What's the average force a cars tires exert against the track to maintain its circular motion in this problem?

A 1200kg car rounds a circular track of a radius of 45 meters in 9 seconds."What is the average force that the car's tires exerts against the track to maintain its circular motion?"



PLEASE show all work/show how to do everything in order to complete the problem.



THANKYOUU.What's the average force a cars tires exert against the track to maintain its circular motion in this problem?Let's start by getting the car's velocity:



Circumference = 2*pi*r = 2*pi*45m = 282.74m



velocity = Circumference/time = 282.74m/9s = 31.416 m/s



The centripetal force can then be calculated as:



Fc = mv^2/r = (1200kg)(31.416 m/s)^2 / 45m = 26318.9 kgm/s^2 = 26318.9 N



This centripetal force is provided by the friction between the tires and the track. The track is exerting this force on the car towards the center of the circle while the car is exerting this force on the track in the opposite direction. 26318.9 N is probably the answer that you want.



However, keep in mind that the car is also exerting a force downward on the track that is equal to the gravitational force on the car (i.e. (1200kg)(9.8 m/s^2) = 11760 N). So, the car is exerting a frictional force on the track of 26318.9 N in the outward radial direction and 11760 N in the downward direction. The magnitude of the net force that the car is exerting on the track is calculated as follows:



Fnet = Sqrt[26318.9^2 + 11760^2] = 28826.8 N



Depending on how detailed your class discussions have been, either one could be the answer that you want.

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