1
00:00:46,699 --> 00:00:54,300
In the last class we have in the main seen
that the differential equations of one
2
00:00:54,300 --> 00:01:00,319
dimensional systems. We could see how these
are obtained from the Lagrangian
3
00:01:00,319 --> 00:01:01,319
equation.
.
4
00:01:01,319 --> 00:01:07,780
Let us now go to one stage higher to the two
dimensional systems. Two dimensional
5
00:01:07,780 --> 00:01:11,140
means two spatial dimensions.
6
00:01:11,140 --> 00:01:12,140
..
7
00:01:12,140 --> 00:01:25,450
Now, let us take as an example a simple 2D
system. Here is the constant a flat plane
8
00:01:25,450 --> 00:01:38,270
with
two walls and there are two masses, they are
9
00:01:38,270 --> 00:01:48,240
linked with a spring and these are also
connected to the walls by means of springs.
10
00:01:48,240 --> 00:01:57,619
And for the present moment we consider
these as frictionless surface. In that case
11
00:01:57,619 --> 00:02:10,290
how we will define the generalized coordinates,
can you suggest.
12
00:02:10,290 --> 00:02:22,700
Simply the positions of this, so this is a
q 1 and this is a q 2 measure from some kind
13
00:02:22,700 --> 00:02:33,670
of a
data, see here. The moment we have written
14
00:02:33,670 --> 00:02:54,609
down these two as the generalized
coordinates. So, let us write now in terms
15
00:02:54,609 --> 00:02:59,160
of these two generalized coordinates we need
to write down the kinetic energy and the potential
16
00:02:59,160 --> 00:03:06,269
energy. So, let us first write down the
T pretty simple, because this has say a mass
17
00:03:06,269 --> 00:03:20,950
M 1 and this has mass M 2. Let this be K 1,
K 2 and K 3. Then the kinetic energy is half
18
00:03:20,950 --> 00:03:31,180
M 1 q 1 dot square for the first one plus
half
19
00:03:31,180 --> 00:03:46,849
M 2 q 2 dot square, simple. The potential
energy V, the potential energy in this spring
20
00:03:46,849 --> 00:04:15,140
will be half K 1 q 1 square, this one is . q
1 minus q 2, any
21
00:04:15,140 --> 00:04:17,790
question?
.
22
00:04:17,790 --> 00:04:42,930
Here there is a question. Let us for the sake
of simplicity assume, that for the this spring
23
00:04:42,930 --> 00:04:49,770
q 1 is measured from it is unstressed position.
And for this spring q 2 is measured form it
24
00:04:49,770 --> 00:04:55,840
is unstressed position. You could assume anything
otherwise what will happen is that,
25
00:04:55,840 --> 00:05:00,910
this will have some constant terms which on
differentiation will vanish. But, this is
26
00:05:00,910 --> 00:05:01,910
easy
27
00:05:01,910 --> 00:05:09,229
.way to write that is why we assume that.
So, her question was where to assume the data,
28
00:05:09,229 --> 00:05:15,209
that we could assume anywhere, but that would
be a convenient way of doing it. So, the
29
00:05:15,209 --> 00:05:32,099
Lagrangian function is T minus V, that would
be half M 1 q 1 dot square plus half M 2 q
30
00:05:32,099 --> 00:05:50,630
2 dot square minus half K 1 q 1 square minus
half K 3 q 2 square minus half, is it visible
31
00:05:50,630 --> 00:06:02,310
K 2 q 1 minus q 2.
.
32
00:06:02,310 --> 00:06:17,810
Now, take the derivatives, the Lagrangian
which is split to say q 1, we will have to
33
00:06:17,810 --> 00:06:23,349
write
the Lagrangian which is split to q 2. We will
34
00:06:23,349 --> 00:06:30,970
have to write the Lagrangian which is split
to q 1 dot and we will have to write it in
35
00:06:30,970 --> 00:06:56,110
terms of do this. From here it is pretty simple
and fair, you can see that the first one is,
36
00:06:56,110 --> 00:07:09,039
. this one would be minus
K 1 . and...
37
00:07:09,039 --> 00:07:33,240
Now, here it would be
minus K 3 q 2 plus here K 2 q 1 minus q 2,
38
00:07:33,240 --> 00:07:40,560
the which respect to
the generalized velocity, it is simple M 1
39
00:07:40,560 --> 00:07:55,440
q 1 dot and M 2, these are the momentum can
you see that q 2 dot. The moment you have
40
00:07:55,440 --> 00:08:03,409
written down this, to remember there would
be two differential equations coming out of
41
00:08:03,409 --> 00:08:07,989
the Lagrangian formulism, one along q 1
another along q 2.
42
00:08:07,989 --> 00:08:08,989
..
43
00:08:08,989 --> 00:08:38,060
So, the differential equation along q 1 would
be d d t of dot minus fine. Now, if you
44
00:08:38,060 --> 00:08:49,240
write down it will be substitute M 1, this
fellow would be q 1 double dot, because you
45
00:08:49,240 --> 00:08:57,010
are taking derivative with respect to time
once again. And then this would be minus let
46
00:08:57,010 --> 00:09:23,250
me copy from here this term plus K 2 q 1 plus
K 2 q 1 minus q 2. Differential equation
47
00:09:23,250 --> 00:09:31,820
along q 2 will be notice the advantage of
this formalism, that simply by substituting
48
00:09:31,820 --> 00:09:36,270
q 1 q
2 you get the accounting differential equation.
