Does it really work? [Answered - No]

[J.C.MacSwell]

 

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?

 

These are very basic classical laws of mechanics. What evidence does not back me up?

[ABV]

Here’s no question about how basic classical mechanics laws is.

The problem is people try to apply wrong model, which brings misunderstanding of Mother Nature.

The good example is rolling resistance. The sliding friction model cannot be use for rolling friction, because rolling body doesn’t have a slippery on a track. The rolling along an incline model is very close to that rolling resistance explanation.

On real world, this rolling resistance is not regular movement on the incline. A precision model must use differential equation for this process full explanation. But this simple model showed a correlation between moment of inertia an acting time, and no “extra spending to nowhere” kinetic energy and momentums.

I understand, regular engineering coefficients may not count this difference, but this is a physics concept which shows a paradox.

Edited June 4, 2009 by ABV

[J.C.MacSwell]

Here’s no question about how basic classical mechanics laws is.

The problem is people try to apply wrong model, which brings misunderstanding of Mother Nature.

The good example is rolling resistance. The sliding friction model cannot be use for rolling friction, because rolling body doesn’t have a slippery on a track. The rolling along an incline model is very close to that rolling resistance explanation.

On real world, this rolling resistance is not regular movement on the incline. A precision model must use differential equation for this process full explanation. But this simple model showed a correlation between moment of inertia an acting time, and no “extra spending to nowhere” kinetic energy and momentums.

I understand, regular engineering coefficients may not count this difference, but this is a physics concept which shows a paradox.

 

No. It does not.

 

If you apply the laws of classical physics exactly, they do not contradict the laws of classical physics. They are consistent with each other.

 

You cannot have an exactly correct model using classical physics laws and prove classical physics laws wrong.

[ABV]

No. It does not.

 

If you apply the laws of classical physics exactly, they do not contradict the laws of classical physics. They are consistent with each other.

 

You cannot have an exactly correct model using classical physics laws and prove classical physics laws wrong.

 

OK.

 

Let's talk about rolling resistance again.

An inelastic rolling body moves without slippery on a track. Correct?

How can you explain rolling resistance for non slippery rolling?

A sliding friction? It’s wrong.

On ideal mode the rolling body hikes on cavity all the time. And this movement let the rolling unit loose energy without slippery. In this case formula must include rolling body moment of inertia. You may not see big difference on solid disk and small radius disk, but it will on thin ring.

 

Any other suggestions?

[J.C.MacSwell]

OK.

 

Let's talk about rolling resistance again.

An inelastic rolling body moves without slippery on a track. Correct?

How can you explain rolling resistance for non slippery rolling?

A sliding friction? It’s wrong.

On ideal mode the rolling body hikes on cavity all the time. And this movement let the rolling unit loose energy without slippery. In this case formula must include rolling body moment of inertia. You may not see big difference on solid disk and small radius disk, but it will on thin ring.

 

Any other suggestions?

 

The higher moment of inertia allows more kinetic energy available at the same speed, but the total momentum (translational momentum) of the disc and ring are the same. That extra energy must be exerted against something external to the system to change the momentum of the system as a whole.

 

Similarly with angular momentum. That extra energy must be exerted against something external to the system to change the angular momentum of the system as a whole. (in this case the ring has a greater angular momentum than a similar system with a solid disc, but neither system can change without an external force)

 

If you were at rest in outer space you could not change the center of gravity of your system no matter what you do, no matter how much energy you expended to do it. If you spat to the left you would move to the right, leaving your previous rest point behind, but the center of gravity of you and your spit would remain at that point. Your system's momentum would remain at zero in your original rest frame.

 

You could swing your arms and contort in what seems like a spin like manner but your angular momentum would remain at zero as well.

 

You need "outside help" to change either one.

[ABV]

The higher moment of inertia allows more kinetic energy available at the same speed, but the total momentum (translational momentum) of the disc and ring are the same. That extra energy must be exerted against something external to the system to change the momentum of the system as a whole.

 

Similarly with angular momentum. That extra energy must be exerted against something external to the system to change the angular momentum of the system as a whole. (in this case the ring has a greater angular momentum than a similar system with a solid disc, but neither system can change without an external force)

There is not necessary to change angular momentum. There is just explanation how extra angular momentum put classical mechanics equations to paradox conditions.

