Circumventing Newton's third law through Euler Inertial Forces

[swansont]

17 hours ago, John2020 said:

 

 

Rotating mass m in outer space.
M: mass of the rotor
m: mass that may move radially
ω: angular velocity
yellow rod: rigid rod

Fcp: centripetal force
Fra: reaction force
Ffr: friction force
Fcf: inertial centrifugal force while Δω ≠ 0

The rotor provides a constant angular velocity while the mass is at distance (r). When the angular velocity increases by Δω within 0.1 radians:
a) How mass (m) will be affected from the change in angular velocity?
b) How the system (M + m) will be affected from the change in angular velocity?

 

Note: Some years ago, I conducted a very simple experiment with a sample of Pb (lead) metal being suspended by a thread hooked under a weight balance (0.001 grams resolution) and applied a DC magnetic field of Nd Magnet that showed something interesting (video available) that depicts the above situation (according to my view).

 

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Could you make the analysis on the last drawing with the rotating mass and to answer on (a) and (b)?

First critique, of course, is that there is no centrifugal force, and no reaction force should appear in a free-body diagram.

The acceleration (i.e. from the net force) is centripetal, and exerted by the rod.   a_c = v^2/r = w^2 r.     (v = wr)

You want to increase r  (from, say, r₁ to r₂) while keeping w constant. We are assuming M>>m so we don't have to worry about the axis of rotation moving (which it will)

That means the new centripetal force is larger, and also means v has to increase, which represents an increase in kinetic energy. Work has to be done, and a centripetal force doesn't do work. A tangential force must be applied by the rod. The final state of the system the rod exerts a larger tension such that a_c = w^2 r₂

The only way for the mass to move would be to release it from the connection to the rod, reducing the effective tension it feels. If it were free to move, it would travel a straight line tangent to its original path, which would increase r. Once you reconnect it to the rod the rod would be doing work on it to get it up to speed. But it's not free to move if there is still some attachment to the rod.

The difference in the KE tells you how much work the rod must do, and that tells you the force it must exert over the path. From that you might be able to solve for the path of the mass, but it will be a messy calculation.

Another thing we know is that if there are no external torques, angular momentum will be conserved. But we know the larger mass is rotating at a constant w and also transferring angular momentum and doing work on the small mass — the only way for that to happen is if it's being driven by an external torque, so angular momentum is not conserved.

3 hours ago, John2020 said:

The right frame rotates clockwise. Seeing the rotation of the circle on the left, shouldn't the right frame rotate counterclockwise in order to agree with the rotation of the circle on the left?

If you're sitting on a carousel that's rotating counter-clockwise, the world looks like it's rotating clockwise.

IOW, if we look at the top of the circle, the carousel is moving left, so it looks like the outside world is moving to the right.

[John2020]

26 minutes ago, swansont said:

First critique, of course, is that there is no centrifugal force, and no reaction force should appear in a free-body diagram.

The acceleration (i.e. from the net force) is centripetal, and exerted by the rod.   a_c = v^2/r = w^2 r.     (v = wr)

You want to increase r  (from, say, r₁ to r₂) while keeping w constant.

Your assertion does not represent the situation above. Maybe I was not clear (I thought it was obvious). In order the mass m to go to r₂, presupposes an increase in angular velocity ω₂ that implies a centrifugal force will rise pushing mass m to r₂ point. This is how I present the situation with the drawing.

26 minutes ago, swansont said:

Another thing we know is that if there are no external torques, angular momentum will be conserved. But we know the larger mass is rotating at a constant w and also transferring angular momentum and doing work on the small mass — the only way for that to happen is if it's being driven by an external torque, so angular momentum is not conserved.

I cannot follow you. The rotor is powered internally being the system in outer space. We start the rotor until it acquires a constant angular velocity and then we leave it. After some secs, there is an internal mechanism that will increase its angular velocity. Moreover, the rod is rigid that means the angular momentum can be transferred to mass m.

Edited October 21, 2020 by John2020

[Ghideon]

58 minutes ago, John2020 said:

, OK. It is like the drawing depicts a situation while being on the circle rotating counterclockwise, the surrounding space turns clockwise.

Correct.

Again: This is crucial to agree upon, otherwise the analysis will be ambiguous or incorrect: It is the same physical situation, governed by same physical laws. The difference is two different coordinate systems to describe the physics and two different viewpoints for the animation. 

