# For a ball falling through a fluid, does the ball do work on the fluid

• Urmi Roy
In summary: The ball does work on the fluid. This is shown by the fact that the level of the fluid doesn't change no matter how much the ball falls. This is because the fluid is being displaced and the ball is applying a force over a distance to do that.3. The energy lost in the interaction is heat.
Urmi Roy
So when a ball falls through a fluid in which it is submerged, it keeps displacing fluid as it descends, but the level of the fluid doesn't rise. So does the ball do work on the fluid?

(Assuming there is no viscosity, the K.E would keep increasing, while potential energy decreases...that's the only energy conversion there's supposed to be...) So even if there is viscosity, the ball reaches a terminal velocity and at that stage, the PE gets converted into heat energy wholly...But in a question we did in class, the teacher said we should also include work done by the ball on the fluid being displaced...I don't get it!

There is a boundary layer of fluid that is carried along with the ball. At the interface between ball and fluid, the fluid is moving along with the ball. There is a force by the ball on the fluid at the interface. The fluid is moving at the interface. So there is work being done by the ball on the fluid at the interface.

The work being done by the ball on the fluid shows up as turbulent energy in the fluid. This eventually damps out and becomes heat energy.

The above view focuses on the boundary between ball and fluid and the fact that at that boundary the ball and the fluid are moving at the same velocity. Because they are moving at the same velocity the work done by the fluid on the ball is equal and opposite to the work done by the ball on the fluid.

There is another viewpoint that can be taken...

If you ignore the details of the interface and pretend that the ball and fluid are non-rotating, rigid objects in relative motion then you could conclude that the fluid is motionless and that, accordingly, no work is being done. [If you have trouble with the notion of a "rigid fluid", think of a fireman sliding down a pole instead].

Taking this viewpoint, the work done by the ball on the fluid and the work done by the fluid on the ball are no longer equal and opposite. If you add the work done by the ball on the fluid to the work done by the fluid on the ball you get a total that is negative. This is an energy deficit in the interaction. But energy is conserved. The "lost" energy in this sort of interaction usually ends up as heat.

Yes, the ball does work on the fluid, because when the ball is at the bottom of the fluid, the center of mass of the fluid is higher than it was when the ball was at the top of the fluid.

This is true even though the level of the fluid doesn't rise, because the fluid changes shape (at first, it had a spherical hole at the top, and by the end, this hole has moved to the bottom).

Urmi Roy said:
So when a ball falls through a fluid in which it is submerged, it keeps displacing fluid as it descends, but the level of the fluid doesn't rise. So does the ball do work on the fluid?

(Assuming there is no viscosity, the K.E would keep increasing, while potential energy decreases...that's the only energy conversion there's supposed to be...) So even if there is viscosity, the ball reaches a terminal velocity and at that stage, the PE gets converted into heat energy wholly...But in a question we did in class, the teacher said we should also include work done by the ball on the fluid being displaced...I don't get it!

Remember that work is force x distance. As the ball drops the fluid is being moved out of the way. So the ball is applying a force over a distance, thus doing work on the fluid.

Your car actually does the same thing to the air. The minimum amount of work your car must do is to push the air to the sides so it can get through. However, some the of air can't get out of the way fast enough and gets "bunched up" and is accelerated before it gets pushed out of the way, which requires more work. This is why cars that are more aerodynamic get better gas mileage. They push the air to the sides and out of the way without letting as much bunch up and get accelerated, thus requiring less work by your engine to keep speed.

There are several things here that need to be kept separate.

1. Assuming the fluid is incompressible (the OP said "submerged" so I assume it was talking about a ball under water), the ball can't move anywhere unless the fluid also moves around it. Because of that, the "effective mass" of the ball in the "F=ma" formula to find its acceleration is greater than its actual mass. If you ignore the viscosity of the fluid, the flow pattern around the ball will remain the same for different velocities and the fluid velocity at any point will be proportional to the ball velocity, so the "effective mass" is a constant.

The above effect reduces the acceleration of the ball, but it doesn't limit the velocity.

2. Almost all real fluids have viscosity, so the fluid flow around the ball converts some of the kinetic energy transferred to the fluid by (1) into heat. The viscosity also changes the flow pattern around the ball. The result is a "drag force" on the ball which is approximately proportional to the square of the the velocity. The drag force depends only on the shape of the ball, not on its mass. This is what limits the ball's velocity to its "terminal velocity", when the drag force is equal to other forces (e.g. gravity) acting on it.

3. The fact that "the fluid gains gravitational potential energy as the ball loses it" is the same as a statement about the buoyancy force acting on the ball (which still exists, though it is not big enough to make the ball float), or the fact that the hydrostatic pressure in the fluid varies with depth, which amounts the same thing.

