A 765-kg car is travelling north and makes a gradual turn to the east at a constant speed of 15 m/s. The radius of the turn is 112 m.

Question

I calculated the angular velocity to be 0.135 m/s and the friction force needed to be 1410.86 N.
What is the smallest radius for which the turn could be designed so that the car doesn't slip at this speed? Assume the coefficient of static friction is 0.65 and the road is level.

Answer 1

omega = v/R = 15 m/s / 112 m = 0.134 radians (not meters) /s
m Ac = m v^2/R = 0.65 * m *g
so at slip
v^2/R = .65 g
225/R = .65 g
R = 225/ .65 g = 225 / (.65*9.81) = 35.3 meters

Answer 2

Thank you so much