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Start today by briefly recapitulating where
we were in the last class, we have seen that
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even though the propositions of Newton. Where
they laid the basic frame work for
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deriving differential equations, there are
a some practical difficulties and we identified
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that difficulties.
.
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Let us enumerate them one by one, the difficult
number 1 was that in a given system
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mainly constraint system there will be constraint
forces, just to recapitulate what the
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constraints, where there were two types of
constraints, holonomic constraints and non
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holonomic constraints. And the holonomic constraints
were those, which can be
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expressed as a algebraic equation, why the
non holonomic constraints are those, which
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can be expressed as in equation.
So, that was what we did and then we are said
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that in order for the body to follow the
constraint or constraint surface, the constraint
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surface has to apply some kind of a force
on it. So, there will always be some force
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acting on the body the constraint force and
in
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.order to write the Newtonian equation for
each body, you will have to take in account
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all
the force acting on it and therefore, the
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constraint forces will have to be considered.
So, number 1 was the constraint forces have
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to be considered, so that was one difficulty,
because the constraint forces are not always
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easily quantifiable. Number 2, difficulty
was
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that in a inter connected system various body
will interact with each other with forces
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and these forces are in general produced by
springs. If two bodies are connected by
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springs, this body will be applying a force
on the other body by means of this spring,
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if
two bodies move against each other by means
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of some kind of frictional element, the one
body will apply a force on the other body.
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So, for each body this forces have to be considered
and all this forces are vectors, so that
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was another difficulty that we the formulation
becomes very complicated. So, all the
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given forces are vectors and we would like
to simplify the matter by writing down the
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differential equations in terms of things
that are scalars, one would be comfortable
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with
that. The third problem was, that we have
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seen that in case of a system with holonomic
constraint for example, the pendulum, it is
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completely stupid to write three differential
equations, because it does not have it does
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not move in all those directions.
So, in general if there are n total number
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of configuration coordinates no not n thrice
n, if
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there are n bodies there will be three times
of n number of configuration coordinates.
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But
it is somewhat unintelligent to write all
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the thrice n equations, rather we should be
able to
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write thrice n minus h number of equation,
where h is the number of holonomic
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constraints. So, the number of equations while
the number actually required, where h is
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the number of holonomic constraints. So, these
where the three major difficulties with
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the Newtonian formalism, that we need to overcome.
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..
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So, let us go one by one, first let us consider
the situation of we consider the situation
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of
the constraint forces one, how to
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and in countering this problem we argued that
even
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though the constraint forces are there in
general they do not work. Because, the actual
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motion is orthogonal to the constraint force,
there are situations somewhat rare
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situations, most situation the constraint
forces do not do any work.
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But, where the holonomic constraint is what
we termed as rheonomic, where the
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holonomic constraint is dependent on time.
Say, there is a bowl in which you have
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released a marble to move and the bowl is
being moved around, so these are the
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situations where you have a holonomic constraint
all right, but that is dependent on time.
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Now, when that happens we need to go into
another stage of abstraction, we said that
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we
will not consider the actual motion, actual
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velocity we will not multiply the force with
the actual velocity.
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Rather, we will multiply the force with a
specific conceptually or emergent velocity,
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emergent motion which is consistent with the
constraint at every instant of time. That
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means, if you have a pendulum with a oscillating
support, then at every point the chord
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has a particular angle and the bob is here.
In that case, the admissible motion at that
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position would be orthogonal to the direction
of the chord.
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So, the moment we conceptually imagine that
kind of a possible motion admissible
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motion. Then, we have a work the moment you
multiply the forces with the, that
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admissible motion we get a work, but that
is what the actual work we done by this
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.system. That is again admissible work, you
might ask, what gain do you make by
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considering this, yes we do will it will be
clear to you later.
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But, the what we did was we say that originally
the equation was, originally the
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Newtonian equation was j is equal to 1 to
N, j is the number of mass points and it was
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m
j r double dot j that is mass into acceleration
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term minus F j this is the applied force or
the given force, this was equal to the constraint
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forces. So, that was our logic that was the
initial starting point, because this was the,
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what Newton said.
