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Dale
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Oops, quite right. I have edited it to fix itjbriggs444 said:I think that you mis-spoke here, @Dale.
Oops, quite right. I have edited it to fix itjbriggs444 said:I think that you mis-spoke here, @Dale.
##F_{rcf}=m_{man}r_{station}\omega_{station}^2##. Note that ##\omega_{station}## is the angular velocity of the station as measured in an inertial frame, so this quantity is frame independent.Aeronautic Freek said:I still don't understand why Ficf and Frcf are not same in magnitude...
can you make simple example with numbers..
Take the example of a rock attached to a string being spun in a circle in space (no air and no gravity). Let the mass be 1 kg, the radius be 2 m, and the tangential velocity be a constant 10 m/s. (all numbers in SI units)Aeronautic Freek said:I still don't understand why Ficf and Frcf are not same in magnitude...
can you make simple example with numbers..
Here start all my problems I didnt have solution in my head that you can spin camera at different ang.velocity than station.But why you will do that!Ibix said:You could set the camera spinning at half the rate the station spins, in which case ##F_{icf}=F_{rcf}/4##.
Thanks ,with your example with numbers and post 37# with camera example, now is clear.Dale said:Take the example of a rock attached to a string being spun in a circle in space (no air and no gravity). Let the mass be 1 kg, the radius be 2 m, and the tangential velocity be a constant 10 m/s. (all numbers in SI units)
In an inertial frame (##\omega=0##):
##F_{icf}=m\omega^2 r=0## acting on the rock
##F_{rcf}=mv^2/r=50## acting on the string
In a co-rotating frame (##\omega=5##):
##F_{icf}=m\omega^2 r=50## acting on the rock
##F_{rcf}=mv^2/r=50## acting on the string
In a frame rotating at double speed (##\omega=10##):
##F_{icf}=m\omega^2 r=200## acting on the rock
##F_{rcf}=mv^2/r=50## acting on the string
It was not invented here, but I also dislike the term. For one thing, the reaction to a centripetal force can be another centripetal force, as in the case of circular orbits in Newtonian gravity.vanhees71 said:This thread shows again that one shouldn't invent unnecessary vocabulary which isn't used in a wider physics community anyway. I've never heard about "reactive centrifugal forces" before, and I do not see what they are in the here discussed example nor what they are good for.
There is the gravitational force of the sun acting on the planet. The third law pair to that force is the gravitational force of the planet acting on the sun. Both forces are centripetal.vanhees71 said:In Newtonian gravity, I guess you refer to the Kepler problem of a planet moving around the Sun, I guess. This problem you treat of course in an inertial frame of reference, and the force you have there is the gravitational force, which of course provides the centripetal force of the planet to keep it on its circular (or elliptic orbit). In this problem there's no other force than the gravitational force and not "another centripetal force", or what should that be?
Dale said:Both forces are centripetal.
Well, that depends on what you take as the "center". If one uses the barycenter then gravity is purely centripetal at all times. If one uses the instantaneous center of curvature of the trajectory then gravity is only purely centripetal at two points, as you point out.weirdoguy said:Only if the orbit is a cricle. In elliptic case force of gravity as a whole is not a centripetal force, only its component perpendicular to ellipse.
Yes. Which is why I said:weirdoguy said:Only if the orbit is a cricle. In elliptic case force of gravity as a whole is not a centripetal force, only its component perpendicular to ellipse.
(emphasis added)Dale said:the reaction to a centripetal force can be another centripetal force, as in the case of circular orbits in Newtonian gravity.
Not in general, only in the most simple scenarios like this one. For example, if the two astronauts in the station were also connected by a rope under tension, then the Frcf would be reduced, and its magnitude no longer equal to Ficf in the co-rotating frame.Aeronautic Freek said:[reference frame] spins at same ang.velocity as station..So Frcf must be equal to Ficf
Reference frames are just abstract constructions, like coordinate systems. Their movement doesn't have to coincide with the movement of any physical object.Aeronautic Freek said:(probelm is i didnt know that reference frame can spins with different ang.velocty than station)
jbriggs444 said:If one uses the barycenter then gravity is purely centripetal at all times.
Dale said:Yes. Which is why I said:
jbriggs444 said:Well, that depends on what you take as the "center". If one uses the barycenter then gravity is purely centripetal at all times. If one uses the instantaneous center of curvature of the trajectory then gravity is only purely centripetal at two points, as you point out.
Yes, ##a=\frac{v^2}{r}## fits the instantaneous center of rotation while ##a=\frac{GM_r}{r^2}## fits the barycenter [here, ##M_r## is the "reduced mass" of the other body].weirdoguy said:But then in that case, can one use the formula for centripetal force ##\frac{mv^2}{r}##? I guess not, since this formula adapts the second convention.
In my book, "centripetal" and "centrifugal" are directions. Literally toward the center and away from the center. But I am not a definition Nazi. Use words as you wish. Just be aware that others may use them differently.etotheipi said:I was under the impression that it is by definition always the latter, i.e. the normal component in intrinsic coordinates, ##\frac{mv^2}{\rho}##.
