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00:00:53,370 --> 00:00:59,030
In the last class, we had basically achieved
the derivation of the Lagrangian equation,
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00:00:59,030 --> 00:01:04,480
so
what actually did it do, if there are n number
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00:01:04,480 --> 00:01:08,830
of bodies in a system each marked as j.
.
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00:01:08,830 --> 00:01:23,369
Then for each equation in the Newtonian way
would be m q double dot is equal to the
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00:01:23,369 --> 00:01:33,800
applied force plus the constant force F c
this would be vector, this would be vector
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00:01:33,800 --> 00:01:38,580
and
all that will have the j subscript, that was
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00:01:38,580 --> 00:01:50,090
the Newton’s way. Now, you can see that
if
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00:01:50,090 --> 00:01:56,350
there are say 25 bodies in a system, you would
need 25 such equations and then you will
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00:01:56,350 --> 00:02:03,069
have to you will get the set of differential
equations as that 25 set.
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00:02:03,069 --> 00:02:12,670
But, that 25 could be in a three dimensional
space, so how many coordinates do you
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00:02:12,670 --> 00:02:20,590
really need three and how many equations do
you really need three, if there are
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00:02:20,590 --> 00:02:27,220
holonomic constants then it even goes to less.
So, that was one problem with the
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00:02:27,220 --> 00:02:36,050
Newton’s equation and from there we had
obtained the Lagrangian equation as d d t
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00:02:36,050 --> 00:02:46,660
of
this is the Lagrangian function with respect
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to q dot here minus Lagrangian function.
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.So, this is the Lagrangian way and these
two forms are mathematically equivalent,
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because they yield the same differential equation,
we had seen that starting from here we
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had arrived at this. Now, these are each equation
will have the subscript i, is i then
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00:03:26,060 --> 00:03:34,939
represents the identifier of the new coordinates
the generalized coordinates. So, if there
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are n number of such generalized coordinates,
which is the total number of necessary
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00:03:41,939 --> 00:03:47,930
configuration variables minus the number of
holonomic constants, this would be that
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00:03:47,930 --> 00:03:53,969
many number.
Now, let us try to understand how this equation
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00:03:53,969 --> 00:04:01,159
can be used, but first what does this
consists of the Lagrangian function is the
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kinetic energy minus the potential energy
that
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what we said. And in the potential energy
we said that at least we have gone up to that
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extent, where we said that at least we understand
how it is for the gravitational potential,
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we understand how it is for the electrostatic
potential. So, that much we understand, but
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in general that could a body could be attached
to other bodies through spring like
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elements, through other kind of elements,
so let us understand that first.
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00:04:30,970 --> 00:04:31,970
.
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Now, suppose you have a spring and this end
is being pulled and from the un stretched
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position it has been pulled by an amount q,
which is the position coordinate, then what
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is
the energy stored in this. So, energy stored
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is half K q square, now notice that this has
the
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00:05:06,670 --> 00:05:17,590
form of the potential energy why, because
if this is the potential then
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00:05:17,590 --> 00:05:26,330
is just K q and K q
is the amount of force that we expect right.
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.So, you get the and why is it minus, because
you are pulling in this direction and the
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force is applied in the opposite direction,
that is why this minus comes has to come.
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00:05:36,569 --> 00:05:43,419
So,
you have the force obtained as the gradient
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of the potential, so the energy stored in
the
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00:05:47,639 --> 00:05:56,340
spring is actually potential energy is that
clear. So, if a system has both springs and
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00:05:56,340 --> 00:06:02,850
gravitational energy stored, so the total
potential energy would be the potential energy
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00:06:02,850 --> 00:06:10,460
stored in the spring as well as the in the
gravitational energy. And this potential energy
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00:06:10,460 --> 00:06:14,870
is
a scalar, just a number it has no direction,
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00:06:14,870 --> 00:06:16,870
so if something is stored here, something
is
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00:06:16,870 --> 00:06:20,669
stored there, something is stored there, just
add them up that is it that gives you the
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00:06:20,669 --> 00:06:25,400
total
potential energy that is one advantage.
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.
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00:06:26,400 --> 00:06:33,720
Let us illustrate with a very, very simple
system the simple oscillator, how this can
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00:06:33,720 --> 00:06:41,430
be
used, suppose there is a spring here connected
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to a mass, simple no complication how
would you obtain the differential equation
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00:06:53,430 --> 00:07:05,550
in this case. Let us start from the simple
system yes, we would first say no before going
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to that let me start from here ((Refer
Time: 07:10)) and illustrate how the process
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00:07:14,830 --> 00:07:20,259
of obtaining the differential equation goes.
In case of the Newtonian equation what did
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you do, for every mass point we identified
the forces acting on it and in order to facilitate
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that we drew the free body diagram, that
you probably learnt in the mechanics class.
