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00:00:54,209 --> 00:01:01,239
Earlier when we were developing the equations
for the Lagrangian' s formalism. We said
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that the generalized forces were expressed
as Q i and that has three components, three
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00:01:08,720 --> 00:01:18,159
possible types of forces. One the forces of
interaction between mass points through
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capacitive elements, so of interaction through
capacitive elements. The second
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possibility is externally impressed forces,
so and third possibility is the forces due
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to
friction. Out of that these two we say that
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these two can be obtained from a suitably
defined potential function, so that is how
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00:02:17,989 --> 00:02:26,689
we proceeded.
And when we deal so we wrote the Q i is minus,
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00:02:26,689 --> 00:02:34,581
so that is how we proceeded. So, this
was just to recapitulate where we were, so
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00:02:34,581 --> 00:02:37,739
far we have only taken into account these
two
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types and we have not yet accounted for the
forces due to friction. So, that is what we
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will try to do today. So, the objective today
will be somehow to include the forces due
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to
friction into the Q i term, but there is a
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problem here.
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.The problem is that while the other forces
could be obtained from a potential function
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by
partially differentiating the potential function
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with respect to a the generalized
coordinates Q i. Obviously that cannot be
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done for the friction why, because the
frictional force say I have something moving
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against say this one, is moving against a
surface like this.
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Then the forces due to friction if it is not
a dry friction, then the frictional force
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is
proportional to what the velocity. Therefore
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there is no way that you can express that
as a
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function of V. Because, after all what you
are what is it is what is dependent on it
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is
dependent on the velocity not the positions,
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so that is the problem. So, in general if
the
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coefficient of friction is say r and the speed
of motion is Q i dot, then I know that the
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force will be R i q i dot.
So, force due to friction where obviously,
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i is the i th generalized coordinate i th
direction in that direction the velocity is
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00:04:38,590 --> 00:04:50,840
q i dot and on that we have to multiply the
frictional coefficient. So, this is the expression,
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obviously that cannot be obtained like
this. But, still we would like to we have
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already seen there are many advantages of
the
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Lagrangian formalism which we do not want
to give up. So, somehow this feat has to be
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obtained that we would not include this into
the Lagrangian formalism.
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.
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Rayleigh proposed that, this can be done by
defining a new type of potential which is
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obviously velocity dependent. So, velocity
dependent potential, potential that would
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be
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.expressed as
half R i q i dot square, along each of these
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00:05:58,560 --> 00:06:09,800
directions. So, total Rayleigh
potential is I will express the R Rayleigh
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00:06:09,800 --> 00:06:14,830
potential, in order to distinguish it from
this
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00:06:14,830 --> 00:06:31,259
force coefficient R by some kind of a script
R. That is equal to summation over a number
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00:06:31,259 --> 00:06:38,099
of generalized coordinates.
That was the proposition of Rayleigh, therefore
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this is called a Rayleigh potential. So,
what are the properties of this potential,
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this potential is firstly not dependent on
the
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coordinate, it is dependent on the velocity.
The second property is that if you
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differentiate this with respect to q i dot
what you get, that is what you wanted. We
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00:07:07,749 --> 00:07:10,289
want
to obtain wanted to obtain the force and that
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is what we have obtained. Which means,
that this Rayleigh potential when partially
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00:07:17,259 --> 00:07:22,699
differentiated with respect to the generalized
velocities give the force.
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00:07:22,699 --> 00:07:23,699
.
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Now, recall how we proceeded to obtain the
Lagrangian function at some stage, if you
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00:07:33,770 --> 00:07:36,930
go
back to the page in which you have written
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00:07:36,930 --> 00:07:48,990
down, where I was deriving the equation,
there was one stage where you have d d t of
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00:07:48,990 --> 00:07:57,830
partial derivative T with respect to q i dot
minus partial derivative of T with respect
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00:07:57,830 --> 00:08:11,819
to q i plus partial derivative of V with respect
to q i. At this stage it was minus Q i equal
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00:08:11,819 --> 00:08:18,080
to 0, and then we had substituted this by
this,
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so this is where the generalized force term
appears. And so long as there is no dissipative
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term, no friction term, no resistive element
in a electrical circuit, this term only contains.
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00:08:31,120 --> 00:08:36,479
.It can be obtained from by differentiating
the potential function V, that is how we
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proceeded. But now, we find that in the new
framework where we need to include the
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resistive things, then you Q i earlier was
expressed as this only. Now, it has to include
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in
addition, so if this is the new term, that
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means this was due to the conservative forces
and
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this is the non conservative forces.
