# Thread: Tilting a spinning wheel

1. Hi guys,

Please watch this video :https://www.youtube.com/watch?v=UZlW1a63KZs, when the girl (at 0:50) pushes her right hand down, making the wheel rotate vertically to her left, the platform turns to her right.
Does this happens because of the 3rd law of motion (as if whe were kicking a football)?

Thanks a lot

2.

3. Originally Posted by whizkid
Hi guys,

Please watch this video :https://www.youtube.com/watch?v=UZlW1a63KZs, when the girl (at 0:50) pushes her right hand down, making the wheel rotate vertically to her left, the platform turns to her right.
Does this happens because of the 3rd law of motion (as if whe were kicking a football)?

Thanks a lot
Well the video explains the effect is due to conservation of angular momentum, i.e. she rotates in a sense opposite to that of the wheel, such that total angular momentum about the vertical axis remans zero.

But do you want an explanation in terms of forces and linear motion? That's quite complicated.

4. Originally Posted by exchemist
do you want an explanation in terms of forces and linear motion? That's quite complicated.
Hi exchemist, nice to hear from you .
I watched a video in which a prof said what is relevant here is 3rd law, and it made sense to me:
I am on a frictionless platform push a football/wheel to the left and start spinning to the right.
Why complicated? : m*v = m1*v1 = (50*.4= 20), if the ball has mass = 1 kg it must be kicked off at v = 20m/s
is that right?

If it is simpler for you ,
suppose the wheel has m=2kg, r =0.318, v =10m/s
the girl-platform m= 50 kg, r = .45, v = 0.40/s

Can you show me how it works with angular momentum?

Thanks

5. Originally Posted by whizkid
Originally Posted by exchemist
do you want an explanation in terms of forces and linear motion? That's quite complicated.
Hi exchemist, nice to hear from you .
I watched a video in which a prof said what is relevant here is 3rd law, and it made sense to me:
I am on a frictionless platform push a football/wheel to the left and start spinning to the right.
Why complicated? : m*v = m1*v1 = (50*.4= 20)

If it is simpler for you ,
suppose the wheel has m=2kg, r =0.318 v =10m/s
the girl-platform m= 50 kg, r = .45, v = 0.40/s
Can you show me how it works with angular momentum?

Thanks
I'll have a go at explaining in terms of linear forces and momentum, then. But I'm going to set to one side your numbers and revert to the video you cited.

The girl holds the wheel with its axis horizontal and it is spun so that the rim that is close to her is moving upward. She then starts to move her right hand downward and her left hand upward.

Think about a segment of the rim close to her. It has linear momentum vertically upward, right? But when she starts to tilt the axis clockwise, as she sees it, this segment is having its direction of motion changed from vertically upward to partly upward but now with a component of motion from left to right. The momentum of the segment of rim in question will resist that change, creating a reaction force to the LEFT.

Now think about a segment of rim on the opposite side. It is initially moving vertically downwards, but as she tilts the axis it is being forced to develop a component of motion to the left. Due its vertical linear momentum, this creates a reaction force to the RIGHT.

So we now have a couple, twisting the axis of the wheel, clockwise as seen from above.

Whereas the spin of the wheel itself, progressively becomes an anticlockwise spin - again seen from above - as its axis is rotated from horizontal to vertical.

That is the root of how gyroscopic forces arise with spinning objects.

Thinking about conservation of angular momentum is a short cut that works, without the tedious necessity for analysing segments of rotating objects all the time.

As a matter of fact what the video ignores is that the turning the axis of rotation will also create another couple, tending to make the girl's head move forward and her feet move backward, so that she starts to tumble, thus conserving the angular momentum in the original direction (when the axis was vertical). Clearly one can't demonstrate that without putting her in a 3D gimbal arrangement.

6. Originally Posted by exchemist
So we now have a couple, twisting the axis of the wheel, clockwise as seen from above.
.
You are fantastic, echemist, it's the first time precession has been clearly explained. (I suppose we can call that, right?)
I'll ask you more later, now, could you be so kind and put the right numbers and formulas into this practical example?, I have no clue as to how to calculate work done by the girl.
If you think my numbers are not plausible (I figured them out from the video) you can change them as you like.

