# Circumventing Newton's third law through Euler Inertial Forces

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2 minutes ago, Ghideon said:

For constant angular velocity there is centripetal acceleration accelerometer will NOT show zero.

This is valid in the inertial frame as I show above but not in the rotating frame.

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2 minutes ago, John2020 said:

This is valid in the inertial frame as I show above but not in the rotating frame.

False

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6 minutes ago, Ghideon said:

I want to know the acceleration of the mass m in the inertial frame of reference, before, during and after the increase the angular velocity. please point out those three accelerations of mass m in your equations for the inertial frame of reference.

Scroll up this page and about 4 posts from here there you may find the expression for the inertial frame.

Just now, Ghideon said:

False

Then show me your version for both frames.

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This part of the equation is incorrect or needs explanation:

Fcp>Ffr

Centripetal force can't be greater than friction as far as I can see in your picture.

16 minutes ago, John2020 said:

Then show me your version for both frames.

No. What you state is physically impossible. Or you have misunderstood centripetal acceleration. A picture of your example may help: The accelerometer is attached to the mass m. It measures the acceleration of mass m. The acceleration is visible on the display (green).

The reading on the display is the same for both observers in the picture. The drawing holds for any of the times before, during, after increased of angular velocity. Mass m can't experience a different acceleration in different frames of reference. The motion of mass m just happen to be described by different coordinates in different frames of reference.

Edited by Ghideon

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14 minutes ago, Ghideon said:

The reading on the display is the same for both observers in the picture. The drawing holds for any of the times before, during, after increased of angular velocity. Mass m can't experience a different acceleration in different frames of reference, it's movement just happen to have different coordinates in different frames of reference.

If you agree with what I wrote about the rotational frame (centrifugal is present and cannot be removed) then I missed something in the inertial frame or vice versa (here you have to justify the centrifugal does not exist but in rotational frames there is always one therefore it has to be taken into consideration).

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3 minutes ago, John2020 said:

If you agree with what I wrote about the rotational frame (centrifugal is present and cannot be removed) then I missed something in the inertial frame or vice versa (here you have to justify the centrifugal does not exist but in rotational frames there is always one therefore it has to be taken into consideration).

Cant comment unless you are more specific about what you wrote.

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21 minutes ago, Ghideon said:

Centripetal force can't be greater than friction as far as I can see in your picture.

You speak about the inertial frame. Well, this is the part I couldn't otherwise justify in the inertial frame in order to agree with what I wrote about the rotating frame (here the centripetal equals to friction at ω1. Afterwards, the centrifugal becomes greater than the friction while up to ω2).

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14 minutes ago, John2020 said:

You speak about the inertial frame. Well, this is the part I couldn't otherwise justify in the inertial frame in order to agree with what I wrote about the rotating frame (here the centripetal equals to friction at ω1. Afterwards, the centrifugal becomes greater than the friction while up to ω2).

No. I speak of real forces, present in any frame. The friction does not change when we change coordinate system. If you start from invalid physics and end up with invalid equations. If two real forces (forces that are not fictitious) are different in two frames of reference we know you made an error.

It sounds like you try to force your misunderstanding of rotation and fictitious forces (for instance centrifugal forces) on the rest of physics. It will not happen.

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59 minutes ago, swansont said:

The radial force isFnet(ω1) and it remains that value, because that's the frictional force the system can supply.

I am not assuming something else. This is the reason I take the difference that equals to Fnet that represents the effective force that applies to mass m. In other words, friction remains where it is but due to the increase in angular velocity, the centrifugal force will be greater than friction, thus it will start to accelerate mass m.

3 minutes ago, Ghideon said:

It sounds like you try to force your misunderstanding of rotation and fictitious forces (for instance centrifugal forces) on the rest of physics. It will not happen.

So you indirectly claim there are no centrifugal forces in rotational frame. Who is violating the laws of physics? You or me.

Could you please show your version for both frames? Just to have a reference for what we are talking about as also how it is correctly expressed mathematically.

Edited by John2020

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9 minutes ago, John2020 said:

Could you please show your version for both frames? Just to have a reference for what we are talking about as also how it is correctly expressed mathematically.

Your mathematics does not match the picture you have provided.  Maybe you should start more simple? Provide the inertial frame of reference first, and describe the steps in the equations more explicitly. Be extra careful with the transition phase.

