I would ideally like to have a
class with maybe just ten
students so that we can have a
lot of dialogue and I can find
out what you're thinking about,
address your questions.
That'll make a lot of fun but
that's not possible in an
introductory course.
But I think you should feel
free to interrupt or ask
questions or discuss any
reasonably related issue that I
didn't get to.
Okay?
But that'll make it interesting
for everybody because I told you
many times the subject is not
new to me.
It's probably--it's new to you.
So what makes it interesting
for me is the fact that it's a
different class with a different
set of people with different
questions.
And trust me,
your ability to surprise me is
limitless because even on a
subject which I thought is
firmly fixed in my mind,
some of you guys come up with
some point of view or some
question that's always of
interest to me,
of people who've been
practicing this business for a
long time.
So I welcome that,
and that'll make it more
lively.
All right, so I think last time
I got off on a rant about all
the different pedagogical
techniques, some of which I
don't endorse fully.
I'm going to go back now 300
years in time,
to one of the greatest laws
that we have because--look at
the power of this law,
right?
Here is the equation,
and all the mechanical
phenomena that we see,
around the world,
can be understood with this
law.
And I was starting to give you
examples on how to put this law
to work, because I think I at
least made you realize that
simply writing down the law does
not give you a good feeling for
how you actually use it.
So, maybe you have understood
it, but I'm going to remind you
one more time on how you're
supposed to use this law.
So, I'm going to take a
concrete example.
The use of any law of physics
is to be able to predict
something about the future,
given something about the
present.
So, all problems that we solve
can be categorized that way.
So I'm going to take a very
simple problem,
a problem to which I will
return in great detail later on;
but let me first start
with--let me start with this
problem, which I will do very
quickly now and we'll come back
and do it more slowly later.
But at least it's a concrete
problem.
The problem looks like this.
So, here is a table and here's
a spring and here's the mass
m.
There's a force constant
k.
I want to pull it by some
amount A,
and let it go.
So that's the knowledge of the
present.
The question is,
when I let it go,
what's this guy going to do?
That's the typical physics
problem.
It can get more and more
complex.
You can replace the mass by a
planet;
you can replace the spring by
the Sun, which is attracting the
planet;
you can put many planets,
you can make it more and more
complicated.
But they all boil down to a
similar situation.
I know some information now and
I want to be able to say what'll
happen next.
So, here I pull the mass,
when I want A and I want
to know what'll it do next.
Remember, when we go back to
the laws of Newton,
the laws of Newton only tell
you this--and we've been talking
about this is useful
information.
The first thing you have to
know in order to use the laws of
Newton, you have to separately
know the left-hand side.
You have to know what force is
going to act on a body.
You cannot simply say,
"Oh, I know the force on the
body, it is m times
a;
ma is not a force acting
on a body;
a is the response to a
force;
you got to have some other
means of finding the force.
And in this case,
the force on the mass is due to
the spring.
So, I pull the spring by
various amounts and I see what
force it exerts.
Now, I think you know now in
practice how I know what force
it exerts, right?
I pull it by some amount,
attach, say,
the one kilogram standard mass.
I see what acceleration it
experiences, and m times
that acceleration,
or 1 times the acceleration,
is the force the spring exerts.
So I pull it by various amounts
and I study the spring.
And I've learned,
by studying the spring,
that the force it exerts is
some number k,
called a force constant,
times the amount by which I
pull it.
If I start off the mass in a
position where the spring is
neither expanded nor contracted,
that's what we like to call
x = 0.
So I pulled it to x =
A.
Now, what I'm told is when you
pull it to any point x,
that's the force the spring
exerts.
So, this is part of an
independent study.
People who work in spring
physics will study springs and
they will find out from you,
find out and tell you that any
time you buy a spring from me
it'll exert this force.
And they have done--they
figured that out by pulling the
spring and attaching it to
various entities and seeing what
acceleration it produces.
There, the masses are taken to
be known, because you can always
borrow the mass from the Bureau
of Standards,
or we discussed last time how
if you have an unknown mass you
can then compare it to this mass
and find out what its value is.
So every object's mass can be
measured.
And then the guys making the
spring have studied what it does
to different masses and figured
this out.
Now, you come with this mass
and you say, what happens when I
connect it?
Well, I'm assuming the mass of
this guy can also be found by
comparisons, the way I described
to you last time.
So we can always find a mass of
any object, as we went into in
some length.
Then, Newton's law says this is
equal to ma,
but I want to write a as
the second derivative of
x.
So, you now go from a physical
law, which is really a
postulate.
There is no way to derive F
= ma.
You cannot just think about it
and get it.
So, whenever I do physics I
will sometimes tell you this is
a law;
that means don't even try to
derive it.
It just summarizes everything
we know in terms of some new
terms, but it cannot be deduced.
On the other hand,
the fate of this mass can now
be deduced by applying Newton's
law to this equation.
Now, this is a new equation,
you may not have seen this
equation before.
For example,
if I told you -- forget the
left-hand side -- if I told you
the right-hand side is 96,
I think you guys know how to
solve that, right?
You have to find a function
whose second derivative is 96
divided by m,
and you all know how to do
that;
it's t^(2) times a
number, and you can fudge the
number so it works.
This is more complicated.
The time derivative of this
unknown function is not a given
number but the unknown function
itself;
in other words,
x itself is a function
of time.
This is called a differential
equation.
A differential equation is an
equation that tells you
something about an unknown
function in terms of its
derivatives.
You can have a differential
equation involving the second
derivative or the first
derivative or the fourteenth
derivative, whatever it is.
You are supposed to find out
what x (t) is,
given this information.
So, one thing is,
you can go to the Math
Department and say,
"Hey look, I got this equation,
what's the solution?"
and they will tell you.
Now, sometimes we have to do
our own work and we can solve
this equation by guessing.
In fact, the only way to solve
a differential equation is by
guessing the answer;
there is no other way.
You can make a lot of guesses
and every time it works you keep
a little table;
then you publish it,
called Table of
Integrals.
So, I have in my office a huge
table, Mark Caprio has got his
own integral,
we don't leave home without our
Table of Integrals.
I got one at home,
I got one at work,
I may want to keep one in the
car because you just don't know
when you will need an integral.
Okay?
So people have tabulated them
over hundreds of years.
But how do they find them?
They're going to find them in
the way I'm going to describe to
you now.
You look at the equation and
you guess the answer.
Let's make our life simple by
taking a case where the forced
constant is just 1,
okay?
It takes 1 Newton per meter to
stretch that spring.
I let the mass of the object be
1 kilogram.
This is just to keep the
algebra simple.
Later on, you can put any
m and k,
and we'll do all that.
Then what am I saying?