7. Originally Posted by whizkid
Originally Posted by exchemist
So we now have a couple, twisting the axis of the wheel, clockwise as seen from above.
.
You are fantastic, echemist, it's the first time precession has been clearly explained. (I suppose we can call that, right?)
I'll ask you more later, now, could you be so kind and put the right numbers and formulas into this practical example?, I have no clue as to how to calculate work done by the girl.
If you think my numbers are not plausible (I figured them out from the video) you can change them as you like.
To be honest I don't think I can do this. The thing is, not only does the girl acquire angular momentum (and thus kinetic energy) about the vertical axis, but she also prevents (by being constrained from toppling forward) the conservation of angular momentum about the horizontal axis from taking place as it should, in a free system. I am not at all sure what overall effect this will have on the final rate of spin of the wheel, compared to its initial rate of spin.

Something else that is nagging at me is that I think I vaguely recollect that angular momentum is only conserved about a common axis. Now the girl and the wheel are not on a common axis: they have parallel axes but different ones. There is, I think, a thing called the parallel axis theorem that may come into it, but I'm not sure.

I think we may need a real physicist to help us understand how to calculate how the initial kinetic energy becomes distributed among the two rotating bodies (wheel and girl) and the various axes of rotation.

8. Originally Posted by exchemist
I think we may need a real physicist .
You have seen in the other thread that are no physicists around. But I started this thread to discuss the issue of no-work-on-normal-direction. This is another example od that issue: both torques right-hand-down and precession-horizontal-clockwise-rotation are perpendicular to the plane of rotation and should require no work, but the girl does work great enough to set the platform moving.

Waiting for a physicist to help us, let's consider the very beginning of the video, I suppose the real case of conservation of L .

Suppose a ball of mass 1kg is rotating on a frictionless surface on a string r = .20 m (that passes through a hole at the centre, and that we hold perpendicular under the surface) at v 10 m/s. L equals to 2 J*s, KE = 50J
Suppose now we pull the string by 0.1 m: L stays 2 J*s, v becomes 20 m/s and therefore KE = 100J 50+50, right?
If that is right, where does the surplus of energy come from?

9. Originally Posted by whizkid
Originally Posted by exchemist
I think we may need a real physicist .
You have seen in the other thread that are no physicists around. But I started this thread to discuss the issue of no-work-on-normal-direction. This is another example od that issue: both torques right-hand-down and precession-horizontal-clockwise-rotation are perpendicular to the plane of rotation and should require no work, but the girl does work great enough to set the platform moving.

Waiting for a physicist to help us, let's consider the very beginning of the video, I suppose the real case of conservation of L .

Suppose a ball of mass 1kg is rotating on a frictionless surface on a string r = .20 m (that passes through a hole at the centre, and that we hold perpendicular under the surface) at v 10 m/s. L equals to 2 J*s, KE = 50J
Suppose now we pull the string by 0.1 m: L stays 2 J*s, v becomes 20 m/s and therefore KE = 100J 50+50, right?
If that is right, where does the surplus of energy come from?
You have to do work against the mv²/r centrifugal force of the ball in order to pull in the string, don't you?

10. Originally Posted by exchemist

You have to do work against the mv²/r centrifugal force of the ball in order to pull in the string, don't you?
That force is fictitious , isn't it? v^2/r makes 500 J, anyway, and, in any case, you are not applying any force tangentially, so you cannot increase velocity, can you?

moreover, if you release the string to 0.4 m, speed falls down to 5 m/s, KE dropping to 12.5J , losing 37.5 J, how that?

Can anyone make an experiment?

11. Originally Posted by whizkid
Originally Posted by exchemist

You have to do work against the mv²/r centrifugal force of the ball in order to pull in the string, don't you?
That force is fictitious , isn't it? v^2/r makes 500 J, anyway, and, in any case, you are not applying any force tangentially, so you cannot increase velocity, can you?

moreover, if you release the string to 0.4 m, speed falls down to 5 m/s, KE dropping to 12.5J , losing 37.5 J, how that?

Can anyone make an experiment?
No Slow down and think harder.

If you hold the string, will you feel a pull from the rotating ball? Yes.

So, if you pull in the string, are you doing work (F x d) ? Yes.

So you ARE putting energy into the system. The only question is how you do the analysis to find out how much and where it is going.

I called the force in question centrifugal for ease of communication, though indeed you are right that classically one is taught to think of it as the force needed to provide the constant centripetal acceleration that the ball requires, in order to keep it moving in a circle. But whatever terminology you use, there is no doubt that you feel a force on the string and if you let go the force will vanish and the circular motion will cease. Right?

