Shankar: Let me start again by reminding you what it is that
was done last time. If you say, “Can you summarize
for me in a few words the main ideas?”
I would like to do that. What I did last time was to
show you how to handle motion in more than one dimension.
I picked for it two dimensions as the standard way to explain
it. By the way, I would like to
make one recommendation. If you guys are coming in a
little late, don’t worry about submitting the homework.
Just come in and settle down, because it’s very hard to
lecture with this amount of traffic.
I do try to start a few minutes late, but I also have to finish
a few minutes early so you can go to your next class.
Also, people who come in early maybe should try to sit in the
middle part of the classroom, so the latecomers can come in
without too much disruption. So, summary of last time.
If you live in two dimensions or more, you’ve got to use
vectors to describe most things. The typical vector is called
V or A or B something with an arrow on it.
The most important vector is the position vector that tells
you where the object is. It’s got components which are
x and y that could vary with time.
I and J are unit vectors in the x and
y directions. You can deal with a vector in
one of two ways. You can either think of it as
an arrow and imagine the arrow, or you can reduce it to a pair
of numbers, x and y.
If you want to add two vectors, you can add the arrows by the
rule I gave you or just add the components of the two guys to
get the component of the sum and likewise for the y.
I mentioned something of increasing importance only
later, which is that you are free to pick another set of
axes, not in the traditional x
and y direction, but as an oblique direction.
If you do, the unit vectors are called I prime and
J prime, the components of the vector
change. You can imagine that if a
vector is viewed from an angle, then its components will vary
with the perspective. So the components of the vector
are not invariant characterizations of the vector.
That is, the vector itself has a life of its own.
The components come in the minute you pick your axis.
It’s not enough to say these are the components of the
vector; you’ve got to tell me,
“I am working with I and J, which I define in the
following manner.” I gave you a law of
transformation of the components;
namely, if the vector has components ax and
ay in one reference frame and ax prime and
ay prime in another reference frame,
how are the two related? They come from writing I
prime and J prime in terms of I and J,
then sticking it in an expression and identifying the
new representation. Somehow, when I told you to
invert the transformation, some of you had some
difficulty. Maybe you didn’t realize that
they are just simultaneous equations that you solve.
Normally, if I tell you 3x + 4y=6,
and 9x + 6y=14, you know how to solve it.
It’s like really the same kind of problem, except that the 3
and 4 will all be placed by sin φ and cos φ but
they are just the numbers. You eliminate them the same way
you eliminate them. I’ve given you a homework
problem where you can try your skills.
When you go today and look at the homework,
you’ll find the problem from the textbook,
one extra problem that deals with all of this.
Then, I gave you one other very important example of a particle
moving in the xy plane. x and y can be
whatever you like, but I picked a very special
example where x looked like this: r times cos
ωt. This times I + r
times sin ωt. You should go back and remember
what we did. If it took you a while to
digest it, I ask you to think harder.
This describes a particle that’s going around in a circle.
We know it’s going around in a circle because if I find the
length of this vector, which is the x-square
part, plus the y-square part, I just get r^(2) at
all times, because sine square plus cosine
square is one. So we know it’s moving on a
circle of radius r. Furthermore,
as time increases, the angle, ωt,
is increasing in this fashion. Omega is called the angle of
velocity. I related it to the time
period, which is the time it takes to go around a full
circle, by saying once you’ve done a
full circle, ωt better be 2π.
So this new quantity ω, which may be new to you,
is related to the time period. How long does it take to go one
round in this fashion? You’re also free to write it in
terms of the frequency. The time period and frequency
are reciprocals. If it takes you 1/60 of a
second to go around once, then you do it 60 times a
second. Omega is really very simple
quantity. It’s related to the frequency
with which you go around the circle but is multiplied by
2π. Why is that?
Frequencies, how many times you go around,
and 2π is the rate at which the angle is being
changed. But if every revolution is
worth 2π radians, then 2πf is the number
of radians per second. f is revolutions per
second and 2πf is radians per second.
It’s called the angle of velocity.
The most important result from last time was that if you took
this r, and you took two derivatives of
this to find the acceleration, d2r over
dt^(2), try to do this in your head.
If you took the two derivatives of this guy, first time it will
become -ω, sin ωt;
second time it will become -ω^(2) cos ωt.
In other words, it will become -ω^(2)
times itself. Same thing there.
The final result is the acceleration is -ω^(2)
times the position. That means the acceleration is
pointing towards the center of the circle and it has a
magnitude a. When I draw something without
an arrow, I’m talking about the magnitude.
It is just ω^(2)r. I have shown you yesterday that
the speed of the particle as it goes around the circle is this
(ωr). Again, you should make sure you
know how to derive this. You can do it any way you like.
You can take one full circle and realize the distance
traveled is 2πr, divide by the time and you will
get this. Another way,
take the derivative of this, get the velocity vector and you
notice his magnitude is a constant and the constant will
be ωr. Whichever way you do it,
you can then rewrite this as v^(2) over r.
This is called the centripetal acceleration. This is the acceleration
directed toward the center. I told you these are very
important results. You’ve got to get this in your
head. Whenever you see a particle
moving in a circle, even if it’s at a constant
speed, it has an acceleration,
v^(2) over r directed towards the center.
This formula doesn’t tell you which way it’s pointing,
because it’s a scaler; it’s not a vector equation.
If you want to write it as a vector equation,
you want to write it as v^(2) over r
minus–I want to say that it’s pointing in the direction toward
the center. So sometimes what we do is we
introduce a little vector here called e_r.
I’ll tell you more about it later.
e_r is a vector at each point of length
one pointing radially away from the center.
It’s like the unit vector I.
Unit vector I points away from the origin in the
x direction. J points away from the
origin in the y direction.
e_r is not a fixed vector.
At each point, e_r is a
different vector pointing in the radial direction of length one.
The advantage of introducing that guy is that if you like,
I can now write an equation for the acceleration as a vector.
The magnitude is v^(2) over r.
The direction is -e_r.
So e_r is a new entity I’ve introduced for
convenience. It plays a big role in
gravitation, in the Coulomb interaction.
It’s good to have a vector pointing in the radial direction
of length one. That’s what it is.
That’s really the heart of what I did last time.
Then we did some projectile problems.
You shoot something, you should know when it will
land, where it will land, with what speed it will land,
how high it will go. I assume that those problems
are not that difficult and I’ve given you a lot of practice.
