Is the concept of reactive centrifugal force valid?

In summary, the article is incorrect in its assertion that the reactive centrifugal force is always centrifugal. It depends on the context.
  • #71
harrylin said:
Einstein's view is not modern anymore. :biggrin: See the physics FAQ: "modern usage demotes the uniform "gravitational" field back to its old status as a pseudo-field"
http://www.desy.de/user/projects/Physics/Relativity/SR/TwinParadox/twin_gr.html

You'll have to explain how your interpretation of this link refutes the point of view that in general, boiled down, there are no special frames of reference, and what is "real" and what is "pseudo" depends on which frames of reference you are comparing and which one you prefer.
 
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  • #72
harrylin said:
See Wikipedia*; I asked you how you derive the path of your inertial object with the use of fictive forces.

*http://en.wikipedia.org/wiki/Fictitious_forces#Rotating_coordinate_systems
Exactly. So in the rotating frame we have:

[tex]m \mathbf{a}= \mathbf{F}_{\mathrm{fict}} +\mathbf{F}_{\mathrm{real}} [/tex]
where
[tex]\mathbf{F}_{\mathrm{fict}} =
- 2 m \boldsymbol\Omega \times \mathbf{v}_{B} - m \boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{x}_B ) - m \frac{d \boldsymbol\Omega }{dt} \times \mathbf{x}_B[/tex]

You have to use the fictitious forces in the rotating frame, otherwise you get the wrong trajectory. It doesn't matter if you do your dynamics in the inertial frame and then transform the result to the rotating frame or if you do your dynamics in the rotating frame directly. Either way the trajectory in the rotating frame has coordinate accelerations that are not attributable to the real forces and require fictitious forces to explain.
 
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  • #73
K^2 said:
Andrew, in this case, the opposing centripetal forces are each-other's reaction forces. Reaction force is just another name for a constraint force. You constrain the clothes to move along the curve. Hence the force on whatever provides the constraint. Nothing else to it.
Exactly. The reaction force is another centripetal force. It is a real force. It is not a centrifugal force.

AM
 
  • #74
Andrew Mason said:
The reaction force is another centripetal force.
If that's the case, then you simply don't use the term "reactive centrifugal force". But in other scenarios the reaction to centripetal force is centrifugal (away from rotation center), so some use the term "reactive centrifugal force".

Andrew Mason said:
It is a real force. It is not a centrifugal force.
"Real" and "centrifugal" are not mutually exclusive. You have been given many examples where the real reaction force to centripetal force is centrifugal (away from rotation center):

O-------->==> +

O : mass
--- : string
>==> : rocket
+ : rotation center of the whole arrangement

Here the rocket exerts a centripetal force on the mass, and the mass exerts a centrifugal reaction force on the rocket . The reactive centrifugal force acting on the rocket is a real force that appears in every frame.

Merry-go-round

Interaction forces that obey Newtons 3rd, and appear in every reference frame:
- centripetal force on the child, by the seat
- centrifugal force on the seat, by the child

Inertial force that doesn't obey Newtons 3rd, and appears only in the rotating reference frame:
- centrifugal force on the child
 
  • #75
Andrew Mason said:
My question was: am I missing something? You are persuading me that I wasn't.

If you put the washing machine in space and put it in spin cycle to spin the clothes, what happens? (let's assume there is electrical power somehow and let's assume that the machine is much heavier than the drum and clothes). The drum containing the clothes spins. But the rotating drum and clothes causes the washing machine to prescribe a circular motion about a point at or very close to the centre of the post on which the drum is centred (depending on how evenly the mass is distributed). In fact, both the drum and the rest of the machine are rotating (differently - at different angular speeds) about a common point. There is no centrifugal reaction force. There are only centripetal forces. The centripetal forces all sum to zero, since there is no net force on the machine as a whole. Where is the centrifugal force (I did not say centrifugal effect)?

Now keep increasing the mass of the washing machine until it has a mass of 6 x 10^24 kg. What, fundamentally, changes? If nothing changes fundamentally, then where does a "real" centrifugal reaction force arise?