And then we said from here, that let us multiply
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both these sides by the admissible
displacement. There by we got this is a vector
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equal to again the right hand side is equal
to 1 to N F c this vector j times, now this
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term will; obviously, vanish because for each
of
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this N bodies, the constraint force times
the admissible motion is 0. So, is equal to
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0 that
is how we eliminated the constraint forces,
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so keep this final equation written in your
copy, we will refer to it later.
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.
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The second problem was the reduction of the
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coordinates and in that what we did we
have do, we said that after all in the thrice
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n dimensional configuration space, each of
the
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holonomic constraints define a surface. And
if there are two holonomic constraints
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imagine for the sake of simplicity in visualization,
the original configuration space is
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three dimensional, in which you have defined
one holonomic constraint that will define
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a
surface.
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.So, that reduces the dimension by one, if
you have another holonomic constraint that
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will
define another surface. So, ultimately the
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body will be constrained to move in
intersection of the two holonomic constraints
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and that reduces the dimension by 2, so
each holonomic constraint is reducing the
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dimension by it is number. So, since ultimately
the body is constrained to move on the intersection
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of holonomic surfaces, there is no
reason why we should still continue to define
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the coordinates in terms of the old x, y,
z
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coordinates.
We said, that now that we have identify a
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subspace, where the body must lie the
configuration point must lie not the body
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really, because we are now talking in terms
of
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configuration space. If there are three bodies,
there will conceptually thrice n three times
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three number of coordinates, where still I
would advise you to still visualize or extend
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the visualization. We cannot visualize in
nine dimensional space all right, but still
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try
extend the visualization, that it is some
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kind of a space in which the moment you have
defined a constraint it defines a conceptually
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a surface.
And the body this the configuration point
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must lie on that surface, must move on that
surface. And therefore, we said that now let
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us define a new set of coordinates on that
constrained surface and these are marked as
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q, the generalized coordinates, so in case
of
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the spherical pendulum what would we do there
is only one body. So, original set of
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coordinates were x, y, z and then since the
body is constrained to lie on the spherical
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surface.
Therefore, on the spherical surface the most
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convenient coordinates are the spherical
coordinates theta and phi. So, we define theta
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and phi as the new set of coordinates
consistent with the constraints and in a system
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there could be some motions that are
constrained to on a spherical surface, some
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motion that are constrained to be on a linear
side. In that case those motions will be we
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would try to define the minimum number of
coordinates consistent with the constraint
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that completely define the positional status
of
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the system and that will be the ultimate our
definition of the generalized coordinates.
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So, we said that if we have a original thing
like this and we know that the behavior is
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constrained on this kind of a surface. Then,
we would define a new set of coordinates
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and then any point which was earlier expressed
in terms of x, y, z, now would have to be
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expressed in terms of q 1 and q 2. Now, extend
that vision this x, y, z could be thrice n
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.and this q 1 and q 2 could be again a large
number of variables, just the idea is that
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we
have now defined a lesser number of variables
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that are consistent with the constraints.
So, q’s are, so suppose a body was there
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at this position, then it is position is given
by
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this r vector in the original coordinate.
If it is given by the r vector, then for this
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body this
r can r was earlier expressed in terms of
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x, y, z, but now it has to be r has to be
expressed
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in terms of q 1 and q 2. New set of coordinates,
similarly there was another body suppose
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here it is coordinate say this would be r
1 and this one would be r 2. And all those
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bodies’ coordinates now would have to be
expressed in terms of the new set of
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coordinates.
And that can be easily done, because we can
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say r 1 is equal to r 1 of the new set of
coordinates q 1, q 2, q n and may be time,
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it might also depend on time, because the
surface itself might depend on time, similarly
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that will go to r n capital N q 1.
.
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So, in short we can write then t j th mass
point this is a vector as r j of q i and time,
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where i goes from 1 to n. Now, once we have
defined the old set of coordinates in terms
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of the new set of the coordinates, we can
also define the velocities, the velocity would
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be
r dot j or d r j d t. So, if r j is dependent
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on all these, when you take the derivative
it will
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be expressed in terms . you will be able to
write and all that plus
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dived depends on time.
