Aristotelian Physics is a Drag

Hey’all! Remember me?

Once again, my teaching duties at the University of Cincinnati got the better of me, and the creation of content for this website took a seat so far back it fell off the edge of the Flat Earth.*

With the onset of Summer, I planned to finally not hold back and finish the many drafts that I have started over the past year and start new posts from the list of topics that I think I have something worth talking about.

Then, at the last minute, I was asked to teach a summer class.

Fortunately, it’s just one class, so in principle, it shouldn’t be as time-consuming as a full load. But it’s an accelerated half-semester course worth three credit hours, meaning the students and I should have six contact-hours per week, which then doubles for me for grading and planning to twelve hours per week for the next seven weeks? That’s still not that bad and should leave plenty of time for writing in addition to all the other things on my summer agenda. And it’s an online class, so student contact, grading and planning all kind of get mashed together it seems, and the course material has already been built. My only job on that front is to keep it working, keep it updated, and perhaps make it better wherever I see fit. But I’ve never taught an online class before, so this is all new to me, and the first few days were complete chaos, because nothing worked. Fortunately, that all got sorted out, and now things seem to be running smoothly.

“So, what is the class,” you may be asking yourself?

It’s called “How Things Work,” and it seems to be a conceptually-based physics survey course covering the same kind of material one would expect to see in the first semester of a traditional physics survey course with a focus on it’s applications to technology. In other words, it covers concepts that explain “how things work,” at least as far a classical mechanics is concerned.

Imagine that!

It also happens to be that for some time now, I have wanted to write a series of posts for this website which will more or less follow along the track of a physics survey course, the reason being to provide a conceptual and theoretical underpinning to help readers understand some of the more complicated things I write about here. It seems like I might be able to kill two birds with one stone if this series of posts bears some resemblance to what I am teaching or vice versa.

So, let’s get on with it!

*No, Internet, the Earth is not flat. It’s a lumpy oblate spheroid. But here the metaphor works, so…

First, I should mention that this specific post isn’t necessarily “foundational.” Moreso, it’s my thoughts on one of the students’ homework questions—one that goes as follows:

In the 9th problem, suppose you dropped a penny exactly at the same time as you threw the other penny towards the bucket.  Which one would hit the ground first?  Suppose you instead dropped a quarter, what then?  Please write a complete argument supporting your conclusions, using concepts you have learned in class and from reading the book.

For reference, the 9th problem is about throwing a penny horizontally off a balcony into a bucket of water on the ground below.

At the expense of giving away the answer, let me say outright that, if gravity is the only factor, both the dropped and thrown pennies hit the ground simultaneously, because gravitational fields act on all objects equally, regardless of their states of motion. (This is even more evident in a uniform gravitational field, such as that near Earth’s surface.) So the dropped and thrown pennies experience the same vertical acceleration and therefore have the same change in vertical velocity and vertical position. Consequently, they travel the vertical distance in the same amount of time and hit the ground simultaneously.

Now, what about the quarter? It has more mass than the penny, and so gravity is going to pull it downward with more force. However, because the quarter has more mass, it also has more inertia or resistance to changes in its state of motion. Thus, the quarter is more resistant to the downward pull of gravity. These two effects wind up balancing each other perfectly, so that the quarter falls at the same rate, and hits the ground at the same time, as the pennies!

Don’t believe me? Check out this sweet demonstration:

That is one of my most favorite experiments of all time. And yes, it actually happened on the Moon.

So why don’t we see hammers and feathers falling at the same rate here on Earth? Well, unlike the Moon, Earth has a pretty substantial atmosphere. (I suppose it’s actually average as far Solar System terrestrial planet atmospheres go, with the exception that it’s full of oxygen.) The matter comprising Earth’s atmosphere is in a fluid state, and when objects move through fluids, they experience a force called “drag.” Drag is an interesting force, in that, unlike gravity, it does depend upon the state of motion of the object experiencing it. If the object is stationary with respect to the fluid, there is no drag. But if the object moves through the fluid, there is drag, and that drag depends upon the object’s speed: the faster moves, the more drag the object experiences.

This relationship is quantified in the drag equation: \begin{equation} F_D = \frac{1}{2} \rho u^2 C_D A. \end{equation}

Here, \(\rho\) is the fluid’s density, \(u\) is the relative speed of the object through the fluid, \(C_D\) is something called the drag coefficient (more on that later), and \(A\) is the object’s projected area—the cross-section of the object as “seen” by the oncoming fluid. The important thing to understand is that the strength of the drag force \(F_D\) is proportional to these quantities. If any one of them is zero, then the drag force is zero. If any one of them gets bigger, then the drag force gets bigger, etc. Of course, most of these quantities are only going to change in special circumstances. The fluid density will only appreciably change if the fluid is a gas and its temperature, pressure, or composition change in a non-isochoric way. The object’s projected area will only change if its size, shape, or orientation relative to the oncoming fluid change. The drag coefficient might change, but that gets complicated fast, so we’re going to assume that it doesn’t.

