# Why does a rotating tire use the static, rather than the dynamic coefficient of friction?

The explanation I have heard of the difference between static and dynamic friction is that static friction is stronger because bonds form when one object is put on top of another object and these have to be overcome to get the movement started.

For a rotating tire, although the point on the ground will be stationary for an instant, it would seem that bonds wouldn't have time to form. So, why isn't the dynamic coefficient of friction used?

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I'm not sure how accurate the "bonds form" explanation is for static friction, but to the extent that it's correct, I don't think that the process would take much time. Also, remember that a real tire isn't quite like the idealized picture we often draw of a circle tangent to a plane at just one point. The tire flattens at the bottom, so that there's a significant contact area. Any one point is in contact with the ground for some time, not just an instant. – Ted Bunn Apr 22 '11 at 0:53

A car drives at 20 m/s. The circumference of the wheel is 2m, so the rotation rate is 10 Hz. A reasonable percentage of the tire is in contact with the ground - maybe around 5%. That would give a contact time of 5*10^-3 s.

This is a pretty long time in molecular terms. The distance between molecules divided by the speed of light is around 10^-18 s, so that's the fastest we can imagine some sort of bonding occurring. That 16 orders of magnitude faster than the contact time.

Real chemical reactions must be slower, but with 10^16 "clicks" of time for the chemistry to sort itself out, there should be plenty of time for the tire to stick.

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You are correct that more than just a point will be touching the ground, but how much of it will be stationary relative to the ground? Perhaps none. – Casebash Apr 22 '11 at 4:33
@Casebash Look at a real tire: it flattens where it contacts the ground. Lets say the contact surface is a rectangle. The front edge of that rectangle is stationary until the tire has rotated enough (and the car moved forward enough) for that edge to become the rear edge and then subsequently lifted. – freespace Apr 22 '11 at 10:11
I've seen simulations of a tire under severe cornering stress, and all sorts of motion was going on. However that was a tire in a severe skid, which differs from the OP. The tire is still deforming considerably during contact, so even in the rolling state, I'd bet the whole contact area isn't stationary. – Omega Centauri Apr 22 '11 at 18:01
It would be great to have a more detailed answer taking into account Omega Centauris comment – student May 25 '11 at 17:08

You'd use the dynamic friction if you slam on your brakes and start skidding. Until then, your tire is rolling with the road and so the tire rubber isn't moving with respect to the road. Hence with tires rolling, you use static friction.

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I mentioned in the question that I knew the tire wasn't moving with respect to the road. The issue is that it is stationary for only an instant which doesn't seem to work with the bond explanation of static friction that I have heard – Casebash Apr 22 '11 at 0:51
@Casebash; Given the weight of the car and the inflation pressure of the tires, you can compute the area of the tire that is in contact with the road. Since the result is not zero, it's not possible for the tire to be in contact with the road for only an "instant". – Carl Brannen Apr 22 '11 at 0:59
@Casebah: replace the tire with a bowling ball, and that argument doesn't work out. – Jerry Schirmer May 26 '11 at 2:37

I am not sure why you are rejecting the static friction on the basis on long the parts are in contact. A "bond" is not a chemical bond that might take time, but rather an interaction between adjacent molecules, or atoms. It propagates at the speed of light, so there is plenty of time for the adjacent molecules to "bond" when sufficiently close enough.

In real life though, pairing down the tire/road contact into a friction coefficient is the wrong approach. It is a non-linear contact, where the higher the normal load the wider the contact patch is and the distribution of contact pressures changes. In addition, some parts have micro sliding as only 1 point in the contact patch is truly stationary.

There is something called the "Pacejka Magic Formula" which is a well established model of a tire contact, and there are newer ones out there which minor and major refinements to it. In the end, it depends on what you want out of it, in order for you to decide what contact/traction model to use.

[ref: Magic Formula]

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It's hard to believe this answer was downvoted because he's the only one who makes the very important point about micro sliding. – Marty Green May 25 '11 at 19:43
Broken link. Also I believe most tire models include an area, rather than a single point, that is in static contact as the tire elastically deforms to keep the tire surface from sliding when the shear stress is less than the normal stress times the static coefficient of friction. The ratio of area in static contact vs. sliding contact partially determines the effective coefficient of friction and is a function of slip rate. – Rick Aug 28 '15 at 12:00

We can theorize about why static or dynamic coefficient of friction is more appropriate, and why. However, the bottom line on why we use the static coefficient is that it has been proven, experimentally, to be the correct one.

As a thought experiment, try this:

You have a car sitting in a flat road. What is the coefficient of friction, static or dynamic? The answer is "Static." Now, roll the car very, VERY slowly (like one inch per hour). What is the coefficient of friction? Still static. Now roll it just a little faster and ask the same question. Now faster.

How fast do you have to go before the answer is that the coefficient of friction is no longer, "static?"

The answer to that question, in my mind, is, "fast enough for something(s) in the chassis/strut/spring/shock/axle/wheel/tire/road system to excite the rubber/concrete interface enough that radial vibration modes of the rubber in the tire were strong enough to lift the wheel off the road and keep it off." ("vibro-planing" if I may be allowed to coin a word). The speed required to do this, on a concrete surface in good repair, is probably an order of magnitude beyond the capability of any street-legal conveyance.

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