This also means that a force having a physical effect in one frame must have the same physical effect in the other frame. If a force of 100N breaks* the string and the string break, then a force of 100N acted upon the string in both frames. Ok? 

Further; fictitious forces, when added to the pictures, will be different in the left and the right picture. Ok? 
If not obvious we can investigate further.

 

35 minutes ago, John2020 said:

Maybe I was not clear

Your examples are not clear, hence I move with more caution this time. 

 

*) Just a random example to illustrate. 

Edited October 21, 2020 by Ghideon

[John2020]

8 minutes ago, Ghideon said:

Further; fictitious forces, when added to the pictures, will be different in the left and the right picture. Ok? 
If not obvious we can investigate further.

In my example it is clear there is centrifugal force at play in the rotating frame of reference. 

[Ghideon]

4 minutes ago, John2020 said:

In my example it is clear there is centrifugal force at play in the rotating frame of reference. 

But is it clear why there is no centrifugal force at play in the non-rotating frame of reference in my example*? 

I am not commenting your advanced example until agreement is reached on how the basics work. Feel free to continue that dialogue with @swansont

 

*) as in any non rotating frame in any similar example. 

Edited October 21, 2020 by Ghideon
clarification which example that is addressed

[John2020]

1 minute ago, Ghideon said:

But is it clear why there is no centrifugal fore at play in the non-rotating frame of reference? 

I am not commenting your advanced example until agreement is reached on how the basics work. 

Yes.

[swansont]

46 minutes ago, John2020 said:

Your assertion does not represent the situation above. Maybe I was not clear (I thought it was obvious). In order the mass m to go to r₂, presupposes an increase in angular velocity ω₂ that implies a centrifugal force will rise pushing mass m to r₂ point. This is how I present the situation with the drawing.

You asked me for analysis, and that's my analysis. There is no centrifugal force, no matter how much you want there to be one. We could just as easily done this with a rope, where there physically could not be a centrifugal force. Using a rod doesn't sneak one into play.

Your analysis is inconsistent with Newton's laws. No force is necessary to move the mass to r₂. It would move there on its own. Look at Ghideon's animations. That shows the path the mass would take if it were able to move freely. See how r increases? No forces are acting on it once it's released.

 

46 minutes ago, John2020 said:

I cannot follow you. The rotor is powered internally being the system in outer space. We start the rotor until it acquires a constant angular velocity and then we leave it. After some secs, there is an internal mechanism that will increase its angular velocity. Moreover, the rod is rigid that means the angular momentum can be transferred to mass m.

You need a motor to keep this thing rotating. You could spin it up and turn the motor off, but the system wouldn't maintain constant w when the mass moved. Also, the motor would be spinning in the opposite direction to conserve angular momentum.

 

[John2020]

Just now, swansont said:

You need a motor to keep this thing rotating. You could spin it up and turn the motor off, but the system wouldn't maintain constant w when the mass moved. Also, the motor would be spinning in the opposite direction to conserve angular momentum.

I think this is already shown on the drawing (see the red and blue direction vectors).

[swansont]

1 minute ago, John2020 said:

I think this is already shown on the drawing (see the red and blue direction vectors).

You said you didn't follow my explanation. Does this explain the situation adequately, or not?

 

(You showed an arrow on your diagram. No mention of a motor, or that it would have its own mass and angular velocity)

[Ghideon]

23 minutes ago, John2020 said:

Yes.

Then we may check the magnitude of the forces. 

Left picture: In this coordinate system the ball follows a circular path*. Newton says that a change of velocity means a force is acting on the ball: F=ma >0. This force is pointing in towards the centre, in the direction of acceleration. No fictitious force exists.

Right picture: In this coordinate system the ball is not moving or accelerating. Newton says that since acceleration a=0 then m*a=0 so F=0. The force point inwards to the circle's center is still present, otherwise the string would have different tension in left and right image (physically impossible). The solution is to add a fictitious force, pointing outwards. This force, blue in the right picture below, is equal and opposite of the red force. This means that, in this coordinate system, the ball is not accelerating; F=ma=0.

Ok?

 

The fictitious force, centrifugal force in this case, does not change the physics.

 

*) (We can neglect the fact that the ball is thrown away at the end of the animation and use the part where ball is attached to string. I had no time to edit the animation)

Edited October 21, 2020 by Ghideon

[John2020]

1 minute ago, swansont said:

(You showed an arrow on your diagram. No mention of a motor, or that it would have its own mass and angular velocity)

The rotor turns counterclockwise (red arrow) and the stator clockwise (blue arrow).