The component toward the barycentre would I guess be towards a centre of some sort, but it wouldn't be what we call "centripetal force"?
Sure, but nowhere (using the usual terminology) is there any centrifugal force.Dale said:There is the gravitational force of the sun acting on the planet. The third law pair to that force is the gravitational force of the planet acting on the sun. Both forces are centripetal.
Indeed. There's at least one seventeen page thread on the topic here. However in the ball on a string example, there is a third law pair to the inward pointing centripetal force on the ball and this force is outward pointing. This would be the "reactive centrifugal force". It isn't present for planets because the third law pairs are both centripetal, since there is no mechanical connection between the planets.vanhees71 said:Sure, but nowhere (using the usual terminology) is there any centrifugal force.
Yes. This is why I dislike the term “reactive centrifugal force”. Sometimes the reaction force to a centripetal force is centripetal, not centrifugal.vanhees71 said:Sure, but nowhere (using the usual terminology) is there any centrifugal force.
Meanwhile, I dislike the "reactive" part. It reinforces the incorrect notion that there is some sort of asymmetry in a third law force pair. Cause and effect, action and reaction rather than simply two facets of the same interaction.Dale said:Yes. This is why I dislike the term “reactive centrifugal force”. Sometimes the reaction force to a centripetal force is centripetal, not centrifugal.
Excellent! Yet another reason to not like the termjbriggs444 said:Meanwhile, I dislike the "reactive" part. It reinforces the incorrect notion that there is some sort of asymmetry in a third law force pair. Cause and effect, action and reaction rather than simply two facets of the same interaction.
What's the normal terminology what you rename "reactive centrifugal force"? There's a string tension here, and it's as centripetal a force as the gravitational force in the planet-Sun (Kepler) example.Ibix said:Indeed. There's at least one seventeen page thread on the topic here. However in the ball on a string example, there is a third law pair to the inward pointing centripetal force on the ball and this force is outward pointing. This would be the "reactive centrifugal force". It isn't present for planets because the third law pairs are both centripetal, since there is no mechanical connection between the planets.
I think the problem is that some sources do use the words "centrifugal force" to mean the thing I've called a "reactive centrifugal force", which is most definitely a proper force in the cases where it's present. This is sloppy terminology - even the Wikipedia article on centrifugal force notes this usage as "deprecated" - and is what's causing all the confusion here.
I also agree that "reactive centrifugal force" is not a term I'd come across before the first time I saw this argument on here. I mostly remember it just to answer people like the OP who've confused it with the inertial force.
action and reaction is not a cause-and-effect relation. It's an artifact of Newtonian physics, where you have action-at-a-distance ineractions, which of course is an approximation to real interactions mediated by fields.jbriggs444 said:Meanwhile, I dislike the "reactive" part. It reinforces the incorrect notion that there is some sort of asymmetry in a third law force pair. Cause and effect, action and reaction rather than simply two facets of the same interaction.
The outward force of rock on string is clearly centrifugal.vanhees71 said:What's the normal terminology what you rename "reactive centrifugal force"? There's a string tension here, and it's as centripetal a force as the gravitational force in the planet-Sun (Kepler) example.
The outward force of rock on string, as I said. That force exists at the interface between rock and string. It has a direction and a location. The direction is unambiguously away from the center of rotation.vanhees71 said:Which outward force?
We do not seem to be communicating well.vanhees71 said:Sure, but it's not a centrifugal force but just a force in an inertial frame. As you say, it's a "contact formce". So don't call something a centrifugal (inertial) force in an inertial frame. By definition there are no inertial forces in inertial frames of reference. The wrong wording leads to utmost confusion of the students, as is demonstrated by this thread which consists of 63 contributions just to eliminate the wrong thinking from this wrong wording again!
It is a force that acts in the centrifugal direction. Taking "centrifugal" here to mean literally "away from the center".vanhees71 said:It's in any case NOT a centrifugal force, because there is no centrifugal force in an inertial frame of reference by definition. In you example the force on the rock is due to the tension in the string. At the end of the string the rock's contact force is opposite and of equal magnitude of course (Newton's 3rd law).
jbriggs444 said:It is a force that acts in the centrifugal direction. Taking "centrifugal" here to mean literally "away from the center".
I don't think that anyone here disagrees with you. The term "reactive centrifugal force" is rather like "relativistic mass". It has a standard known definition that is not from Physics Forums and is not liked by most participants here, but sometimes you see someone who is confused by it and so we have to explain what it means and also why we dislike it.vanhees71 said:I fight against the use of the word "centrifugal" in the context of forces at all as long as we discuss physics from the point of view of an inertial observer, because it's much more economic to avoid misleading language and stick to clear definitions. A centrifugal force is an inertial force and as such can by definition only present in non-inertial (in this case rotating) frames of reference.
I am even more annoyed by the "reactive" part because I dislike the "action/reaction" terminology with respect to Newtons 3rd Law.vanhees71 said:I fight against the use of the word "centrifugal" in the context of forces at all as long as we discuss physics from the point of view of an inertial observer, ...