Posted by Mischa on Thursday, November 20, 2008 at 12:57am.
acceleration = v^2/R
side force = m v^2/R
overturning moment = m v^2 h/R
righting moment = m g d/2
so
m g d/2 = m v^2 h/R
v^2 = g d R /(2h)
v = sqrt [ g d R / (2 h) ]
Consider the moment about the wheels farthest from the center of the track. If the car is about to top over, there will be no weight or friction force on the inside wheels.
Imageine that you al=re lookng at the car head-on and considere the moments acting on it.
The moment due to the car's weight
M g d/2 will be equal to the oppositely directed moment due to the centripetal force acting through the center of mass,
M V^2 h/R
Therefore V^2 = g d R/(2h)
That will tell you the maximum stable velocity, V
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