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In the last class, we are talking about the
Mutual Inductances and we have already
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learned how to model general electrical networks.
Modeling the elements of the network
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and modeling the situations like the mutual
inductance are now understood. In that
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situation, we can develop sort of a general
technique for handling any electrical network.
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So, if there is an electrical network consisting
of many branches loops and no suppose it
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is a very complicated network. Situation like
that often happen in say power systems.
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Then the general technique would be first
to identify the meshes, identifying the open
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windows. And the charges flowing in those
open windows will designate as the
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independent variables.
.
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So, if there is a circuit something like this
or whatever. That all will do is to say that,
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here is a window, here is a window, here is
and these are the independent meshes and we
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say q 1, q 2, q 3 and q 4, thus we will identify.
So, the identification of the independent
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.variables becomes rather trivial. Now, once
we have done. Then, we have to write down
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the kinetic energy term, the potential energy
term and the Ray Leigh function term.
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.
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So, these three will be taking the general
forms like let us first write the potential
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energy
field will be first the energy stored in the
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capacitors. So, it will be half C i q i square
and
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that will have to be integrated over all i’s.
So, i’s are the independent variables. So,
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these
are the very i‘s. So, it will have to be
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integrated over i is equal to 1 to n. Then,
notice that
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these are situations like this, where a capacitor
is placed in basically the charge flowing
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through the capacitor is this invariable.
But, there can also be situations like this,
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where the capacitor shared between two
branches. So, that will have to be written
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as again half, we have to sum up over
something, I will write that later and it
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will be sorry, this would be one by know,
1 by c i
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j. So, this is mutual branch this will be
q i minus q j square and this will be integrate
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from j is equal to 1 to how many it will one
less, because I am not considering this self
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1,
this it is i minus 1.
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So, that exhausts the energy stored in the
capacitors, plus in the potential term, we
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also
have to include the externally applied forces,
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which are the voltage sources, so that will
be included. So, the voltage sources could
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be in such a branch also. So, again we will
have to individually consider them, it will
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be minus, again sum over this is E i q i,
this is
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.i is equal to 1 to n. There will be one more
term, which will where the voltage source
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is
placed in between two branches.
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So, that will have to be written as minus,
again sum over something, I will write that
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later. It will be E i j q i minus q j and
it will have to be wait, wait did I write
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it correctly,
it has also to be summed over i. So, I will
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just insert it has to have a summation over
what i; that means, it will be i is equal
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to 1 to n. Similar thing will have to be done
here.
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So, I will just write it once again, the first
one would be i is equal to 1 to n, j is equal
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to 1
n minus 1 or i minus 1, sorry that completes
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the story about the potential.
Now, comes the kinetic energy T, again has
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to be written in the same fashion. So, you
will have half it will be summed over L i
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q i dot square, so this is the self branches.
So, it
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will be i is equal to 1 to n plus now the
mutual branches, mutual branches means not
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the
mutual inductors, I am saying if a inductor
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is placed somewhere say here. Then, it will
be
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half, again it will need two summations, over
what I will write later L i j q i dot minus
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q j
dot whole square. And this will have to be
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again summed over in the similar way, i is
equal to 1 to n and j is equal to 1 to i minus
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1.
So, that exhausts the self inductances, there
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also be mutual inductances and mutually
inductances will have to written as M q i
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q j, so that will have to be summed over.
So, it
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will be plus half M, M is always i j between
which branch and which branch, q i dot q j
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dot. This will have to be summed over sorry,
again in the same way, not in the same
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way, this will have to be summed over all
i’s and j’s, where i is equal to not equal
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to j.
So, that exhausts all the possibilities of
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the potential energy and the kinetic energy.
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..
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Now, we will have to write the Ray Leigh dissipation
function, R that will have the
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resistances in the self branches and the resistances
in the mutual branches. So, we will
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have to write it as half it will be summed
once i is equal to 1 to n, R i q i dot square.
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These are the self branch resistances, plus
half again it will have to be summed over
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twice, i is equal to 1 to n and j is equal
to 1 to i minus 1, then it will be R i j.
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The current flowing through this branch would
be q i dot minus q j dot square that
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exhaust, the complete thing. So, if you have
a very complicated network, you are will
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still able to write the Lagrangian equation,
exactly in the same way, clear. Now, it is
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true
that often, we do not in our mind we go this
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way. But, actually when we see the circuit
something like this, we identify and then
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and we write down.
For example, if I have write down the kinetic
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energy of this one, we will say this L 1,
half L 1 q 1 dot square plus half a q L 2
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q 4 dot square and so and so forth. Just by
an
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inspection, we can write, you might have notice
that in this formulation I left out the
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current sources that calls for a base based
a question, why? The reason is that the current
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sources need to be treated a little differently.