Now I’m going to move to the really important and central
topic. I guess you can guess what that
is. We’re going to talk about
Newton’s laws. This is a big day in your life.
This is when you learn the laws in terms of which you can
understand and explain a large number of phenomena.
In fact, until we do electricity and magnetism the
next semester, everything’s going to be based
on just the laws of Newton. It’s really amazing that
somebody could condense that much information into a few,
namely three, different laws.
That’s what we’re going to talk about.
Let’s start. Your reaction may be that
you’ve seen Newton’s laws, you applied them in school.
I’ve got to tell you that I realized fairly late in life
they are more subtle than I imagined the first time.
It’s one thing to plug in all the numbers and say,
“I know Newton’s laws and I know how they work.”
But as you get older and you have a lot of spare time,
you think about what you are doing,
which is something I have the luxury of doing right now,
and I realized this is more tricky.
I want to share some of that with you so you can fast forward
and get the understanding it took me much longer to get.
That’s what I’m going to emphasize, more than just
plugging in the numbers. Of course, we have to also know
how to plug in the numbers so we can pass all the tests,
but it’s good to understand the nature of the edifice set up by
Newton. First statement by Newton–I
don’t feel like writing it down. It’s too long and everybody
knows what the law is. It’s called the Law of Inertia.
Let me just say it and talk about it.
The Law of Inertia says that, “If a body has no forces acting
on it, then it will remain at rest if it was at rest to begin
with, or if it had a velocity to
begin with, it will maintain that velocity.”
One way to say it is, every body will continue to
remain in a state of rest or uniform motion in a straight
line. That’s another way of saying
maintaining velocity if it’s not acted upon by a force.
What makes the law surprising is that if I only gave you half
the law, namely every body will remain at rest if it’s not acted
upon by a force, you will say, “That’s fine.
I accept that, because here’s something.
You leave it there, it doesn’t move.
It’s not a big surprise.” People were used to that from
the time of Aristotle. But Aristotle used to think
that if you want something to move, there has to be some
agency making it move. That agency you could call
force. The great discovery that
Galileo and Newton made is that you don’t need a force for a
body to move at constant velocity.
It’s very clear you don’t need a force if something is doing
nothing, just sitting there. The fact that you don’t need a
force for it to move forever at a given speed in a given
direction, that’s not obvious,
because in daily life you don’t see that.
In daily life, everything seems to come to
rest unless you push it or you pull it or you exert some kind
of force. But we all know that the reason
things come to a halt when you push them is,
there eventually is some friction or drag or something
bringing them to rest. Somehow, if you could
manufacture a really smooth frictionless surface,
that if you took a hockey puck or something and an air cushion
and you give it a push, in some idealized world,
it’ll travel forever. So it’s hard to realize that in
the terrestrial situation. But Galileo already managed to
find examples where things would roll on for a very,
very long time. Nowadays, if you go to outer
space, you can check for yourself that if you throw
something out, it just goes on forever without
your intervention. It’s in the nature of things to
go at a constant velocity. They don’t need your help to do
that. You have to be careful that
this first law of Newton is not valid for everybody.
In fact, I’ll give an example in your own life where you will
find that this law doesn’t work. Here is the situation.
You go on an airplane and then after the usual delays,
the plane begins to accelerate down the runway.
At that instant, if you leave anything on the
floor, you know it’s no longer yours.
It’s going to slide down and the guy in the last row is going
to collect everything. Why is that?
Because we find in that plane, when objects are left at what
you think is at rest with no external agency acting on them,
they all slide backwards towards the rear end of the
plane. That happens during takeoff.
That doesn’t happen in flight, but it happens during takeoff.
That is an example of a person for whom the Law of Inertia does
not work. This is something you guys may
not have realized. Newton’s laws are not for
everybody. You have to be what’s called an
“inertial observer.” If you’re an inertial observer,
then in your system of reference, objects left at rest
will remain at rest. The plane that’s ready to take
off or is taking off is not such a system.
The Earth seems to be a pretty good inertial system,
because on the ground, you leave something,
it stays there. It depends on what you leave.
If you leave your iPod, it’s not going to stay there
for very long. But then you can trace it to
some external forces, which are carrying your iPod.
But if you don’t do anything, things stay.
Here is the main point. The point of Newton is,
two things in the Law of Inertia, which one may think is
trivial. First, free velocity,
constant velocity can be obtained for free without doing
anything. There are people for whom this
is true. For example,
in outer space, you’ve got an astronaut.
You send something, you’ll find it goes on forever.
Here’s another thing. If you find one inertial
observer, namely one person for whom this Law of Inertia works,
I can manufacture for you an infinite number of other people
for whom this is true. Who are these other people?
Do you know what I’m talking about?
If I give you one observer for whom the Law of Inertia is true,
I say that others for whom is also true.
[inaudible] Professor Ramamurti
Shankar: Did you hear that? Let me repeat that.
First of all, if the Law of Inertia is valid
for me, it’s valid for other people in the same room at rest
with respect to me. Because if I think it’s not
moving, you think it’s not moving.
That’s just fine. But suppose you are in a train
and you’re moving past me and you look at this piece of chalk.
Of course, everything in my room is going backwards for you.
But things which were at rest will move at a constant
velocity, opposite of the velocity that you have relative
to me. You will find that objects that
are at constant initial velocity maintain the velocity.
If I am an inertial observer, another person moving relative
to me at constant velocity will also be an inertial observer.
Why? Because any velocity I ascribe
to a particle or an object, you will add a certain constant
to it by the law of composition of velocity.
All velocities I see you will add a certain number to get the
velocities according to you. But adding a constant velocity
to objects does not change the fact that those which were
maintaining constant velocity still maintain a constant
velocity. It’s a different constant
velocity. In particular,
the things that I say are at rest, you will say are moving
backwards at the velocity that you have relative to me.
Things that I say are going at 50 miles per hour you may say
are going at 80. But 50 is a constant and 80 is
a constant. Therefore, it’s not that
there’s only one fortunate family of inertial observers.
There’s infinite number of them, but they’re all moving
relative to each other at constant velocity.
If the Earth is an inertial frame of reference,
if you go in a train relative to the Earth at constant
velocity, you’re also inertial. But if you go on a plane which
is accelerating, you’re no longer inertial.
