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John2020

Circumventing Newton's third law through Euler Inertial Forces

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1 hour ago, John2020 said:

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

You have two sets of coordinate systems in your equations and none in your picture. 
You mention kinetic friction above, that is not in your picture.
You have one set of time independent equations (per frame of reference), example speaks of three (at least) different states.
You imply from the equations that there is constant linear motion or that mass m=0 or radius r=0. That is not what the picture shows.
You speak of this dimension of force on the rod. Not in your picture.
Tension is not is your picture.
You have argued that there is outward acceleration* in the inertial frame of reference in your example but now say tension on the rod. that means inward acceleration at all times, Outward acceleration in the inertial frame of reference implies compression of the rod. 

(The above is a short extract)

A short and more to the point way to express how I prefer this discussion to continue:

 

image.png.3c5d5bd209a81c1ec361c66a8723432f.pngcross.png.cc6d3b042469ed377efcec65d8de36de.png

Start with something simpler. You need to be able to draw correct conclusions to be able to write meaningful equations for the example. So far that has not happened. 

The rest of you comment can't be addressed since the seem to describe some situation that does not exist in the example. I'll postpone any answer until more information is provided.

*) https://www.scienceforums.net/topic/123261-circumventing-newtons-third-law-through-euler-inertial-forces/?do=findComment&comment=1157422

 

 

Edited by Ghideon

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5 hours 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)=m(ω+Δω)2(r+Δr)Fcp(ω)=Ffr=μsT=mω2rxAxis:Tx(ω+Δω)Fcp(ω)=Δpm/ΔtΔω=0Δr=0Tx(ω+Δω)=Fcp(ω)=mω2rΔpm/Δt=0Δω0Δr0Tx(ω+Δω)>Fcp(ω)Δpm/Δt0yAxis:Ty(ω+Δω)=m(Δω/Δt)(Δr)Δω=0Δr=0Ty(ω+Δω)=0Δω0Δr0Ty(ω+Δω)0Rotating frameFcf(ω+Δω)=m(ω+Δω)2(r+Δr)Fcp(ω)=Ffr=μsT=mω2rxAxis:Fcf(ω+Δω)Fcp(ω)=Δpm/ΔtΔω=0Δr=0Fcf(ω+Δω)=Fcp(ω)=mω2rΔpm/Δt=0Δω0Δr0Fcf(ω+Δω)>Fcp(ω)Δpm/Δt0yAxis:Ty(ω+Δω)=m(Δω/Δt)(Δr)Δω=0Δr=0Ty(ω+Δω)=0Δω0Δr0Ty(ω+Δω)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

This should be done in a circular coordinate system. You’re already using variables from it.

The frictional force is not u.T

The tension will never be greater than the centripetal force, because it is the centripetal force 

 

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8 minutes ago, John2020 said:

Let us close the discussion here.

Regarding circumventing Newton or regarding this specific example?

 

8 minutes ago, John2020 said:

Well, then I got it all wrong.

 

One example: your deltas = 0 Implying T=0 means the firs equation equals 0. That can only happen when one or more of m, w, r is zero. 

error.png.efcbb3d2b6d43889b6872c3e30af053c.png

Edited by Ghideon

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1 hour ago, John2020 said:

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).

The tension and the frictional force are equal. They are the centripetal force. You are using friction, but you could also just bolt the mass to the rod - it doesn’t matter. The rod isn’t stretching, so there can only be that one force present, regardless of the source.

When you have uniform circular motion, the net force is the centripetal force, however that net force physically manifests itself.

1 hour ago, John2020 said:

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).

It’s from a torque exerted on the rod.

1 hour ago, John2020 said:

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.

Tension is always directed toward the origin. If this were somehow greater than the centripetal force, r would decrease.

When you increase w, you no longer have uniform circular motion. Friction does not supply enough of a force to have that be true. The net force is no longer toward the center of the circle

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13 minutes ago, Ghideon said:

Regarding circumventing Newton or regarding this specific example?

The entire thread.

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15 minutes ago, John2020 said:

Well, then I got it all wrong. Let us close the discussion here.

This a good thing.  Being wrong is not bad but refusing to admit you are wrong is terrible.  All the greatest physicist and scientists were wrong about many things, and learned from their errors.  There is no doubt you have learned a lot of mechanics in this thread, I know I have.  Thanks for the discussion.

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