If we are defining the sources, defining the
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independent variables as the currents flowing
through the meshes, and suppose there is a
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current source appearing somewhere in one
of
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the mesh, what happens? That particular q
i dot immediately becomes equal to that
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current source, it no long as remain as independent
variable.
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.So, the current sources actually offer advantage
and when that happens, when you
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encounter a current source, you would know
that the number of equation that will really
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need would be one less. That is why, we cannot
write down the prescription in the same
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way, but nevertheless let us look at it how
it will happens.
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.
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Suppose, there is a current source
like this, suppose there is a capacitor and
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suppose there
is an inductor and say only this much, then
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according to our prescription, what we will
say, we will say that this will be our q 1
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and this will be our q 2. Now, notice that
q 1 dot
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is nothing but this i and therefore q 1 is
integral of the independent current source.
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So,
this is say I over t, it is the integral of
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the externally imposed current starting from
0 to
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now.
So, it will be integral of say 0 to t, you
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have we will have to define some other variable,
because here I am writing t, it will be I
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of say Epsilon d E. Starting from 0 to now,
whatever current has flown, that is the total
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q 1. Now, once we have defined this, the rest
becomes exactly the same, why. But, because
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in that case we will write the kinetic
energy here as half say L C half L q 2 dot
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square, no problem about it.
We will write the potential energy as stored
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here, it is, why, wait a minute. So, it will
be
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1 by twice c, but now the charge here is q
1 minus q 2 and q 1 is this fellow. So, we
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will
write this whole thing 0 to t I Epsilon d
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Epsilon minus q 2, this whole thing square.
And
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your; obviously, the Ray Leigh term is absent
here, because you are dealing with the
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.conservative system. And then the Lagrangian
function will be half L q 2 dot square
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minus 1 by twice c, this fellow integral,
sorry within bracket 0 to t I Epsilon d Epsilon
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minus q 2 square.
Now, if you differentiate notice that here,
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q 1 does not appear at all and therefore,
there
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will be no equation in terms of q 1 that is
natural, because q 1 has become dependent
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on
an external variable; it is no longer a state
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variable then... So, can you write down the
equation in terms of q 2, then you will have
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to differentiate this, this is a independent
variable, independent of q 2. So, when you
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differentiate this fellow, if it appears alone,
it
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goes off.
So, how will you say, the equation will be
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first d d t of the derivative of Lagrangian
with
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respect to q 2 dot, it will be L. Can you
see, yes L q 2 double dot
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minus now this fellow
just expand it, first it will be this square
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plus this square minus this twice this. So,
it will
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have, whenever it comes to be just this square,
it will cancel off, when you differentiate
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and ultimately what you get, you get one term
with this thing. So, it will be twice q 2
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and
this differentiates with respect to q 2, only
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this remains.
So, it will be 1 by c integral 0 to t I Epsilon
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d Epsilon, there will be another term, which
will be on these only. It will be plus, because
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this was plus, it will be q 2 by c, that is
how it is. Now, notice that the integral appears
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in the differential equation that is natural,
because you have externally imposed the derivative
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of q 1. Therefore, if you want to
obtain q 1 as a state variable, you have to
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have the integral.
So, this appearance a bit uncommon, that is
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why I did not include in the general
formalism, but when you have, then you can
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always do it like this. Now, when we were
talking about the general formalism of electrical
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networks, we sort of took a rather short
cut path. You will later realize, why I say
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it is a short cut path, we said that just
identify
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the meshes and assign state variables to them.
But, as here, I have not bothered to prove
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that, that will be the minimum set of state
variables and in fact in some situations,
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they will not be where they will not be another
thing, we will deal with later. And that is
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exactly, why later, I will also deal with
another
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different method of obtaining differential
equations for specifically for electrical
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networks. Because, there some additional caution
is necessary, but for main ran of
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circuits, it will be done this way.
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.Now, let us come to situations, where you
need to model some kind of a non-linear
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element in a system. For example, let us not
deal with the electrical circuits, now because
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one non-linear element, we have already come
across that is dry fiction.
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.
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So, where you have got a body resting on a
dry surface and if somebody pushes it and
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then it will initially, it will not move,
and then when that threshold force is exceeded,
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it
starts to move and then after that the force
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due to friction remains more or less constant.
So, that is what the character is and we have
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already seen that, it can be represented by
means of a graph something like this, q dot
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here and a force here.
Force due to the friction, I am talking about,
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so that is the character of the, this is
obviously, it is a non-linear function. A
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linear function is where it will be form straight
line dependence between the force and q dot,
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here it is not and therefore, it is a nonlinear
function. How would we represent that, now
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if a system has this kind of elements
or say suppose you have got a circuit, where
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one of the elements is say a saturable
inductor or the capacitance is not really
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a linear capacitance.
