Question

A 810-kg race car can drive around an unbanked turn at a maximum speed of 40 m/s without slipping. The turn has a radius of 120 m. Air flowing over the car's wing exerts a downward-pointing force (called the downforce) of 9200 N on the car. What is the coefficient of static friction between the track and the car's tires? What would be the maximum speed if no downforce acted on the car?

Answer 1

The centripetal force holding the car in a circular path is the inward force of friction f = uN. So from 2nd law;

f = ma

uN = mv^2/R

Where N is the normal force between the car and the road. The vertical forces acting on the car are its downward weight, the downward "downforce" and the upward Normal force exerted by the road. Since there is no motion vertically these forces must add up to zero;

N - mg - 9200 = 0

N = (810)(9.8) + 9200

= 17138 N

This value can now be used in the centripetal force eq. to finf "u"

u = mv^2/NR

= (810)(40)^2/(17138 )(120)

= .630

If there was no "downforce" then N would just equal the car's weight;

N - mg = 0

N = mg

The frictional force, "uN" would be smaller, so max "v" would be smaller;

uN = umg = mv^2/R

v = SqRt[uRg]