A 1500-kg car drives at 30 m/s around a flat circular track 300 m in diameter. Since the net force is equivalent to the force of static friction, your answer to Part D is the magnitude fs. Based on this value, what is the minimum coefficient μs of static friction between the road and the car?
Answer:
The minimum coefficient μs of static friction between the road and the car μs is 0.6116
Explanation:
Here we have
Angular acceleration = v²/r = 30²/150 = 6 rad/s²
Therefore, the force of the car away from its center of motion =
Force, F = Mass × Acceleration = 1500 kg × 6 rad/s² = 9000 N
The centripetal force is balanced by the frictional force so the car does not skid off the track
Frictional force Ff = Normal force (Weight of car) × Coefficient of friction, μs)
Weight of car = Mass × Acceleration due to gravity = 1500 kg × 9.81 m/s²
= 14715 N
Therefore, for equilibrium
Frictional force Ff of car = Centripetal force on car
14715 N × μs = 9000 N
Therefore, μs = 9000 N/(14715 N) = 0.6116
The minimum coefficient μs of static friction between the road and the car μs = 0.6116.