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So, we drew the free body diagram then we
either wrote force is equal to mass into acceleration
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or if we write in the D Alembert's
way, the free body diagram itself contains
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the m q double dot, so just balance of the
forces.
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.In this case, what we do is we have already
achieved the feet of reducing the number of
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differential equations using the holonomic
constants. So, now if looking at a system
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we
find that we can define a set of position
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coordinates, position coordinates remember,
position coordinates that are consistent with
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the constants. The minimum number of
position coordinates necessary to define the
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00:08:19,090 --> 00:08:24,529
position us status of the whole system unitly
that is it.
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.
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So, the first step in the Lagrangian formulation
would be one, identify the generalized
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coordinates. Now, remember this generalized
coordinates there this nomenclature should
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not confuse you, in case of a system like
this . what can be the
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generalized coordinates, here is a body in
the Newtonian way it has x, y, z coordinates
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in
the Lagrangian way I do not need the y and
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z coordinates that is it. We will say, that
from the un stretched position of the spring
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I will measure the variable q that is it just
one necessary.
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..
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So, that is the generalized coordinate in
case of the simple pendulum, what is the
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generalized coordinate, simple . suppose it
is a planar pendulum,
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which means that the first holonomic constant
forces it to move on a sphere and the
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second holonomic constant imposes another
plane on it. So, it is forced to move on an
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arc, if it moves on a arc then what is the
minimum number of position coordinates can
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00:10:02,389 --> 00:10:05,160
I
define, that is consistent with the constant
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just one and that is simple theta.
So, whatever follows from your common sense
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that exactly is the generalized coordinate
it just goes as the name fine .. So, you have
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the number of
generalized coordinates, normally they would
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be denoted as q i that is the way we
normally define. So, even if something is
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theta we can call it q 1, q 2 and all that,
but
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sometimes we might in order to make the identification
easy, continue with the alpha,
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beta, theta and whatever, but always remember
that this is the coordinate.
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Well once we have identified this, then express
the Kinetic energy
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and the Potential
energy
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00:11:21,120 --> 00:11:35,220
in terms of, so we have the position coordinates.
Now, express the Potential
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energy and the Kinetic energy in terms of
this q i and q i dot, then
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that is it, then we just
substitute in the Lagrangian equation. Let
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us illustrate start with this problem,
number one identify the generalized coordinate,
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we have identified only
one yes.
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Student: .
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00:12:29,730 --> 00:12:37,070
.No it will matter, his question is if we
cannot identify the generalized coordinate
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00:12:37,070 --> 00:12:41,880
will it
matter, one clarification might be necessary.
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In a system there might be some generalized
coordinates, that are spherical there might
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be some generalized coordinate that are linear.
And also in electromechanical system there
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can be some generalized coordinate that are
in the mechanical domain, some generalized
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coordinates that are in the electrical domain
it is possible, but you will have to identify,
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that is the first step.
So, that has to be identified and I will work
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out here many problems, so that you will
after sometime be comfortable with it. Once,
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you have done that express the Kinetic
energy and the Potential energy in terms of
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q and q dot, so you can easily do that if
the
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velocity is q dot then T is half m, the Potential
energy is half K q square simple, so the
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Lagrangian is T minus v half m q dot square
minus half K q square. So, the derivative
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of
Lagrangian with respect to q dot is
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and the derivative of the Lagrangian with
respect to q
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is minus K q, once you have written this we
have to substitute here.
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00:14:41,560 --> 00:14:42,560
.
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00:14:42,560 --> 00:15:02,860
So, let us substitute you get d d t of Lagrangian
dou q dot minus substitute this you get m
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00:15:02,860 --> 00:15:15,029
q double dot here, because it is again derivative
with respect to time plus K q equal to 0,
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that is the equation of the system. Pretty
simple really, well you might say that as
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yet the
advantage of the Lagrangian form, Lagrangian
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way of writing equation vis-à-vis the
Newtonian form is not very clear, . because
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it could have done this
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.way, that here if I pull it a bit then the
mass is acted on by a force like K q in the
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direction.
So, K q is the force acting on it and that
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is mass into acceleration that is exactly
what you
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00:15:49,480 --> 00:15:58,269
have written here, yes right from here it
may not be all that clear. Only thing is that,
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in
the Newtonian way logically you will have
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to write the equations for all the directions,
but now you know that they are not moving
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in the all the directions. So, exactly from
this
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one it may not be all that clear, so let us
now handle the this problem which we had done
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earlier, just turn the page where we have
earlier obtained the differential equations
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00:16:23,649 --> 00:16:28,959
by the
Newtonian way ..
