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We're going to discuss today

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resistive forces
and drag forces.

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When you move an object
through a medium,

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whether it's a gas
or whether it's a liquid,

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it experiences a drag force.

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This drag force depends
on the shape of the object,

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the size of the object,

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the medium through
which you move it

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and the speed of the object.

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The medium is
immediately obvious.

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If it's air
and you move through air,

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you feel the wind through
your hair-- that's a drag force.

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If you swim in water,
you feel this drag force.

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In oil, the drag force
would be even larger.

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This drag force, this resistive
force is very, very different

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from the friction
that we have discussed earlier

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when two surfaces move
relative to each other.

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There, the kinetic friction
coefficient remains constant

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independent of the speed.

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With the drag forces
and the resistive forces,

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they are not at all independent
of the speed.

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In very general terms,

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the resistive force
can be written

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as k1 times the velocity plus
k2 times the velocity squared

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and always
in the opposite direction

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of the velocity vector.

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This v here is the speed,

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so all these signs-- k1, v and
k2, and obviously v squared--

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they all are positive values.

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And the k values depend

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on the shape
and the size of the object

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and on the kind of medium
that I have.

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Today I will restrict myself
exclusively to spheres.

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And when we deal with spheres,
we're going to get

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that the force,
the magnitude of the force--

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so that's this part--

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equals C1 times r
times the speed

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plus C2 times r squared
times v squared.

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And again, it's always opposing
the velocity vector.

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C1 in our unit is kilograms
per meters per second

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and C2 has the dimension

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of density kilogram
per cubic meters.

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We call this the viscous term,

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and we call this
the pressure term.

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The viscous term has to do with
the stickiness of the medium.

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If you take,
for instance, liquids--

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water and oil and tar--

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there is a huge difference
in stickiness.

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Physicists also refer to that
as viscosity.

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If you have a high viscosity,
it's very sticky,

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then this number, C1,
will be very high.

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So this we call
the viscous term,

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and this we call
the pressure term.

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The C1 is a strong function
of temperature.

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We all know that if you take tar
and you heat it

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that the viscosity goes down.

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It is way more sticky
when it is cold.

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C2 is not very dependent
on the temperature.

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It's not so easy to see

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why this pressure term here
has a v square.

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Later in the course when we deal
with transfer of momentum,

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we will understand why there is
a v-square term there.

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But the r square is
very easy to see,

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because if you have a sphere
and there is some fluid--

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gas or liquid--
streaming onto it,

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then this has
a cross-sectional area

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which is proportional
to r squared,

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and so it's easy to see

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that the force that
this object experiences--

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we call it the pressure term--

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is proportional to r square,
so that's easy to see.

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Two liquids with the very
same density would have...

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they could have
very different values for C1.

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They could differ by ten...

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not ten, by four or five
orders of magnitude.

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But if they have the same
density, the liquids,

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then the C2 is
very much the same.

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C2 is almost the density rho
of the liquid--

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not quite but almost--
but there is

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a very strong correlation
between the C2 and the density.

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If I drop an object
and I just let it go,

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I take an object
and I let it fall--

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we're only dealing
with spheres today--

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then what you will see,
I have a mass m,

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and so there is a force mg--
that is gravity.

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And as it picks up speed,
the resistive force will grow,

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and it will grow
and it will grow,

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and there comes a time...
because the speed increases,

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so the resistive force
will grow,

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and there comes a time
that the two are equal.

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And when the two are equal, then
there is no longer acceleration,

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so the object has
a constant speed,

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and we call that
the terminal velocity,

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and that will be the case

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when mg equals C1 r v plus
C2 r squared v squared.

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And then we have here
terminal velocity.

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If you know what m is, the mass
of an object, the radius,

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and you know
the values for C1 or C2

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of that medium
in which you move it,

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then you can calculate
what the terminal velocity is.

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It is a quadratic equation,
so you get two solutions

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of which one of them is
nonphysical,

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so you can reject that one.

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Very often will we work
in a domain, in a regime

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whereby this viscous term
is dominating.

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I call that regime one,
but it also happens--

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and I will show you examples
today-- that we're working

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in a regime where really
this force is dominating.

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I call that regime two.

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Where one and two are the same--

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where the force
due to the viscous force

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and the pressure force
are the same--

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we can make
these terms the same,

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so you get C1 r v equals
C2 r square v squared,

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and that velocity we call
the critical velocity,

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even though there is
nothing critical about it.

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It's not critical at all;
 

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it's simply the speed at which
the two terms are equal.

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That's all it means.

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And that, of course, then equals
C1 divided by C2 divided by r.

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Now we're going to make
a clear distinction

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between the domains one and two,
the regimes one and two.

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Regime one is when the speed is
much, much less

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than the critical velocity.

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So we then have that
mg equals C1 r v terminal,

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and therefore the
terminal velocity equals

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mg divided by C1 r.

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If you take objects
of the same material--

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that means they have
the same density,

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the density of the objects
that you drop in the liquid

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or that you drop in the gas--

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so that m equals
4/3 pi rho r cubed--

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this is now the rho,
the density of the object;

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it's not the density
of the medium--

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then you can immediately see,
since you get an r cubed here,

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that this is proportional
to the square of the radius

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if you drop objects in there
with the same density.

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Regime two is the case when v is
much larger than v critical,

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so then mg equals
C2 r squared v squared

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if this is
the terminal velocity.

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So the terminal velocity is then

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the square root of mg
divided by C2 r squared...

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mg divided by C2 r squared.

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And if you take objects
with the same density

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and you compare their radii,
m is proportional to r cubed

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so this is now proportional
to the square root of r.

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So this separates
these two regimes,

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and we will see examples of that

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that sometimes you really work
exclusively in one

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and sometimes you work
in the other.

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I have for you a view graph
that is on the Web

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so you do not have to copy it.

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It summarizes what
I have just told you.

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It has all the key equations.

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You see there on top
the resistive force,

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the magnitude
of the resistive force.

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You see then
the critical velocity.

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There's nothing critical
about it;

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it's just the speed
at which this term has
 

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the same magnitude as this term.

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Then you see the condition here,

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which I call equation one
for terminal speed,

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and then we have regime one,

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whereby the speed is way less
than the critical speed

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and then you get
the terminal velocity,

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as you see on the blackboard,

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which then is proportional
to r squared if you only look

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at objects which have
a particular given density.

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And if the velocity,

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if the speed is way larger
than the critical speed,

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you are in regime two

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and then you have a dependence
with the square root of r.

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I'm going to do a demonstration
and some measurements

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with ball bearings,

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which have very precise
radii-- very well known--

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which I'm going
to drop into syrup.

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And we have chosen for that
Karo light corn syrup.

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It may interest you that
two tablespoons is 180 calories.

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I needed to know more
for this demonstration

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and so I had to do
my own homework on it,

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but at least you see here what
this Karo syrup can do for you.

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You see your 180 calories
per two tablespoons...

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or per tablespoon.

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It's very low fat,

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and the rest may interest you
before you use it.

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I had to know C1, which
I calculated; I measured it.

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In fact,

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the kind of demonstration
you and I will be doing today,

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you can derive it,

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but it's very strongly
temperature-dependent.

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It could be different
yesterday from today.

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I measured C2
to a reasonable accuracy.

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Notice that the density
of the syrup

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in terms of kilograms
per cubic meter

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is very close to C2--
I mentioned that earlier.

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They're very close,
not exactly but very close.

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These steel ball bearings
have a density

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of about 7,800 kilograms
per cubic meter.

