We're going to discuss today resistive forces
and drag forces. When you move an object
through a medium, whether it's a gas
or whether it's a liquid, it experiences a drag force. This drag force depends
on the shape of the object, the size of the object, the medium through
which you move it and the speed of the object. The medium is
immediately obvious. If it's air
and you move through air, you feel the wind through
your hair-- that's a drag force. If you swim in water,
you feel this drag force. In oil, the drag force
would be even larger. This drag force, this resistive
force is very, very different from the friction
that we have discussed earlier when two surfaces move
relative to each other. There, the kinetic friction
coefficient remains constant independent of the speed. With the drag forces
and the resistive forces, they are not at all independent
of the speed. In very general terms, the resistive force
can be written as k1 times the velocity (v) plus
k2 times the velocity squared and always
in the opposite direction of the velocity vector. This v here is the speed, so all these signs-- k1, v and
k2, and obviously v squared-- they all are positive values. And the k values depend on the shape
and the size of the object and on the kind of medium
that I have. Today I will restrict myself
exclusively to spheres. And when we deal with spheres,
we're going to get that the force,
the magnitude of the force-- so that's this part-- equals C1 times r
times the speed plus C2 times r squared
times v squared. And again, it's always opposing
the velocity vector. C1 in our unit is kilograms
per meters per second and C2 has the dimension of density kilogram
per cubic meters. We call this the viscous term, and we call this
the pressure term. The viscous term has to do with
the stickiness of the medium. If you take,
for instance, liquids-- water and oil and tar-- there is a huge difference
in stickiness. Physicists also refer to that
as viscosity. If you have a high viscosity,
it's very sticky, then this number, C1,
will be very high. So this we call
the viscous term, and this we call
the pressure term. The C1 is a strong function
of temperature. We all know that if you take tar
and you heat it that the viscosity goes down. It is way more sticky
when it is cold. C2 is not very dependent
on the temperature. It's not so easy to see why this pressure term here
has a v square. Later in the course when we deal
with transfer of momentum, we will understand why there is
a v-square term there. But the r square is
very easy to see, because if you have a sphere
and there is some fluid-- gas or liquid--
streaming onto it, then this has
a cross-sectional area which is proportional
to r squared, and so it's easy to see that the force that
this object experiences-- we call it the pressure term-- is proportional to r square,
so that's easy to see. Two liquids with the very
same density would have... they could have
very different values for C1. They could differ by ten order not ten, by four or five
orders of magnitude. But if they have the same
density, the liquids, then the C2 is
very much the same. C2 is almost the density rho
of the liquid-- not quite but almost--
but there is a very strong correlation
between the C2 and the density. If I drop an object
and I just let it go, I take an object
and I let it fall-- we're only dealing
with spheres today-- then what you will see,
I have a mass m, and so there is a force mg--
that is gravity. And as it picks up speed,
the resistive force will grow, and it will grow
and it will grow, and there comes a time...
because the speed increases, so the resistive force
will grow, and there comes a time
that the two are equal. And when the two are equal, then
there is no longer acceleration, so the object has
a constant speed, and we call that
the terminal velocity, and that will be the case when mg equals C1 r v plus
C2 r squared v squared. And then we have here
terminal velocity. If you know what m is, the mass
of an object, the radius, and you know
the values for C1 or C2 of that medium
in which you move it, then you can calculate
what the terminal velocity is. It is a quadratic equation,
so you get two solutions of which one of them is
nonphysical, so you can reject that one. Very often will we work
in a domain, in a regime whereby this viscous term
is dominating. I call that regime one,
but it also happens-- and I will show you examples
today-- that we're working in a regime where really
this force is dominating. I call that regime two. Where one and two are the same-- where the force
due to the viscous force and the pressure force
are the same-- we can make
these terms the same, so you get C1 r v equals
C2 r square v squared, and that velocity we call