49
00:09:36,270 --> 00:09:48,250
It will be just substitute this with change
this to the equation with respect to q 2 dot
50
00:09:48,250 --> 00:10:00,450
minus Lagrangian with respect to q 2 is equal
to 0. Substitute you get M 2 q 2 double dot
51
00:10:00,450 --> 00:10:19,750
plus K 3 q 2 minus K 2 q 1 minus q 2, so this
is the first equation and this is the second
52
00:10:19,750 --> 00:10:26,770
equation. Along two different detections simple,
we did not have to bother about
53
00:10:26,770 --> 00:10:32,450
individual bodies. And you can easily see
that, if there where n number of bodies still
54
00:10:32,450 --> 00:10:36,790
we
could do this same procedure pretty simply.
55
00:10:36,790 --> 00:10:47,380
Let us take another, we have already started
doing 1, 2, 3 problems the last day, where
56
00:10:47,380 --> 00:10:54,589
if let us see do we have I lost the paper.
57
00:10:54,589 --> 00:10:55,589
..
58
00:10:55,589 --> 00:11:06,670
So, that day the problem was we took a spring
pendulum. And our variables were theta
59
00:11:06,670 --> 00:11:16,510
and if the total when it hangs vertically,
if the position is like this, then this distance
60
00:11:16,510 --> 00:11:29,260
was
a. And here this distance was a plus r, that
61
00:11:29,260 --> 00:11:34,070
is how we have defined r and theta are the
coordinates. And then we have already written
62
00:11:34,070 --> 00:11:47,270
down the Lagrangian function. Just recall
the Lagrangian function in this case was half
63
00:11:47,270 --> 00:12:10,020
m r dot square plus half m a plus r whole
square theta dot square minus half K r plus
64
00:12:10,020 --> 00:12:16,680
m g by K
We have all done these earlier, so I am not
65
00:12:16,680 --> 00:12:19,030
explaining, I am just writing down, so that
we
66
00:12:19,030 --> 00:12:25,500
can start up from there. Plus I just illustrate
something that students often make mistake,
67
00:12:25,500 --> 00:12:37,240
that is why I am doing this problem to it
is completion. Cos theta minus m g a, that
68
00:12:37,240 --> 00:12:39,970
is
what just check.
69
00:12:39,970 --> 00:12:42,370
.
Third term.
70
00:12:42,370 --> 00:12:59,500
Student: R plus m g the whole square.
This is a square, that would be a square.
71
00:12:59,500 --> 00:13:02,840
So, now you start writing the derivatives
with
72
00:13:02,840 --> 00:13:08,290
respect to the two coordinates. The first
derivative that would write is Lagrangian
73
00:13:08,290 --> 00:13:21,470
with
respect to r. When you write in terms of r,
74
00:13:21,470 --> 00:13:25,990
this has to be differentiated, this has to
be
75
00:13:25,990 --> 00:13:32,250
differentiated, this has to be differentiated.
So, how do you do it, you write it carefully
76
00:13:32,250 --> 00:13:41,870
2
comes forward, so this m theta dot square
77
00:13:41,870 --> 00:13:46,620
a plus r differentiated, so it should be a
plus r.
78
00:13:46,620 --> 00:14:02,589
.Then, because this also this one comes to
this. Then this one it will be minus K, this
79
00:14:02,589 --> 00:14:23,010
thing will be different r plus m g by K
and then from here plus m g cos theta. M g
80
00:14:23,010 --> 00:14:28,080
a
basically ., so this is with respect to r.
81
00:14:28,080 --> 00:14:40,600
Then with respect to r dot,
what do you have these remains, so m r dot
82
00:14:40,600 --> 00:14:48,910
enough. So, you can see from these two I can
write down the first equation.
83
00:14:48,910 --> 00:14:49,910
.
84
00:14:49,910 --> 00:15:08,770
And that would be
d d t of dou Lagrangian dou r dot minus dou
85
00:15:08,770 --> 00:15:16,620
Lagrangian it would be r
equal to 0, so substitute the first term will
86
00:15:16,620 --> 00:15:25,120
come to m r double dot, no problem about it.
Then the next term simply substitute it, minus
87
00:15:25,120 --> 00:15:36,560
m theta dot square a plus r plus K r plus
m
88
00:15:36,560 --> 00:15:49,230
g by K minus cos theta equal to 0 done, first
problem completed.
89
00:15:49,230 --> 00:15:50,230
..
90
00:15:50,230 --> 00:15:58,990
Second we write the same things with respect
to theta, so Lagrangian doe theta will be
91
00:15:58,990 --> 00:16:07,330
well this was the form. . This vanishes, this
vanishes, this vanishes
92
00:16:07,330 --> 00:16:24,240
this is only here, so it is pretty simple.
What remains only this term, it would be minus
93
00:16:24,240 --> 00:16:47,060
because cos to sin m g a plus r sin theta
that does it. Then this term is put a dot
94
00:16:47,060 --> 00:16:58,529
[FL]
simple, so now we have to write the second
95
00:16:58,529 --> 00:17:14,179
equation will be d d t of
dot minus doe theta
96
00:17:14,179 --> 00:17:28,039
equal to 0. Now, this has to be differentiated,
if you differentiate what do you have?
97
00:17:28,039 --> 00:17:34,450
.
