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 --> 00:00:06,080 Your support will help MIT OpenCourseWare 4 00:00:06,080 --> 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 --> 00:00:17,325 at ocw.mit.edu. 8 00:00:22,084 --> 00:00:25,800 PROFESSOR: And I have put on the second board 9 00:00:25,800 --> 00:00:31,260 just a quick review of the key equations that we 10 00:00:31,260 --> 00:00:34,690 use to do the direct method. 11 00:00:34,690 --> 00:00:40,140 So straight from Newton's second law, the first one, 12 00:00:40,140 --> 00:00:45,850 second one-- the summarization of torques about some point a. 13 00:00:45,850 --> 00:00:51,370 a may be g, in which case things get much simplified, g 14 00:00:51,370 --> 00:00:53,270 being the center of mass. 15 00:00:53,270 --> 00:00:59,210 And if point a is moving, then you have this extra term. 16 00:00:59,210 --> 00:01:01,150 And this is a new equation over here, 17 00:01:01,150 --> 00:01:05,060 which I got frustrated with this equation a few days ago 18 00:01:05,060 --> 00:01:10,170 and derive something, which I think is easier to use. 19 00:01:10,170 --> 00:01:15,260 So I'll do a problem today using that expression. 20 00:01:15,260 --> 00:01:17,530 So I won't talk more about it now-- show it to you 21 00:01:17,530 --> 00:01:18,610 in a minute. 22 00:01:18,610 --> 00:01:22,450 And then finally, it's convenient at times 23 00:01:22,450 --> 00:01:24,800 to express angular momentum with respect 24 00:01:24,800 --> 00:01:28,400 to a point a as angular momentum about g 25 00:01:28,400 --> 00:01:36,580 plus the position vector from a to g times the linear momentum. 26 00:01:36,580 --> 00:01:39,340 So that's kind of the covered on the quiz. 27 00:01:39,340 --> 00:01:41,200 And we're going to use these equations. 28 00:01:41,200 --> 00:01:47,250 Today, what I'm going to do is look at a few problems 29 00:01:47,250 --> 00:01:50,730 from the point of view both equations of motion, 30 00:01:50,730 --> 00:01:52,800 by the direct method and the Lagrange method, 31 00:01:52,800 --> 00:01:59,150 and kind of talk about and which one's easier, when should 32 00:01:59,150 --> 00:02:01,580 you choose this one, when to choose that one. 33 00:02:01,580 --> 00:02:04,370 That's the nature of today's lecture to sort of get 34 00:02:04,370 --> 00:02:06,460 you prepped for the quiz. 35 00:02:09,400 --> 00:02:11,355 And a little bit more on generalized forces. 36 00:02:23,467 --> 00:02:25,800 You have questions about what's going to be on the quiz? 37 00:02:29,744 --> 00:02:31,979 AUDIENCE: What is that last word-- balancing of? 38 00:02:31,979 --> 00:02:33,270 PROFESSOR: Balancing of rotors. 39 00:02:37,070 --> 00:02:41,060 There's a rotor-- any thing that we spin about an axis. 40 00:02:41,060 --> 00:02:42,980 A rigid body spinning about an axis, 41 00:02:42,980 --> 00:02:45,040 you can call a rotor of some kind. 42 00:02:45,040 --> 00:02:50,050 And balancing is usually-- the axes are fixed. 43 00:02:50,050 --> 00:02:54,970 And so is this one-- this axis passes 44 00:02:54,970 --> 00:02:59,365 through the center of mass-- so is 45 00:02:59,365 --> 00:03:04,771 this rotor statically balanced? 46 00:03:04,771 --> 00:03:05,270 Right. 47 00:03:05,270 --> 00:03:07,790 Is this rotor dynamically balance? 48 00:03:07,790 --> 00:03:11,120 Would you expect to when this is spinning, 49 00:03:11,120 --> 00:03:15,335 it's direction of spin is vertical, right. 50 00:03:18,930 --> 00:03:22,820 Is the direction of the angular momentum of this 51 00:03:22,820 --> 00:03:28,000 object in the same direction as the angle of rotation? 52 00:03:28,000 --> 00:03:28,500 No. 53 00:03:28,500 --> 00:03:32,420 And so that's an indication of dynamic imbalance. 54 00:03:32,420 --> 00:03:34,370 Angular momentum is not in the same direction 55 00:03:34,370 --> 00:03:36,720 as angular rotation, right. 56 00:03:36,720 --> 00:03:38,090 All right. 57 00:03:38,090 --> 00:03:40,170 And, well, while we're on that subject, 58 00:03:40,170 --> 00:03:41,730 we'll just hit it briefly here. 59 00:03:44,740 --> 00:03:47,115 One other prop I need. 60 00:03:50,180 --> 00:03:54,960 So another rigid body, and I don't know 61 00:03:54,960 --> 00:03:56,755 where g is on this rigid body. 62 00:04:04,260 --> 00:04:07,760 So does-- this is rotating about an axis. 63 00:04:11,010 --> 00:04:14,050 Is this object statically balanced, 64 00:04:14,050 --> 00:04:15,246 just doing this experiment? 65 00:04:15,246 --> 00:04:16,329 We're doing an experiment. 66 00:04:16,329 --> 00:04:19,089 We're sticking this on an axis, and we're 67 00:04:19,089 --> 00:04:20,130 letting it do it's thing. 68 00:04:20,130 --> 00:04:22,540 And I'm asking you is it statically balance. 69 00:04:22,540 --> 00:04:23,290 How do you know? 70 00:04:23,290 --> 00:04:27,736 AUDIENCE: Because if you were to flip it upside down and do it, 71 00:04:27,736 --> 00:04:29,220 it would still come to a stop. 72 00:04:29,220 --> 00:04:31,240 PROFESSOR: So it goes to a low point, right. 73 00:04:31,240 --> 00:04:35,190 And that tells you what about the position of this axis? 74 00:04:35,190 --> 00:04:37,686 AUDIENCE: [INAUDIBLE] 75 00:04:37,686 --> 00:04:39,060 PROFESSOR: The center of mass has 76 00:04:39,060 --> 00:04:42,200 got gravity, puts a force at the center of mass, 77 00:04:42,200 --> 00:04:45,610 creates a torque which brings it to the low point. 78 00:04:45,610 --> 00:04:54,100 Does this object have any symmetries 79 00:04:54,100 --> 00:04:55,230 that you could point out? 80 00:04:55,230 --> 00:04:59,960 Or in other words, can you without doing any math tell me 81 00:04:59,960 --> 00:05:02,400 one principal axis of this object? 82 00:05:11,824 --> 00:05:15,098 AUDIENCE: Where the dotted line is, that's a plane [INAUDIBLE]. 83 00:05:15,098 --> 00:05:17,480 PROFESSOR: So you say I've got a dotted line around here 84 00:05:17,480 --> 00:05:18,920 and that's a plane of what? 85 00:05:18,920 --> 00:05:19,570 AUDIENCE: Plane of symmetry. 86 00:05:19,570 --> 00:05:21,180 PROFESSOR: So there's a plane of symmetry cutting 87 00:05:21,180 --> 00:05:22,310 this thing through here. 88 00:05:22,310 --> 00:05:25,840 So if you can identify one plane of symmetry, 89 00:05:25,840 --> 00:05:28,350 then where can you tell me that for sure you 90 00:05:28,350 --> 00:05:33,090 know that you can make a principal axis? 91 00:05:33,090 --> 00:05:36,550 He says perpendicular to that plane. 92 00:05:36,550 --> 00:05:41,390 Any place on that plane perpendicular to it? 93 00:05:41,390 --> 00:05:43,240 What do you think? 94 00:05:43,240 --> 00:05:46,570 For now, we've identified it has a plane of symmetry 95 00:05:46,570 --> 00:05:57,100 and that an axis perpendicular to it-- a principle axis should 96 00:05:57,100 --> 00:05:58,440 be perpendicular to that plane. 97 00:05:58,440 --> 00:06:00,430 But the question is where do I put it? 98 00:06:00,430 --> 00:06:02,750 Where, perpendicular to that plane 99 00:06:02,750 --> 00:06:05,385 can this axis be and be a principal axis? 100 00:06:13,320 --> 00:06:14,530 She says through g. 101 00:06:14,530 --> 00:06:17,160 Do you agree with that? 102 00:06:17,160 --> 00:06:18,510 So certainly through g. 103 00:06:18,510 --> 00:06:22,490 Let's say that that really is g and I put it through here. 104 00:06:22,490 --> 00:06:24,647 And if it was exactly through g, it 105 00:06:24,647 --> 00:06:25,980 wouldn't rotate to the low spot. 106 00:06:25,980 --> 00:06:26,900 This is pretty close. 107 00:06:26,900 --> 00:06:28,760 It struggles to get there. 108 00:06:28,760 --> 00:06:32,350 That's definitely a principal axis, but are there more? 109 00:06:32,350 --> 00:06:34,390 This is kind of the real point of the question. 110 00:06:34,390 --> 00:06:37,900 Are there other allowable principal axes 111 00:06:37,900 --> 00:06:38,675 in this direction? 112 00:06:41,770 --> 00:06:43,520 So we haven't talked about this very much. 113 00:06:43,520 --> 00:06:46,710 Any axis-- if this is a principal axis, 114 00:06:46,710 --> 00:06:51,200 any axis parallel to it is a principal axis. 115 00:06:51,200 --> 00:06:52,860 It's no longer through g. 116 00:06:52,860 --> 00:06:56,430 It's no longer statically balanced about that point. 117 00:06:56,430 --> 00:06:58,460 But it's still a principal axis. 118 00:06:58,460 --> 00:07:01,830 And one measure of whether or not something 119 00:07:01,830 --> 00:07:06,360 is a principal axis-- if you know the principal axes 120 00:07:06,360 --> 00:07:10,820 of an object and you write the mass moment of inertia 121 00:07:10,820 --> 00:07:14,410 matrix of it with respect to g, what 122 00:07:14,410 --> 00:07:18,150 does that matrix look like? 123 00:07:18,150 --> 00:07:21,190 Where does it have 0s and non-0s? 124 00:07:21,190 --> 00:07:23,870 It's a diagonal matrix. 125 00:07:23,870 --> 00:07:26,800 So if i about g is diagonal, that 126 00:07:26,800 --> 00:07:30,420 means you're computed the moment, mass moment 127 00:07:30,420 --> 00:07:32,980 properties with respect to three principal axes. 128 00:07:32,980 --> 00:07:36,800 If it's diagonal, and those axes all pass through g, 129 00:07:36,800 --> 00:07:39,950 if I now take one of those axes and say, 130 00:07:39,950 --> 00:07:42,560 I want to move it over here, and you've 131 00:07:42,560 --> 00:07:47,480 re-computed the inertia matrix, would it still be diagonal? 132 00:07:50,110 --> 00:07:55,580 So you've moved just one axis to another spot, parallel 133 00:07:55,580 --> 00:07:57,090 to its original one. 134 00:07:57,090 --> 00:07:59,040 You know this one is a principal axis. 135 00:07:59,040 --> 00:08:01,480 I'm going to move it just parallel. 136 00:08:01,480 --> 00:08:04,200 Only that one axis-- the other two are saying where they are. 137 00:08:04,200 --> 00:08:06,600 I've just moved it parallel. 138 00:08:06,600 --> 00:08:09,410 Is that also a principal axis, or another way of saying it, 139 00:08:09,410 --> 00:08:13,990 does that mass matrix, mass moment of inertia matrix stay 140 00:08:13,990 --> 00:08:16,130 diagonal? 