A falling object with no initial velocity with mass $m$ is influenced by a gravitational force $g$ and the drag (air resistance) which is proportional to the object's speed. By Newton´s laws this can be written as:
- $mg-kv=ma$ (for low speeds)
- $mg-kv^2=ma$ (for high speeds).
I assume that $k$ is a positive constant that depends on the geometry of the object and the viscosity. But how can one explain that the air resistance is proportional to the velocity? And to the velocity squared in the second equation?
Answer
One's naive expectation would be that as the object moves through the medium, it collides with molecules at a rate proportional to $v$. The volume swept out in time $t$ is $A v t$, where $A$ is the cross-sectional area, so the mass with which it collides is $\rho A v t$. The impulse in each collision is proportional to $v$, and therefore the drag force should be proportional to $\rho A v^2$, with a constant of proportionality $C_D$ (the drag coefficient) of order unity.
In reality, this is only true for a certain range of Reynolds numbers, and even in the range of Reynolds numbers for which it's true, the independent-collision picture above is not what really happens. At low Reynolds numbers you get laminar flow and $C_D\propto 1/v$, while at higher Reynolds numbers there's turbulence, and you get $C_D$ roughly constant.
No comments:
Post a Comment