So, differentiate this with respect to d d
98
00:17:34,450 --> 00:17:41,809
t, you take d d t, since these are both variables,
so you have to do it carefully, that is what
99
00:17:41,809 --> 00:17:51,350
I want you to point out. So, you have by
differentiating this, you have
100
00:17:51,350 --> 00:18:05,129
.
M a plus r whole square theta double dot plus,
101
00:18:05,129 --> 00:18:12,210
then you differentiate this one, so you
have twice m theta dot a plus r a plus r into
102
00:18:12,210 --> 00:18:21,679
r dot into dot r, minus then becomes plus
m g
103
00:18:21,679 --> 00:18:33,210
a plus r sin theta equal to 0. Now, this is
what students often I have found in exams
104
00:18:33,210 --> 00:18:37,320
miss,
here that both these are time variables. And
105
00:18:37,320 --> 00:18:43,440
therefore, when you differentiate with
respect to time, both have to be differentiated,
106
00:18:43,440 --> 00:18:47,970
that is what often people miss. And I have
seen hundreds of answer scripts saying that
107
00:18:47,970 --> 00:18:50,370
this times theta double dot.
108
00:18:50,370 --> 00:18:56,690
.No it is not, that is one point that is why
I did this problem to its completion. So,
109
00:18:56,690 --> 00:19:00,880
these
two are the differential equations of this
110
00:19:00,880 --> 00:19:09,190
system. So, second equation is this first
equation
111
00:19:09,190 --> 00:19:17,580
was this, notice that these equations are
hopelessly non-linear. Can you see these are
112
00:19:17,580 --> 00:19:20,419
not
linear equations really. So, in such simple
113
00:19:20,419 --> 00:19:32,669
systems also you are having non-linear
equation that we will deal with later.
114
00:19:32,669 --> 00:19:40,840
Now, we need to talk about electrical circuits
because, so far we are dealing with
115
00:19:40,840 --> 00:19:46,340
mechanical systems. But, we started with saying
in our earlier class, that the electrical
116
00:19:46,340 --> 00:19:49,980
systems are equivalent to mechanical systems,
you can have almost exact equivalence.
117
00:19:49,980 --> 00:19:56,919
So, you might ask, then it should be possible
to do solve write equations for electrical
118
00:19:56,919 --> 00:20:00,600
circuits exactly the same way, yes that is
true.
119
00:20:00,600 --> 00:20:10,779
But, in that case we what the steps were,
step number 1 was that define the generalized
120
00:20:10,779 --> 00:20:16,279
coordinates that are consistent with the constants.
In a mechanical system we easy to see
121
00:20:16,279 --> 00:20:23,669
the constants a hanging ball with the wire
I can easily see the constant. In the electrical
122
00:20:23,669 --> 00:20:29,640
circuit do you see the constant, where is
the constant, no does not supply voltage is
123
00:20:29,640 --> 00:20:32,730
not
the constant. Supply voltage is similar to
124
00:20:32,730 --> 00:20:38,720
a force, that is not a constant.
The constant is essentially the where the
125
00:20:38,720 --> 00:20:41,851
circuit is wired up, circuit is connected.
Some
126
00:20:41,851 --> 00:20:45,070
things are connecting series means the two
branch currents are forced to be the same
127
00:20:45,070 --> 00:20:48,850
that
is the constant. Two lines are connected in
128
00:20:48,850 --> 00:20:56,409
parallel the voltages are forced to be the
same, that is the constant. So, the point
129
00:20:56,409 --> 00:21:00,830
is that in electrical circuits the constants
are the
130
00:21:00,830 --> 00:21:12,929
way the circuit is connected, that impose
the constants. But, now our job is to define
131
00:21:12,929 --> 00:21:16,279
the
minimum number of position coordinates.
132
00:21:16,279 --> 00:21:20,070
That is what we did in mechanical system and
what is the equivalent of position in
133
00:21:20,070 --> 00:21:27,609
electrical circuits charge, so we have to
deal with charge, charge is equivalent to
134
00:21:27,609 --> 00:21:31,779
the
position. So, charge becomes the configuration
135
00:21:31,779 --> 00:21:39,630
coordinate in electrical circuit, but charge
wire. Again the rule was that I have to define
136
00:21:39,630 --> 00:21:44,279
the position coordinates, the minimum
number of position coordinates that uniquely
137
00:21:44,279 --> 00:21:49,450
define the position less status of the system.
That was what we did, in this case also we
138
00:21:49,450 --> 00:21:55,240
have to think in terms of the circuit is
connected this way. What are the different
139
00:21:55,240 --> 00:22:02,549
charges that I need to consider as the
minimum number of coordinates.
140
00:22:02,549 --> 00:22:03,549
..
141
00:22:03,549 --> 00:22:13,559
Say suppose the circuit is like this, there
is an inductor, there is another inductor,
142
00:22:13,559 --> 00:22:21,049
there is
a capacitor, there is a capacitor and there
143
00:22:21,049 --> 00:22:27,429
is another capacitor. Then there is a charge
flowing here, there is a charge flowing here,
144
00:22:27,429 --> 00:22:34,960
there is a charge flowing here and what are
the minimum number. Now, there anything could
145
00:22:34,960 --> 00:22:39,679
be taken as the minimum number, but
from your 1st year knowledge of electrical
146
00:22:39,679 --> 00:22:44,090
circuits course you probably have learnt,
that
147
00:22:44,090 --> 00:22:49,020
one simple possible solution to this problem
is simply to consider the charges flowing
148
00:22:49,020 --> 00:22:53,390
in
the loops like so.