AM
I thought that what you were missing are the definitions of "centrifugal" and "force". However, now it seems that what you miss is more fundamental, related to the meaning of Newton's third law and the way he applied it.
Amazingly, you seem to deny that the clothes push against the drum! Perhaps it's useful if you explain what you think that Newton meant with the citations that I gave you, for your convenience here they are again (Motte's translation):

"To every action there is always opposed an equal reaction; or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts."
[..]
"This is the centrifugal force, with which the body impels the circle; and to which the contrary force, wherewith the circle continually repels the body towards the centre, is equal."
Harald
 
  • #76
DaleSpam said:
Exactly. So in the rotating frame we have:

[tex]m \mathbf{a}= \mathbf{F}_{\mathrm{fict}} +\mathbf{F}_{\mathrm{real}} [/tex]
where
[tex]\mathbf{F}_{\mathrm{fict}} =
- 2 m \boldsymbol\Omega \times \mathbf{v}_{B} - m \boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{x}_B ) - m \frac{d \boldsymbol\Omega }{dt} \times \mathbf{x}_B[/tex]

You have to use the fictitious forces in the rotating frame, otherwise you get the wrong trajectory. It doesn't matter if you do your dynamics in the inertial frame and then transform the result to the rotating frame or if you do your dynamics in the rotating frame directly. Either way the trajectory in the rotating frame has coordinate accelerations that are not attributable to the real forces and require fictitious forces to explain.

The starting point for you is the result of my derivation from a coordinate transformation without the use of fictitious forces: it's
aB in the derivation on the page that I mentioned.

This coordinate acceleration is the result of (and attributed to) the rotation of the frame; it really doesn't make any sense to attribute them to fiction. And as we already have the equation of motion, there is also no reason to next introduce forces - let alone fictitious ones - for the determination of the trajectory. :wink:
The only thing to do is not forget that your frame rotates; else what you get is magic and fiction.Harald
 
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  • #77
JeffKoch said:
You'll have to explain how your interpretation of this link refutes the point of view that in general, boiled down, there are no special frames of reference, and what is "real" and what is "pseudo" depends on which frames of reference you are comparing and which one you prefer.

I didn't make the claims that you put in my mouth. In contrast, your claim concerning the use of non-inertial frames, that you "also look forward to his refutation of Einstein's modern view" sounds very much in disagreement with demoting Einstein's uniform "gravitational" field back to its old status as a pseudo-field.Harald
 
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  • #78
A.T. said:
If that's the case, then you simply don't use the term "reactive centrifugal force". But in other scenarios the reaction to centripetal force is centrifugal (away from rotation center), so some use the term "reactive centrifugal force".


"Real" and "centrifugal" are not mutually exclusive. You have been given many examples where the real reaction force to centripetal force is centrifugal (away from rotation center):
The centrifugal effect in those examples is the result of a body's inertia. It is not a real force since it produces no acceleration.

The problem here may be with the concept of a "fixed" centre of rotation. The only fixed centre of rotation is a centre of mass. The reaction force to centripetal acceleration of mass on one side of that centre of mass is the centripetal force on the mass on the opposite side. That has to be the case in every rotation. There is no "real" centrifugal force or "reactive centrifugal force". A real centrifugal force would provide an acceleration away from the centre of mass of rotation. There is no acceleration away from the centre of mass.

AM
 
  • #79
Andrew Mason said:
It is not a real force since it produces no acceleration.
Acceleration of an object depends on the sum of all forces, not just the reactive centrifugal force, acting here on the rocket / seat. The forces on the rocket(by the mass) and on the seat(by the child) are real interaction forces, as defined in Newtons 3rd law. The net acceleration of rocket / seat is completely irrelevant for this consideration.

Andrew Mason said:
A real centrifugal force would provide an acceleration away from the center of mass of rotation.
Only if it was the only force acting, which is not the case in the scenarios.
 
  • #80
Andrew Mason said:
The problem here may be with the concept of a "fixed" centre of rotation.
You could have a inertial (non-rotating) frame of reference with it's origin at center of rotation of a rotating frame of reference, assuming the center of the rotating frame does not have a linear component of acceleration.

