Mr. P: Every year when we do projectile motion
we neglect air resistance and students
always want to know how do we include
the force of drag in our calculations?
Numerical modeling otherwise knows as
the Euler method is the answer.
Bo: Hi guys.
Billy: Hey Bo.
Bobby: Hi bo.
♫ (lyrics) Flipping physics ♫
Mr. P: In a previous
lesson, we determined
that we would need to drop a ball
1.6883 meters in front of a bucket
if we are moving at a horizontal
10.0 miles per hour and holding the ball
0.70 meters above the bucket.
I stated that, "Air resistance
decreases the displacement
"in the x-direction of the ball
"by less than 1 centimeter."
Today we're going to prove that statement
by introducing numerical modeling
or the Euler method.
Now, Leonherd Euler was a mathematician
who was around during the 18th century
and he devised a numerical method
for solving ordinary
differential equations
which is what we're going to do today.
But before we can really
even talk about that
we need to first discuss the
concept of air resistance.
Bo, remind me, what are the conditions
necessary for an object to
be in projectile motion?
Bo: It has to be flying through
the air in two dimensions.
Billy: Not the air, the vacuum
that you can breathe.
Bo: Which gets me to the second condition,
we have to ignore air resistance.
Mr.P: Then Bobby, if we
include air resistance,
is the object in projectile motion?
Bobby: Well no, if there's air resistance
then it's not projectile motion.
Mr. P: Correct, when we include air resistance
that means that the object is no longer
in projectile motion which means
the acceleration in the y-direction
is no longer negative 9.81
meters per second squared
and the velocity in the x-direction
is no longer constant.
Alright, now let's draw
a free body diagram
of the forces acting on the ball.
The force of gravity acts straight down
on the object at it's center of mass
and the force of drag
actually acts opposite
the direction of the
velocity of the object
however, I broke the force
of drag into it's components.
The force of drag in the x-direction
and the force of drag in the y-direction
just to make it easier to work with.
Now a typical equation
for the force of drag is,
The force of drag is equal to
1/2 times row times V squared
times D times A.
Where p is the density of the medium
through which the object is traveling,
V is the velocity of the object,
D is the drag coefficient of the object
which I'll talk about in just a moment
and A is the cross-sectional
area of the object.
Now, the cross-sectional
area of the object
is the area of the object's normal
to the direction of it's travel.
In other words, Bo, when
you look at the ball
traveling straight at
you, what two dimensional
shape do you see?
Bo: I see a circle, so pi r squared
would be the cross-sectional
area of the ball.
Mr. P: Luckily there are
published specifications
for any Lacrosse ball which is what
we are going to use here.
According to
the National Operating
Committee on Standards
for Athletic Equipment
or NOCSAE, the weight value of any
Lacrosse ball must be
within 5.0 to 5.25 ounces
and the circumference value must be
within 7.75 to 8 inches.
The radius of the Lacrosse ball is about
0.031835 meters and it's
mass is 0.14529 kilograms.
What I did there is I used the values
in the middle of those ranges.
I converted from English to metric units
and I went from circumference to radius.
I'm not going to walk
through that right now.
I'm actually going to walk through that
when we go through the spreadsheet
which is posted on my
web page for this video.
What I'd like
to do now is talk about
the drag coefficient D.
According to NASA, the drag coefficient
is a number which aerodynamicists use
to model all of the complex dependencies
of drag on shape, inclination
and some flow conditions.
In other words, it is an experimentally
determined number that helps determine
the drag on an object
and that drag coefficient
changes depending on
the shape of the object,
the type of fluid flow around the object
and the speed of the object.
Also according to NASA,
the drag coefficient
of a smooth sphere is about 0.5
and a Lacrosse ball is a smooth sphere
so that is the number that
we are going to use today.
And please notice that .5 only has
one significant digit so we will only
know our answer to one sig fig.
Now we need the density of the medium
through which the ball is traveling
which of course is air so we need
the density of air which is dependent on
the temperature and pressure of that air
and you can find all
sorts of tables online
that lists the density of air.
We are going to use
About.com's published number
of 1.275 kilograms per meter cubed
for the density of air at sea level
at 15 degrees Celsius
which is pretty close
to the conditions under which
this experiment was performed.
Now, let's talk about the
velocity of the object.