149
00:22:53,390 --> 00:23:03,840
At least it is guaranteed that they would
be the minimum number of independent
150
00:23:03,840 --> 00:23:10,269
coordinates, that is what actually matters
for us. So, let us say this is q 1 is here
151
00:23:10,269 --> 00:23:17,780
and q 2 is
here simple. Q 1 is here q 2 is here, say
152
00:23:17,780 --> 00:23:22,120
this is L 1, this is L 2, this is exactly
why we did
153
00:23:22,120 --> 00:23:29,389
not write the Lagrangian as L as in written
in many of the physics text books. We wrote
154
00:23:29,389 --> 00:23:36,220
it with a script L, because we wanted to distribute
only inductance, so that students do
155
00:23:36,220 --> 00:23:54,190
not get mixed up. C 1, C 2, C 3, in this case
also we will have to write down the kinetic
156
00:23:54,190 --> 00:23:59,289
energy and the potential energy. What is the
equivalent of the kinetic energy in a
157
00:23:59,289 --> 00:24:01,779
electrical circuit?
.
158
00:24:01,779 --> 00:24:07,000
No not the charge, the energy stored in the
inductor that is the equivalent to the kinetic
159
00:24:07,000 --> 00:24:12,240
energy and the energy stored in the capacitor
is the equivalent to potential energy,
160
00:24:12,240 --> 00:24:17,789
because the capacitor has the character of
being a complaint element equivalent to a
161
00:24:17,789 --> 00:24:23,489
.spring, that is why we have to do it that
way. So, if that is so then you have your
162
00:24:23,489 --> 00:24:36,749
T in this
case half L 1 q 1 dot square, q 1 dot is nothing
163
00:24:36,749 --> 00:24:53,450
but, the means current plus half L 2 q 2
dot square
164
00:24:53,450 --> 00:25:00,450
V the potential energy, notice since we are
writing the potential energy as V all the
165
00:25:00,450 --> 00:25:07,049
potentials are applied electromotive forces
we always we will write as E not V. Because,
166
00:25:07,049 --> 00:25:16,190
we do not want to have duplicity of notations
V, V is the charge energy stored here plus
167
00:25:16,190 --> 00:25:20,309
energy stored here plus the energy stored.
If my energy is here, how much is the energy
168
00:25:20,309 --> 00:25:30,929
here, 1 by 2 C 1 notice, here the C is downstairs
in the denominator. 1 by twice C 1 into
169
00:25:30,929 --> 00:25:44,750
q 1 square plus 1 by twice C 2, how much is
the charge flowing to this q 1 minus q 2.
170
00:25:44,750 --> 00:26:01,981
Q 1 minus q 2 square, how much is the energy
stored here, 1 by 2 C 3 q 2 square. When
171
00:26:01,981 --> 00:26:13,159
we have written the T and V this way, we can
write the Lagrangian as half L 1 q 1 dot
172
00:26:13,159 --> 00:26:29,239
square plus half L 2 q 2 dot square minus
T minus V 1 by 2 C 1 q 1 square minus 1 by
173
00:26:29,239 --> 00:26:38,440
2
C 2 q 1 minus q 2 square minus 1 by 2 C 3
174
00:26:38,440 --> 00:26:50,460
q 2 square. Now, differentiate same
procedure, strangely you will notice that
175
00:26:50,460 --> 00:26:56,320
it is yielding exactly the same equation as
this,
176
00:26:56,320 --> 00:27:07,190
see here my can you see all of them ..
Here my T is half L 1 q 1 dot square half
177
00:27:07,190 --> 00:27:10,479
M 1 q 1 dot square, half L 2 q 2 dot square
half
178
00:27:10,479 --> 00:27:17,799
M 2 q 2 dot square exactly the same, only
M and L are interchanged. Therefore, the
179
00:27:17,799 --> 00:27:23,740
kinetic energy in this case is not only are
conceptually the same, even in magnitude are
180
00:27:23,740 --> 00:27:33,600
the same, if provided that you were expressing
M and L in the same quantities, whatever
181
00:27:33,600 --> 00:27:47,779
the units are. V half q 1 square 1 by 2 C
1 q 1 square half K 3 q 2 square 1 by 2 C
182
00:27:47,779 --> 00:27:52,749
3 q 2
square exactly the same, only K 1 is equivalent
183
00:27:52,749 --> 00:27:58,480
to 1 by C 1, K 2 is equivalent to 1 by C 2
and all that.
184
00:27:58,480 --> 00:28:04,990
And therefore, even without writing the differential
equations here I can tell you pretty
185
00:28:04,990 --> 00:28:09,730
blindly, that it will be the exactly the same
differential equations as this ((Refer Time:
186
00:28:09,730 --> 00:28:16,080
28:10)) that we have already obtained. So,
I do not really need to write, because these
187
00:28:16,080 --> 00:28:19,999
two systems are equivalent of each other.
188
00:28:19,999 --> 00:28:20,999
..
189
00:28:20,999 --> 00:28:27,090
These two systems are equivalent of each other.
