Coriolis force caused by tangential velocity

[Swamp Thing]

From another recent thread I learned that you see a Coriolis force if an object in a rotating reference frame moves along a tangent at some velocity v. (I was already familiar with the case where the velocity is radial).

I still find it a little counter-intuitive that the force has the same magnitude irrespective of whether the velocity is directed radially or tangentially. When you move radially towards the center of rotation, you are dashing headlong from a "fast lane" into a "slow lane", so you will soon see a growing mismatch between your own tangential velocity and the local velocity of the surface that you are running over. Fair enough.

But when you move along the tangent, you are following approximately the same direction that the local surface is following anyway, only faster. So you are merely sidling or drifting, in an incidental manner, from a "fast lane" into a "superfast lane" where the local surface velocity is trying to leave you behind.

So intuitively, the Coriolis force in the second example should be an order smaller than in the first. Now obviously, my intuition is wrong -- but is there a simple, correct intuitive picture that will make it at once clear why the two cases produce exactly the same Coriolis force?

 

[Anachronist]

It isn't a force, it's an effect that has the appearance of a force to an observer in the rotating reference frame. To an observer outside that reference frame, no forces are acting on the object other than the usual (centripetal, gravity).

If you sit on a rotating disk and roll a frictionless ball toward the center, you'll see a deflection in its path because your point of view changes with respect to the straight path of the ball. If you roll the ball in the direction of rotation, you'll also see a deflection in its path, because your path diverges from the straight path of the ball. In either case (and everything in between), the deflection appears as if a force were acting on the ball. It's an effect of the reference frame, not an actual force.

In the case of vortices in our atmosphere, a low pressure region attracts air toward it. As the air moves toward it (assuming the pressure region is relatively stationary with respect to the rotation of the planet), it doesn't head directly for the low pressure region but deflects to the right (when viewed from above in the northern hemisphere) no matter what direction from which the air approaches the low pressure region. Air converging in all directions deflected to the right will create a whorl to the left (counterclockwise) around the low-pressure zone.

 

[Dale]

my intuition is wrong -- but is there a simple, correct intuitive picture that will make it at once clear why the two cases produce exactly the same Coriolis force?

Since your intuition is wrong I don’t think that a correct picture will possibly be intuitive. So I think you should focus on simple and correct, discarding intuitive.

The simplest correct explanation is simply the formula for the Coriolis force: $-2m \mathbf{\Omega} \times \mathbf v$. Since this cross product is the same in both cases the acceleration is the same in both cases.

 

[A.T.]

So you are merely sidling or drifting, in an incidental manner,...

The reason for the radial Coriolis force has nothing to do with sidling or drifting.

One intuitive way to think about the radial Coriolis force, is as a velocity dependent modification of the centrifugal force. In fact you could order the inertial force terms in a rotating frame by direction, and lump the radial Coriolis force with radial centrifugal force. But for mathematical and historical reasons we separate them as position dependent term (centrifugal) and a velocity dependent term (Coriolis).

 

[Swamp Thing]

One intuitive way to think about the radial Coriolis force, is as a velocity dependent modification of the centrifugal force.

Ok, that's a nice way to make sense of it.

 

[Swamp Thing]

And I see how I mis-analyzed the second scenario. The acceleration / force due to "lane changing" is not only an order of magnitude less than Case-1, it's actually zero. But a different phenomenon has now kicked in, namely "modified centrifugal force" which can be interpreted as a new force with the same magnitude, different direction.

What is cool is that the same elegant cross product formula encompasses both of these phenomena (which, after all, belong to the same general class of inertial forces).

 

[FactChecker]

Something going tangentially is moving toward an increasing height and simultaneously falling behind the radius that it would have been at if it was traveling in a circle. It's not clear to me how the two effects can be intuitively summed, but it may not be so surprising that the total magnitude of the deviation is the same as the example that you intuitively accept.

 

[Dr.D]

The Coriolis "force" is not at all. It is an "inertial reaction" term, that is a mass x acceleration term.

Recall Newton's 3rd law about every FORCE having an equal and opposite reaction force? Well, where is the reaction to the "Coriolis Force"? It cannot be found because it does not exist.

 

[jbriggs444]

Try this intuition on for size...

If you have a fixed velocity that you are measuring against rotating directions then that velocity will appear to rotate anti-spinward. That rotation amounts to an acceleration at right angles to the current direction. The higher the velocity, the greater the required acceleration. The higher the rotation rate, the greater the required acceleration.