So, this is nothing but, sigma i is equal
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to 1 to n delta r j vector delta q i and this
is q dot i
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plus r j. So, this is how the velocity would
be expressed velocity of each body fine,
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.notice in the left hand side what we have,
this body is moving .
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bodies velocity expressed in terms of the
new set of coordinate, we will use that. Now,
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if
I ask you how does the r dot j this is, so
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this is r dot j how does this fellow depend
on the
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generalized coordinates.
So, if i ask you del r dot j del q dot i what
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you get, suppose you want to express this
derivative, look at the right hand side this
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is a generalized velocity in one direction
i. So,
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if we differentiate it with respect to another
generalized coordinate you get 0, because
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it
general coordinate is a independent quantity.
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So, the moment you want to do this
derivative with respect to q dot i, then you
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will find only that term will remain where
this
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and that are the same.
All the others will vanish and this will also
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vanish, because this tells how does the
constraint surface vary with time and that
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also does not depends on q dot i. So, ultimately
what remains is the set clear, so we keep
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this for future reference, we have argued
that
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this would be valid and we keep this for future
reference fine. So, there are two things we
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are keeping for future reference, one is this
equation another is this equation we will
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come back to that.
Let us keep somewhere, so that we can again
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refer to that fine . now
let us get back to this picture, this was
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the position r of a body, if I ask you what
is it is
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admissible motion it is admissible motion
is on the surface.
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.
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.So, your admissible motion delta r j this
now has to be on the surface that now has
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to be
expressed in terms of the new coordinates
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and how would you expressed that in terms
of
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the new coordinates. Exactly the same way
we will say that this is delta r j del q i
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times
delta q i sum over i is equal to 1 to n correct
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I am not writing the broken up form,
because that will be the same for each body
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j the derivative of that r j in terms of the
ith
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coordinate times the admissible motion along
the ith coordinate.
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So, this has to be summed up, so this is another
thing that we keep for future reference,
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getting too complicated know what you have
essentially done is, that what are we
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interested in we are interested in the transformation
of the coordinates ((Refer Time:
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22:37)) and who do we transfer. First this
vector to the derivative of the vector; that
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means, the velocity position and the velocity
how do they transform into the new
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coordinates. And then the admissible motion
that is the three things we wanted to express
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in terms of new coordinates, that is what
we have written now fine ((Refer Time:
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23:11)).
Now, let us start with this equation that
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we wrote, we had multiplied by the generalized
by the admissible motion and obtained the
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right hand side 0. Let us start from there
. at this time our objective is number 2 all
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the given forces were
vectors. So, everything in the Newtonian framework
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has to be expressed in terms of
vectors, and then in any given system what
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is the absolutely general scalars that we
have
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the energies energy is a scalar.
So, we would like to express that was the
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ingenuity of Lagrange, that he said that
everything can be expressed in terms of energies.
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So, let us try to frame the whole set of
equations in terms of energies, which are
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scalars and what are the energies kinetic
energy
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and the potential energy. So, that is what
we are trying to arrive at, but you we will
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have
to go through a route, because the way Lagrange
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himself did and the way you are to
understand, they are a bit different. Because,
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now we have the hindsight of how this
thing can be derived far easily, that is why
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we are going this way.
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..
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So, we start from here j there are n number
of mass points, so N m j r double dot j minus
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F j I am just copying this, this is a vector,
this is a vector delta r j is equal to 0 that
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is
where we were. Now, we have obtained something
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about how the generalized admissible
motion transforms, we . that and that is what
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we need to substitute
here the first step. So, we substitute what
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we have, you can also substitute on your
copies, you will see N m j r double dot j
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minus F j is a vector times all this stuff
i is equal
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to 1 to n I have just substituted this.
Now, there are two summations here and is
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immaterial in which order you sum them, so
we bring this summation over i forward. So,
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let us do that we bring this summation over
i
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forward, what remains at this term, so you
have the summation over j, summation over
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i
just to recall what is summation over i the
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number of generalized coordinates and
summation over j is the number of bodies two
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things.
