Centrifugal force and Newton 3rd law

[tiny-tim]

The Coriolis term is intuitive and logical. There seems to be unanimous agreement that there is no actual "Coriolis force" knocking the missile eastward, i.e. Coriolis is "virtual". But in the course of the missile flight, where does "centrifugal" come into play? No one seems to produce the origin of this force, but are too quick to defend its significance.

Hi Claude!

Centrifugal force comes in as part of g.

It depends only on position, just like gravitational force, and so it's just part of what we measure as g (which isn't exactly "vertical" anyway, because of mountains etc).

Also, no one has yet explained the origin of cf in the moon orbit question I raised earlier. Is anybody going to attampt to tackle that one?

Which post was that?

 

[Doc Al]

Begging to differ, I don't quite follow you when you say Coriolis is not a correction term, but a term which "appears" in rotating ref frames. I guess different physicists, including the ones that taught me at my university, can view things differently. Call me a Newtonian curmudgeon if you please, but I don't see how an *accelerated" ref frame can be 1 to 1 "mapped" or "transformed" into a stationary one.

Let's take Coriolis. If we were to assume that the Earth is NOT rotating, i.e. a stationary ref frame, while it actually IS rotating, what are the consequences? We err by not accounting for the rotation. So, we introduce a correction term in the kinetics/kinematics equations to account for this. But by doing so we are acknowledging the existence of rotation and an accelerated frame of ref. We say that we refer to Earth as a frame of ref, but we do so with a priori knowledge that it is rotating. Thus we include the Coriolis term, twice the cross product of omega and u, to account for the rotation.

I suppose you can consider analyzing motion that takes into account the Earth's rotation as a "correction" to an analysis that ignores the Earth's rotation. But that's not what is being discussed here. We are talking about analyzing things from an inertial, non-rotating frame (in which the Earth rotates, of course!) versus analyzing things from a non-inertial, rotating frame (in which the Earth is at rest). We are not talking about ignoring the Earth's rotation versus taking it into account. (Both frames take rotation into account!)

When we say that coriolis and centrifugal forces are artifacts of using a non-inertial, rotating frame that does not mean that new physical effects magically appear when using a rotating frame of reference. The physical effects exist even in an inertial frame--they are just much harder to analyze! (In an inertial frame, the effects are "simply" due to the fact that while a projectile goes "straight", the Earth rotates.) If you do your analysis from an inertial frame, there is no need to introduce "forces" such as centrifugal or coriolis. (But good luck carrying out that analysis!)

Also, no one has yet explained the origin of cf in the moon orbit question I raised earlier. Is anybody going to attampt to tackle that one?

I thought I answered that in https://www.physicsforums.com/showpost.php?p=1785157&postcount=51".

 

[George Jones]

I thought I answered that in https://www.physicsforums.com/showpost.php?p=1785157&postcount=51".

I, too, thought that you did.

 

[cmos]

The Coriolis term is intuitive and logical. There seems to be unanimous agreement that there is no actual "Coriolis force" knocking the missile eastward, i.e. Coriolis is "virtual". But in the course of the missile flight, where does "centrifugal" come into play? No one seems to produce the origin of this force, but are too quick to defend its significance.

For this specific example, no one has defended its significance. The reason is that, relative to gravity, Coriolis force, and force due to burning of fuel; the centrifugal force is usually negligible.

Please don't use this as an argument to refute its existence. There are many times when something exist but is negligible. In analyzing the hydrogen atom, you care only about the Coulomb potential. Gravity is negligible in this case, but it is still there physically.

Also, no one has yet explained the origin of cf in the moon orbit question I raised earlier. Is anybody going to attampt to tackle that one? BR.

Yes, Doc Al already explained this. But also, this is the reason I presented the geosynchronous satellite example. It is a simpler example but generalizes to the moon.

 

[George Jones]

Yes, Doc Al already explained this. But also, this is the reason I presented the geosynchronous satellite example. It is a simpler example but generalizes to the moon.

https://www.physicsforums.com/showthread.php?p=1783976#post1783976".

 

[tiny-tim]

For this specific example {a missile}, no one has defended its significance. The reason is that, relative to gravity, Coriolis force, and force due to burning of fuel; the centrifugal force is usually negligible.

Hi cmos!

I don't think that's right:

Coriolis force = $-2m\,\mathbf{\Omega} \times \mathbf{v}_{rel}$

Centrifugal force = $m\,\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r})\,=\,m\,\mathbf{\Omega} \times \mathbf{v}_{rot}$ where $\mathbf{v}_{rot}$ is the velocity of rotation.

So centrifugal force is usually much larger than Coriolis force:

[tex]\frac{|centrifugal|}{|Coriolis|}\,=\,\frac{v_{rot}}{2v_{rel}}[/itex]

 

[cmos]

Hi cmos!

I don't think that's right:

Coriolis force = $-2m\,\mathbf{\Omega} \times \mathbf{v}_{rel}$

Centrifugal force = $m\,\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r})\,=\,m\,\mathbf{\Omega} \times \mathbf{v}_{rot}$ where $\mathbf{v}_{rot}$ is the velocity of rotation.

