Centrifugal forces + pressure and CG

In summary, the conversation discusses the effects of rotation on a torus with solid and liquid parts of equal density. The rotation causes forces from pressure and changes the center of mass without external forces. The question of whether the solid or liquid gives pressure is brought up, as well as the idea of using ropes to limit the pressure. The expert suggests starting with simpler versions of the problem, such as a rotating pipe with a viscous fluid, to gain understanding before tackling the more complex torus problem. The expert also explains that the material properties do not affect the stress distribution, and that boundary conditions are key to determining the stress distribution.
  • #1
Gh778
421
0
Take half torus with solid part of density 1, other half torus with liquid of density 1. The torus turn at W rd/s enough for maintain liquid at external circle. Now, for 0 to 180° we accelerate torus, this increase forces (blue in drawing) from pressure of water and from 180° to 360° we deccelerate torus. Rotationnal speed change from W to W+w to W. It's like torus change the CG without external forces. I thought solid give pressure too but I can imagine 2 liquids with different density, it's the same. I have some questions:

1/ water don't give pressure so no force exist ?
2/ solid give same pressure ?
3/ when we rotate the forces from rotation, this cancel pressure forces ?

Edit: I add another drawing for show different positions
 

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  • #2
Maybe my questions are not precise enough ?
When I drawn solid in black, it's possible to imagine it maintain from center of circle with solid like this no pressure can appear from solid (imagine a lot of smal ropes for maintain all part of solid). So if solid don't give pressure, it is liquid that don't give pressure ?
 
  • #3
Does this attachment help?

There is certainly a tangential ring pressure developed in a rotating ring. In a solid it is a stress and in a liquid it is pressure. Of course these are equal in your ring because the densities are equal. For unequal densities you would have to relocate the centre of mass.
 

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  • #4
I don't understand exactly what you are doing here, but one difference between the solid and liquid parts is that there must be a container that holds the liquid, and the container will apply a force to the liquid. If you ignore that fact, you will get the wrong forces and stresses.
 
  • #5
Does this attachment help?
I'll try to understand this evening. That I can't understand: if I imagine solid half torus like small parts, if each part is attached by a rope, the pressure could be limited a lot I think ?

and the container will apply a force to the liquid
ok, but this force (the force the container apply) will put on support that take system ?
 
  • #6
This is actually a very complicated problem. Based on lots of experience in fluid and solid mechanics modelling, I can tell you that the best way to approach a problem like this is to start with some much simpler versions of the problem, and solve them first, and gradually add complexity until you get to the actual problem you are trying to solve. Why? First of all, if you can't solve the simpler versions of the problem, you certainly won't be able to solve the more complex problem. Secondly, you will gain understanding along the way from the solutions to the simpler problems.

In the system you describe, neither the solid nor the liquid will behave as a rigid body. Both will experience transient non-homogeneous deformations. The liquid will exhibit viscous flow that has to be accounted for, and the solid will exhibit elastic deformation that has to be accounted for. I suggest starting out with the following very simple problem:

Consider an infinitely long straight pipe containing a viscous fluid. Initially the pipe and the fluid are at rest. At time zero, the pipe is set in motion with a steady velocity V along its axis. What is the velocity and pressure distribution within the fluid as a function of time and radial position? If you can't solve this problem, you will never be able to do the torus problem.

As a second version of the problem, consider the same pipe and fluid, but this time make the imposed pipe velocity a function of time.

As a third problem, consider the case where there are rigid plugs within the pipe, and a finite length of fluid between the plugs. The plugs move with the pipe velocity.
 
  • #7
This is actually a very complicated problem.
With W = constant, it's a complicated problem too ? or it's because W change in time ?

With W=constant, are there forces from liquid ? I think forces from solid can't be near 0 if there is a lot of parts (imagine infinite number of parts and ropes). Even if there are forces from liquid is strange because even on a round forces cancel themselves if we place this system on another object in a part of time (60° for example), this change the CG of object I think.

For change velocity, we can change the rotation of the system or place a lot of stems for help solid and liquid to gain velocity very quickly. In theory, if velocity is apply on each atom of liquid and solid, forces are centripetal forces only.
 
