1 00:00:00,070 --> 00:00:02,430 The following content is provided under a Creative 2 00:00:02,430 --> 00:00:03,810 Commons license. 3 00:00:03,810 --> 00:00:06,050 Your support will help MIT OpenCourseWare 4 00:00:06,050 --> 00:00:10,140 continue to offer high quality educational resources for free. 5 00:00:10,140 --> 00:00:12,690 To make a donation or to view additional materials 6 00:00:12,690 --> 00:00:16,590 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,590 --> 00:00:17,260 at ocw.mit.edu. 8 00:00:22,861 --> 00:00:25,560 PROFESSOR: OK, what I thought I would do 9 00:00:25,560 --> 00:00:32,549 is I'm going to go quickly through the problems that 10 00:00:32,549 --> 00:00:33,745 were assigned for practice. 11 00:00:33,745 --> 00:00:35,513 There's seven or eight of them. 12 00:00:35,513 --> 00:00:36,846 I'm not going to do all of them. 13 00:00:36,846 --> 00:00:38,304 I'm going to go through and kind of 14 00:00:38,304 --> 00:00:43,570 talk about what some of the key issues are with each problem 15 00:00:43,570 --> 00:00:51,000 and maybe make some points about identifying 16 00:00:51,000 --> 00:00:55,410 how you do problems, not necessarily specifically focus 17 00:00:55,410 --> 00:00:57,620 on the exact question that was asked here. 18 00:00:57,620 --> 00:01:01,670 So here was this first-- this was a quiz from last year. 19 00:01:01,670 --> 00:01:07,290 And you were asked to find an equation of motion 20 00:01:07,290 --> 00:01:09,530 for this thing. 21 00:01:09,530 --> 00:01:12,190 And it can be a bit of a puzzling problem, 22 00:01:12,190 --> 00:01:15,570 so how many degrees of freedom, what are the constraints, what 23 00:01:15,570 --> 00:01:16,920 equation should you use. 24 00:01:19,510 --> 00:01:21,660 So first of all, I look at a problem like this. 25 00:01:21,660 --> 00:01:24,070 Is it planar motion or not? 26 00:01:24,070 --> 00:01:26,970 OK, so maximum three degrees of freedom. 27 00:01:26,970 --> 00:01:29,463 What are the constraints, though? 28 00:01:34,411 --> 00:01:37,700 STUDENT: So G is [INAUDIBLE]. 29 00:01:37,700 --> 00:01:39,970 PROFESSOR: So G is R/2 from the center. 30 00:01:39,970 --> 00:01:43,380 So that's true. 31 00:01:43,380 --> 00:01:49,520 And so what you're really saying is it's fixed in radial motion, 32 00:01:49,520 --> 00:01:50,290 right? 33 00:01:50,290 --> 00:01:51,730 And you're right. 34 00:01:51,730 --> 00:01:52,820 So that's one constraint. 35 00:01:52,820 --> 00:01:54,780 What's another constraint in this problem? 36 00:01:57,760 --> 00:02:01,050 How many degrees of freedom do you expect to end up with? 37 00:02:01,050 --> 00:02:02,920 How many equations do you need? 38 00:02:02,920 --> 00:02:04,370 How many coordinates? 39 00:02:04,370 --> 00:02:05,368 One, right? 40 00:02:05,368 --> 00:02:07,243 So how do we articulate the other constraint? 41 00:02:10,650 --> 00:02:12,280 So it can't move radially. 42 00:02:12,280 --> 00:02:13,310 We figured that out. 43 00:02:26,780 --> 00:02:29,370 So this is actually kind of an important point 44 00:02:29,370 --> 00:02:30,220 in this problem. 45 00:02:30,220 --> 00:02:32,480 Because if you can get this down, 46 00:02:32,480 --> 00:02:34,515 then the problem becomes very simple. 47 00:02:37,930 --> 00:02:39,090 So talk to a neighbor. 48 00:02:39,090 --> 00:02:43,060 How would you describe the other constraint here? 49 00:02:43,060 --> 00:02:45,130 I'm going to hold it a second. 50 00:02:45,130 --> 00:02:47,070 STUDENT: [INAUDIBLE] 51 00:02:47,070 --> 00:02:47,820 PROFESSOR: Pardon? 52 00:02:47,820 --> 00:02:48,560 STUDENT: Gravity is just [INAUDIBLE]. 53 00:02:48,560 --> 00:02:49,670 PROFESSOR: Well, there is gravity, yes. 54 00:02:49,670 --> 00:02:51,260 But gravity is not a constraint. 55 00:02:51,260 --> 00:02:53,555 So this is a pendulum, actually, of sorts. 56 00:02:56,830 --> 00:02:58,360 So talk to a neighbor. 57 00:02:58,360 --> 00:03:00,275 How would you describe the other constraint? 58 00:03:41,870 --> 00:03:46,310 OK, how would you describe it? 59 00:03:46,310 --> 00:03:49,250 What's the other constraint here? 60 00:03:57,770 --> 00:04:00,840 So sometimes if something is pinned, 61 00:04:00,840 --> 00:04:02,895 we say, well, then it's fixed in x and y. 62 00:04:02,895 --> 00:04:04,680 And that's two constraints. 63 00:04:04,680 --> 00:04:06,290 It's only left to rotate. 64 00:04:06,290 --> 00:04:08,290 But this one doesn't seem to-- it's kind of hard 65 00:04:08,290 --> 00:04:12,140 to understand where this one is pinned. 66 00:04:12,140 --> 00:04:14,057 You know this thing is constrained in x and y. 67 00:04:14,057 --> 00:04:16,390 Because you want to use theta for the coordinate, right? 68 00:04:16,390 --> 00:04:18,070 But how do you say that it's constrained 69 00:04:18,070 --> 00:04:18,903 in that other thing? 70 00:04:18,903 --> 00:04:20,264 Yeah. 71 00:04:20,264 --> 00:04:22,650 STUDENT: [INAUDIBLE] 72 00:04:29,860 --> 00:04:31,985 PROFESSOR: OK, so he's saying that you can define x 73 00:04:31,985 --> 00:04:34,185 and y in terms of r and theta. 74 00:04:34,185 --> 00:04:36,176 And that's true. 75 00:04:36,176 --> 00:04:38,050 And I'm going to say it slightly differently. 76 00:04:38,050 --> 00:04:39,591 So the way I think about this problem 77 00:04:39,591 --> 00:04:42,040 is we've said that it's constrained 78 00:04:42,040 --> 00:04:43,440 in the radial direction. 79 00:04:43,440 --> 00:04:44,880 So that's one. 80 00:04:44,880 --> 00:04:48,040 It can move in the tangential direction, right? 81 00:04:48,040 --> 00:04:50,060 This problem is really exactly the same 82 00:04:50,060 --> 00:04:54,090 as the skateboard problem in a bowl. 83 00:04:54,090 --> 00:04:55,759 You shorten that thing a little bit, 84 00:04:55,759 --> 00:04:57,050 and it looks like a skateboard. 85 00:04:57,050 --> 00:04:59,190 And we're ignoring the inertia of the wheels. 86 00:04:59,190 --> 00:05:01,500 So it's just a stick sliding up and down. 87 00:05:01,500 --> 00:05:06,370 So what's the relationship between the tangential motion 88 00:05:06,370 --> 00:05:07,860 and theta? 89 00:05:07,860 --> 00:05:10,690 Is it a fixed relationship? 90 00:05:10,690 --> 00:05:15,370 So how far does it move in the theta? 91 00:05:15,370 --> 00:05:17,350 If you have a little motion, delta theta, 92 00:05:17,350 --> 00:05:20,290 how far does it move tangentially? 