1 00:00:00,080 --> 00:00:02,430 The following content is provided under a Creative 2 00:00:02,430 --> 00:00:03,810 Commons license. 3 00:00:03,810 --> 00:00:06,050 Your support will help MIT OpenCourseWare 4 00:00:06,050 --> 00:00:10,140 continue to offer high quality educational resources for free. 5 00:00:10,140 --> 00:00:12,690 To make a donation or to view additional materials 6 00:00:12,690 --> 00:00:16,600 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,600 --> 00:00:17,258 at ocw.mit.edu. 8 00:00:25,830 --> 00:00:29,820 PROFESSOR: So today we're going to do two things in particular. 9 00:00:29,820 --> 00:00:33,540 One is finish off with the discussion 10 00:00:33,540 --> 00:00:36,830 of this device, a shaker. 11 00:00:36,830 --> 00:00:39,640 This, by the way, this is a commercial thing. 12 00:00:39,640 --> 00:00:44,120 And out of the catalog, this is the littlest one. 13 00:00:44,120 --> 00:00:45,770 This is a 50 pound shaker. 14 00:00:45,770 --> 00:00:48,900 At full speed it actually puts out 50 pounds. 15 00:00:48,900 --> 00:00:51,902 All it is is masses inside going around and around. 16 00:00:51,902 --> 00:00:53,636 AUDIENCE: What is its commercial purpose? 17 00:00:53,636 --> 00:00:55,510 PROFESSOR: Ah, what's its commercial purpose? 18 00:00:55,510 --> 00:01:03,410 Well, the big ones that are maybe 100 pounds of moving mass 19 00:01:03,410 --> 00:01:07,650 are they bolt them to the floor in nuclear power plants 20 00:01:07,650 --> 00:01:09,390 and test them. 21 00:01:09,390 --> 00:01:13,745 Shake the buildings to represent earthquake kind 22 00:01:13,745 --> 00:01:15,770 of loads and things like that. 23 00:01:15,770 --> 00:01:18,690 And the smaller ones, you can buy these for. 24 00:01:18,690 --> 00:01:20,860 This kind is actually if you're running 25 00:01:20,860 --> 00:01:23,330 an operation like in a flour mill 26 00:01:23,330 --> 00:01:26,600 and you've got particulate stuff trying to get 27 00:01:26,600 --> 00:01:28,310 it to slide through chutes. 28 00:01:28,310 --> 00:01:32,376 Does stuff slide down chutes easier if the things 29 00:01:32,376 --> 00:01:33,500 are vibrating a little bit? 30 00:01:33,500 --> 00:01:34,600 Have you ever banged on something 31 00:01:34,600 --> 00:01:35,870 to get stuff to come loose? 32 00:01:35,870 --> 00:01:38,660 You can just stick one of these on the side and let it run. 33 00:01:38,660 --> 00:01:39,620 Nothing sticks. 34 00:01:39,620 --> 00:01:43,150 So there's lots and lots of purposes for shakers like this. 35 00:01:43,150 --> 00:01:48,340 So we were in the process of analyzing how one of these 36 00:01:48,340 --> 00:01:48,840 works. 37 00:01:55,020 --> 00:01:57,300 And I want to finish that. 38 00:01:57,300 --> 00:02:00,670 And then part two today is we're going to-- we've only 39 00:02:00,670 --> 00:02:03,760 really talked about angular momentum with respect 40 00:02:03,760 --> 00:02:06,010 to particles, individual particles. 41 00:02:06,010 --> 00:02:09,240 And even in your physics classes you 42 00:02:09,240 --> 00:02:11,352 did things with mass moment of inertia. 43 00:02:11,352 --> 00:02:13,060 And so we're going to make the connection 44 00:02:13,060 --> 00:02:17,260 today between particles, mass moment of inertia, 45 00:02:17,260 --> 00:02:18,520 unbalanced shakers. 46 00:02:18,520 --> 00:02:24,180 It all comes together in the second part of today's lecture. 47 00:02:24,180 --> 00:02:29,230 So the problem we are analyzing, literally that little shaker, 48 00:02:29,230 --> 00:02:31,100 can be modeled. 49 00:02:31,100 --> 00:02:35,480 Well, in that one particular application 50 00:02:35,480 --> 00:02:37,365 we can find that thing on rollers. 51 00:02:40,850 --> 00:02:43,280 This is the problem we are discussing. 52 00:02:43,280 --> 00:02:48,920 It has inside of it an unbalanced rotating mass 53 00:02:48,920 --> 00:02:51,110 with an arm that's E long. 54 00:02:51,110 --> 00:02:54,380 It's called the eccentricity in the trade. 55 00:02:54,380 --> 00:02:57,580 And it has some mass m. 56 00:02:57,580 --> 00:03:00,440 And this body that it's in, we'll call it mass of the body, 57 00:03:00,440 --> 00:03:02,870 mb. 58 00:03:02,870 --> 00:03:05,270 And this thing's going around and round. 59 00:03:05,270 --> 00:03:10,050 So this is some angle theta which is described as omega t. 60 00:03:10,050 --> 00:03:13,500 And they're constant rotation rate devices. 61 00:03:13,500 --> 00:03:17,535 So theta dot equals omega and that's a constant. 62 00:03:20,697 --> 00:03:22,280 That's how they're basically designed. 63 00:03:24,790 --> 00:03:27,660 And we label this point A. Over here we 64 00:03:27,660 --> 00:03:34,820 have a inertial coordinate system, xy. 65 00:03:34,820 --> 00:03:38,493 This point a, this point we've called b in our analysis. 66 00:03:42,580 --> 00:03:45,740 And we set out to find the equation 67 00:03:45,740 --> 00:03:48,180 of motion of this thing in the x direction. 68 00:03:48,180 --> 00:03:49,905 Has no movement in the y. 69 00:03:49,905 --> 00:03:51,860 It's confined in the y. 70 00:03:51,860 --> 00:03:54,090 It puts out lots of force in the y direction. 71 00:03:54,090 --> 00:03:56,940 You really have to restrain it to keep it from moving. 72 00:03:56,940 --> 00:03:59,200 But it doesn't move in the y direction, 73 00:03:59,200 --> 00:04:03,426 but it will move in the x. 74 00:04:03,426 --> 00:04:03,925 OK. 75 00:04:09,390 --> 00:04:11,140 So we came to the conclusion that we 76 00:04:11,140 --> 00:04:14,690 could write for the main body the summation 77 00:04:14,690 --> 00:04:20,269 of the external forces on mb. 78 00:04:20,269 --> 00:04:25,062 It's mb times its acceleration. 79 00:04:25,062 --> 00:04:26,520 And it's acceleration is completely 80 00:04:26,520 --> 00:04:27,985 defined by this coordinate. 81 00:04:35,502 --> 00:04:48,660 And if we draw a free body diagram of this mass, 82 00:04:48,660 --> 00:04:55,630 you're going to have a normal force I'll call n 83 00:04:55,630 --> 00:04:58,280 in the y direction upwards. 84 00:04:58,280 --> 00:05:04,275 You're going to have its weight downwards. 85 00:05:07,020 --> 00:05:11,540 And you're going to have some force exerted 86 00:05:11,540 --> 00:05:16,720 on it through this shaft that comes from the little mass. 87 00:05:16,720 --> 00:05:19,750 So we're accounting for everything the little mass, 88 00:05:19,750 --> 00:05:22,700 all its influence on this big block 89 00:05:22,700 --> 00:05:25,670 by the forces that are passed through that rod, which 90 00:05:25,670 --> 00:05:27,574 is hinged at the center. 91 00:05:27,574 --> 00:05:28,870 OK. 92 00:05:28,870 --> 00:05:34,750 And I'm going to call that f mb. 93 00:05:34,750 --> 00:05:37,640 OK, the force from that little link. 94 00:05:37,640 --> 00:05:44,190 Now, that happens to be equal to minus the forces on the mass 95 00:05:44,190 --> 00:05:45,465 that this rod exerts. 96 00:05:45,465 --> 00:05:46,965 It must exert some force on the mass 97 00:05:46,965 --> 00:05:49,630 to make it go around and around. 98 00:05:49,630 --> 00:05:51,960 And because of Newton's third law, 99 00:05:51,960 --> 00:05:54,280 those two forces have to be equal and opposite 100 00:05:54,280 --> 00:05:56,730 because they're operating on the same massless 101 00:05:56,730 --> 00:06:01,345 shaft for the purpose of this example. 102 00:06:04,670 --> 00:06:11,030 OK, in order find that equation of motion, 103 00:06:11,030 --> 00:06:12,810 the sum of the external forces, these 104 00:06:12,810 --> 00:06:14,420 are both in the y direction. 105 00:06:14,420 --> 00:06:17,970 So we just need to find the horizontal component 106 00:06:17,970 --> 00:06:21,240 of this force and we'll be able to complete that equation. 107 00:06:21,240 --> 00:06:22,665 So the point of the exercise here 108 00:06:22,665 --> 00:06:26,030 is just is to find this horizontal component. 109 00:06:26,030 --> 00:06:33,030 So to do that, let's move on to thinking about the little mass, 110 00:06:33,030 --> 00:06:37,070 small mass, and what its free body diagram looks like. 111 00:06:41,690 --> 00:06:46,380 So viewed from the here's our rod. 112 00:06:46,380 --> 00:06:47,540 Here's the small mass. 113 00:06:47,540 --> 00:06:48,690 This is a side view. 114 00:06:48,690 --> 00:06:51,350 So there's the point it's rotating about A. 115 00:06:51,350 --> 00:06:56,140 But I want to draw a free body diagram of this rod. 116 00:06:56,140 --> 00:06:58,950 The rod puts a force on this mass, 117 00:06:58,950 --> 00:07:03,090 which will have a vertical component, fm. 118 00:07:03,090 --> 00:07:05,190 And I'll just call it y. 119 00:07:05,190 --> 00:07:11,680 And it'll put a force that's in horizontal component fm x. 120 00:07:11,680 --> 00:07:15,760 And I'm drawing them both positive, 121 00:07:15,760 --> 00:07:17,709 because I don't know which direction they act. 122 00:07:17,709 --> 00:07:19,500 And if the answer turns out to be positive, 123 00:07:19,500 --> 00:07:20,360 then I guessed right. 