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Wheel dynamics: sliding and skidding 
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#1
Nov3012, 03:10 AM

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Hi all.
Reading on several books and papers, I found that the motion of a wheel moving on a flat surface are given by (assume that positive torques are counterclockwise and the positive direction of the motion is toward the right direction): m \dot v=F J \dot ω=rF+T for T<0, where m is the wheel mass, J is the wheel inertia, F is the tractive force due to the sliding friction, r is the wheel radius and \dot z indicates the time derivative of z and T is the applied torque to the wheel, and m \dot v=F J \dot ω=rF+T for T>0. Now assume to start with the same initial condition v0=ω0 and assume to apply a sufficiently large (in absolute value) T<0 for t0≤t<t1. The system evolves accordingly to the first equations, and, at time t=t1, I would have v(t1)<rω(t1). Now, assume to apply a sufficiently small torque T>0. The system now evolves accordingly to the second dynamics. However, there will be a time t=t1 such that v(t1)=0 and rω(t1)≠0. This means that the wheel stop to translate but it doesn't stop to roll. Does it make sense physically? I struggled a lot to think about an everyday example which exhibit this phenomenon but I cannot find one, and I wonder if there is some mistake in my reasoning. 


#2
Nov3012, 04:45 AM

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If it is not translating but still "rolls" then wheel edge must be sliding on the surface  a real life situation that fits this description is when you accelerate too hard from a standing start in your car  the tires squeal  or when your car is stuck in mud (one wheel spins in the mud spraying your passengers  who have volunteered to get out and push.)
Rolling is actually quite complicated  as any snooker or pool player will tell you: http://www.jimloy.com/billiard/phys.htm Note: for rolling without slipping from left to right, accelerating, the torque that does this will be clockwise wouldn't it? It kinda helps thinking if a positive angular acceleration corresponds to a positive translational acceleration. 


#3
Nov3012, 05:21 AM

P: 9

Regarding the torque I have considered that the rotation axis is "entering" the screen, therefore you have negative torque > acceleration and positive torque > braking (I agree that is less intuitive). I also understand that when you are on the mud you "roll" but you don't translate since you may start with a zero translating speed (standstill) and then you start roll without moving, but my point is that if you start with a nonzero translating speed (and the wheel is rolling very fast), can it happen that when you brake the longitudinal speed goes to zero before the rotational speed? 


#4
Nov3012, 07:45 AM

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Wheel dynamics: sliding and skidding
Well ... since you are braking by applying a retarding torque to the wheel  your brakes are actually slowing the rotation and relying on friction to slow the translation as a consequence.
This means, if you brake hard enough, you can have the wheels spin backwards while translating forwards. In that situation, the translation slows to a stop but the wheels are picking up speed in reverse. But you are talking about soft braking. So the wheel spins forward, and is momentarily stationary ... keep braking and it goes backwards while still spinning forward ... that correct? Considering the mechanism for slowing down, that doesn't seem to make sense. You'd expect the wheel to stop turning before the friction stops the translation every time. I'm uncertain about some aspects of the model you have outlined  i.e. the direction of the friction force depends on the relative motion of the surfaces. Initially the wheel is turning clockwise and translating to the right, an anticlock torque is applied so the rotation slows down ... but it is still, initially, rotating clockwise. If I read the equations correctly, you flip the friction direction immediately the applied torque changes direction. I'm also unsure about the speed dependence of friction in the model. friction is usually described as ##f=\mu mg## between horizontal surfaces ... ##\mu## does has a small speed dependence. But there is a big change between static and kinetic friction ... the situation where ##v=r\omega## which is your initial condition, static friction comes into play doesn't it? So, for "sufficiently small" torques, the model may not apply? 


