I was going to do a brand new
topic but I didn’t want to go without making sure everyone
understands the loop-the-loop problem.
Because that’s one problem where if you didn’t learn
Newton’s laws and apply them properly, you just couldn’t
figure it out, okay?
All the other things we do, you know, blocks sliding down
planes, friction, pulleys,
you have an intuitive feeling for that, but this one really is
very non-intuitive right? Here is a trolley at the top of
the track. The forces on the trolley,
both gravity and the force of the track, are both pushing
down. And the question is,
“Why isn’t that falling down?” Everything is pushing down.
Why isn’t it falling down? That’s the mystery,
so what’s the answer you’d give to somebody who says,
“Why isn’t it falling down?” Yes?
Student: [inaudible] Professor Ramamurti
Shankar: Yes, so they might say,
“Well, if you say it’s falling down, how come it doesn’t end up
here to the center of the circle?” Yes, somebody have an answer to
that, yes? Student: [inaudible]
Professor Ramamurti Shankar: Pardon me?
Student: [inaudible] Professor Ramamurti
Shankar: No, there are no forces in the
horizontal direction; the forces are all–yes?
Student: [inaudible] Professor Ramamurti
Shankar: It’s not negligible.
There is an acceleration that’s producing a change in velocity.
But when we say an apple has a downward acceleration,
you let it go and it falls down;
that’s what we think is downward acceleration.
Here is something which is downward acceleration,
but it’s not falling down. Yes?
Student: Force changes the direction of velocity
vectors over time, acceleration times time,
it just changes the direction of the velocity vector.
Professor Ramamurti Shankar: Ok.
Yes, I’ll take one more answer, yes?
Student: The downward component and the horizontal
component of the velocities are independent of each other so at
the same time that the cart can be falling down it can be moving
so much in one direction [inaudible]
Professor Ramamurti Shankar: Alright.
So, let me summarize all the different answers.
First of all, one thing you should remember,
there are no horizontal forces here.
The forces are what I’ve drawn; that’s why you should be very,
very careful. Just because something has a
tendency to move to the left doesn’t mean there’s a force to
the left. That’s the whole point of
Newton’s law, force is connected to
acceleration not to velocity. So yes, this will have an
acceleration, you cannot avoid it.
F=ma will tell you T + mg is ma,
and the a that you want in this problem,
if it’s moving in a circle, is mv^(2)/R.
So, it is accelerating, and the difference between this
one accelerating, let us say–If I go to the
loop-the-loop and I drop an apple,
forget the track it’s got mg acting,
it’s accelerating too. What’s the difference between
these two? Here, we start with zero
velocity; you accelerate downwards.
That means you pick up a downward velocity.
As it accelerates you keep falling down and picking up more
and more speed. The same acceleration’s
experienced by the cart also, and a little more because
T is also pushing down. But there the change in
velocity over a tiny amount of time is not added to zero
velocity, but to this large velocity it has at that instant.
And the tiny increment in velocity if added this way will
produce a velocity in that direction,
and what that really means if you wait a small amount of time,
this thing is going along the new velocity direction.
So, that’s the difference between–So, the acceleration is
the same for a falling apple as for the loop-the-loop cart.
But in the apple’s case, it is added to zero initial
velocity so the acceleration and the velocity are the same,
namely down. If you have a huge velocity to
which you add a tiny change in velocity, the downward direction
you rotate the velocity vector that just means that as it goes
around the circle, the velocity vector is
constantly tangent to the circle.
There’s another way in which people understand this.
If you are on the Earth and you are standing,
say, on a building and you fire a gun,
it goes like that and it hits the ground because of
acceleration due to gravity. If you fire it a little faster
it’ll go there. There’ll be a certain speed at
which it keeps falling but doesn’t get any closer to the
center. That is in fact how you launch
a bullet into orbit. So, what’s the first thing you
should do when you fire this gun? Move away because it’s going to
come back and get you from behind, okay [students
laughing]. Alright that’s a very useful
lesson. So, you might think,
“Hey I shot it in that direction, it’s not my problem.”
Like everything else it’ll come back and get you when you least
expect it. This is how you have–;So,
is this thing falling or not? It is constantly falling but
the Earth is falling under it so it’s endless search to come to
the Earth. Whereas if you drop the bullet,
it’ll just fall and it’ll hit the ground;
that’s what we always think as drop.
Okay, so this is something I did in a rush,
and I didn’t want that because this is really the part of
physics which is not intuitive, and I think we all know what
the main point is. The main point is velocity is a
vector and its value can change due to change in direction,
not change in magnitude. Yes sir?
Student: That negative or that downward change in
velocity, what is that equal to? Professor Ramamurti
Shankar: Okay, if you wait a certain time
Δt, then the acceleration is
v^(2)/r, put the v,
put the r, get some number,
multiply by the Δt, 1/10^(th) of a second.
That’ll be the change in that time.
So, in that time also the guy would have moved over in the
circle a little bit, just precisely so that at that
point your new velocity vector is tangent to your new location.
So this little Δv I drew is not a fixed number;
it depends on how long you want to wait.
Calculus says wait an infinitesimal amount of time,
in practice wait 1/10^(th) of a second and it’ll be fine.
Okay, today there is a very, very important concept I want
to introduce. It’s a very robust and powerful
concept; it has to do with energy. It is powerful because after
the laws of quantum mechanics were discovered in the sub
atomic world, we basically gave up on many
cherished notions. You know, you must have heard
that particles cannot have a definite position and definite
velocity at a given time. They don’t have any
trajectories. You see a particle here now,
then you see it there. That’s true because they’re
observed events. What happens in between?
You might think they have a trajectory but they don’t.
The forces are not well defined because if you don’t know where
the guy is, you don’t know what the velocity is,
you don’t know what the acceleration is.
So most of the ideas of Newtonian mechanics are
basically surrendered, but the notion of energy turns
out to be very robust and survives all the quantum
revolution. In fact, there is a period when
people are studying nuclear reactions and the energy you
began with didn’t seem to be the energy you ended up with.
So something was missing. So Neils Bohr,
father of the atom said, “Maybe the Law of Conservation
of Energy is not valid.” Then Pauli said,
“I think I put my money on the Law of Conservation of Energy.
