Solving Wheel Coming to a Stop: FBD, Friction & Hysteresis

In summary, the wheel has an initial speed and it will come to a stop eventually due to the two sources of friction.
  • #1
Juanda
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TL;DR Summary
I realized I can't confidently draw the FBD of a wheel coming to a stop.
Simple question. Let's say a solid cylinder has an initial speed ##v_o## and it's rotating on infinitely hard ground without air resistance.
The cylinder will come to a stop eventually. There are two sources of friction.
  1. Since the wheel/cylinder is deformed at the contact patch, there is some sliding friction that will consume energy.
  2. Materials are not really elastic so hysteresis happens. Especially with rubber-like materials typically used for tires. Due to the viscoelastic behavior of the wheel, the rotational energy of the wheel will be transformed into heat because of the internal friction. I believe this to be the main friction source since the sliding friction could not even happen if the wheel is sticky and can deform enough.
1690711670110.png
That picture is from this video. My concern is that, although he doesn't mention it directly, he's assuming the wheel has no angular acceleration to obtain a formula for the rolling resistance.

So I'm trying to draw a FBD of the cylinder and make the equations work but I'd love to have confirmation of my thought process to make sure it's right.

1690713203732.png


Vertical forces. The normal force must necessarly exist and be equal to the weight because there is no vertical acceleration.
$$-mg+N=0 \rightarrow N = mg$$

Horizontal forces. The force ##F## must exist because we know the acceleration in ##x## of the body will decrease with time. I'll consider the value to be initially unknown.
$$-F=ma_x$$

Torques. Due to the hysteresis, the normal force is not alligned with the weight. Let's assume for now that ##d## is known since we should be able to approximate it from the properties of the material. Also, due to the deformation of the wheel, the force ##F## has a smaller leverage ##r## instead of ##R##.
$$-Fr+Nd=I \alpha$$
Where ##I \approx \frac{1}{2}mR^2##

Mechanical link. The rotation of the wheel and the horizontal displacements are related. A positive turn of the wheel moves it in the negative horizontal direction so there will be a minus sign in there. I have doubts about using ##r## or ##R##. I believe ##r## is more accurate.
$$\alpha r \approx -a_x$$

We know everything from the first equation so we are left with 3 equations and 3 unknowns ##F, a_x, \alpha##. This is solvable.
After clearing the unknowns I get the following.
$$\alpha = \frac{Nd}{I+mr^2}$$
$$F = \frac{Ndmr}{I+mr^2}$$
$$a_x = - \frac{Ndr}{I+mr^2}$$

From that, I can get the linear and angular velocities and how they change with time so I can obtain the power being lost. However, I feel I should be able to obtain the same result using ##P=Fv## or ##P= \tau \omega## but I wouldn't even know which force or torque to input in those equations.

To me, it feels like the approach is correct. However, it bothers me it looks so different when compared with the usual ##F_f= \mu N##. How would you solve a problem like this?
 
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  • #2
Here's how I look at this problem. Assume that the sketch is a round wheel (I'm a bad artist) with a flat contact patch. The contact patch is divided into two parts, with the dividing line vertically below the center of the wheel. The total vertical downforce (wheel weight plus load) is mg. That downforce is equal to the total up force at the contact patch. Due to hysteresis, the moment due to force ##F_1## is greater than the moment due to force ##F_2##. Forces ##F_1## and ##F_2## sum to the total downforce mg.

If viscoelastic effects were large enough, force ##F_2## could approach zero. In that case, the entire contact patch would be to the right of the center of the wheel, and rolling resistance would be quite large. If hysteresis and viscoelastic losses were zero, ##F_1## would be equal to ##F_2##, both of those forces would be the same horizontal distance from the downforce, and rolling resistance would be zero.

Rolling friction.jpg
 
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  • #3
jrmichler said:

I can't see the point of dividing the normal force ##N## into ##F_1## and ##F_2##.
Just to clarify, I did put ##N## at the center of pressure which is moved to the right due to the viscoelastic effect. It already encapsulates the effect from ##F_1## and ##F_2## you described. In the picture I posted you can see the pressure distribution at the contact patch and the equivalent normal force due to that pressure distribution.

As a last bonus point, I'm not even certain that the wheel wouldn't come to a stop even if it's 100% elastic. I feel that if we meshed the body and make it completely elastic the kinetic energy would still end up as vibrational energy of the nodes of the wheel which is the transformation of the kinetic energy of the body in a rise in temperature. However, that is beyond the scope of the original question. My main concern still is: why do the equations I posted feel OK but they look so different to what's usually seen regarding rolling resistance?
 
