Circumventing Newton's third law through Euler Inertial Forces

[Ghideon]

5 minutes ago, John2020 said:

No. In the inertial frame the accelerometer will show an inward (e.g. minus) and in the rotating frame an outward (plus).

Then back up that invalid claim with evidence. Do not use more invalid arguments or new invalid claims. 

Or check any introduction to physics, the above seems to be based on a common misunderstanding among physics student.

You are claiming that a accelerometer, for instance with a digital display showing the acceleration, attached to the mass m, will read one number wen seen from rotational frame of reference while simultaneously reads another value seen from the inertial frame of reference. I fail to see why you would claim that.

Edited October 23, 2020 by Ghideon

[John2020]

8 minutes ago, Ghideon said:

Then back up that invalid claim with evidence. Do not use more invalid arguments or new invalid claims. 

I speak always for the case of increase of angular velocity and not while the angular velocity is constant (radial acceleration would be then zero). Inside a car you are influenced by the centrifugal (outwards acceleration), otherwise you wouldn't be pushed against the door of the car.

[Ghideon]

5 minutes ago, John2020 said:

speak always for the case of increase of angular velocity and not while the angular velocity is constant (radial acceleration would be then zero). Inside a car you are influenced by the centrifugal (outwards acceleration), otherwise you wouldn't be pushed against the door of the car.

False. I speak of your example. Both at the time(s) with constant angular velocity and the time of accelerating angular velocity. Read previous answers.

The measured acceleration may vary and is towards the centre. The acceleration may look different from different frames of reference such as a rotating or an inertial frame of reference. 

The fact that you misinterpreted your experience in a car (and many beginners have done so) have no effect on the laws of physics.

 

Edited October 23, 2020 by Ghideon

[John2020]

9 minutes ago, Ghideon said:

False. I speak of your example. Both at the time(s) with constant angular velocity and the time of accelerating angular velocity. Read previous answers.

 

9 minutes ago, Ghideon said:

The fact that you misinterpreted your experience in a car (and many beginners have done) have no effect on the laws of physics.

The same applies for my example and I am still speaking for the rotating frame. Nobody said the laws of physics change because of that. When you say the acceleration will be always towards the center in the rotating frame, it is like you deny the existence of the centrifugal force in the rotating frame (of course there is none in the inertial frame).

Again, I speak for the transition time to a greater angular velocity.

 

Edited October 23, 2020 by John2020

[Ghideon]

15 minutes ago, John2020 said:

Again, I speak for the transition time to a greater angular velocity.

As I have told you numerous times, my explanations covers all the times, including transitions and both frames of reference. 

Do you really know what an accelerometer measures? 

15 minutes ago, John2020 said:

Nobody said the laws of physics change because of that.

You claim that. You may have to read my answers in more detail. You are claiming that an accelerometer, for instance with a digital display showing the acceleration, attached to the mass m, will read one number when seen from rotational frame of reference while simultaneously reads another value seen from the inertial frame of reference. 

What kind of physical principle allows you to have a display on an accelerometer attached to mass m show for instance -2G in one frame of reference and at the same time showing the value of +3G when seen from another frame of reference? Nature does not work that way as far as I know. 

At least we seem to have found the core, were the invalid analyses comes from. 

Edited October 23, 2020 by Ghideon
spelling

[John2020]

15 minutes ago, Ghideon said:

What kind of physical principle allows you to have a display on an accelerometer show for instance -2G and at the same time showing the value of +3G? Nature does not work that way as far as I know. 

Actually, I don't know how you assume an acceleration from the inertial frame perspective since the acceleration can be measured only when the accelerometer is in the rotating frame. I am again confused with the reference frames.

In the rotating frame is for instance +2G (outwards) and in the inertial frame is -2G (inwards) again during the transition. For constant angular velocity there is no radial acceleration (accelerometer will show zero). This is how I understand it. 

Edited October 23, 2020 by John2020

[Ghideon]

7 minutes ago, John2020 said:

Actually, I don't know how you assume an acceleration from the inertial frame perspective since the acceleration can be measured only when the accelerometer is in the rotating frame. I am again confused with the reference frames.

The accelerometer is at all times attached to mass m and hence moves with mass m wherever mass m is going. Again: Do you know what an accelerometer measures? Do you know that the acceleration the mass m experience is absolute and does not depend on frame of reference? 

I am not assuming anything. I read the measurement of the accelerometer attached to the mass m. The accelerometer is not physically moved. We look at the one and the same accelerometer that is attached to the mass m at all times. 