 

If you were at rest in outer space you could not change the center of gravity of your system no matter what you do, no matter how much energy you expended to do it. If you spat to the left you would move to the right, leaving your previous rest point behind, but the center of gravity of you and your spit would remain at that point. Your system's momentum would remain at zero in your original rest frame.

 

You could swing your arms and contort in what seems like a spin like manner but your angular momentum would remain at zero as well.

 

You need "outside help" to change either one.

 

The document shows how it possible. Centripetal force does not count angular momentum, but rolling friction does. Please read.

 

P.S.

The site is updating all the time. Please take a look.

[J.C.MacSwell]

There is not necessary to change angular momentum. There is just explanation how extra angular momentum put classical mechanics equations to paradox conditions.

 

 

 

The document shows how it possible. Centripetal force does not count angular momentum, but rolling friction does. Please read.

 

P.S.

The site is updating all the time. Please take a look.

 

In Figure 5.

 

1. Force Fr. Where is the equal but opposite force? (Don't just say rolling resistance, that's from an engineering convention to direct it opposite the velocity and at center point, at the axle or hub. That's a simplification/approximation where the difference is not relevant - here it is very relevant)

 

2. Force N. Here presumably there can be an equal but opposite force on the track due to the proximity of the track but it is directed slightly behind the center point. Is that your intention? (that is OK if it is, but read on)

 

If those are exactly the forces on the wheel, the wheel would decelerate but at the same time accelerate it's rotation, which means it would be slipping. The disc or ring must decelerate it's velocity and decelerate it's rotation in exact proportions in order to not slip.

 

3. There is a missing force similar to Fr in figure 6. a static friction force forward at track level(and slightly upwards where there is deformation of the track. This force will be greater for a decelerating ring than for a decelerating disc, since it will counteract any tendency for a decelerating wheel to slip (slip backwards, not the frontwards slipping tendency typical in braking)

 

To get it right for a disc or ring freely wheeling on a track with rolling resistance:

 

1. Get rid of Fr in the location of the center point

 

2. Force N is OK or can be combined with force Ff below.

 

3. Add Ff (for friction so as not to confuse it with rolling resistance, or if you prefer Ftr for traction not track), forward at track level and slightly upwards, greater in the case of the ring and lesser in the case of the disc. Or combine it with N

 

The net force from the track to the ring or disc Ftk will be directed ahead of the centerpoint but applied at a point even further ahead of that the wheel is slowed and at the same time the rotation.

Edited June 5, 2009 by J.C.MacSwell

[ABV]

In Figure 5.

 

1. Force Fr. Where is the equal but opposite force? (Don't just say rolling resistance, that's from an engineering convention to direct it opposite the velocity and at center point, at the axle or hub. That's a simplification/approximation where the difference is not relevant - here it is very relevant).

 

2. Force N. Here presumably there can be an equal but opposite force on the track due to the proximity of the track but it is directed slightly behind the center point. Is that your intention? (that is OK if it is, but read on)

 

If those are exactly the forces on the wheel, the wheel would decelerate but at the same time accelerate it's rotation, which means it would be slipping. The disc or ring must decelerate it's velocity and decelerate it's rotation in exact proportions in order to not slip.

 

There is NO SLIPPING. No sliding friction on rolling friction model. Rolling friction base on loosing a kinetic energy from a rolling body lifting.

Please try to understand, sliding friction force much bigger than rolling friction force.If you'll apply sliding frction on rolling, you won't move any cart at shopping mall

NO SLIDING.

 

 

3. There is a missing force similar to Fr in figure 6. a static friction force forward at track level(and slightly upwards where there is deformation of the track. This force will be greater for a decelerating ring than for a decelerating disc, since it will counteract any tendency for a decelerating wheel to slip (slip backwards, not the frontwards slipping tendency typical in braking)

 

To get it right for a disc or ring freely wheeling on a track with rolling resistance:

 

1. Get rid of Fr in the location of the center point

 

2. Force N is OK or can be combined with force Ff below.

 

3. Add Ff (for friction so as not to confuse it with rolling resistance), forward at track level and slightly upwards, greater in the case of the ring and lesser in the case of the disc. Or combine it with N

 

Actually this is a tipical example from classical mechanics book. The link a few posts up. This Fr let disk move without slipping. That's all. This example does not count own rolling friction. The disk just loosing kinetic energy from lifting, which is a good model for rolling resistance.

