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A car at the Indianapolis 500 accelerates uniformly from the pit area, going from rest to 260 km/h in a semicircular arc with a...Asked by asher
A car at the Indianapolis-500 accelerates uniformly from the pit area, going from rest to 290 km/h in a semicircular arc with a radius of 200 m.
Determine the tangential acceleration of the car when it is halfway through the turn, assuming constant tangential acceleration.
Correct: Your answer is correct.
5.16 m/s2
Determine the radial acceleration of the car at this time.
Correct: Your answer is correct.
16.22 m/s2
If the curve were flat, what would the coefficient of static friction have to be between the tires and the roadbed to provide this acceleration with no slipping or skidding?
Only struggling with this question ^^
an explanation would be great, thanks
Determine the tangential acceleration of the car when it is halfway through the turn, assuming constant tangential acceleration.
Correct: Your answer is correct.
5.16 m/s2
Determine the radial acceleration of the car at this time.
Correct: Your answer is correct.
16.22 m/s2
If the curve were flat, what would the coefficient of static friction have to be between the tires and the roadbed to provide this acceleration with no slipping or skidding?
Only struggling with this question ^^
an explanation would be great, thanks
Answers
Answered by
That's Mr. Idontknowhwhatimdoing to you sir!
Since we have tangential and radial acceleration our total acceleration is going to be:
a = sqrt(tangential^2 + radial^2)
Since the car doesn't slip or skid the total force on the car is zero:
fmg = ma <-- where f is friction force
Mass cancels, so our final f is
f = a/g
a = sqrt(tangential^2 + radial^2)
Since the car doesn't slip or skid the total force on the car is zero:
fmg = ma <-- where f is friction force
Mass cancels, so our final f is
f = a/g
Answered by
asher
thank you very much but is this only the Ff correct? I would still need to find the coefficient of Ff, or no?
Answered by
asher
Nevermind I figured it out. Thanks again!
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