Will a falling object minimize air drag?

AI Thread Summary
A falling object experiences air drag, which can influence its motion and orientation. The final configuration of the object may not minimize drag, as it can stabilize in a position that increases drag, depending on its shape and mass distribution. Objects with an orientation-independent center of pressure tend to have their center of mass below this point, affecting their fall. A real-world example is jets entering a flat spin, which is a stable but high-drag state. The discussion raises questions about whether configurations like a "flat penny" truly represent the least drag scenario.
skazis
Messages
8
Reaction score
0
Hi,

An object is falling near the surface of the Earth. We assume there is a drag force acting on it due to the air molecules. It's initial conditions are such that it rotates and is in a chaotic motion.
Will the object will fall in the end in such a way to minimize air drag? And if yes - how can it be explained? I was thinking of least action principle as smaller drag force will make sure object reaches surface sooner.
 
Physics news on Phys.org
Nope - it may end up in a stable configuration (which is not necessarily, or even often, the least drag configuration), or it may continue tumbling. It mainly depends on the shape of the object and its mass distribution. Specifically, if the object has a relatively orientation-independent center of pressure, it will eventually end up with the center of mass below the center of pressure.
 
cjl said:
Nope - it may end up in a stable configuration (which is not necessarily, or even often, the least drag configuration), or it may continue tumbling. It mainly depends on the shape of the object and its mass distribution. Specifically, if the object has a relatively orientation-independent center of pressure, it will eventually end up with the center of mass below the center of pressure.

A real world example: this happens often - and with lethal results - when jets flame out and go into a flat spin.

The flat spin is quite stable (the pilot is unable to escape from it), but at the same time is the highest drag configuration.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top