1 00:00:00,100 --> 00:00:02,450 The following content is provided under a Creative 2 00:00:02,450 --> 00:00:03,830 Commons license. 3 00:00:03,830 --> 00:00:06,070 Your support will help MIT OpenCourseWare 4 00:00:06,070 --> 00:00:10,170 continue to offer high quality educational resources for free. 5 00:00:10,170 --> 00:00:12,710 To make a donation or to view additional materials 6 00:00:12,710 --> 00:00:16,620 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,620 --> 00:00:17,276 at ocw.mit.edu. 8 00:00:20,757 --> 00:00:21,340 PROFESSOR: OK. 9 00:00:21,340 --> 00:00:22,465 We're going to get started. 10 00:00:22,465 --> 00:00:26,016 The homework-- does everybody have a copy of the handout? 11 00:00:26,016 --> 00:00:27,900 If not, there's some on the steps there. 12 00:00:27,900 --> 00:00:29,610 There's Muddy Cards on the steps. 13 00:00:29,610 --> 00:00:32,049 And we're going to do three things today. 14 00:00:32,049 --> 00:00:33,500 That's this complex problem I want 15 00:00:33,500 --> 00:00:36,432 to talk about, a problem on center of percussion. 16 00:00:36,432 --> 00:00:37,890 And then as we have time, I'm going 17 00:00:37,890 --> 00:00:39,740 to summarize some summarising statements 18 00:00:39,740 --> 00:00:44,590 about imbalances, which we've talked a lot about off and on. 19 00:00:44,590 --> 00:00:49,310 So I want you restrain looking at the notes I've handed out. 20 00:00:49,310 --> 00:00:53,141 The notes I've handed out are this complex problem. 21 00:00:53,141 --> 00:00:54,890 And I'll let you look at them in a minute. 22 00:00:54,890 --> 00:00:57,348 But I want to get you to think about some things before you 23 00:00:57,348 --> 00:00:58,412 see what's on the notes. 24 00:00:58,412 --> 00:00:59,995 But the notes are intended so that you 25 00:00:59,995 --> 00:01:02,650 don't have to spend a lot of time writing down messy stuff. 26 00:01:02,650 --> 00:01:04,720 You can think and listen. 27 00:01:04,720 --> 00:01:07,650 And they'll also be sent out on the web, 28 00:01:07,650 --> 00:01:11,120 posted, so you don't have to grab a copy for your friends. 29 00:01:11,120 --> 00:01:14,690 Because I only made 100 copies and there are 122 of you. 30 00:01:14,690 --> 00:01:15,755 OK. 31 00:01:15,755 --> 00:01:17,130 So let's start with this problem. 32 00:01:17,130 --> 00:01:25,490 This is basically a complex system, a mass, a pendulum. 33 00:01:25,490 --> 00:01:29,940 And when you-- I've made up one for you. 34 00:01:29,940 --> 00:01:32,110 So here's the pendulum. 35 00:01:32,110 --> 00:01:33,930 It's on an axle stuck into my cart. 36 00:01:33,930 --> 00:01:36,470 The cart's got springs connecting to it. 37 00:01:36,470 --> 00:01:38,980 And it naturally has some damping. 38 00:01:38,980 --> 00:01:43,910 So this is a realization of what's been drawn there. 39 00:01:43,910 --> 00:01:45,850 So let's just give it a little bump. 40 00:01:49,670 --> 00:01:51,670 Cart moves back and forth. 41 00:01:51,670 --> 00:01:54,020 Pendulum swings back and forth. 42 00:01:54,020 --> 00:02:00,050 If I start it this way, it'll act a lot more crazy. 43 00:02:00,050 --> 00:02:02,210 It'll have a more chaotic looking kind of motion. 44 00:02:02,210 --> 00:02:04,150 The reason for that has to do with, this 45 00:02:04,150 --> 00:02:06,440 is a multiple degree of freedom system, has 46 00:02:06,440 --> 00:02:10,500 more than one natural frequency, has more than one frequency 47 00:02:10,500 --> 00:02:12,330 that responds, all mixed together. 48 00:02:12,330 --> 00:02:15,300 And that's why it'll do kind of crazy things. 49 00:02:15,300 --> 00:02:18,060 Like that'll stop, almost stop, and then start up again, 50 00:02:18,060 --> 00:02:19,150 stuff like that. 51 00:02:19,150 --> 00:02:22,620 All completely natural. 52 00:02:22,620 --> 00:02:24,490 But if I give it a nice, gentle start, 53 00:02:24,490 --> 00:02:26,290 it actually mostly vibrates in what 54 00:02:26,290 --> 00:02:29,701 I call one mode of vibration. 55 00:02:29,701 --> 00:02:30,200 OK? 56 00:02:30,200 --> 00:02:33,115 So we want to get the equations of motion of this system. 57 00:02:35,650 --> 00:02:40,290 So here, I've drawn it. 58 00:02:40,290 --> 00:02:42,350 And the first question about a system 59 00:02:42,350 --> 00:02:43,980 like this when you go to analyze it 60 00:02:43,980 --> 00:02:46,473 is, how many degrees of freedom does it have. 61 00:02:46,473 --> 00:02:47,800 All right? 62 00:02:47,800 --> 00:02:54,590 So I'm going to claim this to be a planar motion problem. 63 00:02:54,590 --> 00:02:58,980 It's confined to the board and confined to rotation 64 00:02:58,980 --> 00:03:01,010 perpendicular to the board. 65 00:03:01,010 --> 00:03:01,810 OK? 66 00:03:01,810 --> 00:03:08,660 Any time that happens, each object has, 67 00:03:08,660 --> 00:03:13,640 at most, three degrees of freedom, x y, and a rotation. 68 00:03:13,640 --> 00:03:18,150 So we said no rotations around x or around y. 69 00:03:18,150 --> 00:03:21,520 So three possible ones for this mass, three possible ones 70 00:03:21,520 --> 00:03:23,620 for this mass. 71 00:03:23,620 --> 00:03:30,490 And then we start looking for the constraints in addition. 72 00:03:30,490 --> 00:03:32,090 Well, this one is constrained. 73 00:03:32,090 --> 00:03:34,739 It can't move out of the-- let's see, 74 00:03:34,739 --> 00:03:36,030 we've already constrained that. 75 00:03:36,030 --> 00:03:36,613 Wait a second. 76 00:03:39,228 --> 00:03:40,990 How do I want to say this? 77 00:03:40,990 --> 00:03:44,930 This one is certainly constrained in the y because 78 00:03:44,930 --> 00:03:46,960 of the track here. 79 00:03:46,960 --> 00:03:48,390 So that's one constraint. 80 00:03:48,390 --> 00:03:50,390 We're starting off with six possible. 81 00:03:50,390 --> 00:03:55,000 We've got one because it's constrained in the y. 82 00:03:55,000 --> 00:03:57,120 But it can certainly move in the x. 83 00:03:57,120 --> 00:04:00,050 Can it rotate in the z, this big mass? 84 00:04:00,050 --> 00:04:01,190 So that's two. 85 00:04:01,190 --> 00:04:03,890 So out of the three possible, the top one only 86 00:04:03,890 --> 00:04:07,460 has one possible degree of freedom, x. 87 00:04:07,460 --> 00:04:11,890 This one also has three possible degrees of freedom. 88 00:04:11,890 --> 00:04:18,643 It's pinned at A. What does that do for constraint? 89 00:04:22,270 --> 00:04:24,850 So this one, take 30 seconds and talk to your neighbor. 90 00:04:24,850 --> 00:04:27,750 How many constraints are caused by the pin at A? 91 00:05:12,660 --> 00:05:14,740 All right. 92 00:05:14,740 --> 00:05:19,420 So at least-- how many believe that there's one constraint 93 00:05:19,420 --> 00:05:20,515 provided by the pin at A? 94 00:05:20,515 --> 00:05:22,390 Let's see the hands that believe we have one. 95 00:05:22,390 --> 00:05:24,560 Get them up high, up high now. 96 00:05:24,560 --> 00:05:26,311 How about two? 97 00:05:26,311 --> 00:05:26,810 OK. 98 00:05:26,810 --> 00:05:29,101 How about three? 99 00:05:29,101 --> 00:05:29,600 All right. 100 00:05:29,600 --> 00:05:31,460 A little uncertainty here. 101 00:05:31,460 --> 00:05:33,110 All right. 102 00:05:33,110 --> 00:05:36,590 Do you know the motion at A? 103 00:05:36,590 --> 00:05:40,260 Have you prescribed the motion at A in any way? 104 00:05:40,260 --> 00:05:40,940 In what way? 105 00:05:40,940 --> 00:05:42,566 I see you nodding your head. 106 00:05:42,566 --> 00:05:44,483 AUDIENCE: It has to move with the other block. 107 00:05:44,483 --> 00:05:46,190 PROFESSOR: It has to move with the block. 108 00:05:46,190 --> 00:05:48,550 But we've already given the other block a coordinate. 109 00:05:48,550 --> 00:05:49,050 What is it? 110 00:05:49,050 --> 00:05:50,260 AUDIENCE: In x. 111 00:05:50,260 --> 00:05:53,447 PROFESSOR: So if you know x, do you know the motion of A? 112 00:05:53,447 --> 00:05:53,946 OK. 113 00:05:53,946 --> 00:05:57,650 So you have established the motion of A. 114 00:05:57,650 --> 00:06:01,280 So if you know A-- you don't need any additional information 115 00:06:01,280 --> 00:06:02,400 about A do you? 116 00:06:02,400 --> 00:06:04,750 You've already chosen a coordinate for it. 117 00:06:04,750 --> 00:06:08,540 So if A is prescribed, that means 118 00:06:08,540 --> 00:06:14,700 x1 and y1, the motion-- the coordinates 119 00:06:14,700 --> 00:06:17,050 describing this thing-- and I've given 120 00:06:17,050 --> 00:06:19,443 started at a center of mass, a little x1 that way, 121 00:06:19,443 --> 00:06:20,630 and a little y1 that way. 122 00:06:20,630 --> 00:06:23,916 Those are my possible displacement coordinates. 123 00:06:23,916 --> 00:06:25,290 And it has a rotation coordinate. 124 00:06:25,290 --> 00:06:27,410 Three possible degrees of freedom, right? 125 00:06:27,410 --> 00:06:31,410 x and y are prescribed at this point. 126 00:06:31,410 --> 00:06:35,280 If you prescribe the motion-- fix the motion-- 127 00:06:35,280 --> 00:06:38,300 at any point on a rigid body, what 128 00:06:38,300 --> 00:06:41,810 does that say about translational motion 129 00:06:41,810 --> 00:06:45,715 any place else on the rigid body? 130 00:06:45,715 --> 00:06:48,510 Remember, this is back to that subtle definition 131 00:06:48,510 --> 00:06:51,890 of what we mean by translation, what we mean by rotation. 132 00:06:51,890 --> 00:06:53,830 AUDIENCE: It's a parallel motion. 133 00:06:53,830 --> 00:06:55,425 PROFESSOR: Parallel motion. 134 00:06:55,425 --> 00:06:59,510 The translation part of this-- every point on the object 135 00:06:59,510 --> 00:07:02,300 moves parallel to every other point. 136 00:07:02,300 --> 00:07:06,840 So if you've prescribed the translation of any one point 137 00:07:06,840 --> 00:07:09,370 you have prescribed the translation for all. 138 00:07:09,370 --> 00:07:12,180 So this second body, basically, its only translation 139 00:07:12,180 --> 00:07:15,820 that it can have is x of the main body. 140 00:07:15,820 --> 00:07:18,270 So you have confined it in x and y. 141 00:07:18,270 --> 00:07:19,890 It has no x and y possibilities. 142 00:07:19,890 --> 00:07:24,940 Those are two constraints leaving you with-- one. 143 00:07:24,940 --> 00:07:25,440 Right? 144 00:07:25,440 --> 00:07:26,890 One degree of freedom. 145 00:07:26,890 --> 00:07:28,530 And then we pick a coordinate for it. 146 00:07:28,530 --> 00:07:32,960 And the natural one to use for that is the angle here. 147 00:07:32,960 --> 00:07:34,350 So we need two. 148 00:07:34,350 --> 00:07:39,200 Two coordinates completely describe the motion. 149 00:07:39,200 --> 00:07:41,490 It will yield two equations of motion. 150 00:07:44,232 --> 00:07:45,150 OK? 151 00:07:45,150 --> 00:07:49,420 So on quizzes, you say, how many independent degrees of freedom 152 00:07:49,420 --> 00:07:50,050 are there. 153 00:07:50,050 --> 00:07:52,680 That is the same question as saying, 154 00:07:52,680 --> 00:07:57,730 how many equations of motion are required to completely describe 155 00:07:57,730 --> 00:07:59,100 this system. 156 00:07:59,100 --> 00:08:01,120 Or how many independent coordinates 157 00:08:01,120 --> 00:08:03,630 are required to completely describe the motion of system? 158 00:08:03,630 --> 00:08:05,254 They're, all three, the same questions. 