  • #8
Chestermiller said:
In the system you describe, neither the solid nor the liquid will behave as a rigid body. Both will experience transient non-homogeneous deformations. The liquid will exhibit viscous flow that has to be accounted for, and the solid will exhibit elastic deformation that has to be accounted for. I suggest starting out with the following very simple problem:

Compresibiitiy or elasticity doesn't have any effect on the stress field, unless you want to model this as a large displacement nonlinear problem. The body forces that cause the centripetal acceleration are known from the geometry and density, and the body forces are also the gradient of the stress tensor. The stresses are independent of the material properties, for a given set of the boundary conditions.

The boundary conditions are the key to the actual stress distrubition. For a rotating solid thin ring or cylinder with no radial constraints imposed, the main stress compoent is tension around the circumference of the ring. But a fluid can't sustain any significant tensile stress (or negative hydrostatic pressure if you prefer to call it that), so there must be some radial constraint forces (for example the fluid pressure on the solid structure that is containing it) to make the situation physically realistic.
 
  • #9
AlephZero said:
Compresibiitiy or elasticity doesn't have any effect on the stress field, unless you want to model this as a large displacement nonlinear problem. The body forces that cause the centripetal acceleration are known from the geometry and density, and the body forces are also the gradient of the stress tensor. The stresses are independent of the material properties, for a given set of the boundary conditions.

The boundary conditions are the key to the actual stress distrubition. For a rotating solid thin ring or cylinder with no radial constraints imposed, the main stress compoent is tension around the circumference of the ring. But a fluid can't sustain any significant tensile stress (or negative hydrostatic pressure if you prefer to call it that), so there must be some radial constraint forces (for example the fluid pressure on the solid structure that is containing it) to make the situation physically realistic.

If the angular velocity is constant, then, as you indicate, then liquid and the solid can be treated as rigid, and the system will be statically determinate. However, if the angular velocity varies with time, then you have to consider the deformational mechanics of both the solid and the liquid.
 
  • #10
Gh778 said:
With W = constant, it's a complicated problem too ? or it's because W change in time ?

With W=constant, are there forces from liquid ? I think forces from solid can't be near 0 if there is a lot of parts (imagine infinite number of parts and ropes). Even if there are forces from liquid is strange because even on a round forces cancel themselves if we place this system on another object in a part of time (60° for example), this change the CG of object I think.

For change velocity, we can change the rotation of the system or place a lot of stems for help solid and liquid to gain velocity very quickly. In theory, if velocity is apply on each atom of liquid and solid, forces are centripetal forces only.

If W is constant, then, as AlephZero points out, the analysis is much simpler than if W varies with time. But this is not the problem that you posed.
 
  • #11
Could any of these forces apply to the Earth's core?
 

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  • #12
1/ I would like to know if there are forces from pressure of liquid when W = constant ? AlephZero said that container would give the same force but that force come from the support of the system, right ? So the system oscillate around a point, the CG move ?

2/ If W varies with time, it's easy to change the system for accelerate/deccelerate not the support of the system but a lot of parts of solid and liquid for reduce problem from deformation I think.

@Studiot:
There is certainly a tangential ring pressure developed in a rotating ring.
Where come from this tangential force ? Could you explain, I don't understand ?

@Blind light:
Could any of these forces apply to the Earth's core?
Do you know an effect not explain on Earth and forces can explain ?
 
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  • #13
...but that force come from the support of the system, right?

I don't think AZ mentioned the system support, however you did in post#5.

AZ did say

I don't understand exactly what you are doing here

and I concur exactly.

So let us take things one step at a time.

Do you understand the idea that any rotating object will be subject to internal forces that can be resolved in two directions at right angles - radially and tangentially?
 
  • #14
Do you understand the idea that any rotating object will be subject to internal forces that can be resolved in two directions at right angles - radially and tangentially?
No, I don't understand that. I see only centripetal forces. Could you explain ?

and I concur exactly.
you don't understant the system or the goal of the system ? If it's the goal, it's only for study it. I know the CG can't move but I don't see that.
 
  • #15
Which direction do you think angular momentum acts?
 