93 00:05:20,290 --> 00:05:27,400 R delta theta is the delta r tangential, 94 00:05:27,400 --> 00:05:29,210 the distance it moves. 95 00:05:29,210 --> 00:05:32,980 So this is the way to say the other constraint, 96 00:05:32,980 --> 00:05:36,610 is if you know theta, you know how far it's 97 00:05:36,610 --> 00:05:39,520 moved in the radial, in the tangential direction. 98 00:05:39,520 --> 00:05:42,180 So that's a way of saying that other constraint. 99 00:05:42,180 --> 00:05:45,080 OK, then we can say we completely describe 100 00:05:45,080 --> 00:05:49,060 this problem by one coordinate. 101 00:05:49,060 --> 00:05:51,660 And that's theta. 102 00:05:51,660 --> 00:05:54,440 And the second you can do that, this problem then 103 00:05:54,440 --> 00:05:59,270 has a center of rotation right in the center of the circle. 104 00:05:59,270 --> 00:06:03,440 And anytime you have a body which 105 00:06:03,440 --> 00:06:07,350 rotates around a central point, then 106 00:06:07,350 --> 00:06:09,960 you know that you can describe the angular 107 00:06:09,960 --> 00:06:13,300 momentum of the body as some I with respect 108 00:06:13,300 --> 00:06:16,250 to A times-- and it's planar motion. 109 00:06:16,250 --> 00:06:22,520 So then it's just omega z, as long as this is a principal, 110 00:06:22,520 --> 00:06:26,760 as long as you have the mass moments of inertia in terms 111 00:06:26,760 --> 00:06:30,410 of principal axes here. 112 00:06:30,410 --> 00:06:33,460 And it's nice when you can write it that simply. 113 00:06:33,460 --> 00:06:37,070 The second you can identify a fixed point about which 114 00:06:37,070 --> 00:06:39,510 something rotates, then the angular momentum 115 00:06:39,510 --> 00:06:41,532 simplifies to that. 116 00:06:41,532 --> 00:06:48,370 OK, and then this problem, of course 117 00:06:48,370 --> 00:06:51,580 the sum of the torques with respect to the center 118 00:06:51,580 --> 00:06:59,990 here has got to be d H with respect to A dt plus vAO 119 00:06:59,990 --> 00:07:02,490 cross PGO. 120 00:07:02,490 --> 00:07:06,960 But this is-- what's the velocity at point A? 121 00:07:09,820 --> 00:07:12,380 0, so you don't have to worry about this term. 122 00:07:12,380 --> 00:07:14,580 And it's just that. 123 00:07:14,580 --> 00:07:17,580 And you know that the sum of the external torques 124 00:07:17,580 --> 00:07:22,930 in this problem comes from gravity. 125 00:07:22,930 --> 00:07:26,202 So you put in the gravity term and compute the torques 126 00:07:26,202 --> 00:07:26,910 about the center. 127 00:07:29,640 --> 00:07:32,330 So our object really looks like that. 128 00:07:32,330 --> 00:07:38,885 And here's G. And in the problem, the way it was posed, 129 00:07:38,885 --> 00:07:42,340 the theta is drawn from this line. 130 00:07:42,340 --> 00:07:47,530 So this is theta. 131 00:07:47,530 --> 00:07:49,702 So what's the moment arm? 132 00:07:49,702 --> 00:07:55,510 It looks like whatever this distance is, which is R/2. 133 00:07:55,510 --> 00:07:58,490 It's given. 134 00:07:58,490 --> 00:08:04,910 R/2 cosine theta is the length of this side. 135 00:08:04,910 --> 00:08:11,391 So the moment that gravity makes is some Mg. 136 00:08:16,860 --> 00:08:20,980 And that's going to be equal to some Izz 137 00:08:20,980 --> 00:08:23,990 about A theta double dot. 138 00:08:23,990 --> 00:08:25,910 OK, good. 139 00:08:25,910 --> 00:08:33,370 Let's take a look at the next problem. 140 00:08:33,370 --> 00:08:34,969 And now so my intention here, I'm 141 00:08:34,969 --> 00:08:36,760 going to go kind of quickly one by another. 142 00:08:36,760 --> 00:08:38,843 And I'm just trying to hit the important concepts. 143 00:08:38,843 --> 00:08:41,789 So if you have a question about the concepts, ask. 144 00:08:41,789 --> 00:08:44,610 If I've left you wondering, I really 145 00:08:44,610 --> 00:08:47,060 want this to be kind of a conversation here. 146 00:08:47,060 --> 00:08:48,180 So that's the whole point. 147 00:08:48,180 --> 00:08:53,610 I've given you the essence of what makes these problems work. 148 00:08:53,610 --> 00:08:57,930 OK, this one, we just find a location of the center of mass. 149 00:08:57,930 --> 00:09:01,790 I think you're pretty good at that sort of thing. 150 00:09:01,790 --> 00:09:05,710 Well, remember just a couple points about center of mass. 151 00:09:05,710 --> 00:09:07,820 This one it says, find it. 152 00:09:07,820 --> 00:09:10,455 And you're given the two objects. 153 00:09:10,455 --> 00:09:12,210 The important point is that you can 154 00:09:12,210 --> 00:09:15,520 pick any coordinate at all in order 155 00:09:15,520 --> 00:09:18,430 to use it to compute the center of mass. 156 00:09:18,430 --> 00:09:22,050 So I pick S here, the M1. 157 00:09:22,050 --> 00:09:24,930 And some distance down here is where the center of mass 158 00:09:24,930 --> 00:09:26,060 is that we're looking for. 159 00:09:26,060 --> 00:09:30,200 And so this is the G we're looking for. 160 00:09:30,200 --> 00:09:38,426 So M1 plus M2 is the total mass times the position of G. 161 00:09:38,426 --> 00:09:43,050 I guess that ought to be a capital-- SG here 162 00:09:43,050 --> 00:09:48,610 must be equal to the sum of the parts times their positions, M1 163 00:09:48,610 --> 00:09:54,680 SG1 plus M2 SG2. 164 00:09:54,680 --> 00:09:56,430 And you can solve for SG. 165 00:09:56,430 --> 00:09:58,090 And that tells you where you're at. 166 00:09:58,090 --> 00:10:01,320 So you can pick any coordinate at all to calculate it. 167 00:10:01,320 --> 00:10:04,600 And once you find it, you know where it is. 168 00:10:04,600 --> 00:10:08,360 So you just solve here for SG. 169 00:10:08,360 --> 00:10:12,600 It's that divided by the total mass, obviously. 170 00:10:12,600 --> 00:10:21,370 OK, the next question was, draw principal axes. 171 00:10:21,370 --> 00:10:27,400 So does this object have some planes of symmetry? 172 00:10:27,400 --> 00:10:28,510 Tell me one. 173 00:10:28,510 --> 00:10:29,714 STUDENT: This one. 174 00:10:29,714 --> 00:10:31,630 PROFESSOR: Slice down through it this way, OK. 175 00:10:31,630 --> 00:10:34,210 And that means you have a principal axis where? 176 00:10:34,210 --> 00:10:36,000 Perpendicular to every plane of symmetry. 177 00:10:36,000 --> 00:10:39,360 So you have a principal axis coming out of the board. 