So centrifugal force is usually much larger than Coriolis force:

[tex]\frac{|centrifugal|}{|Coriolis|}\,=\,\frac{v_{rot}}{2v_{rel}}[/itex]

tiny-tim,

I think you are missing a term in your centrifugal force. Regardless, my counter-argument:

I was thinking more in terms of the centrifugal force and the force due to gravity. At the equator, the ratio of the magnitudes of the centrifugal to gravitational force is 0.0034413. This ratio tends to zero as we approach either the north or south poles. So for these two forces, I would say the centrifugal is negligible.

Regarding your assertion, in your final equation, you are missing at least one sine term (there may be more). So in some cases, the centrifugal force will still be negligible. But I do have to say, I could imagine some trajectory in which this would not be the case.

Well played.

 

[cmos]

I was thinking more in terms of the centrifugal force and the force due to gravity. At the equator, the ratio of the magnitudes of the centrifugal to gravitational force is 0.0034413. This ratio tends to zero as we approach either the north or south poles. So for these two forces, I would say the centrifugal is negligible.

I just want to add, we are talking here about the specific example of an intercontinental ballistic missile traversing the surface of the Earth.

The ratio I mentioned, which monotonically decreases as we approach the poles, helps to explain the phenomenon known as equatorial budge. Clearly not necessarily negligible (depending on the problem).

 

[cabraham]

So what is the consensus? I think we can say that those who legitimize cf do so as follows. In a rotating ref frame, cf is necessary to cancel another force, say gravity, in order to satisfy the sum of forces going to zero.

With the moon orbiting the earth, if we adopt the moon as a ref frame, then from that perspective the moon is attracted to the Earth via gravity. But it doesn't accelerate towards the earth, leaving one to conclude that there must be an equal and opposite force to that of gravity to maintain static equilibrium, i.e. forces summing to zero. Hence, cf is conceptualized.

But who says that the gravitational force is "real" in the moon's ref frame? Here in NE Ohio, there is a city called Sandusky, with an amusement park named "Cedar Point". Maybe you know of it. A ride known as the "Demon Drop" features a platform that the passengers are loaded on to. Then the platform literally drops in free fall, the platform and its passengers accelerate downward at near 1 g.

Let us adopt the free falling platform as our ref frame, an accelerated one at that. If I hold a ball in my hand and release it when the platform is at rest at the top, the ball falls to the floor. Gravity is real and present in this static ref frame. But if the platform commences free fall, and I release the ball in descent, in my accelerated frame of ref, the ball floats in mid-air and does not fall to the floor. Of course to a stationary observer on the ground, the platform, passengers, and ball are all falling downward at 1 g and gravity is active. But in the accelerated ref frame the ball hangs in mid-air. In order to cancel gravity, there must be a "counter gravity force" present to account for the non-motion of the ball. Thus can we conceptualize a "counter gravity" force present in the accel ref frame of the platform?

Or, is it that in the accel ref frame, there is no gravity? The passengers are weightless, and the ball hangs in mid-air. In this frame, gravity is NOT present. Hence there is no counter gravity present either. There is no need to balance the force of gravity as it does not exist in the platform frame of ref.

With the moon or geosat orbiting the earth, many presume that since the geosat does not accelerate towards the Earth and that gravity is present which would force it to do so, that there MUST BE another force opposing gravity, hence centrifugal. But, if I'm on the geosat, and let go of a ball while orbiting the earth, it does not "fall" in my ref frame. Hence there is no gravity in my accel frame of ref. Since there is no gravity, there is no counter force to gravity, the net sum of the forces on the ball in this ref frame is already zero. No gravity, no centrifugal force.

Centrifugal force does not exist.

 

[cmos]

With the moon or geosat orbiting the earth, many presume that since the geosat does not accelerate towards the Earth and that gravity is present which would force it to do so, that there MUST BE another force opposing gravity, hence centrifugal. But, if I'm on the geosat, and let go of a ball while orbiting the earth, it does not "fall" in my ref frame. Hence there is no gravity in my accel frame of ref. Since there is no gravity, there is no counter force to gravity, the net sum of the forces on the ball in this ref frame is already zero. No gravity, no centrifugal force.

Centrifugal force does not exist.

Wrong again. If you are standing on the geosynchronous satellite, then we can say that it is the gravitational force between you and the satellite that keeps you on the satellite (obviously this will be balanced by the normal force since you are not accelerating with respect to the satellite).

So now you take a ball out of your pocket and let it go. I will thus fall to the satellite due to gravity.

Do you still seriously believe that there is no centrifugal force at all or was your last sentence relative to the specific example you erroneously were talking about?

 

[D H]

So what is the consensus? I think we can say that those who legitimize cf do so as follows. In a rotating ref frame, cf is necessary to cancel another force, say gravity, in order to satisfy the sum of forces going to zero.

No. Nobody is saying the centrifugal force is a "real" force. It is a pseudo-force that appears when doing physics from the perspective of a rotating frame and trying to ascribe Newton's second law to describe motion as observed from the perspective of that frame.

From the very end of your post,

Centrifugal force does not exist.

It is a fictitious force. That does not mean it does not exist. It is a force that depends only on the observer, and fictitious forces can appear to be very real to those observing things from the perspective of a rotating frame. This is, I think, why you are stuck on the Coriolis force being a "real" force. It certainly appears to be real! It is not a real force. It, like the centrifugal force, arises solely from the perspective of the observer rather than from the physical envifornment.

On to gravity ...

Let us adopt the free falling platform as our ref frame, an accelerated one at that.

This is your key mistake. The free-falling platform is an inertial frame from the perspective of general relativity. You may not like that definition, but using it gives more accurately describes orbits (particularly Mercury's orbit) than does Newtonian physics.