A system where there is a spring, but the
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spring is not really a linear spring; it is
a hard
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spring or soft spring. In any case, it is
slightly bend, the characteristic slightly
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bend not
really straight line. What you will do, then
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the general technique is first assumed that
it is
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a linear thing. There by what will happen,
we will write some kind of for example this
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one, we will write some kind of a R.
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.If it had been a viscous friction, we will
write R, we will write the equation that way
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and
then say, no, no my R is not then a fixed
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quantity, it is a variable quantity, we just
dependence on these, these, theses, these.
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Then, we can write it, we can substitute that
dependence into the equation, the moment you
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do that the non-linearity is represented,
clear.
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.
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So, let us do a problem with a dry friction
element. Suppose, you have surface, there
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is a
spring connected to a mass and where it is
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pulled with a force F and this fellow is a
dry
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friction. This surface is a dry friction surface.
Again, in this case the convenient
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definition of the state variable would be
the q, where q is measured from the unstretched
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position of the spring that will be convenient.
So, once we do that, we will right the equation
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as the Lagrangian, we can directly right,
because we have already written that once.
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It will be half m q dot square, for this 1
minus
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half k q square, for this 1, plus F q. And
the Ray Leigh function will be half R q dot
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square and notice that, here we are assuming
initially that this fellow is some number
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and
accordingly we are proceeding to derive the
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equations.
There is only one state variable q, therefore
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we can write just one equation, the first
one
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would be in terms of this and it will be,
so I am writing the equation directly. It
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will be m
q double dot, due to this minus derivative
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of the Lagrangian with respect to Q. So, this
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.will be plus, it will be plus k q, for this,
it will be minus F plus R q dot equal to 0.
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So far,
so good, it is a simple stuff.
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Now, assume that the characteristic of this
thing is represented as something like this.
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So,
here is q dot and here is what, suppose you
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are moving it, it will feel a reaction of
m g
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and there will be a coefficient say mu, so
mu m g will be the force created. So, this
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fellow will be mu m g and this is the minus
mu m g convinced. So, in reality it will be
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slightly up here and then it drops. But, let
us assume that it more less that this is not
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bad
assumption, so mu m g and minus mu m g here.
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Now, this needs to be somehow represented
here, somehow this fellow has to represent
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that. So, how to do that, tell me, how would
you do that, R has to represent instead of,
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when I write R, I am assuming a function like
this, linear function. Instead of that, I
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have
to somehow tell this equation, that my dependence
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is like this. So, how will you do that,
this can be done simply by, because what is
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happening, mu m g is the magnitude and it
is
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plus or minus. So, under certain condition,
it will have to take a plus value, some under
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certain condition, it will take a take a minus
value. That can easily be represented by
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means of a signum function.
.
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So, we can write s g n signum function of
x dot, I have to write in terms of x dot is
202
00:25:21,929 --> 00:25:30,070
x dot
divided by the magnitude of x dot. So, the
203
00:25:30,070 --> 00:25:37,590
magnitude cancels of it is only the sign that
remains. So, that will be obviously, plus
204
00:25:37,590 --> 00:25:44,509
1 for x dot greater than 0 and minus 1, for
x dot
205
00:25:44,509 --> 00:25:57,009
.less than 0. So, in terms of that, then we
can write the equation that we have derived
206
00:25:57,009 --> 00:26:08,820
as it
was m q double dot plus k q minus F, all that
207
00:26:08,820 --> 00:26:14,370
remains the same.
But, now here R times q dot that has to be
208
00:26:14,370 --> 00:26:34,390
substituted to say plus mu m g, the magnitude
times x dot by x dot magnitude that is it,
209
00:26:34,390 --> 00:26:46,690
q dot and q dot yes, so this would be
that gives
210
00:26:46,690 --> 00:26:53,990
the equation. So, in general that tells us,
that in general if you encounter a non-linear
211
00:26:53,990 --> 00:26:58,799
element, whatever method of obtaining the
differential equation, will still retain exactly
212
00:26:58,799 --> 00:27:03,679
the same method.
We will initially assume that this term is
213
00:27:03,679 --> 00:27:08,340
R L C, whatever it is and then substitute
the
214
00:27:08,340 --> 00:27:14,779
dependence of the R L C whatever it is on
the corresponding state variable that is how
215
00:27:14,779 --> 00:27:24,980
we can obtain the equations for the non-linear
system also fine. Now, the last day I was
216
00:27:24,980 --> 00:27:32,399
actually planning to do it as a tutorial,
but since we missed it. Let me give you some
217
00:27:32,399 --> 00:27:41,210
problem, which you do at home, but they it
is to be submitted as an assignment, tutorial
218
00:27:41,210 --> 00:27:46,490
normally I prefer to do here, so that I can
supervise. But, since we missed one tutorial
219
00:27:46,490 --> 00:27:52,559
day, we will I will give you the problem do
it and submit, in a separate copy, remember
220
00:27:52,559 --> 00:27:54,170
that, fine.