Then, we this term comes m j r double dot
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j minus F j this and then this term delta
r j
208
00:26:55,539 --> 00:27:13,419
delta q i. So, this term and then you have
to sum over fine, so this term is summed over
209
00:27:13,419 --> 00:27:16,440
j
and that term after that it is summed over
210
00:27:16,440 --> 00:27:23,020
i let us put this term inside, so what do
we have
211
00:27:23,020 --> 00:27:36,220
is this part visible yes i is equal to 1 to
n just put this inside you have sum j is equal
212
00:27:36,220 --> 00:27:45,640
to 1
to N m j r vector double dot j times r j delta
213
00:27:45,640 --> 00:28:13,330
q i I have multiplied it here minus F j vector
r delta q i, just keep check that I am doing
214
00:28:13,330 --> 00:28:22,620
it right fine done.
So, we have this term we need to put a bracket
215
00:28:22,620 --> 00:28:29,330
here, because the whole the summation is
over the whole thing. Now, let us see what
216
00:28:29,330 --> 00:28:36,999
each term means, here we have r j that is
the
217
00:28:36,999 --> 00:28:47,760
.position of the nth or jth body is double
dot times this derivative, what does it means
218
00:28:47,760 --> 00:28:50,540
physically just look at it.
.
219
00:28:50,540 --> 00:29:05,820
First suppose we have we take r dot j delta
r j these are all vectors, suppose we start
220
00:29:05,820 --> 00:29:12,720
with
this term and take a d d t of this can you
221
00:29:12,720 --> 00:29:18,429
do it, there are two terms here we will go
by the
222
00:29:18,429 --> 00:29:23,879
one the standard way of differentiation. So,
we first differentiate this, so that would
223
00:29:23,879 --> 00:29:34,279
be
this is dot r dot here r double dot j times
224
00:29:34,279 --> 00:29:47,559
this plus this remains now and this term is
differentiated with respect to time.
225
00:29:47,559 --> 00:30:04,500
So, we have r dot, so del r dot you can do
it this way, so we have just written the
226
00:30:04,500 --> 00:30:11,670
derivative like this. Now, you notice why
did we do all this, because this term is this
227
00:30:11,670 --> 00:30:22,639
term, so this term is expressible as r this
is what we are interested in expressible as
228
00:30:22,639 --> 00:30:35,720
this is
not there r j del q i is equal to this minus
229
00:30:35,720 --> 00:30:46,309
this r dot j derivative of j with respect
to q i
230
00:30:46,309 --> 00:31:05,529
minus r j is a vector dot. So, we have understood
the meaning of the first term in terms of
231
00:31:05,529 --> 00:31:12,149
two terms, which we are yet to understand
leave it at this stage, but let us substitute
232
00:31:12,149 --> 00:31:18,690
it
here, let us substitute it here.
233
00:31:18,690 --> 00:31:19,690
..
234
00:31:19,690 --> 00:31:25,240
What do we have after substitution we have,
let us start from further left hand side,
235
00:31:25,240 --> 00:31:34,999
because it will be rather large i is equal
to 1 to small n, then we have a bracket then
236
00:31:34,999 --> 00:31:40,799
it is
in terms of j is equal to 1 to capital N,
237
00:31:40,799 --> 00:31:45,159
this is what we are writing m j then this
is to
238
00:31:45,159 --> 00:32:12,940
substituted. So, it will be m j d d t of r
dot j dou r j dou q i fine minus let us break
239
00:32:12,940 --> 00:32:19,210
up the
summations again j is equal to 1 to capital
240
00:32:19,210 --> 00:32:37,230
N, the next term m j remains now comes this
term r dot no this is vector dot j dou r dot
241
00:32:37,230 --> 00:32:44,570
j dou q i.
So, that substitution is over, but there was
242
00:32:44,570 --> 00:32:50,009
another term remaining here, that has to be
written. So, we will again write it as a separate
243
00:32:50,009 --> 00:33:04,361
summation I will write here minus j is
equal to 1 to N it remains as it is F j dou
244
00:33:04,361 --> 00:33:12,169
r j dou q i, now this bracket will be closed
times
245
00:33:12,169 --> 00:33:26,200
your generalized coordinate generalize admissible
motion is equal to 0. Now, after all
246
00:33:26,200 --> 00:33:39,570
these business we have come to three terms,
one term, second term and the third term
247
00:33:39,570 --> 00:33:47,960
three terms.