Then these together give you the Q i term,
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00:09:16,610 --> 00:09:30,250
substitute it here you have d d t of, here
we
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have the and here will be ..
And then we had argued that since the v
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does not depend on q i dot, therefore I can
also write this as the Lagrangian term. So,
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00:10:02,560 --> 00:10:05,990
we
will continue with that argument, we will
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say here is the Lagrangian minus here is the
Lagrangian, this is q i plus the Rayleigh
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term with respect to q i dot.
So, this will now be the modified Lagrangian
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00:10:30,130 --> 00:10:38,450
equation that takes into account, the
dissipative forces here this, this is the
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additional thing. By the way before we proceed
I
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would draw your attention to something that
we have said, which sort of one tends to
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take for granted. We said that this to this
is possible only when, I will write it here
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00:11:01,610 --> 00:11:09,920
only
when, because actually it was the partial
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derivative of T with respect to q i dot. But,
we
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had replaced it by in order to simplify the
notations by the Lagrangian with respect to
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q i
dot.
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And that assumes that the other term is 0,
now naturally these will be valid when this
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00:11:26,470 --> 00:11:31,010
is
true. I will site you some examples where
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00:11:31,010 --> 00:11:34,580
this is not true there can be situations like
that.
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Now, when that is observed, then you will
have to take this equation into consideration
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00:11:41,330 --> 00:11:50,930
not this, just this pointer just keep in mind.
Fine let us now see a few examples of how
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this could be done, this can be done in a
practical system.
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..
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00:11:56,560 --> 00:12:17,050
First consider, we had already considered
this system a two dimensional oscillator,
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there
are two position coordinates q 1 and q 2.
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We had said this was K 1 this was K 2 and
this
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was K 3 and this is the valve, this is the
valve and this is the base. Earlier when we
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00:12:41,070 --> 00:12:45,750
are
solving this problem, we said that these two
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surfaces are perfectly frictionless surfaces,
but now let there be friction. Let us see
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how would you do it. Yes…
.(Refer Time: 12:59))
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Here.
.(Refer Time: 13:16)
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So, his question let me repeat for the sake
of the microphone. His question would be was
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that, since if you differentiate it you get
this, therefore should it not be plus, is
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that the
question yes. Now, you notice that ultimately
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we want to get the signs right, signs right
means here is the equation in which the force
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must appear in the proper direction. Now,
in order to have this force appear in the
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proper direction, since this was negative,
therefore I had to have this as positive,
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this as negative.
So, that ultimately we get it positive, this
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will be clear, when we actually solve the
problem, and then you will be able to check
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whether this particular force term has the
right sense or not. If it has then this is
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if it does not have it is wrong, let us check
that. So,
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00:14:27,950 --> 00:14:35,270
.we had come to this problem and since we
had already done this except for these two,
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suppose here the frictional coefficient R
1 and here the frictional coefficient is R
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2.
Then we had solved the problem except these
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two, so just recall that, and then let us
add
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otherwise we will unnecessarily lose time
writing things. So, we had already written
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the
kinetic energy, the potential energy, they
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remain unchanged. Only what is added is the
Rayleigh term, which is half R 1, R 1 is acting
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along the q 1 dot direction, so q 1 dot
square plus half R 2 q 2 dot square. So, that
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the dou R dou q 1 dot is R 1 q 1 dot and dou
R dou q 2 dot is equal to R 2 q 2 dot.
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Then after that we just recall what was our
expression for the T and V you already have
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it, let me write down, so that we can derive
the whole thing. T was half this was M 1,
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this
was M 2, M 1 q one dot square plus half M
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2 q 2 dot square that was the expression for
the kinetic energy. The potential energy expression
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was, potential would be for this
spring, this spring, this spring and no gravitational
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potential. So, it was half K 1 q 1
square plus half K 2 q 1 minus q 2 square
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plus half K 3 q 2 square.
This would be the potentials due to the three
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springs, at that time we had considered
there is a application of force where was
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it on this mass. So, there was the force,
that
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force will be since it is in the same direction
as q 1, therefore I would have to write it
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minus F q 1, that is how it was. And now you
have included this new term and now do
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the proper differentiations and write these
equations, do this. This will have to be written
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in both q 1 direction as well as the q 2 direction.