178 00:10:39,360 --> 00:10:40,670 That's one. 179 00:10:40,670 --> 00:10:43,127 And give me another plane of symmetry. 180 00:10:43,127 --> 00:10:43,960 STUDENT: [INAUDIBLE] 181 00:10:43,960 --> 00:10:45,043 PROFESSOR: Yeah, this one. 182 00:10:45,043 --> 00:10:46,850 So that means you've got a plane of axis, 183 00:10:46,850 --> 00:10:48,910 principal axis going that way. 184 00:10:48,910 --> 00:10:52,680 And so then the third one has to be perpendicular to that. 185 00:10:52,680 --> 00:10:57,640 So you have one this way, this way, and this way. 186 00:11:09,369 --> 00:11:17,220 So I'd asked for-- so if this is G, then principal axes 187 00:11:17,220 --> 00:11:20,850 with respect to G, one there, one coming out of the board, 188 00:11:20,850 --> 00:11:23,250 and one going up like that. 189 00:11:28,300 --> 00:11:31,050 So what is it? 190 00:11:31,050 --> 00:11:35,840 Let's remind ourselves what it means to be a principal axis. 191 00:11:35,840 --> 00:11:41,370 So if you know the principal axes of an object, 192 00:11:41,370 --> 00:11:46,620 if you rotate, and you've chosen them to go through G, 193 00:11:46,620 --> 00:11:48,140 do they have to go through G? 194 00:11:48,140 --> 00:11:51,150 Do all principal axes have to go through G? 195 00:11:51,150 --> 00:11:52,800 No, not at all. 196 00:11:52,800 --> 00:11:56,650 The principal axes just have to give you, when you work it out, 197 00:11:56,650 --> 00:11:58,635 a diagonal inertia matrix. 198 00:12:04,730 --> 00:12:07,630 But what does it mean to have body 199 00:12:07,630 --> 00:12:09,920 coordinates that we know are principal axes? 200 00:12:09,920 --> 00:12:17,730 Well, one of the things we know, that if you rotate about G, 201 00:12:17,730 --> 00:12:21,340 rotate about a principal axis passing through G, 202 00:12:21,340 --> 00:12:23,440 if you rotate about any one of them, 203 00:12:23,440 --> 00:12:24,775 is it dynamically balanced? 204 00:12:27,880 --> 00:12:30,010 Guaranteed. 205 00:12:30,010 --> 00:12:33,760 You must rotate about the principal axis, though. 206 00:12:33,760 --> 00:12:35,550 Now, I'll ask you a second question. 207 00:12:35,550 --> 00:12:38,840 If you rotate about any axis passing through G, 208 00:12:38,840 --> 00:12:40,730 is it statically balanced? 209 00:12:40,730 --> 00:12:44,530 Right, if you rotate about any axis passing through G, 210 00:12:44,530 --> 00:12:46,960 is it dynamically balanced? 211 00:12:46,960 --> 00:12:48,140 Maybe, right? 212 00:12:48,140 --> 00:12:49,880 Not necessarily. 213 00:12:49,880 --> 00:12:53,600 It might be dynamically balanced if you pass through G, 214 00:12:53,600 --> 00:12:59,100 and it is a principal axis, right? 215 00:12:59,100 --> 00:13:07,350 OK, can you have a principal axis not going through G 216 00:13:07,350 --> 00:13:11,560 and rotate about that axis and have 217 00:13:11,560 --> 00:13:13,388 it be dynamically balanced? 218 00:13:13,388 --> 00:13:18,990 Yeah, OK, so we'll talk about an example in a second. 219 00:13:18,990 --> 00:13:25,860 OK, so I think the second question here said, 220 00:13:25,860 --> 00:13:30,530 OK, we want to rotate this about this point G2. 221 00:13:30,530 --> 00:13:35,317 Now, G2 is right in the middle at this body. 222 00:13:35,317 --> 00:13:36,775 That's where its center of mass is. 223 00:13:36,775 --> 00:13:39,150 And there's a G1 up here that's the center of mass 224 00:13:39,150 --> 00:13:40,430 of the upper body. 225 00:13:40,430 --> 00:13:42,350 And there's a composite center of mass right 226 00:13:42,350 --> 00:13:45,710 here, which we just computed what it was. 227 00:13:45,710 --> 00:13:51,740 And we want-- there's going to be an axis of rotation passing 228 00:13:51,740 --> 00:13:53,125 through G2. 229 00:13:53,125 --> 00:13:56,210 And we're rotating this body around it. 230 00:13:56,210 --> 00:13:59,475 So first question is, this axis coming through G2, 231 00:13:59,475 --> 00:14:01,333 is it a principal axis? 232 00:14:01,333 --> 00:14:02,280 Yes. 233 00:14:02,280 --> 00:14:05,255 If we rotate about that axis only, 234 00:14:05,255 --> 00:14:08,520 do you expect it to be dynamically balanced? 235 00:14:08,520 --> 00:14:09,450 Right. 236 00:14:09,450 --> 00:14:11,520 Do you expect it to be statically balanced? 237 00:14:11,520 --> 00:14:12,020 No. 238 00:14:25,820 --> 00:14:46,900 So we're spinning it around this G2, going like that around G2 239 00:14:46,900 --> 00:14:50,270 at some omega z. 240 00:14:50,270 --> 00:14:52,390 We know that this is G2. 241 00:14:52,390 --> 00:14:56,650 We know that G for the object is here. 242 00:14:56,650 --> 00:14:58,720 That's the center of mass of the whole thing. 243 00:14:58,720 --> 00:15:03,185 So there's some distance between those two, which I don't know. 244 00:15:03,185 --> 00:15:04,310 But we could figure it out. 245 00:15:04,310 --> 00:15:05,760 We'll call it e. 246 00:15:05,760 --> 00:15:10,730 So what's the force required at this pin 247 00:15:10,730 --> 00:15:12,370 that its axle is going around? 248 00:15:12,370 --> 00:15:15,890 What's the force required to allow this thing 249 00:15:15,890 --> 00:15:20,070 to spin around, to hold it, essentially, the force required 250 00:15:20,070 --> 00:15:22,620 to hold it in place? 251 00:15:22,620 --> 00:15:26,570 And you ought to intuitively have an idea of what it is. 252 00:15:26,570 --> 00:15:28,250 Why is there a force? 253 00:15:28,250 --> 00:15:30,260 That thing will spin around this central axle. 254 00:15:30,260 --> 00:15:32,210 And there's a force required to keep 255 00:15:32,210 --> 00:15:33,720 that thing from moving away. 256 00:15:33,720 --> 00:15:36,142 What is it? 257 00:15:36,142 --> 00:15:39,130 STUDENT: Centrifugal force. 258 00:15:39,130 --> 00:15:41,650 PROFESSOR: I hear centrifugal. 259 00:15:41,650 --> 00:15:45,130 So you're accelerating the center of mass of that thing. 260 00:15:45,130 --> 00:15:46,960 You're making it go in a circle. 261 00:15:46,960 --> 00:15:48,800 Mass times acceleration is a force. 262 00:15:48,800 --> 00:15:51,269 And that's the tension in the string when you're 263 00:15:51,269 --> 00:15:52,310 swinging the ball around. 264 00:15:52,310 --> 00:15:55,550 This is no different from swinging the ball on a string. 