141 00:08:16,130 --> 00:08:19,910 Do any of the terms in it change? 142 00:08:19,910 --> 00:08:23,500 Let's say this is the z principal axis and I move away. 143 00:08:23,500 --> 00:08:29,380 So now you have this new mass moment inertia matrix. 144 00:08:29,380 --> 00:08:31,680 Does the Izz term change? 145 00:08:36,580 --> 00:08:39,564 Yeah, by how much? 146 00:08:39,564 --> 00:08:42,059 AUDIENCE: [INAUDIBLE] 147 00:08:42,059 --> 00:08:44,340 PROFESSOR: He said ml squared, l being what? 148 00:08:44,340 --> 00:08:46,300 AUDIENCE: The length you moved. 149 00:08:46,300 --> 00:08:47,370 PROFESSOR: Here. 150 00:08:47,370 --> 00:08:51,690 And that's this thing though as parallel axis there, right? 151 00:08:51,690 --> 00:08:57,920 If you move one axis, any axis parallel to one of those axes 152 00:08:57,920 --> 00:08:59,940 is also a principal axis. 153 00:08:59,940 --> 00:09:01,360 So that's a subtle point. 154 00:09:01,360 --> 00:09:03,390 We don't often talking about it. 155 00:09:03,390 --> 00:09:06,730 Any axis parallel to a principal axis is also principal axis. 156 00:09:16,460 --> 00:09:18,230 Well, let's think about a problem. 157 00:09:18,230 --> 00:09:19,760 We've done this problem before. 158 00:09:19,760 --> 00:09:22,090 But now we have all the full tools. 159 00:09:22,090 --> 00:09:25,870 We have Lagrange, we have direct method. 160 00:09:25,870 --> 00:09:31,835 So that's this problem of the wheel on a horizontal plane. 161 00:09:37,690 --> 00:09:42,080 Kind of a complicated distribution of mass 162 00:09:42,080 --> 00:09:51,320 for which we know Izz about g is some mass radius and gyration 163 00:09:51,320 --> 00:09:51,820 squared. 164 00:09:54,590 --> 00:09:57,540 Maybe we measured it rather than computed it. 165 00:09:57,540 --> 00:10:03,630 So this thing is sliding on ice, and it has cord 166 00:10:03,630 --> 00:10:05,190 wrapped around it. 167 00:10:05,190 --> 00:10:08,360 I'm pulling with a force F, call it F1. 168 00:10:12,050 --> 00:10:15,815 I'll provide the inertial frame. 169 00:10:21,910 --> 00:10:29,600 And it's got some radius R. And we'll 170 00:10:29,600 --> 00:10:31,233 say the surface is frictionless. 171 00:10:44,530 --> 00:10:46,200 So I don't have a string wrapped around. 172 00:10:46,200 --> 00:10:47,730 But we're really talking about this problem. 173 00:10:47,730 --> 00:10:49,425 I'm pulling on the string and this thing 174 00:10:49,425 --> 00:10:51,070 can slide and rotate. 175 00:10:59,360 --> 00:11:00,270 Take a second. 176 00:11:00,270 --> 00:11:01,980 Think about how many different ways 177 00:11:01,980 --> 00:11:07,810 can you think of solving for the equations of motion of this? 178 00:11:07,810 --> 00:11:10,180 So if this on a quiz, I want you to find 179 00:11:10,180 --> 00:11:13,057 the equations of motion. 180 00:11:13,057 --> 00:11:14,140 What method would you use? 181 00:11:16,870 --> 00:11:17,640 And what's this? 182 00:11:17,640 --> 00:11:19,624 And then you're confronted with, well, 183 00:11:19,624 --> 00:11:20,790 I know more than one method. 184 00:11:20,790 --> 00:11:22,280 What's the simplest method? 185 00:11:26,483 --> 00:11:28,290 And before that, you need to decide 186 00:11:28,290 --> 00:11:30,399 how many coordinates you're going to need, 187 00:11:30,399 --> 00:11:32,690 how many equations you're going-- so let's start there. 188 00:11:32,690 --> 00:11:36,470 How many independent coordinates do you need to do this problem? 189 00:11:42,390 --> 00:11:44,220 We'll back up one more question, then. 190 00:11:44,220 --> 00:11:47,680 Is this a planar motion problem? 191 00:11:47,680 --> 00:11:49,430 So it's confined to one plane. 192 00:11:49,430 --> 00:11:50,860 I see a lot of heads nodding. 193 00:11:50,860 --> 00:11:55,250 And it rotates only with one degree of freedom of rotation, 194 00:11:55,250 --> 00:11:57,100 axis perpendicular to the plane. 195 00:11:57,100 --> 00:11:58,599 So this is a planar motion problem. 196 00:11:58,599 --> 00:12:00,140 So in general, planar motion problems 197 00:12:00,140 --> 00:12:04,420 have found many degrees of freedom per rigid body? 198 00:12:04,420 --> 00:12:06,580 So I see a bunch of 3's. 199 00:12:06,580 --> 00:12:09,259 Does this one have any constraints on those three 200 00:12:09,259 --> 00:12:10,050 degrees of freedom? 201 00:12:10,050 --> 00:12:11,860 So this has three degrees of freedom. 202 00:12:11,860 --> 00:12:15,680 You conceivably need three equations of motion. 203 00:12:19,060 --> 00:12:21,267 What are the choices for getting them? 204 00:12:21,267 --> 00:12:23,100 I want you to think about this for a minute. 205 00:12:23,100 --> 00:12:25,360 You decide what way you would do, and let's 206 00:12:25,360 --> 00:12:26,900 take a poll in a minute. 207 00:12:26,900 --> 00:12:28,650 You think about it, and then let's take 208 00:12:28,650 --> 00:12:30,865 a poll for how people would do this problem. 209 00:13:14,020 --> 00:13:17,390 So what's your favorite? 210 00:13:17,390 --> 00:13:19,720 How would you do it? 211 00:13:19,720 --> 00:13:22,440 He'd use the direct method. 212 00:13:22,440 --> 00:13:25,780 Do you need a torque equation? 213 00:13:25,780 --> 00:13:27,630 About what point? 214 00:13:27,630 --> 00:13:28,630 He'd use center of mass. 215 00:13:28,630 --> 00:13:30,979 So he'd say, direct method and for the torque equation, 216 00:13:30,979 --> 00:13:32,270 do it about the center of mass. 217 00:13:32,270 --> 00:13:36,340 Anybody have a different way they would do this problem? 218 00:13:36,340 --> 00:13:39,400 Is there another way to do it? 219 00:13:39,400 --> 00:13:41,610 He wants to do Lagrange. 220 00:13:41,610 --> 00:13:44,960 Everybody agree it could be done that way? 221 00:13:44,960 --> 00:13:50,620 So I agree, and let's quickly do this problem. 222 00:13:50,620 --> 00:14:00,100 And so direct method-- I'm going to make it easier on myself 223 00:14:00,100 --> 00:14:00,600 here. 224 00:14:09,610 --> 00:14:12,970 So the three coordinates, you're going to need an x, a y, 225 00:14:12,970 --> 00:14:16,070 and a theta is what I've chosen for the three coordinates 226 00:14:16,070 --> 00:14:18,000 to do this direct method. 227 00:14:18,000 --> 00:14:19,810 And to make it easy on myself, I'm 228 00:14:19,810 --> 00:14:24,120 going to align the force with my coordinates. 229 00:14:24,120 --> 00:14:26,085 So I'm going to essentially make theta 0. 230 00:14:39,593 --> 00:14:43,820 Essentially aligning F1, my force F1 with the x-axis. 231 00:14:43,820 --> 00:14:49,770 And if I do that, then I can say the sum of the forces in the y, 232 00:14:49,770 --> 00:14:54,440 the external forces in the y direction are nonexistent, 0. 233 00:14:54,440 --> 00:14:59,820 So that must be my double dot, so y double dot is 0. 234 00:14:59,820 --> 00:15:03,000 So that simplifies the problem. 235 00:15:03,000 --> 00:15:07,040 Summation of forces in the x direction, external forces 236 00:15:07,040 --> 00:15:07,660 are F1I. 237 00:15:10,930 --> 00:15:15,350 And they must equal to the mass times the acceleration 238 00:15:15,350 --> 00:15:16,280 in the x direction. 239 00:15:16,280 --> 00:15:19,080 And I'm basically done with that equation. 240 00:15:19,080 --> 00:15:20,876 AUDIENCE: Is that a rotating [INAUDIBLE] 241 00:15:26,288 --> 00:15:28,730 PROFESSOR: Well, I'm doing this. 242 00:15:28,730 --> 00:15:30,230 I'm saying this is my inertia frame. 243 00:15:38,690 --> 00:15:40,950 You add an inertial frame? 244 00:15:40,950 --> 00:15:43,000 Sure. 245 00:15:43,000 --> 00:15:46,670 Now I can-- I don't have to break this force 246 00:15:46,670 --> 00:15:52,255 into components, F1 cosine theta in the x direction 247 00:15:52,255 --> 00:15:55,350 and F1 sine theta in the y direction 248 00:15:55,350 --> 00:16:00,810 and get two-- get a finite amount, a nonzero amount 249 00:16:00,810 --> 00:16:03,710 for y double dot and non-0 amount for x double dot. 250 00:16:03,710 --> 00:16:05,860 I end up with three equations I have to work with. 251 00:16:05,860 --> 00:16:09,670 Just by realigning it, I can make this problem even simpler. 252 00:16:09,670 --> 00:16:11,660 So just to do it quickly, that's what I did. 253 00:16:14,650 --> 00:16:17,800 So I've got two of the three equations done. 254 00:16:17,800 --> 00:16:23,340 And finally, the sum of the torques about g. 255 00:16:32,750 --> 00:16:35,390 And we know the angular momentum of this 256 00:16:35,390 --> 00:16:41,120 would be some izz g times omega z 257 00:16:41,120 --> 00:16:43,190 would be the angular momentum. 258 00:16:43,190 --> 00:16:51,150 And it's d by dt of that izz about g, omega z dot, 259 00:16:51,150 --> 00:16:53,790 which is theta double dot, right? 260 00:16:53,790 --> 00:16:55,240 And what are the external torques? 261 00:17:02,852 --> 00:17:03,560 So figure it out. 262 00:17:03,560 --> 00:17:05,440 What's the external torque in this problem? 263 00:17:05,440 --> 00:17:07,190 What is its size and what's its direction? 264 00:17:28,410 --> 00:17:33,460 So, somebody give me-- what's the external torque 265 00:17:33,460 --> 00:17:36,200 with respect to where, to start with? 266 00:17:36,200 --> 00:17:40,870 So we're summing torques about g, the center of mass. 267 00:17:40,870 --> 00:17:44,355 R cross F, right, which comes out in what direction? 268 00:17:49,830 --> 00:17:51,170 k direction, right? 269 00:17:51,170 --> 00:17:54,150 And minus or plus? 270 00:17:54,150 --> 00:17:55,210 Do your right hand rule. 271 00:17:55,210 --> 00:17:58,040 And notice I've drawn x and y such 272 00:17:58,040 --> 00:18:00,250 that positive z is into the board. 273 00:18:03,460 --> 00:18:04,710 So this is positive. 274 00:18:04,710 --> 00:18:07,250 I put a positive z into the board on purpose 275 00:18:07,250 --> 00:18:13,660 so that this would just work out R F1 in the k direction. 276 00:18:13,660 --> 00:18:22,360 And that's equal to Izz about g theta double dot. 277 00:18:22,360 --> 00:18:24,380 And since it's all with respect to k, 278 00:18:24,380 --> 00:18:27,000 you can drop the unit vectors here. 279 00:18:27,000 --> 00:18:31,170 And so we have one equation. 