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And I'm going to drop in that
Karo syrup four ball bearings,

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and they have diameters
of an eighth of an inch,

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5/32, 3/16 and
a quarter of an inch.

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And what I calculated was
the terminal velocity

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as a function of radius
of these ball bearings.

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All of this is on the Web.

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And so what you see here,
it is a logarithmic plot--

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this is a log scale
and this is a log scale.

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Here you see the speed,

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and here you see the radius
in meters of the ball bearings.

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And this is my solution
to equation number one

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when I substitute
various values of r in there.

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I get the terminal speed
like this.

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And this is the critical speed,

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which has a 1-over-r
relationship.

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If you look
at this black dot here,

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then the terminal speed is
ten times larger here

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than the critical speed.

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And so notice that when
you are at speeds above that

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that you are exclusively
in domain two

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and your terminal speed
is proportional

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to the square root of the radius
of the ball bearings.

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This black dot is a factor
of ten below the critical speed,

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and so you see when
you are at lower speeds,

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when you work here, again,
you see that you fall into...

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exclusively in domain one,
and you see

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that the terminal speed is
proportional to r square.

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This slope here is plus two
in this diagram

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and this slope here
is plus one-half.

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Our ball bearings are all here,

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and so we are exclusively
operating in regime one

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where the viscous term
dominates.

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Now you could say,

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"Well, what is the meaning
of this critical speed here

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if they never reach
that speed anyhow?"

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Well, this critical speed
for a small ball bearing

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would be some hundred
meters per second.

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That's about 200 miles per hour.

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There is nothing wrong
with injecting a ball bearing

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with 400 miles per hour
into this syrup,

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in which case, if you injected
it with 400 miles per hour,

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you would be above
the critical speed

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and so for a short while

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would the motion be controlled
by the pressure term.

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But of course
when gravity takes over,

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then you ultimately end up
in regime one.

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So that's the meaning
of the critical velocity.

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If you could give the ball
bearing such a high speed,

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then the two terms are equal.

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That's all it means.

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Very well.

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Now we are going to look
at the various ball bearings,

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the various sizes, and I'll show
you how we do the experiment.

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You will shortly see
on the screen there seven marks

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which are one centimeter apart.

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They are in the liquid--

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one, two, three, four,
five, six, seven.

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So here is the liquid,

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and the ball bearings
are dropped from above.

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When it reaches this line,
I will start my timer.

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And when it crosses
one, two, three, four...

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when it crosses this line,
I will stop my timer.

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And each mark is
about one centimeter apart,

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so this is a journey
of about four centimeters.

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And we will measure
the time that it takes

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to go from here to here.

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And the terminal velocity
is given.

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It's clearly regime one,

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so you see the terminal velocity
right there.

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Now, the time that it will take
is of course this distance--

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let me call it h--
that the ball bearings travel

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divided by
the terminal velocity,

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and that is proportional,
since you're in regime one,

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by 1 over r squared.

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And now I will give you the...

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not the radii
but I will give you

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the diameters
of these ball bearings.

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That's the way they come.

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So we're going
to get a list here

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of the diameters
of the ball bearings,

286
00:16:08 --> 00:16:14
and the diameters is in inches.

287
00:16:11 --> 00:16:17
My smallest one has a diameter
of an eighth of an inch.

288
00:16:14 --> 00:16:20
Then I have 5/32, I have 3/16--

289
00:16:20 --> 00:16:26
all these things come in inches; 
that's one of those things--

290
00:16:22 --> 00:16:28
and I have
one-quarter-inch diameter.

291
00:16:25 --> 00:16:31

292
00:16:28 --> 00:16:34
If I plot here a plot,

293
00:16:30 --> 00:16:36
if I give you here the diameter
in terms of 1/32s of an inch,

294
00:16:35 --> 00:16:41
then this is four, five, six,
and eight-- easy numbers.

295
00:16:41 --> 00:16:47
Clearly if the time
that it takes

296
00:16:43 --> 00:16:49
is proportional
to 1 over r squared,

297
00:16:45 --> 00:16:51
it will also be proportional
to 1 over d squared--

298
00:16:48 --> 00:16:54
of course, that's the same.

299
00:16:52 --> 00:16:58
What I'm going to plot is
not 1 over d squared,

300
00:16:56 --> 00:17:02
but to get some nice numbers

301
00:16:58 --> 00:17:04
I'm going to plot for you
100 over d squared,

302
00:17:03 --> 00:17:09
whereby d is then in these
units, 1/32 of an inch,

303
00:17:06 --> 00:17:12
and that gives me
some nice numbers.

304
00:17:09 --> 00:17:15
Then I get a 6.25 here,
I get a 4.00 here.

305
00:17:14 --> 00:17:20
You can see 100 divided by 25
is exactly four.

306
00:17:19 --> 00:17:25
I get 2.78 here, and
my last number is 1.56.

307
00:17:25 --> 00:17:31
And now I'm going to time it,
and my timing uncertainty

308
00:17:29 --> 00:17:35
is of course dictated
by my reaction time.

309
00:17:33 --> 00:17:39
That should be
at least 0.1 seconds.

310
00:17:37 --> 00:17:43
However, you will see

311
00:17:38 --> 00:17:44
when I reach
the quarter-inch ball bearing

312
00:17:41 --> 00:17:47
that it goes so fast

313
00:17:43 --> 00:17:49
that my error could well be
2/10 of a second.

314
00:17:46 --> 00:17:52
It goes in a flash.

315
00:17:47 --> 00:17:53
So I would allow here
2/10 of a second,

316
00:17:51 --> 00:17:57
and here I really don't know--

317
00:17:53 --> 00:17:59
maybe 1/10,
maybe 2/10 of a second.

318
00:17:55 --> 00:18:01
You may ask me,

319
00:17:57 --> 00:18:03
"Why didn't you give us
the error in the diameter?"

320
00:17:59 --> 00:18:05
which ultimately,
of course, translates

321
00:18:02 --> 00:18:08
into the error in the mass.

322
00:18:03 --> 00:18:09
The reason is that these are
so precise the way you buy them

323
00:18:07 --> 00:18:13
that the uncertainty is
completely negligible

324
00:18:09 --> 00:18:15
compared to the timing error
that I make,

325
00:18:12 --> 00:18:18
so I won't even take that
into account.

326
00:18:16 --> 00:18:22
All right, so now we can start
the demonstration,

327
00:18:22 --> 00:18:28
and I'll have to switch
to this unit here.

328
00:18:27 --> 00:18:33
Here is this container
with the Karo syrup.

329
00:18:36 --> 00:18:42
It is very sticky indeed.

330
00:18:41 --> 00:18:47
You see there are seven marks,
and just for my own convenience,

331
00:18:45 --> 00:18:51
I have put there two black marks
so that I can easily see

332
00:18:51 --> 00:18:57
the moment that I have
to start my timer

333
00:18:54 --> 00:19:00
and the moment that I stop it.

334
00:18:56 --> 00:19:02
There are so many lines, I may
get confused if I don't do that.

335
00:18:59 --> 00:19:05
And we're going
to time this together,

336
00:19:03 --> 00:19:09
and we'll see
how these objects...

337
00:19:07 --> 00:19:13
how long it takes for them
to go through.

338
00:19:09 --> 00:19:15
I will start with my
one-eighth-of-an-inch diameter.

339
00:19:14 --> 00:19:20
I have one here-- tweezer.

340
00:19:16 --> 00:19:22

341
00:19:19 --> 00:19:25
I release it at zero, three...
oh, oh, you can't see that.

342
00:19:24 --> 00:19:30
You should see that:
three, two, one, zero.