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00:16:28,959 --> 00:16:40,600
There the tension appear and it had to appear,
because that is the force acting on the body
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it had to appear. But, now in the Lagrangian
way we will not bother about it at all why,
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because that is a constant force and we have
eliminated the constant force already that
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from the formulation. So, we will forget about
it, so the generalized coordinate
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is what
theta, the generalized velocity remember one
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00:17:18,740 --> 00:17:22,450
thing, the generalized velocity is simply
the
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00:17:22,450 --> 00:17:26,570
derivative of theta, derivative of the generalized
coordinate not the actual velocity of the
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00:17:26,570 --> 00:17:32,620
body, the actual velocity of the body is...
Student: .
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00:17:32,620 --> 00:17:43,860
So, this is l then l theta dot but we are
not talking about that, your generalized velocity
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00:17:43,860 --> 00:17:49,350
is
simply the derivative of this. Fine, the next
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00:17:49,350 --> 00:17:57,350
step is to obtain the kinetic energy and the
potential energy, the Kinetic energy is what
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is the velocity of this fellow, it is say
this
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00:18:01,210 --> 00:18:18,460
length is l, so it is l theta dot is the velocity,
so half m l theta dot square. And your
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00:18:18,460 --> 00:18:29,910
potential energy is suppose I measure from
here, how much it is below, then if this is
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00:18:29,910 --> 00:18:45,000
theta then this height is l cos theta, so
negative of that minus l cos theta m g l,
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00:18:45,000 --> 00:18:57,250
so yes l
cos theta m g, m g has to be there.
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00:18:57,250 --> 00:19:09,730
So, your Lagrangian function is T minus v
half m l square theta dot square plus m g
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00:19:09,730 --> 00:19:18,059
l cos
theta. So, your derivative of the Lagrangian
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00:19:18,059 --> 00:19:31,940
with respect to theta dot is m l square theta
dot, the derivative of the Lagrangian with
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00:19:31,940 --> 00:19:43,360
respect to theta is do it, it will be m g
l cos
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00:19:43,360 --> 00:19:53,590
theta minus sin theta. The moment you have
this, you can easily is there space here yes
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00:19:53,590 --> 00:19:58,940
you can write down the differential equation,
derivative of this with respect to time
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00:19:58,940 --> 00:20:16,270
which is m l square theta double dot minus
of this, so it is plus m g l sin theta equal
151
00:20:16,270 --> 00:20:17,400
to 0
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00:20:17,400 --> 00:20:31,840
.m cancels off l one l cancels off, so you
get theta double dot is equal to minus g by
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00:20:31,840 --> 00:20:39,980
l sin
theta that is the equation.
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00:20:39,980 --> 00:20:46,830
So, that is how we write it and you see that
we do not need to write three equations, we
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00:20:46,830 --> 00:20:55,640
do not need to consider the tension it becomes
simple and that is the main advantage of
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00:20:55,640 --> 00:21:04,510
the Lagrangian form. If the advantage is salient
even for such a simple system for more
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00:21:04,510 --> 00:21:12,889
complicated system, it will be far more salient
let us try to illustrate some of that.
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00:21:12,889 --> 00:21:13,889
.
159
00:21:13,889 --> 00:21:24,020
Now, you have while talking about the constants
we had considered this kind of a
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00:21:24,020 --> 00:21:40,250
situation, we were considered this. So, let
us obtain the differential equation of this,
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00:21:40,250 --> 00:21:43,250
here
you would notice that here is a tension, here
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00:21:43,250 --> 00:21:51,590
is a tension and we had argued that this
tension does work, this tension also does
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00:21:51,590 --> 00:21:57,890
work, but put together they do no work.
Anyway that is not important, what is important
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00:21:57,890 --> 00:22:02,360
is that written in the Newtonian way,
you will have to considered this two tensions
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00:22:02,360 --> 00:22:07,580
and written in the Lagrangian way we can
write at the beginning forget about them.
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00:22:07,580 --> 00:22:16,120
So, first question how many generalized coordinates
do we need in this case only one,
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00:22:16,120 --> 00:22:22,210
because the moment it goes up it has to go
that amount down. So, both this positions
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00:22:22,210 --> 00:22:25,580
are
given by just one body was one coordinate
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00:22:25,580 --> 00:22:46,169
and let that be starting from say this position,
we can say this is q, if the total length
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00:22:46,169 --> 00:23:04,120
here is say l then this becomes l minus q.
So, the
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00:23:04,120 --> 00:23:24,380
generalized coordinate is consistent with
the constant is q, now write the kinetic energy
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00:23:24,380 --> 00:23:26,320
and the potential energy.
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00:23:26,320 --> 00:23:35,120
.So, what will be the potential energy, potential
energy is the potential of this body and
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00:23:35,120 --> 00:23:46,870
the potential of that body taken together.
