Difference between centrifugal force vs reactive centrifugal force?

[A.T.]

Centrifugal force is not a real, or more precisely an interaction force, ...

Part of the problem is using the "real" label for interaction forces. It tends to turn technical discussions into philosophical musings.

 

[Aeronautic Freek]

The magnitude of Ficf depends on the rotation rate of the chosen reference frame.
Their magnitudes are not the same in general, just in some reference frames.

I still don't understand why Ficf and Frcf are not same in magnitude...
can you make simple example with numbers..

 

[jbriggs444]

By Newton’s 3rd law there is an outward-pointing interaction force from the rock acting on the string. This is the reactive centrifugal force. Note that it is an interaction force. An accelerometer on the rock detects the inward acceleration produced by this force.

I think that you mis-spoke here, @Dale. The phrase "this force" refers to the reactive centrifugal force from the rock acting on the string. There is no inward acceleration of the rock produced by "this force". First, because it acts outward and second because it acts on the string. The inward acceleration of the rock is, of course, produced by the centripetal force.

 

[A.T.]

I still don't understand why Ficf and Frcf are not same in magnitude...
can you make simple example with numbers..

In the inertial frame Ficf is zero, while Frcf is not.

 

[Dale]

show me equation for magnitude for Ficf and Frcf

A vector has more than just a magnitude. However, here are the equations in an inertial frame:
$|F_{icf}|=0$
$|F_{rcf}|=|F_{cf}|$

 

[Dale]

I think that you mis-spoke here, @Dale.

Oops, quite right. I have edited it to fix it

 

[Ibix]

I still don't understand why Ficf and Frcf are not same in magnitude...
can you make simple example with numbers..

$F_{rcf}=m_{man}r_{station}\omega_{station}^2$. Note that $\omega_{station}$ is the angular velocity of the station as measured in an inertial frame, so this quantity is frame independent.

$F_{icf}=m_{man}r_{station}\omega_{frame}^2$. The angular velocity of the frame can be anything. If you imagine the diagrams A. T. drew to be photos taken with a camera, $\omega_{frame}$ is the angular velocity of the camera. This has no effect on the situation being filmed. It could be zero, in which case $F_{icf}=0$ and you have an inertial frame. You could set the camera spinning at half the rate the station spins, in which case $F_{icf}=F_{rcf}/4$. In the case that the camera spins at the same rate as the station, $F_{icf}=F_{rcf}$.

The point is that $F_{icf}$ is a force you have to introduce if you want to use Newton's laws in their simple form to describe motions as recorded by a spinning camera. $F_{rcf}$ is a real force the man exerts on the station floor.

 

[Dale]

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)

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

 

[Aeronautic Freek]

You could set the camera spinning at half the rate the station spins, in which case $F_{icf}=F_{rcf}/4$.

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!

I allways thinking when talking about rotating frame ,that camera is attached on station,so now it is part of station ,so normaly it spins at same ang.velocity as station..So Frcf must be equal to Ficf

 

[Aeronautic Freek]

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

Thanks ,with your example with numbers and post 37# with camera example, now is clear.
Whish you write this at biginning of topic! :)

(probelm is i didnt know that reference frame can spins with different ang.velocty than station)
I allways have only two rafrerence frame in my mind.
inertial-when I watch merrygoaround from outside
and roating (non-inertial)when I am in merry-go around so I spin with him at same rate...

 

[vanhees71]

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.

The usual way the expression "centrifugal force" is used is that of a specific part of the socalled inertial forces in a rotating reference frame in Newtonian mechanics (rotating against inertial reference frames of course). The inertial forces come from bringing parts of the covariant time derivative defining acceleration as the 2nd time derivative of the position vector when expressed in components wrt. the rotating basis of the rotating reference frame to the right-hand side such as to make the equation look like in an inertial frame with additional "inertial forces" on the right-hand side. This is sometimes customary if you want to think about what's going on in a rotating reference frame. We live in one, because the Earth rotates once a day around its axis, and for some physics we cannot ignore this rotation against the inertial frames (e.g., the physics of the Foucault pendulum, which is treated in any introductory physics lecture on classical mechanics, the rotation direction in cylones and anticyclones on both sides of the equator,...).

It is a very good practice, hammered into us by good teachers in high school and professors at universities to talk about the centrifugal force if and ONLY if one refers to a rotating reference frame and the one specific part of the inertial forces in such a rotating frame: $\vec{F}_{\text{centrifug}}=-m \vec{\omega} \times (\vec{\omega} \times \vec{r})$. Here $m$ is the mass of the particle, $\vec{\omega}$ the momentary angular velocity of the rotating basis against the inertial basis, and $\vec{r}$ the position vector, all given as components with respect to the rotating basis.

 

[Dale]

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.

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]

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]

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?

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.

 

[weirdoguy]

Both forces are centripetal.

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.

 

[jbriggs444]

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.

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.

 

[Dale]

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:

the reaction to a centripetal force can be another centripetal force, as in the case of circular orbits in Newtonian gravity.

(emphasis added)

 

[A.T.]

[reference frame] spins at same ang.velocity as station..So Frcf must be equal to Ficf

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.

(probelm is i didnt know that reference frame can spins with different ang.velocty than station)

Reference frames are just abstract constructions, like coordinate systems. Their movement doesn't have to coincide with the movement of any physical object.

 

[weirdoguy]

If one uses the barycenter then gravity is purely centripetal at all times.

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.

 

[weirdoguy]

Yes. Which is why I said:

Sorry, I missed that post

 

[etotheipi]

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.

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"?

 

[jbriggs444]

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.

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].

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"?

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.

 

[vanhees71]

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.

Sure, but nowhere (using the usual terminology) is there any centrifugal force.

 

[Ibix]

Sure, but nowhere (using the usual terminology) is there any centrifugal force.

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.

 

[Dale]

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.

 

[jbriggs444]

Yes. This is why I dislike the term “reactive centrifugal force”. Sometimes the reaction force to a centripetal force is centripetal, not centrifugal.

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]

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.

Excellent! Yet another reason to not like the term

 

[vanhees71]

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.

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.

Also third law refers always to interactions, i.e., to the forces acting on two different bodies or, for the most general case of general $N$-body forces, it boils down to the center-of-mass motion, which for a closed system is always uniform and thus finally to global momentum conservation for closed systems as a consequence of homogeneity of space according to Noether's theorem.

 

[vanhees71]

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.

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]

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 is clearly centrifugal.

Of course, strictly speaking, string tension is not a force at all. It is a condition in the string. A component of a stress tensor. It is more akin to a force pair rather than to a single member of a force pair.