I would postulate that some other tiny particles that you
guys cannot see, that’s carrying away all the
missing energy.” That was a very radical thing
to do in those days when people did not lightly postulate new
particles. Nowadays, if you don’t
postulate a particle you don’t get your PhD in particle physics
[students laugh]. In those days it was very
radical, and the particle was not seen for many,
many years. It was called a neutrino.
Nowadays, neutrinos are one of the most exciting things one
could study. There’s a lot of mysteries in
the universe connected with neutrinos.
Okay, so energy is a concept you’re going to learn.
Here’s again a place where I ask you, don’t memorize the
formulas and start plugging them in.
Try to follow the logic by which you are driven to this
notion, because then it’ll be less luggage for you to carry in
your head. And I’ll try to make the notion
as natural as possible rather than saying, “This is my
definition of energy, this is the definition of
kinetic energy.” Let’s ask where does all that
come from. How many people have seen
kinetic and potential energies in your past?
Okay, if you haven’t seen it that’s fine, this is not
something you’re supposed to know.
If you know it and you learned it properly it’ll be helpful,
but you don’t have know that. You just have to know what we
have done so far. So let’s go back and ask a
simple question. When a force acts on a body,
what is its main effect? What does it do?
Yes? Student: It changes its
motion. Professor Ramamurti
Shankar: Be more precise by it when you say “changes its
motion.” What does it mean to change its
Student: It accelerates the object [inaudible]
Professor Ramamurti Shankar: Right,
that’s from Newton’s law; that means it changes the
velocity. So let’s for a moment,
in fact for the entire lecture today, go back to one dimension.
Then, we will do the whole business in higher dimensions.
In one dimension changing the velocity simply means speeding
up. So bodies are going to speed up
and forces act on them. That we know.
We’re going to find the relation between how much speed
is accumulated when a force acts on a body;
we’re going to derive that. So, let us take an example
where the force is in the x direction and it is
some constant. “Constant” meaning it’s not
varying with time; somebody just has to just apply
10 Newtons constantly to a body. What does that do?
That produces an acceleration which is F/m,
right, that’s Newton’s law. That’s a constant acceleration.
So, let’s go back to lecture number one.
Lecture number one we learned that if a body has constant
acceleration, then the velocity at the end is
the velocity squared at the beginning plus 2 times a
times the distance traveled. That’s what we learned long
back, this is just kinematics. In those days we didn’t ask,
“Why is it having a constant acceleration?”
You were just told, “It’s a constant acceleration,
just go ahead.” Now that we learned dynamics,
we know acceleration has a cause, namely a certain force.
So, I’m going to write here as you can imagine F/m in
the place of a, but I’m going to make one more
cosmetic change in notation. In all these energy formulas,
rather than calling the initial velocity or initial anything
with subscript zero and the final is no subscript,
we are going to use a notation where the final quantities are
denoted with a subscript 2, initial quantities are denoted
by a subscript 1. In other words,
initially, the body had a velocity of
v_1, and then it had an acceleration
a for a distance d,
then the relation between final initial velocities is this.
So let’s put that F/m here and let’s put a d. Since we are interested in the
effect of the force, we want to see what the force
did to the velocity. What you find here is that if
you want to find the change in the velocity,
of course you should take this guy to the left-hand side,
right? Take it to the other side and
you want to isolate the force, so bring m/2 to the
left-hand side. So, do that in your head and
what do you get? m/2V
2^(2) – m/2v 1^(2) is F times d. This is a very important
concept. It says, when the force acts on
a body it changes the velocity, and it depends on how far the
force has been acting, how many meters you’ve been
pushing the object. It’s clear that if you didn’t
push it at all, then even the forces acting,
velocity hasn’t had time to change.
So, it seems to depend on how far the force acted,
and the change is not simply in velocity but in velocity
squared. That’s what comes out;
then, you realize that it’s the most natural thing in the world
to give this combination a name because that’s what comes out of
this. That combination is called
kinetic energy, and I’m going to call it
K_2, meaning K at the end
minus K at the beginning is F times d.
So F times d is denoted by the symbol W,
and that’s called the work done by F.
That’s how we introduce a notion of a force times distance
as the definition of work. This is how people–you can
say, “Why did anyone think of taking the combination F
times d?” Well, this is how it happens.
So, you need to have a unit for this, which is actually Newton
meters, but we are tired of calling everything a Newton
meter, so we’re going to call it a
joule. So joule or J,
is just a shorthand for Newton meters. Alright, so let us now write
down this following expression here.
ΔK, Δ in anything we do is the change in any
quantity, it’s called delta of that quantity,
ΔK is a change in K is equal to the
F times the distance traveled.
Let the distance traveled be some tiny amount Δx.
This formula is valid even if d is 1km,
as long as force is constant. But I’m saying,
let’s focus on a problem, whether the force is constant
or not, over a tiny interval in which the body moves the
distance Δx. Then, we find that the change
in kinetic energy is force times Δx.
Now, what if there are 36 forces acting on the body?
Which one should I use? I’m pulling and you’re pushing
and…. It’s got to be the net force
because Newton’s law connects the net force to the
acceleration. If you and I have a tug of war
and we cancel each other out to zero then there’s no
acceleration. So, you’ve got to go back to
the definition; this is very, very important.
This F better be the total force.
I should really use a subscript T for total,
but I don’t want to do that and we write F=ma;
it is understood F is everything.
We’ll come back later to the notion of what happens if
F is made up of two parts.
So, maybe I’ll just give you a simple example.
Here’s a body, right, and I’m pushing it this
way and you’re pushing it that way, let’s say with equal force.
And the body could be at rest or if the body had initial
velocity it’ll maintain the initial velocity because
velocity is for free. Then, you can ask,
“What does this theorem say?” This is called a Work Energy
Theorem. The Work Energy Theorem says,
“The change in energy is equal to the work done by all the
forces.” In this case,
there is no change in energy, but there are two forces at
work that cancel. Then, it turns out that it is
sensible to define the work done by me, which is equal to
F times the distance traveled,
and the work done by you will be in fact given a minus sign.
So, when do we attach a plus sign and when do we attach a
minus sign? If you go back to the whole
derivation, it was understood that a was a positive
quantity. Then everything works.
