1 00:00:00,100 --> 00:00:02,450 The following content is provided under a Creative 2 00:00:02,450 --> 00:00:03,830 Commons license. 3 00:00:03,830 --> 00:00:06,070 Your support will help MIT OpenCourseWare 4 00:00:06,070 --> 00:00:10,170 continue to offer high quality educational resources for free. 5 00:00:10,170 --> 00:00:12,710 To make a donation or to view additional materials 6 00:00:12,710 --> 00:00:16,620 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,620 --> 00:00:17,270 at ocw.mit.edu. 8 00:00:22,010 --> 00:00:26,250 PROFESSOR: OK, let's talk about the important thoughts 9 00:00:26,250 --> 00:00:27,625 for the week, important concepts. 10 00:00:30,442 --> 00:00:31,150 What do you have? 11 00:00:33,928 --> 00:00:35,317 AUDIENCE: Energy method. 12 00:00:35,317 --> 00:00:36,706 PROFESSOR: Energy method. 13 00:00:36,706 --> 00:00:40,745 I'm going to say this work energy theorem. 14 00:00:54,820 --> 00:01:01,053 And from that, these turn out to be our potential energies. 15 00:01:16,481 --> 00:01:20,610 And this minus the work done by the conservative forces 16 00:01:20,610 --> 00:01:23,890 we call V2 minus V1. 17 00:01:23,890 --> 00:01:26,870 That's the change in the potential energy of the system. 18 00:01:26,870 --> 00:01:28,380 So that's the work energy theorem. 19 00:01:28,380 --> 00:01:30,250 That's an important one for the week. 20 00:01:30,250 --> 00:01:30,750 What else? 21 00:01:41,420 --> 00:01:43,370 AUDIENCE: Different types of problems. 22 00:01:43,370 --> 00:01:45,870 PROFESSOR: Yeah, we did several example problem 23 00:01:45,870 --> 00:01:47,120 particularly emphasizing what? 24 00:01:52,669 --> 00:01:54,335 AUDIENCE: With respect to moving points? 25 00:01:54,335 --> 00:01:57,202 PROFESSOR: Yeah, or points that look like they're moving. 26 00:01:57,202 --> 00:02:00,020 But you can use special considerations 27 00:02:00,020 --> 00:02:03,360 in calling them instantaneous centers of rotation and things 28 00:02:03,360 --> 00:02:07,560 like that, so summation of the torques with respect to points 29 00:02:07,560 --> 00:02:18,030 sometimes moving defined with respect to ICRs, Instantaneous 30 00:02:18,030 --> 00:02:24,250 Centers of Rotation, and also moving points. 31 00:02:27,365 --> 00:02:29,410 We definitely did some of that. 32 00:02:29,410 --> 00:02:30,320 How about a third? 33 00:02:45,950 --> 00:02:48,820 I'd say one more thing that we got started on 34 00:02:48,820 --> 00:03:07,710 is expressions-- 1/2 i, 1/2 omega dot h plus 1/2 mv, v dot 35 00:03:07,710 --> 00:03:09,730 v, those kind of terms for this. 36 00:03:09,730 --> 00:03:14,229 And for potential energy, if it's purely mechanical devices, 37 00:03:14,229 --> 00:03:15,770 we have two kinds of potential energy 38 00:03:15,770 --> 00:03:17,020 that we'll be dealing with. 39 00:03:17,020 --> 00:03:24,000 And they are associated with springs and gravity. 40 00:03:24,000 --> 00:03:25,970 So that's the two we'll deal with. 41 00:03:25,970 --> 00:03:27,836 Are there others? 42 00:03:27,836 --> 00:03:30,210 What's an example of a different kind of potential energy 43 00:03:30,210 --> 00:03:31,643 that you might run into? 44 00:03:31,643 --> 00:03:32,550 AUDIENCE: Magnetic. 45 00:03:32,550 --> 00:03:33,985 PROFESSOR: Magnetic fields, yeah. 46 00:03:33,985 --> 00:03:36,850 We can do work pulling something out of a magnetic field. 47 00:03:36,850 --> 00:03:37,420 What else? 48 00:03:37,420 --> 00:03:38,100 AUDIENCE: Electric fields. 49 00:03:38,100 --> 00:03:39,570 PROFESSOR: Electric fields, sure. 50 00:03:39,570 --> 00:03:41,230 But they do have their complications. 51 00:03:41,230 --> 00:03:45,370 So that's pretty good for the week. 52 00:03:47,890 --> 00:03:52,990 Now, we're going to talk about this problem. 53 00:03:52,990 --> 00:03:54,770 I'll give him a chance to shoot it. 54 00:03:54,770 --> 00:03:56,210 This is a good diagram of it. 55 00:03:56,210 --> 00:04:00,570 You guys had a homework problem that instead of a uniform rod, 56 00:04:00,570 --> 00:04:06,820 you had a concentrated mass on the end of a massless rod. 57 00:04:06,820 --> 00:04:09,690 So this is almost identical to that. 58 00:04:09,690 --> 00:04:15,530 But we now have actual mass, distributed mass, in the stick. 59 00:04:15,530 --> 00:04:20,579 I've got a little demo here, which will work, 60 00:04:20,579 --> 00:04:21,760 works after a fashion. 61 00:04:25,440 --> 00:04:27,030 Put a little tension on these springs, 62 00:04:27,030 --> 00:04:29,740 so now I've got my cart on a spring. 63 00:04:29,740 --> 00:04:31,390 The damping is natural in the system, 64 00:04:31,390 --> 00:04:34,154 comes from the wheels and everything else. 65 00:04:34,154 --> 00:04:36,070 I've got to get this kind of close to the edge 66 00:04:36,070 --> 00:04:36,880 so it'll work. 67 00:04:40,600 --> 00:04:44,020 This thing has two natural frequencies actually. 68 00:04:44,020 --> 00:04:46,790 They're kind of hard to isolate them. 69 00:04:46,790 --> 00:04:51,300 There's one-- no, can't do it. 70 00:04:51,300 --> 00:04:54,590 In one natural mode, the thing swings forward 71 00:04:54,590 --> 00:04:56,540 and moves forward together. 72 00:04:56,540 --> 00:04:59,135 And in the other natural mode, this'll go aft, that way, 73 00:04:59,135 --> 00:05:00,760 while the other one's going frontwards. 74 00:05:00,760 --> 00:05:03,220 So let's see if I can-- that's higher frequency. 75 00:05:03,220 --> 00:05:07,480 That's the motion of what's called the second mode, 76 00:05:07,480 --> 00:05:08,300 opposite. 77 00:05:08,300 --> 00:05:11,720 And the first mode is lower frequency than that. 78 00:05:25,490 --> 00:05:28,010 It's hard to get a perfect start to it. 79 00:05:28,010 --> 00:05:31,010 So this is the kind of motion we're talking about. 80 00:05:31,010 --> 00:05:33,530 And we're going to have you help me solve. 81 00:05:33,530 --> 00:05:37,910 We want to obtain equations of motion for this system. 82 00:05:37,910 --> 00:05:41,890 And let's quickly run through the exercise of-- you 83 00:05:41,890 --> 00:05:43,814 can do this intuitively first, it's so simple. 84 00:05:43,814 --> 00:05:44,980 Plus you've already done it. 85 00:05:44,980 --> 00:05:48,240 How many independent coordinates does it take to do it? 86 00:05:48,240 --> 00:05:57,150 Two, and how many rigid bodies are there in this problem? 87 00:05:57,150 --> 00:05:59,140 How many rigid bodies in the problem? 88 00:06:02,932 --> 00:06:03,806 AUDIENCE: Two. 89 00:06:03,806 --> 00:06:05,430 PROFESSOR: There's gotta be two, right? 90 00:06:05,430 --> 00:06:07,250 There's a block that moves back and forth, 91 00:06:07,250 --> 00:06:08,970 and there's this rod that swings. 92 00:06:08,970 --> 00:06:11,380 So the number of degrees of freedom, we'll call it d, 93 00:06:11,380 --> 00:06:15,260 is 6 times the number of bodies. 