Good morning. If I drop an ball in the vacuum you can breathe, the object is in free fall.
However, the reality is that air actually does exist, which is helpful for us as humans.
However, it does mean the ball will actually not be in free fall. Billy, why not?
♪ Flipping Physics ♪
The only force acting on an object in free fall is the force of gravity. If the object
is moving through air, then there is also a force of drag acting on it and it's not really in free fall.
Correct. Because there is a force of drag or a resistive force pushing on the object
and caused by the fluid the object is moving through, the object is no longer in free fall.
Therefore, this ball is actually not quite in free fall. … The free body diagram of the forces acting
on an object after it is released, at rest, in the air has an upward force of drag and a downward
force of gravity. … And I do want to clarify that the magnitude of the force of drag in the free
body diagram has been greatly exaggerated here. If the force of drag were shown at a length which
were proportionally correct relative to the length of the force of gravity, you would not be able to see it in
this free body diagram. I will show proportionally correct drag force magnitudes later. … Bobby,
please sum the forces on the object as it is falling and determine its acceleration.
Okay. Uh. … The net force in the y-direction equals the force of drag
minus the force of gravity. That also equals mass times acceleration in the y-direction.
The equation for the force of drag is … one half times the drag coefficient times the
density of the medium times the cross-sectional area times speed of the object squared. … The
equation for the force of gravity is mass times acceleration due to gravity. … All
you need to do is divide by mass to get the acceleration in the y-direction of the ball.
Thank you Bobby. Now, notice that, in the absence of air resistance,
this term cancels out and the acceleration in the y-direction of the object equals negative g.
Because, without air resistance, the object would be in free fall
and have an acceleration equal to negative g. Exactly. And the same is true right
at the moment the object is dropped. … Oh. Because the initial velocity equals
zero and therefore the initial drag force equals zero. So, the initial acceleration of the ball
equals negative 9.81 meters per second squared. Correct Bo. And Billy, what happens
as the object movers faster? Well, … the velocity of the ball
increases in magnitude, causing the force of drag to increase in magnitude, causing the acceleration
of the ball to get closer and closer to zero. Yes Billy. And when the force of drag has
increased to the point where it is equal in magnitude to the force of gravity,
the acceleration of the object equals zero, the velocity of the object is now constant,
and this constant velocity is called the terminal velocity of the object, v sub t.
I thought v sub t was tangential velocity. Right. Unfortunately, it is the same symbol.
Hopefully there will never be a situation with both tangential and terminal velocities
in it. … Okay, Bo, please solve for the terminal velocity of a dropped object.
Sure. … We can solve for the terminal velocity of the object by setting the acceleration of
the object in the y-direction equal to zero. … Add acceleration due to gravity to both sides. … Solve
for terminal velocity squared … and then take the square root of the whole equation to get. … Terminal
velocity equals the square root of 2 times mass times acceleration due to gravity all divided
by the quantity drag coefficient times density of the medium times the cross-sectional area.
Great. Thank you Bo! … Please, do not memorize this equation for terminal velocity.
It is easy to derive and you will be much better off
if you understand how to derive it rather than memorize the equation. … Okay,
let’s look at the position as a function of time of 5 coffee filters stacked on top of one another.
Looking at the position as a function of time graph, you can see the initial slope of the
best fit line of the data equals zero because the coffee filters are initially at rest. After
the coffee filters are dropped, the best fit line slope decreases until it reaches a value
of roughly negative 3.3 m/s. 3.3 meters per second is the terminal velocity of the 5 coffee filters
stacked on top of one another. The velocity is negative because the coffee filters are
going down. … Bobby, tell me what you see when you look at the velocity as a function of time graph.
Well … The velocity as a function of time graph starts out at zero, just like the slope
of the position graph. … The velocity then decreases in value and then seems to level
out and becomes constant at roughly negative 3.3 m/s which is the terminal velocity we saw
in the position as a function of time graph. Thank you Bobby. … Billy, what about the
acceleration as a function of time graph. What do you see there?