the critical velocity, even though there is
nothing critical about it. It's not critical at all; it's simply the speed at which
the two terms are equal. That's all it means. And that, of course, then equals
C1 divided by C2 divided by r. Now we're going to make
a clear distinction between the domains one and two,
the regimes one and two. Regime one is when the speed is
much, much less than the critical velocity. So we then have that
mg equals C1 r v terminal, and therefore the
terminal velocity equals mg divided by C1 r. If you take objects
of the same material-- that means they have
the same density, the density of the objects
that you drop in the liquid or that you drop in the gas-- so that m equals
4/3 pi rho r cubed-- this is now the rho,
the density of the object; it's not the density
of the medium-- then you can immediately see,
since you get an r cubed here, that this is proportional
to the square of the radius if you drop objects in there
with the same density. Regime two is the case when v is
much larger than v critical, so then mg equals
C2 r squared v squared if this is
the terminal velocity. So the terminal velocity is then the square root of mg
divided by C2 r squared... mg divided by C2 r squared. And if you take objects
with the same density and you compare their radii,
m is proportional to r cubed so this is now proportional
to the square root of r. So this separates
these two regimes, and we will see examples of that that sometimes you really work
exclusively in one and sometimes you work
in the other. I have for you a view graph
that is on the, on the Web so you do not have to copy it. It summarizes what
I have just told you. It has all the key equations. You see there on top
the resistive force, the magnitude
of the resistive force. You see then
the critical velocity. There's nothing critical
about it; it's just the speed
at which this term has the same magnitude as this term. Then you see the condition here, which I call equation one
for terminal speed, and then we have regime one, whereby the speed is way less
than the critical speed and then you get
the terminal velocity, as you see on the blackboard, which then is proportional
to r squared if you only look at objects which have
a particular given density. And if the velocity, if the speed is way larger
than the critical speed, you are in regime two and then you have a dependence
with the square root of r. I'm going to do a demonstration
and some measurements with ball bearings, which have very precise
radii-- very well known-- which I'm going
to drop into syrup. And we have chosen for that
Karo light corn syrup. It may interest you that
two tablespoons is 180 calories. I needed to know more
for this demonstration and so I had to do
my own homework on it, but at least you see here what
this Karo syrup can do for you. You see your 180 calories
per two tablespoons... oh per tablespoon. It's very low fat, and the rest may interest you
before you use it. I had to know C1, which
I calculated; I measured it. In fact, the kind of demonstration
you and I will be doing today, you can derive it, but it's very strongly
temperature-dependent. It could be different
yesterday from today. Ehm I measured C2
to a reasonable accuracy. Notice that the density
of the syrup in terms of kilograms
per cubic meter is very close to C2--
I mentioned that earlier. They're very close,
not exactly but very close. These steel ball bearings
have a density of about 7,800 kilograms
per cubic meter. And I'm going to drop in that
Karo syrup four ball bearings, and they have diameters
of an eighth of an inch, 5/32, 3/16 and
a quarter of an inch. And what I calculated was
the terminal velocity as a function of radius
of these ball bearings. All of this is on the Web. And so what you see here,
it is a logarithmic plot-- this is a log scale
and this is a log scale. Here you see the speed, and here you see the radius
in meters of the ball bearings. And this is my solution
to equation number one when I substitute
various values of r in there. I get the terminal speed
like this. And this is the critical speed, which has a 1-over-r
relationship. If you look
at this black dot here, then the terminal speed is
ten times larger here than the critical speed. And so notice that when
you are at speeds above that that you are exclusively
in domain two and your terminal speed
is proportional to the square root of the radius
of the ball bearings. This black dot is a factor
of ten below the critical speed, and so you see when
you are at lower speeds, when you work here, again,
you see that you fall into... exclusively in domain one,
and you see that the terminal speed is
proportional to r square. This slope here is plus two
in this diagram and this slope here