Gravity is real and present in this static ref frame.

Your "static ref frame" is not an inertial frame.

 

[cabraham]

The gravity you refer to is that between the ball and the geosat. The mass of the geosat is tiny vs. that of the earth. The ball will be attracted to the geosat, but the force is very very small. The ball does not feel the Earth's gravity. Of course the gravity between me and the geosat is there, as well as the normal force. The normal force balances the local geosat gravity. The forces sum to zero WITHOUT introducing centrifugal force.

I believe this sums it up. The Earth's gravity is not felt on the geosat ref frame. There is a local gravity field, but not the earth's. The downward force of Earth gravity in the geosat frame of ref is ZERO. The upward centrifugal force relative to the geosat frame of ref is ZERO.

There is no centrifugal force to an observer on the geosat moving with it.

There is no centrifugal force to a stationary observer.

There is no centrifugal force.

 

[rcgldr]

Getting back to the OP:

Lets say I have an object who moves in a circular path, we know the object is accelerating because the velocity is constantly changing direction, the acceleration is towards the center.

Now, where does the centrifugal force comes to play?

Instead of centrifugal force, call it "reaction force to centripetal force", since the meaning of "centrifugal force" was changed sometime in the past.

Is it equal to f = m*w^2*R = m*v^2/r ?
Does it comes from Newton 3rd law to the centripetal force?

Yes, it's simply Newtons 3rd law about action + reaction forces. Something applies a force to an object, and the object responds with a reaction force applied to that "something". Doesn't matter what the direction of the force is (parallel to path, perpendicular to path or some combination).

 

[jostpuur]

Now, where does the centrifugal force comes to play?

Instead of centrifugal force, call it "reaction force to centripetal force", since the meaning of "centrifugal force" was changed sometime in the past.

This was not a good reply. What is the "it" supposed to mean? If it meant the centrifugal force is modern sense in the original post, then you shouldn't call it a reaction force. The truth probably is, that asi123 didn't know what he was meaning with the centrifugal force, and that's why he was asking. But we should respond by explaining what the term "centrifugal force" means (with modern conventions), and not by what something else means.

 

[cmos]

The gravity you refer to is that between the ball and the geosat. The mass of the geosat is tiny vs. that of the earth. The ball will be attracted to the geosat, but the force is very very small. The ball does not feel the Earth's gravity. Of course the gravity between me and the geosat is there, as well as the normal force. The normal force balances the local geosat gravity. The forces sum to zero WITHOUT introducing centrifugal force.

This is more or less what I said, ok. But the relative masses can be whatever you want them to be, within reason I suppose. Also, the ball not feeling the Earth's gravity, I just want to point out that this would be an approximation. Valid in the same sense that I can approximate that I moon's gravity does not directly affect me.

I believe this sums it up. The Earth's gravity is not felt on the geosat ref frame. There is a local gravity field, but not the earth's. The downward force of Earth gravity in the geosat frame of ref is ZERO. The upward centrifugal force relative to the geosat frame of ref is ZERO.

This is completely false. Go study Newton's law of gravitation.

There is no centrifugal force to an observer on the geosat moving with it.
There is no centrifugal force to a stationary observer.
There is no centrifugal force.

At this point, are you just doing this to waste people's time?

 

[jostpuur]

Let us adopt the free falling platform as our ref frame, an accelerated one at that. If I hold a ball in my hand and release it when the platform is at rest at the top, the ball falls to the floor. Gravity is real and present in this static ref frame. But if the platform commences free fall, and I release the ball in descent, in my accelerated frame of ref, the ball floats in mid-air and does not fall to the floor. Of course to a stationary observer on the ground, the platform, passengers, and ball are all falling downward at 1 g and gravity is active. But in the accelerated ref frame the ball hangs in mid-air. In order to cancel gravity, there must be a "counter gravity force" present to account for the non-motion of the ball. Thus can we conceptualize a "counter gravity" force present in the accel ref frame of the platform?

The formula for the effective force in a non-inertial frame is this:

$$\vec{F}_{\textrm{eff}}\; =\; \vec{F} \;-\; m\ddot{R} \;-\; m\dot{\vec{\omega}}\times\vec{r} \;-\; m\vec{\omega}\times(\vec{\omega}\times\vec{r}) \;-\; 2m\vec{\omega}\times\vec{v}$$

(copied from the page 392 of the Classical Dynamics of Particles and Systems, fifth edition, by Thorton and Marion.)

Now the term $$-m\ddot{R}$$ is the relevant one for your question. It means that if the coordinate set itself is accelerating, there is a corresponding pseudo force in the opposite direction. So if you choose to think the gravity as a force (so that inertial frames are the ones which are in rest with the surface of the earth), and then choose to use an accelerating frame that falls in a free fall, then yes, there is a pseudo force that you chose to call "counter gravity".

Or, is it that in the accel ref frame, there is no gravity? The passengers are weightless, and the ball hangs in mid-air. In this frame, gravity is NOT present. Hence there is no counter gravity present either. There is no need to balance the force of gravity as it does not exist in the platform frame of ref.

This is the case if you choose to think about the gravity as not a force, but as an alternative definition of the inertial frames.

