Question

travels along a perfectly flat (no banking) circular track of radius 532 m. The car increases its speed at uniform rate of at ≡ d |v| dt = 2.96 m/s 2 until the tires start to skid. If the tires start to skid when the car reaches a speed of 36.1 m/s, what is the coefficient of static friction between the tires and the road?

Answer 1

Answer:

The coefficient of static friction = 0.392

Explanation:

When the tire starts to skirt, The net force on the car is equal to the maximum friction.

F = μR................ Equation 1

R = mg.............. Equation 2

F = ma ............ Equation 3

Substituting equation 2 and 3 into Equation 1

ma = μmg.

a = μg

μ = a/g................................... Equation 4

Where F = frictional Force, μ = Coefficient of frictional force, R = normal reaction. a = total acceleration of the car. g = acceleration due to gravity.

Note: The tangential acceleration  is perpendicular to the radius. The total acceleration is

a = √(a₁² + a₂²)........................ Equation 4

Where a₁ = tangential acceleration of the tire, a₂ = centripetal acceleration of the car.

a₁ = 2.96 m/s²

a₂ = v²/R..................... Equation 5

Where v = speed of the car = 36.1 m/s, R = radius of the circular part traveled by the car = 532 m

Substituting into equation 5

a₂ = 36.1²/532

a₂ = 1303.21/532

a₂ = 2.45 m/s².

Therefore,

a = √(2.45²+2.96²)

a = √(6.0025 + 8.7612)

a = √14.764

a = 3.84 m/s².

Substituting into equation 4

μ = 3.84/9.8

μ = 0.392.

Thus the coefficient of static friction = 0.392