On the other hand, the relative speed of the object and the fluid is almost guaranteed to change as a consequence of drag. That’s because drag is a force acting on the object, forces cause acceleration, and an object’s acceleration is by definition the change of its velocity over time. So in other words, drag changes the object’s speed relative to the fluid, and this change in speed changes the drag. (That’s called a feedback loop, which is a common feature in dynamics.) But, there is one assumption in what I just said—it’s that drag is the only force acting on the object, or at least it’s the only force that’s not balanced out by other forces. That’s because drag always points in the direction opposite the object’s motion through the fluid, and so left to its own devices, it has the effect of slowing the object down until it comes to a stop. At that point, the object is no longer moving relative to the fluid, and so the drag force disappears. An example of this would be a neutrally buoyant submarine gliding through water without running its motor. In that case, drag is the only unbalanced force, and it will eventually bring the sub to a stop.

But what if drag isn’t the only unbalanced force? Or what if the drag force itself is balanced out by something else? Well, as I said above, whenever the forces on an object are unbalanced, that object accelerates, meaning its velocity changes over time. Just as we saw above, this process will continue until the drag force disappears—or, is no longer unbalanced.

As an example, let us return to the submarine, but say now it restarts its motor. This will ultimately cause a thrust force which pushes the sub through the water. If the ship starts from rest, there is no drag, and so nothing to balance out this thrust. Thus the craft accelerates forward. However, once the sub begins moving, the drag force appears, and the faster the ship moves, the bigger this drag becomes. Additionally, if the sub started from rest, then it will be moving in the direction the thrust force points—forward. Drag, which again opposes motion, will, therefore, point backward—opposite the thrust force. When two forces point in opposite directions, they have the effect of canceling each other out to some degree. If they point in opposite directions and have the same strength, then they cancel each other out or balance each other out, completely. (Think of a stalemate in tug ‘o war.) If the thrust and drag acting on the sub do not perfectly balance out, then there is an unbalanced force, and the sub accelerates—its speed changes—and so the drag force changes. Again, these changes continue until there are no more unbalanced forces. Before, this happened when the velocity and drag force both vanished. Now, it occurs when the drag on the ship becomes as strong as the thrust, and the two forces perfectly balance out. At that point, since there is no unbalanced force, there is no acceleration, and because there is no acceleration, there is no change in velocity. But there is still a velocity—it just doesn’t change! This final, constant velocity corresponding to a drag force that balances out all other forces acting on an object is called the terminal velocity, because it’s the final velocity the object achieves before it stops accelerating and its velocity stops changing. At that point, our submarine cruises along at a constant speed, so long as its motor produces thrust.

Or airplane, or car, or bicycle, or person running, or anything that moves through a fluid, as most things on Earth’s surface do, thanks to Earth’s non-negligible atmosphere or hydrosphere, depending on where on the planet you happen to be.

So, for us Earthlings, drag has a ubiquitous effect on our lives—one that is so common it easily goes unnoticed as its own thing. Instead, it’s lumped in with other phenomena, which is why brilliant people like Aristotle kind of missed the mark when it came to moving objects.

Kind of.

I mean, his laws of motion are not universally correct. But, for systems in which drag is present, they’re actually more or less right!

To see this, let’s imaging acting on some general object at rest in a fluid with a force of strength \(F\). As with the submarine starting from rest, the object will begin to move in the direction the force acts and a drag force acting in the opposite direction will immediately manifest. Very quickly, that drag force will grow sufficiently large to balance out the applied force at which point the object stops accelerating and then moves at a constant terminal velocity. However, despite its name, we can actually increase this velocity if we increase the strength of \(F\). That’s because when the object moves at the terminal speed, “the forces acting on it balancing each other out” means those forces’ magnitudes must be the same: \begin{equation} F = F_ D = \frac{1}{2} \rho v^2 C_D A. \label{151919deqeq} \end{equation}

Or, in other words, \begin{equation} F \propto v^2. \label{05192019v2} \end{equation}

(FYI, the funny \(\propto\) symbol means that the quantities on either side are proportional to each other. Also note that I have subtly exchanged the symbol for the object’s speed relative to the fluid \(u\) with one representing the speed of the terminal velocity, which I’m calling \(v\). When the object is moving through the fluid at terminal velocity, those two things are the same.)

