A NASCAR racecar rounds one end of the Martinsville Speedway. This end of the tr

Question

A NASCAR racecar rounds one end of the Martinsville Speedway. This end of the track is a turn with radius approximately 57.0 m. If we approximate the track to be completely flat and the racecar is traveling at a constant 29.5 m /s (about 66 mph) around the turn, what is the racecar's centripetal (radial) acceleration? Number m/s What is the force responsible for the centripetal acceleration in this case? O friction gravity normal O weight To keep from skidding into the wall on the outside of the turn what is the minimum coefficient of static friction between the racecar's tires and track? Number

Explanation / Answer

The centripetal acceleration is given by:

a = v2/r

=> a = 29.52/57 = 15.267 m/s2

The force responsible for centripetal acceleration is the Friction Force as it acts radially outwards.

To prevent the car from skidding during the turn, the centripetal force must be equal to the friction force.

usmg = mv2/r

=> usg = v2/r

=> us = 15.267/9.8 = 1.558.

This is the minimum coefficient of static friction required.

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