124 00:07:20,360 --> 00:07:22,526 If it's negative, it means it's going the other way. 125 00:07:24,870 --> 00:07:27,440 And what other forces that are acting on this? 126 00:07:27,440 --> 00:07:31,720 Well, there is certainly is an mg downwards 127 00:07:31,720 --> 00:07:32,750 acting on that mass. 128 00:07:36,690 --> 00:07:37,190 OK. 129 00:07:37,190 --> 00:07:42,090 And there's no forces in and out of the page on it. 130 00:07:42,090 --> 00:07:43,720 And this is operating in the plane. 131 00:07:43,720 --> 00:07:47,900 So this is a planar motion problem. 132 00:07:47,900 --> 00:07:56,549 And we note that in here r dot equals r double dot equals 0. 133 00:07:56,549 --> 00:07:58,340 This thing doesn't change in length at all. 134 00:07:58,340 --> 00:08:02,350 It's just going round and round fixed length. 135 00:08:02,350 --> 00:08:06,290 So we can write, then, that the summation 136 00:08:06,290 --> 00:08:10,572 of the external forces on this little mass 137 00:08:10,572 --> 00:08:17,800 had better equal its mass times the acceleration of point A 138 00:08:17,800 --> 00:08:20,854 with respect to the inertial frame. 139 00:08:20,854 --> 00:08:22,440 And whoops. 140 00:08:22,440 --> 00:08:25,020 Not A, but what? 141 00:08:30,660 --> 00:08:34,370 B. The acceleration of this point. 142 00:08:34,370 --> 00:08:35,280 This is B. 143 00:08:35,280 --> 00:08:38,360 We need to figure out what the acceleration of that point 144 00:08:38,360 --> 00:08:40,900 is in the inertial frame. 145 00:08:40,900 --> 00:08:42,730 But we've done enough of these problems, 146 00:08:42,730 --> 00:08:44,290 so this should be pretty easy. 147 00:08:44,290 --> 00:08:48,430 This is the mass times the acceleration 148 00:08:48,430 --> 00:08:51,930 of point A with respect to O plus the mass 149 00:08:51,930 --> 00:08:59,020 times the acceleration of B with respect to A. B and A. 150 00:08:59,020 --> 00:09:05,006 These are all vectors until I break them down 151 00:09:05,006 --> 00:09:06,255 into their x and y components. 152 00:09:15,200 --> 00:09:17,140 So what's the acceleration of A with respect 153 00:09:17,140 --> 00:09:22,190 to O in the coordinate systems that we have written here? 154 00:09:29,569 --> 00:09:31,610 So that's just kind of our generic representation 155 00:09:31,610 --> 00:09:34,140 of acceleration, right? 156 00:09:34,140 --> 00:09:38,592 But we've chosen some coordinates here. 157 00:09:38,592 --> 00:09:40,050 Specifically have a coordinate that 158 00:09:40,050 --> 00:09:42,300 describes the motion of the main mass, right? 159 00:09:42,300 --> 00:09:43,950 What is that? 160 00:09:43,950 --> 00:09:46,620 So what's the acceleration of point A? 161 00:09:46,620 --> 00:09:47,770 x double dot. 162 00:09:47,770 --> 00:09:57,060 So we know that this then is m x double dot and plus. 163 00:09:57,060 --> 00:10:01,190 Now, it's easiest to describe this in terms 164 00:10:01,190 --> 00:10:04,030 of cylindrical coordinates. 165 00:10:04,030 --> 00:10:06,600 And we can then write that, well, then this 166 00:10:06,600 --> 00:10:18,600 must be a mass times the terms in the r hat direction. 167 00:10:26,105 --> 00:10:28,030 r hat. 168 00:10:28,030 --> 00:10:34,840 And then terms over here in the theta hat direction. 169 00:10:34,840 --> 00:10:39,830 Theta double dot plus 2r dot theta dot. 170 00:10:42,800 --> 00:10:45,440 Now, which of these are 0? 171 00:10:45,440 --> 00:10:48,645 Does that arm change length? 172 00:10:48,645 --> 00:10:49,145 No. 173 00:10:49,145 --> 00:10:52,710 So this is 0. 174 00:10:52,710 --> 00:10:56,700 Is the angular acceleration constant? 175 00:10:56,700 --> 00:10:58,710 So this is 0. 176 00:10:58,710 --> 00:11:00,310 The arm doesn't change length. 177 00:11:00,310 --> 00:11:01,300 The Coriolis is 0. 178 00:11:01,300 --> 00:11:04,370 So there's no Coriolis force, no [INAUDIBLE] force, 179 00:11:04,370 --> 00:11:07,480 no radial acceleration, only a single term. 180 00:11:07,480 --> 00:11:09,550 Just the centrifugal. 181 00:11:09,550 --> 00:11:15,592 So this becomes a pretty simple expression. 182 00:11:22,770 --> 00:11:27,250 So the summation of the external forces on our little mass, 183 00:11:27,250 --> 00:11:35,092 then, we can write as mx double dot in the i hat direction. 184 00:11:35,092 --> 00:11:37,710 I'm going to break it into its vector components here. 185 00:11:40,810 --> 00:11:43,570 Minus m. 186 00:11:43,570 --> 00:11:47,210 And I know that r here equals e. 187 00:11:47,210 --> 00:11:48,630 That's the eccentricity. 188 00:11:48,630 --> 00:11:50,630 I'm going to start using these terms. 189 00:11:50,630 --> 00:11:54,380 Minus me omega squared. 190 00:11:59,140 --> 00:12:02,460 r hat. 191 00:12:02,460 --> 00:12:06,450 But I'm going to break that r. 192 00:12:06,450 --> 00:12:07,680 Goes round and round. 193 00:12:07,680 --> 00:12:09,970 I need to break it into x and y components, 194 00:12:09,970 --> 00:12:13,060 but we've done that many times before. 195 00:12:13,060 --> 00:12:18,340 That looks like a cosine omega t in the i hat 196 00:12:18,340 --> 00:12:25,484 direction plus a sine omega t in the j hat direction. 197 00:12:25,484 --> 00:12:27,150 So as this thing goes around and around, 198 00:12:27,150 --> 00:12:29,699 it has a cosine term and a sine term. 199 00:12:29,699 --> 00:12:31,740 And this is in the x direction, this is in the y. 200 00:12:36,840 --> 00:12:42,150 So we're really interested in the equation of motion 201 00:12:42,150 --> 00:12:47,110 on small mass m in the x direction. 202 00:12:47,110 --> 00:12:49,900 So we just need to pull out the x components from this. 203 00:12:49,900 --> 00:13:03,420 So we have an m x double dot i hat minus me 204 00:13:03,420 --> 00:13:08,390 omega squared cosine omega t i hat. 205 00:13:08,390 --> 00:13:10,990 And we can drop the i hats now because we just have 206 00:13:10,990 --> 00:13:12,400 one single component equation. 207 00:13:15,780 --> 00:13:19,540 And this is this quantity I called fm. 208 00:13:24,080 --> 00:13:29,090 And fm x then is the x term in my little free body diagram. 209 00:13:29,090 --> 00:13:31,810 And the force that it exerts on the main mass 210 00:13:31,810 --> 00:13:33,640 is in the x direction. 211 00:13:39,020 --> 00:13:40,950 So this is the force that the rod 212 00:13:40,950 --> 00:13:43,680 places on that little mass in the x direction. 213 00:13:43,680 --> 00:13:45,480 What's the force that the rod places 214 00:13:45,480 --> 00:13:48,420 on the big mass in the x direction? 215 00:13:48,420 --> 00:13:49,580 Minus that. 216 00:13:49,580 --> 00:13:57,750 So this is minus f mb in the x direction. 217 00:13:57,750 --> 00:13:59,060 That's what we're after. 218 00:13:59,060 --> 00:14:03,890 We need that force so we can go back now and we'll finish out 219 00:14:03,890 --> 00:14:05,310 are the equation of motion that we 220 00:14:05,310 --> 00:14:09,080 were after for the main mass. 221 00:14:09,080 --> 00:14:09,670 This up here. 222 00:14:09,670 --> 00:14:12,990 We need to sum the external forces to get that, 223 00:14:12,990 --> 00:14:14,820 to fill out that expression. 224 00:14:14,820 --> 00:14:17,470 But while I'm here, just to have it, 225 00:14:17,470 --> 00:14:22,446 the summation of the forces on the small mass in the y 226 00:14:22,446 --> 00:14:22,945 direction. 227 00:14:25,470 --> 00:14:27,250 Look at our free body diagram. 228 00:14:27,250 --> 00:14:29,780 It has a minus mg. 229 00:14:32,510 --> 00:14:43,520 And it then has this term, minus me omega squared sine omega t. 230 00:14:43,520 --> 00:14:46,320 So just for completeness, we have also the y component 231 00:14:46,320 --> 00:14:50,020 of the force that the rod places on the small mass. 232 00:14:50,020 --> 00:14:53,422 And minus this amount is what it places on the main mass 233 00:14:53,422 --> 00:14:54,380 that it's connected to. 234 00:15:00,790 --> 00:15:03,540 So now let's go back to our equation up here, 235 00:15:03,540 --> 00:15:08,650 the summation of the forces on the main body. 236 00:15:13,410 --> 00:15:16,425 In the x direction. 237 00:15:16,425 --> 00:15:19,170 This is going to be the main body. 238 00:15:19,170 --> 00:15:21,080 x double dot. 239 00:15:21,080 --> 00:15:28,440 And it's now the x direction forces. 240 00:15:28,440 --> 00:15:32,830 There's only x component of this force. 241 00:15:32,830 --> 00:15:34,700 And that's what we have right here. 242 00:15:34,700 --> 00:15:35,835 It's minus that. 243 00:16:02,260 --> 00:16:06,010 And that's our equation of motion. 244 00:16:06,010 --> 00:16:09,520 We can rearrange it a little bit and it remarkably simplifies, 245 00:16:09,520 --> 00:16:10,540 actually. 246 00:16:10,540 --> 00:16:13,750 You end up, if you collect the motion terms involving 247 00:16:13,750 --> 00:16:32,510 x on the left hand side equals an external excitation 248 00:16:32,510 --> 00:16:34,160 on the right hand side. 249 00:16:34,160 --> 00:16:38,560 And I've been kind of following the commentaries in mb. 250 00:16:38,560 --> 00:16:40,290 Little confusion about some questions. 251 00:16:40,290 --> 00:16:43,400 When you're asked to find an equation of motion, 252 00:16:43,400 --> 00:16:46,680 is that the same thing as meaning 253 00:16:46,680 --> 00:16:47,965 solve the equation of motion? 