  • #4
Juanda said:
As a last bonus point, I'm not even certain that the wheel wouldn't come to a stop even if it's 100% elastic. I feel that if we meshed the body and make it completely elastic the kinetic energy would still end up as vibrational energy of the nodes of the wheel which is the transformation of the kinetic energy of the body in a rise in temperature.
100% elastic means that the macroscopic KE can be fully retrieved, not that you have 100% elastic interactions somewhere on the microscopic level.
 
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  • #5
A.T. said:
100% elastic means that the macroscopic KE can be fully retrieved, not that you have 100% elastic interactions somewhere on the microscopic level.
Hmmm. I had no idea about that distinction but it makes sense to classify it like that.
 
  • #6
Juanda said:
Torques. [...]
$$-Fr+Nd=I \alpha$$
This is wrong because this:
Juanda said:
$$F = \frac{Ndmr}{I+mr^2}$$
doesn't make sense. The force ##F## cannot depend on the inertia or mass of the wheel. By definition ##FR = Nd##. See the video from your post at 11:29 for the correct use of the reaction pair.

You took the moment about the center of the wheel but since ##FR = Nd##, this is not helpful.

You need to take the moments at the contact patch centerline. Thus:
$$I\alpha + ma_xR = Nd\ (= FR)$$
Or
$$\alpha = \frac{FR}{I+mR^2}$$
Juanda said:
However, I feel I should be able to obtain the same result using ##P=Fv## or ##P= \tau \omega## but I wouldn't even know which force or torque to input in those equations.
##F= \frac{Nd}{R}## as already defined. ##\tau = FR## such that ##P=Fv = F(\omega R) = (FR) \omega = \tau \omega##.

Juanda said:
To me, it feels like the approach is correct. However, it bothers me it looks so different when compared with the usual ##F_f= \mu N##. How would you solve a problem like this?
The friction here is not sliding but just static because the velocity at the contact patch with respect to the ground is zero.

This will be true as long as ##F \le \mu N##. Above that value, your wheel is spinning and you are not in pure rolling anymore.

Juanda said:
Also, due to the deformation of the wheel, the force ##F## has a smaller leverage ##r## instead of ##R##.

[...]

I have doubts about using ##r## or ##R##. I believe ##r## is more accurate.
Note that I have used ##R## everywhere. This is true with a tire as the rolling circumference of a tire is made of a belt that always keeps the same length per revolution even when deformed. The rolling radius of a tire is the equivalent radius of that belt length. (NOT the outside radius of the tire tread. The effective radius of the belt is ##\frac{wheel\ diameter}{2} + \frac{aspect\ ratio}{100} * tire\ width## with the proper units, of course.)

5.0Tires101_5.2Engineering_StructureDiagram.jpg

tiresizes.jpg
 
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  • #7
First of all, thanks for the replies. It's definitely helpful.
jack action said:
doesn't make sense. The force ##F## cannot depend on the inertia or mass of the wheel. By definition ##FR = Nd##. See the video from your post at 11:29 for the correct use of the reaction pair.
Is there a reason the force ##F## cannot depend on the inertia or mass? The value of the normal force can be known because the wheel is not accelerating in the vertical direction so ##N=mg##. That would be a force dependent on the mass and I see no reason to forbid it.

Also, you mention that we must accept ##FR = Nd## by definition but I don't understand why. I think this is the key point in the argument since there is a lot of information built on top of it.
In the original post I mentioned how, although the video doesn't explicitly say it, I believe he's forcing a situation where the wheel is not accelerating which is the underlying reason he can use the equality ##FR = Nd##. That's why I was concerned about a scenario where ##\alpha \neq 0##. After considering ##F## and ##\alpha## to be unknown and imposing some additional equations (Newtons ↑,→, torque, mechanical link due to rotation) those are the values I obtained.

jack action said:
You took the moment about the center of the wheel but since ##FR = Nd##, this is not helpful.

You need to take the moments at the contact patch centerline. Thus:
$$I\alpha + ma_xR = Nd\ (= FR)$$
Or
$$\alpha = \frac{FR}{I+mR^2}$$
I was under the impression that if we're talking about a system where ##\alpha \neq 0## then it is necessary to solve the problem around either the CoM or the instantaneous center of rotation and both methods should yield the same result. Is that not correct? I wasn't sure if the instantaneous center of rotation would be in the centerline of the contact patch or the point of the resultant force ##N## so I decided to use the CoM instead which seemed to be a way to avoid taking that decision. I know sometimes choosing the convenient center of rotation for the equation of moments makes things significantly easier because you can make an unknown variable disappear but in this case, there were enough equations to find all the unknowns anyways.

jack action said:
##F= \frac{Nd}{R}## as already defined. ##\tau = FR## such that ##P=Fv = F(\omega R) = (FR) \omega = \tau \omega##.The friction here is not sliding but just static because the velocity at the contact patch with respect to the ground is zero.