159 00:08:05,254 --> 00:08:07,860 And that threw a couple people on the last quiz. 160 00:08:07,860 --> 00:08:08,470 OK. 161 00:08:08,470 --> 00:08:11,030 Let's move on to free body diagrams. 162 00:08:11,030 --> 00:08:14,124 So now we know we've got two coordinates, theta and x. 163 00:08:14,124 --> 00:08:15,665 And now you're free to look at your-- 164 00:08:15,665 --> 00:08:17,122 you can be free to look at things. 165 00:08:17,122 --> 00:08:18,830 And I want you to spend most of your time 166 00:08:18,830 --> 00:08:21,690 thinking and listening and not having to write. 167 00:08:21,690 --> 00:08:24,260 But make notes as you get insights about things. 168 00:08:24,260 --> 00:08:26,770 Free body diagram then, two of them. 169 00:08:26,770 --> 00:08:31,530 The pendulum piece is actually pretty simple. 170 00:08:31,530 --> 00:08:33,610 You've got Mg down. 171 00:08:33,610 --> 00:08:37,000 And you have two possible forces at this point of rotation. 172 00:08:37,000 --> 00:08:38,960 And I've just named them F1 and F2. 173 00:08:38,960 --> 00:08:41,299 And I've drawn them-- not arbitrarily, 174 00:08:41,299 --> 00:08:43,659 but I've picked the direction to draw them in. 175 00:08:43,659 --> 00:08:47,670 I don't know what direction they're in. 176 00:08:47,670 --> 00:08:50,040 And that's my complete free body diagram. 177 00:08:50,040 --> 00:08:52,110 If I've missed anything, tell me. 178 00:08:52,110 --> 00:08:55,501 Or if you have any questions about it, ask me. 179 00:08:55,501 --> 00:08:56,000 OK. 180 00:08:56,000 --> 00:08:59,530 The free body diagram for this guy-- this 181 00:08:59,530 --> 00:09:03,310 is capital M. This one's M1. 182 00:09:03,310 --> 00:09:06,500 This is mass M2 just to keep it straight. 183 00:09:06,500 --> 00:09:08,470 Lots of possible forces in this thing. 184 00:09:13,070 --> 00:09:14,730 Reaction forces through the wheels, 185 00:09:14,730 --> 00:09:16,390 there are only vertical. 186 00:09:16,390 --> 00:09:18,385 I've left out friction, ignored friction. 187 00:09:21,100 --> 00:09:23,760 But also, these F1 and F2 act at to pin. 188 00:09:23,760 --> 00:09:25,680 And notice I've drawn them exactly equal 189 00:09:25,680 --> 00:09:27,290 and opposite to these. 190 00:09:27,290 --> 00:09:28,610 Kind of a key thing to do. 191 00:09:28,610 --> 00:09:29,680 And why do we do that? 192 00:09:32,770 --> 00:09:35,655 There's a law. 193 00:09:35,655 --> 00:09:37,475 What's the law? 194 00:09:37,475 --> 00:09:38,820 Newton's third. 195 00:09:38,820 --> 00:09:41,491 You've bot to do that or else it won't work. 196 00:09:41,491 --> 00:09:41,990 OK? 197 00:09:41,990 --> 00:09:44,980 These are common unknowns. 198 00:09:44,980 --> 00:09:47,330 But they're equal and opposite at this point. 199 00:09:47,330 --> 00:09:49,930 And we have an M1g hanging down. 200 00:09:49,930 --> 00:09:51,560 The spring force resists. 201 00:09:51,560 --> 00:09:54,680 Any positive motion, a spring pulls back. 202 00:09:54,680 --> 00:09:58,690 Any positive velocity, the damper holds back. 203 00:09:58,690 --> 00:10:01,784 And that's all of the forces on this thing. 204 00:10:01,784 --> 00:10:03,700 It's going to be necessary to be able to break 205 00:10:03,700 --> 00:10:04,600 these things down. 206 00:10:04,600 --> 00:10:08,080 Because I'm going to sum things. 207 00:10:08,080 --> 00:10:12,580 Going to need to have bits and pieces of F2 and F1, 208 00:10:12,580 --> 00:10:14,410 so cosine thetas and sine thetas. 209 00:10:21,490 --> 00:10:26,160 Now resist looking at your paper for a second. 210 00:10:26,160 --> 00:10:29,740 Next concept question. 211 00:10:29,740 --> 00:10:31,220 We talked a lot in the last couple 212 00:10:31,220 --> 00:10:36,010 of the lectures about the best approach to do problems, 213 00:10:36,010 --> 00:10:37,965 right, especially using angular momentum. 214 00:10:37,965 --> 00:10:39,881 Do you think you're going to need some angular 215 00:10:39,881 --> 00:10:42,240 momentum to solve this problem? 216 00:10:42,240 --> 00:10:45,622 An approach using torques and angular momentum? 217 00:10:45,622 --> 00:10:46,330 More than likely. 218 00:10:46,330 --> 00:10:49,560 Anytime things are rotating, more than likely. 219 00:10:49,560 --> 00:10:53,550 So what's the best approach here? 220 00:10:53,550 --> 00:10:56,490 Are we going to compute angular momentum with respect 221 00:10:56,490 --> 00:10:58,890 to A, that pivot point? 222 00:10:58,890 --> 00:11:01,295 With respect to g, the center of mass? 223 00:11:03,900 --> 00:11:05,940 And it's not too obvious. 224 00:11:05,940 --> 00:11:09,430 So if you were starting out this problem, 225 00:11:09,430 --> 00:11:10,540 how would you begin it? 226 00:11:10,540 --> 00:11:15,070 Would you decide, I'm going to sum my torques around point A 227 00:11:15,070 --> 00:11:17,810 or, I'm going to sum my torques around g? 228 00:11:17,810 --> 00:11:18,700 Did I mark g? 229 00:11:18,700 --> 00:11:21,790 Yeah, it's right at the center. 230 00:11:21,790 --> 00:11:22,850 All right. 231 00:11:22,850 --> 00:11:24,580 Think about that for a second. 232 00:11:24,580 --> 00:11:26,352 Got a question? 233 00:11:26,352 --> 00:11:27,234 All right. 234 00:11:27,234 --> 00:11:29,450 I'm going to ask you this. 235 00:11:29,450 --> 00:11:30,970 Any questions about the question? 236 00:11:30,970 --> 00:11:33,030 I want a real show of hands here. 237 00:11:33,030 --> 00:11:35,505 I want you to just-- what would you do to start with? 238 00:11:35,505 --> 00:11:39,710 So how many would sum torques, compute angular momentum 239 00:11:39,710 --> 00:11:43,490 with respect to point A? 240 00:11:43,490 --> 00:11:44,080 OK. 241 00:11:44,080 --> 00:11:47,390 How about g? 242 00:11:47,390 --> 00:11:48,510 Hmm, interesting. 243 00:11:48,510 --> 00:11:50,782 And another way? 244 00:11:50,782 --> 00:11:52,240 Some of you are holding back on me. 245 00:11:52,240 --> 00:11:54,130 Not everybody raised their hands here. 246 00:11:54,130 --> 00:11:55,040 OK. 247 00:11:55,040 --> 00:11:55,540 All right. 248 00:11:55,540 --> 00:11:57,920 Most people would do it around A. 249 00:11:57,920 --> 00:11:58,700 A would work. 250 00:12:02,070 --> 00:12:03,470 g would work too. 251 00:12:03,470 --> 00:12:06,030 And in fact, when I sat down to do this problem, 252 00:12:06,030 --> 00:12:08,310 I did it with A first. 253 00:12:08,310 --> 00:12:09,850 And then I went and did it with g. 254 00:12:09,850 --> 00:12:11,945 And it turns out that doing it with respect to g 255 00:12:11,945 --> 00:12:13,230 is just slightly easier. 256 00:12:18,910 --> 00:12:19,410 OK. 257 00:12:19,410 --> 00:12:23,720 So the approach we need to find two equations of motion. 258 00:12:23,720 --> 00:12:25,090 We have two bodies. 259 00:12:25,090 --> 00:12:28,560 We're going to use Newton's and Euler's laws to go after them. 260 00:12:28,560 --> 00:12:32,840 So starting out, first one, sum of the forces 261 00:12:32,840 --> 00:12:35,625 in the x direction on this body. 262 00:12:38,470 --> 00:12:39,850 There's a lot of unknowns. 263 00:12:39,850 --> 00:12:41,610 So I'm going to end up-- or start off-- 264 00:12:41,610 --> 00:12:43,640 with more than two equations. 265 00:12:43,640 --> 00:12:45,350 Because I've got how many unknowns? 266 00:12:45,350 --> 00:12:53,850 One, two, three, four, x, the motion x, five, 267 00:12:53,850 --> 00:12:56,170 and the angle theta, six. 268 00:12:56,170 --> 00:13:01,070 I could need as many as six equations to start with. 269 00:13:01,070 --> 00:13:03,670 If I sum forces in the vertical direction I can-- 270 00:13:03,670 --> 00:13:06,700 it turns out that N1 and N2 two here, I never actually 271 00:13:06,700 --> 00:13:09,990 have to deal with. 272 00:13:09,990 --> 00:13:11,250 So don't start there. 273 00:13:11,250 --> 00:13:12,810 If you're thinking you might not have to mess with them, 274 00:13:12,810 --> 00:13:13,580 don't start there. 275 00:13:13,580 --> 00:13:14,940 You're going to waste a lot of time. 276 00:13:14,940 --> 00:13:16,350 I don't think I'm going to have to deal with them. 277 00:13:16,350 --> 00:13:18,058 And in fact, I'm only going to need four. 278 00:13:18,058 --> 00:13:23,516 I'm going to find four equations involving F1, F2, x, and theta. 279 00:13:23,516 --> 00:13:24,016 OK. 280 00:13:30,266 --> 00:13:32,140 So here's how I'm saving a little time today. 281 00:13:32,140 --> 00:13:36,150 I've written it down for you and I've written it out. 282 00:13:36,150 --> 00:13:37,670 Same thing on your paper. 283 00:13:37,670 --> 00:13:39,800 Let's talk about the first one. 284 00:13:39,800 --> 00:13:44,990 This is a sum of forces in this capital X direction, which 285 00:13:44,990 --> 00:13:51,650 is our inertial frame, X, on the main cart on mass 1. 286 00:13:51,650 --> 00:13:54,340 So it's got to be the mass times the acceleration. 287 00:13:54,340 --> 00:13:55,700 And this is not [INAUDIBLE]. 288 00:13:55,700 --> 00:13:59,590 I don't how that sneaked in there. 289 00:13:59,590 --> 00:14:02,400 And I've called it-- in capital, I had direction. 290 00:14:02,400 --> 00:14:03,020 OK? 291 00:14:03,020 --> 00:14:04,210 Mass times acceleration. 292 00:14:04,210 --> 00:14:06,540 I just sum up all the forces on the cart. 293 00:14:06,540 --> 00:14:07,910 Spring force holds back. 294 00:14:07,910 --> 00:14:12,780 The direction of the arrow on my free body diagram 295 00:14:12,780 --> 00:14:14,450 tells you the sense of it. 296 00:14:14,450 --> 00:14:19,290 Minus KX minus bx dot plus F2 cosine theta 297 00:14:19,290 --> 00:14:23,620 plus F1 sine theta, all in the X I hat direction. 298 00:14:23,620 --> 00:14:24,550 OK? 299 00:14:24,550 --> 00:14:29,415 Some of the forces on my little bar here in the x1 direction. 300 00:14:29,415 --> 00:14:34,180 Now x1 is down here. 301 00:14:34,180 --> 00:14:36,040 And y1 is off that way. 302 00:14:38,930 --> 00:14:42,440 So sum of the forces in x1, which is now on mass 2, 303 00:14:42,440 --> 00:14:45,935 must its mass times its acceleration. 304 00:14:45,935 --> 00:14:50,834 And I've just called it-- it doesn't matter where I do my-- 305 00:14:50,834 --> 00:14:52,250 what point I'm going to do angular 306 00:14:52,250 --> 00:14:53,900 momentum with respect to. 307 00:14:53,900 --> 00:14:55,330 This is Newton's law. 308 00:14:55,330 --> 00:14:57,480 And it's the acceleration of the center 309 00:14:57,480 --> 00:15:00,420 of the gravity with respect to an inertial frame. 310 00:15:00,420 --> 00:15:02,600 But this is in the little i hat direction. 311 00:15:02,600 --> 00:15:04,710 So it's the complement in that direction. 312 00:15:04,710 --> 00:15:08,850 I have a minus F1i in that direction. 313 00:15:08,850 --> 00:15:15,460 And I have a plus M2g cosine theta i in the i hat direction. 