I'll stop here to check we agree and then go further once this is established.

12. Originally Posted by exchemist
So, if you pull in the string, are you doing work (F x d) ? Yes. So you ARE putting energy into the system.
you feel a force on the string and if you let go the force will vanish and the circular motion will cease. Right?
.
I am doing work the same as when I lift a book and put it on a shelf.
When I pull the string I am contrasting a radial force. When I lift a book and put it on a shelf I am not putting energy into the system. Energy spent has vanished into PE. I hope you won't say that centrifugal force is real and has PE.
Anyway, if for some technical reason it is as you say (?) the energy is spent radially and there is no way energy can be put tangentially into the system. Lastly, the value of v^2/r does not even fit your hypothesis

Then there is the reverse case when I loosen the string : no work done, same as when I let a book drop to the ground.
Do you agree?

13. Originally Posted by whizkid
Originally Posted by exchemist
So, if you pull in the string, are you doing work (F x d) ? Yes. So you ARE putting energy into the system.
you feel a force on the string and if you let go the force will vanish and the circular motion will cease. Right?
.
I am doing work the same as when I lift a book and put it on a shelf.
When I pull the string I am contrasting a radial force. When I lift a book and put it on a shelf I am not putting energy into the system. Energy spent has vanished into PE. I hope you won't say that centrifugal force is real and has PE.
Anyway, if for some technical reason it is as you say (?) the energy is spent radially and there is no way energy can be put tangentially into the system. Lastly, the value of v^2/r does not even fit your hypothesis

Then there is the reverse case when I loosen the string : no work done, same as when I let a book drop to the ground.
Do you agree?
Whizkid yes you do work when you put a book on the shelf. W = F x d. You are increasing the gravitation potential energy of the book and this requires work to be done. So you ARE putting energy into the "system". Conservation of energy demands this. It is just that it is not kinetic energy in this case.

When you pull the string in your scenario, you pull against a force. F x d = W, ergo you do work. Which means some form of energy in the system must increase.

Now you said yourself that conservation of angular momentum tells you the kinetic energy of the system is higher after you have pulled in the string. So, what do you think has happened?

As to the idea that the energy is put in radially that is clearly not so if you stop and think about it for a second. Your ball is initially circling at radius r = 0.2m. As you start to pull the string, the path taken by the ball becomes not a circle but a spiral, because the radius is now progressively decreasing. That means there is a component of force that is not acting at right angles to the ball's direction of travel. This will give it angular acceleration.

As for what happens when you loosen the string again you allow the system to do work on you, by allowing it to move your hand a distance , d, against the restraining force, F, that you continue to exert to stop the whole thing flying off tangentially. F x d = W. So it is simply the reverse process.

As for when you drop your book to the ground, gravity accelerates it, converting the potential energy into kinetic, which is then released as sound and heat when the book hits the floor. This last bit is very, very basic, Whizkid. If you are still not understanding this level of basic mechanics, it calls into question the value of our previous discussions.

Note added later: I've found an identical explanation to the one I've given you, for a skater pulling in her arms, here: http://physics.stackexchange.com/que...tia-is-changed

14. Originally Posted by exchemist

That means there is a component of force that is not acting at right angles to the ball's direction of travel. This will give it angular acceleration.
You should consider that if you give a moving body momentum/acceleration/Ke in a normal direction you are accelerating it and changing the direction:
if a flying ball is hit perpendicularly by another identical ball you can calculate the result by a parallelogram. Same for a cannonball fired horizontally from a height. (yet you are not altering its horizontal velocity, no acceleration in that direction)
..
When you work against a force you do not give moventum/v/KE in that direction: when you lift a book, the book stops when you stop lifting. You do not accelerate it horizontally, if you are running.
The moon is flying above the earth: you work against Centrifugal-force by G, pulling it towards the earth. It returns into the orbit but does not accelerate , its speed does not increase: you are not putting any energy in the system, let alone tangentially.