Equivalent is what sense, equivalent in
190
00:28:27,090 --> 00:28:31,869
the following sense, that here what will happen
if you say move this one and release, it
191
00:28:31,869 --> 00:28:37,989
will oscillate, it will transfer the oscillation
here, if this will also oscillate, so there
192
00:28:37,989 --> 00:28:41,460
will
be some dynamics. And the dynamics you will
193
00:28:41,460 --> 00:28:45,409
learn how to actually obtain the dynamics
and plot it, but it is not difficult to see
194
00:28:45,409 --> 00:28:48,029
that there will be dynamics q 1 and q 2 will
vary
195
00:28:48,029 --> 00:28:56,700
and you can draw the waveforms q 1 and q 2.
In this case you are not really moving something
196
00:28:56,700 --> 00:29:04,909
and releasing, what would be the
equivalent of moving M 1 and releasing. What
197
00:29:04,909 --> 00:29:05,909
is the equivalent here?
.
198
00:29:05,909 --> 00:29:11,119
Yes, position is changed which means the K
1 is moved, K 1 is moved means some
199
00:29:11,119 --> 00:29:17,119
energy is stored initially in q 1, which is
equivalent to some charge being stored initially
200
00:29:17,119 --> 00:29:24,249
in C 1. So, store some charge and switch it
on you will see that this system will also
201
00:29:24,249 --> 00:29:29,179
oscillate. The currents and voltages will
also oscillate, exactly the same way as this
202
00:29:29,179 --> 00:29:33,580
one
and then if the L 1, L 2, L 3 and C 1, C 2,
203
00:29:33,580 --> 00:29:42,070
C 3 are numerically the same as these, they
will be the exactly the same dynamics.
204
00:29:42,070 --> 00:29:48,909
When it is excursion q as excursion reaches
the maximum point, this current through the
205
00:29:48,909 --> 00:29:56,440
inductor it is or charge through to the inductor.
It is excursion also reaches the same
206
00:29:56,440 --> 00:30:00,989
point, exactly the same dynamics. That is
why the electrical and mechanical system
207
00:30:00,989 --> 00:30:14,419
.would be same to be system wise not only
element wise, system wise equivalent. So,
208
00:30:14,419 --> 00:30:21,309
far
we were considering systems that had only
209
00:30:21,309 --> 00:30:26,359
one type of potential, either it was a we
have
210
00:30:26,359 --> 00:30:30,370
considered the system with gravitational and
potential. We have considered systems
211
00:30:30,370 --> 00:30:35,710
where the potential is represented by a spring
and the energies in spring. What about
212
00:30:35,710 --> 00:30:44,730
applied force, if there is an applied force
can it still be is still a conservative system,
213
00:30:44,730 --> 00:30:46,289
what
was the definition of conservative system
214
00:30:46,289 --> 00:30:47,289
we said thatâ€¦
.
215
00:30:47,289 --> 00:31:02,031
If you can write the force as minus the derivative
of potential function with respect to, if
216
00:31:02,031 --> 00:31:07,239
you can do that then we say that it is a conservative
system.
217
00:31:07,239 --> 00:31:08,239
..
218
00:31:08,239 --> 00:31:18,739
Now, if you imagine here on this, suppose
I am applying some kind of a force here F,
219
00:31:18,739 --> 00:31:22,259
that force could be a sinusoidal forcing function
or something like that. But, some would
220
00:31:22,259 --> 00:31:31,820
be applying a force here. Can that force be
somehow included in this potential, so that
221
00:31:31,820 --> 00:31:40,169
. this is valid, that is true. That means
even if there is a externally
222
00:31:40,169 --> 00:31:46,849
impressed force, then also the system is conservative.
In the sense that it can be expressed in this
223
00:31:46,849 --> 00:31:51,710
form but how, in that case the rule is that
you
224
00:31:51,710 --> 00:32:03,219
have to add to the potential function. Force
times the direction of the generalized
225
00:32:03,219 --> 00:32:10,820
coordinate, that particular generalized coordinates
along with it is applied. So, in this
226
00:32:10,820 --> 00:32:30,610
case we will write V is equal to F times q
1 with a negative sign, why you will see that
227
00:32:30,610 --> 00:32:32,090
it
is necessary in order to set the sign, if
228
00:32:32,090 --> 00:32:39,469
you write it this way the V due to the applied
force
229
00:32:39,469 --> 00:32:49,479
is this much. There we have already written
the rest of the V function, we will only add
230
00:32:49,479 --> 00:32:53,340
if this is applied, let us do it separately.
231
00:32:53,340 --> 00:32:54,340
..
232
00:32:54,340 --> 00:33:08,460
Now, we are talking about the system where
233
00:33:08,460 --> 00:33:11,409
and there is a force applied here, here this
q
234
00:33:11,409 --> 00:33:26,450
1 here is q 2. Then what, we have already
written the rest of it, so let me write V
235
00:33:26,450 --> 00:33:37,519
is equal
to K 1 q 1 square plus half K 3 q 2 square
236
00:33:37,519 --> 00:33:44,870
plus half K 2 q 1 minus q 2, this is that
and this
237
00:33:44,870 --> 00:33:55,559
remains square, this remains. What gets additionally
added to this potential function, so
238
00:33:55,559 --> 00:34:05,359
that the total force is still obtainable from
that potential function by simple derivative.
239
00:34:05,359 --> 00:34:17,230
What I said is that here you have to add minus
F times q 1 that is it, if you do so let us
240
00:34:17,230 --> 00:34:23,250
see what happens to the Lagrangian equation.