This will be true as long as ##F \le \mu N##. Above that value, your wheel is spinning and you are not in pure rolling anymore.
The reason I believe there is some sliding friction is that rotation can only happen around one point or one axis so there should be a line in the contact patch that's really static but, shouldn't the rest of the contact patch have to slide on the ground during the rotation? Shouldn't that consume some energy too? According to what I could find on the internet most of the energy lost is due to the wheels not being completely round and the hysteresis but I was curious about that possible sliding.

jack action said:
Note that I have used ##R## everywhere. This is true with a tire as the rolling circumference of a tire is made of a belt that always keeps the same length per revolution even when deformed. The rolling radius of a tire is the equivalent radius of that belt length. (NOT the outside radius of the tire tread. The effective radius of the belt is ##\frac{wheel\ diameter}{2} + \frac{aspect\ ratio}{100} * tire\ width## with the proper units, of course.)
I had no idea about that. I will take it into account when working with these kinds of tires. Is there an equivalency for something like a solid steel cylinder? That's the scenario I initially was thinking of because I imagined it would be simpler. In that case, would you just use the original radius of the wheel when it isn't loaded?
 
  • #8
Juanda said:
Is there a reason the force F cannot depend on the inertia or mass?
Juanda said:
Also, you mention that we must accept FR=Nd by definition but I don't understand why.
F is not a reaction force. F is the equivalent horizontal force that produces the same torque N and mg produce. F only appears when you align N and mg. F and d cannot both be present.

I guess FR≡Nd would be more appropriate.

And that is where I misguided you in my post. If you take the moment about the center of rotation, it is either:
Iα+maxR=Nd
Or
Iα+maxR=FR
And since FR≡Nd both will yield the same result.

Juanda said:
The reason I believe there is some sliding friction is that rotation can only happen around one point
That is true if the wheel is round, but it is now deformed. Imagine a hexagon "rolling" instead of a circle: Every time it lands on a flat surface, all points on that surface have the same velocity, i.e. zero.

The losses really happen when it compresses (in front of the contact patch) and the decompression (at the rear of the contact patch) which do not exhibit the same amount of energy.

Juanda said:
Is there an equivalency for something like a solid steel cylinder?
Well, the value of ##d## for steel wheel is about ##0.0015R## so I guest the difference between ##R## and ##r## would be unoticeable.
 
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  • #9
Juanda said:
Is there a reason the force ##F## cannot depend on the inertia or mass?
F depends on mass because N does. I can also imagine, that d depends on the mass distribution (moment of inertia).

Juanda said:
Also, you mention that we must accept ##FR = Nd## by definition but I don't understand why.
Both sides of this equation are just coarse proxy models of the same complex underlying effect. They express the same net resistance in different ways, and this equation states their equivalence. You don't put both in the same dynamic analysis, but choose one of the representations.

Juanda said:
The reason I believe there is some sliding friction is that rotation can only happen around one point or one axis
Pure rotation is a rigid body motion, but the wheel is continuously deforming. Consider tracked vehicles, which have an even longer base of support, without any slippage needed (when moving straight).
 
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  • #10
jack action said:
F is not a reaction force. F is the equivalent horizontal force that produces the same torque N and mg produce. F only appears when you align N and mg. F and d cannot both be present.

I guess FR≡Nd would be more appropriate.

And that is where I misguided you in my post. If you take the moment about the center of rotation, it is either:
Iα+maxR=Nd
Or
Iα+maxR=FR
And since FR≡Nd both will yield the same result.

I'm not sure I'm following your train of thought. Are you saying ##F## doesn't really "exist"? I think it is necessary to add ##F## so there is some force responsible for diminishing the horizontal acceleration as expected. Simultaneously, ##Nd## was necessary to add a torque responsible for diminishing the angular acceleration.

jack action said:
That is true if the wheel is round, but it is now deformed. Imagine a hexagon "rolling" instead of a circle: Every time it lands on a flat surface, all points on that surface have the same velocity, i.e. zero.
I feel like in the hexagon only the leading edge in contact with the ground has 0 velocity. The rest is rotating around it and therefore has non-zero speed. At the point where the flat surface contacts the ground, the instantaneous center of rotation instantly jumps to the leading edge and the flat surface changes its speed in no time. I'm aware that implies infinite acceleration but that's how it should be in an ideal world for the hexagon. The point I'm trying to make is that the flat surface doesn't have 0 speed.
EDIT. After reading the example from A.T. about deformed bodies and tracked vehicles I think I'm ready to believe the contact patch does have no speed so there's no dynamic friction with it.

jack action said:
Well, the value of ##d## for steel wheel is about ##0.0015R## so I guest the difference between ##R## and ##r## would be unoticeable.
I should have expected a very low value. After all, that's one of the reasons things like trains use steel wheels on steel tracks. At that point, it's probably more useful to focus on other sources of friction like air resistance.

