Why is rolling easier than sliding?

[UNForces_885]

I learned that rolling involves the coefficient of static friction unlike sliding that involves the coefficient of kinetic friction. It's known that the coefficient of static friction is always higher than the coefficient of kinetic friction. This should result in rolling to be more difficult than sliding as it involves higher frictional force, which is not the case in real life.
Could someone please help elaborate??

 

[Lnewqban]

The static friction happens between wheel and surface.
Between wheel and axis there is dynamic friction.

 

[kuruman]

Without any friction, a cylinder on an incline will slide but not roll. Static friction does not dissipate energy; it converts translational energy of the center of mass to rotational energy about the CM. There is rolling friction (look it up) which dissipates some energy but not as much as kinetic friction.

 

[jbriggs444]

Consider, for instance, a wagon wheel. It is about one meter in diameter and has a hub with a hole in it which is about five centimeters in diameter. Inside the hole sits the axle about which the wheel rotates.

The drag from the rolling wheel relates to kinetic friction at the axle. The drag on a wheel with the brakes locked up relates to kinetic friction at the road. Static friction is not involved.

The reason that a wagon wheel rolls easily is due to mechanical advantage. (And grease). You have the same load on the wheel bearing (1/4 of the wagon weight) as there is on the road. So the force of kinetic friction is nominally the same. But with a 20 to 1 mechanical advantage, rolling is easier than sliding. Grease improves that further. And reduces wear.

It is worth noting that the coefficient of static friction is the maximum frictional force that can exist between two mating surfaces that are not yet slipping. Just because the surfaces are not yet slipping that doesn't mean that the force has to be that high. Two surfaces with no tangential force between them can succeed in not slipping just fine! Like a book sitting on a table.

 

[PeroK]

I learned that rolling involves the coefficient of static friction unlike sliding that involves the coefficient of kinetic friction. It's known that the coefficient of static friction is always higher than the coefficient of kinetic friction. This should result in rolling to be more difficult than sliding as it involves higher frictional force, which is not the case in real life.
Could someone please help elaborate??

Rolling doesn't involve any friction to continue. Rolling continues through conservation of linear and angular momentum. Imagine a wheel rolling off a surface into space - it will keep rotating.

Static friction is necessary to initiate rolling from a surface (and to brake!). It's the static friction that provides the torque for angular acceleration. But, as the friction is static (no slipping), there is no displacement between the wheel and surface while they are in contact and no work done.

Also, if you slide a wheel along a rough surface, then kinetic friction will a) reduce the kinetic energy of the wheel and b) translate some of its linear motion to rotational motion. The wheel then reaches an equilibrium where it's rolling without slipping and friction is no longer required.

 

[UNForces_885]

Thanks for your answers. But I am lost indeed in those details you mentioned. Let me rephrase the question.
Assume a single car tire on a horizontal surface in two situations not attached to anything:
1- It's rolling (µs is involved)
2- It's sliding (µk is involved)
since Fsmax=µsN (where Fsmax is the maximum static friction and µs is the coefficient of static friction), Fk=µkN (where Fk is the kinetic friction and µk is the coefficient of kinetic friction), and µs > µk, I can assume that the tire will experience higher frictional force while rolling than while sliding.
This conclusion is totally counterintuitive to me.

Additional Info:
Please put my question in context with the following quote of my physics teacher.
"If you lock your wheels driving down the road on dry concrete if they are sliding, or skidding, you will have less friction than if they are rolling. (µs > µk)
This is in theory the idea of antilock breaking systems (ABS) in cars; they cause intermittent lockage of breaks to keep the wheeling rolling intermittently to prevent sliding and thus provide higher friction force (stoppage force) using μs instead of μk."

 

[PeroK]

Thanks for your answers. But I am lost indeed in those details you mentioned. Let me rephrase the question.
Assume a single car tire on a horizontal surface in two situations not attached to anything:
1- It's rolling (µs is involved)

This is wrong. There is no friction involved in rolling. See post #5.

 

[jbriggs444]

since Fsmax=µsN (where Fsmax is the maximum static friction and µs is the coefficient of static friction), Fk=µkN (where Fk is the kinetic friction and µk is the coefficient of kinetic friction), and µs > µk, I can assume that the tire will experience higher frictional force while rolling than while sliding.