94 00:06:15,260 --> 00:06:18,270 n is the number bodies minus the number of constraints. 95 00:06:18,270 --> 00:06:21,330 So this is 6 times 2 minus the constraints. 96 00:06:21,330 --> 00:06:23,870 So we've got 12 minus c. 97 00:06:23,870 --> 00:06:25,860 And you know the answers. 98 00:06:25,860 --> 00:06:29,150 The answer, how many independent coordinates to completely 99 00:06:29,150 --> 00:06:30,237 describe this motion? 100 00:06:30,237 --> 00:06:30,820 AUDIENCE: Two. 101 00:06:30,820 --> 00:06:31,819 PROFESSOR: Just the two. 102 00:06:31,819 --> 00:06:34,135 So you know the answer is two. 103 00:06:34,135 --> 00:06:36,380 So that means you've got to come up with 10, 104 00:06:36,380 --> 00:06:37,940 a list of 10 of these things. 105 00:06:37,940 --> 00:06:42,250 So let's see if you can write down quickly your own list. 106 00:06:42,250 --> 00:06:47,370 How would you eliminate through coming up with constraints? 107 00:06:47,370 --> 00:06:49,420 Write down the 10 constraints. 108 00:06:49,420 --> 00:06:52,720 And I recommend you do it by taking one body at a time. 109 00:06:52,720 --> 00:06:56,824 So constraints on mass one, the first block, what do you got? 110 00:06:56,824 --> 00:06:59,610 AUDIENCE: y, y dot, and y double dot equal 0. 111 00:06:59,610 --> 00:07:02,200 PROFESSOR: OK, so that says no motion in the y direction. 112 00:07:02,200 --> 00:07:03,815 Because we're on rollers. 113 00:07:03,815 --> 00:07:06,670 It's fixed that way, so no motion in the y. 114 00:07:06,670 --> 00:07:07,629 What else? 115 00:07:07,629 --> 00:07:08,670 AUDIENCE: No motion in z. 116 00:07:08,670 --> 00:07:09,030 PROFESSOR: No motion in z. 117 00:07:09,030 --> 00:07:11,430 We're telling it it can't come in and out of the board. 118 00:07:11,430 --> 00:07:15,190 OK, any other translational constraints on that one? 119 00:07:15,190 --> 00:07:21,685 So no y or z translation. 120 00:07:24,260 --> 00:07:26,470 And so that gives us two constraints there. 121 00:07:26,470 --> 00:07:28,030 What else on that body? 122 00:07:28,030 --> 00:07:29,620 AUDIENCE: No rotations. 123 00:07:29,620 --> 00:07:31,370 PROFESSOR: No rotations in what direction? 124 00:07:31,370 --> 00:07:32,210 AUDIENCE: x, y, or z. 125 00:07:32,210 --> 00:07:33,251 PROFESSOR: In x, y, or z. 126 00:07:33,251 --> 00:07:39,980 The thing can only translate, so no x, y, z rotations. 127 00:07:42,920 --> 00:07:43,540 That's three. 128 00:07:43,540 --> 00:07:45,510 So that's a total of five. 129 00:07:45,510 --> 00:07:46,620 But we have one left. 130 00:07:46,620 --> 00:07:48,600 It's got to be able to move in the x. 131 00:07:48,600 --> 00:07:51,710 And how about M2? 132 00:07:51,710 --> 00:07:55,786 What are the constraints on the second one? 133 00:07:55,786 --> 00:07:58,030 AUDIENCE: [INAUDIBLE] 134 00:07:58,030 --> 00:07:59,400 PROFESSOR: So one at a time. 135 00:07:59,400 --> 00:08:00,275 AUDIENCE: [INAUDIBLE] 136 00:08:02,950 --> 00:08:04,840 PROFESSOR: So omega which? 137 00:08:04,840 --> 00:08:06,010 AUDIENCE: x and y. 138 00:08:06,010 --> 00:08:09,986 PROFESSOR: No rotations about which axes? 139 00:08:09,986 --> 00:08:10,820 AUDIENCE: x and y. 140 00:08:10,820 --> 00:08:13,683 PROFESSOR: So you're saying no rotations of this about the x. 141 00:08:13,683 --> 00:08:15,340 So it can't roll over. 142 00:08:15,340 --> 00:08:17,960 About the y, it can't turn. 143 00:08:17,960 --> 00:08:20,260 And about the z. 144 00:08:20,260 --> 00:08:22,756 But we're doing this one. 145 00:08:22,756 --> 00:08:24,380 So it's no spinning about-- let's do it 146 00:08:24,380 --> 00:08:25,550 in its body coordinates. 147 00:08:25,550 --> 00:08:30,070 x1-- no spinning about x1. 148 00:08:30,070 --> 00:08:33,120 No spinning about y1. 149 00:08:33,120 --> 00:08:35,679 But can it rotate around z1? 150 00:08:35,679 --> 00:08:47,310 OK, so for the second mass, we have no x1, y1 rotations. 151 00:08:47,310 --> 00:08:48,100 That's two. 152 00:08:48,100 --> 00:08:50,180 And what about translations on the second body? 153 00:08:53,916 --> 00:08:54,850 Hmm? 154 00:08:54,850 --> 00:08:56,730 AUDIENCE: No translations x, y, and z. 155 00:08:56,730 --> 00:08:58,380 PROFESSOR: No translations x, y, and z. 156 00:08:58,380 --> 00:09:01,320 So z, pretty obvious-- we're not letting it do this. 157 00:09:01,320 --> 00:09:07,136 But why can you argue there's no translation in x and y? 158 00:09:07,136 --> 00:09:10,608 AUDIENCE: Because it's strictly rotation. 159 00:09:10,608 --> 00:09:13,856 PROFESSOR: You say it's strictly rotation. 160 00:09:13,856 --> 00:09:17,994 AUDIENCE: For it to translate [INAUDIBLE]. 161 00:09:17,994 --> 00:09:19,785 PROFESSOR: OK, so you're saying translation 162 00:09:19,785 --> 00:09:22,680 is strictly described as parallel motion 163 00:09:22,680 --> 00:09:24,980 of all points on it. 164 00:09:24,980 --> 00:09:27,915 But where is it constrained in x and y? 165 00:09:27,915 --> 00:09:30,532 AUDIENCE: [INAUDIBLE] 166 00:09:30,532 --> 00:09:31,990 PROFESSOR: It's pinned to the cart. 167 00:09:31,990 --> 00:09:34,240 And if you fix x, which you're allowed to do, 168 00:09:34,240 --> 00:09:36,712 a test you're allowed to do, you have set this 169 00:09:36,712 --> 00:09:37,920 as an independent coordinate. 170 00:09:37,920 --> 00:09:41,230 If you fix it, this point is fixed. 171 00:09:41,230 --> 00:09:44,780 And you've constrained it in x and y. 172 00:09:44,780 --> 00:09:48,100 And if any point in the object is constrained in x and y, 173 00:09:48,100 --> 00:09:54,590 then all points are constrained in x and y pure translation. 174 00:09:54,590 --> 00:10:02,035 So we have in this one no x, y, and z translations. 175 00:10:04,610 --> 00:10:05,520 So that's three. 176 00:10:05,520 --> 00:10:06,880 You add them up, you get 10. 177 00:10:06,880 --> 00:10:08,590 OK, good. 178 00:10:08,590 --> 00:10:12,720 Now, the next thing I want you to do, 179 00:10:12,720 --> 00:10:16,320 another key step in doing problems like this, 180 00:10:16,320 --> 00:10:17,930 is our free body diagrams. 181 00:10:17,930 --> 00:10:20,850 And how many do you need? 182 00:10:20,850 --> 00:10:27,525 As many as you have rigid bodies. 183 00:10:27,525 --> 00:10:30,400 You need one for every rigid body. 184 00:10:30,400 --> 00:10:32,030 So we got two rigid bodies. 185 00:10:32,030 --> 00:10:41,770 Draw for me your free body diagram for the cart 186 00:10:41,770 --> 00:10:44,265 and for the rod. 187 00:11:16,930 --> 00:11:20,830 OK, what group wants to volunteer here, 188 00:11:20,830 --> 00:11:23,319 tell me what I ought to do? 189 00:11:23,319 --> 00:11:24,610 How about you guys in the back? 190 00:11:24,610 --> 00:11:26,330 You're quite today. 