Bubble in a water column

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On Earth. Theoretical study. With a perfect liquid (no viscosity). A bubble in a column of liquid is moving up, the column is on a table, I have some questions:

1/ the force on the table increase/decrease when bubble is moving up ?
2/ the center of gravity of {column+liquid} is always the same when the bubble moving up ?
 
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  • #2
tiny-tim
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Hi Gh778! :smile:
2/ the center of gravity of {column+liquid} is always the same when the bubble moving up ?

How can that be? :confused: When the bubble is at the bottom, the centre of gravity is at the centre of the liquid. When the bubble is at the top, the centre of gravity is at the centre of the liquid. But the centre of the liquid is lower.
1/ the force on the table increase/decrease when bubble is moving up ?

What do you think? :smile:
 
  • #3
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Hi tiny-tim,

2/ great, I'm agree and understand that point :)

Even there is another strange thing: look at images please: if center of gravity change, I can do this cycle, step by step the column move up ? Or is it like second image ?

So it is for that I ask 1/

1/ I call "weight" the force from pressure and quantity of mouvement.

I explain with V=100 m/s for all the system. If quantity of mouvement give force, this force Fm give work while time dt, so energy giving is Fm*V*dt, it's bigger than at 10 m/s.

For me, the weight is constant, the center of gravity is constant, but quantity of mouvement give a force, this force give a lot of work if velocity is not 0. I would like to understand where I'm wrong.
 

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  • #4
tiny-tim
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Hi Gh778! :smile:
Even there is another strange thing: look at images please: if center of gravity change, I can do this cycle, step by step the column move up ? Or is it like second image ?

Sorry, I don't understand your diagrams at all. :redface:

The bubble starts at the bottom (as in step 2), and finishes at the top (as in step 1) … what are all the other steps for? :confused:
1/ I call "weight" the force from pressure and quantity of mouvement.

I explain with V=100 m/s for all the system. If quantity of mouvement give force, this force Fm give work while time dt, so energy giving is Fm*V*dt, it's bigger than at 10 m/s.

For me, the weight is constant, the center of gravity is constant, but quantity of mouvement give a force, this force give a lot of work if velocity is not 0. I would like to understand where I'm wrong.

(btw, it's "movement" :wink:

and, in britain, it's also "centre"!)


the centre of gravity is not constant, it is moving down

look at the forces on the bubble … which way are the forces on it? where do they come from?

so what are the forces on the liquid? where do they come from?

and so what are the forces on the table?
 
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Sorry, I wrote bubble move up, but is it possible to study when bubble move down (with a spring on the column for example) ?

I'm starting at step 1 and after it is step2, after it is step 3, etc. I would like to know how column is at each step. Vacuum object is at top first and spring move down the vacuum. At step 2 I "forget" vacuum (I can: it's vacuum) and I restart with vacuum at top (I can: it's vacuum). Etc.

thanks for help in my poor english...

I have a general question: when bubble move up or move down (with a spring), quantity of movement give force on the table ?

Image: P=weight, dp=additionnal weight due to the vacuum in liquid, -dp from static spring, F1=force when spring pull (dynamic), this give F1' to the bottom, and spring -F1 to, the total weight is always P.
 

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  • #6
tiny-tim
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Hi Gh778! :smile:

(just got up :zzz:)
I have a general question: when bubble move up or move down (with a spring), quantity of movement give force on the table ?

the bubble can only move up if there is a net force acting on it

that net force has to come from somewhere

except for its own weight, it has to come from the liquid

there has to be an equal and opposite force on the liquid from the bubble

the liquid is also moving, so you can calculate F = ma for the liquid

alternatively, you can treat the liquid-and-bubble as a single body, and since that single body is moving (its centre of mass is moving down), you can do F = ma for the liquid-and-bubble

the problem is the same as for a person standing on a weighing scale …
if he bends his knees (so he moves down), what happens to the reading on the scale while he is moving? :wink:
 
  • #7
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The reading on the scale increase. And it's true: a standard problem with masses it's the same than with bubble and liquid. It's easier to understand :approve:

When a mass move down at constant velocity 100 m/s (image) with its spring. A fixed system can control speed at 100 m/s (red). If spring move up a little mass, the spring give a force to the fixed system, and this force works on a great distance due to the velocity (100 m/s). So, if I recover this energy, where I lost energy ? The weight of mass is reduce of the spring force so the total is the same weight ?
 