That’s the main point. The point is that there are
inertial frames of reference. You must know the Earth is not
precisely inertial. The Earth has an acceleration. Can you tell me why I’m sure
the Earth has an acceleration? Yes?
Student: Because it’s moving in a
circle. Professor Ramamurti
Shankar: It’s going around the Sun.
Let’s imagine it’s a circular orbit.
Then we’ve just shown here, it’s an accelerated frame of
reference. It just turns out that if you
put the v^(2) and you put the r,
and r is 93 million miles, you will find the
acceleration is small enough for us to ignore.
But there are effects of the Earth’s acceleration,
which we’ll demonstrate. The Focault pendulum is one
example where you can see that the Earth is rotating around its
own axis. Then, the fact that the Earth
is going around the Sun. All of them mean it’s really
not inertial, but it’s approximately
inertial. But if you go to outer space
nowadays, you can find truly inertial frames of reference.
That’s the first law. The first law,
if you want, if you want to say,
“Okay what’s the summary of all of this?”
The summary is that constant velocity doesn’t require
anything. The reason it looks like a
tautology, because you look around, nothing seems to have
its velocity forever. Then you say,
“Oh, that’s because there’s a force acting on it.”
It looks like a tautology because you’re never able to
show me something that moves forever at a constant velocity,
because every time you don’t find such a thing,
I give an excuse, namely, a force is acting.
But it’s not a big con, because you can set up
experiments in free space far from everything,
where objects will, in fact, maintain their
velocity forever. That’s a possibility.
It’s a useful concept on the Earth, because Earth is
approximately inertial. Now, we have come to the second
law, which is “the law.” This is the law that we all
memorize and learn. It says that,
“If a body has an acceleration, then you need a force and the
relation of the force to acceleration is this thing:
F=ma” Now, I have to say a few words about
units. Acceleration is measured in
meters per second squared. Mass is measured in kilograms.
So, the way to measure force is in kilogram meters per second
squared. But we get tired of saying that
long expression, so we’re going to call that a
Newton, right? If you invented it,
we’d call it whatever your name is, but this is the guy who
invented it, so it’s called a Newton, usually denoted by a
capital N. A typical problem that you may
have done in your first pass at Newton’s law,
someone tells you a force of 36 Newtons is acting on a mass of
whatever, 4 kilograms; what’s the acceleration?
You divide and you find it’s 9 and you say, “Okay,
I know what to do with Newton’s laws.”
That’s where I want to tell you that it’s actually more
complicated than that. Let’s really look at this
equation. Take yourself back to
1600-whatever, whenever Newton was inventing
these laws. You don’t know any of these
laws. You have an intuitive
definition of force. You sort of know what force is.
Somebody pushes you or pulls you.
That’s a force. Suddenly, you are told there is
a law. Are you better off in any way?
“Can you do anything with this law?”
is what I’m asking you. What can you do with this law?
I give you Newton’s law and say, “Good luck.”
What will you do? What does it help you predict?
Can you even tell if it’s true? Here’s a body that’s moving,
right? I want you to tell me,
is Newton right? How are we going to check that?
Well, you want to measure the left-hand side and you want to
measure the right-hand side. If they’re equal,
maybe you will say the law is working.
What can you measure in this equation?
Professor Ramamurti Shankar: Pardon me?
Student: Force and acceleration.
Professor Ramamurti Shankar: All right.
Let’s start with acceleration. What’s your plan for measuring
acceleration if some little thing is moving?
What do you need to measure it? Student:
The change in velocity over the change [inaudible]
Professor Ramamurti Shankar: Right,
but what instruments will you need to measure it?
You are supposed to really measure it.
What will you ask me for? Student: A watch.
Professor Ramamurti Shankar: A watch?
[inaudible] Professor Ramamurti
Shankar: And what else? Student:
[inaudible] Professor Ramamurti
Shankar: A ruler? Okay.
That’s right. You don’t mean Queen Elizabeth,
right? You mean– Very good.
What you really want–a ruler may not be enough,
but maybe it’s enough. So here’s the long ruler and
here’s this thing moving, right?
You ask for a Rolex so here’s your watch.
It’s telling time. Tell me exactly what you want
to do to measure acceleration. What do you have to do?
I want the acceleration now. What will you measure?
Okay, you go ahead. You can try.
I would start with the object at rest and then [inaudible]
Professor Ramamurti Shankar: It may not be at
rest to begin with. It’s doing its thing.
It’s going at some speed. Yes?
Student: Measure the distance it travels
over a constant interval of time?
Professor Ramamurti Shankar: That’ll give you
the velocity. Student:
Well, if you do distance after one second versus distance over
the next second versus distance over a third second,
you see how it increases. Professor Ramamurti
Shankar: Good, in principle.
Let me repeat what he said. He said, first,
let it go a little distance, take the distance over time.
That gives you the velocity now. Let it go a little more,
that gives you the velocity later.
Take the difference of the two velocities and divide by the
difference of the two times, and you’ve got the
acceleration. Of course, you have allowed it
to move a finite distance in a finite time.
What you should imagine doing is making these three
measurements more and more quickly.
You need three positional measurements.
Now, a little later, and a little later later,
because between the first and second,
you get a velocity, the second and third you get
another velocity. Their difference divided by the
difference in times is going to be the acceleration.
But if you imagine making these measurements more and more and
more quickly, in the end, you can measure
what you can say is the acceleration now.
That’s the meaning of the limit in calculus.
You take Δx and you take Δt.
What’s the meaning of Δx and Δt going
to 0? It means, measure them as
quickly as you can. In the real world,
no one’s going to measure it instantaneously,
but we can make the difference as small as we like.
Mathematically, we can make it 0.
In that limit, we can measure velocity right
now. That means we can also measure
velocity slightly later and make the slightly later come as close
to right now as we want. Then, that ratio will give you
acceleration. Acceleration is the easiest
thing to measure of these three quantities.
We all have a good intuitive feeling for what acceleration
is. You want to test if what Newton
told you is right. You see an object in motion,
you find a and you give a a certain numerical
value, 10 meters per second squared.
But that’s not yet testing the equation, because you’ve got to
find both sides. What about the mass?
What’s the mass of this object? Anybody want to try from this
section here? Yes?
Student: You could use some sort of a
standard unit of mass and then a balance to measure the mass of
something else, like put it at a certain
distance from the fulcrum on a scale and figure out the
relative mass. Professor Ramamurti
Shankar: Okay. His idea was the following.