 

[FactChecker]

Try this intuition on for size...

If you have a fixed velocity that you are measuring against rotating directions then that velocity will appear to rotate anti-spinward. That rotation amounts to an acceleration at right angles to the current direction. The higher the velocity, the greater the required acceleration. The higher the rotation rate, the greater the required acceleration.

This is my favorite way of looking at the difference between the velocity derivative in a rotating frame versus acceleration due to a force. Suppose there is a coordinate system (XYZ) attached to an airplane pilot's head, where X is straight forward from his head and Y is out his left ear. When he looks forward, the velocity can be Mach 1 in the X direction. When he turns his head to look right, the X velocity becomes 0 and the Y velocity becomes Mach 1. That can happen in a second, so the velocity derivatives are huge. But there is no force and no physical acceleration. This example shows how important it is to distinguish between physical acceleration due to a force and the derivative of velocity.

 

[zanick]

The Coriolis "force" is not at all. It is an "inertial reaction" term, that is a mass x acceleration term.

Recall Newton's 3rd law about every FORCE having an equal and opposite reaction force? Well, where is the reaction to the "Coriolis Force"? It cannot be found because it does not exist.

actually, you could have a force due to Coriolis, that would be the force in opposition in slowing tangentially, a north bound object from the equator. there would have to be a force to slow the object down, or it would break a conservation law. without the "force" there would be an apparent deflection , called Coriolis.

 

[Dr.D]

Every force has a reaction. Where is the reaction in your example, @zanick ?

 

[zanick]

Every force has a reaction. Where is the reaction in your example, @zanick ?

assume have a 1000mph bullet train going north bound from the equator.
the force of the "train" acting on the east side of the rail and the rail acting back on the train. suddenly, a fictitious force or deflection is not so "apparent" anymore. ;) similar to centrifugal force of your head hitting the side window of a car in circular path. centripetal force (proper) is the window... centrifugal force ( head hitting window) inertial force/fictitious force. or you can look at it from a conservation of angular momentium... the train going north spins up the earth, requiring a force

 

[FactChecker]

Due to the Coriolis tendency to drift, there may be a force required to prevent an object from drifting, just like a centrifugal force has an associated centripetal force to keep an object "in place" in the rotating coordinate system. The "equal and opposite" would be seen in the infinitesimal change in the rotation of the Earth.

 

[Swamp Thing]

... there would have to be a force to slow the object down, or it would break a conservation law.

... there may be a force required to prevent an object from drifting ... The "equal and opposite" would be seen in the infinitesimal change in the rotation of the Earth.

These remarks dovetail very nicely with something that I've come to understand in the last couple of days, as a result of thinking about Veritasium's video on the Dzhanibekov Effect.

If we have a free rigid body that is rotating with no external torque, then of course angular momentum has to be conserved. However, there can be "internal" off-axis torques due to unbalanced centrifugal forces, akin to the dynamic balancing problem -- e.g. when balancing crankshafts and vehicle wheels. These torques will try to shift the orientation of the object with respect to the angular momentum vector. When that happens, the moment of inertia around the direction of angular momentum could (e.g.) increase, which would require a decrease in angular velocity in order to conserve angular momentum.

Moreover, an individual component of the system may happen to move towards the angular momentum axis, while another may move away from it, requiring a redistribution of energy/momentum from one element to another.

So in a rigid freely evolving system, Coriolis forces are merely the mechanisms that mediate the change in angular velocity and the internal redistribution of energy described above. The reaction forces -- Dr D asked about those -- those are just the structural rigidity of the system trying to prevent elastic distortions within the system. If we imagine the system as a massless wireframe with point masses attached to certain nodes with strong rubber bands, then the Coriolis reaction forces are the rubber band forces that remain after you factor out the "conventional" centrifugal forces.

Through those rubber bands, the point masses are all constantly "negotiating" with each other (sending forces/torques to each other) to collectively determine how the system is going to move, such that each element has to swerve as little as possible from its constant-velocity motion, while still respecting Newton's laws and global conservation of angular momentum. A heavy point mass has more "say" in the negotiation that a light one, for example. This negotiation aspect is reflected in certain simulation algorithms by solving nonlinear simultaneous equations to determine what has to happen during a small time step.

Edit: I am not too happy with the way I have expressed this:

However, there can be "internal" off-axis torques due to unbalanced centrifugal forces...

but it will have to do for the moment.