So, you have s minus let us see let this be
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00:23:46,870 --> 00:23:54,610
M 1
and let this be M 2, so you have minus M 1
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00:23:54,610 --> 00:24:08,450
g and q for this one and minus M 1, M 2 g
l
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00:24:08,450 --> 00:24:17,470
minus q for this one. So, that is the total
potential energy and the kinetic energy is
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00:24:17,470 --> 00:24:22,270
kinetic
energy they of course, move together.
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00:24:22,270 --> 00:24:27,159
So, if this fellow moves with a velocity q
dot that fellow also moves with a velocity
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00:24:27,159 --> 00:24:33,400
q
dot, we can simply add them up. So, half M
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00:24:33,400 --> 00:24:43,409
1 plus M 2 q dot square, so the Lagrangian
is
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00:24:43,409 --> 00:25:24,070
half M 1 plus M 2 q dot square plus M 1 g
q plus M 2 g will it go no l minus q, now
183
00:25:24,070 --> 00:25:31,470
let
us differentiate. So, is M 1 plus M 2 q dot
184
00:25:31,470 --> 00:25:40,290
and the derivative with respect to q is let
us
185
00:25:40,290 --> 00:25:56,200
see this term M 1 g this fellow cancels off
this fellow vanishes, so minus M 2 g.
186
00:25:56,200 --> 00:26:03,360
So, these are two parts one the fixed part
another the variable part that is what remains,
187
00:26:03,360 --> 00:26:10,340
so ultimately since there is space here we
can write, the derivative of this with respect
188
00:26:10,340 --> 00:26:18,520
to
time M 1 plus M 2 q dot it becomes q double
189
00:26:18,520 --> 00:26:39,880
dot minus this. So, minus M 1 g plus M 2
that is the equation, one thing might be worth
190
00:26:39,880 --> 00:26:45,290
mentioning . that in
the last problem that we solve the simple
191
00:26:45,290 --> 00:26:53,990
pendulum, where written the potential as minus
m g l cos theta, taking this point as the
192
00:26:53,990 --> 00:26:57,440
datum.
Suppose, some of you say no, no I do not want
193
00:26:57,440 --> 00:27:03,980
to take that as the data, after all this
fellow comes here that is the minimum position
194
00:27:03,980 --> 00:27:09,649
I will take that as the datum from
measure it from there, yes you can do that.
195
00:27:09,649 --> 00:27:15,200
In that case this total length is here l and
how
196
00:27:15,200 --> 00:27:23,600
much is the elevation l minus this do that
no problem l minus l cos theta, but then what
197
00:27:23,600 --> 00:27:32,440
happens in that case just do it, it will be
l minus l cos theta this whole thing m g,
198
00:27:32,440 --> 00:27:36,470
you
would notice that this will have two components,
199
00:27:36,470 --> 00:27:42,649
one a constant component m g l minus
m g l cos theta when you take the derivative
200
00:27:42,649 --> 00:27:47,720
this fixed term vanishes. And therefore, that
does not appear, so it does not really matter
201
00:27:47,720 --> 00:27:55,580
where you take the datum is that clear, so
that is up to you wherever you want you can
202
00:27:55,580 --> 00:27:56,690
take.
203
00:27:56,690 --> 00:27:57,690
..
204
00:27:57,690 --> 00:28:13,610
Now, let us go to a slightly more involved
problem think of this, suppose you have a
205
00:28:13,610 --> 00:28:28,220
pendulum as before, but here the point of
suspension is moving like this, if we actually
206
00:28:28,220 --> 00:28:34,460
simulate and see it is behavior such a thing,
you would find that it has a very, very
207
00:28:34,460 --> 00:28:37,880
complicated behavior very, very complicated
motion. So, let us try to write down the
208
00:28:37,880 --> 00:28:49,260
differential equation that can capture that
motion, so this length is l here is your m,
209
00:28:49,260 --> 00:28:53,940
we
would need the x coordinate and y coordinate,
210
00:28:53,940 --> 00:28:56,269
because this fellow is moving in the x
coordinate.
211
00:28:56,269 --> 00:29:30,120
So, here is the x and here is the y, now x
is if this is theta x is l sin theta no let
212
00:29:30,120 --> 00:29:33,539
us redefine
because things are happening in this direction,
213
00:29:33,539 --> 00:29:42,990
let us call it y and let us call this x, then
y
214
00:29:42,990 --> 00:29:58,399
is equal to l sin theta and x is equal to
l cos theta. Now, your additional thing is
215
00:29:58,399 --> 00:30:04,029
adding to
the y, suppose that suspension’s movement
216
00:30:04,029 --> 00:30:13,220
is called y s and this fellows’ position
is y, if
217
00:30:13,220 --> 00:30:23,260
you have this if this fellow is not moving,
then you can say your x dot is minus l sin
218
00:30:23,260 --> 00:30:44,130
theta
and y dot is l cos theta into theta dot.