 

[cabraham]

This is more or less what I said, ok. But the relative masses can be whatever you want them to be, within reason I suppose. Also, the ball not feeling the Earth's gravity, I just want to point out that this would be an approximation. Valid in the same sense that I can approximate that I moon's gravity does not directly affect me.



This is completely false. Go study Newton's law of gravitation.



At this point, are you just doing this to waste people's time?

Not at all, I assure you. Here is a summary.

For a geosat orbiting earth, let's use 2 frames of ref. A stationary observer in ref frame S, and an observer on the geosat revolving around the earth, ref frame R. In the early part of this thread, I was rebuked because I persistently used the stationary S frame for my reference. I was told that there are instances when it is more appropriate to consider ucm from the moving frame R. Also it was emphasized that in the R frame, the geosat feels the force of the Earth's gravity but does not accelerate towards the earth. Therefore it was reasoned that an equal and opposite counter force to gravity must account for the forces on the geosat summing to zero in its own ref frame R. Hence centrifugal force is conceptualized. It actually sounds logical. If Earth gravity pulls the geosat down, yet it stays put in the R frame of ref, centrifugal force must be countering gravity so that the net forces acting on the geosat are zero in ref frame R.

But in an accelerated ref frame, R in this case, the force of Earth gravity is not present. With the "demon drop" amusement park ride, a platform full of passengers is in free fall. They are weightless and objects released from their grip float weightless in space along with them in ref frame R, the free falling platform. In the R frame of ref, no gravity is present, hence the forces on the ball released by a passenger are zero without any force to counter gravity since it is zero.

In the geosat frame of ref R, the ball, the observer, and the geosat have the same speed of zero in frame R. If the observer releases the ball it will float weightless with him and the geosat. In frame R, the forces sum to zero as the Earth gravity force is zero downward, and there is zero upward force as well.

In frame S, the geosat does feel the gravity force. It has an acceleration towards the earth. The net force on the geosat is equal to the centripetal acceleration times the mass in frame S.

Back to frame R, yes cmos the geosat has ITS OWN gravity. It pullsdown on the observer, and a normal force counters this local gravity. The forces sum to zero, local gravity and normal. All is balanced without centrifugal.

There is no centrifugal force in frame R.

There is no centrifugal force in frame S.

There is no centrifugal force period.

Regarding Newton's law of gravitation, of course the Earth attracts the geosat **in frame S**. Newton's law is not being challenged. Just like the free falling platform, there is no gravitation **in frame R**. Hence no counter force. That is the crux.

If there was a centrifugal force, even if only in one of the ref frames, my physics prof Dr. M would have taught me so. Ditto for the other physics profs and ME profs as well. I trust them. They know their stuff. I trust them more than myself. The reason I have such confidence in "my position" is because it is NOT my position, but that of others more learned than me regarding these issues.

Thanks to every one ESPECIALLY THOSE who opposed me. These debates keep all of us sharp and I've been well educated. I know much more now than a few days ago. To those who disagree with me, I extend my hand in peace and friendship. You've helped me immensely. Have a great day!

Claude

 

[D H]

The formula for the effective force in a non-inertial frame is this:

$$\vec{F}_{\textrm{eff}}\; =\; \vec{F} \;-\; m\ddot{R} \;-\; m\dot{\vec{\omega}}\times\vec{r} \;-\; m\vec{\omega}\times(\vec{\omega}\times\vec{r}) \;-\; 2m\vec{\omega}\times\vec{v}$$

(copied from the page 392 of the Classical Dynamics of Particles and Systems, fifth edition, by Thorton and Marion.)

I already said this in https://www.physicsforums.com/showthread.php?p=1783888#post1783888".

 

[rcgldr]

What is the "it" supposed to mean? If it meant the centrifugal force is modern sense in the original post, then you shouldn't call it a reaction force. The truth probably is, that asi123 didn't know what he was meaning with the centrifugal force, and that's why he was asking. But we should respond by explaining what the term "centrifugal force" means (with modern conventions), and not by what something else means.

Since the OP specifically mentioned Newtons 3rd law, it should be clear that he's interested in reaction force, not the more complicated modern meaning of centrifugal force. Plenty of othres have already given the new modern definition.

I would have preferred that centripetal and centrifugal were used to describe action / reaction pair forces, but I'm not in charge of physics terminology, and there's already a thread about this.

When did centrifugal force die?

 

[jostpuur]

cabraham, frames that move along orbiting planets are either examples of accelerating frames, with the direction of acceleration changing, or examples of frames that are both accelerating and rotating. Using them requires good understanding of pseudo forces. You need to learn pseudo forces with simpler examples first, instead of trying to jump into complicated examples in attempt to find confusion so that you could prove pseudo forces inconsistent.

 

[Doc Al]

Regarding Newton's law of gravitation, of course the Earth attracts the geosat **in frame S**. Newton's law is not being challenged. Just like the free falling platform, there is no gravitation **in frame R**. Hence no counter force. That is the crux.

Of course you are challenging Newton's law of gravity!

Whether you realize it or not, you are assuming the perspective of general relativity, which makes gravity a pseudoforce that goes away in a free falling reference frame (which is an inertial frame in GR). Somehow you have no problem doing away with gravity, but that other pseudoforce--centrifugal force--seems to really throw you.

If you mean to do that, fine. But if you take the ordinary Newtonian view of gravity as a "real" force, then that force exists in every frame.

This is basic stuff described in any mechanics book.

 

[jostpuur]

I already said this in https://www.physicsforums.com/showthread.php?p=1783888#post1783888".

