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What exactly is centrifugal force

by hms.tech
Tags: centrifugal, force
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sophiecentaur
#37
Jan29-13, 09:51 AM
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Quote Quote by A.T. View Post
It is really not that confusing if you make clear which reference frame you consider.
We didn't 'do' astronauts when I was at school (they hadn't been invented - except for Dan Dare) and the term 'reference frame' was University stuff at the time.

Having tried those ideas on teenagers, not long ago, I did find it worked with some. It didn't work with others but, so what? They'll have ended up in banking or as 'TV personalities'.
A.T.
#38
Jan29-13, 09:53 AM
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Quote Quote by A.T. View Post
In the rotating frame the centrifugal reaction force can push an object outwards, just like the inertial centrifugal force can.
Quote Quote by sophiecentaur View Post
But will 'flee' along a tangent and not a radius, when you cut the string.
I was talking about about the rotating frame, where it flees along the radius.
sophiecentaur
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Jan29-13, 09:54 AM
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Quote Quote by A.T. View Post
I was talking about about the rotating frame, where it flees along the radius.
See my post - just above.
stevendaryl
#40
Jan29-13, 10:48 AM
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Quote Quote by A.T. View Post
Yeah, you can always replace a simple approach with a mathematically equivalent, but more complicated one.
That's what introducing the idea of "centrifugal force" does. It makes things unnecessarily complicated and confusing.
stevendaryl
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Jan29-13, 10:57 AM
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Quote Quote by A.T. View Post
Yeah, you can always replace a simple approach with a mathematically equivalent, but more complicated one.
Really, you have things completely backwards here. The SIMPLE approach is to use vector equations:
  1. In the absence of any forces, [itex]\frac{\stackrel{\rightarrow}{dU}}{dt} = 0[/itex]
  2. In the presence of an external force [itex]m \frac{\stackrel{\rightarrow}{dU}}{dt} = \stackrel{\rightarrow}{F}[/itex]
  3. If one object exterts a force [itex]\stackrel{\rightarrow}{F}[/itex] on another, then the second object exerts a force [itex]-\stackrel{\rightarrow}{F}[/itex] on the first.

There is nothing confusing about these rules, and there are no exceptions for "fictitious forces" or "noninertial frames". The only complicated thing is that you need to realize that a vector can change for two reasons: (1) the components change, or (2) the basis vectors change. It might be simpler to pretend that (2) never happens, but it's a falsehood, and you're doing physics in a crippled way when you do it.
stevendaryl
#42
Jan29-13, 11:08 AM
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Quote Quote by stevendaryl View Post
That's what introducing the idea of "centrifugal force" does. It makes things unnecessarily complicated and confusing.
Actually, different people disagree about what is "complicated". I consider it complicated when you have lots of ad hoc rules that apply in specific circumstances: If you are using inertial Cartesian coordinates, do this, if you're using curvilinear coordinates, do that, if you're using noninertial coordinates, do that. I'd rather have a fundamental set of principles that apply in all those circumstances, even if working out the details might be complicated.
ZealScience
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Jan29-13, 11:33 AM
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It appears because of choice of coordinates and reference frame, and yes it does not exist in inertial reference frame.
Andrew Mason
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Jan29-13, 12:33 PM
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Quote Quote by A.T. View Post
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)
Ok. But it gets confusing to students.

Example: there is no centrifugal reaction force to the weight of a car sitting on a road on the equator, even though the car is whipping around at 1000 mph. as the earth turns and even though it's weight bears on the earth surface. Ques. What about the normal force? Do we call that a centrifugal force? Answer: no. Ques. But it is directed away from the centre of rotation (ie. up). Answer: Yes, but we do not call it centrifugal. Ques. Why not? Answer: My answer would be: We don't call it centrifugal because it does not cause anything to flee the centre of rotation. I am not sure what your answer would be. Should we call it centrifugal?

This describes the potential effects of the inertial centrifugal force as seen in the rotating frame.
Yes, but the point is that it means "fleeing the centre".

Wrong. In the rotating frame the centrifugal reaction force can push an object outwards, just like the inertial centrifugal force can.
But that is a fictitious centrifugal force, not the "centrifugal reaction force". Give us an example - just one example where the reaction force to a centripetal force causes matter to flee the centre of rotation.

AM
rcgldr
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Jan29-13, 01:09 PM
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Quote Quote by Andrew Mason View Post
There is no centrifugal reaction force to the weight of a car sitting on a road on the equator, even though the car is whipping around at 1000 mph.
A more generalized version of this would be 2 body system in outer space, free of external forces, each object orbiting about a common center of mass due to gravity or opposite charge. The Newton third law pair of forces are the two attractive forces between the objects and directed towards a common center of mass. If the orbits are circular, then the two attractive forces are also centripetal forces. In this situation, there are no reactive centrifugal forces.

