is gravity, and that affects the bucket and water equally. This is a physics problem; cows are spherical, and cannons are massless. Idealized behavior can be assumed.

[Bignose]

But if the buckets have any kind of shape to them, the inertia of the fluid will cause some rotation, which then will lead to centrifugal or Coriolis forces in the fluid. These things aren't really all that negligible.

 

I might buy it if the fluid weren't water. If you had something very viscous, like glycerin or similar, then the viscosity would dampen all these disturbances out very quickly -- and virtually little to no sloshing, too! -- and I would feel that the simplifying assumptions would be more valid.

 

But, water just isn't viscous enough for those assumptions to be valid. Eddies in water form too easily and last too long to be ignored. The non-laminar, rotating, sloshing flow that real water would experience will not fit with the simplifying assumptions that are needed to keep the problem from getting into fluid mechanics and a more detailed analysis.

[max.yevs]

despite of where gravity is at what point, the centrifugal force always pushes the contents outwards, thus the one swinging will empty first.

[Dudde]

I don't know guys - once the water gets to a certain level, it'll start staying mainly on the sides of the bucket and avoiding the hole. How big a hole are we talking here?

[CaptainPanic]

Partial solution follows at the end. First I have to myth-bust some bad answers.

I don't want a complex answer involving fluid mechanics.

 

Too late. You posted on the scienceforums.

 

If it is proportional to the force, it would be faster for the swinging pendulum, since on average the vertical force will be the same, and there are additional components to the force in that case.

 

It’s proportional to the square root of the force, as I’ll explain below…

 

I don't know guys - once the water gets to a certain level, it'll start staying mainly on the sides of the bucket and avoiding the hole. How big a hole are we talking here?

Water will not stay at the sides unless you spin the bucket around its vertical axis (which we don't do). Let's ignore the very last drops, shall we?

Furthermore, the size of the hole is irrelevant, assuming it's the same size hole in both buckets.

We're studying the difference between the two buckets, and we're looking for a qualitative answer. In this question, all that matters is that there is a hole, and that it's at the bottom of both buckets.

 

Bernoulli

As I said before in post 2: you want to study the flow through an orifice (flow through a hole)... which is a simplified version of Bernoulli's equation.

 

The formula clearly shows that the volumetric flowrate is proportional to the square root of the pressure difference.

 

Pressure is force per area. Area does not change, so pressure is proportional to the force of the water pressing down.

 

Since this force is caused by acceleration, the pressure is proportional to the acceleration (gravity and acceleration caused by swinging).

 

Conclusion so far: volumetric flowrate is proportional to the square root of the acceleration.

 

Now we have to determine the acceleration which acts on the water, and relate that to the motion of the bucket. If the acceleration of the water is equal to that of the bucket (as it would be in free fall), then no water will come out.

 

I'm not 100% sure at this point. The bucket moving up and down would have its acceleration average out. But since the square root of the maximum and minimum is not the same as twice the square root of the average, the flowrates would not average out.

 

However, there is also a sideways motion which causes additional acceleration. I'm not 100% sure how to deal with this.

 

Anyway, I hope that this helped.

[swansont]

despite of where gravity is at what point, the centrifugal force always pushes the contents outwards, thus the one swinging will empty first.

 

If it's swinging in a circle, what's the net force at the top?

[YT2095]

my Experimental evidence suggests the swinging one will be a LITTLE bit faster, roughly a 44:47 ratio faster.

 

I just tested this using:

 

like this:

 

when still it averaged 47 seconds to empty, and in motion it averaged 44 seconds in 5 trials of each (10 total experiments).

 

hope that helps a little

Edited March 26, 2009 by YT2095

[J.C.MacSwell]

Would it? The bucket and water have the same acceleration due to gravity. If the line were not taught, no water would leak out at all, so the additional pressure is dependent on the tension, which is varying sinusoidally in time.

 

The time-average of sine oscillating between 0 and 1 is 1/2. It's possible (I haven't actually worked it out) that the effects cancel, and the buckets take the same amount of time (to within one period of oscillation) to empty.

 

My guess (haven't worked it out either) is that the vertical component cancels out exactly, and because there is an additional horizontal component to the force, always adding to the tension except when the bucket is at the bottom of it's swing, then it would, on average, have more tension and therefore empty faster.

[Sisyphus]

My guess (haven't worked it out either) is that the vertical component cancels out exactly, and because there is an additional horizontal component to the force, always adding to the tension except when the bucket is at the bottom of it's swing, then it would, on average, have more tension and therefore empty faster.

 

I don't think that's right. I'm thinking the tension would be greatest at the bottom of the swing, and so would the flow. If you think about it, if the pendulum swung up to a full 90 degrees, the tension (and flow) would be zero at that point, as both bucket and water would be in freefall.