191 00:11:26,330 --> 00:11:29,460 So pick this first one. 192 00:11:29,460 --> 00:11:33,880 AUDIENCE: You have the spring force kx to the left. 193 00:11:33,880 --> 00:11:35,002 PROFESSOR: kx. 194 00:11:35,002 --> 00:11:36,970 AUDIENCE: Also bx dot. 195 00:11:36,970 --> 00:11:38,710 PROFESSOR: bx dot. 196 00:11:38,710 --> 00:11:40,570 AUDIENCE: Normal force up. 197 00:11:40,570 --> 00:11:41,487 PROFESSOR: N. 198 00:11:41,487 --> 00:11:46,844 AUDIENCE: You have M1g down, and then the force of the rod 199 00:11:46,844 --> 00:11:48,305 is down and to the right. 200 00:11:48,305 --> 00:11:52,155 PROFESSOR: Oh, so you've made-- ahh, colinear with the rod? 201 00:11:52,155 --> 00:11:52,780 AUDIENCE: Yeah. 202 00:11:52,780 --> 00:11:54,490 PROFESSOR: OK, so you're saying we've 203 00:11:54,490 --> 00:11:59,000 got a vector force like that, some F, right? 204 00:11:59,000 --> 00:12:02,240 OK, how about a little help on this. 205 00:12:02,240 --> 00:12:06,994 Anybody have a different result here for this one? 206 00:12:06,994 --> 00:12:10,930 AUDIENCE: I broke up my normal into two [INAUDIBLE]. 207 00:12:10,930 --> 00:12:13,950 PROFESSOR: So you have two pieces here? 208 00:12:13,950 --> 00:12:17,712 So in what direction do you draw them? 209 00:12:17,712 --> 00:12:20,060 AUDIENCE: Reaction force of the first failed reaction? 210 00:12:20,060 --> 00:12:22,750 PROFESSOR: I guess which coordinate system? 211 00:12:22,750 --> 00:12:27,260 We need two coordinate systems to define this problem. 212 00:12:27,260 --> 00:12:29,140 AUDIENCE: Just in the regular x, y. 213 00:12:29,140 --> 00:12:34,100 PROFESSOR: So I've got an inertial frame here. 214 00:12:34,100 --> 00:12:37,720 And that gives me my X, capital X, motion for this guy. 215 00:12:37,720 --> 00:12:40,000 But we need, because if we want to be able to moments 216 00:12:40,000 --> 00:12:43,550 of inertia and things like that, a coordinate system attached 217 00:12:43,550 --> 00:12:45,440 to this rigid body, right? 218 00:12:45,440 --> 00:12:48,990 So I've called it little x1, y1, and it moves with the body. 219 00:12:48,990 --> 00:12:52,130 OK, so you've picked your two reaction 220 00:12:52,130 --> 00:12:57,094 forces aligned with these two or aligned in this system? 221 00:12:57,094 --> 00:12:59,360 AUDIENCE: With the inertial. 222 00:12:59,360 --> 00:13:01,300 PROFESSOR: OK, you've done it this way. 223 00:13:01,300 --> 00:13:03,010 So you're coming up with a pair of forces 224 00:13:03,010 --> 00:13:07,198 that have a piece like this and a piece like that? 225 00:13:07,198 --> 00:13:11,011 AUDIENCE: No, I'm talking about for the wheels. 226 00:13:11,011 --> 00:13:12,260 PROFESSOR: Oh, for the wheels. 227 00:13:12,260 --> 00:13:13,930 Oh, OK. 228 00:13:13,930 --> 00:13:15,750 You came up with individual ones here? 229 00:13:15,750 --> 00:13:17,530 Yeah, OK. 230 00:13:17,530 --> 00:13:20,390 I'll call that N1 and N2. 231 00:13:20,390 --> 00:13:23,520 And that's probably an improvement. 232 00:13:23,520 --> 00:13:26,210 Because that's what keeps it from rolling over. 233 00:13:26,210 --> 00:13:28,780 OK, that's true. 234 00:13:28,780 --> 00:13:29,990 And what else? 235 00:13:29,990 --> 00:13:32,892 So you've got a reaction force here like that. 236 00:13:32,892 --> 00:13:33,850 Can we improve on that? 237 00:13:36,670 --> 00:13:39,960 AUDIENCE: You can break it up into tangential and radial. 238 00:13:39,960 --> 00:13:41,470 PROFESSOR: Yeah, but I understood. 239 00:13:41,470 --> 00:13:44,570 The intention here was it was drawn colinear 240 00:13:44,570 --> 00:13:47,290 with the axis of the thing. 241 00:13:47,290 --> 00:13:50,060 And that suggested just that one part. 242 00:13:50,060 --> 00:13:53,630 You could break that into a horizontal in the master frame, 243 00:13:53,630 --> 00:13:54,130 yeah. 244 00:13:54,130 --> 00:13:58,090 But does anybody have something different about that, 245 00:13:58,090 --> 00:13:59,520 another way to do this? 246 00:13:59,520 --> 00:14:01,450 Do you agree with this? 247 00:14:01,450 --> 00:14:03,510 Is this correct? 248 00:14:03,510 --> 00:14:04,010 Pardon? 249 00:14:07,080 --> 00:14:09,160 You're treating it like a string. 250 00:14:09,160 --> 00:14:09,966 Is it a string? 251 00:14:12,840 --> 00:14:15,935 Can that shaft transmit shear forces? 252 00:14:20,111 --> 00:14:20,610 Hmm? 253 00:14:25,330 --> 00:14:27,730 If I grab this thing and go left and right on it, 254 00:14:27,730 --> 00:14:30,020 it's putting forces in that direction perpendicular 255 00:14:30,020 --> 00:14:32,650 to this. 256 00:14:32,650 --> 00:14:35,150 So is it conceivable that there are 257 00:14:35,150 --> 00:14:37,915 forces on this axle caused by that 258 00:14:37,915 --> 00:14:42,120 that are in that direction as well as that direction? 259 00:14:42,120 --> 00:14:44,000 What do you think? 260 00:14:44,000 --> 00:14:45,800 Because I can't make an argument for sure 261 00:14:45,800 --> 00:14:48,169 that would eliminate either one of those. 262 00:14:48,169 --> 00:14:49,960 Pretty sure there's certainly a radial one, 263 00:14:49,960 --> 00:14:52,585 because that's centripetal, and there's got to be some of that. 264 00:14:52,585 --> 00:14:56,100 But might there be some in the other direction? 265 00:14:56,100 --> 00:15:01,270 OK, so an improvement on this, a necessary improvement, I think, 266 00:15:01,270 --> 00:15:05,370 is that rather than just one, we'd better have two. 267 00:15:05,370 --> 00:15:08,360 And we're going to use my system of naming 268 00:15:08,360 --> 00:15:11,110 so that the answers I've got here will work out. 269 00:15:11,110 --> 00:15:16,640 I'm going to call that one F2 and this one F1. 270 00:15:16,640 --> 00:15:20,900 So now the rod is placing on the cart two reaction forces 271 00:15:20,900 --> 00:15:22,180 which you don't know. 272 00:15:22,180 --> 00:15:26,160 Now, is that the complete thing for the cart? 273 00:15:26,160 --> 00:15:28,390 OK, now let's do this one. 274 00:15:28,390 --> 00:15:33,210 Another group, tell me how to do the second one, free body 275 00:15:33,210 --> 00:15:36,165 diagram. 276 00:15:36,165 --> 00:15:39,116 AUDIENCE: So you have F1 and F2, but opposite. 277 00:15:39,116 --> 00:15:40,240 PROFESSOR: Ahh, good point. 278 00:15:40,240 --> 00:15:46,950 You've got an F2 this way and an F1 like that, 279 00:15:46,950 --> 00:15:48,200 is what you're saying? 280 00:15:48,200 --> 00:15:50,330 AUDIENCE: Yeah, and an Mg comes down. 281 00:15:50,330 --> 00:15:53,863 PROFESSOR: And an Mg, yeah, M2g. 282 00:15:53,863 --> 00:15:55,530 OK, good. 283 00:15:55,530 --> 00:16:00,340 Real important point-- Newton's third law. 284 00:16:00,340 --> 00:16:02,810 If you draw F1 and F2 here, those on that side 285 00:16:02,810 --> 00:16:07,300 have to be equal and opposite in order for all this 286 00:16:07,300 --> 00:16:08,890 to work out without problems. 