280 00:18:31,170 --> 00:18:35,100 We have two equations. 281 00:18:35,100 --> 00:18:38,730 And we have three equations. 282 00:18:38,730 --> 00:18:40,461 And that's all we need to complete this, 283 00:18:40,461 --> 00:18:42,960 is three degrees of freedom to completely define the motion. 284 00:18:42,960 --> 00:18:43,459 Yes? 285 00:18:43,459 --> 00:18:46,965 AUDIENCE: Do you want us to have a trivial case? 286 00:18:46,965 --> 00:18:51,240 PROFESSOR: Do you mean the my double dot equals 0? 287 00:18:51,240 --> 00:18:57,600 Well, in general-- when I started this one, when 288 00:18:57,600 --> 00:19:01,952 I didn't align them, then you're forced to use the three, right? 289 00:19:01,952 --> 00:19:03,410 So the only difference between what 290 00:19:03,410 --> 00:19:09,390 I did and this probe making a problem slightly more messy, 291 00:19:09,390 --> 00:19:14,120 is if I had let theta, this angle, be other than 0, I 292 00:19:14,120 --> 00:19:16,940 would have had to have come up with three 293 00:19:16,940 --> 00:19:22,400 non-zero acceleration equations. 294 00:19:22,400 --> 00:19:28,110 So I'd say technically, or I'd just say I think it's prudent. 295 00:19:28,110 --> 00:19:32,280 If you're ever not sure, if you can 296 00:19:32,280 --> 00:19:36,630 look at it and by inspection write down, my double dot is 0, 297 00:19:36,630 --> 00:19:39,230 you've essentially just written an equation of motion. 298 00:19:39,230 --> 00:19:42,720 Rather than say let's just ignore it, 299 00:19:42,720 --> 00:19:45,630 I'd keep it in your-- I'd just put it on the paper, 300 00:19:45,630 --> 00:19:47,670 my double dot equals 0. 301 00:19:47,670 --> 00:19:50,100 And that's your first trivial equation of motion. 302 00:19:50,100 --> 00:19:52,880 But it is an equation of motion, because this body 303 00:19:52,880 --> 00:19:55,480 has a true degree of freedom in that direction. 304 00:19:57,911 --> 00:19:58,910 That's really the point. 305 00:19:58,910 --> 00:20:02,080 Don't confuse trivial results because there's 306 00:20:02,080 --> 00:20:06,860 no external forces with constraints. 307 00:20:06,860 --> 00:20:09,260 It's unconstrained in the x direction, 308 00:20:09,260 --> 00:20:12,190 or in the y direction the way I've set it up. 309 00:20:12,190 --> 00:20:17,640 So I think you're safer if you do just-- 310 00:20:17,640 --> 00:20:20,195 say the torques or the forces in that direction are 0 311 00:20:20,195 --> 00:20:22,640 and you write it down. 312 00:20:22,640 --> 00:20:25,390 Let's look at this problem from the point of view of Lagrange. 313 00:20:25,390 --> 00:20:28,930 It's also equally simple, the Lagrange, 314 00:20:28,930 --> 00:20:33,370 except for the generalized forces. 315 00:20:33,370 --> 00:20:36,102 Actually, Lagrange is harder because you 316 00:20:36,102 --> 00:20:38,310 have to think your way through the generalized forces 317 00:20:38,310 --> 00:20:39,420 in this problem. 318 00:20:39,420 --> 00:20:41,690 And I got myself in a pickle with this problem 319 00:20:41,690 --> 00:20:44,110 and I spent hours trying to figure out 320 00:20:44,110 --> 00:20:47,210 a good explanation of a conundrum I 321 00:20:47,210 --> 00:20:49,800 ran into with the generalized forces. 322 00:20:49,800 --> 00:20:50,644 So let's look. 323 00:20:50,644 --> 00:20:52,310 I'll tell you what that was in a second. 324 00:20:52,310 --> 00:20:53,268 Let's look at Lagrange. 325 00:21:00,480 --> 00:21:07,310 So you need T. And we're pretty good at this now, 326 00:21:07,310 --> 00:21:10,900 I with respect to g zz, theta dot squared. 327 00:21:10,900 --> 00:21:13,220 That's your rotational kinetic energy. 328 00:21:13,220 --> 00:21:19,430 1/2 mx dot squared plus y dot squared. 329 00:21:21,940 --> 00:21:23,950 Translational kinetic energy associated 330 00:21:23,950 --> 00:21:25,080 with the center of mass. 331 00:21:25,080 --> 00:21:27,360 Notice this is with respect to the center 332 00:21:27,360 --> 00:21:29,260 of mass. 333 00:21:29,260 --> 00:21:30,290 How about v? 334 00:21:35,096 --> 00:21:36,470 Potential energy in this problem? 335 00:21:39,520 --> 00:21:40,710 Any? 336 00:21:40,710 --> 00:21:41,810 It's all 0's, right? 337 00:21:41,810 --> 00:21:42,810 There's no springs. 338 00:21:42,810 --> 00:21:46,820 There's no gravity in this plane, operating in this plane. 339 00:21:46,820 --> 00:21:48,337 So this is 0. 340 00:21:48,337 --> 00:21:50,295 This problem's going to be particularly simple. 341 00:22:17,550 --> 00:22:23,340 So in the theta direction here, this derivative 342 00:22:23,340 --> 00:22:28,760 with respect to Q dot gives us Iz theta dot 343 00:22:28,760 --> 00:22:34,910 and the time derivative of that zz theta double dot. 344 00:22:34,910 --> 00:22:40,930 T with respect to theta-- nothing. 345 00:22:40,930 --> 00:22:42,850 V with respect to theta? 346 00:22:42,850 --> 00:22:43,675 Nothing. 347 00:22:43,675 --> 00:22:50,710 And on the right hand side, we need to get Q theta here. 348 00:22:50,710 --> 00:22:58,940 And the other equation, it's the x equation. 349 00:22:58,940 --> 00:23:00,930 Well, actually, we can have x and y. 350 00:23:00,930 --> 00:23:02,930 So what about the x equation? 351 00:23:02,930 --> 00:23:04,570 Derivative with respect to x dot? 352 00:23:04,570 --> 00:23:08,170 I get an mx dot, time derivative, mx double dot. 353 00:23:15,100 --> 00:23:16,420 And with respect to y. 354 00:23:16,420 --> 00:23:20,900 The thing about Lagrange, if you can do Lagrange, just write 355 00:23:20,900 --> 00:23:24,140 down the total expression and just crank it out, 356 00:23:24,140 --> 00:23:27,380 because if you say, oh, well, I know this is trivial, 357 00:23:27,380 --> 00:23:29,410 then you're actually employing some information 358 00:23:29,410 --> 00:23:30,554 from the direct method. 359 00:23:30,554 --> 00:23:33,220 If you're saying, I know the sum of the forces in this direction 360 00:23:33,220 --> 00:23:34,310 is 0. 361 00:23:34,310 --> 00:23:37,440 So when you straight out in applying Lagrange, 362 00:23:37,440 --> 00:23:41,340 you can do it without any reference at all to Newton. 363 00:23:41,340 --> 00:23:45,830 So in this case, we just blindly clunk along and we say, well, 364 00:23:45,830 --> 00:23:49,190 what's the derivative now with respect to y dot? 365 00:23:49,190 --> 00:23:55,810 Will I get an my dot time derivative y double dot. 366 00:23:55,810 --> 00:23:57,490 This derivative would be 0. 367 00:23:57,490 --> 00:23:58,910 This derivative is 0. 368 00:23:58,910 --> 00:24:02,760 And over here I get Q1. 369 00:24:02,760 --> 00:24:06,620 Now I'm left-- so the left hand side of Lagrange equations-- 370 00:24:06,620 --> 00:24:08,855 remember, you just add these terms together. 371 00:24:13,790 --> 00:24:16,225 And these are going to be our three equations of motion 372 00:24:16,225 --> 00:24:19,430 that result. And now the remaining work 373 00:24:19,430 --> 00:24:21,640 is to get these three generalized forces. 374 00:24:28,450 --> 00:24:30,980 Let's go back to our picture. 375 00:24:30,980 --> 00:24:35,480 If we had lined it up like this, and now we 376 00:24:35,480 --> 00:24:39,470 give it a general little virtual displacement delta 377 00:24:39,470 --> 00:24:42,940 y, how much work gets done? 378 00:24:42,940 --> 00:24:43,680 None. 379 00:24:43,680 --> 00:24:47,780 And so you suddenly realize, oh, I went to a lot of work 380 00:24:47,780 --> 00:24:48,680 for nothing here. 381 00:24:48,680 --> 00:24:50,900 This equation is trivial. 382 00:24:50,900 --> 00:24:51,980 Nothing happens there. 383 00:24:51,980 --> 00:24:56,370 And so now it's reduced to doing the Qx and Qy. 384 00:24:59,400 --> 00:25:03,700 So just to give us some practice, 385 00:25:03,700 --> 00:25:10,350 let's do Q theta and Qx the rigorous kinematic way. 386 00:25:37,050 --> 00:25:39,420 I'm going to draw the general picture here, 387 00:25:39,420 --> 00:25:42,225 because this is where I got myself into trouble. 388 00:25:42,225 --> 00:25:43,600 You're working along on a problem 389 00:25:43,600 --> 00:25:46,310 and then something doesn't quite work for you. 390 00:25:46,310 --> 00:25:49,647 I had set this problem to do generalized-- to compute 391 00:25:49,647 --> 00:25:50,605 the generalized forces. 392 00:25:50,605 --> 00:26:00,030 And I set it up looking like set it up like this so theta is 0. 393 00:26:00,030 --> 00:26:02,540 It's lined up like that. 394 00:26:02,540 --> 00:26:05,430 And I started thinking about, then generalized, 395 00:26:05,430 --> 00:26:08,470 my little virtual displacements, and I 396 00:26:08,470 --> 00:26:11,426 got into a conceptual problem that I couldn't sort out 397 00:26:11,426 --> 00:26:11,925 for a while. 398 00:26:11,925 --> 00:26:17,699 And I'll work it into the explanation here. 399 00:26:17,699 --> 00:26:19,490 So I'm going to set this up more generally. 400 00:26:19,490 --> 00:26:23,290 I have an angle theta here, and this is my F1. 401 00:26:23,290 --> 00:26:25,400 So to do this by this kinematic way, 402 00:26:25,400 --> 00:26:28,020 I need to find the position vector. 403 00:26:28,020 --> 00:26:30,720 Here's a point-- I'm going to call it D, 404 00:26:30,720 --> 00:26:35,800 and this is my point G, and here's O. 405 00:26:35,800 --> 00:26:39,580 So I have a position vector going to here-- 406 00:26:39,580 --> 00:26:44,490 it's RG; and a position vector going to here-- that's 407 00:26:44,490 --> 00:26:54,000 R of D with respect to G; and a position vector from here, 408 00:26:54,000 --> 00:27:07,000 and that's RD in O. And RD in O is RG in O plus RD with respect 409 00:27:07,000 --> 00:27:08,639 to G. I can write that. 410 00:27:08,639 --> 00:27:09,305 They're vectors. 411 00:27:12,820 --> 00:27:15,510 So I'm going to need those. 412 00:27:15,510 --> 00:27:17,830 So what is RG in O? 413 00:27:17,830 --> 00:27:19,900 Well, I have this coordinate system. 414 00:27:19,900 --> 00:27:27,560 It's going to be some XI plus YJ in the inertial frame. 