is plus one-half. Our ball bearings are all here, and so we are exclusively
operating in regime one where the viscous term
dominates. Now you could say, "Well, what is the meaning
of this critical speed here if they never reach
that speed anyhow?" Well, this critical speed
for a small ball bearing would be some hundred
meters per second. That's about 200 miles per hour. There is nothing wrong
with injecting a ball bearing with 400 miles per hour
into this syrup, in which case, if you injected
it with 400 miles per hour, you would be above
the critical speed and so for a short while would the motion be controlled
by the pressure term. But of course
when gravity takes over, then you ultimately end up
in regime one. But that's the meaning
of the critical velocity. If you could give the ball
bearing such a high speed, then the two terms are equal. That's all it means. Very well. Now we are going to look
at the various ball bearings, the various sizes, and I'll show
you how we do the experiment. You will shortly see
on the screen there seven marks which are one centimeter apart. They are in the liquid-- one, two, three, four,
five, six, seven. So here is the liquid, and the ball bearings
are dropped from above. When it reaches this line,
I will start my timer. And when it crosses
one, two, three, four... when it crosses this line,
I will stop my timer. And each mark is
about one centimeter apart, so this is a journey
of about four centimeters. And we will measure
the time that it takes to go from here to here. And the terminal velocity
is given. It's clearly regime one, so you see the terminal velocity
right there. Now, the time that it will take
is of course this distance-- let me call it h--
that the ball bearings travel divided by
the terminal velocity, and that is proportional,
since you're in regime one, by 1 over r squared. And now I will give you the... not the radii
but I will give you the diameters
of these ball bearings. That's the way they come. So we're going
to get a list here of the diameters
of the ball bearings, and the diameters is in inches. My smallest one has a diameter
of an eighth of an inch. Then I have 5/32, I have 3/16-- all these things come in inches;
that's one of those things-- and I have
one-quarter-inch diameter. If I plot here a plot, if I give you here the diameter
in terms of 1/32s of an inch, then this is four, five, six,
and eight-- easy numbers. Clearly if the time
that it takes is proportional
to 1 over r squared, it will also be proportional
to 1 over d squared-- of course, that's the same. What I'm going to plot is
not 1 over d squared, but to get some nice numbers I'm going to plot for you
100 over d squared, whereby d is then in these
units, 1/32 of an inch, and that gives me
some nice numbers. Then I get a 6.25 here,
I get a 4.00 here. You can see 100 divided by 25
is exactly four. I get 2.78 here, and
my last number is 1.56. And now I'm going to time it,
and my timing uncertainty is of course dictated
by my reaction time. That should be
at least 0.1 seconds. However, you will see when I reach
the quarter-inch ball bearing that it goes so fast that my error could well be
2/10 of a second. It goes in a flash. So I would allow here
2/10 of a second, and here I really don't know-- maybe 1/10,
maybe 2/10 of a second. You may ask me, "Why didn't you give us
the error in the diameter?" which ultimately,
of course, translates into the error in the mass. The reason is that these are
so precise the way you buy them that the uncertainty is
completely negligible compared to the timing error
that I make, so I won't even take that
into account. All right, so now we can start
the demonstration, and I'll have to switch
to this unit here. Here is this container
with the Karo syrup. It is very sticky indeed. You see there are seven marks,
and just for my own convenience, I have put there two black marks
so that I can easily see the moment that I have
to start my timer and the moment that I stop it. There are so many lines, I may
get confused if I don't do that. And we're going
to time this together, and we'll see
how these objects... how long it takes for them
to go through. I will start with my
one-eighth-of-an-inch diameter. I have one here-- tweezer. I release it at zero, three...
oh, oh, you can't see that. You should see that:
three, two, one, zero. Look how beautiful
it's working it's way through! You see, it's building up. You see that nice air bubble
at the top? It's going very slowly, but just wait when it has broken
through the surface. There it goes. Now! One centimeter, two centimeter,
three centimeter. Now! Okay, what is that? (students responding ) LEWIN:
Nothing. What happened? STUDENT:
5.93. LEWIN:
I didn't see it. I want to do another one. Did I... did I clean...
