Why is rolling easier than sliding?

[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

 

[thetrellan]

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??

I'm no scientist. I had to Google the subject. But I think you're thinking of friction in general as something that happens between moving surfaces. That's only the case with kinetic friction. Static friction is between objects that are not moving in relation to each other. If they do move- say if a wheel slips (spins)- it's kinetic. Don't confuse heat with friction. Heat is a byproduct. If I understand this right, static friction is the force that grips, or keeps things from sliding. It has to be strong to do that. If it's not, it starts sliding and turns kinetic. https://en.wikipedia.org/wiki/Friction#Static_friction

 

[haruspex]

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.

What do you mean when you say rolling is "less difficult" than sliding? Doesn't it depend on what is considered constant?

If a ball rolls down a slope, from rest, it does so at a certain acceleration. If I reduce the friction so much that it slides instead the acceleration will be greater, so sliding must be 'easier' in this sense.

Perhaps, as @jbriggs444 and @rbelli1 suggest, you are comparing sliding with a subcritical rolling state, i.e. one in which the static friction force is less than its maximum. If I push a heavy ball from rest I can get it to move with very little force (ignoring rolling resistance), but I have to push it much harder to get it to slide. But that has nothing to do with $\mu_s$.

Rolling doesn't involve any friction to continue

No, in general it does involve friction. If applying the brakes of a vehicle there is friction required between road and tyre to maintain rolling contact. Similarly for a car or ball rolling on a slope. Only in a few idealised contexts is it not required.

 

[Mister Perfessor]

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??

I should've spotted this mistake sooner!

I think you have considered the static friction keeping a tires contact patch from slipping across the road surface, as being the frictional forces which are inside the wheel bearings? Where a rolling tire involves the coefficient of static friction is located at the contact patch where that wheel touches the road. The bearing do not experience this so much

When driving on black ice, this coefficient is is very low and slippage becomes a concern. Nobody likes to unexpectedly find themselves driving on black ice, so it's a great thing that the dry road elsewhere has a very high coefficient of friction, which prevents the car from losing control and sliding sideways

So in a condition that could be referred to as 'standard', where the road surface has a constant coefficient of friction, no bumps or potholes to complicate matters and a few other things I will omit here, the roadbeds coefficient of friction has nothing at all to do with any part of the car which may add other sources of friction

But it's a bit over-simplified to always say that a freely rolling passenger car tire only experiences static friction loads when moving in a straight line, not being acted on to either speed up or slow down or to perform any other activity other than just simply rolling along. Car tires have sidewalls that greatly influence the outer edges of the contact patch. If you were to define a boundary line where the sidewall ends and the tread begins, then consider that the outer edge of this defined sidewall has a rate of speed that mostly relates to the diameter of the tire at any point other than where the contact patch is located combined with how fast that tire is rotating. Now due to how pneumatic tires work, and the weight of the car pushing that tire downwards into the roadbed, the area where the contact patch is located has a much greater area than the more simple model would hint at. Basically, the tire is 'squished'. So the contact patch is at shorter radius from the center of the wheel, than any other point of the tread. This is why there's a term called 'rolling radius' of a car tire

So the truth is, a car tire has a small amount of kinetic friction at the outer edges of it's contact patch being mixed into a much greater amount of static friction. This ratio can be made much greater by reducing the air pressure inside the tire, and/or changing the tire for one which is much wider

A interesting thing to do is to mount a thermal imaging camera above the tire and aimed downwards at the tread. The thermal distribution across the tire is always in a state of flux when outside forces act to change the conditions acting on the contact patch, things such as turning through corners or braking hard

But we are supposed to only be discussing an artificially simplified set of conditions? So yeah sure, there's only the coefficient of static friction to prevent the coefficient of kinetic friction from establishing itself as the dominant factor!

ps - don't forget the static friction keeping the tire attached to the wheel rim is always supposed to be greater than the static friction of the roadbed

 

[FactChecker]

If a ball rolls down a slope, from rest, it does so at a certain acceleration. If I reduce the friction so much that it slides instead the acceleration will be greater, so sliding must be 'easier' in this sense.

It seems to me that any nonzero force on the bottom would start a rotation that we would call rolling. So there can be no sliding with nonzero friction without also rolling. I don't see how that can be considered to make sliding "easier" than rolling. On the other hand, it is possible to roll an object with no sliding at the bottom contact point.

 

[FactChecker]

One could compare the rates of linear motion to the rate of rotation and see if there is more linear motion than allowed by rolling. If the excess (call it sliding) is greater than the rolling velocity, then it can be considered to be sliding easier than it is rolling. It is easy to think of an object designed with a large moment of inertia so that there would be more strictly linear acceleration than the component due to rotational acceleration. But that says more about the moments of inertia than it does about the friction.