254 00:16:47,965 --> 00:16:51,374 No, asking find the equation of motion means get this far. 255 00:16:51,374 --> 00:16:53,540 Now, if I wanted to know a solution for this, pretty 256 00:16:53,540 --> 00:16:56,540 trivial in this case, it's going to look like cosine omega t, 257 00:16:56,540 --> 00:17:01,210 but then I'd say solve that equation of motion. 258 00:17:01,210 --> 00:17:04,630 OK, now let's see. 259 00:17:04,630 --> 00:17:06,599 We also know-- let's just finish this-- 260 00:17:06,599 --> 00:17:11,869 that the summation of the forces on this main body in the y 261 00:17:11,869 --> 00:17:15,770 direction must be 0 because it can't move. 262 00:17:15,770 --> 00:17:16,479 No acceleration. 263 00:17:19,329 --> 00:17:22,660 And from the free body diagram for that, 264 00:17:22,660 --> 00:17:40,350 we can write that this is n and y minus mmbg 265 00:17:40,350 --> 00:17:55,400 minus mg from the little mass plus me 266 00:17:55,400 --> 00:18:00,302 omega squared sine omega t. 267 00:18:00,302 --> 00:18:02,700 That's the other phase of this. 268 00:18:02,700 --> 00:18:10,860 And the interesting thing here, then, is to solve for the force 269 00:18:10,860 --> 00:18:14,870 that it takes to hold this thing in place. 270 00:18:14,870 --> 00:18:25,288 So you get mb plus m times g plus or minus. 271 00:18:25,288 --> 00:18:26,252 All right, yeah. 272 00:18:36,297 --> 00:18:37,380 All right, so what's that? 273 00:18:37,380 --> 00:18:39,810 So just kind of step back and look at these things 274 00:18:39,810 --> 00:18:41,800 and say what's it telling us. 275 00:18:41,800 --> 00:18:43,830 So first of all, just to keep this thing 276 00:18:43,830 --> 00:18:49,040 from moving up and down, there's a force on it 277 00:18:49,040 --> 00:18:50,740 that has to support its weight. 278 00:18:50,740 --> 00:18:52,540 And it's the combined weight of whatever's 279 00:18:52,540 --> 00:18:54,350 inside that container. 280 00:18:54,350 --> 00:18:57,340 The weight of the rotating mass and the weight of the object. 281 00:18:57,340 --> 00:19:00,320 They have to be supported by a normal force, which 282 00:19:00,320 --> 00:19:03,240 this is a constant term. 283 00:19:03,240 --> 00:19:05,220 Weight down, normal force up. 284 00:19:05,220 --> 00:19:10,700 And around that constant force is an oscillating force. 285 00:19:10,700 --> 00:19:13,222 me omega squared sine omega t. 286 00:19:13,222 --> 00:19:15,520 e omega squared you should recognize 287 00:19:15,520 --> 00:19:18,530 as a centripetal acceleration. 288 00:19:18,530 --> 00:19:21,180 Mass times acceleration to force. 289 00:19:21,180 --> 00:19:26,370 And because it goes round and round, when 290 00:19:26,370 --> 00:19:28,240 it's like this it's pulling up and when 291 00:19:28,240 --> 00:19:30,740 it's like this it's pulling down and when it's like this is, 292 00:19:30,740 --> 00:19:32,150 it's only going to the sides. 293 00:19:32,150 --> 00:19:35,850 So sine omega t for the vertical parts, cosine omega 294 00:19:35,850 --> 00:19:36,935 t for the horizontal. 295 00:19:41,680 --> 00:19:44,550 And that's actually all there is to the shake. 296 00:19:44,550 --> 00:19:47,610 That's all there is to the shakers. 297 00:19:47,610 --> 00:19:50,740 The rotating mass inside. 298 00:19:50,740 --> 00:19:55,330 Now, in the homework, from the second homework where 299 00:19:55,330 --> 00:20:00,090 you had this thing, this ball running around inside, 300 00:20:00,090 --> 00:20:01,640 where I posed the question in a way 301 00:20:01,640 --> 00:20:03,075 I didn't really quite intend. 302 00:20:07,510 --> 00:20:19,280 But I asked here's the track. 303 00:20:19,280 --> 00:20:23,790 And you had this roller going around inside. 304 00:20:23,790 --> 00:20:29,900 And I asked to find the normal force that the track exerts 305 00:20:29,900 --> 00:20:32,420 on the roller. 306 00:20:32,420 --> 00:20:34,830 So it's an unknown. 307 00:20:34,830 --> 00:20:46,380 And there must also be a tangential force on this thing. 308 00:20:46,380 --> 00:20:51,600 And there's also going to be this thing certainly has 309 00:20:51,600 --> 00:20:52,915 weight mg. 310 00:21:02,140 --> 00:21:05,460 And so that's the complete free body diagram. 311 00:21:05,460 --> 00:21:08,370 Now, let's if this is frictionless, which it won't 312 00:21:08,370 --> 00:21:11,190 be in reality, but for the purposes of analysis, 313 00:21:11,190 --> 00:21:14,100 let's say it's frictionless, it's only a normal force. 314 00:21:14,100 --> 00:21:16,800 Where does this tangential force come from? 315 00:21:16,800 --> 00:21:20,890 Why's it there in this problem? 316 00:21:20,890 --> 00:21:23,010 There's a key piece of information you're told, 317 00:21:23,010 --> 00:21:25,400 and that is that the angular acceleration of this thing 318 00:21:25,400 --> 00:21:26,090 is constant. 319 00:21:26,090 --> 00:21:28,005 It's constant speed going around. 320 00:21:28,005 --> 00:21:31,770 If you had a ball rolling around there at constant speed, 321 00:21:31,770 --> 00:21:33,610 would it go constant if you just pushed 322 00:21:33,610 --> 00:21:35,679 it and it started rolling? 323 00:21:35,679 --> 00:21:38,220 It would slowdown going up and it would speed up coming down. 324 00:21:38,220 --> 00:21:39,920 Why? 325 00:21:39,920 --> 00:21:41,330 AUDIENCE: [INAUDIBLE]. 326 00:21:41,330 --> 00:21:42,420 PROFESSOR: Gravity, right? 327 00:21:42,420 --> 00:21:45,490 So there must be something that has to overcome gravity 328 00:21:45,490 --> 00:21:48,140 going uphill and holding it back coming downhill. 329 00:21:48,140 --> 00:21:50,010 So the way these things actually work 330 00:21:50,010 --> 00:21:52,999 is they've got ports pushing compressed air in here. 331 00:21:52,999 --> 00:21:54,790 And this is driven around by compressed air 332 00:21:54,790 --> 00:21:56,844 and there's a pressure difference 333 00:21:56,844 --> 00:21:58,510 between this side and that side and that 334 00:21:58,510 --> 00:22:01,120 generates the necessary tangential force to make 335 00:22:01,120 --> 00:22:02,960 the thing go around and around. 336 00:22:02,960 --> 00:22:04,300 But they're really easy to make. 337 00:22:04,300 --> 00:22:05,980 You can imagine very few moving parts. 338 00:22:05,980 --> 00:22:08,130 Just hook up a compressed air hose to that 339 00:22:08,130 --> 00:22:10,139 and it's just pushing the ball around inside. 340 00:22:10,139 --> 00:22:11,180 You get the same outcome. 341 00:22:17,130 --> 00:22:20,280 On this ball, on this roller, if there 342 00:22:20,280 --> 00:22:29,190 is a-- the problem we just solved is we found fm 343 00:22:29,190 --> 00:22:34,980 in the y and fm in the x. 344 00:22:34,980 --> 00:22:38,400 And this problem said yeah, but why can't we get the same thing 345 00:22:38,400 --> 00:22:46,240 but have those coordinates be f normal and f tangential? 346 00:22:46,240 --> 00:22:49,270 And sure, that's just a coordinate rotation. 347 00:22:49,270 --> 00:22:51,960 So what can you say about these forces? 348 00:22:51,960 --> 00:22:59,000 Well, one thing you could say is fn squared plus ft squared 349 00:22:59,000 --> 00:23:06,230 had better be equal to fmx squared plus fmy squared, 350 00:23:06,230 --> 00:23:08,040 right? 351 00:23:08,040 --> 00:23:10,410 And then just like converting from polar 352 00:23:10,410 --> 00:23:15,000 to Cartesian coordinates, you can do these conversions. 353 00:23:15,000 --> 00:23:17,100 And you could find out, for example, 354 00:23:17,100 --> 00:23:28,600 that fn is-- keep my notation consistent 355 00:23:28,600 --> 00:23:42,820 here. fn will be fm in the x cosine omega t plus fm in the y 356 00:23:42,820 --> 00:23:45,320 sine omega t. 357 00:23:45,320 --> 00:23:46,620 And so there's the answer. 358 00:23:46,620 --> 00:23:49,056 This is what you're asked for in that problem set. 359 00:23:59,968 --> 00:24:02,470 OK. 360 00:24:02,470 --> 00:24:07,170 So all you need to know about shakers. 361 00:24:07,170 --> 00:24:10,170 If you're ever confronted with something like this, 362 00:24:10,170 --> 00:24:14,410 what's the magnitude of the force that the shaker puts out? 363 00:24:22,825 --> 00:24:25,049 AUDIENCE: [INAUDIBLE]. 364 00:24:25,049 --> 00:24:26,090 PROFESSOR: Little louder? 365 00:24:26,090 --> 00:24:27,780 AUDIENCE: Is it mr omega squared? 366 00:24:27,780 --> 00:24:30,540 PROFESSOR: mr omega squared, but substitute 367 00:24:30,540 --> 00:24:35,920 for r the actual eccentricity. 368 00:24:35,920 --> 00:24:38,960 It's whatever that mass in length 369 00:24:38,960 --> 00:24:40,960 out there that's spinning around. 370 00:24:40,960 --> 00:24:44,300 me omega squared is the magnitude of the force 371 00:24:44,300 --> 00:24:46,710 and it's going to oscillate up and down 372 00:24:46,710 --> 00:24:49,070 and it's going to have gravity that it adds to. 373 00:24:49,070 --> 00:24:51,910 But the important part is me omega squared 374 00:24:51,910 --> 00:24:54,480 is the magnitude of the force. 375 00:24:54,480 --> 00:24:58,730 OK, now we're going to move on to the next topic. 376 00:24:58,730 --> 00:25:03,210 The next topic is mass moments of inertia. 377 00:25:03,210 --> 00:25:05,360 And it has a strong connection to these. 