did I erase it? (student answers ) LEWIN:
How much was it? STUDENTS:
5.93. LEWIN:
5.93. Keep that in mind. It's nice to see whether
they reproduce, actually. Okay, there it goes. Now! One centimeter, two centimeter,
three centimeter. Now! 5.66-- that shows you
the uncertainty in my timing. So we had a 5.93,
and we have now a 5.66. It's not so bad, 5.9, 5.7-- timing error
a tenth of a second. My timing error could be a little bit larger
than a tenth of a second. You don't have very much time. So now we go to the 5/32. Okay? 5/32. It takes some time to break
through the surface. Isn't that funny? Because a thin film has formed
on the surface of the syrup due to its exposure to air. It's wonderful--
that lets us wait patiently. But now it goes--
there it goes. Now! One, two, three. Now! 3.80. Is that what you have? It's going to be tougher
and tougher for me. 3/16 of an inch. It's actually a good thing that it stays for a while
at the surface so that I can get ready. Really, that really helps,
doesn't it? If you do this in water,
it goes (whoosh ). You don't even see it. Come on... There we go. Now! Now! So you see,
that's very hard for me, and so I could easily have
a substantial error. 2.69. And now we have the
quarter-inch, the real big one. I have to do that again. I don't trust this at all. It went through the surface
too fast. (timet button clicking) (timer button clicking ) I... can't do it
very accurately. What was the first number,
by the way? (students respond ) LEWIN:
One point...? (student responds ) LEWIN:
Six eight. And this is 1.40. 1.68 and 1.40. So you see I wasn't kidding when I said that my uncertainty
could easily be .2. Now comes the acid test. And the acid test is
that if I'm... if the measurements
were done correctly and if we really work
in that regime, then if I plot 100 divided
by d squared versus t on linear paper, then
it should be a straight line. All right. Here I have a plot
which I prepared, and I'm going to put
these numbers in there. So first we're going to get
six point... five point... let's put in 5.8 seconds
for the smallest ball bearing. This is the smallest one. Don't be misled, because this is
100 divided by d squared, so this is
the smallest one-- 5.8. So we are here on this line, and
we are at 5.8-- somewhere here. That's it. Notice the point is lower
than where I expected it, and the reason is the
temperature went up. And if the temperature went up, then the viscosity goes down
and they go faster. But that's okay,
that doesn't worry me. The next one, 3.80. Four, 3.80. Ahhhh! You see that? I predicted that--
straight line. Isn't that a straight line? (students respond ) LEWIN:
It's not a straight line? (students respond ) LEWIN:
What's wrong? Okay, we'll put in
a third point. 2.69. Two point... This is 2.7-- 2.69. I can hardly put in
the error of the timing, because it is not much larger
than the size of my dots. And now we have the last one. One point... let's take
the average-- 1.55. 1.55... with an error of
about 2/10 of a second, and this has an error
of about 2/10 of a second. All right, there we go. Now... is this a straight line
or is it not? A gorgeous straight line. And so you see you are really working here in the regime where the terminal
velocity is proportional to the radius squared. Okay, we'll give you
your lights back. Now comes a question which is
relevant to this experiment, and that is, how long does it take for the
terminal speed to be reached? Well, the object has
a certain mass, so there's
a gravitational force on it, and the gravitational force
equals mg. And then there is a resistive
force, which in the case-- because we are operating
in regime one exclusively-- that resistive force equals
C1 r v in terms of magnitude, C1 r v, because we deal
with regime one. And so if I call this
the increasing value of y, the second... Newton's
Second Law would give me ma equals mg minus C1 r times v,
and this equals m dv dt, so I have here
a differential equation in v, and that can be solved. And you're going to solve it
on your assignment number four. What you're going to see is that
the speed as a function of time is going to build up
to a maximum value... this is the... to a maximum value, which
is the terminal velocity-- or you may want to call it
terminal speed-- and it's going to build up
in some fashion and then it's going
to asymptotically approach the terminal speed. And this is what I'm asking you
on your third... on your fourth assignment,
to calculate that. If there were no drag force
at all, I hope you realize that the velocity
would increase linearly, so you would get
something like this. So there's no drag. So the behavior is extremely
different due to the drag. And I calculated already something that is part
of your assignment-- how long does it take for
the quarter-inch ball bearing... how long does it take in time to reach a speed which is
about 99% of the terminal speed? And I calculated that, and you will go through
that calculation for yourself. That is only nine milliseconds. In other words, once it has
broken through the surface-- that takes a while
because of the thin film-- then in nine milliseconds will I already be
at 99% of the terminal speed, and so there was
no problem at all; when I waited for the object
to cross the first mark, it was already clearly going
at the terminal speed. So that was fine. Now I want to turn to air. Air, of course, behaves
in an extremely different way. The principle is the same, but the values for C1 and C2