That conclusion is overstated. The tire can experience higher frictional force while rolling than sliding.$$F_\text{s} \leq F_\text{smax} = \mu_sN > \mu_kN = F_\text{k}$$The actual force of static friction can be either greater or less than that of kinetic friction.

Modern ABS systems do not actually achieve an increase in friction above locked-up skidding. But they do preserve some measure of control while braking with maximum pedal force. Hence their utility.

Edit: It seems that modern controllers are smarter than the ones I'd understood from yesteryear and can improve traction above kinetic.

 

[PeroK]

I can assume that the tire will experience higher frictional force while rolling than while sliding.
This conclusion is totally counterintuitive to me.

If you have a bicycle, you can try the following experiment:

1) Lock the brakes and try to push the bike along the ground. It's very difficult. It can take quite a force to get it moving at all. That is static friction.

2) Once it is moving, it's not so hard to keep it going. That's kinetic friction, which is less than static friction.

3) Release the brakes so that the wheels can roll. Now, as you push static friction gets the wheels rotating. But, as the bike moves there is almost no resistance. And, once the bike is moving at constant speed static friction no longer plays a part.

When talking about static friction it's imporant to distinguish case 1) from case 3)!

 

[Lnewqban]

...
Additional Info:
Please put my question in context with the following quote of my physics teacher.
"If you lock your wheels driving down the road on dry concrete if they are sliding, or skidding, you will have less friction than if they are rolling. (µs > µk)
This is in theory the idea of antilock breaking systems (ABS) in cars; they cause intermittent lockage of breaks to keep the wheeling rolling intermittently to prevent sliding and thus provide higher friction force (stoppage force) using μs instead of μk."

That explanation is correct, except that the ABS releases the brake pressure from the locked tire and immediately re-applies the brake.
Each time the tire stops rolling while the car is still moving, the ABS repeats those steps.

That achieves two important things: better deceleration and keeping steering (in case of front wheels).
What seems confusing to you?

 

[UNForces_885]

This is wrong. There is no friction involved in rolling. See post #5.

How then can you explain the following quote?

"If you lock your wheels driving down the road on dry concrete if they are sliding, or skidding, you will have less friction than if they are rolling. (µs > µk)
This is in theory the idea of antilock breaking systems (ABS) in cars; they cause intermittent lockage of breaks to keep the wheeling rolling intermittently to prevent sliding and thus provide higher friction force (stoppage force) using μs instead of μk."

 

[UNForces_885]

That explanation is correct, except that the ABS releases the brake pressure from the locked tire and immediately re-applies the brake.
Each time the tire stops rolling while the car is still moving, the ABS repeats those steps.

That achieves two important things: better deceleration and keeping steering (in case of front wheels).
What seems confusing to you?

If the explanation is correct, it's a fact that the static friction of rolling wheels is higher than the kinetic friction of sliding wheels, which is not experienced when I roll or slide a wheel on its own as it's much easier to roll the wheel probably because less friction is involved.

 

[PeroK]

How then can you explain the following quote?

Braking is very different from rolling. The brakes induce the static friction to activate. Until the brakes are applied there is no friction (static or otherwise) involved in rolling - that's why it's so efficient and you can coast along without power for so long.

PS I didn't notice that you were talking about braking until the "additional info" in post #6.

 

[jbriggs444]

If the explanation is correct, it's a fact that the static friction of rolling wheels is higher than the kinetic friction of sliding wheels, which is not experienced when I roll or slide a wheel on its own as it's much easier to roll the wheel probably because less friction is involved.

It's because you are not applying the brakes. Static friction (of tires on road) can be greater than kinetic friction (of tires on road) if you are applying the brakes very hard, just short of breaking loose and going into a skid.

 

[A.T.]

I learned that rolling involves the coefficient of static friction unlike sliding that involves the coefficient of kinetic friction. It's known that the coefficient of static friction is always higher than the coefficient of kinetic friction. This should result in rolling to be more difficult than sliding as it involves higher frictional force, which is not the case in real life.