I noticed it earlier, and my first though was to quote your post now, but you had left the $$-m\ddot{R}$$ term out, and only included the rotational terms, so I copied the equation again.

btw. I'm not sure I would recommend Thorton & Marion. I didn't like their way of deriving the expression of pseudo forces. Although on the other hand I don't know a better source at the moment... A little knowledge with Lie groups would allow one to do more easily understandable derivation.

 

[cmos]

btw. I'm not sure I would recommend Thorton & Marion. I didn't like their way of deriving the expression of pseudo forces. Although on the other hand I don't know a better source at the moment... A little knowledge with Lie groups would allow one to do more easily understandable derivation.

I'm not sure if it has been mentioned already, but I swear by Fowles and Cassiday.

 

[cmos]

I would have preferred that centripetal and centrifugal were used to describe action / reaction pair forces, but I'm not in charge of physics terminology, and there's already a thread about this.

Sigh... we all should have just joined a thread arguing about what and what not to call a planet...

 

[cmos]

But in an accelerated ref frame, R in this case, the force of Earth gravity is not present. With the "demon drop" amusement park ride, a platform full of passengers is in free fall. They are weightless and objects released from their grip float weightless in space along with them in ref frame R, the free falling platform. In the R frame of ref, no gravity is present, hence the forces on the ball released by a passenger are zero without any force to counter gravity since it is zero.

As jostpuur mentioned, you are taking a relativistic viewpoint. From the classical viewpoint, clearly there is gravity (otherwise the "demon drop" would just be "demon"... without the drop). The fact that the ball floats in the air is represented by the fact that gravity is balanced by a fictious inertial force. No special name here, just some fictious inertial force (Fint, or Ffict, or Fwithwhateversubscriptyouwanttogiveit if you will).

 

[D H]

I noticed it earlier, and my first though was to quote your post now, but you had left the $$-m\ddot{R}$$ term out, and only included the rotational terms, so I copied the equation again.

Well, dang. That was a f*up, pure and simple, particular since I said accelerating frame.

 

[cabraham]

Good grief! In my early posts I was being "too Newtonian" and needed to look at GR. Now I'm to GRian and need to study Newton! If there was no gravity the demon wouldn't drop! I know that! But you are all now doing what I did earlier, i.e. considering only the static S ref frame and ignoring the accelerated frame R.

No one is disputing gravity! Or Newton. Or GR. You folks are trying to have it both ways. You acknowledge gravity in the S frame, then copy and paste it to the R frame and then create a fictitious or "inertial" force to balance it out. ON the demon drop, passengers in the R ref frame are weightless and the ball floating in mid air is not inertial force cancelling g force, but rather the forces acting on said ball do not balance. Since there is no force to counter gravity, the ball incurs an acceleration sum F = ma. Is that Newtonian enough for you? No GR here. But in the R frame is there gravity? The passengers are weightless and objects released just float in mid air. In the R frame there is no gravity/g-force. Of course to a staionary observer on the ground in the S frame there is gravity. That is why the platform drops. To a ground observer in the S frame the force of gravity acts on the platform, passengers, and ball WITH NO COUNTER FORCE. Thus the system accelerates downward.

If we wish to stay in the S frame, fine, we'll use S. If you prefer R, be my guest. Let us be careful not to mix them.

Regarding Doc Al's "gravity must exist in every frame" per Newton, does it follow that centrifugal must exist in every frame as well, as it surely does not in the S frame? Just wondering if the rules you state are consistent. Best regards and thanks to all for the mentally stimulating and exciting friendly discussion.

 

[jostpuur]

Good grief! In my early posts I was being "too Newtonian" and needed to look at GR. Now I'm to GRian and need to study Newton! If there was no gravity the demon wouldn't drop! I know that! But you are all now doing what I did earlier, i.e. considering only the static S ref frame and ignoring the accelerated frame R.

No one is disputing gravity! Or Newton. Or GR. You folks are trying to have it both ways. You acknowledge gravity in the S frame, then copy and paste it to the R frame and then create a fictitious or "inertial" force to balance it out. ON the demon drop, passengers in the R ref frame are weightless and the ball floating in mid air is not inertial force cancelling g force, but rather the forces acting on said ball do not balance. Since there is no force to counter gravity, the ball incurs an acceleration sum F = ma. Is that Newtonian enough for you? No GR here. But in the R frame is there gravity? The passengers are weightless and objects released just float in mid air. In the R frame there is no gravity/g-force. Of course to a staionary observer on the ground in the S frame there is gravity. That is why the platform drops. To a ground observer in the S frame the force of gravity acts on the platform, passengers, and ball WITH NO COUNTER FORCE. Thus the system accelerates downward.

If we wish to stay in the S frame, fine, we'll use S. If you prefer R, be my guest. Let us be careful not to mix them.

We can think that the gravity acts as a force, and use a stationary frame S (which is inertial).

We can think that the gravity acts as a force, and use the free fall frame R (which is not inertial).

We can think that the gravity acts through an alternative definition of the inertial frame, and use the stationary frame S (which is not inertial).

We can think that the gravity acts through an alternative definition of the inertial frame, and use the free fall frame R (which is inertial).

You see, there is four alternatives, in this sense. The choice of the frame does not yet fix the way we think about the gravity. It does not make sense to merely ask question like "does the gravitation force exist in the frame R". The question doesn't have unique answer. There is nobody mixing these two frames, but you have mixed these four alternatives into smaller amount.