.
221
00:27:54,170 --> 00:28:18,020
So, let us take this problem, how will you
do this, it is a double pendulum, assume for
222
00:28:18,020 --> 00:28:24,600
the sake of simplicity that these two l’s
are the same. Because, I am more interested
223
00:28:24,600 --> 00:28:28,919
in
you are being able to derive it, than the
224
00:28:28,919 --> 00:28:35,250
unnecessary complications. So, let these two
masses to be the same that will make it simpler.
225
00:28:35,250 --> 00:28:37,850
So, how will you define the state
226
00:28:37,850 --> 00:28:46,630
.variables, they coordinates the two thetas,
two angles, that uniquely define the positional
227
00:28:46,630 --> 00:28:56,210
status of the system.
So, this will be theta 1 and then you will
228
00:28:56,210 --> 00:29:01,142
have to write the potential energy and the
kinetic energy, it is a non dissipative system.
229
00:29:01,142 --> 00:29:07,811
So, you do not have to write the Ray Leigh
term. Suppose there is system like this, so
230
00:29:07,811 --> 00:29:25,330
this problem number 1, problem number 2. A
mass is hanging from the roof by means of
231
00:29:25,330 --> 00:29:38,840
a spring and then here you have a spring and
a damper connected to another mass and this
232
00:29:38,840 --> 00:30:01,620
mass is been pulled downwards with a
force F of t.
233
00:30:01,620 --> 00:30:07,590
So, how will you adapt this problem, how will
you conveniently define the variables q 1,
234
00:30:07,590 --> 00:30:15,120
q 2, but that q 1, q 2 how do you define?
There are two ways; obviously, you allow it
235
00:30:15,120 --> 00:30:17,740
to
hang vertically. And then wherever these two
236
00:30:17,740 --> 00:30:21,799
masses are call it q 1, q 2 that is one
possibility. The other possibility is no,
237
00:30:21,799 --> 00:30:25,080
no do not do that, you raise this fellow with
your
238
00:30:25,080 --> 00:30:30,270
hand, so that ultimately, it goes to a position,
where this unstretched.
239
00:30:30,270 --> 00:30:34,590
Hold it there and now raise this one, so that
this one also goes to unstretched position
240
00:30:34,590 --> 00:30:37,320
and
call that, these two positions of the masses
241
00:30:37,320 --> 00:30:41,919
as q 1 and q 2 or as the 0 positions of q
1 and
242
00:30:41,919 --> 00:30:47,350
q 2 and measure from there. Both these possibilities
are there and you define according
243
00:30:47,350 --> 00:30:51,860
to what you find convenient, both are the
same actually, both are equivalent actually.
244
00:30:51,860 --> 00:30:53,870
In
one, you will have one additional term, which
245
00:30:53,870 --> 00:30:59,030
will cancel off, when you differentiate.
.
246
00:30:59,030 --> 00:31:20,080
.How about this problem 3, there are two masses,
this one is connected by a spring
247
00:31:20,080 --> 00:31:28,740
damper arrangement. This one is also connected
by a spring damper arrangement and
248
00:31:28,740 --> 00:31:41,691
these two are connected by means of a spring
and say there is a force F, pulling this
249
00:31:41,691 --> 00:31:56,570
mass. In that case, how will you define the
state variables, again there are two masses;
250
00:31:56,570 --> 00:32:01,559
their positions uniquely define the systems,
positional status.
251
00:32:01,559 --> 00:32:12,009
Then, how will you conveniently define that,
may not be possible, in this case that may
252
00:32:12,009 --> 00:32:15,870
not be possible, in the earlier case it was
possible, but here since these are two rigid
253
00:32:15,870 --> 00:32:23,820
walls, this may not be possible to bring this
springs to unstretched positions. So, in that
254
00:32:23,820 --> 00:32:30,659
case what, you will bring two of them to unstretched
position and allow one to stretch.
255
00:32:30,659 --> 00:32:37,710
When you start it, just define a q; that means,
is already stretched to this extent and then
256
00:32:37,710 --> 00:32:46,080
start from there.
And once you define the states variables,
257
00:32:46,080 --> 00:32:48,289
it will not be very difficult thing to write
down
258
00:32:48,289 --> 00:32:56,769
the kinetic energy and potential energy, this
step itself is not very complicated. Let us
259
00:32:56,769 --> 00:34:15,730
gives some circuits, call them k 1. Now, take
a circuit, so here you have L 1, L 2, C 1,
260
00:34:15,730 --> 00:34:24,950
C
2. Now, you can say E 1 and E 2, what will
261
00:34:24,950 --> 00:34:36,389
you define as state variables. Let us add
some more complication; there is mutual inductance
262
00:34:36,389 --> 00:34:39,879
between these two, oriented like
that with the dots.