287 00:16:08,890 --> 00:16:11,110 OK, those look pretty good to me. 288 00:16:14,010 --> 00:16:15,401 Now let's move on. 289 00:16:15,401 --> 00:16:16,650 This is a complicated problem. 290 00:16:16,650 --> 00:16:20,480 And to get a lot with it in an hour is a challenge. 291 00:16:20,480 --> 00:16:22,490 And you've done a bunch of this problem already. 292 00:16:22,490 --> 00:16:24,615 You did the one with the particle on it. 293 00:16:24,615 --> 00:16:27,050 So there's not a great deal of difference here. 294 00:16:27,050 --> 00:16:29,690 So I'm emphasizing a few of the nuances. 295 00:16:29,690 --> 00:16:32,020 What I now want to do is to talk about, 296 00:16:32,020 --> 00:16:36,870 how do we get equations of motion for this problem? 297 00:16:36,870 --> 00:16:38,956 How many will there be to start with? 298 00:16:38,956 --> 00:16:39,789 AUDIENCE: Two. 299 00:16:39,789 --> 00:16:41,580 PROFESSOR: Got to get two, two rigid bodies 300 00:16:41,580 --> 00:16:43,831 and two degrees of freedom. 301 00:16:43,831 --> 00:16:46,330 And you're going to end up-- it's the two degrees of freedom 302 00:16:46,330 --> 00:16:48,705 that tell you you're going to get two equations of motion 303 00:16:48,705 --> 00:16:49,260 at the end. 304 00:16:49,260 --> 00:16:53,210 OK, how many-- and I've written three up there, 305 00:16:53,210 --> 00:16:55,730 but it's not necessarily true. 306 00:16:55,730 --> 00:16:58,490 You've been taught to this point two different methods 307 00:16:58,490 --> 00:17:02,990 for getting at these equations of motion. 308 00:17:02,990 --> 00:17:05,079 Somebody describe one for me. 309 00:17:05,079 --> 00:17:08,400 How would you personally go about this problem? 310 00:17:08,400 --> 00:17:12,730 Just pick a method and in words describe it. 311 00:17:12,730 --> 00:17:14,821 And think of what you did in the last homework. 312 00:17:14,821 --> 00:17:19,640 AUDIENCE: Sum of external force equals [INAUDIBLE]. 313 00:17:19,640 --> 00:17:22,894 PROFESSOR: So sum of external forces is-- 314 00:17:22,894 --> 00:17:23,819 AUDIENCE: [INAUDIBLE] 315 00:17:23,819 --> 00:17:28,460 PROFESSOR: OK, so on method one here, I'll call it, 316 00:17:28,460 --> 00:17:33,191 you're going to say sum of the forces on what? 317 00:17:33,191 --> 00:17:34,232 AUDIENCE: The rigid body. 318 00:17:34,232 --> 00:17:36,145 PROFESSOR: So you've got two bodies here. 319 00:17:36,145 --> 00:17:38,011 So pick one. 320 00:17:38,011 --> 00:17:39,454 Huh? 321 00:17:39,454 --> 00:17:42,650 All right, so I'm going to put this on M1 for sure. 322 00:17:42,650 --> 00:17:44,360 That one is translating. 323 00:17:44,360 --> 00:17:46,260 And you're going to get an equation that 324 00:17:46,260 --> 00:17:49,040 involves its x double dot. 325 00:17:49,040 --> 00:17:50,870 It's got to be able to move back and forth. 326 00:17:50,870 --> 00:17:53,200 You're going to need an equation like this. 327 00:17:53,200 --> 00:17:55,270 So the sum of the external forces on that mass 328 00:17:55,270 --> 00:18:04,900 has got to be equal to M1 times-- right? 329 00:18:04,900 --> 00:18:11,390 And we can write an equation for that in terms of this free body 330 00:18:11,390 --> 00:18:12,790 diagram. 331 00:18:12,790 --> 00:18:15,310 In the i direction, you've got this. 332 00:18:15,310 --> 00:18:16,750 You've got this. 333 00:18:16,750 --> 00:18:21,330 You have the normal forces gravity don't contribute. 334 00:18:21,330 --> 00:18:28,700 And components of this and this have to be added in. 335 00:18:28,700 --> 00:18:32,650 You have to do an F2, probably cosine theta, and an F1 sine 336 00:18:32,650 --> 00:18:33,960 theta, and add them in. 337 00:18:33,960 --> 00:18:35,460 Because they're the external forces. 338 00:18:35,460 --> 00:18:37,710 So you've got all these summation of forces 339 00:18:37,710 --> 00:18:39,390 on that side. 340 00:18:39,390 --> 00:18:41,560 That gives you one equation. 341 00:18:41,560 --> 00:18:46,530 This will give you an equation for the motion 342 00:18:46,530 --> 00:18:47,900 of the first mass. 343 00:18:47,900 --> 00:18:49,950 And it'll have some unknowns in it. 344 00:18:49,950 --> 00:18:50,700 And what are they? 345 00:18:53,510 --> 00:18:56,205 Describe them. 346 00:18:56,205 --> 00:18:57,580 So you're going to get a function 347 00:18:57,580 --> 00:19:00,640 over here of a couple unknowns. 348 00:19:00,640 --> 00:19:03,491 This is going to be a function of what that you don't know? 349 00:19:03,491 --> 00:19:04,892 AUDIENCE: [INAUDIBLE] 350 00:19:04,892 --> 00:19:08,840 PROFESSOR: F1 and F2, and then also x 351 00:19:08,840 --> 00:19:12,366 double dot and thetas and theta dots and so forth, 352 00:19:12,366 --> 00:19:13,740 all the variables in the problem. 353 00:19:13,740 --> 00:19:15,114 But these are two unknowns you've 354 00:19:15,114 --> 00:19:16,800 got to ultimately get rid of. 355 00:19:16,800 --> 00:19:18,870 OK, what else are you going to do in method one? 356 00:19:22,132 --> 00:19:24,090 You've got to deal with the second mass, right? 357 00:19:24,090 --> 00:19:25,300 What would you do then? 358 00:19:25,300 --> 00:19:26,175 AUDIENCE: [INAUDIBLE] 359 00:19:31,420 --> 00:19:34,360 PROFESSOR: So some of the force is now on M2, for sure. 360 00:19:34,360 --> 00:19:37,120 And that's going to give you a couple equations. 361 00:19:37,120 --> 00:19:40,020 What else do you do in your method? 362 00:19:40,020 --> 00:19:42,241 AUDIENCE: [INAUDIBLE] 363 00:19:42,241 --> 00:19:43,990 PROFESSOR: [INAUDIBLE] torques about what? 364 00:19:47,490 --> 00:19:50,232 AUDIENCE: [INAUDIBLE] 365 00:19:50,232 --> 00:19:51,440 PROFESSOR: Have I named them? 366 00:19:51,440 --> 00:19:53,820 Look at that diagram there. 367 00:19:53,820 --> 00:19:56,460 So you're going to do your torques, sum 368 00:19:56,460 --> 00:20:00,020 of the torques, about A, OK? 369 00:20:00,020 --> 00:20:06,690 And if you sum torques about A, what are they? 370 00:20:06,690 --> 00:20:10,470 I'll let you guys pursue that a little bit, figure out, 371 00:20:10,470 --> 00:20:17,890 what are the torques about this point in this problem? 372 00:20:17,890 --> 00:20:20,040 OK, what do you get? 373 00:20:20,040 --> 00:20:21,670 Sum of the torques about A? 374 00:20:26,090 --> 00:20:29,430 z is in which-- which positive z? 375 00:20:29,430 --> 00:20:31,960 x, y, positive z coming out of the board. 376 00:20:31,960 --> 00:20:36,060 What produces torque about this point in this problem? 377 00:20:36,060 --> 00:20:36,940 Just what? 378 00:20:36,940 --> 00:20:37,690 AUDIENCE: Gravity. 379 00:20:37,690 --> 00:20:38,690 PROFESSOR: Just gravity. 380 00:20:38,690 --> 00:20:40,822 And is it positive or negative? 381 00:20:40,822 --> 00:20:44,390 And it's a component of gravity perpendicular to this. 382 00:20:44,390 --> 00:20:49,340 So it's probably sine theta M2g sine theta in the minus. 