 

[Doc Al]

Good grief! In my early posts I was being "too Newtonian" and needed to look at GR. Now I'm to GRian and need to study Newton! If there was no gravity the demon wouldn't drop! I know that! But you are all now doing what I did earlier, i.e. considering only the static S ref frame and ignoring the accelerated frame R.

Good grief, indeed! You are all over the map.

No one is disputing gravity! Or Newton. Or GR. You folks are trying to have it both ways. You acknowledge gravity in the S frame, then copy and paste it to the R frame and then create a fictitious or "inertial" force to balance it out.

Realize that Newton's law of gravity is not revoked depending upon what frame you view things from. In the Newtonian view, gravity is a "real" force--meaning: it has an agent--not a pseudoforce that only exists in non-inertial frames.

ON the demon drop, passengers in the R ref frame are weightless and the ball floating in mid air is not inertial force cancelling g force, but rather the forces acting on said ball do not balance. Since there is no force to counter gravity, the ball incurs an acceleration sum F = ma. Is that Newtonian enough for you? No GR here. But in the R frame is there gravity? The passengers are weightless and objects released just float in mid air. In the R frame there is no gravity/g-force. Of course to a staionary observer on the ground in the S frame there is gravity. That is why the platform drops. To a ground observer in the S frame the force of gravity acts on the platform, passengers, and ball WITH NO COUNTER FORCE. Thus the system accelerates downward.

Don't confuse the colloquial expression "weightless" with meaning "no gravity". Jump out the window and (ignoring air resistance) you are briefly "weightless", but the force of gravity still acts. (The unfortunate term "weightless" means that your apparent weight is zero, not your actual weight. Your apparent weight is what a scale would read if you stood on it. Your actual weight is the force of gravity, as given by Newton's law of gravity.)

If we wish to stay in the S frame, fine, we'll use S. If you prefer R, be my guest. Let us be careful not to mix them.

No one is "mixing frames" but you. You seem to treat gravity as a real force when it suits you, and as a pseudoforce when it doesn't.

Regarding Doc Al's "gravity must exist in every frame" per Newton, does it follow that centrifugal must exist in every frame as well, as it surely does not in the S frame? Just wondering if the rules you state are consistent.

The rules are consistent and simple: real forces (forces that have agents, not pseudoforces) exist in all frames. In the Newtonian view, gravity is a real force. The ball is pulled by the earth. On the other hand, pseudoforces--like centrifugal force--only exist in non-inertial frames.

 

[cabraham]

Good grief, indeed! You are all over the map.

Realize that Newton's law of gravity is not revoked depending upon what frame you view things from. In the Newtonian view, gravity is a "real" force--meaning: it has an agent--not a pseudoforce that only exists in non-inertial frames.

Don't confuse the colloquial expression "weightless" with meaning "no gravity". Jump out the window and (ignoring air resistance) you are briefly "weightless", but the force of gravity still acts. (The unfortunate term "weightless" means that your apparent weight is zero, not your actual weight. Your apparent weight is what a scale would read if you stood on it. Your actual weight is the force of gravity, as given by Newton's law of gravity.)

No one is "mixing frames" but you. You seem to treat gravity as a real force when it suits you, and as a pseudoforce when it doesn't.

The rules are consistent and simple: real forces (forces that have agents, not pseudoforces) exist in all frames. In the Newtonian view, gravity is a real force. The ball is pulled by the earth. On the other hand, pseudoforces--like centrifugal force--only exist in non-inertial frames.

Well then based on what you just said, gravity is real, and centrifugal is pseudo, your words verbatim. In ucm, such as the moon or geosat orbiting the earth, the centripetal is the gravity, which you include in both R and S ref frames. But you admit the the centrifugal is a pseudo force. So then, can I assume that you give gravity, and centripetal which in this example is due to gravity, more "weight" (no pun intended) or better yet "recognition" than centrifugal. In other words centripetal is an active force included in all frames of ref, and it shows up in free body diagrams. The same cannot be said for centrifugal. That is what I read into your words.

Regarding the demon drop, if g force is present in the R frame, should we then conclude that an equal but opposite force counters it, known as "inertial force"? I'm just trying to summarize what has been stated. I'm not trying to be difficult or play "devils advocate" but the basis for recognizing inertial force is to counter g force so that the forces sum to zero in the R frame.

So now I think we have this entire issue down to one concept. In the R frame is there a g force? The answer to that question determines whether or not there is a need for an "inertial" or "centrifugal" force to counter it so that the summation of forces is zero.

Not to annoy anyone by rehashing the demon drop, but we haven't addressed normal force. When I'm standing on the platform before the free fall, the floor exerts a upward normal force on me, countering the downward g force. In free fall, the normal force is gone. During free fall if the g force is present in the R frame, without the normal force, then what keeps me balanced? Let me guess, the inertial force? So the normal force disappeared and was replaced with inertial force. I'm only trying to summarize all that has been stated. All I ask is that the scenario just mentioned be explained in terms of all forces in both R and S frames of ref.

My questions are only put forward with the intention of rigorously examining the physics involved in these scenarios, and not to annoy or enrage. The responses I've been getting are pretty hostile. BR.