314 00:15:15,460 --> 00:15:17,970 Pretty straightforward Newton's second law. 315 00:15:17,970 --> 00:15:20,030 Again, the other component for Newton's 316 00:15:20,030 --> 00:15:25,180 in the Y1 direction, mass times the acceleration in the j hat 317 00:15:25,180 --> 00:15:28,900 direction, minus F2j-- because my F2 happens 318 00:15:28,900 --> 00:15:31,570 to be in the minus direction in the free body 319 00:15:31,570 --> 00:15:38,020 diagram-- minus M2g sine theta in the j hat. 320 00:15:38,020 --> 00:15:41,510 And finally, I'm going to sum my moments 321 00:15:41,510 --> 00:15:45,680 about the center of mass. 322 00:15:45,680 --> 00:15:49,760 It would also work-- if I did it around A, 323 00:15:49,760 --> 00:15:54,380 I would-- the reason you normally do it around A, which 324 00:15:54,380 --> 00:15:56,870 is why I started there, is why? 325 00:15:56,870 --> 00:15:58,960 What's the advantage of doing it around A? 326 00:15:58,960 --> 00:16:01,824 The potential advantage of going to A? 327 00:16:01,824 --> 00:16:04,534 AUDIENCE: So you can eliminate forces at the pin. 328 00:16:04,534 --> 00:16:05,200 PROFESSOR: Yeah. 329 00:16:05,200 --> 00:16:08,474 F1 and F2 don't generate moments at the pin. 330 00:16:08,474 --> 00:16:09,890 And there's a hope that, then, you 331 00:16:09,890 --> 00:16:11,920 can get what you need to know without ever 332 00:16:11,920 --> 00:16:13,730 having to evaluate F1 and F2. 333 00:16:13,730 --> 00:16:14,690 Right? 334 00:16:14,690 --> 00:16:20,884 And in this case, they're going to pop up in these equations. 335 00:16:20,884 --> 00:16:23,550 And you're going to have to deal with them anyway, it turns out. 336 00:16:23,550 --> 00:16:26,120 So it doesn't actually give you much of an advantage. 337 00:16:26,120 --> 00:16:29,390 So it's easier because the torque equation's 338 00:16:29,390 --> 00:16:33,610 easier in terms of the number of terms it has. 339 00:16:33,610 --> 00:16:35,970 That's what makes it slightly easier in this problem. 340 00:16:35,970 --> 00:16:38,260 You've got to deal with F1 and F2 anyway. 341 00:16:38,260 --> 00:16:40,670 So it's just the time rate of change of the angular 342 00:16:40,670 --> 00:16:43,580 momentum with respect to g. 343 00:16:43,580 --> 00:16:45,010 And that's external torques. 344 00:16:45,010 --> 00:16:49,480 And the only external torque is caused by F2. 345 00:16:49,480 --> 00:16:52,200 And I left out a minus sign. 346 00:16:52,200 --> 00:16:54,945 You have it on your paper, I think, here. 347 00:16:58,600 --> 00:16:59,610 No, it's correct. 348 00:16:59,610 --> 00:17:00,790 It's positive. 349 00:17:00,790 --> 00:17:05,707 So you have, F2 acts on a moment arm about g, L2. 350 00:17:05,707 --> 00:17:07,540 And it's going to be in the k hat direction. 351 00:17:07,540 --> 00:17:10,200 That's the external torque on the system. 352 00:17:10,200 --> 00:17:16,050 Its gravity causes no torque because it's acting at g. 353 00:17:16,050 --> 00:17:17,069 All right? 354 00:17:17,069 --> 00:17:19,364 And this has got to be the time rate of change of h. 355 00:17:24,280 --> 00:17:28,349 I claim in this problem-- you don't 356 00:17:28,349 --> 00:17:31,290 have to do this-- but in this problem, 357 00:17:31,290 --> 00:17:34,540 I claim I can write h as a mass moment of inertia 358 00:17:34,540 --> 00:17:42,760 matrix times a vector that tells you what the rotation rate is. 359 00:17:42,760 --> 00:17:45,740 You, in fact, can always do this about g. 360 00:17:45,740 --> 00:17:48,925 This is I with respect to g. 361 00:17:48,925 --> 00:17:53,210 If you know what it is, this I-- I 362 00:17:53,210 --> 00:17:55,900 have chosen a set of coordinates that pass 363 00:17:55,900 --> 00:17:59,090 through the center of mass. 364 00:17:59,090 --> 00:18:03,150 x1 that way, y1 perpendicular to it, are they principal coord-- 365 00:18:03,150 --> 00:18:06,100 and z coming out of the board-- are they principal coordinates? 366 00:18:06,100 --> 00:18:07,130 Sure, you know that. 367 00:18:07,130 --> 00:18:09,080 This is just a uniform rod. 368 00:18:09,080 --> 00:18:12,230 But just symmetry, immediately, should tell you that they are. 369 00:18:12,230 --> 00:18:13,092 Yes. 370 00:18:13,092 --> 00:18:16,290 AUDIENCE: I thought you said that IMA could only work 371 00:18:16,290 --> 00:18:18,996 for a stationary rotation axis. 372 00:18:18,996 --> 00:18:21,702 I thought IMA could only work for a stationary axis 373 00:18:21,702 --> 00:18:22,450 of rotation. 374 00:18:22,450 --> 00:18:26,290 PROFESSOR: And rotation about the center of mass. 375 00:18:26,290 --> 00:18:29,361 You can always do things with the center of mass. 376 00:18:29,361 --> 00:18:29,860 OK? 377 00:18:29,860 --> 00:18:35,580 But you could just work this out by the basic definition 378 00:18:35,580 --> 00:18:37,280 of angular momentum. 379 00:18:37,280 --> 00:18:40,731 Don't do it the hard way, r cross p's and those kind 380 00:18:40,731 --> 00:18:41,230 of things. 381 00:18:41,230 --> 00:18:43,870 And you end up in the same place. 382 00:18:43,870 --> 00:18:47,940 This one, because it's 0 0 omega z, 383 00:18:47,940 --> 00:18:50,230 when you do the multiplication, this is a diagonal. 384 00:18:50,230 --> 00:18:52,877 The only term that matters is this one. 385 00:18:52,877 --> 00:18:54,100 Right? 386 00:18:54,100 --> 00:18:57,980 So that's going to give us an IzzG theta double dot, which 387 00:18:57,980 --> 00:19:01,211 is omega z, k hat direction. 388 00:19:01,211 --> 00:19:02,460 And it's not theta double dot. 389 00:19:02,460 --> 00:19:04,270 The h gives you theta dot. 390 00:19:04,270 --> 00:19:05,790 Omega z is theta dot. 391 00:19:05,790 --> 00:19:07,330 And we've taken the time derivative, 392 00:19:07,330 --> 00:19:10,230 which gets us theta double dot. 393 00:19:10,230 --> 00:19:13,330 And we know what the mass moment of inertia 394 00:19:13,330 --> 00:19:16,440 about a uniform stick is with respect 395 00:19:16,440 --> 00:19:19,600 to G, ML squared over 12. 396 00:19:19,600 --> 00:19:20,700 Theta dot-- OK. 397 00:19:20,700 --> 00:19:23,600 So there's our four equations that we have to work with. 398 00:19:23,600 --> 00:19:26,980 And do they involve N1 or N2? 399 00:19:26,980 --> 00:19:27,589 Not at all. 400 00:19:27,589 --> 00:19:29,380 That's why I said we had, potentially, six. 401 00:19:29,380 --> 00:19:31,850 We really only have four unknowns. 402 00:19:31,850 --> 00:19:39,962 And now you use these two to solve for F1 and F2. 403 00:19:39,962 --> 00:19:43,210 And once you get expressions for F1 and F2 404 00:19:43,210 --> 00:19:47,230 you can eliminate them from here and here. 405 00:19:47,230 --> 00:19:49,060 And you're done. 406 00:19:49,060 --> 00:19:52,360 But there's a bit of work left to do that. 407 00:19:52,360 --> 00:19:54,092 But that's the approach. 408 00:19:54,092 --> 00:19:57,860 You use these two to isolate F1 and F2 409 00:19:57,860 --> 00:19:59,500 and plug them into these two to get 410 00:19:59,500 --> 00:20:00,890 your final equations of motion. 411 00:20:04,600 --> 00:20:08,140 But we have a couple of things we don't know yet 412 00:20:08,140 --> 00:20:11,000 that we need in here, the acceleration of that center 413 00:20:11,000 --> 00:20:14,870 of mass in the i hat-- and break into two 414 00:20:14,870 --> 00:20:17,310 components, i hat and j hat. 415 00:20:17,310 --> 00:20:20,700 But that's what we've been studying kinematics for. 416 00:20:34,290 --> 00:20:38,310 So we need to know the velocity of G with respect to O. 417 00:20:38,310 --> 00:20:40,410 And we need to know the acceleration of G 418 00:20:40,410 --> 00:20:42,380 with respect to O. OK? 419 00:20:46,700 --> 00:20:49,580 So the velocity is pretty straightforward. 420 00:20:49,580 --> 00:20:50,890 We've done this many times. 421 00:20:50,890 --> 00:20:53,680 So the velocity of A with respect-- 422 00:20:53,680 --> 00:20:58,890 Remember, you pick things you know to work from. 423 00:20:58,890 --> 00:21:02,970 And you try to make as few as possible things you don't know. 424 00:21:02,970 --> 00:21:04,740 Or put them in forms that we know 425 00:21:04,740 --> 00:21:06,250 how to go about getting it. 426 00:21:06,250 --> 00:21:08,720 These are vectors. 427 00:21:08,720 --> 00:21:10,812 Do we know the velocity of A with respect to O? 428 00:21:15,740 --> 00:21:16,410 What's A? 429 00:21:16,410 --> 00:21:20,170 A is the place where the pin is. 430 00:21:20,170 --> 00:21:23,124 What's its velocity at that point? 431 00:21:23,124 --> 00:21:23,980 AUDIENCE: x dot. 432 00:21:23,980 --> 00:21:24,830 PROFESSOR: x dot. 433 00:21:24,830 --> 00:21:26,682 And in what direction? 434 00:21:26,682 --> 00:21:27,807 AUDIENCE: Capital I hat. 435 00:21:27,807 --> 00:21:28,390 PROFESSOR: OK. 436 00:21:28,390 --> 00:21:33,700 So this is x dot capital I hat. 437 00:21:33,700 --> 00:21:36,970 And now this term, the velocity of G with respect to A, 438 00:21:36,970 --> 00:21:44,320 this is a rigid body which is rotating and translating. 439 00:21:44,320 --> 00:21:46,480 And we've run into this before, right? 440 00:21:46,480 --> 00:21:50,630 And basically, the equation for such things 441 00:21:50,630 --> 00:21:55,150 is the motion of the translational velocity 442 00:21:55,150 --> 00:22:01,430 of the object plus the velocities within the object, 443 00:22:01,430 --> 00:22:03,870 including anything contributed by rotation. 444 00:22:03,870 --> 00:22:10,192 So this term is the velocity of the point G, with respect to A, 445 00:22:10,192 --> 00:22:14,590 to the center-- velocity of the center G with respect 446 00:22:14,590 --> 00:22:15,780 to point A. 447 00:22:15,780 --> 00:22:20,650 And that is-- I'll write it consistently here. 448 00:22:20,650 --> 00:22:26,500 This is the velocity of G with respect to A evaluated 449 00:22:26,500 --> 00:22:30,640 if you had no rotation. 450 00:22:30,640 --> 00:22:32,300 If you're sitting-- the other way 451 00:22:32,300 --> 00:22:34,380 the books often say it is this is 452 00:22:34,380 --> 00:22:37,370 what you would see if you were on the object sitting at A 453 00:22:37,370 --> 00:22:39,150 looking and G. Is it moving? 454 00:22:39,150 --> 00:22:40,660 You'd say no. 455 00:22:40,660 --> 00:22:42,760 Well, another way of saying that, that's 456 00:22:42,760 --> 00:22:46,288 the velocity of this thing if there were no rotation. 457 00:22:46,288 --> 00:22:47,180 All right? 458 00:22:47,180 --> 00:22:49,750 So this term happens to be 0. 459 00:22:49,750 --> 00:22:53,890 Plus omega-- now we're kind of doing this on purpose. 460 00:22:53,890 --> 00:22:58,300 We know this is omega z in the k direction. 461 00:22:58,300 --> 00:23:00,685 But reminding you, make it always 462 00:23:00,685 --> 00:23:05,060 with respect to the inertial frame-- cross 463 00:23:05,060 --> 00:23:12,460 RGA, the position vector from between the two points. 464 00:23:12,460 --> 00:23:15,810 And so this case, we end up with an x 465 00:23:15,810 --> 00:23:31,860 dot I hat plus-- this is going to be-- omega zk cross L/2 466 00:23:31,860 --> 00:23:34,010 little i hat. 467 00:23:34,010 --> 00:23:36,264 k cross i is j. 468 00:23:36,264 --> 00:23:42,380 So x dot I hat. 469 00:23:42,380 --> 00:23:47,530 This is in the little j moving coordinate system direction. 470 00:23:47,530 --> 00:23:50,650 Omega z is theta dot. 471 00:23:50,650 --> 00:23:57,520 Theta dot L over 2 j hat. 472 00:24:04,220 --> 00:24:06,520 I didn't even leave enough room here. 473 00:24:06,520 --> 00:24:08,680 Let me see this. 474 00:24:08,680 --> 00:24:13,104 x dot I plus. 475 00:24:13,104 --> 00:24:13,980 OK? 476 00:24:13,980 --> 00:24:16,620 Familiar, our omega term. 477 00:24:16,620 --> 00:24:20,260 So that's your velocity of G with respect to O. 478 00:24:20,260 --> 00:24:27,920 We need to find an acceleration of that same point. 