383 00:20:49,340 --> 00:20:57,220 So it's minus M2g L over 2-- you need the moment arm-- sine 384 00:20:57,220 --> 00:21:01,030 theta in the k hat direction. 385 00:21:01,030 --> 00:21:17,840 And that's going to be equal to the angular dH, dT about A. 386 00:21:17,840 --> 00:21:20,310 Sum of the torques, time rate of change of the angular 387 00:21:20,310 --> 00:21:21,960 momentum about A-- because that's what 388 00:21:21,960 --> 00:21:23,580 you've chosen to work about. 389 00:21:23,580 --> 00:21:27,010 And you always have this term you have to check on. 390 00:21:30,460 --> 00:21:32,770 OK, so that's the method. 391 00:21:32,770 --> 00:21:34,530 This is method one. 392 00:21:34,530 --> 00:21:37,730 I'll call this the tau with respect to A method. 393 00:21:40,940 --> 00:21:42,560 You have to write this equation. 394 00:21:42,560 --> 00:21:47,700 You have two unknowns that pop up in it. 395 00:21:47,700 --> 00:21:49,120 You go to the second mass. 396 00:21:49,120 --> 00:21:51,180 You write the F equals ma for that, 397 00:21:51,180 --> 00:21:54,150 and you're going to get two equations. 398 00:21:54,150 --> 00:21:56,520 If you do it in the local coordinate system, 399 00:21:56,520 --> 00:21:59,590 in the rotating one, you get mass times acceleration 400 00:21:59,590 --> 00:22:01,210 in the i direction. 401 00:22:01,210 --> 00:22:04,970 You get F1 and a component of gravity. 402 00:22:04,970 --> 00:22:08,610 And in the j direction, m2 acceleration in j, 403 00:22:08,610 --> 00:22:11,730 you get F2 and a piece that's acceleration of gravity. 404 00:22:16,610 --> 00:22:18,600 You don't know this yet. 405 00:22:18,600 --> 00:22:20,310 But you can find it. 406 00:22:20,310 --> 00:22:24,520 You know how to find those velocities now. 407 00:22:24,520 --> 00:22:25,650 You've done it before. 408 00:22:25,650 --> 00:22:30,440 So the velocity of G with respect to O, 409 00:22:30,440 --> 00:22:32,230 probably going to have an X double 410 00:22:32,230 --> 00:22:38,930 dot I hat plus something that will be like L/2 theta dot, 411 00:22:38,930 --> 00:22:40,995 and in what direction? 412 00:22:40,995 --> 00:22:41,495 Hmm? 413 00:22:44,080 --> 00:22:49,130 It's rotating like this-- j hat, little j. 414 00:22:49,130 --> 00:22:50,350 So there's your velocity. 415 00:22:50,350 --> 00:22:54,030 You can take the time derivative of that to get an acceleration, 416 00:22:54,030 --> 00:22:56,370 X double dot-- that's not true. 417 00:22:56,370 --> 00:22:58,170 Down here, you get an X double dot. 418 00:22:58,170 --> 00:22:59,580 In this one, you get two terms. 419 00:22:59,580 --> 00:23:01,580 Because that's a rotating piece. 420 00:23:01,580 --> 00:23:06,200 You get a centripetal term and an Euler term 421 00:23:06,200 --> 00:23:07,410 that come out of that. 422 00:23:07,410 --> 00:23:10,660 So then now you know velocities and accelerations. 423 00:23:10,660 --> 00:23:13,320 You can put the accelerations on this side. 424 00:23:13,320 --> 00:23:15,060 You can solve this first equation 425 00:23:15,060 --> 00:23:18,645 for F1, second equation for F2. 426 00:23:18,645 --> 00:23:21,340 You take those and put them in here. 427 00:23:21,340 --> 00:23:25,110 And you finally have one of your equations of motion. 428 00:23:25,110 --> 00:23:32,590 Down here, the external torque only involves gravity. 429 00:23:32,590 --> 00:23:36,150 So do your F1 and F2 appear-- do you have to mess with those 430 00:23:36,150 --> 00:23:43,450 in this final equation or not? 431 00:23:43,450 --> 00:23:47,800 This is the reason you use point A. This equation does not 432 00:23:47,800 --> 00:23:49,220 involve F1 or F2. 433 00:23:49,220 --> 00:23:51,860 There's no moments created by F1 and F2. 434 00:23:51,860 --> 00:23:55,120 The stuff on the right hand side is only kinematics and mass 435 00:23:55,120 --> 00:23:57,890 moments of inertia. 436 00:23:57,890 --> 00:24:00,260 This thing will give you an i with respect-- 437 00:24:00,260 --> 00:24:03,260 you can think of this as it's going to give you 438 00:24:03,260 --> 00:24:07,450 an i with respect to G theta double dot 439 00:24:07,450 --> 00:24:10,210 plus some other terms. 440 00:24:10,210 --> 00:24:12,145 And I want to talk about this. 441 00:24:12,145 --> 00:24:13,410 Does this term go to 0? 442 00:24:17,200 --> 00:24:20,210 Is the velocity of A 0 in this problem? 443 00:24:20,210 --> 00:24:20,960 No way. 444 00:24:20,960 --> 00:24:23,800 And in what direction is it? 445 00:24:23,800 --> 00:24:27,450 Horizontal, right, capital I hat. 446 00:24:27,450 --> 00:24:29,570 P, the momentum of the second thing, 447 00:24:29,570 --> 00:24:33,170 has these velocity components. 448 00:24:33,170 --> 00:24:36,330 The P is equal to m times that. 449 00:24:36,330 --> 00:24:38,520 And you have velocity components in the I, 450 00:24:38,520 --> 00:24:39,960 which is the same as vA. 451 00:24:39,960 --> 00:24:43,440 But you have another velocity component that's not the same. 452 00:24:43,440 --> 00:24:45,890 So is that cross product going to be 0? 453 00:24:45,890 --> 00:24:46,930 No way. 454 00:24:46,930 --> 00:24:49,140 You have to deal with that term. 455 00:24:49,140 --> 00:24:53,780 It happens that you can express H with respect 456 00:24:53,780 --> 00:25:02,440 to A as H with respect to G plus rG A cross P. 457 00:25:02,440 --> 00:25:07,440 And you know P, because you just found v. It's mv. 458 00:25:07,440 --> 00:25:14,140 And r, rG A, is just L over 2 j, little j, 459 00:25:14,140 --> 00:25:16,700 which is from here to here. 460 00:25:16,700 --> 00:25:21,896 It's L over 2 little-- no i, excuse me. 461 00:25:21,896 --> 00:25:23,920 So that's r, L over 2 i. 462 00:25:23,920 --> 00:25:27,239 So you can crank out this cross product. 463 00:25:27,239 --> 00:25:29,530 Then you have to take the time derivative of this right 464 00:25:29,530 --> 00:25:31,560 here, this whole thing. 465 00:25:31,560 --> 00:25:38,340 Every time when you take the time derivative of this part, 466 00:25:38,340 --> 00:25:41,000 you will get a P. So you get multiple pieces. 467 00:25:41,000 --> 00:25:43,150 But one of them will be minus that. 468 00:25:45,920 --> 00:25:48,940 This thing creates a piece that's exactly minus that, 469 00:25:48,940 --> 00:25:50,300 and they'll cancel. 470 00:25:50,300 --> 00:25:54,290 But you've got to go through the cranking it out 471 00:25:54,290 --> 00:25:55,930 till you get to that point. 472 00:25:55,930 --> 00:25:58,050 OK, that's one method of doing it. 473 00:25:58,050 --> 00:26:01,370 Let's talk briefly about a second method. 474 00:26:01,370 --> 00:26:03,620 What is the other way to go about doing this problem? 475 00:26:07,564 --> 00:26:09,540 AUDIENCE: [INAUDIBLE] 476 00:26:09,540 --> 00:26:12,330 PROFESSOR: That's method three. 