378 00:25:05,360 --> 00:25:07,640 And I'm going to use this kind of analysis 379 00:25:07,640 --> 00:25:11,470 as the transition to talking about moments of inertia. 380 00:25:11,470 --> 00:25:14,170 Moments of inertia and products of inertia. 381 00:25:14,170 --> 00:25:17,830 So any final questions about this before we go on? 382 00:25:17,830 --> 00:25:18,901 Yeah? 383 00:25:18,901 --> 00:25:21,727 AUDIENCE: Your summation of [INAUDIBLE], 384 00:25:21,727 --> 00:25:24,082 why did you not include mg? 385 00:25:24,082 --> 00:25:25,530 PROFESSOR: Why didn't I include? 386 00:25:25,530 --> 00:25:26,480 AUDIENCE: Mg. 387 00:25:26,480 --> 00:25:28,600 PROFESSOR: Mg. 388 00:25:28,600 --> 00:25:29,890 in the. 389 00:25:29,890 --> 00:25:31,875 AUDIENCE: First summation. 390 00:25:31,875 --> 00:25:32,374 [INAUDIBLE] 391 00:25:35,080 --> 00:25:37,380 PROFESSOR: Oh. 392 00:25:37,380 --> 00:25:39,000 Yeah, you're right. 393 00:25:42,900 --> 00:25:44,335 And where's my free body diagram? 394 00:25:48,650 --> 00:25:50,150 Has it on it, right? 395 00:25:50,150 --> 00:25:52,660 Just didn't get it down into the-- and what direction's 396 00:25:52,660 --> 00:25:53,160 it in? 397 00:25:57,570 --> 00:25:59,265 Because then we did get it back in. 398 00:26:02,134 --> 00:26:04,370 AUDIENCE: [INAUDIBLE]. 399 00:26:04,370 --> 00:26:08,540 PROFESSOR: Back in the last line for the y component. 400 00:26:08,540 --> 00:26:11,334 Now, does it appear in this one, this equation at all? 401 00:26:11,334 --> 00:26:13,000 It has absolutely nothing to do with it. 402 00:26:13,000 --> 00:26:15,470 Gravity is in the j hat direction. 403 00:26:15,470 --> 00:26:18,460 This is a force equation in i hat. 404 00:26:18,460 --> 00:26:21,390 But it does appear in that normal tangential expression 405 00:26:21,390 --> 00:26:24,550 when you go look at the solution for that problem, 406 00:26:24,550 --> 00:26:29,390 because it has compounds in both of the i and j directions. 407 00:26:29,390 --> 00:26:30,960 And so it'll show up. 408 00:26:30,960 --> 00:26:34,730 Gravity will show up in this expression. 409 00:26:34,730 --> 00:26:37,030 Right through this term. 410 00:26:37,030 --> 00:26:38,925 Yeah? 411 00:26:38,925 --> 00:26:42,725 AUDIENCE: In the bottom equation on that middle board, 412 00:26:42,725 --> 00:26:48,340 you have my minus mbg minus mg plus. 413 00:26:48,340 --> 00:26:51,274 I don't understand where that last plus came from. 414 00:26:51,274 --> 00:26:53,719 Because in your equation on the left 415 00:26:53,719 --> 00:26:57,142 you're using the force of little f, correct? 416 00:26:57,142 --> 00:26:59,910 And you have two negatives there. 417 00:26:59,910 --> 00:27:01,510 PROFESSOR: This is 0. 418 00:27:01,510 --> 00:27:03,720 I left the n where it was and moved everything 419 00:27:03,720 --> 00:27:05,840 to the other side. 420 00:27:05,840 --> 00:27:08,330 So that plus becomes a minus. 421 00:27:08,330 --> 00:27:13,070 AUDIENCE: No, from the board to the left to the middle board. 422 00:27:13,070 --> 00:27:13,960 PROFESSOR: OK. 423 00:27:13,960 --> 00:27:15,421 AUDIENCE: So down. 424 00:27:15,421 --> 00:27:18,343 So you have the summation of the force 425 00:27:18,343 --> 00:27:25,585 on the little mass as negative mg minus m-- yep, that one. 426 00:27:25,585 --> 00:27:28,495 And from what I can understand, you just 427 00:27:28,495 --> 00:27:32,375 moved that force over to the large force, 428 00:27:32,375 --> 00:27:34,315 but you [INAUDIBLE], correct? 429 00:27:43,510 --> 00:27:47,000 PROFESSOR: It should be minus this thing. 430 00:27:47,000 --> 00:27:48,180 The summation here. 431 00:27:48,180 --> 00:27:52,340 This force is minus the little mass force. 432 00:27:52,340 --> 00:27:55,440 So that ought to become a plus and a plus, right? 433 00:27:55,440 --> 00:27:58,085 And so if I do that carefully. 434 00:28:19,155 --> 00:28:25,660 To this one is OK. 435 00:28:25,660 --> 00:28:28,815 But this one appears to have a sign problem, right? 436 00:28:32,110 --> 00:28:36,470 But these two terms have got to be the same. 437 00:28:36,470 --> 00:28:38,080 And so I've got a mistake somewhere. 438 00:28:38,080 --> 00:28:42,790 And rather than spend 10 minutes fixing it on the fly, 439 00:28:42,790 --> 00:28:44,060 I'll take note of that. 440 00:28:50,520 --> 00:28:51,770 This should be OK. 441 00:28:51,770 --> 00:28:53,736 AUDIENCE: Yeah, intuitively makes sense too, 442 00:28:53,736 --> 00:28:55,235 I just don't understand [INAUDIBLE]. 443 00:29:03,170 --> 00:29:05,680 PROFESSOR: Ah, wait a second. 444 00:29:08,540 --> 00:29:11,630 No, I'm not going to try to fix it right now. 445 00:29:11,630 --> 00:29:14,120 I made a slip in my notes somewhere. 446 00:29:14,120 --> 00:29:15,490 But I will repair that. 447 00:29:15,490 --> 00:29:16,466 Yeah? 448 00:29:16,466 --> 00:29:18,450 AUDIENCE: Why do we need mg at all? 449 00:29:18,450 --> 00:29:22,418 Because doesn't this force the angular acceleration 450 00:29:22,418 --> 00:29:23,410 is constant? 451 00:29:23,410 --> 00:29:26,386 Or the angular velocity is constant, right? 452 00:29:26,386 --> 00:29:28,122 So the centrifugal acceleration is 453 00:29:28,122 --> 00:29:32,834 going to be constant, which means that the part that's 454 00:29:32,834 --> 00:29:34,818 driven by the motor is going to be changing 455 00:29:34,818 --> 00:29:36,802 to account for gravity. 456 00:29:36,802 --> 00:29:39,778 So isn't gravity taking into account 457 00:29:39,778 --> 00:29:44,738 that we have a constant force or [INAUDIBLE]? 458 00:29:50,000 --> 00:29:53,410 PROFESSOR: Yeah, you're asking if gravity is not 459 00:29:53,410 --> 00:29:58,520 taken into account somehow by that rotating. 460 00:29:58,520 --> 00:30:05,530 The gravitational force that is on the main mass that 461 00:30:05,530 --> 00:30:07,900 comes from the little mass certainly 462 00:30:07,900 --> 00:30:11,690 has to pass through the rod. 463 00:30:11,690 --> 00:30:17,535 It's got to be contained in the forces in the connecting rod. 464 00:30:17,535 --> 00:30:19,960 So it's definitely there. 465 00:30:19,960 --> 00:30:24,510 But the force that causes the centripetal acceleration 466 00:30:24,510 --> 00:30:29,670 of that rotating mass is completely 467 00:30:29,670 --> 00:30:31,870 independent of gravity. 468 00:30:31,870 --> 00:30:34,700 With or without gravity, it takes a particular force 469 00:30:34,700 --> 00:30:37,640 to make that thing travel in a circular path. 470 00:30:37,640 --> 00:30:42,340 And that's m minus mr theta dot squared always. 471 00:30:42,340 --> 00:30:42,840 Yep? 472 00:30:42,840 --> 00:30:45,877 AUDIENCE: So doesn't that mean that on your first expression 473 00:30:45,877 --> 00:30:49,919 on that board, there should be no mb? 474 00:30:49,919 --> 00:30:51,210 PROFESSOR: On which expression? 475 00:30:51,210 --> 00:30:52,127 AUDIENCE: That one. 476 00:30:52,127 --> 00:30:52,960 PROFESSOR: This one. 477 00:30:52,960 --> 00:30:55,593 OK, this is the total forces on the little mass. 478 00:30:55,593 --> 00:30:56,509 AUDIENCE: [INAUDIBLE]. 479 00:31:05,990 --> 00:31:08,520 PROFESSOR: We need to back up to here. 480 00:31:08,520 --> 00:31:13,160 The total forces are mass times the acceleration 481 00:31:13,160 --> 00:31:17,180 of the main body it's connected to plus the mass 482 00:31:17,180 --> 00:31:19,940 times the acceleration of B with respect to A. 483 00:31:19,940 --> 00:31:22,950 So we have to have that term. 484 00:31:22,950 --> 00:31:25,960 And we then go into our four terms 485 00:31:25,960 --> 00:31:29,190 here and find there's only one left. 486 00:31:29,190 --> 00:31:34,580 So that's the force exerted on the small mass by the rod. 487 00:31:34,580 --> 00:31:41,980 And that is positive mx double dot minus mr theta dot squared. 488 00:31:41,980 --> 00:31:46,040 So we sum the forces on that little mass. 489 00:31:46,040 --> 00:31:50,529 It has got to be equal to-- ah, I 490 00:31:50,529 --> 00:31:51,820 know where we made the mistake. 491 00:31:51,820 --> 00:31:54,020 So we've just discovered our mistake. 492 00:31:54,020 --> 00:31:56,640 This has got to be able to mass times acceleration. 493 00:31:56,640 --> 00:32:01,450 And what are the forces? 494 00:32:01,450 --> 00:32:04,150 The summation the forces is mass times acceleration. 495 00:32:04,150 --> 00:32:08,220 So the acceleration is this plus this. 496 00:32:08,220 --> 00:32:09,555 But the sum of the forces. 497 00:32:50,040 --> 00:32:56,260 The problem here is I've used a notation where 498 00:32:56,260 --> 00:33:00,820 this is very similar looking to the forces 499 00:33:00,820 --> 00:33:02,440 that I've noted here. 500 00:33:02,440 --> 00:33:09,260 So this is the actual force in the y direction, j hat, 501 00:33:09,260 --> 00:33:16,302 plus the actual force in the x direction, i hat, minus mgi. 502 00:33:20,860 --> 00:33:23,670 So when I solve for the i component, 503 00:33:23,670 --> 00:33:28,340 I'm going to get the i pieces of that plus mgi. 504 00:33:32,019 --> 00:33:33,310 I mean, excuse me, j component. 505 00:33:39,128 --> 00:33:40,535 Should it be like that? 506 00:33:44,580 --> 00:33:48,990 The j component will have this piece times sine omega 507 00:33:48,990 --> 00:33:51,710 t with a minus. 