415 00:27:27,560 --> 00:27:33,380 And then RD with respect to G-- I 416 00:27:33,380 --> 00:27:37,500 need to have some kind of a coordinate system that 417 00:27:37,500 --> 00:27:40,200 rotates with the body. 418 00:27:40,200 --> 00:27:47,240 So I'll call this y1, and this x1-- how 419 00:27:47,240 --> 00:27:49,380 it's rotating with the body. 420 00:27:49,380 --> 00:27:58,170 So now, this one down here is minus R in the j1 direction. 421 00:28:02,370 --> 00:28:07,400 So that's the position of this point D with respect to O, 422 00:28:07,400 --> 00:28:09,040 written as the sum of two vectors. 423 00:28:25,150 --> 00:28:27,690 And I'm eventually going to need to be able to express 424 00:28:27,690 --> 00:28:33,480 this j1 in an inertial frame. 425 00:28:33,480 --> 00:28:37,500 So here's just this little vector j1, 426 00:28:37,500 --> 00:28:46,810 and I need its components in the J and I directions. 427 00:28:46,810 --> 00:28:51,220 So it's made up of a piece like that, a piece like this, 428 00:28:51,220 --> 00:28:57,120 and this angle here is theta. 429 00:28:57,120 --> 00:29:13,365 So little j1-- so I can converge. 430 00:29:13,365 --> 00:29:17,790 I can move from this unit vector system to that one 431 00:29:17,790 --> 00:29:19,239 by this transformation. 432 00:29:19,239 --> 00:29:20,697 I'm going to need that in a minute. 433 00:29:26,898 --> 00:29:29,020 Actually, I'll invoke it right now. 434 00:29:29,020 --> 00:29:34,555 So my position vector here is-- the x component is x. 435 00:29:48,190 --> 00:29:49,270 Can't read my plus. 436 00:29:49,270 --> 00:29:52,620 I crossed out a plus and minus, so I got to sort this out 437 00:29:52,620 --> 00:29:53,575 carefully here. 438 00:30:04,890 --> 00:30:05,545 Is this right? 439 00:30:15,077 --> 00:30:17,701 What do you think? 440 00:30:17,701 --> 00:30:18,950 Laura, what have I done wrong? 441 00:30:22,950 --> 00:30:25,480 So this is the plus I direction. 442 00:30:25,480 --> 00:30:32,400 This piece here is-- so minus sine 443 00:30:32,400 --> 00:30:37,015 theta in the I plus cosine theta in the J are the components. 444 00:30:37,015 --> 00:30:39,441 So I'm missing a minus sign there. 445 00:30:39,441 --> 00:30:39,940 OK. 446 00:30:42,640 --> 00:30:45,070 So I want to collect the I pieces together. 447 00:30:45,070 --> 00:30:48,250 So I have an X in the I, and I have 448 00:30:48,250 --> 00:30:52,690 a minus R minus sine theta. 449 00:30:52,690 --> 00:31:00,660 So, plus R sine theta-- this is the I contribution. 450 00:31:00,660 --> 00:31:09,600 And then over here I have a Y and a minus R cosine 451 00:31:09,600 --> 00:31:15,720 theta in the J. Then I have this vector all worked out 452 00:31:15,720 --> 00:31:16,780 in my inertial frame. 453 00:31:20,410 --> 00:31:29,810 So now I can say that-- remember, Qj dot 454 00:31:29,810 --> 00:31:38,290 dR-- it's a sort of virtual deflection for the jth 455 00:31:38,290 --> 00:31:39,920 contribution. 456 00:31:39,920 --> 00:31:44,965 So in this case, I'm interested in, say, the 1 in the-- which 457 00:31:44,965 --> 00:31:48,195 one should we do first-- The x direction. 458 00:31:52,260 --> 00:32:02,530 So this is the F1 in the I dot derivative of RD with respect 459 00:32:02,530 --> 00:32:08,900 to x delta x. 460 00:32:08,900 --> 00:32:11,100 But now I have an-- I can work this out. 461 00:32:11,100 --> 00:32:13,150 I can figure this one out directly now, 462 00:32:13,150 --> 00:32:17,690 because I have an expression for R-- 463 00:32:17,690 --> 00:32:22,780 that vector-- in terms of one set of unit vectors, 464 00:32:22,780 --> 00:32:26,160 and I can take the derivative with respect to x. 465 00:32:26,160 --> 00:32:27,812 And the only x that shows up in here 466 00:32:27,812 --> 00:32:29,645 is this, so the derivative is pretty simple. 467 00:32:29,645 --> 00:32:31,490 It's just 1. 468 00:32:31,490 --> 00:32:38,710 So this is F1I dot I delta x. 469 00:32:38,710 --> 00:32:45,670 So this tells us that Qx equals F1. 470 00:32:45,670 --> 00:32:46,610 No great surprise. 471 00:32:46,610 --> 00:32:49,920 We knew the answer to that from before. 472 00:32:49,920 --> 00:32:57,760 And I-- oh, I made a mistake. 473 00:33:01,740 --> 00:33:04,620 What didn't I-- I was mixing the two. 474 00:33:04,620 --> 00:33:07,085 I did a simplification when I did the direct one. 475 00:33:07,085 --> 00:33:08,186 Is F1 in the I direction? 476 00:33:17,090 --> 00:33:20,390 So F1 has components. 477 00:33:20,390 --> 00:33:33,420 Vector F1 has a magnitude F1 times cosine theta I plus sine 478 00:33:33,420 --> 00:33:37,000 theta J. And so down here, I need 479 00:33:37,000 --> 00:33:40,020 to put those in and take the dot product. 480 00:33:40,020 --> 00:33:44,980 So the only dot product that will matter is the I component. 481 00:33:44,980 --> 00:33:50,372 This is going to be F1 cosine theta I 482 00:33:50,372 --> 00:33:54,460 dot this, which we know this gives us just and I. 483 00:33:54,460 --> 00:33:56,830 So I'm leaving out the J piece. 484 00:33:56,830 --> 00:34:02,010 So I end up here with F1 cosine theta. 485 00:34:02,010 --> 00:34:04,630 Does that look right? 486 00:34:04,630 --> 00:34:06,990 So here is your theta. 487 00:34:06,990 --> 00:34:10,680 Here is your-- this is F1 cosine theta. 488 00:34:13,360 --> 00:34:18,159 So for our coordinate system that's squared up like this, 489 00:34:18,159 --> 00:34:21,730 it's only that term that's in the F1 direction, 490 00:34:21,730 --> 00:34:24,610 and this would come out correctly. 491 00:34:24,610 --> 00:34:25,149 Now 492 00:34:25,149 --> 00:34:28,790 And if we let-- if we now reorient our coordinate system 493 00:34:28,790 --> 00:34:36,199 so that theta is 0, then this would just turn out to be F1. 494 00:34:36,199 --> 00:34:38,719 If you can see that this is F1 cosine theta without going 495 00:34:38,719 --> 00:34:42,489 through all the work, what's Qy for this system? 496 00:34:47,090 --> 00:34:47,900 Sure. 497 00:34:47,900 --> 00:34:50,359 F1 sine theta. 498 00:34:50,359 --> 00:34:51,734 We won't go through all the work, 499 00:34:51,734 --> 00:34:55,300 but you go through the same taking 500 00:34:55,300 --> 00:34:57,500 the derivative-- in this case, the derivative of R 501 00:34:57,500 --> 00:35:01,880 with respect to y, and then dot it with F, 502 00:35:01,880 --> 00:35:05,900 and you get F1 sine theta in the J direction. 503 00:35:05,900 --> 00:35:08,020 So there's Qy. 504 00:35:08,020 --> 00:35:22,540 Now, the problem I ran into, conceptually, 505 00:35:22,540 --> 00:35:25,570 is around this other piece. 506 00:35:25,570 --> 00:35:31,000 So Q theta, the virtual work done in the theta direction, 507 00:35:31,000 --> 00:35:34,360 we know has to do with rotation. 508 00:35:34,360 --> 00:35:41,460 And that is going to be sum F1 dot-- now 509 00:35:41,460 --> 00:35:46,421 this is the derivative of RD with respect to theta delta 510 00:35:46,421 --> 00:35:46,920 theta. 511 00:35:53,620 --> 00:36:02,632 So RD is here, and it's indeed a function of theta, right? 512 00:36:02,632 --> 00:36:04,840 And so I can take a derivative with respect to theta, 513 00:36:04,840 --> 00:36:08,010 and I'll get derivative of that R sine theta term, 514 00:36:08,010 --> 00:36:10,996 and I'll get a derivative of the R cosine theta term. 515 00:36:10,996 --> 00:36:12,620 But here's how I got myself in trouble. 516 00:36:12,620 --> 00:36:14,190 It took me-- and this is the sort of thing 517 00:36:14,190 --> 00:36:15,273 that happens to all of us. 518 00:36:15,273 --> 00:36:18,407 You're working a problem, throw in a simplification, 519 00:36:18,407 --> 00:36:19,740 and then something doesn't work. 520 00:36:19,740 --> 00:36:31,290 Well, what if you had oriented the axis, initially, 521 00:36:31,290 --> 00:36:38,720 so that the force is aligned with x and perpendicular to y? 522 00:36:38,720 --> 00:36:45,780 Then this theta angle is 0, right? 523 00:36:45,780 --> 00:36:48,670 And you know, when you set it up, 524 00:36:48,670 --> 00:36:53,980 you don't end up with-- the cosine of 0 is 1, and sine of 0 525 00:36:53,980 --> 00:36:55,672 is 0. 526 00:36:55,672 --> 00:36:58,260 You don't end up with the cosine and sines in there 527 00:36:58,260 --> 00:37:00,140 to take derivatives of. 528 00:37:00,140 --> 00:37:02,450 You just jump to the simplification, 529 00:37:02,450 --> 00:37:13,060 and you just find out that you just have R in the J. 530 00:37:13,060 --> 00:37:14,480 There's no R sine theta term. 531 00:37:14,480 --> 00:37:17,380 It's just vanished on you. 532 00:37:17,380 --> 00:37:21,400 And now you need to take that derivative to do this step. 533 00:37:26,300 --> 00:37:29,010 And I do count with it as huh? 534 00:37:29,010 --> 00:37:31,600 What have I done wrong? 535 00:37:31,600 --> 00:37:33,550 In order to do this method, you need 536 00:37:33,550 --> 00:37:37,040 to leave it in the sort of general formulation, 537 00:37:37,040 --> 00:37:40,010 and then let theta be 0 at the end, if you want to do that. 538 00:37:40,010 --> 00:37:45,340 So if I had then finished it, I can 539 00:37:45,340 --> 00:37:47,690 do this for this general problem, 540 00:37:47,690 --> 00:37:50,520 and then let theta go to 0, and the answer 541 00:37:50,520 --> 00:37:51,520 will come out all right. 542 00:37:51,520 --> 00:37:53,140 So should we do this? 543 00:37:53,140 --> 00:37:56,460 So this is F1 dot this. 544 00:37:56,460 --> 00:38:09,920 So, F1 dotted with the derivative of R 545 00:38:09,920 --> 00:38:12,010 with respect to theta. 546 00:38:12,010 --> 00:38:21,957 So I get an R cosine theta in the I, 547 00:38:21,957 --> 00:38:23,915 and the derivative of this term will be-- well, 548 00:38:23,915 --> 00:38:26,150 the cosine theta is minus sine theta 549 00:38:26,150 --> 00:38:36,550 times a minus is a plus R sine theta in the J delta theta. 550 00:38:36,550 --> 00:38:40,810 And I do this dot product, so I get 551 00:38:40,810 --> 00:38:46,715 the I times the I's, and the J terms times the J terms. 552 00:39:11,980 --> 00:39:13,510 And it all works out nicely. 553 00:39:13,510 --> 00:39:16,090 Sine squared plus cosine squared-- you get 1. 