A.T. said:
Both sides of this equation are just coarse proxy models of the same complex underlying effect. They express the same net resistance in different ways, and this equation states their equivalence. You don't put both in the same dynamic analysis, but choose one of the representations.
I'm having trouble to understand why I can't put both in the same dynamic analysis. If I don't have both of them I can't express the change in horizontal acceleration due to ##F## because it's the only force acting in that direction and the change in angular acceleration due to ##Nd## because without it there'd be a torque on the wheel caused by ##FR## that keeps accelerating it forward instead of decreasing its movement.

A.T. said:
Pure rotation is a rigid body motion, but the wheel is continuously deforming. Consider tracked vehicles, which have an even longer base of support, without any slippage needed (when moving straight).
I think this is one of the main sources of my confusion. I'm trying to see it from the lens of pure rotation of a rigid body because it is what I know and it simply isn't. The tracked vehicle example helped me visualize how the contact patch can have no velocity.
My main concern now is understanding why ##FR## and ##Nd## can't be used simultaneously.
 
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  • #11
Juanda said:
I'm having trouble to understand why I can't put both in the same dynamic analysis. If I don't have both of them I can't express the change in horizontal acceleration due to ##F## because it's the only force acting in that direction and the change in angular acceleration due to ##Nd## because without it there'd be a torque on the wheel caused by ##FR## that keeps accelerating it forward instead of decreasing its movement.
I think you are right here. In the accelerating case, the linear and angular accelerations must be matched to avoid slippage. So the net torque from F and N must provide the right angular acceleration to match the linear acceleration from F, thus the relationship between F and N depends on the MoI (unless changes in d exactly cancel the effect of MoI changes).
 
  • #12
Juanda said:
If I don't have both of them I can't express the change in horizontal acceleration due to ##F## because it's the only force acting in that direction and the change in angular acceleration due to ##Nd## because without it there'd be a torque on the wheel caused by ##FR## that keeps accelerating it forward instead of decreasing its movement.
I think I figure out the problem here. The ##F## in your equations is not the same ##F## as presented in the video, the one referred to as "rolling resistance". Your ##F## is the reaction friction force coming from the ground.

Imagine setting ##d## to ##0##. It is still possible to have ##a_x \ne 0## and ##\alpha \ne 0##. There would still be a ##F## at the ground where ##F = ma_x##. You could even add a ##F_a## at the center of rotation - a horizontal force coming from an axle - or an added braking wheel torque and they would all increase the magnitude of your ##F##.

Rolling Resistance - let's call it ##F_{RR}## is defined as ##\frac{d}{R}N## where ##\frac{d}{R}## has been observed to be a constant. It is thus more practical to apply ##F_{RR}## at the contact patch than using ##N## at some distance ##d##; a distance that must be reevaluated every time ##R## changes, if it is known at all.
 
  • #13
jack action said:
I think I figure out the problem here. The ##F## in your equations is not the same ##F## as presented in the video, the one referred to as "rolling resistance". Your ##F## is the reaction friction force coming from the ground.
Yes, I think it's key to distinguish those two force terms. They are equal only in special cases, like rolling against rolling resistance at constant speed propelled by a force that doesn't create a torque around the CoM. But in the un-propelled slowing down case those force terms aren't equal, then their relationship depends on the MoI.
 