508 00:33:51,710 --> 00:34:01,250 And you move the mgj to that side and it becomes a plus. 509 00:34:01,250 --> 00:34:02,200 All right. 510 00:34:02,200 --> 00:34:02,965 That makes sense. 511 00:34:05,970 --> 00:34:10,710 The rod has to hold up the weight of that little mass, 512 00:34:10,710 --> 00:34:12,560 right? 513 00:34:12,560 --> 00:34:13,760 The weight's down. 514 00:34:13,760 --> 00:34:19,340 But the rod has to push up on it in the y direction. 515 00:34:19,340 --> 00:34:23,060 So the force the rod puts on the little mass 516 00:34:23,060 --> 00:34:26,880 has got to be equal to the weight of the small mass 517 00:34:26,880 --> 00:34:29,350 minus this me omega squared term, which 518 00:34:29,350 --> 00:34:37,530 is the force necessary to create the centripetal acceleration. 519 00:34:37,530 --> 00:34:40,120 OK, so we've got this now fixed. 520 00:34:45,530 --> 00:34:48,450 This term is OK. 521 00:34:56,449 --> 00:35:01,730 And the minus that force is then the force on the main body. 522 00:35:01,730 --> 00:35:02,545 So minus. 523 00:35:08,721 --> 00:35:09,220 Plus. 524 00:35:11,950 --> 00:35:15,020 Now I've got to figure out what I did wrong here. 525 00:35:15,020 --> 00:35:17,620 You're doing what I said I wasn't going to do. 526 00:35:17,620 --> 00:35:21,200 We're on the fly trying to figure out where the. 527 00:35:21,200 --> 00:35:24,200 AUDIENCE: [INAUDIBLE]. 528 00:35:24,200 --> 00:35:25,200 PROFESSOR: OK. 529 00:35:25,200 --> 00:35:26,250 You're happy now. 530 00:35:26,250 --> 00:35:28,250 Good. 531 00:35:28,250 --> 00:35:29,294 OK. 532 00:35:29,294 --> 00:35:30,710 All right, we're going to move on. 533 00:35:39,474 --> 00:35:40,390 AUDIENCE: [INAUDIBLE]. 534 00:35:43,960 --> 00:35:45,460 PROFESSOR: What about the summation? 535 00:35:45,460 --> 00:35:47,043 AUDIENCE: It's not really a summation. 536 00:35:47,043 --> 00:35:51,340 It's just the force that arm is [INAUDIBLE]. 537 00:35:55,760 --> 00:35:57,230 PROFESSOR: Right. 538 00:35:57,230 --> 00:35:58,614 Fair enough. 539 00:35:58,614 --> 00:35:59,114 Yep. 540 00:36:03,600 --> 00:36:06,850 This is just minus f. 541 00:36:06,850 --> 00:36:08,570 This is on the little mass. 542 00:36:08,570 --> 00:36:10,950 This is the force on the little mass from the free body 543 00:36:10,950 --> 00:36:11,670 diagram. 544 00:36:11,670 --> 00:36:12,400 Right. 545 00:36:12,400 --> 00:36:16,660 And that helps. 546 00:36:16,660 --> 00:36:22,050 And that's different from the summation one here. 547 00:36:22,050 --> 00:36:23,110 OK. 548 00:36:23,110 --> 00:36:23,610 All right. 549 00:36:23,610 --> 00:36:26,310 I think we've got it sorted out. 550 00:36:26,310 --> 00:36:34,170 Now, I've put on the Stellar website 551 00:36:34,170 --> 00:36:36,370 under readings a little one page thing called 552 00:36:36,370 --> 00:36:37,750 "Moments of Inertia." 553 00:36:37,750 --> 00:36:42,360 It's two pages of information taken from the Williams 554 00:36:42,360 --> 00:36:44,120 textbook on dynamics. 555 00:36:44,120 --> 00:36:49,186 And it's going to show some of what 556 00:36:49,186 --> 00:36:51,060 I'm going to put on the board, and especially 557 00:36:51,060 --> 00:36:52,910 the detailed stuff you don't have to copy. 558 00:36:52,910 --> 00:36:55,410 OK, we're going to come up with some expressions for angular 559 00:36:55,410 --> 00:36:59,050 momentum in terms of particles and their positions. 560 00:36:59,050 --> 00:37:03,060 And this is now the subject of mass and moments 561 00:37:03,060 --> 00:37:04,945 of inertia and products of inertia. 562 00:37:04,945 --> 00:37:07,320 And I'm going to put some of these equations on the board 563 00:37:07,320 --> 00:37:10,068 and you don't have to copy them all. 564 00:37:10,068 --> 00:37:13,730 All these expressions become the definitions 565 00:37:13,730 --> 00:37:16,760 of mass moments of inertia and products of inertia. 566 00:37:16,760 --> 00:37:21,130 And if you just drop down one last little bit, 567 00:37:21,130 --> 00:37:24,050 we come up with an expression for angular momentum. 568 00:37:24,050 --> 00:37:30,240 Three vector components look like ixx omega x plus ixy omega 569 00:37:30,240 --> 00:37:33,540 y and so forth. 570 00:37:33,540 --> 00:37:38,400 These compounds in terms of particle masses and positions 571 00:37:38,400 --> 00:37:41,630 are defined in these final equations. 572 00:37:41,630 --> 00:37:44,370 So I'm going to tell you what I'm going to tell you. 573 00:37:44,370 --> 00:37:45,990 We're going to make the transition 574 00:37:45,990 --> 00:37:55,000 from dealing with particles and angular momentum of particles 575 00:37:55,000 --> 00:38:00,200 to angular momentum of rigid bodies. 576 00:38:00,200 --> 00:38:01,200 OK? 577 00:38:01,200 --> 00:38:07,450 And in my own experience this is something 578 00:38:07,450 --> 00:38:09,860 that is generally done badly. 579 00:38:09,860 --> 00:38:11,710 And I'm going to try to do it well. 580 00:38:11,710 --> 00:38:17,240 I'm going to try to give you an intuitive understanding of why 581 00:38:17,240 --> 00:38:23,020 we have these diagonal terms called the moments of inertia 582 00:38:23,020 --> 00:38:25,420 and what they're useful for and why 583 00:38:25,420 --> 00:38:28,850 these off diagonal terms call products of inertia turn up 584 00:38:28,850 --> 00:38:30,470 and what they actually mean. 585 00:38:30,470 --> 00:38:31,970 When I was taught the stuff, I never 586 00:38:31,970 --> 00:38:34,370 got a gut feeling for why or what 587 00:38:34,370 --> 00:38:36,180 the off diagonal terms meant. 588 00:38:36,180 --> 00:38:38,494 You don't know it, but we've been using them. 589 00:38:38,494 --> 00:38:39,910 And then I'll tell you the answer. 590 00:38:39,910 --> 00:38:50,060 The answer is that when we have a problem like well, 591 00:38:50,060 --> 00:38:56,000 the motorcycle problem were talking about is this. 592 00:38:56,000 --> 00:39:00,140 Basically here's the motorcycle wheel spinning around 593 00:39:00,140 --> 00:39:00,740 and round. 594 00:39:00,740 --> 00:39:02,920 And it has these two masses. 595 00:39:02,920 --> 00:39:05,650 This is set up B. So one little mass 596 00:39:05,650 --> 00:39:08,920 was off to the side of the rim a bit. 597 00:39:08,920 --> 00:39:11,390 And the other mass was off to the side. 598 00:39:11,390 --> 00:39:13,030 On the picture it looked like this. 599 00:39:13,030 --> 00:39:14,680 Here's the axle, motorcycle, and forks 600 00:39:14,680 --> 00:39:16,350 would be coming down like this. 601 00:39:16,350 --> 00:39:19,090 And these two little masses. 602 00:39:19,090 --> 00:39:21,610 Equal distance but opposite sides from one another. 603 00:39:21,610 --> 00:39:24,880 And if this spins, it puts a heck of a wobble 604 00:39:24,880 --> 00:39:25,560 into this thing. 605 00:39:28,930 --> 00:39:34,140 And this puts a moment about this point. 606 00:39:34,140 --> 00:39:36,342 It tries to make this thing rock back and forth 607 00:39:36,342 --> 00:39:37,050 as it's spinning. 608 00:39:37,050 --> 00:39:39,560 It's really hard to hold. 609 00:39:39,560 --> 00:39:41,996 You hold the axle there. 610 00:39:41,996 --> 00:39:43,370 And you got to do it so you don't 611 00:39:43,370 --> 00:39:45,870 get hit by the-- there you go. 612 00:39:45,870 --> 00:39:47,940 Now tell me if you feel a moment. 613 00:39:47,940 --> 00:39:50,281 It's really hard to keep that thing straight, right? 614 00:39:50,281 --> 00:39:52,780 Well that's what it's trying to do to that motorcycle wheel. 615 00:39:56,460 --> 00:39:57,600 OK. 616 00:39:57,600 --> 00:40:02,870 For this problem, those off diagonal terms, 617 00:40:02,870 --> 00:40:07,110 those products of inertia are not 0. 618 00:40:07,110 --> 00:40:10,440 The product of inertia terms cause these things 619 00:40:10,440 --> 00:40:13,380 called dynamic imbalances. 620 00:40:13,380 --> 00:40:17,940 It causes there to be angular-- makes the angular momentum 621 00:40:17,940 --> 00:40:20,280 terms instead of the angular momentum 622 00:40:20,280 --> 00:40:25,672 being aligned with the axis of rotation, the rotation vector, 623 00:40:25,672 --> 00:40:27,130 it's pointed off in this direction. 624 00:40:30,270 --> 00:40:34,160 Anytime the angular momentum vector and the rotation vector 625 00:40:34,160 --> 00:40:37,280 are not aligned, you have off diagonal terms 626 00:40:37,280 --> 00:40:40,060 and you will have dynamic imbalance. 627 00:40:40,060 --> 00:40:42,100 So there's a physical consequence 628 00:40:42,100 --> 00:40:44,000 of those off diagonal terms. 629 00:40:44,000 --> 00:40:46,580 And they explain the dynamic imbalance. 630 00:40:46,580 --> 00:40:52,904 So let's see if we can't make some headway on that. 631 00:40:52,904 --> 00:40:53,945 So you've seen the rotor. 632 00:40:58,360 --> 00:40:59,540 Let's look at two cases. 633 00:41:06,170 --> 00:41:08,830 One that looks like that, which I just had set up a second ago. 634 00:41:20,910 --> 00:41:22,180 One that looks like that. 635 00:41:22,180 --> 00:41:29,710 Call this A, B. And in both cases, 636 00:41:29,710 --> 00:41:32,990 the rotation is around the vertical axis 637 00:41:32,990 --> 00:41:36,455 and it's constant at omega. 