554 00:39:16,090 --> 00:39:19,390 And you find out that the torque doesn't care about the angle. 555 00:39:19,390 --> 00:39:20,839 And does that make sense? 556 00:39:20,839 --> 00:39:21,755 AUDIENCE: [INAUDIBLE]? 557 00:39:25,434 --> 00:39:26,350 PROFESSOR: Absolutely. 558 00:39:26,350 --> 00:39:30,160 I did this on purpose to give us a little more practice using 559 00:39:30,160 --> 00:39:33,330 this kinematic grind it out method. 560 00:39:33,330 --> 00:39:36,870 But, for sure, the intuitive method 561 00:39:36,870 --> 00:39:41,390 would have yielded result a lot faster here, right? 562 00:39:41,390 --> 00:39:45,500 We would've said, let's have a little deflection delta x. 563 00:39:45,500 --> 00:39:46,870 What's the virtual work, then? 564 00:39:46,870 --> 00:39:50,700 Well, would have been F1 cosine theta delta x. 565 00:39:50,700 --> 00:39:53,740 It would be Qx delta x. 566 00:39:53,740 --> 00:40:00,140 Qy delta y would be F1 delta y sine theta, and we're done. 567 00:40:00,140 --> 00:40:04,660 And the rotation one-- we would have looked at this 568 00:40:04,660 --> 00:40:08,130 and said, how much-- in a little motion delta theta-- how 569 00:40:08,130 --> 00:40:09,370 much does this move? 570 00:40:09,370 --> 00:40:15,565 It moves R delta theta crossed with the F 571 00:40:15,565 --> 00:40:19,440 that it moves through, or dot product with that distance, 572 00:40:19,440 --> 00:40:22,550 would have given you F1 R delta theta equals 573 00:40:22,550 --> 00:40:24,130 Q theta delta theta. 574 00:40:24,130 --> 00:40:26,215 So we could have done it in a minute. 575 00:40:28,850 --> 00:40:30,840 And sometimes, doing it the hard way 576 00:40:30,840 --> 00:40:32,360 is what will get you in trouble. 577 00:40:32,360 --> 00:40:35,869 I didn't have the cosine theta and sine thetas 578 00:40:35,869 --> 00:40:38,160 working out in this and I couldn't take the derivative. 579 00:40:38,160 --> 00:40:40,540 What did I do wrong? 580 00:40:40,540 --> 00:40:43,084 And I finally realized, I made the problem too simple, 581 00:40:43,084 --> 00:40:46,840 simplified it too soon. 582 00:40:46,840 --> 00:40:50,810 So I wanted to use this as a stepping stone 583 00:40:50,810 --> 00:40:55,147 to something I meant to cover perhaps weeks ago. 584 00:40:55,147 --> 00:40:56,605 And there's a general little lesson 585 00:40:56,605 --> 00:40:59,860 that we can learn from this, which you've probably 586 00:40:59,860 --> 00:41:02,990 been taught before, but I'm going to remind you. 587 00:41:02,990 --> 00:41:05,840 Here's a rigid body. 588 00:41:05,840 --> 00:41:08,440 Got some center of mass. 589 00:41:08,440 --> 00:41:13,840 And I've got a force acting on it here, 590 00:41:13,840 --> 00:41:15,880 just some arbitrary position and angle. 591 00:41:19,600 --> 00:41:22,410 So that force has a line of action, 592 00:41:22,410 --> 00:41:25,680 and perpendicular to that line of action 593 00:41:25,680 --> 00:41:28,825 is a distance, which I'll call d here. 594 00:41:31,450 --> 00:41:34,580 And we know that in R cross F, we 595 00:41:34,580 --> 00:41:44,290 can-- this force exerts a moment about G that would be R cross 596 00:41:44,290 --> 00:41:47,160 F, where R goes from here to here, 597 00:41:47,160 --> 00:41:50,830 but it's only the component perpendicular to it. 598 00:41:50,830 --> 00:41:54,060 So we know that this is going to create a moment. 599 00:41:54,060 --> 00:41:56,570 And I won't draw pictures yet here for a moment. 600 00:41:56,570 --> 00:42:00,555 This diagram is the same as-- well, 601 00:42:00,555 --> 00:42:03,510 I could draw the same peanut here. 602 00:42:03,510 --> 00:42:09,740 It's the same as the following-- a force here, that's F, 603 00:42:09,740 --> 00:42:12,540 and another force at G, that's F. Equal and opposite, 604 00:42:12,540 --> 00:42:15,340 so it's like adding 0 to the problem. 605 00:42:15,340 --> 00:42:19,310 So I can do that with impunity, and I still 606 00:42:19,310 --> 00:42:23,970 have my force F down here. 607 00:42:23,970 --> 00:42:25,882 So this problem's equal to that problem. 608 00:42:29,530 --> 00:42:36,260 But now, this force and this force, equal and opposite, 609 00:42:36,260 --> 00:42:37,860 cancel one another, but create what's 610 00:42:37,860 --> 00:42:42,440 called a couple-- a moment that we can compute about my point G 611 00:42:42,440 --> 00:42:43,340 here. 612 00:42:43,340 --> 00:42:48,110 So these two forces create just a pure torque, 613 00:42:48,110 --> 00:42:51,670 leaving this as a force. 614 00:42:51,670 --> 00:42:53,550 They're equal and opposite, so no net force. 615 00:42:53,550 --> 00:42:56,280 But they create a torque in this direction, 616 00:42:56,280 --> 00:42:59,910 and there's a remaining force on the center of mass. 617 00:42:59,910 --> 00:43:04,110 So this whole thing is also, then, 618 00:43:04,110 --> 00:43:12,220 equal to the equivalent of a torque 619 00:43:12,220 --> 00:43:16,400 around the center of mass and a force 620 00:43:16,400 --> 00:43:19,620 applied at the center of mass. 621 00:43:19,620 --> 00:43:23,520 And in this case, the torque would be dF. 622 00:43:28,310 --> 00:43:30,750 This problem is easier to do if you're 623 00:43:30,750 --> 00:43:34,200 trying to compute generalized forces than the one we started 624 00:43:34,200 --> 00:43:39,600 with, because now, the position vector that we need 625 00:43:39,600 --> 00:43:41,035 goes only to the center of mass. 626 00:43:45,630 --> 00:43:50,890 And the general rule for this-- you have a body, 627 00:43:50,890 --> 00:43:59,590 you have several forces-- F1, F2, F3. 628 00:44:03,410 --> 00:44:15,800 That's equal to the body with G, with a F total on it, 629 00:44:15,800 --> 00:44:20,700 and a moment-- a torque-- total. 630 00:44:20,700 --> 00:44:27,430 And F total-- it's a vector-- is a summation of the Fi's. 631 00:44:27,430 --> 00:44:32,890 And T, the torque total, is the summation 632 00:44:32,890 --> 00:44:34,630 of each of these guys. 633 00:44:34,630 --> 00:44:36,920 Here's G. This has an R1. 634 00:44:39,560 --> 00:44:40,490 Here is an R2. 635 00:44:43,800 --> 00:44:49,020 Here's an Ri to the ith force. 636 00:44:49,020 --> 00:44:55,380 So this is a summation of the Ri cross Fi's. 637 00:44:55,380 --> 00:44:58,240 So if you sum all the forces and compute the equivalent torque 638 00:44:58,240 --> 00:45:00,555 about the center of mass, sum all the forces 639 00:45:00,555 --> 00:45:03,030 and apply it at the center of mass, 640 00:45:03,030 --> 00:45:07,440 then this problem is equal to that problem. 641 00:45:07,440 --> 00:45:09,540 So this can come in handy when you're 642 00:45:09,540 --> 00:45:10,785 trying to simplify problems. 643 00:45:22,330 --> 00:45:38,800 And for this particular problem, let's draw our hockey puck. 644 00:45:46,745 --> 00:45:51,400 Going to put my force in the simplest direction here. 645 00:45:51,400 --> 00:45:53,260 But there's my force. 646 00:45:53,260 --> 00:46:00,520 This problem is equal, doing the same thing here. 647 00:46:00,520 --> 00:46:04,925 It's the same thing as if I had drew forces like this. 648 00:46:12,570 --> 00:46:15,060 And that gives me an equivalent problem, 649 00:46:15,060 --> 00:46:22,420 where I have-- this couple here produces a torque, 650 00:46:22,420 --> 00:46:28,200 and there is a net force acting on the center of mass. 651 00:46:28,200 --> 00:46:32,980 So this problem is equal to this problem now. 652 00:46:32,980 --> 00:46:36,410 A torque and a force both acting at the center of mass, 653 00:46:36,410 --> 00:46:43,250 and the torque is RF1 in the k. 654 00:46:43,250 --> 00:46:50,530 And the force is just F1 in the positive I direction. 655 00:46:53,620 --> 00:46:58,500 And R-- but now, what is the little R vector? 656 00:46:58,500 --> 00:47:02,650 If we're going to do this the kinematic way? 657 00:47:02,650 --> 00:47:08,210 From here to G, this is R1, I'll call it. 658 00:47:08,210 --> 00:47:09,470 And that's it. 659 00:47:09,470 --> 00:47:13,700 I only-- now I have everything acting here. 660 00:47:13,700 --> 00:47:26,740 And Qx delta x is F1 dot derivative of R 661 00:47:26,740 --> 00:47:30,610 with respect to x delta x. 662 00:47:30,610 --> 00:47:33,375 But what is R? 663 00:47:38,690 --> 00:47:41,610 It's XI plus YJ. 664 00:47:41,610 --> 00:47:49,310 The derivative of R1 with respect to x is just 1 665 00:47:49,310 --> 00:47:52,940 in the I, and the derivative of R with respect to Y 666 00:47:52,940 --> 00:47:58,460 is just J. So this becomes a trivial calculation. 667 00:47:58,460 --> 00:48:07,060 F1I dot I-- I get Qx equals F1 delta x. 668 00:48:07,060 --> 00:48:15,910 Then I get Qy is F1, which is in the I direction dot J is 0. 669 00:48:15,910 --> 00:48:19,600 And I get Q theta-- ah, Q theta. 670 00:48:19,600 --> 00:48:23,270 Now, do I use this expression? 671 00:48:23,270 --> 00:48:27,100 So, I'm not dealing with little changes in distance anymore. 672 00:48:27,100 --> 00:48:29,400 When you can put rotation-- when you put moments 673 00:48:29,400 --> 00:48:31,500 about the centers of mass, it's just easier. 674 00:48:31,500 --> 00:48:34,790 This one-- you call intuitive, but this is just 675 00:48:34,790 --> 00:48:38,960 going to be the sum delta theta is 676 00:48:38,960 --> 00:48:46,070 equal to the total torques acting 677 00:48:46,070 --> 00:48:53,070 at the center of mass dot delta theta. 678 00:48:53,070 --> 00:48:56,110 And in this case, the torque is in the k direction, 679 00:48:56,110 --> 00:48:58,210 delta theta's in the k direction, 680 00:48:58,210 --> 00:49:12,910 and this is RF1k dot k delta theta, which just gives us 681 00:49:12,910 --> 00:49:16,740 RF1 equals Q theta. 682 00:49:16,740 --> 00:49:20,252 So the torque equations you can derive straight out. 683 00:49:20,252 --> 00:49:21,710 You don't have to take derivatives. 684 00:49:21,710 --> 00:49:24,140 It doesn't have anything to do with change in position. 685 00:49:24,140 --> 00:49:26,554 It's just a little rotation, delta theta. 