295 km/h =295000/3600=81.9 m/s
s=2πR/2
a(τ) = v²/2s = v²/2s = v²2/2•2 πR= v² /2•πR=81.9²/2•3.14•190=5.62 m/s²
v1= sqrt(2a( τ) •s1) = sqrt[2•5.62•(2•π•R/4)] =
=57.91 m/s
a(n) = v1²/R=57.91²/190= 17.68 m/s².
ma=F(fr)
m• v²/R =μ•mg
μ=v²/Rg =81.9²/190•9.8 =1.8
A car at the Indianapolis-500 accelerates uniformly from the pit area, going from rest to 295 km/h in a semicircular arc with a radius of 190 m.
Determine the tangential acceleration of the car when it is halfway through the turn, assuming constant tangential acceleration.
5.6247 m/s2
Determine the radial acceleration of the car at this time.
17.6898 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?
1.889
Obviously the answers are there I just don't know how to get to those answers. Help!!!>>
2 answers
Don't use *just* the centripetal component for μ. The real equation is:
μ = a / 9.8
Do sqrt(a_c^2 + a_t^2) to find the magnitude of a.
μ = a / 9.8
Do sqrt(a_c^2 + a_t^2) to find the magnitude of a.
0 answers