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What exactly is centrifugal force

by hms.tech
Tags: centrifugal, force
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Andrew Mason
#19
Jan29-13, 05:48 AM
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Quote Quote by rcgldr View Post
An "inertial" frame of reference means the frame of reference is not accelerating (no linear acceleration and no rotation). Newton's laws of physics apply in an inertial frame without any modification.

If the frame was accelerating, such as a rocket in space, then to an observer inside the rocket, there would be an apparent force on any object within the spacecraft, similar to gravity on earth.

The wiki article on this goes into more detail:

http://en.wikipedia.org/wiki/Inertia...e_of_reference



http://www.answers.com/topic/centrifugal-force

In your diagram of the car in a turn, there is a Newton third law pair of forces, an inwards centripetal force exerted by the road onto the car, and an equal in magnitude but opposing outwards reaction force that the car applies to the road. That outwards force is real, the issue for some people is using the term "reactive centrifugal force" to describe that outwards force that the car exerts to the road.
The centripetal force produces centripetal acceleration. The force that causes centripetal acceleration produces mechanical tensions (ie chains of electromagnetic forces at the atomic level). These mechanical tensions operate radially. The net sum of these tensions is the centripetal force that produces the centripetal acceleration.

Somebody had the idea that we should call the reaction to a centripetal force that is the result of the application mechanical forces a "centrifugal reaction force". As far as I can tell, this is a concept that serves no purpose and simply causes confusion. It can produce no acceleration so it is hardly a centre-fleeing force as the term "centrifugal" would imply. It is a mechanical tension so it does not exist where the centripetal force is applied by a force operating at a distance, such as gravity or an electric field.

Calling it a centripetal reaction force also confuses the physics. The true reaction to a centripetal force is another centripetal force. One does not appreciate this in the situation of a tethered ball rotating about a pole fixed to the earth but that is only because we do not take into account the movement of the earth. In the case of two balls tethered to each other by a rope and rotating in space about their centre of mass, the rope has a tension. Each ball pulls on the rope and, through the rope, pull on each other ie toward the centre. The ball pulls on the rope but produces no acceleration of the rope radially outward. Rather this pull is applied through the rope, via tensions, to accelerate the balls and rope toward the centre.

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A.T.
#20
Jan29-13, 06:20 AM
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Quote Quote by Andrew Mason View Post
"centrifugal reaction force". As far as I can tell, this is a concept that serves no purpose
Because you are considering only the most trivial cases, like:
Quote Quote by Andrew Mason View Post
two balls tethered to each other by a rope and rotating in space about their centre of mass,
In the other thread we gave you plenty examples where a reactive centrifugal force exists.

Quote Quote by Andrew Mason View Post
It can produce no acceleration so it is hardly a centre-fleeing force as the term "centrifugal" would imply.
The term "centrifugal" has nothing to do with producing acceleration. It simply means that the force points away from the center of rotation. Acceleration is a matter of net force, not just one centrifugal force.

Quote Quote by Andrew Mason View Post
The true reaction to a centripetal force is another centripetal force.
Sometimes it is, and sometimes it is not.
stevendaryl
#21
Jan29-13, 06:53 AM
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Quote Quote by A.T. View Post
Because you are considering only the most trivial cases, like:

In the other thread we gave you plenty examples where a reactive centrifugal force exists.
If a string pulls on an object, the object pulls back on the string. That's always the case, by Newton's third law. In the special case where the force of the string on the object is radially inward, the force of the object on the string is radially outward.
A.T.
#22
Jan29-13, 07:05 AM
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Quote Quote by stevendaryl View Post
If a string pulls on an object, the object pulls back on the string. That's always the case, by Newton's third law. In the special case where the force of the string on the object is radially inward, the force of the object on the string is radially outward.
Yes, for local interactions the reaction to a centripetal force is always a centrifugal force. It's only when we invoke action at a distance (like Newton's gravity or a mass-less string that is not considered an object itself, just a means of force transmission) that we have a Newtons 3rd force pair of two centripetal forces.
stevendaryl
#23
Jan29-13, 07:07 AM
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To me, "fictitious forces" are very ugly, because they spoil Newton's laws of motion. In a rotating coordinate system, there are "centrifugal forces" acting on absolutely every object in the universe. Where's the equal and opposite forces? There are none.

