The exam on Wednesday will cover our first five lectures and the
first two homework assignments. So I list here the topics
the way we discussed them. Of course, it is not possible
to discuss all of them today but I will make a selection. I recall
that we discussed scaling and we used the interesting
example of Galileo Galilei-- an animal,
and the animal has legs. And we defined the overall size
of the animal as yea big-- we called that "s." And then we said,
well, there is here the femur and the femur has length l
and thickness d. It was completely reasonable
to say well... that l will have
to be proportional to S. If an animal is
ten times larger than another its legs will be typically
ten times longer. Since the mass of the animal must be proportional to its size
to the power three it will also be proportional
to the length of the femur to the power three, and then came in
this key argument-- namely, you don't want
the bones to be crushed. Which is called "yielding"
in physics. If I take a piece of concrete,
a block of concrete, and I put too much pressure
on it, it starts to crumble. And that's what Galileo Galilei
may have had in mind. And in order to protect animals who get bigger
and bigger and bigger against this crushing, we argued-- and I will not go through that argument
now anymore-- that the mass will have to be
proportional to d squared, which is the cross-section
of the femur. And so, you see immediately that d squared has to be
proportional to l to the third so d must be proportional
to the length of the femur to the power one and a half. So this would mean that if you
compare an elephant with a mouse the elephant's overall size is about 100 times larger
than a mouse. You would expect the femur
to be about 100 times larger, which is true. But you would then expect
the femur to be about 1,000 times thicker and that turns out to be
not true, as we have seen. In fact, the femur
of the elephant is only 100 times thicker,
so it scales just as the size. So that was a scaling argument. And let's now talk
about dot products. If I look there... I scan it a little bit
in a random way over my topics, so let's now talk
about dot products. I have a vector A... Ax times x roof, which is the
unit vector in the x direction, plus Ay y roof plus Az Z roof. So these are
the three unit vectors in the x, y and z direction. And these are the x components,
y and the z component of the vector A. I have another vector, B. B of x, x roof, B of y, y roof, B of z, z roof. Now, the dot product... A dot B-- also called
the scalar product-- is the same as B dot A and it is defined as Ax Bx
plus Ay By plus Az Bz. And it's a number. It is a scalar,
it is a simple number. And so this number can be larger
than zero-- it can be positive-- it can be equal to zero, it can
also be smaller than zero. They're just dumb numbers. There is another way
that you can define... You can call this method number
one, if you prefer that. There is another way that
you can find the dot product. It gives you exactly
the same result. If you have a vector A
and you have the vector B and the angle
between them is theta, then you can project B on A-- or A on B, for that matter,
it makes no difference-- and that projection...
the length of this projection is then, of course,
B cosine theta. And so A dot B... and that is exactly the same. You may want to go
through a proof of that. It is the length of A
times the length of B times cosine of theta. And that will give you
precisely the same result. What is interesting about this
formulation, which this lacks, that you can immediately see that if the two are
at 90-degree angles or 270 degrees, for that matter,
then the dot product is zero. So that's an insight
that you get through this one which you lack
through that other method. Let us take a down-to-earth
example of a dot product. Suppose A equals 3x and B equals 2x plus 2y, and I am asking you,
what is the dot product? Well, you could use method
number one, which, in this case is by far the fastest,
believe me. Ax is 3 and Bx is 2,
so that gives me a 6. There is no Ay, there is no Az,
so that's the answer. It's just 6--
that's the dot product. You could have done it that way. It's a little bit
more complicated but I certainly want
to show you that it works. If this is the x direction
and this is the y direction-- we don't have to look
into the z direction because there is
no z component-- then this would be vector A and this point would be at 3. B... this would be 2,
and this would be 2 and so this would be
the vector B. And it's immediately clear now
that this angle... 45 degrees. That follows
from the 2 and the 2. So if we now apply
method number two, A dot B. First the length of A, that's 3, times the length of B,
that is 2, times the square root of 2-- this is 2, this is 2,
this is 2... square root 2-- times the cosine of 45 degrees,
which is one-half square root 2, and the answer is 6. Notice that
this square root of 2 and this square root of 2
equal just 2, and you get 6. You get the same answer,
of course. But it would be
a dumb thing to do it since it can be done
so much easier. On cross products... I don't want to go
through the formalism of cross products the way we
did that with the determinant. I just want to remind you that if you have a cross product
of two vectors, that is minus B cross A, and that the magnitude of C is the length of A
times the length of B times the sine
of the angle between them. The vector C, the dot product...