343
00:19:28 --> 00:19:34
Look how beautiful
it's working it's way through!

344
00:19:31 --> 00:19:37
You see, it's building up.

345
00:19:32 --> 00:19:38
You see that nice air bubble
at the top?

346
00:19:35 --> 00:19:41
It's going very slowly,

347
00:19:36 --> 00:19:42
but just wait when it has broken
through the surface.

348
00:19:39 --> 00:19:45
There it goes.

349
00:19:41 --> 00:19:47
Now!

350
00:19:43 --> 00:19:49
One centimeter, two centimeter,
three centimeter.

351
00:19:47 --> 00:19:53
Now!

352
00:19:48 --> 00:19:54
Okay, what is that?

353
00:19:50 --> 00:19:56
(students responding )

354
00:19:51 --> 00:19:57
LEWIN:
Nothing.

355
00:19:52 --> 00:19:58
What happened?

356
00:19:54 --> 00:20:00
STUDENT:
5.93.

357
00:19:55 --> 00:20:01
LEWIN:
I didn't see it.

358
00:19:57 --> 00:20:03
I want to do another one.

359
00:19:58 --> 00:20:04

360
00:19:59 --> 00:20:05
Did I... did I clean...
did I erase it?

361
00:20:02 --> 00:20:08
(student answers )

362
00:20:04 --> 00:20:10
LEWIN:
How much was it?

363
00:20:05 --> 00:20:11
STUDENTS:
5.93.

364
00:20:06 --> 00:20:12
LEWIN:
5.93.

365
00:20:08 --> 00:20:14

366
00:20:11 --> 00:20:17
Keep that in mind.

367
00:20:12 --> 00:20:18
It's nice to see whether
they reproduce, actually.

368
00:20:14 --> 00:20:20

369
00:20:17 --> 00:20:23
Okay, there it goes.

370
00:20:19 --> 00:20:25
Now!

371
00:20:20 --> 00:20:26
One centimeter, two centimeter,
three centimeter.

372
00:20:25 --> 00:20:31
Now!

373
00:20:26 --> 00:20:32
5.66-- that shows you
the uncertainty in my timing.

374
00:20:30 --> 00:20:36
So we had a 5.93,
and we have now a 5.66.

375
00:20:36 --> 00:20:42
It's not so bad, 5.9, 5.7--

376
00:20:39 --> 00:20:45
timing error
a tenth of a second.

377
00:20:41 --> 00:20:47
My timing error could be

378
00:20:42 --> 00:20:48
a little bit larger
than a tenth of a second.

379
00:20:44 --> 00:20:50
You don't have very much time.

380
00:20:46 --> 00:20:52
So now we go to the 5/32.

381
00:20:49 --> 00:20:55

382
00:20:53 --> 00:20:59
Okay?

383
00:20:54 --> 00:21:00
5/32.

384
00:20:56 --> 00:21:02

385
00:21:00 --> 00:21:06
It takes some time to break
through the surface.

386
00:21:02 --> 00:21:08
Isn't that funny?

387
00:21:04 --> 00:21:10
Because a thin film has formed
on the surface of the syrup

388
00:21:06 --> 00:21:12
due to its exposure to air.

389
00:21:09 --> 00:21:15
It's wonderful--
that lets us wait patiently.

390
00:21:12 --> 00:21:18
But now it goes--
there it goes.

391
00:21:13 --> 00:21:19
Now!

392
00:21:14 --> 00:21:20
One, two, three.

393
00:21:17 --> 00:21:23
Now!

394
00:21:19 --> 00:21:25
3.80.

395
00:21:21 --> 00:21:27
Is that what you have?

396
00:21:24 --> 00:21:30

397
00:21:27 --> 00:21:33
It's going to be tougher
and tougher for me.

398
00:21:29 --> 00:21:35
3/16 of an inch.

399
00:21:33 --> 00:21:39
It's actually a good thing

400
00:21:34 --> 00:21:40
that it stays for a while
at the surface

401
00:21:36 --> 00:21:42
so that I can get ready.

402
00:21:39 --> 00:21:45

403
00:21:42 --> 00:21:48
Really, that really helps,
doesn't it?

404
00:21:44 --> 00:21:50
If you do this in water,
it goes (whoosh ).

405
00:21:47 --> 00:21:53
You don't even see it.

406
00:21:49 --> 00:21:55

407
00:21:51 --> 00:21:57
Come on... There we go.

408
00:21:53 --> 00:21:59
Now!

409
00:21:54 --> 00:22:00

410
00:21:56 --> 00:22:02
Now!

411
00:21:57 --> 00:22:03
So you see,
that's very hard for me,

412
00:21:58 --> 00:22:04
and so I could easily have
a substantial error.

413
00:22:00 --> 00:22:06
2.69.

414
00:22:02 --> 00:22:08

415
00:22:04 --> 00:22:10
And now we have the
quarter-inch, the real big one.

416
00:22:10 --> 00:22:16

417
00:22:14 --> 00:22:20
I have to do that again.

418
00:22:16 --> 00:22:22
I don't trust this at all.

419
00:22:18 --> 00:22:24
It went through the surface
too fast.

420
00:22:21 --> 00:22:27

421
00:22:27 --> 00:22:33
(timer button clicking )

422
00:22:29 --> 00:22:35
I... can't do it
very accurately.

423
00:22:30 --> 00:22:36
What was the first number,
by the way?

424
00:22:32 --> 00:22:38
(students respond )

425
00:22:33 --> 00:22:39
LEWIN:
One point...?

426
00:22:34 --> 00:22:40
(student responds )

427
00:22:35 --> 00:22:41
LEWIN:
Six eight.

428
00:22:36 --> 00:22:42
And this is 1.40.

429
00:22:38 --> 00:22:44
1.68 and 1.40.

430
00:22:42 --> 00:22:48
So you see I wasn't kidding

431
00:22:44 --> 00:22:50
when I said that my uncertainty
could easily be .2.

432
00:22:47 --> 00:22:53
Now comes the acid test.

433
00:22:50 --> 00:22:56
And the acid test is
that if I'm...

434
00:22:54 --> 00:23:00
if the measurements
were done correctly

435
00:22:55 --> 00:23:01
and if we really work
in that regime,

436
00:22:57 --> 00:23:03
then if I plot 100 divided
by d squared versus t

437
00:23:02 --> 00:23:08
on linear paper, then
it should be a straight line.

438
00:23:06 --> 00:23:12

439
00:23:10 --> 00:23:16
All right.

440
00:23:12 --> 00:23:18
Here I have a plot
which I prepared,

441
00:23:16 --> 00:23:22
and I'm going to put
these numbers in there.

442
00:23:19 --> 00:23:25
So first we're going to get
six point... five point...

443
00:23:24 --> 00:23:30
let's put in 5.8 seconds
for the smallest ball bearing.

444
00:23:31 --> 00:23:37
This is the smallest one.

445
00:23:33 --> 00:23:39
Don't be misled, because this is
100 divided by d squared,

446
00:23:37 --> 00:23:43
so this is
the smallest one-- 5.8.

447
00:23:42 --> 00:23:48
So we are here on this line, and
we are at 5.8-- somewhere here.

448
00:23:51 --> 00:23:57
That's it.

449
00:23:53 --> 00:23:59
Notice the point is lower
than where I expected it,

450
00:23:56 --> 00:24:02
and the reason is the
temperature went up.

451
00:23:59 --> 00:24:05
And if the temperature went up,

452
00:24:00 --> 00:24:06
then the viscosity goes down
and they go faster.

