# centripetal and centrifugal force

by monty37
Tags: centrifugal, centripetal, force
 Sci Advisor PF Gold P: 11,380 Sorry to hear about the Plymouth. Insurance companies just don't understand about emotional attachment. "Beyond economic repair" is a phrase which they often apply to priceless objects. Look, we're only arguing about a name. Because of Newton 3 we can always say that these forces appear in pairs. Mostly, we only consider and name one of the pair and I agree that centrifugal force is not the way to go in explaining what makes things go in a curve. My fave car was a Lotus Super Seven, which was a bit like a roller skate. It could do 0 - 60 in less than 6 seconds (soft bits and all) with a 1500 Ford Cosworth engine. It woudn't do more than 100 and even at that it was screaming. Great at traffic lights in town until some brute in a Corvette Stingray left me behind in a cloud of dust. Really cut me down to size - then I bought a Ford Escort. . . .
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 Quote by sophiecentaur But I FEEL a force pushing my soft bits towards my back- 'outwards'. That force is away from the centre. Centrifugal, in fact.
You don't feel forces directly. You feel deformations of your body. Theses deformations occur because the centripetal force is applied non-uniformly to parts of you body. So what you feel indirectly is the centripetal force, not the centrifugal force. You feel a force pushing your back inwards against your soft bits.

If the centripetal force is suddenly gone and you fly off tangentially, the centrifugal force in the rotating frame is still there and accelerates you away from the center, but you don't feel any load. This shows that you cannot feel the centrifugal force.
 Sci Advisor PF Gold P: 11,380 OK - say we've got a conker on a string, whirling around. There MUST be two forces on each element of the string, keeping it taught. Those constitute the 3law pair I'm talking about. The must both exist or the string would be slack. Yes, 'everyone' (at least you and I) knows the conker will fly off tangentially. That's not the issue. If I am on the conker, that's the frame I'm interested in and I guess I can change my interest when the string is cut - but I'm not going to cut the string. What would you call the force pulling outwards on the string, then? (Bearing in mind that it does exist and is directed away from the centre and it needs a name). I don't think you realise that we are arguing about semantics and not Physics.
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P: 40,905
 Quote by sophiecentaur What would you call the force pulling outwards on the string, then? (Bearing in mind that it does exist and is directed away from the centre and it needs a name).
Why does every force need a special name? Note that you are talking about an outward force on the string, not on the object. (Viewed from a rotating frame, centrifugal force acts on the object.) The string and object exert forces on each other. If you want to call that force on the string something, don't call it the 'centrifugal force'--that term already has a specific meaning in physics. (Some use the term 'reactive centrifugal force'--use that at the risk of confusing people.)
PF Gold
P: 1,767
 Quote by monty37 so how do you conclude the nature of this force,as to where which acts,centripetal or centrifugal,i mean generally,as in books especially in engineering concepts,it is highly unclear.
I always kept them straight by remembering the 'f' in centrifugal stands for 'flee' so it is the apparent radial outward force in a rotating frame while a centripetal force is the force typically opposing the centrifugal and effecting the circular motion. (The latin root fugere means 'to flee' also use to define fugacity in thermodynamics.)
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P: 14,468
 Quote by sophiecentaur What would you call the force pulling outwards on the string, then?
Tension.

 I don't think you realise that we are arguing about semantics and not Physics.
Misusing names can lead to confusion. The term centrifugal force has a specific meaning.
PF Gold
P: 11,380
 Quote by D H Tension.
Doesn't tension operate in both directions in the string? I think that's a cop-out.
 Misusing names can lead to confusion. The term centrifugal force has a specific meaning.
I guess that's the best argument against using it when it's not appropriate. So what can we call it? Tension is just not specific enough - because it doesn't describe the force of my body against the seat on the ride.
 HW Helper P: 6,929 Wiki has updated it's "reactive centrifugal force" article: http://en.wikipedia.org/wiki/Reactive_centrifugal_force Perhaps it's not common usage in physics, but it is common usage in English, which is a larger audience, and I doubt the term centrifugal force is confusing to anyone in physics, who would know that it's a reactive force in a standard inertial frame.
 Sci Advisor PF Gold P: 11,380 So they now have "reactive centrifuges" in Biology labs? Fair enough. Do you think it will catch on?
P: 3,551
 Quote by sophiecentaur What would you call the force pulling outwards on the string, then? (Bearing in mind that it does exist and is directed away from the centre and it needs a name).
That is the reactive centrifugal force (a 'real' interaction force that exists in every frame):
http://en.wikipedia.org/wiki/Reactive_centrifugal_force