the cross product is always perpendicular to
both A and perpendicular to B. In other words,
it's perpendicular to the plane of the two vectors. Now, if it's perpendicular
to the plane, then in that case, it's
perpendicular to the blackboard. You have two choices: it's either coming
at you perpendicular or it's coming right
straight into the blackboard. And now everyone has
his own way of doing it. I taught you what's called
"the right-hand corkscrew" rule. You take the first one that is
mentioned-- in this case, A-- and you rotate it
over the shortest angle to B. When you do that,
you rotate your corkscrew-- seen from where you're sitting--
counterclockwise. Then the corkscrew comes to you. And so the direction
of the vector is such that you will see
the tip of the vector so it's coming
straight out of the blackboard. And so that gives you,
then, the direction. Now I will give you
the position x of an object as a function of time and then we're going to ask
ourselves a lot of questions about velocities, accelerations, sort of everything
you can think of, everything we have covered--
speeds... And I will cover here
four seconds of time. So this is the time axis
in seconds and we will cover four seconds. So let this be
one, two, three, four. And let the object be
at position plus six. This is my x-axis, this is where
the object is actually moving, and this is three,
and here is minus three and this, let's say,
is in meters. Let's make a little grid so that's easier for me
to put in the curve. All right, so now comes
x as a function of t. The time t equals t seconds. And this part is a parabola and this parabola
here is horizontal. It's important,
you have to know that, so this is a parabola and this, here, is horizontal. So the object goes
from plus six to three, then it goes to minus three, then it stays there
for one second and then it goes back
in one second to plus six. It's a one-dimensional problem. The motion is only in the
x-axis, along the x direction. Let's analyze all these
different seconds that occur. Let's first take the first
second, during the first second. Since this is a parabola, you know that
the acceleration is constant so I hope
you will conclude immediately that a must be a constant. If a is a constant, the
position x as a function of time should change as follows: x zero plus v zero t
plus one-half a t squared. I expect you to know
this equation. Very often do I give you
equations at the exam and that may well happen during
the second and the third exam, but it will not be the case
this time. The equations are
all very fundamental and you have to make them
part of your world. So this is an equation
that you will have to remember. All right,
what is the velocity here? The velocity starts out
to be zero and the velocity here is
not zero anymore. If I look at time t equals one, then I have here x zero or six. It starts out with
velocity zero-- that's a given. And I get plus one-half
times a t squared but this is only one second, and so when it is at three, I have six plus one-half a
times one squared, and so you find that a equals minus six meters
per second squared. So during this first second the acceleration is minus six
meters per second squared. And the velocity, v,
as a function of time, is the derivative of this one,
is v zero plus at. V zero was zero, and so
that is minus six times t. So the velocity is changing
in a linear fashion. What do I know about
the end of the first second? Well, I can say that x is three. What do I know
about the velocity? Well, the velocity is minus six. What do I know about a? I don't... I don't know about a. It's true that
during this first second a is minus six meters
per second squared, but it changes abruptly
at this point so it's ill-defined
at this point. In fact, it's
actually nonphysical. So I really don't know
exactly at the end what the acceleration is. Let's now go
to the second second and let's see
what happens there. The second second. And first let's look during, and then we'll look
at the situation at the end. During the second second,
it is clear-- since this is a straight line-- that the velocity
remains constant and it remains
minus six meters per second. That is exactly what it was
at this point at the end. You can see it go six meters-- from plus three
to minus three-- in one second so the velocity is
minus six meters per second. The acceleration is
therefore zero. You see that the acceleration
changes abruptly from minus six meters
per second squared to zero so I can't tell you what it is
exactly at this moment in time. So that's the situation
during the second second. And what is the situation at the end of the second
section... second second? At the end, I know
that x equals minus three. What is the velocity? I don't know, because
it changes abruptly here from minus six to zero, so I don't know exactly
what it is at that point. It's a nonphysical thing,
it's a very abrupt change. And the acceleration, yeah,
that's also a very tricky thing, because if the velocity is minus six on this side
of the two seconds and here becomes zero, and if
that happens in a split second, there must be
a huge acceleration just at that point