I'm saying, find me a function
whose second derivative is minus
that function.
Now, as a word problem maybe it
rings a bell,
right?
Do you know such a function?
Student: Exponential.
Professor Ramamurti
Shankar: Exponential is
good.
Another one?
Yes?
Student: Any trig
functions.
Professor Ramamurti
Shankar: Trigonometric
functions are good.
So I'm going to temporarily
dismiss the exponential.
I will tell you why.
An exponential function is
actually a very good guess.
You can say x = e^(t),
for example.
The trouble is,
this is growing exponentially,
and we know this spring mass
system, the x is not
growing exponentially.
Or you might say,
well, let me fix that by making
it minus t;
then it quickly comes to rest
after a time and doesn't move.
We sort of feel we want this
thing to be jiggling around for
a long time.
So, the exponential function is
even better than this function
because it says,
take one derivative and I
reproduce myself.
But that's too much;
we only want a function that
reproduces itself when you take
two derivatives,
and that's where the
trigonometric functions come in.
So, I make a guess,
x (t) is cos t,
and you can check if you take
two derivatives,
it obeys the equation,
the second derivative of
x is –x itself.
But do I like this solution or
is there something not fully
satisfactory?
Yes?
Student: But you can't
help the negative when you
[inaudible]
Professor Ramamurti
Shankar: Pardon me?
Student: [inaudible]
Professor Ramamurti
Shankar: Oh it will be
negative x.
So, let's do that.
So let's take dx/dt = -sin
(t).
Then, take one more derivative,
d²x over dt²,
derivative of sine is cosine.
So, if this is –cos
(t), but x was cos
(t);
so it's really minus of the
x I started with.
Yes?
So you have a question or--?
Student: I was thinking,
you wrote it up there
[inaudible]
Professor Ramamurti
Shankar: Oh,
this one?
Student: Yes.
Professor Ramamurti
Shankar: Caught on film.
Yes, that's a mistake.
Okay?
That should be minus.
Very good.
Look, it makes me immensely
happy when you guys catch me.
Of course, it becomes an
addiction, you try to catch me;
sometimes it doesn't work,
but keep trying.
This time you're absolutely
right.
That's why you were looking
confused.
Anything else?
Everyone follow the way this
very elementary problem is
solved?
It's solved by making a guess.
So, you're right.
So, the minus sign was wrong
but now you agree everything's
working.
This is the answer.
But no, I meant something else
is not quite okay with this.
Yes?
Student: Your function
also works if you added a factor
of the sine, also?
Professor Ramamurti
Shankar: Yes.
But first of all,
I can add some sin (t)
and it'll still work.
But more importantly,
if I put t = 0,
I get x = 1.
Why should it be true that I
pulled it by exactly one meter?
I could have pulled it by 2 or
3 or 9 meters.
I want to be able to tell how
much I pulled it by at t
= 0.
Suppose it was pulled to 5
meters and released.
I want x (0) to be 5.
The point is,
you can easily fix that,
by putting a 5 here.
Does it screw up everything?
It doesn't because the 5 just
comes along for the ride,
everywhere.
And don't think it's 5x,
because this 5 is part of
x.
So you can multiply your answer
by any number you like,
and you pick that number,
that reproduces your initial
displacement.
So if you pulled it by 5,
you put a 5.
So, now we're very happy to
describe this problem.
It is true, you can put sin
(t) but I will,
in this problem we don't need
it.
Here's an answer that does
everything you want it to do.
At the initial time,
it gives you 5 times cos
0, which is 5;
and that's what you told me was
the initial displacement.
Initial velocity,
if you take a dx/dt,
it's proportional to sin
(t) and a t = 0,
that vanishes,
that's correct;
you pulled it and you let it go.
So, the instant you released
it, it had no velocity.
And it satisfies Newton's laws,
and that's your answer.
I want to do this simple
example in totality because this
is the paradigm.
This is the example after which
everything else is modeled.
Yes?
Student: How do you know
that it's not sin (t) at
the end?
Professor Ramamurti
Shankar: Ah.
You need a t = 0,
if the answer was sin
(t);
well sin (t) vanishes a
t = 0.
And cos (t), it doesn't
vanish, has a robust value of 1.
But I want a value of 5,
I put a 5 in front.
Look, you guys may have a few
more things to say about this
problem because this is not the
final and most general solution
to the problem.
It is a solution that certainly
works for the one example I had,
which is a mass being pulled to
5 meters and released.
So, carry this in your mind.
This is really to me the most
important thing I can tell you
right now.
The way F = ma is
applied in real life is half the
world is working on the
left-hand side,
finding the forces that act on
bodies under various conditions.
In this example,
the force is due to a spring
and by playing with a spring you
can study the force it exerts
the various displacements of
various amounts by which you
pull it,
and in this example if you pull
the spring by an amount
x, it exerts the force
-kx,
and you can measure k
and you put the label on your
product and you sell the spring
with a given k.
Other people are measuring
masses of various objects and
they've decided the mass of this
is whatever, maybe 1 kilogram.
You put the two sides together
and you find the response of the
system for all future times.
And what this tells you then,
you can now plot this.
This is a graph that looks like
this;
it goes from 5 to -5 and
repeats itself and it's very
nice.
That's what we think the spring
would do.
Now, you think this is a real
spring or am I missing
something?
This is the whole story?
Yes?
Student: There's
friction.
Professor Ramamurti
Shankar: There's friction
because we know if you pull a
real spring it's not going to
oscillate forever.
It's going to slowly stop
shaking and pretty much come to
rest.
That just means--Who do you
want to blame for this equation?
Do we conclude now there's
something wrong with Newton's
laws?
Or what's the problem?
Student: The k
coefficients should not be
[inaudible]
Professor Ramamurti
Shankar: Pardon me?
Student: The k
coefficient is wrong.
Professor Ramamurti
Shankar: The fault is not
with the k.
Yes?
Student: We didn't take
into account all the force
that's friction and the amount
of force.
Professor Ramamurti
Shankar: That is correct.
So, we will say we have missed
something.
There's another force acting on
this mass, besides the spring;
that's the force of friction
that'll oppose the motion of the
mass.
So, you can say one of two
things.
Either you can say something is
wrong with Newton's laws,
or you can say we've not
applied Newton's laws properly
because we haven't identified
all the forces.
So, if you never knew about
friction, you do this and you
find, hey, it doesn't work.
The mass comes to rest,
whereas the theory says it
should oscillate forever.
So, you are at the fork.
Either you can say Newton's
laws are right;
I'm missing some forces,
I'm going to look around and
find out what they are.
Then, you will derive friction
from that.
Or maybe it's time to give up
Newton's laws.