Uh, it looks really weird. … The acceleration should be zero while you are holding the coffee
filters, then jump to negative 9.81 meters per second squared when you
drop them and then increase to zero. That’s what it should do, but I do not really see that here.
Billy, you are correct about what we should see on the acceleration graph.
Realize the only data we actually collected about the 5 stacked coffee filters was position as a
function of time. That blue item on the ground is a motion sensor which detected the position of the
coffee filters 50 times per second. The velocity data was calculated from the position data and the
acceleration data was calculated from the velocity data. In other words, with each calculation there
is more and more “noise” in the data. That is why the position data looks great, the velocity
data looks okay, and the acceleration data looks … pretty scattered. However, the acceleration does
actually follow, pretty okay what you described. The average initial acceleration is roughly zero.
Clearly, I was unable to hold the coffee filters still. The acceleration then drops suddenly. It
does not reach negative 9.81 meters per second squared, however, it does reach about negative
7.5 meters per second squared and then increase in value until it reaches zero acceleration.
Sure. I guess I can see that. It is pretty scattered though.
Now, I just want to make it abundantly clear than this is not uniformly accelerated motion.
Right. You can clearly see on the graph that the acceleration is not constant.
And the force of drag changes with velocity causing the net
force and acceleration to not be constant. I agree. However, I still sometimes get students
who want to apply the uniformly accelerated motion equations to this situation. You cannot do that.
Right. Let’s take a moment
to look at the forces acting on the filters as they fall. Bobby, tell me what you see.
Uh … Before you drop the filters, you are applying a force upward on the filters. …
As soon as you let go of the filters, that force applied disappears, and the initial speed of the
filters is zero, so there is no force of drag right when you drop them. … As the filters fall,
their speed increases and therefore the force of drag increases … until the magnitudes of the
force of drag and force of gravity are equal, right before the coffee filters hit the ground.
Very nice Bobby. …. Okay, we have already determined the terminal speed of 5 stacked
coffee filters to be 3.3 meters per second; let’s also determine the drag coefficient of
the coffee filters. Bo, please go back to our original Newton’s Second Law equation and solve
for the drag coefficient of the coffee filters. Sure. … The acceleration in the y-direction at
terminal velocity is zero, therefore the force of drag equals the force of gravity. … We can plug
in equations for both of those. … And solve for the drag coefficient which equals 2 times mass
times acceleration due to gravity all divided by the quantity density times cross sectional
area … actually let’s put in the area of a circle or pi times radius squared there because we know
the cross sectional area of the coffee filters is a circle … times terminal speed squared.
Thanks Bo. Billy, what variables do we need in order to solve for
the drag coefficient of the coffee filters? We need the mass of the 5 coffee filters.
(The total mass of the 5 coffee filters is 7.8 grams.)
We need the density of … the coffee filters? No, that’s the density of the medium
the coffee filters are going through. Oh, that’s right. We need the density of air.
(The density of air at this temperature is roughly 1.2 kilograms per meter cubed.)
We also need the radius of the coffee filters. (The diameter of the coffee filters is
17.4 cm. That makes the radii of the coffee filters 0.087 meters.)
And we already have the terminal speed of the 5 coffee filters. It is 3.3 meters per second.
Now we can calculate that the drag coefficient of the 5 stacked coffee
filters equals 0.492489 or 0.49. That’s cool and all, but why did we
drop coffee filters? We could have done something more interesting.
Like what? I don’t know. A baseball?
The terminal velocity of a baseball is roughly 90 miles per hour.
So?
So, a major league pitcher can throw a baseball faster than its terminal velocity?
Woah. … But still,
why couldn’t we have used a baseball instead of 5 coffee filters. That would have been more fun.
… Does anybody have an answer for Bo? … You seem to drop the coffee filters
from kind of a short height off the ground. Yeah, the terminal velocity of the 5 coffee
filters is 3.3 meters per second, but a baseball has a terminal velocity of roughly 42 meters per second.