Remember this means that the drag force
on the ball is dependent on
the velocity of the ball.
So, as the velocity of the ball changes
the drag force on the
ball is going to change
which will change the
net force on the ball
which will then change the
acceleration on the ball
which will mean, as the velocity changes
the net force will change which will
change the acceleration which will change
the velocity which will
change the net force
which will change the acceleration
which will change the velocity
which will change the net
force which will change
the acceleration and I think
you can see the problem.
Bobby: That is definitely not
uniformly accelerated motion.
Mr. P: No, it is definitely not UAM
however, if you split this motion
into a whole bunch of
very short lasting parts
you can assume that that acceleration
is approximately constant which is exactly
what numerical modeling
or the Euler method does.
We can take this motion and split it
into a whole bunch of parts.
Let's say that they last for only
1/100th of a second and for each
1/100th of a second part, we can assume
that that acceleration is constant
and we can use the UAM equations.
Again, just for that 1/100th of a second.
Bo: That sounds like a lot
of calculations to me.
Mr. P: Which is exactly why we're going to
harness the power of Excel.
Billy: Okay.
Bobby: What?
Mr. P: Basically we're going
to use a spreadsheet
to do all the calculations for us.
But we need to first determine
what those calculations need to be.
So Billy, could you please look at
the free body diagram and sum the forces
in the x-direction?
Billy: Sure, the net force in
the x-direction equals
actually well there's only one force
in the x-direction and that is of course
drag in the x-direction
which is to the left
so it is negative and the net force equals
the mass of the ball
times the acceleration
of the x-direction.
We have our equation for the force of drag
so we can substitute in 1/2 times
the density of air times
the initial velocity
and the x-direction squared times
the drag coefficient times the
cross-sectional section of the ball
dividing by the mass of the ball
and substituting in pi r squared
for the cross-sectional area give us
that the acceleration in the x-direction
equals the negative of the density of air
times the initial velocity
in the x-direction squared
times the drag coefficient
times pi r squared
that whole quantify divided by two times
the mass of the ball.
We have all those numbers so it works out,
actually, no wait, we need to convert
the initial velocity to
meters per second first
so the velocity initial in the x-direction
equals 10.0 miles per hour times
one hour over 3600 seconds
to cancel out hours
and multiply by 1609 meters over one mile
to cancel out miles and you get--
Bobby: 4.4694444 meters per second.
Billy: For the initial velocity
in the x-direction
now we can plug numbers in
to get our acceleration.
Negative 1.275 times 4.4694444 squared
times 0.5 times pi times 0.031835 squared
that quantity divided by two times 0.14529
which give us the acceleration in
the x-direction as--
Bo: negative 0.13953 meters
per second squared.
Billy: Yeah.
Mr. P (voice-over): I just want to take this moment
to apologize, the density of air
as written on the board
is in kilograms per meter
squared which is incorrect.
It should be in kilograms per meter cubed.
I'm sorry.
Mr. P: Now that we have the acceleration
in the x-direction of the ball
for that first 1/100th of a second,
we can assume that that is a constant
acceleration we can use the UAM equations
to figure out the final velocity again,
after the 1/100th of a second.
Bo, please find the final velocity
in the x-direction of the ball.
Bo: We can use the final velocity
in the x-direction equals
the initial velocity
in the x-direction plus the acceleration
in the x-direction times
the change in time.
We have all those numbers
so the final velocity
in the x-direction just equals 4.469444
plus negative 0.13953 times 0.01
or 4.46805 meters per second.
Mr. P: So notice that after 1/100th of a second
the velocity of the
ball in the x-direction
has decreased by
approximately 1.4 thousandths
of a meter per second.
Now, Bobby could you please go through
and figure out the final
position of the ball
in the x-direction after
1/100th of a second?
Bobby: We can use the UAM
equation displacement equals
hold up, we're trying to
find the final position
so let's make delta x
equal to the final position
minus the initial position which equals
the velocity initial in the x-direction
times a change in time plus 1/2
times the acceleration in the x-direction
times a change in times squared.
Add position initial to both sides to get
the final position equals
the initial position
plus the initial velocity
in the x-direction
times a change in time plus 1/2 times
the acceleration in the x-direction
times the change in times squared.
Let's let the initial position be zero.
