A couple weeks ago, I was in the car with Steve and we passed a truck broken down on the side of the road. It had lost a wheel, and Steve commented that if that happens to a car moving fast enough, it can often flip over. I was curious exactly how fast the car would need to be going, so I decided to model the situation.
When a front wheel falls off a car, it will tip over slightly, since most of the weight is under the hood. Suppose the car's axles are a height
h off the ground, and the center of mass of the car is a distance
d forward from its geometrical center.
The offset of the center of mass will create a torque when the wheel falls off
where
w and
l are the width and length of the car, and
m is its mass. We're interested in the downward force this creates where the axle touches the road,
The friction from this contact point will tend to slow that corner of the car, but not all of that force will be transferred to the center of mass. Because the center of mass is higher than the road surface, some of the force will cause the car to lift (think of a pole vaulter turning their forward momentum into lifting power). The deceleration caused by the friction will be
Some amount of this acceleration will contribute a torque to the center of mass, lifting the car. If the angle between the axle and the road is
θ, then the angular acceleration is
Notice that both
θ and
h are time-dependent. By expressing
h in terms of
θ, we get a differential equation.
where
Unfortunately, there's no easy way to solve this exactly, but by using some typical car dimensions, we can come up with a numerical solution. I got this diagram from a parking garage manufacturer's
site:
Plugging in these numbers, along with a typical
coefficient of friction [Link broken.
Replacement f
rom Arthur Kirkby], we find for
θ(
t)
but we're only interested in θ up to 90°, since we know at that point the car will flip over. Solving for this point, we find t = 0.66 s. That means for the car to flip over, it must still be moving after 0.66 s. The velocity of the car will be given by
where
v0 is its velocity when the wheel falls off, and
ax is the acceleration along the road surface. Using
with the
a and
θ found above, and solving for
v0, we find that the car must be moving at least 4.46 m/s, or about 10 mph. Surprisingly slow, so maybe some of my assumptions have driven it down from the real speed, but I don't think this is too far off. In any case, a stern warning to make sure your wheels are in good condition...
I think that assuming that the car body is perfectly rigid is a big factor in the low speed you're seeing. The first few things to hit the ground should crumple and increase drag.
ReplyDeleteOn the other hand, human pole vaulters run at less than 10m/s with a pole as long as your car and all the mass at the far end, so maybe this isn't too far off.