But if the body is moving to the right, and I’m pushing to
the right, then the work done by me is positive.
And if you were pushing to the left and the body still moved to
the right, the work done by you is negative.
In other words, if you get your way,
namely, things move the way you’re pushing,
the work done by you is positive.
If people are pushing you counter to your will,
in the opposite direction to your force, work done by you is
negative. See, here is a simple example.
I have a piece of chalk and I’m lifting it at constant velocity
from the ground. Its kinetic energy is not
changing. That means the total work done
on it is zero, but it’s not because there are
no forces on it. There is gravity acting down
and there is me countering gravity with exactly the right
amount I’m applying the force mg.
So, work done by me would be positive because I want it to go
up, and it goes up. So, if it goes up by an amount
h work done by me is mg times h.
Work done by gravity in the meantime is minus mg
times h, so the total work done is zero.
So, this formula says this is the total force,
but I’m telling you if you take the work done and call it
ΔW it maybe a result of many works,
work done by me, work done you,
work done by somebody else, some are positive,
some are negative, you add them all up.
So, that is the notion of work done by one of many forces
acting on a body. And the total work done is the
algebraic sum of the work done by all the forces.
Alright, so we’ve got this result;
let’s do the following. Let us imagine all of this
occurs in a time Δt. Then ΔK/Δt,
which will eventually become dK/dt if you take all the
proper limits becomes F times dx/dt,
which is force times velocity and that’s called power.
So, power is the rate at which work is done.
For example, if I climb a 12-story building
I’ve done some work; my m times g
times the height of the building.
I can climb the building in one minute, I can climb the building
in one hour, the work done is the same,
but the power is a measure of how rapidly work is done.
That’s why it’s the product of force and velocity.
So, the units for power are joules per second;
now, that also has a new name, which is watts or just
W. You may use a kilowatt,
which is kW, little k then W is thousand
watts. So, if you’ve got a 60-watt
bulb, it’s consuming energy at the rate of 60 joules per
second. That’s the meaning of a 60-watt
bulb. Okay, now I’m going to do the
next generalization, which is when the force is not
a constant but varies with x. So, do we know any example of a
force that varies with location? Student: A spring.
Professor Ramamurti Shankar: A spring is one
example. Even gravity I think we know is
not a secret, the force of gravity is
mg near the Earth, but if you go sufficiently far
you will notice gravity itself is getting weaker.
It’ll still look like mg but g won’t be 9.8,
g will become 9.6 and 9.2 and 9.1, it’ll decrease.
But the variation of gravity in terrestrial experiments is
pretty small so we won’t worry about it.
But the force of a spring definitely varies with x
in a noticeable way. The force of a spring is one
concrete example; it’s what you all seem to
recognize; it’s this one [(-kx)].
So then, we have to ask what is the Work Energy Theorem when the
forces itself vary? So, let’s draw ourselves a
force which is varying, here is x,
here is my force; it does something.
F=kx or –kx is a particular graph,
I’m just taking any function of x that I want.
Now, the acceleration is not constant because the force is
not a constant. So, if you wait for one hour
the change in kinetic energy is not simply force times distance
because what force are you going to use?
It was something now, it is something later,
it is positive, now it’s negative later.
So, we cannot apply this formula because F is not
a constant. So, the usual trick in calculus
is find me an interval in x, which is so narrow,
and I’ll make it as narrow as you insist, so that during that
period I’ll think that F is a constant and the constant
value of F is just the function F at that time
x, at that time where the location
is x. So, for that tiny interval I
can still say that F is constant;
F over m is some constant over the tiny interval,
then the change in kinetic energy will be then F(x)
times Δx. Then, you wait a little longer,
you go to a new place, then you’ve got a new value of
the force, maybe somewhere here. Then, that times Δx is
the work done in that region. But geometrically,
F(x) times Δx is the area of the little rectangle
whose base is Δx and whose height is the function
F at that time. So, if you eventually went from
here to here [distance on graph], let me call that point
as x_1 and let me call the ending point as
x_2, then the work done by you is
really given by the area under that graph.
And in every segment you pick up the change in kinetic energy,
you add it all up. When we say add it all up on
the left-hand side, all the ΔKs will add up
to give me the final minus initial kinetic energy.
This one is what is written in the notation of calculus as
F(x)dx from x_1 to
x_2. This is called the integral of
the function F(x). Now again, in a course at this
level I’m assuming that you guys know what an integral is.
Even if you’ve never heard of an integral, if I give you a
function and you never heard of calculus, you can still deal
with this problem. Because you’ll come to me and
say, “Give me your function.” I’m going to plot it in some
kind of graph paper with a grid on it, and I’m just going to
count the number of tiny squares enclosed here.
That’s the area and that’s the change in kinetic energy.
So, integration is finding the area under graphs bounded by the
function of the top, x axis below,
and two vertical lines at the starting and ending points. Now, a little digression.
Even though you’re supposed to know this, I want to make
sure–part of my “No Child Left Behind” program,
that you all follow this. If you never heard calculus or
integral calculus this is going to be a two-minute,
three-minute introduction. There is a great secret,
which is that if you give me a function and you tell me to find
the area under this from x_1 to
x_2, I can go to the graph paper and
draw that, or there is quite often a trick.
You don’t have to draw anything, you know.
And I will tell you a little more about what it is.
If you give me a function F(x), the area under this
graph happens to be the following.
There is another function G(x) that will be
specified in a moment. And you just have to find
G at the upper limit minus G at the lower
limit, and that will be the area.
And who is this mystery function G?
It is any function whose derivative is the given function
F. So, the whole business of
finding area which was solved around Newton’s time,
is if you have a function they want the area under it,
you go and think of a function whose derivative is the given
function. So, it is the opposite of
taking derivatives. Differentiation is an
algorithmic process; you take F(x),
you add a Δx to the x, you find the change in
F divide by Δx. Given any function you can do
that. This is the opposite;
this is guessing the function G whose derivative is
F. So, if I say F(x) equals
x^(3), then G(x) is that
function whose derivative is x^(3).
So, I know I’ve got to start with x to the 4^(th)
because when I take derivatives it’ll become an x^(3),
but with the 4 in it. So, I fight that by putting a 4
downstairs and that’s the function G.