Andrew Mason said:
There is no acceleration away from the centre of mass.
From an inertial frame of reference, the center of mass of a closed system can't accelerate, for example a rocket and it's expelled mass (spent fuel). If the rocket were in space free of external forces, and using it's engine to move in a circular path, then the rocket would experience centripetal force, and the exhaust, centrifugal force. In a frame of reference centered at the center of the circular path and rotating at the same speed as the rocket, the rocket would appear to be at rest, and the exhaust (and center of mass of rocket and exhaust) would appear to be moving "outwards".
 
  • #81
Andrew Mason said:
The centrifugal effect in those examples is the result of a body's inertia. It is not a real force since it produces no acceleration.

Only if you aren't in the rotating frame. If you are sitting on the merry-go-round, there is indeed a force, you will accelerate if the force is not balanced e.g. by you clinging to the pole that's impaling your wooden horsey, and if you do move then the force will have done work. If it looks like a duck, walks like a duck, and quacks like a duck...
 
  • #82
JeffKoch said:
Only if you aren't in the rotating frame.
No, we are not talking about the inertial centrifugal force in the in the rotating frame. We are talking about the reactive centrifugal force that acts on a different object and appears in every frame.

See scenarios here:
https://www.physicsforums.com/showpost.php?p=3470239&postcount=74
 
  • #83
harrylin said:
The starting point for you is the result of my derivation from a coordinate transformation without the use of fictitious forces: it's
aB in the derivation on the page that I mentioned.
Obviously. Nobody is saying that there are fictitious forces in the inertial frame.

harrylin said:
This coordinate acceleration is the result of (and attributed to) the rotation of the frame; it really doesn't make any sense to attribute them to fiction.
I don't know what you are talking about. They are named "fictitious forces" they aren't attributed to fiction. I don't even know how you could attribute a force to a body of literature. Perhaps you just don't like the term "fictitious forces" and would prefer the term "inertial forces" or "pseudo forces". My preference is "inertial forces".

harrylin said:
And as we already have the equation of motion, there is also no reason to next introduce forces - let alone fictitious ones - for the determination of the trajectory.
The point is that the equation of motion is not f=ma, there are additional terms, and those additional terms are the fictitious forces. It doesn't matter if you start with the equations of motion in the non-inertial frame, or if you transform from the inertial frame into the non-inertial frame. Once you get to the non-inertial frame there are fictitious forces. You cannot remove those fictitious forces and correctly determine the trajectory.
 
  • #84
rcgldr said:
You could have a inertial (non-rotating) frame of reference with it's origin at center of rotation of a rotating frame of reference, assuming the center of the rotating frame does not have a linear component of acceleration.
Provided the centre of rotation was also the centre of mass of the rotating system. For example in the earth/moon system, the inertial frame of reference (ignoring the orbit around the sun and gravitational effects of other planets) would be the centre of mass of the earth/moon system which is between the centre of the Earth and centre of the moon.

From an inertial frame of reference, the center of mass of a closed system can't accelerate, for example a rocket and it's expelled mass (spent fuel). If the rocket were in space free of external forces, and using it's engine to move in a circular path, then the rocket would experience centripetal force, and the exhaust, centrifugal force. In a frame of reference centered at the center of the circular path and rotating at the same speed as the rocket, the rocket would appear to be at rest, and the exhaust (and center of mass of rocket and exhaust) would appear to be moving "outwards".
How would you get the rocket to prescribe a circular path without using tangental rocket thrust?

AM
 
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  • #85
Andrew Mason said:
How would you get the rocket to prescribe a circular path without using tangental rocket thrust?
Circular motion at constant speed requires only a centripetal acceleration which acts radially. No tangential thrust is needed once the rocket has the right linear and angular velocity.
 
  • #86
A.T. said:
Circular motion at constant speed requires only a centripetal acceleration which acts radially. No tangential thrust is needed once the rocket has the right linear and angular velocity.
That is not a problem if the rotation is about the centre of mass. How can the rocket rotate about a point other than its centre of mass and still have angular momentum being conserved here? You have to take into account the rocket gases flying out the rear of the rocket.