The Lagrangian would be the kinetic
241
00:34:23,250 --> 00:34:31,620
energy remains the same, so I will directly
write the Lagrangian. It was half M 1 q 1
242
00:34:31,620 --> 00:34:39,400
dot
square plus half M 2 q 2 dot square minus
243
00:34:39,400 --> 00:34:55,659
half K 1 q 1 square minus half K 3 q 2 square
minus half K 2 q 1 minus q 2 square plus F
244
00:34:55,659 --> 00:35:09,420
q 1. Because, this is minus it becomes plus.
Now, when we write the Lagrangian with respect
245
00:35:09,420 --> 00:35:27,369
to q 1, what you had now, here minus
K 1 q 1 here, what was it earlier, let me
246
00:35:27,369 --> 00:35:42,700
just refer to that. Here earlier it was this
minus K
247
00:35:42,700 --> 00:35:57,790
2 q 1 minus q 2. Now, this will be added to
that, so it would be plus F, but that will
248
00:35:57,790 --> 00:36:03,869
not
happen
249
00:36:03,869 --> 00:36:09,440
if you differentiate with respect to q 2,
so that remains the same minus K 3 q 2
250
00:36:09,440 --> 00:36:23,470
plus K 2 q 1 minus q 2. Now, you write the
first equation first differential equation.
251
00:36:23,470 --> 00:36:35,450
These two remain the same as earlier.
252
00:36:35,450 --> 00:36:36,450
..
253
00:36:36,450 --> 00:36:46,319
It will be M 1 q 1 double dot as usual, but
then this one has to be written, it will be
254
00:36:46,319 --> 00:36:59,920
a plus
K 1 q 1 plus K 2 q 1 minus q 2 minus F equal
255
00:36:59,920 --> 00:37:07,849
to 0. See that the signs have been right,
because this is mass into acceleration, this
256
00:37:07,849 --> 00:37:14,589
is the force applied on it by the springs
and
257
00:37:14,589 --> 00:37:21,700
this is the force applied externally. So,
that is how the equation is written which
258
00:37:21,700 --> 00:37:26,380
is right,
the second equation remains exactly the same.
259
00:37:26,380 --> 00:37:44,599
It would be M 2 q 2 double dot plus K 3 q
2 minus K 2 q 1 minus q 2 without any difference.
260
00:37:44,599 --> 00:37:53,589
So, it is trivial to take into account the
external forces. What will happen if we want
261
00:37:53,589 --> 00:37:55,880
to do
the same thing in an electrical circuit, external
262
00:37:55,880 --> 00:38:02,890
force is equivalent to a better externally
applied EMF. And what would be the equivalent
263
00:38:02,890 --> 00:38:07,210
of this ., a force
being applied in the q 1 direction.
264
00:38:07,210 --> 00:38:08,210
..
265
00:38:08,210 --> 00:38:10,240
What is the equivalent to that?
.
266
00:38:10,240 --> 00:38:19,750
A source notice here, here is the M 1 which
is equivalent to the L 1 and what is the
267
00:38:19,750 --> 00:38:28,799
character of this force, this force share
the same velocity with M 1. And in order for
268
00:38:28,799 --> 00:38:31,080
that
to happen in the electrical circuit you have
269
00:38:31,080 --> 00:38:36,400
to apply a battery here, that is it. So, if
we
270
00:38:36,400 --> 00:38:44,550
really have a battery here and it is applied
voltage is E, in that case the equation and
271
00:38:44,550 --> 00:38:46,279
the
process of derivation of the equation will
272
00:38:46,279 --> 00:38:51,329
exactly would be the same. So, in that case
what we will do, we will argue you that.
273
00:38:51,329 --> 00:38:52,329
..
274
00:38:52,329 --> 00:38:57,549
Here E is applied in the direction of the
generalized coordinate q 1. And therefore,
275
00:38:57,549 --> 00:39:03,910
with
the V we will have minus E times q 1 that
276
00:39:03,910 --> 00:39:21,829
is it, so that is how we will write the
differential equation. Let us solve a relatively
277
00:39:21,829 --> 00:39:28,119
difficult problem which has an applied
force on it, and which is the very practically
278
00:39:28,119 --> 00:39:35,660
relevant problem. Slowly we are going into
engineering problems, have you seen ever launching
279
00:39:35,660 --> 00:39:39,039
of a spacecraft, a rocket for
example.
280
00:39:39,039 --> 00:39:46,730
You must have seen in the in the TV, so how
does it happen, the spacecraft is launched
281
00:39:46,730 --> 00:39:51,810
alright, but initially it has to be held there.
And it is held there with the help of braces,
282
00:39:51,810 --> 00:39:54,890
if
the braces are there and it fires the whole
283
00:39:54,890 --> 00:40:01,450
thing will break off. So, that cannot be allowed,
so the braces have to come off, if it comes
284
00:40:01,450 --> 00:40:09,450
off then the rocket will fall. So, how to
prevent rocket from falling, what happens
285
00:40:09,450 --> 00:40:13,049
is that down there is actually a vehicle like
thing.
