In this video, we’re going to learn about drag and terminal velocity. We’ll learn what these terms mean, how they affect object motion, and how to work with them practically.
To start out, imagine that you are part of a competition called Earth First. The way the competition works is simple. Each participant designs an object light enough and small enough that they’re able to carry. The objects are loaded onto a plane and released from the plane as it flies at the same moment. The first object to reach the ground wins, hence the name of the competition. In order to have a chance at this year’s prize, you’re going to want to know something about drag and terminal velocity.
When we talk about drag, we’re speaking of a force that’s caused by resistance to motion through a fluid, whether a liquid or a gas. Drag is something we’re all familiar with. If you’ve ever put your hand out the window of a moving car, you can feel the resistance of the air against your arm. That’s the force of drag. And the interesting thing is you can feel that as the car speeds up, the force increases. In fact, let’s take this example of a car driving along the road to understand the equation or functional form of this drag force. We’ll call this force 𝐹 sub 𝐷. And to find out what it is, let’s consider some of what’s going on as a car drives along the road.
Perhaps the first thing we might think of is that the car is not driving through a vacuum. But it’s driving through air of density 𝜚. This air is what supplies the air molecules which create the resistance to the car’s motion and therefore the drag. We could expect that the drag force is proportional to the density of the fluid we’re moving through. As the car moves through the air, we can also consider its area, specifically its cross-sectional area, as it faces the directional travel. If this area were much bigger or a good bit smaller, we would expect that to affect the drag of the car. So it also seems to make sense that drag force would be proportional to cross-sectional area 𝐴.
Now let’s think about the shape of our area or the shape of our object in motion through a fluid. What if our car had a flat front like this, same cross-sectional area but a different shape? Or what if it was designed like this, so that air hitting the front of the car would very easily slide over it and move back? In each of these three cases, we have the same cross-sectional area 𝐴, but very different performance when it comes to the drag force. There’s a parameter that summarizes these differences in shape. And that’s the drag coefficient. We write it 𝐶 sub 𝑑. This coefficient tells us just how aero- or hydrodynamic a shape is. The lower the drag coefficient is, the less drag is experienced.
We mentioned earlier that an object’s drag goes up the faster it’s moving. So if we were guessing the functional form of the drag force, we might put in a factor of 𝑣. This is on the right track. But from experimental evidence, we find that the true form of the drag force is better approximated by putting a factor of 𝑣 squared and a factor of one-half in our expression. This relationship, though not true in all scenarios with all motion, is a helpful rule of thumb for the drag force acting on an object.
Because velocity is squared in this expression, it tells us that if we’re moving through a fluid and want to double our speed, our drag force will go up by a factor of four. So to go faster, we need to push that much harder, say with our engines or other power supply. And in fact, as we speed up and up, there gets to be a point where the drag force becomes too strong to overcome. This happens whenever an object is in free fall long enough. We know that every object near Earth surface experiences a gravitational force we can call 𝐹 sub 𝑔. When an object is also in motion through a fluid, such as this ball that we’ve dropped, it experiences a drag force, 𝐹 sub 𝐷. And this drag force, as we’ve seen, goes up with the velocity of the object.
After an object has fallen far enough and continued to speed up and up and up, its drag force increases to the point where it’s equal to the gravitational pull on that object. When this happens, that is, when the magnitude of the drag force equals the magnitude of the gravitational force on a falling object, it will stop falling faster. It will stay at the same speed as it descends. At this point, the object has reached what we call terminal velocity. It’s the highest speed an object can have while it’s in free fall. When an object reaches terminal velocity, it’s still falling, and usually quite fast. But it’s just not speeding up anymore. Let’s get some practice calculating drag and terminal velocity through a couple of examples.
A car has a frontal area of 0.722 meters squared and a drag coefficient of 0.216. The car drives through air of density 1.225 kilograms per cubic meter. The car drives at a velocity of 62.1 kilometers per hour. What is the magnitude of the drag force on the car? The car drives at a velocity of 90.4 kilometers per hour. What is the magnitude of the drag force on the car?
We want to solve for the drag force on the car when it’s moving at two different speeds. We’ll call this drag force 𝐹 sub 𝐷. To help us solve for it, we’re told the car’s frontal area, we’ll name that capital 𝐴; its drag coefficient with the air it’s in, 0.216, which we’ll call 𝐶 sub 𝐷; and the density of the air it drives through, which we’ll call 𝜚. We want to solve for the car’s drag force at two different speeds. We’ve called them 𝑣 one and 𝑣 two.