479 00:24:27,920 --> 00:24:32,540 And we're going to take the derivative of this to get it. 480 00:24:32,540 --> 00:24:37,800 A point to-- we haven't talked about this in a while. 481 00:24:37,800 --> 00:24:41,110 This has mixed unit vectors. 482 00:24:41,110 --> 00:24:41,610 Right? 483 00:24:41,610 --> 00:24:44,670 It's got unit vectors in the inertial frame and the unit 484 00:24:44,670 --> 00:24:45,950 vectors in the rotating frame. 485 00:24:45,950 --> 00:24:49,080 Is that allowed? 486 00:24:49,080 --> 00:24:50,680 No problem. 487 00:24:50,680 --> 00:24:53,060 It's perfectly legitimate, right? 488 00:24:53,060 --> 00:24:54,665 You have to reduce it, eventually, 489 00:24:54,665 --> 00:24:56,740 to get a workable equation. 490 00:24:56,740 --> 00:25:00,270 But that's just fine at this intermediate stage. 491 00:25:00,270 --> 00:25:01,200 OK. 492 00:25:01,200 --> 00:25:05,900 The next thing you want to do is find this a with respect 493 00:25:05,900 --> 00:25:10,760 to G-- aG with respect to O, the acceleration. 494 00:25:10,760 --> 00:25:17,790 And we know that's just a derivative, remember, 495 00:25:17,790 --> 00:25:20,735 with respect to the inertial frame-- that's 496 00:25:20,735 --> 00:25:24,590 why we have to deal with this rotation business-- of velocity 497 00:25:24,590 --> 00:25:28,660 of G with respect to O. But the derivative of this-- does I 498 00:25:28,660 --> 00:25:29,960 change direction? 499 00:25:29,960 --> 00:25:30,460 Nope. 500 00:25:30,460 --> 00:25:32,020 So this is just x double dot. 501 00:25:35,710 --> 00:25:38,170 And we have a single term here that we 502 00:25:38,170 --> 00:25:39,390 have to take a derivative of. 503 00:25:39,390 --> 00:25:41,935 So we get a theta double dot L/2 j. 504 00:25:48,730 --> 00:25:51,460 This is a little lowercase j. 505 00:25:51,460 --> 00:25:53,250 But now, does j rotate? 506 00:25:56,010 --> 00:25:56,775 Yeah? 507 00:25:56,775 --> 00:25:59,910 And so this is the derivative of a rotating vector. 508 00:25:59,910 --> 00:26:06,810 The derivative of j is minus theta dot I. 509 00:26:06,810 --> 00:26:21,690 So minus theta dot squared L over 2 i hat. 510 00:26:25,298 --> 00:26:26,200 All right? 511 00:26:26,200 --> 00:26:28,870 And that's all there is to getting G. 512 00:26:28,870 --> 00:26:35,185 Now we could have done-- oops, I want that one. 513 00:26:41,990 --> 00:26:51,780 I don't want you to be afraid of using the big, kind of hairy 514 00:26:51,780 --> 00:26:58,450 looking 3D vector equation for acceleration. 515 00:26:58,450 --> 00:27:01,230 Acceleration and this gets called-- this is A with respect 516 00:27:01,230 --> 00:27:03,870 to O, putting it in terms of this problem, 517 00:27:03,870 --> 00:27:07,490 plus the acceleration of G with respect 518 00:27:07,490 --> 00:27:23,230 to A evaluated with no rotation plus omega dot cross RGA 519 00:27:23,230 --> 00:27:39,310 plus omega cross omega cross RGA plus 2 omega cross 520 00:27:39,310 --> 00:27:46,610 velocity GA omega equals 0. 521 00:27:46,610 --> 00:27:50,870 And I should have one, two, three, four, five terms. 522 00:27:50,870 --> 00:27:52,980 There's always five potential terms 523 00:27:52,980 --> 00:27:58,280 when you're evaluating the acceleration of a point 524 00:27:58,280 --> 00:28:02,640 on a rigid body which is translating and rotating. 525 00:28:02,640 --> 00:28:07,950 This is translating and rotating reference frames, 526 00:28:07,950 --> 00:28:10,090 attached to the rigid body. 527 00:28:10,090 --> 00:28:13,470 And then you just go in and fill it in. 528 00:28:13,470 --> 00:28:15,570 This is the acceleration of the rigid body, 529 00:28:15,570 --> 00:28:17,050 the translational acceleration. 530 00:28:17,050 --> 00:28:20,550 In this problem, what is that? 531 00:28:20,550 --> 00:28:21,542 Louder. 532 00:28:21,542 --> 00:28:22,500 AUDIENCE: X double dot. 533 00:28:22,500 --> 00:28:23,083 PROFESSOR: OK. 534 00:28:23,083 --> 00:28:26,286 This guy is capital X double dot I hat. 535 00:28:26,286 --> 00:28:28,780 All right? 536 00:28:28,780 --> 00:28:32,000 This is the acceleration of point G relative to A 537 00:28:32,000 --> 00:28:33,900 if you were in the object, on the object. 538 00:28:33,900 --> 00:28:34,933 It is? 539 00:28:34,933 --> 00:28:35,432 AUDIENCE: 0. 540 00:28:35,432 --> 00:28:36,307 PROFESSOR: All right. 541 00:28:36,307 --> 00:28:37,430 0. 542 00:28:37,430 --> 00:28:41,560 This is the Eulerian term. 543 00:28:41,560 --> 00:28:46,310 This is theta double dot crossed with the distance 544 00:28:46,310 --> 00:28:47,720 from between the two points. 545 00:28:47,720 --> 00:28:49,240 Is this 0? 546 00:28:49,240 --> 00:28:50,290 No, not necessarily. 547 00:28:50,290 --> 00:28:53,524 So this is theta double dot. 548 00:28:56,842 --> 00:28:59,740 This is the L/2. 549 00:28:59,740 --> 00:29:01,180 That's this. 550 00:29:01,180 --> 00:29:05,130 And when you do the cross product of k and I, you get j. 551 00:29:05,130 --> 00:29:07,280 That's that term. 552 00:29:07,280 --> 00:29:11,380 Omega cross omega cross RGA, this is the centripetal term. 553 00:29:11,380 --> 00:29:13,180 Would you think it's going to be 0? 554 00:29:13,180 --> 00:29:14,120 Nope. 555 00:29:14,120 --> 00:29:18,402 So this is k cross I. That's j. k cross j is 556 00:29:18,402 --> 00:29:22,070 I minus-- this is the term that gives you minus-- theta 557 00:29:22,070 --> 00:29:27,880 dot squared L/2 i. 558 00:29:27,880 --> 00:29:30,440 And this is our Coriolis term. 559 00:29:30,440 --> 00:29:33,735 It requires motion of that point, G, relative to A. 560 00:29:33,735 --> 00:29:34,920 Is that moving? 561 00:29:34,920 --> 00:29:35,420 Nope. 562 00:29:35,420 --> 00:29:37,250 So this term just goes to 0. 563 00:29:37,250 --> 00:29:39,197 Get the same answer? 564 00:29:39,197 --> 00:29:39,696 All right. 565 00:29:39,696 --> 00:29:40,990 So don't be afraid of this. 566 00:29:40,990 --> 00:29:42,350 This thing, just lay it down. 567 00:29:42,350 --> 00:29:46,850 Just plug the things in and the right things will fall out. 568 00:29:46,850 --> 00:29:47,350 All right. 569 00:29:47,350 --> 00:29:54,160 On the paper, I break down the acceleration of G with respect 570 00:29:54,160 --> 00:29:58,680 to O into its little i and little j components. 571 00:30:03,000 --> 00:30:07,200 And I end up with acceleration of G with respect 572 00:30:07,200 --> 00:30:09,940 to O. I group the terms. 573 00:30:09,940 --> 00:30:15,871 The I capital X double dot sine theta-- 574 00:30:15,871 --> 00:30:17,370 I actually have to break this-- see, 575 00:30:17,370 --> 00:30:20,570 this is not in the direction of little i or little j. 576 00:30:20,570 --> 00:30:28,030 So I know that I can express capital I as a lowercase i sine 577 00:30:28,030 --> 00:30:30,090 theta plus j cosine theta. 578 00:30:30,090 --> 00:30:31,110 And I use that. 579 00:30:31,110 --> 00:30:35,220 So it's X double dot sine theta minus L/2 580 00:30:35,220 --> 00:30:40,090 theta dot squared-- it is in the little i hat direction. 581 00:30:40,090 --> 00:30:48,610 So this is the i hat term-- plus X double dot cosine theta 582 00:30:48,610 --> 00:30:51,610 plus L/2 theta double dot. 583 00:30:51,610 --> 00:30:53,900 And this is the j hat term. 584 00:30:53,900 --> 00:30:54,820 OK? 585 00:30:54,820 --> 00:30:55,590 Yeah. 586 00:30:55,590 --> 00:30:58,085 AUDIENCE: Why would you choose to put everything in terms 587 00:30:58,085 --> 00:31:02,004 of the rotating inertial frame? 588 00:31:02,004 --> 00:31:03,420 PROFESSOR: You have to always have 589 00:31:03,420 --> 00:31:04,760 these same decisions to make. 590 00:31:04,760 --> 00:31:06,960 And it's what you're comfortable with, 591 00:31:06,960 --> 00:31:09,160 what you think is going to lead to the least work. 592 00:31:09,160 --> 00:31:11,980 You have no idea how much time I spent working on this problem 593 00:31:11,980 --> 00:31:13,520 to put it in a form that I thought 594 00:31:13,520 --> 00:31:14,650 I could teach it to you. 595 00:31:14,650 --> 00:31:19,220 I spent a lot of time on it, the point being, occasionally, you 596 00:31:19,220 --> 00:31:22,820 have to spend a lot of time working them out, going down 597 00:31:22,820 --> 00:31:26,090 a path that doesn't pay off, backing up, going down 598 00:31:26,090 --> 00:31:27,240 the next one. 599 00:31:27,240 --> 00:31:31,410 On quizzes, on homework, most of what 600 00:31:31,410 --> 00:31:34,927 stops you from getting to the final right answer 601 00:31:34,927 --> 00:31:35,510 is confidence. 602 00:31:38,080 --> 00:31:39,490 You've got to believe that you've 603 00:31:39,490 --> 00:31:41,510 learned these things well enough that you know 604 00:31:41,510 --> 00:31:43,710 this is the right thing to do. 605 00:31:43,710 --> 00:31:46,740 And it'll get you there if you just do the arithmetic right. 606 00:31:46,740 --> 00:31:47,647 OK? 607 00:31:47,647 --> 00:31:48,980 And we all make little mistakes. 608 00:31:48,980 --> 00:31:51,120 And you'll all have to back up and do it again. 609 00:31:51,120 --> 00:31:53,161 But that's why the fundamentals and understanding 610 00:31:53,161 --> 00:31:54,980 the basic concepts are so important. 611 00:31:54,980 --> 00:31:57,140 If you've got the concepts down cold, 612 00:31:57,140 --> 00:32:00,730 you'll have confidence that your method's going to work. 613 00:32:00,730 --> 00:32:01,630 OK. 614 00:32:01,630 --> 00:32:06,810 So this is my i direction term. 615 00:32:06,810 --> 00:32:08,740 This is my j direction term. 616 00:32:08,740 --> 00:32:17,290 I can take those and take this bit and plug it in here. 617 00:32:17,290 --> 00:32:18,650 Let's just give it a name. 618 00:32:18,650 --> 00:32:25,694 Let's call this A and this piece here B. 619 00:32:25,694 --> 00:32:27,240 And your A goes right here. 620 00:32:30,730 --> 00:32:35,899 That's A. And this is B. OK? 621 00:32:35,899 --> 00:32:37,690 And not that you've made that substitution, 622 00:32:37,690 --> 00:32:40,310 you can solve for F1. 623 00:32:40,310 --> 00:32:42,880 And they're all in one direction now. 624 00:32:42,880 --> 00:32:45,170 They're all in the little i hat direction. 625 00:32:45,170 --> 00:32:48,100 You make the other substitution, everything's in little j hat. 626 00:32:48,100 --> 00:32:49,450 And you can drop them. 627 00:32:49,450 --> 00:32:51,530 You no longer have to carry it along. 628 00:32:51,530 --> 00:32:57,910 You now have scalar equations you can solve for F1 and F2. 629 00:32:57,910 --> 00:33:00,870 So that's done. 630 00:33:00,870 --> 00:33:04,020 And you take those two expressions-- 631 00:33:04,020 --> 00:33:13,440 so this implies F1 equals-- and this one implies F2 equals-- 632 00:33:13,440 --> 00:33:14,740 and you take those. 633 00:33:14,740 --> 00:33:23,150 And you put F1 and F2-- you need F1 and F2. 634 00:33:23,150 --> 00:33:25,690 And you plug it in here. 635 00:33:25,690 --> 00:33:29,125 And in this equation, you only need an F2. 636 00:33:37,739 --> 00:33:42,210 And when you do that, you get your two equations of motion. 637 00:33:42,210 --> 00:33:43,360 Now a couple things happen. 638 00:33:43,360 --> 00:33:49,920 It turns out, when you do this-- when 639 00:33:49,920 --> 00:33:54,970 you make the substitution in here-- you end up with an MX 640 00:33:54,970 --> 00:34:00,420 double dot cosine squared theta and an M2 X double dot sine 641 00:34:00,420 --> 00:34:01,920 squared theta. 