The losses in rolling are determined by the coefficient of rolling resistance:
https://en.wikipedia.org/wiki/Rolling_resistance#Rolling_resistance_coefficient

The coefficient of static friction determines whether you get rolling or sliding, but has nothing to do with the efficiency of rolling.

 

[Lnewqban]

If the explanation is correct, it's a fact that the static friction of rolling wheels is higher than the kinetic friction of sliding wheels,...

That is a fact.

... which is not experienced when I roll or slide a wheel on its own as it's much easier to roll the wheel probably because less friction is involved.

That is an error of perception.
You cannot push a tire (bracing yourself on something solid) and make it roll if there is no friction at the contact patch: you need that resisting force to create the torque that will induce the tire rotation.

When you perceive that your tire easily rolls, what you have is a succession of contact patches that grip hard against the surface, one little area after the other.
When each little section of contact patch finishes its job, it is lifted and separated from the surface by the geometry of the wheel, while the following one is beginning to land and establishes new grip.
None of those little areas slide or skid under normal conditions of pure rolling.

For that reason, a car can happily roll forward while is cornering hard: the persistent and strong static friction of each little area of successive contact patches prevents the car from sliding out of the curve.
If the cornering force is strong enough to overwhelm the static friction, a slide starts (dynamic sideways friction) and does not stop unless some other condition changes, because its value is always smaller than the value of static friction.

Think of a rack and pinion mechanism: you have smooth rolling with tenacious grip (think huge static friction) between wheel and linear gear:

 

[zinq]

"There is no friction involved in rolling."

Maybe not once a wheel has already begun to roll on a perfectly level ideal frictionless track.

But certainly friction is necessary for rolling something uphill, even slightly uphill.

 

[PeroK]

"There is no friction involved in rolling."

Maybe not once a wheel has already begun to roll on a perfectly level ideal frictionless track.

But certainly friction is necessary for rolling something uphill, even slightly uphill.

This is wrong.

You can rotate a wheel uphill by any torque. The same way you can rotate a wheel that is not touching any surface.

The track does not need to be frictionless for there to be no friction.

 

[Dullard]

A bacon and egg breakfast 'involves' a pig and a chicken. Like sliding and rolling, their involvement isn't the same.

 

[UNForces_885]

This is wrong.

You can rotate a wheel uphill by any torque. The same way you can rotate a wheel that is not touching any surface.

The track does not need to be frictionless for there to be no friction.

The more details you introduce the more I get confused. :(
Does a rolling car wheel on its own experience more friction than the same wheel sliding? If yes, why is it easier to roll the wheel than to slide it? [Please keep the answer as simple as possible]

 

[PeroK]

The more details you introduce the more I get confused. :(
Does a rolling car wheel on its own experience more friction than the same wheel sliding? If yes, why is it easier to roll the wheel than to slide it? [Please keep the answer as simple as possible]

I would ignore this digression. Concentrate on rolling and braking.

 

[UNForces_885]

I would ignore this digression. Concentrate on rolling and braking.

Could you please answer the question?
Does a rolling car wheel on its own experience more friction than the same wheel sliding? If yes, why is it easier to roll the wheel than to slide it? [Please keep the answer as simple as possible]

 

[PeroK]

Could you please answer the question?
Does a rolling car wheel on its own experience more friction than the same wheel sliding? If yes, why is it easier to roll the wheel than to slide it? [Please keep the answer as simple as possible]

What part of "friction is not involved in rolling" do you not understand?

 

[UNForces_885]

What part of "friction is not involved in rolling" do you not understand?

Actually, rolling involves static friction. That's why I am confused.

Giancoli-Physics-Principles-with-Applications-7th (2014) Page 204

 

[PeroK]

Actually, rolling involves static friction. That's why I am confused.
View attachment 259204
Giancoli-Physics-Principles-with-Applications-7th (2014) Page 204

Well, one of us is wrong!

 

[UNForces_885]

Well, one of us is wrong!

Most probably, neither of you is wrong and I only misunderstand the whole concept that's why I asked for elaboration.

 

[PeroK]

Most probably, neither of you is wrong and I only misunderstand the whole concept that's why I asked for elaboration.

If I was you I would trust the guy that wrote the book! It's a bit of a problem if he is wrong, though.

