3, A car is on a circular track of radius r = 120 m. It accelerates from rest to

Question

3, A car is on a circular track of radius r = 120 m. It accelerates from rest to a speed of 20 m/s in 8 seconds with constant acceleration. At the end of 8 seconds the car is next to a marker in the track. a) Find the ratio of the tangential acceleration to the centripetal acceleration as the car reaches the marker. (10 points) b) Find the minimum coefficient of friction between the tires and the track's pavement so that the car does not go out of the track at the marker. (10 points) c) At the marker the tires are rolling without slipping. The radius of the tires is 0.25m. Find the angular velocity of the tire at the marker. Find also the linear speed of the point of contact of the tire with the pavement at the marker. (10 points)

Explanation / Answer

tangential acceleration atan = (vf-vi)/t = (20-0)/8 = 2.5 m/s^2

centripetal acceleration ac = v^2/r = 20^2/120 = 3.33 m/s^2

ratio = atan/ac = 2.5/3.33 = 0.75 <<<-------------ANSWER

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b)

friction force f = uk*m*g

f = Fc

uk*m*g = m*v^2/r

uk = v^2/(rg)

uk = 20^2/(120*9.81)

uk = 0.34 <<<----ANSWER

===========================

c)

angular velocity w = v/rtire = 20/0.25 = 80 rad/s

speed of the point of contact = 0

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