[CaptainPanic]

My guess (haven't worked it out either) is that the vertical component cancels out exactly, and because there is an additional horizontal component to the force, always adding to the tension except when the bucket is at the bottom of it's swing, then it would, on average, have more tension and therefore empty faster.

 

Please, please, please, please - if you join a science forum, at least READ THE INFORMATION THAT IS AVAILABLE. I provided quite an essential bit of information which is called

***Bernoulli's equation***.

You may disagree with it, fine - explain me why. I'm interested. But I'm a bit tired of being ignored, and then reading what people are guessing.

 

I have linked not once, but twice before, in two separate posts, to the formula which clearly shows that: the vertical component does not cancel out. The accelerations cancel out perhaps, if you like to say that, but flowrate is proportional to the frickin' square root of the pressure difference between the bottom of the bucket and the outside atmospheric pressure.

The hydrostatic pressure will change linearly with the acceleration of the bucket, and therefore the pressure difference changes linearly with the acceleration, and therefore the flowrate changes with the square root of the pressure difference. And that means that it does not cancel out.

 

[math]Q = A_2\;\sqrt{\frac{2\;(P_1-P_2)/\rho}{1-(A_2/A_1)^2}}[/math] <-- formula. Read it. Source.

Pressure difference (hydrostatic pressure, caused by the weight of the water) pushes the water out. The velocity of that water is proportional to the square root of the pressure difference.

That means that flowrate Q is proportional to the square root of the pressure difference. It's derived from Bernoulli's equation, which is standard textbook material. I may have applied it the wrong way, but so far I'm the only one here to put up a formula... so can we please discuss that?

 

Please note that while it's true that the average of 0 and 2 equals the average of 1 and 1, it's not true that their square roots are equal: [math]\sqrt{2}+\sqrt{0}<\sqrt{1}+\sqrt{1}[/math]

 

This was the 3rd time I try to bring up this equation. I believe it's a vital part of the answer. Can we please use it, or find out why it cannot be applied? "Guessing" just seems a bit silly here.

 

Thank you for your attention.

 

Apologies for using big red letters, but I believe that YT's pictures have distracted people from my links to things that I think are important. Mods can make it less flashy if they want to.

Edited March 26, 2009 by CaptainPanic

[Dudde]

Water will not stay at the sides unless you spin the bucket around its vertical axis (which we don't do). Let's ignore the very last drops, shall we?

 

Oh yeah, I don't know what my brain was visualizing... I think I was using a much faster swing as a rationale, as well as wider than 45 to each side.

 

I think YT's experiment helped at least to demonstrate that the swinging bucket will empty faster? That was the original question, after all - and simple, per the OP's request

[swansont]

Please, please, please, please - if you join a science forum, at least READ THE INFORMATION THAT IS AVAILABLE. I provided quite an essential bit of information which is called

***Bernoulli's equation***.

You may disagree with it, fine - explain me why. I'm interested. But I'm a bit tired of being ignored, and then reading what people are guessing.

 

I have linked not once, but twice before, in two separate posts, to the formula which clearly shows that: the vertical component does not cancel out. The accelerations cancel out perhaps, if you like to say that, but flowrate is proportional to the frickin' square root of the pressure difference between the bottom of the bucket and the outside atmospheric pressure.

The hydrostatic pressure will change linearly with the acceleration of the bucket, and therefore the pressure difference changes linearly with the acceleration, and therefore the flowrate changes with the square root of the pressure difference. And that means that it does not cancel out.

 

[math]Q = A_2\;\sqrt{\frac{2\;(P_1-P_2)/\rho}{1-(A_2/A_1)^2}}[/math] <-- formula. Read it. Source.

Pressure difference (hydrostatic pressure, caused by the weight of the water) pushes the water out. The velocity of that water is proportional to the square root of the pressure difference.

That means that flowrate Q is proportional to the square root of the pressure difference. It's derived from Bernoulli's equation, which is standard textbook material. I may have applied it the wrong way, but so far I'm the only one here to put up a formula... so can we please discuss that?

 

Please note that while it's true that the average of 0 and 2 equals the average of 1 and 1, it's not true that their square roots are equal: [math]\sqrt{2}+\sqrt{0}<\sqrt{1}+\sqrt{1}[/math]

 

This was the 3rd time I try to bring up this equation. I believe it's a vital part of the answer. Can we please use it, or find out why it cannot be applied? "Guessing" just seems a bit silly here.

 

Thank you for your attention.

 

Apologies for using big red letters, but I believe that YT's pictures have distracted people from my links to things that I think are important. Mods can make it less flashy if they want to.

