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Foolishly leaving the window too widely open, earlier today my bathroom door slammed shut with the wind. How could I theoretically work out the time it takes to close?

Would it be impossible to reach an answer based on the conservation of energy, or does the nature of the wind's force ruin that method?

That was my question, what follows are just hunches.

I'm not sure how to approach the problem- the angular velocity is $w= \frac{d \theta}{dt}$, so the force on a point on the door is $\propto (u-wrsin\theta)^2$ (if you measure the angle so that it is 0 when the door is parallel to the wind's motion, u is the wind's speed and r the distance from the door's pivot). I assume a stationary thing hit b the wind is analogous to a moving thing hitting the stationary wind.

Am I right in thinking you would then integrate to get the torque on the entire door, and from that, knowing the moment of inertia of the door, find the time for $\theta=90$? I am unfamiliar to most of angular mechanics, if that helps formulate a good answer.

Meow
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I don't think your "drag on each point of the door is the same as on a free particle in the wind" is going to be a good simulation of what goes on when your door slams. Actually, with your model, the door would never start moving.

If you wanted a back of the envelope estimate, I would base it on the Venturi effect: the air is stalled on one side of the door, flowing with a speed $v$ on the other one, so there is a pressure difference of $\rho v^2/2$ between both sides, where $\rho$ is the air's density. The force created by this pressure difference is perpendicular to the door's surface, and you could consider it applied at the center of the door.

You could refine this in several ways, taking into account how the flow through the door changes while it closes, or have better model of how the air affects the door once it is half way closed using thin airfoil theory. But its probably not worth the effort.

Jaime
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