It turns out,
even if there is no friction
this is not the correct law of
physics.
The correct law is given by
Einstein's relativistic
dynamics.
This does not satisfy the
relativistic dynamics;
by that I mean,
if you really pull a real mass
by 5 meters or whatever and
release it, its motion will not
be exactly what I said;
or it'll be off by a very,
very, very tiny amount that you
probably will not discover in
most laboratories.
But if the mass begins to move
at a velocity that is comparable
to that of light,
then this equation will be
wrong.
So, people say,
oh, what kind of business are
you in?
Every once in a while somebody
is wrong.
I'll tell you right now.
Everything we know is wrong,
you know?
And it's no secret;
you can publish it,
you can tell the whole world,
this practitioner said
everything is wrong.
But it's wrong in this limited
sense.
Is Newton right or wrong?
Well, Newton didn't try to
describe things moving at speeds
comparable to light.
He dealt with what problem he
could deal with at that time.
So, it's a law that has a
limited domain of validity.
You can always push the
frontiers of observation until
you come to a situation where
the law doesn't work.
But the specialty of relativity
doesn't also work all the time.
If the mass becomes very tiny,
it becomes of atomic
dimensions, then you need the
laws of quantum mechanics.
That's wrong too.
So, things work in a certain
domain and sometimes you abandon
the formalism;
but don't rush to do that.
In this case,
the problem is not with
formalism but due to friction.
So, you have to make sure you
got all the forces.
Once you got all the forces,
it still doesn't work,
then of course the
experimentalists are very happy
because there's something wrong
with the theory;
then you got to find the new
theory, till that works.
Okay, so that's the cycle.
So it's an ongoing process;
physics is still a work in
progress.
This is work as of 300 years
ago.
And no one's found anything
wrong with Newton's laws,
provided you don't violate two
conditions.
You don't deal with objects
moving at speeds comparable to
that of light,
and you don't deal with objects
which are very,
very tiny.
The notion of what's very tiny
will be very clear when you
learn quantum mechanics next
term.
Right now, there is something
rough, called "tiny," there's
things like atomic dimensions.
All right.
So now we are going to do more
concrete problems,
but I wanted to write that up
for you as a simple,
solvable, worked out -- an
example that displays everything
that happens in physics.
All right.
So, by the way,
people are always looking for
new forces, the electrical
forces.
You find a charge on a table,
it's not connected to a spring;
I bring another charge next to
it, in this case it starts
moving.
What do I do?
A body accelerating without any
force… I can say Newton is
wrong or I can say maybe there's
a new force between that body
and this body,
and that will turn out to be
the electrical force.
Electrical force occurs
whenever you take something,
you take a cat and you take a
piece of amber or you rub it,
you do various things,
then you put them next to other
things and they're attracted or
repelled,
and you discover you're
charging things,
and that's how electricity was
discovered.
But you don't have to have a
cat;
if you have a buffalo,
you can take it with a buffalo,
it still works.
Then, you got to find out what
you really need to find electric
forces.
But you find there is a new
force and you know you've got a
whole new force to study;
then you got to find out what's
the cause of the force.
It turns out,
it's not the mass of the
object, it's something else
called electric charge.
Then, how does the force vary
with distance?
Well, it varies like
1/r².
You know that.
So that's how you slowly study
and identify new forces.
So, the forces were found in
the order of gravitation first,
and then electricity,
magnetism and now of course
there are new forces called
"strong forces."
There are weak forces.
They all act on the subnuclear
level;
so that's why that business is
not done.
We are not finished with the
left-hand side.
We're still looking for new
forces.
Okay, so now I'm going to start
doing problems in two
dimensions.
Everything I did last time was
1D.
You're pulling something,
you're pushing something,
you got one block connected to
a second block to a third block
and you pull it.
We're done all that.
So we're going to elevate this
to high dimensions.
And as you know,
in high dimensions,
acceleration is a vector,
and so force better be a vector
because when you take a vector
and multiply it by a number you
get something else.
And there, by that process,
you've gone to higher
dimensions.
So let's take again simple
problems and try to make them
more and more complex.
And you cannot tell me
eventually you agree the problem
is complex enough,
or I will crank it up as we go
along.
If at some point you plead for
mercy, then we'll stop.
But there is no limit to how
difficult mechanics problems can
be.
If you go back and read
Cambridge University exams,
in 1600 and 1700,
there were really difficult
problems.
That's why they invented
quantum mechanics,
which is a lot easier than
those problems from Newtonian
mechanics.
You can make it as hard as you
like.
So, I'm just going to do a few,
then once you got the idea,
you stop.
But again, we're going to start
with easy one.
Easy one is a table on which
this mass is sitting,
doing absolutely nothing.
And we want to study that guy
and understand it in terms of
Newton's laws.
Because we're in two
dimensions, you got to have two
axes.
So here is my x and here
is my y.
So, the first rule is if F =
ma, we're really saying
i times fx +
j times fy,
is miax + mjay.
It's a vector equation.
The force has to be a
combination of i and
j because nobody--By the
way, I'm working in the
xy plane.
I told you I don't need to go
to 3D, for a long time.
In 3D, in every vector there's
some amount of i and some
amount of j,
that's, what's true for
f, that's true for
a.
You write it out.
And I told you,
if two vectors are equal,
then the x components
have to match and the y
parts also have to match.
You can see that.
If I draw two arrows on the
plane and I tell you this arrow
is the same as that arrow,
it's the same.
It's got the same horizontal
part and the same vertical part.
So, this is really two
equations, and I'm going to
apply them to this block.
So, when I look at this block,
in the x direction,
it's not moving at all.
There are no known forces
acting in the x
direction, and therefore it's a
case of 0 = 0.
Did I miss something?
You guys look a little worried.
Yes?
Okay, now I look in the
y direction.
If you look in the y
direction, I take this block and
I draw its free body diagram,
and I write all the forces on
it.
So, there is the force of
gravity, mg;
however, this M is a big
M or a small m?
Well, let's change the figure
to small m,
so that the rest of the
equations are right.
The little mg is acting
down.
If that's all you had,
the block would fall through
the table;
it's not doing that,
so the table we know is
exerting a force.
The standard name for that is
N, and N stands
for normal.
And normal is a mathematical
term for perpendicular.
You say this vector is normal
with that vector,
you mean they are
perpendicular,
and here we mean this force is
perpendicular to the table.
There are the forces in the
y direction.
In the y direction,
N is a positive force,
it's with a plus sign;
mg is a negative force,
and that's equal to m
times ay.
It's interesting that in this
application of the Newtonian
equation, I'm going to say that
I know the right-hand side.
I know the right hand side is 0
because I know this block is
neither sinking into the table
nor flying off the table.
