A car is rounding a curve with a radius of 12.0 m at 8.2 m/s when it hits a slippery patch. What is the minimum coefficient of static friction between the car and the slick patch of road that will allow the car to round the corner safely?
To determine the minimum coefficient of static friction required for the car to round the corner safely, we need to consider the forces acting on the car as it rounds the curve.
First, we need to calculate the acceleration of the car. The centripetal acceleration of an object moving in a circular path is given by the equation:
a = v² / r
where v is the velocity of the car and r is the radius of the curve.
In this case, the velocity of the car is given as 8.2 m/s and the radius of the curve is given as 12.0 m. Plugging these values into the equation, we can calculate the acceleration:
a = (8.2 m/s)² / 12.0 m
a ≈ 5.62 m/s²
The centripetal acceleration is caused by the net force acting towards the center of the curve. This force is given by the equation:
F = m * a
where m is the mass of the car. However, in this case, we're only concerned with the static friction force that prevents the car from slipping, which is equal to the centripetal force acting on the car.
So, the static friction force can be written as:
f_friction = m * a
Now, the maximum static friction force can be calculated by multiplying the coefficient of static friction (μ_s) by the normal force (F_normal):
f_friction = μ_s * F_normal
In this case, the normal force is equal to the weight of the car (m * g), where g is the acceleration due to gravity. Since the car is not accelerating vertically, the normal force is equal to the weight of the car.
Combining these equations, we can solve for the coefficient of static friction (μ_s):
μ_s * (m * g) = m * a
μ_s = a / g
Now, we can substitute the values we have:
μ_s = (5.62 m/s²) / (9.8 m/s²)
Calculating this, we find:
μ_s ≈ 0.573
Therefore, the minimum coefficient of static friction required between the car and the slick patch of road to round the corner safely is approximately 0.573.