Does it really work? [Answered - No]

[ABV]

It depends on a lot of factors. I was hoping you would draw the force vectors and see the difference between the disc and ring and to see what it meant to have the "same" rolling resistance for decelerating discs or rings. The resultant force vectors cannot have the same magnitude, direction and application point in each case (unless at least one of them is skidding to some extent)

 

I agree. Real rolling resistance depends on many factors.

But I disagree if we talk about ideal models.

Anyway rolling resistance is just a force. Even if increase force, acting time for deceleration period going to be short. But deceleration acting time depends from rolling body kinetic energy, which will be spend to this deceleration (i.e work).

In this case no matter what value of force will apply to rolling body, net momentum will depend from work which rolling body does at deceleration period.

Edited May 29, 2009 by ABV

[J.C.MacSwell]

I agree. Real rolling resistance depends on many factors.

But I disagree if we talk about ideal models.

Anyway rolling resistance is just a force. Even if increase force, acting time for deceleration period going to be short. But deceleration acting time depends from rolling body kinetic energy, which will be spend to this deceleration (i.e work).

In this case no matter what value of force will apply to rolling body, net momentum will depend from work which rolling body does at deceleration period.

 

A force is a vector quantity.

 

You have chosen to model the ring with the same mass as the disc but higher moment of inertia. They have different ratios of mass to moment of inertia. If they roll without slipping while decelerating the decelerating force cannot be applied the same way. Not in the real world and not in an idealized model.

 

You can simplify and say they are approximately the same, but don't point to the difference later and say "Paradox". When forces don't balance or momentums don't balance in an isolated system that is a sign that you have made a mistake, not a sign that there is an exception to a physical law.

Edited May 30, 2009 by J.C.MacSwell

[ABV]

A force is a vector quantity.

 

You have chosen to model the ring with the same mass as the disc but higher moment of inertia. They have different ratios of mass to moment of inertia. If they roll without slipping while decelerating the decelerating force cannot be applied the same way. Not in the real world and not in an idealized model.

The rolling resistance applies identically for thin ring and solid disk. It just depend on mass and rolling resistance coefficient. A different ratio of mass gives different acting time to same velocity destination.

 

 

You can simplify and say they are approximately the same, but don't point to the difference later and say "Paradox". When forces don't balance or momentums don't balance in an isolated system that is a sign that you have made a mistake, not a sign that there is an exception to a physical law.

 

This is a main point. The deceleration momentums are different in these two cases.

That because I’m calling this PARADOX.

[J.C.MacSwell]

The rolling resistance applies identically for thin ring and solid disk.

 

So you can prove this by drawing it on a free body diagram.

 

Draw me a force, that will slow them down as you describe.

[ABV]

So you can prove this by drawing it on a free body diagram.

 

Draw me a force, that will slow them down as you describe.

 

Please look on few posts up to wikipedia link. You'll see there a generic rolling resistance diagram. Doesn't matter thin ring or solid disk, rolling resistance depend just on rolling body mass and rolling resistance coefficient. No any description about a body shape.

[Bignose]

Why don't you just post a drawing so that everyone can see clearly what you're talking about? I've read through this thread, and I personally cannot tell what exactly you are talking about. J.C. has been a saint in this thread, most people would have given up a long time ago.

 

The big thing, ABV, is that the onus is on you to demonstrate the issue, since you are the one claiming it. The onus is not on any of us to go to wikipedia and use their generic picture in your situation. If you want to convince us of the issue, draw us a diagram of your specific situation, put in all the forces, and show us what the issue is. We shouldn't have to, and most of us aren't going to piece together all the parts of the puzzle for you -- you should have the puzzle completely together before you start, and give that to us. Then, we can look at the whole picture instead of it piece by piece.

[ABV]

Why don't you just post a drawing so that everyone can see clearly what you're talking about? I've read through this thread, and I personally cannot tell what exactly you are talking about. J.C. has been a saint in this thread, most people would have given up a long time ago.

 

The big thing, ABV, is that the onus is on you to demonstrate the issue, since you are the one claiming it. The onus is not on any of us to go to wikipedia and use their generic picture in your situation. If you want to convince us of the issue, draw us a diagram of your specific situation, put in all the forces, and show us what the issue is. We shouldn't have to, and most of us aren't going to piece together all the parts of the puzzle for you -- you should have the puzzle completely together before you start, and give that to us. Then, we can look at the whole picture instead of it piece by piece.