Here's a question for you and Giancoli, though. If a rolling wheel reaches a big hole in the road, does it keep spinning? Or, because it no longer has static friction to keep it spinning, does it stop spinning? If so, what happened to conservation of angular momentum?

Or more pointedly, what force does it take to keep a spinning wheel spinning? By conservation of angular momentum, no force is required to keep it spinning; only to start it spinning, speed it up or slow it down.

 

[jbriggs444]

Possibly we should shift to a simpler situation.

A 1 kg book is sitting in the middle of a horizontal table. The coefficient of static friction between book and table is 0.2. What is the force of static friction between book and table?

 

[UNForces_885]

Possibly we should shift to a simpler situation.

A 1 kg book is sitting in the middle of a horizontal table. The coefficient of static friction between book and table is 0.2. What is the force of static friction between book and table?

My whole point was to address the confusion related to static and kinetic friction while a wheel is rolling or sliding thus my question;
Does a rolling car wheel on its own experience more friction than the same wheel sliding? If yes, why is it easier to roll the wheel than to slide it?

 

[jbriggs444]

My whole point was to address the confusion related to static and kinetic friction while a wheel is rolling or sliding thus my question;
Does a rolling car wheel on its own experience more friction than the same wheel sliding? If yes, why is it easier to roll the wheel than to slide it?

The fact that you ask this question suggests that you do not yet grasp the basics. So please answer my question.

 

[UNForces_885]

Possibly we should shift to a simpler situation.

A 1 kg book is sitting in the middle of a horizontal table. The coefficient of static friction between book and table is 0.2. What is the force of static friction between book and table?

Without applying any force on the book the force of static friction is zero N

Assuming that you apply some force on the book, the static friction increases with increasing the applied force until the maximum static friction is reached, beyond which the book will start moving.

Fsmax=µsN
In this case the force of normal equals to the weight of the book, thus
Fsmax=µsmg = (0.2)(1kg)(10m/s2)
The maximum static frictional force on the book is 2 N

 

[A.T.]

If yes, why is it easier to roll the wheel than to slide it?

Because the rolling resistance is usually smaller than kinetic friction.

 

[FactChecker]

Static friction is what stops a ball from just sliding down a slope, but it is not the force that must be overcome for the ball to roll. For a ball to roll, only the rolling friction must be overcome. That can be very low, much lower than either static or dynamic friction.

 

[russ_watters]

Without applying any force on the book the force of static friction is zero N

Assuming that you apply some force on the book, the static friction increases with increasing the applied force until the maximum static friction is reached,

Good, so you know that the static friction force is equal to the applied force at the contact point and can be much less than the theoretical maximum available static friction. So when an object is rolling, what causes a force to be applied at the contact point?

 

[FactChecker]

Consider the case of a cogged wheel rolling down a cogged straight rail, with the cog teeth meshing. Although the meshed cogs do not fit the definition of static friction, they can illustrate the essential concepts. It would be very hard for the wheel to slide down the rail without rolling; it might require breaking the cog teeth. That illustrates the concept of static friction. On the other hand, there is very little to stop the wheel from rolling down the rail. That illustrates the concept of rolling friction. In general, static friction is much greater than rolling friction.

 

[Lnewqban]

I believe that your confusion comes from mixing in one question:
1) Static friction and kinematic friction.
2) ABS brakes.
3) The simple machine of wheel and axle.

Let's go back to the explanation of your professor, which involves only #1 and #2.
Please, go back to post #16 and then watch the following video:

 

[kuruman]

It seems that @UNForces_885 thinks of static and kinetic friction as having similar effects on a rolling object having misjudged the importance of the following statement out of Giancolli, post #24, "The friction is static because the rolling object's point of contact with the ground is at rest at each moment." (emphasis mine)
To @UNForces_885:
The effects are not similar. Consider the difference between the following two cases.
1. You push a bicycle on a flat surface so it moves forward with the wheels rolling.
2. You push a bicycle on a flat surface gripping the brakes so it moves forward while the wheels slide and scrape on the floor.

Why do you have to push harder in case 2 than in case 1 to get the bike to move forward at the same speed?
Answer: In case 1 the energy per unit time (power) that your muscles spend is not lost; it is divided into moving the center of mass of the wheel faster and into getting the wheel to spin faster about its axis. Static friction is what divides the power. If the floor were frictionless, it would be just as easy to move the bike forward, although the wheels would not roll without static friction.