are vastly different. If we take air
at one atmospheres, and we take it
at room temperature, then C1 is about 3.1 times ten
to the minus four in our units and C2 is about 0.85. This is very close
to the density of air, which is about one kilogram
per cubic meter, which I told you earlier, C2 and
rho are very strongly related. And so the critical speed,
which is C1 divided by C2 by r, is about 3.7 times
ten to the minus four
divided by r meters per second. And that is
about 400 times lower than the critical speed
in syrup, in the Karo syrup for the same value of r. So if I compare
the quarter-inch ball bearing and I drop it in the Karo syrup, then the terminal velocity
in the Karo syrup is way below the critical
velocity of the Karo syrup. The critical velocity
of the Karo syrup would be 100 miles per hour
for a quarter-inch ball bearing. So it's way below. Here, in air,
the critical velocity is something like
11 centimeters per second. This is 11 centimeters
in one second, and we know when you drop a one-quarter-inch
ball bearing in air that the speed is way larger,
and therefore in the case of air, a quarter-
inch ball bearing would have a speed way
above the critical speed, and so you are now
exclusively in regime two. That's the regime two. Almost all spheres that you drop
in air operate exclusively in regime two. Whether it is a raindrop or whether it is a baseball
that you hit, or a golf ball,
or even a beach ball, or you throw a pebble
off a high building, or whether you jump
out of an airplane, with or without a parachute,
makes no difference, you're always dominated
by the pressure term, by the v-square term,
and you always are in a range whereby the terminal speed
is proportional to the square root of the radius for a given density
of the object. If you take a pebble with
a radius of about one centimeter and you throw it
off a high building, it will reach a speed which will
not exceed 75 miles per hour because of the air drag. If you jump out of a plane
and you have no parachute and I make the assumption that your mass is
about 70 kilograms-- I want rough numbers--
if I can approximate you by a sphere with a radius
of about 40 centimeters-- that's also an approximation,
you're not really like a sphere, but I want to get
some rough numbers-- then the terminal velocity is
150 miles per hour. So if you jump out of a plane
and you have no parachute, you will not go much faster
than 150 miles per hour. I just read an article yesterday about sky divers
who jump out of planes, and they want to open
the parachute at the very last possible, and they reach
terminal velocities of 120 miles per hour,
which doesn't surprise me. It's very close to this rough
number that I calculated. Of course, then
they open the parachute and then the air drag
increases enormously and then they slow down
even further. I told you a raindrop... almost all raindrops operate
in regime two when they fall. So the terminal velocity is
dictated by the v-square term. However, if you make the drops
exceedingly small, there comes a time that
you really enter regime one. And on assignment number four I've asked you to calculate
where that happens, and I can't do it for water because the radius of that
water drop will be so small that it would
evaporate immediately, so I chose oil for that. So I'm asking you in assignment
four, take an oil drop, make it smaller and smaller
and smaller and smaller, and there comes a time that
you begin to enter regime one, and I want you to calculate
where that crossover is between these two domains. I have here a ball-- you may call it a balloon,
but I call it a ball because there's no helium in it. And this ball-- ooh!--
weighs approximately 34 grams. So let me erase some here,
because I want, yeah... So we know the mass
and we know the radius. The mass is about 34 grams and the radius is
about 35 centimeters. It's about
70 centimeters across. I can calculate what
the terminal velocity is-- a better name would be
terminal speed. I know I'm definitely going
to be in this regime, so I know the mass, I know C2,
which in air is 0.85, I know the radius and I know g, and so I find that I find
about 1.8 meters per second. So if I drop it from
a height of three meters, which I'm going to do,
then you would think that the time that it takes
to hit the floor would be about my three meters divided by 1.8 meters
per second, which is about 1.7 seconds. That's not bad. That's not a bad approximation. However, it will,
of course, take longer. and the reason why
it will take longer is that the terminal velocity, the terminal speed is not
achieved instantaneously. With the ball bearings, it was
within nine milliseconds. I can assure you that here
it will take a lot longer. Now, if you want to calculate
the time that it takes to get close to terminal speed,
that is not an easy task, because you are going to end up with a nasty
differential equation. You're going to get mg. You're going to get
the acceleration which is the result of... Let's go with
the equation we have. You see, we have ma equals mg and then we get minus
the resistive force. And the resistive force has
a term in v and has a term in v squared. You see, v and v squared, and so this cannot be
solved analytically. But I've asked my graduate
student Dave Pooley, who is one of your instructors, to solve this for me
numerically. And I'm going to show you
the results. that he prepared...