which is nonphysical. So I would also put a question
mark at the a... I don't know what the a is. So we'll go to the
third second... this part. Let's first look during
the third second. Well, the object
isn't going anywhere, it's just sitting there. x remains minus three and the velocity is zero
and a is zero-- We can agree on that. What is the situation
at the end of the third second? That means t equals three. Well, all I know is
that x is minus three. That's nonnegotiable. What the velocity is,
I don't know, because it's changing abruptly
from zero to a positive value. So that's ill-defined and the same is true
for the acceleration. There is a sudden change
in velocity. That means there must be
a huge acceleration. It's unknown, ill-defined because this curve is,
of course, not very physical. Let's now look
at the last second. This is the fourth second. First, during. Well, it's going
from minus three to plus six and it's a straight line,
so the velocity is constant. If the velocity is constant then you can immediately
conclude that a is zero-- there is no acceleration-- and it goes nine meters
in a time span of one second. But it's now plus--
plus nine meters per second. So the object first went
from positive values of x to zero
and to negative values for x. During all that time,
the velocity was negative by our sign convention and now the velocity,
it goes back to plus six. The velocity becomes
plus nine meters per second. What is the story at the end
of the fourth second? Well, all I can say is
that x equals plus six. I don't know much more. I don't know
what the velocity is. Neither do I know
what the acceleration is. The plot stops there, anyhow. Now, I would think that it is reasonable
to ask the following question: What is the average velocity,
for instance, between time zero and time four? Average velocity. We define average velocity as the position
at time four seconds minus the position at time zero,
divided by four. That is our definition. At zero, it is at plus six,
at four, it is at plus six. So the upstairs is zero, so the average velocity during
this four-second trip is zero. You may not like that, it may go
against your intuition. Of course!
I couldn't agree more with you but that's the way
we define velocity. Speed is defined differently. Speed is the magnitude
of the velocity vector and the speed, therefore,
always has a positive value. And I will show you now what is the average speed
between time zero and four. That is the distance that it has traveled
in these four seconds. Well, let's first go
through the first second. It goes from plus six
to plus three. So it already travels
three meters. Then in the second second it goes from plus three
to minus three so it travels
another six meters. And then in the third second it's lazy,
it doesn't do anything, so the distance traveled
is zero. And then in the last one second it gets very active
and it travels nine meters. Notice you only see
plus signs here. There are no minus signs,
it would make no sense. And this occurs in four seconds, so that is 4.5 meters
per second. So the average speed is
4.5 meters per second, but the average velocity
is zero. We could now make a plot of the
velocity as a function of time. Let me put here the 4.5. I just have enough room here
to make the velocity plot as a function of time. I'll make a new one. This is my time axis,
and this is the velocity. This is zero. One second, two seconds,
three seconds, four seconds. And this velocity is
in meters per second. I go up here to plus ten and here is minus five,
here is minus six. So, what do I do now? I know that the velocity during
the first second is minus six t so it's linear. And so during the first second this is the velocity
as a function of time. It starts at zero,
you can see that, and when it is here it has a velocity
of minus six meters per second. During the second second it remains
minus six meters per second. So during the second second,
the velocity is not changing. It stays there. During the third second the velocity jumps
all of a sudden to zero-- you see
how nonphysical that is. And so all of a sudden,
during the third second it becomes zero. So there has to be somehow a connection, of course,
between the two to make this physical. So in a very small amount
of time that will have to occur. That's why you get a huge
acceleration here at that point. Of course, you also get an
acceleration here at this point, because there's also
a change in velocity. And then,
during the fourth second, the velocity is
plus nine meters per second, and so we jump up. Let's make this plus nine. And so we have here
during the last second... And again, this is nonphysical, so there has to be
somehow a transition. And so here you see the velocity
as a function of time. Now comes
an interesting question. Is it possible,
if I gave you this-- so this is a given,
you can't see that-- could you convert
this back to that? And the answer is yes, provided that I tell you
what the position is at t zero. At t equals zero,
x equals plus six and that is sufficient for you to use this information
and to reconstruct that. It's an interesting thing to do,