It's sitting on the table;
it has no velocity,
it has no acceleration,
vertically.
So that has to be 0 because I
know the right-hand side.
So, in Newton's laws,
quite often either I will know
the left or I will know the
right;
either I will know the force or
I will know the acceleration.
So, if this is 0,
I come to the conclusion N =
mg.
This is a very simple example,
but this is how you figure out
how the Newton's laws work in
this particular problem.
Okay, now I'm ready to
introduce another force,
and that's the force of
friction.
As you all seem to be aware,
there is generally a force of
friction between this mass
m, and this table.
So, how do we learn there is a
force of friction?
Suppose someone said,
how do you know there is
friction?
Here is a mass.
What experiment will you do
that tells you,
hey, there is another force
called friction?
Okay, anybody tell me something
you would do?
Yes?
Student: Maybe put the
mass on an incline.
Professor Ramamurti
Shankar: Okay,
but suppose it's on this table.
But you are around,
you can do what you want.
Yes?
Go ahead.
Student: [inaudible]
constant v;
in order to move with a
constant v,
you'd have to apply force.
Professor Ramamurti
Shankar: That's possible.
Student: But moving
forward, over the constant
v, there's no force.
Professor Ramamurti
Shankar: That's a very good
answer.
One answer is,
you find that to keep it moving
at constant velocity,
you have to apply a force.
That means the force you're
applying is cancelled by
somebody else,
because there is zero
acceleration,
there's got to be zero force.
Yes?
Student: Just apply
force to it and push it on the
table and it'll stop eventually,
and that's how you think there
must be a force,
that's out there.
Professor Ramamurti
Shankar: Very good.
But even before that,
even before it starts moving,
you find that if you push it,
it doesn't move.
Right?
I try to push the podium,
well, I don't know my own
strength, but I'm going to
imagine, I push it,
it's not moving.
But if I push hard enough it
will move.
What's happening before it
moves?
I'm applying a force and I'm
getting nothing in return for
it.
So I know there is another
force opposing what I do,
and that's called a force of
static friction.
So, static friction is when the
body is not moving.
So, I'm applying a force
F.
Let me give it some other name,
little f.
No, that's also bad.
Little f was the name
for friction.
So, I'm going to apply a force
called F_me.
You can label these things any
way you like.
Okay?
This particular notation has
not caught on because people are
not as interested in what I
think as I am.
But right now I'm going to call
it F sub-me.
I invite you to invent any
notation you want because
there's nothing sacred about a
notation, at least I want to
convey to you that notion.
So, I call it
F_me,
I'm applying to the right,
then I know that it's not
moving;
therefore, there has to be
another force that cancels it.
But how much is the other force?
Yes?
Student: Equal to
F_me.
Professor Ramamurti
Shankar: That's correct.
So, it is not a fixed force.
Static friction is not a fixed
force;
it's whatever it takes to keep
it from moving.
It will not be less than what I
apply, because then it'll move;
it cannot be more than what I
apply because then it'll start
moving backwards.
So, it will do what it takes to
keep it from moving.
So, static friction is a force
which has a range from 0 to some
maximum.
So, let's call it the force of
friction is less than or equal
to some maximum.
And that maximum turns out to
be a number called the
coefficient of static friction
times the normal force.
The force of friction seems to
depend on how much normal force
the table is exerting on the
object,
which in our example,
if you want in our example,
not always, this is always
true,
but in our example you may
replace it by
μ_s times
mg,
because N = mg.
So the force of friction seems
to depend on how heavy the
object is that's sitting on the
table.
But what does it not depend on
that it could depend on?
Can you think of something else
it could depend on but doesn't?
Yes?
Student: The amount of
surface area.
Professor Ramamurti
Shankar: It doesn't depend
on the area of contact.
In other words,
if this was a rectangular
block, it doesn't seem to matter
whether it's doing that or
whether it's doing that.
You might think,
hey, there's going to be more
friction because there's more
contact.
But it doesn't.
So, it's interesting not only
what it is but what it's not.
Okay, this is called a
coefficient of static friction.
So you guys should remember
that the force that friction
applies is equal to the force
that I apply,
up to a maximum of this.
And once I beat the maximum,
it'll give up and it'll start
moving.
Once it starts moving,
you're going to have a
situation like this.
I apply F_me
this way, and friction will
apply a force which now is
called new μ sub-kinetic
times N.
So, the force of friction is
different when it's moving as
compared to when it's not
moving, it's slightly less,
once a body is in motion.
So, it's always true for all
situations;
μ sub-kinetic is less
than μ sub-static.
Okay?
So, this is how you
characterize friction.
You study various objects and
you find out friction certainly
depends on the nature of the two
surfaces.
Given that constant,
it seems to depend only on the
normal force and not on the
area.
Yes?
Student: So is μ
sub-kinetic less than the
maximum [inaudible]
Professor Ramamurti
Shankar: Remember,
μ sub-kinetic is a
fixed number.
Student: Right.
Professor Ramamurti
Shankar: Maybe .2.
μ sub-static is a fixed
number, .25.
If it's .25 for static,
it means if the normal force is
some mg,
up to a quarter of the weight
is the frictional force it can
apply to prevent its motion.
Suppose I apply a force on a
body, which is originally
1/10^(th) of its weight;
it won't move;
2/10^(th) of its weight it
won't move;
a quarter of its weight it's a
tie;
.26 of its weight it'll start
moving.
Once it starts moving,
frictional force will be .2
times its weight,
not .25, because the kinetic
friction is less.
This is just the summary of
friction.
Okay, so far even though I used
vectors, most of the action is
just in one or the other
direction.
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: No,
it again seems to--It can
actually depend on many things.
Friction is something we don't
understand fully.
But it depends only on the
normal force,
just the way I described it.
Okay.
Now, we are going to do the one
problem that has sent more
people away from physics than
anything else.
Do you know what it is?
You haven't heard about the
problem that sends people away?
Okay, this is called the
inclined plane.
A lot of people may not
remember where they were during
the Kennedy assassination,
but they say,
"I remember the day I saw the
inclined plane;
that's the day I said I'm not
going into physics."
Because this is very bad
publicity for our field.
You come into a subject hearing
about relativity and quantum
mechanics and then--or maybe
gravitation and astrophysics,
and we hit you with this.
So why am I doing this?
Well, if you can do this and
you find it boring,
you can move on.
But if you couldn't do this,
you have no right to complain
because this is the entry ticket
into the business.
You've got to be able to do
this.
Only then I can tell you other
things.
Okay, so we have to.
We'll do this quickly.
I'm watching the clock.
I won't linger over this,
but we have to do this problem.
Okay?
So let's quickly finish this
problem.
