Professor Ramamurti 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
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. Yes?
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?
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
[inaudible] 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? Yes.
Professor Ramamurti Shankar: And what else?
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. Student:
[inaudible] 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.
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?
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. Yes?
Student: 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. Yes?
Student: 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,
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? Yes?
Student: 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. Yes?
Professor Ramamurti Shankar: What do I learn
from the period? Student:
[inaudible] Professor Ramamurti
Shankar: How? Student:
[inaudible] 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?
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.
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
Student: 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
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
What's the difference between saying F = -kx and F =
ma? They are saying very different
Student: 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. Then?
Student: 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
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? Yes?
Student: 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. Student:
[inaudible] Professor Ramamurti
Shankar: And how big is that?
Student: 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
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 problem.
[inaudible] 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? Student:
[inaudible] 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