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I see it explained as each rolling wheel experiences a centripetal force in the direction toward a turn center (axis or rotation) as in Fig 1 below. This is the definition of circular motion.

But what if that reference frame is accelerated as in Fig 2 where the turning/cornering car is on a rotating turntable? The car still travels along a curved path on the turntable though there are no longer centripetal forces to guide it along that circular path.

How does it still travel along that circular contact path? enter image description here enter image description here

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  • $\begingroup$ Does this answer your question? Why do cars turn by turning the front wheel? $\endgroup$
    – Solar Mike
    Commented Feb 24 at 19:53
  • $\begingroup$ Simplify it by considering the case where the car has no speed relative to the surface it is sitting on. $\endgroup$ Commented Feb 24 at 21:43
  • $\begingroup$ Thanks @Solar Mike. I would say that is a different way to change direction than what a typical car does. And if the car had just two wheels (very wide wheels so it doesn't tip over) it would still travel along a circular path. $\endgroup$
    – Matt Zusy
    Commented Feb 25 at 15:36

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Okay, you have three reference frames for this - Car, Track, and inertial - the latter I'm calling Earth.

The car's steering is fixed so that if it traverses, it will always trace an arc on the track. But the car's trajectory relative to Earth (at four contact points) requires us to combine the velocity vector or the track relative to Earth, and the car's velocity relative to Track (at those contact points). The car's acceleration vector in Earth can vary wildly, and so will the forces at the contact points. The forces are whatever is needed to realize the car's acceleration relative to Earth - not relative to Track.

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  • $\begingroup$ Thanks for the answer. I can agree with most of it but the first sentence is what my question is all about. “The car's steering is fixed so that if it traverses, it will always trace an arc on the track.” Why is this so? $\endgroup$
    – Matt Zusy
    Commented Feb 25 at 15:27
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The car track is locked by the geometry of the wheels, the forces created or referenced frame accelartion have nothing to do with the track, they will be incidental to to track of the car and frame of refrence.

If one creates a small scale model of the car with battery and scaled wiehgt and friction on a board and put the board on a roller coaster the track of the car will be the same! As long as the accelerations of the roller coaster are not going to toss the car off track.

I f we were tto study the forces on the track and tires we would have to solve a tensor with the path of roller coaster as a metric.

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  • $\begingroup$ Thanks @kamran. Through the principle of circular motion we have a circular track. The axis is the turn center and with the centripetal force at each wheel along with the forward motion of the car, there is a circular track left behind. I agree with how you explain it except at the first line “The car track is locked by the geometry of the wheels”. Why? This is my question. We use the principle of circular motion to explain it but I show that couldn't be the reason because there are no longer circular track centripetal forces in an accelerated reference frame $\endgroup$
    – Matt Zusy
    Commented Feb 25 at 22:28
  • $\begingroup$ Why a train turns with the train tracks. The wheels are mechanically trapped in the tracks. Car tires are mechanically trapped in the track. Unless one is a stunt driver and intentionally forces the car to skid or burn the tires for making them stick. Or as it was our case at the Mammoth there’s black ice on the road. $\endgroup$
    – kamran
    Commented Feb 25 at 23:08
  • $\begingroup$ A train travels a curved path on a curved track because the oncoming track does not allow the straight path motion it would otherwise travel. The ball on a roulette wheel does the same for the same reason. Car wheels are mechanically trapped in place, at any instant only, by friction. Why does it change direction at the next instant? $\endgroup$
    – Matt Zusy
    Commented Feb 26 at 1:21

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