 

[Dale]

the force of the "train" acting on the east side of the rail and the rail acting back on the train.

Neither of those are the Coriolis force. @Dr.D is correct, inertial forces do not follow Newton’s 3rd law.

 

[zanick]

I agree... and also, I didnt say that both of them were "Coriolis force".. that's how you know you are in a non -inertial reference frame, as things don't follow Newton mechanics . I think we would agree, that Coriolis is an apparent deflection not a force for the reason you mentioned. so, what would that force be called if you prevented the apparent deflection with a force?

 

[Dale]

what would that force be called if you prevented the apparent deflection with a force?

Depends on the nature of the real force. If it were a contact force then I would call it a contact force. If it were a tension then I would call it a tension. If it were an electromagnetic force then I would call it an electromagnetic force. It doesn’t have a special name.

I didnt say that both of them were "Coriolis force"..

You sort of did. You listed them as an answer to the question about what is the reaction to the Coriolis force. Since those two forces are reaction forces to each other then it sure seems like you are describing one of them as the Coriolis force.

 

[zanick]

Due to the Coriolis tendency to drift, there may be a force required to prevent an object from drifting, just like a centrifugal force has an associated centripetal force to keep an object "in place" in the rotating coordinate system. The "equal and opposite" would be seen in the infinitesimal change in the rotation of the Earth.

agreed... for example, an airliner has to decal laterally only 1.5mph per min while traveling north bound.. very small force guiding the plane trimmed to the west, but still a force.. technically, what would you call that force... as you mentioned, centripical force could be analogous to this...… "anti Coriolis force" ? ;)

 

[jbriggs444]

technically, what would you call that force

Why must it have a name?

 

[Swamp Thing]

Why must it have a name?

Indeed! "The Force that can be named is not the True Force"

 

[zanick]

Why must it have a name?

to help explain it best. for example... centrifugal force is not a true force, but its called "centrifugal"... sure we call these forces , inertial , or pseudo, 'not true force" , etc. the term" Coriolis " force is used on occasion, can we feel good about using it here?

 

[zanick]

Depends on the nature of the real force. If it were a contact force then I would call it a contact force. If it were a tension then I would call it a tension. If it were an electromagnetic force then I would call it an electromagnetic force. It doesn’t have a special name.

You sort of did. You listed them as an answer to the question about what is the reaction to the Coriolis force. Since those two forces are reaction forces to each other then it sure seems like you are describing one of them as the Coriolis force.

Thanks Dale... I just want to be as correct as possible , especially when discussing the roots of terms with my son who is in his first year of physics now. I like your comment. …. its more of a "reaction" to Coriolis and could be a" contact force" in my example. but, as I have always understood it, it is an apparent deflection in the non inertial reference frame. the point was, could it be called" corilos", similarly to how centrifugal is used.

 

[jbriggs444]

to help explain it best. for example... centrifugal force is not a true force, but its called "centrifugal"... sure we call these forces , inertial , or pseudo, 'not true force" , etc. the term" Coriolis " force is used on occasion, can we feel good about using it here?

But what is it that you want to explain?

You want to explain the nature of the force that an airplane pilot needs to command in order to keep his plane on a "straight" course? Surely that comes under the heading of aerodynamics and control systems rather than under the heading of physics.

 

[vanhees71]

I'd not call inertial forces "fictitious", but that's another story...

 

[zanick]

But what is it that you want to explain?

You want to explain the nature of the force that an airplane pilot needs to command in order to keep his plane on a "straight" course? Surely that comes under the heading of aerodynamics and control systems rather than under the heading of physics.

not really, the reaction to those forces would be what a pilot controls . the cause could be Coriolis or it could be the jet stream... coincidentally those "forces" would be pretty similar in magnitude and direction on the aircraft flying north bound from the equator. the question of what we are trying to explain, IS the cause and effect. what's wrong with being as accurate and descriptive as possible for a more complete understanding? just the term "pseudo force", or "non force" is not well understood . its not always easy to explain a force that is reference frame dependent.

 

[zanick]

I'd not call inertial forces "fictitious", but that's another story...

do tell... can you summarize... I don't usually , but when I hear it ,I don't disagree. should I? :)

 

[Dr.D]

assume have a 1000mph bullet train going north bound from the equator.
the force of the "train" acting on the east side of the rail and the rail acting back on the train. suddenly, a fictitious force or deflection is not so "apparent" anymore. ;)

If there is a load on the wheel flange, it is a constraint force.