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  • #14
jack action said:
I think I figure out the problem here. The ##F## in your equations is not the same ##F## as presented in the video, the one referred to as "rolling resistance". Your ##F## is the reaction friction force coming from the ground.
I posted the source from the video as an attempt to make the thread clearer and by changing the letters I think I accomplished the opposite. I'll take it into account in future posts.

jack action said:
Imagine setting ##d## to ##0##. It is still possible to have ##a_x \ne 0## and ##\alpha \ne 0##. There would still be a ##F## at the ground where ##F = ma_x##. You could even add a ##F_a## at the center of rotation - a horizontal force coming from an axle - or an added braking wheel torque and they would all increase the magnitude of your ##F##.
If ##d## is set to ##0## it can either be because the wheel is perfectly rigid and round or because the wheel doesn't show hysteresis. I believe you are referring to the first option. If that's the case and we want to have ##a_x \ne 0## and ##\alpha \ne 0## then it's not that we COULD add a force but that we NEED to add an additional force (maybe pushing the axle) or torque (something like braking) besides what's going on at the contact point with the ground.
As counterintuitive as it may seem, the Newton equation says that if a perfectly round and rigid wheel is rotating without slipping on a surface then the frictional force must be 0. That's why it just keeps moving in the horizontal direction without changing its linear speed. Also, the weight and gravity don't contribute to the rotation because they're aligned with the CoM (##d=0##).
For example, it could be rolling on the asphalt where ##\mu \neq 0## and then transitioning to an ideal surface where ##\mu = 0## and nothing would change for the wheel because the tangential force at the contact point with the ground is 0 in both cases. To accelerate/decelerate said wheel without slipping it'd be necessary to add a force or torque and the no slipping condition would only be maintained as long as the necessary force to keep the rotation going doesn't surpass the limit static friction force.

jack action said:
Rolling Resistance - let's call it ##F_{RR}## is defined as ##\frac{d}{R}N## where ##\frac{d}{R}## has been observed to be a constant. It is thus more practical to apply ##F_{RR}## at the contact patch than using ##N## at some distance ##d##; a distance that must be reevaluated every time ##R## changes, if it is known at all.
And here lies what I believe is our main disagreement. You point the force I named ##F## or, maybe more conveniently, you named ##F_{RR}## is defined as ##\frac{d}{R}N##.
As I commented earlier, the derivation from the video implicitly uses ##\alpha=0## to obtain that result.
$$-F_{RR}R+Nd=I \alpha \rightarrow \left \{ \alpha=0 \right \} \rightarrow F_{RR}=\frac{d}{R}N$$

On the other hand, what I'm trying to say here is that that expression is not a general result. For example, in the exercise I showed, I feel that force would be ##F_{RR} = \frac{NdmR}{I+mR^2}## (I renamed it ##F_{RR}## and used ##R## instead of ##r## when compared with the OP to keep track of what we discussed in the thread so far).
I'll admit though that the force being dependent on the inertia feels weird but that's what I get from some equations I believe to be true (Newton and rotation).
As a last point, I still believe it's necessary to consider both ##F_{RR}R## and ##Nd## in the problem. Even if they were equal (when ##\alpha=0##), they're still both acting on the wheel after all so I can't understand why we should ignore one of them.
 
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  • #15
Juanda said:
And here lies what I believe is our main disagreement. You point the force I named ##F## or, maybe more conveniently, you named ##F_{RR}## is defined as ##\frac{d}{R}N##.
As I commented earlier, the derivation from the video implicitly uses ##\alpha=0## to obtain that result.
That's because ##F_{RR}## (roiling resistance) is supposed to represent the resistance due to deformation only, and not due to inertia. It's a mere convention that is useful, not something that is wrong or right. When it's determined experimentally, it is also usually done at constant speed.

But it is also messy, because depending on the application it lumps different effects together (the road can deform too).

Juanda said:
I feel that force would be ##F_{RR} = \frac{NdmR}{I+mR^2}##
I think that would be ##F_{fric}##, the actual physical frictional force, which results from both: deformation and inertia combined.

Juanda said:
I'll admit though that the force being dependent on the inertia feels weird but that's what I get from some equations I believe to be true (Newton and rotation).
It's not weird that ##F_{fric}## depends on inertia here. But ##F_{RR}## should not depend on inertia, per definition. However, it cannot be excluded, that the mass distribution affects how the wheel deforms, so the coefficient of rolling resistance would be acceleration dependent.

Juanda said:
As a last point, I still believe it's necessary to consider both ##F_{RR}R## and ##Nd## in the problem.
##F_{fric}## and ##N## are both real forces, that act together on the wheel, and have both to be considered.

But ##F_{RR}## is just a term you can compute from ##N## and ##d##, which in special cases equals ##F_{fric}##, but otherwise just represents a "component" of ##F_{fric}##.
 
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  • #16
I'm glad that you are stubborn @Juanda , because I just found that we are both right acceleration-wise but not on the definition of ##F_{RR}##. And I'll try to explain it some other way.

