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Centripetal and centrifugal force 
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#37
Nov2009, 06:24 AM

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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. . . . 


#38
Nov2009, 07:07 AM

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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. 


#39
Nov2009, 09:11 AM

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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. 


#40
Nov2009, 09:28 AM

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#41
Nov2009, 09:34 AM

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#42
Nov2009, 10:10 AM

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#43
Nov2009, 10:19 AM

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#44
Nov2009, 10:34 AM

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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. 


#45
Nov2009, 11:49 AM

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So they now have "reactive centrifuges" in Biology labs? Fair enough. Do you think it will catch on?



#46
Nov2009, 12:42 PM

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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 


#47
Nov2009, 02:55 PM

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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"? 


#48
Nov2009, 03:14 PM

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Stop that, Jeff!



#49
Nov2109, 09:21 AM

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#50
Nov2109, 11:39 AM

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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: [tex]\vec{a}= \ddot{\vec{r}} = (\ddot{r}r\omega^2) \hat{r} + (r\dot{\omega} +2\dot{r}\omega)\hat{\theta}[/tex] [tex](\omega = \dot{\theta})[/tex] We get two sets of terms. The coordinate accelerations: [tex]\vec{a}_{cord}= \ddot{r}\hat{r} + r\dot{\omega}\hat{\theta}[/tex] and the components emerging from rates of change of our basis (local frame): [tex]\vec{a}_{frame} = r\omega^2 \hat{r} + 2\dot{r}\omega\hat{\theta}[/tex] We can then express Newton's 2nd law in two forms: [tex] \vec{F} = m\vec{a}[/tex] or [tex] \vec{F}_{eff} = m\vec{a}_{coord}[/tex] where the l.h.s. is an "effective force" which is the "physical force" plus the "pseudoforces" or "fictional forces" we get by subtracting out mass times the frame accelerations. They are the coriolis force: [tex]\vec{F}_{cori}= 2m\dot{r}\omega\hat{\theta}[/tex] and the centrifugal force: [tex]\vec{F}_{cnf} = mr\omega^2\hat{r}[/tex] 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 nonrectangular coordinates we will again need to resolve out coordinate and "local frame" components of acceleration while also taking into account the effects of the overall 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 "pseudoforces" 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 KaluzaKlein 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. 


#51
Nov2109, 12:48 PM

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#52
Nov2109, 04:06 PM

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#53
Nov2109, 07:12 PM

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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. 


#54
Nov2109, 07:34 PM

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