So some people would take this to mean "Newton's laws only apply in an inertial frame". I don't like that conclusion. If you view them as vector equations, then they apply in all circumstances, not just inertial frames.

If [itex]\stackrel{\rightarrow}{U}[/itex] is the velocity vector, you write
[itex]m \frac{\stackrel{\rightarrow}{dU}}{dt} = \stackrel{\rightarrow}{F}[/itex]

this equation is always valid, regardless of what coordinate system you are using, and the force on an object always has an equal and opposite force. What changes when you go to a noninertial or curvilinear coordinate system is not Newton's laws, but the idea that you can compute the time derivative of a vector by only taking the derivative of its components. That is, it is not the case that

[itex](\frac{\stackrel{\rightarrow}{dU}}{dt})^j = \frac{
dU^j}{dt}[/itex]
A.T.
#24
Jan29-13, 07:19 AM
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Quote Quote by stevendaryl View Post
To me, "fictitious forces" are very ugly, because they spoil Newton's laws of motion.
They extend the applicability of Newton's 1nd & 2nd to non-inertial frames. Which is a very useful thing, when you are actually using physics to compute something, not just muse about the beauty of laws.
Quote Quote by stevendaryl View Post
Where's the equal and opposite forces? There are none.
Newton's 3rd doesn't apply to inertial forces, because momentum is not conserved in non-inertial frames. So we cannot extend Newton's 3rd to them.
stevendaryl
#25
Jan29-13, 07:34 AM
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Quote Quote by A.T. View Post
They extend the applicability of Newton's 1nd & 2nd to non-inertial frames.
No, they don't. As vector equations, Newton's laws hold in any coordinate system or frame. It's always the case that in the absence of external forces, the velocity vector [itex]\stackrel{\rightarrow}{U}[/itex] satisfies

[itex]\frac{\stackrel{\rightarrow}{dU}}{dt} = 0[/itex]

It's always the case that in presence of a (real, non-fictitious) force [itex]\stackrel{\rightarrow}{F}[/itex], the velocity satisfies

[itex]m \frac{\stackrel{\rightarrow}{dU}}{dt} = \stackrel{\rightarrow}{F}[/itex]

It's always the case that if one object exerts a (real, non-fictitious) force [itex]\stackrel{\rightarrow}{F}[/itex] on another, then the second exerts a force [itex]-\stackrel{\rightarrow}{F}[/itex] on the first.

Newton's laws, as vector equations are not changed at all by using noninertial or curvilinear coordinates. Fictitious forces are not needed to extend Newton's laws to noninertial frames. What's needed is a more sophisticated notion of what

[itex]\frac{\stackrel{\rightarrow}{dU}}{dt}[/itex]

means when the basis vectors themselves are nonconstant.
stevendaryl
#26
Jan29-13, 07:36 AM
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Quote Quote by A.T. View Post
Newton's 3rd doesn't apply to inertial forces, because momentum is not conserved in non-inertial frames. So we cannot extend Newton's 3rd to them.
That's not true. Momentum as a vector quantity is conserved in non-inertial frames. What isn't conserved are the components of momentum.
stevendaryl
#27
Jan29-13, 07:55 AM
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Quote Quote by stevendaryl View Post
That's not true. Momentum as a vector quantity is conserved in non-inertial frames. What isn't conserved are the components of momentum.
Actually, I should clarify this. There are two different departures from good old inertial Cartesian coordinates: (1) Using curvilinear coordinates and (2) using noninertial coordinates.

With curvilinear coordinates, the basis vectors become functions of position. With noninertial coordinates, the basis vectors become functions of time. Curvilinear coordinates can be handled with the usual Newton's laws of motion, viewed as vector equations. Noninertial coordinates can also be handled with the usual Newton's laws of motion, but only if you adopt a 4-D spacetime view.
A.T.
#28
Jan29-13, 08:01 AM
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Quote Quote by stevendaryl View Post
Fictitious forces are not needed to extend Newton's laws to noninertial frames. What's needed is a more sophisticated notion ... basis vectors themselves are nonconstant... adopt a 4-D spacetime view...
Yeah, you can always replace a simple approach with a mathematically equivalent, but more complicated one.
Studiot
#29
Jan29-13, 08:04 AM
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hms.tech

My advice is to stick with one the two methods I outlined for you, either will always get you the right answer, D'Alembert's often leads to rather less computational effort.