So, is it visible yes, I can see that in the
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q 1 direction I will write the Lagrangian
differentiated with respect to q 1 dot will
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be only this. So, M 1 q 1 double dot as the
first
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00:18:26,600 --> 00:18:37,370
term, the second term would be this, where
I will be interested in, let me write down
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the
Lagrangian else it becomes problematic. This
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minus this half M 1 I will write it later,
so
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that was the Lagrangian function and Rayleigh
function is there.
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..
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So, if you now differentiate in the first
direction you will get, in the q 1 direction
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you
have this, K 1 q 1 plus K 2 no. And the Rayleigh
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terms derivative was this ((Refer Time:
20:12)). So, my final expressions becomes
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M 1 q 1 double dot plus K 1 q 1 plus K 2 q
1
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minus q 2 minus F, then I have to include
this term plus R 1 q 1 dot equal to 0. Now,
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00:20:43,269 --> 00:20:46,660
let
us see is it right, first let us write down
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00:20:46,660 --> 00:20:56,470
the second equation also.
The second equation would be M 2 q 2 double
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dot minus K 2 q 1 minus q 2 plus K 3 q 2
this will not be there, then plus R 2 q 2
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dot. So, you see these are the expressions
and let
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us get back to the picture can you see. Force,
if you take the other side force is acting
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00:21:48,249 --> 00:21:51,110
in
this direction and this is moving in this
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direction, so force is equal to mass into
acceleration here. Force will have F in this
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00:21:59,399 --> 00:22:11,320
direction minus K 1 q 1 pulling in this
direction K 2 q 1 minus q 2 in that direction.
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And this fellow, so if you bring it to this
side F will be positive and this fellow is
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negative, so this is acting in the direction
opposite to the force. This fellow is moving
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in
this direction, so this is also acting in
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this in this direction, so that has got the
sense. So,
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these are the final two equations for this
coordinates.
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I would later consider how to handle this,
tackle this, how to solve those things we
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00:22:55,170 --> 00:22:58,029
will
handle later. For at this stage we are only
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obtaining the differential equation, so this
is
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one good example. Let us for this system can
you visualize what will be the electrical
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00:23:07,659 --> 00:23:18,830
.equivalent, we have already seen there will
be two resistances. So, we have already seen
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00:23:18,830 --> 00:23:29,889
that the representation of this would be the
two masses will be represented by inductors.
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00:23:29,889 --> 00:23:30,889
.
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The springs will be represented by capacitors
and this force will be represented by
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00:23:41,320 --> 00:23:54,289
battery, and there is a resistance that shares
the same motion with this mass the inductor.
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00:23:54,289 --> 00:24:06,020
So, you have, so this is the equivalent of
this mechanical system. If you obtain the
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00:24:06,020 --> 00:24:11,360
equations for this electrical system you will
got exactly the same equations, only the K
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00:24:11,360 --> 00:24:23,409
and 1 by C will have to be interchange, that
is all. And let us solve one typical electrical
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00:24:23,409 --> 00:24:37,150
circuit problem to illustrate how this may
that helps in obtaining the electrical circuits.
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00:24:37,150 --> 00:24:38,150
.
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00:24:38,150 --> 00:24:47,960
.Let there be a resistance here, let there
be another resistance here, let there be
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00:24:47,960 --> 00:25:05,360
capacitance here and let there be a voltage
source here, the
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00:25:05,360 --> 00:25:11,139
differential equations of this
system that is what we would like to obtain.
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00:25:11,139 --> 00:25:17,419
This is our E, this is L, this is C 1, this
is C
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2, this is R 1, this is R 2, first how would
you decide what are the generalized
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coordinates. The minimum number of potentials,
potential coordinates that we can
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00:25:40,970 --> 00:25:46,970
identify, no not potential charge.
Because the equivalent of the position is
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charge, so I want to identify what are the
independent directions of flow of charge.
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And from the first circuit theory course you
have learnt, that if there are there is a
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circuit like this with you have to first identify
the
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windows or mesh. And the minimum number of
variables necessary would be the
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number of meshes, so it will be very convenient
to identify the three meshes like this.