Isn't it so?
If you think I am wrong, start calculating the energy you spend to pull the string from r =.2 to r =.1 m, and see if it fits

15. Originally Posted by whizkid
Originally Posted by exchemist

That means there is a component of force that is not acting at right angles to the ball's direction of travel. This will give it angular acceleration.
You should consider that if you give a moving body momentum/acceleration/Ke in a normal direction you are accelerating it and changing the direction:
if a flying ball is hit perpendicularly by another identical ball you can calculate the result by a parallelogram. Same for a cannonball fired horizontally from a height. (yet you are not altering its horizontal velocity, no acceleration in that direction)
..
When you work against a force you do not give moventum/v/KE in that direction: when you lift a book, the book stops when you stop lifting. You do not accelerate it horizontally, if you are running.
The moon is flying above the earth: you work against Centrifugal-force by G, pulling it towards the earth. It returns into the orbit but does not accelerate , its speed does not increase: you are not putting any energy in the system, let alone tangentially.

Isn't it so?
If you think I am wrong, start calculating the energy you spend to pull the string from r =.2 to r =.1 m, and see if it fits
No, Whizkid. Rather than start introducing new scenarios I strongly suggest you try first to get your head round the one you originally proposed (the ball on the string). That is an excellent scenario for understanding the process (for which I congratulate you) but adding further examples at this stage is just a recipe for confusion. So I'm going instead to slow right down and go step by step.

If you don't like my explanation of your ball on a string scenario please read the link about the skater. It's the same problem and the explanation is essentially the same as the one I have given you but in different words, so maybe it will be easier for you.

And then answer the question I posed to you earlier, namely if:

a) as you yourself calculated, conservation of angular momentum implies the kinetic energy of the ball increases as you pull the string and if

b) when you pull the string you are exerting a force through a distance and are thus doing work, since F x d = W……………..

………...what do you think might be happening?

Let's agree this first. Then we can consider the dynamics that lead to it.

16. Originally Posted by exchemist
………...what do you think might be happening? .
I do not imagine, echemist, I am too ignorant. I am just a reflexive student that browses in the prairies of physics and is perplexed when he thinks there are some contradictions such as: hearing that changing the direction of (L) a spinning wheel requires no work/energy and then seeing that a girl just by doing that sets a platform of 50 kg spinning.

It may be as well as you say, I am just offering you arguments (I hope not so stupid) to your reflexion.
I cannot imagine what happens until you tell me why the moon should behave differently from the ball.
The parallel seems adequate, perfectly fitting to me, you show me why it is not, what makes the difference and that'll make me understand faster.
I hope you don't mind.

17. Originally Posted by whizkid
Originally Posted by exchemist
………...what do you think might be happening? .
I do not imagine, echemist, I am too ignorant. I am just a reflexive student that browses in the prairies of physics and is perplexed when he thinks there are some contradictions such as: hearing that changing the direction of (L) a spinning wheel requires no work/energy and then seeing that a girl just by doing that sets a platform of 50 kg spinning.

It may be as well as you say, I am just offering you arguments (I hope not so stupid) to your reflexion.
I cannot imagine what happens until you tell me why the moon should behave differently from the ball.
The parallel seems adequate, perfectly fitting to me, you show me why it is not, what makes the difference and that'll make me understand faster.
I hope you don't mind.
But who says changing the direction of a spinning wheel takes no energy? I assume you are now jumping back to the scenario of the girl and the wheel. Surely all that has been asserted in that case is that angular momentum does not change. I don't recall anyone saying no energy was expended by the girl - unless I made a typo through replying too fast or something).

On the contrary, I suspect energy is expended, in that a gyroscope resists having its orientation changed and that implies a force has to be applied through a distance (F x d = W) in order to make it happen.

As we've seen so elegantly in your nice simple scenario of the ball on the string, conservation of angular momentum does NOT imply no energy change to the system.

Do you now understand this? If so we can move on to the moon etc BUT NOT BEFORE.

18. Originally Posted by exchemist
But who says changing the direction of a spinning wheel takes no energy?
...conservation of angular momentum does NOT imply no energy change to the system.
You are right, you never said that, but that's what they say, and what you and the 'physicists' implied when you refuted that when tilting the axis of the magnetic moment mu you must also account for the work done to tilt the direction of L, the mechanical angular momenum, too
but we can postpone this.

You said in the other thread that it takes no energy/work to change the direction of a moving body.
That is the case of the moon. The ball on a string/spoke seem similar to me. In the other thread I mentioned a man jumping on an asteroid.
you can choose what you think best.
In the case of the ball , if that can avoid misunderstandings, consider it on a telescopic spoke , so that there is no doubt the force is applied absolutely perpendicular to the motion and there is no torque whatsoever.