 

[Dr.D]

could it be called" corilos", similarly to how centrifugal is used

When applying Newton's Second Law (Sum F = m*a), neither a "coriolis force" nor a "centrifugal force" belong on the left side of the equation. They belong on the right side.

 

[zanick]

If there is a load on the wheel flange, it is a constraint force.

which is a variant of a "contact" force , correct? and, could or would you call it "due to Coriolis"?

 

[Dr.D]

I would simply call a constraint force by that name and not try to attribute it to any particular source other than the fact that it constrains (guides) the motion.

 

[zanick]

When applying Newton's Second Law (Sum F = m*a), neither a "coriolis force" nor a "centrifugal force" belong on the left side of the equation. They belong on the right side.

yes, I understand, and is the reason why I though using that naming convention logic that calling it" Coriolis force" would be appropriate. Using the example of the train and the "constraining " force caused by Coriolis, there would be no net force as there would be no acceleration or velocity to the east traveling north. Unconstrained... in the non inertial frame of refence, you would see the "m" and an "a" to the east, following Newton, wouldn't there have to be an "F"?

 

[vanhees71]

do tell... can you summarize... I don't usually , but when I hear it ,I don't disagree. should I? :)

Well, in a non-inertial frame you have inertial forces. They are defined by writing $\vec{F}=m \vec{a}$ also in the non-inertial frame. Then in addition to the "true forces" (i.e., some mediated by the electroweak or strong interaction) you have to add the inertial forces on the left-hand side of this equation. They are then not fictitious in any way.

If you work in a general covariant way, of course, no such forces belong to the left-hand side but occur on the right-hand side of the equation, but I don't know any textbook that takes this point of view. All I know lump the inertial forces to the left-hand side and thus they call them forces.

You can argue about gravity. In Newtonian physics you consider it a "true force", in relativistic physics you have to use general relativity, and there gravitational (local!) effects on a test particle are usually reinterpreted as the effect of free motion in curved spacetime. In this sense you cannot distinguish between gravitational interaction of a test particle and inertial forces in a sufficiently small spacetime region. Would you then call gravity a "fictitious forces", because you do so for the inertial forces in Newtonian physics (or special relativity) in accelerated reference frames?

This is of course all no problem, because you always end up with the same equations of motion to solve.

 

[zanick]

I would simply call a constraint force by that name and not try to attribute it to any particular source other than the fact that it constrains (guides) the motion.

Thank you... sorry it took so long to get to this, but if that is the best naming convention, ill use that from now on to avoid error and/or confusion. I've been using "apparent deflection due to Coriolis " and rarely "Coriolis force" ill stop using the later. Now the challenge is to describe how the "apparent" deflection is actually real, if you are looking at it in the non inertial reference frame. ;)

 

[zanick]

Well, in a non-inertial frame you have inertial forces. They are defined by writing $\vec{F}=m \vec{a}$ also in the non-inertial frame. Then in addition to the "true forces" (i.e., some mediated by the electroweak or strong interaction) you have to add the inertial forces on the left-hand side of this equation. They are then not fictitious in any way.

If you work in a general covariant way, of course, no such forces belong to the left-hand side but occur on the right-hand side of the equation, but I don't know any textbook that takes this point of view. All I know lump the inertial forces to the left-hand side and thus they call them forces.

You can argue about gravity. In Newtonian physics you consider it a "true force", in relativistic physics you have to use general relativity, and there gravitational (local!) effects on a test particle are usually reinterpreted as the effect of free motion in curved spacetime. In this sense you cannot distinguish between gravitational interaction of a test particle and inertial forces in a sufficiently small spacetime region. Would you then call gravity a "fictitious forces", because you do so for the inertial forces in Newtonian physics (or special relativity) in accelerated reference frames?

This is of course all no problem, because you always end up with the same equations of motion to solve.

GREAT, thanks for that ^^^^^^ it reinforces the correct way of looking at the relationships.

 

[jbriggs444]

the reaction to those forces would be what a pilot controls

We usually reserve the term "reaction" for 3rd law force pairs. These are not third law force pairs. These are "second law force pairs" -- equal and opposite forces that result in zero acceleration for a test body.

A common instance of a "second law force pair" would be your weight and the support force from your bathroom scale.

 

[zanick]

that makes sense as well

 

[Dale]

I just want to be as correct as possible , especially when discussing the roots of terms with my son who is in his first year of physics now.