265 00:15:55,550 --> 00:15:57,100 And you're essentially calculating 266 00:15:57,100 --> 00:16:00,970 the tension required to keep that thing from flying off. 267 00:16:00,970 --> 00:16:02,580 And we could probably just guess. 268 00:16:02,580 --> 00:16:10,990 We can almost guess what it is, minus the total mass times 269 00:16:10,990 --> 00:16:14,174 the acceleration, which is inward. 270 00:16:14,174 --> 00:16:15,590 That's where the minus comes from. 271 00:16:15,590 --> 00:16:16,320 And what is it? 272 00:16:19,122 --> 00:16:20,530 STUDENT: [INAUDIBLE] 273 00:16:20,530 --> 00:16:21,648 PROFESSOR: Louder. 274 00:16:21,648 --> 00:16:23,239 STUDENT: [INAUDIBLE] 275 00:16:23,239 --> 00:16:24,280 PROFESSOR: Omega squared. 276 00:16:24,280 --> 00:16:25,680 I hear an omega z squared. 277 00:16:25,680 --> 00:16:27,540 We need something else. 278 00:16:27,540 --> 00:16:28,440 Pardon? 279 00:16:28,440 --> 00:16:29,872 STUDENT: [INAUDIBLE] 280 00:16:29,872 --> 00:16:30,830 PROFESSOR: Yeah, the r. 281 00:16:30,830 --> 00:16:32,577 And the r is this little e here. 282 00:16:32,577 --> 00:16:33,993 It's called the eccentricity, when 283 00:16:33,993 --> 00:16:36,230 you have something unbalanced. e omega squared, 284 00:16:36,230 --> 00:16:39,720 and it's in the r hat direction. 285 00:16:39,720 --> 00:16:40,910 We'll call it inwards. 286 00:16:40,910 --> 00:16:42,850 And the inwards is the minus. 287 00:16:42,850 --> 00:16:44,080 So that's the acceleration. 288 00:16:44,080 --> 00:16:46,430 Mass times acceleration is the force. 289 00:16:46,430 --> 00:16:53,950 Also, you could compute this by computing d by dt of P of G 290 00:16:53,950 --> 00:16:59,350 with respect to O. And that's then 291 00:16:59,350 --> 00:17:07,270 M1 plus M2 times v of G. You have to take the d by dt of it. 292 00:17:07,270 --> 00:17:10,380 And that gives you that, which is the acceleration. 293 00:17:10,380 --> 00:17:16,750 So the forces you get directly from computing dP/dt, 294 00:17:16,750 --> 00:17:18,730 if you need to remember a way to do that. 295 00:17:18,730 --> 00:17:26,950 All right, so conceptually, if I don't know what e is, 296 00:17:26,950 --> 00:17:29,080 then I don't want to have to figure it out. 297 00:17:33,380 --> 00:17:36,090 So I'm going to describe to you a way in which you could 298 00:17:36,090 --> 00:17:37,870 get directly at this answer, which 299 00:17:37,870 --> 00:17:41,320 reminds us something about angular 300 00:17:41,320 --> 00:17:43,530 momentum and these bodies. 301 00:17:43,530 --> 00:17:48,830 This object is made up of two objects. 302 00:17:51,480 --> 00:17:55,970 And we're spinning it about the center of one of them. 303 00:17:55,970 --> 00:17:58,890 So let's take them one at a time. 304 00:17:58,890 --> 00:18:09,350 How much force is required to keep this spinning 305 00:18:09,350 --> 00:18:14,170 about that axle that goes right through its center of mass? 306 00:18:14,170 --> 00:18:15,670 None, right? 307 00:18:15,670 --> 00:18:18,270 Because there's no r omega squared. 308 00:18:18,270 --> 00:18:19,520 There's no acceleration. 309 00:18:19,520 --> 00:18:21,980 You're not making the center of mass of this object 310 00:18:21,980 --> 00:18:23,010 go in a circle. 311 00:18:23,010 --> 00:18:28,340 So this piece doesn't enter into the solution. 312 00:18:28,340 --> 00:18:32,320 The actual force, the mass, this mass here, 313 00:18:32,320 --> 00:18:35,080 is M2 in the problem. 314 00:18:35,080 --> 00:18:39,350 M2 doesn't actually have anything to do with the answer. 315 00:18:39,350 --> 00:18:40,820 Because it's perfectly statically 316 00:18:40,820 --> 00:18:42,630 balanced around that point. 317 00:18:42,630 --> 00:18:45,500 All of this force making the thing go in a circle 318 00:18:45,500 --> 00:18:49,730 is making this piece go in a circle. 319 00:18:49,730 --> 00:18:54,380 So in fact, this force up here is minus M1. 320 00:18:56,910 --> 00:19:02,790 And now we need the distance from the axis of rotation 321 00:19:02,790 --> 00:19:06,240 to its center of mass. 322 00:19:06,240 --> 00:19:07,840 Well, that's some r. 323 00:19:07,840 --> 00:19:09,390 We need to know what that r is. 324 00:19:09,390 --> 00:19:13,970 And I think that's basically-- this was L/2. 325 00:19:13,970 --> 00:19:17,250 And this was b, so plus b/2. 326 00:19:17,250 --> 00:19:19,860 So the distance from this rotation point 327 00:19:19,860 --> 00:19:34,280 to that center of mass is L/2 plus b/2, omega z squared. 328 00:19:34,280 --> 00:19:37,430 Mass-- this is acceleration, and it's inwards. 329 00:19:37,430 --> 00:19:38,740 And that's the total force. 330 00:19:41,270 --> 00:19:43,020 All right, let's look at the next problem. 331 00:19:47,580 --> 00:19:49,320 Vicente, can you pull up the next one? 332 00:20:06,380 --> 00:20:08,530 OK, this problem-- this problem and the next 333 00:20:08,530 --> 00:20:09,470 are rather similar. 334 00:20:09,470 --> 00:20:11,859 This is the elevator problem with the pendulum 335 00:20:11,859 --> 00:20:12,525 in the elevator. 336 00:20:16,604 --> 00:20:18,580 There's a couple subtleties here. 337 00:20:18,580 --> 00:20:21,580 The elevator, you're told that it's moving upwards 338 00:20:21,580 --> 00:20:23,740 at some particular rate of acceleration, 339 00:20:23,740 --> 00:20:29,220 some y double dot. 340 00:20:29,220 --> 00:20:30,140 That's given. 341 00:20:30,140 --> 00:20:32,963 So how many degrees of freedom does this problem have? 342 00:20:36,810 --> 00:20:38,230 OK, two or one? 343 00:20:38,230 --> 00:20:39,530 I hear one. 344 00:20:39,530 --> 00:20:42,230 I see a two. 345 00:20:42,230 --> 00:20:43,960 How many coordinates is it going to take 346 00:20:43,960 --> 00:20:46,485 to completely describe the motion, 347 00:20:46,485 --> 00:20:48,390 or how many equations of motion do 348 00:20:48,390 --> 00:20:49,681 you think you're going to need? 349 00:20:52,280 --> 00:20:53,400 I see ones and twos. 350 00:20:53,400 --> 00:20:55,950 So there's a little not-- two, OK. 351 00:20:59,340 --> 00:21:04,290 If we know the acceleration of the thing-- we start 352 00:21:04,290 --> 00:21:06,110 with some initial conditions, at time 0, 353 00:21:06,110 --> 00:21:08,140 it's sitting on the ground, and y equals 0, 354 00:21:08,140 --> 00:21:11,770 and now it takes off from the first floor-- we 355 00:21:11,770 --> 00:21:15,300 know for all time its position. 