 

Hey, CP, I'm not ignoring you. I hadn't had time to work out details, so I wasn't sure by inspection if there was an exponent problem or not.

 

The solution to a harmonic oscillator is that [math]\theta=sin\omega t[/math], (for small angles) and the tension will be the centripetal force, which is [math]ml({\frac{d\theta}{dt}})^2[/math], so you end up with a force (and thus pressure) that depends on the square of the trig function. Combine this with the flow rate being dependent on the square root of the pressure, and we're back to averaging cosine and getting 1/2.

 

BUT: the actual results are going to depend on the initial conditions, because you will get slower draining at the peak and faster at the bottom of the path, so it matters where you start the experiment. If you start draining at the minimum level, but with the pendulum in motion (you start it swinging and pull a plug when it's at the lowest point), it will drain slightly faster than if it starts draining at the maximum amplitude.

 

It's not completely certain that YT's discrepancy is statistically significant (~6% is a small difference for 5 trials) and if it is, whether it's due to deviations from ideal motion (e.g. sloshing) or from the solution for large angles being slightly different than the small-angle case. But I think the period increases for increasing angle, which means there should be more time spent at the lower flow rate, which is opposite of the observed effect.

[YT2095]

I used roughly a Litre for the tests, but they we all the same amount because of a marker line/fill level.

 

the stopper for the Swinging tests was pulled at it`s greatest height before letting the bottle go.

 

the hole was done with a 1/4 inch drill bit.

 

I have no idea of the exact angle as a number but the bottle was raised to a specific height for all tests.

[J.C.MacSwell]

I don't think that's right. I'm thinking the tension would be greatest at the bottom of the swing, and so would the flow. If you think about it, if the pendulum swung up to a full 90 degrees, the tension (and flow) would be zero at that point, as both bucket and water would be in freefall.

 

That should be correct, for the tension. The flow would lag somewhat behind.

 

But that doesn't change the fact that there is no horizontal component adding to the tension at that point in the swing. The tension is all vertical at that point.The tension at that point is greatest there because the centripetal acceleration is greatest at that point.

[YT2095]

Hmmm... here`s a simple test someone could try, you`ll need a similar setup as I did, but a regular weight will do.

then you`ll need to hook the string up to some Spring scales, the sort fishermen use.

I would do this myself but I don`t have any.

when you swing the weight, This Time you look at the scales reading, if what is being said in here is true, I predict the weight should be greatest when the mass is at it`s lowest.

 

if no one tries this before me, I`ll have a look at what junk I have laying around and build something similar if I get time.

[J.C.MacSwell]

Hey, CP, I'm not ignoring you. I hadn't had time to work out details, so I wasn't sure by inspection if there was an exponent problem or not.

 

The solution to a harmonic oscillator is that [math]\theta=sin\omega t[/math], (for small angles) and the tension will be the centripetal force, which is [math]ml({\frac{d\theta}{dt}})^2[/math], so you end up with a force (and thus pressure) that depends on the square of the trig function. Combine this with the flow rate being dependent on the square root of the pressure, and we're back to averaging cosine and getting 1/2.

 

BUT: the actual results are going to depend on the initial conditions, because you will get slower draining at the peak and faster at the bottom of the path, so it matters where you start the experiment. If you start draining at the minimum level, but with the pendulum in motion (you start it swinging and pull a plug when it's at the lowest point), it will drain slightly faster than if it starts draining at the maximum amplitude.

 

It's not completely certain that YT's discrepancy is statistically significant (~6% is a small difference for 5 trials) and if it is, whether it's due to deviations from ideal motion (e.g. sloshing) or from the solution for large angles being slightly different than the small-angle case. But I think the period increases for increasing angle, which means there should be more time spent at the lower flow rate, which is opposite of the observed effect.

 

Isn't the period constant for an ideal pendulum?

[max.yevs]

nice, im glad someone decided to actually do it...

 

what i said before is not completely correct, if you're on some planet with very high gravity, centrifugal force might be irrelevant...

[Sisyphus]

That should be correct, for the tension. The flow would lag somewhat behind.

 

But that doesn't change the fact that there is no horizontal component adding to the tension at that point in the swing. The tension is all vertical at that point.The tension at that point is greatest there because the centripetal acceleration is greatest at that point.

 

Why would the flow lag behind? The force of the flow will be directly proportional to the tension on the rope.

 

nice, im glad someone decided to actually do it...

 

what i said before is not completely correct, if you're on some planet with very high gravity, centrifugal force might be irrelevant...

 

No, the situation would be the same. If you increase the gravity, then the pendulum swings faster, and the centrifugal effect increases.