Here is how I would solve the problem (it seems overly complex here, but it simplifies more complex machines):

First I use the concept of equivalent mass ##m_e## to take into account the rotational inertia:
$$\frac{1}{2}m_ev^2 =\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $$
$$\frac{1}{2}m_ev^2 =\frac{1}{2}mv^2 + \frac{1}{2}I\frac{v^2}{R^2}$$
Or:
$$m_e = m +\frac{I}{R^2}$$
Using this concept I get:
$$N = mg$$
$$F_{RR} = m_e a$$
Note that I ignore both ##d## and ##\alpha## here and I assume ##F_{RR}R = Nd## such that:
$$a = \frac{F_{RR}}{m_e} = \frac{mg\frac{d}{R}}{m +\frac{I}{R^2}}$$
$$a = \frac{\frac{d}{R}g}{1+\frac{I}{mR^2}}$$
Or:
$$\frac{a}{g} = \frac{C_{RR}}{1+\frac{I}{mR^2}}$$
Where ##C_{RR} = \frac{d}{R} = \frac{F_{RR}}{N}## and it is know as the rolling resistance coefficient.

I know this to be true and valid.

But, where you assume ##FR \ne Nd##, you get:
$$a = \frac{NdR}{I+mR^2} = \frac{mgdR}{I+mR^2}$$
$$a = \frac{\frac{mgdR}{mR^2}}{\frac{I+mR^2}{mR^2}}$$
$$a = \frac{\frac{d}{R}g}{1+\frac{I}{mR^2}}$$
Or:
$$\frac{a}{g} = \frac{C_{RR}}{1+\frac{I}{mR^2}}$$
Which is the exact same result that I got! So all of your weirdly written equations are actually right ... like mines are.

But I say:
$$F_{RR} = N\frac{d}{R} = m_e a = \left(1+\frac{I}{mR^2}\right)ma$$
Note that you would get the same result by solving ##F_{RR}R = maR + I\alpha## (which is what I did in post #6).

And you say:
$$F= ma = \frac{N\frac{d}{R}}{1+\frac{I}{mR^2}}$$
This could be rewritten as:
$$F = \frac{F_{RR}}{1+\frac{I}{mR^2}}$$
I think this is really the friction force needed from the ground to compensate for ##F_{RR}## minus the effect of the rotational inertia like @A.T. said:
A.T. said:
I think that would be ##F_{fric}##, the actual physical frictional force, which results from both: deformation and inertia combined.

If somehow the radius of gyration would be zero, then ##F=F_{RR}##.

You must see that ##N\frac{d}{R}## is some force acting at radius ##R## with the equivalent moment of ##Nd##. What would be the name of that force if not the rolling resistance?

To come back with your question about power:
Juanda said:
From that, I can get the linear and angular velocities and how they change with time so I can obtain the power being lost. However, I feel I should be able to obtain the same result using ##P=Fv## or ##P= \tau \omega## but I wouldn't even know which force or torque to input in those equations.
If you want to know how much power is lost due to rolling resistance, you should imagine adding a wheel torque that will maintain a constant velocity ##v##. This means no acceleration, thus no horizontal forces. Therefore the necessary torque will clearly be ##Nd = \frac{d}{R}NR = C_{RR}NR## (or litterally the equivalent ##F_{RR}R##) and thus the power loss will be ##(C_{RR}NR)\omega## or ##(C_{RR}N)v##.
 
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  • #17
jack action said:
I'm glad that you are stubborn @Juanda
I get called stubborn somewhat often although people usually aren't happy about it 🤣 . In my defense, I'll say that rather than being stubborn I prefer to call it being skeptical. I try to have solid foundations built for the little I know so when I encounter something that contradicts them it's an opportunity to either revalidate them or demolish them if they're proved to be wrong. Demolishing them typically takes more effort though.

jack action said:
because I just found that we are both right acceleration-wise but not on the definition of ##F_{RR}##. And I'll try to explain it some other way.
That was an interesting ride I must say. Having two methods yielding the same answer definitely gives some confidence in the results obtained.

jack action said:
But, where you assume ##FR \ne Nd##, you get:
$$a = \frac{NdR}{I+mR^2} = \frac{mgdR}{I+mR^2}$$
$$a = \frac{\frac{mgdR}{mR^2}}{\frac{I+mR^2}{mR^2}}$$
$$a = \frac{\frac{d}{R}g}{1+\frac{I}{mR^2}}$$
Or:
$$\frac{a}{g} = \frac{C_{RR}}{1+\frac{I}{mR^2}}$$
Which is the exact same result that I got! So all of your weirdly written equations are actually right ... like mines are.
I wouldn't say I assume ##FR \ne Nd##. I just did not assume ##FR = Nd## because I couldn't understand why I should do it. You obtained the same expression for ##\frac{a}{g}## in both cases (both cases being: assuming ##F_{RR}R = Nd## with ##m_e## and considering ##F## to be unknown and deriving it from Newton and rotation as in the OP). This confuses me a little but I think the point is that ##F_{RR} = N\frac{d}{R}## is a convenient entity that allows for a simpler method to solve the problem although in my opinion it feels more abstract which makes it harder to follow.

