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When you turn the front wheels to the right for example, your car will tend to continue in its original forward direction. However, there is friction between the wheel and the ground that prevents it from continuing in the forward direction. But if we are turning slowly, the wheel is not sliding against the ground (sliding along the axis of rotation of the wheel). Hence we should not be causing kinetic energy loss through kinetic friction. However, when the car comes out of the turn, it is inevitably much slower. My question is, where does the original kinetic of energy of the car go if there was no kinetic friction? All of the above assumes turning at the same height with no change in grav. potential energy.

We can assume that this is happening in a vacuum with no air resistance. With standard gravity. The car turning in such a way that it is not sliding. Also assume that the car engine is off so take away the complication with idle engines. Also assume that the car transmission has a differential so the two wheels can spin independently at different rates.

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First, even when a car is idle, the engine is still turning and it's possible for it to do work. So if you're going slow enough, you'll come out of the turn faster than when you entered it.

There is no kinetic friction between the tire and the road, but that does not mean there is no friction. The tire and road exert static friction between them. If there were no friction, the car would keep going straight, even with the tires at an angle. But static friction is not necessarily where the energy is lost either, since it could just as easily be doing positive work to the car, like when you speed up.

The real loss of kinetic energy comes from air resistance and friction/deformation losses within the car. A tire is not a perfect circle, it is slightly flatter on the bottom than the top. As the tire rotates, the rubber changes shape and heats up in the process. Similar deformations occur in the axle, and any other moving part of the engine. Anywhere that moving pieces touch, such as the axle touching its housing or the pistons against the cylinder wall will have kinetic friction, even if lubricated.

Edit in response to your update: All that's changed is you removed one source energy loss: air resistance. All other sources of loss are still there, internal friction and deformation.

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  • $\begingroup$ The fact that thermal losses due to rubber deformation are significant is the reason why merely having under-inflated tires will cause gas mileage to drop measurably: under-inflated tires will deform "a bit more". Note also that even when turning slowly there is always a bit of grinding (kinetic friction) as the extended contact surface between all four tires and the ground has to rotate, no matter how slowly you turn. Without that rotation, the front tires would go in their own separate way, away from the back tires. The car structural integrity prevents that, forcing that rotation. $\endgroup$ – Euro Micelli Sep 6 '18 at 6:23
  • $\begingroup$ Not necessarily. If the turn is large enough, it could be achieved through stretching of the rubber in contact with the road alone. It would have to be nearly straight, but it could be possible. We're already dealing with an air resistance free situation with no engine, so why not use another idealization? $\endgroup$ – Johnathan Gross Sep 6 '18 at 6:27
  • $\begingroup$ Good point. It's conceivable, but it probably requires different materials and structure. I strongly suspect (but don't claim to prove) that tire rubber is rigid enough (especially to lateral flexing where the rim contributes significantly to rigidity) that the contact surface will transition from static to kinetic well before any significant deformation will contribute to the rotation. In reality there might a combination of both effects, but I think a rigid model is likely a better approximation. $\endgroup$ – Euro Micelli Sep 6 '18 at 12:27
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Besides the deformation that the rubber tyre necessarily undergoes as it turns (already referred to in the earlier answers), I have always assumed that an important loss mechanism in the turn is that all four tyres have finite width, and for none of those four tyres is the required road speed the same across the width of the tyre. At some point across the width of the tyre, hopefully near the middle, the road speed matches the wheel speed nicely, but for every other point there's a mismatch. The side of the tyre on the inside of the turn may be turning too fast, and on the outside too slow, so for most of the width of the tyre you're dragging rubber over road. You can hear something that suggests this is going on too, with your car windows open.

Secondly, the geometry of the steering might change (some cars show lots of camber in the steering wheels when they turn) so there might be other loss mechanisms there.

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