Let's begin with the linear drag force. From fluid mechanics, a constant known as the viscosity, \( \eta \), determines how resistant the air (or whatever fluid we are moving through) is to being deformed; larger \( \eta \) means that the force exerted by adjacent layers of fluid on each other is larger. Without going into too much detail, \( \eta \) has units of \( N \cdot s / m^2 \) (see Taylor problem 2.2.)
This is a low-speed effect which treats the air like a smooth, continuous medium. To see the dependence on the other properties of our moving object, consider this sketch:
From the frame of reference of the moving object, the air is flowing across it at speed \( v \). Near the object itself, the moving air has its path deformed around the object; the size of the deformation is proportional to the linear size of the object, \( D \). The complete formula for a spherical object is
\[ \begin{aligned} f_{\rm lin}(v) = 3\pi \eta D v. \end{aligned} \]
where \( D \) is specifically the diameter of the sphere. For air at STP and a spherical object, Taylor goes on to give the result
\[ \begin{aligned} b = \beta D,\ \ \beta = 1.6 \times 10^{-4}\ N \cdot s / m^2 \end{aligned} \]
which is convenient for talking about air resistance in particular.
Next, we have the quadratic drag force. This is a more violent effect, arising at higher speeds. Basically, if the speed of our object is large enough, the air doesn't have enough time to flow smoothly out of the way, and instead the individual air molecules are swept up and kicked out of the path of our object:
(This is an example of turbulent flow, as opposed to the smooth laminar flow in the linear regime.) We can think of there being two different \( v \)-dependent effects here. First, the speed at which the air molecules have to get out of the way will increase proportional to \( v \). Second, for larger \( v \) our object will simply run into more air molecules per unit time. So indeed, we expect a quadratic dependence \( v^2 \) for this force.
Two other factors influence how many air molecules are swept up per unit time. First, the cross-sectional area \( A \) of our object - the depth of the object doesn't matter because they're just scattering out of its path completely. Second, the density of the air, \( \rho \). (If we're counting molecules it's the number density \( \rho/m_{\rm air} \) that matters, but we're interested in the drag force, which puts the factor of \( m_{\rm air} \) back in.) Thus, the final result is
\[ \begin{aligned} f_{\rm quad}(v) = \frac{1}{2} C_D \rho A v^2 \end{aligned} \]
This introduces one more physical quantity, the dimensionless number \( C_D \) which is known as the drag coefficient. This number is usually between 0 and 1, and determines how streamlined the object is (for a smoother object, the air doesn't have to be deflected completely out of the way as soon as the front of the object hits it.) For a sphere, \( C_D \approx 0.5 \); a reasonably aerodynamic car has \( C_D \approx 0.25 \), and a 747 jumbo jet has \( C_D \approx 0.03 \).
Once again, we can specialize to air at STP and a spherical object, and Taylor gives the result
\[ \begin{aligned} c = \gamma D^2,\ \ \gamma = 0.25\ N \cdot s^2 / m^4. \end{aligned} \]
This is handy, but do keep in mind that it assumes a sphere; if you want to study air resistance for an object that you know the drag coefficient for, you should adjust the formula, multiplying by \( C_D / (1/2) = 2C_D \). (The extra 2 deals with the fact that \( C_D = 1/2 \) for a sphere.)
Unfortunately, the general drag force \( f(v) = bv + cv^2 \) is still slightly too complicated to work with analytically for solving the motion. We will talk about that general case, but first we'll try to understand two restricted cases. In the linear drag regime, the \( bv \) term is relatively large and we can ignore the quadratic term completely; this is obviously where \( v \) is very small. In the other limit where \( v \) is very large, we can ignore the linear term and solve in the quadratic drag regime.
Of course, we can't just say "\( v \) is really big", we have to be quantitative - the size of \( b \) and \( c \) matters too! We should really just compare the forces directly:
\[ \begin{aligned} \frac{f_{\rm quad}}{f_{\rm lin}} = \frac{cv^2}{bv} = \frac{C_D \rho Av^2}{2 \eta D v} = \left( \frac{C_D A}{D^2} \right) \frac{D \rho v}{\eta} \end{aligned} \]
The combination \( C_D A / D^2 \) is some dimensionless number which is generally not very far from 1, while the other fraction is another dimensionless combination known as the Reynolds number,
\[ \begin{aligned} R \equiv \frac{D \rho v}{\eta}. \end{aligned} \]
So if \( R \ll 1 \), the linear force is definitely dominant and we can ignore the quadratic term. If \( R \gg 1 \), the opposite is true and we just study the quadratic term. If \( R \) is sort of close to 1, then we probably can't ignore either effect and need to be careful.
It's helpful to know the ratio of these forces for air and some example objects. In example 2.1, Taylor estimates that \( f_{\rm quad} / f_{\rm lin} \sim 600 \) for a baseball moving at 5 m/s, \( \sim 1 \) for a raindrop at 0.5 m/s, and \( \sim 10^{-7} \) at \( 5 \times 10^{-5} \) m/s in the Millikan oil drop experiment. From the way the Reynolds number depends on \( D \) and \( v \), this means that air resistance for everyday objects is almost always dominantly quadratic.