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Then, we will say this is our q 1 direction
of flow, this is our q 2 and this is our q
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00:26:30,720 --> 00:26:38,100
3. Once
we have done that we can then proceed to write
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00:26:38,100 --> 00:26:43,880
the kinetic energy, and the potential
energy of this circuit. What will they have
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00:26:43,880 --> 00:26:52,650
first they kinetic energy T, kinetic energy
will
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00:26:52,650 --> 00:27:00,370
consist of the energy stored in the inductor
and the energy stored in this inductor. And
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00:27:00,370 --> 00:27:06,450
the charges flowing are q 1 and q 2, therefore
the current flowing through this is q 1 dot
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00:27:06,450 --> 00:27:15,640
minus q 3 dot, so this is what produces the
stored charge.
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00:27:15,640 --> 00:27:28,730
So, you have half L q 1 dot minus q 3 dot
square and that is the only place where kinetic
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00:27:28,730 --> 00:27:39,220
energy could be stored, the potential energy
is the energy stored in the two capacitors.
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00:27:39,220 --> 00:27:46,990
So, let us take this one first, it will be
half 1 by 2 C, C is in the denominator 1 by
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2 C this
is C 1, charge flowing is q 1 minus q 2, so
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q 1 minus q 2 square. In this one it is C
2 it is
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00:28:05,999 --> 00:28:16,940
q 3 plus there is something here, this is
the externally applied force which is acting
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00:28:16,940 --> 00:28:19,399
in the
positive direction of q 1. Therefore, you
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00:28:19,399 --> 00:28:33,840
will have to say minus q 1 E.
So, that is the potential energy and the Rayleigh
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00:28:33,840 --> 00:28:44,359
function would be here, for this one it
would be half R 1, then I have to identify
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00:28:44,359 --> 00:28:48,169
what are the currents flowing through this.
It is
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q 2 dot minus q 3 dot, q 2 dot is flowing
this way and q 3 dot is flowing in the opposite
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way, so q 2 dot minus q 3 dot square. And
this one plus half R 2 q 2 dot square, the
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00:29:16,989 --> 00:29:25,649
moment you have written this the rest is trivial,
so all you have to do is to differentiate
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these three functions.
So, let the write their Lagrangian, it is
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00:29:30,059 --> 00:29:35,809
I will write fast half L q 1 dot minus q 3
dot
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00:29:35,809 --> 00:29:50,739
square minus 2 C 1 q 1 minus q 2 square minus
2 C 2 q 3 square plus q 1 E and Rayleigh
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00:29:50,739 --> 00:30:20,590
.is this. So, the first direction q 1, in
the q 1 coordinate, what do you have minus
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00:30:20,590 --> 00:31:04,620
q 1 by C
1q 1 minus q 2 by C 1, is L into q 1 dot q
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00:31:04,620 --> 00:31:09,429
3 dot.
.
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00:31:09,429 --> 00:31:25,989
And your Rayleigh function is what is your
Rayleigh function, yes it is independent of
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00:31:25,989 --> 00:31:30,490
q
1 0. So, now if you write the equation it
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00:31:30,490 --> 00:31:36,929
will be, so the equation is d d t of this
which
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00:31:36,929 --> 00:31:52,759
means L q 1 double dot minus L q 3 double
dot. Then this one negative of this one q
207
00:31:52,759 --> 00:32:02,950
1
minus q 2 by C minus E, this term Rayleigh
208
00:32:02,950 --> 00:32:10,220
term is 0, therefore this is ok. Let us do
it the
209
00:32:10,220 --> 00:32:22,850
direction of q 2 also, because that is that
illustrates the use of the Rayleigh term.
210
00:32:22,850 --> 00:32:40,649
In the q 2 coordinate, what we have q 2 is
this one we are differentiating this one,
211
00:32:40,649 --> 00:32:43,379
we are
can you see yes, this one we are differentiating
212
00:32:43,379 --> 00:32:56,409
with respect to q 2, so you have plus
and
213
00:32:56,409 --> 00:33:14,960
nothing. Lagrangian with respect to q 2 dot
is 0 and delta R with respect to is, it was
214
00:33:14,960 --> 00:33:24,289
here
yes it will be, it will be R 1 q 1 dot minus
215
00:33:24,289 --> 00:33:50,119
q 3 dot minus. And the other term will also
remain, now if you write down the equation
216
00:33:50,119 --> 00:33:57,650
it will take the form. This is 0, therefore
d d
217
00:33:57,650 --> 00:33:59,779
t of this does not exist.
218
00:33:59,779 --> 00:34:00,779
..
