Fcentripetal = m Ac = m v^2/r = m v^2/26
toppling moment = height Fc = (.6/26) m v^2
Righting moment = m g (1.4/2)
start to tip when toppling moment = righting moment
(.6/26)v^2 = 9.8 (1.4/2)
The left and right wheels of an automobile are separated by a transverse distance of 1.40 m. The center of mass of this automobile is .6 m above the ground. If the automobile is driven around a flat (no banking) curve a radius of 26.0 m with an excessive speed, it will topple over sideways. What is the speed at which it will begin to topple? Assume that the wheels do not skid.
3 answers
The car will topple over if the centripetal force, applied at the center of mass, reaches a value such that the net torque about the wheels on the OUTSIDE of the curve is zero. Under these conditions, the centripetal-force torque about a line through the outer wheel's contact points with the ground will be equal and opposite to the gravity-force torque. At that speed, the inner wheels will have no ground force on them, and will start to lift. The car will topple.
The equation to solve(for V) is
(M V^2/R) * y = M g L
where y = 0.6 m , L = 1.40 m and R = 26.0 m. Note that M cancels out. Long wheelbase (high L) and low CG (low value of y) tend to increase the toppling velocity.
The equation to solve(for V) is
(M V^2/R) * y = M g L
where y = 0.6 m , L = 1.40 m and R = 26.0 m. Note that M cancels out. Long wheelbase (high L) and low CG (low value of y) tend to increase the toppling velocity.
The max velocity is
Sqrt(Rgl/2h)
R = 26 m
g = 9.8 m/s^2
l = 1.4 m
h = 0.6 m
Sqrt(Rgl/2h)
R = 26 m
g = 9.8 m/s^2
l = 1.4 m
h = 0.6 m
1 answer