he has a nice view graph, and you can see the effect
of time on the ball if you drop it
from three meters. Here it is. This is on the Web,
so don't copy anything. You have the values for C1
and C2 are given at the top. You may not be able to see them
from your seat, but they are there,
and what you see here is the height above the ground
as a function of time-- this is one second,
this is 1½ seconds; this is the three-meter mark. If there were no air drag-- remember we dropped an apple
early on from three meters-- it will hit the floor
at about 780 milliseconds. However, with the air drag, it
will be about one second later, more like 1.8 seconds. So the 1.7 wasn't bad,
as you see, but if you look here at how the speed builds up
as a function of time, you see it takes about three-,
four-tenths of a second to build up
to that terminal speed, which is the 1.8 meters per
second that you have there. And needless to say, of course, that the acceleration
due to gravity does not remain constant
but goes down very quickly, and when the acceleration
reaches, approaches zero, then you have terminal speed, and then there is no longer
any change in the velocity. So let's try this. I'll give more light. And we're going to throw
this object, and I don't think you're
going to get 1.8 seconds. You may get something that
is larger than 1.8 seconds, and the reasons are
the following. Number one, this is
not a perfect sphere, and only for spheres
do these calculations hold-- that's number one. Number two,
this thing is very springy, so the moment that I let it go, it probably goes
in some kind of oscillation. That doesn't help either. That will probably
also slow it down, because what causes, of course,
this slowdown in regime two is really turbulence. Turbulence is extremely hard
to understand and predict. And so almost anything I do,
I will only add turbulence, and therefore I predict that the time that it will take
from three meters will be probably
larger than 1.8 seconds, but it will be
substantially larger than the 780 milliseconds, which is what you would have
seen if you drop an apple. So let's see how close we are. I have to turn on this timer. Make sure I zero it-- I did. And... It's not so easy
to release it, by the way, and start the timer
at the same moment. And it's not even so easy for me
to see when it hits the ground, so there's a huge uncertainty
in this experiment. Okay. Three, two, one, zero. What do we see? Did I see something? 2.0-- that's not bad. See? The prediction was 1.8;
you get 2.0. That's not bad, so this takes
the air drag into account, and it is not even
an approximation. It uses the entire term linear
in v as well as in v square. But it's almost exclusively
dominated by the v-square term. I also asked Dave
to show me what happens when I throw a pebble
off the Empire State Building. And the pebble that we chose was
at a radius of one centimeter-- it's the kind of pebble
that all of us could find-- I know roughly
the density of pebbles, and when we throw it
off the Empire State Building, we reach a terminal speed
of about 75 miles per hour. Without the air drag, we would
have reached 225 miles per hour. So I want to show you that, too. So now you see this
Empire State Building, which has a height
of 475 meters, so that's where you start,
at t0; this is one second, five seconds... ten seconds,
five seconds, 15 seconds, and if there had been
no air drag, it would hit the ground
a little less than ten seconds, but now it will hit the ground
more like 16, 17 seconds. And you see that
the terminal speed builds up in about five, six seconds. It's very close
to the final value, and if there had been
no air drag, then the speed at which
it would hit the ground it would, of course, grow linearly, and when it hits the ground,
it would be somewhere here, which is 225 miles per hour. So you see that it's even a pebble
you wouldn't expect to be... to have a very large effect
on air drag, it is huge, provided that you
throw it from a high building. Now, you may remember that we dropped an apple
from three meters and that we calculated
the gravitational acceleration given the time
that it takes to fall. That was one of your... one of the things you did
in your assignment. We had 781 milliseconds,
I think. And out of that you can
calculate g, right? Because you know that three
meters is one-half g t squared, so I give you the three,
with an uncertainty, I give you the time-- 781 milliseconds with an
uncertainty of two milliseconds, which we allowed. Out pops g. So I asked Dave, "What is the effect of air drag
on this apple? Is it... Was it a responsible thing
for us to ignore that?" The apple has
a mass of 134 grams. It's easy to weigh, of course. So this was our apple during our
first lecture-- m is 134 grams. It's almost a sphere, really--
not quite but almost a sphere-- and the radius is
about three centimeters. And that leads
to a terminal velocity which you can calculate
if you want to using the v-square term, but I was not interested
in that. I wanted to know
how many milliseconds is the touchdown delayed
because of the air drag. And Dave made the calculations, and he found that that is two
milliseconds from three meters. From 1½ meters
it's almost nothing, and the reason
why it's almost nothing for 1½ meters-- you see, when you throw an apple in air,
it's really in regime two, so you're really dominated
by the speed squared, and the first 1½ meters it doesn't get
at very high speed yet. The speed grows linearly,
and so it is the last portion where you really get hit by the
air drag, by the v-square term. Two milliseconds from three
meters, so if h is three meters, there is a two-millisecond...