219
00:34:00,779 --> 00:34:18,429
So, you will have minus of q 1 minus q 2 by
C plus R 1 q 2 dot minus q 3 dot plus R 2
220
00:34:18,429 --> 00:34:25,790
q
2 dot equal to 0, similarly you obtain the
221
00:34:25,790 --> 00:34:50,220
equation in third direction. And it will actually
take the form L q 1 double dot minus q 3 double
222
00:34:50,220 --> 00:35:11,170
dot plus 1 by C 2 q 3 plus R 1 q 2 dot
minus q 3 dot. So, three equations are this,
223
00:35:11,170 --> 00:35:16,510
this and...
.(Refer Time: 35:13))
224
00:35:16,510 --> 00:35:21,579
Where?
.(Refer Time: 35:16))
225
00:35:21,579 --> 00:35:40,460
This one, let us check sure yes, wait this
term since you are differentiating this there
226
00:35:40,460 --> 00:35:49,329
will
be a minus term here and the third term will
227
00:35:49,329 --> 00:36:04,920
also minus. See in this case no, I should
actually write explicitly otherwise this is
228
00:36:04,920 --> 00:36:07,699
creating problems.
229
00:36:07,699 --> 00:36:08,699
..
230
00:36:08,699 --> 00:36:18,809
So, the Lagrangian with respect to q 3 dot
is, notice the Lagrangian it will be L times
231
00:36:18,809 --> 00:36:27,250
this, but now this has to be differentiated,
so it will be minus. So, minus L q 1 dot minus
232
00:36:27,250 --> 00:36:47,289
q 3 dot, the derivative of L with respect
to q 3 is q 3 only here, minus q C 2 q 3.
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00:36:47,289 --> 00:36:57,960
And,
then
234
00:36:57,960 --> 00:37:09,250
here again it will lead to a minus, so R 1
q 2 dot minus q 3 dot and the second it will
235
00:37:09,250 --> 00:37:20,539
lead to a minus sign. So, minus R 1 q 2 dot
minus q 3 dot and this is 0.
236
00:37:20,539 --> 00:37:28,900
So, now the equation becomes minus L q 1 double
dot minus q three double dot
237
00:37:28,900 --> 00:37:48,589
derivative of this with respect to time this
plus 1 by C 2 q 3 plus or minus, minus R 1
238
00:37:48,589 --> 00:37:56,880
q 2
dot minus q 3 dot. So, that is the equation,
239
00:37:56,880 --> 00:38:24,460
then so you have the three equations, where
are they three equations, this is the q 2
240
00:38:24,460 --> 00:38:29,069
equation, this is the q 3 equation. And where
was
241
00:38:29,069 --> 00:38:39,540
the q 1 equation here, now notice one thing
what is this equation actually, notice
242
00:38:39,540 --> 00:38:52,849
carefully what does this say look at the circuit,
no.
243
00:38:52,849 --> 00:39:00,520
Can you see these two now . no, any way look
at this circuit and
244
00:39:00,520 --> 00:39:06,990
you have already written down, so tell you
what does it say L q 1 double dot minus L
245
00:39:06,990 --> 00:39:13,170
q 3
double dot, so it is actually L d i d t, where
246
00:39:13,170 --> 00:39:25,849
i is q 1 dot minus q 3 dot. Here is the voltage
across the capacitor q by C and here is E,
247
00:39:25,849 --> 00:39:32,660
so this is nothing but, the KVL. Which means
though we have not derived the Lagrangian
248
00:39:32,660 --> 00:39:35,039
equation using the Kirchhoff’s laws, it
is
249
00:39:35,039 --> 00:39:38,340
actually consistent with the Kirchhoff’s
laws.
250
00:39:38,340 --> 00:39:45,180
So, when we obtain a set of differential equations
we can also start from the Kirchhoff’s
251
00:39:45,180 --> 00:39:50,299
law and do it, no problem. And you will not
be asked forced to do it this way, but you
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00:39:50,299 --> 00:39:56,470
.will find that for many systems since I do
not have to worry about the direction of this
253
00:39:56,470 --> 00:40:01,059
for example, because it is squared. I do not
really care which direction is positive, which
254
00:40:01,059 --> 00:40:06,619
direction is negative, it is far more advantageous
in terms of writing down the equations
255
00:40:06,619 --> 00:40:12,539
correctly.
