1
00:00:03,640 --> 00:00:05,170
So now we want to
find the center
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00:00:05,170 --> 00:00:06,820
of mass of a uniform rod.
3
00:00:06,820 --> 00:00:09,930
And we have the result for
a continuous body, which
4
00:00:09,930 --> 00:00:15,730
is that integral over the body
of dmr to that mass element dm
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00:00:15,730 --> 00:00:19,250
divided by an interval.
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00:00:19,250 --> 00:00:21,170
Now our goal is
to figure out how
7
00:00:21,170 --> 00:00:24,210
to apply this
result, specifically,
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00:00:24,210 --> 00:00:26,140
to real physical objects.
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00:00:26,140 --> 00:00:29,060
And the key, as always, is
choosing a coordinate system.
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00:00:29,060 --> 00:00:31,250
So now I'll draw
the object, again.
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00:00:31,250 --> 00:00:34,230
And the first thing I'll
do is choose an origin.
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00:00:34,230 --> 00:00:36,534
I can pick my origin
anywhere I want.
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00:00:36,534 --> 00:00:37,700
I can pick it in the middle.
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00:00:37,700 --> 00:00:38,825
I can put it in the middle.
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00:00:38,825 --> 00:00:39,950
I could put it at this end.
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00:00:39,950 --> 00:00:41,050
I could put it that end.
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00:00:41,050 --> 00:00:44,990
I could put it down here,
but I'll choose it over here.
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00:00:44,990 --> 00:00:46,890
Because the object
is linear, this
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00:00:46,890 --> 00:00:49,450
is a very Cartesian system.
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00:00:49,450 --> 00:00:51,580
I'm only doing a one
dimensional object.
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00:00:51,580 --> 00:00:54,890
So I choose my
coordinate system plus x.
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00:00:54,890 --> 00:00:57,380
That's step one.
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00:00:57,380 --> 00:01:01,150
Now my origin-- now here
comes the crucial thing.
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00:01:01,150 --> 00:01:06,150
In this argument, dm is the
infinitesimal mass element.
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00:01:06,150 --> 00:01:09,970
And I want to pick that at an
arbitrary place in the object.
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00:01:09,970 --> 00:01:11,760
I don't want to pick
it at the origin.
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00:01:11,760 --> 00:01:13,800
I don't want to
pick it at the end.
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00:01:13,800 --> 00:01:16,660
Note down here
this is x equals L.
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00:01:16,660 --> 00:01:20,660
So I'll arbitrarily pick
an infinitesimal element.
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00:01:20,660 --> 00:01:24,289
I'll shade it in dm.
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00:01:24,289 --> 00:01:26,640
That represents--
this is what I'm
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00:01:26,640 --> 00:01:30,050
going to make my summation
over when I do my integral.
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00:01:30,050 --> 00:01:33,440
I'm going to add
up all these dm's.
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00:01:33,440 --> 00:01:36,300
And the point is that the
dm's are different distances
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00:01:36,300 --> 00:01:37,630
from the origin.
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00:01:37,630 --> 00:01:40,120
So the vector-- and
here's the next step--
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00:01:40,120 --> 00:01:44,200
is I draw a picture
of my vector rdm.
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00:01:44,200 --> 00:01:48,630
So now I have these terms, at
least, explained in my diagram.
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00:01:48,630 --> 00:01:52,250
The next step is to turn-- is
to introduce an integration
40
00:01:52,250 --> 00:01:55,060
variable for both
of these quantities.
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00:01:55,060 --> 00:01:59,950
So step one was the
coordinate system.
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00:02:04,270 --> 00:02:13,381
Step two, was the
identification of dm.
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00:02:13,381 --> 00:02:15,500
And step three and
I think this is
44
00:02:15,500 --> 00:02:19,610
absolutely the crucial one is
to introduce the integration
45
00:02:19,610 --> 00:02:21,420
variable.
46
00:02:21,420 --> 00:02:25,370
Now you'll see that will
come in two different cases.
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00:02:25,370 --> 00:02:28,445
So this is the quantity,
the distance from dm
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00:02:28,445 --> 00:02:30,000
to the origin that's changing.
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00:02:30,000 --> 00:02:33,740
You can see for each of these
little elements, that changes.