and if you feel like it I would say, give it a shot. All right, so far, about speeds and average velocities
and accelerations. Let's now go to trajectories,
three-dimensional trajectories. Trajectories, thank goodness, are almost
never three-dimensional. They're always two-dimensional, because the trajectory itself
is in a vertical plane and so we normally... When we throw up an object
in a gravitational field, you have the trajectory
in a plane. So we're going to have
one trajectory. Let this be the x direction
and let this be the y direction. Increasing values of y,
increasing values of x. I take an object and I throw it up with
an initial velocity v zero. And what is the object
going to do? You're going to get a parabola
under the influence of gravity and it comes down here again. And where we have
this kind of a problem we will decompose it
in two one-dimensional motions, one in the x direction
and one in the y direction. We already decompose
right away the velocity at time t equals zero into a component
which I call v zero x and that, of course, is v zero
times the cosine of alpha if the angle is alpha. And the velocity
in the y direction at time t equals zero-- I will call that
v zero in the y direction and that is v zero
times the sine of alpha. And now I have to know how the object moves in the x
direction as a function of time and how it behaves as a function
of time in the y direction. So here come the equations
for the x direction. x as a function of time equals
x zero plus v zero x times t. That's all--
there is no acceleration. The velocity in the x direction
as a function of time is simply v zero x--
it never changes. So that's the x direction. Now we take the y direction. y as a function of time equals y zero plus v zero y t
plus one-half at squared. My g value that I'm going
to use is always positive-- either 9.8 meters
per second squared or sometimes I make it
easy to use it, 10-- but mine is always positive. And since in this case I have chosen this to be
the increasing value of y, that's the only reason why I would now have to put in
minus one-half gt squared-- not, as some of you think, because the acceleration
is down. That's not a reason. Because I could have called
this direction increasing y. Then it would have been
plus one-half gt squared. So the consequence
of my choosing this the direction
in which y increases... Therefore, the
plus one-half at squared that you would normally see, I'm going to replace that now
by minus one-half gt squared. Then the velocity in the y
direction as a function of time would be this derivative,
that is, v zero y minus gt and the acceleration
equals minus g. So these are the three equations that govern the motion
in the y direction. This only holds
if there is no air drag, no friction of any kind. That is very unrealistic
if we are near Earth, but when we are
far away from Earth, as we were with the KC-135-- which was flying at an altitude
of about 30,000 feet-- that, of course, is
a little bit more realistic. And therefore the example that I have picked
to throw up an object is the one whereby the KC-135, at an altitude somewhere
around 25,000 or 30,000 feet, comes in at a speed of 425 miles
per hour, turns the engines off and then, for
the remaining whatever it was-- about 30 seconds-- everyone, including
the airplane, has no weight. That's the case that I now want to work out
quantitatively with you. In the case of the KC-135, we will take an angle for alpha
of 45 degrees and we will take v zero, which
was about 425 miles per hour. You may remember that
from that lecture. 425 miles per hour translates into about 189 meters
per second. And so that means that the
velocity v zero y and v zero x are both the same
because of the 45-degree angle, and that is, of course, the 189
divided by the square root of 2. And that is
about 133 meters per second. Both are positive--
keep that in mind because this is what I call
the increasing value for y and this is
the increasing value of x. They are both positive values. Signs do matter. But this is a given now. And now comes the first question
that I could ask you on an exam. When is the plane at its highest
point of its trajectory and how high is it
above the point where it started
when it turned the engines off when it went into free fall? So when is it here
and what is this distance? Well, when is it there? That's when the velocity
in the y direction becomes zero. It is positive. It gets smaller and smaller because of
the gravitational acceleration, comes to a halt
and becomes zero. So I ask this equation,
when are you zero? This is the one I pick and so I say, zero equals
plus 133 minus 10 times t. You may think that the
gravitational acceleration at an altitude of 30,000 feet
could be substantially less than the canonical number of 10. It is a little less because you're a little bit
further away from the Earth, but it's only 0.3 percent less,
and so we'll just accept the 10. It's easy to work with. And so when is it
at the highest point? That is when t
equals 13.3 seconds. So that's about how long it
takes to get there. When I gave
the lecture last time, I said it's about 15 seconds, because I made the numbers...