And later we will also illustrate a more suitable
256
00:40:12,539 --> 00:40:16,990
method for specifically devoted to
electrical circuits, but you can easily see
257
00:40:16,990 --> 00:40:23,840
that the set of equation that we develop using
mechanical systems as example are equally
258
00:40:23,840 --> 00:40:26,310
applicable to electrical circuits. And they
are
259
00:40:26,310 --> 00:40:32,549
not contradicting the Kirchhoff’s law which
to be correct. Similarly, you can check that
260
00:40:32,549 --> 00:40:36,680
the second equation will be actually the KVL
equation Kirchhoff’s voltage law equation
261
00:40:36,680 --> 00:40:48,090
in the q 2 loop and the third one is along
the q 3 loop.
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00:40:48,090 --> 00:40:57,040
Only one pointer, that is I do not know if
whether you know this. That we had started
263
00:40:57,040 --> 00:41:00,760
by
the assumption, that we can define a minimum
264
00:41:00,760 --> 00:41:05,760
number of necessary coordinates by
simply looking at the windows and put in the
265
00:41:05,760 --> 00:41:13,099
q 1, q 2, q 3 along them, this is in general
true. But, there are situations where this
266
00:41:13,099 --> 00:41:19,869
is not true, there are situations where this
procedure does not give the minimum set of
267
00:41:19,869 --> 00:41:26,240
coordinates. That gives you can always
define that way, it would not be wrong.
268
00:41:26,240 --> 00:41:30,600
But that will not give you the minimum set
of coordinates and how to write down the
269
00:41:30,600 --> 00:41:35,279
minimum set of coordinates that is a problem,
we will take up later. Just a pointer that
270
00:41:35,279 --> 00:41:49,079
this works and what you get is, but that is
not the minimum thing, let us carry on. So,
271
00:41:49,079 --> 00:41:53,710
we
have learnt how to use it can you just quickly
272
00:41:53,710 --> 00:41:59,079
write it for this system.
273
00:41:59,079 --> 00:42:00,079
..
274
00:42:00,079 --> 00:42:14,751
Say there is a mass, there is a spring, there
is a damper, there is another mass, there
275
00:42:14,751 --> 00:42:22,010
is a
spring, there is a damper and now it is connected
276
00:42:22,010 --> 00:42:33,100
it is resting on the ground. And now
you apply a force, can you write down the
277
00:42:33,100 --> 00:42:39,109
differential equations for this system. Do
this
278
00:42:39,109 --> 00:42:44,099
first you will need to define the coordinates.
How will you define a coordinates, position
279
00:42:44,099 --> 00:42:53,210
coordinates of these two masses. And if you
want to define the position coordinates of
280
00:42:53,210 --> 00:42:57,230
these two masses, you have to specify what
is their 0.
281
00:42:57,230 --> 00:43:06,920
So, you cannot simply say that here is my
position, and then this is say q 1 and this
282
00:43:06,920 --> 00:43:12,740
is say
q 2. You can say that, but the moment you
283
00:43:12,740 --> 00:43:16,800
will have to say where is my 0, now let that
0
284
00:43:16,800 --> 00:43:23,119
position be where these two springs are relaxed,
no energy stored in the springs. So, the q
285
00:43:23,119 --> 00:43:29,039
1 and q 2 will directly give the amount of
energy stored in the spring. So, you can easily
286
00:43:29,039 --> 00:43:33,579
see that if you do not apply any force let
it rest, then two springs will come to a
287
00:43:33,579 --> 00:43:38,270
compressed position at rest.
And that is not what I am defining as the
288
00:43:38,270 --> 00:43:44,300
0 why, because in that case how much has it
been compressed initially and all that we
289
00:43:44,300 --> 00:43:47,339
have to consider. We say that let these two
be
290
00:43:47,339 --> 00:43:52,170
measured from their zero position, where in
the zero position there is no energy stored
291
00:43:52,170 --> 00:44:04,470
in
the springs. Quickly M 1, M 2, so let this
292
00:44:04,470 --> 00:44:14,069
be M 1 and let this be M 2. Now, if you
proceed to write first you will write the
293
00:44:14,069 --> 00:44:18,440
kinetic energy which is somewhat trivial half
M
294
00:44:18,440 --> 00:44:36,010
1 q 1 dot square plus half M 2 q 2 dot square,
potential will be the potential stored.