219
00:30:44,130 --> 00:30:53,840
So, these two are the velocities of this fellow
and we have to add to that the velocity
220
00:30:53,840 --> 00:31:01,960
here, in order to obtain the actual velocity
of the bob, it is if this were not moving
221
00:31:01,960 --> 00:31:08,010
whatever the velocity would be plus the velocity
that is imposed on it, that will be the
222
00:31:08,010 --> 00:31:14,590
easy way of looking at it that is what we
will do. So, ultimately when we have obtained
223
00:31:14,590 --> 00:31:20,190
this then you are the kinetic energy is half
m v square.
224
00:31:20,190 --> 00:31:32,860
.So, we have to say half m and v square is
x square plus y square x dot square plus y
225
00:31:32,860 --> 00:31:39,919
dot
square, it would be then x dot there is nothing
226
00:31:39,919 --> 00:31:54,270
happening in the x dot direction x dot
square plus y dot plus y s dot square, what
227
00:31:54,270 --> 00:32:00,950
you have done in this direction the movement
the velocity was y dot added to that you have
228
00:32:00,950 --> 00:32:06,480
got some velocity here. So, this fellow will
actually move with this velocity in the y
229
00:32:06,480 --> 00:32:10,400
direction and x square plus y square is the
actual
230
00:32:10,400 --> 00:32:19,460
velocity square that is what we need.
So, let us break it up you have half m x dot
231
00:32:19,460 --> 00:32:35,289
square plus y dot square plus y s dot square
plus twice y dot y s dot. Now, why we know
232
00:32:35,289 --> 00:32:41,769
what these guys are in terms of theta,
because that is what we want to write it in,
233
00:32:41,769 --> 00:32:55,100
so we substitute m x dot is x dot square minus
goes off l square sin square theta theta dot
234
00:32:55,100 --> 00:33:00,070
square, we have already done it plus this
is l
235
00:33:00,070 --> 00:33:17,740
square cos square theta theta dot square plus
y dot s square plus yes twice y this comes
236
00:33:17,740 --> 00:33:28,210
down twice y is this fellow y dot l cos theta
theta dot times y s dot is it enough this
237
00:33:28,210 --> 00:33:46,990
screen, let us move it a bit then you will
be able to see yes twice l cos theta theta
238
00:33:46,990 --> 00:33:55,720
dot y s
dot right happy. What can we do from here,
239
00:33:55,720 --> 00:34:03,409
wait, wait, so it is done essentially these
two
240
00:34:03,409 --> 00:34:10,120
can be put together, because sin square theta
cos square theta it is one, so that can be
241
00:34:10,120 --> 00:34:11,120
simplified.
.
242
00:34:11,120 --> 00:34:27,600
So, T is half m l square theta dot square
plus these two terms they remain plus y s
243
00:34:27,600 --> 00:34:38,500
dot
square plus twice l cos theta theta dot y
244
00:34:38,500 --> 00:34:44,990
s dot happy, look the procedure is pretty
straight
245
00:34:44,990 --> 00:34:48,540
forward. What we have done, . we have identified
what the velocity
246
00:34:48,540 --> 00:34:53,899
.would be and accordingly we have written
in terms of the actual generalized coordinate,
247
00:34:53,899 --> 00:35:01,860
these are this theta and of course, y is also
has to be one coordinate.
248
00:35:01,860 --> 00:35:10,280
So, we have written the T and then we have
to write V the potential energy and the
249
00:35:10,280 --> 00:35:14,850
potential energy is pretty simple, because
the amount of potential energy is in no way
250
00:35:14,850 --> 00:35:21,520
different from for the simple pendulum, simple
pendulum we had written minus m g l
251
00:35:21,520 --> 00:35:33,690
cos theta, so minus m g l cos theta. So, your
Lagrangian term becomes this whole thing
252
00:35:33,690 --> 00:35:51,750
half m let me multiply let me get rid of this
bracket, so half m l square theta dot square
253
00:35:51,750 --> 00:36:08,950
plus half m y dot s square plus no two and
half of m l cos theta theta dot y dot s that
254
00:36:08,950 --> 00:36:18,330
does
it T minus V, so plus m g l cos theta.