OK, so to be as correct as possible:
1) The Coriolis force is an inertial force.
2) Inertial forces are also called fictitious forces, although like @vanhees71 I avoid the term "fictitious force" and use the term "inertial force" as a general rule.
3) Inertial forces exist only in non-inertial reference frames, in this case in a rotating reference frame
4) Inertial forces do not follow Newton's 3rd law
5) Inertial forces do follow Newton's 2nd law and are necessary to describe dynamics in the inertial frame
6) Inertial forces are always proportional to the mass
7) Inertial forces cannot be detected by an accelerometer
8) Inertial forces may (sometimes) have an associated potential energy in the non-inertial reference frame

 

[vanhees71]

We usually reserve the term "reaction" for 3rd law force pairs. These are not third law force pairs. These are "second law force pairs" -- equal and opposite forces that result in zero acceleration for a test body.

That's a common misconception and should be avoided from the very beginning. The 3rd law states

If two bodies are interacting, then if the force on body 1 due to the presence of body 2 is $\vec{F}_{12}$ then there acts also a force on body 2 due to the presence of body 1, which is $\vec{F}_{21}=-\vec{F}_{12}$.

If you have two bodies with only interaction forces acting you have
$$m_1 \vec{a}_1=\vec{F}_{12}, \quad m_2 \vec{a}_2=\vec{F}_{21}$$
and thus the total momentum is conserved since
$$m_1 \vec{a}_1 + m_2 \vec{a}_2=\dot{\vec{p}}_1 + \dot{\vec{p}}_2=\vec{F}_{12}+\vec{F}_{21}=0.$$
Thus the total momentum
$$\vec{P}=\vec{p}_1+\vec{p}_2$$
is conserved, i.e., its center of mass
$$\vec{R}=\frac{m_1 \vec{x}_1+m_2 \vec{x}_2}{m_1+m_2}$$
moves like a free particle according to the 1st law uniformly, and in this sense and only this sense there's 0 net acceleration.

On each of the points the interaction force due to the other is acting, and each point thus is acclerated.

Mostly discussed example: The Sun and the Earth interacting due to Newtonian gravity. Neglecting all other heavenly bodies around that's the Kepler problem: The Sun and the Earth go around their common center of mass on ellipses, i.e., both are accerated. The center of mass goes with constant velocity, i.e., is unaccelerated. By clever choice of the reference frame you can make it stay at rest.

 

[Dale]

When applying Newton's Second Law (Sum F = m*a), neither a "coriolis force" nor a "centrifugal force" belong on the left side of the equation. They belong on the right side.

You are free to write any equation with any number of the terms on either side. That is basic algebra.

 

[vanhees71]

OK, so to be as correct as possible:
1) The Coriolis force is an inertial force.
2) Inertial forces are also called fictitious forces, although like @vanhees71 I avoid the term "fictitious force" and use the term "inertial force" as a general rule.
3) Inertial forces exist only in non-inertial reference frames, in this case in a rotating reference frame
4) Inertial forces do not follow Newton's 3rd law
5) Inertial forces do follow Newton's 2nd law and are necessary to describe dynamics in the inertial frame
6) Inertial forces are always proportional to the mass
7) Inertial forces cannot be detected by an accelerometer
8) Inertial forces may (sometimes) have an associated potential energy in the non-inertial reference frame

Ad 7) It depends on the accelerometer, or what are you having in mind? Fix a spring with some mass on an axis and let it rotate. It will get extended from its equilibrium position. Watching this from the point of view of co-rotating observer, he'll trace this back to the centrifugal force.

Also there are nice experiments done with a smart phone using its accelerometer. An example is here:

https://phyphox.org/experiment/centrifugal-acceleration/

 

[Dale]

Ad 7) It depends on the accelerometer, or what are you having in mind?

I had in mind an ideal 6 degree of freedom accelerometer. Also called an inertial measurement unit (IMU). Ideal meaning that it is of negligible size and mass, and can measure any acceleration to infinite accuracy and precision.

Any real accelerometer will of course deviate from the ideal behavior, but that is common to all measurement devices.

Fix a spring with some mass on an axis and let it rotate. It will get extended from its equilibrium position. Watching this from the point of view of co-rotating observer, he'll trace this back to the centrifugal force.

Actually, this detects the real centripetal forces at the axis, not the inertial centrifugal forces. The extensions of the springs are in the wrong direction to be attributed to the centrifugal force.