You take a standard mass and you maybe go to a seesaw.
You put the standard mass here and you put some other fellow at
the mass you’re looking at there.
You add just the lengths and when it balances,
you can sort of tell what this mass is, right?
But suppose you were in outer space.
There’s no gravity. Then the seesaw will balance,
even if you put a potato on one side and an elephant on the
other side. You cannot tell the mass,
because what you are doing now is appealing to the notion of
mass as something that’s related to the pull of the Earth on the
object. But Newton’s law is–You see,
you’ve got to go back and wipe out everything you know.
If this is what you have, there is no mention of the
Earth in these equations. Yet, the notion of mass is
defined. It’s not talking about the
gravitational pull of the Earth on an object.
If you had an object that you say has a certain mass that you
don’t know and then you took one that’s the same density,
but say twice the size and you can see if it slows the force.
Professor Ramamurti Shankar: But do you know
what the force on the body is? We don’t know how to measure
that either; do you agree?
We don’t know what the force is, because we are hoping to say
force is going to be m times a.
We’re just getting on to measuring m.
It looks like a circular definition right now.
Do you mean what material this object is made of and its
density [inaudible] Professor Ramamurti
Shankar: Ah, but density is mass over volume.
But we don’t know the mass, right?
Student: But if you use the density of
that [inaudible] Professor Ramamurti
Shankar: How is anyone going to give you a density of
anything? We’re just asking what’s the
mass of any object? We have not yet found a
satisfactory answer to what’s the mass of an object?
Are we still operating in outer space?
Professor Ramamurti Shankar: Yes.
This cannot depend on the planet Earth.
Student: Could we have,
by any chance, a spring?
Professor Ramamurti Shankar: Yes.
Student: Whose spring constant and the
definition [inaudible] Professor Ramamurti
Shankar: I’m sorry. What did you just say?
You can have a spring. Student:
Yeah, you could have a spring and you could compare how fast
that objects will travel when the spring is compressed and
they’re placed against it and released, but you’d only have a
comparison then. Professor Ramamurti
Shankar: Okay. Let me repeat what he said.
He said, take a spring to outer space and we’ll hook up some
objects to them and see how fast they move and do a comparison.
That’s fairly close to what I had in mind.
But it’s not the word perfect answer.
I just want to take some time thinking about it.
[inaudible] Professor Ramamurti
Shankar: What do I learn from the period?
Professor Ramamurti Shankar: How?
Professor Ramamurti Shankar: Remember,
you cannot peek into chapters 6 and 7, because you’ve seen it
before. I’m asking you if somebody
wrote apparently Shakespeare’s plays for him;
it’s one of the rumors, right? Suppose Newton comes to you and
says, “I have this great law, but I don’t want to publish it
under my name. I’m going to give it to you.”
You have got this new law, but how are you going to sell
this? You’ve got to tell people how
to use it. You realize it’s very subtle,
because the very first thing in that equation,
which is m, has not yet been defined.
He gave an answer which is fairly close.
It doesn’t rely on gravity. It doesn’t rely on the planet.
You cannot say to me, “Take a force,
due to a spring, and see what force it applies
and divide by the acceleration and get the mass,”
because we haven’t defined force either.
You’ve got to realize that. Let me ask you something.
How do we decide how long a meter is?
Can you tell me, how do you know how long a
meter is? Student:
It’s just sort of arbitrary. Professor Ramamurti
Shankar: Right. Student:
You can just pick an arbitrary mass, as well,
like one object, which is an arbitrary mass.
Professor Ramamurti Shankar: And we’ll call it.
First thing you’ve got to do is, realize that some of these
things are not God-given. A meter, for example,
is not deduced from anything. Napoleon or somebody said,
“The size of my ego is one meter.”
That’s a new unit of length. You take a material like silver
and put it in a glass case and that’s the definition of a
meter. It’s not right or wrong.
Then I ask you, “What is two meters and what is
three meters?” We have ways of doing that.
I take the meter and put it next to the meter,
that’s two meters. I cut it in half,
I’ll use some protractors and dividers and compasses,
you can split the meter into any fraction you like.
Likewise for mass, we will take a chunk of some
material and we will call it a kilogram.
I should give you some hint. That kilogram,
I don’t expect you to deduce. That is a matter of convention.
Just like one second is some convention we use and one meter
is some convention we use. I’m going to give you a little
help. I’m going to give you a glass
case and in the glass case is an entity with this as one
kilogram, by definition. Then I give you another object,
an elephant. Here’s an elephant.
I’m telling you, “What’s the mass of the
elephant?” How do you find this mass?
You got to take the hint he gave.
I give you a spring and an elephant.
What should we do? Yes?
Student: Measure how far,
if you hung an elephant from a spring [inaudible]
Professor Ramamurti Shankar: No.
Remember, when we hang the elephant in outer space,
nothing is going to happen. Student: Okay.
You push the elephant and measure how far– what the
distance from the unstretched spring is that the elephant
travels in either direction. Professor Ramamurti
Shankar: Okay. Student:
Then, you also do that with one kilogram.
It should be proportional to how far from the resting state
of the spring. That should be the proportion
that the one kilogram object is to the other.
Professor Ramamurti Shankar: When you say,
“push the elephant,” you want to push it in a particular way?
What do you want to do? Student:
Towards the [inaudible] to compress the spring or to
[inaudible] Professor Ramamurti
Shankar: That is correct, but you don’t want to give it a
definite push. Yes?
Student: Do you push it radially along
the axis of the spring, because you know that the
centripetal force is mv^(2) is over t.
Professor Ramamurti Shankar: Yeah.
Maybe that’s an interesting thing.
That’s correct. You can do that, too.
Let me now put you out of your suspense.
I think I’ve heard bits and pieces of the answer everywhere,
so I don’t want to wait until we get it word-perfect.
The point is, the one kilogram is a matter of
convention. We want to know what is the
mass of the elephant. We can do the seesaw
experiment, you suggested, but the seesaw experiment
requires gravity, so we don’t want to do that.
A spring will, on the other hand,
exert a force. We don’t know what the force of
the spring is. If you assume that,
you are not playing by the rule, because we don’t know what
force it exerts. We do know it exerts a force,
so here’s what you do. You hook one end of the spring
to a wall and you pull it from rest by some amount and you
attach the one kilogram mass to it.