686 00:49:29,210 --> 00:49:31,920 So it simplifies this problem and, actually, totally avoids 687 00:49:31,920 --> 00:49:35,180 the conundrum I got into. 688 00:49:35,180 --> 00:49:37,060 If I work everything about the center of mass 689 00:49:37,060 --> 00:49:40,010 of that rigid body, the R gets easier, 690 00:49:40,010 --> 00:49:42,580 the derivatives get easier, and the torque 691 00:49:42,580 --> 00:49:45,050 becomes obvious how it applies. 692 00:50:16,950 --> 00:50:21,330 We'll talk about-- I have a pendulum in kind 693 00:50:21,330 --> 00:50:22,060 of a weird shape. 694 00:50:29,260 --> 00:50:34,400 But the pendulum has the following properties. 695 00:50:34,400 --> 00:50:38,760 So that weird shape-- that may be this. 696 00:50:38,760 --> 00:50:45,100 It's got one plane of symmetry like this does. 697 00:50:45,100 --> 00:50:49,650 And I'm going to put-- what did I do with my axle? 698 00:50:49,650 --> 00:50:53,780 I'm going to put the axis of rotation perpendicular 699 00:50:53,780 --> 00:50:57,510 to that plane of symmetry, somewhere not at G, 700 00:50:57,510 --> 00:51:01,280 not at the center of mass-- so, just what I've done here. 701 00:51:01,280 --> 00:51:07,790 I can make any object that has a plane of symmetry-- 702 00:51:07,790 --> 00:51:13,940 if I have it rotate about an axis that's 703 00:51:13,940 --> 00:51:16,030 perpendicular to that plane of symmetry, 704 00:51:16,030 --> 00:51:19,440 it becomes a simple pendulum. 705 00:51:19,440 --> 00:51:27,210 And you can write down with my inspection 706 00:51:27,210 --> 00:51:28,930 what the equation of motion is. 707 00:51:28,930 --> 00:51:33,720 So, how many degrees of freedom does this system have? 708 00:51:37,120 --> 00:51:40,870 So first of all, then, is it planar motion? 709 00:51:40,870 --> 00:51:45,760 So it has, at most, 3, and how many-- 710 00:51:45,760 --> 00:51:47,900 and it has a pin through it that doesn't allow it 711 00:51:47,900 --> 00:51:50,590 to move in x or y. 712 00:51:50,590 --> 00:51:51,945 How many constraints is that? 713 00:51:51,945 --> 00:51:52,599 AUDIENCE: 2. 714 00:51:52,599 --> 00:51:53,140 PROFESSOR: 2. 715 00:51:53,140 --> 00:51:55,240 So how many you got left? 716 00:51:55,240 --> 00:51:55,740 AUDIENCE: 1. 717 00:51:55,740 --> 00:51:56,281 PROFESSOR: 1. 718 00:51:56,281 --> 00:51:59,580 So this is a single degree of freedom problem, 719 00:51:59,580 --> 00:52:04,070 and I've chosen to have it move about an axis that 720 00:52:04,070 --> 00:52:08,450 is-- is this axis a principal axis of this body? 721 00:52:08,450 --> 00:52:10,390 Absolutely. 722 00:52:10,390 --> 00:52:13,330 And so, if I ask you, what's the mass moment of inertia of this? 723 00:52:13,330 --> 00:52:18,217 If I gave you Izz about G for this thing, and this distance, 724 00:52:18,217 --> 00:52:19,800 what would you tell me the mass moment 725 00:52:19,800 --> 00:52:21,380 of inertia about the pivot is? 726 00:52:26,330 --> 00:52:27,925 I'll give you this. 727 00:52:34,534 --> 00:52:35,450 What's the rest of it? 728 00:52:38,670 --> 00:52:41,155 Not a G. You've got some distance here, 729 00:52:41,155 --> 00:52:46,860 which you're given-- L, we'll call it. 730 00:52:46,860 --> 00:52:47,390 Correct. 731 00:52:47,390 --> 00:52:50,060 So, parallel axis-- remember, ML squared. 732 00:52:50,060 --> 00:52:55,350 So we know that that's true, and the G is over here someplace, 733 00:52:55,350 --> 00:53:00,120 and this is this distance L. And what's the equation of motion 734 00:53:00,120 --> 00:53:00,990 for this? 735 00:53:00,990 --> 00:53:03,860 And gravity acts. 736 00:53:03,860 --> 00:53:05,660 No damping for now. 737 00:53:05,660 --> 00:53:08,204 What's the equation of motion? 738 00:53:08,204 --> 00:53:09,370 Right off the top your head. 739 00:53:09,370 --> 00:53:10,602 You've done enough of these. 740 00:53:14,670 --> 00:53:16,960 I'll give you one minute think about it. 741 00:53:16,960 --> 00:53:19,690 We'll figure out the equation of motion for this problem. 742 00:53:30,050 --> 00:53:32,205 I highly recommend you don't do the Lagrange. 743 00:54:35,420 --> 00:54:37,170 Somebody give me their equation of motion. 744 00:54:54,090 --> 00:54:58,210 So, what method would you choose to use? 745 00:54:58,210 --> 00:55:01,440 Simplest one you could think of. 746 00:55:01,440 --> 00:55:02,130 Direct? 747 00:55:02,130 --> 00:55:03,576 About what point? 748 00:55:03,576 --> 00:55:04,390 AUDIENCE: Around A. 749 00:55:04,390 --> 00:55:05,848 PROFESSOR: About A. And what do you 750 00:55:05,848 --> 00:55:09,940 say is-- what equation do you apply? 751 00:55:09,940 --> 00:55:11,487 General equation. 752 00:55:11,487 --> 00:55:16,927 AUDIENCE: [INAUDIBLE] It's just AIzz [INAUDIBLE]. 753 00:55:16,927 --> 00:55:19,010 PROFESSOR: Yeah, that would be part of the answer. 754 00:55:19,010 --> 00:55:21,468 And the equation-- remember, I wrote a list of them-- 1, 2, 755 00:55:21,468 --> 00:55:22,770 3 over there. 756 00:55:22,770 --> 00:55:23,860 Which one do you use? 757 00:55:27,077 --> 00:55:28,660 AUDIENCE: You just use the second one. 758 00:55:28,660 --> 00:55:30,250 PROFESSOR: Just a second one, right. 759 00:55:30,250 --> 00:55:33,086 And do the nuisance terms go away? 760 00:55:33,086 --> 00:55:34,410 Point A's not moving. 761 00:55:34,410 --> 00:55:37,010 So some of the torque is equal to dHdt, 762 00:55:37,010 --> 00:55:38,890 and it's H with respect to A, right? 763 00:55:57,740 --> 00:55:59,670 So that's H, right? 764 00:56:03,170 --> 00:56:12,880 d by dt Izz with respect to A theta double dot k 765 00:56:12,880 --> 00:56:14,650 equals the sum of the external torques. 766 00:56:14,650 --> 00:56:18,320 And the rest of the problem is finding the external torques. 767 00:56:18,320 --> 00:56:20,130 What's the free body diagram look like? 768 00:56:24,730 --> 00:56:39,400 So, here's the object, possibly A and Rx, and an Ry, and an Mg. 769 00:56:39,400 --> 00:56:41,650 Those are your possible-- that will set your free body 770 00:56:41,650 --> 00:56:43,280 diagram, right? 771 00:56:43,280 --> 00:56:47,530 So here's the Mg down. 772 00:56:47,530 --> 00:56:51,800 What's the torque that gravitational force 773 00:56:51,800 --> 00:56:55,570 puts on this object with respect to point A? 774 00:56:58,100 --> 00:56:59,500 R cross F, right? 775 00:57:02,340 --> 00:57:05,582 R cross F into the board. 776 00:57:05,582 --> 00:57:06,790 Is that positive or negative? 777 00:57:10,090 --> 00:57:11,510 And what's the link to this R? 778 00:57:16,830 --> 00:57:19,520 Well, from here to here is, I guess-- oh, that 779 00:57:19,520 --> 00:57:23,395 is L. So the length of this side and that triangle? 780 00:57:26,350 --> 00:57:28,820 L sine theta? 781 00:57:28,820 --> 00:57:31,890 L sine theta cross Mg. 782 00:57:31,890 --> 00:57:36,260 And it is positive or negative-- what did we decide? 783 00:57:36,260 --> 00:57:42,790 Looks like it's-- this is positive theta xy system. 784 00:57:42,790 --> 00:57:51,180 It gets negative minus MgL sine theta, and we're done. 785 00:57:51,180 --> 00:57:59,780 So, generically-- and this is also in the k direction, 786 00:57:59,780 --> 00:58:01,200 so we can drop the k's now. 787 00:58:01,200 --> 00:58:08,680 We have our equation of motion Izz theta double dot plus MgL 788 00:58:08,680 --> 00:58:10,530 sine theta. 789 00:58:15,843 --> 00:58:24,520 Is there any then single degree of freedom pendulum made out 790 00:58:24,520 --> 00:58:30,320 of a rigid body, and rotating about an axis 791 00:58:30,320 --> 00:58:35,161 that is a principal axis that has that equation of motion? 792 00:58:40,240 --> 00:58:43,400 Any of them, no matter what the shape. 793 00:58:43,400 --> 00:58:46,090 If you can rotate it about a principal axis, 794 00:58:46,090 --> 00:58:48,790 and not through G, because if you put-- 795 00:58:48,790 --> 00:58:53,680 what's the natural frequency if you run the axis through G? 796 00:58:53,680 --> 00:58:55,516 What's the torque? 797 00:58:55,516 --> 00:58:56,015 0. 798 00:58:56,015 --> 00:58:58,999 Nothing happens, right? 799 00:58:58,999 --> 00:59:00,290 It doesn't want to do anything. 800 00:59:00,290 --> 00:59:03,980 But as soon as it's not through G, then the things oscillates. 801 00:59:03,980 --> 00:59:06,860 And that's because this is now a differential equation. 802 00:59:06,860 --> 00:59:11,320 And I need to say this equals 0 here. 803 00:59:11,320 --> 00:59:12,340 There's no other forces. 804 00:59:12,340 --> 00:59:14,548 If you've got a damping force, then it would show up. 805 00:59:14,548 --> 00:59:18,060 But this is a generic, undamped equation 806 00:59:18,060 --> 00:59:22,250 of motion for any single degree of freedom pendulum 807 00:59:22,250 --> 00:59:27,400 rotating about one of its principal axes. 808 00:59:27,400 --> 00:59:29,980 And if the body you're shown or given 809 00:59:29,980 --> 00:59:33,280 has a single plane of symmetry, then you 810 00:59:33,280 --> 00:59:35,010 can figure out one way, immediately, 811 00:59:35,010 --> 00:59:38,790 that you know this will-- that equation applies. 812 00:59:38,790 --> 00:59:42,540 An axis perpendicular to that plane of symmetry is a pendulum 813 00:59:42,540 --> 00:59:45,210 and it is that equation. 814 00:59:45,210 --> 00:59:50,970 And the i respect to A you can get from simple parallel axis, 815 00:59:50,970 --> 00:59:54,960 if you're told what i with respect to G is. 816 01:00:00,221 --> 01:00:00,720 Questions? 817 01:00:13,490 --> 01:00:16,670 I'm going to put up two additional problems. 818 01:00:16,670 --> 01:00:18,659 And I'm not going to solve them. 819 01:00:18,659 --> 01:00:19,950 We're going to talk about them. 820 01:00:19,950 --> 01:00:22,366 You're going to tell me how you would go about doing them. 821 01:00:22,366 --> 01:00:24,930 And they wouldn't be bad problems for you 822 01:00:24,930 --> 01:00:25,980 to practice on. 823 01:00:35,590 --> 01:00:40,200 And you've actually seen both of them before. 824 01:00:40,200 --> 01:00:51,935 So, the first one-- pulley rotating about that fixed axis. 825 01:00:54,800 --> 01:00:59,895 And this is that thing we call Atwood's Machine. 826 01:01:02,825 --> 01:01:04,075 How did I draw my coordinates? 