638 00:41:39,590 --> 00:41:44,500 And I just mean these to be two different cases. 639 00:41:44,500 --> 00:41:46,750 I'll make it lowercase so I don't confuse it 640 00:41:46,750 --> 00:41:50,520 with my coordinate system notation. 641 00:41:50,520 --> 00:41:55,320 This is going to be point A in both of these problems. 642 00:41:55,320 --> 00:41:58,560 And it's going to be the origin of a coordinate system. 643 00:42:09,830 --> 00:42:14,144 So if you cause this to spin, these both have-- 644 00:42:14,144 --> 00:42:16,615 did I write these masses as m over 2? 645 00:42:22,850 --> 00:42:26,220 For a moment, let's just think of these as being equal masses. 646 00:42:26,220 --> 00:42:30,860 If you do this problem, do you think this one will wobble? 647 00:42:30,860 --> 00:42:34,860 No, it's perfectly balanced. 648 00:42:34,860 --> 00:42:38,010 And it'll just spin nice and smoothly. 649 00:42:38,010 --> 00:42:43,010 It has angular momentum around the z-axis, the omega axis. 650 00:42:43,010 --> 00:42:46,490 It has angular momentum in that direction, 651 00:42:46,490 --> 00:42:48,570 certainly, when calculated. 652 00:42:48,570 --> 00:42:54,240 This one has same mass, same distance away from this axis, 653 00:42:54,240 --> 00:42:55,830 but now one up and one down. 654 00:42:55,830 --> 00:42:57,870 This one wobbles. 655 00:42:57,870 --> 00:43:00,450 But this one has a component of angular momentum 656 00:43:00,450 --> 00:43:03,380 in this direction, which is exactly equal to this one. 657 00:43:03,380 --> 00:43:08,340 But this also has a component that's in this direction. 658 00:43:08,340 --> 00:43:10,570 And we're going take a look and see what that is. 659 00:43:15,100 --> 00:43:18,678 So we're going to do this problem here. 660 00:43:18,678 --> 00:43:29,450 We're going to analyze B. This case B. And here's the goal. 661 00:43:29,450 --> 00:43:35,540 The goal is to show you that the angular 662 00:43:35,540 --> 00:43:42,850 momentum of this system with respect to this point 663 00:43:42,850 --> 00:43:47,500 can be written as a matrix with constants in it, 664 00:43:47,500 --> 00:43:51,260 which you can call the mass moment of inertia matrix. 665 00:43:51,260 --> 00:44:01,390 Times the vector components of the rotation rate. 666 00:44:01,390 --> 00:44:06,060 Now, this problem, the z-axis will be 667 00:44:06,060 --> 00:44:10,440 upwards and will only have one component one non 0 component. 668 00:44:10,440 --> 00:44:15,380 But in general, we want to be able to express the angular 669 00:44:15,380 --> 00:44:19,462 momentum as a product of this inertia matrix. 670 00:44:19,462 --> 00:44:21,170 And these are the inertias we'll find out 671 00:44:21,170 --> 00:44:28,360 with respect to A. Times the vector of angular velocities. 672 00:44:28,360 --> 00:44:30,860 We've got to be very careful about some definitions. 673 00:44:42,720 --> 00:44:45,770 So we're going to do this specific problem, 674 00:44:45,770 --> 00:44:50,380 but we're going to use methods that are completely general. 675 00:44:50,380 --> 00:44:53,410 So I want to describe the general problem. 676 00:44:53,410 --> 00:45:07,420 Here is a inertial coordinate system fixed. 677 00:45:07,420 --> 00:45:13,430 Here's a body out here in space. 678 00:45:13,430 --> 00:45:26,450 And it is rotating about some point A. So point one 679 00:45:26,450 --> 00:45:30,230 and the rotation vector, the angular 680 00:45:30,230 --> 00:45:32,260 rotation and some omega. 681 00:45:32,260 --> 00:45:36,100 And it's just in some direction. 682 00:45:36,100 --> 00:45:38,880 And that omega is with respect. 683 00:45:38,880 --> 00:45:41,670 We always in these angular momentum problems 684 00:45:41,670 --> 00:45:45,750 define rotation rate with respect to an inertia 685 00:45:45,750 --> 00:45:46,540 coordinate system. 686 00:45:49,440 --> 00:45:54,910 Now, this point A. So first carefully 687 00:45:54,910 --> 00:46:00,865 define A is a fixed point. 688 00:46:04,867 --> 00:46:06,075 So is that an inertial point? 689 00:46:09,710 --> 00:46:10,940 Yeah. 690 00:46:10,940 --> 00:46:12,780 You can do Newton's laws from this point 691 00:46:12,780 --> 00:46:15,370 just as well as you could any fixed point 692 00:46:15,370 --> 00:46:18,420 in this inertial reference frame is an inertial point 693 00:46:18,420 --> 00:46:20,040 and you can use Newton's laws. 694 00:46:20,040 --> 00:46:21,230 So this is a fixed point. 695 00:46:21,230 --> 00:46:22,970 I'm defining it that way. 696 00:46:22,970 --> 00:46:30,540 This body is rotating about that point with this angular rate. 697 00:46:33,630 --> 00:46:45,220 But attached to the body is a coordinate system 698 00:46:45,220 --> 00:46:46,630 that rotates with the body. 699 00:46:46,630 --> 00:46:50,480 So this would be some a xyz coordinate 700 00:46:50,480 --> 00:46:52,640 system attached to the body. 701 00:46:52,640 --> 00:46:56,500 So it's like this problem where I've 702 00:46:56,500 --> 00:46:58,870 got a coordinate system attached to my wheel. 703 00:46:58,870 --> 00:47:01,780 There's x, here's y, z coming out of it. 704 00:47:01,780 --> 00:47:03,160 And in a really simple case, it's 705 00:47:03,160 --> 00:47:05,130 rotating around the z-axis. 706 00:47:05,130 --> 00:47:08,920 But I can make it rotate around some other axis. 707 00:47:08,920 --> 00:47:10,997 I pushed a nail through here and I'm 708 00:47:10,997 --> 00:47:12,330 trying to hold it constant here. 709 00:47:12,330 --> 00:47:16,520 And now it's rotating about a different axis, right? 710 00:47:19,440 --> 00:47:23,070 Same rotation rate, but it doesn't 711 00:47:23,070 --> 00:47:25,130 have to be lined up in any pretty way. 712 00:47:25,130 --> 00:47:29,200 If I make that thing rotate around that other axis, 713 00:47:29,200 --> 00:47:31,530 it looks weird, but we can define it. 714 00:47:31,530 --> 00:47:33,390 And that's what we're talking about here. 715 00:47:33,390 --> 00:47:37,870 So this body is rotating around, has some rotation 716 00:47:37,870 --> 00:47:42,800 rate with respect to a reference frame attached to the body. 717 00:47:46,710 --> 00:47:58,260 So A xyz is a frame that can't-- going to make this go up. 718 00:48:01,810 --> 00:48:02,500 Come on. 719 00:48:05,210 --> 00:48:11,770 This is attached to the body. 720 00:48:16,860 --> 00:48:19,530 And I've drawn them at kind of funny angles here, 721 00:48:19,530 --> 00:48:23,620 just to emphasize that they're not necessarily 722 00:48:23,620 --> 00:48:26,240 lined up with these. 723 00:48:26,240 --> 00:48:28,024 And it's going to rotate. 724 00:48:28,024 --> 00:48:28,524 OK. 725 00:48:46,885 --> 00:48:47,385 Omega. 726 00:49:13,900 --> 00:49:14,890 Just to emphasize. 727 00:49:14,890 --> 00:49:16,500 It's always in the inertial frame. 728 00:49:21,020 --> 00:49:26,220 The last point may be confusing to start with. 729 00:49:26,220 --> 00:49:37,770 Omega measured with respect to O can 730 00:49:37,770 --> 00:49:49,445 be expressed in terms of the axyz unit vectors. 731 00:49:56,870 --> 00:49:57,870 We're going to do that. 732 00:49:57,870 --> 00:50:00,440 It turns out it vastly simplifies the problem 733 00:50:00,440 --> 00:50:04,100 to express the rotation in the unit 734 00:50:04,100 --> 00:50:07,600 vectors of the frame attached to the body. 735 00:50:07,600 --> 00:50:10,190 Remember, that frame is still fit. 736 00:50:10,190 --> 00:50:14,520 Its origin A is at a fixed point in the inertial frame. 737 00:50:14,520 --> 00:50:18,360 So it's just the system's going around and inside 738 00:50:18,360 --> 00:50:22,090 of that system you have a rotation 739 00:50:22,090 --> 00:50:24,990 and you can break it down into xyz components. 740 00:50:24,990 --> 00:50:27,520 Just a vector and you can express it in those components. 741 00:50:27,520 --> 00:50:28,728 That's all we're saying here. 742 00:50:35,970 --> 00:50:39,920 Now I want to do the motorcycle problem. 743 00:50:39,920 --> 00:50:43,470 I'm going to just turn it on its side. 744 00:50:43,470 --> 00:50:45,730 And the reason I'm going to do this specific example, 745 00:50:45,730 --> 00:50:47,580 the hope here is to actually now give you 746 00:50:47,580 --> 00:50:50,160 a physical feeling for what's going on. 747 00:50:50,160 --> 00:50:53,250 We've done a lot of illustrations of it. 748 00:50:53,250 --> 00:50:55,890 And you know that it produces imbalances. 749 00:50:55,890 --> 00:51:01,530 So here's my z-axis and my rotation rate. 750 00:51:04,280 --> 00:51:10,610 Omega with respect to O is some omega in the k hat 751 00:51:10,610 --> 00:51:14,370 direction in the fixed frame. 752 00:51:14,370 --> 00:51:19,540 And in this case, it's going to be simpler 753 00:51:19,540 --> 00:51:23,760 than the general case, so that we can do it 754 00:51:23,760 --> 00:51:26,660 in a reasonable length of time. 755 00:51:26,660 --> 00:51:28,195 So actually here's my rod. 756 00:51:30,820 --> 00:51:40,190 Here's my point A. This is my coordinate system axyz. 757 00:51:43,440 --> 00:51:47,970 So this is now attached to the body. 758 00:51:47,970 --> 00:51:53,102 My rigid body is a massless rod with two masses on it. 759 00:51:56,060 --> 00:52:00,035 And this distance, this is the x. 760 00:52:02,860 --> 00:52:06,310 Going that way will be a y, which we have little use of. 761 00:52:06,310 --> 00:52:09,540 There's nothing happening in that direction. 