286
00:40:13,049 --> 00:40:20,319
There is actually a vehicle like thing which
is allowed to move and that moves and there
287
00:40:20,319 --> 00:40:27,079
by keeps it vertical, it is like a inverted
pendulum. The pendulum is vertical and here
288
00:40:27,079 --> 00:40:29,950
this
point is allowed to move and it is actually
289
00:40:29,950 --> 00:40:36,690
a control system enabling which you keep it
vertically and then you fire. That why it
290
00:40:36,690 --> 00:40:39,900
remains vertical and then it goes off that
is how
291
00:40:39,900 --> 00:40:47,520
the rocket launching is normally done. And
to model such a thing let us model the
292
00:40:47,520 --> 00:40:49,699
inverted pendulum problem.
293
00:40:49,699 --> 00:40:50,699
..
294
00:40:50,699 --> 00:41:02,670
So, inverted pendulum problem is where you
have the base along with the wall, and there
295
00:41:02,670 --> 00:41:21,640
is a car with wheels. And the pendulum is
like this, say normally this could be a you
296
00:41:21,640 --> 00:41:27,240
know mass could be uniformly distributed.
But, let us assume for our simplicity, that
297
00:41:27,240 --> 00:41:32,500
this
is a mass concentrated at the top and here
298
00:41:32,500 --> 00:41:36,829
is another mass, this fellow also has mass.
So,
299
00:41:36,829 --> 00:41:46,300
let us this be M 1 and this fellow be M 2.
And then you have to apply force here, in
300
00:41:46,300 --> 00:41:50,230
order
for this to move and balance it.
301
00:41:50,230 --> 00:42:00,089
So, that force is F, well in this case how
would you define the generalized coordinates,
302
00:42:00,089 --> 00:42:07,030
there are two generalized coordinates. One
position of this fellow which has to be
303
00:42:07,030 --> 00:42:18,839
measured simply from the wall, it can be measured
from the wall. And this fellow
304
00:42:18,839 --> 00:42:31,630
position can be measured in terms of the angle.
Either you call this as q 1 and this as q
305
00:42:31,630 --> 00:42:36,260
2,
or if you want to retain the physical things
306
00:42:36,260 --> 00:42:39,740
theta, then let us call it x, then it would
be is x
307
00:42:39,740 --> 00:42:43,330
and theta.
But, in our case it is essentially the q 1
308
00:42:43,330 --> 00:42:46,369
and q 2 the 2 positional coordinates. Now,
we
309
00:42:46,369 --> 00:42:50,609
have all set to write down the differential
equations, so the generalized coordinates
310
00:42:50,609 --> 00:43:15,539
in
this case x and theta. Now, the kinetic energy
311
00:43:15,539 --> 00:43:22,049
T, kinetic energy would be a combination
of the two kinetic energies, kinetic energy
312
00:43:22,049 --> 00:43:25,210
of this one and the kinetic energy of this
one.
313
00:43:25,210 --> 00:43:40,720
For this one it is simple, half m 1, then
x 1 dot square x dot square. For this one
314
00:43:40,720 --> 00:43:44,369
it is a bit
complicated, because it has a circumferential
315
00:43:44,369 --> 00:43:49,740
motion, there is no radial motion, but that
circumferential motion has to be broken into
316
00:43:49,740 --> 00:43:53,049
two components x and y. And x has this
317
00:43:53,049 --> 00:44:02,190
.additional x dot adequate, there are two
components really. So, the horizontal component
318
00:44:02,190 --> 00:44:15,349
of this motion, if this is L, then L theta
dot is the circumferential motion. L theta
319
00:44:15,349 --> 00:44:18,640
dot cos
theta would be
320
00:44:18,640 --> 00:44:25,280
.
No, horizontal component and sin theta would
321
00:44:25,280 --> 00:44:31,289
be the vertical component. The vertical
component is unaffected by this motion, the
322
00:44:31,289 --> 00:44:38,520
horizontal component has this motion erect
to it, so that way we can write down the T.
323
00:44:38,520 --> 00:44:52,790
So, how what will it be a horizontal
component would be plus half m 2, it would
324
00:44:52,790 --> 00:45:00,130
be l theta dot cos theta, l theta dot cos
theta
325
00:45:00,130 --> 00:45:25,050
plus x dot. L theta dot cos theta is plus
x we add up square plus half m 2 l theta dot
326
00:45:25,050 --> 00:45:43,990
sin
theta l theta dot sin theta square.
327
00:45:43,990 --> 00:45:51,220
This is the kinetic energy completed, potential
energy potential energy there are because
328
00:45:51,220 --> 00:46:00,859
of two components one, because of gravitational
energy in m 2. This does not have
329
00:46:00,859 --> 00:46:05,250
gravitational energies moves horizontally,
but also there will be another component of
330
00:46:05,250 --> 00:46:24,299
the potential due to F. So, what will be this
quantity, it would be m g l cos theta, now
331
00:46:24,299 --> 00:46:28,490
F is
being applied in the direction of x. X increases
332
00:46:28,490 --> 00:46:40,359
in this direction, F is also applied in this
direction, so it will be minus F x. What?
333
00:46:40,359 --> 00:46:52,849
.
M 2, now there will be a term here that will
334
00:46:52,849 --> 00:46:55,779
vanish, there will be one term coming out
of
335
00:46:55,779 --> 00:47:00,829
this that we have cos square theta and here
sin square theta and they will add up to one.
336
00:47:00,829 --> 00:47:09,270
So, this can be simplified, just simplify
that, so let us write down the Lagrangian.
337
00:47:09,270 --> 00:47:10,270
..