453
00:24:02 --> 00:24:08
But that's okay,
that doesn't worry me.

454
00:24:05 --> 00:24:11
The next one, 3.80.

455
00:24:10 --> 00:24:16
Four, 3.80.

456
00:24:13 --> 00:24:19
Ah!

457
00:24:14 --> 00:24:20
You see that?

458
00:24:16 --> 00:24:22
I predicted that--
straight line.

459
00:24:18 --> 00:24:24

460
00:24:20 --> 00:24:26
Isn't that a straight line?

461
00:24:22 --> 00:24:28
(students respond )

462
00:24:24 --> 00:24:30
LEWIN:
It's not a straight line?

463
00:24:26 --> 00:24:32
(students respond )

464
00:24:27 --> 00:24:33
LEWIN:
What's wrong?

465
00:24:30 --> 00:24:36

466
00:24:31 --> 00:24:37
Okay, we'll put in
a third point.

467
00:24:34 --> 00:24:40
2.69.

468
00:24:35 --> 00:24:41

469
00:24:37 --> 00:24:43
Two point...

470
00:24:41 --> 00:24:47
This is 2.7-- 2.69.

471
00:24:46 --> 00:24:52
I can hardly put in
the error of the timing,

472
00:24:48 --> 00:24:54
because it is not much larger
than the size of my dots.

473
00:24:52 --> 00:24:58
And now we have the last one.

474
00:24:54 --> 00:25:00
One point... let's take
the average-- 1.55.

475
00:25:00 --> 00:25:06
1.55...

476
00:25:03 --> 00:25:09
with an error of
about 2/10 of a second,

477
00:25:07 --> 00:25:13
and this has an error
of about 2/10 of a second.

478
00:25:10 --> 00:25:16
All right, there we go.

479
00:25:12 --> 00:25:18
Now... is this a straight line
or is it not?

480
00:25:19 --> 00:25:25

481
00:25:21 --> 00:25:27
A gorgeous straight line.

482
00:25:23 --> 00:25:29
And so you see

483
00:25:24 --> 00:25:30
you are really working here
in the regime of 1 over...

484
00:25:30 --> 00:25:36
in the regime where the terminal
velocity is proportional

485
00:25:35 --> 00:25:41
to the radius squared.

486
00:25:37 --> 00:25:43
Okay, we'll give you
your lights back.

487
00:25:41 --> 00:25:47
Now comes a question which is
relevant to this experiment,

488
00:25:45 --> 00:25:51
and that is,

489
00:25:46 --> 00:25:52
how long does it take for the
terminal speed to be reached?

490
00:25:52 --> 00:25:58
Well, the object has
a certain mass,

491
00:25:56 --> 00:26:02
so there's
a gravitational force on it,

492
00:26:01 --> 00:26:07
and the gravitational force
equals mg.

493
00:26:05 --> 00:26:11
And then there is a resistive
force, which in the case--

494
00:26:10 --> 00:26:16
because we are operating
in regime one exclusively--

495
00:26:14 --> 00:26:20
that resistive force equals
C1 r v in terms of magnitude,

496
00:26:20 --> 00:26:26
C1 r v, because we deal
with regime one.

497
00:26:25 --> 00:26:31
And so if I call this
the increasing value of y,

498
00:26:29 --> 00:26:35
the second... Newton's
Second Law would give me

499
00:26:32 --> 00:26:38
ma equals mg minus C1 r times v,
and this equals m dv dt,

500
00:26:43 --> 00:26:49
so I have here
a differential equation in v,

501
00:26:47 --> 00:26:53
and that can be solved.

502
00:26:48 --> 00:26:54
And you're going to solve it
on your assignment number four.

503
00:26:52 --> 00:26:58
What you're going to see is that
the speed as a function of time

504
00:26:58 --> 00:27:04
is going to build up
to a maximum value...

505
00:27:02 --> 00:27:08
this is the...

506
00:27:03 --> 00:27:09
to a maximum value, which
is the terminal velocity--

507
00:27:08 --> 00:27:14
or you may want to call it
terminal speed--

508
00:27:11 --> 00:27:17
and it's going to build up
in some fashion

509
00:27:14 --> 00:27:20
and then it's going
to asymptotically approach

510
00:27:18 --> 00:27:24
the terminal speed.

511
00:27:19 --> 00:27:25
And this is what I'm asking you
on your third...

512
00:27:23 --> 00:27:29
on your fourth assignment,
to calculate that.

513
00:27:25 --> 00:27:31
If there were no drag force
at all, I hope you realize

514
00:27:29 --> 00:27:35
that the velocity
would increase linearly,

515
00:27:34 --> 00:27:40
so you would get
something like this.

516
00:27:36 --> 00:27:42
So there's no drag.

517
00:27:40 --> 00:27:46
So the behavior is extremely
different due to the drag.

518
00:27:45 --> 00:27:51
And I calculated already

519
00:27:46 --> 00:27:52
something that is part
of your assignment--

520
00:27:50 --> 00:27:56
how long does it take for
the quarter-inch ball bearing...

521
00:27:54 --> 00:28:00
how long does it take in time

522
00:27:56 --> 00:28:02
to reach a speed which is
about 99% of the terminal speed?

523
00:28:02 --> 00:28:08
And I calculated that,

524
00:28:03 --> 00:28:09
and you will go through
that calculation for yourself.

525
00:28:05 --> 00:28:11
That is only nine milliseconds.

526
00:28:07 --> 00:28:13
In other words, once it has
broken through the surface--

527
00:28:10 --> 00:28:16
that takes a while
because of the thin film--

528
00:28:12 --> 00:28:18
then in nine milliseconds

529
00:28:14 --> 00:28:20
will I already be
at 99% of the terminal speed,

530
00:28:16 --> 00:28:22
and so there was
no problem at all;

531
00:28:19 --> 00:28:25
when I waited for the object
to cross the first mark,

532
00:28:23 --> 00:28:29
it was already clearly going
at the terminal speed.

533
00:28:27 --> 00:28:33
So that was fine.

534
00:28:30 --> 00:28:36
Now I want to turn to air.

535
00:28:33 --> 00:28:39
Air, of course, behaves
in an extremely different way.

536
00:28:37 --> 00:28:43
The principle is the same,

537
00:28:39 --> 00:28:45
but the values for C1 and C2
are vastly different.

538
00:28:43 --> 00:28:49
If we take air
at one atmospheres,

539
00:28:47 --> 00:28:53
and we take it
at room temperature,

540
00:28:50 --> 00:28:56
then C1 is about 3.1 times ten
to the minus four in our units

541
00:28:57 --> 00:29:03
and C2 is about 0.85.

542
00:29:01 --> 00:29:07
This is very close
to the density of air,

543
00:29:03 --> 00:29:09
which is about one kilogram
per cubic meter,

544
00:29:05 --> 00:29:11
which I told you earlier, C2 and
rho are very strongly related.

545
00:29:09 --> 00:29:15
And so the critical speed,
which is C1 divided by C2 by r,

546
00:29:18 --> 00:29:24
is about 3.7 times
ten to the minus four
divided by r meters per second.

547
00:29:25 --> 00:29:31
And that is
about 400 times lower

548
00:29:29 --> 00:29:35
than the critical speed
in syrup, in the Karo syrup

549
00:29:34 --> 00:29:40
for the same value of r.

550
00:29:36 --> 00:29:42
So if I compare
the quarter-inch ball bearing

551
00:29:41 --> 00:29:47
and I drop it in the Karo syrup,

552
00:29:45 --> 00:29:51
then the terminal velocity
in the Karo syrup

553
00:29:48 --> 00:29:54
isway below the critical
velocity of the Karo syrup.