477 00:26:12,330 --> 00:26:15,452 We're going to hurry so I can get to it. 478 00:26:15,452 --> 00:26:17,785 What's a second way to do this problem, a close relative 479 00:26:17,785 --> 00:26:18,750 of the first way? 480 00:26:21,956 --> 00:26:24,630 But I want you to have all these different ways 481 00:26:24,630 --> 00:26:26,320 of doing it in your tool kit. 482 00:26:26,320 --> 00:26:29,490 Because some problems are easy, more easily done. 483 00:26:29,490 --> 00:26:33,050 There's another way than this. 484 00:26:33,050 --> 00:26:37,090 This is perfectly appropriate for this type of problem. 485 00:26:37,090 --> 00:26:44,890 If the problem we were doing involved this, 486 00:26:44,890 --> 00:26:46,520 would you do this method? 487 00:26:46,520 --> 00:26:48,460 What would you do? 488 00:26:48,460 --> 00:26:49,400 AUDIENCE: [INAUDIBLE] 489 00:26:49,400 --> 00:26:50,900 PROFESSOR: No, can't use energy yet. 490 00:26:50,900 --> 00:26:51,920 That's next week. 491 00:26:51,920 --> 00:26:52,860 AUDIENCE: [INAUDIBLE] 492 00:26:52,860 --> 00:26:54,405 PROFESSOR: Ahh, right. 493 00:26:54,405 --> 00:26:57,270 So the second method is basically 494 00:26:57,270 --> 00:26:59,900 this is the same, this piece. 495 00:26:59,900 --> 00:27:04,200 But the second part-- and this is the same, 496 00:27:04,200 --> 00:27:05,740 sum of the forces about M2. 497 00:27:05,740 --> 00:27:07,600 You do both of those forces. 498 00:27:07,600 --> 00:27:11,460 You do sum forces for sure, M1 and M2. 499 00:27:11,460 --> 00:27:17,838 But the third step is you do sum of torques about g. 500 00:27:17,838 --> 00:27:26,600 If you sum torques about g, which is here, 501 00:27:26,600 --> 00:27:29,720 what appears in the sum torques? 502 00:27:29,720 --> 00:27:31,343 Does Mg contribute? 503 00:27:31,343 --> 00:27:35,021 No, what are the external torques about g? 504 00:27:35,021 --> 00:27:36,452 AUDIENCE: [INAUDIBLE] 505 00:27:36,452 --> 00:27:43,550 PROFESSOR: F2, L/2, positive F2 L/2 k. 506 00:27:43,550 --> 00:27:47,390 So that equation, summing torques about g, 507 00:27:47,390 --> 00:27:51,030 will give you an F2 L/2 k. 508 00:27:51,030 --> 00:27:53,310 And now you've got that unknown popping up 509 00:27:53,310 --> 00:27:54,830 in your torque equation. 510 00:27:54,830 --> 00:27:58,020 And you know that it also pops up in these two equations. 511 00:27:58,020 --> 00:28:01,140 Because we solved them. 512 00:28:01,140 --> 00:28:07,030 So now why does this method have a slight advantage 513 00:28:07,030 --> 00:28:09,162 over this method? 514 00:28:09,162 --> 00:28:10,940 AUDIENCE: [INAUDIBLE] 515 00:28:10,940 --> 00:28:13,641 PROFESSOR: Well, you don't get F2 or F1 popping up 516 00:28:13,641 --> 00:28:14,640 in your torque equation. 517 00:28:14,640 --> 00:28:16,680 You immediately get to the answer for that one. 518 00:28:16,680 --> 00:28:18,013 So it's a little more efficient. 519 00:28:24,490 --> 00:28:32,800 So is there a way to get directly 520 00:28:32,800 --> 00:28:36,870 at the two equations we're after? 521 00:28:36,870 --> 00:28:41,950 The second equation we get from this torque thing. 522 00:28:41,950 --> 00:28:45,880 The first equation we get from resolving this. 523 00:28:45,880 --> 00:28:50,510 This is a mass times-- M1 X double dot. 524 00:28:50,510 --> 00:28:52,680 And you get this function of F1 and F2. 525 00:28:52,680 --> 00:28:54,920 And you have to substitute in for these F1's 526 00:28:54,920 --> 00:28:58,060 and F2's until you finally get that equation. 527 00:28:58,060 --> 00:29:00,570 So that's kind of tedious. 528 00:29:00,570 --> 00:29:05,230 So is there a way to get directly at this without having 529 00:29:05,230 --> 00:29:07,700 to mess with F1 and F2 at all? 530 00:29:07,700 --> 00:29:10,670 And that's what is the key to what you're talking about. 531 00:29:10,670 --> 00:29:11,980 Let's look at something here. 532 00:29:11,980 --> 00:29:20,360 Let's do the vector sum of the forces on M1. 533 00:29:24,270 --> 00:29:29,170 We get an F. Let's look at our free body diagram, 534 00:29:29,170 --> 00:29:47,920 F2j plus F1i minus kx minus bx dot and a minus M1g. 535 00:29:47,920 --> 00:29:49,920 Well, those are-- well, we'll put them in there, 536 00:29:49,920 --> 00:29:52,600 at least throw them in there. 537 00:29:52,600 --> 00:30:04,370 This is in my capital J plus N1 plus N2 capital J hat here. 538 00:30:04,370 --> 00:30:07,320 Is that all the forces, all the vector forces in that problem? 539 00:30:12,630 --> 00:30:13,860 Did I miss anything? 540 00:30:13,860 --> 00:30:16,400 I've got spring, the dashpot, the forces 541 00:30:16,400 --> 00:30:19,614 caused by the rod, gravity, and the normal forces. 542 00:30:19,614 --> 00:30:32,490 OK, and we could figure out the component of this, 543 00:30:32,490 --> 00:30:37,150 the sum of the forces on M1 in the capital X direction. 544 00:30:37,150 --> 00:30:39,200 We could figure that out from-- this 545 00:30:39,200 --> 00:30:41,450 has component in that direction. 546 00:30:41,450 --> 00:30:43,160 It just involves thetas. 547 00:30:43,160 --> 00:30:44,195 These go away, then. 548 00:30:44,195 --> 00:30:46,570 You don't have to deal with this stuff in the x equation. 549 00:30:49,950 --> 00:30:52,217 But there's our sum of the forces on the first body. 550 00:30:52,217 --> 00:30:54,300 Let's do the sum of the forces on the second body. 551 00:30:57,870 --> 00:30:58,930 What are they? 552 00:31:04,056 --> 00:31:05,460 AUDIENCE: Gravity. 553 00:31:05,460 --> 00:31:10,800 PROFESSOR: OK, so you've got a minus M2g on it. 554 00:31:10,800 --> 00:31:12,120 And what else? 555 00:31:15,992 --> 00:31:17,882 AUDIENCE: [INAUDIBLE] 556 00:31:17,882 --> 00:31:19,590 PROFESSOR: But with opposite sign, right? 557 00:31:19,590 --> 00:31:25,710 F2 little j minus F1 little i, and anything else? 558 00:31:28,330 --> 00:31:29,570 Nothing, right? 559 00:31:29,570 --> 00:31:30,700 Let's add the two together. 560 00:31:33,570 --> 00:31:39,020 Oh, and this one has got to be equal to the mass, M1, 561 00:31:39,020 --> 00:31:47,720 times the acceleration, the x component I plus M1 562 00:31:47,720 --> 00:31:53,380 acceleration in the-- I can't write over here-- ay 563 00:31:53,380 --> 00:31:58,250 in the capital J hat direction. 564 00:31:58,250 --> 00:31:59,940 That gives us mass times acceleration, 565 00:31:59,940 --> 00:32:02,790 is what all the sums of the forces is equal to. 566 00:32:02,790 --> 00:32:09,180 The same thing here-- this is equal to the little mass, M2, 567 00:32:09,180 --> 00:32:11,250 times its acceleration. 568 00:32:11,250 --> 00:32:19,500 And it'll have components in the-- this is mass 2. 569 00:32:19,500 --> 00:32:23,680 And it'll have components we can put in the I hat direction, 570 00:32:23,680 --> 00:32:33,720 and plus M2 acceleration 2-- maybe I'll do it like this. 571 00:32:37,380 --> 00:32:40,430 a2y, and this is in the j direction. 