And these just by looking at these terms it
248
00:33:47,960 --> 00:33:50,360
is very difficult to physically understand
what
249
00:33:50,360 --> 00:33:56,309
they are. So, let us try to physically understand
what they are, we will come back to this,
250
00:33:56,309 --> 00:34:03,940
so let us keep this here and let us write
down individually, we said that we want to
251
00:34:03,940 --> 00:34:06,239
express things in terms of energies.
252
00:34:06,239 --> 00:34:07,239
..
253
00:34:07,239 --> 00:34:10,252
A first energy is kinetic energy, what is
the kinetic energy total kinetic energy, kinetic
254
00:34:10,252 --> 00:34:19,720
energy expressed written as T. So, kinetic
energy is half m v square for each body, so
255
00:34:19,720 --> 00:34:41,389
half m 1 that is r dot 1 square plus half
m 2 r 2 vector dot square plus, so that is
256
00:34:41,389 --> 00:34:49,720
the total
kinetic energy, that is half no I will keep
257
00:34:49,720 --> 00:34:52,820
the summation outside half this is summed
over
258
00:34:52,820 --> 00:35:10,520
j is equal to 1 to N, N number of bodies half
m j r dot j square fine do we have this term
259
00:35:10,520 --> 00:35:14,930
around here no not really not exactly this
term.
260
00:35:14,930 --> 00:35:20,430
But, we have what could possibly be obtained
from it is derivative, so let us take the
261
00:35:20,430 --> 00:35:33,110
derivative of this term, this kinetic energy.
So, this is the kinetic energy, so the derivative
262
00:35:33,110 --> 00:35:38,320
of the kinetic energy, kinetic energy is after
all dependent on the velocity, so we will
263
00:35:38,320 --> 00:35:46,800
derive take the derivative in terms of the
generalized velocity T del q dot i that is
264
00:35:46,800 --> 00:35:51,320
the
generalized velocity.
265
00:35:51,320 --> 00:35:56,650
If you take what do you have in the right-hand
side, we have the summation first j is
266
00:35:56,650 --> 00:36:05,440
equal to 1 to capital N, when you take this
derivative this two comes forward and gets
267
00:36:05,440 --> 00:36:22,480
cancelled with the half. So, it will be m
j r dot j and then the derivative of r j del
268
00:36:22,480 --> 00:36:34,710
dot j dot
q i, this is how would you express fine, so
269
00:36:34,710 --> 00:36:38,980
we are better off because we can identify
this
270
00:36:38,980 --> 00:36:45,880
term somewhere here. So, we know that here
this term was actually talking about the
271
00:36:45,880 --> 00:36:50,520
derivative of the kinetic energy with respect
to the generalized coordinate good at least
272
00:36:50,520 --> 00:37:06,190
one thing is settled.
Now, where did I keep the expressions here,
273
00:37:06,190 --> 00:37:12,100
. see we have already
obtained that the derivative of the velocities
274
00:37:12,100 --> 00:37:13,570
would be equal to the derivative of the
275
00:37:13,570 --> 00:37:19,980
.coordinates, we had already done this we
had kept it for future reference that is where
276
00:37:19,980 --> 00:37:26,270
we
will use it. This term is then equal to m
277
00:37:26,270 --> 00:37:33,931
j r dot j, but now we do not need to write
the
278
00:37:33,931 --> 00:37:52,850
velocities just the positions, so this is
something that we will keep and then let us
279
00:37:52,850 --> 00:37:56,370
also
differentiate the kinetic energy with respect
280
00:37:56,370 --> 00:37:57,370
to the position.
.
281
00:37:57,370 --> 00:38:05,190
So, we have derivative of the kinetic energy
with respect to the position not q dot i it
282
00:38:05,190 --> 00:38:08,850
is q
i, then we will again have to write is as
283
00:38:08,850 --> 00:38:14,580
the same summation of partial derivatives.
So, it
284
00:38:14,580 --> 00:38:25,820
will be summation of j is equal to 1 to capital
N and in that case what will remain m j see
285
00:38:25,820 --> 00:38:31,240
this is what we are trying to express or take
a derivative of in terms of the q i. So, it
286
00:38:31,240 --> 00:38:41,760
will
be same way m j r dot j times the derivative
287
00:38:41,760 --> 00:38:53,220
of r dot j times the derivative i, you can
possibly identify that this is this term.