255
00:36:18,330 --> 00:36:28,780
So, that is the Lagrangian, so see I am doing
absolutely algorithmically I know that I
256
00:36:28,780 --> 00:36:34,041
have identified the coordinates and after
that I am proceeding algorithmically. All
257
00:36:34,041 --> 00:36:38,900
these
are simple algebra, no complication really
258
00:36:38,900 --> 00:36:48,550
even though this may look a bit longish, so
you have with respect to with respect to what
259
00:36:48,550 --> 00:36:58,120
will you differentiate theta. So, you have
theta dot and we will have to differentiate
260
00:36:58,120 --> 00:37:05,970
with respect to theta, now this becomes m
l
261
00:37:05,970 --> 00:37:19,390
square theta dot, this goes off this remains,
so this was what plus this plus this is not
262
00:37:19,390 --> 00:37:24,750
there.
So, you have if you differentiate it will
263
00:37:24,750 --> 00:37:39,140
be plus m l cos theta y dot s this goes off,
now
264
00:37:39,140 --> 00:37:45,630
when we differentiate with respect to theta,
this is this goes, this goes this remains
265
00:37:45,630 --> 00:37:53,220
yes m
l theta dot y dot s and then cos theta, so
266
00:37:53,220 --> 00:37:58,130
you have...
Student: .
267
00:37:58,130 --> 00:38:18,140
Sin theta minus yes no I am talking about
this term here minus.
268
00:38:18,140 --> 00:38:25,970
Student: .
Yes fine, now we have to differentiate these
269
00:38:25,970 --> 00:38:32,680
two and that is it, so can you write down
now starting from here.
270
00:38:32,680 --> 00:38:33,680
..
271
00:38:33,680 --> 00:38:54,800
Now, we have to write down the d d t let me
explicitly write it, so you have to say d
272
00:38:54,800 --> 00:39:06,290
d t
of this term m l square theta dot I am doing
273
00:39:06,290 --> 00:39:39,550
this especially, because people make a
mistake at this stage right. Now, you differentiate
274
00:39:39,550 --> 00:39:47,820
this, if you differentiate this what you
get firstly, you get m l square theta double
275
00:39:47,820 --> 00:39:55,450
dot, now here are two quantities that are
depending on time and therefore, you have
276
00:39:55,450 --> 00:39:58,940
to be careful in differentiating this term,
you
277
00:39:58,940 --> 00:40:19,600
have to say plus m l cos theta y double dot
s plus m l y dot s sin theta [FL] into theta
278
00:40:19,600 --> 00:40:24,070
dot.
And this is the crucial thing that people
279
00:40:24,070 --> 00:40:35,910
miss, so plus m l theta dot y dot sin theta
plus m
280
00:40:35,910 --> 00:41:00,470
g l sin theta equal to 0 done that is what
I am looking for what cancels.
281
00:41:00,470 --> 00:41:14,960
Student: .
These two cancel off no yes, so these things
282
00:41:14,960 --> 00:41:30,540
three things remain, then we can cancel
certain things. So, things become simpler
283
00:41:30,540 --> 00:41:44,980
theta double dot plus g by l sin theta these
things were there in the simple pendulum that
284
00:41:44,980 --> 00:41:50,490
is why I wanted to write it this way plus
1
285
00:41:50,490 --> 00:42:04,880
by l cos theta y double dot yes equal to 0,
that becomes the equation. Can you now write
286
00:42:04,880 --> 00:42:13,600
down on your copy, if this y yes happens to
be a sinusoidal oscillation, then what will
287
00:42:13,600 --> 00:42:17,970
this equation be.
Student: .
288
00:42:17,970 --> 00:42:23,440
No that is the externally imposed motion,
that is why it is not a independent degree
289
00:42:23,440 --> 00:42:25,450
of
freedom of the system, that is why it is not
290
00:42:25,450 --> 00:42:30,120
necessary at this stage, her question is that
291
00:42:30,120 --> 00:42:37,570
.would not this become another coordinate
. no it is not necessary,
292
00:42:37,570 --> 00:42:45,450
because it has no a internal degree of freedom
it is moved by somebody y s is being
293
00:42:45,450 --> 00:42:51,210
imposed by somebody.
So, the system has no freedom in choosing
294
00:42:51,210 --> 00:42:54,800
that, that is why you do not need that as
a
295
00:42:54,800 --> 00:43:00,000
coordinate that is imposed by somebody, even
without solving the differential equation
296
00:43:00,000 --> 00:43:10,700
you know y s should be at any point of time
that is why. So, you need a coordinate if
297
00:43:10,700 --> 00:43:12,540
in
order to find out what that coordinate value
298
00:43:12,540 --> 00:43:15,770
will be at a specific point of time, you need
to solve the differential equation, but in
299
00:43:15,770 --> 00:43:20,900
this case you do not need to…
Student: .
300
00:43:20,900 --> 00:43:28,200
Which one this fellow.
Student: If it had a mass then kinetic energy
301
00:43:28,200 --> 00:43:33,630
you would have considered a term.
