A car at the Indianapolis 500 accelerates uniformly from the pit area, going fro

Question

A car at the Indianapolis 500 accelerates uniformly from the pit area, going from rest to 320 km/h in a semicircular arc with a radius of 210 m

Part A:

Determine the tangential and radial acceleration of the car when it is halfway through the arc, assuming constant tangential acceleration.

Express your answers using three significant figures separated by a comma.

Part B:

If the curve were flat, what would the coefficient of static friction would be necessary between the tires and the road to provide this acceleration with no slipping or skidding?

Express your answer using three significant figures.

Explanation / Answer

ans:-r = 210 m , v1=0 km/s=0m/s , v2=330 km/h=91.66m/s = 92 m/s

tangeital acceleration is related to rotational acceleration or alpha.

since

a = r

and we know

2^2 =1^2 + 2(b) where 1 , and 1 are angular velocities one and two, and is angular acceleration, and b is the angle in radans . so since we have v1, and v2 and we know that :

v = r

we can find 2, and 1

1 = v1 / r = 0 radan/second
2 = v2/ r = 92/210 = .438 radan/s

since the car moves in a semicircular arc it is 180 degrees so half way is a quarter of the circle or almost 90 degree which is equal to /2 radan so b=/2

so we have

.438^2 = 0^2 + 2(/2)

so = .0610 radan/s/s

now that we have we find a

a = r = 210 * .0610 = 12.83 m/s^2

if by radial acceleration u mean angular acceleration then we have : = .0610 radan/s/s


since we also have a Centripetal acceleration we need a centripetal force


F = mv^2/r

in this case the F or force is the force of friction between tires and the asfalt since there is no tilt to the path.

so we have

F(friction) = mv^2/r

and we know that F = mg

so

mg = m v^2/r

so

g = v^2 /r

so

= v^2 /r*g since the maximum speed reached by the car in our interval is v =330 km/g = 91.66 m/s we have

= (91.66)^2 / 9.8 * 210 = 4.08

As you can clearly see this quoficient is UNREALISTIC and so the only meaningful conclusion is that the car would go off track if it was to drive at 91.66 m/s around a curve of 210 meter radius.. Either the car has to slow down, or the curvature has to be bigger or else this car is up to some really nasty crash.

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