288
00:38:53,220 --> 00:39:06,250
So, we are slowly homing on to what we wanted
to express, . finally
289
00:39:06,250 --> 00:39:16,280
if there is a force acting on this body, that
force if now we expressed in terms of the
290
00:39:16,280 --> 00:39:20,000
new
coordinates how what shape will it take, in
291
00:39:20,000 --> 00:39:30,730
the new coordinates suppose that generalized
force is called Q i will be expressed as summation
292
00:39:30,730 --> 00:39:47,770
of again j is equal to 1 to N F is the
vector j r vector j. So, all we are doing
293
00:39:47,770 --> 00:39:51,450
we had the actual forces applied F j, now
we are
294
00:39:51,450 --> 00:39:55,490
converting into transforming into the new
set of coordinates Q i.
295
00:39:55,490 --> 00:40:02,780
So, these are the forces generalized forces
that are applied in the direction of the
296
00:40:02,780 --> 00:40:11,780
generalized coordinates, so this will be the
force. Now, suddenly we see that all the terms
297
00:40:11,780 --> 00:40:16,940
that we wanted to know, . they are all clear
what is this, this is
298
00:40:16,940 --> 00:40:23,570
.actually the original forces acting on the
bodies converted into the generalized forces,
299
00:40:23,570 --> 00:40:32,330
what is this term, this term we have just
indentified as this term.
300
00:40:32,330 --> 00:40:44,820
So, this term is same as this term, so d d
t of this is d d t of this clear, so and this
301
00:40:44,820 --> 00:40:48,900
term is
already we have derived. So, all we can now
302
00:40:48,900 --> 00:40:59,750
express this equation in a much simpler and
elegant form as i is equal to 1 to small n
303
00:40:59,750 --> 00:41:07,800
and then we had written, now the all the
summation go away, because this term summed
304
00:41:07,800 --> 00:41:12,750
together is some simple term, this term
summed together is some simple term, this
305
00:41:12,750 --> 00:41:14,870
term summed is sum together is some simple
term.
306
00:41:14,870 --> 00:41:28,730
So, all these summation go away over j and
we get d d t of del t q dot i, because that
307
00:41:28,730 --> 00:41:33,920
is
what we had written, this term were expressed
308
00:41:33,920 --> 00:41:40,390
as this and that is what we have in that
particular term. So, this is the first term
309
00:41:40,390 --> 00:41:44,790
minus the second term here is what we wanted
to
310
00:41:44,790 --> 00:41:59,770
express, so we have del T q i dou Y dou q
i that is the second term and the third term
311
00:41:59,770 --> 00:42:04,910
is
minus simply Q i, this whole thing simple
312
00:42:04,910 --> 00:42:16,380
it has to be multiplied by the admissible
motion is equal to 0.
313
00:42:16,380 --> 00:42:23,090
So, true that we went through a lot of you
know partial derivates and messy stuff, but
314
00:42:23,090 --> 00:42:27,660
ultimately the objective was to simplify stuff
and here is the simple thing. And you
315
00:42:27,660 --> 00:42:33,321
notice, that here is something that is multiplied
by the admissible motion and the
316
00:42:33,321 --> 00:42:38,040
admissible motion along the first coordinate
and the admissible motion under second
317
00:42:38,040 --> 00:42:44,610
coordinate are all independent quantities.
And this term summed over i would be 0 only
318
00:42:44,610 --> 00:42:53,920
if this term is 0 always can you see that.
So, here is the term for the ith thing it
319
00:42:53,920 --> 00:42:58,970
is multiplied by the ith admissible motion
and all
320
00:42:58,970 --> 00:43:05,120
that summed over will has to be 0. And that
can be 0 only if since this cannot vanish,
321
00:43:05,120 --> 00:43:07,420
this
is something that we have assumed that it
322
00:43:07,420 --> 00:43:15,820
can be just any motion, therefore this term
must vanish. Not only that this term must
323
00:43:15,820 --> 00:43:19,660
vanish for all the generalized coordinates,
not
324
00:43:19,660 --> 00:43:25,080
the summation, just individually along all
the generalized coordinates this term must
325
00:43:25,080 --> 00:43:26,080
vanish.
