Video Transcript
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.