Claude

 

[cmos]

Well then based on what you just said, gravity is real, and centrifugal is pseudo, your words verbatim. In ucm, such as the moon or geosat orbiting the earth, the centripetal is the gravity, which you include in both R and S ref frames. But you admit the the centrifugal is a pseudo force. So then, can I assume that you give gravity, and centripetal which in this example is due to gravity, more "weight" (no pun intended) or better yet "recognition" than centrifugal. In other words centripetal is an active force included in all frames of ref, and it shows up in free body diagrams. The same cannot be said for centrifugal. That is what I read into your words.

This is more or less correct (see more in my next post). Whether we are analyzing a non-inertial frame or an inertial frame, we still take the stance that we are "enlightened" and we do know all the forces that are prevailing. So when we analyze the non-inertial frame, we realize it is a non-inertial frame and we remember that there are forces that prevail that are inherent to the inertial frame. See the third quote for this to make more sense.

Regarding the demon drop, if g force is present in the R frame, should we then conclude that an equal but opposite force counters it, known as "inertial force"? I'm just trying to summarize what has been stated. I'm not trying to be difficult or play "devils advocate" but the basis for recognizing inertial force is to counter g force so that the forces sum to zero in the R frame.

Yes, conclude it as stated. Correct!

So now I think we have this entire issue down to one concept. In the R frame is there a g force? The answer to that question determines whether or not there is a need for an "inertial" or "centrifugal" force to counter it so that the summation of forces is zero.

Again, we are "enlightened" so we know there is a gravitational force. Being "enlightened" tough, we realize that in the non-inertial frame, it will seem as if there is no gravitational force (note: we assume that the people in the demon drop cannot see outside the box and that they are in fact plummeting to the Earth). So to make this work, we balance the gravitational force by a fictious force.

I think this is they key you are missing. In doing a mechanics problem you are, in a sense, God. You know what is truly happening, and you write the equations of motion for whatever frame you want to analyze.

Not to annoy anyone by rehashing the demon drop, but we haven't addressed normal force. When I'm standing on the platform before the free fall, the floor exerts a upward normal force on me, countering the downward g force. In free fall, the normal force is gone. During free fall if the g force is present in the R frame, without the normal force, then what keeps me balanced? Let me guess, the inertial force? So the normal force disappeared and was replaced with inertial force. I'm only trying to summarize all that has been stated. All I ask is that the scenario just mentioned be explained in terms of all forces in both R and S frames of ref.

This is pretty much correct. The normal force occurs because the Earth's gravity is pulling you downward but the drop box won't let you accelerate through the floor. One the box falls, you no longer feel the effect of the gravitational force because it is balanced by the inertial pseudo-force. So the normal force disappears.

My questions are only put forward with the intention of rigorously examining the physics involved in these scenarios, and not to annoy or enrage. The responses I've been getting are pretty hostile. BR.

In due fairness, I would say that is how many of your post come off as; that's what set me off a few pages back. If you have a question, ask the question. At many times, instead of asking a straightforward question, you come off as stating false physics as fact. Then end with a 3-4 line tyrade which repeats "there is no centrifugal force." This comes off as somewhat childish.

Please don't take offense to this. I'm trying to help everyone out here to make the discussions more scholarly. Also think about the casual websurfer who might happen upon this thread, see you have the last post, and think you are stating physical fact rather than asking a question?

 

[cmos]

Well then based on what you just said, gravity is real, and centrifugal is pseudo, your words verbatim. In ucm, such as the moon or geosat orbiting the earth, the centripetal is the gravity, which you include in both R and S ref frames. But you admit the the centrifugal is a pseudo force. So then, can I assume that you give gravity, and centripetal which in this example is due to gravity, more "weight" (no pun intended) or better yet "recognition" than centrifugal. In other words centripetal is an active force included in all frames of ref, and it shows up in free body diagrams. The same cannot be said for centrifugal. That is what I read into your words.

I just wanted to throw some caution in your claim to the "ever-prevalent" centripetal force. The centripetal force, as well as the normal force, are more of geometrical terms. So yes, in uniform circular motion, there is a "centripetal" (center seeking) acceleration. So for a circular orbit, we must conclude that this centripetal force is in fact the gravitational force.

For the case of a car though, it is the engine that allows for this acceleration; or rather your engine and how your front axle is turned. Turn the axle forward and there is no force acting on the car (except of gravity of course ).

This "geometric" argument is also what gives the centrifugal force its name. The force tends to push a body outward in a rotating frame of reference, hence "centrifugal" (center fleeing). The Coriolis force, on the other hand, deflects the trajectory of a projectile; we might as well have called this the deflecting force (sorry Dr. Coriolis).

 

[D H]

cabraham, why are you harping so on centrifugal force? Everyone has said it is a pseudo-force. A recap:

• Centrifugal force is a pseudo-force that appears only because a rotating observer (i.e., an observer in a non-inertial frame) wants to use Newton's second law to describe observed motions.
  The centrifugal force does not appear only when there is a need to balance forces. This is the freshman physics point-of-view. The centrifugal force arises in a reference frame whenever that reference frame is rotating.
• The Coriolis force is also a pseudo-force. The Coriolis force arises in a rotating reference frame when the object being observed has a non-zero velocity in that frame.
• Gravity is a real force in Newtonian mechanics.
• Gravity is a pseudo-force in general relativity. Newtonian mechanics and general relativity disagree on the underlying concept of what constitutes an inertial reference frame and on the spatial extent of an inertial reference frame.