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  • #8
tiny-tim
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When a mass move down at constant velocity 100 m/s (image) with its spring. A fixed system can control speed at 100 m/s (red).

if the speed is constant, then there is no extra force (because there is no acceleration)

if the speed of the bubble was constant, again it would make no difference to the force

but in fact (if we ignore viscosity) the bubble does accelerate
If spring move up a little mass, the spring give a force to the fixed system, and this force works on a great distance due to the velocity (100 m/s). So, if I recover this energy, where I lost energy ? The weight of mass is reduce of the spring force so the total is the same weight ?

when you say (i think) that the potential energy lost by the mass is the same as the potential energy gained by the spring, you are assuming that the system is in equilibrium, and that the kinetic energy is zero (or constant)

but in fact the mass will accelerate (or decelerate), so the total potential energy of the mass and of the spring won't be the same :smile:
 
  • #9
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Could you help me on this point:

1/ If I take a system of 2 masses, even the system move down (free fall) due to gravity, the centre never moved if I apply forces from one mass to another ?
2/ If I take a column of water with a bubble (free fall), in this case centre can move ?

If it's possible I consider the velocity fixed if forces are extremely quick.
 
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  • #10
tiny-tim
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1/ If I take a system of 2 masses, even the system move down (free fall) due to gravity, the centre never moved if I apply forces from one mass to another ?

sorry, i don't understand :redface:

if they're in free fall, ie with no external forces other than gravity,

then internal forces make no difference …​

the movement of the centre of gravity (centre of mass) is only governed by the external forces

2/ If I take a column of water with a bubble (free fall), in this case centre can move ?

yes, if the bubble is moving up or down, then the centre of mass of the water-and-bubble will move …

isn't this obvious? :wink:
I consider the velocity fixed if forces are extremely quick.

i don't understand, but that looks very wrong :frown:
 
  • #11
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yes, if the bubble is moving up or down, then the centre of mass of the water-and-bubble will move …

isn't this obvious? :wink:


No, it's not obvious for me...in one part it's not possible to change the centre of mass without external force but the centre of mass of the water-and-bubble will move if the bubble moving up or down (in the same time free fall)
 
  • #12
tiny-tim
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No, it's not obvious for me...in one part it's not possible to change the centre of mass without external force …

the external force is gravity

no internal force (eg the reaction force between the water and the bubble) can alter the movement of the centre of mass
 
  • #13
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great !!! ok like that it's obvious

Another question, a bubble move up (in the same time the system is free falling) in a column, the centre of the system move and I lost/win energy because potential energy is mgh ? Even the bubble is in water, the weight is always the same I think.
 
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  • #14
tiny-tim
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a bubble move up (in the same time the system is free falling) in a column, the centre of the system move and I lost/win energy because potential energy is mgh ? Even the bubble is in water, the weight is always the same I think.

m of the bubble and m of the water are different,

so the mgh for the bubble increases, but the mgh for the water decreases by a different amount (when they exchange places) …

so the centre of gravity moves, and the total mgh changes, as expected :smile:
 
  • #15
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but mass of bubble is 0 (in theory) for me, no ?
 
  • #16
tiny-tim
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but mass of bubble is 0 (in theory) for me, no ?

even if it was, the mgh of the bubble would be 0, and so it certainly wouldn't balance the mgh lost by the water :smile:
 
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and where goes the mgh lost of the water ?
 
  • #18
tiny-tim
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the increased kinetic energy of the accelerating bubble and of the accelerating water :smile:
 
  • #19
Baluncore
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So, what happens to the pressure at the bottom of the column as a bubble moves from the bottom to the top?
The bubble will increase in size as it rises, (due to reduced hydrostatic pressure), which will progressively increase the liquid level in the column until the bubble reaches the top at which point there will be a step change downwards to below the original liquid level.
 
  • #20
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the increased kinetic energy of the accelerating bubble and of the accelerating water
so if I repeat cycle, more and more with a new "bubble" at bottom at each time, the system liquid-bubble accelerate more than gravity give ?
 
  • #21
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if the speed is constant, then there is no extra force (because there is no acceleration)

That's not strictly correct because the bubble may change its volume due to the change of pressure with depth
 

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