let's call it delay. So we were on the hairy edge
of being lucky and unlucky. If you really want to recalculate the gravitational
acceleration using our data, you should really subtract the
two milliseconds from the time. On the other hand, since we allowed a two-
millisecond uncertainty, we really weren't too far off. Now comes my last part, and that is, how does air drag
influence trajectories? And that is also part
of your assignment, but I'm going to help you
a little bit with that. In your assignment number four,
I'm asking you to evaluate quantitatively the
motion of an object in liquid, but I give that object an initial speed
in the x direction, and then there's
the liquid below. Then gravity is there, and
there is that initial speed. If there were no drag, then this, of course,
would be a parabola, and the horizontal velocity
would always be the same. There would never be any change. But that's not the case now. Because of the resistive forces, because of the drag
by the liquid, the object is going to get
a velocity in this direction, so there's going to be a component of the
resistive force opposing it. It has a speed
in this direction, so there's also going to be a
component of the resistive force in this direction. And that will
decrease the speed-- this component
in the x direction. And so you can already see that the curve that you're going
to see is a very different one. It's going to look
something like this. And then ultimately, there is
nothing left of this component, and ultimately,
when you go vertically down, you have the terminal speed
that you can find from dropping an object just
into the liquid vertically. So that's something
you're going to deal with in your assignment number four, and this is exclusively done
in regime one, because we have
an object with liquid, and with liquid, you almost
always work with regime one. Suppose I take a tennis ball and I throw up
a tennis ball in 26.100. There is air drag
on that tennis ball. In the absence of any air drag,
I would get a nice parabola which will be
completely symmetric. I throw it up with
a certain initial speed, v-- call it v0, I don't care-- and the horizontal component
would never change. This would be v0x;
it would always be the same. But now with air drag,
you're going to see that there's going to be
a force, an air drag force in the y direction. If the object goes up in this
direction, then there will be a resistive force component
in the y direction, and since it has a speed
in this direction, there will also be a resistive
force in the x direction, so this speed is going to be
eaten up in the same way that this component was going
to be eaten up. This component is going
to suffer. It will not stay
constant throughout, and as a result of that, you're going to get a trajectory
that looks more like this. It's asymmetric. Clearly you don't reach the highest point that you would
have reached without air drag, for the reason
that this resistive force
in the y direction will not allow it to go
as high-- that's obvious. You don't go as far
as you would without air drag, for obvious reasons
that this resistive force is going to kill this speed, but you also will get asymmetry
in the curvature, and I want you to see that. I call this point O, this point
P, and let's call this point S. So what I will do is
I will throw up a tennis ball and then I will throw up
a Styrofoam ball, and the Styrofoam ball has
very closely the same radius as the tennis ball. That means the resistive force
is the same on both, because the resistive force
is only dictated by r squared and by v--
r squared v squared, remember? However, this has a way
larger mass than this one, and even though the resistive
forces will be closely the same if I throw them up
with the same initial speed, it has a way larger effect
on a smaller mass than on a larger mass. F = ma, right? So on a very large mass, the resistive force will have
a much lower effect than on the smaller mass, even though the resistive forces
are about the same. So, try to see that the tennis ball is very
close to an ideal parabola. You will not even see
any effect of asymmetry. It will not be the case
for the Styrofoam, though. So, look
at the tennis ball first. I should really do it here. Did it look
more or less symmetric? Okay, now I'll try this one. Did it look asymmetric? Could you see it? Are you just saying yes,
or you really saw it? Let me do it once more;
I can throw it back. So now it should curve like this and then sort of come down
like that. You ready? You see the asymmetry? Okay, now comes
my last question. I'm going to ask you
the following, and there's
a unique answer to that. I want you to think about it
and I want you to be able to give that answer
with total, 100% confidence. When this object goes
from O to P, that takes
a certain amount of time. When it goes from P to S,
back to the ground, that takes also
a certain amount of time. Is this time
the same as this time? Or is this time longer than this
time, or is this time shorter? Think about it. See you Friday.