 

[haruspex]

It seems to me that any nonzero force on the bottom would start a rotation that we would call rolling. So there can be no sliding with nonzero friction without also rolling. I don't see how that can be considered to make sliding "easier" than rolling. On the other hand, it is possible to roll an object with no sliding at the bottom contact point.

The usage with which I am familiar defines rolling as not sliding. It can be rotating and sliding.

 

[FactChecker]

The usage with which I am familiar defines rolling as not sliding. It can be rotating and sliding.

Ok. I'll buy that. I don't know if there is any official definition, but your's will make me drop my objection.

 

[skystare]

The main stopping benefit of ABS brakes is to avoid the steep temperature rise at the skidding footprint, which softens the rubber and allows it to slough off, thus lubricating the interface, leaving the skid mark and lowering the available stopping force. On ice, the reverse occurs; the surface of the ice under the tire melts in a skid, also best avoided for the shortest stop. On gravel ABS makes little difference, except maintenance of control (important in all three situations).
At least that's how my aircraft maintenance instructor explained it all those years ago.

 

[haruspex]

The main stopping benefit of ABS brakes is to avoid the steep temperature rise at the skidding footprint, which softens the rubber and allows it to slough off, thus lubricating the interface, leaving the skid mark and lowering the available stopping force. On ice, the reverse occurs; the surface of the ice under the tire melts in a skid, also best avoided for the shortest stop. On gravel ABS makes little difference, except maintenance of control (important in all three situations).
At least that's how my aircraft maintenance instructor explained it all those years ago.

Interesting... but irrelevant to the question in post #1.

 

[PeroK]

No, in general it does involve friction. If applying the brakes of a vehicle there is friction required between road and tyre to maintain rolling contact. Similarly for a car or ball rolling on a slope. Only in a few idealised contexts is it not required.

There is conservation of angular momentum in the absence of any external torque. If you consider conservation of angular momentum for a wheel to be valid in "only a few idealised contexts", then you ought be be consistent and apply that to everything: no parabolic projectile motion (air resistance); no frictionless pulleys (friction); no massless strings or ropes (mass); etc. All of classical mechanics practically is an idealised context.

Newton's first law only applies in a few idealised contexts as well.

 

[haruspex]

There is conservation of angular momentum in the absence of any external torque. If you consider conservation of angular momentum for a wheel to be valid in "only a few idealised contexts", then you ought be be consistent and apply that to everything: no parabolic projectile motion (air resistance); no frictionless pulleys (friction); no massless strings or ropes (mass); etc. All of classical mechanics practically is an idealised context.

Newton's first law only applies in a few idealised contexts as well.

I wrote "in a few" idealised contexts. I should have omitted "idealised". You can take idealised versions of a ball rolling down a slope (no drag, no rolling resistance) and my point still stands. The only case that springs to mind where there is in principle no friction on a frictional surface is rolling on the level at constant speed.
I interceded because it appeared to be confusing the OP.

 

[PeroK]

I wrote "in a few" idealised contexts. I should have omitted "idealised". You can take idealised versions of a ball rolling down a slope (no drag, no rolling resistance) and my point still stands. The only case that springs to mind where there is in principle no friction on a frictional surface is rolling on the level at constant speed.
I interceded because it appeared to be confusing the OP.

The argument went along the following lines (to summarise):

Staring Premise: Rolling without slipping (even at constant speed on a level surface requires static friction)
Static friction is greater than kinetic friction, so rolling should be "harder" than sliding against kinetic friction

One rebuttal is that the static friction need not be the maximum possible; so perhaps only a small force of static friction is required to maintain rolling - whereas, kinetic friction is constant.

A second point is that, by conservation of angular and linear momentum, it takes no external torque or force to maintain constant rolling motion. The required force of static friction is zero in this case. I.e. even on a frictionless surface rolling without slipping may continue (theoretically at least).

To alter the state of motion (either accelerate or decelerate the wheel) while maintaining rolling requires an external force and external torque in the correct proportions. This can be achieved on a rough surface with static friction. This cannot be achieved on a frictionless surface.

 

[haruspex]

The argument went along the following lines (to summarise):
Staring Premise: Rolling without slipping (even at constant speed on a level surface requires static friction)

Up to the point in question, this was the exchange:

I learned that rolling involves the coefficient of static friction

Rolling doesn't involve any friction to continue.