263
00:34:39,879 --> 00:34:51,690
So, how will you define, simple q 1, q 2,
q 3 and then follow the same logic that I
264
00:34:51,690 --> 00:35:04,630
showed, this exactly can be done by that,
fine. Do you any other circuit problem or
265
00:35:04,630 --> 00:35:08,140
this
more or less gives you practices. You asked
266
00:35:08,140 --> 00:35:16,970
me a question, that day, that in case of the
pendulums tip is being or the suspension is
267
00:35:16,970 --> 00:35:23,750
being moved, then does not they define the
another state variable. For that system, where
268
00:35:23,750 --> 00:35:28,780
they want that was considering at that time,
it was independently been moved. And therefore,
269
00:35:28,780 --> 00:35:33,410
that does not become a state variable,
but suppose that is not a dependent one and
270
00:35:33,410 --> 00:35:39,120
take a situation like this.
271
00:35:39,120 --> 00:35:40,120
..
272
00:35:40,120 --> 00:36:04,710
So, here you have a F, and then from here
then what, so you have got a mass in a simple
273
00:36:04,710 --> 00:36:08,890
harmonic motion. But, it will not remain a
simple harmonic motion, because there will
274
00:36:08,890 --> 00:36:12,400
be another thing, pulling it or pushing it
or whatever it is. So, it will be another
275
00:36:12,400 --> 00:36:17,200
application of force, but can you see that,
it becomes then quite complicated stuff. But,
276
00:36:17,200 --> 00:36:20,400
if
you try to write down the differential equations,
277
00:36:20,400 --> 00:36:25,240
it will not remain quite complicated, it
will be simple, but how do you do that; let
278
00:36:25,240 --> 00:36:32,950
me see.
In this case, what will be the state variables;
279
00:36:32,950 --> 00:36:41,140
yes q may be position of this one. Now, you
can define it as the deviation from the unstretched
280
00:36:41,140 --> 00:36:50,530
position, so this becomes a state
variables say q 1. And or maybe you can call
281
00:36:50,530 --> 00:36:56,460
it q 2, which is theta, does it uniquely
define the positional status of the system,
282
00:36:56,460 --> 00:37:04,830
is it does. Notice that, one is a translational
variable; another is a rotational variable,
283
00:37:04,830 --> 00:37:07,420
generalized coordinate that is the concept
of
284
00:37:07,420 --> 00:37:11,210
generalized coordinate. Whatever, you want
to define as the minimum number of
285
00:37:11,210 --> 00:37:15,910
coordinates that defines the positional status
of the system, fine.
286
00:37:15,910 --> 00:37:20,920
Now, can you write down now the kinetic energy
and the potential energy of this fellow
287
00:37:20,920 --> 00:37:34,430
with respect to this q 1 and q 2 try this,
it will be good? So this was problem 5. Let
288
00:37:34,430 --> 00:37:43,099
us
think of a mechanism, where you have a rigid
289
00:37:43,099 --> 00:37:53,510
wall to a spring, but at this point, there
is a
290
00:37:53,510 --> 00:38:09,451
wheel, which is able to move in between this
and there is another guide. So, there is a
291
00:38:09,451 --> 00:38:15,869
guide here, which forces this thing to move
this way only and there is another guide
292
00:38:15,869 --> 00:38:18,640
which forces this thing to move this way only.
293
00:38:18,640 --> 00:38:32,750
.And suppose, there is another wheel and this
two fellows are connected by
294
00:38:32,750 --> 00:38:41,819
and then you
have got, say a mass hanging here. You see,
295
00:38:41,819 --> 00:38:50,579
the mass is hanging here and you can apply
a some kind of an initial force to this one
296
00:38:50,579 --> 00:38:53,790
to see what will happen to it. As you pull
it,
297
00:38:53,790 --> 00:38:59,740
this will come down, this will come this way,
the spring will stretch and then it will go
298
00:38:59,740 --> 00:39:04,000
through a some kind of a motion.
How do you write the differential equations
299
00:39:04,000 --> 00:39:10,809
for that, what are the state variables, q
1 is
300
00:39:10,809 --> 00:39:27,000
what, assume that these two are mass less
and then what
301
00:39:27,000 --> 00:39:32,740
displacement of the mass. Say
and then
302
00:39:32,740 --> 00:39:41,450
.
Do, you really need that…
303
00:39:41,450 --> 00:39:48,599
.
