1 00:00:03,520 --> 00:00:05,890 I'd like to now calculate the angular momentum of a two 2 00:00:05,890 --> 00:00:07,220 particle system. 3 00:00:07,220 --> 00:00:08,380 So here's one particle. 4 00:00:08,380 --> 00:00:09,730 Call that 1. 5 00:00:09,730 --> 00:00:11,300 And here's the other particle 2. 6 00:00:11,300 --> 00:00:15,730 And particle 1 has a momentum in the downward direction. 7 00:00:15,730 --> 00:00:19,430 And particle 2 has a momentum in the upper direction. 8 00:00:19,430 --> 00:00:23,410 Now, in this case, I want to make some of the momentum-- 9 00:00:23,410 --> 00:00:27,070 so this momentum of the system is 0. 10 00:00:27,070 --> 00:00:29,560 And now I want to calculate the angular momentum 11 00:00:29,560 --> 00:00:31,370 about two different points. 12 00:00:31,370 --> 00:00:34,150 So let's just draw an axis here. 13 00:00:34,150 --> 00:00:39,220 And let's choose one point A. And we'll call this distance r, 14 00:00:39,220 --> 00:00:40,540 and this distance r. 15 00:00:40,540 --> 00:00:42,460 And let's choose another point B, 16 00:00:42,460 --> 00:00:44,650 and we'll call that distance d. 17 00:00:44,650 --> 00:00:48,380 Now, in order to calculate angular momentum of the system, 18 00:00:48,380 --> 00:00:51,820 I need to draw my vectors. 19 00:00:51,820 --> 00:00:54,010 Let's start with the angular momentum 20 00:00:54,010 --> 00:00:59,140 about the point A. That's going to be a vector from A to object 21 00:00:59,140 --> 00:01:06,070 1, cross the momentum of 1, plus the vector from A to object 2, 22 00:01:06,070 --> 00:01:08,510 cross the momentum of 2. 23 00:01:08,510 --> 00:01:11,440 Now what's crucial here, is at this point, 24 00:01:11,440 --> 00:01:13,640 things are a little bit of a abstract, 25 00:01:13,640 --> 00:01:17,710 but we want to draw these vectors, rA1 and rA2. 26 00:01:17,710 --> 00:01:21,340 So we're now constructing our angular momentum diagrams. 27 00:01:21,340 --> 00:01:26,440 So there's rA1, and there rA2. 28 00:01:26,440 --> 00:01:29,060 but again, in order to calculate this, 29 00:01:29,060 --> 00:01:31,510 we could do it geometrically with right hand rules, 30 00:01:31,510 --> 00:01:34,280 but I now want to introduce unit vectors, 31 00:01:34,280 --> 00:01:37,600 so I can do vector decomposition at every point. 32 00:01:37,600 --> 00:01:42,460 So I'm going to choose a unit vector, i-hat, j-hat, 33 00:01:42,460 --> 00:01:44,550 and k-hat. 34 00:01:44,550 --> 00:01:47,800 And now I can decompose all of these vectors 35 00:01:47,800 --> 00:01:49,420 in terms of my unit vectors. 36 00:01:49,420 --> 00:01:52,250 This is the angular momentum of the system. 37 00:01:52,250 --> 00:01:55,020 So now it's not hard at all to write these out. 38 00:01:55,020 --> 00:02:01,620 rA1 is minus r in the i-hat direction. 39 00:02:01,620 --> 00:02:04,030 Cross P1 is down. 40 00:02:04,030 --> 00:02:07,540 That's minus P1 in the j-hat direction, 41 00:02:07,540 --> 00:02:09,820 where P1 is the magnitude . 42 00:02:09,820 --> 00:02:13,370 And rA2 is in the plus i-hat direction. 43 00:02:13,370 --> 00:02:16,090 So that's r i-hat. 44 00:02:16,090 --> 00:02:19,480 And I'm crossing that with P2, which 45 00:02:19,480 --> 00:02:24,650 I'll write right now as P2 in the j-hat direction. 46 00:02:24,650 --> 00:02:32,650 Now remember, P2 has the same magnitude as P1, 47 00:02:32,650 --> 00:02:35,930 but they're pointing in opposite directions. 48 00:02:35,930 --> 00:02:37,780 So when I take the cross project, 49 00:02:37,780 --> 00:02:41,930 i-hat cross j-hat is k-hat minus sign minus sign. 50 00:02:41,930 --> 00:02:44,650 So I get rP1 k-hat. 51 00:02:44,650 --> 00:02:49,510 And now over here, P2 is equal to P1. 