Not to be confused with centrifugal force (a 'fictional' inertial force that exists only in the rotating frame):
http://en.wikipedia.org/wiki/Centrif...rence_frame%29
 HW Helper P: 6,929 In some cases which forces are real or reactive get a bit fuzzy. Here is an example of a radio control glider dynamic soaring in circles at speeds up to 375 mph (the mentioned 392 mph pass wasn't caught on video): http://www.youtube.com/watch?v=WaQB16ZaNI4&fmt=18 While circling, the air exerts a centripetal force on the glider, causing the glider to accelerate inwards, following a circular path. This coexists with the glder exerting a centrifugal force on the air, causing the air to accelerate outwards in a spiraling path. Here the glider's outwards reactive force coincides with the outward force the glider exerts onto the air, and the air's inwards reactive force coincides with the inwards force the air exerts onto the glider. Similarly imagine a rocket in space void of gravitational effect, using it's thrust to follow a circular path. The spent fuel exerts a centripetal force on the rocket, causing the rocket to accelerate inwards, following a circular path. This coexists with the rocket exerting a centrifugal force on the spent fuel, causing the spent fuel to accelerate outwards in a spiraling path. Here the rocket's outwards reactive force coincides with the outward force the rocket exerts onto the spent fuel, and the spent fuel's inwards reactive force coincides with the inwards force the spent fuel exerts onto the rocket. update - so which of the forces in these examples are "fictitious"?
 Mentor P: 14,468 Stop that, Jeff!
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 Quote by Jeff Reid Here the glider's outwards reactive force coincides with the outward force the glider exerts onto the air, and the air's inwards reactive force coincides with the inwards force the air exerts onto the glider.
They coincide because they are one and the same. Same is the case with the rocket example too. When you hit a wall, there is only one force exerted by you on the wall. You seem to think there are two- namely the force exerted by you on the wall and the reactive force you exert on the wall. This isn't true. We cannot absolutely classify a force as action or reaction. Its just that forces observed from an inertial frame occur in pairs of action and reaction.