572 00:32:40,430 --> 00:32:45,090 And those two things we can add together. 573 00:32:45,090 --> 00:32:53,148 And so if I add them together, and I put M1 plus M2 times-- 574 00:32:53,148 --> 00:32:57,151 I'll just call it the total acceleration. 575 00:32:57,151 --> 00:32:58,850 Well, I don't want to do that. 576 00:32:58,850 --> 00:33:00,780 I don't want to say this. 577 00:33:07,362 --> 00:33:09,320 I'm hurrying because we're running out of time. 578 00:33:09,320 --> 00:33:17,850 M1 times its acceleration times M2 plus its acceleration 579 00:33:17,850 --> 00:33:20,740 is equal to-- and now we sum up everything on the right hand 580 00:33:20,740 --> 00:33:22,610 side. 581 00:33:22,610 --> 00:33:25,440 And the key thing that happens here 582 00:33:25,440 --> 00:33:26,780 is what happens at F1 and F2? 583 00:33:26,780 --> 00:33:27,610 AUDIENCE: Drop out. 584 00:33:27,610 --> 00:33:29,766 PROFESSOR: They completely drop out. 585 00:33:33,080 --> 00:33:40,630 So on the right hand side, it's not a function of F1 and F2. 586 00:33:47,050 --> 00:33:51,880 The equating of motion that we were trying to get here 587 00:33:51,880 --> 00:33:54,270 was just in the I direction. 588 00:33:54,270 --> 00:33:56,810 That's all we need for that big cart. 589 00:33:56,810 --> 00:33:59,460 So if we go down here and extract 590 00:33:59,460 --> 00:34:06,830 the I component of this total sum, it'll be an equation. 591 00:34:06,830 --> 00:34:08,710 It'll involve M1 and M2. 592 00:34:08,710 --> 00:34:15,760 And it'll have cosines and sines and thetas and things like 593 00:34:15,760 --> 00:34:17,860 that, but no F1's, no F2's. 594 00:34:17,860 --> 00:34:24,090 You will get directly to this, the final result that you got 595 00:34:24,090 --> 00:34:26,250 here where you had to go in and substitute 596 00:34:26,250 --> 00:34:29,376 in for F1's and F2's. 597 00:34:29,376 --> 00:34:30,709 So it'll get it to you directly. 598 00:34:33,790 --> 00:34:36,420 What do you need to solve it, though? 599 00:34:36,420 --> 00:34:38,540 You need to know the acceleration of mass 1. 600 00:34:41,070 --> 00:34:41,900 You know that. 601 00:34:41,900 --> 00:34:43,340 That's trivial. 602 00:34:43,340 --> 00:34:47,530 You need to know the acceleration of mass 2. 603 00:34:47,530 --> 00:34:52,400 And that you did by taking the time 604 00:34:52,400 --> 00:34:53,580 derivatives of these terms. 605 00:34:53,580 --> 00:34:56,210 You know the acceleration of that mass. 606 00:34:56,210 --> 00:34:58,870 So you need that expression. 607 00:34:58,870 --> 00:35:02,190 And you plug that in here. 608 00:35:02,190 --> 00:35:04,770 And everything else in the right hand side you know. 609 00:35:04,770 --> 00:35:11,022 It's kx dot, bx dot, minus Mg, M1g minus M2g, and so forth. 610 00:35:11,022 --> 00:35:12,480 All the rest of the stuff is known. 611 00:35:12,480 --> 00:35:12,770 Yeah. 612 00:35:12,770 --> 00:35:15,170 AUDIENCE: Does this give you three equations of motion? 613 00:35:15,170 --> 00:35:17,003 PROFESSOR: Well, it gives you two equations. 614 00:35:18,980 --> 00:35:22,120 At this point, you'd break this into capital I and capital 615 00:35:22,120 --> 00:35:26,280 J. The one in the capital J direction is equal to 0. 616 00:35:26,280 --> 00:35:27,770 There's no motion in that direction 617 00:35:27,770 --> 00:35:30,490 for the main mass, the cart. 618 00:35:30,490 --> 00:35:36,990 There's only motion in the inertial frame x direction. 619 00:35:36,990 --> 00:35:39,417 So that's all. 620 00:35:39,417 --> 00:35:40,500 This is a vector equation. 621 00:35:40,500 --> 00:35:46,535 It has two pieces to it, I and J. The J is a static equation. 622 00:35:46,535 --> 00:35:48,310 The I is your dynamic one. 623 00:35:48,310 --> 00:35:52,250 And it's this one that you would have found in this problem, 624 00:35:52,250 --> 00:35:55,630 but without ever having to solve for F1 and F2. 625 00:35:55,630 --> 00:36:00,700 But you do have to go out and find the acceleration 626 00:36:00,700 --> 00:36:02,910 of both masses. 627 00:36:02,910 --> 00:36:06,050 Mass times acceleration, what this is really saying-- this 628 00:36:06,050 --> 00:36:08,380 is what you asked about the system. 629 00:36:08,380 --> 00:36:09,940 This is sort of the system approach. 630 00:36:09,940 --> 00:36:15,230 If you think of this whole thing as a system, 631 00:36:15,230 --> 00:36:20,080 draw a box around it, the sum of the external forces 632 00:36:20,080 --> 00:36:24,910 on the system is equal to the mass, 633 00:36:24,910 --> 00:36:30,650 the total mass, times actually the acceleration of the center 634 00:36:30,650 --> 00:36:32,620 of mass of the system. 635 00:36:32,620 --> 00:36:35,250 But this acceleration times the center 636 00:36:35,250 --> 00:36:38,480 of mass of the system, you can break this into two pieces. 637 00:36:38,480 --> 00:36:45,040 It is this term plus this term. 638 00:36:45,040 --> 00:36:49,080 And then that's the total system. 639 00:36:49,080 --> 00:36:52,170 And all you have to put on this side 640 00:36:52,170 --> 00:36:56,170 is the summation of the external forces. 641 00:36:56,170 --> 00:37:00,060 But the external forces are not-- the internal forces 642 00:37:00,060 --> 00:37:00,919 don't count. 643 00:37:00,919 --> 00:37:02,210 They're not part of the system. 644 00:37:02,210 --> 00:37:04,100 So you only have to put the external forces 645 00:37:04,100 --> 00:37:10,040 on it-- Mg's springs, dashpots, normal forces, outside forces, 646 00:37:10,040 --> 00:37:13,820 the only ones that go over here. 647 00:37:13,820 --> 00:37:16,450 The reason we don't usually think of doing this 648 00:37:16,450 --> 00:37:18,470 is because when you think about the system, 649 00:37:18,470 --> 00:37:23,240 you write it as the acceleration of the center of mass. 650 00:37:23,240 --> 00:37:25,270 But you have to realize that you can 651 00:37:25,270 --> 00:37:29,680 get that expression as the sum of the parts, the sum 652 00:37:29,680 --> 00:37:32,970 of each bit times its acceleration, 653 00:37:32,970 --> 00:37:35,110 and sum them all up, is the same answer 654 00:37:35,110 --> 00:37:37,900 as the sum of the masses times the acceleration 655 00:37:37,900 --> 00:37:41,240 of the center of mass of the whole system. 656 00:37:41,240 --> 00:37:44,275 So we've still got a couple of minutes. 657 00:37:44,275 --> 00:37:46,400 Last hour, I ran over, and we couldn't get to this. 658 00:37:46,400 --> 00:37:50,010 So there is, in a way, yet another thing you can do. 659 00:37:50,010 --> 00:37:52,770 And that's think of the thing as a whole system. 660 00:37:52,770 --> 00:37:54,520 And sometimes that'll give you an equation 661 00:37:54,520 --> 00:37:56,540 without having to use F1, without having 662 00:37:56,540 --> 00:37:59,110 to solve for internal forces-- pretty cool. 663 00:38:04,170 --> 00:38:09,980 All right, so questions, thoughts? 