To get started on our solution, we can recall that drag force, 𝐹 sub 𝐷, is equal to one-half an object’s front-facing area, 𝐴, multiplied by the density of the fluid it’s moving through, 𝜚, times the drag coefficient, 𝐶 sub 𝐷, which depends on the shape of the object, multiplied finally by its speed, 𝑣, squared. Applying this relationship to our scenario, we want to solve for the drag force when our speed is equal to 𝑣 one and when our speed is 𝑣 two. We’re given the car’s area 𝐴, that air density 𝜚, and the drag coefficient 𝐶 sub 𝐷. Before we plug in, we’ll want to convert the units of 𝑣 one into meters per second so that the units of all the values in our expression are consistent.
In general, if we take an arbitrary speed in units of kilometers per hour and multiply that speed by 1000 meters in one kilometer and then by one hour in 3600 seconds, we’ll see that the units of hours cancel out, as do the units of kilometers. And we’re left with units of meters per second. This tells us that, for a given speed 𝑣 in kilometers per hour, if we divide that speed by 3.6, then it will be expressed in meters per second. So when we plug in for our various values, inserting 𝑣 one, we have 62.1 over 3.6 meters per second. When we calculate this value, we find that, to three significant figures, it’s 28.4 newtons. That’s the drag force on the car when it’s moving at 62.1 kilometers per hour.
Next, we wanna solve for the drag force when the car is moving, not at 𝑣 one, but at 𝑣 two, 90.4 kilometers per hour. The only change we need to make is to replace 62.1 with 90.4 in the numerator of this fraction. When we calculate this drag force, we find it’s equal to 60.2 newtons. So we see that, by increasing our velocity by about 50 percent, we’ve increased our drag force by over 100 percent. Now let’s look at an example involving terminal velocity.
A skydiver with a mass of 80.0 kilograms falls through air that has a uniform density of 1.23 kilograms per cubic meter. The skydiver has a surface area of 0.140 meters squared and a drag coefficient of 0.690. What is the skydiver’s terminal velocity?
We can call this terminal velocity 𝑣 sub 𝑇. And we’ll also write down the skydiver’s mass, cross-sectional area, drag coefficient, and the density of the air the skydiver falls through. As the skydiver falls down to Earth, there’s a gravitational force pulling the skydiver in that direction. Along with the gravitational force, there’s the drag force resisting the skydiver’s fall. If we recall that drag force is proportional to the square of an object’s speed, at the start of their fall, the skydiver has very little drag force. But as they fall faster and faster, this force increases. And eventually, we get to a point where the drag force is equal to the gravitational force on the skydiver. At this point, they are no longer speeding up. They’ve reached terminal velocity.
We can rewrite 𝐹 sub 𝐷 in terms of the parameters it relies on. And 𝐹 sub 𝑔, we know, is equal to the mass of the skydiver times 𝑔, where 𝑔, the acceleration due to gravity, is 9.8 meters per second squared. So one-half the skydiver’s area times the air density times the drag coefficient times the terminal velocity squared is equal to the skydiver’s mass times 𝑔. Or 𝑣 sub 𝑇, the skydiver’s terminal velocity, is the square root of two 𝑚𝑔 divided by 𝐴, the skydiver’s area, times the density, 𝜚, multiplied by the drag coefficient. When we plug in for all these values and solve for 𝑣 sub 𝑇, we find it’s equal to 115 meters per second. That’s the maximum speed that this skydiver will achieve. Let’s summarize what we we’ve learned about drag and terminal velocity.
We’ve seen that drag is a force that’s caused by a fluid’s resistance to an object moving through it. Written as a function, the drag force connects several parameters of the object moving through the fluid and the fluid itself. The drag force magnitude depends on the object’s cross-sectional area 𝐴. It depends on the object’s drag coefficient, 𝐶 sub 𝐷. It depends on the density of the fluid the object is moving through, 𝜚. And it also depends on the object’s speed, 𝑣, squared. This functional form tells us that if an object’s speed is doubled, then the drag force is quadrupled. And finally, when the force of gravity and the drag force balance on a freely falling object, that object has reached its maximum speed, also called its terminal velocity. In idealized exercises, we often neglect the drag force. But in real-life scenarios, it has an important effect on object motion.