Yes? Student: Shouldn’t there
be a c? Professor Ramamurti
Shankar: Yes, that is the plus c.
The c is, of course, velocity of light
[students laugh]. Now, why is this plus c
there? She’s quite right,
because if you’ve got a function whose derivative is
something, adding a constant doesn’t change the derivative.
Then you can say, “Well, we are in trouble now
because you told me to find the function G,
and you’re telling me the world cannot agree on what G is
because I’ve got one G, you’ve got another one with a
different constant.” But the beauty is that when you
take this difference, I may not put the constant,
and you may put the constant, but it’s going to cancel out.
So, most of the time, people don’t bother with the
constant. Sometimes it’s very important
to keep the constant in place — it plays a special role — but
we don’t want that here. We’ll just call this the
integral. So, maybe it’s again worth
asking for a second, why is this true,
you know. Why is G the function
whose derivative is F? Why is that the answer to the
problem? Well, one way to think about
it, this is really more for your completeness than anything else.
Suppose there is a function F, and I’m finding the
area, let me find it up to the point that I’m going to call
x_2. For the starting point let me
just pick zero. This is the area I’m looking
for, and I write it as F(x)dx from zero to the
point x_2. You agree that as I slide
x_2 back and forth, the area is varying with
x_2, so it’s legitimate to call that
a function of x_2. A function is anything that
depends on something else, and we all agree that if the
point actually moves, the area moves.
This is my shorthand for the area.
The starting point is nailed at zero;
this is G(x_2). The left asks what is
G(x_2+Δ). That is the area to this point,
that we denoted as F(x)dx from zero to
x_2 + Δ, that’s
G(x_2+Δ). So, what is the difference
between the two? The difference between the two
is G(x_2 + Δ) – G(x_2).
If you look at that graph, the extra area that I have is
this thing, and that is clearly equal to F at the point
x_2 times Δx.
In other words, by how much is the area
changing when I change the upper limit?
Well, it’s changing by the height of the function at the
very last moment times the change in x.
Well, we know what to do now, divide by Δx and take
the limit, and you can see that F(x_2) is
dG/dx at the point x_2.
But the point x_2 is whatever
point you want. So, we drop the label,
and we just say that F as a function of x is
dG/dx. This is the origin of the
result that the area can be found if you can find the
inverse derivative of the function.
So, let’s keep a couple of popular examples in mind.
One, if F(x) equals the constant, then the G(x)
is equal to constant times x.
If F(x)=x then, G(x) is equal to
x^(2) over 2. As you know,
the integral of x^(n) is x^(n) +1 over n + 1.
There are integrals for all kinds of functions,
and you check them by taking derivatives and make sure you
get back your F. Alright, suppose this is what
we learned from the math guys, then go back to the Work Energy
Theorem and see what it says, K_2-K
_1. Oh, by the way,
I didn’t complete one story; let me finish that.
This was, if G(x_2) was the
area from zero to the point x_2,
G(x_1) is the area from zero to the point
x_1. That is perhaps clear to you,
if I only wanted the area between the point
x_1 and the point x_2,
I can take the area all the way up to x_2,
subtract from it the area all the way up to
x_1, and that’s of course the claim
that this integral F(x)dx from x_1 to
x_2 is G(x_2) –
G(x_1). That’s the reason the integral
is given by a difference of this G.
So, let me use a shorthand notation and call
G(x_2) as simply G_2,
G(x_1) is G_1.
You should understand, G_2 means
G evaluated at the final point, G_1 to
G evaluated initial point.
So, you have got K_2 –
K_1 is G_2 –
G_1, but G is some definite
function of x. You’ve got to understand that.
For example, it could be
x_2^(2) over 2 – x_1^(2) over
2 in the problem where F is equal to x.
Then, let’s rearrange this thing to read K_2 –
G_2 is equal to K_1 –
G_1. So, the Work Energy Theorem
says that, “If a force is acting on a body, a variable force
associated with the force is the function G,
a variable function of x, whose derivative is
the given force.” Then, at the beginning,
if you take the difference of kinetic energy minus the value
of G_2, it’ll be the same.
I’m sorry, K_1 – G_1,
it’ll be the same as K_2 –
G_2. Now, we don’t like the form in
which this is written. We have to make a little
cosmetic change, and the cosmetic change is to
introduce the function U(x), which is minus this
function G(x). So, once you do that,
you’ve got to remember now the force is equal to minus the
derivative of this function, because of the minus sign
relating G to U. Then, we can write it as
K_2 + U_2=K_1 +
U_1. And that combination,
if you call that as E_2 and you
call this combination as E_1,
we are saying, E_1=
E_2, and that’s called the
Conservation of Energy. So, what does conservation mean?
Conservation of Energy in physics has a totally different
meaning from conservation of energy in daily life.
Here it means, when a body’s moving under the
effect of this force F(x),
you agree it’s gaining speed or losing speed or doing all kinds
of things. So, as it moves around it’s
speeding up and slowing down, so its speed is definitely a
variable, but a certain magic combination
connected to the velocity of the object and to its location,
U depends on x, that does not change with time.
And it’s very useful to know that does not change with time,
because if you knew the value of E,
even if E were to equal some fixed number E at
one time you know E at all of the times.
So, let’s take a simple example. So, we take a rock and we drop
it. We know it’s picking up speed;
we know it’s losing height. So, you may think maybe there
is some combination of height and speed, which does not change
in this exchange, and we find the combination by
this rule now. In the case of gravity,
F=-mg; therefore, U=mg times
y, because –du/dy will now be my
F. Then, we may conclude that
½mv^(2) + mgy at the initial time will be equal
to ½mv_2^(2) + mgy_2 at the
final time. And that is the–And the whole
thing can be evaluated any time. You take any random time you
want. The sum of these two numbers
doesn’t change. And that’s a very,
very powerful result. So, I’m going to give you one
more example later when you can see the power of this.
But in the case of gravity, this is what it means.
So, this agrees with the intuitive notion that when you
lose height you gain speed, when you lower your y
you gain speed, but it’s much more precise than
that. Instead of saying vaguely,
“you lose and gain,” ½mv^(2) is the part
connected with the speed and mgy is the part connected
with the height and that sum does not change with time.
So, let’s go back to the mass and spring system.
