To test the performance of its tires, a car

travels along a perfectly flat (no banking) circular track of radius 120 m. The car increases
its speed at uniform rate of
at ≡
d |v|
dt
= 2.99 m/s
2
until the tires start to skid.
If the tires start to skid when the car reaches
a speed of 18.2 m/s, what is the coefficient of
static friction between the tires and the road?
The acceleration of gravity is 9.8 m/s
2

Ac = v^2/r = 18.2^2/120 = 2.76 m/s^2

at ≡d |v|/dt= 2.99 m/s2
sqrt(2.76^2+2.99^2)=4.069 m/s^2
μs=(|a|max)/(g)=4.069/9.8=.4152

To find the coefficient of static friction between the tires and the road, we need to analyze the forces acting on the car when it starts to skid.

1. Centripetal Force (Fc):
The car is traveling along a circular track, so it experiences a centripetal force towards the center of the circle. The formula for centripetal force is given by Fc = m * ac, where m is the mass of the car and ac is the centripetal acceleration.
Given that the radius of the circular track is 120 m and the car's speed at the point of skidding is 18.2 m/s, we can calculate the centripetal acceleration using the formula ac = (v^2) / r, where v is the speed and r is the radius.
ac = (18.2 m/s)^2 / 120 m = 2.7827 m/s^2

2. Maximum Static Friction Force (Fsf):
When the tires start to skid, the static friction force between the tires and the road reaches its maximum value. This force provides the centripetal force needed to keep the car moving in a circular path.
The formula for the maximum static friction force is Fsf = μsf * N, where μsf is the coefficient of static friction and N is the normal force acting on the car.
Since the car is on a flat surface with no banking, the normal force is equal to the weight of the car, which is given by N = m * g, where g is the acceleration due to gravity (9.8 m/s^2).
Substituting the value of N into the formula gives Fsf = μsf * m * g.

3. Equating Centripetal and Friction Forces:
At the point of skidding, the maximum static friction force equals the centripetal force, so we can equate them:
Fsf = Fc

Substituting the respective formulas for Fsf and Fc, we get:
μsf * m * g = m * ac

We can cancel out the mass of the car (m) from both sides of the equation:
μsf * g = ac

Now we can solve for the coefficient of static friction (μsf):
μsf = ac / g

Substituting the given values:
μsf = 2.7827 m/s^2 / 9.8 m/s^2

μsf ≈ 0.2835

Therefore, the coefficient of static friction between the tires and the road is approximately 0.2835.

To find the coefficient of static friction between the tires and the road, we can use the concept of centripetal acceleration.

Centripetal acceleration is the acceleration that keeps an object moving in a circular path. It is given by the formula:

a_c = v^2 / r

Where:
a_c is the centripetal acceleration
v is the velocity of the car
r is the radius of the circular track

In this case, the car is increasing its speed at a uniform rate, so we can calculate the centripetal acceleration when the tires start to skid.

Given:
v = 18.2 m/s
r = 120 m

Using the formula, we can find the centripetal acceleration:

a_c = (18.2)^2 / 120

Next, we need to calculate the net force acting on the car. The net force is equal to the product of mass and acceleration (Newton's second law):

F_net = m * a_c

At the point of skidding, the maximum force of static friction is reached, which is given by:

F_max(static friction) = μ_s * m * g

Where:
μ_s is the coefficient of static friction
m is the mass of the car
g is the acceleration due to gravity

Now, we equate the net force and the maximum force of static friction:

F_net = F_max(static friction)

Substituting the values, we have:

m * a_c = μ_s * m * g

The mass of the car cancels out:

a_c = μ_s * g

Finally, we rearrange the equation to solve for the coefficient of static friction:

μ_s = a_c / g

Substituting the known values:

μ_s = (18.2^2 / 120) / 9.8

Calculating this expression will give us the coefficient of static friction between the tires and the road.

a = 2.99 m/s^2 is irrelevant I suspect

Ac = v^2/r = 18.2^2/120 = 2.76 m/s^2
m * Ac = friction force = m g mu
2.76 = 9.8 mu
mu = 2.76/9.8 = .282