Compare this with what Aristotle says about moving objects: things which move “unnaturally” (i.e., they are moved by something else, as opposed to moving “via their material nature”), only move when acted upon by a force. The stronger the force, the faster the object moves.

Or, in other words, \begin{equation} F \propto v. \label{05192019v} \end{equation}

Again, for systems with drag, he’s on the right track! Of course, it’s important to note that I cannot say with any certainty that Aristotle would choose \eqref{05192019v} over \eqref{05192019v2} because as far as I know, he never actually wrote anything down using mathematical notation. All we have are his words, which say that a larger force is needed to move an object faster—a statement that is consistent with both \eqref{05192019v2} and \eqref{05192019v}.

So, is there anything else we can say about an object moving through a fluid at the terminal speed?

Of course! We can calculate what that speed actually is by taking \eqref{151919deqeq} and solving for \(v\): \begin{equation} v = \sqrt{\frac{2 F}{\rho C_D A}}. \label{05212019vter} \end{equation}

Or, in other words, \begin{equation} v^2 \propto \frac{F}{\rho}. \end{equation}

Compare that with what Wikipedia says Aristotle says in the Physics (215a25):

Aristotle effectively states a quantitative law, that the speed, v, of a falling body is proportional to its weight, W, and inversely proportional to the density, ρ, of the fluid in which it is falling.

Or, in other words, \begin{equation} v \propto \frac{W}{\rho}, \end{equation}where weight, the force of gravity, take the place of our general force \(F\).

Again, up to a possible power of 2, Aristotle is not wrong.

So I think drag’s omnipresence in our Earthly lives is why it seems counterintuitive to us that on the Moon, where there is effectively no atmospheric drag, a hammer and a feather fall a the same rate and hit the ground at the same time. And why it was so surprising when Galileo claimed that all objects fall at the same rate, regardless of their weight.

Perhaps that also explains why some of the more ambitious responses to the homework question above have sought to elucidate the effects of drag on the falling penny vs. quarter. The one that stands out to me and ultimately caused this post was a response which said that because the quarter has a larger projected area as it falls (assuming the planes of both coins’ faces are more or less oriented horizontally as they fall), it will feel more drag and therefore lag behind the penny, thus hitting the ground later.

That’s partially true.

Yes, the larger quarter feels more drag, but it’s also more massive than the penny. This increased mass results in an increased weight acting against that drag, and greater inertia resisting the effects of all forces, and these factors together will determine which coin reaches the ground first. So, a more detailed analysis is required.

Enter Sir Isaac Newton.

Newton’s second law of motion states that an object’s acceleration \(a\) is proportional to the total force \(F_{net}\) acting on it and inversely proportional to the object’s mass \(m\).

Or, in other words, \begin{equation} a = \frac{F_{net}}{m}. \end{equation}

For an object being propelled through a resistive fluid by some force \(F\), the net force acting on it is \begin{equation} \begin{aligned} F_{net} & = F - F_D \\[1ex] & = F - \frac{1}{2} \rho u^2 C_D A. \end{aligned} \end{equation}

The object’s acceleration is therefore \begin{equation} a = \frac{F}{m} - \frac{\rho u^2}{2} \frac{C_D A}{m}. \label{05212019acc} \end{equation}

Note that here I am using a sign convention where the object’s velocity vector—i.e., the direction in which is it moving—defines the positive direction. Therefore the force \(F\), which causes the motion, will always be positive and the drag \(F_D\), which resists the motion, will always be negative. The force \(F_{net}\) can either be positive or negative, depending on whether \(F\) or \(F_D\) is stronger. Consequently, \(a\) will either be positive or negative, which in this case indicates that the object is speeding up or slowing down, respectively. Or \(F_{net}\) and \(a\) can be zero when \(F\) and \(F_D\) have equal strength and balance each other out so that the object is neither speeding up nor slowing down. This, of course, is the condition that occurs when the object reaches the terminal velocity. I use this scheme now because I think students find it intuitive.

However, in practice, this is not the best way to do things! Better it is by far to nail your coordinate system down and accept the fact that your object can move in both positive and negative directions, and have both positive and negative accelerations, and that the relative sign between the object’s motion and acceleration determines whether it is speeding up or slowing down. That’s because, in this scheme, vectors which, say, point up and right are always positive, and those that point down and left are always negative. That makes things a lot easier later on. Meanwhile, in the object-velocity-defines-positive scheme, vectors will flip signs depending on whether the object is traveling in the direction they point, in which case the vector will be positive, or the opposite, in which case the vector will be negative. If the object changes direction, then all the vector signs change along with it! Fortunately, I don’t expect the object to change direction here. But it could. (That said, even in a fixed coordinate system, the sign of the drag vector will change whenever the object’s direction of motion changes, but that’s one sign-changing vector versus many.)