In the case of the mass and spring system,
if you have a mass connected to a spring,
then we say here ½mv^(2) +
½kx^(2) equal to E is a constant.
That means, find it at any initial time one,
find it at any final time two, add these two numbers,
the total will not change. So, let me put that to work.
Let me use that in a calculation.
So, I say here is a spring; here is where it likes to be.
I’m going to pull it to a new location a.
x=0 is where the spring likes to be.
I’m going to pull it by an amount A and I’m going to let it
go. And I want to know how fast the
guy is moving when it comes back to, say, x=0. If you go back to Newton’s
laws, it’s a pretty complicated problem.
You have got to think about why it’s a complicated problem.
Because you start with a mass at rest, if you pull it by an
amount A, a force acting on it,
it would be –k times A;
that will produce a tiny acceleration,
will produce an acceleration, right, kA/m,
which will give it a new velocity by the time it comes
here. But once it comes to the new
location, a little later, a different force is acting on
it. Because the x is now
different, so the acceleration during this interval is
different, and the gain and the velocity
of that interval will be different from that in the first
interval. You’ve got to add all these
changes to find the velocity here.
That’s a difficult proposition, but with the Law of
Conservation of Energy you’ll do that in one second.
In fact, let’s do it not just to be at the mid point here,
but at any point x measured from equilibrium.
Because we will say initially ½mv_1^(2) +
½kx_1²=½m v_2^(2) +
½kx_2². This is the truth;
this is a constant that does not change with time.
Now, to get some mileage out of this, you’ve got to know what
the constant value is. Well, we know what it is in the
beginning because when you pull this guy, at the instant we knew
it had no velocity, and we knew it was sitting at
the point A, so we know ½k
A_1² is the total energy of this mass and spring
system. That’s not going to change with
time. We like to say that at the
initial time all the energy is potential energy;
there is no kinetic energy because there is no motion.
At any subsequent time, this number has to be equal to
what you get. Now, let me drop the subscript
to there [pointing to equation]. Let it stand for some generic
instant in the future. This much we know.
Now, do you see that we can solve for the velocity on any
location x? You pick the x;
if you want x to be zero that’s very easy,
you kill that term, you balance these two,
you cancel the half and you can see that’s a formula for
v^(2). v^(2) is going to be
ka^(2) over m; that is going to be the
velocity when it swings by this midpoint.
But you can pick any x you like, if the x was
not zero but .1; well, put the .1,
take it to the other side and solve for v.
This is one example of the Law of Conservation of Energy.
So, the power of the Law of Conservation of Energy is if you
knew the initial energy, you don’t have to go through
Newton’s laws, and you’ll find it in all the
problem sets and many exams. When a problem is given to you
and you are told to find the velocity of something,
you should try whenever possible to use the Law of
Conservation of Energy rather than going to Newton’s laws.
So, if you want to know what have we done,
how come finding the velocity seemed so difficult when I
described it to you earlier, and would we have done it,
the answer is, if you tell me the force of the
spring is a function of x,
I have done that integral in the Work Energy Theorem,
once and for all. The integral of the spring
force has been integrated once and for all and that’s what the
potential energy is doing. The potential energy comes from
integrating that force. Alright, so now lets take
another problem where there are two forces acting.
Yes? Student: Could you
explain really quick how you got the ½ in front of the
kx^(2) for the energy? Professor Ramamurti
Shankar: Let me see, here?
Student: No, down at the bottom of the
center, yeah. Professor Ramamurti
Shankar: Here? Student: Is it just
because of the integration? Professor Ramamurti
Shankar: Yeah, for a spring I think we said
here, if the force is x,
the potential energy is G=x^(2)/2,
U will be -x^(2)/2.
So, the force is –kx, then you have to find the
U so that –du/dx=–kx and the answer to
that is U is kx^(2)/2.
So, if you knew the force, you just find the function
U by asking, “Whose derivative gives me
F with a minus sign?” Then you are done.
That’s the potential energy for that problem.
Now, lets take another problem, a mass is hanging from the
ceiling. Then, it’s got two kinds of
potential energy, one is just a gravitational
potential energy, and the other is a spring
potential energy, and that’s going to come
because K_2 – K_1 will be the
integral of F(x)dx. I’m sorry, let’s call it
dy and F(y)dy from y_1 to
y_2. And this F(y) will have
two parts, the force of gravity — I’m tired of writing the
limits — then, the force of the spring,
from 1 to 2 [adding the limits to the integrals].
Right, there are two forces acting on it.
This one will be mgy_2 –
mgy_1, this one will be
½ky_2^(2) – ½ky_1^(2).
So, every force acting on the body will turn into a potential
energy, because the force–Newton’s law says Work
Energy Theorem applies to the total force.
So, the total force is two parts, and you can do two
integrals; each integral is an upper limit
minus lower limit of some function and this is what you
get [pointing to formula]. Oh, I think I made one mistake
here. I think this should give me,
yes, this overall sign is wrong in these equations.
Really, there’s a minus in that and minus in that.
In other words, the work done by the force,
if there’s only one force acting, is really U(1) –
(U2). You see why I got the sign
wrong? When you do the integral in
calculus, integral of F is G_2 –
G_1, but G and U are
minuses of each other. So, integral of force will give
U initially -U finally.
So, it should really be + mgy_1 –
mgy_2. So, the Law of Conservation of
Energy for this object will be that, ½mv^(2) + mgy +
½ky^(2) is the total energy and will not change.
So, this mass can bump up and down and go back and forth.
If you knew the total of these three numbers at one instant,
you know the value at any other instant.
For example, if you pulled it from its
equilibrium by .2 centimeters and you released it,
and you say what’s the speed when it’s .1 centimeters away,
it’s very easy because you know the initial energy was all
spring energy and gravitational energy.
At any later time if you knew the y,
if you knew the position, you know the potential energy;
that’s the whole point. Potential energy depends on
where you are; kinetic energy depends on how
fast you’re moving. And some combination of where
you are and how fast you’re moving is invariant,
does not change with time. So, if you tell me where you
are, I will tell you how fast you’re moving.
By the way, remember the kinetic energy enters as
v^(2). You can imagine,
when you solve for v you’re always going to get two
answers. Why is that?
