Now we're going to analyze a more complicated example
of drag forces, where we have an object falling
in a gravitational field with gravity.
We have a resistive force.
And this is an object in air.
And so our model will be for the resistive force
that it's going to be proportional to the velocity
squared.
Now to get its direction right, opposing the motion, j hat.
So this will turn out to be a more complicated analysis.
But first again, let's think about the units
of our coefficient beta.
It has the units of force divided by velocity squared.
So we write the units of force, kilogram meters
per second squared, and the units
of velocity squared, meter squared per second squared.
And so we see that we have units of kilogram per meter
for our coefficient beta.
Now, we'll apply as usual Newton's second law,
F equals m a, to get our equation of motion.
We're looking at the j direction.
And our forces are gravity minus the velocity
squared resistive drag force.
And that's equal to the derivative of the velocity
dv/dt.
In this example, it's a one dimensional motion.
So I'm dropping any mention of y direction for simplicity.
Now we can rewrite this equation as dv/dt.
Let's divide through by m is g minus beta over m v square.
And this is a linear-- it's a first order differential
equation, dv/dt.
It's a non-linear equation because the velocity
term appears here as a square and there's a constant term.
But we can still apply our technique
of separation of variables.
And so when we write this equation as dv/dt
we'll separate out dv's and t's, so we have d d times
g minus beta over and m v squared is equal to dt.
Now, I'm going to do two things just to clean this up
for algebra a little bit later.
I'm going to multiply both sides by a minus sign.
And I'm going to pull the g out.
So I have 1 minus beta over m g v squared.
And on the other side, I have minus dt.
And now I can get this equation in the form
that I'd like to integrate, which
is minus dv times 1 minus beta over m g v squared equals
minus gdt.
Now, the trouble here is this integral is a little bit
complicated.
So I'd like to make a change of variable.
And my change of variable will be
u equals the square root of beta m g times v. And that implies
that du is beta m g dv.
And the limits, if we start our object at rest,
so if u0 equals 0, then because v0 equal 0,
that's our first limit.
Now we have to be a little bit careful
because if we drop this object at rest, initially
it will be moving very slowly.
And so our resistive model doesn't actually apply.
However, we're going to neglect that effect even
though if we were to do a more complicated analysis,
we would have to change our model as the object is
following, so it would be a multi-stage motion.
First, at the beginning with our only velocity
dependent resistant.
And then as it gets some initial speed and it's going faster,
we change our model.
That's why the actual problem can be quite complicated.
But we're just trying to keep things simple here.
And then u of t is square root of beta mg d of t.
And then with this change of variable, my integration,
remember I have a dv, so I have to multiply the left side by mg
over beta.
And I'm integrating with a minus sign du times 1
minus u squared from 0 to this final value u of t
And over here I'm just integrating minus g dt.
Now again, for simplicity, I'm going
to bring this term, the beta over mg over to the other side.
So I'll use the magic of our light board
by just erasing that and bringing it
to the other side, which makes my life a little easier.
And now, this interval can be done by the method
of integration by parts.
It's a nice problem in calculus.
And you can verify for yourself that the result
is one half natural log of 1 plus u over 1
minus u evaluated at our limits.
And over here I have minus g square root of beta over mg.
Now, once again, for a little bit of simplicity,
I'm going to bring the 2 over to the other side.
And now, I evaluate my limits.
Now recall that when you have a minus log,
we're flipping, because log of b over a equals minus
log over ab, so when I put out my limits in,
I have natural log of-- now remember, what are our limits?
We have 1-- I'm flipping-- minus u is beta mg times v of t.
And I have the 1 plus beta mg v of t.
And that's equal to minus as 2g square root of beta mg.
Now, again, we'll use the fact that e to log of x is x.
And so if I exponentiate both sides,
I end up with 1 minus the square root of beta mg d of t over 1
plus square root of beta mg v of t
is equal to exponential minus 2g beta mg times t.
And we'll just move that.
OK, now this is a little bit of algebra.
I want to solve for v of t.
If I bring this side over to there,
I'll just do that to make the first step a little simpler
to see.
So we have 1 minus square root of beta mg v of t
equals 1 plus square root of beta over mg v of t times
e to the minus this factor 2g square root of beta mg t.
Now this is a lot of stuff to carry around.
I'd like to introduce a constant here,
tao, which I'm going to find to be square root of mg beta 1
over 2g.
And so this whole term is going to just be e to the minus tao.
It's a nice example for you to work out that the units of tao
are the units of seconds.
And that's a little exercise to work out.
Now, I just have to collect my terms.
And what I'll do is I'll collect the T
terms on the right and the terms that don't have vt on the left.
So I have 1 minus e to the minus t over tao
on the left is equal to 1 plus e to the minus t over tao
on the right times beta mg v of t.
And so I get my solution, v of t equals
the square root of mg over beta times 1 minus
e to the minus t over tau over 1 plus e to the minus t over tao.
Well, it's not a simple solution at all.
But let's examine when you have a case like this-- again,
it would be a nice exercise to graph this out.
But right now we're going to consider the limit
as t goes to infinity.
And remember that e minus t over t
goes to 0 when t goes to infinity.
So we just have 1 over 1.
And what we get as t goes to infinity
is the quantity mg over beta.
And this is what we call the terminal velocity.
Now what does terminal velocity mean?
Well, when object is falling and there's a resistive force,
as the object falls faster and faster,
the resistive force gets greater and greater
until if we go back to Newton's second law and look at it,
as v gets faster and faster, eventually
these two terms are equal.
And when these two terms are equal,
that's the statement that the right-hand side has to be zero.
So what we mean by terminal velocity
is it's the velocity is no longer changing in time.
And then we can immediately check our work
by going to Newton's second law and see what
that case is if we set this quantity equal to zero.
In other words, when we set mg minus beta v
squared terminal equal to 0 we can solve for v terminal,
and we get a square root of mg divided by beta.
And that agrees with our lengthy calculation.
So we think we're on the right track.