I rounded them off. It's about 30.3 seconds. And what is this distance h now? Ah! Now I have to go
to this equation. I say h equals zero, because
I'm going to define the point where the plane
starts its trajectory. I call that y zero zero,
I'm free to do that. h equals zero plus 133--
that is the speed-- times 13.3 seconds minus
one-half times g-- that is 5-- times 13.3 squared. That is what h must be. And that turns out to be
about 885 meters. I think I told you last time
it's about 900, close enough. So we know now
how long it takes to reach p and we know
what the vertical distance is. And the whole trip
back to this starting point-- if we call this sort
of a starting point, starting altitude-- this whole trip will take
twice the amount of time. To get back to this point
when the engines are restarted is about 26.5, 27 seconds. How far has the plane traveled,
then, in horizontal direction? Well, now I go
back to this equation. So now I say, aha! x then,
when it is back at this point, must be x zero-- which I
conveniently choose zero-- plus 133 meters per second, which is the velocity in the x
direction, which never changes. When the plane is here,
that velocity in the x direction is the same 133 meters
per second as it was here, which, by the way,
is about 300 miles per hour. That never changes if there is no air drag
or air friction of any kind. So we get plus 133
times the time and the whole trip
takes 26.6 seconds, and that, if you convert
that to kilometers is about 3.5 kilometers. Now, you could ask yourself
the question: What is the velocity
of that plane when it is at that point s? And now... you may want
to abandon now this one-dimensional idea
of x and y. You may say, "Well, look. "This is a parabola and
it is completely symmetric. "If the plane comes up here "with 425 miles per hour
at an angle of 45 degrees, "then obviously it comes down
here at an angle of 45 degrees and the speed must again be
425 miles per hour." And you would score 100 percent,
of course-- it's clear. I want you
to appreciate, however, that I could continue
to think of this as two one-dimensional motions. And I can therefore calculate what the velocity
in the x direction is at s and what the velocity
in the y direction is at s. So what is the velocity
in the x direction at point s? I go to equation...
the second equation there. That is v zero x, that is
plus 133 meters per second. What is the velocity
in the y direction? Ah, I have to go
to this equation now. v zero y minus gt. So I get plus 133 minus 10 times the 26.6 seconds
to reach that point s. And what do I find? Minus 133 meters per second. The velocity in the y direction
started off plus 133, but now it is minus 133. You see, this is sign-sensitive. This is wonderful. That's the great thing
about treating it that way. So you now know that it comes
in with a velocity of 133 in the x direction--
positive-- 133 in the minus y direction, and so what is the net,
the sum of the two vectors? That, of course, is this vector and no surprise,
this angle is 45 degrees and this one is
the square root of 2 times 133 and that, of course, gives you
back your 189 meters per second. 189 meters per second, and
that is 425 miles per hour. I'm not recommending that
you would do this, of course. It is perfectly reasonable to immediately come
to that conclusion because of the symmetry
of the parabola. Let's now turn
to uniform circular motion. Uniform circular motion occurs when an object goes around
in a circle and when the speed
never changes. If the speed doesn't change, then the velocity,
of course, does change because the direction changes
all the time, but the speed does not. So here we have our circle. Let this be radius r, and at this moment in time,
the object is here. It has a certain velocity. This is 90 degrees. And later in time, the object is here,
the speed is the same, but the direction
has changed, 90 degrees. So these vectors,
they have the same length. In a situation like this that we have
uniform circular motion-- so it's uniform... circular motion-- we first identify what we call
the period T in seconds. That's the time to go around. Then we identify what we call
the frequency, that is, how many times
it goes around per second. I prefer the letter f, but our
book uses the Greek letter nu. I find the nu
often very confusing with the v of velocity. That's why I prefer the f. It is one over T, and so
the units are seconds minus one but most physicists
would call that "hertz." Ten hertz means to go
ten times around per second. And then we identify omega,
the angular velocity. Omega, which is
in radians per second. Since it takes T seconds
to go around two pi radians, omega is two pi divided by T. Now, then we have the speed, which we can also think
of as a linear velocity. How many meters per second
is linear, as opposed to
how many radians per second, which is angular velocity. So this is a linear velocity,
this is an angular velocity. And that linear velocity, which, in this case,
is really your speed, is of course
