Question

A car travels at a constant speed of 32.5 mi/h (14.5 m/s) on a level circular turn of radius 49.0 m, as shown in the bird's-eye view in figure a. What minimum coefficient of static friction, μ_s, between the tires and the roadway will allow the car to make the circular turn without sliding?

1 ) make the circular turn without sliding?

2 ) At what maximum speed can a car negotiate a turn on a wet road with coefficient of static friction 0.215without sliding out of control? The radius of the turn is 25.5 m.
m/s

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Part 1

Mass of the car = M

Speed of the car = V = 32.5 mi/h = 14.5 m/s

Radius of the circular turn = R = 49 m

Coefficient of static friction =

Maximum friction force that can act on the car = f_max

f_max = Mg

The friction force acting on the car provides the necessary centripetal force for the circular motion.

Minimum coefficient of static friction = 0.437

Part 2

Mass of the car = M

Maximum speed of the car = V_m

Radius of the turn = R = 25.5 m

Coefficient of static friction on wet road = = 0.215

Friction force on the car = f

f = Mg

The friction force acting on the car provides the necessary centripetal force for the circular motion.

f = MV_m²/R

Mg = MV_m²/R

g = V_m²/R

(0..215)(9.81) = V_m²/(25.5)

V_m = 7.33 m/s

Maximum speed at which the car can take turn without slipping = 7.33 m/s