338
00:47:10,270 --> 00:47:25,910
Lagrangian is half m 1 x dot square, let us
breaks that that up half m 2 l square theta
339
00:47:25,910 --> 00:47:37,460
dot
square cos square theta plus half m 2 x dot
340
00:47:37,460 --> 00:47:50,380
square. Plus half goes off m 2 l theta dot
cos
341
00:47:50,380 --> 00:47:59,359
theta x dot. I have broken up just check it
if I have written correctly, I have just expanded
342
00:47:59,359 --> 00:48:13,680
this square. Plus half m 2 l theta dot, I
will not write it l square theta square sin
343
00:48:13,680 --> 00:48:24,869
square
theta minus m 2 g l cos theta plus F x, that
344
00:48:24,869 --> 00:48:29,180
is the total Lagrangian.
Now, you will notice that this fellow and
345
00:48:29,180 --> 00:48:37,349
this fellow, this term gets common cos square
plus sin square, so I will just drop this
346
00:48:37,349 --> 00:48:43,819
term and drop this term, I can do that. So,
it
347
00:48:43,819 --> 00:48:50,549
becomes relatively simplified expression for
the Lagrangian. Now, we will lead to take
348
00:48:50,549 --> 00:49:02,280
the derivative of the Lagrangian with respect
to x, it will be this goes off, this goes
349
00:49:02,280 --> 00:49:19,660
off,
this goes off. Now, you have m 1 x dot from
350
00:49:19,660 --> 00:49:25,579
here, this goes off here this remains, so
plus
351
00:49:25,579 --> 00:49:39,900
m 2 x dot. This one we remains, but this is
a complicated stuff differentiate it properly
352
00:49:39,900 --> 00:50:02,319
plus m 2 l theta dot cos theta. So, let us
write the other one before proceeding, no
353
00:50:02,319 --> 00:50:05,219
let me
write the equation from here.
354
00:50:05,219 --> 00:50:06,219
..
355
00:50:06,219 --> 00:50:18,130
The equation would be then d d t of Lagrangian
with respect to x dot minus Lagrangian
356
00:50:18,130 --> 00:50:30,190
with respect to x equal to 0. Now, when we
expand it when we expand it would be write
357
00:50:30,190 --> 00:50:40,589
the d d t of this correctly. Now, I writing
same thing, because most students making
358
00:50:40,589 --> 00:51:02,609
mistake here, write the d d t of this correctly.
It will be m 1 plus m 2 x double dot plus
359
00:51:02,609 --> 00:51:15,650
m
2 l cos theta theta double dot theta double
360
00:51:15,650 --> 00:51:26,820
dot minus. Now, we have to differentiate this
m 2 l theta dot...
361
00:51:26,820 --> 00:51:36,890
Student: Sin theta.
Sine theta theta dot square
362
00:51:36,890 --> 00:51:46,170
.
So, this is the derivative of this. Now, we
363
00:51:46,170 --> 00:52:04,700
have to write minus this, so that would be
minus F minus F equal to 0 done. Next one
364
00:52:04,700 --> 00:52:12,744
derivative of Lagrangian with respect to theta
is if you do it with respect to theta, can
365
00:52:12,744 --> 00:52:14,200
you see the expression, yes. (Refer Time:
52:15)
366
00:52:14,200 --> 00:52:22,340
this goes off, this goes off, this goes off,
this remains this remains, so take the derivative
367
00:52:22,340 --> 00:52:43,961
correctly, it will be minus m 2 l theta dot
sin theta x dot plus m 2 g l sin theta. And
368
00:52:43,961 --> 00:52:50,910
with
respect to theta dot is with respect to theta
369
00:52:50,910 --> 00:53:01,270
dot is this term remains m 2 l square theta
dot
370
00:53:01,270 --> 00:53:15,299
plus m 2 l cos theta x dot.
Now, again take the derivatives properly when
371
00:53:15,299 --> 00:53:36,700
you write this, write it properly here we
have this. The first term is simple m 2 l
372
00:53:36,700 --> 00:53:39,869
square theta double dot, no problem here.
Next
373
00:53:39,869 --> 00:54:03,500
.term will have problem plus m 2 l cos theta
x double dot minus m 2 l dot sin theta dot.
374
00:54:03,500 --> 00:54:18,950
Then plus this one, so plus m 2 l theta dot
sin theta x dot minus m 2 g l sin theta equal
375
00:54:18,950 --> 00:54:21,410
to
0, does anything cancel off.
376
00:54:21,410 --> 00:54:29,190
.
No third and the fourth, they cancel, m 2
377
00:54:29,190 --> 00:54:39,359
l cancels off, so it is the actually simpler
equation theta double dot l theta double dot
378
00:54:39,359 --> 00:54:52,619
plus cos theta x double dot minus g sin theta
equal to 0, does the other equation simplify,
379
00:54:52,619 --> 00:54:58,819
no so that is the equation then good. So,
this
380
00:54:58,819 --> 00:55:04,700
is how the equations are to be written. Remember
this if you try with any other method
381
00:55:04,700 --> 00:55:10,309
would be enormously complicated, this particular
problem. So, that is why the strength
382
00:55:10,309 --> 00:55:16,680
of the Lagrangian approach becomes salient
in this kind of problems.
383
00:55:16,680 --> 00:55:18,089
Thank you.
384
00:55:18,089 --> 00:55:18,089
.