295
00:44:36,010 --> 00:44:42,579
.For now let us ignore gravity because ultimately
if you consider gravity, it will lead to
296
00:44:42,579 --> 00:44:52,150
constant terms that will be differentiated
away, so who cares I do not want that. So,
297
00:44:52,150 --> 00:44:56,069
V
will be due to the two springs and the force,
298
00:44:56,069 --> 00:45:03,359
so it will be half, let this be K 1 and let
this
299
00:45:03,359 --> 00:45:16,279
be K 2 and this is be R 1, R 2. Half K 1,
then how will you write no q 1 minus q 2,
300
00:45:16,279 --> 00:45:20,640
q 1
minus q 2 is the amount of displacement experienced
301
00:45:20,640 --> 00:45:30,890
by this spring, so q 1 minus q 2
square plus half K 2 q 2 square.
302
00:45:30,890 --> 00:45:50,339
Now, here you see q 1 is up and F is down,
so we will say plus no just q 1 is acting
303
00:45:50,339 --> 00:45:57,279
in the
direction of q 1. Then your Lagrangian becomes
304
00:45:57,279 --> 00:46:23,220
I will just quickly write, so that we can
differentiate fast what I
305
00:46:23,220 --> 00:46:31,240
will write. Now, the Rayleigh function it
will be what is the
306
00:46:31,240 --> 00:46:42,029
relative velocity experienced by this R, it
is q 1 dot minus q 2 dot. So, it is half R
307
00:46:42,029 --> 00:46:51,780
1 q 1
dot minus q 2 dot square and how much is the
308
00:46:51,780 --> 00:46:57,059
it is only q 2 dot, for R 2 it is only q 2
dot
309
00:46:57,059 --> 00:47:19,260
So, it is plus half R 2 q 2 dot square.
From here do I really need to the rest is
310
00:47:19,260 --> 00:47:21,569
trivial, the rest is trivial and therefore,
I will
311
00:47:21,569 --> 00:47:29,039
leave it to you know how to do it. See you
did you notice that, the actual thing that
312
00:47:29,039 --> 00:47:32,420
you
have to do is to write these two nothing else.
313
00:47:32,420 --> 00:47:37,349
The rest is so algorithmic that you do not
really need to bother about it. You only have
314
00:47:37,349 --> 00:47:44,619
to do the differentiations right, which I
suppose you learnt in school, so this is rather
315
00:47:44,619 --> 00:47:54,799
simple way of doing it. By the way this
actually represents physical cushioning system,
316
00:47:54,799 --> 00:48:03,990
cushioning system where cars and motor
cycles they have a cushion.
317
00:48:03,990 --> 00:48:11,790
So, that you do not directly feel the vibration
of the these are somewhat like that, you
318
00:48:11,790 --> 00:48:22,730
may have seen this arrangement around the
back wheel of the motor cycles, you get a
319
00:48:22,730 --> 00:48:30,470
cylinder in which the frictional spring arrangement
is there. So, exactly this kind of
320
00:48:30,470 --> 00:48:40,660
arrangements in order to dampen or cushion
your the riders motion. Now, let us come to
321
00:48:40,660 --> 00:48:45,099
another issue that you might notice that you
have, where was the circuit I was dealing
322
00:48:45,099 --> 00:48:54,319
with circuit. It is not necessary we can again
do the circuit, there may be a number of
323
00:48:54,319 --> 00:48:58,260
inductances in the circuit.
And if there are number of inductances, so
324
00:48:58,260 --> 00:49:03,701
there may also be mutual coupling be
between the inductances. What happens, then
325
00:49:03,701 --> 00:49:08,299
if there are two mutual inductances, then
there are two inductances there is a mutual
326
00:49:08,299 --> 00:49:13,720
coupling between them. Then how do you,
how does it get reflected in the equations
327
00:49:13,720 --> 00:49:18,200
and through what, there is a question let
us deal
328
00:49:18,200 --> 00:49:23,980
with. So, let us take one example a circuit
simple one.
329
00:49:23,980 --> 00:49:24,980
..
330
00:49:24,980 --> 00:49:34,460
This is the positive direction of the voltage
E, here say an inductor and here say another
331
00:49:34,460 --> 00:49:44,230
inductor. And the rest of it can be connected
to resistance and capacitive branches. So,
332
00:49:44,230 --> 00:49:55,180
you have L 1, L 2, C, R, but now in addition
to this there is a mutual coupling between
333
00:49:55,180 --> 00:50:04,180
them. Now, whenever there was a mutual coupling
between two things what you do, you
334
00:50:04,180 --> 00:50:14,540
always put these blobs what does it mean.
It means that if the current is flowing into
335
00:50:14,540 --> 00:50:17,650
the
blob, in these two directions, then they would
336
00:50:17,650 --> 00:50:19,299
be additive, magnetic fields would be
additive.