554
00:29:51 --> 00:29:57
The critical velocity
of the Karo syrup

555
00:29:53 --> 00:29:59
would be 100 miles per hour
for a quarter-inch ball bearing.

556
00:29:56 --> 00:30:02
So it's way below.

557
00:29:57 --> 00:30:03
Here, in air,
the critical velocity is

558
00:30:00 --> 00:30:06
something like
11 centimeters per second.

559
00:30:03 --> 00:30:09
This is 11 centimeters
in one second, and we know

560
00:30:06 --> 00:30:12
when you drop a one-quarter-inch
ball bearing in air

561
00:30:09 --> 00:30:15
that the speed is way larger,
and therefore

562
00:30:12 --> 00:30:18
in the case of air, a quarter-
inch ball bearing would have

563
00:30:16 --> 00:30:22
a speed way
above the critical speed,

564
00:30:18 --> 00:30:24
and so you are now
exclusively in regime two.

565
00:30:23 --> 00:30:29
That's the regime two.

566
00:30:26 --> 00:30:32
Almost all spheres that you drop
in air operate exclusively

567
00:30:33 --> 00:30:39
in regime two.

568
00:30:35 --> 00:30:41
Whether it is a raindrop

569
00:30:37 --> 00:30:43
or whether it is a baseball
that you hit,

570
00:30:39 --> 00:30:45
or a golf ball,
or even a beach ball,

571
00:30:42 --> 00:30:48
or you throw a pebble
off a high building,

572
00:30:45 --> 00:30:51
or whether you jump
out of an airplane,

573
00:30:48 --> 00:30:54
with or without a parachute,
makes no difference,

574
00:30:51 --> 00:30:57
you're always dominated
by the pressure term,

575
00:30:54 --> 00:31:00
by the v-square term,
and you always are in a range

576
00:30:59 --> 00:31:05
whereby the terminal speed
is proportional

577
00:31:02 --> 00:31:08
to the square root of the radius

578
00:31:04 --> 00:31:10
for a given density
of the object.

579
00:31:08 --> 00:31:14
If you take a pebble with
a radius of about one centimeter

580
00:31:12 --> 00:31:18
and you throw it
off a high building,

581
00:31:14 --> 00:31:20
it will reach a speed which will
not exceed 75 miles per hour

582
00:31:18 --> 00:31:24
because of the air drag.

583
00:31:20 --> 00:31:26
If you jump out of a plane
and you have no parachute

584
00:31:25 --> 00:31:31
and I make the assumption

585
00:31:28 --> 00:31:34
that your mass is
about 70 kilograms--

586
00:31:32 --> 00:31:38
I want rough numbers--
if I can approximate you

587
00:31:35 --> 00:31:41
by a sphere with a radius
of about 40 centimeters--

588
00:31:38 --> 00:31:44
that's also an approximation,
you're not really like a sphere,

589
00:31:41 --> 00:31:47
but I want to get
some rough numbers--

590
00:31:43 --> 00:31:49
then the terminal velocity is
150 miles per hour.

591
00:31:51 --> 00:31:57
So if you jump out of a plane
and you have no parachute,

592
00:31:56 --> 00:32:02
you will not go much faster
than 150 miles per hour.

593
00:32:00 --> 00:32:06
I just read an article yesterday

594
00:32:02 --> 00:32:08
about sky divers
who jump out of planes,

595
00:32:04 --> 00:32:10
and they want to open
the parachute

596
00:32:05 --> 00:32:11
at the very last possible,

597
00:32:07 --> 00:32:13
and they reach
terminal velocities

598
00:32:09 --> 00:32:15
of 120 miles per hour,
which doesn't surprise me.

599
00:32:11 --> 00:32:17
It's very close to this rough
number that I calculated.

600
00:32:14 --> 00:32:20
Of course, then
they open the parachute

601
00:32:16 --> 00:32:22
and then the air drag
increases enormously

602
00:32:19 --> 00:32:25
and then they slow down
even further.

603
00:32:22 --> 00:32:28

604
00:32:25 --> 00:32:31
I told you a raindrop...
 

605
00:32:27 --> 00:32:33
almost all raindrops operate
in regime two when they fall.

606
00:32:32 --> 00:32:38
So the terminal velocity is
dictated by the v-square term.

607
00:32:37 --> 00:32:43
However, if you make the drops
exceedingly small,

608
00:32:41 --> 00:32:47
there comes a time that
you really enter regime one.

609
00:32:45 --> 00:32:51
And on assignment number four

610
00:32:47 --> 00:32:53
I've asked you to calculate
where that happens,

611
00:32:50 --> 00:32:56
and I can't do it for water

612
00:32:52 --> 00:32:58
because the radius of that
water drop will be so small

613
00:32:56 --> 00:33:02
that it would
evaporate immediately,

614
00:32:58 --> 00:33:04
so I chose oil for that.

615
00:32:59 --> 00:33:05
So I'm asking you in assignment
four, take an oil drop,

616
00:33:02 --> 00:33:08
make it smaller and smaller
and smaller and smaller,

617
00:33:05 --> 00:33:11
and there comes a time that
you begin to enter regime one,

618
00:33:09 --> 00:33:15
and I want you to calculate
where that crossover is

619
00:33:13 --> 00:33:19
between these two domains.

620
00:33:15 --> 00:33:21

621
00:33:16 --> 00:33:22
I have here a ball--

622
00:33:18 --> 00:33:24
you may call it a balloon,
but I call it a ball

623
00:33:22 --> 00:33:28
because there's no helium in it.

624
00:33:25 --> 00:33:31
And this ball-- ooh!--
weighs approximately 34 grams.

625
00:33:31 --> 00:33:37

626
00:33:34 --> 00:33:40
So let me erase some here,
because I want, yeah...

627
00:33:40 --> 00:33:46

628
00:33:42 --> 00:33:48
So we know the mass
and we know the radius.

629
00:33:47 --> 00:33:53
The mass is about 34 grams

630
00:33:51 --> 00:33:57
and the radius is
about 35 centimeters.

631
00:33:54 --> 00:34:00
It's about
70 centimeters across.

632
00:33:57 --> 00:34:03
I can calculate what
the terminal velocity is--

633
00:34:01 --> 00:34:07
a better name would be
terminal speed.

634
00:34:04 --> 00:34:10
I know I'm definitely going
to be in this regime,

635
00:34:08 --> 00:34:14
so I know the mass, I know C2,
which in air is 0.85,

636
00:34:12 --> 00:34:18
I know the radius and I know g,

637
00:34:14 --> 00:34:20
and so I find that I find
about 1.8 meters per second.

638
00:34:20 --> 00:34:26
So if I drop it from
a height of three meters,

639
00:34:23 --> 00:34:29
which I'm going to do,
then you would think

640
00:34:25 --> 00:34:31
that the time that it takes
to hit the floor

641
00:34:28 --> 00:34:34
would be about my three meters

642
00:34:30 --> 00:34:36
divided by 1.8 meters
per second,

643
00:34:34 --> 00:34:40
which is about 1.7 seconds.

644
00:34:36 --> 00:34:42
That's not bad.

645
00:34:38 --> 00:34:44
That's not a bad approximation.

646
00:34:40 --> 00:34:46
However, it will,
of course, take longer,

647
00:34:42 --> 00:34:48
and the reason why
it will take longer

648
00:34:44 --> 00:34:50
is that the terminal velocity,

649
00:34:46 --> 00:34:52
the terminal speed is not
achieved instantaneously.