642 00:34:01,920 --> 00:34:04,740 And sine squared plus cosine squared 643 00:34:04,740 --> 00:34:07,110 conveniently equals-- 1. 644 00:34:07,110 --> 00:34:09,159 And that collapses and goes away. 645 00:34:09,159 --> 00:34:19,300 So you end up with the two final expressions 646 00:34:19,300 --> 00:34:21,260 here after you've made those combinations. 647 00:34:21,260 --> 00:34:23,464 So from four over here-- I'll call it four prime-- 648 00:34:23,464 --> 00:34:25,130 you get one of your equations of motion. 649 00:35:01,760 --> 00:35:03,010 That's one equation of motion. 650 00:35:03,010 --> 00:35:07,500 It's primarily about the translation of-- I 651 00:35:07,500 --> 00:35:09,260 mean rotation-- of the system. 652 00:35:09,260 --> 00:35:13,320 It derives from all those substitutions in this equation. 653 00:35:13,320 --> 00:35:15,910 Curiously, these two bits go together, 654 00:35:15,910 --> 00:35:18,790 which, if you had worked around A and you used, 655 00:35:18,790 --> 00:35:20,980 dangerously, perhaps, a parallel axis theorem, 656 00:35:20,980 --> 00:35:23,160 you would have ended up with ML squared over 3. 657 00:35:29,842 --> 00:35:31,300 But that's your first equa-- that's 658 00:35:31,300 --> 00:35:32,640 one of your equations of motion. 659 00:35:32,640 --> 00:35:37,150 And from the first one, it comes from the sum of the forces 660 00:35:37,150 --> 00:35:43,990 on the main mass, M1, plus M2 X double dot 661 00:35:43,990 --> 00:35:48,640 plus b X dot plus KX. 662 00:35:48,640 --> 00:35:51,490 All the usual, just things for mass spring oscillator. 663 00:35:51,490 --> 00:35:54,540 But now you've got these additional forces 664 00:35:54,540 --> 00:35:57,180 that are exerted on that because it's got this pendulum 665 00:35:57,180 --> 00:35:58,420 flagging back and forth. 666 00:36:18,670 --> 00:36:22,548 And if I made a mistake on the board, believe the paper. 667 00:36:22,548 --> 00:36:25,320 I think I've got it all transcribed right. 668 00:36:25,320 --> 00:36:28,810 This is your force equation on the main mass. 669 00:36:28,810 --> 00:36:31,520 If you didn't have the pendulum hanging there, 670 00:36:31,520 --> 00:36:35,560 there's your equation of motion for a cart going 671 00:36:35,560 --> 00:36:37,460 back and forth with springs. 672 00:36:37,460 --> 00:36:40,570 Then you have this pendulum hanging on it, which 673 00:36:40,570 --> 00:36:42,740 puts additional forces on it. 674 00:36:42,740 --> 00:36:45,690 It comes through those F1 and F2 terms. 675 00:36:45,690 --> 00:36:48,410 And you can see they have to do with having to accelerate 676 00:36:48,410 --> 00:36:50,770 things down there. 677 00:36:50,770 --> 00:36:51,270 OK? 678 00:36:55,270 --> 00:36:56,900 Now these two equations are actually 679 00:36:56,900 --> 00:36:59,350 quite easy to linearize. 680 00:36:59,350 --> 00:37:01,910 So linearize means you can always 681 00:37:01,910 --> 00:37:04,630 linearize around the equilibrium position. 682 00:37:04,630 --> 00:37:06,560 This is its equilibrium position. 683 00:37:06,560 --> 00:37:07,465 Small motions. 684 00:37:10,010 --> 00:37:12,390 That can be described, what you see there, 685 00:37:12,390 --> 00:37:14,790 by linearized equations of motion. 686 00:37:14,790 --> 00:37:25,590 And to linearize-- so for theta small sine theta 687 00:37:25,590 --> 00:37:28,240 is approximately equal to theta. 688 00:37:28,240 --> 00:37:30,560 Cosine theta is approximately equal to 1. 689 00:37:30,560 --> 00:37:32,880 And you just have sines and cosines in here. 690 00:37:32,880 --> 00:37:34,380 You let this guy go to 1. 691 00:37:34,380 --> 00:37:36,710 You let this go to theta. 692 00:37:36,710 --> 00:37:39,090 Here, 1. 693 00:37:39,090 --> 00:37:40,880 Here, theta. 694 00:37:40,880 --> 00:37:42,740 And you have linearized equations of motion. 695 00:37:45,360 --> 00:37:48,340 Now other problems can be harder to linearize. 696 00:37:48,340 --> 00:37:51,200 But this is particularly simple. 697 00:37:51,200 --> 00:37:53,992 Two linear equations, solvable. 698 00:37:53,992 --> 00:37:55,450 If you solve them, they'll give you 699 00:37:55,450 --> 00:38:03,970 two natural frequencies for the system and two vibration modes. 700 00:38:03,970 --> 00:38:04,470 All right. 701 00:38:04,470 --> 00:38:07,910 That's the end of this problem. 702 00:38:07,910 --> 00:38:10,170 Any last questions about it? 703 00:38:10,170 --> 00:38:12,800 I'm going to move on to this topic of center percussion. 704 00:38:28,470 --> 00:38:28,970 OK. 705 00:38:28,970 --> 00:38:30,178 This next one is kind of fun. 706 00:38:30,178 --> 00:38:32,220 It has a real practical purpose in life. 707 00:38:32,220 --> 00:38:35,870 I didn't bring my tennis racket or my baseball bat. 708 00:38:35,870 --> 00:38:38,300 But if any of you play sports that 709 00:38:38,300 --> 00:38:42,640 use things that hit things, you know, when hit the baseball 710 00:38:42,640 --> 00:38:44,660 and you hit it down on the handle, 711 00:38:44,660 --> 00:38:45,900 it really stings your hands. 712 00:38:45,900 --> 00:38:47,840 Right? 713 00:38:47,840 --> 00:38:49,930 Or the tennis racket, if you don't hit it right, 714 00:38:49,930 --> 00:38:51,850 you feel a lot of forces in your hands. 715 00:38:51,850 --> 00:38:55,425 And you hit it really sweet, you feel almost no force at all. 716 00:38:55,425 --> 00:38:58,010 How many have had that experience? 717 00:38:58,010 --> 00:38:58,650 Really common. 718 00:38:58,650 --> 00:38:59,149 OK. 719 00:38:59,149 --> 00:39:01,665 So is there a right place to hit it? 720 00:39:01,665 --> 00:39:02,820 The answer's yes. 721 00:39:02,820 --> 00:39:08,020 And we're going to go through that right now. 722 00:39:08,020 --> 00:39:10,760 So here's our-- I'm looking down. 723 00:39:10,760 --> 00:39:12,250 Here's my baseball bat. 724 00:39:12,250 --> 00:39:14,170 z is in this direction. 725 00:39:14,170 --> 00:39:16,750 This is kind of the top view. 726 00:39:16,750 --> 00:39:18,490 So here's my handle of my bat. 727 00:39:21,920 --> 00:39:22,890 Ball's coming in. 728 00:39:27,710 --> 00:39:29,200 Put some force on it. 729 00:39:29,200 --> 00:39:33,400 I think I put it lowercase in the notes. 730 00:39:33,400 --> 00:39:35,060 OK? 731 00:39:35,060 --> 00:39:38,060 And I want to know-- and I want to minimize that force 732 00:39:38,060 --> 00:39:39,124 at my hands. 733 00:39:39,124 --> 00:39:40,540 I'm going to call this place where 734 00:39:40,540 --> 00:39:47,685 it hits P. I'm going to say here's its center of mass at G. 735 00:39:47,685 --> 00:39:49,730 The point about which you're holding it 736 00:39:49,730 --> 00:39:57,100 and it's rotating is A. And my coordinate system attached 737 00:39:57,100 --> 00:40:01,240 to the bat-- we always have this coordinate system attached 738 00:40:01,240 --> 00:40:03,520 to the bad in these rotational problems 739 00:40:03,520 --> 00:40:06,670 so that we can divine things like moments of inertia. 740 00:40:06,670 --> 00:40:09,540 So it's a coordinate system attached to the bat. 741 00:40:09,540 --> 00:40:11,770 And here's my y direction. 742 00:40:11,770 --> 00:40:13,110 And z's coming out of the board. 743 00:40:16,490 --> 00:40:21,940 And this distance is going to be important to me, RGA. 744 00:40:25,770 --> 00:40:30,990 And this distance here, I'm going to call q. 745 00:40:30,990 --> 00:40:31,950 It's my unknown. 746 00:40:31,950 --> 00:40:35,490 And I want to know, where's the sweet spot. 747 00:40:35,490 --> 00:40:40,180 What do I want q to be so that I minimize the force at A, 748 00:40:40,180 --> 00:40:41,380 where I'm hanging onto it. 749 00:40:45,970 --> 00:40:47,440 So how many degrees of freedom? 750 00:40:47,440 --> 00:40:51,210 How many independent coordinates do I need to do this problem? 751 00:40:51,210 --> 00:40:54,160 This problem is complicated, because it's a-- you know, 752 00:40:54,160 --> 00:40:56,270 a real life situation is probably complicated. 753 00:40:56,270 --> 00:40:59,346 And I'm going to do a lot of simplification and claim 754 00:40:59,346 --> 00:41:01,220 that the answer-- and then look at the answer 755 00:41:01,220 --> 00:41:02,480 and say, does it makes sense. 756 00:41:02,480 --> 00:41:05,590 Did my simplifications make sense? 757 00:41:05,590 --> 00:41:08,850 So I'm going to do a number of simplifications. 758 00:41:08,850 --> 00:41:12,210 I'm going to argue that I think I can get at most of the answer 759 00:41:12,210 --> 00:41:16,240 by essentially saying, right here where I'm hitting it, 760 00:41:16,240 --> 00:41:18,430 my wrists are like a hinge point. 761 00:41:18,430 --> 00:41:21,970 That's point A. I'm going to assume it rotates about a fixed 762 00:41:21,970 --> 00:41:25,522 point at that moment. 763 00:41:25,522 --> 00:41:26,880 OK? 764 00:41:26,880 --> 00:41:29,600 Pretty gross simplification of the real thing going on. 765 00:41:29,600 --> 00:41:32,090 You've really got muscles, you're putting moments on it, 766 00:41:32,090 --> 00:41:34,280 your wrists actually are moving some. 767 00:41:34,280 --> 00:41:37,450 But I'm just going to say rotating only about that point. 768 00:41:37,450 --> 00:41:39,010 It stays in this plane. 769 00:41:39,010 --> 00:41:40,360 It's a planar motion problem. 770 00:41:40,360 --> 00:41:43,010 At most, three possible degrees of freedom, 771 00:41:43,010 --> 00:41:45,620 x, y, and some theta. 772 00:41:45,620 --> 00:41:46,904 Right? 773 00:41:46,904 --> 00:41:48,570 And I'm going to make the simplification 774 00:41:48,570 --> 00:41:50,150 that it's pinned at A. 775 00:41:50,150 --> 00:41:53,665 So if it's pinned at A, how many constraints is that? 776 00:41:53,665 --> 00:41:54,455 AUDIENCE: Two. 777 00:41:54,455 --> 00:41:55,080 PROFESSOR: Two. 778 00:41:55,080 --> 00:41:56,160 One left. 779 00:41:56,160 --> 00:42:00,660 So I only need-- I'm going to pick one coordinate. 780 00:42:00,660 --> 00:42:03,220 That's going to be my theta. 781 00:42:03,220 --> 00:42:06,210 So even though I've made these gross simplifications, 782 00:42:06,210 --> 00:42:09,470 will the answer actually turn out to be meaningful? 783 00:42:09,470 --> 00:42:12,940 And this G is my center of mass. 784 00:42:12,940 --> 00:42:22,850 And let's assume that we know Izz with respect 785 00:42:22,850 --> 00:42:29,020 to G, that is, mass moment of inertia for rotation in z. 786 00:42:29,020 --> 00:42:29,950 OK? 787 00:42:29,950 --> 00:42:31,085 Assume that we know that. 788 00:42:31,085 --> 00:42:31,650 It's given. 789 00:42:35,110 --> 00:42:41,480 So for this problem, when you have fixed 790 00:42:41,480 --> 00:42:44,430 rotation points-- this problem here, 791 00:42:44,430 --> 00:42:46,330 that A point wasn't fixed. 792 00:42:46,330 --> 00:42:48,230 It was moving, accelerating. 793 00:42:48,230 --> 00:42:50,980 It could have had lots of complicated terms 794 00:42:50,980 --> 00:42:54,910 in its expression for torque if you'd done it that way. 795 00:42:54,910 --> 00:42:57,070 This is fixed at A. It makes a lot of sense 796 00:42:57,070 --> 00:43:01,790 to compute moments about A. All right? 