Why does a car skid during circular motion when its velocity is too high?

I have come into this approach as follows, but I don't know whether it is correct or not.

By $F_{\text{net}} = \frac{\text{Mass} \times \text{Tangential velocity}^2}{\text{Radius}}$, since the kinetic friction is fixed, the radius of the circular motion will need to increase in order to compensate the increased velocity. Therefore, the car skids because of a circular motion with a larger radius.


Cars' tires when in circular motion keep on the road because the centrifugal instantaneous force is balanced by the frictional forces,which generate a centripetal, and thus a circle can be maintained. If this balance is not zero, skidding will result across the tangent to the circle, which is the instantaneous direction of velocity.

This can happen when going over a patch of water. When the velocity is very high the frictional forces between tires and road can diminish due to the heat generated in contact with the road. At some point the tires could melt, as seen by the dark traces on roads where accidents have happened. Thus the centrifugal would overcome the centripetal and there will be a skid on the tangent.

For curves in general the same mathematics holds, except the radius of the circle changes according to the curvature. Putting bends on roads , having the force of gravity in the centripetal direction, helps in allowing to keep the velocity constant without skidding, up to a point in velocity.


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