650
00:34:51 --> 00:34:57
With the ball bearings, it was
within nine milliseconds.

651
00:34:54 --> 00:35:00
I can assure you that here
it will take a lot longer.

652
00:34:58 --> 00:35:04
Now, if you want to calculate
the time that it takes

653
00:35:02 --> 00:35:08
to get close to terminal speed,
that is not an easy task,

654
00:35:07 --> 00:35:13
because you are going to end up

655
00:35:09 --> 00:35:15
with a nasty
differential equation.

656
00:35:12 --> 00:35:18
You're going to get mg.

657
00:35:13 --> 00:35:19
You're going to get
the acceleration

658
00:35:16 --> 00:35:22
which is the result of...

659
00:35:17 --> 00:35:23
Let's go with
the equation we have.

660
00:35:19 --> 00:35:25
You see, we have ma equals mg

661
00:35:26 --> 00:35:32
and then we get minus
the resistive force.

662
00:35:31 --> 00:35:37
And the resistive force has
a term in v

663
00:35:34 --> 00:35:40
and has a term in v squared.

664
00:35:36 --> 00:35:42
You see, v and v squared,

665
00:35:39 --> 00:35:45
and so this cannot be
solved analytically.

666
00:35:41 --> 00:35:47
But I've asked my graduate
student Dave Pooley,

667
00:35:44 --> 00:35:50
who is one of your instructors,

668
00:35:46 --> 00:35:52
to solve this for me
numerically.

669
00:35:49 --> 00:35:55
And I'm going to show you
the results.

670
00:35:52 --> 00:35:58
In fact, he prepared...
he has a nice view graph,

671
00:35:55 --> 00:36:01
and you can see the effect
of time on the ball

672
00:36:01 --> 00:36:07
if you drop it
from three meters.

673
00:36:05 --> 00:36:11

674
00:36:07 --> 00:36:13
Here it is.

675
00:36:09 --> 00:36:15
All the numbers are there.

676
00:36:10 --> 00:36:16
This is on the Web,
so don't copy anything.

677
00:36:13 --> 00:36:19
You have the values for C1
and C2 are given at the top.

678
00:36:16 --> 00:36:22
You may not be able to see them
from your seat,

679
00:36:19 --> 00:36:25
but they are there,
and what you see here

680
00:36:21 --> 00:36:27
is the height above the ground
as a function of time--

681
00:36:27 --> 00:36:33
this is one second,
this is 1½ seconds;

682
00:36:29 --> 00:36:35
this is the three-meter mark.

683
00:36:31 --> 00:36:37
If there were no air drag--

684
00:36:33 --> 00:36:39
remember we dropped an apple
early on from three meters--

685
00:36:37 --> 00:36:43
it will hit the floor
at about 780 milliseconds.

686
00:36:41 --> 00:36:47
However, with the air drag, it
will be about one second later,

687
00:36:45 --> 00:36:51
more like 1.8 seconds.

688
00:36:47 --> 00:36:53
So the 1.7 wasn't bad,
as you see, but if you look here

689
00:36:52 --> 00:36:58
at how the speed builds up
as a function of time,

690
00:36:56 --> 00:37:02
you see it takes about three-,
four-tenths of a second

691
00:37:00 --> 00:37:06
to build up
to that terminal speed,

692
00:37:02 --> 00:37:08
which is the 1.8 meters per
second that you have there.

693
00:37:04 --> 00:37:10
And needless to say, of course,

694
00:37:06 --> 00:37:12
that the acceleration
due to gravity

695
00:37:08 --> 00:37:14
does not remain constant
but goes down very quickly,

696
00:37:11 --> 00:37:17
and when the acceleration
reaches, approaches zero,

697
00:37:15 --> 00:37:21
then you have terminal speed,

698
00:37:17 --> 00:37:23
and then there is no longer
any change in the velocity.

699
00:37:21 --> 00:37:27
So let's try this.

700
00:37:24 --> 00:37:30
We'll give more light.

701
00:37:27 --> 00:37:33
And we're going to throw
this object,

702
00:37:29 --> 00:37:35
and I don't think you're
going to get 1.8 seconds.

703
00:37:32 --> 00:37:38
You may get something that
is larger than 1.8 seconds,

704
00:37:36 --> 00:37:42
and the reasons are
the following.

705
00:37:38 --> 00:37:44
Number one, this is
not a perfect sphere,

706
00:37:41 --> 00:37:47
and only for spheres
do these calculations hold--
 

707
00:37:44 --> 00:37:50
that's number one.

708
00:37:45 --> 00:37:51
Number two,
this thing is very springy,

709
00:37:48 --> 00:37:54
so the moment that I let it go,

710
00:37:50 --> 00:37:56
it probably goes
in some kind of oscillation.

711
00:37:52 --> 00:37:58
That doesn't help either.

712
00:37:53 --> 00:37:59
That will probably
also slow it down,

713
00:37:55 --> 00:38:01
because what causes, of course,
this slowdown in regime two

714
00:37:58 --> 00:38:04
is really turbulence.

715
00:38:00 --> 00:38:06
Turbulence is extremely hard
to understand and predict.

716
00:38:03 --> 00:38:09
And so almost anything I do,
I will only add turbulence,

717
00:38:07 --> 00:38:13
and therefore I predict

718
00:38:09 --> 00:38:15
that the time that it will take
from three meters

719
00:38:12 --> 00:38:18
will be probably
larger than 1.8 seconds,

720
00:38:14 --> 00:38:20
but it will be
substantially larger

721
00:38:16 --> 00:38:22
than the 780 milliseconds,

722
00:38:18 --> 00:38:24
which is what you would have
seen if you drop an apple.

723
00:38:23 --> 00:38:29
So let's see how close we are.

724
00:38:25 --> 00:38:31

725
00:38:30 --> 00:38:36
So, turn on this timer.

726
00:38:32 --> 00:38:38
Make sure I zero it-- I did.

727
00:38:36 --> 00:38:42
And...

728
00:38:37 --> 00:38:43

729
00:38:45 --> 00:38:51
It's not so easy
to release it, by the way,

730
00:38:48 --> 00:38:54
and start the timer
at the same moment.

731
00:38:50 --> 00:38:56
And it's not even so easy for me
to see when it hits the ground,

732
00:38:53 --> 00:38:59
so there's a huge uncertainty
in this experiment.

733
00:38:58 --> 00:39:04
Okay.

734
00:39:00 --> 00:39:06

735
00:39:02 --> 00:39:08
Three, two, one, zero.

736
00:39:03 --> 00:39:09

737
00:39:07 --> 00:39:13
What do we see?

738
00:39:08 --> 00:39:14

739
00:39:11 --> 00:39:17
Did I see something?

740
00:39:12 --> 00:39:18
2.0-- that's not bad.

741
00:39:15 --> 00:39:21
See? The prediction was 1.8; 
you get 2.0.

742
00:39:18 --> 00:39:24
That's not bad, so this takes
the air drag into account,

743
00:39:21 --> 00:39:27
and it is not even
an approximation.

744
00:39:23 --> 00:39:29
It uses the entire term linear
in v as well as in v square.

745
00:39:30 --> 00:39:36
But it's almost exclusively
dominated by the v-square term.

746
00:39:35 --> 00:39:41

747
00:39:37 --> 00:39:43
I also asked Dave
to show me what happens

748
00:39:40 --> 00:39:46
when I throw a pebble
off the Empire State Building.