356 00:21:15,300 --> 00:21:17,570 So that's actually a given. 357 00:21:17,570 --> 00:21:19,950 It's a specified thing. 358 00:21:19,950 --> 00:21:22,610 And we don't have to write a separate equation of motion 359 00:21:22,610 --> 00:21:26,540 for that that we then will have to solve. 360 00:21:26,540 --> 00:21:27,940 So it's actually given. 361 00:21:27,940 --> 00:21:30,580 So the actually only dynamic equation 362 00:21:30,580 --> 00:21:32,730 of motion we have to write is about the pendulum. 363 00:21:32,730 --> 00:21:34,540 Now, will it involve y double dot? 364 00:21:34,540 --> 00:21:35,610 Absolutely. 365 00:21:35,610 --> 00:21:36,657 But it's a given number. 366 00:21:36,657 --> 00:21:37,990 It's just a number you're given. 367 00:21:37,990 --> 00:21:40,239 You don't have to write a separate equation of motion. 368 00:21:40,239 --> 00:21:42,571 So actually you've got just one degree of freedom. 369 00:21:42,571 --> 00:21:43,071 Yeah. 370 00:21:43,071 --> 00:21:45,230 STUDENT: [INAUDIBLE] 371 00:21:45,230 --> 00:21:51,680 PROFESSOR: OK, so put up the next problem-- that one. 372 00:21:51,680 --> 00:21:55,090 So this problem has how many degrees of freedom? 373 00:21:55,090 --> 00:21:56,540 This is definitely two. 374 00:21:56,540 --> 00:21:57,850 Because you don't know. 375 00:21:57,850 --> 00:22:01,570 The motion in the up and down of the mass 376 00:22:01,570 --> 00:22:04,104 and the slider with a spring on it is unknown. 377 00:22:04,104 --> 00:22:06,520 You're going to have to write an equation of motion, which 378 00:22:06,520 --> 00:22:11,290 would have to be solved to get it, so a distinction, 379 00:22:11,290 --> 00:22:12,790 a subtlety that can trip you up. 380 00:22:12,790 --> 00:22:15,390 When the motion is actually specified, 381 00:22:15,390 --> 00:22:17,280 and you know what it is for all time, 382 00:22:17,280 --> 00:22:20,890 you don't need a separate equation for that. 383 00:22:20,890 --> 00:22:23,060 So this problem requires two. 384 00:22:23,060 --> 00:22:25,150 The behavior of the pendulum part 385 00:22:25,150 --> 00:22:29,390 is essentially the same in both. 386 00:22:29,390 --> 00:22:31,840 When you write the equation-- well, let's talk about it. 387 00:22:31,840 --> 00:22:34,650 How will we go about solving this problem? 388 00:22:34,650 --> 00:22:36,630 We need two equations, two coordinates. 389 00:22:36,630 --> 00:22:39,293 What would you pick for your coordinates? 390 00:22:39,293 --> 00:22:40,126 STUDENT: [INAUDIBLE] 391 00:22:43,920 --> 00:22:45,630 PROFESSOR: Theta, and then a coordinate. 392 00:22:45,630 --> 00:22:47,950 This one I guess is called y, so y 393 00:22:47,950 --> 00:22:50,520 for the square block sliding up and down, 394 00:22:50,520 --> 00:22:53,510 and a theta for the pendulum, two equations of motion. 395 00:22:53,510 --> 00:22:56,670 And to get the y equation of motion, 396 00:22:56,670 --> 00:22:58,600 you would use-- we're not using Lagrange here. 397 00:22:58,600 --> 00:23:00,725 We're going to do a little review of direct method. 398 00:23:00,725 --> 00:23:05,670 How would you write the equation of motion using-- up there 399 00:23:05,670 --> 00:23:08,380 I kind of wrote the key to everything 400 00:23:08,380 --> 00:23:11,740 that we do with the direct method, some of the forces, 401 00:23:11,740 --> 00:23:14,280 some of the torques. 402 00:23:14,280 --> 00:23:19,730 Every rigid body has at max how many degrees of freedom? 403 00:23:19,730 --> 00:23:21,530 Six for rigid body. 404 00:23:21,530 --> 00:23:23,910 In planar motion, this is reduced to three. 405 00:23:23,910 --> 00:23:27,640 But every rigid body in 3D space has 406 00:23:27,640 --> 00:23:32,180 six degrees of freedom-- three positions, three rotations. 407 00:23:32,180 --> 00:23:35,070 So you write some of the forces on the body. 408 00:23:35,070 --> 00:23:37,590 That is equal to the mass times acceleration. 409 00:23:37,590 --> 00:23:40,810 That gives you three equations, one in each vector, component 410 00:23:40,810 --> 00:23:41,710 direction. 411 00:23:41,710 --> 00:23:43,990 You write the sum of the torques. 412 00:23:43,990 --> 00:23:45,490 That's also a vector equation. 413 00:23:45,490 --> 00:23:47,780 You can get as many as three equations out of it 414 00:23:47,780 --> 00:23:49,060 if you need it. 415 00:23:49,060 --> 00:23:50,960 Then we keep reducing down how many 416 00:23:50,960 --> 00:23:54,390 equations we need to get by figuring out the constraints 417 00:23:54,390 --> 00:23:58,170 until we get down to the number of actual degrees of freedom. 418 00:23:58,170 --> 00:24:01,440 And that's the remaining equations, the number 419 00:24:01,440 --> 00:24:03,150 of equations you actually need. 420 00:24:03,150 --> 00:24:06,060 So this one requires two. 421 00:24:06,060 --> 00:24:09,000 And if you're going to write an equation of motion 422 00:24:09,000 --> 00:24:14,910 about the main mass that slides up and down, what law would 423 00:24:14,910 --> 00:24:15,480 you use? 424 00:24:15,480 --> 00:24:16,440 STUDENT: [INAUDIBLE] 425 00:24:16,440 --> 00:24:18,010 PROFESSOR: Yeah, sum of the forces 426 00:24:18,010 --> 00:24:20,219 equal to the mass times acceleration. 427 00:24:20,219 --> 00:24:21,760 Then we need to get another question. 428 00:24:21,760 --> 00:24:23,936 What law would you use for the second one? 429 00:24:23,936 --> 00:24:24,810 STUDENT: Torque. 430 00:24:24,810 --> 00:24:26,660 PROFESSOR: Torque about where? 431 00:24:26,660 --> 00:24:27,500 STUDENT: A. 432 00:24:27,500 --> 00:24:30,310 PROFESSOR: Torque about A. So let's look at that. 433 00:24:30,310 --> 00:24:32,790 So that's the second equation up there. 434 00:24:32,790 --> 00:24:36,530 Torque about A is the time derivative 435 00:24:36,530 --> 00:24:38,710 of the angular momentum with respect to A plus 436 00:24:38,710 --> 00:24:43,750 that vA cross momentum term, which is 437 00:24:43,750 --> 00:24:46,780 a little annoying to work out. 438 00:24:46,780 --> 00:24:50,360 So the simpler version is the third equation. 439 00:24:50,360 --> 00:24:53,120 It's a pretty quick derivation to go from that second equation 440 00:24:53,120 --> 00:24:54,290 to the third. 441 00:24:54,290 --> 00:24:57,970 And it's much less work, generally, 442 00:24:57,970 --> 00:25:01,270 to get that mass times acceleration 443 00:25:01,270 --> 00:25:04,160 term, a lot less work actually. 