664 00:38:09,980 --> 00:38:10,650 Yeah. 665 00:38:10,650 --> 00:38:13,548 AUDIENCE: Is it safe to say that if we 666 00:38:13,548 --> 00:38:17,090 have two more rigid bodies, we should probably consider it 667 00:38:17,090 --> 00:38:19,830 as a whole system [INAUDIBLE]? 668 00:38:19,830 --> 00:38:22,930 PROFESSOR: Well, I can't generalize on that. 669 00:38:22,930 --> 00:38:25,980 I haven't done enough problems like that myself to even 670 00:38:25,980 --> 00:38:30,330 have enough experience to know just how often is 671 00:38:30,330 --> 00:38:31,455 that going to help you out. 672 00:38:34,780 --> 00:38:37,500 This one, there's a pretty obvious equation 673 00:38:37,500 --> 00:38:40,950 that you're going to write in this direction on that mass. 674 00:38:40,950 --> 00:38:45,360 It's constrained to a single motion and a single coordinate 675 00:38:45,360 --> 00:38:46,930 you're assigning to it. 676 00:38:46,930 --> 00:38:48,870 And probably in cases like that where 677 00:38:48,870 --> 00:38:51,730 you can isolate one of the rigid bodies, 678 00:38:51,730 --> 00:38:54,436 there might be some advantage in doing something like this. 679 00:38:54,436 --> 00:38:56,102 AUDIENCE: --more difficult to figure out 680 00:38:56,102 --> 00:38:58,440 what the accelerations for both rigid bodies was, 681 00:38:58,440 --> 00:39:01,210 then it would be better [INAUDIBLE]. 682 00:39:01,210 --> 00:39:04,890 PROFESSOR: Yeah, this one, all of these methods 683 00:39:04,890 --> 00:39:08,880 you're going to end up having to compute the velocities 684 00:39:08,880 --> 00:39:11,720 and accelerations of each piece. 685 00:39:11,720 --> 00:39:13,400 And if you've got that information, 686 00:39:13,400 --> 00:39:17,770 you may as well use it if there's 687 00:39:17,770 --> 00:39:24,020 something to be gained by doing a clever step like this. 688 00:39:24,020 --> 00:39:24,827 Yeah. 689 00:39:24,827 --> 00:39:27,326 AUDIENCE: What confuses me is that method one, method three, 690 00:39:27,326 --> 00:39:29,410 there's no-- so you said there's going 691 00:39:29,410 --> 00:39:30,535 to be two things in motion. 692 00:39:30,535 --> 00:39:31,680 But why is it-- 693 00:39:31,680 --> 00:39:33,410 PROFESSOR: So this, this method three, 694 00:39:33,410 --> 00:39:37,940 this shows you a different way to get that equation. 695 00:39:37,940 --> 00:39:39,900 You still need this. 696 00:39:39,900 --> 00:39:42,260 AUDIENCE: Oh, this is just for the first-- 697 00:39:42,260 --> 00:39:45,160 PROFESSOR: This is a clever way to get just this first equation 698 00:39:45,160 --> 00:39:46,810 down. 699 00:39:46,810 --> 00:39:49,240 You still need a second equation. 700 00:39:49,240 --> 00:39:50,890 And you still are going to have to-- 701 00:39:50,890 --> 00:39:53,780 and then you don't want to have to mess with F1 and F2. 702 00:39:53,780 --> 00:39:55,950 So we've just found a way to never have 703 00:39:55,950 --> 00:39:59,010 to mess with F1 and F2 in at least this problem. 704 00:39:59,010 --> 00:40:00,790 And that is get the first equation 705 00:40:00,790 --> 00:40:04,370 by using this trick, the system equation. 706 00:40:04,370 --> 00:40:07,500 And it's a sum forces expression. 707 00:40:07,500 --> 00:40:11,790 And then the second equation you need is this moment equation. 708 00:40:11,790 --> 00:40:15,070 And you have two choices, about G or about A. 709 00:40:15,070 --> 00:40:17,400 And which one can you do it without having 710 00:40:17,400 --> 00:40:19,760 to find F1 and F2? 711 00:40:19,760 --> 00:40:22,784 I'm doing it about A. So there's a way 712 00:40:22,784 --> 00:40:24,200 that you can do this whole problem 713 00:40:24,200 --> 00:40:28,130 and never have to address what F1 and F2 are. 714 00:40:28,130 --> 00:40:32,490 And that's kind of a combination of these two pieces here. 715 00:40:32,490 --> 00:40:34,740 The traditional way, the textbook way, you open it up, 716 00:40:34,740 --> 00:40:36,510 the textbook will teach you this one. 717 00:40:36,510 --> 00:40:39,776 And it'll teach you about A and teach you about G. 718 00:40:39,776 --> 00:40:44,560 But you don't often run into-- there's sometimes 719 00:40:44,560 --> 00:40:49,890 a direct way of doing it without finding the F1 and F2. 720 00:40:52,704 --> 00:40:54,120 Unfortunately, there are thousands 721 00:40:54,120 --> 00:40:55,161 of problems in the world. 722 00:40:55,161 --> 00:41:01,400 And it's a little hard to say in advance what method to use 723 00:41:01,400 --> 00:41:03,310 and what's going to work the easiest. 724 00:41:03,310 --> 00:41:05,990 But there's some insight. 725 00:41:05,990 --> 00:41:07,820 There's some generalizations that you 726 00:41:07,820 --> 00:41:13,020 can get from looking at things in these different ways. 727 00:41:13,020 --> 00:41:18,140 If you have forces at a pivot point, 728 00:41:18,140 --> 00:41:20,360 taking the moments of the pivot point, 729 00:41:20,360 --> 00:41:23,070 unknowns of the pivot point, taking the moments 730 00:41:23,070 --> 00:41:25,290 about that point will get rid of them for you. 731 00:41:25,290 --> 00:41:27,400 That's the one generalization that you can usually 732 00:41:27,400 --> 00:41:28,170 take to the bank. 733 00:41:31,720 --> 00:41:35,310 But beyond that-- and if the thing's unconstrained. 734 00:41:35,310 --> 00:41:38,960 So if it's fixed axis rotation, then you 735 00:41:38,960 --> 00:41:42,100 want to use about the axis of fixed axis. 736 00:41:42,100 --> 00:41:44,190 This is a moving axis. 737 00:41:44,190 --> 00:41:48,999 But it is still rotating about a fixed point that's moving. 738 00:41:48,999 --> 00:41:50,790 There's still some advantage of doing this. 739 00:41:54,270 --> 00:41:59,400 If it's the eraser problem, this thing, 740 00:41:59,400 --> 00:42:02,330 where you have no fixed points of rotation-- 741 00:42:02,330 --> 00:42:06,070 if you take a body, and you do this with it, 742 00:42:06,070 --> 00:42:08,270 where is it spinning about? 743 00:42:08,270 --> 00:42:10,295 Where is its axis of rotation? 744 00:42:10,295 --> 00:42:11,170 AUDIENCE: [INAUDIBLE] 745 00:42:11,170 --> 00:42:12,890 PROFESSOR: Always, right? 746 00:42:12,890 --> 00:42:15,130 That you can take to the bank. 747 00:42:15,130 --> 00:42:19,780 If there's no forced pivot, bodies on their own 748 00:42:19,780 --> 00:42:24,000 only rotate about their centers of gravity, centers of mass. 749 00:42:24,000 --> 00:42:26,530 And then when you have free bodies that 750 00:42:26,530 --> 00:42:30,250 don't have constraints that are forcing them to move 751 00:42:30,250 --> 00:42:32,800 with respect to something, if there's no constraints, 752 00:42:32,800 --> 00:42:36,350 then you want to take the torques about G for sure. 753 00:42:40,190 --> 00:42:47,140 Good, all right, see you guys next Tuesday.