I'm sure you can see that a by product of introducing reference frames is to cause experts to squabble amongst themselves as a result of a great increase in complexity.
Andrew Mason
#30
Jan29-13, 08:28 AM
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Quote Quote by stevendaryl View Post
If a string pulls on an object, the object pulls back on the string. That's always the case, by Newton's third law. In the special case where the force of the string on the object is radially inward, the force of the object on the string is radially outward.
That is quite true, but you are only looking at part of the picture. If there was only the object and the string you could not have a centripetal force acting on the object. There are necessarily other forces on the string.

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sophiecentaur
#31
Jan29-13, 08:34 AM
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Words like "fictitious force" and "particle" should have a warning sign on them when they're used for public consumption. If you can feel the effect then it's not really fictitious so a different word should really be used to describe it. Likewise, calling a photon a particle is just asking for the sort of misinterpretations we read every day. A term like quasi-force or quasi-particle would at least ring warning bells for people who might want to rush off and think in terms of conventional meanings.
A.T.
#32
Jan29-13, 08:41 AM
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Quote Quote by Andrew Mason View Post
That is quite true, but you are only looking at part of the picture.
"Part of the picture" is fully sufficient for Newtons 3rd Law and local interactions. You don't have to consider all forces acting on a object or the net acceleration of the object, to tell that an individual force acting on it is centrifugal (points away from the rotation center) and forms a 3rd Law pair with a centripetal force acting on a different object.
Andrew Mason
#33
Jan29-13, 08:46 AM
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Quote Quote by A.T. View Post
Because you are considering only the most trivial cases, like:
So where is the concept of a centrifugal reaction force used? It cannot ever produce a centrifugal acceleration. So all it does is explain the tension.

In the other thread we gave you plenty examples where a reactive centrifugal force exists.
My quibble is not with the force per se but with the name "centrifugal" in conjunction with the term "force". A force is something that is capable of producing an acceleration of an object. The acceleration that this reaction force produces is always centripetal ie. opposite to the direction of the force.

The term "centrifugal" has nothing to do with producing acceleration. It simply means that the force points away from the center of rotation. Acceleration is a matter of net force, not just one centrifugal force.
From Wikipedia:
Centrifugal force (from Latin centrum, meaning "center", and fugere, meaning "to flee") is the apparent outward force that draws a rotating body away from the center of rotation.

There is nothing about the "centrifugal reaction force" that causes anything to flee from the centre. Nothing. The only thing that causes a rotating object and rotating rope to "flee" the centre is inertia, not a force.

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sophiecentaur
#34
Jan29-13, 08:48 AM
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Quote Quote by A.T. View Post
"Part of the picture" is fully sufficient for Newtons 3rd Law and local interactions. You don't have to consider all forces acting on a object or the net acceleration of the object, to tell that an individual force acting on it is centrifugal (points away from the rotation center) and forms a 3rd Law pair with a centripetal force acting on a different object.
Using the term 'Centrifugal Force' when I was at school was verboten. It was regarded as worse than talking about sex in front of your parents.
A.T.
#35
Jan29-13, 09:27 AM
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Quote Quote by Andrew Mason View Post
So where is the concept of a centrifugal reaction force used?
It's not a concept. The concept here is Newtons 3rd Law: A pair of equal but opposite force acting at the interface of two objects: one inwards (centripetal) the other outwards (centrifugal)

Quote Quote by Andrew Mason View Post
From Wikipedia:
Centrifugal force (from Latin centrum, meaning "center", and fugere, meaning "to flee") is the apparent outward force that draws a rotating body away from the center of rotation.
This describes the potential effects of the inertial centrifugal force as seen in the rotating frame.

Quote Quote by Andrew Mason View Post
There is nothing about the "centrifugal reaction force" that causes anything to flee from the centre.
Wrong. In the rotating frame the centrifugal reaction force can push an object outwards, just like the inertial centrifugal force can.
sophiecentaur
#36
Jan29-13, 09:44 AM
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Quote Quote by A.T. View Post

Wrong. In the rotating frame the centrifugal reaction force can push an object outwards, just like the inertial centrifugal force can.
But will 'flee' along a tangent and not a radius, when you cut the string. That was why the word was not 'permitted'. Of course, they failed to mention that a tangent takes you further from the centre, too.


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