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 30.5 m / s (about 68 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 turn what is the minimum coefficient of static friction between the racecar's tires and track?

Explanation / Answer

m = mass

u = coefficient of f iction

a) v^2/r = 30.5^2/57

= 16.32m/s^2

b) friction

c) v^2 /r = mu*g

30.5^2/57 = mu * 9.8

mu = 1.67

the minimum coefficient of static friction between the race cars tires and track is 1.67

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.