the circumference of the circle divided by how many seconds
it takes to go around. And that is also omega r and that is now
in meters per second. All this is only possible
if there is an acceleration, and the acceleration is called
the centripetal acceleration. It is always pointed
towards the center: "a" centripetal, "a" centripetal. And the centripetal
acceleration-- the magnitude-- is v squared divided by r, which is therefore
also omega squared r, and that, of course,
is in meters per second squared. I want to work out
a specific example, and the example that I have
chosen is the human centrifuge that is used by NASA in Houston
for experiments on humans to see how they deal
with strong accelerations. This is that centrifuge. The radius
from the axis of rotation-- the axis of rotation is here-- and the distance
from here to here, though you may not think so,
is about 15 meters. So the astronauts go in here
and then the thing goes around. And so I would like to work out
this with some numbers. The radius r--
I'll give your light back because it may
be nicer for you... The radius is 15 meters. It depends, of course,
a little bit on where the person
is located in that sphere. It goes around
24 revolutions per minute and that translates
into 0.4 hertz. So the period to go around
for one rotation is 2.5 seconds. The thing goes around
once in 2.5 seconds. So the angular velocity omega,
which is two pi divided by T... If you take two pi
and divide it by 2.5, it just comes out
to be roughly 2.5. (chuckling):
It's a purely accident,
that's the way it is. Don't ever think that that
has to be the same, of course. It just happens to come out
that way for these dimensions. So omega is about 2.5 radians
per second. And the speed, linear speed-- tangential speed, if you
want to call it-- is omega r. That comes out to be about 35...
37.7 meters per second, and that translates into
about 85 miles per hour, so it's a sizable speed. What, of course,
the goal is for NASA: What is the centripetal
acceleration-- that is omega squared r-- or, if you prefer to take
v squared divided by r, you'll find, of course,
exactly the same answer if you haven't made a slip, and that is 95 meters
per second squared. And that is about ten times the gravitational acceleration
on Earth, which is really phenomenal,
if you add, too, the fact that the direction is changing
all the time when you go around, so you feel the 10 g
in this direction and then you feel it
in a different direction. I can't imagine how people
can actually survive that-- I mean, not faint. Most people, like you and me, if we were to be accelerated
along a straight line, not even a circle,
where the direction changed, but along a straight line, most of us faint
when we get close to 6 g. And there is a reason for that. You get problems
with your blood circulation and not enough oxygen goes
to your brains, and that's why you faint. How these astronauts
can do it at 10 g and the direction changing
all the time, it beats me. If you take a Boeing 747,
it takes 30 seconds from the moment
that it starts on the runway until it takes off. You should time that,
when you get a chance. It's very close to 30 seconds, and by that time
the plane has reached a speed of about 150 miles per hour. And if you calculate, if you assume that the
acceleration is constant-- it's an easy calculation--
it turns out that the acceleration is only
two meters per second squared. That is only one-fifth of
the gravitational acceleration. Feels sort of good, right? It's very comfortable,
when you're taking off. It's only 2 meters
per second squared. These poor people,
men and women, 95 meters per second squared. I would like
to address something that is not part of the exam, but that is something
that I want you to think about, something that is fun, and it's always nice
to do something that is fun. It has to do
with my last lecture. I have to clean my hands first
for it to work quite well. I have a yardstick here, and I
am going to put the yardstick on my hands, on my two fingers,
which I hold in front of me. Here it is. It's resting on my two fingers, and I'm going to move my
two fingers towards each other. One of them begins
to slide first, of course. I can't tell you which one. But something
very strange will happen. If this one starts to slide
first, it comes to a stop and then the other one
starts to slide and it comes to a stop. And then this one starts
to slide and so on. And that is very strange. This is something
you should be able to explain, certainly after the lecture
we had last time. Look at this. Did you see the alternation?
I'll do it a little faster. Left is going, right is going,
left is going, right is going, left is going. Once more... look at it. Left is going, right is going,
left is going, right is going. They alternate. Give this some thought, and you know,
PIVoT has an option that you can discuss
problems with other students, so make use
of this discussion button and see whether you can
come to an explanation. Good luck on your exam. See you next Friday.
