ABV

The main idea was to understand if little human transfer momentums (surface to rolling body and backwards) differently or not, for rolling bodies with same initial velocity and different moment of inertia.

I clearly understand what I want, but it hard to explain it to you. I’m sorry. I try to do abstract in a few forces and use simple conditions. Unfortunately, during this discussion I did many changes into doc. This may be confusing everyone a lot. But I’m not going to break any law without facts or logical explanation. I try to explain my vision to you. And if it’s wrong – understand why. I have to put back my doc close to original and start new topics again. Sorry, but I have to find a truth. And I’m not going to use any pseudo theories.

The second one – transferred momentum from rolling body to surface through human. Second is yes. Good. Otherwise. The human will transfer different surface momentums in these cases to a rolling body to reach velocity V. And different momentums will transfer back during deceleration time. And pair acceleration/deceleration for these rolling bodies must be identical. I1>I2 - moment of inertia Pa1>Pa2 and Pd1>Pd2 Pa1=Pd1 and Pa2=Pd2 Correct? Sorry about my language (: Merged post follows: Consecutive posts merged This is normal a physics questions. Why pseudoscience?

And again. The initial condition is no slippery. With slippery - yes. Each linear momentum frame will decrease a rolling body momentum and do slipping for reducing a angular velocity and momentum. But in this case this is continuous stopping force and no slippery at all.

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.

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

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?

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.

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. Merged post follows: Consecutive posts merged 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. Merged post follows: Consecutive posts mergedFor 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?

I added a few more necessary forces into diagrams. Please take a look. Merged post follows: Consecutive posts mergedOK. 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.

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.

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. 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.

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.

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?

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.

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.

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.

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 mergedAnother site with good rolling friction explanation. http://webphysics.davidson.edu/faculty/dmb/py430/friction/rolling.html Merged post follows: Consecutive posts mergedThis is very interesting document http://arxiv.org/ftp/physics/papers/0606/0606254.pdf Merged post follows: Consecutive posts mergedThe 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.

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. Is rolling resistance depends on velocity? 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. 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 (please use same rolling resistance on a straight track) 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?

What do you mean? What kind of difference would be? Would it be big hole on a disks center?

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. no comments Merged post follows: Consecutive posts merged 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. 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 mergedWell, looks like no explanation, what is going on in this paradox.

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 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 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.

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.

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.

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. This is a main point. The deceleration momentums are different in these two cases. That because I’m calling this PARADOX.