 

The rolling resistance is very simple thing. This is enough open sources like (wikipedia for example) to show this force does not depend on rolling body shape.

 

 

I don't understand, why this simple thing is going to puzzle.

 

 

Therefore, just keep in mind, doesn’t matter what force value it would, acting time depends on force value and work (kinetic energy difference). Base on these two values multiplication (deceleration force and acting time) we’ll get momentum which gains to track. The thin ring momentum is higher than solid disk momentum, because thin ring discharges more kinetic energy for same velocity differentiation than solid disk.

[J.C.MacSwell]

The rolling resistance is very simple thing. This is enough open sources like (wikipedia for example) to show this force does not depend on rolling body shape.

 

 

I don't understand, why this simple thing is going to puzzle.

 

 

Therefore, just keep in mind, doesn’t matter what force value it would, acting time depends on force value and work (kinetic energy difference). Base on these two values multiplication (deceleration force and acting time) we’ll get momentum which gains to track. The thin ring momentum is higher than solid disk momentum, because thin ring discharges more kinetic energy for same velocity differentiation than solid disk.

 

If you draw it, you might see the difference and begin to understand.

 

It is not the shape, it's the way the force vector must be applied in each case, to slow down the disc and ring without slipping.

 

They have the same amount of momentum.

 

They have different amounts of angular momentum.

 

This affects the way any force vector must be applied to slow down the disc or ring without slipping.

 

You might then see how it applies in the simplest case of a fixed straight track. If you cannot do this how can you possibly understand the more complicated isolated circuit track.

---

Merged post follows:

Consecutive posts merged

Why don't you just post a drawing so that everyone can see clearly what you're talking about? I've read through this thread, and I personally cannot tell what exactly you are talking about. J.C. has been a saint in this thread, most people would have given up a long time ago.

 

The big thing, ABV, is that the onus is on you to demonstrate the issue, since you are the one claiming it. The onus is not on any of us to go to wikipedia and use their generic picture in your situation. If you want to convince us of the issue, draw us a diagram of your specific situation, put in all the forces, and show us what the issue is. We shouldn't have to, and most of us aren't going to piece together all the parts of the puzzle for you -- you should have the puzzle completely together before you start, and give that to us. Then, we can look at the whole picture instead of it piece by piece.

 

Thanks BN. The curious thing is, that he offered to do just that, fix any errors or omissions if I pointed them out, but he seems to balk at doing it. From his link in the first post he is obviously capable of drawing, he just needs to slow down and do it.

 

I would be happy if he at least gets this one part right for now, the deceleration of the disc and ring on a fixed straight track. I suspect he will not get it right the first time he tries it, since he thinks they are the same, but it will be a start.

[navigator]

ABV: What the experiment shows is that by changing the shape of the mass in motion, compressing the angular momentum to a more usable place, it develops more force.

 

Lets take the experiment one step further and change the shape a little more as well as the path it travels.

 

Instead of having a constant overall diameter. Make the ring cone shaped so the longest diameter is in the very center and gradually decreases until at its widest point it reaches the open center of the ring.

 

Instead of using a flat track, use 2 rails with the exact same height, oval size etc. The rails at the top will be wide enough so the ring rides the widest part of the ring, which is also the smallest diameter of the cone shape, but doesn't fall through. The rails run parallel until it reaches the bottom of the track and begins to enter the loop. At this point the rails begin to narrow and continue until at the 9:00 position, it is narrowest and the cone shaped ring is intersecting the rails at its longest diameter.

 

Not only are we taking advantage of angular momentum, we have added an increase to the radial momentum.

 

We have again developed more force with the same input. The only change was the shape and path of the object in motion.

[Bignose]

The rolling resistance is very simple thing. This is enough open sources like (wikipedia for example) to show this force does not depend on rolling body shape.

 

 

I don't understand, why this simple thing is going to puzzle.

 

 

Therefore, just keep in mind, doesn’t matter what force value it would, acting time depends on force value and work (kinetic energy difference). Base on these two values multiplication (deceleration force and acting time) we’ll get momentum which gains to track. The thin ring momentum is higher than solid disk momentum, because thin ring discharges more kinetic energy for same velocity differentiation than solid disk.

 

If you want us to understand, why don't you do what we've asked to help us understand? If you really want to get your point across, shouldn't you be very interested in trying to make it as clear to others exactly what your point is? Why the resistance to try to make it easy for the rest of us to see?