797 00:43:01,790 --> 00:43:04,440 So my first equation here, that I 798 00:43:04,440 --> 00:43:11,350 want, is the sum of the moments, torques with respect to A. 799 00:43:11,350 --> 00:43:13,790 And these are-- well, they will all turn out 800 00:43:13,790 --> 00:43:17,460 to be in the z direction here. 801 00:43:17,460 --> 00:43:27,280 And they are time rated change of the angular momentum 802 00:43:27,280 --> 00:43:31,180 with respect to G, which, in these planar motion problems, 803 00:43:31,180 --> 00:43:35,600 always then boils down to the mass moment of inertia 804 00:43:35,600 --> 00:43:40,250 for the axis you're rotating it about times the angular 805 00:43:40,250 --> 00:43:41,250 acceleration. 806 00:43:41,250 --> 00:43:43,890 That's that term. 807 00:43:43,890 --> 00:43:50,450 Oops, not G. A. 808 00:43:50,450 --> 00:43:50,950 Excuse me. 809 00:43:50,950 --> 00:43:51,783 We picked our point. 810 00:43:51,783 --> 00:43:53,170 We've got to stick with it here. 811 00:43:53,170 --> 00:43:53,669 OK. 812 00:43:53,669 --> 00:43:56,680 I need that. 813 00:43:56,680 --> 00:44:03,690 And this is in the k hat direction. 814 00:44:03,690 --> 00:44:07,420 This is the only term in our torque equation. 815 00:44:10,390 --> 00:44:13,560 What are the external torques? 816 00:44:13,560 --> 00:44:17,830 Well now, z is out of the board like this. 817 00:44:17,830 --> 00:44:21,920 I've got a force acting on a moment arm 818 00:44:21,920 --> 00:44:24,780 giving me a torque in the other direction. 819 00:44:24,780 --> 00:44:30,448 So this is minus fq-- also k hat. 820 00:44:39,074 --> 00:44:40,490 Now I might have done this sooner, 821 00:44:40,490 --> 00:44:43,530 but we should look at a free body diagram. 822 00:44:43,530 --> 00:44:47,830 So here's a little stick figure of my bat. 823 00:44:47,830 --> 00:44:50,850 And I potentially have an unknown force 824 00:44:50,850 --> 00:44:56,260 in the y direction, Ry, and another unknown force 825 00:44:56,260 --> 00:45:00,040 in the x direction here at A, that pin point 826 00:45:00,040 --> 00:45:02,552 where it's rotating it about. 827 00:45:02,552 --> 00:45:05,920 Out here is G. 828 00:45:05,920 --> 00:45:11,100 And at G, is there an mG term? 829 00:45:11,100 --> 00:45:12,610 Well, it's gravity, right? 830 00:45:12,610 --> 00:45:15,300 But it's acting-- it's in the minus z direction. 831 00:45:15,300 --> 00:45:17,420 And I'm looking down on it. 832 00:45:17,420 --> 00:45:18,480 This is a top view. 833 00:45:21,340 --> 00:45:25,040 So gravity would create a moment in the y direction. 834 00:45:25,040 --> 00:45:27,360 And I don't have to deal with that. 835 00:45:27,360 --> 00:45:30,710 My only moment-- this is my torque equation-- 836 00:45:30,710 --> 00:45:33,310 only has k hat terms in it. 837 00:45:33,310 --> 00:45:36,490 Gravity creates a moment in the y, 838 00:45:36,490 --> 00:45:39,841 which, in fact, you have to-- that's a static equilibrium 839 00:45:39,841 --> 00:45:41,590 problem-- you do have to provide that when 840 00:45:41,590 --> 00:45:43,440 you're holding the bat or else the bat would droop. 841 00:45:43,440 --> 00:45:43,939 Right? 842 00:45:43,939 --> 00:45:46,610 It doesn't have anything to do with the dynamics, actually. 843 00:45:46,610 --> 00:45:47,110 OK. 844 00:45:47,110 --> 00:45:50,100 Here's G. There's no gravity term that you can see. 845 00:45:50,100 --> 00:45:52,400 It's pointing into the board. 846 00:45:52,400 --> 00:45:57,750 And out here is my point P. And here's f. 847 00:45:57,750 --> 00:46:00,510 OK. 848 00:46:00,510 --> 00:46:05,460 So I need an equation-- I have an equation for moment 849 00:46:05,460 --> 00:46:06,590 equilibrium. 850 00:46:06,590 --> 00:46:08,960 I need some dynamic equation. 851 00:46:08,960 --> 00:46:13,320 I have two possible equations for force equilibrium. 852 00:46:13,320 --> 00:46:14,701 How many unknowns do I have? 853 00:46:20,960 --> 00:46:22,270 But no others. 854 00:46:22,270 --> 00:46:23,270 Three possible unknowns. 855 00:46:23,270 --> 00:46:24,645 I'm going to need three equations 856 00:46:24,645 --> 00:46:26,840 to get rid of these two terms. 857 00:46:26,840 --> 00:46:27,900 Right? 858 00:46:27,900 --> 00:46:30,040 And I am looking ahead. 859 00:46:30,040 --> 00:46:32,929 My objective is to make that term go to 0. 860 00:46:32,929 --> 00:46:34,720 So I both have to find an expression for it 861 00:46:34,720 --> 00:46:36,511 and then figure out how to make it go to 0. 862 00:46:36,511 --> 00:46:38,630 That's how you do this problem. 863 00:46:41,140 --> 00:46:41,640 OK. 864 00:46:41,640 --> 00:46:46,510 So let's do sum of the forces in the y. 865 00:46:51,220 --> 00:47:02,106 So you have an Ry j hat minus f, also in the j hat direction. 866 00:47:02,106 --> 00:47:03,855 And that's basically all there is to that. 867 00:47:03,855 --> 00:47:07,140 But that must be the mass of the bat 868 00:47:07,140 --> 00:47:09,760 times the acceleration of G with respect 869 00:47:09,760 --> 00:47:13,460 to O in the y direction. 870 00:47:13,460 --> 00:47:15,800 So that's our j hat component. 871 00:47:15,800 --> 00:47:19,000 This is just the component of acceleration in the j hat 872 00:47:19,000 --> 00:47:19,870 direction. 873 00:47:19,870 --> 00:47:25,770 And the sum of the forces in x, we look at those. 874 00:47:25,770 --> 00:47:28,850 Well, we have an Rx in the i. 875 00:47:28,850 --> 00:47:31,630 And there are no others. 876 00:47:31,630 --> 00:47:34,180 And that must be, then, the mass of the bat 877 00:47:34,180 --> 00:47:38,480 times the acceleration, G, with respect to O, of the bit 878 00:47:38,480 --> 00:47:40,160 in the x direction. 879 00:47:40,160 --> 00:47:44,230 And that'll be in the i hat. 880 00:47:44,230 --> 00:47:46,640 So I've got two unknown accelerations, now, 881 00:47:46,640 --> 00:47:48,410 that I have to deal with. 882 00:47:48,410 --> 00:47:51,260 But we know lots of kinematics now. 883 00:48:03,100 --> 00:48:06,330 So let's do it this way just to give you the practice. 884 00:48:06,330 --> 00:48:10,060 Acceleration of G with respect to O. Acceleration of A 885 00:48:10,060 --> 00:48:15,890 with respect to O plus the acceleration of G with respect 886 00:48:15,890 --> 00:48:32,940 A omega equals 0 plus omega dot cross RGA plus 2 omega 887 00:48:32,940 --> 00:48:46,160 cross VGA omega equals 0, and finally, plus omega 888 00:48:46,160 --> 00:48:51,215 cross omega cross RGA. 889 00:48:54,350 --> 00:48:59,150 One, two, three, four, five terms. 890 00:48:59,150 --> 00:49:02,230 Our vector 3D equation is figured out quickly. 891 00:49:02,230 --> 00:49:02,970 What's this term? 892 00:49:06,218 --> 00:49:08,080 AUDIENCE: 0. 893 00:49:08,080 --> 00:49:08,900 PROFESSOR: 0? 894 00:49:08,900 --> 00:49:09,716 Why? 895 00:49:09,716 --> 00:49:10,590 AUDIENCE: It's fixed. 896 00:49:10,590 --> 00:49:12,048 PROFESSOR: Fixed point of rotation, 897 00:49:12,048 --> 00:49:13,090 that's our assumption. 898 00:49:13,090 --> 00:49:14,270 This guy goes to 0. 899 00:49:14,270 --> 00:49:15,100 OK. 900 00:49:15,100 --> 00:49:18,662 The velocity of G with respect to A? 901 00:49:18,662 --> 00:49:19,162 AUDIENCE: 0. 902 00:49:19,162 --> 00:49:19,745 PROFESSOR: OK. 903 00:49:19,745 --> 00:49:21,580 Another 0. 904 00:49:21,580 --> 00:49:22,890 Omega dot cross RGA? 905 00:49:27,920 --> 00:49:29,861 So that could-- you know, if there's 906 00:49:29,861 --> 00:49:31,980 an angular acceleration, which there might be, 907 00:49:31,980 --> 00:49:35,130 cross R, that's a perfectly legitimate term. 908 00:49:35,130 --> 00:49:38,600 And that'll give you a-- this will give you 909 00:49:38,600 --> 00:49:42,260 a theta double dot, just before. 910 00:49:42,260 --> 00:49:49,590 It's in the k direction, cross RGA in the i. 911 00:49:49,590 --> 00:49:52,020 That's this term right here. 912 00:49:52,020 --> 00:49:52,520 OK. 913 00:49:52,520 --> 00:49:54,820 This velocity of G with respect to A 914 00:49:54,820 --> 00:49:56,140 on the bat, that went to 0. 915 00:49:56,140 --> 00:49:58,140 The Coriolis, that goes away. 916 00:49:58,140 --> 00:50:01,200 So at this point, I've only got one term from there. 917 00:50:01,200 --> 00:50:02,910 And this term is our centripetal term. 918 00:50:02,910 --> 00:50:04,993 You think there's going to be a centripetal force? 919 00:50:07,694 --> 00:50:14,580 Common sense should be any time you have a rotational moment, 920 00:50:14,580 --> 00:50:18,570 something rotating about a point, if the center of mass 921 00:50:18,570 --> 00:50:24,700 is not the point of rotation, you 922 00:50:24,700 --> 00:50:27,310 will be forcing a mass to move in a circle. 923 00:50:27,310 --> 00:50:31,270 And that always requires-- that centripetal acceleration-- 924 00:50:31,270 --> 00:50:32,520 always requires a force. 925 00:50:32,520 --> 00:50:35,040 This, for sure, will give you a term 926 00:50:35,040 --> 00:50:43,057 which is a minus omega z squared type term times the distance 927 00:50:43,057 --> 00:50:43,556 RGA. 928 00:50:46,180 --> 00:50:47,660 That's its radius. 929 00:50:47,660 --> 00:50:48,840 And it's inward. 930 00:50:48,840 --> 00:50:52,010 So in this direction, it will be i hat. 931 00:50:52,010 --> 00:50:53,710 Three terms, same as before. 932 00:50:53,710 --> 00:50:54,960 We now have our accelerations. 933 00:51:05,190 --> 00:51:05,750 OK. 934 00:51:05,750 --> 00:51:07,790 So from our top equation up there 935 00:51:07,790 --> 00:51:11,650 from the summation of forces in the y, 936 00:51:11,650 --> 00:51:13,770 we'll now substitute these in. 937 00:51:13,770 --> 00:51:21,940 We have M RGA j hat. 938 00:51:21,940 --> 00:51:28,750 The acceleration term, we ended up with the-- the k cross i 939 00:51:28,750 --> 00:51:29,800 term gives you j. 940 00:51:29,800 --> 00:51:33,110 So the Eulerian term is going to show up here. 941 00:51:33,110 --> 00:51:38,490 So that, we need a theta double dot. 942 00:51:38,490 --> 00:51:39,810 And this is in the j direction. 943 00:51:39,810 --> 00:51:44,680 That comes from the acceleration we figured out. 944 00:51:44,680 --> 00:51:50,390 And that's equal to minus f plus Ry. 945 00:51:50,390 --> 00:51:52,881 And this is all little j hat. 946 00:52:03,690 --> 00:52:05,570 Let me write the other one first here. 947 00:52:05,570 --> 00:52:10,330 Here's the summation in the x direction. 948 00:52:10,330 --> 00:52:16,550 It's mass times our acceleration in the i hat direction. 949 00:52:16,550 --> 00:52:19,790 And we only ended up with one term doing that, 950 00:52:19,790 --> 00:52:21,870 and that's our centripetal term. 951 00:52:21,870 --> 00:52:36,460 So we get minus RGA theta dot squared i hat. 952 00:52:36,460 --> 00:52:39,430 That's the centripetal acceleration times the mass. 953 00:52:39,430 --> 00:52:42,950 And the only term that it is equated to 954 00:52:42,950 --> 00:52:45,788 is the other reaction force. 955 00:52:45,788 --> 00:52:47,150 OK? 956 00:52:47,150 --> 00:52:51,290 So here's the key step in this problem. 957 00:52:51,290 --> 00:52:56,910 We want what to be 0 so that we're 958 00:52:56,910 --> 00:53:00,032 hitting it at the sweet spot? 959 00:53:00,032 --> 00:53:01,282 What's the original objective? 960 00:53:05,220 --> 00:53:07,730 We want Ry to go away. 961 00:53:07,730 --> 00:53:08,710 All right? 