749
00:39:44 --> 00:39:50
And the pebble that we chose had
a radius of one centimeter--

750
00:39:48 --> 00:39:54
it's the kind of pebble
that all of us could find--

751
00:39:51 --> 00:39:57
I know roughly
the density of pebbles,

752
00:39:53 --> 00:39:59
and when we throw it
off the Empire State Building,

753
00:39:55 --> 00:40:01
we reach a terminal speed
of about 75 miles per hour.

754
00:40:00 --> 00:40:06
Without the air drag, we would
have reached 225 miles per hour.

755
00:40:05 --> 00:40:11
So I want to show you that, too.

756
00:40:08 --> 00:40:14

757
00:40:13 --> 00:40:19
So now you see this
Empire State Building,

758
00:40:16 --> 00:40:22
which has a height
of 475 meters,

759
00:40:19 --> 00:40:25
so that's where you start,
at t0; this is one second, 

760
00:40:22 --> 00:40:28
five seconds... ten seconds,
five seconds, 15 seconds,

761
00:40:25 --> 00:40:31
and if there had been
no air drag,

762
00:40:28 --> 00:40:34
it would hit the ground
a little less than ten seconds,

763
00:40:31 --> 00:40:37
but now it will hit the ground
more like 16, 17 seconds.

764
00:40:35 --> 00:40:41
And you see that
the terminal speed builds up

765
00:40:38 --> 00:40:44
in about five, six seconds.

766
00:40:39 --> 00:40:45
It's very close
to the final value,

767
00:40:41 --> 00:40:47
and if there had been
no air drag,

768
00:40:44 --> 00:40:50
then the speed at which
it would hit the ground

769
00:40:46 --> 00:40:52
would, of course, grow linearly,

770
00:40:48 --> 00:40:54
and when it hits the ground,
it would be somewhere here,

771
00:40:52 --> 00:40:58
which is 225 miles per hour.

772
00:40:53 --> 00:40:59
So you see

773
00:40:54 --> 00:41:00
that it's even a pebble
you wouldn't expect to be...

774
00:40:57 --> 00:41:03
to have a very large effect
on air drag,

775
00:40:59 --> 00:41:05
it is huge, provided that you
throw it from a high building.

776
00:41:04 --> 00:41:10
Now, you may remember

777
00:41:06 --> 00:41:12
that we dropped an apple
from three meters

778
00:41:10 --> 00:41:16
and that we calculated
the gravitational acceleration

779
00:41:16 --> 00:41:22
given the time
that it takes to fall.

780
00:41:17 --> 00:41:23
That was one of your...

781
00:41:19 --> 00:41:25
one of the things you did
in your assignment.

782
00:41:21 --> 00:41:27
We had 781 milliseconds,
I think.

783
00:41:23 --> 00:41:29
And out of that you can
calculate g, right?

784
00:41:27 --> 00:41:33
Because you know that three
meters is one-half g t squared,

785
00:41:30 --> 00:41:36
so I give you the three,
with an uncertainty,

786
00:41:33 --> 00:41:39
I give you the time--

787
00:41:34 --> 00:41:40
781 milliseconds with an
uncertainty of two milliseconds,

788
00:41:37 --> 00:41:43
which we allowed.

789
00:41:39 --> 00:41:45
Out pops g.

790
00:41:40 --> 00:41:46
So I asked Dave,

791
00:41:41 --> 00:41:47
"What is the effect of air drag
on this apple?

792
00:41:45 --> 00:41:51
Was it a responsible thing
for us to ignore that?"

793
00:41:50 --> 00:41:56
The apple has
a mass of 134 grams.

794
00:41:56 --> 00:42:02
It's easy to weigh, of course.

795
00:41:58 --> 00:42:04
So this was our apple during our
first lecture-- m is 134 grams.

796
00:42:05 --> 00:42:11
It's almost a sphere, really--
not quite but almost a sphere--

797
00:42:09 --> 00:42:15
and the radius is
about three centimeters.

798
00:42:12 --> 00:42:18
And that leads
to a terminal velocity

799
00:42:15 --> 00:42:21
which you can calculate
if you want to

800
00:42:18 --> 00:42:24
using the v-square term,

801
00:42:21 --> 00:42:27
but I was not interested
in that.

802
00:42:23 --> 00:42:29
I wanted to know
how many milliseconds

803
00:42:27 --> 00:42:33
is the touchdown delayed
because of the air drag.

804
00:42:31 --> 00:42:37
And Dave made the calculations,

805
00:42:33 --> 00:42:39
and he found that that is two
milliseconds from three meters.

806
00:42:37 --> 00:42:43
From 1½ meters
it's almost nothing,

807
00:42:39 --> 00:42:45
and the reason
why it's almost nothing

808
00:42:41 --> 00:42:47
from 1½ meters-- you see,

809
00:42:43 --> 00:42:49
when you throw an apple in air,
it's really in regime two,

810
00:42:48 --> 00:42:54
so you're really dominated
by the speed squared,

811
00:42:52 --> 00:42:58
and the first 1½ meters

812
00:42:54 --> 00:43:00
it doesn't get
at very high speed yet.

813
00:42:56 --> 00:43:02
The speed grows linearly,
and so it is the last portion

814
00:42:59 --> 00:43:05
where you really get hit by the
air drag, by the v-square term.

815
00:43:04 --> 00:43:10
Two milliseconds from three
meters, so if h is three meters,

816
00:43:10 --> 00:43:16
there is a two-millisecond...
let's call it delay.

817
00:43:15 --> 00:43:21
So we were on the hairy edge
of being lucky and unlucky.

818
00:43:19 --> 00:43:25
If you really want

819
00:43:20 --> 00:43:26
to recalculate the gravitational
acceleration using our data,

820
00:43:24 --> 00:43:30
you should really subtract the
two milliseconds from the time.

821
00:43:28 --> 00:43:34
On the other hand,

822
00:43:29 --> 00:43:35
since we allowed a two-
millisecond uncertainty,

823
00:43:32 --> 00:43:38
we really weren't too far off.

824
00:43:35 --> 00:43:41

825
00:43:37 --> 00:43:43
Now comes my last part,

826
00:43:39 --> 00:43:45
and that is, how does air drag
influence trajectories?

827
00:43:45 --> 00:43:51
And that is also part
of your assignment,

828
00:43:48 --> 00:43:54
but I'm going to help you
a little bit with that.

829
00:43:51 --> 00:43:57
In your assignment number four,
I'm asking you

830
00:43:56 --> 00:44:02
to evaluate quantitatively the
motion of an object in liquid,

831
00:44:05 --> 00:44:11
but I give that object

832
00:44:07 --> 00:44:13
an initial speed
in the x direction,

833
00:44:11 --> 00:44:17
and then there's
the liquid below.

834
00:44:13 --> 00:44:19
Then gravity is there, and
there is that initial speed.

835
00:44:16 --> 00:44:22
If there were no drag,

836
00:44:17 --> 00:44:23
then this, of course,
would be a parabola,

837
00:44:20 --> 00:44:26
and the horizontal velocity
would always be the same.

838
00:44:24 --> 00:44:30
There would never be any change.

839
00:44:26 --> 00:44:32
But that's not the case now.

840
00:44:28 --> 00:44:34
Because of the resistive forces,

841
00:44:31 --> 00:44:37
because of the drag
by the liquid,

842
00:44:34 --> 00:44:40
the object is going to get
a velocity in this direction,

843
00:44:38 --> 00:44:44
so there's going to be

844
00:44:39 --> 00:44:45
a component of the
resistive force opposing it.