Yes, only q 1 will satisfy the requirement,
304
00:39:48,599 --> 00:39:51,630
because this position is uniquely defined
by q
305
00:39:51,630 --> 00:39:56,800
1. So, the task is to define the minimum number
of variables necessary in order to define
306
00:39:56,800 --> 00:40:03,140
the positional status and that can be done
only with one variable in this kind of system.
307
00:40:03,140 --> 00:40:08,260
So, once you have written down that, then
you have to q 1 may be the displacement
308
00:40:08,260 --> 00:40:14,579
something like going like this q. And the
on the basis of that, you have to specify
309
00:40:14,579 --> 00:40:19,880
this as
it.
310
00:40:19,880 --> 00:40:24,930
Try in a Newtonian way, it will become enormously
complicated, try in a Lagrangian
311
00:40:24,930 --> 00:40:36,960
way, it will become enormously simple. Similarly,
you had considered a pendulum,
312
00:40:36,960 --> 00:40:42,109
where this fellow is moving, this way. Can
you imagine a system, where this fellow is
313
00:40:42,109 --> 00:40:47,440
moving that way; yes you can always do that
say like this.
314
00:40:47,440 --> 00:40:48,440
..
315
00:40:48,440 --> 00:40:59,790
You have got a spring and there is a say guide
and this well as coming, so that this is not
316
00:40:59,790 --> 00:41:06,160
allowed to move this way, this only allowed
moving that way and here you have a point
317
00:41:06,160 --> 00:41:20,559
of suspension from which
the pendulum is hung. So, this is a guide
318
00:41:20,559 --> 00:41:31,260
that allows it to move
only up and down. In that case, how many variables
319
00:41:31,260 --> 00:41:47,650
would you need, think 2, which 2?
.
320
00:41:47,650 --> 00:41:56,480
So, the vertical displacement of not the mass
of this point, because mass is here. This
321
00:41:56,480 --> 00:42:00,970
is
just a point, where is being held. So, point
322
00:42:00,970 --> 00:42:03,420
of suspension and this point of suspension
are
323
00:42:03,420 --> 00:42:09,460
going up and down. So, you can say this position
is one state variable; the other is
324
00:42:09,460 --> 00:42:16,770
obviously, theta. So, you can say, if it is
a x coordinate, it is a y coordinate, then
325
00:42:16,770 --> 00:42:23,550
this can
be simply y from here and it is possible,
326
00:42:23,550 --> 00:42:32,160
then to define y 0 as the unstretched position
of
327
00:42:32,160 --> 00:42:46,609
the spring, this would be problem number 7.
You have done a problem with the spring pendulum;
328
00:42:46,609 --> 00:43:01,329
let us complicate your life a little
bit. Here, you have the spring pendulum, but
329
00:43:01,329 --> 00:43:20,770
at this point, what is hanging is a bar, so
yes
330
00:43:20,770 --> 00:43:27,650
you will have to consider, it is centre of
mass and then you need only two variables.
331
00:43:27,650 --> 00:43:40,420
Though, it is a far complicated system, two
variables, no, because if this were not there,
332
00:43:40,420 --> 00:43:45,520
then because it is a spring, you would have
to consider the radial motion, as well as
333
00:43:45,520 --> 00:43:46,540
the
angular motion.
334
00:43:46,540 --> 00:43:52,170
.So, for this point you need two and again
for this one, there is another additional
335
00:43:52,170 --> 00:43:57,799
angle,
three variables, you need three variables.
336
00:43:57,799 --> 00:44:02,650
So, you can say theta 1, theta 2 and again
the
337
00:44:02,650 --> 00:44:08,680
same way you have to define. As I told you,
that there are many possible ways of
338
00:44:08,680 --> 00:44:16,510
defining this, how it define is your business
in the problem, that I did probably I had
339
00:44:16,510 --> 00:44:20,650
taken the vertically hanging downward position
as the data. So, that is the one possible
340
00:44:20,650 --> 00:44:28,540
way. So, there has to be a say r direction
in this side, but how you define, it is your
341
00:44:28,540 --> 00:44:30,000
business.
.
342
00:44:30,000 --> 00:44:56,240
How, about this sum of the clocks, have it
like this a spring, another spring, so there
343
00:44:56,240 --> 00:45:00,130
are
two springs and it is able to move like this
344
00:45:00,130 --> 00:45:12,589
and there is a tortional spring that connects
a
345
00:45:12,589 --> 00:45:32,319
rod. So, this is a tortional spring that also
gives a springy effect. So, how will you define
346
00:45:32,319 --> 00:45:37,400
in this case, you can say this is k 1, this
is this one is k 2 and this is a tortional
347
00:45:37,400 --> 00:45:44,270
spring say
k r rotational spring and then it has a mass
348
00:45:44,270 --> 00:45:51,790
m length l.