So here is the problem.
There is a mass,
m, sitting on its inclined
plane, it'll make some angle
here.
You want to know what it's
going to do.
All right, we know it's going
to slide down the hill,
but we want to be more precise,
and the whole purpose of
Newton's Laws is to quantify
things for which you already
have an intuition.
So the only novel thing about
the inclined plane is that for
the first time we are going to
pick our x and y
axes not along the usual
directions,
but along and perpendicular to
the incline.
That's going to be my x
and that's going to be my
y.
And I ask, what are the forces
on this mass?
Well, I've told you,
first deal with contact forces.
But the only thing in contact
with the mass is the plane.
The inclined plane can exert a
force, in general,
along its own surface and
perpendicular to its own
surface.
But I'm going to take a case
where there is no friction.
If there is no friction,
by definition it cannot exert a
force along its own length.
So it can only exert a force in
this direction and we're going
to call that N.
Then, there's only one other
force;
that's the force of gravity
which we agreed we have to
remember.
Even though the Earth is not
touching this block,
the Earth is somewhere down
here, it's able to reach out and
pull this block down.
So, let me show you the forces.
So here is N,
and here is mg,
these are your forces.
And with no friction, that's it.
That's the key to doing any
problem I give you,
in any exam.
No matter how stressful,
I guarantee you'll get it from
Newton's laws,
provided you write every force
that is there and don't write
any force that is not there.
So, this is it.
These are the two forces,
and the mass will do what these
forces tell it to do.
So, what we want to do then,
since my axes are in this
direction, I have to take this
mg and write it as the
sum of two vectors,
one which is pointing in this
direction and one which is
pointing in that direction.
So, if I add these two arrows,
I get mg.
That's called resolving the
force into some other direction.
Now, the key to all of this is
to know that this angle,
here, is the same as this angle
here.
In other words,
if mg is acting down,
that angle θ is the
same as this angle θ
here.
This is going to be used all
the time, so I'll tell you a
little trick that tells you how
I figured that out.
θ is the angle between
the horizontal and the incline.
Now, you've got to agree that
if I draw perpendicular to the
horizontal and I draw
perpendicular to the incline,
the angle between those two
perpendiculars will be the same
as the angle between the two
lines I began with,
because perpendicular means
rotate by 90.
If you rotate both lines by 90,
the angle between the rotated
lines is the same.
The vertical is perpendicular
to this horizontal,
and this is the perpendicular
to the plane.
Once you got that,
the rest is very easy.
So, in the x direction,
I'm going to write the
following equation:
mg times sin θ.
So, you've got to understand
that this part here will be
mg times sin θ,
and this part here will be
mg times cos θ.
You've got--This is something
you better get really used to.
If a vector is like this and
you want the force in that
direction, the angle between is
θ, then you put the
cosine.
The component that's adjacent
to the vector,
is the cosine,
this one here is a sin
θ.
So, I got the horizontal part
in the x direction,
I got mg sin θ,
and that's it.
There's nothing else.
That's got to be equal to
m times ax.
Remember ax is not
horizontal, ax is along
the incline, downhill.
In the y direction I
write N - mg cos θ =
m times ay.
That's simply writing down
Newton's laws as two laws,
one along x and one
along y,
and putting in whatever we
know.
What happens next is up to what
else we know.
Well, we know that this block
is sliding down the hill,
it's not going into the block,
nor is it flying out of the
block;
it's moving along the block.
That's the reason we chose a
coordinate to be perpendicular
to the block,
because that coordinate,
the y coordinate,
it's not changing.
Even though the block is
sliding down the hill,
when it's sliding down the hill
the usual x and y
coordinates,
if you define them this way,
they're all changing.
But by this clever choice,
it's always at y = 0.
So, it has no y
coordinate, there's no y
velocity, there's no y
acceleration.
So, we know this ay = 0.
So, this is an example where if
we know the acceleration,
by this kind of argument,
from that you deduce that
N has to be mg cos
θ.
Okay?
Then, you come to this fellow
here.
There we find out that
ax, you cancel the
m, and you find it's
equal to g sin θ.
So, that's the big result.
It'll slide down the hill with
an acceleration g sin θ.
This is a good way to measure
g, by the way,
because if you drop something,
it falls too fast for you to
time it.
But if you let it go down an
incline, by making θ
very small so sin θ is
very small and the grading is
very low,
the body can accelerate with a
much slower acceleration
downhill, because it's reduced
by this factor,
sin θ.
Okay, here's another thing I
should tell you right now.
Most of us, when we become
professionals,
don't put in numbers till the
very end.
So, if you are told the incline
is an angle of 37 degrees,
g is 9.8,
don't start putting numbers
into the first equation.
We don't do that and I know for
some of you it's pretty
traumatic to work with symbols,
and sometimes on exams they
say, oh, you had all these
problems with symbols.
But I want to sell you another
idea.
It's much better to work with
symbols than with numbers till
the end.
Why is that?
First of all,
if you put numbers in,
and I suddenly tell you,
"Hey, I was wrong about the
slope,
it's really 39 degrees and not
37," you're going to do the
whole calculation again.
But this way,
you come to the formula and you
say, "What's your θ?"
You change your mind,
okay, that's the new θ.
You change the value of
g, you made a better
measurement, that's the answer.
Then, you can see if your
problem had some mistakes in it.
Suppose you got g² of
sin θ.
You will know it's wrong
because this is an acceleration
and that's an acceleration.
The units have to match and
sin θ,
of course, has no units.
But maybe you got the
trigonometry wrong.
Maybe it's really g cos
θ that should be the right
answer.
How do I know it's sin
θ?
You can do a test.
We know, for example,
as the incline becomes less and
less inclined,
you got to get less and less
acceleration,
because it's going downhill
less and less,
and when θ goes to
small values,
ax also diminishes.
Or, if you take an incline and
you really make it almost
vertical, and in the end you
make it completely vertical,
the block is just falling under
gravity.
So, when θ is 90,
a becomes g.
So, you can test your results.
So, be prepared for that part,
if you are in this course.
I will give you some homework
problems in which numbers are
not put in till the very end,
or I may not even ask you to
put the numbers in,
because the thing is to get
this formula because anyone can
take the numbers and put them in
the calculator.
So, that's going to be part of
our agreement that we will work
with symbols.
And if you look at any
calculation we do,
if you caught me in my office,
I'm not putting numbers in very
early.
I mean, I cannot wait to find
out what the answer is but I
keep the symbols till the end,
then I put the numbers.
Then, you can vary different
numbers.
You can say,
"Well, what if gravity was
weaker?
I go up to Denver and do the
experiment.
Will I get a different answer?"
You will.