Does it make any sense? Yes?
Student: Because you’re going to have the same potential
energy [Inaudible] problem if you’re oscillating
[inaudible] Professor Ramamurti
Shankar: Or, if a spring is going back and
forth at a given point x, you can have two velocities
because you could be going away from the center,
or it can be going back towards the center.
So, this tells you the velocity on those two will be related by
a minus sign, right?
If you pull the spring this way, when you go through the
origin you say, “What’s my velocity?”
I’ll get two answers because I could be going to the origin or
I could go the other way and on the way back again pass the
origin. So, every x you’ll get
two velocities. And that agrees well with the
fact that v^(2) is what you will solve for.
And because it’s quadratic you will get two answers and you
should get two answers; you should be expecting that.
Okay, now, we are going to find one bad boy, bad girl,
bad person, which is going to ruin this whole thing.
There is–you know what I’m talking about here,
yes? Student: Friction
[inaudible] Professor Ramamurti
Shankar: So, friction is a force acting on a
body, so what’s the thing I should try to do?
Let’s take a body with only friction acting on it.
Let’s not worry about anything else, or maybe friction and a
spring. So, I try as usual,
to get a Law of Conservation of Energy when friction is acting,
and you will see that I will not succeed.
And it looks like everything here is water-tight,
right? Everything looks good.
You tell me the force acting on a body.
I integrate the force from start to finish and I call that
the difference of potential energies;
then I’ve got my result. Try to follow this logic:
K_2 – K_1 is integral to
force, integral to G(x_2)
– G(x_1) or which is U(x_1) –
U(x_2); you can rearrange it to get
K_2 + U_2=
K_1 + U_1.
Looks like you can always get Law of Conservation of Energy.
And you want to know why that fails when there is friction.
Okay, why does it fail when there is friction?
So, when there is friction–Let me try to do it again.
K_2 – K_1 is the force
due to the spring, which is whatever –kx.
Let me do that integral plus the force of friction dx,
from x_1 to x_2.
This one is from x_1 to
x_2 and we saw what this is.
Let me write it one more time, ½kx_1^(2)
if you didn’t have friction you are done.
Because you can get kinetic plus potential for spring,
is kinetic plus potential. Why don’t we just integrate the
force of friction and then turn that into a U and then
we’re done? What’s going to be the problem?
Let me just give some–anybody else want–yes?
Student: It’s always against the direction of motion
so it would be always changing direction force [inaudible]
Professor Ramamurti Shankar: Okay what was your
answer? Student: You can’t
associate a potential function with friction because it depends
on the path, not only the initial position
and final position. The anti-derivative only
considers the final and the initial position.
Professor Ramamurti Shankar: Right,
but let’s suppose I’m somewhere and I’m pulling a block and
there is some friction, there is some force F at
the location, there’s my F.
Why don’t I just use the value of F?
Yes? Student: Energy is lost
due to heat. Professor Ramamurti
Shankar: I know but see, you are saying something else
which is speaking, I had into the book,
but right now what’s the problem with integrating the
force of friction? Student: I was just
going to say the energy of the force of friction is no longer
within the system [inaudible] Professor Ramamurti
Shankar: No, but you’re all telling me why
there is a loss of mechanical energy but I’m just a
mathematician. I’m saying, give me a force,
find its integral. What’s the problem?
Yes. Student: Is it because
the function to the force of friction depends on the
velocity? Professor Ramamurti
Shankar: Yes, I think somebody else said that
earlier also. The force of friction is not a
function of x. You might say,
“What do you mean?” I’m pushing this guy,
I know how hard it’s pushing me back, but I’m telling you now,
“Push it the opposite way!” Then, you’ll find the force of
friction at the very same location is pointing in a
different direction whereas the force of a spring is the force
of the spring. It’s -kx,
whether you’re going towards the spring or you’re going away
from the spring; try to think about that.
Same with gravity. Gravity is pulling you down
with a force -mg and it doesn’t care whether the chalk
is going up or coming down. On the way up it’s pulling it
down; on the way down it’s also
pulling it down, because at a given location
there is a fixed force. So, the problem is not the
force varies with x or y, but the force is a
function only of x. But if the function of x
and velocity then you also have to know the velocity,
and no one’s giving you then a definite value for the velocity
because in the vibrating mass and spring,
velocity can have either sign. Therefore, we do this integral
but we don’t do this. Even though I write it this
way, mathematically it’s wrong because the force of
friction–Is it really a function of x?
I’ll tell you in what limited sense it is a function and what
technical sense it is not. Let’s defer that for now.
Take all these guys to the left-hand side,
so this is K_2,
this’ll become U_2,
so you’ll get (K_2 + U_2) –
(K_1+U _1) is equal to
this last thing, work done by friction from
x_1 to x_2 and the
question is, “Can I do the integral?”
The answer is, if you just tell me
x_1 and x_2,
then I know that’s the point you’re making;
I cannot just do an integral because I will have to know the
whole story, but in limited cases I can do this.
Suppose I pulled a mass by an amount A and I let it go,
but now there is friction. I think we all know that if it
came to the center it won’t be going quite as fast.
And we also know, when it overshoots the other
side, it will not go back to –A;
normally, it will go back to -A.
Everyone knows why it goes back to -A,
right? When you go the other extremity
without friction, your energy is all potential.
It was ½kA^(2) to begin with, it’s going to be
½kA^(2) at the end, except A can go to
-A without changing the answer;
that’s why it’ll go from A to -A.
But when you’ve got friction, I cannot say the energies are
equal because I’ve got to do this integral.
For this part of the journey from here to here,
during that part of the journey, the force is a
well-defined function. Namely, it is a constant value
-µ times mg and it is acting this way.
So, for this part of the trip, I can do the integral.
It is very easy to find out what it is;
it’s -mg times µ kinetic, times the distance
traveled. Of course, it’s a little more
complicated now. You have to tell me where you
want your velocities. You might say,
find the velocity at x=0.
If you want me to do that, the distances traveled is the
original displacement. That’s what is the stuff on
that side. And here, K_2
=½mv^(2) which is what I’m trying to find.
I am at the origin so it’s ½k times 0^(2),
no potential energy. Initially, there is no kinetic
energy but there was potential energy ½kA^(2).