[ABV]

If you draw it, you might see the difference and begin to understand.

 

It is not the shape, it's the way the force vector must be applied in each case, to slow down the disc and ring without slipping.

 

They have the same amount of momentum.

 

They have different amounts of angular momentum.

 

This affects the way any force vector must be applied to slow down the disc or ring without slipping.

 

You might then see how it applies in the simplest case of a fixed straight track. If you cannot do this how can you possibly understand the more complicated isolated circuit track.

---

Merged post follows:

Consecutive posts merged

 

 

Thanks BN. The curious thing is, that he offered to do just that, fix any errors or omissions if I pointed them out, but he seems to balk at doing it. From his link in the first post he is obviously capable of drawing, he just needs to slow down and do it.

 

I would be happy if he at least gets this one part right for now, the deceleration of the disc and ring on a fixed straight track. I suspect he will not get it right the first time he tries it, since he thinks they are the same, but it will be a start.

 

There is no error. Please pay attention a disk and a ring same round shape. But ring has hole on center. This is just difference between them. Then they contact to track they do track deformation. Base on this deformation rolling body gets rolling resistance. Both bodies have a same shape which contact to the track. This means they get same force.

 

P.S.

Please do not suspect. Better if we follow a classical mechanic 

---

Merged post follows:

Consecutive posts merged

If you want us to understand, why don't you do what we've asked to help us understand? If you really want to get your point across, shouldn't you be very interested in trying to make it as clear to others exactly what your point is? Why the resistance to try to make it easy for the rest of us to see?

 

I hope the person who reviews the system knows elementary classical mechanic laws.

But I should not complicate full diagram with useless details. All forces are present there.

 

I pointed to open source wikipedia, where clearly explains rolling resistance. If it’s not clear, let’s use another source. I’m OK with that. A lot of other sources which explain how these forces come. Let discuss on existing models. I hope you understand it.

---

Merged post follows:

Consecutive posts merged

We have again developed more force with the same input. The only change was the shape and path of the object in motion.

 

For getting better result I would redesign test system in a future. Better makes a new experiment series with solid disk and very thin ring.

I’ll do it when I’ll get a chance.

Edited May 31, 2009 by ABV
Consecutive posts merged.

[J.C.MacSwell]

Both bodies have a same shape which contact to the track. This means they get same force.

 

.

 

...and yet, one loses proportionally more momentum than the other, and the other proportionally more angular momentum.

 

...and of course, you can't be wrong so...

 

Paradox!

[Bignose]

Is it really that hard to draw up a diagram as asked? I don't think anybody is looking for a professional-level diagram. Something scratched together in MS Paint would probably get the point across. Heck, something scratched on a cocktail napkin and scanned in would probably get the point across. I just think that you could really help your case.

 

And, I don't appreciate the insulting tone I perceive in your response. I am showing you plenty of respect, as is everybody else in this thread; we deserve the same from you.

[ABV]

...and yet, one loses proportionally more momentum than the other, and the other proportionally more angular momentum.

Paradox!

Please try to understand. This is a rolling body, just ONE body, which has ONE kinetic energy. During deceleration period this energy spends to gains momentum to track. Value of this kinetic energy discharge will be different for thin ring and solid disk. Just one force is applying to rolling body during this deceleration period. This is rolling resistance force. No matter how big is this force (I’m using identical in both cases for better understanding) acting time depends on energy discharge. Amount of this kinetic energy discharge (i.e. work) will be different for the thin ring and the solid disk. Therefore these rolling bodies gain different momentums to the track.

 

...and of course, you can't be wrong so...

no comments

---

Merged post follows:

Consecutive posts merged

Is it really that hard to draw up a diagram as asked? I don't think anybody is looking for a professional-level diagram. Something scratched together in MS Paint would probably get the point across. Heck, something scratched on a cocktail napkin and scanned in would probably get the point across. I just think that you could really help your case.

Yea, but a lot resources on internet, which explains better this force. Why we need useless "cut and paste". Just put a link and give some comments, please.

 

And, I don't appreciate the insulting tone I perceive in your response. I am showing you plenty of respect, as is everybody else in this thread; we deserve the same from you.

 

I apologize if my previous post hurts you. But I would rather give/get real help than many times talk to about new rolling resistant picture.

---

Merged post follows:

Consecutive posts merged

Well, looks like no explanation, what is going on in this paradox.

Edited May 31, 2009 by ABV
Consecutive posts merged.