444 00:25:04,160 --> 00:25:07,340 And if I've got time, I'll go back 445 00:25:07,340 --> 00:25:09,440 and we'll work out a problem like this. 446 00:25:09,440 --> 00:25:11,160 And we'll do that part. 447 00:25:11,160 --> 00:25:14,110 But let's go on to the other questions. 448 00:25:14,110 --> 00:25:19,220 OK, this problem-- two rigid bodies. 449 00:25:19,220 --> 00:25:21,810 Is it planar motion? 450 00:25:21,810 --> 00:25:22,492 STUDENT: Yes. 451 00:25:22,492 --> 00:25:23,200 PROFESSOR: Right? 452 00:25:23,200 --> 00:25:27,180 How many possible degrees of freedom for each rigid body? 453 00:25:27,180 --> 00:25:31,970 Times 2, 6, now how many constraints? 454 00:25:31,970 --> 00:25:33,880 So let's take them one rigid body at a time. 455 00:25:33,880 --> 00:25:36,551 It's a big roller. 456 00:25:36,551 --> 00:25:37,050 Pardon? 457 00:25:37,050 --> 00:25:37,730 STUDENT: No y. 458 00:25:37,730 --> 00:25:38,850 PROFESSOR: No y, she says. 459 00:25:38,850 --> 00:25:41,120 OK, that's one constraint in it-- two left, 460 00:25:41,120 --> 00:25:42,940 two possible things left. 461 00:25:42,940 --> 00:25:43,720 STUDENT: No slip. 462 00:25:43,720 --> 00:25:46,780 PROFESSOR: Ahh, so I think this one's no slip. 463 00:25:46,780 --> 00:25:49,970 So how does a no slip then give you a constraint? 464 00:25:49,970 --> 00:25:54,570 STUDENT: So x dot equals r omega of the wheel turning. 465 00:25:54,570 --> 00:25:56,920 PROFESSOR: So r, there's an r. 466 00:25:56,920 --> 00:25:59,530 X equals r theta, or minus r theta 467 00:25:59,530 --> 00:26:02,570 depending on how you define the theta, right? 468 00:26:02,570 --> 00:26:08,239 So x is not independent of theta. 469 00:26:08,239 --> 00:26:09,530 And that's a second constraint. 470 00:26:09,530 --> 00:26:14,340 So you only need one equation, one generalized coordinate, 471 00:26:14,340 --> 00:26:15,872 to describe the motion of the wheel. 472 00:26:15,872 --> 00:26:16,830 And what would you use? 473 00:26:19,620 --> 00:26:20,495 STUDENT: Theta? 474 00:26:20,495 --> 00:26:21,370 PROFESSOR: OK, theta. 475 00:26:21,370 --> 00:26:22,203 You could use theta. 476 00:26:22,203 --> 00:26:23,850 Or you could use x. 477 00:26:23,850 --> 00:26:27,290 All right, and now the other object, how many constraints 478 00:26:27,290 --> 00:26:30,075 does it have? 479 00:26:30,075 --> 00:26:31,142 And what are they? 480 00:26:31,142 --> 00:26:32,725 What are the constraints on the T bar? 481 00:26:36,032 --> 00:26:39,860 STUDENT: Can't translate the y. 482 00:26:39,860 --> 00:26:41,490 PROFESSOR: OK, that's true. 483 00:26:41,490 --> 00:26:44,940 Let me give you a little hint about how I process 484 00:26:44,940 --> 00:26:46,500 this when I'm looking at it. 485 00:26:46,500 --> 00:26:53,230 Remember when you're picking generalized coordinates, 486 00:26:53,230 --> 00:26:55,530 they need to be independent. 487 00:26:55,530 --> 00:26:57,880 And independent means if you freeze all but one, 488 00:26:57,880 --> 00:26:59,810 that last one can still move. 489 00:26:59,810 --> 00:27:01,360 OK, so we've picked one. 490 00:27:01,360 --> 00:27:03,410 You said theta or x. 491 00:27:03,410 --> 00:27:05,710 Let's freeze it. 492 00:27:05,710 --> 00:27:09,154 What motion is left of that T bar? 493 00:27:09,154 --> 00:27:10,819 STUDENT: [INAUDIBLE] 494 00:27:10,819 --> 00:27:12,360 PROFESSOR: It can only rotate, right? 495 00:27:12,360 --> 00:27:15,800 And that means it has a pin, a point 496 00:27:15,800 --> 00:27:20,170 about at which it rotates, which you have just fixed. 497 00:27:20,170 --> 00:27:21,210 So it's constrained. 498 00:27:21,210 --> 00:27:24,005 That pin constrains it in how many directions? 499 00:27:24,005 --> 00:27:25,250 STUDENT: Two. 500 00:27:25,250 --> 00:27:26,940 PROFESSOR: Two, right off the bat. 501 00:27:26,940 --> 00:27:29,050 You have the two constraints right at the pin. 502 00:27:29,050 --> 00:27:31,840 But it's not so obvious until you say, let's freeze 503 00:27:31,840 --> 00:27:33,594 at other coordinates. 504 00:27:33,594 --> 00:27:35,010 And then we see that, ahh, there's 505 00:27:35,010 --> 00:27:37,470 only one possible motion left, the rotation. 506 00:27:37,470 --> 00:27:40,970 OK, so we have a rotation and a translation, or two rotations. 507 00:27:40,970 --> 00:27:44,720 That's the way we could write equations of motion. 508 00:27:44,720 --> 00:27:47,910 If we did translation of the main disk, 509 00:27:47,910 --> 00:27:51,710 what equation would you use to write the equation of motion 510 00:27:51,710 --> 00:27:55,140 for the main disk? 511 00:27:55,140 --> 00:27:56,600 STUDENT: Force. 512 00:27:56,600 --> 00:27:57,950 PROFESSOR: Force, yeah. 513 00:27:57,950 --> 00:28:01,190 Newton's second law, sum of the forces in what direction? 514 00:28:03,985 --> 00:28:05,607 It can only move in-- 515 00:28:05,607 --> 00:28:06,440 STUDENT: [INAUDIBLE] 516 00:28:06,440 --> 00:28:07,630 PROFESSOR: Horizontal, yeah. 517 00:28:07,630 --> 00:28:09,740 So sum of the forces in the x direction, 518 00:28:09,740 --> 00:28:12,120 sum of external forces in the x direction, 519 00:28:12,120 --> 00:28:14,160 equals mass times acceleration. 520 00:28:14,160 --> 00:28:16,870 Now this, though, has some problematic external forces, 521 00:28:16,870 --> 00:28:18,300 right? 522 00:28:18,300 --> 00:28:21,499 What are they, the difficult ones? 523 00:28:21,499 --> 00:28:22,397 STUDENT: [INAUDIBLE] 524 00:28:22,397 --> 00:28:23,480 PROFESSOR: Well, friction. 525 00:28:23,480 --> 00:28:24,521 There's a friction force. 526 00:28:24,521 --> 00:28:31,020 And there's also the internal forces at the pin. 527 00:28:31,020 --> 00:28:33,310 So you're going to have to sort out how 528 00:28:33,310 --> 00:28:34,920 to work your way through that. 529 00:28:34,920 --> 00:28:37,710 If you don't want to mess with external forces at the pin, 530 00:28:37,710 --> 00:28:42,133 you have to sum rotations at the-- 531 00:28:42,133 --> 00:28:42,966 STUDENT: [INAUDIBLE] 532 00:28:46,626 --> 00:28:48,000 PROFESSOR: Yeah, so this problem, 533 00:28:48,000 --> 00:28:50,220 there's no escaping something messy. 534 00:28:53,290 --> 00:28:59,850 So let's say one equation will be 535 00:28:59,850 --> 00:29:03,010 the sum of the forces on that main roller 536 00:29:03,010 --> 00:29:04,610 in the horizontal direction. 