326
00:43:26,080 --> 00:43:27,080
..
327
00:43:27,080 --> 00:43:42,840
So, that gives a very simple equation d d
t of dou T dou q i dot minus doh T doh q i
328
00:43:42,840 --> 00:43:53,180
minus the generalized forces equal to 0. Now,
this is the Lagrangian equation in the most
329
00:43:53,180 --> 00:44:14,610
generalized form, in the most general form,
you might wonder that, so far we were
330
00:44:14,610 --> 00:44:19,930
initially talking about energies, energies
would be kinetic as well as potential. But,
331
00:44:19,930 --> 00:44:22,230
so far
we have talked out only kinetic energy, some
332
00:44:22,230 --> 00:44:27,290
where does the potential energy come or
where does the potential energy go.
333
00:44:27,290 --> 00:44:33,940
Now, you notice that here we have got a generalized
force, what does that mean, it
334
00:44:33,940 --> 00:44:41,170
means the forces on the bodies in the direction
of the generalized coordinates. So, in case
335
00:44:41,170 --> 00:44:48,220
of the pendulum, if you apply a force in the
direction of the theta, so that will be the
336
00:44:48,220 --> 00:44:53,850
generalized force. Now, what does this force
consist of what could be the different types
337
00:44:53,850 --> 00:45:00,810
of forces acting on a body just think, these
are the forces, so these are the forces that
338
00:45:00,810 --> 00:45:06,670
we
need to consider.
339
00:45:06,670 --> 00:45:28,000
What kind of forces could it be, it could
be one forces
340
00:45:28,000 --> 00:45:35,380
like the gravitational force it could
be force applied by one fellow on another
341
00:45:35,380 --> 00:45:55,630
fellow by means of a spring in between. So,
a
342
00:45:55,630 --> 00:46:01,470
force could be also externally applied somebody
pushes a body, so externally applied
343
00:46:01,470 --> 00:46:13,270
force
344
00:46:13,270 --> 00:46:29,990
and there could be also the force due to friction.
So, there could be all these types
345
00:46:29,990 --> 00:46:34,510
of forces have we taken it into account everything
is there any other type of force that
346
00:46:34,510 --> 00:46:41,050
you can imagine, that can be applied on a
body, no we have taken everything in.
347
00:46:41,050 --> 00:46:55,060
So, if you drop this the last one, then all
the other types of forces are actually derivable
348
00:46:55,060 --> 00:47:03,550
from a suitable defined potential energy I
will show that. But, . this
349
00:47:03,550 --> 00:47:10,310
.will have to be separately treated, but other
forces gravitational force we know that there
350
00:47:10,310 --> 00:47:14,710
is something called a gravitational potential
and the gravitational force is the gradient
351
00:47:14,710 --> 00:47:19,800
of
that potential you learned in school. But,
352
00:47:19,800 --> 00:47:22,140
yes you may not know how that can be done
for
353
00:47:22,140 --> 00:47:25,900
a force applied through a spring or the externally
applied forces.
354
00:47:25,900 --> 00:47:32,100
But, for now take it for granted I will illustrate
all of these separately, take it for granted
355
00:47:32,100 --> 00:47:38,391
that all these forces can be treated the same
way we treat the gravitational force. That
356
00:47:38,391 --> 00:47:43,530
means, in the gravitational force we said
that there is a gravitational potential and
357
00:47:43,530 --> 00:47:47,190
the
force that is applied is actually the derivative
358
00:47:47,190 --> 00:47:50,330
or the gradient of the potential.
.
359
00:47:50,330 --> 00:48:09,430
So, the potential energy is expressed as v,
so the generalized force can be obtained as
360
00:48:09,430 --> 00:48:19,150
the
gradient or negative gradient clear and that
361
00:48:19,150 --> 00:48:29,560
satisfies your concept of the gravitational
force. But, in general in order to visualize
362
00:48:29,560 --> 00:48:32,960
imagine in this way, the potential is like
a
363
00:48:32,960 --> 00:48:39,270
potential surface, say suppose you have got
a potential surface like this and the body.