AM
 
  • #87
JeffKoch said:
Only if you aren't in the rotating frame. If you are sitting on the merry-go-round, there is indeed a force, you will accelerate if the force is not balanced e.g. by you clinging to the pole that's impaling your wooden horsey, and if you do move then the force will have done work. If it looks like a duck, walks like a duck, and quacks like a duck...
I am not trying to suggest that the centrifugual effect is imagined. It is quite apparent. It is just that it is not a real force.

But my point is that the reaction force to the centripetal force that is applied to the merry-go-round rider is not a centrigufal reaction force. Rather it is always an equal and opposite centripetal force toward the centre of mass of the rotating system. If the centre of rotation of the merry-go-round is its centre of mass, then there is no force on the Earth and the reaction force to the centripetal force on the rider is the centripetal force keeping mass on the opposite side of the merry-go-round rotating.

AM
 
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  • #88
Andrew Mason said:
How can the rocket rotate about a point other than its centre of mass and still have angular momentum being conserved here? You have to take into account the rocket gases flying out the rear of the rocket.
Yes, if you take into account the rocket gases the total momentum is conserved, while the rocket rotates around some point, other than the rockets COM. And the force applied to the gases by the rocket is called reactive centrifugal force, because it points away from that center of rotation.
 
  • #89
Andrew Mason said:
But my point is that the reaction force to the centripetal force that is applied to the merry-go-round rider is not a centrigufal reaction force. Rather it is always an equal and opposite centripetal force toward the centre of mass of the rotating system.
Would you not agree that the merry-go-round must exert a centripetal force on the rider? If so, then the rider exerts and equal and opposite (in this case, outward) force on the merry-go-round. Simple as that. (Call that force what you will.)
 
  • #90
Andrew Mason said:
But my point is that the reaction force to the centripetal force that is applied to the merry-go-round rider is not a centrigufal reaction force.
The force on the seat, by the rider is centrifugal (= away from the center of rotation).
Andrew Mason said:
Rather it is always an equal and opposite centripetal force toward the centre of mass of the rotating system.
The specifiers centripetal & centrifugal refer to the center of rotation, not to the center of mass of some closed system for which momentum is conserved.

But if you are so obsessed with the center of mass, just put a second rider of equal weight on the opposite side. Both riders will exert reactive centrifugal forces on their seats. These forces point away from :
- the center of rotation
- the center of mass
You can't get more centrifugal than that.
 
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  • #91
Doc Al said:
Would you not agree that the merry-go-round must exert a centripetal force on the rider? If so, then the rider exerts and equal and opposite (in this case, outward) force on the merry-go-round. Simple as that. (Call that force what you will.)
All the molecules in the merry-go-round and rider exert forces on each other. But the net force on each particle is a centripetal force. Those centripetal forces sum to zero if the mass is distributed symmetrically in the merry-go-round (ie. the centre of rotation is the centre of mass). (They do not sum to zero if the centre of rotation is not the centre of mass, in which case there is a net force on the Earth which has to be included.) But from Newton's third law, we know that all action/reaction pairs must sum to zero. Therefore, the centripetal forces on all the molecules in the merry-go-round must include all action and reaction pairs. So the reaction force to a centripetal force must be another centripetal force. There can be no centrifugal reaction force. If you add in a "real" centrifugal force they do not sum to zero.

The confusion arises if the merry-go-round is not balanced so that the (apparent) centripetal forces about the centre of the merry-go-round do not sum to zero. In order to make these apparent centripetal forces sum to 0 one has to postulate a centrifugal force on the centre of rotation. But that is a mistake because the centre of the merry-go-round is not an inertial frame in that case. There is a force on the Earth that accelerates the earth, and the merry-go-round centre, towards the centre of mass of the earth/merry-go-round system. In order to see this, one has to think of the Earth as being much less massive than it is.

