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Say you have a car with perfect wheels and a perfect surface with so much friction that it never skids.

When the car is traveling forward, the wheels INSTANTANEOUSLY turn 90 degrees.

Does the car come to a complete stop or does it change direction and continue traveling left/right at the same speed?

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  • $\begingroup$ For the car to move sideways, there would have to be a sideways force. Where do you think that would come from? $\endgroup$ – ACuriousMind Aug 2 '14 at 19:49
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    $\begingroup$ Think about it: You posit a physically impossible surface, an a physically impossible maneuver, and then ask what happens? Physics cannot answer your question. Your question is meaningless. Which runs faster: a unicorn or pegasus? $\endgroup$ – garyp Aug 2 '14 at 19:52
  • $\begingroup$ @ACuriousMind I think the force for the car to move sideways comes from the surface itself, like a train that runs on tracks $\endgroup$ – maze-cooperation Aug 2 '14 at 20:34
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    $\begingroup$ @garyp I don't think the surface is impossible, it just needs to exert "so much friction that it never skids" which is not infinite or something. I think the crucial point is to define what is meant by instantaneaous $\endgroup$ – maze-cooperation Aug 2 '14 at 20:37
  • $\begingroup$ @garyp So? We often ignore friction, which would then be meaningless by the same argument. $\endgroup$ – Bernhard Aug 3 '14 at 8:22
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Let's rework this problem slightly into something more realistic so that people who are complaining this problem can't exist are somewhat satisfied . Lets say the car went airborne while traveling forward, and then the wheels turned to 90 degrees while it was airborne. It lands in one of 3 ways:

  1. Front wheels first

  2. Back wheels first

  3. All wheels simultaneously

In all cases one should think of a rigid body and apply forces at the respective areas (like in engineering statics).

In all cases the line of actions for the reaction force are pointing towards the rear of the car and create a moment tipping the rear upwards. Now whether or not there is enough force to tip the car over depends on the energetics involved.

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The car comes to a complete stop - after flipping tail over front.

The wheels at 90 degrees do not permit the car to go forward. For the car to turn, there would have to be a lateral force on the front wheels. But since the wheels point at 90 degrees, and are "perfect", there is no source for this force.

The car coming to an "instant stop" requires "infinite deceleration", but with the center of gravity above the ground - since you have a wheel - the car would have to flip over.

Better not turn so sharply.

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    $\begingroup$ haha, thanks Floris! I accepted Skyler's answer for imagining a realistic scenario which might help future readers, but I gave you an up-vote. $\endgroup$ – Tyler Langan Aug 27 '14 at 21:34
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If the angle of the tires is not exactly 90 degrees to the original line of motion, the car will follow the new tire heading. Ignoring flexing of the car and its suspension, an instantaneous change in tire angle will produce a very short spike of infinite acceleration sideways, as well as a spike of infinite deceleration along the original line of motion. Again ignoring the effects of the suspension, this will cause the vehicle to temporarily rise so that only one tire is touching the ground. For low speeds the center of gravity will not rise far enough to cause the car to flip, and it will come back down moving on a new course. This ignores the effects of the torque when the car is only on one wheel.

Actually, the description of the tires needs some tweaking. Tires with infinite friction will not allow turning, since this always involves a certain amount of differential skidding between the inside and the outside areas of the turn. That is, during a turn, (using a turn to the left for example) the tire's right side has to travel farther than the left side, and infinite friction won't allow this.

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  • $\begingroup$ When you say differential skidding are you talking about the difference in speed between left and right wheel (covered by the differential drive of the car) or speed withing parts of the same tire (for perfect tires, infinitesimal contact point, so no problem). I think you are focusing on the wrong problem. $\endgroup$ – Floris Aug 3 '14 at 19:43
  • $\begingroup$ I'm talking about speed within parts of the same tire. And no, appealing to "infinitesimal contact point" does not solve the problem. The OP specified as much friction as necessary. And if a point of contact is assumed, normal operation provides an infinitesimally wide contact area across the width of the tires, resulting in an infinitesimal contact area. And I'm not really focusing on it, just pointing out that the idea of essentially infinite friction has consequences which are, shall we say, inconvenient. $\endgroup$ – WhatRoughBeast Aug 3 '14 at 20:09
  • $\begingroup$ Fair enough. These thought experiments often run into inconvenient alleys. $\endgroup$ – Floris Aug 3 '14 at 20:11
  • $\begingroup$ This is a really interesting comment. Thank you! I never thought about friction within the wheel like that! $\endgroup$ – Tyler Langan Aug 27 '14 at 21:35
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The part of the question about perfect wheels and perfect surface (with as much friction as needed and no skid) is not that important. One can have an example in existing roadsways (albeit with good asphalt).

Now, friction is a force component which depends on velocity (or relative velocity between adjacent bodies or surfaces), moreover depends non-linear way. In the case of a car, the relative velocity is the velocity of the car itself.

One can do sample computations, but the intuition is based on the previous paragraph (which backs it up).

So, for a certain range of car velocities (small velocity), the car will actually turn, as expected, when the wheels do the instantaneous 90-degree turn.

As the car velocity grows larger, the car will not turn that smoothly, at some point it cannot turn because the non-linearity of the velocity-dependent friction will create such force components which will prevent the turn.

Also the part of car inertia (and car momentum) enters the picture and it is possible the car will do a complete u-turn (even in the opposite direction) or even turn upside-down.

Hope this answers the question.

The following was my answer for what happens without friction (misread the question too fast). However i wil leave it as it is still relevant to the whole answer.

Under these circumstances the car will not turn and also will not move (except if sliding due to another external force). Since both turning and moving the car (using the force transfered from the engine to the wheels) an amount of friction is needed. Isnt this sth, friction is needed for car movement :)

The mechanics of car movement is based intimately on friction between the surface of the "road" and the wheels. To move the car, the wheels (using friction) exert a force on the underlying surface (force generated from the car engine) and the surface, in its turn, exerts an opposite force which moves the car. Similar mechanics hold for turning the car, etc..

To have a vivid example, see a car trying to move on ice (sth similar happens also on sand, although seems counterfactual), the car will not move (except if it will slide by means of another force), since the wheels will have no friction (or minimum friction) to "hold on" to the underlying surface and thus receive the necessary opposite force which will move the car. Effectively the wheels push the underlying surface to one direction (and through an amount of friction) the underlyng surface pushes the car to the other direction, and thus movement (of the car).

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  • $\begingroup$ In the question it says that there is friction. As much friction as needed. $\endgroup$ – maze-cooperation Aug 2 '14 at 20:27
  • $\begingroup$ @maze-cooperation, oops! read it too fast, will edit the answer $\endgroup$ – Nikos M. Aug 2 '14 at 20:28
  • $\begingroup$ and why the downvote? i did not hear you. Maybe next time the downvoter will actually have sth to say in response. $\endgroup$ – Nikos M. Aug 3 '14 at 21:37
  • $\begingroup$ I did not downvote ... here, have an upvote $\endgroup$ – maze-cooperation Aug 4 '14 at 8:30
  • $\begingroup$ @maze-cooperation, thanx, i do not know who the downvoter is btw $\endgroup$ – Nikos M. Aug 4 '14 at 15:59

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