827 01:01:10,240 --> 01:01:21,250 So I have my Izz about G. It's sum M3 kappa squared. 828 01:01:21,250 --> 01:01:23,790 You're given the radius of gyration of this pulley. 829 01:01:23,790 --> 01:01:27,090 So that's its mass moment of inertia about the center. 830 01:01:27,090 --> 01:01:34,755 And you know R. You could specify a theta. 831 01:01:38,780 --> 01:01:48,500 And we have a pair of masses-- start off like this, M1, M2. 832 01:01:48,500 --> 01:01:51,440 And this, now-- so I've going to say, no slip. 833 01:01:54,550 --> 01:01:59,390 And this rope goes over the pulley, no slip, initially 834 01:01:59,390 --> 01:02:00,425 stationary. 835 01:02:00,425 --> 01:02:02,580 I let go. 836 01:02:02,580 --> 01:02:05,310 I want an equation of motion for this system. 837 01:02:05,310 --> 01:02:07,490 So how many is it-- first of all, 838 01:02:07,490 --> 01:02:10,800 is it a planar motion problem? 839 01:02:10,800 --> 01:02:11,530 OK. 840 01:02:11,530 --> 01:02:12,500 How many rigid bodies? 841 01:02:15,280 --> 01:02:17,980 How many potential degrees of freedom? 842 01:02:17,980 --> 01:02:18,750 AUDIENCE: None. 843 01:02:18,750 --> 01:02:20,360 PROFESSOR: None. 844 01:02:20,360 --> 01:02:21,810 How many do you think they really 845 01:02:21,810 --> 01:02:23,270 are going to end up with here? 846 01:02:27,982 --> 01:02:29,590 How many do you expect to find? 847 01:02:29,590 --> 01:02:34,330 How many necessary coordinates are you 848 01:02:34,330 --> 01:02:36,870 going to need to completely describe the motion? 849 01:02:36,870 --> 01:02:38,220 I see some 1's going up. 850 01:02:38,220 --> 01:02:40,660 Everybody believe that? 851 01:02:40,660 --> 01:02:42,040 Lots of constraints. 852 01:02:42,040 --> 01:02:43,870 We're not going to let this-- we're 853 01:02:43,870 --> 01:02:46,895 constraining this to move in-- only allowing 854 01:02:46,895 --> 01:02:48,000 it to move up and down. 855 01:02:48,000 --> 01:02:52,060 These can't rotate, so there's one for each. 856 01:02:52,060 --> 01:02:57,870 These can't go in this direction, so two more. 857 01:02:57,870 --> 01:03:00,990 This can only rotate, no translations. 858 01:03:00,990 --> 01:03:03,660 So you end up with 1. 859 01:03:03,660 --> 01:03:05,800 And finally, there's this no slip condition, 860 01:03:05,800 --> 01:03:11,015 which means that R theta equals x, 861 01:03:11,015 --> 01:03:14,650 and that's one final one that you'd need to write down. 862 01:03:14,650 --> 01:03:17,780 So if you had that, gotten this far-- 863 01:03:17,780 --> 01:03:19,550 find the equation of motion. 864 01:03:19,550 --> 01:03:21,110 What method would you use? 865 01:03:21,110 --> 01:03:31,335 So, your choices are here-- direct method, Lagrange. 866 01:03:31,335 --> 01:03:31,960 Think about it. 867 01:03:31,960 --> 01:03:34,479 How would you-- what's the easiest way for you to 868 01:03:34,479 --> 01:03:35,145 do this problem? 869 01:03:42,400 --> 01:03:46,110 Maybe an ancillary question is, how many ways can you 870 01:03:46,110 --> 01:03:48,110 think of that you could do this problem? 871 01:03:48,110 --> 01:03:50,340 How many different approaches could you use? 872 01:04:15,510 --> 01:04:22,585 So, how many ways can you think of doing this problem, Kristen? 873 01:04:27,660 --> 01:04:29,010 I'm just picking on her, but-- 874 01:04:29,010 --> 01:04:30,093 AUDIENCE: I'm not Kristen. 875 01:04:32,911 --> 01:04:34,410 PROFESSOR: Give me an answer anyway. 876 01:04:34,410 --> 01:04:34,951 AUDIENCE: OK. 877 01:04:38,258 --> 01:04:40,522 I think you could do two, but I would choose Lagrange. 878 01:04:40,522 --> 01:04:42,480 PROFESSOR: You would want to do it by Lagrange. 879 01:04:42,480 --> 01:04:44,625 Helen, how else could you do it? 880 01:04:44,625 --> 01:04:48,270 If you did it by Lagrange, got the answer, and said, 881 01:04:48,270 --> 01:04:50,635 I want to check it, what would you do next? 882 01:04:50,635 --> 01:04:52,610 AUDIENCE: The direct method. 883 01:04:52,610 --> 01:04:53,360 PROFESSOR: Direct. 884 01:04:53,360 --> 01:04:55,990 But where would do apply? 885 01:04:55,990 --> 01:04:57,220 Would you use torques? 886 01:04:57,220 --> 01:04:58,710 Would you use forces? 887 01:04:58,710 --> 01:05:00,170 How would you go about doing this? 888 01:05:00,170 --> 01:05:03,800 AUDIENCE: I would use torques about the center of the pulley. 889 01:05:03,800 --> 01:05:05,300 PROFESSOR: Torques about the center. 890 01:05:05,300 --> 01:05:06,120 We did that. 891 01:05:06,120 --> 01:05:07,620 I mean, I worked that-- I actually 892 01:05:07,620 --> 01:05:11,830 did this problem by that method earlier in the term. 893 01:05:11,830 --> 01:05:13,980 Torques about A. Let's have the pivot. 894 01:05:13,980 --> 01:05:14,730 Works fine. 895 01:05:14,730 --> 01:05:17,950 Actually, it's pretty efficient. 896 01:05:17,950 --> 01:05:22,380 So, Lagrange or torques about A. You only 897 01:05:22,380 --> 01:05:24,260 need-- there's one degree of freedom, right? 898 01:05:24,260 --> 01:05:28,920 How many equations do you need to write to do this problem? 899 01:05:28,920 --> 01:05:29,420 Just one. 900 01:05:29,420 --> 01:05:32,870 So the sum of the torques about A will give you the answer. 901 01:05:32,870 --> 01:05:36,330 Lagrange equation will give you the answer. 902 01:05:36,330 --> 01:05:40,425 That's a good problem to practice on. 903 01:05:40,425 --> 01:05:46,815 Another problem-- this is not very big. 904 01:05:49,810 --> 01:05:51,420 It's basically this problem. 905 01:05:51,420 --> 01:05:52,891 This is just-- you've seen this. 906 01:05:52,891 --> 01:05:55,390 I'm sure you've been shown this in physics and stuff before. 907 01:05:55,390 --> 01:05:59,170 This is the falling stick problem. 908 01:05:59,170 --> 01:06:02,620 You can't set this up without having some friction. 909 01:06:02,620 --> 01:06:05,150 So there's definitely friction on the table. 910 01:06:05,150 --> 01:06:07,930 But until it hits, it's doing some things. 911 01:06:07,930 --> 01:06:10,020 And so, there's two problems that you 912 01:06:10,020 --> 01:06:12,170 could set up and try to do. 913 01:06:12,170 --> 01:06:19,070 One is the problem with no friction-- frictionless table. 914 01:06:19,070 --> 01:06:22,940 And then, you could allow friction. 915 01:06:22,940 --> 01:06:24,990 Tricky thing about allowing friction 916 01:06:24,990 --> 01:06:28,130 is-- you think there's a normal force. 917 01:06:28,130 --> 01:06:30,920 So let's say our usual friction model is-- the friction force 918 01:06:30,920 --> 01:06:33,940 is mu times the normal force. 919 01:06:33,940 --> 01:06:38,687 Does the normal force change with time in this problem? 920 01:06:43,360 --> 01:06:46,000 So, standing up, what's the normal force? 921 01:06:48,640 --> 01:06:51,060 Mg, right? 922 01:06:51,060 --> 01:06:54,010 And the sum of the forces in the vertical direction 923 01:06:54,010 --> 01:06:58,700 is mass times acceleration in the vertical direction. 924 01:06:58,700 --> 01:07:00,870 And what are the-- if you draw a free body 925 01:07:00,870 --> 01:07:03,495 diagram of this problem-- so let's draw it now. 926 01:07:03,495 --> 01:07:08,070 Here's my-- if there's no friction force, 927 01:07:08,070 --> 01:07:09,445 I still-- without friction force, 928 01:07:09,445 --> 01:07:12,260 I still have a normal force, and I still 929 01:07:12,260 --> 01:07:15,350 have Mg pulling down here. 930 01:07:15,350 --> 01:07:19,180 So the sum of the forces in the y 931 01:07:19,180 --> 01:07:26,160 certainly have a minus Mg plus N equals 932 01:07:26,160 --> 01:07:31,440 the mass times the acceleration in the y direction. 933 01:07:31,440 --> 01:07:44,020 And if you solve for N, do you-- then the issue-- 934 01:07:44,020 --> 01:07:47,110 my question is, does this remain constant? 935 01:07:47,110 --> 01:07:51,930 Depends on whether or not ay remains constant, right? 936 01:07:51,930 --> 01:07:55,310 Do you think the acceleration in the y direction of this thing 937 01:07:55,310 --> 01:08:00,670 will change as it goes more and more horizontal? 938 01:08:00,670 --> 01:08:02,970 There's some nods, up and down, and left and rights. 939 01:08:02,970 --> 01:08:04,178 I think it's going to change. 940 01:08:04,178 --> 01:08:07,840 For sure, you can't assume that it won't change. 941 01:08:07,840 --> 01:08:09,530 So you have to assume it'll change. 942 01:08:09,530 --> 01:08:11,820 And that means N-- this normal force becomes 943 01:08:11,820 --> 01:08:15,400 a function of time, which makes certain ways of doing 944 01:08:15,400 --> 01:08:18,410 this problem a little harder. 945 01:08:18,410 --> 01:08:22,080 So how many ways? 946 01:08:22,080 --> 01:08:38,310 Let's say the friction-- let's do a, no friction, and b, 947 01:08:38,310 --> 01:08:39,729 with friction. 948 01:08:39,729 --> 01:08:44,630 No friction-- first of all, is it a planar motion problem? 949 01:08:44,630 --> 01:08:47,029 Yeah, we can do that. 950 01:08:47,029 --> 01:08:48,729 So, at most three degrees of freedom. 951 01:08:48,729 --> 01:08:51,939 How many does this one have? 952 01:08:51,939 --> 01:08:53,229 Work this one out. 953 01:08:53,229 --> 01:08:55,970 Figure out how many degrees of freedom this problem has. 954 01:08:55,970 --> 01:08:58,712 How many separate equations do you need to come up with? 955 01:09:42,380 --> 01:09:44,600 So, you decide what your-- let's say 956 01:09:44,600 --> 01:09:46,120 we're going to use Lagrange. 957 01:09:46,120 --> 01:09:48,200 What would your generalized coordinates 958 01:09:48,200 --> 01:09:49,425 be to do this problem? 959 01:10:21,140 --> 01:10:24,930 So, problem a-- no friction. 960 01:10:24,930 --> 01:10:26,914 How many generalized coordinates? 961 01:10:26,914 --> 01:10:28,080 How many degrees of freedom? 962 01:10:28,080 --> 01:10:29,955 How many generalized coordinates do you need? 963 01:10:33,390 --> 01:10:34,275 I heard a three. 964 01:10:36,800 --> 01:10:38,388 I hear one. 965 01:10:38,388 --> 01:10:39,730 Somebody give me two. 966 01:10:39,730 --> 01:10:41,340 I got a two. 967 01:10:41,340 --> 01:10:43,250 All right. 