Change this 2 body system to one where there are no attractive forces, and the two objects are connected by a string and rotate in a circular path about a common center of mass. Both objects exert a reactive centrifugal force on the ends of the string (assuming that the common center of mass is not located within one of the objects, in which case only one end of the string experiences a reactive centrifugal force).
Andrew Mason
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Jan29-13, 05:46 PM
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Quote Quote by stevendaryl View Post
Actually, different people disagree about what is "complicated". I consider it complicated when you have lots of ad hoc rules that apply in specific circumstances: If you are using inertial Cartesian coordinates, do this, if you're using curvilinear coordinates, do that, if you're using noninertial coordinates, do that. I'd rather have a fundamental set of principles that apply in all those circumstances, even if working out the details might be complicated.
The principles and the laws of physics are the same. The problem is that nature makes a distinction between non-inertial and inertial frames of reference.

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Andrew Mason
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Jan29-13, 05:55 PM
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Quote Quote by rcgldr View Post
Change this 2 body system to one where there are no attractive forces, and the two objects are connected by a string and rotate in a circular path about a common center of mass. Both objects exert a reactive centrifugal force on the ends of the string (assuming that the common center of mass is not located within one of the objects, in which case only one end of the string experiences a reactive centrifugal force).
And how would the objects produce a centrifugal acceleration of the ends of the string?

AM
A.T.
#48
Jan30-13, 03:44 AM
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Quote Quote by stevendaryl View Post
It might be simpler to pretend that (2) never happens,
It is simpler, thatís why inertial forces are widely used.
Quote Quote by stevendaryl View Post
but it's a falsehood, and you're doing physics in a crippled way when you do it.
It leads to the same quantitative predictions, which is all that matters in physics. The rest is philosophy.
dextercioby
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Jan30-13, 03:48 AM
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In classical mechanics we treat the Coriolis force, the translation inertial force and the centrifugal force on equal footing.

I wonder why only the centrifugal force is poorly understood by people, since it's the only inertial force generating multi-page debates o PF...
stevendaryl
#50
Jan30-13, 06:09 AM
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Quote Quote by A.T. View Post
It is simpler, thatís why inertial forces are widely used.
No, inertial forces are NOT simpler. They may seem simpler to people who prefer to memorize formulas instead of understanding them.
stevendaryl
#51
Jan30-13, 06:34 AM
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Quote Quote by stevendaryl View Post
No, inertial forces are NOT simpler. They may seem simpler to people who prefer to memorize formulas instead of understanding them.
Can someone write down what are the equations for inertial forces, and what are the rules for using them? It seems to me that it amounts to this:
  1. Write down the equations of motion using inertial Cartesian coordinates.
  2. Transform to curvilinear, noninertial coordinates.
  3. Note that there are extra terms in the equations of motion that were not present in the inertial Cartesian case.
  4. Call these extra terms "inertial forces".

Surely, the last step isn't doing anything for you. Calling them "forces" doesn't help anything. They are different from other forces you're likely to encounter, because they don't depend on the substance an object is made of, and they don't have an equal and opposite reactive force. Calling them forces is a confusion--it's not a simplification. There is nothing that becomes simpler because of that choice of names.

The real confusion that is at the heart of discussions of "inertial forces" is the assumption that, if [itex]\stackrel{\rightarrow}{U}[/itex] is a vector (say, a velocity vector) with components [itex]U^i[/itex], then [itex]\frac{\stackrel{\rightarrow}{dU}}{dt}[/itex] must be a vector with components [itex]\frac{dU^i}{dt}[/itex]. That's just bad mathematics. It's just not true. It's true for inertial Cartesian coordinates, but not for other coordinates. It's not a "simplification" to assume something that is provably false.
A.T.
#52
Jan30-13, 07:39 AM
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Quote Quote by stevendaryl View Post
No, inertial forces are NOT simpler...
Then I'm sure you will soon have convinced anyone to stop using them. Let us know when all the books have been revised.
Quote Quote by stevendaryl View Post
That's just bad mathematics. It's just not true...
It leads to the same quantitative predictions, which is all that matters in physics. The rest is philosophy.
rcgldr
#53
Jan30-13, 08:34 AM
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Quote Quote by rcgldr View Post
Change this 2 body system to one where there are no attractive forces, and the two objects are connected by a string and rotate in a circular path about a common center of mass. Both objects exert a reactive centrifugal force on the ends of the string (assuming that the common center of mass is not located within one of the objects, in which case only one end of the string experiences a reactive centrifugal force).
Quote Quote by Andrew Mason View Post
And how would the objects produce a centrifugal acceleration of the ends of the string?
They wouldn't. The net force on each object is inwards. The objects (and the ends of the string) are accelerated "inwards" by the tension in the string. The outwards force exerted by the objects onto the ends of the string is a reaction (to acceleration) force, not a net force, equal in magnitude and opposing the tension at the ends of the string.
sophiecentaur
#54
Jan30-13, 08:34 AM
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Quote Quote by A.T. View Post
Then I'm sure you will soon have convinced anyone to stop using them. Let us know when all the books have been revised.

It leads to the same quantitative predictions, which is all that matters in physics. The rest is philosophy.
This thread reads like something out of Gulliver's Travels, actually. Anyone would think that there is some actual 'reality' in it all. People don't acknowledge that Science is the pragmatic business of predicting things - all the rest is faith.


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