(side note: This thread has now reached the third place in the unofficial competition "speculations thread with the most answers". )

Edited by Ghideon

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1 minute ago, Ghideon said:

Your mathematics does not match the picture you have provided.  Maybe you should start more simple? Provide the inertial frame of reference first, and describe the steps in the equations more explicitly. Be extra careful with the transition phase.

This is all I can. It will be the same from wherever I start. I would suggest to present your version for both frames and to show me where I am wrong.

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23 minutes ago, John2020 said:

This is all I can. It will be the same from wherever I start. I would suggest to present your version for both frames and to show me where I am wrong.

Please explain what the point is. I will provide the mathematics, you will interprete it with your personal version of physics and state that calculations are wrong or have some other issue. Note that @swansont already provided the matehamtics for the inertial frame of reference, you may start by commenting on that.

I think the problem is that you require your personal interpretation to be correct. It looks like you need/want fictitious centrifugal forces have physical effect and adopt the equations to that.

I will provide mathematics as soon as you are past the initial step of understating the basics of centripetal acceleration. Feel free to provide a simpler example if needed. There must be many example that have the same basic properties but that are less complicated.
Edited by Ghideon

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18 minutes ago, Ghideon said:

I will provide mathematics as soon as you are past the initial step of understating the basics of centripetal acceleration. Feel free to provide a simpler example if needed. There must be many example that have the same basic properties but that are less complicated.

OK. I will find a simpler example to explain what is going on in the inertial frame. A little bit later.

I will present a little bit later the simple classical case of the centripetal acceleration.

21 minutes ago, Ghideon said:

Please explain what the point is. I will provide the mathematics, you will interprete it with your personal version of physics and state that calculations are wrong or have some other issue. Note that @swansont already provided the matehamtics for the inertial frame of reference, you may start by commenting on that.

OK. I answered to his post, however wrongly because I assumed that he was speaking for the rotating frame.

Edited by John2020

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2 hours ago, John2020 said:

This is valid in the inertial frame as I show above but not in the rotating frame.

An accelerometer is a real device and will not register fictitious forces. It doesn't matter what frame you are in. If your connection to the mass was a spring, it would extend until it was supplying the required centripetal force (because springs follow F = -kx). It will have this extension regardless of which frame you analyze the motion in. It's a real event, and everybody agrees on it.

1 hour ago, John2020 said:

I am not assuming something else. This is the reason I take the difference that equals to Fnet that represents the effective force that applies to mass m. In other words, friction remains where it is but due to the increase in angular velocity, the centrifugal force will be greater than friction, thus it will start to accelerate mass m.

There is no centrifugal force. This is an inertial frame analysis.

Quote

This is all I can. It will be the same from wherever I start. I would suggest to present your version for both frames and to show me where I am wrong.

One approach might have been not to do any analysis in a rotating frame, which people have been pointing out for basically this entire thread. You misunderstand basic physics, and are trying to do analysis that's more difficult (IMO) Basically you're adding non-Newtonian physics and your misconceptions, and it's unclear whether your misconceptions are with Newtonian or the non-Newtonian parts

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4 minutes ago, swansont said:

There is no centrifugal force. This is an inertial frame analysis.

This is my fault, I will come again on this later.

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I have been following this thread with interest and I would like thank the participants.  I never really got the idea of a fictitious force, but thanks to this thread the concept now seems clear to me.

It also looks like John2020 has long ago reached the point where he should have admitted his idea was wrong, if he was ever going to.

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19 minutes ago, Bufofrog said:

It also looks like John2020 has long ago reached the point where he should have admitted his idea was wrong, if he was ever going to.

Not yet. I have still some tricks under my sleeve.

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32 minutes ago, John2020 said:

Not yet. I have still some tricks under my sleeve.

Not ones that will work. We already looked at the end of the story. Newton wins.

The ONLY way to refute an existing (i.e. experimentally tested and mathematically self-consistent) theory is with experimental evidence. Until such time as you build a device that moves on its own, you will not have shown Newtonian physics to be wrong.

All you can offer is analysis that is flawed in some way, and all of your efforts have either been to try and hide the flaws, or by baldly asserting that they aren't flaws (which requires an experiment; see above)

53 minutes ago, Bufofrog said:

It also looks like John2020 has long ago reached the point where he should have admitted his idea was wrong, if he was ever going to.

This is the playbook that we see over and over again.

1. Convince yourself that your idea is right.

2. Analyze your idea with physics

3. Contort the analysis to reach the conclusion you desire

4. If the flaw is obvious, obfuscate by making the example more complex

Whereas in reality, if your analysis reaches a conclusion that violates laws of physics, you know you made a mistake someplace. The math is internally consistent, so this just points to a math error.