Obviously, you can define 1 coordinate theta,
349
00:45:51,790 --> 00:46:12,980
what other you need, so the position of
this. So, you can say, can you write down
350
00:46:12,980 --> 00:46:19,280
the kinetic energy and potential energy in
terms of this would be a little tough, but
351
00:46:19,280 --> 00:46:38,760
nevertheless, not it not a very formidable
challenge. What problem 9? I will give you
352
00:46:38,760 --> 00:46:54,309
10 problems; the tenth problem is where you
have, say do you know, how cars are balanced
353
00:46:54,309 --> 00:47:00,079
on the suspension.
Suppose, you have got a car, which I am representing
354
00:47:00,079 --> 00:47:07,540
by means of a very badly drawn
rectangle, no one would like to get into such
355
00:47:07,540 --> 00:47:10,010
a car, but nevertheless imagine that is a
car.
356
00:47:10,010 --> 00:47:15,920
.And I will not draw wheels, because all I
am tried to depict is that, while the wheels
357
00:47:15,920 --> 00:47:22,770
move, they are on a bumpy ground. So, the
bumpy ground is giving some kind of a force
358
00:47:22,770 --> 00:47:27,530
this way and this fellow is moving, so that
is what you feel all the time, when we move
359
00:47:27,530 --> 00:47:32,970
in a car.
And these act at specific positions, where
360
00:47:32,970 --> 00:47:35,700
this can be represented this suspension can
be
361
00:47:35,700 --> 00:47:52,150
represented as springs. At this stage, I will
not make it complicated by saying that this
362
00:47:52,150 --> 00:47:56,790
two position are being pushed up and down.
What, I say is, just let it rest and then
363
00:47:56,790 --> 00:48:00,380
give a
push somewhere here, it will oscillate and
364
00:48:00,380 --> 00:48:04,020
move. Now, it has the point is that you have
to
365
00:48:04,020 --> 00:48:13,020
find that how many degrees of freedom it has.
Obviously, I will need to mention, where is
366
00:48:13,020 --> 00:48:22,200
the center of gravity, I will need to mention
where these two suspensions are, fine, let
367
00:48:22,200 --> 00:48:33,640
that be that be given. So, the total length
is l,
368
00:48:33,640 --> 00:48:47,190
say out of that these two are symmetrically
placed, this is l by 5 and this is also l
369
00:48:47,190 --> 00:48:53,230
by 5
and the whole thing has a mass m. You will
370
00:48:53,230 --> 00:48:57,839
also need the rotational motion; there will
be
371
00:48:57,839 --> 00:49:05,810
a rotational motion another degree of freedom.
So, you will also need the moment of
372
00:49:05,810 --> 00:49:10,260
inertia there.
So, what are the degrees of freedom and how
373
00:49:10,260 --> 00:49:16,569
is it allow to move, just one mass
remember, but the same mass will now have
374
00:49:16,569 --> 00:49:24,460
two degrees of freedom, it can go up and
down, it can also rock like this. So, in order
375
00:49:24,460 --> 00:49:30,349
to define the positional status of the system
uniquely, how many variables do you need,
376
00:49:30,349 --> 00:49:36,770
2. One represents this position and
represents that and the elongation of the
377
00:49:36,770 --> 00:49:41,799
springs will have to be measured from both,
it is
378
00:49:41,799 --> 00:49:47,559
not just this.
So, if there is a tilt here, so this fellow
379
00:49:47,559 --> 00:49:53,260
this particular spring is compressed and this
particular spring is elongated, so that will
380
00:49:53,260 --> 00:49:59,069
have to be taken into account in order to
measure the potential energy in the springs.
381
00:49:59,069 --> 00:50:03,070
So, the problem is that just a bar resting
on
382
00:50:03,070 --> 00:50:09,950
springs and if you press one, it will go on
doing like this and that is what we are trying
383
00:50:09,950 --> 00:50:15,000
to
find out later. But, first job is to obtain
384
00:50:15,000 --> 00:50:17,730
the differential equations, clear, do you
able, fine
385
00:50:17,730 --> 00:50:20,470
wonderful.
So, I have given you 10 problems and that
386
00:50:20,470 --> 00:50:23,190
10 problems solution have to be submitted
in
387
00:50:23,190 --> 00:50:35,130
the next day, no not tomorrow, next week,
I will give you the week and in between.
388
00:50:35,130 --> 00:50:41,690
Now, you have noticed, that whenever we were
trying to use the Lagrangian method to
389
00:50:41,690 --> 00:50:47,070
.obtain the differential equations, it always
yielded second order differential equations.
390
00:50:47,070 --> 00:50:54,440
Obviously, m q double dot that was one term.
