A car traveling on a flat circular track accelerates uniformly from rest with a

Question

A car traveling on a flat circular track accelerates uniformly from rest with a tangential acceleration of 1.7 m/s^2. The car makes it one-quarter of the way around the track before it skids off. Determine the coefficient of kinetic friction between the car and truck.

Explanation / Answer

In order for the car to stay within the circular track, the static friction between the tires of the car and the road must provide the centripetal acceleration. Just when the car begins to skid: F(friction,static) = F(centripetal) ==> u(s)mg = mv^2/r ==> v^2 = u(s)rg. The car skids once it travels a distance of (1/4)(2pr) = pr/2, where r is the radius of the circle. At a constant acceleration a, we see that the speed of the car after it has traveled a distance of (1/2)pr is: v^2 = 2ad = 2a(pr/2) = par. Equating the values of v^2 gives: par = u(s)rg ==> u(s) = pa/g = (3.14)(1.7m0/s^2)/(9.8 m/s^2) = 0.5446.

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