 so which of the forces in these examples are "fictitious"?
Terming a force as fictitious depends on the reference frame. If you have observed the above two events from an inertial frame, then all the forces you measure are real.
 Sci Advisor PF Gold P: 1,767 If you want to argue semantics then lets make it clear what the debate is. *** In one context "centripetal" and "centrifugal" are just qualifiers for resp. negative radial and positive radial directions and you can also speak of centripetal velocity or centrifugal displacements as well as further qualify with "reactive" or "applied" or whatever. In such context the two words are redundant (which isn't necessarily bad) as we could as easily speak of positive centrifugal and negative centrifugal or similarly negative centripetal and positive centripetal. *** In another context we resolve the vector acceleration of a particle in polar coordinates: $$\vec{a}= \ddot{\vec{r}} = (\ddot{r}-r\omega^2) \hat{r} + (r\dot{\omega} +2\dot{r}\omega)\hat{\theta}$$ $$(\omega = \dot{\theta})$$ We get two sets of terms. The coordinate accelerations: $$\vec{a}_{cord}= \ddot{r}\hat{r} + r\dot{\omega}\hat{\theta}$$ and the components emerging from rates of change of our basis (local frame): $$\vec{a}_{frame} = -r\omega^2 \hat{r} + 2\dot{r}\omega\hat{\theta}$$ We can then express Newton's 2nd law in two forms: $$\vec{F} = m\vec{a}$$ or $$\vec{F}_{eff} = m\vec{a}_{coord}$$ where the l.h.s. is an "effective force" which is the "physical force" plus the "pseudo-forces" or "fictional forces" we get by subtracting out mass times the frame accelerations. They are the coriolis force: $$\vec{F}_{cori}= -2m\dot{r}\omega\hat{\theta}$$ and the centrifugal force: $$\vec{F}_{cnf} = mr\omega^2\hat{r}$$ Now when considering systems where the coordinate acceleration is zero the centrifugal force must be canceled by a radial component of the "physical force" which component we call the "centripetal force". *** Now we can also adopt a third context not totally distinct from the second one wherein we consider a uniformly rotating frame which itself may be resolved in rectangular or polar coordinates or some nastier system. We may even use an origin distinct from the center of rotation. In this context when using non-rectangular coordinates we will again need to resolve out coordinate and "local frame" components of acceleration while also taking into account the effects of the over-all time dependency of the rotating coordinate system. It is thus useful (and consistent with Einstein's equivalence principle) to treat the coriolis and centrifugal forces due to the global frame rotation as "physical" e.g. a form of gravity. We will also have as before local frame components of the acceleration which we can again refer to as "pseudo-forces" if that is our inclination. Typical examples in this last context are ballistics and the dynamics of a hurricane given the rotation of the Earth. Now I think it is silly to argue over "fictitious" vs. "real" forces especially given GR where all gravitational forces are just as "fictitious" as the coriolis and centrifugal ones here. The important point is correct bookkeeping. We must be sure that Newton's 2nd law (F - ma =0) gets transformed correctly when we change coordinate systems. Personally I am for dropping the "fictitious" or "pseudo" qualifiers in physics. (Though they may still be appropriate in engineering.) We should properly recognize that Einstein's equivalence principle goes both ways and such distinctions are meaningless. After all in Kaluza-Klein theory EM forces end up being "fictitious" as well. Consider also covariant momenta and gauge transformations when we define forces as rates of change of momenta.
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 Quote by Jeff Reid In some cases which forces are real or reactive get a bit fuzzy. Here is an example ... While circling, the air exerts a centripetal force on a glider, causing the glider to accelerate inwards, following a circular path. This coexists with the glder exerting a centrifugal force on the air, causing the air to accelerate outwards in a spiraling path. Here the glider's outwards reactive force coincides with the outward force the glider exerts onto the air, and the air's inwards reactive force coincides with the inwards force the air exerts onto the glider.
Exactly, centrifugal/centripetal is just action/reaction between 2 bodies. After all, it's not possible to have action/reaction with just one, - the 2 always pull or push on each other. Which is the action and which the reaction is just an argument of semantics.

 Quote by Jeff Reid Similarly imagine a rocket in space void of gravitational effect, using it's thrust to follow a circular path. The spent fuel exerts a centripetal force on the rocket, causing the rocket to accelerate inwards, following a circular path. This coexists with the rocket exerting a centrifugal force on the spent fuel, causing the spent fuel to accelerate outwards in a spiraling path. Here the rocket's outwards reactive force coincides with the outward force the rocket exerts onto the spent fuel, and the spent fuel's inwards reactive force coincides with the inwards force the spent fuel exerts onto the rocket. update - so which of the forces in these examples are "fictitious"?
both, it's all down to individual perception, ie which frame of reference you're using. And we could probably argue that all forces are fictitious if we follow Einstein's lead. And don't forget that Newton's first attempt at putting circular motion into a mathematical formulation involved the centrifugal not centripetal force.
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 Quote by YellowTaxi Exactly, centrifugal/centripetal is just action/reaction between 2 bodies. After all, it's not possible to have action/reaction with just one, - the 2 always pull or push on each other. Which is the action and which the reaction is just an argument of semantics.
Centrifugal force (rather than the archaic "reactive centrifugal force" that Jeff likes) does *not* obey Newton's third law. It is not just a matter of semantics.
HW Helper
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 Quote by D H Centrifugal force (rather than the archaic "reactive centrifugal force" that Jeff likes) does *not* obey Newton's third law.
In the examples I mentioned with circling glider and rocket, where is Newtons third law being violated?

In a sense, virtually all forces involving accelerations are "reactive". It's just that in most cases, one or more of the objects involved is attached some massive object (such as the earth) where that massive object is treated as if it had infinite inertia, and the momentum effects on that massive object are ignored.
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 Quote by Jeff Reid In the examples I mentioned with circling glider and rocket, where is Newtons third law being violated?
Are you intentionally missing D H's point? Your examples have nothing to do with centrifugal force (as used in physics), only 'reactive centrifugal force'. (At this point I can only think that you are yanking our chains by purposely leaving off the label 'reactive'.)

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