Uniform circular motion: A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding?
A) 1.23
B) 0.816
C) 0.736
D) 0.952
E) 0.662

Question

Quizplus · Accepted Answer

The centripetal force required to keep the car moving in a circle is provided by the frictional force between the tires and the road. Therefore, we can use the formula for centripetal force:

Fc = mv²/r

where Fc = centripetal force, m = mass of the car, v = velocity, and r = radius of the curve.

We can rearrange this equation to solve for the coefficient of static friction (μ):

μ = Ff/N = Fc/mg

where Ff = frictional force, N = normal force, and g = acceleration due to gravity.

Substituting the values given in the problem:

Fc = mv²/r = (m)(40.0 m/s)²/(200 m) = 4000m
N = mg = (m)(9.81 m/s²)
μ = Fc/N = (4000m)/[(m)(9.81 m/s²)] = 408.2/9.81

μ = 0.816 (rounded to three significant figures)

Therefore, the minimum coefficient of static friction needed for the car to travel at this speed without sliding is 0.816. The correct answer is B.

Uniform Circular Motion: a Car Travels at a Steady 40