How is static friction the centripetal force during a car turning?

[Richard R Richard]

The distance between two points on a circle is a chord. The arc of circumference between these two points is always longer than the chord. When the tire hits the ground, the circumference is elastically compressed, to reduce its length from the arc measure to the chord measure.
If we define a forward direction and turn to the left, the compression on the left side of the tire will be less than on the right side, just enough to absorb the difference in turning radius.
As long as the area or weight differential over the area differential, multiplied by the mean coefficient of friction, does not exceed the maximum static friction force, integral over the tread area of the mean pressure per unit, there will be static frictional force with a component perpendicular to the speed of the car, that will be the maximum centripetal force without slip.
If the tire demands more, with tight curves, adding more load by weight or by aerodynamics, etc., then, there may be "also" a dynamic friction force, which opposes the relative sliding of the tire with respect to the ground, whose direction is a combination of the speed of the car and the tangential speed of the tire (it can be higher if you accelerate, the same if you only turn, zero if you skid, or retrograde as very few drivers know how to achieve).
To pretend that all the centripetal force is caused by the sliding friction of the tire, in my opinion is an error, easily demonstrable.
In the 500 miles more than half of the trip, the vehicle is turning, although there is cant, the tire forces in a centripetal direction during each turn, thus lasting approximately 100 miles. But can you imagine how long a tire lasts while sliding it 100 miles, even though I changed the speed of the vehicle very little? They are different demands.
A city car will be doubling 2% of the route with a duration of 60000km, it would be the equivalent of dragging it 1200km… and it is clear that this last wear will be higher than the previous one.
That is my point of view, static friction is always one or two orders greater than dynamic, and only in severe conditions of use, is it the main cause of wear, together with those derived from acceleration or braking. A vehicle on the road where there are curves that are longer, it wears less than in the city, where the turns are short and slower. But on the road more length is covered by doubling than in the city, so the dynamic friction would not correspond to the wear.

 

[sophiecentaur]

To pretend that all the centripetal force is caused by the sliding friction of the tire, in my opinion is an error, easily demonstrable.

Is anyone suggesting that? Not me (intentionally).

I am suggesting that there must be sliding somewhere on the edge of the footprint because the normal force is not great enough to produce static friction. Good tyres, inflated to a proper value will have little of this effect, compared with the static friction but when there's something wrong - say underinflation, the slip angle will increase and the distortion can be extreme in regions where the contact pressure is low.
In the examples I have already given, there is significant slip yet the car or motorcycle still corners.

 

[hutchphd]

To pretend that all the centripetal force is caused by the sliding friction of the tire, in my opinion is an error, easily demonstrable.

No one is making that claim.
But my instinct here is that any sliding anywhere on the contact patch will likely lead to a cascade where the entire patch will break free. And then you have only sliding friction.
I admit I had not considered the simple fact that the chord is shorter...gives the tire a lot more ways to stretch and conform.

 

[Richard R Richard]

Is anyone suggesting that? Not me (intentionally).

No one is making that claim.

Hi ! Maybe I have gone very to the extreme and it was not the intention.
I'm just trying to give an alternative point of view to

Cornering only happens when there is slip in some form.

which in my opinion is the opposite in meaning.

The lateral slip of a tire is the angle between the direction it is moving and the direction it is pointing. This can occur, for instance, in cornering, and is enabled by deformation in the tire carcass and tread. ... Sliding may occur, starting at the rear of the contact patch, as slip angle increases.

I have already agreed that there may be some dynamic friction, but not as the main cause for the centripetal force in all cases, if in some cases already analyzed very evident, I continue in my attempt to explain to OP how to relate "friction" vs "centripetal force"

So some slip occurs and that, of course, can't be where there's static friction and it is not the major factor in normal cornering.

I anticipated that I disagree, the same slippage in meters occurs for a curve with a wide radius as for a curve with a small radius, it is only a function of the total turning angle and the width of the tire, therefore the same slippage of the same tire in the same speed conditions cannot explain the two different radii of gyration. The different turning radii with their associated centripetal force need something else to be explained, and my point of view is that the cause is static friction, parallel to the wheel axis, which has a component in the direction of the center of the curve.

Regards.