Plug in 4.4694444 times 0.01 plus
1/2 times negative 0.13953
times 0.01 squared and we get--
0.04469 meters.
Mr. P: Now, we have figured out for the first
1/100th of a second in the x-direction
the acceleration, the final velocity
and the final position.
Now, for the next 1/100th of a second
we need to again go through and figure out
the acceleration, the final velocity
and the final position.
The final velocity becomes
the initial velocity
for the second part and the final position
becomes the initial position again
for that second 1/100th of a second.
Let us now see how that
works by harnessing
the power of Excel.
Let's start by looking at the information
for the Lacrosse ball.
The weight of the Lacrosse ball
was in the range between
0.5 and 5.25 ounces
so we'll take 5.125 ounces as right
in the middle of that range.
And the mass, now to convert from ounces
to kilograms, we need to remember
that on planet earth
there are 35.27396 ounces
in one kilogram so we
take our weight in ounces
and we multiply by one kilogram divided by
35.27396 ounces and we get
our mass of 0.14529 kilograms.
Billy: Why did he say on planet earth?
Bobby: Remember weight changes depending
on location, depending on the
acceleration due to gravity.
Billy: Oh yeah, that's right.
Mr. P: And you can see I've already plugged in
our values for our drag coefficient
and the density of air.
Circumference was anywhere in the range
of 7.75 to 8 inches so
we'll use a value of
7.875 inches as a good average.
In order to convert from inches to meters
we multiply our inches by one meter
divided by 39.37 inches.
That give us out circumference in meters
because the circumference equals two pi
times the radius we can divide
the circumference by two pi to get
the radius so we take that circumference,
we divide it by two times.
Now, in order to get pi in Excel,
you do pi, left parenthesis,
right parenthesis.
That will give you the value for pi
and you can see we have here our radius.
In order to figure out the area,
we take and we multiply pi again,
pi, left parenthesis, right parenthesis.
We multiply that by the radius squared
and that will give us the area.
So let's take a look at the x-direction.
You can see our initial time is zero.
Our initial position is zero.
Our initial velocity was equal to 4.469
with a repeating four and the drag force
was equal to negative 1/2 multiplied
by the density of air.
Now, I need that density of air
to be an absolute reference.
I need to add the dollar signs.
The easiest way to do that on my Mac
is to press command t and then I multiply
by the velocity which
is a relative reference
and that is squared multiplied by
the drag coefficient absolute reference
and then multiplied by
the cross-sectional area
again, an absolute reference.
The easiest way to get
an absolute reference
on a Window's machine is to press F4
just so you know and we
end up with our drag force.
Bobby: What's an absolute reference?
Billy: When you copy and past from one cell
to another, the cell referred to
from the original cell stays the same
in an absolute reference.
Bobby: Okay.
Mr. P: So that drag force is the only force
acting the x-direction and therefore
it is also the net
force in the x-direction
which is equal to mass
times the acceleration
in the x-direction.
So to figure out the acceleration in
the x-direction, we simply divide
that drag force by the mass.
So that's going to be
equal to the drag force
divided by the mass.
Again, mass needs to be
an absolute reference
and that gives us our acceleration
negative 0.13953 meters
per second squared,
which is exactly what we got when
we did it on the board and now we can
go through and figure
out our final velocity
in the x-direction.
Alright so we need to
add 1/100th of a second
to our time and our final velocity is
equal to our initial velocity plus
our acceleration times the change in time.
That give us our final velocity
which is 4.46805 meters per second.
Exactly the same again as what
we got on the board.
Our final position is equal to
our initial position
plus our initial velocity
times again, the change in time plus
1/2 times the acceleration times
the change in time squared.
And again, we end up with the exact
same position, 0.04469 meters.
Bo, why don't we go back now to the board
and figure out the information
for the y-direction please?
Bo: Sure, the net force in the y-direction
equals the force of
drag in the y-direction
minus the force of gravity because
the force of gravity is down.
So that equals 1/2
times the density of air
times the initial velocity
in the y-direction
squared times the drag coefficient times
the cross-sectional area minus
the mass times the
acceleration due to gravity.
The initial velocity in
the y-direction is zero
so the drag force cancel out and--
Hey wait a minute.
You can cancel out mass
so the acceleration
of the y-direction equals negative 9.81
meters per second squared.