So, how things depend on the
input parameters is interesting.
Here's another interesting
result.
What is missing in this formula
that could have been there?
Student: Mass.
Professor Ramamurti
Shankar: Pardon me?
The mass.
It doesn't seem to matter what
the object is.
That's an interesting property
of the result.
Right?
And that's because m
cancelled out of the two sides
of that equation.
So, these are all interesting
features you would not know if
you kept numbers everywhere and
you got a = .62 meters
per second squared.
To me, that number doesn't tell
me as much.
At the end, I want to know the
number but the formula is very
interesting.
All right.
So, now we are going to make it
a little more complicated.
We are going to now add
friction to this.
So, here is the block but now
there's a coefficient,
the friction,
and let's say the block
is--Let's ask the following
simple question.
I think it was related to what
one of you guys said.
Let the block be at rest and
θ is an angle I can
vary.
Maybe there is a little thingy
here, you can turn it back and
forth.
Let me crank up θ and
see where it'll start slipping.
In other words,
there is a coefficient of
static friction between the
block and the plane,
and I want to know,
like parking a car in San
Francisco, what angle can it
take before it starts sliding
down.
So, let's write the forces and
let's write F = ma.
So, the forces are again here.
Let me cut to the chase and
write mg cos θ here,
N here,
mg sin θ here.
What shall I write for friction?
Student: The first one.
Professor Ramamurti
Shankar: Do not write
μ_s times
N,
because the frictional force is
anything you need,
up to μ_s
times N.
In other words,
if θ was very small and
mg sin θ is very small,
the frictional force will be in
fact mg sin θ.
So, it'll equal whatever it
takes to keep it from slipping.
But if I've cranked up the
angle to that maximum angle,
beyond which it cannot stay
there, let's give a name to that
angle that is sin θ*.
Well, at the maximum angle
friction is doing the maximum
thing it can do;
then, we can certainly say
mg sin θ is then equal
to the static friction times
N,
which is now
μ_s times
mg cos θ.
Again, the mass cancels.
So, it doesn't matter what kind
of car you parked on the slope,
it only depends on the
following.
But g cancels too.
So, whether you park your car
on the Earth or park it on
another planet,
this is going to be the same
restriction,
which is that tan θ has
to be less than or equal to
μ_s;
equal to μ_s
is the critical value,
less than that is acceptable.
Okay, that's a simple problem.
If you tilt it to more than
that it'll topple over.
So, now let's take another
problem where the block has
started moving;
it's going down the hill.
Now, what formula do I write?
You notice that I'm not writing
every step, every time.
Each time I deal with it I'm
going to do some things faster.
For example,
I already resolved gravity to
two parts, without going through
the whole song and dance,
because you know how to do
that.
Now, I put in the friction.
So now, I've got μ sub
kinetic not equal to 0,
mass is moving.
You can ask,
"What's the acceleration now?"
So, the horizontal equation
along x is going to be
mg sin θ minus μ
kinetic times a normal force,
but allow me to write for the
normal force mg cos θ,
and that's got to be equal to
m times ax.
Now, can you do this?
That's the test.
It is not enough to watch me do
this and say,
okay, he seems to know what
he's doing.
The question is,
can you do this?
Would you have thought about
this?
Do you agree that I'm not doing
anything beyond Newton's laws
when I do this?
Yes, each one of you should ask.
If you cannot do that,
you have to deal with that now
because it's only going to get
compounded.
On the other hand,
if you can do this,
you know everything I know
about Newton's laws.
This is all there is to
Newton's laws and how to apply
them.
So, you cancel the m.
So, notice here,
this is the normal force,
but I've taken the y
equation that says normal force
is mg cos θ.
Put that here.
Then, I find acceleration is
g times sin θ –
μ sub-kinetic cos
θ.
Okay, here is the result:
ax is what I'm calling
a now.
I want to study this in various
limits.
The first limit I take is no
coefficient of kinetic friction,
I get back to my sin θ.
Well, let me try the following
interesting limits.
I crank up the coefficient of
kinetic friction,
more and more and more.
Look what happens here:
sin θ and cos θ
are fixed numbers.
If this becomes larger than
some amount, than say 10,
what happens to the expression?
a is negative.
And what does that mean?
Student: It's going
uphill.
Professor Ramamurti
Shankar: The block is moving
uphill.
And do you buy that?
So, what's wrong?
I mean, I took the equation,
right?
I told you all the time how you
can study various limits.
Something is wrong.
It says, if kinetic friction is
very large, the block will move
uphill.
Student: [inaudible]
Professor Ramamurti
Shankar: That's not the
reason.
Yes?
Student: It can't be
greater than the static one.
Professor Ramamurti
Shankar: No,
that's not the reason.
I mean, you can also vary the
angles, maybe keeping k
at some value,
vary the angles so that this
number times that can beat this.
Then what happens?
Yes?
Student: Where do you
find the direction of sin
θ downhill?
Professor Ramamurti
Shankar: That is the point.
Okay?
Let me repeat his answer;
that's the correct answer.
You're right about kinetic
being less than static and so
on.
But you can always imagine
keeping that number constant but
changing the angle to the cosine
gets bigger and sine gets
smaller.
At some point,
you agree, cosine will beat the
sine, because sine is going
towards 0 and cosine's going
towards 1.
But the correct answer is,
in writing this forced law,
I'm assuming the force of
friction points to the left,
uphill.
That is correct only if the
body is moving downhill.
But if you got an answer where
a was negative,
therefore it's going uphill,
the starting premise is wrong.
So, one very useful lesson to
learn here is when you apply
formula, don't forget the
conditions under which you
derived it,
and don't apply the result to a
case where the answer does not
belong to the domain that you
assumed you're applying it to.
So, the trouble with friction
is the following.
Friction is not a definite
force with a definite magnitude
or direction.
The magnitude is definite,
but direction is not definite.
It is uphill if you're going
downhill;
it's downhill if you're going
uphill.
It's like the constant
opposition to motion.
There is force of gravity,
the normal force,
everything else is a fixed
direction;
you can draw it once and for
all.
So in a formula,
where the forces are what they
are in any context,
you can apply the formula in
any limit.
But if the frictional force
assumed a downward motion,
don't apply it when the block
is going up.
Okay, so I'm going to do one
more of these inclined planes
and then pretty much done with
the inclined plane.
Okay, this is a problem of two
masses.
There's a rope that goes over
the pulley, let's call this
m, let's call that
M, and this angle again,
it's θ,
and no friction.
You can mix and match various
things but let's keep friction
out because I'm trying to do
something different here.
So, you want to know what these
guys will do now.
So first guess is,
well, M looks so
impressive compared to small
m, so maybe it'll go
downhill.