AM
 
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  • #92
Andrew Mason said:
All the molecules in the merry-go-round and rider exert forces on each other.
Well, the ones in contact do, but OK.
But the net force on each particle is a centripetal force.
True.
Those centripetal forces sum to zero if the mass is distributed symmetrically in the merry-go-round (ie. the centre of rotation is the centre of mass). (They do not sum to zero if the centre of rotation is not the centre of mass, in which case there is a net force on the Earth which has to be included.) The forces include all action and reaction pairs.
Sure, the net force is zero. So what? Third law pairs are actual, individual forces, not 'net forces'.
So the reaction force to a centripetal force must be another centripetal force.
Nonsense.
There can be no centrifugal reaction force.
:confused:

Answer this question: Does the section of the merry-go-round in contact with the rider exert a force on the rider?

And this: What direction is that force?

And this: Do you agree that Newton's 3rd law applies?
 
  • #93
Doc Al said:
Andrew Mason said:
But my point is that the reaction force to the centripetal force that is applied to the merry-go-round rider is not a centrigufal reaction force.
Would you not agree that the merry-go-round must exert a centripetal force on the rider? If so, then the rider exerts and equal and opposite (in this case, outward) force on the merry-go-round. Simple as that. (Call that force what you will.)
I agree with Andrew here: The equal but opposite reaction to a real centripetal force is a real centripetal force. It is directed inward, not outward. You have to look at it all squinty-eyed and such to see the reaction force as outward.

A different example: Consider a binary star system; call the two stars A and B. The gravitational force exerted on Star A by star B is directed toward star B and thus toward the system center of mass. This force makes star A's trajectory curve inward toward the center of mass. It is a centripetal force. The gravitational force exerted on Star B by star A is an equal but opposite force and causes star B's trajectory curve inward toward the center of mass. It is also a centripetal force.

The same goes for your merry-go-round. The force on the merry-go-round when viewed from the perspective of an inertial frame is centripetal, not centrifugal.

I submit that there is such a thing as a reactive centrifugal force. The equal but opposite reaction to a real centrifugal force is a real centrifugal force. For example, the interactions between two positively charged objects are equal but opposite centrifugal forces.

This however is not what that nonsense wikipedia article is talking about.
 
  • #94
D H said:
The same goes for your merry-go-round. The force on the merry-go-round when viewed from the perspective of an inertial frame is centripetal, not centrifugal.
Obviously the net force on some segment of the merry-go-round must be centripetal. But the reaction to the centripetal force that the merry-go-round exerts on the rider must be an outward 'centrifugal' force on the merry-go-round.

No squinting involved!
 
  • #95
Doc Al said:
Obviously the net force on some segment of the merry-go-round must be centripetal. But the reaction to the centripetal force that the merry-go-round exerts on the rider must be an outward 'centrifugal' force on the merry-go-round.

No squinting involved!
Lots of squinting involved!

Rhetorical question: What is it that makes the merry-go-round+rider situation different from the binary star situation? The answer is that the merry-go-round is attached to the Earth. That attachment is just a distraction. So, let's put the merry-go-round out in space. The result is an astronaut in a rotating space station. The astronaut feels something very much like a gravitational force due to the rotation of the space station.

What's happening to the rotating space station? The answer is that it too is undergoing a uniform circular motion. The center of mass of the rotating space station is orbiting about the space station + astronaut center of mass. The curvature is inward, not outward. The space station is therefore subject to a centripetal force, just as is the astronaut.
 
  • #96
D H said:
What's happening to the rotating space station? The answer is that it too is undergoing a uniform circular motion. The center of mass of the rotating space station is orbiting about the space station + astronaut center of mass. The curvature is inward, not outward. The space station is therefore subject to a centripetal force, just as is the astronaut.
Again you confuse the net force on an element of the space station with the force on it due to the astronaut.
 
  • #97
Doc Al said:
Again you confuse the net force on an element of the space station with the force on it due to the astronaut.
What net force? Assume the space station is in deep space; the only force acting on it is the force exerted on it by the astronaut.
 
  • #98
Andrew Mason said:
But the net force on each particle is a centripetal force.
We are not interested in the net force on the merry-go-round-particle. We are interested in the force that rider-particles exert on merry-go-round-particles. That force is called "reactive centrifugal force".

The fact that the net force on the merry-go-round-particles is centripetal does not change the fact that the force from rider-particles on merry-go-round-particles is centrifugal.