One problem that we’d have, though, is if we’re in outer
space, where are we going to find a wall that won’t move?
Professor Ramamurti Shankar: A wall that will
not move? Student:
Actually, if you hook it up to any wall in outer space
[inaudible] Professor Ramamurti
Shankar: You’re right. What can actually happen is
that if you’re in an enormous laboratory, which is made up of
an enormous object, then you will find that you
can, in fact, attach it to the wall.
You can go in outer space– Student:
It won’t move very much? Professor Ramamurti
Shankar: No, it won’t move very much.
Once we know enough dynamics, we can answer your question.
Does outer space even rob you of something to which you can
anchor a spring? The answer is, “no.”
You can anchor things to objects in outer space.
They just won’t act the way gravity does.
But you can nail it to the wall and pull one end.
So you pull one end. We don’t know what force it
exerts. But it exerts some force.
Now, I tie this one kilogram mass to it and let it go.
I find the acceleration. That is the force which I do
not know in magnitude. But this is the acceleration of
the one kilogram mass. Then, I bring the elephant and
I pull the spring by the same amount and I find the
acceleration of the elephant and the denominator is obviously the
mass of the elephant. The force is not known,
but it’s the same force. So, when I divide these two
numbers, I’m going to find a_1 over
a_E is equal to m_E over
m, which is the one kilogram mass.
What we needed was some mechanism of exerting some fixed
force. We didn’t have to know its
magnitude. But the acceleration it
produces on the elephant and on the mass, are in an inverse
ratio of their masses. If you knew this was one
kilogram, then the acceleration of the elephant,
which will be some tiny number, maybe 100^(th) of what this guy
did; the mass of the elephant is
then 100 kilograms. Note there are,
again, subtleties even here. If you think harder,
you can get worried about other things.
For example, how do I know that when I pull
the spring the first time for the mass,
it exerted the same force when I pulled the spring the second
time for the elephant? After all, springs wear out.
That’s why you change your shock absorbers in your car.
After a while, they don’t do the same thing.
First, we got to make sure the spring exerts a fixed force
every time. You can say,
“How am I going to check that? I don’t have the definition of
force yet.” But we do know the following.
If I pull the one kilogram mass and I let it go,
it does something, some acceleration.
Then, I pull it again by the same amount and let it go;
I do it 10 times. If every time I get the same
acceleration, I’m convinced this is a
reliable spring that is somehow producing the same force under
the same condition. On the eleventh time,
I pull the mass. I will put the elephant in.
With some degree of confidence, I’m working with a reliable
spring and then I will get the mass of the elephant.
Why is it so important? It’s important for you guys to
know that everything you write down in the notebook or
blackboard as a symbol is actually a measured quantity.
You should know at all times how you measure anything.
If you don’t know how to measure anything,
you are doing algebra and trigonometry.
You are not doing physics. This also tells you that the
mass of an object has nothing to do with gravitation.
Mass of an object is how much it hates to accelerate in
response to a force. Newton tells you forces cause
acceleration. But the acceleration is not the
same on different objects. Certain objects resist it more
than others. They are said to have a bigger
mass. We can be precise about how
much bigger by saying, “If the acceleration of a body
to a given force is ten times that of a one kilogram mass,
then this mass is one-tenth of one kilogram.”
This is how masses can be tabulated using a spring.
Imagine then from now on, we can find the mass of any
object, right? We know now with the same
spring, by this comparison, we will find.
All objects now can be attributed a mass.
Then we may, from this equation,
say a certain force is acting in a given situation by
multiplying the m times the a.
Then, here is what we actually do.
Now, we go back to the spring. We go back to the spring and we
want to learn something about the spring.
We want to know how much force it exerts when I pull it by a
certain amount. Now I can measure that,
because I pull it by one centimeter and I find the
acceleration it exerts on a known mass.
That m times a is the force the spring is
exerting. Then I pull it by 1.1 cm,
and I find ma. I find 1.2, I find ma.
I draw a graph here of the amount by which I pull the
spring versus the force it exerts.
It will typically look like this and the formula we say is
F=-kx, where k is called “the
force constant.” You got to understand what the
minus sign is doing here. This is the force exerted by
the spring on the mass. It says, if you pull it to the
right, so that x is positive, the spring will exert
a force which is in the negative direction;
that’s why you have a minus sign.
Then, all springs will do something like this.
Further out they can do various things.
The force may taper off, the force may not be given by a
straight line, but for modest deformations,
every spring will have a linear regime in which the force is
linearly proportional to the stretching.
It also tells you that if you compress the spring,
compress it means x is now–x is measured from
this position, where the spring is neither
compressed nor expanded. So x is not really the
coordinate of the end point. You’ve got to understand that.
Springs have a natural length; x is measured from that
length. If it’s positive,
it means you’ve stretched it, if it’s negative,
it means compressed. This equation is telling you if
you compress it, namely if x is negative,
F will be then positive, because it’s pushing you
outwards. Therefore, what we have done
now is, we can take all kinds of springs and we can calibrate the
force they will exert under various conditions.
Namely, if you pull it by so much, that’s the force it will
exert. I want you to think for a
second about two equations. One equation says F=ma.
Other equation says F=-kx. What’s going on?
Is one of them Newton’s law? Then what’s the other one?
Maybe F=-kx should be called Newton’s law?
Why is F=ma called Newton’s law?
Then, what’s the meaning of this expression? What’s the difference between
saying F=-kx and F=ma?
They are saying very different things.
F=ma is a universal law.
Professor Ramamurti Shankar: Right.
Let me repeat. F=ma is universally
true, independent of the nature of the force acting on a body.
F=-kx is only describing how the spring is.
Professor Ramamurti Shankar: Very good.
That’s the whole point. The cycle of Newtonian dynamics
has two parts. First one says,
if you knew the force acting on any body, without going into
what caused the force, then you may set that force
equal the mass times acceleration of the body.
We think of force as the cause and a as the result or
the effect. Force causes acceleration and
this is a precise statement. There, Newton doesn’t tell you
what forces are going to be acting on a body in a given
situation. If you leave the body alone,
maybe there’s no force acting on it.
If you connect the body to a spring, which is neither
compressed nor extended, there’s no force acting on it.
If you pull the spring, there is a force acting on it.
Newton is not going to come and tell you what force the spring
will exert when it’s pulled by some amount.
That is another part of your assignment.