Now, let us be a bit more specific, and say that our object is falling through a resistive fluid in a uniform gravitation field. In that case, the generic force \(F\) is replaced by the object’s gravitational weight minus the effect of the object’s buoyancy within the fluid. (I add that emphasis because I almost wrote this entire post without considering buoyancy before I realized my oversight and convinced myself it’s worth including since I’m dealing with solids immersed in fluids. Plus it gives me a reason to mention Archimedes. Eureka!) The object’s weight is the product of its mass and the local gravitational acceleration \(g\) and the buoyant force is the weight of the fluid that the object displaces. Both of these weights can be expressed as products of \(g\), the object’s volume \(V\), and the object and fluid densities, respectively: \begin{equation} \begin{aligned} W_o & = m_o g = \rho_o V g, \\[1ex] W_f & = m_f g = \rho_f V g. \end{aligned} \end{equation}

From here on out, subscripts \(o\) ascribe a quantity to the object, \(f\) to the fluid, and a lack of subscript indicates the quantity is common to both.

With this, the net force on the object becomes \begin{equation} \begin{aligned} F_{net, \ o} & = W_o - W_f - F_{D, \ o} \\[1ex] & = m_o g - m_f g - \frac{1}{2} \rho_f u^2 C_D A. \end{aligned} \label{05222019netf} \end{equation}

And the object’s acceleration is \begin{equation} \begin{aligned} a_o & = g - \frac{m_f}{m_o} g - \frac{\rho_f u^2}{2} \frac{C_D A}{m_o} \\[1ex] & = g - \frac{\rho_f}{\rho_o} g - \frac{u^2}{2} \frac{\rho_f}{\rho_o} \frac{C_D A}{V}. \end{aligned} \label{05222019neta} \end{equation}

Let’s unpack that, shall we?

First, as I’ve written it, the object’s acceleration has three terms corresponding to the three forces acting on it. The first term is the gravitational acceleration. The second term is the contribution of buoyancy, which depends on gravity (without which there is no buoyancy!) and the relative densities of the object and fluid. The third term is the contribution of drag, which, like buoyancy, depends on the object/fluid density relationship but also their geometric relationship encoded within the factor \(C_D A / V\). And of course, it depends on the relative speed object and fluid \(u\), too.

And there is one more thing. Remember when I said that sign-changing vectors will come back to bite us? Well, here it is. Equations \eqref{05222019netf} and \eqref{05222019neta}, as they are currently written, are true if the object is denser than the fluid and therefore moves downwards, in the direction of gravity, at the terminal speed when the forces acting on it are balanced. If instead, the object is less dense than the fluid and moves upwards, in the direction of buoyancy, at the terminal speed when the forces acting on it are balanced, then the signs on the gravitational and buoyancy terms have to switch. Fortunately, however, in this schema, drag is always negative, so at least we have that. But, with this in mind, \eqref{05222019neta} should really say \begin{equation} a_o = \mp \left( 1 - \frac{\rho_f}{\rho_o} \right) g - \frac{u^2}{2} \frac{\rho_f}{\rho_o} \frac{C_D A}{V}. \label{05222019mess} \end{equation}

In equation \eqref{05222019mess}, we take the minus sign if the object is traveling up and the plus sign if it’s going down. I’ve arranged things so that this can be remembered by the fact that the minus sign is above (up) while the plus sign is below (down). I’ve also taken the liberty of factoring \(g\) out of the gravitational and buoyancy terms both to satisfy my OCD tendencies and to drive home the fact that the \(\mp\) sign has the job of keeping the numerical value of \( \left( 1 - \rho_f / \rho_o \right) \) positive, which is very important.


Well, let’s take a look at the terminal velocity, when \(a_o = 0\): \begin{equation} \begin{aligned} 0 = & \mp \left( 1 - \frac{\rho_f}{\rho_o} \right) g - \frac{v^2}{2} \frac{\rho_f}{\rho_o} \frac{C_D A}{V} \\[1ex] & \Longrightarrow \mp \left( 1 - \frac{\rho_f}{\rho_o} \right) g = \frac{v^2}{2} \frac{\rho_f}{\rho_o} \frac{C_D A}{V} \\[1ex] & \Longrightarrow v = \sqrt{ \mp \left(1 - \frac{\rho_f}{\rho_o} \right) \frac{\rho_o}{\rho_f} \frac{2 g V}{C_D A}} \\[1ex] & \Longrightarrow v = \sqrt{ \mp \left(1 - \frac{\rho_f}{\rho_o} \right) \frac{2 m_o g}{\rho_f C_D A}}. \end{aligned} \label{05222019vmess} \end{equation}