537 00:29:04,610 --> 00:29:06,880 And that's going to force us to deal with a friction 538 00:29:06,880 --> 00:29:11,720 force and an internal, two reaction forces at the pin. 539 00:29:11,720 --> 00:29:13,360 Well, you just write them down. 540 00:29:13,360 --> 00:29:18,200 Then go and write the same expression, sum of the forces 541 00:29:18,200 --> 00:29:19,580 on the other object. 542 00:29:19,580 --> 00:29:22,809 And it'll also show up with those two internal forces. 543 00:29:22,809 --> 00:29:24,350 You add those two equations together. 544 00:29:24,350 --> 00:29:25,183 Those forces cancel. 545 00:29:27,700 --> 00:29:30,190 And that gives you the first equation. 546 00:29:30,190 --> 00:29:33,880 And it'll involve acceleration of both masses. 547 00:29:33,880 --> 00:29:36,220 And you'll have to figure out the acceleration 548 00:29:36,220 --> 00:29:39,164 of that second mass around its own. 549 00:29:39,164 --> 00:29:40,330 But you know how to do that. 550 00:29:40,330 --> 00:29:41,288 That's just kinematics. 551 00:29:44,340 --> 00:29:46,780 And then you've got to do the torque equation. 552 00:29:46,780 --> 00:29:50,300 And you'd probably do it about A. 553 00:29:50,300 --> 00:29:56,752 And does this involve-- what about the velocity of A 554 00:29:56,752 --> 00:29:57,680 in this problem? 555 00:29:57,680 --> 00:29:59,610 Is it 0? 556 00:29:59,610 --> 00:30:02,400 No, so you may have to deal with that second term. 557 00:30:02,400 --> 00:30:04,740 And again, I would go use the third expression. 558 00:30:04,740 --> 00:30:08,460 Because you need to find the accelerations anyway. 559 00:30:08,460 --> 00:30:10,480 And I would use that. 560 00:30:10,480 --> 00:30:11,302 Yeah. 561 00:30:11,302 --> 00:30:14,719 STUDENT: Could you also take the sum of the torques about d? 562 00:30:14,719 --> 00:30:16,260 PROFESSOR: Some of the torques about? 563 00:30:16,260 --> 00:30:16,560 STUDENT: d? 564 00:30:16,560 --> 00:30:18,310 PROFESSOR: Then d is the point of contact. 565 00:30:22,480 --> 00:30:27,291 Yeah, and that would give you a way of getting at, 566 00:30:27,291 --> 00:30:29,040 I think, an equation of motion that mostly 567 00:30:29,040 --> 00:30:30,450 describes the big roller. 568 00:30:36,840 --> 00:30:38,910 I don't know. 569 00:30:38,910 --> 00:30:41,880 In terms of the other one, I'm not-- 570 00:30:41,880 --> 00:30:45,270 summing it, then talking about the second rigid body? 571 00:30:45,270 --> 00:30:47,460 I don't think I would do it. 572 00:30:47,460 --> 00:30:49,146 It gets a little complicated. 573 00:30:49,146 --> 00:30:50,020 But it's interesting. 574 00:30:50,020 --> 00:30:53,057 It might be worth a try. 575 00:30:53,057 --> 00:30:55,390 I'd probably do it about A, is the point I would choose, 576 00:30:55,390 --> 00:30:57,980 and deal with the fact that you need 577 00:30:57,980 --> 00:31:01,070 to know the acceleration of A. But in this problem, 578 00:31:01,070 --> 00:31:02,320 that's pretty straightforward. 579 00:31:02,320 --> 00:31:04,100 What's the acceleration of point A? 580 00:31:04,100 --> 00:31:08,380 And let's say our horizontal coordinate is x. 581 00:31:08,380 --> 00:31:10,520 What's the acceleration of point A? 582 00:31:10,520 --> 00:31:13,310 x double dot in the I direction-- pretty easy 583 00:31:13,310 --> 00:31:17,210 to stick that in this third expression up here. 584 00:31:17,210 --> 00:31:20,710 And you know Rg with respect to A 585 00:31:20,710 --> 00:31:25,290 is just the length of the distance down 586 00:31:25,290 --> 00:31:28,730 to the center of mass of that T. So that term's 587 00:31:28,730 --> 00:31:29,930 pretty easy to figure out. 588 00:31:29,930 --> 00:31:32,810 And then the lead term is just everything with respect to g. 589 00:31:32,810 --> 00:31:35,360 And that's pretty straightforward. 590 00:31:35,360 --> 00:31:38,240 OK, good, how are we doing on time? 591 00:31:38,240 --> 00:31:39,180 Not a lot left. 592 00:31:39,180 --> 00:31:41,880 And is that the last problem? 593 00:31:41,880 --> 00:31:42,519 Yeah. 594 00:31:42,519 --> 00:31:43,518 STUDENT: Quick question. 595 00:31:43,518 --> 00:31:46,018 The previous problem, we don't need to worry about friction, 596 00:31:46,018 --> 00:31:48,427 do we? 597 00:31:48,427 --> 00:31:50,010 STUDENT: If it was slipping, we would. 598 00:31:50,010 --> 00:31:53,410 PROFESSOR: Yeah, will friction ultimately 599 00:31:53,410 --> 00:31:57,565 end up in the answer to this one? 600 00:31:57,565 --> 00:31:58,065 No. 601 00:32:01,073 --> 00:32:01,906 STUDENT: [INAUDIBLE] 602 00:32:06,790 --> 00:32:08,880 PROFESSOR: Well, speaking in Lagrange terms, 603 00:32:08,880 --> 00:32:10,256 does it do any work? 604 00:32:10,256 --> 00:32:11,994 STUDENT: [INAUDIBLE] 605 00:32:11,994 --> 00:32:14,160 PROFESSOR: I know, but you know about it now, right? 606 00:32:14,160 --> 00:32:16,350 Constraints that do no work usually 607 00:32:16,350 --> 00:32:19,045 don't end up in the final answer. 608 00:32:27,220 --> 00:32:31,050 I don't have a simple way of explaining why it doesn't. 609 00:32:31,050 --> 00:32:35,115 But if in the first case, instead of doing Newton's law 610 00:32:35,115 --> 00:32:38,730 on the first one, we had done torques about d, 611 00:32:38,730 --> 00:32:40,850 it would have given us an equation of motion 612 00:32:40,850 --> 00:32:43,040 for the roller. 613 00:32:43,040 --> 00:32:44,540 And then you could have done torques 614 00:32:44,540 --> 00:32:46,629 about A, a different equation. 615 00:32:46,629 --> 00:32:48,920 And we'd have got an equation of motion largely dealing 616 00:32:48,920 --> 00:32:51,910 with T. You still have some internal forces 617 00:32:51,910 --> 00:32:54,470 and torques that you're going to have to eliminate. 618 00:32:54,470 --> 00:32:59,280 So this is not a simple problem, but straightforward if you 619 00:32:59,280 --> 00:33:04,080 know what laws to apply to some of the forces, some 620 00:33:04,080 --> 00:33:04,820 of the torques. 621 00:33:04,820 --> 00:33:08,360 And be careful with the extra term. 622 00:33:08,360 --> 00:33:11,680 If this object were just given an initial deflection, 623 00:33:11,680 --> 00:33:14,275 like the T bar is picked up and let go, and it's just there, 624 00:33:14,275 --> 00:33:16,800 and it's just doing its thing, what 625 00:33:16,800 --> 00:33:20,225 can you say about the center of mass of the system? 