962 00:53:13,610 --> 00:53:15,127 Just make it go away. 963 00:53:18,330 --> 00:53:37,010 So f, then-- I'll do it as an intermediate step. 964 00:53:37,010 --> 00:53:39,240 Just a second. f is then-- I'm going to leave Ry 965 00:53:39,240 --> 00:53:40,560 for just a second longer. 966 00:53:40,560 --> 00:53:48,270 Ry minus M RGA theta double dot. 967 00:53:48,270 --> 00:53:50,910 And since this is my-- I can drop the j hats. 968 00:53:50,910 --> 00:53:53,380 This is now a scalar equation. 969 00:53:53,380 --> 00:53:55,200 I can find an expression for this force 970 00:53:55,200 --> 00:53:56,950 that-- I don't even know what the force is 971 00:53:56,950 --> 00:53:59,190 that the ball exerts on the bat. 972 00:53:59,190 --> 00:54:01,060 But it exists. 973 00:54:01,060 --> 00:54:02,990 I can say this must be true. 974 00:54:02,990 --> 00:54:07,000 And my objective is that this should be 0. 975 00:54:07,000 --> 00:54:10,330 So that says, when that's true, f 976 00:54:10,330 --> 00:54:14,010 is minus M RGA theta double dot. 977 00:54:14,010 --> 00:54:17,056 And I'm going to use that in just a moment. 978 00:54:17,056 --> 00:54:22,660 So I'm going to substitute this into our first equation 979 00:54:22,660 --> 00:54:26,640 up there for moment, equation number one. 980 00:54:26,640 --> 00:54:35,870 So this implies that f is minus M RG with respect 981 00:54:35,870 --> 00:54:38,570 to A theta double dot. 982 00:54:38,570 --> 00:54:46,590 And this goes into 1, which I have in fq. 983 00:54:46,590 --> 00:54:49,891 So I do that. 984 00:54:53,249 --> 00:54:54,040 Where are my notes? 985 00:55:17,640 --> 00:55:19,610 This is now also a scalar equation. 986 00:55:19,610 --> 00:55:21,190 I can drop the k hats. 987 00:55:21,190 --> 00:55:28,520 So Izz with respect to A theta double dot equals minus fq. 988 00:55:28,520 --> 00:55:32,680 But f is minus, so the minuses cancel. 989 00:55:32,680 --> 00:55:42,916 So I get M RGA theta double dot times q. 990 00:55:48,580 --> 00:56:10,370 Now Izz with respect to A can be expressed as M times some thing 991 00:56:10,370 --> 00:56:14,240 we call the radius of gyration squared with respect-- 992 00:56:14,240 --> 00:56:16,250 and this has got to be-- with respect to A. 993 00:56:16,250 --> 00:56:20,676 The radius of gyration for rotation about A, 994 00:56:20,676 --> 00:56:23,440 I can find an expression like this 995 00:56:23,440 --> 00:56:26,310 so that it's equal to Izz with respect to A. 996 00:56:26,310 --> 00:56:28,670 And all the radius of gyration means 997 00:56:28,670 --> 00:56:33,210 is that if I took all the mass and put it that distance away, 998 00:56:33,210 --> 00:56:36,450 the mass moment of inertia of that concentrated point 999 00:56:36,450 --> 00:56:38,240 mass with respect to the point of rotation 1000 00:56:38,240 --> 00:56:41,410 is the same as in the real object. 1001 00:56:41,410 --> 00:56:44,640 So this is just a convenience that we use. 1002 00:56:44,640 --> 00:56:47,370 So that's M Kappa squared. 1003 00:56:47,370 --> 00:57:00,840 And that says, then, that M Kappa squared A equals M RGA 1004 00:57:00,840 --> 00:57:03,450 theta double dot-- I have a theta double dot here too. 1005 00:57:03,450 --> 00:57:06,310 Sorry-- times q. 1006 00:57:06,310 --> 00:57:08,380 And I can solve for q. 1007 00:57:08,380 --> 00:57:11,000 The theta double dots go away. 1008 00:57:11,000 --> 00:57:13,540 The M's go away. 1009 00:57:13,540 --> 00:57:21,797 And q is just Kappa squared A divided by RGA. 1010 00:57:33,870 --> 00:57:42,610 And that's the answer to this center of percussion problem. 1011 00:57:42,610 --> 00:57:43,370 Here's your bat. 1012 00:57:47,900 --> 00:57:59,686 Here's A. Here's G. Here's P. The q is about here. 1013 00:58:02,560 --> 00:58:04,280 Whoops, excuse me. 1014 00:58:04,280 --> 00:58:12,530 q is the distance to P. It's always 1015 00:58:12,530 --> 00:58:16,811 outside of G. It will always be outside of G. 1016 00:58:16,811 --> 00:58:19,060 We'll have to think about, maybe, the reason for that. 1017 00:58:19,060 --> 00:58:21,290 So here's the center of mass. 1018 00:58:21,290 --> 00:58:21,790 Here's RGA. 1019 00:58:26,440 --> 00:58:27,870 and here's this point at which you 1020 00:58:27,870 --> 00:58:29,840 want to hit at the sweet spot. 1021 00:58:29,840 --> 00:58:35,390 And it's always Kappa squared A over RGA. 1022 00:58:35,390 --> 00:58:38,540 Now how do you get to Kappa squared A? 1023 00:58:38,540 --> 00:58:41,250 Well, you need to evaluate Izz with respect to A. 1024 00:58:41,250 --> 00:58:45,770 But Izz with respect to A, you can do parallel axis theorem, 1025 00:58:45,770 --> 00:58:51,883 Izz with respect with respect to G plus M RGA squared. 1026 00:58:51,883 --> 00:58:52,630 Right? 1027 00:58:52,630 --> 00:58:53,860 And then you have that. 1028 00:58:53,860 --> 00:58:56,400 So if we were to do this-- I'll give you a quick example. 1029 00:58:56,400 --> 00:59:13,980 If this were a uniform rod, Izz G equals ML squared over 12. 1030 00:59:13,980 --> 00:59:27,960 Izz with respect to A ML squared over 12 plus MRG 1031 00:59:27,960 --> 00:59:31,840 with respect to A squared. 1032 00:59:31,840 --> 00:59:39,410 That is ML squared over 12 plus ML squared-- 1033 00:59:39,410 --> 00:59:41,830 this is L/2 to RGA, half the length, 1034 00:59:41,830 --> 00:59:44,130 if I put A right at the end. 1035 00:59:44,130 --> 00:59:46,090 So I'm doing the simple rod. 1036 00:59:46,090 --> 00:59:50,040 I'm putting A right here, G right in the middle. 1037 00:59:50,040 --> 00:59:56,830 So this is L/2-- ML squared over 2 squared, 4. 1038 00:59:56,830 --> 01:00:03,440 You add those two together, you get ML squared over 3 1039 01:00:03,440 --> 01:00:08,150 equals M Kappa A squared. 1040 01:00:08,150 --> 01:00:10,566 So Kappa squared A is L squared over 3. 1041 01:00:15,880 --> 01:00:24,940 And q equals L squared over 3 divided by L/2. 1042 01:00:24,940 --> 01:00:33,080 So you get 2/3 L. 2/3 of the length puts you out here. 1043 01:00:33,080 --> 01:00:35,650 Or at 1/2 the length-- so here's L/2. 1044 01:00:35,650 --> 01:00:39,640 Here's P at 2L/3. 1045 01:00:45,970 --> 01:00:51,650 So anytime you run into these center of percussion problems-- 1046 01:00:51,650 --> 01:00:56,810 tennis racquet, baseball bats, whatever-- the right place 1047 01:00:56,810 --> 01:01:01,671 to hit it is away from the point of rotation. 1048 01:01:01,671 --> 01:01:03,920 Now do you think this is-- now how good is this model? 1049 01:01:12,100 --> 01:01:14,220 You know, you start thinking about, well, 1050 01:01:14,220 --> 01:01:18,000 what are the things-- if the bat handles really moving. 1051 01:01:18,000 --> 01:01:21,605 It's still actually remarkably good. 1052 01:01:21,605 --> 01:01:23,390 It's remarkably good. 1053 01:01:23,390 --> 01:01:25,504 In fact, the fact these other things 1054 01:01:25,504 --> 01:01:26,920 are happening, like you're putting 1055 01:01:26,920 --> 01:01:30,160 some moment on with your wrists, the fact that you still 1056 01:01:30,160 --> 01:01:32,610 have some speed down here, all you're doing 1057 01:01:32,610 --> 01:01:36,680 is kind of maybe changing the point it's rotating around. 1058 01:01:36,680 --> 01:01:40,237 But you're still having to exert forces with your hands 1059 01:01:40,237 --> 01:01:41,820 at that point where you're holding it. 1060 01:01:41,820 --> 01:01:44,070 And you just want those forces to go to 0. 1061 01:01:44,070 --> 01:01:45,842 All of these other complications, 1062 01:01:45,842 --> 01:01:48,300 you're still going to find out, to make that force go to 0, 1063 01:01:48,300 --> 01:01:51,350 it's approximately that. 1064 01:01:51,350 --> 01:01:53,660 Really a pretty good answer. 1065 01:01:53,660 --> 01:01:54,180 All right. 1066 01:01:58,440 --> 01:02:03,710 Ah, let's go back. 1067 01:02:03,710 --> 01:02:05,799 And we have another piece of information 1068 01:02:05,799 --> 01:02:08,090 that we developed in this problem that we haven't used. 1069 01:02:11,960 --> 01:02:13,430 We had another equation. 1070 01:02:13,430 --> 01:02:17,080 We had the acceleration in the i direction. 1071 01:02:17,080 --> 01:02:19,763 And it's just equal to RX. 1072 01:02:19,763 --> 01:02:24,015 The RX force is M times the acceleration in the i 1073 01:02:24,015 --> 01:02:24,515 direction. 1074 01:02:28,810 --> 01:02:42,510 And that's minus RG with respect to A theta dot squared. 1075 01:02:42,510 --> 01:02:46,770 That's our old centripetal force term, right? 1076 01:02:46,770 --> 01:02:48,655 Centrifugal force, centripetal acceleration. 1077 01:02:48,655 --> 01:02:50,910 Here's the centripetal acceleration times 1078 01:02:50,910 --> 01:02:56,070 the mass is the force to make the bat go in the circle. 1079 01:02:56,070 --> 01:02:58,763 You have to provide that force. 1080 01:02:58,763 --> 01:02:59,263 OK? 1081 01:03:06,010 --> 01:03:11,420 So I want to finish up by a loose end 1082 01:03:11,420 --> 01:03:15,524 that many people have-- continue to have-- 1083 01:03:15,524 --> 01:03:16,440 a little trouble with. 1084 01:03:16,440 --> 01:03:20,250 Because I get questions on Muddy cards about this. 1085 01:03:20,250 --> 01:03:21,800 And it was on this last homework. 1086 01:03:21,800 --> 01:03:26,340 I don't know how that wheel with the little bit missing-- 1087 01:03:26,340 --> 01:03:28,240 if you all got that sorted out. 1088 01:03:28,240 --> 01:03:30,684 If you're all perfectly clear in your mind about that then 1089 01:03:30,684 --> 01:03:32,100 I don't need to say what I'm going 1090 01:03:32,100 --> 01:03:33,350 to talk about for a second. 1091 01:03:33,350 --> 01:03:36,040 But I thought I would just tie that 1092 01:03:36,040 --> 01:03:37,920 up, the last little bit here. 1093 01:03:37,920 --> 01:03:40,280 This problem is a nice lead to it. 1094 01:03:46,090 --> 01:03:54,380 This last force that I computed is 1095 01:03:54,380 --> 01:03:58,760 the force required to swing this thing in a circle. 1096 01:03:58,760 --> 01:04:00,400 And I'm putting that force right here. 1097 01:04:00,400 --> 01:04:03,280 And I'm saying it's-- actually the point of rotation is right 1098 01:04:03,280 --> 01:04:04,780 here. 1099 01:04:04,780 --> 01:04:09,680 And that force is the mass times the acceleration of that point. 1100 01:04:09,680 --> 01:04:14,350 And that point is a distance from my point of rotation 1101 01:04:14,350 --> 01:04:16,000 to the center of mass. 1102 01:04:16,000 --> 01:04:18,230 No accident. 1103 01:04:18,230 --> 01:04:21,180 You're accelerating the center of mass. 1104 01:04:21,180 --> 01:04:25,440 That times the rotation rate squared 1105 01:04:25,440 --> 01:04:28,680 is the acceleration, the centripetal acceleration. 1106 01:04:28,680 --> 01:04:30,220 This is the force. 1107 01:04:30,220 --> 01:04:46,250 So any time you have an object-- maybe it's a wheel, 1108 01:04:46,250 --> 01:04:51,690 and maybe it has, stuck on it, a little extra mass 1109 01:04:51,690 --> 01:04:53,800 you don't know about, a rock stuck in your tire. 1110 01:04:57,760 --> 01:04:59,600 This is the center of rotation. 1111 01:04:59,600 --> 01:05:01,520 Your axle on your wheel doesn't move. 1112 01:05:01,520 --> 01:05:03,470 Tire doesn't move relative to the axle. 1113 01:05:03,470 --> 01:05:05,980 You've got this little bit of mass missing 1114 01:05:05,980 --> 01:05:08,880 or added, doesn't much matter. 