So long as the object is less dense than the fluid \(\left( \rho_o < \rho_f \right)\) and we take the upper minus sign because less dense objects float upward in equilibrium, we’re good. Or, if the object is denser than the fluid \(\left( \rho_o > \rho_f \right)\) and we take the lower plus sign because denser objects sink downward in equilibrium, we’re good. But if we mix those two situations up, either way, we get a negative value under the radical, and thus an imaginary terminal speed. Not good—at least if you ever hope to measure that speed.

That’s what we get for considering buoyancy—that thing which allows things to float. Will all this effort ultimately be worth it when we finally return to the original problem of dropping coins through air?

No, probably not.

That’s because metal coins are way denser than air, in which case \(\left( \rho_o \gg \rho_f \right)\) and \eqref{05222019vmess} very well approximates to \begin{equation} v_c \approx \sqrt{ \frac{2 m_c g}{\rho_f C_D A} } = \sqrt{ \frac{2 W_c}{\rho_f C_D A} }. \label{05222019vapp} \end{equation}

That’s just equation \eqref{05212019vter} with the coin’s weight, \(W_c\) taking the place of the generic force \(F\). The subscript \(c\) of course indicates that we are dealing with a dense coin instead of an arbitrary object, and I have taken the liberty of using the plus sign from \eqref{05222019vmess} since coins in equilibrium most certainly fall through air.

A similar approximation can be made for the coin’s acceleration by simply dropping the buoyancy term in \eqref{05222019neta} and rearranging: \begin{equation} a_c \approx g - \frac{\rho_f C_D A}{2 m_c} u^2. \label{05222019aapp} \end{equation}

I leave it as an exercise to the reader to verify that \eqref{05222019vapp} follows from \eqref{05222019aapp} when \(a_c = 0\), but I’ll make it easy for you by pointing out that \eqref{05222019aapp} can be rewritten as \begin{equation} a_c \approx g \left( 1- \frac{u^2}{v_c^2} \right). \label{05222019aappvc} \end{equation}

From these expressions, we see that any coin dropped from rest will start out motionless and with an acceleration due entirely to gravity. Then, as the coin falls and picks up speed, drag manifests in opposition to gravity, and the coin’s acceleration reduces, ultimately reaching zero as the coin’s velocity approaches the terminal value. The coin then continues traveling at this speed until it hits the ground, assuming that didn’t happen already. Or, say the coin is thrown at the ground with an initial speed that is greater than \(v_c\). In that case, the coin’s acceleration will be negative, which again, due to our sign convention, just means that it’s slowing down. It will continue slowing until it reaches the terminal speed or hits the ground, whichever happens first. The timing of all this and the highest speed which the coin can attain is determined not just by the coin’s projected area, but instead, as I alluded to above, by the ratio of the coin’s mass to that area, \(m_c / A\). The greater that ratio is, the higher the coin’s terminal velocity, and the less acceleration reduction due to drag at a given speed the coin experiences. Thus this coin should take less time to hit the ground than a coin for which that ratio is smaller.

Which makes sense, right? After all, that’s why people who jump out of airplanes wear parachutes. By strapping a parachute to themselves, they’re increasing their mass, and so in the Aristotelian model, they should fall faster. Note that before chute deployment, our model actually says the same thing, since the jumper’s \(m / A\) ratio has increased! However, when the parachute deploys, the parachutist’s projected area increases much more than their mass did. This decreases their \(m / A\) ratio, thereby reducing their terminal speed to something which is not so “terminal” upon impact with the ground. I wonder how Aristotle would explain that?

Okay, so a coin with a larger \(m_c / A\) ratio seems likely to hit the ground in less time than a coin with a smaller ratio dropped from the same height. But can we be sure?

Of course!

Equation \eqref{05222019aappvc} is the tool we need to figure out any physical question we could ask of our system, which is why it’s called an equation of motion. Mathematically speaking \eqref{05222019aappvc} is a non-linear first order ordinary differential equation for an object’s speed through fluid, ignoring buoyancy (which is actually easy to recover by simply rescaling \(g \to \left( 1 - \rho_f / \rho_c \right) g\).) Moreover, it’s one of the very few non-linear first order ordinary differential equations that can actually be solved analytically!