337
00:50:19,299 --> 00:50:29,260
If the currents flow in the opposite directions
that means, in one it goes through into the
338
00:50:29,260 --> 00:50:37,109
blob and the other one it goes this way that
means, it goes out of the blob, then they
339
00:50:37,109 --> 00:50:40,970
are
subtractive. After all that addition or subtraction
340
00:50:40,970 --> 00:50:48,430
does what it adds to or subtracts from
the total kinetic energy that is stored in
341
00:50:48,430 --> 00:50:52,290
these two. So, this change should be reflected
in
342
00:50:52,290 --> 00:51:00,849
some kind of a change in the kinetic energy.
So, the kinetic energy if this is not there,
343
00:51:00,849 --> 00:51:04,619
the
kinetic energy would be written as half, let
344
00:51:04,619 --> 00:51:24,210
us designate the two coordinates.
Then half L 1 q 1 dot square plus half L 2,
345
00:51:24,210 --> 00:51:31,529
here it is q 1 minus q 2, q 1 dot minus q
2 dot
346
00:51:31,529 --> 00:51:43,430
square, in the absence of this mutual inductance.
Now, if the mutual inductance is there,
347
00:51:43,430 --> 00:51:50,559
if the mutual inductance is additionally there,
then you see due to q 1 it is going out of
348
00:51:50,559 --> 00:52:02,810
this blob and the q 1 minus q 2 is going into
the blob. So, how will you designate this,
349
00:52:02,810 --> 00:52:13,380
the amount of additional energy stored in
the inductor is half M, the two i, i 1, i
350
00:52:13,380 --> 00:52:15,200
2 you
know that.
351
00:52:15,200 --> 00:52:32,359
.So, the additional kinetic energy would be
plus half mutual inductance half M q 1 dot
352
00:52:32,359 --> 00:52:47,369
into q 1 dot minus q 2 dot. Now, here I will
first put plus minus, and then do the choice
353
00:52:47,369 --> 00:52:53,010
this is the amount, and then we will do the
choice. Now, you see because I put the dots
354
00:52:53,010 --> 00:52:59,230
here and here, q 1 goes out through the dot
and q 1 minus q 2 goes into the dot.
355
00:52:59,230 --> 00:53:04,109
Therefore, these two currents are subtractive,
if it is subtractive I have to take the minus
356
00:53:04,109 --> 00:53:16,050
sign, the rest remains the same.
All you need to do is to do this, now if I
357
00:53:16,050 --> 00:53:18,390
change the position of the dot say, this one
is
358
00:53:18,390 --> 00:53:22,769
moved to here obviously, you can see that
this will become additive you have to take
359
00:53:22,769 --> 00:53:28,829
plus. And in when you are actually doing this,
in problems then you have to take that into
360
00:53:28,829 --> 00:53:32,640
account somewhat minutely. So, that many people
I have seen that making mistakes in
361
00:53:32,640 --> 00:53:38,440
the signs here, because this is see if these
are two independent coordinates there is no
362
00:53:38,440 --> 00:53:40,930
problem.
But, because these are two related q 1 minus
363
00:53:40,930 --> 00:53:47,320
q 2, that is why I do not why, but students
tend to make mistakes. So, be careful about
364
00:53:47,320 --> 00:53:51,069
that if you are taking q 1 minus q 2 it is
going into, if you taking q 2 minus q 2 it
365
00:53:51,069 --> 00:54:05,210
is going out, then you can continue with this.
Mechanical equivalent of a mutual inductance,
366
00:54:05,210 --> 00:54:15,289
no mechanical equivalent of a mutual
inductance tough question, But, there can
367
00:54:15,289 --> 00:54:19,619
be, but there can be and that this question
I will
368
00:54:19,619 --> 00:54:26,940
come to about a month later, when they required
background has been built up.
369
00:54:26,940 --> 00:54:31,559
At this moment I will not get into these,
because that will require a lot of things
370
00:54:31,559 --> 00:54:36,589
to be
told I will come to that later. But, keep
371
00:54:36,589 --> 00:54:43,410
this question in mind and when the time comes
raise it again and we will see where it comes.
372
00:54:43,410 --> 00:54:48,701
So, what is the mechanical equivalent of a
mutual inductance, we will come to that.
373
00:54:48,701 --> 00:54:49,701
Thank you very much today.
374
00:54:49,701 --> 00:54:49,701
.