762 00:52:09,540 --> 00:52:16,770 So this distance here I'll call x1. 763 00:52:16,770 --> 00:52:20,700 This distance here is z1. 764 00:52:20,700 --> 00:52:28,769 Over here, this is z2 and x2. 765 00:52:28,769 --> 00:52:30,310 Now we're going to make this problem. 766 00:52:30,310 --> 00:52:31,680 We'll substitute a number. 767 00:52:31,680 --> 00:52:33,380 So this is symmetric. 768 00:52:33,380 --> 00:52:37,150 So x2 is going to be minus x1 and so forth. 769 00:52:37,150 --> 00:52:40,400 But we want to keep them separate for the moment 770 00:52:40,400 --> 00:52:43,620 so you see what happens to different terms. 771 00:52:43,620 --> 00:52:45,055 OK, so that defines a problem. 772 00:52:52,710 --> 00:52:55,080 So the coordinates. 773 00:52:55,080 --> 00:52:57,975 And we'll call this mass m1. 774 00:52:57,975 --> 00:52:59,950 I'll keep this a little general for a moment. 775 00:52:59,950 --> 00:53:01,750 And this is m2. 776 00:53:01,750 --> 00:53:14,800 So m1 is at the coordinates x1, i, 0, and z1 k. 777 00:53:14,800 --> 00:53:24,640 And m2 is at x2, i, 0, and z2 k. 778 00:53:29,740 --> 00:53:32,280 Just points in a plane. 779 00:53:32,280 --> 00:53:39,040 And I want now to compute the-- I 780 00:53:39,040 --> 00:53:46,980 want to find the angular momentum of this object 781 00:53:46,980 --> 00:53:49,960 with respect to point a. 782 00:53:49,960 --> 00:53:52,620 Remember we compute angular momentum in respect to points. 783 00:53:52,620 --> 00:53:55,120 So I'm going to do it with respect to point A. 784 00:53:55,120 --> 00:53:57,640 And that's going to be the sum of the angular 785 00:53:57,640 --> 00:54:00,990 momentum of mass 1 with respect to A, 786 00:54:00,990 --> 00:54:31,850 plus the angular momentum of mass 2 with respect to A. 787 00:54:31,850 --> 00:54:37,410 So the angular momentum of any particle i with respect to A 788 00:54:37,410 --> 00:54:39,340 is r cross p. 789 00:54:39,340 --> 00:54:41,570 r cross the linear momentum. 790 00:54:41,570 --> 00:54:50,080 So it's r i with respect to A cross p i. 791 00:54:50,080 --> 00:54:53,520 Now the p, this is the momentum of the particle. 792 00:54:53,520 --> 00:54:57,170 That's always with respect to what kind of frame. 793 00:54:57,170 --> 00:54:59,820 When you compute angular momentum. 794 00:54:59,820 --> 00:55:01,370 Must be the inertial frame, right? 795 00:55:01,370 --> 00:55:04,460 So technically to start with, just remind you of that, 796 00:55:04,460 --> 00:55:05,950 we'd say oh. 797 00:55:05,950 --> 00:55:08,760 But we've already said our A is a fixed 798 00:55:08,760 --> 00:55:10,430 point in an inertial frame. 799 00:55:10,430 --> 00:55:16,840 So it's OK to write r i with respect to A cross, 800 00:55:16,840 --> 00:55:20,610 in this case, p i with respect to A. They're the same thing. 801 00:55:20,610 --> 00:55:22,450 These two things are exactly the same thing. 802 00:55:22,450 --> 00:55:24,230 The momentum measured at any two fixed 803 00:55:24,230 --> 00:55:26,422 points in an inertial frame is the same. 804 00:55:26,422 --> 00:55:28,338 Doesn't matter where you're measuring it from. 805 00:55:32,541 --> 00:55:33,040 OK. 806 00:55:40,390 --> 00:55:45,580 And we know that p i with respect to A 807 00:55:45,580 --> 00:55:51,940 now, we'll call it, is the mass i times the velocity of i 808 00:55:51,940 --> 00:55:56,650 with respect to A. That's just ordinary linear momentum. 809 00:56:01,890 --> 00:56:06,700 So I need an expression for the velocity of i with respect 810 00:56:06,700 --> 00:56:08,860 to A. Any point. 811 00:56:12,930 --> 00:56:14,130 So these are fixed now. 812 00:56:14,130 --> 00:56:16,400 These are fixed length things. 813 00:56:16,400 --> 00:56:21,500 The velocity of a moving point is just the derivative 814 00:56:21,500 --> 00:56:23,940 of the position vector. 815 00:56:23,940 --> 00:56:28,170 But you have this equation some people call a transport 816 00:56:28,170 --> 00:56:29,172 equation. 817 00:56:29,172 --> 00:56:31,130 So the length of this thing's not changing any, 818 00:56:31,130 --> 00:56:32,750 so it's just going to have one term in it. 819 00:56:32,750 --> 00:56:33,625 So what's a velocity? 820 00:56:36,980 --> 00:56:43,328 In vector notation, omega cross. 821 00:56:43,328 --> 00:56:43,828 All right. 822 00:56:51,650 --> 00:56:54,400 Right. 823 00:56:54,400 --> 00:56:55,970 And this could also. 824 00:56:55,970 --> 00:56:57,790 All right, these are vectors. 825 00:57:03,290 --> 00:57:10,350 And because I can say that, then I can say hi with respect to A 826 00:57:10,350 --> 00:57:19,835 is mi riA cross omega with respect 827 00:57:19,835 --> 00:57:28,090 to O cross ri with respect to A. OK. 828 00:57:28,090 --> 00:57:28,680 All vectors. 829 00:57:33,030 --> 00:57:36,472 So any rigid body. 830 00:57:36,472 --> 00:57:37,430 So here's the link now. 831 00:57:37,430 --> 00:57:44,010 Here's the jump from points particles to rigid bodies. 832 00:57:44,010 --> 00:57:47,480 Any rigid body is made up of the whole mess of particles, 833 00:57:47,480 --> 00:57:49,410 connected rigidly together. 834 00:57:49,410 --> 00:57:50,560 No relative motion. 835 00:57:50,560 --> 00:57:52,100 But a whole mass of particles. 836 00:57:52,100 --> 00:58:00,630 So I can compute the total momentum of a rigid body 837 00:58:00,630 --> 00:58:06,321 as the summation over all the little particles in it. 838 00:58:06,321 --> 00:58:06,820 mi. 839 00:58:10,270 --> 00:58:17,680 riA cross omega with respect to O cross riA. 840 00:58:17,680 --> 00:58:19,030 Just sum them all up. 841 00:58:21,740 --> 00:58:23,790 And when you have continuous bodies, 842 00:58:23,790 --> 00:58:28,110 these summations turn into integrals. 843 00:58:28,110 --> 00:58:30,400 So you'll find definitions for like there's 844 00:58:30,400 --> 00:58:34,060 a mass moment of inertia about this axis of this wheel. 845 00:58:34,060 --> 00:58:36,750 It's mr squared over 2. 846 00:58:36,750 --> 00:58:42,960 And it comes from the-- and that's 847 00:58:42,960 --> 00:58:45,190 the number that you have to multiply by omega 848 00:58:45,190 --> 00:58:47,140 to get the angular momentum. 849 00:58:47,140 --> 00:58:49,880 So it comes from summing up all these little particles 850 00:58:49,880 --> 00:58:53,700 in this thing is the total momentum, angular momentum, 851 00:58:53,700 --> 00:58:55,530 of the object. 852 00:58:55,530 --> 00:58:57,310 All right, let's do that. 853 00:58:57,310 --> 00:59:00,940 We're going to do that for our too little masses 854 00:59:00,940 --> 00:59:07,265 here and see what kind of things result. Oops. 855 00:59:17,580 --> 00:59:21,180 I want to get my h with respect to A 856 00:59:21,180 --> 00:59:26,630 is the sum of h1 with respect to A plus h2 with respect to A. 857 00:59:26,630 --> 00:59:28,340 And I'm just going to use that formula. 858 00:59:34,420 --> 00:59:36,730 So it's m1. 859 00:59:36,730 --> 00:59:39,890 So if I were just work out that little vector products there. 860 00:59:42,460 --> 00:59:43,660 m1. 861 00:59:43,660 --> 00:59:45,270 here's riA. 862 00:59:45,270 --> 01:00:02,375 It's x1 i times z1k cross omega zk cross x1i plus z1k. 863 01:00:06,120 --> 01:00:10,380 And then I have a second term, the m2 term. 864 01:00:10,380 --> 01:00:27,650 x2i plus z2k omega zk x2i plus z2k. 865 01:00:27,650 --> 01:00:30,740 So just a lot of little vector terms. 866 01:00:30,740 --> 01:00:35,290 That is that expression for our two little particles. 867 01:00:35,290 --> 01:00:40,020 With their specific positions at x1 and z1 and x2 and z2. 868 01:00:43,130 --> 01:00:46,630 So if I multiply all that out, then I'll 869 01:00:46,630 --> 01:00:52,445 get the following result. An h with respect to A here. 870 01:00:56,210 --> 01:01:13,890 It's m1 x1 squared omega zk minus m1 x1 z1 omega 871 01:01:13,890 --> 01:01:25,100 z in the i hat direction plus an m2 x2 squared omega 872 01:01:25,100 --> 01:01:38,040 zk minus m2 x2 z2 omega z in the i direction. 873 01:01:38,040 --> 01:01:41,110 So this is the angular momentum of particle one. 874 01:01:41,110 --> 01:01:42,940 This is angular momentum of a particle two. 875 01:01:47,610 --> 01:01:49,340 And I'm going to do a special case. 876 01:01:54,088 --> 01:01:59,950 And the special case I'm going to let 877 01:01:59,950 --> 01:02:07,280 m1 equal m2 equal m over 2. 878 01:02:07,280 --> 01:02:09,790 So they'll do sum to m. 879 01:02:09,790 --> 01:02:18,380 And x1 equals minus x2 and z1 equals minus z2. 880 01:02:18,380 --> 01:02:24,190 So they're nice and symmetrically opposite 881 01:02:24,190 --> 01:02:25,960 like drawn in the picture. 882 01:02:25,960 --> 01:02:28,600 That I'm making an equal masses in equal distances 883 01:02:28,600 --> 01:02:30,435 on either side of the origin. 884 01:02:30,435 --> 01:02:32,810 And that's going to make this thing simplify quite a bit. 885 01:03:37,330 --> 01:03:39,870 This is of the form. 886 01:03:39,870 --> 01:03:42,480 This angular momentum vector is of the form 887 01:03:42,480 --> 01:03:44,740 has three vector components. 