50
00:02:33,740 --> 00:02:38,260
So what I'll write [INAUDIBLE]
as a vector is x prime,
51
00:02:38,260 --> 00:02:40,690
which will be my integration
variable in the i hat
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00:02:40,690 --> 00:02:41,900
direction.
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00:02:41,900 --> 00:02:44,250
So the integration
variables x prime.
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00:02:44,250 --> 00:02:47,410
That's the first place that
I introduced the integration
55
00:02:47,410 --> 00:02:48,150
variable.
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00:02:48,150 --> 00:02:52,065
And x prime, you
can see, will vary.
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00:02:54,850 --> 00:03:00,020
And it varies from x prime
equals 0 to x prime equals L.
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00:03:00,020 --> 00:03:03,760
And that will show up in terms
of the limits of my integral.
59
00:03:03,760 --> 00:03:08,270
Now the second place that the
integration variable comes in
60
00:03:08,270 --> 00:03:09,320
is dm.
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00:03:09,320 --> 00:03:13,700
I want to express
in terms of x prime,
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00:03:13,700 --> 00:03:16,410
which is a measure of
where this object is.
63
00:03:16,410 --> 00:03:20,579
And that's how, if we
choose this length here
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00:03:20,579 --> 00:03:25,430
to be dx prime, notice in terms
of the integration variable,
65
00:03:25,430 --> 00:03:30,600
then I have a relationship
between and dm and dx prime.
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00:03:30,600 --> 00:03:33,150
dm is mass in this
little element.
67
00:03:33,150 --> 00:03:35,350
dx prime is the
length of the element.
68
00:03:35,350 --> 00:03:40,890
And if the whole object is a
uniform rod with a mass capital
69
00:03:40,890 --> 00:03:48,280
M and a length L, then its just
given by M over L dx prime.
70
00:03:48,280 --> 00:03:52,590
And this quantity M
over L is an example
71
00:03:52,590 --> 00:03:55,940
of a mass, linear
mass density, which we
72
00:03:55,940 --> 00:03:57,790
have a scale challenge about.
73
00:03:57,790 --> 00:04:01,530
So I have two places,
where my integration
74
00:04:01,530 --> 00:04:04,520
variable has been introduced.
75
00:04:04,520 --> 00:04:09,280
And now I can write up every
piece in this interval.
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00:04:09,280 --> 00:04:13,580
So let's now indicate
that we're integrating
77
00:04:13,580 --> 00:04:16,839
from x prime equals
0 to x prime equals
78
00:04:16,839 --> 00:04:27,390
L. dm is M over L dx prime, and
our vector is x prime i hat.
79
00:04:27,390 --> 00:04:31,730
And downstairs, it's
just M over L dx
80
00:04:31,730 --> 00:04:37,710
prime from x prime equals
0 to x prime equals L.
81
00:04:37,710 --> 00:04:40,700
And that's how I
set up the integral
82
00:04:40,700 --> 00:04:42,480
for the center of mass.
83
00:04:42,480 --> 00:04:46,490
Both of these integrals are
now not difficult to do.
84
00:04:46,490 --> 00:04:48,670
Notice, it's x prime dx prime.
85
00:04:48,670 --> 00:04:52,110
So this integral is
x squared over 2.
86
00:04:52,110 --> 00:04:58,370
And I get 1/2 M
over L, L squared.
87
00:04:58,370 --> 00:05:02,840
And downstairs, dx prime
from 0 to is just L.
88
00:05:02,840 --> 00:05:08,610
So the downstairs
integral is just M over L.
89
00:05:08,610 --> 00:05:12,380
And when you have
M over L's cancel,
90
00:05:12,380 --> 00:05:17,860
we just are left
with a-- this is--
91
00:05:17,860 --> 00:05:23,520
I'm sorry-- this is just M not
M over L, dimension incorrect.
92
00:05:23,520 --> 00:05:26,400
So we get for the
position of the center
93
00:05:26,400 --> 00:05:28,080
of mass, the M's cancel.
94
00:05:28,080 --> 00:05:29,660
One of the L's cancel.
95
00:05:29,660 --> 00:05:33,040
And we have an i hat
in this expression
96
00:05:33,040 --> 00:05:38,460
so our answer is r equals
L over 2 I hat, which
97
00:05:38,460 --> 00:05:40,360
is exactly what we expected.
98
00:05:40,360 --> 00:05:46,508
We expected the center of mass
to be half way down the rod.