The statement you mentioned from OpenStax textbook is correct. When a car goes faster around a turn, the coefficient of static friction between the car's tires and the road needs to be larger to prevent the tires from slipping. This is because the centripetal force required to keep the car moving in a curved path increases with the square of the speed.
To understand why the coefficient of friction needs to be larger, let's consider the forces acting on a car moving in a turn. When a car turns, there are two primary forces at play: the frictional force and the centripetal force.
Frictional force: The friction between the car's tires and the road provides the necessary force to change the direction of the car and keep it on the curved path. This force is directed inward, toward the center of the turn, and is responsible for providing the centripetal force required for circular motion.
Centripetal force: The centripetal force is the net force acting on an object moving in a circular path. In the case of a car turning, the centripetal force is provided by the frictional force between the tires and the road. It is given by the equation:
Fc = m * v^2 / r
where Fc is the centripetal force, m is the mass of the car, v is its velocity, and r is the radius of the turn.
As the car's speed increases, the centripetal force required to keep the car moving in a turn increases. The maximum frictional force that can be exerted by the tires depends on the coefficient of static friction (μs) between the tires and the road. The maximum frictional force is given by:
Ff_max = μs * N
where Ff_max is the maximum frictional force, μs is the coefficient of static friction, and N is the normal force between the tires and the road (equal to the car's weight in this case).
For the car to maintain its turn without slipping, the frictional force (Ff) must provide the required centripetal force (Fc). Therefore, we can write:
Fc = Ff
m * v^2 / r = μs * N
Since the mass and the radius of the turn remain constant, the equation becomes:
v^2 = μs * N / m
v^2 = μs * g
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
From this equation, we can see that the speed of the car squared is directly proportional to the coefficient of static friction. Therefore, if the car goes faster around a turn, the coefficient of static friction needs to be larger to provide the necessary centripetal force and prevent the tires from slipping.