jack action said:
To come back with your question about power:

If you want to know how much power is lost due to rolling resistance, you should imagine adding a wheel torque that will maintain a constant velocity ##v##. This means no acceleration, thus no horizontal forces. Therefore the necessary torque will clearly be ##Nd = \frac{d}{R}NR = C_{RR}NR## (or litterally the equivalent ##F_{RR}R##) and thus the power loss will be ##(C_{RR}NR)\omega## or ##(C_{RR}N)v##.
That makes sense. Because the input torque is enough to match the rolling resistance then the horizontal acceleration is 0 which in turn makes the sum of forces in the horizontal direction 0 as well so the only forces left to worry about are the vertical ones.
My concern left about that scenario is the following: is it that the horizontal force to the left at the contact patch ##F## or ##F_{RR}## became 0 or that the input torque is creating an opposite force to the right at the same point that neglects the first? Is there a physical difference in those scenarios or, in terms of Newton's equations, I am just describing the same thing with different words?
I feel like the scenarios are different. For example, a bar can be in external equilibrium but compressed by two forces. However, here the force would be applied at the same point so I don't know what to make of it.
 
  • #18
Juanda said:
My concern left about that scenario is the following: is it that the horizontal force to the left at the contact patch ##F## or ##F_{RR}## became 0 or that the input torque is creating an opposite force to the right at the same point that neglects the first?
Oh, you will love the discussion from "Plate sliding on ice with friction (Physics competition question)" starting at post #26:
jbriggs444 said:
jack action said:
So if there's only rolling resistance, you are saying that a wheel-driven vehicle rolling without accelerating requires no friction force from the ground?
Yes, that is what I am saying.

Rolling resistance manifests as an external torque, not an ordinary external force. The engine's forward torque on the wheel goes into cancelling the external torque from rolling resistance. Under pure rolling resistance, there is no external linear force to resist and no static friction required to resist it.
And from post #29:
haruspex said:
Consider the pure rolling resistance scenario, no drag.
The tyre has a greater elastic constant during deformation than during relaxation. That displaces the normal force ##N## some distance ##s## ahead of the axle. Combined with the bike's load on the axle, that produces a retrograde torque ##Ns## on the wheel.

In the steady motion of the bike, the motor provides a balancing torque to the driving wheel. ##\tau_{motor}=Ns##. The reaction torque on the bike is matched by a slight shift in the distribution of weight between the two axles. At this point, no horizontal forces are involved, so there is no frictional force in the direction of travel (but some lateral to stay upright, no doubt).
To overcome an external retarding force ##F## instead, the motor would need to provide axle torque ##Fr##, where r is the wheel radius. Thus the external force equivalent, as far as the motor is concerned, is ##Ns/r##.

Now consider the idling wheel. With no motor to provide torque, the rolling resistance torque would tend to slow the wheel down. ##I_{wheel}\alpha=Ns##. To keep the wheel turning at the rolling speed, there will be a backwards frictional force from the ground. ##Ns=rF_{fric}##. That is transmitted through the axle to the bike and to the driving wheel. As the motor works to maintain speed, the driving wheel therefore experiences an equal forward frictional force.

Adding these up, the torque the motor has to supply is ##2Ns##, whence ##2Ns/r=400N##. The frictional forces the wheels will exert on the plate are ##Ns/r## forwards at the front wheel and ##Ns/r## backwards at the rear wheel.
 
  • #19
Juanda said:
My concern left about that scenario is the following: is it that the horizontal force to the left at the contact patch ##F## or ##F_{RR}## became 0 or that the input torque is creating an opposite force to the right at the same point that neglects the first?
The horizontal friction force is 0, which you can decompose in infinitely many ways into multiple forces that add up to 0. But this is pure math and has no physical relevance.
 
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  • #20
Juanda said:
it bothers me it looks so different when compared with the usual ##F_f= \mu N##.
jack action said:
I think this is really the friction force needed from the ground
Yes, but it is static friction, so ##F_f= \mu N## is not right. Should be ##F_{fs}\leq \mu_sN##.
The retrograde torque created by the forward displacement of the normal force is tending to slow the rotation rate, which would make it less than ##v/r## (r being the height of the axle from the road). As a result, static friction arises, reducing the linear velocity while maintaining a matching rotational velocity.
 