[J.C.MacSwell]

 

Yea, but a lot resources on internet, which explains better this force. Why we need useless "cut and paste". Just put a link and give some comments, please.

 

 

Well, looks like no explanation, what is going on in this paradox.

 

If you give a link showing a diagram I will be happy to comment on it.

 

This link:

 

http://en.wikipedia.org/wiki/Rolling_resistance

 

...we already know does not represent your case.

[ABV]

If you give a link showing a diagram I will be happy to comment on it.

 

This link:

 

http://en.wikipedia.org/wiki/Rolling_resistance

 

...we already know does not represent your case.

 

What do you mean? What kind of difference would be?

Would it be big hole on a disks center?

[J.C.MacSwell]

What do you mean? What kind of difference would be?

Would it be big hole on a disks center?

 

The force vector must be placed in such a way as to slow the ring down without slipping. Nature does this automatically, as She reduces the momentum and angular momentum of the ring. This will not be the same for the disc, as the momentum and angular momentum are proportionally different. But neither of these cases match the diagram of the Wiki link.

 

At constant velocity, where the disc and ring have a driving force to match the rolling resistance, there is no difference, the rolling resistance force vectors can be the same in each case, assuming the driving force is applied the same in each case. This is not the case you are describing as you do not have constant velocity.

 

An analogy would be the force of friction between the rock and the ground, if you are pushing on it, or a small squirrel is pushing on it- if it doesn't move, then it doesn't move, but the force is not the same. Until it budges, the rock "knows" how much and in what direction to push back.

 

And in all the cases you describe, the track "knows" how much to push back as well. Nature does the accounting, and leaves no "paradox".

[ABV]

The force vector must be placed in such a way as to slow the ring down without slipping. Nature does this automatically, as She reduces the momentum and angular momentum of the ring. This will not be the same for the disc, as the momentum and angular momentum are proportionally different. But neither of these cases match the diagram of the Wiki link.

This is no arguing about rolling resistance difference. Wiki shows how to a rolling resistance works. And this is correct concept. The rolling resistance force depends from the other forces. You should not use momentums for that. A force goes to forces, a momentum goes to momentums. Do not mix up.

 

 

At constant velocity, where the disc and ring have a driving force to match the rolling resistance, there is no difference, the rolling resistance force vectors can be the same in each case, assuming the driving force is applied the same in each case. This is not the case you are describing as you do not have constant velocity.

 

Is rolling resistance depends on velocity?

 

 

An analogy would be the force of friction between the rock and the ground, if you are pushing on it, or a small squirrel is pushing on it- if it doesn't move, then it doesn't move, but the force is not the same. Until it budges, the rock "knows" how much and in what direction to push back.

 

You’re absolutely incorrect.

There is no model like rock and ground for rolling friction. This model explains friction for linear movement.

A normal generic model is rolling body hikes to mount all the time. Because rolling body deforms a track and this deformation is making a rolling resistance.

Rolling resistance does not depends on velocity and shape. Extra angual momentum gives more roling capacity to rolling body, wich will gains more momentum to the track. Energy should not be discharge to nowhere.

 

And in all the cases you describe, the track "knows" how much to push back as well. Nature does the accounting, and leaves no "paradox".

 

The ideal thin ring and solid disk has no rolling resistance difference. This means this Isolated System has a PARADOX for classical mechanics.

---

Merged post follows:

Consecutive posts merged

One more question. Will thin ring get longer ride than solid disk with same mass and velocity? Yes or No?

(please use same rolling resistance on a straight track)

 

Yes.

 

At few posts before I received answer about thin rink longest ride.

And latter rolling resistance explanation from you was completely opposite.

Would physics laws change by situation?

Edited June 1, 2009 by ABV

[J.C.MacSwell]

 

 

At few posts before I received answer about thin rink longest ride.

And latter rolling resistance explanation from you was completely opposite.

Would physics laws change by situation?

 

The laws don't change. The application of the laws changes.

[ABV]

The laws don't change. The application of the laws changes.

 

But we still look through classical mechanic for these cases.

I understand all classical mechanics laws must work. But this particular case shows a paradox.