I bet you have to drop the baseball from pretty high up to get it to reach its terminal velocity.
Oh, right. I get it. Compared to the baseball, the coffee filters have a smaller mass and a
larger radius and therefore have a much smaller terminal velocity, and therefore will reach
terminal velocity after having only fallen that small distance. … How far does a dropped baseball
have to go before reaching its terminal velocity? Well, actually, it is important to realize a
dropped object will never actually reach its terminal velocity.
What? Right. The terminal velocity is
an ideal limit the object will never quite reach. The acceleration gets closer and closer to zero,
but will never technically reach zero. Oh.
Okay. Technically it would take an infinite amount of time for the acceleration to reach exactly zero.
Which is not going to ever happen. So then, I’ll rephrase my question.
How far does a dropped baseball have to go before getting close to its terminal velocity?
(Hold up Mr P.) Mr. Workenaour...
Do you really mean to tell me that an object is never going to reach its terminal velocity?
Yes. In an ideal world no object will ever reach its terminal velocity because
its acceleration will never reach exactly zero. Right, I agree that the math model approaches a
limit of the terminal velocity, but how close do two numbers have to be before we just call them
the same number? After all, our measurements can only be so precise. If we measure the speed of the
coffee filter to be 3.2 m/s instead of 3.3 m/s, terminal velocity is still within our measurement
uncertainty. And even if we made this measurement a full minute into the coffee filter’s movement,
we would inevitably make measurements a little bit below and above the terminal velocity,
just because of measurement uncertainty. Right, physics uses mathematical
equations to model motion which have uncertainty in their measurements.
But, ultimately it is a measurement based science. That formula for the drag
force is full of simplifying assumptions. Okay, I see what you are getting at. These
equations do not take into account turbulent air flow, or that the temperature, and therefore the
density, of the air is actually not constant. Exactly. Even the gravitational field of the
planet is not constant. … And, eventually, even before the baseball hits the ground, we should
just go ahead and say it is at terminal velocity. Well, not this baseball.
Right, okay, this baseball does not have enough height to reach terminal velocity.
Yeah, so I think our difference here is that you are looking at it from an actual
measurement standpoint and I am looking at it as an ideal mathematical model. Can we agree that,
using an ideal mathematical model, a dropped baseball will never reach its terminal velocity,
however, the reality is that at some point it will get close enough to that terminal velocity, that,
as far as measurement uncertainty is concerned, the baseball has reached terminal velocity.
We can agree on that, right? I can agree with that.
Okay, well then, thank you for your help Mr. Workenaur.
You are welcome mr.p. Gentelmen. Thanks. Woah. Uh.
Who was that? That's Mr. Workenaour. ... Okay.
I calculated that, in order for a dropped baseball to get very close to its
terminal velocity and have an acceleration less than 0.1 meters per second squared,
the baseball will have to have been dropped from a height of more than 400 meters and will
have fallen for almost 13 seconds. I get it. You would have to drop
a baseball from about a quarter mile off the ground for it to get close to its
terminal velocity of 90 miles per hour. I don’t think his ladder is that tall.
It’s pretty close. … But,
how did you do those calculations? As you said, this is not uniformly accelerated motion.
I used numerical modeling. What’s that?
Numerical modeling is what we are going to go over next time. I’ve included an Excel
file of the calculations on the webpage for this lecture if you are interested.
You will understand the calculations much better if you take a look at them after our next video
about numerical modelling. We are going to go back to the projectile motion problem where I
dropped a ball from a moving car, and it landed in a bucket. Only this time, we are going to solve
it using numerical modelling so we can include the force of drag. We will be able to predict
how much difference it made to assume the force of drag was negligible in the problem.
Cool! Sounds good. Alright. Glad you are looking forward to
it. … Thank you very much for learning with me today, I enjoyed learning with you.