 

Both pictures are mostly a copies from other sites. Plese take a look.

Edited June 5, 2009 by ABV

[J.C.MacSwell]

There is NO SLIPPING. No sliding friction on rolling friction model. Rolling friction base on loosing a kinetic energy from a rolling body lifting.

Please try to understand, sliding friction force much bigger than rolling friction force.If you'll apply sliding frction on rolling, you won't move any cart at shopping mall

NO SLIDING.

 

No slipping? Look at the diagram for diagram #5. What force is slowing the rotation? (hint:there is none) It needs to be corrected.

 

[

Actually this is a tipical example from classical mechanics book. The link a few posts up. This Fr let disk move without slipping. That's all. This example does not count own rolling friction. The disk just loosing kinetic energy from lifting, which is a good model for rolling resistance.

 

Both pictures are mostly a copies from other sites. Plese take a look.

 

Which site has # 5, and which has #6?

[ABV]

No slipping? Look at the diagram for diagram #5. What force is slowing the rotation? (hint:there is none) It needs to be corrected.

No slipping. Fr - result force from loosing a kinetic energy. It's force F(dc) (fig. 6) projection to movement direction.

 

Just keep in mind, sliding force much bigger than rolling resistance. Back to shopping cart example. Try to lock all this cart wheels and see what is happen. I hope you know result of this experiment.

 

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

http://cnx.org/content/m14312/latest/ or http://203.158.100.140/physics/charud/scibook/Physics-for-Scientists-and-Engineers-Serway-Beichner-4/11%20-%20Rolling%20Motion%20and%20Angular%20Momentum.pdf

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The site is updated. Please take a look.

Edited June 5, 2009 by ABV

[ABV]

I added a few more necessary forces into diagrams.

Please take a look.

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OK.

Forget about this paradox.

Let's build a concept.

Take a look on this.

http://knol.google.com/k/alex-belov/an-antigravity-engine-concept/1xmqm1l0s4ys/8#

 

I'll do description later today.

[J.C.MacSwell]

I added a few more necessary forces into diagrams.

Please take a look.

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OK.

Forget about this paradox.

Let's build a concept.

Take a look on this.

http://knol.google.com/k/alex-belov/an-antigravity-engine-concept/1xmqm1l0s4ys/8#

 

I'll do description later today.

 

Absolutely. No problem at all building an anti-gravity device from the same ideas. Just turn the thing at 90 degrees and use the same false assumptions and the same faulty arguments.

[ABV]

Absolutely. No problem at all building an anti-gravity device from the same ideas. Just turn the thing at 90 degrees and use the same false assumptions and the same faulty arguments.

 

No faulty arguments.

A rolling body along an incline kinematics equation depends on moment of inertia. It's standard example. But moment of inertia won’t include into equation with centripetal force.

This concept does not include a rolling resistance and gravity force. An electrostatic field opposite charge holds a thin ring on track. Electromagnets forces (arrows shows a direction of this force) emulate a local gravity force. Electromagnets are ON just in one time, when the thin rings move on deceleration part of track. Thin rings are running synchronously.

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One more question about straight track.

Let say external friction force Ffr, no rolling resistance, initial velocity V0 moment of inertia I,

Let's make a kinematic equation

 

[math]\vec{ma} = -\vec{F_{fr}} - \vec{I\frac {a} {R^2}}[/math]

 

follow to

 

[math]a = -\frac {F_{fr}}{m+\frac {I} {R^2}}[/math]

 

and

 

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

 

finally

 

[math]\Delta t = \Delta V\frac {m+\frac {I} {R^2}}{F_{fr}}[/math]

 

A solid disk's moment of inertia

 

[math] I = m\frac{R^2}{2}[/math]

 

A ring's moment of inertia

 

[math] I = m\frac{R_{1}^2+R_{2}^2}{2}[/math]

for thin ring R1 ~ R2

[math] I = mR^2[/math]

 

 

Time for solid disk is

 

[math]\Delta t = \Delta V\frac {1\frac {1} {2}m}{F_{fr}}[/math]

 

 

Time for thin ring is

 

[math]\Delta t = \Delta V\frac {2m}{F_{fr}}[/math]

 

 

 

 

I't means, force is same but time frame is different. Correct?

If no, please provide a correct a kinematic equation and time frame formula.