626 00:33:23,342 --> 00:33:26,150 STUDENT: [INAUDIBLE] 627 00:33:26,150 --> 00:33:30,215 PROFESSOR: Well, it's not that it doesn't necessarily move. 628 00:33:35,500 --> 00:33:36,780 Well, let's put it this way. 629 00:33:36,780 --> 00:33:39,830 What if I gave this thing a push to start with, 630 00:33:39,830 --> 00:33:41,751 and now it's going to roll along? 631 00:33:41,751 --> 00:33:42,250 Huh? 632 00:33:42,250 --> 00:33:43,600 STUDENT: [INAUDIBLE] 633 00:33:43,600 --> 00:33:46,160 PROFESSOR: OK, in that case, you're 634 00:33:46,160 --> 00:33:50,550 saying that the center of mass moves with constant velocity. 635 00:33:50,550 --> 00:33:52,550 And Newton would agree with you. 636 00:33:52,550 --> 00:33:53,159 Why? 637 00:33:53,159 --> 00:33:53,992 STUDENT: [INAUDIBLE] 638 00:33:57,349 --> 00:33:59,390 PROFESSOR: Because once it gets rolling, is there 639 00:33:59,390 --> 00:34:00,300 any friction force? 640 00:34:00,300 --> 00:34:01,012 STUDENT: No. 641 00:34:01,012 --> 00:34:03,262 PROFESSOR: So you've got to draw the free body diagram 642 00:34:03,262 --> 00:34:06,100 and decide whether or not there are any forces on the system. 643 00:34:06,100 --> 00:34:08,500 If there are no external forces on the system, 644 00:34:08,500 --> 00:34:13,710 Newton says no acceleration, no change in momentum. 645 00:34:13,710 --> 00:34:15,659 So once you get it rolling, there's 646 00:34:15,659 --> 00:34:17,070 actually no friction force. 647 00:34:17,070 --> 00:34:20,790 Because it isn't trying to either speed up or slow down 648 00:34:20,790 --> 00:34:21,870 the rolling. 649 00:34:21,870 --> 00:34:23,449 So it goes actually to 0. 650 00:34:23,449 --> 00:34:25,086 Is there slip? 651 00:34:25,086 --> 00:34:26,610 No, it still isn't slipping. 652 00:34:26,610 --> 00:34:28,250 But it's just happily rolling along. 653 00:34:28,250 --> 00:34:30,020 There are no external forces acting on it. 654 00:34:30,020 --> 00:34:33,010 It means the mass times the acceleration of the center 655 00:34:33,010 --> 00:34:34,469 of mass of the system is zero. 656 00:34:34,469 --> 00:34:36,625 So the center of mass does not accelerate. 657 00:34:36,625 --> 00:34:38,790 That means it has constant velocity. 658 00:34:38,790 --> 00:34:41,639 Now, if I contrive to have the initial velocity of the center 659 00:34:41,639 --> 00:34:46,760 of mass be 0, just cause that thing to swing back and forth, 660 00:34:46,760 --> 00:34:51,290 what would you see the center of mass do? 661 00:34:51,290 --> 00:34:54,159 It has no linear momentum. 662 00:34:54,159 --> 00:34:56,120 The linear momentum of the system is 0. 663 00:34:56,120 --> 00:35:01,470 So the center of mass sits still. 664 00:35:01,470 --> 00:35:02,850 Its velocity now is 0. 665 00:35:02,850 --> 00:35:04,170 There's still no acceleration. 666 00:35:04,170 --> 00:35:07,410 It's just this T bar is rocking back and forth. 667 00:35:07,410 --> 00:35:09,696 So what must the roller be doing? 668 00:35:09,696 --> 00:35:11,010 STUDENT: [INAUDIBLE] 669 00:35:11,010 --> 00:35:13,810 PROFESSOR: So the roller has to move to the left 670 00:35:13,810 --> 00:35:15,740 when the T bar is going to the right. 671 00:35:15,740 --> 00:35:17,940 So the two things are going like this. 672 00:35:17,940 --> 00:35:21,970 In the center of mass, wherever it is, it's just sitting still. 673 00:35:21,970 --> 00:35:24,240 All right, is there a last problem? 674 00:35:26,820 --> 00:35:31,150 This forces you to go back to the fundamental definition 675 00:35:31,150 --> 00:35:33,060 of angular momentum. 676 00:35:33,060 --> 00:35:39,440 It's R cross, this little summation of each RI cross PI. 677 00:35:39,440 --> 00:35:42,030 And add them up, and you get the angular momentum with respect 678 00:35:42,030 --> 00:35:47,790 to the rotational point, 1 degree of freedom, 679 00:35:47,790 --> 00:35:48,691 has a center of mass. 680 00:35:48,691 --> 00:35:50,440 Where is the center of mass of the system? 681 00:35:50,440 --> 00:35:51,273 Can you describe it? 682 00:35:54,437 --> 00:35:55,820 STUDENT: Right in between-- 683 00:35:55,820 --> 00:35:57,440 PROFESSOR: Somewhere in between the two masses. 684 00:35:57,440 --> 00:35:59,065 There are two point masses, center mass 685 00:35:59,065 --> 00:36:00,610 has to be someplace in between. 686 00:36:00,610 --> 00:36:02,390 So once you figure out where that is, then 687 00:36:02,390 --> 00:36:06,440 you can concentrate all the mass there and go from there. 688 00:36:06,440 --> 00:36:08,400 Does this have mass moment of inertia? 689 00:36:13,810 --> 00:36:15,970 Can you write the equation of motion? 690 00:36:15,970 --> 00:36:16,840 This is a pendulum. 691 00:36:16,840 --> 00:36:25,580 Can you write I about A theta double dot equals minus Mg 692 00:36:25,580 --> 00:36:29,570 something sine theta? 693 00:36:29,570 --> 00:36:30,880 This is a pendulum. 694 00:36:30,880 --> 00:36:34,670 The restoring torque on it comes from gravity. 695 00:36:34,670 --> 00:36:37,150 And you'll end up with an expression 696 00:36:37,150 --> 00:36:39,000 that looks like some I with respect 697 00:36:39,000 --> 00:36:41,050 to the point of rotation. 698 00:36:41,050 --> 00:36:42,760 So basically, you just need to figure out 699 00:36:42,760 --> 00:36:48,290 what is I with respect to A. And they're particles. 700 00:36:48,290 --> 00:36:52,500 So each particle has an I with respect to its center of mass. 701 00:36:52,500 --> 00:36:59,020 It's equal to, for a particle, concentrated point mass? 702 00:36:59,020 --> 00:36:59,994 STUDENT: Mr squared? 703 00:36:59,994 --> 00:37:01,410 PROFESSOR: Yeah, but there's no r. 704 00:37:01,410 --> 00:37:06,890 So I for a particle about its center mass is 0. 705 00:37:06,890 --> 00:37:10,700 So this only has parallel axis theorem components. 706 00:37:10,700 --> 00:37:12,960 It has M1 times this distance squared 707 00:37:12,960 --> 00:37:15,480 from the center plus M2 times this distance squared 708 00:37:15,480 --> 00:37:16,210 from the center. 709 00:37:16,210 --> 00:37:19,180 And that's the total Izz. 710 00:37:19,180 --> 00:37:20,960 So then you can write it out. 711 00:37:20,960 --> 00:37:26,460 OK, so I think we've run out of time.