Coriolis force caused by tangential velocity

[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.

 

[Dale]

Now the physicist interprets the equilibrium condition r′=r′0=r0 as the point where no force acts on the pearl, i.e., in his interpretation the elongation of the spring measures the centrifugal force.

Remove the centripetal force and the elongation disappears even though the centrifugal force remains. So much for that interpretation.

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.

It is not as complicated as you are making it. My point 7 is correct, but instead of simply learning something new you want to argue. You and I have been down this road before and I did not enjoy it last time and am not willing to do it again. You are free to learn something new here or not. I will leave with one last post explaining the concept to the best of my ability and then I am done.

Suppose that we have an accelerometer involving a mass sliding on a frictionless rod with a spring oriented along the rod. At the equilibrium point the rod is marked with a 0.

We know the spring constant and the mass so we also mark the rod with a pair of 1’s on either side of the 0 for 1 g and a pair of 2’s for 2 g and so on, each indicating where the mass is for a given magnitude of acceleration, but at this time we do not mark directions. So now we have a device that can measure the magnitude of the acceleration along the axis but does not indicate the direction.

Now, to set the directions we experiment with our device. We place it horizontally on the table with the axis along the left-right direction. We note that the mass is at the 0 mark. We accelerate it to the right at 1 g and note that the mass is at the left 1 mark, so we mark an arrow pointing right next to the left 1 mark. We accelerate it to the left at 1 g and note that the mass is at the right 1 mark, so we mark an arrow pointing left next to the right 1 mark. And repeat.

In the end, we note that all of the arrows point in towards the 0 on the rod. A rightward acceleration is indicated by a leftward deflection.

Now, we put the accelerometer in a centrifuge on our table, oriented with the rod along the radial direction. We turn it on to 1 g and note that the mass deflects outward to the 1 g mark. An outward deflection indicates an inward acceleration, and sure enough if we look at the arrow it points inward, towards the 0 mark. This is consistent with the real inward centripetal force.

Now, an ambitious student seeks to analyze it in the rotating frame. In the rotating frame the real centripetal force still exists and a centrifugal force is added. The addition of the centrifugal force does not change the reading, so it should be immediately apparent that the accelerometer does not detect the centrifugal force, but the student wants to consider all possibilities.

The student enumerates all four possibilities. A) the accelerometer detects both forces. B) the accelerometer detects neither force. C) the accelerometer detects only the real centripetal force. D) the accelerometer detects only the fictitious centrifugal force. This is an exhaustive and mutually exclusive set of possibilities.

A) if it detects both then the reading should be 0. The reading is 1 g inward so A) is inconsistent with observation.

B) if it detects neither then the reading should be 0. The reading is 1 g inward so B) is inconsistent with observation.

C) if it detects only the real centripetal force then the reading should be 1 g inward. The reading is 1 g inward so C) is consistent with observation.

D) if it detects only the fictitious centrifugal force then the reading should be 1 g outward. The reading is 1 g inward so D) is inconsistent with observation.

The only possibility consistent with observation is that the accelerometer detects only the real centripetal force and does not detect the inertial centrifugal force.

Another way that it should be obvious that accelerometers do not detect inertial forces is to consider other non-inertial frames. There is nothing physically privileged about the co rotating frame. In other non inertial frames the inertial force will be different. By judicious choice of frame we can make it have any size and direction desired. Regardless of the size or direction of the inertial force the reading of the accelerometer is the same and the magnitude and direction are both given by the real force. Therefore, the accelerometer can not detect any inertial force in any non-inertial frame. The observation is always consistent only with the inertial forces being undetected by the accelerometer.

That is it. I am done here with you. You are free to learn or not as you choose, but I lack the desire to fight you on this. My statement 7 is correct, whether you will accept it or not.

 

[Swamp Thing]

This is how Wikipedia defines an accelerometer:

An accelerometer is a device that measures proper acceleration. Proper acceleration, being the acceleration (or rate of change of velocity) of a body in its own instantaneous rest frame, is not the same as coordinate acceleration, being the acceleration in a fixed coordinate system. For example, an accelerometer at rest on the surface of the Earth will measure an acceleration due to Earth's gravity, straight upwards (by definition) of g ≈ 9.81 m/s2. By contrast, accelerometers in free fall (falling toward the center of the Earth at a rate of about 9.81 m/s2) will measure zero.

And this is how the article explains how they work:

Conceptually, an accelerometer behaves as a damped mass on a spring. When the accelerometer experiences an acceleration, the mass is displaced to the point that the spring is able to accelerate the mass at the same rate as the casing. The displacement is then measured to give the acceleration.

Maybe the disagreement about Dale's statement

7) Inertial forces cannot be detected by an accelerometer

has a lot to do with which acceleration one is talking about. I think Dale's Statement 7 is true about "coordinate acceleration" -- or, to coin an alternative term, "geometric acceleration". An accelerometer won't really sense inertial forces that are associated with coordinate acceleration.

When it comes to proper acceleration on the other hand... By definition an accelerometer measures proper acceleration, and the way it does this is to first measure some effect (e.g. deflection) of the inertial force arising from the proper acceleration, then scale the result based on the known elasticity and mass to get acceleration. So in this case we necessarily measure inertial force in order to measure acceleration.

 

[Dale]

So in this case we necessarily measure inertial force in order to measure acceleration.

An accelerometer never measures inertial forces as shown exhaustively above.

the inertial force arising from the proper acceleration

I get the feeling from your comment that you do not understand what an inertial force is. “Inertial force” is another name for “fictitious force”. They arise from analyzing a scenario in a non inertial frame; they do not arise from proper acceleration.

https://en.m.wikipedia.org/wiki/Fictitious_force

 

[Swamp Thing]

Ok, I think I now see what you mean. My sentence...

The way (the accelerometer works) is to first measure some effect (e.g. deflection) of the inertial force arising from the proper acceleration, then scale the result based on the known elasticity and mass to get acceleration.

... should be corrected to read : "We allow a spring to provide the required centripetal force, then calibrate the spring's deformation to read out the acceleration." In this picture, clearly, the centrifugal force is not (and can not) be measured.