A car at the Indianapolis-500 accelerates uniformly from the pit area, going from rest to 340 km/h in a semicircular arc with a radius of 196 m.
Determine the tangential acceleration of the car when it is halfway through the turn, assuming constant tangential acceleration.
Determine the radial acceleration of the car at this time.
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?
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To determine the tangential acceleration of the car when it is halfway through the turn, we can use the formula for tangential acceleration:
a_t = v^2 / r
where:
a_t = tangential acceleration
v = velocity
r = radius of the circle
First, we need to convert the velocity from km/h to m/s:
v = 340 km/h * (1000 m/1 km) * (1 h/3600 s) = 94.4 m/s
Next, we can substitute the values into the formula:
a_t = (94.4 m/s)^2 / 196 m
a_t ≈ 45.5 m/s^2
Therefore, the tangential acceleration of the car when it is halfway through the turn is approximately 45.5 m/s^2.
To determine the radial acceleration of the car at this time, we can use the formula for centripetal acceleration:
a_r = v^2 / r
where:
a_r = radial acceleration
Substituting the values:
a_r = (94.4 m/s)^2 / 196 m
a_r ≈ 44.9 m/s^2
Therefore, the radial acceleration of the car at this time is approximately 44.9 m/s^2.
If the curve were flat, the coefficient of static friction (μ_s) between the tires and the roadbed can be calculated using the formula:
μ_s = a_c / g
where:
a_c = centripetal acceleration
g = acceleration due to gravity (9.8 m/s^2)
Substituting the values:
μ_s = 44.9 m/s^2 / 9.8 m/s^2
μ_s ≈ 4.59
Therefore, the coefficient of static friction between the tires and the roadbed would have to be approximately 4.59 to provide this acceleration with no slipping or skidding.