Andrew Mason said:
But that is a mistake because the centre of the merry-go-round is not an inertial frame in that case. There is a force on the Earth that accelerates the earth,

Nonsense. I told you already how to balance the merry-go-round:

Just put a second rider of equal mass on the opposite side. Both riders will exert reactive centrifugal forces on their seats. These forces point away from :
- the center of rotation
- the center of mass

You can put the whole thing into space. Considering the interaction with the Earth is not needed here, unless you want to continue to confuse yourself.
 
  • #99
D H said:
What net force? Assume the space station is in deep space; the only force acting on it is the force exerted on it by the astronaut.
Forget the astronaut for a moment. You have a gigantic space station spinning around its center of mass. Do you not agree that an element of the space station has forces acting on it due to the surrounding material?
 
  • #100
This thread has taught me one important thing..., not to be embarrassed so much, when I'm not able to understand some simple physics 101 principles.
I wonder if a thought example might help me get a clear picture of what seems to be the point of such confusion.

If on one of the spokes of the merry-go-round, anywhere toward the outer edge, a vertical stand post (rigid) that has a pivot point for a bar or tube which is perfectly balanced and free to rotate (like a prop) assume no wind interference. If a strain gauge was in position to measure any pressure top or bottom, inside and outside, what measurement would show and where ?
If some out of balance be needed, where would it need to be placed to show any thing being discussed ?

Ron

P.S. Just noticed this is post 666:eek::rolleyes:
 
  • #101
D H said:
The center of mass of the rotating space station is orbiting about the space station + astronaut center of mass. The curvature is inward, not outward. The space station is therefore subject to a centripetal force, just as is the astronaut.
Your first two sentences are correct, but the third sentence does not follow. The first two sentences correctly establish the direction of the force, but to determine whether or not a given force is centripetal or centrifugal requires not only specification of the direction of the force, but also it's point of application.

Assuming that the center of rotation is the origin and that a force is pointing in the positive x direction then the force is centrifugal if applied on the positive x axis, centripetal if applied on the negative x axis, and tangential if applied on the y axis.

In this scenario the force is applied at the location of the astronaut and winds up being centrifugal, despite the fact that it is causing uniform circular motion of the center of mass. I.e. It is not applied at the center of mass so the motion of the center of mass is not sufficient.
 
  • #102
D H said:
I submit that there is such a thing as a reactive centrifugal force. The equal but opposite reaction to a real centrifugal force is a real centrifugal force. For example, the interactions between two positively charged objects are equal but opposite centrifugal forces.
I agree. That is an example of a real centrifugal force - and it does not require, or have anything really to do with, rotation.
This however is not what that nonsense wikipedia article is talking about.
That was my take - complete nonsense. For students of classical mechanics, better to stick to a good textbook. But even good textbooks can be misleading or confusing. Even Newton appears to be misleading (incorrect) in this aspect of rotational motion. The animated discussion on this thread highlights the challenge and difficulty in analysing the physics of rotation.

AM
 
  • #103
Doc Al said:
Sure, the net force is zero. So what? Third law pairs are actual, individual forces, not 'net forces'.
The fact is that there is acceleration always occurring. These are not balanced static forces. The third law pairs are forces that are each causing acceleration.

Answer this question: Does the section of the merry-go-round in contact with the rider exert a force on the rider?

And this: What direction is that force?
The merry-go-round exerts a force on the rider and it is toward the centre of rotation.

And this: Do you agree that Newton's 3rd law applies?
Of course.

AM
 
  • #104
Andrew Mason said:
The third law pairs are forces that are each causing acceleration.
Wrong. There is nothing in Newtons 3rd Law about the acceleration of the interacting objects. When I lean against the wall, there are two equal opposite forces acting on me and the wall respectively. But neither me nor the wall is accelerating.


That is Newtons 3rd Law:
Andrew Mason said:
The merry-go-round exerts a force on the rider and it is toward the centre of rotation, ...
...while the rider exerts a force on the merry-go-round and it is away from the centre of rotation.
 
  • #105
Doc Al said:
Forget the astronaut for a moment.
Or make it 2 astronauts, on opposite sides. Now we have no confusion about center of mass vs. center of rotation and momentum conservation

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