  • #16
(I want to understand first when W = constant) it's tangential but I don't see how forces come from on your drawing at message #3
 
  • #18
Maybe for me it's easier to understand with small solid deformable material. I imagine a part of sector like drawing show. The material is solid but deformable. The deformation put pressure at right and at left, where is the C force in your message #3 ? The deformation is more and more high at external circle.
 

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  • #19
Your last sketch is not far off, so we are getting somewhere. I have drawn something similar in fig4 of the new attachment.

Starting with the static situation of a ring or section of a cylinder under internal pressure, Fig1 shows the internal pressure tending to burst apart the ring. This would break it across some diameter, say AA.

The bursting pressures and forces act radially.

Fig2 shows how this bursting is resisted. There must be a forces F-F acting across the section AA, holding the ring together.
These forces exactly equal the bursting forces and are equivalent to the forces F in my earlier diagram in post#3

Fig 3 shows how the forces F-F must be the same all the way round the ring to preserve equilibrium.

The resisting forces are therefore tangential or circumferential.
They are called hoop stresses or forces, after the iron hoops used to hold wooden barrels together.

Now in Fig4 we proceed to a rotating situation. Here the internal pressure is replaced by the a stress due to the increasing radius in the inner and outer faces of a small segment. This is like your last sketch.
The circumferential stresses are considered constant and equal to the hoop stresses described before.

(Note I have moved from forces to stresses the force is obviously stress time cross sectional area.)

Now a rotating disk is not in equilibrium.

However we may use equilibrium analysis if we introduce the fictitious centrifugal force to balance the real centripetal one. This is equal and opposite to the centripetal force and the technique is known as D'Alembert's Principle.

Alternatively we may analyse the motion by Newton's laws of motion using the centripetal acceleration.

I have done this in the previous attachment and called the centrifugal force C.
This equals mass times distance to the centre of mass of the rotating half ring times the square of the angular velocity.

Using D'Alembert allows me to then use the laws of equilibrium to equate this to the two hoop forces acting across the cut line forming the diameter of the half ring.
 

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  • #20
Ok, like that I understand :)

If I change liquid and take gas, gas can pressure liquid at 2 surfaces of half torus but gas can't pressure itself because temperature prevent this, how does it works in this case ?
 
  • #21
gas can't pressure itself because temperature prevent this, how does it works in this case ?

Swirling gasses are still subject to rotational forces, but please keep to the step by step approach as I am having trouble understanding where you are heading.

I envisage your original question to apply to something like say a petri dish, containing a liquid and with some plasticine or bluetack sausage pressed round the inside of half the rim. If you centred the dish on a turntable and rotated it this would form the half solid half liquid ring you talk about and the forces would be as we have discussed.

In principle the same would apply to a gas, however the speeds required to move the gas aswy form the centre and collect towards the rim would be enormous. You should always expect a significant density distribution radially.
 
  • #22
Yes, I speak always with my dish with solid/liquid or solid/gas.

And now, if speed varies, the problem come from the speed of sound in liquid or gas ?
 
  • #23
the problem come from the speed of sound in liquid or gas ?

What problem? You didn't specify.

Of course, with a gas, you may be moving away from a petri dish and Aleph Zero's confining structures may be forces (magnetic, electric, gravitational).

You haven't mentioned the scale of your thoughts either. Spiral galaxies are formed by processes like these.
 
  • #24
The scale is 1 m or 10 meters only, a human mechanical system.

The problem I don't understand is when the system accelerate, the delay for transmit energy is not zero, it depends of the sound in material. I think it's that Chestermiller would like to say in its messages. See drawing
 

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  • #25
The problem I don't understand is when the system accelerate, the delay for transmit energy is not zero, it depends of the sound in material

Sorry I don't understand.

The system is always accelerating if it is rotating, even if the angular velocity ω is constant. If ω is constant then the acceleration is constant.

What energy are you trying to transmit and what does this have to do with the rotation?
 
  • #26
I increase W, the gas form a shape, if I take the same shape for solid (see last drawing), I see a force that move cog, what compensate this ?
 
  • #27
I see a force that move cog,

I have no idea what you mean by this.
 
