On the same unbanked 500 m curve at 25.0 m/s, static friction alone supplies the centripetal force on the 900 kg car. Taking g = 9.80 m/s^2, what minimum coefficient of static friction between tires and road is needed?

1. On level ground the normal force equals the car's weight, N = mg, and friction supplies the centripetal force, so Fc = f = mu_s * m * g. 2. Setting this equal to the centripetal force expression m*v^2/r and cancelling the mass gives mu_s = v^2 / (r*g). 3. Substituting the known values, mu_s = (25.0 m/s)^2 / [(500 m)(9.80 m/s^2)]. 4. The numerator is 625 m^2/s^2 and the denominator is 4900 m^2/s^2, so mu_s = 625/4900. 5. That division gives mu_s = 0.13 to two significant figures, matching the approximate precision expected of a friction coefficient. 6. 1.28 and 2.55 come from failing to square the speed before dividing, and 0.026 comes from dividing by ten times the correct denominator. _Source: OpenStax College Physics (CC BY 4.0), Ch 6 "Uniform Circular Motion and Gravitation", section 6.3 Centripetal Force_