---

Merged post follows:

Consecutive posts merged

Another site with good rolling friction explanation.

http://webphysics.davidson.edu/faculty/dmb/py430/friction/rolling.html

---

Merged post follows:

Consecutive posts merged

This is very interesting document

http://arxiv.org/ftp/physics/papers/0606/0606254.pdf

---

Merged post follows:

Consecutive posts merged

The rolling friction for inelastic bodies can be described like “rolling along an incline”.

http://cnx.org/content/m14312/latest/

 

This model is reversible and as it shown acceleration depends on rolling body’s moment of inertia I.

 

a=[math]\frac{g*sin(\alpha)}{(1+\frac{I}{m*R^2})}[/math]

 

For small angles sin([math]\alpha[/math]) is equivalent to just [math]\alpha[/math].

[math]\alpha[/math][math]\rightarrow[/math]0 then sin([math]\alpha[/math])[math]\approx[/math][math]\alpha[/math]

 

In our case deceleration is:

–a = -[math]\frac{g*\alpha}{(1+\frac{I}{m*R^2})}[/math]

Where [math]\alpha[/math] is rolling friction coefficient

 

Rolling body has initial velocity V. And link with deceleration and time frame is:

 

[math]\Delta[/math]V=a*[math]\Delta[/math]t

 

The time frame for stoping rolling body from initial velocity V is:

 

[math]\Delta[/math]t=V*[math]\frac{(1+\frac{I}{m*R^2})}{g*\alpha}[/math]

 

Otherwise the rolling body as thin ring has a longest ride than the solid disk, because the thin ring has bigger moment of inertia I than the solid disk.

Base on these formulas, equal conditions (mass, gravity, rolling friction coefficient) and ride difference, the thin ring gains more momentum to the track than solid disk does.

Edited June 3, 2009 by ABV
Consecutive posts merged.

[J.C.MacSwell]

, the thin ring gains more momentum to the track than solid disk does.

 

Hi ABV, I haven't had time to check all this out and respond but if they both roll until they stop, they will both impart the same amount of momentum to the track, even though the ring will roll further. So for the same distance rolled the ring will have imparted less momentum to the track than the disc. (the ring will have more momentum, as the disc slows down more quickly)

 

I realize that may not make sense at a glance but it is true.

 

I looked at the first link and it is interesting. He has applied the rolling resistance as a force at the center point opposite the direction of motion.

 

There is a convention to use it this way in engineering, applying it at axle height to a vehicle. It is a simplification. He then points out additional forces that appear when it is applied in this manner. Nothing wrong with that if he makes sure everything balances out so that the net result is the same.

Edited June 3, 2009 by J.C.MacSwell

[ABV]

Hi ABV, I haven't had time to check all this out and respond but if they both roll until they stop, they will both impart the same amount of momentum to the track, even though the ring will roll further. So for the same distance rolled the ring will have imparted less momentum to the track than the disc.

 

I realize that may not make sense at a glance but it is true.

 

There is a convention to use it this way in engineering, applying it at axle height to a vehicle. It is a simplification. He then points out additional forces that appear when it is applied in this manner. Nothing wrong with that if he makes sure everything balances out so that the net result is the same.

 

For engineering may be “it’s ok”, because a wheel is not so thin ring. But on a theory, riding tine should difference. The sliding friction should not apply to rolling bodies, because if just linear momentum gains to the track then system breaks the law of angular momentum conservation. This is wrong.

Therefore better model for rolling resistance is "Rolling along an incline". I did some formulas on previous post.

[J.C.MacSwell]

For engineering may be “it’s ok”, because a wheel is not so thin ring. But on a theory, riding tine should difference. The sliding friction should not apply to rolling bodies, because if just linear momentum gains to the track then system breaks the law of angular momentum conservation. This is wrong.

Therefore better model for rolling resistance is "Rolling along an incline". I did some formulas on previous post.

 

Believe me, angular momentum is conserved as well.

 

Would you believe that the angular momentum is the same for the system, track and ring, at the start, as it is at the end, when the ring comes to a halt? (as well as all points in between)

[ABV]

Believe me, angular momentum is conserved as well.

 

Would you believe that the angular momentum is the same for the system, track and ring, at the start, as it is at the end, when the ring comes to a halt? (as well as all points in between)

 

Is someone did experiment about it? As I saw on docs, rolling friction has a lot of surprises. I just represent my simple one, which is showng paradox.

 

Thank you for all your comments anyway.

[navigator]

Believe me, angular momentum is conserved as well.

 

 

You must not be aware the your placing your reputation on the line with only theory and speculation, the majority of which the supporting evidence is not conclusive, backing you up.

 

How many experiments have you actually performed?