circular motion?

Question

A 2.20 103 kg car rounds a circular turn of radius 14.0 m. If the road is flat and the coefficient of static friction between the tires and the road is 0.7, how fast can the car go without skidding?
m/s

Answer 1

V=sqr(r*mu*g)
V= sqr(14.0m*0.7*9.8)
V=96.04
V=9.8m/s

Answer 2

f = ma = Fc - Ff; where f = net force on the car in the curve, m = car mass, a = acceleration along the radius of the turn, Fc = centrifugal (centripetal) force = mv^2/R, R = turn radius, and Ff = frictional force = kN, where k is the friction coefficient and N the normal weight of the car N = mg on a flat road.

The car will not move sideways if and only if f = ma = 0; so a = 0. Thus, we have Fc = Ff as the necessary and sufficient condition for not skidding out of control and ruining a perfectly good holiday. So we have Fc = mv^2/R = kmg = Ff; so that v^2/R = kg and v^2 = kgR or v = sqrt(kgR) = the max speed (tangential velocity) while in the turn and not skidding. (You have the numbers to plug in, you can do the math.)

Physics lesson: v = sqrt(kgR); note the velocity max does not depend on the weight or mass of the car. In fact, nothing but the coefficient of friction is related to the car and k depends on the nature and material of the tires as well as of the road. What do we gather from this observation...it could be a sports car or an 18 wheeler, it makes no difference, the max speed is still v max = sqrt(kgR).