8 minutes ago, Ghideon said:

Right picture: In this coordinate system the ball is not moving or accelerating. Newton says that since acceleration a=0 then m*a=0 so F=0. The force point inwards to the circle's center is still present, otherwise the string would have different tension in left and right image (physically impossible). The solution is to add a fictitious force, pointing outwards. This force, blue in the right picture below, is equal and opposite of the red force. This means that, in this coordinate system, the ball is not accelerating; F=ma=0.

Ok?

It is not so clear to me. Anyway, proceed further by analyzing my example to see if it fits to the above description. I have to go and I will be back a little bit later.

[Ghideon]

6 minutes ago, John2020 said:

It is not so clear to me.

Ok, Ask a more precise question and I'll add more explanations. 

 

6 minutes ago, John2020 said:

Anyway, proceed further by analyzing my example to see if it fits to the above description

Once we fully agree on the very basics of rotation and frames of reference.

[John2020]

1 hour ago, swansont said:

See how r increases? No forces are acting on it once it's released.

I think you misinterpret the drawing. Initially the ball is at r₁ when the rotor turns with constant angular velocity. r does not increases by its own. We have a fixed length of a rigid rod where mass m is allowed to slide over it but it is held in r₁ position due to friction (static friction should be better to say). On increasing the angular velocity of the rotor, the inertial centrifugal force has to overcome the static friction and push mass m (acceleration) upto position r₂. Up until the r₂ point the angular velocity increases thus the centrifugal too. Assuming being accelerated all the way down to r₂.

Edited October 21, 2020 by John2020

[John2020]

2 hours ago, Ghideon said:

Ok, Ask a more precise question and I'll add more explanations. 

In your example the system is in equilibrium , thus no acceleration and no centrifugal force is being developed on the rotating frame (assuming the ball is not thrown). That is all OK. My example is different since the system is not in equilibrium because of the change in angular velocity. Consequently, in my example there is an inertial centrifugal at play. Is it clear for you?

2 hours ago, Ghideon said:

Once we fully agree on the very basics of rotation and frames of reference.

All clear. We may proceed now.

Edited October 21, 2020 by John2020

[Ghideon]

43 minutes ago, John2020 said:

In your example the system is in equilibrium, thus no acceleration and no centrifugal force is being developed on the rotating frame (assuming the ball is not thrown).

The above is a misunderstanding that has to be addressed before moving on to your example. 
In uniform circular motion, that is moving with constant speed along a circular path, an object such as the ball experiences an acceleration resulting from the change of the direction of the velocity vector, while velocity vector magnitude remains constant. There is always a centrifugal force in the rotating frame, and it is constant when angular velocity, mass and radius is constant and greater than zero. 

 

Notes:
I asked about this your previous examples; why you never mentioned the centrifugal forces when angular velocity w was constant

There are other errors in your response but focusing on the above may solve the other ones

Edited October 21, 2020 by Ghideon

[John2020]

8 minutes ago, Ghideon said:

There is always a centrifugal force in the rotating frame, and it is constant when angular velocity, mass and radius is constant and greater than zero. 

For constant angular velocity there is an equilibrium in your example where the centripetal equals to the centrifugal force. The same conditions apply for my example when the angular velocity is constant.

 

8 minutes ago, Ghideon said:

Notes:
I asked about this your previous examples; why you never mentioned the centrifugal forces when angular velocity w was constant

Forget about the previous examples, I made some errors there. Let us focus on the example with the rotating mass. Is the drawing clear for you? Do you have any questions?

Edited October 21, 2020 by John2020

[Ghideon]

1 minute ago, John2020 said:

For constant angular velocity there is an equilibrium in your example where the centripetal equals to the centrifugal force.

Yes, in the rotating frame of reference centripetal force is equal and opposite to the fictitious centrifugal force
No, in the non-rotating frame of reference there is no equilibrium. centripetal force is zero (it does not exist)

Clear?

 

 

[John2020]

1 minute ago, Ghideon said:

Yes, in the rotating frame of reference centripetal force is equal and opposite to the fictitious centrifugal force
No, in the non-rotating frame of reference there is no equilibrium. centripetal force is zero (it does not exist)

Clear?