In the inertial frame there is only an inward real force, the mass deflects outward, correctly indicating the inward acceleration.

In the co-rotating frame there is an inward real force and an outward inertial force, the mass deflects outward, incorrectly indicating an inward acceleration when there is no acceleration. The accelerometer thus gives a wrong reading because it does not detect the inertial force.

 

[A.T.]

We usually reserve the term "reaction" for 3rd law force pairs. These are not third law force pairs. These are "second law force pairs" -- equal and opposite forces that result in zero acceleration for a test body.

The whole "action/reaction" terminology should be avoided. It wrongly implies a cause/effect relationship. But the 3rd Law says nothing about that. If we would call it 3rd-Law-pair, the confusion with force balance per 2nd Law would be less likely.

 

[A.T.]

You are free to write any equation with any number of the terms on either side. That is basic algebra.

In fact, the whole point of the mass factor in the inertial force terms, is to be able, to add them to the other forces as part of F_net in F_net = m*a.

 

[vanhees71]

I had in mind an ideal 6 degree of freedom accelerometer. Also called an inertial measurement unit (IMU). Ideal meaning that it is of negligible size and mass, and can measure any acceleration to infinite accuracy and precision.

Any real accelerometer will of course deviate from the ideal behavior, but that is common to all measurement devices.

Actually, this detects the real centripetal forces at the axis, not the inertial centrifugal forces. The extensions of the springs are in the wrong direction to be attributed to the centrifugal force.

In the inertial frame there is only an inward real force, the mass deflects outward, correctly indicating the inward acceleration.

In the co-rotating frame there is an inward real force and an outward inertial force, the mass deflects outward, incorrectly indicating an inward acceleration when there is no acceleration. The accelerometer thus gives a wrong reading because it does not detect the inertial force.

Then, how do you explain the correct functioning of the accelerometer measurements using a Smartphone, I quoted? These are precisely using the principle of the spring.

Of course you are right, seen from the inertial frame the spring shows the centripetal force needed to keep the rotating "point mass" on its circular orbit. Seen in the rotating frame, however, you can interpret the elongating of the spring as due to the centrifugal force.

Another nice example is the experiment with the free-falling Smartphone. When at rest relative to Earth the Smartphone accelorometer correctly shows the gravitational acceleration (i.e., the elongation of the "spring" due to the action of the gravitational force of the Earth, compensated by the elastic force of the spring). Letting the phone freely fall makes the shown acceleration 0 in accordance with the (weak) equivalence principle: a freely falling reference frame is a local inertial frame.

 

[Dale]

Then, how do you explain the correct functioning of the accelerometer measurements using a Smartphone, I quoted? These are precisely using the principle of the spring.

The video didn't play for me, but if the accelerometers are using the spring principle then as I already explained the deflection points in the wrong direction to be measuring the fictitious centrifugal force. It points in the direction to measure the real centripetal force.

Seen in the rotating frame, however, you can interpret the elongating of the spring as due to the centrifugal force.

No, you cannot. First, the deflection is in the wrong direction. Second, if the centrifugal force were causing deflection then it would counteract the deflection from the real centripetal force leading to no net deflection.

The magnitude and the direction of the deflection can only be explained in the rotating frame by asserting that the accelerometer can not detect the inertial centrifugal force.

 

[A.T.]

Of course you are right, seen from the inertial frame the spring shows the centripetal force needed to keep the rotating "point mass" on its circular orbit. Seen in the rotating frame, however, you can interpret the elongating of the spring as due to the centrifugal force.

The real centripetal force is still present in the rotating frame. So why would you need a different explanation?

 

[vanhees71]

The video didn't play for me, but if the accelerometers are using the spring principle then as I already explained the deflection points in the wrong direction to be measuring the fictitious centrifugal force. It points in the direction to measure the real centripetal force.

No, you cannot. First, the deflection is in the wrong direction. Second, if the centrifugal force were causing deflection then it would counteract the deflection from the real centripetal force leading to no net deflection.

The magnitude and the direction of the deflection can only be explained in the rotating frame by asserting that the accelerometer can not detect the inertial centrifugal force.

To be precise, it measures the constraining force of a motion under constraints. Maybe we can agree on that, though I think the discussion of inertial forces are usually so overcomplicated that it comes to unnecessary discussions like this. One should only keep in mind that inertial forces occur only in non-inertial reference frames, i.e., in an inertial frame there can never be Coriolis and/or centrifugal forces. Whether or not you interpret the inertial forces as forces depends simply on which side of the equation of motion you put the corresponding terms.