All of the above applies to compressible fluids as well, but it's a bit harder to visualize exactly what is going on.

Note that the effect in (1) is usually small, and often ignored, unless the fluid flow is restricted in some way - for example if a long cylinder is moving down a tube with only a small clearance for the fluid to flow past it. In that situation, the effect of the fluid viscosity is also increased, but there are practical devices where you need to take (1) into account to make an accurate model of the dynamics of the system.

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I think AlepZero's statement (below) stresses on a point I've been most perplexed about:
AlephZero said:
3. The fact that "the fluid gains gravitational potential energy as the ball loses it" is the same as a statement about the buoyancy force acting on the ball (which still exists, though it is not big enough to make the ball float), or the fact that the hydrostatic pressure in the fluid varies with depth, which amounts the same thing.

So if you were to write the first law for the system comprising the ball and the fluid, how would you include the change in potential energy of the fluid?

jbriggs444 said:
If you ignore the details of the interface and pretend that the ball and fluid are non-rotating, rigid objects in relative motion then you could conclude that the fluid is motionless and that, accordingly, no work is being done. [If you have trouble with the notion of a "rigid fluid", think of a fireman sliding down a pole instead].

That's how I've been thinking about it so far and this view implies that if we set aside the fact that there is a layer of fluid stuck to the the ball (boundary layer), the ball doesn't do any other form of work (in displacing the fluid continuously)...is that right?

jbriggs444 said:
Taking this viewpoint, the work done by the ball on the fluid and the work done by the fluid on the ball are no longer equal and opposite. If you add the work done by the ball on the fluid to the work done by the fluid on the ball you get a total that is negative. This is an energy deficit in the interaction. But energy is conserved. The "lost" energy in this sort of interaction usually ends up as heat.

The ball, in this case isn't supposed to be doing any work on the fluid, right?

So are AlephZero and jbriggs444 saying 2 different things?

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Urmi Roy said:
So are AlephZero and jbriggs444 saying 2 different things?

To the extent that AlephZero and I are saying contradictory things, you would be wise to listen to AlephZero and not to me.

I was ignoring, for instance, buoyancy. The fluid is pushed away from the falling ball at its bottom and closes back in on its top. The outward motion at the bottom takes place under higher pressure than the inward motion at the top. Thus work is performed on the fluid. If you ignore the interface details and treat the fluid as a motionless rigid body, you are ignoring this fact.

If you hand-wave viscosity away, decide that the ball leaves no turbulence in its wake and disregard the kinetic energy of the water being displaced near to the ball (which displacement can indeed be made arbitrarily small by changing the shape of the ball) it seems to me that all that is left is buoyancy. Ignore that and you're ignoring the heart of the problem.

jbriggs444 said:
To the extent that AlephZero and I are saying contradictory things, you would be wise to listen to AlephZero and not to me.

I was ignoring, for instance, buoyancy. The fluid is pushed away from the falling ball at its bottom and closes back in on its top. The outward motion at the bottom takes place under higher pressure than the inward motion at the top. Thus work is performed on the fluid. If you ignore the interface details and treat the fluid as a motionless rigid body, you are ignoring this fact.

If you hand-wave viscosity away, decide that the ball leaves no turbulence in its wake and disregard the kinetic energy of the water being displaced near to the ball (which displacement can indeed be made arbitrarily small by changing the shape of the ball) it seems to me that all that is left is buoyancy. Ignore that and you're ignoring the heart of the problem.

One remaining question in regard to this: how would we incorporate the work done on the fluid (just due to the effect of buoyancy and ignoring all other effects) in the formulation of the first law of thermodynamics for the system comprising the ball and the fluid?

Let M=mass of ball
m=mass of displaced fluid
C=heat capacity of ball
Then we have M*g*(h2-h1)+(1/2)*M*(v2^2-v1^2)+ M*C*(T2-T1)+m*g*(h2-h1)=0 (no heat/shaft work no the system)

But the mass of fluid being displaced, though constant in magnitude, is at different parts of the fluid everytime!

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jbriggs444 said:
To the extent that AlephZero and I are saying contradictory things, you would be wise to listen to AlephZero and not to me.

I was ignoring, for instance, buoyancy.

FWIW I was going to criticize the "fireman and pole" analogy, because it magics away the point of the OP's question (as I saw it). The sliding fireman is causing the air to move, even though the forces that generates will probably be negligible compared with other forces acting on him.

But then I decided to write a longer post instead!

But the mass of fluid being displaced, though constant in magnitude, is at different parts of the fluid everytime!