  • #21
I was gone doing the dishes which gave the time to think more about that.

Rolling resistance is a torque. Everywhere they talk about this force, but that is misguiding at best because it is a fictitious force.

Let's talk about a rolling resistance torque instead.

This torque ##T_{RR}## is ##Nd##. and ##d## is known to be proportional to ##R##, so ##d=C_{RR}R## or:
$$T_{RR} = NC_{RR}R$$
Where ##C_{RR}## is the rolling resistance coefficient. There is no real need to define a fictitious force for ##C_{RR}N##.

Let's also put vertical and horizontal forces ##F_{ay}## and ##F_{ax}## acting at the axle and some wheel torque input ##T_w## for completeness of the general case. Subscript ##w## is for wheel characteristics. ##a_x## is still equal to ##\alpha_wR##.

The equations then become:
$$N= m_wg+F_{ay}$$
$$m_wa_x = F_{ax} + F_f$$
$$I_w\alpha_w + T_w = NC_{RR}R - F_f R$$
From this:
$$I_w\alpha_w = - T_w + NC_{RR}R+ F_{ax}R - m_wa_x R$$
$$I_w\alpha_w = - T_w + NC_{RR}R+ F_{ax}R - m_w\alpha_w R^2$$
Or:
$$\alpha_w =\frac{NC_{RR}R - T_w - F_{ax}R}{I +m_wR^2}$$
And:
$$a_x = \frac{NC_{RR} - \frac{T_w}{R} - F_{ax}}{m_w \left(1 +\frac{I_w}{m_wR^2}\right)}$$
And:
$$F_f = m_wa_x -F_{ax}$$
Or:
$$F_f = \frac{NC_{RR} - \frac{T_w}{R} - F_{ax}}{1 +\frac{I_w}{m_wR^2}} -F_{ax}$$
So the friction force ##F_f## can be equal to zero and depends on a lot more than the rolling resistance torque ##T_{RR}##.

Let's see some special cases where there is no acceleration:
  1. If ##T_w = NC_{RR}R## and ##F_{ax} = 0## then:
    • ##\alpha_w = 0##
    • ##F_f = 0##
  2. If ##T_w = 0## and ##F_{ax} = NC_{RR}## then:
    • ##\alpha_w = 0##
    • ##F_f = - NC_{RR}##
  3. If ##T_w = 2NC_{RR}R## and ##F_{ax} = - NC_{RR}## then:
    • ##\alpha_w = 0##
    • ##F_f = NC_{RR}##
Aaah! Sweet endorphins! Now my mind is at ease.
 
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  • #22
jack action said:
Rolling resistance is a torque.
If only it was that simple....

Once the ground deforms, the normal force at the slope that the wheel is continuously climbing has a horizontal component, which also becomes part of rolling resistance.

220px-Rolling_Resistance_2.png


As already said, rolling resistance is a combination of various effects, with application dependent composition. It's a useful engineering tool rather than some clean piece of physics.

The rolling resistance coefficient is defined based on forces, not torques. It's based on the force you need to pull something at constant speed and no other propulsion. But you could define your own coefficient, for example based on the torque you need to provide via the shaft for constant speed.
 
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  • #23
A.T. said:
If only it was that simple....

Once the ground deforms, the normal force at the slope that the wheel is continuously climbing has a horizontal component, which also becomes part of rolling resistance.

View attachment 330768

As already said, rolling resistance is a combination of various effects, with application dependent composition. It's a useful engineering tool rather than some clean piece of physics.

The rolling resistance coefficient is defined based on forces, not torques. It's based on the force you need to pull something at constant speed and no other propulsion. But you could define your own coefficient, for example based on the torque you need to provide via the shaft for constant speed.

This thread specifies rigid ground, deformed wheel, so the normal forces remain vertical. For this case, I stand by my post #20.

I do agree that it is unsatisfactory to deal in torque alone, as post #21 tries to do. It can’t slow without a horizontal force to provide the acceleration. Indeed, this seems to be the purpose of thread, to explain what horizontal force causes the slowing.

For the rigid wheel, deformed ground, case, I agree that the key is the redirected normal force. I presume from your diagram that the motion is to the left.
You haven’t said what ##F## is, but it works for me if that is static friction. The angled normal force tends to slow the linear motion without directly affecting the rotation. Consequently, static friction acts in a forward direction. That creates a torque slowing the rotation, while partly countering the linear acceleration from the normal force.
Note how this a reversal of the process in the rigid ground case.
 
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