Understanding the Mechanics of a Falling Bucket of Water: A Conceptual Question

[tyco05]

Hi everyone, this is my first post so please be gentle.

Just a conceptual question:


We have a bucket of water with a hole in the bottom (see attached picture for clarification). If it is dropped from a sufficient height as to reach terminal velocity (assume no rotation), at what point (if any) will the water begin to flow out of the bucket? (The hole is unplugged as it is dropped)


My understanding is this:

1/ With no wind resistance we get no forces other than g, and the water will fall with the bucket (ie. be weightless) for the duration of the fall.

2/ As soon as the bucket begins falling there is a wind resistance force (upwards) which will be proportional to some function of the bucket's velocity. This force acts on the bucket (and water in the bucket experiences a force from the bucket, ie. it has weight) , but not on the water leaving the hole. Therefore the water will begin leaving the bucket as soon as it is dropped. The RATE of the flow will increase until it reaches terminal velocity, at which point it is a maximum (like if you were just standing on Earth holding a bucket?)


Is my understanding correct? Is there more to it? Any thoughts would be great.

Cheers

 

[HallsofIvy]

Air resistance does not act on the water that is "shielded" by the bucket but does act on the water trying to come out of the hole. I suspect that some small amount of water would get out of the hole but not much.

 

[Gonzolo]

If I understand correctly, I suspect the high upward airflow will create low pressure near the hole "pulling" the water out of the hole faster than if the bucket was just standing on earth, until a small vacuum is created above the water in the bucket due to it falling. This vacuum would slow down the rate, but should not stop it. At some point, there will be no water left.

I think the shape of the bucket could affect the results. A drop-shaped bucket for example would cause greater water flow. Aerodynamics could be relevant here.

 

[pervect]

If I understand correctly, I suspect the high upward airflow will create low pressure near the hole "pulling" the water out of the hole faster than if the bucket was just standing on earth, until a small vacuum is created above the water in the bucket due to it falling. This vacuum would slow down the rate, but should not stop it. At some point, there will be no water left.

I think the shape of the bucket could affect the results. A drop-shaped bucket for example would cause greater water flow. Aerodynamics could be relevant here.

In order for the bucket to be deaccelerated aerodynamicallyk, I think that the pressure in front of the bucket has to be higher than the pressure behind the bucket. So I don't think there will be a low pressure region in front of the bucket to pull water out, this is not consistent with the bucket slowing down.

I think the situation will be like having the bucket suspended vertically, with a hole in, and a large fan blowing upwards.

I'm not sure if the fan will keep all the water in the bucket, I have an intuitive feeling that the air-water interface will look rather chaotic, with the water breaking up into drops and slowly leaking out. But that's just a guess.

I don't think the low pressure region behind the bucket will be strong enough to drag water upwards out the rear of the bucket, either, BTW.

 

[broomfieldjay]

In order for the bucket to be deaccelerated aerodynamicallyk, I think that the pressure in front of the bucket has to be higher than the pressure behind the bucket. So I don't think there will be a low pressure region in front of the bucket to pull water out, this is not consistent with the bucket slowing down.

I agree, I think Bernoulli's equation supports this idea too after the bucket reaches terminal velocity.

The equation for wind resistance assuming resistance is proporional to velocity squared is:
$$F = .5 c_d \rho v^2 A$$
The only parameter that might be different for the bucket and the water is the drag coefficient $$c_d$$ . ρ is the density of air, A is the projected area.

Now wait a minute, that may be where the diffence is, the projected area. The water has a smaller projected area! I suspect that because of air resistance there may be a difference.

 

[Gonzolo]

In order for the bucket to be deaccelerated aerodynamicallyk, I think that the pressure in front of the bucket has to be higher than the pressure behind the bucket. So I don't think there will be a low pressure region in front of the bucket to pull water out, this is not consistent with the bucket slowing down.

I think the situation will be like having the bucket suspended vertically, with a hole in, and a large fan blowing upwards.

I'm not sure if the fan will keep all the water in the bucket, I have an intuitive feeling that the air-water interface will look rather chaotic, with the water breaking up into drops and slowly leaking out. But that's just a guess.

I don't think the low pressure region behind the bucket will be strong enough to drag water upwards out the rear of the bucket, either, BTW.

Don't forget that the hole is on the side of the bucket, not the bottom (check tyco05's drawing). A hole on the side surely means water shooting out because of lower pressure.

