Ball Fountain Physics: Understand Equilibrium & Water Force

[krickon]

Hey! I have been looking over all these threads and have learned a lot. I was curious if anyone here could help me understand the physics of the ball fountain. I was thinking about the forces of the equilibrium with the weight of the granite ball and the water shooting up. I believe their is a thin film layer of water, maybe someone can help me better understand that. Thanks!

 

[marlon]

Hey! I have been looking over all these threads and have learned a lot. I was curious if anyone here could help me understand the physics of the ball fountain. I was thinking about the forces of the equilibrium with the weight of the granite ball and the water shooting up. I believe their is a thin film layer of water, maybe someone can help me better understand that. Thanks!

With all do respect, there is a reason why your question has not been answered yet :

1) specify your problem : what are you studying EXACTLY ?
2) What are your input data ? What data have been given ?
3) What are your strategies toward solving the problem ?
4) Which formula's are you going to use to do so ?
5) Show us some attempts and/or show us CLEARLY where you are having problems

good luck

greets marlon

 

[FredGarvin]

It operates on the same premise as a hydrodynamic or fluid film bearing. The rotation of the granite ball creates a very thin layer of higher pressure under the rotating stone.

http://www.efunda.com/DesignStandards/bearings/bearings_fluid_film.cfm

 

[krickon]

With all do respect, there is a reason why your question has not been answered yet :

1) specify your problem : what are you studying EXACTLY ?
2) What are your input data ? What data have been given ?
3) What are your strategies toward solving the problem ?
4) Which formula's are you going to use to do so ?
5) Show us some attempts and/or show us CLEARLY where you are having problems

good luck

greets marlon

I am looking for any physics concepts that are involved with the weight of the ball being supported by the pressure of water. I decided to make one for myself. Here is all the information that I have for you.

Weight of ball- 11.5 lbs
Diameter of ball-8.270"
Ring is perfectly is shaped to ball
Water pump, pumps 1,521,360 ml per hour.
- I apoligize for the lack of information. I just would like to know the physics concepts that go along with the ball fountain.

Thanks again! Posts are greatly appreciated!

 

[cesiumfrog]

Ring is perfectly is shaped to ball

That seems quite unquantitative, after you decided to tell us the ball diameter to the thousandth of your unit. You ought to read the other ball fountain threads on this forum, for your explanation of the physics.

 

[berkeman]

I haven't read much of the other threads, but it seems like one of the key items is to be sure that the pump can generate enough pressure to lift the ball. To figure that out, use the diameter of the ring and the weight of the ball to calculate the water pressure needed. Having a flow rate for the pump is not enough to know what pressure it can sustain.

 

[FredGarvin]

Does anyone have a picture of what we are talking about here?

 

[berkeman]

Does anyone have a picture of what we are talking about here?

Here you go. Pretty neat the first time you see it and go give it a roll:

http://new.photos.yahoo.com/qara0/album/576460762329653158/photo/294928803416373096/18

 

[krickon]

I have created a ball fountain, it works. I just would like to know what physics are behind it. What physics concepts are acting on this project. (equilibrium, gravity etc.)

 

[berkeman]

I have created a ball fountain, it works. I just would like to know what physics are behind it. What physics concepts are acting on this project. (equilibrium, gravity etc.)

I think in the basic analysis, it is just the water pressure pushing on the underside surface of the ball that balances the weight of the ball (which comes from gravity pulling the ball down). One subtlety is that the pressure acts on each little part of the surface in a normal direction (each little tiny force vector points toward the center of the ball). So if you understand how to integrate vectors along a surface, you see that all the horizontal components cancel out, and you are left just with the vertical component balancing the weight. If you haven't learned how to integrate vectors over a surface yet, don't worry about it. That's why I said in a previous post to use the diameter of the ring in your area calculation of the required pressure. Then:

F = weight = mass of ball * acceleration of gravity = m * 9.8m/s^2

F = water pressure * area of ring

Set the two equal to determine the water pressure required to balance the ball. A little excess pressure lifts it enough for a little water to spill out of the seal around the ring, and presto. Disneyland!

 

[berkeman]

Oh, and be careful to check your units in those calculations. m/s^2 is not directly compatible with PSI, for example. You need to do the appropriate unit conversions.