I’m not going to walk through the solution, because believe it or not, that’s actually beyond the scope of this post, but here’s what the solution is: \begin{equation} u_c(t) = v_c \tanh \left[\frac{g}{v_c} (t+ t_0) \right]. \end{equation}

The function \(\tanh\) is the hyperbolic tangent and \(t_0\) is a “phase constant” defined by \begin{equation} t_0 = \frac{v_c}{g} \text{arctanh}{\left(\frac{u_0}{v_c}\right)}. \end{equation}

This allows for the possibility that instead of being dropped from rest, our coin is thrown down towards the ground with an initial speed \(u_0\). (Note that the function \(\text{arctanh}\) is the inverse hyperbolic tangent.)

So, what does this mean? Well, as they say, a picture is worth a thousand words.

Plots of \(u_c(t)\) for different values of \(u_0\) but the same value of \(v_c\).

Plots of \(u_c(t)\) for different values of \(u_0\) but the same value of \(v_c\).

The graph above shows \(u_c(t)\) plotted for different values of \(u_0\) but the same value of \(v_c\). Notice how regardless of whether we drop the coin from rest or throw it down at an arbitrary speed, it eventually winds up falling at the terminal speed \(v_c\). This is characteristic of any object falling through a fluid with drag.

It’s also worth taking a look at the coin’s acceleration function. Since, once again, an object’s acceleration is the time rate of change of its velocity, the acceleration function can be obtained by calculating the derivative of \(u_c(t)\) with respect to \(t\). Since, as was the case with solving differential equations, I don’t expect the general reader to know how to do that, here it is: \begin{equation} a_c(t) = \frac{d}{d t} u_c(t) = g \, \text{sech}^2{\left[\frac{g}{v_c} (t+t_0)\right]}. \end{equation}

That’s the hyperbolic secant, in case you were wondering. Here’s what the function looks like for the same values of \(u_0\) and \(v_c\) used in the plots above:

Plots of \(a_c(t)\) for the same values of \(u_0\) and \(v_c\) as above.

Plots of \(a_c(t)\) for the same values of \(u_0\) and \(v_c\) as above.

As we expected, the coin that is dropped from rest has an initial acceleration of \(g\). Coins that are thrown down have different initial accelerations, possibly with negative values, if they were thrown at speeds greater than \(v_c\). In any case, as time goes on and the coin’s velocity approaches \(v_c\), the coin’s acceleration goes to zero—just as we predicted.

But do you remember what our original question was? We were trying to figure out whether a penny or a quarter dropped from the same height through air hits the ground first. To answer this, we need to know the positions of the coins as functions of time. Fortunately, we have all we need to get that, too! Just like an object’s acceleration is the change in its velocity over time, an object’s velocity is its change in position over time. This means we can obtain the position function from the velocity function similarly to how we derived the acceleration function from the velocity function. Only instead of taking a derivative, where we see how something changes, we have to integrate, which lets us predict what something causes.

Think of it this way: say you know where something is at a particular moment in time. Say you also know how that object is moving. Then, you can predict where the object will be in the future!

Sweet, right?!

It is, for whether you realize it or not, you are doing calculus. Therefore, let me present a slightly more detailed description of what’s going on:

Imagine you have an object and you know its velocity function \(u(t)\). Remember, this tells you where the object is going at a given moment in time—which direction and how fast. Thus, you can use the function to determine where the object will be in the near future. So, you move to that location at that time, and then you recheck the velocity function. It now tells you how the object is moving at this moment, and just like before, you can use that information to move to where the object will be in the next instant. Rinse and repeat. As you do this over and over again, the path that you trace out will be the path that the object takes through space. Who knew that integral calculus is just that easy?!

Well, it’s easy in principal, but the actual techniques used to perform these calculations can be a bit complex. So, once again, I’ve done it on your behalf: \begin{equation} y_c(t) = \int u_c(t) \, dt +C = \frac{v_c^2}{g} \ln \left\{ \cosh \left[ \frac{g}{v_c} (t+t_0) \right] \right\} + C. \end{equation}

Okay, it is getting a bit complicated, so let me parse it out: \(\ln\) is the natural logarithm function, \(\cosh\) is the hyperbolic cosine, and—perhaps most importantly—\(C\) is something called an integration constant. These things appear whenever one performs an indefinite integral like the one above, and in physics, they often relate to a system’s initial conditions. In this case, it has something to do with the initial speed of our coin \(u_0\), and the height from which we drop it. Let’s figure that relationship out.