845
00:44:43 --> 00:44:49
It has a speed
in this direction,

846
00:44:45 --> 00:44:51
so there's also going to be a
component of the resistive force

847
00:44:49 --> 00:44:55
in this direction.

848
00:44:51 --> 00:44:57
And that will
decrease the speed--

849
00:44:53 --> 00:44:59
this component
in the x direction.

850
00:44:55 --> 00:45:01
And so you can already see

851
00:44:57 --> 00:45:03
that the curve that you're going
to see is a very different one.

852
00:45:01 --> 00:45:07
It's going to look
something like this.

853
00:45:06 --> 00:45:12
And then ultimately, there is
nothing left of this component,

854
00:45:10 --> 00:45:16
and ultimately,
when you go vertically down,

855
00:45:13 --> 00:45:19
you have the terminal speed
that you can find

856
00:45:17 --> 00:45:23
from dropping an object just
into the liquid vertically.

857
00:45:23 --> 00:45:29
So that's something
you're going to deal with

858
00:45:25 --> 00:45:31
in your assignment number four,

859
00:45:27 --> 00:45:33
and this is exclusively done
in regime one,

860
00:45:30 --> 00:45:36
because we have
an object with liquid,

861
00:45:33 --> 00:45:39
and with liquid, you almost
always work with regime one.

862
00:45:39 --> 00:45:45
Suppose I take a tennis ball

863
00:45:42 --> 00:45:48
and I throw up
a tennis ball in 26.100.

864
00:45:45 --> 00:45:51
There is air drag
on that tennis ball.

865
00:45:48 --> 00:45:54
In the absence of any air drag,
I would get a nice parabola

866
00:45:53 --> 00:45:59
which will be
completely symmetric.

867
00:45:56 --> 00:46:02
I throw it up with
a certain initial speed, v--

868
00:46:03 --> 00:46:09
call it v0, I don't care--

869
00:46:05 --> 00:46:11
and the horizontal component
would never change.

870
00:46:10 --> 00:46:16
This would be v0x;
it would always be the same.

871
00:46:12 --> 00:46:18
But now with air drag,
you're going to see

872
00:46:16 --> 00:46:22
that there's going to be
a force, an air drag force

873
00:46:20 --> 00:46:26
in the y direction.

874
00:46:22 --> 00:46:28
If the object goes up in this
direction, then there will be

875
00:46:26 --> 00:46:32
a resistive force component
in the y direction,

876
00:46:29 --> 00:46:35
and since it has a speed
in this direction,

877
00:46:32 --> 00:46:38
there will also be a resistive
force in the x direction,

878
00:46:37 --> 00:46:43
so this speed is going to be
eaten up in the same way

879
00:46:41 --> 00:46:47
that this component was going
to be eaten up.

880
00:46:44 --> 00:46:50
This component is going
to suffer.

881
00:46:47 --> 00:46:53
It will not stay
constant throughout,

882
00:46:49 --> 00:46:55
and as a result of that,

883
00:46:51 --> 00:46:57
you're going to get a trajectory
that looks more like this.

884
00:46:55 --> 00:47:01

885
00:46:57 --> 00:47:03
It's asymmetric.

886
00:46:59 --> 00:47:05
Clearly you don't reach

887
00:47:00 --> 00:47:06
the highest point that you would
have reached without air drag,

888
00:47:05 --> 00:47:11
for the reason
that this resistive force
in the y direction

889
00:47:09 --> 00:47:15
will not allow it to go
as high-- that's obvious.

890
00:47:12 --> 00:47:18
You don't go as far
as you would without air drag,

891
00:47:15 --> 00:47:21
for obvious reasons
that this resistive force

892
00:47:18 --> 00:47:24
is going to kill this speed,

893
00:47:20 --> 00:47:26
but you also will get asymmetry
in the curvature,

894
00:47:24 --> 00:47:30
and I want you to see that.

895
00:47:26 --> 00:47:32
I call this point O, this point
P, and let's call this point S.

896
00:47:31 --> 00:47:37
So what I will do is
I will throw up a tennis ball

897
00:47:34 --> 00:47:40
and then I will throw up
a Styrofoam ball,

898
00:47:37 --> 00:47:43
and the Styrofoam ball has
very closely the same radius

899
00:47:41 --> 00:47:47
as the tennis ball.

900
00:47:43 --> 00:47:49
That means the resistive force
is the same on both,

901
00:47:47 --> 00:47:53
because the resistive force
is only dictated

902
00:47:50 --> 00:47:56
by r squared and by v--
r squared v squared, remember?

903
00:47:56 --> 00:48:02
However, this has a way
larger mass than this one,

904
00:48:00 --> 00:48:06
and even though the resistive
forces will be closely the same

905
00:48:03 --> 00:48:09
if I throw them up
with the same initial speed,

906
00:48:06 --> 00:48:12
it has a way larger effect
on a smaller mass

907
00:48:08 --> 00:48:14
than on a larger mass.

908
00:48:10 --> 00:48:16
F = ma, right?

909
00:48:11 --> 00:48:17
So on a very large mass,

910
00:48:13 --> 00:48:19
the resistive force will have
a much lower effect

911
00:48:16 --> 00:48:22
than on the smaller mass,

912
00:48:18 --> 00:48:24
even though the resistive forces
are about the same.

913
00:48:21 --> 00:48:27
So, try to see

914
00:48:23 --> 00:48:29
that the tennis ball is very
close to an ideal parabola.

915
00:48:27 --> 00:48:33
You will not even see
any effect of asymmetry.

916
00:48:30 --> 00:48:36
It will not be the case
for the Styrofoam, though.

917
00:48:33 --> 00:48:39
So, look
at the tennis ball first.

918
00:48:36 --> 00:48:42

919
00:48:38 --> 00:48:44
I should really do it here.

920
00:48:41 --> 00:48:47
Did it look
more or less symmetric?

921
00:48:43 --> 00:48:49
Okay, now I'll try this one.

922
00:48:45 --> 00:48:51

923
00:48:46 --> 00:48:52
Did it look asymmetric?

924
00:48:48 --> 00:48:54
Could you see it?

925
00:48:50 --> 00:48:56
Are you just saying yes,
or you really saw it?

926
00:48:53 --> 00:48:59
Let me do it once more; 
I can throw it back.

927
00:48:55 --> 00:49:01
So now it should curve like this

928
00:48:58 --> 00:49:04
and then sort of come down
like that.

929
00:49:00 --> 00:49:06
You ready?

930
00:49:01 --> 00:49:07

931
00:49:04 --> 00:49:10
You see the asymmetry?

932
00:49:06 --> 00:49:12
Okay, now comes
my last question.

933
00:49:08 --> 00:49:14
I'm going to ask you
the following,

934
00:49:09 --> 00:49:15
and there's
a unique answer to that.

935
00:49:12 --> 00:49:18
I want you to think about it
and I want you to be able

936
00:49:15 --> 00:49:21
to give an answer
with total, 100% confidence.

937
00:49:18 --> 00:49:24
When this object goes
from O to P,

938
00:49:21 --> 00:49:27
that takes
a certain amount of time.

939
00:49:23 --> 00:49:29
When it goes from P to S,
back to the ground,

940
00:49:27 --> 00:49:33
that takes also
a certain amount of time.

941
00:49:30 --> 00:49:36
Is this time
the same as this time?

942
00:49:34 --> 00:49:40
Or is this time longer than this
time, or is this time shorter?

943
00:49:39 --> 00:49:45
Think about it.

944
00:49:41 --> 00:49:47
See you Friday.

945
00:49:43 --> 00:49:49

946
00:49:47 --> 00:49:53.000