Why is rolling easier than sliding?

[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.

 

[Mister Perfessor]

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.

I agree with that, as the coefficient of friction between the tire's contact patch and the road surface is a variable, and that very aggressive braking will alter that state dramatically. Temperature rise inside the tire & the road surface play a very important role

Just consider the brake pads & rotor on that wheel in two different cases. In a static condition with the car at rest, the brakes applied to a preset hydraulic pressure, and with a long lever temporarily attached to the wheel - measure the amount of torque it takes to get the brake rotor to slip across the brake pads holding it in place. Then compare that torque to what is available in a kinetic case, where the car is moving before the brakes are applied. That adds a thermal effect to both the brake rotor and the pads. The pad material will exhibit very interesting behaviors, one of which is the generation of gases that act to lubricate the contact point between the pads and the rotor. So the standard rotor should have two versions, one a plain rotor and another as a vented type. My point though, is that you will always find a much higher braking force is possible when the car is at rest before applying the brakes

If for example, a tilt table were to be built, large enough to carry the car in question. And that this table were to be covered with the same exact material as used on a test track located at the same facility. And then install in the test car a simple pendulum with a damping mechanism to steady it's position. Park the car on top of the tilting table and while holding the brakes at a set pressure, tilt the table forwards until the car slips (safety straps would be used to limit the slip to a safe distance, say 4") Now take this car out on the test track and perform a stopping distance test. No matter how small you can get that stopping distance, the pendulum will never reach as high up it's scale as when the car was on the tilt table. This is in my opinion mostly due to the effects of thermal rise at various points, which can be viewed in a simple manner as the exchange of energy required to slow the car

But the OP wasn't asking for real-world conditions. I would only suggest adding the effects of temperature rise to the hypothetical 'rubber tire rolling down an incline' model

 

[A.T.]

...why is it easier to roll the wheel than to slide it?

Because the rolling resistance is usually smaller than kinetic friction.

To see that the above is only usually true, one might consider a counter example:

A very soft wheel on a hard surface can be easier to slide, than to roll.

This is because in rolling both the wheel and the surface are being continuously deformed. In sliding only the surface is continuously deformed, while the wheel adopts an approximately constant deformation. So if you make the kinetic friction and the surface deformation low, then sliding becomes more efficient.