364
00:48:39,270 --> 00:48:48,120
So,
this is the q q i coordinate and here is the
365
00:48:48,120 --> 00:48:50,460
body sail, so body’s position in q i is
this much
366
00:48:50,460 --> 00:48:56,470
from here to here, what will be the force
acting on the body the gradient.
367
00:48:56,470 --> 00:49:05,560
And you might imagine a three dimensional
surface and in that if the body position is
368
00:49:05,560 --> 00:49:10,760
such, then the force acting on the body is
nothing but, the gradient of the potential
369
00:49:10,760 --> 00:49:19,460
surface. So, potential is conceptually a surface,
you have got the q 1 coordinate, q 2
370
00:49:19,460 --> 00:49:24,960
coordinate generalized coordinate and in that
it is a surface and the position of the body,
371
00:49:24,960 --> 00:49:30,670
if you know then depending on the position
it applies a force which is a gradient.
372
00:49:30,670 --> 00:49:35,590
.And how it also applies to other kind of
forces I will come to, but that at least satisfies
373
00:49:35,590 --> 00:49:46,670
your concept of the gravitational potential
does it not. Gravitational potential concept
374
00:49:46,670 --> 00:49:51,290
was exactly this, even electro static potential’s
concept is exactly the same, what was the
375
00:49:51,290 --> 00:49:55,610
concept of electrostatic potential, that if
you have a particular body at a particular
376
00:49:55,610 --> 00:49:58,700
place
how much force is applied and the force is
377
00:49:58,700 --> 00:50:01,650
obtained as the gradient of that potential.
So,
378
00:50:01,650 --> 00:50:06,380
always the potential’s concept is the same
that the force is a gradient of the potential
379
00:50:06,380 --> 00:50:14,570
negative of the gradient like here, how it
applies I will to the other kinds of forces
380
00:50:14,570 --> 00:50:17,130
I will
come to that later.
381
00:50:17,130 --> 00:50:18,130
.
382
00:50:18,130 --> 00:50:28,880
But, once you assume this, then this can be
substituted from there we will substitute
383
00:50:28,880 --> 00:50:44,090
and
write d d t i dot minus here we substitute
384
00:50:44,090 --> 00:50:53,190
minus dou v dou Q i, so this will be plus.
Now,
385
00:50:53,190 --> 00:51:05,440
Lagrange defined a new term called, now it
is called Lagrangian we write it as script
386
00:51:05,440 --> 00:51:12,180
L
which is the kinetic energy minus the potential
387
00:51:12,180 --> 00:51:21,250
energy. So, this can easily be seen that
this is a derivative of the Lagrangian with
388
00:51:21,250 --> 00:51:24,370
respect to q i, can you see that this term
the
389
00:51:24,370 --> 00:51:28,820
whole term.
But, this term well here you were you are
390
00:51:28,820 --> 00:51:31,750
taking a derivative of the kinetic energy
with
391
00:51:31,750 --> 00:51:38,350
respect to the velocity and the potential
is normally not depend on the velocity. So,
392
00:51:38,350 --> 00:51:44,180
we
can very simply write it as d d t of derivative
393
00:51:44,180 --> 00:51:53,610
of the Lagrangian with respect to minus
derivative of the Lagrangian with respect
394
00:51:53,610 --> 00:51:58,970
to, so simple after all this day’s labour,
we
395
00:51:58,970 --> 00:52:00,960
have obtained such a simple equation.
396
00:52:00,960 --> 00:52:07,030
.That is the actual form of the Lagrangian
equation for a non dissipative system,
397
00:52:07,030 --> 00:52:12,980
because we have yet not considered this term,
how to consider this term
398
00:52:12,980 --> 00:52:24,730
we will come to that later. But, this is the
Lagrangian equation
399
00:52:24,730 --> 00:52:39,150
for non dissipative system
for conservative system. Let us stop here
400
00:52:39,150 --> 00:52:44,260
and we will illustrate how to apply this to
physical system in the next class.
401
00:52:44,260 --> 00:52:44,260
.