Question

To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 130 m. The car increases its speed at uniform rate of at ≡ d |v| dt = 5.22 m/s 2 until the tires start to skid. If the tires start to skid when the car reaches a speed of 25.6 m/s, what is the coefficient of static friction between the tires and the road?

Answer 1

Answer:

0.739

Explanation:

If we treat the four tire as single body then

W ( weight of the tyre ) =  mass × acceleration due to gravity (g)

the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r

where v is speed 25.6 m/s and r is the radius of the circle

centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²

net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²

coefficient of static friction between the tires and the road = frictional force / force of normal

frictional force = m × net acceleration / m×g

where force of normal = weight of the body in opposite direction

coefficient of static friction = (7.2567 × m) / (9.81 × m)

coefficient of static friction = 0.739