Yes her question then is if this fellow were
302
00:43:33,630 --> 00:43:40,590
a mass, then somebody would not be able to
move it just like that, it would be application
303
00:43:40,590 --> 00:43:54,160
of force and this will then have it is own
dynamics, in that case yes you need to…
304
00:43:54,160 --> 00:43:56,720
Student: .
Yes in that case y s becomes an independent
305
00:43:56,720 --> 00:44:01,760
coordinate of this system an independent
degree of freedom, but in this case the way
306
00:44:01,760 --> 00:44:08,620
we have defined the problem as yet it does
not become, yes if this fellow had a mass
307
00:44:08,620 --> 00:44:15,760
you are applying a force and then the mass
moves depending on the force and it is own
308
00:44:15,760 --> 00:44:21,500
dynamics, because this being pulled by this
fellow also. So, in that case you do not know
309
00:44:21,500 --> 00:44:27,840
beforehand what the position of mass is at
any point of time, you need to solve the differential
310
00:44:27,840 --> 00:44:34,630
equation for that and for that it
becomes a independent coordinate.
311
00:44:34,630 --> 00:44:43,090
So, in this case as I told you in case y s
is a sinusoidal function, say y s is equal
312
00:44:43,090 --> 00:44:48,950
to say A
cos omega t, can you see yes in that case
313
00:44:48,950 --> 00:44:55,890
what is the equation write down yourself,
because most logical way of moving this fellow
314
00:44:55,890 --> 00:45:01,330
would be a sinusoidal oscillation. So, if
that happens then what will be the equation,
315
00:45:01,330 --> 00:45:09,360
this will just become a omega square cos
becomes sin with a negative sign here and
316
00:45:09,360 --> 00:45:19,710
that is it, only I wanted to you to know that
this is a double derivative.
317
00:45:19,710 --> 00:45:31,530
.So, you will need to take a derivative twice
that is it, but if it is a unknown quantity
318
00:45:31,530 --> 00:45:34,440
and
something is imposed from outside, then it
319
00:45:34,440 --> 00:45:48,440
is simply this equation. Now, are you now
comfortable with this idea, how the equations
320
00:45:48,440 --> 00:45:57,430
are derived, note one thing that you are
getting second order differential equations
321
00:45:57,430 --> 00:46:03,060
along each position coordinate in a, so far
we
322
00:46:03,060 --> 00:46:05,490
have considered problems with just one coordinate.
.
323
00:46:05,490 --> 00:46:12,130
Let us now take the next step, let us take
a simple next step to a system, where there
324
00:46:12,130 --> 00:46:14,600
are
more than one position coordinate, say again
325
00:46:14,600 --> 00:46:18,580
let us take the pendulum, but let us make
it
326
00:46:18,580 --> 00:46:26,680
a little more complicated. Suppose here it
is, but instead of just being connected to
327
00:46:26,680 --> 00:46:35,471
a here
is a spring and then you connect it [FL] what
328
00:46:35,471 --> 00:46:47,130
has happened, because of this spring, there
is another degree of freedom in this direction.
329
00:46:47,130 --> 00:46:53,980
Now, you really have to apply your brain a
bit in order to identify the generalized
330
00:46:53,980 --> 00:47:05,870
coordinates [FL] here is your theta. Now,
you would notice that there are many ways
331
00:47:05,870 --> 00:47:09,680
of
defining the position coordinates theta is
332
00:47:09,680 --> 00:47:15,690
a natural choice alright, but how will you
in the
333
00:47:15,690 --> 00:47:18,810
radial direction there is another degree of
freedom I can see. But, how do you define
334
00:47:18,810 --> 00:47:21,660
that
there are various ways of doing it, you might
335
00:47:21,660 --> 00:47:26,120
say that distance is bigger from here that
is
336
00:47:26,120 --> 00:47:33,310
it, you might say that no let it hang freely
here in this position, then it takes a particular
337
00:47:33,310 --> 00:47:35,910
length of this.
Remember in this range in this position it
338
00:47:35,910 --> 00:47:39,240
is not un stretched it is already stretched,
but
339
00:47:39,240 --> 00:47:44,460
that let that be the datum and from there
I will measure the difference, you could say
340
00:47:44,460 --> 00:47:45,460
that
341
00:47:45,460 --> 00:47:52,850
.and they will actually make no difference.
So, in order to do that let us do it that
342
00:47:52,850 --> 00:47:56,930
way, so
what did we say, we said that when it freely
343
00:47:56,930 --> 00:48:02,930
hangs vertically, then your position is the
datum position.