It’s actually pretty simple. All we have to do is say that when \(t=0\), the position of our coin is some value of our choosing. I’m going to call it \(y_0\), because, why not? This means that \begin{equation} y_0 = y_c(0) = \frac{v_c^2}{g} \ln \left\{ \cosh \left[ \frac{g}{v_c} t_0 \right] \right\} + C. \end{equation}

From this, one can immediately see that \begin{equation} C = y_0 - \frac{v_c^2}{g} \ln \left\{ \cosh \left[ \frac{g}{v_c} t_0 \right] \right\}. \end{equation}

Therefore, \begin{equation} \begin{aligned} y_c(t) &= \frac{v_c^2}{g} \ln \left\{ \cosh \left[ \frac{g}{v_c} (t + t_0) \right] \right\} - \frac{v_c^2}{g} \ln \left\{ \cosh \left[ \frac{g}{v_c} t_0 \right] \right\} + y_0 \\[1ex] &= \frac{v_c^2}{g} \ln \left\{ \frac{ \cosh \left[ \frac{g}{v_c} (t + t_0) \right] }{ \cosh \left[ \frac{g}{v_c} t_0 \right] } \right\} + y_0. \end{aligned} \end{equation}

At \(t = 0\), the logarithm vanishes, leaving behind \(y_0\). Let us see what this function does after that, again plotting values of \(u_0\) and \(v_c\) used above.

Plots of \(y_c(t)\) for the same values of \(u_0\) and \(v_c\) as above.

Plots of \(y_c(t)\) for the same values of \(u_0\) and \(v_c\) as above.

Note that the graph above does not show the actual trajectories of our coins. The path of a coin that is dropped or thrown straight down is purely in the vertical direction—not diagonal as it appears above. Therefore, graphs of these trajectories through space would just be vertical lines, which isn’t super informative. The figure above is, however, for it shows us quite clearly how the position of each coin changes over time. It also shows us which coin reaches the ground first. The graph’s horizontal axis represents the ground. Its vertical axis represents \(t=0\) when each coin is at the starting position \(y_0\). As time increases to the right, each coin’s position at that time is found by following the coin’s position function down from \(y_0\) to the ground. A coin reaches the ground when its position function reaches the horizontal axis, and the farther to the left this intersection is, the sooner the coin has completed its journey. It should come as no surprise that the coin which is thrown towards the ground fastest reaches the ground first, while the coin which is merely dropped from rest reaches the ground last.

But in the original question, coins aren’t thrown, they are dropped, and the thing which differs between them is the \(m_c/A\) ratio, which determines their terminal speed. So, let us see what effect that has.

Accelerations over time for coins with different \(m_c/A\) ratios dropped from the same height.

Accelerations over time for coins with different \(m_c/A\) ratios dropped from the same height.

Speeds over time for coins with different \(m_c/A\) ratios dropped from the same height.

Speeds over time for coins with different \(m_c/A\) ratios dropped from the same height.

Positions over time for coins with different \(m_c/A\) ratios dropped from the same height.

Positions over time for coins with different \(m_c/A\) ratios dropped from the same height.

The three graphs above acceleration, speed, and position over time for three coins with different \(m_c/A\) ratios. Instead of assigning specific values, I set one coin (gold) to be a reference with some arbitrary \(m_c/A\). I then set the other coins to have twice and half that value (green and blue, respectively.) Note that each line terminates when the coin it represents reaches the ground, and the impact times and terminal speeds of each coin are labeled with subscripts equal to the relative value of their \(m_c/A\) ratio. It should come as no surprise that the coin with the largest of these values, and thus the largest terminal speed, hits the ground first.

Alright, so we’re finally getting somewhere! All we have to do is figure out \(m_c/A\), and then we can answer the question that started this whole mess.

But wait!

For the sake of completeness, I want to explain how I figured out when to terminate the lines in the graphs above by calculating specifically when a coin hits the ground. Don’t worry, it’s not hard, but it does require you to look at a few more equations. To figure this out, all we have to do is set our position equation equal to zero (the ground) and solve for what I’m calling \(t_G\): \begin{equation} \begin{aligned} y_c(t_G) &= 0 \\[1ex] \implies t_G &= \frac{v_c}{g} \text{arccosh}{\left[ e^{- y_0 g / v_c^2} \cosh\left(\frac{g}{v_c} t_0\right) \right]} -t_0. \end{aligned} \end{equation}

Alright, that’s all the equations I’ve got (for now). But before I proceed with answering the question only one or two people have been asking, I think it’s interesting to note that when \(y_0 = 0\) above, meaning the coin starts on the ground, the \(\text{arccosh}\) (inverse hyperbolic cosine) term collapses to \(t_0\), which then annihilates the second term. All that is to say, I have proven mathematically that a coin which starts on the ground takes zero time to reach the ground. I’m glad the equation did not suggest otherwise.