You follow that logic now? This is the difference in
energies due to the conserved forces, the conserved energies,
but the difference E_2 –
E_1 is no longer 0 but is given by this number.
So, I can calculate it for this part.
I can even ask, — this is one of the homework
problems, I think — you’re asked, “How far does it go to
the left?” Well, what do you do?
Let’s call the place where it goes to.
The left most extremity is A prime.
Then we don’t know, so there when you go to
K_2 + U_2 – K_1+
U_1, you will go back to this
formula. First of all,
the distance that the force of friction acts on is how much–is
the distance A to come here and another A prime
to go to the left. A prime is a magnitude
of how far you go to the left. And likewise,
on the left-hand side you’ve got to have the total kinetic
plus potential minus kinetic plus potential.
At the end of the day, you only have potential energy
½k A prime squared, no kinetic and initially,
again, you only have potential energy ½kA^(2).
You will see that A prime now satisfies a quadratic equation;
you can solve it. But what is the point?
The point is you cannot do this integral once and for all and
say the answer depends on the end points.
In this particular problem, if you told me you are doing
this part of the journey, during that part of the journey
because for the entire trip, the velocity had a particular
sign, namely to the left,
force of friction had a definite magnitude and sign,
namely to the right, and then of course it’s as good
as any legitimate function and you can do the integral.
In fact, the forces are constant;
you can pull it out of the integral, and the integral
dx is just the distance traveled.
Okay, so work done by friction–it will always come
out negative and you should keep track of that. Okay, so the final bottom line
for Law of Conservation of Energy that you guys should
remember is the following. You take all the forces acting
on a body. So, lets do it one more time in
our heads, so you know where everything comes from.
Everything comes from this result K_2 –
K_1, which is the work done,
is integral of all the forces. Divide the forces into forces
that depend on location only and not on anything else,
like velocity. Let me just divide them into
two forces, one is a good force, F_g dx,
and one is a force of friction, dx.
This guy will turn into a potential energy difference.
Let me just take one good force and the force of friction.
If you had two, each will generate its own
potential when you do the integral.
Then, we may then say E_2 –
E_1; that’s the same as
(K_2 + U_2) – (K_1
+U_1) is equal to the work done by friction.
Work done by friction is the shorthand for that integral
which you can do only; you must divide the motion
involved into pieces. For example,
if you have a mass that swung to the left then turns around
and goes back to the right, of course, again,
it won’t come back to the original, it’ll come to a
smaller number. That’s because there’s more
loss of energy on the return trip.
For the return trip, you must do the problem
separately. On the return trip,
you start from this A prime that you found,
then start moving to the right, the friction will act to the
left. So, you must divide the motion
into segments. During each segment the force
should have a definite direction and magnitude so you can do the
integral. The point is,
there is no universal formula for F(x) due to friction.
So, in one dimension that’s all it is, bringing the good and bad
forces, those which lead to potential, those which don’t;
the difference is the good forces are functions of
x, bad forces are functions of x and
something else, so that at a given x
it’s possible to have two values for that function.
Mathematically, we don’t call it a function,
that’s why you cannot do the integral once and for all.
So, this is the Law of Conservation of Energy.
Okay? Alright now,
what I have to do here for the remaining time is a little more
of some mathematical preparation for what I’m going to do next.
What I’m going to do next is to go to higher dimensions and
we’ve got to make sure we all know some of the mathematical
ideas connected to the higher dimensions.
So, I’m going to start off with some things and this is like a
mathematical interlude. Again, I told you,
if you’re not strong on some of these things you should read
this math book I mentioned to you on Basic Training;
it’ll certainly cover these topics.
But you don’t have to have that, I’m going to tell you now
how to do it. So, the math idea that I’m
going to use, see this is really a
mathematical digression to make sure everybody knows,
they are very simple things, but we do it all the time in
physics so I want to make sure you guys can also do that.
So, let’s take some function F(x) that looks like
this. This is F(x),
this is x [pointing to graph]. And let’s say you start here,
at some point x. And I’m going to go to a nearby
point, x + Δx. Δx is exaggerated so
you can see it. So, actually in the end it’s
going to be a very small number. The question is,
“What is the change in the function in going from here to
here?” So, we all know what it is;
that is the change in the function. We are going to need
approximations to the change in the function.
The useful approximation is to say, let us pretend the function
is a straight line and keeps going with the slope it had to
begin with. You can see that if the
function is not curving up or down too much,
you almost got the right change in F.
And what is the change in F?
That change in F is a derivative of the function at
the point x times the change in x.
But it is not an equality; it is not an equality because
there is this little guy on the top that you have missed.
Because the function is not following the same slope that
you have to begin with, it’s curving up.
So, let’s take a concrete example.
Suppose F(x)=x^(2), then F(x + Δx )=x^(2)
+ 2xΔx + Δx^(2). Therefore, the change in
F — you got to take it to the other side — change in
2xΔx, 2x is,
of course, what you realize is that the derivative of the
function at that point. And that’s giving you this part
but there’s a tiny portion on the top, which is Δx^(2)
that you’re not getting. If the function is
x^(3), then x to the fourth,
then hyperbolic cos (x) there’ll be more errors.
But the point is, the change in the function has
a leading piece that’s proportional Δx,
and then the corrections that we see mathematically are of
higher order in Δx. So, if Δx is made very
small but not zero, then it’s a very useful formula
to know that the change in function and going from one
point to another point is this rate of change times the change.
I don’t think anything could be more natural than that.
But we should be comfortable by saying it’s naively people say,
“Well, you should do that because in calculus you’re
canceling this Δx with that dx.”
That’s not what we’re really doing.
We all know that this stands for a certain limit,
but the limit stands literally for the slope which is at that
point. So, if functions had only one
slope and they never changed their slope and you knew that
slope, then if you know the value to
begin with, you can find the value a mile away from where you
are because it’s going at the same clip.
But most functions of course will change the slope,
and for that reason there will be an error in the formula you
get if you use this. But if Δx is very,
very small and it’s your intention in the end to make
very, very tiny Δxs then we
say we will use this approximation. That’s very simple but it’s
very useful to know. So, let’s take an example of
this function, of this rule,
because I’m going to use it all the time. So, one example is F(x)
=(1 + x)^(n). Well, F(1) is just one.