1115 01:05:08,880 --> 01:05:12,370 The center of mass of this system 1116 01:05:12,370 --> 01:05:15,230 is no longer at the center of rotation. 1117 01:05:15,230 --> 01:05:17,180 So I'll call the center of rotation here A. 1118 01:05:17,180 --> 01:05:19,560 That's what it's rotating about. 1119 01:05:19,560 --> 01:05:23,280 This addition or loss of a piece of mass 1120 01:05:23,280 --> 01:05:26,940 makes the center of mass of the system move a little bit. 1121 01:05:26,940 --> 01:05:30,100 So I have an extra little bit of mass out here. 1122 01:05:30,100 --> 01:05:34,150 The new actual center gravity of this system 1123 01:05:34,150 --> 01:05:36,430 is a little distance away. 1124 01:05:36,430 --> 01:05:41,093 And I'll call that distance e, eccentricity. 1125 01:05:43,840 --> 01:05:46,990 So now I have a system whose center of mass 1126 01:05:46,990 --> 01:05:49,560 is not at the center of rotation. 1127 01:05:49,560 --> 01:06:01,260 And its distance, RGA, I'm calling the eccentricity. 1128 01:06:01,260 --> 01:06:04,960 And as this rotates-- and I'll put a coordinate system 1129 01:06:04,960 --> 01:06:05,620 on here. 1130 01:06:05,620 --> 01:06:11,330 i, x, y, rotating with the system. 1131 01:06:11,330 --> 01:06:18,820 The force, it's going to appear here on this axle. 1132 01:06:18,820 --> 01:06:26,390 It's going to have a force that I'll call Rx. 1133 01:06:26,390 --> 01:06:36,310 Rx equals the mass times the acceleration of the center 1134 01:06:36,310 --> 01:06:39,040 of gravity of the system. 1135 01:06:39,040 --> 01:06:41,000 And the acceleration of the center 1136 01:06:41,000 --> 01:06:52,090 of gravity in this system is minus e theta dot squared. 1137 01:06:55,480 --> 01:06:58,790 And it's inward in the i hat direction. 1138 01:06:58,790 --> 01:07:01,180 Therefore, the minus sign. 1139 01:07:01,180 --> 01:07:05,040 So the force-- this is the centripetal acceleration 1140 01:07:05,040 --> 01:07:07,940 times mass-- the force required to make it 1141 01:07:07,940 --> 01:07:14,750 do that is an inward force minus M e theta dot squared i hat. 1142 01:07:17,280 --> 01:07:18,080 OK? 1143 01:07:18,080 --> 01:07:20,370 I drew Rx in a positive direction. 1144 01:07:20,370 --> 01:07:21,530 The answer comes out minus. 1145 01:07:21,530 --> 01:07:23,054 It says it's going the other way. 1146 01:07:28,780 --> 01:07:31,760 So this is always the case. 1147 01:07:31,760 --> 01:07:35,620 If you have a system that, for some reason, 1148 01:07:35,620 --> 01:07:40,450 does not rotate about its true center of mass, 1149 01:07:40,450 --> 01:07:44,440 but in fact the center of mass is off a little bit or a lot, 1150 01:07:44,440 --> 01:07:47,700 you will have to provide a force that's inwardly 1151 01:07:47,700 --> 01:07:50,554 directed as that thing spins. 1152 01:07:50,554 --> 01:07:51,850 OK? 1153 01:07:51,850 --> 01:07:57,650 Now a subtlety I want you to know, to really go away with, 1154 01:07:57,650 --> 01:08:02,900 is that we've talked quite a lot about unbalance 1155 01:08:02,900 --> 01:08:06,640 due to these masses that aren't concentrated 1156 01:08:06,640 --> 01:08:10,690 at the center, these unbalanced masses. 1157 01:08:10,690 --> 01:08:16,960 There are two kinds of unbalance the engineers have 1158 01:08:16,960 --> 01:08:20,430 chosen to describe the world with, two kinds. 1159 01:08:20,430 --> 01:08:26,410 One is static imbalance and the other is dynamic imbalance. 1160 01:08:26,410 --> 01:08:34,180 A statically imbalanced system is one in which you-- 1161 01:08:34,180 --> 01:08:45,609 I'm looking at a side view of-- an edge view of this thing 1162 01:08:45,609 --> 01:08:51,680 as it's spinning, maybe like this. 1163 01:08:51,680 --> 01:08:54,510 And the center of rotation is here. 1164 01:08:54,510 --> 01:08:57,189 And the G is here. 1165 01:08:57,189 --> 01:09:00,029 That force is inward on it. 1166 01:09:00,029 --> 01:09:04,520 But really, it's perfectly symmetric. 1167 01:09:04,520 --> 01:09:08,270 So this force that I'm having to provide 1168 01:09:08,270 --> 01:09:12,029 to keep this thing from flying off 1169 01:09:12,029 --> 01:09:13,779 is perfectly inward directed. 1170 01:09:13,779 --> 01:09:16,979 Does it generate any moments about this point? 1171 01:09:16,979 --> 01:09:17,710 None. 1172 01:09:17,710 --> 01:09:19,359 This is called static imbalance. 1173 01:09:22,290 --> 01:09:26,729 And here's an example of static imbalance. 1174 01:09:26,729 --> 01:09:28,340 This is my rotor. 1175 01:09:28,340 --> 01:09:30,370 I'm going to put it on its side. 1176 01:09:30,370 --> 01:09:32,597 Gravity says, I want this to hang down. 1177 01:09:32,597 --> 01:09:34,680 Because where's the center of mass of this system? 1178 01:09:34,680 --> 01:09:36,350 Up here? 1179 01:09:36,350 --> 01:09:38,280 Along here? 1180 01:09:38,280 --> 01:09:41,270 It's down here somewhere, right? 1181 01:09:41,270 --> 01:09:44,540 That center of mass, you do a statics calculation, MG sine 1182 01:09:44,540 --> 01:09:48,017 theta, it's going to hang down. 1183 01:09:48,017 --> 01:09:49,850 So that's why they call it static imbalance. 1184 01:09:49,850 --> 01:09:53,050 And the way you can correct a statically imbalanced system 1185 01:09:53,050 --> 01:09:54,570 is just do the test. 1186 01:09:54,570 --> 01:09:57,630 Put the thing-- hang the thing on an axle. 1187 01:09:57,630 --> 01:09:59,810 And do I have a handy axle here today? 1188 01:10:05,630 --> 01:10:08,770 Put it on an axle and see if it rotates. 1189 01:10:08,770 --> 01:10:12,292 And if it always rotates to some point hanging down, 1190 01:10:12,292 --> 01:10:14,125 you know this thing's statically imbalanced. 1191 01:10:19,668 --> 01:10:21,910 Look in my kit of parts here. 1192 01:10:21,910 --> 01:10:23,718 Just maybe do this. 1193 01:10:28,558 --> 01:10:30,910 I don't know if this is heavy enough to do the job. 1194 01:10:30,910 --> 01:10:34,330 But now I've put a little bit of mass on here. 1195 01:10:34,330 --> 01:10:41,270 And I presume that if I do this with it, OK, it goes down. 1196 01:10:41,270 --> 01:10:43,140 This system is statically imbalanced. 1197 01:10:43,140 --> 01:10:46,450 This system, now, is also dynamically imbalanced. 1198 01:10:46,450 --> 01:10:46,950 Why? 1199 01:10:51,620 --> 01:10:54,180 It kind of depends on where I do the calculation. 1200 01:10:54,180 --> 01:10:59,020 But if I say that I want to know-- this 1201 01:10:59,020 --> 01:11:03,520 is when put a force, a mass, off-center here. 1202 01:11:03,520 --> 01:11:10,950 And if I compute moments about a point that's perfectly lined up 1203 01:11:10,950 --> 01:11:15,590 with it, this system will create no moments about that point. 1204 01:11:15,590 --> 01:11:17,860 This system is perfectly balanced now. 1205 01:11:17,860 --> 01:11:21,170 The mass that I've stuck on here is there. 1206 01:11:21,170 --> 01:11:22,940 This system is perfectly balanced. 1207 01:11:22,940 --> 01:11:24,610 It generates no moments as it stands, 1208 01:11:24,610 --> 01:11:26,520 no matter where you calculate this point. 1209 01:11:26,520 --> 01:11:33,270 But this extra little bit-- this is 1210 01:11:33,270 --> 01:11:36,320 A. If I compute H with respect to A 1211 01:11:36,320 --> 01:11:38,700 here and do the derivatives of it, 1212 01:11:38,700 --> 01:11:42,240 I get no torques except in the z direction. 1213 01:11:42,240 --> 01:11:44,720 But as soon as I move away from there, 1214 01:11:44,720 --> 01:11:48,500 and there is a distance here, and I compute moments 1215 01:11:48,500 --> 01:11:51,280 about that, do I get some static moments 1216 01:11:51,280 --> 01:11:53,310 not in the direction of spin? 1217 01:11:53,310 --> 01:11:54,170 Yes. 1218 01:11:54,170 --> 01:11:58,360 So when you do that, this system-- if this is my point A, 1219 01:11:58,360 --> 01:12:01,250 let's call it, over here-- this system, 1220 01:12:01,250 --> 01:12:03,690 is it statically imbalanced? 1221 01:12:03,690 --> 01:12:04,190 Yep. 1222 01:12:04,190 --> 01:12:06,700 Is it dynamically imbalanced? 1223 01:12:06,700 --> 01:12:08,010 Yeah. 1224 01:12:08,010 --> 01:12:11,320 So how would you balance it? 1225 01:12:11,320 --> 01:12:14,340 Final question of the term for me. 1226 01:12:17,250 --> 01:12:20,696 How would you put this system into balance? 1227 01:12:20,696 --> 01:12:25,088 AUDIENCE: Put another mass on the other side. 1228 01:12:25,088 --> 01:12:26,552 PROFESSOR: All right. 1229 01:12:26,552 --> 01:12:29,720 So if I put another mass here, will that statically 1230 01:12:29,720 --> 01:12:30,350 balance it? 1231 01:12:33,098 --> 01:12:36,550 Statically balance, meaning if I let go, 1232 01:12:36,550 --> 01:12:39,520 will it have to rotate around and find a low point. 1233 01:12:39,520 --> 01:12:41,570 Is there any low point-- if I put 1234 01:12:41,570 --> 01:12:46,720 one equal and opposite, equal distance away, opposite side, 1235 01:12:46,720 --> 01:12:48,303 would it be statically balanced? 1236 01:12:48,303 --> 01:12:48,886 AUDIENCE: Yes. 1237 01:12:48,886 --> 01:12:51,219 PROFESSOR: Yeah, because gravity pulls down on this one, 1238 01:12:51,219 --> 01:12:53,240 equal and opposite down on this one. 1239 01:12:53,240 --> 01:12:54,740 So it would be statically balanced. 1240 01:12:54,740 --> 01:12:57,480 Would it be dynamically balanced? 1241 01:12:57,480 --> 01:13:00,345 The system, as you spin, will try to twist like this. 1242 01:13:00,345 --> 01:13:01,720 So it's not dynamically balanced. 1243 01:13:01,720 --> 01:13:02,886 What if I move it over here? 1244 01:13:06,670 --> 01:13:08,042 Is it statically balanced? 1245 01:13:08,042 --> 01:13:08,625 AUDIENCE: Yes. 1246 01:13:08,625 --> 01:13:10,221 PROFESSOR: Is it dynamically balanced? 1247 01:13:10,221 --> 01:13:10,804 AUDIENCE: Yes. 1248 01:13:13,444 --> 01:13:14,069 PROFESSOR: Hmm. 1249 01:13:17,450 --> 01:13:18,740 Compute the torques. 1250 01:13:18,740 --> 01:13:19,820 You know? 1251 01:13:19,820 --> 01:13:24,050 Let's put A right in the center, compute the angular momentum, 1252 01:13:24,050 --> 01:13:25,500 take its time derivative, and you 1253 01:13:25,500 --> 01:13:28,350 will find that this one, as it spins, 1254 01:13:28,350 --> 01:13:30,940 tries to create a moment like that. 1255 01:13:30,940 --> 01:13:34,170 And this one, as it spins tries to create a moment like that. 1256 01:13:34,170 --> 01:13:37,360 And they're exactly equal and opposite. 1257 01:13:37,360 --> 01:13:41,143 This is dynamically balanced and statically balanced. 1258 01:13:45,080 --> 01:13:45,580 Hmm. 1259 01:13:45,580 --> 01:13:47,204 AUDIENCE: When there was just one mass, 1260 01:13:47,204 --> 01:13:50,855 was it dynamically balanced about A, the axis of rotation. 1261 01:13:50,855 --> 01:13:52,480 PROFESSOR: This is not dynamic-- if you 1262 01:13:52,480 --> 01:13:55,740 put A right in the usual center-- the original center 1263 01:13:55,740 --> 01:13:59,610 of gravity, we'll call A, at this wheel-- this system is not 1264 01:13:59,610 --> 01:14:02,380 dynamically balanced with respect to an A that's 1265 01:14:02,380 --> 01:14:03,990 right on the axle and on the center. 1266 01:14:03,990 --> 01:14:06,770 Because it has this little offset. 1267 01:14:06,770 --> 01:14:08,320 OK?