19. Try backing up a bit and think that the forces in the rotating body are balanced. If you want to you can consider the revolving body as traveling in a straight line according to its frame of reference.

Maybe this makes it a bit easier to understand why forcing it to change direction (tilting the wheel) means you are applying a force to it, what direction you apply the force in, and why it is at an angle to the axis of the rotation.

20. Originally Posted by dan hunter
Maybe this makes it a bit easier to understand why forcing it to change direction (tilting the wheel) means you are applying a force to it, what direction you apply the force in, and why it is at an angle to the axis of the rotation.
I am convinced of that, others are not.

21. Originally Posted by whizkid
Originally Posted by exchemist
But who says changing the direction of a spinning wheel takes no energy?
...conservation of angular momentum does NOT imply no energy change to the system.
You are right, you never said that, but that's what they say, and what you and the 'physicists' implied when you refuted that when tilting the axis of the magnetic moment mu you must also account for the work done to tilt the direction of L, the mechanical angular momenum, too
but we can postpone this.

You said in the other thread that it takes no energy/work to change the direction of a moving body.
That is the case of the moon. The ball on a string/spoke seem similar to me. In the other thread I mentioned a man jumping on an asteroid.
you can choose what you think best.
In the case of the ball , if that can avoid misunderstandings, consider it on a telescopic spoke , so that there is no doubt the force is applied absolutely perpendicular to the motion and there is no torque whatsoever.
Whizkid, I agree we should NOT now jump to quantum theory of the atom. Let's deal with this macroscopic classical mechanics issue first. There is ample scope for me or another respondent to have inadvertently confused you in some way - that is the imperfect nature of these forums. One reacts quickly, in writing, without seeing one's interlocutor. This is why I prefer to go slowly, one issue at a time, when something is apparently not clear.

What I said, I think, is that there is no work done to keep a body moving in a circle. That is what happens if you hold the string, in your ball on string scenario, but do not pull it in or let it out. In that situation, no work is done because the force does not act through a distance in the direction of the force. So d = 0 and hence F x d = 0. This is why the Earth does no work on the moon to keep it in orbit. The point is that, in both cases, though a net force acts on the circulating object in both cases, that force is always perpendicular to the direction of motion. So the force does not act through a distance. A spiral is different though. As the string pulls the ball in it is forcing it to deviate from its circular path, inward, i.e. it is giving it a component of motion in the direction of the string - which is the direction of action of the force . So the force starts to act through a distance - and work begins to be done. If you replace the string by a telescopic spoke it makes no difference, you are still pulling it inward, which means you are making the force act through a distance. So work is done.

22. Originally Posted by exchemist
I agree we should NOT now jump to quantum theory of the atom.
I prefer to go slowly, one issue at a time,.
Right, one issue at a time.
1) the issue I raised in the 1H thread had nothing to do with QM.
I maintained that when you twist a spinning electron you must win the resistance both of mu and L. If you say that an electron has intrinsic (mechanical) angular momentum then you must grant that if you try to rotate mu you are rotating L, too, and so doing extra work.

If you now agree on that, that is settled : " It takes work/energy to change the direction of angular momentum".
If you think, for some obscure reason, that that rule does not apply to orbiting electrons, please explain.

2) if you (and mainstream) say(s) that " it takes no work to change the direction of linear momentum (in any context)" I am prepared to give you sound counter-arguments. If you agree, there is no dispute.

23. Originally Posted by whizkid
Originally Posted by exchemist
I agree we should NOT now jump to quantum theory of the atom.
I prefer to go slowly, one issue at a time,.
Right, one issue at a time.
1) the issue I raised in the 1H thread had nothing to do with QM.
I maintained that when you twist a spinning electron you must win the resistance both of mu and L. If you say that an electron has intrinsic (mechanical) angular momentum then you must grant that if you try to rotate mu you are rotating L, too, and so doing extra work.

If you now agree on that, that is settled : " It takes work/energy to change the direction of angular momentum".
If you think, for some obscure reason, that that rule does not apply to orbiting electrons, please explain.

2) if you (and mainstream) say(s) that " it takes no work to change the direction of linear momentum (in any context)" I am prepared to give you sound counter-arguments. If you agree, there is no dispute.
Neither point relates to my previous post, so I will not respond.