 

Edited October 23, 2020 by Ghideon

[John2020]

1 minute ago, Ghideon said:

I am not assuming anything. I read the measurement of the accelerometer attached to the mass m. The accelerometer is not physically moved. We look at the one and the same  accelerometer that is attached to the mass m at all times. 

Then we have just one reading that should be outwards.

[Ghideon]

2 minutes ago, John2020 said:

Then we have just one reading that should be outwards.

False. For the reasons already stated. But now at least your claim can be addressed mathematically.

We have agreed that the absolute acceleration of the mass m is independent of frame of reference, OK? It's just that you have misunderstood the direction.

Edited October 23, 2020 by Ghideon

[John2020]

1 minute ago, Ghideon said:

We have agreed that the absolute acceleration of the mass m is independent of frame of reference, OK? It's just that you have misunderstood the direction.

OK.

[Ghideon]

2 minutes ago, Ghideon said:

We have agreed that the absolute acceleration of the mass m is independent of frame of reference, OK?

 

1 minute ago, John2020 said:

OK.

Then you realise that since the acceleration is absolute we may do the mathematics for the example in any frame of reference?  

[John2020]

3 minutes ago, Ghideon said:

Then you realise that since the acceleration is absolute we may do the mathematics for the example in any frame of reference?  

OK. Let us see what may come out from this now.

Edited October 23, 2020 by John2020

[Ghideon]

2 minutes ago, John2020 said:

OK. Let us see what may come out from this now.

Yes. Please provide your formulas and equations so I can comment. Inertial frame of reference first.

 

Edited October 23, 2020 by Ghideon

[John2020]

1 minute ago, Ghideon said:

Yes. Please provide your formulas and equations so I can comment.

Show me yours first.

[Ghideon]

3 minutes ago, John2020 said:

Show me yours first.

Again; this is your speculative thread, where you have a possibility to argue for your claims.

 

 

[John2020]

1 minute ago, Ghideon said:

Again; this is your speculative thread, where you have a possibility to argue for your claims.

OK, I will present them within the next 30 min or so. 

[Ghideon]

5 minutes ago, John2020 said:

OK, I will present them within the next 30 min or so. 

Ok. Please remember to use the inertial frame of reference and that centrifugal force does not exist there. 

[John2020]

49 minutes ago, Ghideon said:

Ok. Please remember to use the inertial frame of reference and that centrifugal force does not exist there. 

I will show a solution for each frame.

Please see below:

\[\text{rotating frame} \\
\sum F_{net(\omega))} = F_{cf} - F_{cp} = F_{cf} - F_{fr} = m \omega^2r - F_{fr} \\ \\
\text{Initially: } \omega = \omega_1 \Rightarrow r = r_1 \Rightarrow  F_{cf} = F_{fr} \Rightarrow \\
\sum F_{net(\omega_1))} = 0  \Rightarrow a = \sum F_{net(\omega_1))} / m = 0 \\ \\
\text{Later: } \omega = \omega_2 \Rightarrow r = r_2 \Rightarrow  F_{cf} > F_{fr} \Rightarrow \\
\sum F_{net} = \sum F_{net(\omega_2))} -  \sum F_{net(\omega_1))} = m (\omega_2^2r_2 - \omega_1^2r_1) > 0 \Rightarrow \\ 
a = \sum F_{net} / m > 0 \\ 
\\ \\ \\
\text{inertial frame} \\
\sum F_{net(\omega))} = F_{cp} = F_{fr} = m \omega^2r \\ \\
\text{Initially: } \omega = \omega_1 \Rightarrow r = r_1 \Rightarrow  F_{cp} = F_{fr} \Rightarrow \\
\sum F_{net(\omega_1))} > 0  \Rightarrow a = \sum F_{net(\omega_1))} / m > 0 \\ \\
\text{Later: } \omega = \omega_2 \Rightarrow r = r_2 \Rightarrow  F_{cp} > F_{fr} \Rightarrow \\
\sum F_{net} = \sum F_{net(\omega_2))} -  \sum F_{net(\omega_1))} = m (\omega_2^2r_2 - \omega_1^2r_1) > 0 \Rightarrow \\ 
a = \sum F_{net} / m > 0 \\ 
\]

 

Edited October 23, 2020 by John2020

[swansont]

I'm not sure what summing the net forces is supposed to tell you, and you change the sum to a difference in the middle of that last set of equations.

The net force at t2 is mw₂²r² if the object is moving in a circle at that point and you've declared that the frictional force (the actual force that is acting as the centripetal force) is smaller than this, so this equation will not actually give you the net force. It's only valid for circular motion.

What this tells you is the motion is not circular at t2 (or any time ofter you increased the angular speed)

[John2020]

11 minutes ago, swansont said:

I'm not sure what summing the net forces is supposed to tell you, and you change the sum to a difference in the middle of that last set of equations.

