Posted by Dwight on Sunday, November 29, 2009 at 2:20pm.
Let
g = acceleration due to gravity = 9.8 m/s²
r = the radius in m,
m = mass of car in kg
v = tangential velocity = 24 m/s
θ = superelevation angle = 30°
horizontal component, H
= mv²/r
vertical component, V
= mg
If the superelevation is perfectly matched to the speed of the car, then
tan(θ) = V/H
solve for r.
I get about 34 m.
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