In case 2 the tires rub and slide across the floor and unlike case 1, a lot of heat is generated at the point where the tires touch the floor. Where does the power to generate this heat come from? Your muscles that must now work harder.

In case 1 where static friction is present, hardly any muscle power goes into generating heat; in case 2 where kinetic friction is present, quite a bit of muscle power generates heat. This is because static friction does not convert energy into heat; it is not a dissipative force. By contrast, kinetic friction does convert energy into heat and is a dissipative force.

 

[Lnewqban]

Regarding #3 above:
In this case, we have both types of friction again, but we also have mechanical advantage involved, which is the easiness you feel about using a wheel.
The following link tells us about the reason for which pushing/pulling a wheeled vehicle requires less effort than dragging the same wheel-less vehicle:

https://en.wikipedia.org/wiki/Wheel#Mechanics_and_function

In the particular case of your detached wheel-tire, or any free rolling disc or coin on a flat surface and following a straight trajectory, that mechanical advantage does not apply.
The static friction between both surfaces in contact does not need to be big, but there is some of it if there is no relative movement between the surfaces.
If the trajectory becomes curvilinear, then some additional lateral friction force needs to appear, in order to compensate for the cent

 

[anorlunda]

Wow. What a spat over what I believe is just semantics.

Just this morning I rode my bicycle through a patch of soft sand. It took all my strength to pedal. Also when the tire pressure is very low, the resistance to pedaling becomes bigger and the tire sidewalls become hot to the touch. @A.T. 's link below mentions all those things, but it is called "resistance" rather than "friction". The link even shows a table of resistance for a number of cases of hard/soft wheels on hard/soft surfaces.

The losses in rolling are determined by the coefficient of rolling resistance:
https://en.wikipedia.org/wiki/Rolling_resistance#Rolling_resistance_coefficient

If we avoid the word friction, is there still a debate?

 

[PeroK]

What a spat over what I believe is just semantics.

I tried to explain braking to a friend of mine a few years ago and explain the difference between controlled braking (static point of contact with the road) and skidding (locked wheels and burning rubber) and that's what he said: that's just semantics!

 

[jbriggs444]

If we avoid the word friction, is there still a debate?

I think there is still room for a debate on what question the OP is trying to ask.

My take on it is: "how can it possibly be easier to push a bike with the brakes off than with the brakes on when my teacher [seems to have] said different".

 

[kuruman]

I tried to explain braking to a friend of mine a few years ago and explain the difference between controlled braking (static point of contact with the road) and skidding (locked wheels and burning rubber) and that's what he said: that's just semantics!

My explanation to a friend under similar circumstances was, "Imagine trying to prevent a falling weight by grabbing a rope to which it is tied. You can grab and hold the rope until it stops sliding through your palm (not recommended) or you can grab and let go in very rapid succession (recommended if you can do it)."

 

[PeroK]

My explanation to a friend under similar circumstances was, "Imagine trying to prevent a falling weight by grabbing a rope to which it is tied. You can grab and hold the rope until it stops sliding through your palm (not recommended) or you can grab and let go in very rapid succession (recommended if you can do it)."

I always use one of these:

https://www.petzl.com/INT/en/Sport/Belay-devices-and-descenders/REVERSO

 

[kuruman]

I always use one of these:

https://www.petzl.com/INT/en/Sport/Belay-devices-and-descenders/REVERSO

To explain controlled braking or to go up and down vertical cliffs? This is the first time I've seen one of these, and probably the last.

 

[etotheipi]

It's probably not my place to post here, as I don't want to risk adding more fuel to the fire. Though I found rolling a little confusing as well, so I thought I'd share a few pointers.

When something is rolling, it just implies the translational speed of the wheel wrt the ground, $v_t$, equals the linear speed of the wheel wrt the COM: $v_t = r\omega$. If something is rolling constantly, then $a_{t} = r\alpha$. Crucially, the rotational acceleration times $r$ has to equal the linear acceleration.