888 01:03:44,740 --> 01:03:48,410 In this particular case, this one's 0. 889 01:03:48,410 --> 01:03:53,900 And we call the first component, this one here will be hx. 890 01:03:53,900 --> 01:03:58,274 And this one here is clearly hz, the component 891 01:03:58,274 --> 01:03:59,065 in the z direction. 892 01:04:02,690 --> 01:04:06,420 And if we draw, here's our system. 893 01:04:11,000 --> 01:04:12,500 Here's our coordinate system. 894 01:04:17,860 --> 01:04:20,070 The coordinate system attached to the body. 895 01:04:22,640 --> 01:04:28,165 It has a z component of angular momentum positive upwards. 896 01:04:30,750 --> 01:04:36,650 And it has an x component of angular momentum in the minus 897 01:04:36,650 --> 01:04:40,140 direction like that. 898 01:04:40,140 --> 01:04:43,650 When you add them together, you get that. 899 01:04:43,650 --> 01:04:50,053 So this is h with respect to A. This is hz, this is hx. 900 01:04:53,720 --> 01:04:55,702 Now, we found this before. 901 01:04:55,702 --> 01:04:58,160 We didn't talk anything about moments of inertia, anything. 902 01:04:58,160 --> 01:05:01,350 We just deal in particles earlier as we did problems. 903 01:05:01,350 --> 01:05:04,830 We found out that when you have this kind of unbalance, 904 01:05:04,830 --> 01:05:07,790 the direction of the angular momentum vector 905 01:05:07,790 --> 01:05:11,140 is not in the same direction as the rotation vector. 906 01:05:11,140 --> 01:05:14,060 In this case, the rotation doesn't make a zk. 907 01:05:14,060 --> 01:05:15,200 It's like that. 908 01:05:15,200 --> 01:05:17,470 The vector is going around it. 909 01:05:17,470 --> 01:05:18,230 Angular momentum. 910 01:05:46,300 --> 01:05:49,576 Now, in general you would write hx. 911 01:05:52,830 --> 01:05:54,017 General case. 912 01:05:54,017 --> 01:05:56,350 And this is what you can pull off, this little two sheet 913 01:05:56,350 --> 01:05:59,120 handout that you can download and you 914 01:05:59,120 --> 01:06:00,710 don't have to copy everything. 915 01:06:00,710 --> 01:06:04,500 This is going to look like an ixx omega 916 01:06:04,500 --> 01:06:13,505 x plus ixy omega y plus ixz omega z. 917 01:06:18,280 --> 01:06:21,210 So if we look at that and we look 918 01:06:21,210 --> 01:06:30,780 at this, this particular case the hx term is this, right? 919 01:06:30,780 --> 01:06:46,110 So this is the general expression for hx. 920 01:06:46,110 --> 01:06:48,920 And in this particular case, that 921 01:06:48,920 --> 01:07:00,140 will look like minus m x1 z1 omega z. 922 01:07:00,140 --> 01:07:02,870 And this is the piece that's in the i direction. 923 01:07:02,870 --> 01:07:04,610 That's why we call it hx. 924 01:07:04,610 --> 01:07:13,340 And this is then ixz omega z. 925 01:07:13,340 --> 01:07:19,440 So this piece here is what we call ixz. 926 01:07:19,440 --> 01:07:20,910 It's where it comes from. 927 01:07:25,320 --> 01:07:26,380 And we can write it. 928 01:07:26,380 --> 01:07:29,346 So this is our particular case. 929 01:07:39,560 --> 01:07:45,784 Get this result. And we find there's h in the h. 930 01:07:45,784 --> 01:07:48,350 y is 0. 931 01:07:48,350 --> 01:08:02,290 And hz is mx1 squared omega z omega z. 932 01:08:02,290 --> 01:08:07,050 And that's got to be of the form izz omega z. 933 01:08:10,710 --> 01:08:15,250 Now, how do you remember what the subscripts mean? 934 01:08:15,250 --> 01:08:23,050 ixz means this is the h component 935 01:08:23,050 --> 01:08:29,000 and this is the omega component it's multiplied by. 936 01:08:29,000 --> 01:08:35,090 So ixz is the product of inertia for hx. 937 01:08:35,090 --> 01:08:39,970 It's related to rotation in the z component rotation. 938 01:08:39,970 --> 01:08:41,590 That's what the subscripts mean. 939 01:08:52,394 --> 01:08:53,888 Maybe I'll do this. 940 01:09:35,750 --> 01:09:40,770 So in general, if you know what these constants are 941 01:09:40,770 --> 01:09:46,460 for your rigid body and you know your rotation rate, 942 01:09:46,460 --> 01:09:49,680 you instantly know your angular momentum. 943 01:09:49,680 --> 01:09:53,220 These things, the products in moments of inertia, 944 01:09:53,220 --> 01:09:57,080 are basically cataloged-- you'll find them 945 01:09:57,080 --> 01:09:59,330 in the back of your textbook-- for all sorts 946 01:09:59,330 --> 01:10:01,090 of different objects. 947 01:10:01,090 --> 01:10:11,620 So I know that if you have z in this direction and this thing's 948 01:10:11,620 --> 01:10:16,160 rotating around the z, hz is the total mass 949 01:10:16,160 --> 01:10:18,940 of the system times the radius squared 950 01:10:18,940 --> 01:10:24,890 divided by 2. mr squared over 2 would be izz for this object. 951 01:10:24,890 --> 01:10:26,840 And for all sorts of objects. 952 01:10:26,840 --> 01:10:30,150 These are just cataloged values. 953 01:10:30,150 --> 01:10:32,360 And then there's ways of moving the axes, called 954 01:10:32,360 --> 01:10:34,990 parallel axis theorems that you've probably run into, 955 01:10:34,990 --> 01:10:42,370 that allows you then to construct these values from one 956 01:10:42,370 --> 01:10:45,190 known point to moving the point to someplace else 957 01:10:45,190 --> 01:10:46,930 and having it move around that. 958 01:10:46,930 --> 01:10:51,650 So these values are tabulated, calculated, 959 01:10:51,650 --> 01:10:54,600 with respect to the centers of mass. 960 01:10:54,600 --> 01:10:57,130 And if you want to have the mass moments of inertia 961 01:10:57,130 --> 01:10:59,760 with respect to any other point, then you 962 01:10:59,760 --> 01:11:02,750 will use something which we call a parallel axis theorem, which 963 01:11:02,750 --> 01:11:04,520 we'll get to in due course. 964 01:11:08,130 --> 01:11:11,630 Pretty good on timing here. 965 01:11:11,630 --> 01:11:14,700 A note about textbooks. 966 01:11:20,840 --> 01:11:21,895 Textbook conventions. 967 01:11:28,980 --> 01:11:30,820 This I matrix. 968 01:11:30,820 --> 01:11:41,501 In some they write it ixx ixy ixz and so forth. 969 01:11:41,501 --> 01:11:42,000 ix. 970 01:11:45,470 --> 01:11:46,180 No, iyx. 971 01:11:51,560 --> 01:11:53,670 iyy. 972 01:11:53,670 --> 01:11:54,170 iyz. 973 01:11:57,001 --> 01:11:57,500 cx. 974 01:12:06,130 --> 01:12:07,620 Some write it like that. 975 01:12:07,620 --> 01:12:10,897 And others write it with all of these with minus signs 976 01:12:10,897 --> 01:12:11,980 on the off diagonal terms. 977 01:12:17,110 --> 01:12:26,994 So Hibbler uses the minus signs. 978 01:12:29,920 --> 01:12:34,540 Williams does not. 979 01:12:37,110 --> 01:12:40,032 So the diagonal terms are always positive. 980 01:12:40,032 --> 01:12:41,024 Yeah? 981 01:12:41,024 --> 01:12:43,780 AUDIENCE: [INAUDIBLE]. 982 01:12:43,780 --> 01:12:45,840 PROFESSOR: All the off diagonals are negative. 983 01:12:45,840 --> 01:12:47,020 So this is positive. 984 01:12:47,020 --> 01:12:49,290 Positive, positive, positive and then 985 01:12:49,290 --> 01:12:52,752 negative, negative, negative, negative, negative. 986 01:12:52,752 --> 01:12:54,460 Now there are actually negative-- they'll 987 01:12:54,460 --> 01:12:58,040 be negative-- the numbers will pop up negative and so forth. 988 01:12:58,040 --> 01:13:01,250 It's just that in the notation, some authors have adopted 989 01:13:01,250 --> 01:13:03,660 putting the minus signs here. 990 01:13:03,660 --> 01:13:07,690 Others have embedded them in the value itself. 991 01:13:07,690 --> 01:13:15,220 So Williams' notation, he would say that ixz 992 01:13:15,220 --> 01:13:19,390 is minus m x1 z1 for this body. 993 01:13:19,390 --> 01:13:23,530 Hibbler would say it's plus and he'd put the minus sign 994 01:13:23,530 --> 01:13:26,250 in the notation. 995 01:13:26,250 --> 01:13:27,400 So just beware of that. 996 01:13:27,400 --> 01:13:28,430 Because all your life you're going 997 01:13:28,430 --> 01:13:30,430 to run into people saying the product of inertia 998 01:13:30,430 --> 01:13:33,945 of this thing is and you got to know which way they define it. 999 01:13:33,945 --> 01:13:34,445 All right. 1000 01:13:43,630 --> 01:13:44,850 Compute torques. 1001 01:13:44,850 --> 01:13:48,290 You just take time to [INAUDIBLE] angular momentum. 1002 01:13:48,290 --> 01:13:50,590 And we'll do that as a last little step next time. 1003 01:13:50,590 --> 01:13:55,130 But you've got the essence of the movement 1004 01:13:55,130 --> 01:13:57,700 from talking about particles to how we're going 1005 01:13:57,700 --> 01:13:59,620 to talk about rigid bodies. 1006 01:13:59,620 --> 01:14:00,830 So you have muddy cards. 1007 01:14:00,830 --> 01:14:03,600 You have two or three minutes. 1008 01:14:03,600 --> 01:14:05,970 Write down what was tough for you here. 1009 01:14:05,970 --> 01:14:08,010 Write down what wasn't. 1010 01:14:08,010 --> 01:14:11,165 And see you next Tuesday. 1011 01:14:11,165 --> 01:14:16,620 Oh, I must say, so this stuff about-- the mass moment 1012 01:14:16,620 --> 01:14:17,570 of inertia matrix. 1013 01:14:17,570 --> 01:14:20,600 That stuff is not on the exam. 1014 01:14:20,600 --> 01:14:26,960 But knowing about particles and particle moments of inertia is.