Suppose you go to a nearby point which is x away
from the .1; so, this x is really
like your Δx, then the change in the function
will be df/dx at the point 1 times the change in
Student: [inaudible] Professor Ramamurti
Shankar: I’m sorry, ah, ok, that’s right.
F(0) if you like; if x=0, the function has a
value of one. Then, if you go to a
neighboring point–;So, here is what I’m saying,
the function does something which value here is one and I
want to go to a nearby point which I’m calling x.
Then, df/dx you know is n times 1 + x^(n)
-1 times x. But you have evaluated it at
x=0, so this goes away,
then you get nx. So, the change in the function,
this function as you move off x=0 is nx.
So, what we are saying then is 1 + x raised to the
n is 1 + nx + tiny numbers, if x is small. That is a very useful result.
If something looks like 1 + x raised to a power,
and if you’re not going too far from the point 1,
namely if your x is very small, then don’t bother with
doing the whole 1 + x times 1 + xn times,
it’s approximately equal to 1 + nx. So, let me give you one example;
you will use this later on. According to relativity,
a mass of a particle is given by this [writes formula on the
board]. It’s a body sitting still,
it has certain mass. In Newtonian mechanics we say
the mass is the mass and that’s the end of it.
Relativity, you don’t have to know the origin of this;
that’ll be taught to you in due course.
But it just says particle to the velocity have a mass
different from when they are at rest.
And this is the velocity and c is the velocity of
light. This is the exact formula.
Now, most of the time we don’t care about it because v/c
maybe 10^(-10) and the square of that is 10^(-20) and one minus
such a number is negligible and we just put it as one.
Suppose you want an answer that’s good to some accuracy.
I will write this as follows, as m_0 times 1
– v^(2) over c^(2) to the power of -½ ,
do you agree? That’s what a square root in
the bottom is, it’s this power upstairs.
Now, we can write it as follows, m_0,
this is (1 + x)^(n), okay?
So the x that we use is -v^(2)/c^(2) and the
n that we use is -½. That’s what I’m using in this
formula; x is
-v^(2)/c^(2) and n, which is the power to
which I’m raising, is -½;
therefore, this is equal to 1 + v^(2)/2c^(2) plus
corrections, which are negligible if
v^(2)/c^(2) is very small.
This is a very useful result. It tells you the mass variation
is that is the mass at rest plus m_0v^(2)/
2c^(2), it’s a very useful formula.
Sometimes this formula is much more friendly to use than the
exact formula. You cannot beat the exact
formula; it’s the exact truth but if
you’re interested in that exact function for small values of
v/c, it’s much more useful to use this
approximation. And you will have a lot of
problems like this in your homework, now,
later, everywhere else. So, if I’m going to use it all
the time I should tell you what it is.
So, let me summarize what I’ve told you so far.
First result is, if a function is known at a
certain point and you want to know its value at a nearby
point, what are you supposed to do?
That’s not so hard to guess right?
For example, your tuition costs are so much
this year at Yale. And we know how much it’s going
up every year; you can make a guess on what
it’ll be a year from now. Of course in reality,
it’ll be higher than that because there is other terms,
but that’s the best guess you would make.
That’s how we predict the future by looking at past rates
of change; that’s very normal.
The point is that you can predict the tuition in a year
from now maybe, but maybe not five years from
now or ten years from now because there’s inflation on top
of inflation. That’s like saying the rate of
change itself has a rate of change, the slope here is not
the same as the slope there, that’s what the higher
derivatives of the function do for you.
So, I will conclude by telling you one last result which I’m
not going to prove in this course.
And you guys should learn from the math department the
following result. There is some crazy function,
okay, this is f(x), I want to know what it is at
the point x. I know what it is at the origin.
If I know nothing else, how do I predict the future
value of x, meaning for other x I
don’t know. My best guess is,
well, this is the number I knew that’s what it’s going to do.
But if I knew its rate of change here, you know,
its slope here, the next guess is the
derivative of f at the origin times x.
That’s just what I did earlier; instead of Δx I’m
calling it x. If x is small enough
it’ll work. But of course,
x need not be small; if it gets bigger,
it turns out you can do a little better by adding this
[writes formula on board]. This is the second derivative
of the function at the origin times x^(2) and you can
go on adding more and more terms.
I’ll just add one more term, which is 1/3 times
d^(3)x, f/dx^(3) times
x^(3). This goes on forever.
And it says that if you knew every derivative of the function
at the origin you can reconstruct the function.
Not just infinitesimally close to it, but as far as you want to
go, there are some mathematical restrictions on this because
there are infinite number of terms in this thing.
When you add them all up you may get infinity which would be
nonsense. So, you should really sum up
this infinite number of terms if they add up to a finite number,
and if they do that will give you the function here.
What we have done is to just use the first two terms as an
approximation. Or what I’m telling you is that
there is something called The Taylor Expansion of a function
that allows you to know the function even here,
macroscopically, far from here if you knew all
the derivatives. And if you put those
derivatives here and if the sum added up to something finite,
that finite thing is the value of the function at the new
point. Here is an example you guys can
take, f(x)=e^(x). Why don’t you go home and work
out the series for f(x) at all points?
You have got to remember that e^(x) has a beauty,
that the derivative is e^(x) itself.
And its value at the origin of every derivative is simply one.
So, this just becomes 1 + x + x^(2)/2.
And that is, in fact, the expression for
e^(x) and it turns out that no matter how big x
is, the sum in fact adds up to a
finite number that we call e^(x).
And in particular, what is e itself,
e itself e to the 1;
e is e to the 1 so put x=1 you get 1 +
1 + 1^(2) over 2 factorial plus 1^(3) over 3 factorial.
You do all that you get 2 point something, something.
Alright, that’s e. So, this is a very powerful
notion. If you knew all the derivatives
of a function, you can predict its value not
only the tiny neighborhood, but over big distance.
If you knew the first derivative and you were not very
ambitious. Look, the logic is very simple.
If x is a tiny number x^(2) is even tinier,
x^(3) is even tinier. We’ve got no respect for these
terms. We just drop them and we stop
here. Okay, I’ll have something more
to say about this later on. But for now,
these ideas will be useful for what we’re doing next time.
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