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For rolling body which causes external force Ffr, a distance to stop is

[math]L=\frac{1}{F_{fr}}(m\frac{V^2}{2}+I\frac{V^2}{2R^2})[/math]

 

Correct?

A thin ring gets a longest ride?

Edited June 8, 2009 by ABV
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[D H]

This barely qualifies as pseudoscience, let alone real science.

 

Thread moved.

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One more question about straight track.

Let say external friction force Ffr, no rolling resistance, ...

A big part of your problem, ABV, is that you are not specifying the problem clearly. This includes properly drawing system boundaries, which is the chief source of your problems to date.

 

Here you have not fully the specified initial conditions and the initial conditions are contradict your equations of motion. Is the body initially rolling or sliding? You didn't say. Is the object rolling without slipping? I guess that is what "no rolling resistance" means. Is the object rolling rolling up or down an incline or rolling on a flat surface? You didn't say.

 

If the object is initially slipping, the dynamic frictional forces will do two things: It will reduce the object's kinetic energy and it will make the object start rolling.

 

If the object is rolling without slipping, it will keep on rolling forever. Why? The velocity at the contact point is zero. The friction needed to keep the body rolling does zero work. Zero work = zero change in energy.

Edited June 9, 2009 by D H
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[ABV]

I have a few questions about rolling body.

First.

The rolling bodies with different moment of inertia have initial velocity V and move with without slipping.

The rolling bodies cause an external force with opposite to drive direction F.

Whould they have a different distance to stop?

Whould they gains a different momentum?

 

 

My answer. Yes. Because base on energy conservation.

[math]L=\frac{1}{F}(m\frac{V^2}{2}+I\frac{V^2}{2R^2})[/math]

 

If moment of inertea different, then a distance a different.

Different distance gives with same force gices a different time.

Correct?

 

Second.

A rolling bodies with different moment of inertia moves on same curve part of track with radius R, with same velocity V.

Would they cause same Centripetal force?

 

My answer. Yes. Because centripatal force does not depend on rolling body moment of inertia.

[math]F_{c}=-m\frac{V^2}{R}[/math]

 

This formula does not contain the moment of inertia.

It means net momentum from curve part of track does not depend on moment of inertia.

Correct?

 

 

What do you think about it?

Thank you.

[D H]

Stop spamming, please. You are posting multiple threads on the same topic.

 

Thread merged with your other nonsense.

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Well, dang. I don't have the authority to merge a thread with another thread in the pseudoscience sub forum.

[J.C.MacSwell]

I have a few questions about rolling body.

First.

The rolling bodies with different moment of inertia have initial velocity V and move with without slipping.

The rolling bodies cause an external force with opposite to drive direction F.

Whould they have a different distance to stop?

Whould they gains a different momentum?

 

 

My answer. Yes. Because base on energy conservation.

[math]L=\frac{1}{F}(m\frac{V^2}{2}+I\frac{V^2}{2R^2})[/math]

 

If moment of inertea different, then a distance a different.

Different distance gives with same force gices a different time.

Correct?

 

Second.

A rolling bodies with different moment of inertia moves on same curve part of track with radius R, with same velocity V.

Would they cause same Centripetal force?

 

My answer. Yes. Because centripatal force does not depend on rolling body moment of inertia.

[math]F_{c}=-m\frac{V^2}{R}[/math]

 

This formula does not contain the moment of inertia.

It means net momentum from curve part of track does not depend on moment of inertia.

Correct?

 

 

What do you think about it?

Thank you.

 

What is the nature of the external force?

 

If they have the same external force backwards, opposite the direction of motion, applied say at their centre of gravity, it will trigger a lesser external force forward at ground level. The greater the moment of inertia, the greater this triggered force will be, so the net braking force will be reduced. So the greater the moment of inertia, the greater the kinetic energy, and the further it will go before stopping and the longer it will take.

 

However, the amount of momentum that will be transferred to the ground will be independent of the moment of inertia.

 

Second: If they have the same mass then yes, the centripetal force should be the same.

[ABV]

What is the nature of the external force?

 

If they have the same external force backwards, opposite the direction of motion, applied say at their centre of gravity, it will trigger a lesser external force forward at ground level. The greater the moment of inertia, the greater this triggered force will be, so the net braking force will be reduced. So the greater the moment of inertia, the greater the kinetic energy, and the further it will go before stopping and the longer it will take.