I thought you said this
wasn't projectile motion?
Mr. P: Okay Bo, let's take a
look at what happens here.
The initial velocity in the y-direction
is equal to zero therefore, that initial
drag force in the
y-direction is equal to zero
which means that yes, initially
the acceleration of the y-direction
equals negative 9.81
meters per second squared
but it is not constant.
It doesn't stay there.
That velocity in the y-direction
is going to increase in magnitude
which means that that force of drag
is going to increase, which means that
that net force in the y-direction
is going to change and the acceleration
in the y-directions is actually
going to decrease in
magnitude and get closer
and closer to zero.
So while the acceleration to y-direction
starts out at negative 9.81
meters per second squared,
it doesn't stay there.
It is not a constant acceleration.
Okay, let's harness the
power of Excel some more.
Alright, looking at the y-direction.
The first thing we need to do is unhide
all the y-direction
information in those cells
and then zoom out so that we can see
all of the information.
Now, the initial position
is equal to zero.
The initial velocity is equal to zero
in the y-direction.
The drag force actually
uses the same equation
with one exception, the drag force
in the y-direction is up and is therefore
going to be positive.
So we can copy and paste the equation
and we just need to remove the negative.
And of course we have a drag force
initially equal to zero.
The force of gravity is down so
it is negative and it's going to be
negative times the mass which is
an absolute reference times
the acceleration of gravity here
on planet earth, 9.81
meters per second squared.
The net force is just the summation
of the drag force and the force of gravity
and the acceleration, we can again
copy and paste because it's equal
to the net force divided by the mass.
And we do end up with negative 9.81
meters per second squared.
Now, speaking of copy and paste.
We can actually copy and paste
both the position and velocity equations
because they are exactly
the same equations.
We do need to change some of the relative
references though because
we need the times.
We switch all these A's to F's.
We need the time to reference
the correct column and we also need
the acceleration to reference
the correct column as well and that's
going to be in the L column.
So we need to do the same thing
with the velocity so we
need the acceleration
to reference the L column and the time
to reference the A column.
So there we have all of our equations
for position and velocity.
We figure out the final velocity
and final position after
1/100th of a second.
Now, who want to really harness
the power of Excel?
Billy: Me.
Bo: Yeah, I'm ready.
Mr. P: Okay so what we do is
we click and drag copy
all of our force and
acceleration equations
and once we have done
that, we can actually
click and drag copy all of our formulas
and we have now done all
of those calculations.
Just imagine trying to do
all of that without a spreadsheet?
Billy: Very nice Mr. Pete.
Bo: Poor Leonhard Euler.
Bobby: Yeah and he didn't have a calculator.
Mr. P: Alright let's refine this a little bit.
Now, this displacement in the y-direction
should be equal to negative 0.70
but it's not quite.
What we need to do is go through
and change our last time to add
in 0.008064 which is
a number I figured out
beforehand so that this displacement
in the y-direction would work out to be
pretty much exactly negative 0.70 meters.
Now let's take a look at
what's going on in this motion.
You can see the velocity
in the x-direction
is decreasing as a function of time
because the drag force in the x-direction.
The velocity in the y-direction
is increasing in magnitude as
a function in time which is causing
the drag force in the y-direction
to increase in magnitude which is then
causing the acceleration
in the y-direction
to decrease in magnitude
as a function of time
as that ball falls.
Now our goal here was to figure out
the displacement in the
x-direction with drag
but we wanted to compare that to the ideal
x displacement in meters which
was equal to 1.68843.
Now, the difference
caused by the drag force
also in meters, is going to be equal
to the ideal displacement minus
the displacement caused by drag
which was the 1.67984 meters,
which means, the difference caused by
the force of drag is 0.00859 meters
which is roughly nine millimeters
which is just less than one centimeter
as I said in the video.
Bobby: Yeah, that makes sense.
Billy: Very nice.
Bo: Thanks.
Mr. P: You are welcome.
Just so you know, this Excel file
that we have just created together
I have posted on my website along with
the video so that you can download it
and you can make whatever
changes you want to
and you could change the drag coefficient
and see how that effects
the motion for example.
I hope that I provided you a good
introduction to the drag force
and numerical modeling
or the Euler method.
Thank you very much for
learning with me today.
I enjoyed learning with you.
