A NASCAR racecar rounds one end of the Martinville Speedway. This end of the tra
ID: 1343972 • Letter: A
Question
A NASCAR racecar rounds one end of the Martinville Speedway. This end of the track is a turn with radius approxmately 57.0m. 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? 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 tire and track?
Explanation / Answer
A)
In circular motin, Centripital acceleration is given by
Ac = V^2/R
Ac = 29.5^2 / 57
Ac = 15.26 m/s^2
The Centripital acceleration of Race car will be 15.26 m/s^2
B)
The force resposible for Ac is Frictional Force
C)
In circualr motin , Centripital Force is given as
Fc = MV^2/R or M* Ac
M* Ac = Frictional Force between tire and road
15.29 * M = us * Mg
us = 15.29 / g = 15.29 / 9.8
us = 1.56
The Coefficient of Static Friction is 1.56
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