But I realize that's of course
fallacy because M is a
symbol;
the value associated with it
maybe 1/10^(th) that of
m.
But if M really is
bigger than m,
then am I assured it'll go
downhill?
Student: The mass has to
be greater.
[inaudible]
Professor Ramamurti
Shankar: No,
no, I'm saying,
if the big mass M is
bigger than the small mass
m,
is that enough to say it'll go
downhill?
Yes?
Student: No,
it depends on the angle.
Professor Ramamurti
Shankar: It depends on the
angle because part of the mass
is not helpful in going
downhill.
See, if I had a pulley like
this, then you're absolutely
right, the big guy wins.
Now, it's not clear because
this has got all of gravity
pulling it down;
this has got only part of
gravity pulling it down.
So, let's skip this direction,
it's not very interesting.
But let's look at this
direction, for which the
equation is the following.
If I just took that block,
it's got mg sin θ
acting down, and this rope will
exert a certain tension
T, here,
which I don't know.
Under the combined effect of
these two, I can say mg sin -
T = ma, where a is
assumed to be positive going
downhill.
That's one equation.
Then, I come to this fellow
here, this block;
well, it's got mg acting
down and tension acting up.
I'm sorry guys,
I made one mistake here.
I think you know what it is.
These should be M,
that should be M,
sorry for that,
and M (Mg sin θ – T =
Ma).
For m,
the equation is T –
mg.
Now, let's be careful with what
I do next.
I'm going to say m times
a, which is the same
a.
You should all understand why
the acceleration in magnitude is
the same.
We're not saying the direction
is the same.
This is accelerating downhill,
at some angle;
that's accelerating uphill.
These are not equations which
are vectors;
these stand for equations along
certain directions.
This is the up/down direction
here, and that is up the slope,
down the slope direction,
for that one.
But I think you realize if the
rope is inelastic,
if this mass moves down one
inch, that's got to go up one
inch.
Therefore, if it goes one inch
in one second,
so will the other guy,
they'll have the same velocity
and they'll have the same
acceleration.
So, that's why there's only one
unknown acceleration,
a.
Once you realize that you're
done because now we know what to
do.
Whenever you see a –T
and a + T,
you add them and you get
g times M sin θ -
m.
On this side,
it's going to be M + m
times a.
Can you see that?
So, let me short circuit one
step, and divide by M +
m.
So, that's the formula here.
And again, you got to ask,
does it make sense?
Now you notice that for it to
be positive, and to go downhill,
it's not enough if M is
bigger than m.
M sin θ should be
bigger than m;
that's because M sin θ
is the part that's really
pulling downhill.
M cos θ is trying to
ram it into the inclined plane,
and that's being countered by
the normal force.
The second thing to notice is:
can I use this formula when
m is bigger than M sin
θ?
In other words,
can I use this formula when
a becomes a negative?
After all, I did all my
thinking saying it's going down,
but can I use it when it's
going up?
Students: Yes.
Professor Ramamurti
Shankar: Yes,
and the reason?
Anybody want to say why that's
okay here?
Yes?
Student: No friction.
Professor Ramamurti
Shankar: Right.
Student: And friction is
a force, depending upon which
[inaudible]
Professor Ramamurti
Shankar: That's correct.
All the forces I drew here,
they're not going to change
their mind when you change the
body's motion.
Gravity is always going to pull
down.
So, all these forces are fixed
in magnitude and direction,
and therefore once you got a
formula for positive a,
you can apply it to negative
a, but you cannot do it
when you have friction because
in friction,
you assume the direction of
motion.
All right, so now I'm going to
leave the plane.
This is it.
We are finished with the plane.
You are not finished because,
oh don't clap here,
I got a homework problem,
by popular demand.
Two masses, one pulley and
friction.
That's to really say we're
finished with the plane.
Once you can do that,
you've graduated from this boot
camp;
you can handle anything else.
But that's not something I want
to do in class.
That's something you do in the
privacy of your room,
put all these things in and
crank it up.
And you also have enough time.
But that's a very--that's got
some interesting angles you
should think about.
Okay, so now I'm going to do
physics that basically involves
rotational motion.
But let me do one Mickey-Mouse
problem, and let it go,
because I assigned it to you.
The typical problem is:
there's a tree here,
there's a tree here,
and that is--there are two
ropes and there's a backpack
hanging here with some weight.
You are told what that angle is;
you're told what that angle is.
You are supposed to find the
tension on these two ropes.
I think it's fairly simple so I
won't do the details but you
know what you're supposed to do.
You take these two tensions,
break them into the horizontal
parts, with sines and cosines,
and into vertical parts;
horizontal parts that cancel
because of this mass.
This backpack is not going
anywhere horizontally;
the vertical parts of these
two, which will be in fact
additive, should cancel the
weight of the object.
That gives you two simultaneous
equations for T_1 and
T_2,
and you can fiddle with them,
and solve them.
Well, by the way I should tell
you, if you got some equation
like T_1 sin θ
_1 =
T_2,
sin θ _2,
or that – T_2
sin θ is 0,
and another equation involving
T_1 with some
numbers and
T_2.
Some of you seemed to have
trouble the other day when there
were sin θs here;
whereas, if these had been
numbers like 3 and 4,
you knew how to solve these
simultaneous equations.
But I want you to realize that
the solving of the simultaneous
equations for
T_1 and
T_2 involve
forming combinations of these
coefficients to eliminate either
T_1 or
T_2.
So, when you add them,
one of them drops out.
And that doesn't depend on
whether these are trigonometric
functions or any other numbers,
they're eliminated in the same
way.
Likewise on the right hand
side, instead of some numbers
like 3 and 4,
you have the weight,
W, of this mass;
well, if you multiply this
equation by 3,
multiply everything by 3 and
juggle them.
So, you should get used to
eliminating coefficients which
are not simply numbers,
which are symbols,
sines, cosines,
but you know how to--If you had
a sine here and a cosine there,
for example,
how will you combine them?
Maybe you'll multiply this by a
cosine and that by a sine,
and if you subtract it'll drop
out.
You've got to do stuff like
that.
Okay?
But it's important that in a
Physics 200 Level course you are
able to solve those things.
Okay, now I'm coming to the
interesting problems in which
motion involves going in a
circle.
So here's one problem.
This is a string on which there
is a mass and the mass is going
around in a circle.
This is like an amusement park,
so that they have these
dangling little,
tiny,
baby rockets and you sit there
and start spinning,
and instead of being vertical,
it starts lifting up,
and you want to know why it's
lifting up.
So, if this angle here is
θ, and this is some mass
m, and it's going around
in a circle with velocity
v,
and the radius of the circle is
say r;
we want to find some relation
between this angle θ and
these other parameters in the
problem.