52 00:02:49,510 --> 00:02:51,820 i-hat cross j-hat is k-hat. 53 00:02:51,820 --> 00:02:56,320 And I get another rP1 k-hat. 54 00:02:56,320 --> 00:02:59,890 And so I have 2rP1 k-hat. 55 00:02:59,890 --> 00:03:04,300 And that's the angular momentum of the system about point A. 56 00:03:04,300 --> 00:03:08,560 Now I'd like to calculate the angular momentum about B So 57 00:03:08,560 --> 00:03:13,610 let's draw the same diagram, 1, 2. 58 00:03:13,610 --> 00:03:16,300 Here is P2. 59 00:03:16,300 --> 00:03:17,980 Here's P1. 60 00:03:17,980 --> 00:03:22,870 Here is B. And now I'll draw my vectors. 61 00:03:22,870 --> 00:03:30,610 This is rB2 and this is rB1. 62 00:03:30,610 --> 00:03:31,840 This distance was 2r. 63 00:03:31,840 --> 00:03:34,120 This distance was d. 64 00:03:34,120 --> 00:03:36,800 I'm going to use the same unit vectors. 65 00:03:36,800 --> 00:03:41,140 And so I get l for the system about B. Again, 66 00:03:41,140 --> 00:03:42,910 I'll just write everything out. 67 00:03:42,910 --> 00:03:48,310 P1 plus rB2 cross P2. 68 00:03:48,310 --> 00:03:52,910 So I'm going to write P2 as, in this case, 69 00:03:52,910 --> 00:03:54,910 it's equal to minus P1. 70 00:03:54,910 --> 00:03:57,790 And magnitudes P2 equals P1. 71 00:03:57,790 --> 00:04:00,160 In magnitude directions are opposite. 72 00:04:00,160 --> 00:04:09,220 So rB1 is equal 2r plus D i-hat, and it's pointing 73 00:04:09,220 --> 00:04:12,440 in the minus i-hat direction. 74 00:04:12,440 --> 00:04:17,839 Cross P1 is in the minus j-hat direction. 75 00:04:17,839 --> 00:04:22,010 rB2 is also in the minus i-hat direction, 76 00:04:22,010 --> 00:04:24,760 so that's minus D i-hat. 77 00:04:24,760 --> 00:04:31,600 And P2 is in the plus j direction, so that is plus-- 78 00:04:31,600 --> 00:04:33,670 the magnitudes are the same-- 79 00:04:33,670 --> 00:04:35,420 P1 j-hat. 80 00:04:35,420 --> 00:04:39,460 We don't need the plus sign there. 81 00:04:39,460 --> 00:04:42,280 And now I take the cross products. 82 00:04:42,280 --> 00:04:45,790 i-hat cross j-hat is k-hat. 83 00:04:45,790 --> 00:04:47,950 A minus sign minus sign. 84 00:04:47,950 --> 00:04:48,730 That makes plus. 85 00:04:48,730 --> 00:04:52,990 So I have 2r plus D P1 k-hat. 86 00:04:52,990 --> 00:04:57,720 Now notice here I have i-hat cross j-hat, which is k-hat, 87 00:04:57,720 --> 00:04:59,510 but there's a minus sign. 88 00:04:59,510 --> 00:05:03,570 So I have minus D P1 k-hat. 89 00:05:03,570 --> 00:05:07,640 Here I have plus D P1 k-hat minus D1 P k-hat. 90 00:05:07,640 --> 00:05:11,430 And so I get 2rP1 k-hat. 91 00:05:11,430 --> 00:05:20,550 And I have the same result. L system A equals L system B. 92 00:05:20,550 --> 00:05:23,860 Now that's not a coincidence in this problem. 93 00:05:23,860 --> 00:05:26,890 And the reason is that whenever-- 94 00:05:26,890 --> 00:05:28,980 so we can say it this way. 95 00:05:28,980 --> 00:05:35,460 Whenever the momentum of the system is 0, 96 00:05:35,460 --> 00:05:50,260 then LA is independent of the choice of the point A. 97 00:05:50,260 --> 00:05:52,750 So coming back to our example, no matter where 98 00:05:52,750 --> 00:05:55,540 I picked our points A and B-- 99 00:05:55,540 --> 00:05:57,470 any point I could pick, anywhere I want-- 100 00:05:57,470 --> 00:05:59,050 I could pick a point C up there-- 101 00:05:59,050 --> 00:06:01,180 I make this cross product calculation. 102 00:06:01,180 --> 00:06:04,770 I would get exactly the same answer, 2rP1k.