So, whenever you differentiate with respect
391
00:50:54,440 --> 00:51:00,990
to time the derivative of Lagrangian with
respect to q i dot, it obviously, yields a
392
00:51:00,990 --> 00:51:06,140
double derivative. But, then probably you
have
393
00:51:06,140 --> 00:51:10,200
also done some courses on numerical methods,
where you have solved the differential
394
00:51:10,200 --> 00:51:16,099
equations. Have you solved second ordered
differential equations, how did you solve;
395
00:51:16,099 --> 00:51:23,319
you first brought it down to first order.
So, obviously, then after you have obtain
396
00:51:23,319 --> 00:51:28,549
the differential equations this way, you need
first to bring it down to the first order,
397
00:51:28,549 --> 00:51:32,829
if it is a one second order differential equations,
you define two first order differential equation
398
00:51:32,829 --> 00:51:39,549
and so on and so forth. And, then only
you apply those techniques. So, even if you
399
00:51:39,549 --> 00:51:43,099
are trying to solve it analytically, then
also it
400
00:51:43,099 --> 00:51:48,350
is always convenient to have it in the first
order form.
401
00:51:48,350 --> 00:51:52,380
Though, I will not say that techniques do
not exists to directly solve the second order
402
00:51:52,380 --> 00:51:58,250
equations, there are, but nevertheless. Since,
we are very comfortable, we know the
403
00:51:58,250 --> 00:52:02,089
techniques in the first order; the next step
will be to bring the equations down to the
404
00:52:02,089 --> 00:52:07,490
first
order. Normally, how do you do that, you define
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00:52:07,490 --> 00:52:17,050
an additional variable and that
additional variable will can be anything,
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00:52:17,050 --> 00:52:21,990
can be what, what will you normally define
as
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00:52:21,990 --> 00:52:23,079
the additional variable
.
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00:52:23,079 --> 00:52:35,110
Q dot may be, so you have m q double dot,
you say q dot something, you do not do that.
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00:52:35,110 --> 00:52:42,690
But, then that will not prove to be very convenient
algebraically, why because it is
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00:52:42,690 --> 00:52:50,550
always m q double dot. So, if you say m q
dot is a variable, then it will be algebraically
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00:52:50,550 --> 00:52:58,450
far simpler to write it down. So, in general
that is the technique of deriving the first
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00:52:58,450 --> 00:53:03,799
order
equations, we define what are known as momentum.
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00:53:03,799 --> 00:53:10,230
Momentum as the other variable not
the q dot; that is the standard procedure,
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00:53:10,230 --> 00:53:13,080
which will make things very simple, I will
show
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00:53:13,080 --> 00:53:14,080
you.
416
00:53:14,080 --> 00:53:15,080
..
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00:53:15,080 --> 00:53:25,141
So, first we define the p I, notice we had
q i and now we are trying to define p I, what
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00:53:25,141 --> 00:53:31,619
is
p I, notice what is it, the momentum in the
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00:53:31,619 --> 00:53:39,850
generalized i direction, that is nothing but
that
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00:53:39,850 --> 00:54:00,480
is it. Imagine any of the situations that
we have come across, say there is a mass,
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00:54:00,480 --> 00:54:07,390
what
was your Lagrangian, Lagrangian had this term
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00:54:07,390 --> 00:54:17,990
m. You differentiate with respect to q i
dot, it get that is, what p i is.
423
00:54:17,990 --> 00:54:29,820
So, we in general define, the p i as this,
for the trivial cases, where this are you
424
00:54:29,820 --> 00:54:32,770
can easily
identify this no problem. But, you will see
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00:54:32,770 --> 00:54:41,670
that the advantage, when there are coordinates
coupled, I will show you. For example, the
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00:54:41,670 --> 00:54:48,410
mass connected with a spring as a pendulum,
we have seen that there the kinetic energy
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00:54:48,410 --> 00:54:52,630
content, both these things radial as well
as the,
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00:54:52,630 --> 00:54:57,320
in that case, what is the momentum simply.
We already had the Lagrangian differentiate
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00:54:57,320 --> 00:55:04,920
with respect to r, differentiate with respect
to theta, differentiate with respect to r
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00:55:04,920 --> 00:55:10,990
dot and theta dot, so that gives. So, once
you have
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00:55:10,990 --> 00:55:27,760
this, you can write then d d t of p i is nothing
but the first term in the equation that we
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00:55:27,760 --> 00:56:02,510
had, so our equation was R not i. Now, notice
this term becomes, so you can directly
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00:56:02,510 --> 00:56:19,309
write the first order differential equations.
So, finally your equation becomes p i dot
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00:56:19,309 --> 00:56:45,700
minus this, so it has this dotted term, so
you directly obtain the first order differential
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00:56:45,700 --> 00:56:52,940
equation. Now, I will illustrate this with
examples in the next class that is all.
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00:56:52,940 --> 00:56:52,940
.