So what do I do?
I apply F = ma,
that's it;
and I'm going to get everything
by applying F = ma,
to this guy.
So, here is that mass.
The tension on the rope can
only be along the rope.
The rope is not a rigid object.
It cannot exert a force
perpendicular to itself.
It can only pull it along each
direction.
That's the tension on the rope.
There is mg,
and that's it.
Under these things it's going
to obey F = ma.
So, what I do with this tension
is I trade this guy for two
forces.
This is T.
If that angle is θ;
let me see, that angle is
θ;
so this is T sin θ,
and this is T cos θ.
So, you trade that oblique
T for two Ts that
are in the actual simple
directions.
The fact that it's not going up
or down vertically tells me
mg = T sin θ.
Students: [inaudible]
Professor Ramamurti
Shankar: Oh,
I'm sorry, yes.
Okay, then in the other
direction T sin θ is the
horizontal force,
that's got to be mass times
acceleration.
At the instant--but I've shown
the object, because it's going
in a circle, it has an
acceleration,
in this direction,
with size v²/R.
Therefore, if you divide this
equation by that equation you
will get v²/Rg = tan θ,
and that's the angle that the
string will assume.
So, if you know the velocity
you can do this problem.
Okay, then here's another
interesting problem.
When you go on these roads,
if you're going in a circle--so
for this is your car,
I'm looking at you from the
top, you're going in a circle.
Anything wrong?
Going in a circle,
you're accelerating towards the
center;
someone's got to provide the
force, that's what Newton says.
That force is the friction
between your tire and the road.
In fact, it's a case of static
friction.
You might think it's kinetic
because the car is moving,
but it's not moving in this
direction.
So, it's a static friction that
keeps you from slipping.
So, you need a certain static
friction;
so that N times
μ static -- that
N is equal to mg
-- must be equal to
mv²/R.
If you don't have the static
friction, your car will not be
able to make the curve,
it'll fly off.
So, what people have done is to
find a clever way in which you
don't have to have any friction
and you can still make the turn;
that is, to bank your road.
So imagine, now you're going
into the blackboard and you're
turning left,
and the road looks like this,
going away from me.
Now, I don't have to draw this
thing.
This road is a track,
a racetrack going around and
round like this.
And I gave it some angle
θ.
This road has no friction so it
can only exert a force that way.
But let me resolve that force
into two parts,
one like this and one like
this.
Again, if these two have an
angle θ,
that is the same angle
θ there,
and there is the mass of the
car this way.
So, what do I find?
I find N cos θ = mg and
N sin θ = mv²/R.
Therefore, if you again take
the ratios you will find tan
θ = v²/Rg.
I think this may be a little
fast now but you guys can do the
algebra.
What that means is the
following.
If you want the car to go
around the bend,
at a certain speed,
40 miles per hour or so many
kilometers per hour,
and the road itself is part of
a circle of radius r;
it doesn't have to be a full
circle.
At that instant it has to be a
part of a circle.
Then, if you bank your road at
that angle, you don't need any
friction to make the turn.
You got to understand how we
beat the system.
If your road instead of being
flat is tilted,
then even though the road can
only exert a frictionless normal
force,
now part of the normal force is
pointing to the center.
In fact, this picture is not
drawn to scale.
If I drew it that way,
this part of the vector is
horizontal and directed towards
the center of the circle,
on which you orbit.
So, that's how you do the
banking.
In any good system of roads you
will find there is banking.
Okay, I'm going to do the last
problem because it's in the
homework, and maybe I'm a little
late today, I apologize for
that.
The last problem is a very
famous, very important,
and that's the loop-the-loop
problem.
As you know,
the loop-the-loop is when you
come down on a track,
you go on a circle,
and for awhile you're upside
down.
And the famous question is,
"Why don't you fall down?"
What's the trick?
Does it violate Newton's laws,
and so on?
We'll find we can understand it
fully with Newton's laws.
But we'll find out that in
order for this thing to pull it
off, its speed at that instant
has to have a minimum value.
And the application of Newton's
law to this object is very
simple.
Here is gravity.
What about the track on which
it's going?
What's the direction of the
force the track exerts?
Student: Down.
Professor Ramamurti
Shankar: Pardon me?
Down.
Anybody said up?
A track cannot exert a force
upwards, down.
So everybody,
track, mg and they're
all acting down.
So you see, that's when you get
a little worried.
Okay, this guy's up there,
all the forces are down.
So, we got to ask,
so what does that mean?
Well, I have to agree they're
all down.
I say T + mg is the
downward force,
that's got to be mv²/r.
Therefore, T = m times
(v²/R – g).
So, what does that mean?
If this T comes out
negative, you're dead,
because if it's negative,
you want the track to exert an
upward force and it cannot.
In a real amusement park they
have other T-brackets and
so on to support you.
But if you really believe in
the laws of physics,
you don't need any of that,
you just got to make sure this
number is positive.
For it to be positive,
you need v² bigger than
or equal to Rg.
So, you got to make sure when
you're on the top,
you have this minimum velocity.
If you go faster than that,
that is just fine;
if you go faster than that,
T will be some positive
number.
So, here's the interesting
thing.
Have we escaped the pull of
gravity?
How come we are not falling
down?
And the track is not helping
us, it's also pushing us down.
Yes?
Student: The velocity
vector is moving towards the
left and the force;
well, the full force is
changing the direction of the
velocity towards the center,
but it's still going along a
circular path.
Professor Ramamurti
Shankar: Okay.
Student: You can make
that for forced radials today.
Professor Ramamurti
Shankar: Yes.
So, the point is,
if you have an apple,
with two forces,
both pushing down,
that apple is accelerating
down,
and accelerating down means
really falling towards the
ground, towards the center of
the circle here.
In this example,
it is definitely accelerating,
but the acceleration is not
added to zero velocity,
in which case it'll pick up
more and more speed;
it's added to a huge horizontal
velocity.
So, in a tiny time,
you give it a little velocity
like this, your new velocity
points at a new angle.
That just means your body has
come here and has stopped going
like that and started going like
this.
Therefore, downward
acceleration does not mean
coming closer to the ground,
it only means your velocity
vector is changing its
direction.
On the other hand,
if you had no velocity vector,
downward acceleration means
what you think it is.
That is why,
if you drop an apple from here,
it has an acceleration due to
mg;
if the track was pushing it
down even more,
it'll fall towards the ground
and hit the ground.
Here it actually has an
acceleration but it doesn't mean
that you are any closer to the
center.
Going around in a circle is an
example of constantly
accelerating towards the center
but not getting any closer.
That's the way vectors work and
you should think about that.