To summarize, the general drag force takes the form
\[ \begin{aligned} f(v) = bv + cv^2, \end{aligned} \]
where for air at STP and an object with diameter \( D \) we have \( b = \beta D \), \( c = \gamma D^2 \), with
\[ \begin{aligned} \beta = 1.6 \times 10^{-4}\ N \cdot s / m^2,\\ \gamma = 0.25\ N \cdot s^2 / m^4. \end{aligned} \]
However, we know that the real formulas for the drag coefficients are
\[ \begin{aligned} b = 3\pi \eta D \Rightarrow \beta = 3\pi \eta, \\ c = \frac{1}{2} C_D \rho A \Rightarrow \gamma = \frac{1}{2} C_D \rho (A/D^2). \end{aligned} \]
where \( \eta \) and \( \rho \) are the viscosity and density of the medium. We will rarely use these formulas in full, but in many situations it is useful to know how they would scale if we have another situation which we can compare to air at STP. For example, water is about 50 times more viscous than air. So if we are interested in linear drag acting on an object in water, we can just multiply the value of \( \beta \) in air by 50.
Since I mentioned water, there is one more microscopic force that is present for objects moving in a fluid medium: the buoyant force. The buoyant force occurs when we move in a fluid that itself exists within a constant gravitational field \( g \). Roughly speaking, an object immersed in fluid feels the combined weight of all of the fluid molecules above a given point, which results in increased pressure proportional to the depth. The difference in pressure between the bottom and top of a solid object gives rise to a buoyant force pushing against gravity,
\[ \begin{aligned} F_b = -\rho V \vec{g}, \end{aligned} \]
where \( \rho \) is the fluid density and \( V \) is the solid volume of the immersed object. This can be stated in words as "Archimedes' principle": the magnitude of the buoyant force is equal to \( g \) times the mass of the fluid displaced (pushed out of the way) by our solid object.
Taylor doesn't really talk about buoyant force, and for most applications involving air resistance, we can safely ignore it, as long as the density of the objects moving around is much larger than the density of air (so that \( mg \) will be much greater than \( \rho V g \).) However, in physics it's always important to know when it's safe to ignore something. Obviously if we wanted to study, say, the motion of a hot-air balloon with air resistance - or drag forces of relatively light objects underwater - then we'd want to include buoyant force as well.
Clicker Question
Which has the larger quadratic drag coefficient \( c \): a human being moving through air at STP, or a baseball moving through water (which is about 800 times more dense than air?)
A. The baseball has a larger \( c \).
B. The human has a larger \( c \).
C. The \( c \) values are about the same.
Answer: A
The formula for the quadratic drag coefficient is
\[ \begin{aligned} c = \frac{1}{2} C_D \rho A. \end{aligned} \]
In this example we're changing all three quantities, although we might suspect that \( C_D \) isn't incredibly different for a human (estimates I can find give roughly \( C_D \sim 1 \) for a human who isn't trying to curl into a ball, so about double the coefficient of a sphere.) So this is mainly about comparing the changes in density and in cross-sectional area - and we were given the change in density.
Sitting here at my desk, I can easily look up accurate values for the size of a baseball and a human; but first I'll try to deal with the problem the way you would in class, which is using estimation (a very useful physics skill!) I know that the average human is on the order of 2 meters tall and maybe half a meter wide, so let's say that \( A_{\rm human} = 1\ {\rm m}^2 \).
On the other hand, a baseball is pretty small, but not tiny; if I imagine stacking baseballs together near a meter stick, I would say I can probably fit at least 10 of them, so I'll say that a baseball has a diameter of about 10 cm. So the cross-sectional area is
\[ \begin{aligned} A_{\rm baseball} \sim 4\pi (0.05\ {\rm m})^2 \approx 2.5 \times 10^{-2}\ {\rm m}^2. \end{aligned} \]
Some mental math: \( 4\pi \) is about 10, and I can decompose
\[ \begin{aligned} (0.05)^2 = (5 \times 10^{-2})^2 = 25 \times 10^{-4} \end{aligned} \]
which gives me the result above. I've rounded lots of things off very badly, so I don't really trust my accuracy to be much better than order of magnitude (i.e. a factor of 10), but since my initial estimates were probably order of magnitude accuracy at best, that's fine with me!
(Note: in class, many of you skipped straight to the area, simply estimating that \( A_{\rm human} / A_{\rm baseball} \) is less than 800 by imagining stacking baseballs in front of a person. I think this is a good shortcut, and leads to the same answer!)
Now I can compute the ratio of drag coefficients: if I ignore the \( C_D \), I find that
\[ \begin{aligned} \frac{c_{\rm baseball}}{c_{\rm human}} \approx \frac{\rho_{\rm water} A_{\rm baseball}}{\rho_{\rm air} A_{\rm human}} = 800 \frac{0.025}{1} \approx 20. \end{aligned} \]
Although I did lots of pretty rough approximations, I think I can safely conclude that the baseball in water has a larger \( c \).
Now I can use the internet to find more accurate results: the diameter of a regulation MLB baseball is about 7.4 cm, so my estimate was actually pretty close! The cross-sectional area of a skydiver is closer to about \( 0.5 {\rm m}^2 \), since 1 meter wide is definitely a bad overestimate. If I include the fact that the drag coefficient \( C_D \) is about double for a human, I find the improved estimate
\[ \begin{aligned} \frac{c_{\rm baseball}}{c_{\rm human}} \approx 110. \end{aligned} \]
Using more accurate numbers really improves our estimate - but it doesn't change the qualitative answer, which is that \( c \) is larger for the baseball in this situation.
Next time, we'll get to apply our ODE knowledge to some real physics problems! Following the book, we will start by solving the case of linear drag, which is less practically interesting but easier to work with mathematically.