If it were me to decide, I'd never introduce inertial forces to begin with but just covariant time derivatives in a frame-independent formulation of mechanics. Then all these terms are just terms in the equations for the vector components belonging to the acceleration relative to an inertial frame with respect to basis vectors in a frame accelerated with respect to this inertial frame. Unfortunately I guess, that's impossible given the other interpretation of these terms as "inertial forces". Under no circumstances, I'd call the fictitious, because when interpreting these terms as "forces" they are very real and have to be taken properly into account to get the correct equations of motion in an inertial frame!

 

[vanhees71]

The real centripetal force is still present in the rotating frame. So why would you need a different explanation?

Let's do this calculation explicitly to make my point clear.

Take a perl of mass $m$ on a rod (frictionless) fastened with a spring of spring constant $k$. Then let the rod rotate around an axis perpendicular at it with constant angular velocity $\omega$ around the point where the spring is fastened.

Let the rod be in the $(x,y)$ plane of an inertial reference frame and let $r'=r$ the distance of the perl in the corotating frame. It's most simple to work with Lagrange. The coordinates in the inertial frame are
$$x(t)=r(t) \cos(\omega t), \quad y(t)=r(t) \sin(\omega t).$$
The free Lagrangian thus is
$$L_0=\frac{m}{2} (\dot{x}^2+\dot{y}^2)=\frac{m}{2} (\dot{r}^2+r^2 \omega^2).$$
You can now read this equation as valid in the inertial frame. As is explicitly clear, then all the terms belong to the kinetic energy and thus all of the terms from the Euler-Lagrange equation belong to the $m \vec{a}$ terms (acceleration wrt. the inertial frame):
$$\text{LHS=m a term}=m\ddot{x}-m r \omega^2.$$
The potential energy of the "true forces", i.e., the reaction force of the spring is
$$L_{\text{int}}=-\frac{k}{2} r^2.$$
From the point of view of the inertial observer all terms from this part of the Lagrangian belong to the right-hand or "force" side of the equation. Thus we have
$$\text{RHS=force}=-k r.$$
Thus from the point of view of the inertial frame we have
$$m\ddot{r}-m r \omega^2=-k r.$$
Now consider the stationary state, $r=r_0=\text{const}$. In this interpretation, seen from the inertial frame, the spring is elongated from its equilibrium position such as to provide the centripetal force directed inwards. Thus from this point of view the spring measures the constraint force to keep the pearl in uniform circular motion.

Now transform the equation to the corotating frame. This is particularly simple in this example. We just have to write $r'$ everywhere and reinterpret the Lagrangian a bit. Here we interpret only the terms
$$L_0'=\frac{m}{2} (\dot{x}^{\prime '}+\dot{y}^{\prime '}=\frac{m}{2} \dot{r}^{\prime 2}$$
as kinetic energy and the corresponding pieces of the Euler-Lagrange equation as the "ma terms":
$$\text{LHS=m a' terms}=m \ddot{r}'.$$
The rest of the Lagrangian is interpreted as potential energy ans the corresponding pieces of the Euler-Lagrange equations as "forces",
$$L_{\text{int}}'=-\frac{k}{2} r^{\prime 2} + \frac{m}{2} r{\prime 2} \omega^2,$$
and
$$\text{RHS=F' terms}=-k r' + m r' \omega^2.$$
Now the force is due to the reaction force of the string + the "centrifugal force" (which we know to be an inertial force). Now the physicist interprets the equilibrium condition $r'=r_0'=r_0$ as the point where no force acts on the pearl, i.e., in his interpretation the elongation of the spring measures the centrifugal force.

 

[A.T.]

Let's do this calculation explicitly to make my point clear.

I don't think there is any disagreement about the math. It's more about the conceptual interpretation, and associating frame depended "causes" (inertial centrifugal force) to frame independent "effects" (string elongation).

There are two real forces acting directly on the ends of the spring in every frame. They completely explain the elongation of the spring in every frame. There is no need to introduce additional "causes" for the elongation.

...in his interpretation the elongation of the spring measures the centrifugal force.

It's a possible interpretation, but conceptually different from:

...you can interpret the elongating of the spring as due to the centrifugal force.

It's one thing to use one quantity as a proxy measurement of some other quantity, in some special case. It's a different thing to claim causation.