Good morning. If I drop a steel sphere into a tank of water, the steel sphere will eventually
reach its terminal velocity. We are going to start by solving for the drag coefficient of the steel
sphere in terms of variables. Then we are going to determine those variables and solve for the drag
coefficient of the steel sphere.
♪ Flipping Physics ♪
Billy, please start with the free body diagram of the forces acting on the steel sphere
and then solve for its drag coefficient. Absolutely. As the steel sphere moves through the water,
the force of drag acting on the sphere acts upward and the force of gravity acts
downward. At its terminal velocity those two forces are equal, and the acceleration of the
steel sphere is zero. … We can show that is true by summing the forces in the y-direction.
We get force of drag minus force of gravity equals mass times acceleration in the y-direction.
As I already stated, the acceleration in the y-direction at terminal velocity equals zero,
so the force of drag and the force of gravity are then equal in magnitude. … Substitute in equations
for those two forces … and solve for the drag coefficient. We get … the drag coefficient equals
the quantity 2 times mass times acceleration due to gravity all divided by the quantity density
times cross sectional area times the terminal speed squared. … Actually, let’s substitute
in pi times radius squared for cross-sectional area because we know the cross section of the
steel sphere is a circle and pi times radius squared is the area of a circle. … That is the
equation for drag coefficient in this situation. Thanks Billy. Bo, what then do we need to then in order to solve
for the drag coefficient of the steel sphere? We need the mass, … density, … radius, … and
terminal speed of the steel sphere. … Why did you use capital C sub D for
coefficient of drag? I thought the symbol for drag coefficient was just capital D?
I’ve seen both capital D and capital C sub D for drag coefficient, so I thought I would use capital
C sub D this time. (Okay.)
Okay. So, to determine the terminal speed of the steel sphere, I measured that it goes down
a distance of 12.6 cm in a time of 0.056 seconds. That gives us a terminal velocity of magnitude
2.25 meters per second. The mass of the steel sphere is 67.8 grams. Its diameter is 2.55 cm,
which means it has a radius of 0.01275 meters. And the density of steel is 7,800 kilograms per meter
cubed. When we plug all of those numbers into the equation Billy derived for the drag coefficient
in this situation, we get 0.0659628 or 0.066 for the drag coefficient of the steel sphere.
… Uh, that does not make sense. Why not?
According to NASA’s website, the drag coefficient of a smooth sphere is about
0.5. … 0.066 is not close to 0.5. Did you memorize that?
I looked it up after we determined the terminal velocity of a baseball.
Right. Billy,
you are correct. We should get something close to 0.5.
Then we did something wrong. (Yes.)
Dang. Yeah.
We used the wrong density. Uh, what?
The density in the force of drag equation is the density
of the medium the object is moving through. Right. We needed to use the density of water.
1000 kilograms per meter cubed. Yes. This is a very common mistake to make.
All the other variables in the force of drag equation have to do with the object, except the density, which
is the density of the fluid the object is moving through. Please remember that. … And when we
substitute in 1000 kilograms per meter cubed for the density of water, we get 0.514510 or 0.51 for
the drag coefficient of the smooth, steel sphere. The physics works. The physics works. Uh huh,
uh huh. The physics works. The physics works. Uh huh, uh huh. …
Sorry, I have to interrupt the merrymaking and admit to something.
It’s not that the steel sphere never fully reaches terminal velocity is it? That terminal
velocity is the ideal limit and it never quite reaches that? Because we went over that before.
No, that’s not it. But it does have to do with the fact that it takes a long time
and therefore a long distance for an object to get close to its terminal velocity. … You see,
my fish tank is quite small. In order for the sphere to be near its terminal velocity during
the short depth of the fish tank, I calculated what height to drop the sphere from such that it
would be moving at a speed close to its terminal velocity when it landed in the water, therefore,
the sphere is moving near its terminal velocity before it runs into the bottom of the fish tank.
You cheated. A little.
It was to help you learn. Okay, I forgive you. That makes sense. Sure.
Thank you for your understanding. … And realize, if you drop the sphere from higher than I did,
it will start in the water moving faster than its terminal speed,
the drag force will be greater than the force of gravity, and the sphere will be slowing down
on its way to its terminal speed. … And if you drop the sphere from lower than I did,
it will start in the water moving slower than its terminal speed, the drag force will be less
than the force of gravity, and the sphere will be speeding up on its way to its terminal speed. …
And … There is one other thing. We actually neglected a force acting on the steel sphere.
One that is not negligible. Bummer. Yeah. What force?
The force buoyancy. Oh. Ah. The what?
The buoyancy force. The force acting upward on
the sphere. It’s what makes a boat float. Okay. Have we learned about that yet?
No. That is correct. The force buoyancy is not
a part of the AP Physics C: Mechanics curriculum, however, it is not negligible in this situation.
I’ll show the solution which includes force buoyancy and point out that it changes the
drag coefficient to 0.45, which reiterates that force buoyancy is actually not negligible in
this situation. I am not going to walk my way through the whole solution; however,
I will point out that the force buoyancy equals the weight of the fluid displaced by the object.
In other words, it equals the mass of the displaced fluid times acceleration due to
gravity. And, because density equals mass divided by volume, we can substitute the density of water
times the volume of the sphere for the mass of the water displaced by the sphere. And the volume of a
sphere equals four thirds pi times radius cubed. So, there you go. We get the drag coefficient
of this sphere to be 0.45, however, remember this is not a great way to solve for drag coefficient. We would need a much
deeper vessel of water to be able to ensure the steel sphere is moving at its terminal speed.
Thank you very much for learning with me today, I enjoyed learning with you.
Why was the buoyancy force not negligible in this example with the steel sphere moving through water,
but it was when we dropped objects in the air last time?
It’s because of the different media the objects are moving through. What?
This example has the sphere going through water; last time the objects were going through air. ... So?
Water has a higher density than air, so the buoyancy force in water is much larger than the buoyancy force in air.
Right. Water has a density of roughly 1000 kilograms per meter cubed
and air has a density of roughly 1.2 kilograms per meter cubed.
In other words, the buoyancy force acting on an object in water is roughly 800 times larger
than the buoyancy force acting on the same object in air. Oh, yeah. That makes sense. Thanks.
You are welcome.