  • #28
1/ I come back to the torus when W is constant. Full torus with liquid. Now if we increase the radius of the system, we need kinetics energy for additional liquid sure and increase radius (add water). But the C force is like an additionnal force that increase energy of the system. If the radius increase, we can put water in the center, we need energy for rotate water, ok, but water move outside circle more easily due to the C force.
If I compare the potential energy in a rotational solid and a fluid it's not the same ?

2/ Centrifugal forces apply a pressure like gravity can do. The pressure is more at external with C, it's easier to move an empty object at external circle ? See drawing.
 

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  • #29
With the shape I drawing, solid and liquid has the same density, I don't know how forces from pressure (F+C) can cancel themselves ?
 

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  • #30
What makes you think the forces cancel?

It would help if you drew the correct forces.

When analysing a mechanical problem you need to distinguish between applied or external forces (also called loads) and internal forces which appear as stresses.

The response of a liquid is different from the response of a solid to external force.
However the external applied force will be the same in both from the same source.

Please not that Chestermiller has already said this is a very complicated problem (post#6).
In my post 19 I posted a partial differential equation that takes three pages of maths to solve, even after simplifyung to an ordinary differential equation. Are you up for this?
 
  • #31
Like density of liquid and solid are the same, I think centrepetal forces are the same for liquid and solid, so all centripetal forces cancel themselves and I don't show them on drawings.

Now, forces that I drawn on my last drawings are only forces from pressure of liquid, C and F forces like we spoke before. Like W is constant I don't see others forces. In my second drawing if I put forces for cancel all torques, the support of the system must see a force to the left and up. I think it's logical to say sum of torques is zero. For me, the square part of solid adding reduce C forces at external circle.

It's a difficult problem but interresting too, if someone can help me to pose equations I can be resolve it. The difficulty for me it's to understand what's happen in reality.
 
  • #32
If you are using centripetal analysis you must use accelerations and Newtons laws.

If you want to do force analysis you must use D'Alembert's method.

I have noted this before.

I have also noted that I do not understand your diagrams or yours arrows in particular.
 
  • #33
I drawn with more details on forces like I think there are. I think it's not necessary to use Newton or Alembert method because my error is enormous when I see sum of forces. I think I forget something in the system.
In the drawing, red forces are forces from centripetal effect, like solid and liquid have same density, sum of these forces are zero.
Green forces are only pressure of liquid but without the centripetal forces (red color for that). We have F and C forces like we see before. For me torques from Fa, Fb and Fc must be to 0 because this system can't give energy from nothing. So I need anothers forces for compensate sum of Fa+Fb+Fc (in vector) but sum of c1+c2+c3+c4+c5 don't do that, these forces are at left too. For me c4 and c5 are lower because there is less water in a radius length due to the presence of solid, solid which give only centripetal forces not pressure. What another forces are in the system ?
 

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  • #34
So with the shape in the drawing, in a small part of rotation F1>F2, this give a torque ? F2 move more than F1 so how torque can cancel ?
 

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1. What is a centrifugal force?

A centrifugal force is a fictitious force that appears to act on objects moving in a circular path. It is caused by the inertia of the object and its tendency to resist changes in motion.

2. How does centrifugal force affect pressure?

Centrifugal force can cause a decrease in pressure in the center of a rotating object, as the force pushes the particles outward and decreases their density. This effect is known as the centrifugal pressure gradient.

3. What is the relationship between centrifugal force and the center of gravity (CG)?

Centrifugal force and CG are directly related, as the direction of centrifugal force always points away from the center of gravity. Additionally, the magnitude of the centrifugal force is dependent on the distance between the object's center of gravity and the axis of rotation.

4. How does centrifugal force affect objects of different masses?

Centrifugal force affects objects of different masses in the same way, as it is dependent on the object's inertia and not its mass. However, objects with larger masses may require more force to maintain a circular path at a given speed.

5. Can centrifugal force and pressure be used for practical applications?

Yes, centrifugal force and pressure have various practical applications, such as in centrifugal pumps and centrifugal separators. They are also used in centrifugal force fields for particle separation and in centrifugal casting for creating hollow objects.

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