344
00:48:02,930 --> 00:48:20,120
In that position, suppose it came here and
this distance is say a and then this distance
345
00:48:20,120 --> 00:48:23,180
is
the deviation from there. So, distance is
346
00:48:23,180 --> 00:48:27,430
a plus the coordinate that we are talking
about r,
347
00:48:27,430 --> 00:48:38,310
we can do it this way that also uniquely defines
it is position is it not plus theta, theta
348
00:48:38,310 --> 00:48:41,230
has
to be there as one of the coordinates. So,
349
00:48:41,230 --> 00:49:06,050
r and theta, so then generalized coordinates
becomes r and theta, now in terms of this
350
00:49:06,050 --> 00:49:13,080
r and theta what is kinetic energy and what
is
351
00:49:13,080 --> 00:49:16,800
potential energy, that is what we will try
to write.
352
00:49:16,800 --> 00:49:20,190
Kinetic energy will be relatively simple to
write, potential energy will require a bit
353
00:49:20,190 --> 00:49:25,580
of
thought, the kinetic energy is half m v square.
354
00:49:25,580 --> 00:49:32,290
So, half m we will write first and then we
will try to figure out what the v square is,
355
00:49:32,290 --> 00:49:39,400
now is notice that there is a component of
the
356
00:49:39,400 --> 00:49:43,020
motion in the radial direction and there is
a component of the motion in the
357
00:49:43,020 --> 00:49:50,830
circumferential direction, so if you can identify
these two, then it will be simply this
358
00:49:50,830 --> 00:49:57,150
square plus that square.
So, in the radial direction the velocity is...
359
00:49:57,150 --> 00:50:04,730
Student: r dot.
R dot yes, so it is r dot square plus.
360
00:50:04,730 --> 00:50:13,680
Student: .
Yes a plus r theta dot whole square of this,
361
00:50:13,680 --> 00:50:28,620
so both this have the squares done, the
potential energy in this case there are many
362
00:50:28,620 --> 00:50:30,380
sources of the potential energy there is a
fun
363
00:50:30,380 --> 00:50:36,800
of this problem. One, it is contributed by
the gravity, two it is contributed also by
364
00:50:36,800 --> 00:50:40,630
the
spring and they have no difference we will
365
00:50:40,630 --> 00:50:48,930
simply add them up, because they are scalars
that is the advantage fine. So, how do you
366
00:50:48,930 --> 00:50:57,020
go about it, notice let us start the argument
from the position that it is un stretched,
367
00:50:57,020 --> 00:51:00,330
if it is un stretched it has some position,
but from
368
00:51:00,330 --> 00:51:07,060
here it has already being elongated by application
of a force.
369
00:51:07,060 --> 00:51:09,530
Student: .
M g.
370
00:51:09,530 --> 00:51:14,100
.Student: .
So, how much has it already elongated.
371
00:51:14,100 --> 00:51:20,120
Student: m g by K.
M g by K it has already elongated in this
372
00:51:20,120 --> 00:51:26,730
position by m g by k and I am now counting
the
373
00:51:26,730 --> 00:51:34,460
difference from that position here and therefore,
what is the actual elongation m g by K
374
00:51:34,460 --> 00:52:00,710
plus r. So, the actual elongation is let me
start from here actual m g by K plus r that
375
00:52:00,710 --> 00:52:03,961
is the
amount the actual energy, so the potential
376
00:52:03,961 --> 00:52:17,130
energy in the spring V spring is half K into
actual elongation square.
377
00:52:17,130 --> 00:52:35,510
So, you have, so that is the amount of energy
in the spring, when it is hanging vertically
378
00:52:35,510 --> 00:52:42,990
in this position, then the gravitational energy
is 0 and we are taking that as the datum and
379
00:52:42,990 --> 00:52:48,240
counting gravitational energy from there.
So, how much has it gone up from there that
380
00:52:48,240 --> 00:53:01,610
position, this much is a plus r cos theta,
so a minus a plus r cos theta is the amount
381
00:53:01,610 --> 00:53:14,540
by
which it has gone up. So, V gravity is m g
382
00:53:14,540 --> 00:53:26,470
times this amount which is a minus a plus
r
383
00:53:26,470 --> 00:53:34,400
cos theta is the line of argument.
The moment you have identified this two, the
384
00:53:34,400 --> 00:53:36,200
total potential energy is the addition of
this
385
00:53:36,200 --> 00:53:43,520
two and the Lagrangian is t minus that and
the rest is simply algorithmic, you do not
386
00:53:43,520 --> 00:53:50,970
have to apply your head just algebra. So,
you obtain that differential equation starting
387
00:53:50,970 --> 00:54:00,110
from there, there is the rest of the problem
is simply in a routine procedure, you do not
388
00:54:00,110 --> 00:54:07,320
have to really apply your head a lot for the
rest of this problem, bring that in the next
389
00:54:07,320 --> 00:54:10,390
class.
Thank you very much, let us stop here.
390
00:54:10,390 --> 00:54:10,390
.