So, in the race between the penny and the quarter, who wins? Well, a penny has a mass of \(2.5\,\mathrm{g}\) and a diameter of \(19.05\,\mathrm{mm}\) and a quarter has a mass of \(5.67\,\mathrm{g}\) and a diameter of \(24.26\,\mathrm{mm}\). You figure it out.

I kid, of course—here are the numbers. \begin{equation} \begin{array}{r|cccc} \text{Coin} & \text{Mass} \left(\mathrm{g}\right) & \text{Diameter} \left(\mathrm{mm}\right) & m_c / A \left(\mathrm{kg}/\mathrm{m}^2\right) & v_c \left( \mathrm{m}/\mathrm{s}\right) \\ \hline \text{Penny} & 2.50 & 19.05 & 8.77 & 11.19 \\ \text{Quarter} & 5.67 & 24.26 & 12.27 & 13.24 \\ \end{array} \end{equation}

I calculated the terminal speeds using \eqref{05222019vapp} and a drag coefficient of \(1.12\). Remember when I said we would talk later about drag coefficients way up above? Finally, that time is now. Drag coefficients account for the fact that objects with the same projected area moving through a fluid at the same speed feel different amounts of drag if the objects have different shapes. The reasons why are pretty complicated and the utility of the drag coefficient comes from the fact that it distills this complexity down to a single number. The bigger the number is, the more drag an object feels for a given projected area, speed relative to the fluid, and fluid density. Consequently, shapes that we think of as being “aerodynamic” usually have very small drag coefficients. For a short cylinder like a coin moving through a fluid face-on as we assume, a coefficient around \(1.12\) is suggested.

Now for the graphs! Since we are using realistic values for our coins, I figure we ought to drop them from a realistic height of a second story balcony. Since those are usually at least ten feet above the ground, and a person might drop coins from about six feet above that, I figure somewhere around sixteen feet is a good initial height. Therefore, I’m setting \(y_0 = 5\,\mathrm{m}\).

y(t) p vs q.png
u(t) p vs q.png
a(t) p vs q.png

It turns out, perhaps unsurprisingly, that though the quarter does win, both coins hit the ground at almost the same time. This makes sense because they were in the air for such a short time that the differences in drag didn’t really have time to manifest. I mean, they didn’t even reach their terminal speeds! If the coins were dropped from higher up, like say from the top of a skyscraper, then drag would have a more pronounced effect on their motion, something akin to the generic coin plots that I presented above. But alas, that’s not the situation we were given. Note that for comparison I also plotted the position, speed, and acceleration of a coin (or any object) which feels no drag. Even that object hits the ground at about the same time as the penny and the quarter, meaning that ignoring drag altogether is a “good enough” simplification for this specific problem. Again, however, that wouldn’t be the case if the coins were dropped from higher up.

How did I get the “No Drag” functions plotted above? I used the position, speed, and acceleration equations we derived and took their limits as \(v_c \to \infty\), meaning there is no upper bound to the coin’s speed, so it can always fall faster. This is equivalent, by the way, to saying that \(\rho_f \to 0\), as one can see from \eqref{05222019vapp}, and Aristotle, who conflated a falling object’s actual speed with its terminal speed, believed that this meant an object falling through a vacuum would fall infinitely fast, which makes no sense. Apparently, he therefore concluded that a vacuum can not exist. He may or may not have been correct in this conclusion (albeit for the wrong reason), but he was certainly correct that an object cannot have an infinite speed. In any case, when we take those limits, we get the following: \begin{equation} \begin{aligned} & \lim \limits_{v_c \to \infty} y_c (t) = y_0 + u_0 t + \frac{1}{2} g\,t^2, \\[1ex] & \lim \limits_{v_c \to \infty} u_c (t) = u_0 + g\,t, \\[1ex] & \lim \limits_{v_c \to \infty} a_c (t) = g. \end{aligned} \end{equation}

Some readers might recognize these as the basic kinematic equations for an object with an acceleration \(a = g\). Oh, and also, \begin{equation} \lim \limits_{v_c \to \infty} t_G = - \frac{u_0}{g} + \sqrt{ \left( \frac{u_0}{g} \right)^2 - 2 \left( \frac{y_0}{g} \right)}. \end{equation}

So much for not having any more equations.

Finally, you have reached the end, and if you have read this far, I am impressed. This post has turned out to be way longer than I expected and intended it to be. But hey along the way, you maybe kinda sorta accidentally learned the fundamentals of classical mechanics, which underlies many of the things I plan to write about this summer. Therefore, instead of having to explain all the physics in those posts, I can just refer back to this! So hopefully all that stuff will be shorter and maybe a little less technical. We shall see—stay tuned.