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7 hours ago, John2020 said:

I will present a little bit later the simple classical case of the centripetal acceleration.

Ok.

4 hours ago, John2020 said:

Not yet. I have still some tricks under my sleeve.

Some correctly performed application of Newton will do. The tricks you have pulled out of your sleeves so far are not very impressing.

Edited by Ghideon
spelling

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4 hours ago, John2020 said:

Not yet. I have still some tricks under my sleeve.

You mean like magic?

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I admit I didn't backed my arguments properly. This is what I wanted to show you (I hope now you may agree with this):

$\text{Inertial frame} \\ T_{x_{(\omega + \Delta \omega)}} = m (\omega + \Delta \omega )^2(r + \Delta r) \\ F_{cp_{(\omega)}} = F_{fr} = \mu_s\cdot T = m \omega^2r \\ \\ x-Axis: \overbrace{T_{x_{(\omega + \Delta \omega)}}}^{\longrightarrow} \overbrace{- F_{cp_{(\omega)}}}^{\longleftarrow} = \Delta p_m / \Delta t \\ \\ \Delta \omega = 0 \Rightarrow \Delta r = 0 \Rightarrow T_{x_{(\omega + \Delta \omega)}} = F_{cp_{(\omega)}} = m \omega^2r \Rightarrow \Delta p_m / \Delta t = 0 \\ \Delta \omega \neq 0 \Rightarrow \Delta r \neq 0 \Rightarrow T_{x_{(\omega + \Delta \omega)}} > F_{cp_{(\omega)}} \Rightarrow \Delta p_m / \Delta t \neq 0 \\ \\ y-Axis: T_{y_{(\omega + \Delta \omega)}} = m (\Delta \omega / \Delta t)\cdot (\Delta r) \\ \Delta \omega = 0 \Rightarrow \Delta r = 0 \Rightarrow T_{y_{(\omega + \Delta \omega)}} = 0 \\ \Delta \omega \neq 0 \Rightarrow \Delta r \neq 0 \Rightarrow T_{y_{(\omega + \Delta \omega)}} \neq 0 \\ \\ \\ \\ \\ \text{Rotating frame} \\ F_{cf_{(\omega + \Delta \omega)}} = m (\omega + \Delta \omega )^2(r + \Delta r) \\ F_{cp_{(\omega)}} = F_{fr} = \mu_s\cdot T = m \omega^2r \\ \\ x-Axis: \overbrace{F_{cf_{(\omega + \Delta \omega)}}}^{\longrightarrow} \overbrace{- F_{cp_{(\omega)}}}^{\longleftarrow} = \Delta p_m / \Delta t \\ \\ \Delta \omega = 0 \Rightarrow \Delta r = 0 \Rightarrow F_{cf_{(\omega + \Delta \omega)}} = F_{cp_{(\omega)}} = m \omega^2r \Rightarrow \Delta p_m / \Delta t = 0 \\ \Delta \omega \neq 0 \Rightarrow \Delta r \neq 0 \Rightarrow F_{cf_{(\omega + \Delta \omega)}} > F_{cp_{(\omega)}} \Rightarrow \Delta p_m / \Delta t \neq 0 \\ \\ y-Axis: T_{y_{(\omega + \Delta \omega)}} = m (\Delta \omega / \Delta t)\cdot (\Delta r) \\ \Delta \omega = 0 \Rightarrow \Delta r = 0 \Rightarrow T_{y_{(\omega + \Delta \omega)}} = 0 \\ \Delta \omega \neq 0 \Rightarrow \Delta r \neq 0 \Rightarrow T_{y_{(\omega + \Delta \omega)}} \neq 0$

T: Tension of the rigid rod

Fcp: Centripetal force

Fcf: Centrifugal force

Ffr: Static friction force

μs:  Static friction coefficient

Consequently, for Δω≠0 and from the conservation of linear momentum on x-Axis and y-Axis, the system (M+m) will start to accelerate (following a curve) during the transition of angular velocity from ω to ω+Δω

Edited by John2020

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44 minutes ago, John2020 said:

I admit I didn't backed my arguments properly. This is what I wanted to show you (I hope now you may agree with this):

Inertial frameTx(ω+Δω

edit: Can you provide definitions what the variables means?

x-posted; Definitions were added to the post above.