 

However, the amount of momentum that will be transferred to the ground will be independent of the moment of inertia.

 

Second: If they have the same mass then yes, the centripetal force should be the same.

for first:

A simple one.

A rolling body has an initial velocity V on flat surface without rolling resistance and without a slippery.

An External force. Let’s say a small human on the surface holding a rolling body axis and try to stop it with force Ffr.

He tried twice with different rolling bodies. These rolling bodies have a different moment of inertia. It means a different full kinetic energy.

Would these rolling bodies have different distance and time?

Would human gains different momentums to the surface?

[D H]

Would these rolling bodies have different distance and time?

Yes.

 

Would human gains different momentums to the surface?

No.

[J.C.MacSwell]

for first:

A simple one.

A rolling body has an initial velocity V on flat surface without rolling resistance and without a slippery.

An External force. Let’s say a small human on the surface holding a rolling body axis and try to stop it with force Ffr.

He tried twice with different rolling bodies. These rolling bodies have a different moment of inertia. It means a different full kinetic energy.

Would these rolling bodies have different distance and time?

 

If they have the same mass, then Yes. The one with the greater moment of inertia will go further due to the forward traction force at surface level

 

Would human gains different momentums to the surface?

 

Yes, the human would transfer more momentum to the surface for the one with the greater moment of inertia:

 

1. All of the initial momentum

 

Plus

 

2. All of the additional momentum transferred from the surface to the rolling body due to the forward traction force at surface level

 

However, the second component of the momentum transferred by the human to the surface will be totally offset by the amount transferred to the rolling body by the forward traction force at surface level.

 

So:

 

1. the net result for momentum will be the same, regardless of moment of inertia, even though the human will exert the same force for a longer time. (again, the greater traction force will have made up the difference)

 

2. the net result for energy will not be the same, for different of moments of inertia, since the human will exert the same force for a longer distance. (the traction force does not absorb or add energy*)

 

* the traction force does not act through a distance. It is a static force. The very bottom of the wheel is not moving.

[ABV]

Yes.

 

 

No.

 

Why?

The distance is different, time is different. Force is same. Same initial velocity V (just talk about stopping a rolling body).

A Work will be a different (different full kinetic energy)

momentum P = F*t.

F=const, t1>t2

P1>P2

[J.C.MacSwell]

Why?

The distance is different, time is different. Force is same. Same initial velocity V (just talk about stopping a rolling body).

A Work will be a different (different full kinetic energy)

momentum P = F*t.

F=const, t1>t2

P1>P2

 

Same force for the human, but the net force is different due to the traction forces being different.

[D H]

The "yes" answer was to the question "Would these rolling bodies have different [stopping] distance and time?" This is obvious. They two objects have different initial kinetic energy, so different amounts of work are needed to stop the two objects. Work in this case is simply force * stopping distance.

 

The "no" answer was to the question "Would human gains different momentums to the surface?". In both cases the human start with the same initial velocity and end with the same final velocity. The change in the human's momentum is the same in both cases. Once again, this is obvious.

[J.C.MacSwell]

The "yes" answer was to the question "Would these rolling bodies have different [stopping] distance and time?" This is obvious. They two objects have different initial kinetic energy, so different amounts of work are needed to stop the two objects. Work in this case is simply force * stopping distance.

 

The "no" answer was to the question "Would human gains different momentums to the surface?". In both cases the human start with the same initial velocity and end with the same final velocity. The change in the human's momentum is the same in both cases. Once again, this is obvious.

 

I am pretty sure (but I could be wrong) he meant by "Would human gains different momentums to the surface?" to mean transferred momentum to the surface due to the continuous force over the time period. . He is apparently ignoring the forward traction force (a backwards force on the surface) that is triggered by the slowing of the body.

Edited June 9, 2009 by J.C.MacSwell

[ABV]

The "no" answer was to the question "Would human gains different momentums to the surface?". In both cases the human start with the same initial velocity and end with the same final velocity. The change in the human's momentum is the same in both cases. Once again, this is obvious.

Even if he change same velocity frame, acting time is different. And during this time human apply const force Ffr.

The momentum result of is F*dt. And no slippery. Because if counts linear momentum frames P, then anculyar momentum must be lost on slippery. But no sipplery.

Edited June 9, 2009 by ABV