The physicist has to constantly find out what forces act on
bodies. That’s a separate exercise.
In every context in which I place a body,
I’ll have to know what are the forces acting on it.
I’ve got to find them by experimenting,
by putting other bodies and seeing how they react and then
finding out what’s the force that acts on a body when it’s
placed in this or that situation.
Once you’ve got that, then you come back.
In the case of a spring, this is the law that you will
deduce. If it’s something else,
you will have to deduce another law.
For example, we know that if a body is near
the surface of the Earth, the force of gravity and that
object seems to be m times g,
where g is 9.8. That’s something you find out
by experiment. Every time you are finding out
a different force that’s acting on a body with different origin.
One says, leave any body near the Earth, it yields a force.
I know this is the right answer, because if I now find
the acceleration, I find it’s mg divided
by m and I get -g as the answer for all bodies.
By the way, that’s a very remarkable property of the
gravitational force–the cancellation of the two
ms. If you look at the electrical
force, the force of electricity, on the proton and electron or
something, it’s not proportional to the
mass of either object. It’s proportional to the
electric charge of either object.
Therefore, when you divide by the mass to get the
acceleration, the response of different
bodies is inverse to the mass. But gravity has a remarkable
property that the pull of the Earth is itself proportional to
the inertia of the object. So, when you divide by
m, m cancels and everything falls at the same
rate on the surface of the Earth.
In fact, there’s a property of gravitational fields anywhere,
even in outer space, but there is some residual
field between all the planets and all the stars in the
universe, that the force on a body is
proportional to the mass of the body.
So, when you divide by the mass to get the acceleration,
you get the same answer. Everything — gold,
silver, diamonds, particles — everything
accelerates the same way in a gravitational field,
due to this remarkable fact. This was known for a long time,
but it took Mr. Einstein to figure out why
nature is behaving in that fashion.
If I have some time, I’ll tell you later.
But there are two qualities which happen to be equal.
One is inertial mass, which is how much you hate your
velocity to change, how hard you resist
acceleration. That exists far from planets,
far from everything. Other is gravitational mass,
which is the measure of how much you’re attracted to the
Earth. There’s no reason why these two
attributes had to be proportional,
but they are proportional and they are equal by choice of
units and you can ask, “Is this just an accident or is
it part of a big picture?” It turns out,
it’s part of a big picture and all of general relativity is
based on this one great equivalence of two quantities
which are very different attributes.
Why should the amount by which you’re attracted to the Earth be
also a measure of how much you hate acceleration?
Two different features, right? But they happen to be the same.
Anyway, what physicists do is they put bodies in various
circumstances and they deduce various forces.
This is the force of gravity. This is the force of the spring.
Here’s another force you might find.
You put a chunk of wood on a table and you try to move it at
constant speed. Then you find that you have to
apply minimum force. We are moving at constant
velocity. That means the force you’re
applying is cancelled by another force, which has got to be the
force of friction. So force of friction is yet
another force. Then, there are other forces.
You guys know there is the electrical force.
If you bring a plus charge near a plus charge,
if my body m, has a plus charge and another
plus charge is there, it’ll feel a force due to that.
That’s not going to be given by Newton.
So, Newton did not ever tell you what the expression for
force is in a given context. That is a constant study.
Coulomb discovered the Coulomb’s law,
which is a repulsion between charges.
Nowadays we know if you go into the neutron or the proton,
there are quarks. There are forces between the
quarks. You can ask what the force that
this quark will exert and that quark at a certain separation.
That was obviously not known to Newton.
Remember, Newton said F=ma, but didn’t tell you what
value F has in a given context.
He just said whenever there’s an acceleration,
it’s going to be due to some forces and it’s your job to find
what the forces are. To find the force,
what you will do is, suppose somebody says,
“Hey, I’ve got a new force. Every time I go near the
podium, I find I’m drawn to it.” Okay, that’s a new force.
The word gets around and we want to measure the force.
What do I do? I stand near this podium.
I’m drawn to it. I cannot stop.
I tie a spring to my back and I anchor it to the wall and see
how much the spring stretches before the two forces balance.
Then I know that kx is equal to the force this is
exerting at this separation. I move a little closer and I
find the stretching is a different number.
Maybe the force is getting stronger.
That’s how by either balancing the unknown force with a known
force or by simply measuring the acceleration as I fall towards
this podium and multiplying by mass,
you can find the force that exerts on me.
It’s not a cyclical and useless definition.
It’s a very interesting interplay and that’s the
foundation of all of mechanics. We are constantly looking for
values of F and we’re constantly looking for responses
or bodies to a known force. Here’s a simple example of a
complete Newtonian problem. A mass is attached to a spring.
It is pulled by a certain amount x,
and is released. What is it going to do?
We go to Newton. Newton says F=ma,
so to make it a useful result of this problem,
we know the mass of this guy. We did the comparison with the
elephant or something; a is the second
derivative of x and for this problem,
when F is due to a spring,
we know the force is that by studying the spring.
Suddenly, you have a mathematically complete problem.
Mathematically complete problem is that you can find the
function x(t) by saying that the second derivative of
the function is equal to -k over m times
the function. Then, you go to the Math
Department and say, “Please tell me what’s the
answer to this equation?” We don’t have to worry about
how you solve it, but it’s problem in mathematics
and the answer will be–surprise,
it’s going to be oscillating back and forth and that’ll come
out of the wash. This is how you formulate
problems. You can formulate another
problem. Later on, we know about gravity.
Newton finds out there’s a force of gravity acting on
everything. Here’s the Sun.
Here’s your planet. At this instant,
the planet may be moving at that speed.
Then the acceleration of the planet is the force of gravity
between the planet and the Sun, which Newton will tell you is
directed towards the Sun and it depends on how far you are.
Depending on how far the planet is from the Sun and where it’s
located, you will get the left-hand side.
That’s another law. That’s the Law of Universal
Gravitation. Then again, you will find the
evolution of the planetary motion, because the rate of
change of the position is connected to the position.
Again, go to the math guys and say, “What’s the answer to
that?” and they tell you the answer,
which will be some elliptical motion.
Okay. By the way, Mr.
Newton did not have math guys he could go to.
Not only did he formulate laws of gravitation,
he also invented calculus and he also learned how to solve the
differential equation for calculus.
He probably felt that nobody around was doing any work,
because all the thing was given to this one person.
It’s really amazing that what Newton did in the case of
gravity was to find the expression for this.