Nothing there about "constant speed on a level surface". I'm sure that's what you had in mind when you made the remark, but I see no evidence it is what the OP had in mind. The OP appeared, unsurprisingly, somewhat confused.

 

[PeroK]

Up to the point in question, this was the exchange:Nothing there about "constant speed on a level surface". I'm sure that's what you had in mind when you made the remark, but I see no evidence it is what the OP had in mind. The OP appeared, unsurprisingly, somewhat confused.

We'll have to disagree about the reason for the OP's confusion. The textbook he quoted says that "rolling without slipping depends on static friction". That definitely gives me the impression that without friction an object simply cannot roll.

You could say "motion depends on an external force". That's true in general and, of course, in practice. But, you still have Newton's first law.

You do not need a force to have motion; and you do not need a torque to have rolling without slipping.

 

[haruspex]

The textbook he quoted says that "rolling without slipping depends on static friction".

Yes, that is a common error. Let's just say your response needed a little clarification.

 

[zoki85]

You could say "motion depends on an external force". That's true in general and, of course, in practice. But, you still have Newton's first law.

You do not need a force to have motion; and you do not need a torque to have rolling without slipping.

 

[anorlunda]

The OP has not been back on PF since post #32 last Monday, the thread is now up to #50.

The late Jim Hardy was fond of saying "A question well asked is half answered." I would like to challenge all of you to restate the OP question in a way that could have been answered without a big debate.

 

[jbriggs444]

OP has clarified his own confusion in #6 and #12:

In #6, he is speaking of his teacher's claim:

"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 #12, he relates how the teacher's claim seems to run counter to his own experience:

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.

Rephrasing:

You can stop very much more quickly [by a factor of 20, perhaps] with the brakes locked up and the tires skidding than with the brakes not applied and the tires rolling freely as you gently coast to a stop.

You can stop even more quickly if the brakes are applied at a threshold such that the tires are rolling but are just on the verge of skidding.

Rolling involves static friction. Skidding involves kinetic friction. How can static friction result both in less retarding force than kinetic friction and in more retarding force than kinetic friction?

Edit: One wonders whether the PF denizens can spin this out for another +20 responses yet.

 

[kuruman]

The OP has not been back on PF since post #32 last Monday, the thread is now up to #50.

You mean #70.

The late Jim Hardy was fond of saying "A question well asked is half answered." I would like to challenge all of you to restate the OP question in a way that could have been answered without a big debate.

My interpretation of OP's original question is "Why is it easier (less energy required) to accelerate an object to a final speed $V$ when it is allowed to roll (e.g. bicycle with wheels free) than when it is not allowed to roll (e.g. bicycle with wheels rolled)". I responded accordingly with #3.

Rolling involves static friction. Skidding involves kinetic friction. How can static friction result both in less retarding force than kinetic friction and in more retarding force than kinetic friction?

Yes, OP seems to think that the effects of friction are of a similar nature. That was pointed out in #37. OP's confusion appears to stem from the following erroneous reasoning because of lack of understanding of the difference between static and kinetic friction (statement 3):
1. Static friction comes into play when objects are rolling on a surface.
2. Kinetic friction comes into play when objects are sliding on a surface.
3. The motion of an object is similarly affected by static and kinetic friction.
4. The coefficient of static friction is higher than the coefficient of kinetic friction.
5. Therefore it should be harder not easier to roll an object than to slide it, no?

 

[Saw]

I didn't read all posts and don't know what made this thread so long... Really leaving aside the semantic part (what should be called "static" versus "kinetic" and what deserves the name "friction" and what not), I would think that this explanation from Tipler should answer the OP:

"As the car moves down the
highway, the rubber flexes radially inward
where the tread initiates contact with the
pavement, and flexes radially outward where
the tread loses contact with the road. The tire
is not perfectly elastic, so the forces exerted on
the tread by the pavement that flex the tread
inward are greater than those exerted on the
tread by the pavement as the tread flexes back
as it leaves the pavement. This imbalance of
forces results in a force opposing the rolling of
the tire. This force is called a rolling frictional
force. The more the tire flexes, the greater the
rolling frictional force."

... though to be honest I myself have now doubts.

Does Tipler mean that when "the rubber flexes radially inward where the tread initiates contact with the pavement", that is like kinetic friction pushing the car back, but when it "flexes radially outward where the tread loses contact with the road" that is like static friction pushing it forward, so one thing would compensate for the other, as long as the collision were perfectly elastic without energy being lost due to deformation? If that were true the reply to the OP would be as easy as this: in rolling you have the same as sliding but compensated for, though not fully and that is why rolling is easier than sliding. But as I said, in the end I am not sure if this is the right interpretation of Tipler's text.