You are right my mistake (regarding the sum). The difference is the transition from one to another angular velocity.

\[\text{rotating frame} \\ 
    F_{net(\omega))} = F_{cf} - F_{cp} = F_{cf} - F_{fr} = m \omega^2r - F_{fr} \\ \\ 
    \text{Initially: } \omega = \omega_1 \Rightarrow r = r_1 \Rightarrow  F_{cf} = F_{fr} \Rightarrow \\ 
    F_{net(\omega_1))} = 0  \Rightarrow a = F_{net(\omega_1))} / m = 0 \\ \\ 
    \text{Later: } \omega = \omega_2 \Rightarrow r = r_2 \Rightarrow  F_{cf} > F_{fr} \Rightarrow \\ 
    F_{net} = F_{net(\omega_2))} -  F_{net(\omega_1))} = m (\omega_2^2r_2 - \omega_1^2r_1) > 0 \Rightarrow \\  
    a = F_{net} / m > 0 \\  
    \\ \\ \\ 
    \text{inertial frame} \\ 
    F_{net(\omega))} = F_{cp} = F_{fr} = m \omega^2r \\ \\ 
    \text{Initially: } \omega = \omega_1 \Rightarrow r = r_1 \Rightarrow  F_{cp} = F_{fr} \Rightarrow \\ 
    F_{net(\omega_1))} > 0  \Rightarrow a = F_{net(\omega_1))} / m > 0 \\ \\ 
    \text{Later: } \omega = \omega_2 \Rightarrow r = r_2 \Rightarrow  F_{cp} > F_{fr} \Rightarrow \\ 
    F_{net} = F_{net(\omega_2))} -  F_{net(\omega_1))} = m (\omega_2^2r_2 - \omega_1^2r_1) > 0 \Rightarrow \\  
    a = F_{net} / m > 0 \\
\]

11 minutes ago, swansont said:

What this tells you is the motion is not circular at t2 (or any time ofter you increased the angular speed)

Correct. I would say for the time frame of the analysis, from above ω₁ up to ω₂ motion is not circular anymore.

Edited October 23, 2020 by John2020

[Ghideon]

4 minutes ago, John2020 said:

The difference is the transition from one to another angular velocity.

I see two accelerations in the equations, both positive. They are both pointing to the centre of the circle. Have I understood that correct?

 

 

[swansont]

11 minutes ago, John2020 said:

Correct. I would say for the time frame of the analysis, from above ω₁ up to ω₂ motion is not circular anymore.

Even at w2 the motion is not circular, because thr frictional force is equal to mw₁²r₁

Since both r and w have increased, you need a larger force in order to have circular motion. The mass will slide until it hits the stop, which will then provide an additional force

[John2020]

5 minutes ago, Ghideon said:

I see two accelerations in the equations, both positive. They are both pointing to the centre of the circle. Have I understood that correct?

It depends how one interprets it. Normally, I should have placed a negative sign in front of the centripetal force for the inertial frame in order to be consistent with the forces directions in all frames. In the inertial frame, it should appear with an opposite sign from that in the rotating frame. We shouldn't argue about this. The issue here is do we have an acceleration upon the mass m or not?

2 minutes ago, swansont said:

Even at w2 the motion is not circular, because thr frictional force is equal to mw₁²r₁

OK.

Edited October 23, 2020 by John2020

[swansont]

Quote

Fnet=Fnet(ω2))−Fnet(ω1))=m(ω22r2−ω21r1)>0⇒a=Fnet/m>0

 No this is not correct. The radial force isFnet(ω1) and it remains that value, because that's the frictional force the system can supply. But there is also a tangential force, because you are increasing the angular speed, and the mass is moving, so the linear speed in increasing

As long as r is increasing, there will be a tangential force. r keeps increasing until it hits the stop (lets call that radius R)

[Ghideon]

I just saw this:

2 hours ago, John2020 said:

In the rotating frame is for instance +2G (outwards) and in the inertial frame is -2G (inwards) again during the transition. For constant angular velocity there is no radial acceleration (accelerometer will show zero). This is how I understand it. 

Completely false. I asked if you know what an accelerometer and absolute acceleration is and there is no answer.

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

 

7 minutes ago, John2020 said:

It depends how one interprets it. Normally, I should have placed a negative sign in front of the centripetal force for the inertial frame in order to be consistent with the forces directions in all frames. In the inertial frame, it should appear with an opposite sign from that in the rotating frame. We shouldn't argue about this. The issue here is do we have an acceleration upon the mass m or not?

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. 

 

 

Edited October 23, 2020 by Ghideon