 

[broomfieldjay]

Woops! Yep, you're right, I was thinking the hole was on the bottom of the bucket. I do think the water would flow out when the bucket reaches terminal velocity because the weight of the water above the hole forces the water out of the bucket through the hole.

 

[Nenad]

I think you guyes are overthinking the problem with this pressure. I think the question is asking and focusing on friction. The Bucket, as it falls is being slowed down by friction, but the water inside is still being pulled by gravity, causing a leak out of the bucket. And the way it would happen is, at the beginning, the water would leak very slowly, but as the velocity of the bucket increases, so does the leak, until the velocity of the bucket reaches a maximum, (because of friction) causing the leak to be constant and at the same rate.

 

[broomfieldjay]

The Bucket, as it falls is being slowed down by friction, but the water inside is still being pulled by gravity, causing a leak out of the bucket. .

Gravity acts vertically, so it cannot directly cause the leak from the side of the bucket. Indirectly it does through the pressure that the water above the hole exerts on the water at the hole. Pressure = density of water x height of water above the hole x a. a = acceleration of the water = g-(air resistance drag force)/m Note that the air resistance drag force is "transmitted" to the water by the normal force from the bottom of the bucket on the water.

Pressure acts equally in all directions, and is opposed on the sides of the bucket by the bucket walls except at the hole. Thus the water leaks out.

Also, as air moves past the hole, it sucks water out. This is from Bernoulli's principle. Much like air moving across a chimney draws air up. (hot air rising also plays a part in chimney's)

 

[pervect]

Don't forget that the hole is on the side of the bucket, not the bottom (check tyco05's drawing). A hole on the side surely means water shooting out because of lower pressure.

Ooops - I missed this point

 

[tyco05]

Gravity acts vertically, so it cannot directly cause the leak from the side of the bucket. Indirectly it does through the pressure that the water above the hole exerts on the water at the hole. Pressure = density of water x height of water above the hole x a. a = acceleration of the water = g-(air resistance drag force)/m Note that the air resistance drag force is "transmitted" to the water by the normal force from the bottom of the bucket on the water.

Is this saying that at terminal velocity (ie g=drag force) then there will be no flow (acceleration of the water = 0 )?

 

[pervect]

Also, as air moves past the hole, it sucks water out. This is from Bernoulli's principle. Much like air moving across a chimney draws air up. (hot air rising also plays a part in chimney's)

Well, now that I understand the problem correctly - I'm still not so sure about this. After a bit of refreshing my meomry...

Bernoulli's eq says that the sum of the static pressure plus the dynamic pressure is constant along a streamline, where the dynamic pressure is
$$\frac{ \rho V^2}{2}$$

I'll assume the flow velocity is low enough for it's limitations to be met (static, incompressible flow-no turbulence).

If someone can demonstrate that the air is moving faster along a streamline near the bucket than it is along a streamline that's not near the bucket, I'll conceed the point - but I don't see it.

The following link, describing the action of a pitot tube, would seem to treat the case where the bucket does not "flare out". I've quoted a relevant section of text

http://www.grc.nasa.gov/WWW/K-12/airplane/pitot.html

Since the outside holes are perpendicular to the direction of travel, these tubes are pressurized by the local random component of the air velocity. The pressure in these tubes is the static pressure (ps) discussed in Bernoulli's equation.

I do agree that the pressure of the water above the hole will force water through the hole though.

 

[Zhou Yu]

Having the hole at the side of the bucket does indeed change things, and my general feeling is that the water would come out of the side of the bucket.

However my question is, if the hole was at the bottom of the bucket, wouldn't it be more likely that the air would go up through the hole instead of the water coming out?

 

[broomfieldjay]

If someone can demonstrate that the air is moving faster along a streamline near the bucket than it is along a streamline that's not near the bucket, I'll conceed the point - but I don't see it.

Inside the bucket, at the hole, there is no moving air, just water. I may have to look closer at the Bernoulli equation P2 - P1 = .5 (rho H20) v(h20)^2 0- .5(rho air)v(air)^2 more carefully though.

I've considered the terminal velocity a little more. The bucket is no longer in free fall, terminal velocity means that there is no net acceleration for the bucket, just a constant velocity. Therefore, as far as the pressure= water density* g*height of water in bucket, goes, we might as well consider the bucket at rest. We still must consider the Bernoulli effect as though the bucket were moving.

