1 00:00:03,640 --> 00:00:05,170 So now we want to find the center 2 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 5 00:00:15,730 --> 00:00:19,250 divided by an interval. 6 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, 8 00:00:24,210 --> 00:00:26,140 to real physical objects. 9 00:00:26,140 --> 00:00:29,060 And the key, as always, is choosing a coordinate system. 10 00:00:29,060 --> 00:00:31,250 So now I'll draw the object, again. 11 00:00:31,250 --> 00:00:34,230 And the first thing I'll do is choose an origin. 12 00:00:34,230 --> 00:00:36,534 I can pick my origin anywhere I want. 13 00:00:36,534 --> 00:00:37,700 I can pick it in the middle. 14 00:00:37,700 --> 00:00:38,825 I can put it in the middle. 15 00:00:38,825 --> 00:00:39,950 I could put it at this end. 16 00:00:39,950 --> 00:00:41,050 I could put it that end. 17 00:00:41,050 --> 00:00:44,990 I could put it down here, but I'll choose it over here. 18 00:00:44,990 --> 00:00:46,890 Because the object is linear, this 19 00:00:46,890 --> 00:00:49,450 is a very Cartesian system. 20 00:00:49,450 --> 00:00:51,580 I'm only doing a one dimensional object. 21 00:00:51,580 --> 00:00:54,890 So I choose my coordinate system plus x. 22 00:00:54,890 --> 00:00:57,380 That's step one. 23 00:00:57,380 --> 00:01:01,150 Now my origin-- now here comes the crucial thing. 24 00:01:01,150 --> 00:01:06,150 In this argument, dm is the infinitesimal mass element. 25 00:01:06,150 --> 00:01:09,970 And I want to pick that at an arbitrary place in the object. 26 00:01:09,970 --> 00:01:11,760 I don't want to pick it at the origin. 27 00:01:11,760 --> 00:01:13,800 I don't want to pick it at the end. 28 00:01:13,800 --> 00:01:16,660 Note down here this is x equals L. 29 00:01:16,660 --> 00:01:20,660 So I'll arbitrarily pick an infinitesimal element. 30 00:01:20,660 --> 00:01:24,289 I'll shade it in dm. 31 00:01:24,289 --> 00:01:26,640 That represents-- this is what I'm 32 00:01:26,640 --> 00:01:30,050 going to make my summation over when I do my integral. 33 00:01:30,050 --> 00:01:33,440 I'm going to add up all these dm's. 34 00:01:33,440 --> 00:01:36,300 And the point is that the dm's are different distances 35 00:01:36,300 --> 00:01:37,630 from the origin. 36 00:01:37,630 --> 00:01:40,120 So the vector-- and here's the next step-- 37 00:01:40,120 --> 00:01:44,200 is I draw a picture of my vector rdm. 38 00:01:44,200 --> 00:01:48,630 So now I have these terms, at least, explained in my diagram. 39 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. 41 00:01:55,060 --> 00:01:59,950 So step one was the coordinate system. 42 00:02:04,270 --> 00:02:13,381 Step two, was the identification of dm. 43 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. 47 00:02:25,370 --> 00:02:28,445 So this is the quantity, the distance from dm 48 00:02:28,445 --> 00:02:30,000 to the origin that's changing. 49 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 52 00:02:40,690 --> 00:02:41,900 direction. 53 00:02:41,900 --> 00:02:44,250 So the integration variables x prime. 54 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. 56 00:02:48,150 --> 00:02:52,065 And x prime, you can see, will vary. 57 00:02:54,850 --> 00:03:00,020 And it varies from x prime equals 0 to x prime equals L. 58 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. 61 00:03:09,320 --> 00:03:13,700 I want to express in terms of x prime, 62 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 64 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. 66 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. 76 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.