# Kinetic friction and regenerative braking dilemma

1. Apr 7, 2015

### rcgldr

Assume a car has a smart regenerative braking system that optionally locks up the wheels with the tires skidding or allows the wheels to rotate while skidding, using the torque from the rotating and skidding tires to generate power into some type of kinetic energy recovery system. Assume the coefficient of kinetic friction is the same in both cases.

In the first case with the wheels locked up, all of the reduction of kinetic energy of the car during braking is converted into heat. In the second case with the wheels rotating, some of the reduction in kinetic energy is transferred into the kinetic energy recovery system, and the rest of it converted into heat by the skidding tires.

The dilemma here is that it seems the heat produced by kinetic friction would be the same regardless if the wheels are locked up or rotating but still skidding, but in the second case some of the kinetic energy is going into the kinetic energy recovery system.

I'm wondering if the relative velocity between tire surface and road while skidding makes a difference, base on the formula that power equals force times velocity.

2. Apr 7, 2015

### Staff: Mentor

Why would you assume that? If they are locked up completely then the "slipping rate" will be higher than if they are rotating but still skidding. Surely there should be higher frictional losses with a faster slipping rate, all other things equal.

3. Apr 7, 2015

### rcgldr

Which is why I continued with (this was an edit (text went blank while I was trying to compose message), maybe we cross posted)).

Clearly the car velocity matters since KE = 1/2 m v^2, so the same rate of deceleration at a higher speed means more energy consumed per unit of time, but I was wondering if the speed of the tire surface matters in a skidding situation, and apparently it does.

Last edited: Apr 8, 2015