Why is that a problem? Think about all of the fluid. The flluid always contains a hole, of the same volume as the ball. It's easier to think of the fluid without the hole, plus another ball-shaped piece of fluid with density ##-\rho##.

When the ball moves, the "negative density fluid" moves with it. It doesn't matter exactly how the real fluid moves everywhere to make that happen.

There are two different ways of thinking about fluid flow. One way is to look at a fixed region of space, and see what happens to the fluid that is inside the region, and what is flowing into and out of it. The other way is to look at a fixed piece of fluid, and see where it moves. In this case, the first way is hard (as you have found by trying to think about it) because you have to keep track of what happens to the whole of the fluid, and that depends on a lot of irrelevant things like the shape of the container, etc. But for this problem, the second way is easy.

AlephZero said:
...The flluid always contains a hole, of the same volume as the ball. It's easier to think of the fluid without the hole, plus another ball-shaped piece of fluid with density ##-\rho##.

When the ball moves, the "negative density fluid" moves with it. It doesn't matter exactly how the real fluid moves everywhere to make that happen.

So after the ball has entered the fluid, it keeps displacing the same mass of fluid during its descent, thereby doing work.

Now, coming to the application of this, If I'm asked to write energy conservation for the system, how do I include the work being done on the fluid by the ball? (I've elaborated on my thinking in regard to this in my previous post.)

as the ball moves lower, the water that is displaced moves towards the previous position, there can't be much energy being displaced because the potential energy the ball has doesn't change very much,all the heat would then start to be dissipated throughout the water

Ok so in answer to my own question as mentioned in post #10 (as to how to write the first law of thermodynamics for this problem), I figure that if the ball starts off outside the body of the fluid, we need to account for the work done by the ball in lifting the center of gravity of the fluid and the del(h) term (in expression for the change in potential energy) would be the shift in height of the center of gravity . However, if the ball starts off submerged in the fluid (as per requirement of the problem), then since the height of the center of gravity (of the fluid) doesn't shift over time, we also don't account for it in the first law.

The center of mass of the fluid is raised as the ball descends, so the fluid's potential energy increases while the balls potential energy decreases. I doubt this accounts for all of the energy change in real life, but in the idealized case where no heat is generated and no velocity imparted to the fluid, then it seems that total potential energy of ball and fluid should remain constant.

We're dealing with the idealized case here. But as the ball descends, the center of gravity of the ball does not "continuously" change places. It changed right in the beginning when the ball landed in it.After that, the center of gravity of the bulk of fluid doesn't change...

As the ball descends, the center of mass of the ball also descends, and it displaces the surrounding fluid upwards, and the fluid's center of mass is raised. Initially the spherical void in the fluid filled by the ball is at the top, and eventually, the spherical void in the fluid filled by the ball is at the bottom. The void in the fluid moved downwards, so the center of mass of the fluid moved upwards.

1 person
rcgldr, Hmm I see what you mean. So I'm back to my previous question...given that someone asks you to solve a problem in which there is a ball that is falling through a fluid and you need to write the first law of thermo for it, how would you do so?
(please refer to post #10 where I've written out the first law, so it'd be great if you could tell me how to modify it).

I'm not sure about the thermo aspects of this. In real life, vortices are going to be created in the fluid as a wake that trails the ball flows downwards while the surrouding fluid travels upwards. Also when the ball impacts the bottom of the container, some of that energy will be transferred to the earth, assuming the container is resting on the earth, and some of that energy will be converted into heat by the collision.

1 person

## What is the definition of work in relation to a ball falling through a fluid?

Work is the exertion of force over a distance, resulting in the transfer of energy. In the case of a ball falling through a fluid, work is done when the ball's motion results in a change in the fluid's energy.

## How does a ball falling through a fluid result in work?

As the ball moves through the fluid, it experiences resistance from the fluid molecules. This results in a transfer of energy from the ball to the fluid, causing the fluid to move or change in some way. This transfer of energy is considered work.

## Does the amount of work done depend on the speed of the ball?

Yes, the amount of work done is directly proportional to the speed of the ball. This means that the faster the ball falls, the more work it does on the fluid.

## Is the work done by the ball on the fluid the same as the work done by the fluid on the ball?

No, the work done by the ball on the fluid and the work done by the fluid on the ball are not the same. The work done by the ball results in a change in the fluid's energy, while the work done by the fluid results in a change in the ball's energy.

## How does the density of the fluid affect the work done by the ball?

The density of the fluid affects the amount of resistance the ball experiences as it falls through the fluid. A higher density will result in a greater amount of work being done by the ball on the fluid.

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