Clear. I just addressed above the case with the rotating frame. Of course in the non-rotating frame we have no centripetal force.

Edited October 21, 2020 by John2020

[Ghideon]

1 minute ago, John2020 said:

Clear. I addressed the case with the rotating frame.

Ok. Then you realise that Newton holds, there is no room fo any "interpretation" or circumventing. We apply the correct laws to your example and check the outcome. Ok?

[John2020]

Just now, Ghideon said:

Ok. Then you realise that Newton holds, there is no room fo any "interpretation" or circumventing. We apply the correct laws to your example and check the outcome. Ok?

OK!

[swansont]

2 hours ago, John2020 said:

I think you misinterpret the drawing. Initially the ball is at r₁ when the rotor turns with constant angular velocity. r does not increases by its own.

But r can increase on its own. You don't need a radial force for this to happen. If there is no radial force, it travels in a straight line, which increases the distance to the center of the circle.

 

Quote

We have a fixed length of a rigid rod where mass m is allowed to slide over it but it is held in r₁ position due to friction (static friction should be better to say). On increasing the angular velocity of the rotor

I read the part where you said angular velocity was constant.

If you increase it, the centripetal force must increase. If friction does not permit this to happen, then the net force is insufficient to keep it moving in a circle. There's no radial force pushing outward on it - just friction. But you had a_c1 = v1^2/r1 = w1^2 r1 = F_fr1/m

 If you increase to w2, the frictional force can't supply the required centripetal force. The mass will slide because of this. Not because there is an outward force. And it won't stop, because the force needed for circular motion is larger than friction can supply. It continues to slide until it hits the stop.

 

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, the inertial centrifugal force has to overcome the static friction and push mass m (acceleration) upto position r₂. Up until the r₂ point the angular velocity increases thus the centrifugal too. Assuming being accelerated all the way down to r₂.

There is no centrifugal force. Friction is not directed to the outside. There is an inward radial component. That's all there is.

 

[John2020]

44 minutes ago, swansont said:

 If you increase to w2, the frictional force can't supply the required centripetal force. The mass will slide because of this.

From the moment the friction plays the role of the centripetal, before we increase the angular velocity the system is in equilibrium that means the centrifugal equals to the centripetal. Increasing the angular velocity the net force will have an outwards direction that implies centrifugal is greater than centripetal. Of course, the centripetal will increase too, however as long the angular velocity increases, the centrifugal will be greater than the centripetal. 

Moreover the centripetal has a direction towards the center of the rotor and the net force is outwards, otherwise mass m acceleration to the right cannot be justified. Additionally, the rotor is not fixed but in outer space.

Edited October 21, 2020 by John2020

[swansont]

9 minutes ago, John2020 said:

From the moment the static friction plays the role of the centripetal, before we increase the angular velocity the system is in equilibrium that means the centrifugal equals to the centripetal. Increasing the angular velocity the net force will have an outwards direction that implies centrifugal is greater than centripetal. Of course, the centripetal will increase too, however as long the angular velocity increases, the centrifugal will be greater than the centripetal. 

Moreover the centripetal has a direction towards the center of the rotor and the net force is outwards, otherwise mass m acceleration to the right cannot be justified. Additionally, the rotor is not fixed but in outer space.

I am analyzing it in an inertial frame, which is what you indicate in the drawing. The object is rotating, and it clearly labeled as such, so there is no centrifugal force.

If you are in the rotating frame, there is no initial rotation of the object. So you have to pick one. Both cannot exist in the same analysis. Either the object is rotating (inertial frame of reference) or the item is fixed and the reference frame is rotating.

 

[John2020]

2 minutes ago, swansont said:

Either the object is rotating (inertial frame of reference) or the item is fixed and the reference frame is rotating.

Of course the object is rotating. How you are going to make the analysis has nothing to do with me. The goal/challenge is (as we agreed with Ghideon) to analyze this system from an inertial frame of reference as also from a rotating frame of reference and then to compare the results.

[swansont]

3 minutes ago, John2020 said:

Of course the object is rotating. How you are going to make the analysis has nothing to do with me. The goal/challenge is (as we agreed with Ghideon) to analyze this system from an inertial frame of reference as also from a rotating frame of reference and then to compare the results.

And I did the inertial frame. There is no centrifugal force in the inertial frame. There are no fictitous forces, because things in inertial frames follow Newton's laws.

When you do the analysis in the rotating frame, it will start at rest in that frame.