 

[geometer]

We have a bucket of water with a hole in the bottom (see attached picture for clarification). If it is dropped from a sufficient height as to reach terminal velocity (assume no rotation), at what point (if any) will the water begin to flow out of the bucket? (The hole is unplugged as it is dropped)

tyco - A very interesting question and I think we have gotten a little sidetracked here discussing the effect of the air flowing past the hole. In an attempt to answer your question, I conducted a little experiment. I took a standard, empty pop can, punched a hole about 3/4 of an inch up from the bottom of the can, filled the can with water while holding my finger over the hole, then dropped the can from a height of about 7 feet. No water flowed from the hole for the entire duration of the fall. I did only three trials, but the result was the same for all three trials. I'm not sure a drop of 7 feet was enough for the can to reach terminal velocity, but for at least this distance, there was no flow out the hole.

 

[pallidin]

I think tyco is wondering if their is a way to "shield" the effect of falling object velocity.
For some reason, he/she is under the assumption that water in a bucket with a hole behaves differently than if the bucket and water were dropped seperately; as if the water would flow downwards quicker that the bucket.
A general answer: One can not shield the effects of gravity, so the water inside the bucket, even with a really big hole, will move downward at the same rate as the bucket. This is excusing special aerodynamic effects, of course, which has nothing to do with the question as I understand it.

 

[pervect]

Inside the bucket, at the hole, there is no moving air, just water. I may have to look closer at the Bernoulli equation P2 - P1 = .5 (rho H20) v(h20)^2 0- .5(rho air)v(air)^2 more carefully though.

The following two website simulations are worth a look
http://www.ce.utexas.edu/prof/kinnas/319LAB/Applets/Venturi/venturi.html
http://www.efunda.com/formulae/fluids/venturi_flowmeter.cfm

The first one is graphically nicer, the second one has some of the theory. It appears to me from the first simulation, if you make the pipe flat, not only do both "taps" have the same pressure, this pressure goes up as the flowrate increases.

The second webpage has the math, I believe that the pressure is increasing because the flow is not inviscid.

Before I found these web pages, I took a straw and did an impromptu experiment, making a T junction with no flare, and immersing the lower end of the T in a glass of water. I satisfied myself that as I blew through the pipe, the pressure in the T increased, and as I blew harder it increased more, eventually causing air to bubble out through the water.

Another cute webpage I found was

http://www.grc.nasa.gov/WWW/K-12/airplane/wrong3.html

You can change the airfoil shape into a flat plate, then adjust the angle of the plate relative to the flow, and "measure" the pressure with a probe.

This can give some insight as to what the pressure looks like if the bucket walls "flare" and are not at right angles to the flow.

[added]
http://www.grc.nasa.gov/WWW/K-12/airplane/foil2.html
at the same site is also interesting.

 

[broomfieldjay]

The applets for the Bernoulli principle are neat, thanks for the links.

For some reason, he/she is under the assumption that water in a bucket with a hole behaves differently than if the bucket and water were dropped seperately; as if the water would flow downwards quicker that the bucket.
A general answer: One can not shield the effects of gravity, so the water inside the bucket, even with a really big hole, will move downward at the same rate as the bucket. This is excusing special aerodynamic effects, of course, which has nothing to do with the question as I understand it.

Ok, we can leave the Bernoulli principle out as a special aerodynamic effect.

I agree, if the bucket is falling with no air resistance, the water will not leak out of the bucket through the hole because they are both falling with acceleration g.

But the original question asks what happens when the bucket reaches terminal velocity. Under these circumstances, can we agree that water does flow out of the hole as if the bucket was at rest?

 

[BobG]

Okay, I finally have to ask the question I've been too embarrassed to ask.

Once the bucket reaches terminal velocity and is no longer accelerating, the water should try to come out the hole. I would assume the water that makes it out the hole is going to have a lower terminal velocity than the bucket filled with water.

That means it should flow up the side of the bucket?

And, since there is less pressure directly above the bucket (shielded from the wind), the water should move back above the bucket?

And, once above the bucket, where it's shielded from the wind, it will then begin accelerating at 9.8 m/sec^2 or at least will have a much higher terminal velocity than it did when it was on the side of the bucket.

End result would be for the water to fall back in the top of the bucket?