On a flat surface (smooth or rough, however for now no rolling resistance) at a constant speed, we don't need any frictional force. Both the rotational and linear accelerations are zero. A frictional force in either direction would cause the translational speed to increase in one direction and the rotational speed in the opposite sense - the wheel would no longer roll! If it's going down a hill, no frictional force would result in the translational speed increasing (as $g\sin{\theta}$) but the rotational speed remaining constant (since the weight has no torque!). So to keep rolling, we must have a frictional force pointing up the slope.

If we return to the flat surface, now a rough one, and try and accelerate (i.e. moving off in a car), the setup changes again. The axle provides a torque to the wheels in the clockwise direction. If no friction acted, they'd just spin and we'd go nowhere! Now, a frictional force must act pointing forward so that we also get linear acceleration. The upshot is that the torque of the frictional force now opposes the torque provided by the axle, so we need to take this into account as well - but this causes no problems.

Of course, you can keep adding complexity, namely rolling resistance. But this post is now too long and I want to go and watch TV. Thanks for reading my thesis!

 

[jack action]

I learned that rolling involves the coefficient of static friction unlike sliding that involves the coefficient of kinetic friction. It's known that the coefficient of static friction is always higher than the coefficient of kinetic friction. This should result in rolling to be more difficult than sliding as it involves higher frictional force, which is not the case in real life.
Could someone please help elaborate??

The whole problem of your misunderstanding is that the friction force works in opposite directions in your two examples.

Sliding: The friction force works against the applied force. If you push a little, the friction force is equal and opposite and, up to the maximum static friction force, the object doesn't move. If you push harder, you break the static friction and the object begins to move. The kinetic friction force now applies, which is lower than the static force.

Rolling: The friction force works in the direction of motion. In this case, a torque is applied to the wheel. The friction force reacts to the torque, pushing the wheel forward. The friction force is equal to the torque times the wheel radius, up to the maximum static friction force. If the torque increases, then you break the maximum friction force and the wheel begins to slip. The kinetic friction force now applies, which is lower than the static force (which means the wheel could still go forward, even while slipping).

Braking: The wheel torque is now reversed, thus the friction force works against the direction of motion, just like with sliding. But just like with rolling, if the braking torque is small, the friction force is small (less than the maximum static friction force), and there is still rolling. If the torque increases causing the maximum static friction force to be broken, the wheel locks up and sliding occurs (with the kinetic friction force).

 

[Revolucien]

I thought this was interesting and that the original statement you made
"It's known that the coefficient of static friction is always higher than the coefficient of kinetic friction."
I think this is entirely variable based on materials and direction of applied of force.
Materials 1st. If your wheels are rubber and placed on pavement the surface tension created by the contact patch will create a very large friction to overcome and slide, however if your wheels are made of highly polished chrome and placed on an ice surface the wheel will slide before rolling.
Direction of applied force 2nd. If you have the wheel (rubber or chrome) placed on an ice surface and you apply force from one side at the bottom upward, you will release some of the surface tension and the wheel will roll forward. If you push downward on one side with the same force, the wheel will slide forward.

 

[jbriggs444]

I thought this was interesting and that the original statement you made
"It's known that the coefficient of static friction is always higher than the coefficient of kinetic friction."
I think this is entirely variable based on materials and direction of applied of force.
Materials 1st. If your wheels are rubber and placed on pavement the surface tension created by the contact patch will create a very large friction to overcome and slide, however if your wheels are made of highly polished chrome and placed on an ice surface the wheel will slide before rolling.
Direction of applied force 2nd. If you have the wheel (rubber or chrome) placed on an ice surface and you apply force from one side at the bottom upward, you will release some of the surface tension and the wheel will roll forward. If you push downward on one side with the same force, the wheel will slide forward.

All of this is irrelevant. The definitions of the coefficient of kinetic friction and of the coefficient of static friction already specify the directions of the relevant forces. Further, the statement being made assumes that one is comparing the coefficients for the same pair of materials.

The key (for the proposition that rolling is easier than sliding) is that the coefficient of static friction does not determine the force of static friction. It only imposes a maximum.