A few years earlier he had also gotten this law.
By putting the two together, out comes the elliptical motion
of the planets. We’ll come back to that,
but you have to understand the structure of Newtonian
mechanics. Generally, any mechanics will
require knowledge of the force. Now, I’m going to add one more
amendment. You don’t have to write in your
notebook, but you’ve got to remember.
Maybe you’ll but a little T and circle it.
Let me write it here. F_T means the
total force on a body. You’ve got many forces acting
on a body. The acceleration is controlled
by the sum. If I’m now working in one
dimension, it’s obvious because I’m not using any vectors.
Then, you may have forces to the right, forces to the left
pushing, pulling. You add them all up
algebraically, keeping track of their sign,
and that’s the total force. That’s connected to mass times
acceleration. Now, I’ll give you the third
law. The third law says that if
there are two bodies, called one and two,
force of one on two is minus the force of the second on the
first one. This is the thing about action
and reaction. All the laws that anybody knows
have this property. What does it require to be a
successful mechanic, to do all the mechanics
problems? You got to be good at writing
down the forces acting on a body.
That’s what it’s all going to boil down to.
Here is my advice to you. Do not forget the existing
forces and do not make up your own forces.
I’ve seen both happen. Right now at this point in our
course, whenever you have a problem where there is some body
and someone says, “Write all the forces on it”,
what you have to do is very simple.
Every force, with one exception,
can be seen as a force due to direct contact with the body.
Either a rope is pulling, a rope is pushing it,
you are pushing it, you are pulling it.
That’s a contact on the body. If nothing is touching the
body, there are no forces on it, with one exception which is,
of course, gravity. Gravity is one force that acts
on a body without the source of the force actually touching it.
That’s it. Do not draw any more forces.
People do draw other forces. When a body is going around a
circle, they say that’s some centrifugal force acting.
There is no such thing. Be careful.
Whenever there is a force, it can be traced back to a
tangible material cause, which is all the time a force
of contact, with the exception of gravity.
Okay, so with that, if you write the right forces,
you will be just fine. You will be able to solve all
the problems we have in mechanics.
I’m going to now start doing simple problems in mechanics.
They will start out simple and, as usual, they will get
progressively more difficult. Let’s start with our first
triumph will be motion in 1D. Here is some object,
it’s 5 kilograms and I apply 10 Newtons.
Someone says, “What’s the acceleration?”
Everyone knows it is 10 over 5 equals 2.
Now we know how we got all the numbers that go into the very
question. How do we know 10 Newtons is
acting? I think you people know how we
can say that with confidence. How do we know the mass of this
is 5 kilograms? We know how you got that from
an earlier experiment. Now, we know how the numbers
come in. The algebra is,
of course, very trivial here. Then, the next problem is a
little more interesting. Here I got 3 kg and I got 2 kg
and I’m pushing with 10 Newtons and I want to know what happens. One way is to just use your
common sense and realize that if you push it this way,
these two guys are going to move together.
And know intuitively that if they move together,
they will behave like an object of mass 5 and the acceleration
will again be 2. But there’s another way to do
this and I’m going to give you now the simplest example of the
other way, which is to draw free-body diagrams.
By the way, when I say there’s 10 Newtons acting this way,
you might say, “What about gravity?
What about the table?” Imagine that this is in outer
space where there is no gravity for now.
The motion is just along the x axis.
The free-body diagram, it says you can pick any one
body that you like and apply F=ma to it,
provided you identify all the forces acting on that body.
We’ll first pick the body, with mass 3.
Here’s the body of mass 3. What are the forces on it?
This is certainly acting on it. Then, you have to ask,
“What other force is acting?” Here is where you have to think.
Anybody want to guess from the last row what force?
The reactive force of 2 on 3? Professor Ramamurti
Shankar: Right. So let’s give it a name.
Let’s point it that way and call it F2 on 3.
That’s the end of this guy. Let’s look at the other fellow.
Maybe you should complete the force acting on this one.
Can you tell me what it is? Same person in the last row.
Professor Ramamurti Shankar: And how big is
The same as F2, 3.
Professor Ramamurti Shankar: Right.
I don’t want to give it too many different names,
because F2, 3 and F3,
2 are equal and opposite. I’m already showing F2,
3 acting to the left. Let me give it some other name
like f. Then you agree this will be the
same f but pointing that way.
Here is the mistake some people make.
They add to that the 10 Newtons. It’s the 10 Newtons acting on
it, because I know the 10 Newton is pushing me and I’m going to
feel it even if I’m here. That will be a mistake.
That’s an example of adding a force that you really shouldn’t
be adding. The only force acting on this
guy is this little f. That’s, in turn,
because this guy’s being pushed by the 10 Newtons,
but that’s not your problem. Your problem is to only look at
the forces of contact on you, and that is just this f.
Then we do F=ma for these two guys.
For this guy, F=ma will be 10 minus
f is 3 times a. The other one,
it’ll be f=2a. Notice I’m using the same
acceleration for both. I know that if the second mass
moved faster than the first one, then the picture is completely
wrong. If it moved slower than the
first one, it means it’s rammed into this one.
That also cannot happen, so they’re moving with the same
acceleration. There’s only one unknown
a. Once you got this,
you realize what you got to do. This equation is begging you to
be added to this equation. You got a plus f and a
minus f, so you got to do what you got
to do. You add the two,
you get 10=5a and you get a=2.
Once you got a=2, you can go back and realize
here that f=4, so you got really 4 Newtons.
Now, we know the full story. 4 Newtons acting on 2 kg,
gives you an acceleration of 2. On this guy,
I have 10 from the left and I have 4–10 acting this way,
4 acting that way. That is 6 Newtons divided by 3
kg is also an acceleration of 2. This is a simpler example.
A simple example of free body diagrams.
Very simple. If you can do this,
you can do most of the problems you will run into.
Just don’t add stuff that’s not there.
That’s all you have to be careful about.
The stuff that people tend to add sometimes is to keep drawing
the 10 Newtons that’s acting on that. Now, here’s another variation.
The variation looks like this. I got 3 kg and I have a rope.
I got 2 kg and I pull this guy with 10 Newtons. What’s going to happen?
Again, your common sense tells you, “Look, you are pulling
something whose effective mass seems to be 5,
the answer is 2.” Let’s get that systematically
by using free-body diagrams. Now, there are really three
bodies here. Block one and block two and the
rope connecting them. In all these examples,
this rope is assumed to be massless.