Note well: Just because these pseudo-forces are not real (have no causative agent) does not mean they do not exist. They certainly do exist in the eye of the observer.

 

[Doc Al]

I agree with what D H and cmos haved deftly explained (yet again), but here's my response just the same.

Well then based on what you just said, gravity is real, and centrifugal is pseudo, your words verbatim.

Yes. I've practically been shouting this.

In ucm, such as the moon or geosat orbiting the earth, the centripetal is the gravity, which you include in both R and S ref frames. But you admit the the centrifugal is a pseudo force. So then, can I assume that you give gravity, and centripetal which in this example is due to gravity, more "weight" (no pun intended) or better yet "recognition" than centrifugal. In other words centripetal is an active force included in all frames of ref, and it shows up in free body diagrams. The same cannot be said for centrifugal. That is what I read into your words.

Sounds right. But don't get hung up on the term "centripetal". It just means "acting towards the center". Centripetal force is not a separate force unto itself, it's just a catch-all term for whatever is making something move in a circle. In the case of the moon orbiting the earth, gravity provides the centripetal force; in the case of ball on a string being twirled in a horizontal circle, string tension provides the centripetal force. But those forces are "real": they have agents.

Regarding the demon drop, if g force is present in the R frame, should we then conclude that an equal but opposite force counters it, known as "inertial force"? I'm just trying to summarize what has been stated. I'm not trying to be difficult or play "devils advocate" but the basis for recognizing inertial force is to counter g force so that the forces sum to zero in the R frame.

Good!

So now I think we have this entire issue down to one concept. In the R frame is there a g force? The answer to that question determines whether or not there is a need for an "inertial" or "centrifugal" force to counter it so that the summation of forces is zero.

In all frames there is a gravitational force.

Not to annoy anyone by rehashing the demon drop, but we haven't addressed normal force. When I'm standing on the platform before the free fall, the floor exerts a upward normal force on me, countering the downward g force.

Right.

In free fall, the normal force is gone.

Right. You are accelerating.

During free fall if the g force is present in the R frame, without the normal force, then what keeps me balanced? Let me guess, the inertial force? So the normal force disappeared and was replaced with inertial force.

If you choose to analyze your motion from the non-inertial frame of the falling platform, then you must include the inertial forces. Realize that a normal force is a "real" force--either it's there or it isn't in all frames.

I'm only trying to summarize all that has been stated. All I ask is that the scenario just mentioned be explained in terms of all forces in both R and S frames of ref.

Done.

Here's another scenario you might find interesting to analyze. Imagine a railroad car with an object tied to a string hanging from the ceiling of the car. The train accelerates. What angle does the string make with the vertical?

From the inertial frame of the tracks, the object is accelerated. The only forces acting on the object are the string tension and the weight. Both of these forces are "real". Newton's 2nd law applies without modification.

To analyze the situation from the non-inertial frame of the railroad car, one must modify Newton's laws by adding in an inertial pseudoforce. Otherwise you get nonsense. The inertial force is just an artifact of analyzing things from a non-inertial reference frame. (The "real" forces still exist in either frame, of course.)

 

[cabraham]

I understand where you're coming from, but still see a problem. So when the demon drop platform and passengers are in free fall, the normal force disappears, and in Doc Al's words because "You are accelerating". Not to nitpick but that is where I see the problem. I have no issue so far. But, if I am accelerating, then we don't need a pseudo force "inertial" to balance gravity. Were my physics and ME profs wrong when they taught me that the summation of forces goes to zero when a body is in static equilibrium, i.e. at rest. But if the body is "accelerating", which I have no dispute with, why do we need a counter force to gravity? The inertial counter or pseudo force is brought into the picture to balance out the gravity to keep the summation of forces at zero. Again, was I taught wrong that the sum of forces is NONZERO in acceleration?

Yes, Doc Al, I am well aware that "centripetal" is a term indicating the direction of a force being radially inward. The part about the string, friction, or gravity providing the centripetal was something I was about to post when you just did. I agree completely.

Maybe now is a good time to check outside resources. I visited MIT physics site, and found a lot about bodies in ucm. Centripetal is discussed in every lecture note, but centrifugal is not mentioned. I asked 2 ME people here at work, BSME education, about it, and they said that centripetal acts inward, and velocity is tangential, but centrifugal is something they don't use. I'll find a Ph.D. physics prof at the university I go to, but school is out for the summer, so it may take a while.

The only thing I'm sure of is what I've believed for years, that accelerated ref frames are very tricky to deal with.

I'll think about the railroad car with the object dangling from a string. But, the gravity in this case is vertical while the train's acceleration is horizontal. The gravity is of course fully present in both R and S ref frames. But in the demon drop, the acceleration is vertical as is gravity, no coincidence.

I've thought about it. When at rest, the dangling object is pulled down by gravity and the string tension counters the g force keeping the object in static equilibrium. When accelerating, the horizontal force is forward coincident with the acceleration, vertical force is still downward due to gravity, and the tension in the string where it is attached counters the horizontal accelerating force and gravity which is vertical. The string angle, I would expect, would be the result of vector summation. Just an off the cuff 2 minute analysis.

Doc Al, further back in this thread, was you stated that when a rock is twirled on a string, the outward force you feel is something you just call as "tension". That is my position precisely. In the train case, I see it as the tension always acting counter to the force. Maybe I, or others, is just making too much out of it and should let it rest.

Good day to all.