Drag Forces

When an object is sitting at rest, the air around it presses on it fairly equally on all sides, as molecules on one side hit it, on the average, with as much force/area as molecules on the other side and the total cross-sectional area of the object seen from either side is the same (we'll analyze that statement in more detail later).

When the same object is moving with respect to the air (or the air is moving with respect to the object, i.e. - there is a wind) then the molecules hit on the side facing the direction of motion harder, on the average, then molecules on the other side and the recoil of these molecules then exerts, according to Newton's third law, an unbalanced force on the object. This ``frictional force'' of contact with the air is called a drag force, and we must account for it in our force diagrams of object moving in the air whenever the force it exerts is relevant (commensurate with the other forces in play).

Experimentally:

1. $\left\vert \vec{F}_{\rm drag} \right\vert \propto v^n \approx bv^n$. The drag force is approximately proportional to some power of the speed of the object.
2. The drag force of a medium at rest always opposes the velocity; that is, it acts to slow the object down. If the medium is moving (that is, there is a wind or current) then the drag force tends to accelerate the object toward the speed of the medium.
3. At low speeds, $n \approx 1$ and it tends toward $n = 2$ at high speeds.

We thus express the force law for as:

$$\vec{F}_d = -b \frac{\vec{v}}{\left\vert \vec{v} \right\vert} \left\vert \vec{v} \right\vert^n$$ (2.59)

(to be really picky and formally correct). We'll frequently just write this as:

$$F_d = -b v^n$$ (2.60)

and keep track of the direction by hand.

Terminal velocity

One immediate consequence of this is that objects dropped in a graviational field in air do not just keep speeding up ad infinitum.

When first we drop an object it speeds up, but as it does so the drag force on it (opposing the increase caused by gravity) also increases. Eventually the two forces balance and the object drops at a constant speed. We call this speed terminal velocity. We see that (in the negative $y$-direction):

$$F_T = mg - bv_t^n = m\frac{d v}{dt} = 0$$ (2.61)

when terminal velocity is reached or:

$$v_t = \left(\frac{mg}{b}\right)^{1/n}$$ (2.62)

For the simple case $n = 1$ we can actually solve the full equation of motion for at least the velocity as a function of time:

$$mg - bv = m \frac{d v}{dt}$$ (2.63)

or

$$\frac{dv}{dt} = g - \frac{b}{m}v$$ (2.64)

$$\frac{dv}{g - \frac{b}{m}v} = dt$$ (2.65)

$$- \frac{m}{b}\int\frac{-\frac{b}{m}dv}{\left(g - \frac{b}{m}v\right)} = \int dt$$ (2.66)

$$-\frac{m}{b} \ln\left(g - \frac{b}{m}v\right) = t + C$$ (2.67)

$$g - \frac{b}{m}v = C e^{-\frac{b}{m}t}$$ (2.68)

$$v(t) = \frac{mg}{b}\left(1 - e^{-\frac{b}{m}t}\right)$$ (2.69)

or

$$v(t) = v_t\left(1 - e^{-\frac{b}{m}t}\right)$$ (2.70)

(where the last step is for $v(0) = 0$ initial conditions). This equation describes exponential approach to a constant (terminal) velocity. There are various other ways to solve this differential equation (one of which I covered in class), but this one suffices. Note that if $n \ne 1$, the equation is also directly integrable, but the approach to a constant velocity will be algebraic, not exponential.