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I tried looking this question up online but am left a little confused with all the answers, and the diagrams.

When a tire of a car negotiates a turn in the road, there exists friction which is providing the necessary centripetal acceleration. Why must this be towards a center? If someone could please draw the diagram of the wheel, that would be great.

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  • $\begingroup$ If you want to drive a circle there has to be always a centripetal force. $\endgroup$ – EuklidAlexandria Aug 4 at 3:21
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Your confusion is a common one. It applies to all uniform circular motion. Consider a rock tied to a string, and you swing the rock in a circle at a uniform speed around your head.

You feel the rock pull outward on you, and this leads you to think the force is outward. This is correct. That is the force the rock exerts on you.

Another question is what you are doing to the rock. Left to itself, the rock would fly in a straight line. You are pulling on the rock. The direction of the force is along the string. You are pulling it away from a straight line toward yourself.

The two forces - you pulling on the rock, and the rock pulling on you - are equal in strength, and opposite in direction.

For a car, it isn't as easy to see two equal and opposite forces, but they do exist. But we will only consider the force exerted on the car by the road. Left to itself, say if it was sliding on ice, the car would go in a straight line. Turning the tires would not change that.

On a road, if you turn the wheels to the left, they push the car to the left. That is, there is a lot of friction from the road that prevents them from sliding. The road pushes the tires to the left to prevent sliding, and the tires push the car.

If you keep turning, you go in a circle. You see the force to the left is toward the center of the circle.

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One way to answer your question:

Suppose the acceleration was not perpendicular to the "center", then $a$ would have a component in the direction of motion, which would resolve a change in the norm of $v$(speed). However, the speed was unchanged, a contradiction. Thus $a$ must be perpendicular to the direction of $v$.

(Perhaps looking for vector calculus, calculus III.)

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enter image description here

(„METHODE ZUR ERSTELLUNG UND ABSICHERUNG EINER MODELLBASIERTEN SOLLVORGABE FÜR FAHRDYNAMIKREGELSYSTEME Michael Graf“)

Look at this figure at the front wheel the centrifugal force act outwards so the friction force is always opposite to the velocity

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  • $\begingroup$ What a simple diagram! (Why is the car not perpendicular to $r_h$ and $r_v$, what's labelled as $\beta$ and why is there a SP?) $\endgroup$ – wizzwizz4 Aug 4 at 20:56
  • $\begingroup$ $\beta$ ist the so call side slip angle SP is the center of mass. the car is not perpendicular to the $r$ because the side slip angle. This Modell is a “Single Track Modell “ $\endgroup$ – Eli Aug 4 at 21:08
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Because when a body takes a turn... if we consider the body to be our frame of reference... It has a tendency to skid radially outwards due to centrifugal reaction force acting on it... As we know friction opposes motion hence it acts radially inwards...

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  • $\begingroup$ If the body is the frame of reference, there is no acceleration with respect to that frame, and centrifugal force is a fictitious force. From an inertial (non-accelerating) frame of reference, both forces could be considered as reaction forces, since the body accelerates inwards and the earth accelerates a very tiny amount outwards in response to the Newton third law pair of forces (acceleration = force / mass). $\endgroup$ – rcgldr Aug 8 at 2:36
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Let's start with a wheel that is totally locked, it cannot rotate; the brake is applied with full force, it locks the wheel. The wheel acts like a solid block in a simple friction-related experiment, i.e. it generates friction that opposes the movement. Top-down view:

locked

In the above picture the friction $\overrightarrow F$ has two components: $\overrightarrow {F_L}$ "along" the tire and $\overrightarrow {F_T}$ "across". Both components are sliding friction. Note they are applied to the bottom of the wheel, where it contacts the ground; the wheel itself transmits these forces to the axis and then to the rest of the car.

If you unlock the wheel, the $\overrightarrow {F_L}$ component will make it rotate. But then it will no longer be sliding friction. This component will become rolling friction $\overrightarrow {F_R}$ which is far less than $\overrightarrow {F_L}$ was. The next picture underestimates the effect, rolling friction is relatively even smaller:

rotating

The net force $\overrightarrow F$ has changed its direction. It's now pulling the wheel not only backwards but also (mainly!) to the left. Even if other phenomena decrease $\overrightarrow {F_T}$, it won't be as drastic change as from $\overrightarrow {F_L}$ to $\overrightarrow {F_R}$ and the main conclusion will stand: the net force will pull the wheel mainly to the left.

We can decompose $\overrightarrow v$ to "along" and "across" components and think what happens: relatively strong $\overrightarrow {F_T}$ is able to reduce the "across" component quickly, while in the same time relatively weak $\overrightarrow {F_R}$ barely affects the "along" component of the velocity. In effect the velocity vector approaches the plane of the wheel.

If steered wheels are on the front axis, the front of the car will be pulled left. The entire car will yaw left. At this moment the rear (non-steered) wheels will behave like in the second picture, they will contribute to the centripetal force that bends the vehicle path to the left.

If steered wheels are on the the rear axis (like in a forklift or a crop harvester), the rear of the vehicle will be pulled left and the vehicle will yaw right. At this moment the front (non-steered) wheels will behave like in the mirrored second picture, they will pull the front part of the vehicle to the right, assert centripetal force to the right and hopefully bend the vehicle path to the right.

Note in both cases steered wheels affect the yaw in the way the driver wants, but only in the former case they add to the centripetal force the diver (usually) wants. In the latter case they act against the desired centripetal force. Dragging steered wheels behind, if you try to turn too rapidly, the vehicle will swing its steered end and you could even turn it around with only little disturbance to the direction of its translational motion (if you didn't roll over first). Remember these car chases in the movies when the hero drives backward, violently turns the steering wheel and ends up driving forward without changing lanes? This is possible. And it doesn't work so easily from forward to backward because in this case all wheels bend the path in the same direction.

Back to the main point. By allowing steered wheels to rotate, we greatly reduce one component of the friction. If the wheel is skew, the net friction is not in the anti-direction of the translational velocity. This makes the friction change the direction of the velocity. Rotating wheels are crucial for steering control. The purpose of the anti-lock braking system (ABS) is to avoid situations like in the first picture above. Compare this video.

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  • $\begingroup$ Regardless of which wheels are steered (front or back), once in a steady state (and not skidding), the centripetal force exerted by the front and rear tires will correspond to the front and rear weight distribution of the car. If it's a rear engine car with weight ratio 40% front, 60% rear, then the front tires exert 40% of the overall centripetal force, and the rear tires exert 60% of the overall centripetal force. $\endgroup$ – rcgldr Aug 7 at 19:35
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To follow up on mmesser314's answer, there is a Newton 3rd law pair of forces. There is the centripetal force exerted by the road onto the tires, and this coexists with an outwards force exerted by the tires onto the road (which wasn't mentioned in mmesser314's answer).

Why must this be towards a center?

Assuming the current velocity is not zero, then an acceleration perpendicular to the velocity will result in a curved path, and at any point in that path, there will be a radius of curvature, and the perpendicular (centripetal) acceleration will be in the direction of the radius of curvature.

Consider the case of a car traveling at constant speed on a winding road with variable radius turns. The path is not circular, but at any instant in time, the curved path has a radius of curvature and a centripetal acceleration in the direction of the radius of curvature, except when the path is straight.

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