24. Originally Posted by exchemist
This is why the Earth does no work on the moon to keep it in orbit. .
This is not the issue we started in the 1H thread. But if you care:

The moon is not on a string. Any body flying in the sky bends its trajectory when it gets near to another body and gets acceleration (gravitational slingshot) if it is captured in an orbit this process continues unaltered , the only difference that the route correction by g is so precise to take back the body exatly on the right orbit. This is less visible in circular than in elliptical orbits. That is why orbiting ships feel no g, because the body is free to fall. Oscillations may be as little as you wish but they must be there.
That's what they taught me.
But our discussion was about jumping on an asteroid, wasn't it?

Thanks for the discussion , is really nice to talk to you.

25. Originally Posted by whizkid
Originally Posted by exchemist
This is why the Earth does no work on the moon to keep it in orbit. .
This is not the issue we started in the 1H thread. But if you care:

The moon is not on a string. Any body flying in the sky bends its trajectory when it gets near to another body and gets acceleration (gravitational slingshot) if it is captured in an orbit this process continues unaltered , the only difference that the route correction by g is so precise to take back the body exatly on the right orbit. This is less visible in circular than in elliptical orbits. That is why orbiting ships feel no g, because the body is free to fall. Oscillations may be as little as you wish but they must be there.
That's what they taught me.
But our discussion was about jumping on an asteroid, wasn't it?

Thanks for the discussion , is really nice to talk to you.
No. Our discussion was about a ball moving on a frictionless table, constrained by a string to move in a circle, and what happens when you pull the string, in terms of angular momentum and kinetic energy.

This scenario was introduced by you to help understand the video of the girl with the wheel.

I do not want now to start talking about asteroids, or quantum theory, or anything else, while there remain outstanding issues with the ball on the string. I've explained to you what happens when you pull the string and I've sent you a link to another explanation of a skater pulling in her arms, but you have not reacted to the points made in either of those, except by proposing yet more scenarios of various kinds. This in my view will not be productive.

26. Originally Posted by exchemist
. Our discussion was about a ball moving on a frictionless table,
Our discussion originated in the 1H thread, you can check what was the issue.
The ball on the table was a detour to discuss a side aspect, I wanted to get eventually to the H electron falling from n2 to n1 or the like. That example has no relevance in our discussion , as it is patent that moving the ball requires some work, I take your word for the fact work done radially corresponds exactly to the value of KE increase, we can postpone or skip it.

Do you want to discuss "changing the direction of momentum" ?
do you know of any case in which you can change P or L doing no work, spending no energy?

27. Originally Posted by whizkid
Originally Posted by exchemist
. Our discussion was about a ball moving on a frictionless table,
Our discussion originated in the 1H thread, you can check what was the issue.
The ball on the table was a detour to discuss a side aspect, I wanted to get eventually to the H electron falling from n2 to n1 or the like. That example has no relevance in our discussion , as it is patent that moving the ball requires some work, I take your word for the fact work done radially corresponds exactly to the value of KE increase, we can postpone or skip it.

Do you want to discuss "changing the direction of momentum" ?
do you know of any case in which you can change P or L doing no work, spending no energy?
Obviously yes, since the example we have been discussing shows that to conserve momentum, energy changes. So the converse can clearly apply, that if energy does NOT change then momentum can be expected to.

28. Originally Posted by exchemist
Originally Posted by whizkid
Originally Posted by exchemist
I agree we should NOT now jump to quantum theory of the atom.
I prefer to go slowly, one issue at a time,.
Right, one issue at a time.
1) the issue I raised in the 1H thread had nothing to do with QM.
I maintained that when you twist a spinning electron you must win the resistance both of mu and L. If you say that an electron has intrinsic (mechanical) angular momentum then you must grant that if you try to rotate mu you are rotating L, too, and so doing extra work.

If you now agree on that, that is settled : " It takes work/energy to change the direction of angular momentum".
If you think, for some obscure reason, that that rule does not apply to orbiting electrons, please explain.

2) if you (and mainstream) say(s) that " it takes no work to change the direction of linear momentum (in any context)" I am prepared to give you sound counter-arguments. If you agree, there is no dispute.
Neither point relates to my previous post, so I will not respond.
Orbiting electrons are not really orbiting, nor are they really spinning when they are said to spin up or down.
There is no way to actually taste the different flavours of quarks either.
A lot of the terms from QM do not really translate into classical mechanics terms, or vice versa.

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