Question

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m. If the track is completely flat and the race car is traveling at a constant 27.5 m/s (about 62 mph) around the turn, what is the race car's centripetal (radial) acceleration? centripetal acceleration: 12.19 m/s2 Incorrect Which force is responsible for the centripetal acceleration in this case? O weight gravity friction normal What is the minimum coefficient of static friction between the race car's tires and the track necessary to keep the car from skidding into the wall on the outside of the turn? coefficient of static friction: 1.24 Incorrect

Answer 1

V- 27.5W = 27.5m/s R= Radius = 57m u centripetal accelerations 12 G R • mv² R f (27.5)2 m/s² 57 a = 13.27 m/s2 () Centipetal force is directed towards the centre of the curve. therefore, friction is responsible for centripetal acceleration (ii) at equillibrium along redus centripetal force = centrifugal force yng - me Ms: coefficient of static friction = vz (27.5) Rg - 57x9.8 Ms = 1.35