We know there is no thing called a massless rope,
but most ropes have a mass, but maybe negligible compared
to the two blocks you are pulling, so we’ll take the
idealized limit where the mass of the rope is 0.
Here is the deal. 3 kg is being pulled by the
rope on the right with a force that I’m going to call T,
which stands for tension. The rope is being pulled
backwards by this guy, the T.
What is the force on the other side?
What should that be? Who said T?
Why is it definitely T and not something else?
What would happen if it’s something else?
Student: If it was greater,
then the rope would snap. Well, if it was greater
[inaudible] Professor Ramamurti
Shankar: It won’t snap, but something else will be a
Professor Ramamurti Shankar: Not only faster,
but what will be its acceleration?
If the two forces don’t cancel, you have a net force.
What are you going to divide by to get the acceleration?
Professor Ramamurti Shankar: Zero,
right? So, a massless body cannot have
a net force on it, because the acceleration of the
rope cannot be infinite. In fact, it has to be some
finite number, which is the acceleration of
either of these two guys. So, massless bodies will always
have, like a massless rope, equal and opposite forces on
the two ends. That is called the tension on
the rope. When you say the rope is under
tension being pulled from both sides by a certain force,
the tension is not 0 just because this T and that
T cancel. It’s true the net force is 0,
but it doesn’t mean you can ignore it.
Suppose you are being pulled by my favorite animals — the
elephants — from both sides by equal force.
You don’t find any consolation in the fact that these forces
add up to 0. You feel the pain.
That pain is what–This gentleman doesn’t agree.
One of you guys nodding your head.
Do you feel the pain? Do you agree?
Okay. That force is called a tension.
Whenever you’re asked on a problem, “What’s the tension on
the rope?” you’re looking for that equal
and opposite forces acting at two ends of the rope.
We don’t know what it is. We’ll give it a name,
but by linking that T to that T,
we can also figure out the same T must be exerted on this
one by Newton’s third law, because if this block is the
only one that could be pulling this rope with T.
Therefore, the ropes will be pulling the block with T
in the other direction, then I got 10 Newtons here.
Now, you can do F=ma for the three different objects.
There’s nothing to do here, because the forces are 0,
the mass is 0. It doesn’t tell you anything.
The first one tells you T=3a.
The other one tells you 10 – T=2a.
Again, we know what to do. We add these numbers and we get
10=5a. Therefore, we find a=2.
Once you find a=2, you find T=6.
So, tension on the rope is 6 Newtons.
This is very important, because when you buy a rope,
they will tell you how much tension it can take before it’ll
snap. If your plan is to accelerate a
3 kg mass with an acceleration of 2 meters per second,
you better have a rope that can furnish that force and it can
take the tension of 6 Newtons. Now, for the- whoa! I’m going to give the last
class a problem which is pretty interesting, which is what
happens to you when you have an elevator.
Here is a weighing machine and that’s you standing on the
elevator. We’re going to ask,
“What’s the needle showing at different times?”
First, take the case in the elevator is on the ground floor
of some building and completely addressed.
Then, let’s look at the spring. The spring is getting squashed
because you are pushing down and the floor is pushing up.
You are pushing down with the weight mg,
and the floor has got to be pushing up with the md,
because the spring is not going anywhere.
So the spring is being pushed by mg here and mg
here. Therefore, it’ll compress by an
unknown x, which is equal to mg
divided by the force constant of the spring.
By the way, that is a subtle thing people may not have
realized. Even in the case of this
spring, when you pull it, if you pull it to the right by
some force. Remember, the wall is pulling
to the left with the same force. So springs are always pushed or
pulled on either side with the same force.
We focus on one because we are paying for it,
but the wall is doing the opposite.
We don’t pay any attention to that.
You cannot have a spring pulled only at one side,
because then it will then accelerate with infinite
acceleration in that direction. This spring is getting squashed
on either side, and it’ll squash by certain
amount x, that depends on your mass,
and that x will be turned into a motion of a needle
and that’ll read your mass. Now, what happens if the
elevator is accelerating upwards with an acceleration a?
That’s the question. The way I analyze it is,
I say, if I look at me and I write the forces on me,
that is mg acting down, then we use w as the
symbol to represent the force exerted by the spring;
w – mg=ma.
That is, F=ma. When I was not accelerating and
everything was at rest, ma was 0,
w was mg, and w was the reading on
the needle. But if I’m accelerating,
the force exerted by the spring and therefore,
the needle the weighing machine reads is m times g
plus a. It means, when you are
accelerating upwards, as the elevator picks up speed,
the reading on the spring will be more and you will feel heavy.
You feel heavy and it reads more because the poor spring not
only has to support you from falling through the floor,
but also accelerate you counter to what gravity wants to do.
That’s why it is g plus a.
So, you picked up some speed, then you’re coasting along at a
steady speed. Then, a drops out and
you weigh your normal self. As you come to the top of the
building, the elevator has to decelerate, so that it loses its
positive velocity and comes to rest.
So a will be negative and w,
in fact, will be less than mg.
So, you will feel weightless for a short time or you’ll feel
your weight is little. Then, you come to rest and the
opposite happens on the way down.
Let me just briefly look at the ride on the way down.
As you start on the top and go down, your acceleration is
negative. Remember, you’ve got to keep
track of the sign of acceleration,
so if you’re picking up speed towards the ground,
a is negative, so it will be g plus
a, but a is a negative number.
Let me write it as g minus the absolute value of
a. You can see that if a
was equal to g, your downward acceleration is
that of gravity, namely the cable has snapped in
the elevator, then you don’t feel any weight.
You don’t feel any weight because your weight is the
opposition you get to falling through the floor;
but if the floor is giving way and you’re just falling freely,
you feel weightless. It’s wrong to think that you
feel weightless because you escaped the pull of gravity.
We all know that in a falling elevator, you definitely do not
escape the pull of gravity. It’s going to catch up with you
in a few seconds. Likewise, when in outer space,
when you are orbiting the Earth, people are always
floating around in these space stations.
They have not escaped the pull of gravity either.
They have just stopped fighting it.
If you escape the pull of gravity, your spaceship will be
off, won’t be orbiting the Earth.
Anyway, I will return to this next time.
Do your problems. There’s quite a few problems.
You should start them right away.