## 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 fric

Question

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?

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5 months 2021-08-04T19:46:01+00:00 1 Answers 71 views 0

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.