This part looks strange* given the definite of T=tension of rigid rod and F centripetal force:

How can the tension be greater than the centripetal force? It looks like you are mixing forces acting on different objects in the equations making it pretty hard to see if the equations have any physical meaning or not.

Also note that you have added tension in the y direction, you probably mean shear stress.

Note @John2020 It may help drive the discussion if you refer to exactly which example the equations are meant to describe.

*) There are more issues ...

Edited by Ghideon
x-post

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1 hour ago, Ghideon said:

How can the tension be greater than the force applied?

The idea behind this is that the centripetal equals to the static friction force initially that means the mass m is not fixed on the rod. If it was fixed then the tension would be always equal to the centripetal force (like having an infinite friction coefficient) at any angular velocity e.g. ω+Δω. From the moment mass m is not fixed, when the angular velocity increases the centripetal force remains the same during the transition (or better it becomes equal to the kinetic friction (which is much smaller than the static friction, something that is not shown in the equations) we may formulate the tension as being proportional to Δω.

1 hour ago, Ghideon said:

Note @John2020 It may help drive the discussion if you refer to exactly which example the equations are meant to describe.

They describe the drawing with the rotating mass m.

1 hour ago, Ghideon said:

Also note that you have added tension in the y direction, you probably mean shear stress.

I don't know how should I call it. What is certain is, it is not an external force.

Edited by John2020

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1 hour ago, John2020 said:

They describe the drawing with the rotating mass m.

<Sarcasm> Any member who would like to contribute just have to read through the 500+ posts and try to locate which exact example you mean.  </Sarcasm>

If you mean the following, then the equations does not match:

1 hour ago, John2020 said:

The idea behind this is that the centripetal equals to the static friction force initially that means the mass m is not fixed on the rod. If it was fixed then the tension would be always equal to the centripetal force (like having an infinite friction coefficient) at any angular velocity e.g. ω+Δω. From the moment mass m is not fixed, when the angular velocity increases the centripetal force remains the same during the transition (or better it becomes equal to the kinetic friction (which is much smaller than the static friction, something that is not shown in the equations) we may formulate the tension as being proportional to Δω.

That does not describe what happens. Introducing tension along the rod makes this complicated; you need to account for the fact that tension is not constant along the rod.
And maximum tension in the rod can't be less that centripetal force, not in the example i linked.

A friendly suggestion: Start with something simpler instead or study the physics (and in this case engineering) required (doing both would also be fine).

1 hour ago, John2020 said:

I don't know how should I call it. What is certain is, it is not an external force.

Ok, but it makes the analysis tricky when personal terms are introduced without definition. The force may not be external to the system but it is external to the rod, so again the equations are not easy to follow.  But removing all non external (=internal) forces would simplify our analysis of the whole system:

F=ma F=0, a=0. Done (m = total mass, F = total external forces, a=acceleration of the whole system)

Edited by Ghideon

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33 minutes ago, Ghideon said:

<Sarcasm> Any member who would like to contribute just have to read through the 500+ posts and try to locate which exact example you mean.  </Sarcasm>

If you mean the following, then the equations does not match:

Yes this is it. Please show me what doesn't match.

33 minutes ago, Ghideon said:

And maximum tension in the rod can't be less that centripetal force, not in the example i linked.

The first expression (as also in the rotating frame where the tension is labelled as centrifugal (see @swansont comment on centripetal) shows it is equal or greater and never less than the centripetal. When the mass m reaches the stop then, tension again becomes equal to the centripetal (stop=infinite static friction coefficient).

33 minutes ago, Ghideon said:

Ok, but it makes the analysis tricky when personal terms are introduced without definition. It is external to the rod. But removing all non external (=internal) forces would simplify our analysis:

F=ma F=0, a=0. Done (m = total mass, F = total external forces, a=acceleration of the whole system)

How it is external to the rod when nothing applies externally to the rod. The rod has an angular velocity and when this changes then it acquires an angular acceleration. Thus, the tangential acceleration cannot be attributed to an external force but just from the y-component tension of the rod (or call it whatever you like it).

33 minutes ago, Ghideon said:

That does not describe what happens. Introducing tension along the rod makes this complicated; you need to account for the fact that tension is not constant along the rod.

I have already answered this. As I mentioned above since the centripetal is the static friction force, we could express the tension proportional to the angular velocity. At Δω = 0, we have the tension being equal to the centripetal force (uniform circular motion). For  Δω ≠ 0, results  a non-uniform circular motion that implies Tension > Centripetal. That is all.

Edited by John2020

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