968 01:10:43,250 --> 01:10:45,560 Obviously a good question. 969 01:10:45,560 --> 01:10:48,040 So, at most there can be three, because we've 970 01:10:48,040 --> 01:10:50,090 agreed that it's planar motion. 971 01:10:50,090 --> 01:10:51,830 Does this problem have any constraints? 972 01:10:55,730 --> 01:10:56,230 Where? 973 01:10:59,513 --> 01:11:03,130 AUDIENCE: The bottom of the line can't move the y. 974 01:11:03,130 --> 01:11:06,080 PROFESSOR: So it can't move in the y direction. 975 01:11:06,080 --> 01:11:10,860 So if any part of a body can't move in-- translate, 976 01:11:10,860 --> 01:11:13,840 then there's no translation in that direction, 977 01:11:13,840 --> 01:11:17,660 pure translation of the body in that direction. 978 01:11:17,660 --> 01:11:21,290 So there's a constraint in the y, 979 01:11:21,290 --> 01:11:23,550 if we draw a coordinate system here. 980 01:11:28,250 --> 01:11:29,440 Constraint in the y. 981 01:11:29,440 --> 01:11:30,040 True. 982 01:11:30,040 --> 01:11:32,260 So we're down to two. 983 01:11:32,260 --> 01:11:34,877 Are there any other constraints? 984 01:11:34,877 --> 01:11:38,162 AUDIENCE: I would say that, yes, because the upper line-- 985 01:11:38,162 --> 01:11:40,370 PROFESSOR: No, no, it's not leaning against the wall. 986 01:11:40,370 --> 01:11:42,585 This is just a skip. 987 01:11:42,585 --> 01:11:43,450 This is the problem. 988 01:11:50,974 --> 01:11:52,890 So it definitely can't move through the table. 989 01:11:52,890 --> 01:11:54,500 So it's constrained in y. 990 01:11:54,500 --> 01:11:57,215 We've got that so now that leaves us, at most, two. 991 01:12:00,737 --> 01:12:02,070 Are there any other constraints? 992 01:12:07,110 --> 01:12:09,500 How many say no? 993 01:12:09,500 --> 01:12:11,169 How many say yes? 994 01:12:11,169 --> 01:12:13,210 If you say yes, you've got to tell me what it is, 995 01:12:13,210 --> 01:12:14,740 but I don't see any others. 996 01:12:14,740 --> 01:12:15,910 So we're left with two. 997 01:12:15,910 --> 01:12:19,370 We need two coordinates. 998 01:12:19,370 --> 01:12:20,500 What might we pick here? 999 01:12:27,030 --> 01:12:28,500 What would you pick? 1000 01:12:28,500 --> 01:12:30,620 You're now confronted with this problem. 1001 01:12:30,620 --> 01:12:32,036 Do you have rotational-- if you're 1002 01:12:32,036 --> 01:12:34,660 going to use Lagrange do you have rotational kinetic energy 1003 01:12:34,660 --> 01:12:36,500 in this problem? 1004 01:12:36,500 --> 01:12:39,509 You're probably going to need an angle. 1005 01:12:39,509 --> 01:12:40,550 What else would you need? 1006 01:12:40,550 --> 01:12:44,454 So we're going to need an angle for sure. 1007 01:12:44,454 --> 01:12:44,995 Say it again? 1008 01:12:44,995 --> 01:12:47,120 AUDIENCE: The height of the center of mass. 1009 01:12:47,120 --> 01:12:48,150 PROFESSOR: You're going to need the height 1010 01:12:48,150 --> 01:12:49,165 of the center of mass. 1011 01:12:49,165 --> 01:12:51,540 Yeah, you're going to need a potential energy expression, 1012 01:12:51,540 --> 01:12:55,590 but we've decided that y is constrained. 1013 01:12:55,590 --> 01:12:58,410 So you can't have a theta and a y. 1014 01:12:58,410 --> 01:12:59,642 Let's make this our theta. 1015 01:13:03,370 --> 01:13:06,470 What else is there? 1016 01:13:06,470 --> 01:13:06,970 AUDIENCE: x. 1017 01:13:06,970 --> 01:13:07,750 PROFESSOR: x. 1018 01:13:07,750 --> 01:13:09,480 You're going to need an x. 1019 01:13:09,480 --> 01:13:11,320 So our generalized coordinates are 1020 01:13:11,320 --> 01:13:16,740 going to be an x and a theta. 1021 01:13:16,740 --> 01:13:19,370 But you do need to be able to write down a potential energy 1022 01:13:19,370 --> 01:13:22,220 expression, and it involves motion in the y. 1023 01:13:22,220 --> 01:13:23,220 So what do you do? 1024 01:13:26,771 --> 01:13:27,270 Right. 1025 01:13:27,270 --> 01:13:29,670 So what's the height of this thing 1026 01:13:29,670 --> 01:13:37,755 is-- above the ground-- is sum L over 2 cosine theta equals y. 1027 01:13:37,755 --> 01:13:38,880 Something like that, right? 1028 01:13:38,880 --> 01:13:41,260 And I might have a-- depending on whether you make-- 1029 01:13:41,260 --> 01:13:42,630 I haven't thought this through. 1030 01:13:42,630 --> 01:13:45,180 Whether or not it's xy this way or xy that way, 1031 01:13:45,180 --> 01:13:47,490 theta might be plus or minus. 1032 01:13:47,490 --> 01:13:50,760 This could be-- there might be a sine in there. 1033 01:13:50,760 --> 01:13:56,690 But basically, this is-- theta and y are constrained. 1034 01:13:56,690 --> 01:13:58,554 If you know one, you know the other. 1035 01:13:58,554 --> 01:14:00,720 So you do your potential energy expression this way. 1036 01:14:05,680 --> 01:14:07,130 Now we've gotten that far. 1037 01:14:07,130 --> 01:14:09,650 We know we need two coordinates. 1038 01:14:09,650 --> 01:14:12,200 We're going to use x and theta. 1039 01:14:12,200 --> 01:14:14,750 But now you have to decide what method to use. 1040 01:14:14,750 --> 01:14:18,380 Direct method using one of these, or Lagrange? 1041 01:14:18,380 --> 01:14:20,750 What would you do? 1042 01:14:20,750 --> 01:14:21,833 How would you go about it? 1043 01:14:21,833 --> 01:14:22,340 It's a quiz. 1044 01:14:22,340 --> 01:14:26,580 You got 20 minutes to finish, and you 1045 01:14:26,580 --> 01:14:30,153 want to do this in the safest, quickest possible way. 1046 01:14:34,310 --> 01:14:36,270 What would you try? 1047 01:14:36,270 --> 01:14:39,640 And this is the realistic situation, right? 1048 01:14:39,640 --> 01:14:41,680 Next Tuesday. 1049 01:14:41,680 --> 01:14:44,251 What do you trust yourself to do the most? 1050 01:14:44,251 --> 01:14:46,542 It's probably what you ought to do in a quiz situation. 1051 01:14:51,800 --> 01:14:53,800 Anne, how many ways can you think of doing this? 1052 01:14:53,800 --> 01:14:55,657 How many ways could you do this, Rob? 1053 01:14:55,657 --> 01:14:56,240 AUDIENCE: Two. 1054 01:14:56,240 --> 01:14:57,010 PROFESSOR: Two. 1055 01:14:57,010 --> 01:14:58,120 Any way. 1056 01:14:58,120 --> 01:15:00,070 Direct, indirect, what would you use? 1057 01:15:00,070 --> 01:15:02,210 Where would you choose to-- you're 1058 01:15:02,210 --> 01:15:04,570 going to need a torque equation to do this problem, 1059 01:15:04,570 --> 01:15:05,484 if you use direct. 1060 01:15:05,484 --> 01:15:07,150 Where would you take your torques about? 1061 01:15:12,790 --> 01:15:13,784 Think about that. 1062 01:15:13,784 --> 01:15:15,950 For this problem, if you're doing the direct method, 1063 01:15:15,950 --> 01:15:17,390 where are you going to compute the torques? 1064 01:15:17,390 --> 01:15:18,170 About what point? 1065 01:15:23,200 --> 01:15:24,700 So you can do it about that-- you 1066 01:15:24,700 --> 01:15:30,630 could call this point A here. 1067 01:15:30,630 --> 01:15:32,950 And what equation would you-- now 1068 01:15:32,950 --> 01:15:34,790 you've got to use that equation that's got 1069 01:15:34,790 --> 01:15:37,250 the problematic terms in it, right? 1070 01:15:37,250 --> 01:15:38,380 But it's all right. 1071 01:15:38,380 --> 01:15:40,060 You can punch those through and do it. 1072 01:15:40,060 --> 01:15:41,490 That will work. 1073 01:15:41,490 --> 01:15:43,692 And if you use Lagrange? 1074 01:15:43,692 --> 01:15:45,150 You've got to be able to figure out 1075 01:15:45,150 --> 01:15:47,630 the potential and kinetic energy, and so forth. 1076 01:15:56,710 --> 01:16:01,109 So could all of you do this by Lagrange? 1077 01:16:01,109 --> 01:16:02,650 This is a good one to go practice on. 1078 01:16:02,650 --> 01:16:03,661 It's not that hard. 1079 01:16:03,661 --> 01:16:04,160 Do it. 1080 01:16:04,160 --> 01:16:07,860 It's a good practice problem. 1081 01:16:07,860 --> 01:16:11,100 Now add friction. 1082 01:16:11,100 --> 01:16:12,070 So, the b problem. 1083 01:16:12,070 --> 01:16:13,105 It now has friction. 1084 01:16:17,900 --> 01:16:26,334 Can you-- I don't think-- is Lagrange a good choice 1085 01:16:26,334 --> 01:16:27,292 if it now has friction? 1086 01:16:33,130 --> 01:16:37,620 I think it's got a problem, and I'd be careful using Lagrange 1087 01:16:37,620 --> 01:16:39,540 if it had friction. 1088 01:16:39,540 --> 01:16:42,550 Not that-- Lagrange is perfectly-- 1089 01:16:42,550 --> 01:16:43,860 it is certainly valid. 1090 01:16:43,860 --> 01:16:46,430 It's just hard. 1091 01:16:46,430 --> 01:16:47,255 Why is it hard? 1092 01:16:53,040 --> 01:16:55,870 Well, yeah, it might be hard to find. 1093 01:16:55,870 --> 01:16:57,610 What do you need to know to-- so, 1094 01:16:57,610 --> 01:17:00,850 are there any external non-conservative forces 1095 01:17:00,850 --> 01:17:02,550 in the problem with friction? 1096 01:17:02,550 --> 01:17:03,480 AUDIENCE: Friction. 1097 01:17:03,480 --> 01:17:06,010 PROFESSOR: Friction. 1098 01:17:06,010 --> 01:17:09,169 You're going to have to figure out the friction force. 1099 01:17:09,169 --> 01:17:10,710 And to figure out the friction force, 1100 01:17:10,710 --> 01:17:14,900 you're going to have to apply direct method. 1101 01:17:14,900 --> 01:17:16,410 No other way. 1102 01:17:16,410 --> 01:17:19,050 You are going to have to apply some direct method to do 1103 01:17:19,050 --> 01:17:22,530 this problem, no matter what. 1104 01:17:22,530 --> 01:17:26,750 So you can't just say-- Lagrange is not 1105 01:17:26,750 --> 01:17:29,060 going to bail you out and not have 1106 01:17:29,060 --> 01:17:32,400 to solve some of the F equals ma, 1107 01:17:32,400 --> 01:17:33,920 and torque, and those sort of things 1108 01:17:33,920 --> 01:17:36,820 to figure out what friction is. 1109 01:17:36,820 --> 01:17:41,620 So there are problems that Lagrange-- isn't all as simple 1110 01:17:41,620 --> 01:17:44,300 as we sometimes make it out to be. 1111 01:17:44,300 --> 01:17:46,050 All right, I've run a couple minutes over. 1112 01:17:46,050 --> 01:17:46,570 Thank you. 1113 01:17:46,570 --> 01:17:48,820 We'll see you in class on Tuesday. 1114 01:17:48,820 --> 01:17:51,030 We'll do some more review.