 

[Mister Perfessor]

Thanks for your answers. But I am lost indeed in those details you mentioned. Let me rephrase the question.
Assume a single car tire on a horizontal surface in two situations not attached to anything:
1- It's rolling (µs is involved)
2- It's sliding (µk is involved)
since Fsmax=µsN (where Fsmax is the maximum static friction and µs is the coefficient of static friction), Fk=µkN (where Fk is the kinetic friction and µk is the coefficient of kinetic friction), and µs > µk, I can assume that the tire will experience higher frictional force while rolling than while sliding.
This conclusion is totally counterintuitive to me.

Additional Info:
Please put my question in context with the following quote of my physics teacher.
"If you lock your wheels driving down the road on dry concrete if they are sliding, or skidding, you will have less friction than if they are rolling. (µs > µk)
This is in theory the idea of antilock breaking systems (ABS) in cars; they cause intermittent lockage of breaks to keep the wheeling rolling intermittently to prevent sliding and thus provide higher friction force (stoppage force) using μs instead of μk."

I think you may have left out one important detail in the stated quote from your physics teacher: "If you lock your wheels driving down the road on dry concrete if they are sliding, or skidding, you will have less friction than if they are rolling"

I think that what your teacher had meant to say is that the overall braking force of the car will be lower due to a loss of frictional conditions while the wheels are locked-up & skidding as compared to how much braking force is being generated with the brakes fully engaged, yet not quite locking up the wheels. I would think this was the case because the total of all frictional forces while the car is rolling freely without any action from either the drivetrain or the braking system will definitely show much less total friction vs locked-up skidding

I also think you may have made a mistake in how you describe what ABS braking systems are doing: "they cause intermittent lockage of breaks" 'Lockage of brakes' would seem to imply lockage of the tire? But this is definitely not what an ABS system should be doing! It's more like a form of pulse-width modulation being applied to regulate the braking force from the hydraulic pump inside the ABS system. Basically, the ABS is designed to apply a little too much braking effort(without excessive overshoot), and then to use the wheel speed sensor data to form a continously changing pulse-width modulation that acts to reduce the hydraulic pressure from the ABS pump, so that the wheels do not ever come to a complete stop. The hydraulic pressure reduction happens as a series of very rapid on/off cycles that you will feel at the brake pedal as a kind of 'shuddering'. Just keep in mind that what you feel is not actually the tire contact patches, or brake pads, locking up and then breaking free, but is instead the hydraulic pressure rapidly cycling from a high state to a low state

A good ABS system will also control how much pressure is being generated by it's pump. This could for example, be done by changing how fast the pump is running depending on how fast the car was traveling before the brakes were engaged. Basically, a lookup table could be established for that particular car

The loss of braking effect from a skidding tire vs a carefully controlled braking event on that same tire is a well-known hot topic on race tracks and there's plenty of published material just in straight-line braking distances measured on closed courses, showing what happens when the basics: tire pressure & weight distribution are adjusted for either better or worse performance, better being a reduced stopping distance

That sort of closed-course testing is also done showing the differences with ABS completely disabled, working decently and working to it's fullest potential after some fine tuning of the entire car as a closed system

There's a lot being done to teach proper braking technique when a car does not have ABS, such as defensive driving schools that show how to drive an older vehicle in snow. Having a good understanding of the physics involved will help to understand what those schools are trying to explain. Basically, you yourself would be attempting to approximate what an ABS system does, but with the same exact goals in mind

Just an idea, but I think you could disable the ABS on a modern car, then tap into the wheel speed sensors in order to provide a dashboard feedback of some sort to let you know when you are braking too hard

 

[rbelli1]

The key (for the proposition that rolling is easier than sliding) is that the coefficient of static friction does not determine the force of static friction. It only imposes a maximum.

You will also need to know the force holding the wheel down.

Does a rolling car wheel on its own experience more friction than the same wheel sliding?

The rolling wheel will have a larger coefficient of friction than the sliding wheel. The locked wheel will have more force decelerating the car than a freely rolling wheel. Breaking without sliding will have a force decelerating the car than can be smaller or larger than the locked wheel.

The phrase "more friction" is ambiguous. You can see the freely rolling wheel and say it has more friction (larger coefficient of friction) and the locked up wheel decelerating the car has more friction (more force from sliding surfaces due to a lower coefficient of friction) and get confused because the two identical phrases ("more friction" and "more friction") are actually referring to two entirely different things.

BoB