#5
Dec312, 02:16 AM

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Alternatively, I may flip the friction by checking the applied torque every time the condition v=rω is verified. So, indipendently of the applied torque (if it is positive of negative), when v<rω, then the equation of the motion satisfies the following: m \dot v=F J \dot ω=rF+T No matter if T is greater or less than zero. Then, at a certain point, if v=rω is verified, I check which of the two models I should consider. Is it correct? However, even in this case I would have a strange physical behavior. Imagine again that v<rω so that the motion is described by m \dot v=F J \dot ω=rF+T And imagine that I am braking. The longitudinal speed continue to increase. Isn't it strange that the longitudinal speed increases while braking? More in detail, by considering the above scenario, I have that the condition v<rω still holds for a little bit more, and the wheel speed is decreasing. This means that at some point I reach the condition v=rω, and only in that instant I can flip the sign of the friction force. Nevertheless, for all the time in takes to go from v<rω to v=rω, I have an increase of the longitudinal speed. Does it make sense that when you brake you still have an increase of the longitudinal speed, even if it is for small time? This poses the question: should I immediately change the friction direction as I change the direction of the torque`as I did before, or should I wait until ##v=r\omega## and then, depending on the applied torque, change the direction of the friction? In each case I would have a drawback when ##v<r\omega## and when braking: in the first case the longitudinal speed would go to zero before the wheel speed; in the second case I would have an increase of the longitudinal speed, despite I am braking. Which one is correct? 


#6
Dec312, 08:24 PM

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You should make the friction direction explicitly depend on the direction of the slip between the two surfaces that moving against each other.
Note: you will always have to account for the point where static friction takes over. It may be that it erases the artifact you have found anyway. 


#7
Dec412, 03:03 AM

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Welcome to the forum and very good work on this subject so far..At some point I think we need to address the effect of vehicle weight transfer when the brakes are applied. Specially
the effect of vertical tire loading and resultant increased ability of tire to “ grip” the surface. 


#8
Dec412, 03:18 AM

P: 9

m\dot v = F*sgn(rωv) J\dot w = rF*sgn(rωv)+T But in this case it is equivalent to split the dynamics as m\dot v = F J\dot ω = rF+T if rω>v and m\dot v = F J\dot ω = rF+T if rω<v and finally \dot v = k*T/r \dot ω = k*T if rω=v, where k depends on the mass, inertia, wheel radius etc. If we focus on the first two dynamics, then the right solution is the second one I proposed in the previous post, namely:"This poses the question: should I immediately change the friction direction as I change the direction of the torque`as I did before, or should I wait until v=rω and then, depending on the applied torque, change the direction of the friction?", right? 


#9
Dec512, 03:04 AM

P: 166

You may do better to rewrite the equations to track both torque values, since in practice both are applied at all times that the vehicle is in motion. Would this not remove any need to 'flip' things around and switch formulas? 


#10
Dec512, 05:13 AM

P: 9

However, I think I have found the solution. First of all, we have to keep in mind that the sliding friction force is independent of the applied torque and it depends only on the relative speed of the contact point. For instance, the magnitude of such friction force is equal to ##\mu mg##. The direction of such force points to the opposite direction of the relative speed, and therefore it has been modeled with ##sgn(rωv)##. Full stop. The effect of the torque T just changes the derivative of ω (as you can see it is an additional term in the ##\dot ω## equation) but it does not influence the magnitude of the friction force, not in its magnitude, nor in its direction. Therefore, the real dynamics are given by ##m\dot v = Fsgn(rωv)## ##J\dot w = rFsgn(rωv)+T## which can be split into two different dynamics, depending on the ##sgn(rωv)##. Therefore, the switching is completely determined by the term ##rωv##. Do you agree? :) 


#11
Jan914, 08:55 AM

P: 9

After long time I was dealing a little bit more with this issue. I thank all of you again for the support. I am sure that by using the sign function we capture the wheel behavior properly. :)
Next, we may want to include the effect the accelerating/braking actions have on the weight transfer as suggested by Ranger Mike. I was thinking to model it based on the following fact: when accelerating too much (imagine that the wheel is skidding), the wheel tends to detach the ground, while during a steep braking action, the "weight" that (vertically) pushes on the contact point wheel/ground is larger. Therefore, the effects when braking or accelerating can be modeled as a variation of the vertical pressure in the contact point wheel/ground, and whose magnitude depends on the applied torque and (eventually) on the wheel speed. Hence, the vertical force will not be only due to the "weight" mg, but we can add a corrective term that "scales" the term "mg" depending if we are accelerating of braking. What do you think? Furthermore, a component of such a vertical force may also be due the centrifugal force applied to the contact point wheel/ground? I imagine a wheel spinning very fast and not translating that tends to "jump" losing contact with the ground. Is it an effect of the centrifugal force? What do you think? 


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