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I'm trying to numerically integrate the motion of an object (say, a falling vertical cylinder). Here, there's a drag force: the wind "acting" on the body (presumably adding horizontal velocity) and the air itself slowing down the vertical motion.

Is it correct to calculate both forces with the drag equation, $F_D = \frac{1}{2} \rho v^2 C_D A$? Here I suppose the velocity would be the relative velocity between the falling object (i.e. initially $(0, -15)\ m/s$) and the "air" (i.e. $(5, 0)\ m/s$). If that's the case, where would the drag force point towards to? $-\hat{v}_{relative}$?

Thanks a lot for your help in advance.

user1264201
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2 Answers2

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Yes, the force points along the vector of the relative velocity between the object and the air.

Quadratic drag is an interesting phenomenon. You have to calculate the net velocity vector (which includes a horizontal and vertical component) and compute the force along that axis; when you then decompose it into horizontal and vertical components you will find that the vertical drag is greater because of the cross wind. This is not an intuitive result!

Floris
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"The vertical drag is greater" Could this be "lift"?

The vertical and horizontal velocity components do indeed produce a higher net velocity vector, but the vertical and horizontal kinetic energy components also have a Velocity squared function.

Interestingly, if you dropped an unpowered object from a tower in a 2 mph cross wind, it's horizontal component would "slow down" by drag to Velocity=0, relative to the air mass, and continue to plummet vertically to its terminal velocity and impact. The horizontal drag component of the object goes to zero, relative to the air mass, but it will have a Velocity=2 mph horizontal energy component relative to the ground on impact.

If the aerodynamic forces were purely drag vs kinetic energy, we should be able to decompose. Something else may be going on here, which lies at the heart of "when does a falling object become a glider".

Robert DiGiovanni
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