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 24.5 m / s (about 55 mph) around the turn, what is the racecar's centripetal (radial) acceleration?

What is the force responsible for the centripetal acceleration in this case?

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?

Explanation / Answer

a_c = v^2 / r

= 24.5^2 / 57

= 10.5 m/s^2 .....Ans (Centripetal acc.)

friction is the force . ....Ans

f = m a_c

and us = f/N = a_c / g

us = 10.5 / 9.81 = 1.07 ......Ans

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