1 00:00:00,070 --> 00:00:02,430 The following content is provided under a Creative 2 00:00:02,430 --> 00:00:03,800 Commons license. 3 00:00:03,800 --> 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 --> 00:00:17,251 at ocw.mit.edu. 8 00:00:21,040 --> 00:00:23,540 PROFESSOR: Last, kind of under the announcements category, 9 00:00:23,540 --> 00:00:25,990 is I want to talk about the muddy cards. 10 00:00:25,990 --> 00:00:28,510 So I've used those many times in the past. 11 00:00:28,510 --> 00:00:30,870 Last time was the first time I handed them out. 12 00:00:30,870 --> 00:00:32,790 And your comments were great. 13 00:00:32,790 --> 00:00:38,445 This was really a good lecture to have handed them out. 14 00:00:38,445 --> 00:00:42,340 We covered lots of interesting important concepts. 15 00:00:42,340 --> 00:00:45,170 And so I'm going to review a couple of things that 16 00:00:45,170 --> 00:00:47,800 came up in the muddy cards. 17 00:00:47,800 --> 00:00:50,730 A couple of the most positive comments 18 00:00:50,730 --> 00:00:52,840 is people really like the demos, and they really 19 00:00:52,840 --> 00:00:56,535 like the explanations, especially with examples. 20 00:00:59,550 --> 00:01:01,320 People particularly commented it was 21 00:01:01,320 --> 00:01:05,180 really helpful to compute angular momentum from two 22 00:01:05,180 --> 00:01:07,010 different points. 23 00:01:07,010 --> 00:01:08,780 And you get the revelation that you 24 00:01:08,780 --> 00:01:12,030 get two very different answers. 25 00:01:12,030 --> 00:01:16,440 So that's a really important point. 26 00:01:16,440 --> 00:01:19,090 And somebody then asked the question, 27 00:01:19,090 --> 00:01:22,060 said, well, I thought that vectors 28 00:01:22,060 --> 00:01:25,450 were independent of the coordinate system 29 00:01:25,450 --> 00:01:27,270 that you select. 30 00:01:27,270 --> 00:01:28,080 It's true. 31 00:01:28,080 --> 00:01:29,620 A velocity ought to be a velocity 32 00:01:29,620 --> 00:01:32,660 no matter whether it's r theta or x, y, z. 33 00:01:32,660 --> 00:01:33,800 But why? 34 00:01:33,800 --> 00:01:37,990 That seems to kind of violate that notion that vectors should 35 00:01:37,990 --> 00:01:40,770 be independent of coordinates. 36 00:01:40,770 --> 00:01:43,800 And yet we computed an angular momentum with respect 37 00:01:43,800 --> 00:01:45,930 to one place and respect to a different place, 38 00:01:45,930 --> 00:01:47,260 and we got different answers. 39 00:01:51,460 --> 00:01:52,850 How do you resolve that? 40 00:01:55,836 --> 00:01:56,336 Yeah. 41 00:01:56,336 --> 00:01:59,822 AUDIENCE: It's sort of like since it's moving, 42 00:01:59,822 --> 00:02:01,316 the coordinates shouldn't matter. 43 00:02:01,316 --> 00:02:05,300 Like if it was equilibrium, it wouldn't matter where 44 00:02:05,300 --> 00:02:06,794 you would put the [INAUDIBLE]. 45 00:02:14,264 --> 00:02:17,750 PROFESSOR: OK, well, you're getting close. 46 00:02:17,750 --> 00:02:21,380 Here was the problem I think we did. 47 00:02:21,380 --> 00:02:24,700 And we chose this point, which I'll call 2, 48 00:02:24,700 --> 00:02:27,410 and this point, which I'll call 1. 49 00:02:27,410 --> 00:02:34,960 And we computed h1. 50 00:02:34,960 --> 00:02:37,740 I'll just call h1 with respect to point 1. 51 00:02:37,740 --> 00:02:40,970 We'll call this A. We computed with respect to 1. 52 00:02:40,970 --> 00:02:43,510 And we computed the angular momentum of A 53 00:02:43,510 --> 00:02:45,250 with respect to 2. 54 00:02:45,250 --> 00:02:50,250 But in this case, angular momentum of a particle with 55 00:02:50,250 --> 00:02:57,710 respect to some location, origin of a coordinate system, 56 00:02:57,710 --> 00:03:04,440 is defined as r of the partial with respect to the coordinate 57 00:03:04,440 --> 00:03:10,920 system crossed with the linear momentum of the particle that 58 00:03:10,920 --> 00:03:14,095 you're-- we'll call it B here, just the name of the particle. 59 00:03:17,430 --> 00:03:20,520 The definition-- these are both vectors. 60 00:03:20,520 --> 00:03:22,740 You arch changing the vector. 61 00:03:22,740 --> 00:03:24,030 It's a different vector. 62 00:03:24,030 --> 00:03:28,960 Because this changes in the two parts. 63 00:03:28,960 --> 00:03:31,860 So it's not a constant vector at all. 64 00:03:31,860 --> 00:03:33,960 By its definition, it is just something 65 00:03:33,960 --> 00:03:37,180 different when you move to a different place and this piece 66 00:03:37,180 --> 00:03:37,680 changes. 67 00:03:37,680 --> 00:03:40,700 This piece is invariant, but this piece is not. 68 00:03:40,700 --> 00:03:43,390 And that's the answer to that one. 69 00:03:43,390 --> 00:03:46,730 Lots of people were still not clear about Coriolis. 70 00:03:46,730 --> 00:03:50,090 We'll work on that as time goes by. 71 00:03:50,090 --> 00:03:56,530 And people were interested in how 72 00:03:56,530 --> 00:03:59,900 to pick reference frames and so forth. 73 00:03:59,900 --> 00:04:02,487 Somebody made the suggestion, try using some colored chalk. 74 00:04:02,487 --> 00:04:03,070 It would help. 75 00:04:03,070 --> 00:04:04,319 I don't own any colored chalk. 76 00:04:04,319 --> 00:04:06,594 My assistant just walked in with some. 77 00:04:06,594 --> 00:04:08,010 She found some at the last minute. 78 00:04:08,010 --> 00:04:08,980 So I'll try doing that. 79 00:04:08,980 --> 00:04:10,032 That's a good idea. 80 00:04:10,032 --> 00:04:11,990 And someone else says, take a break in an hour. 81 00:04:11,990 --> 00:04:13,850 And that's a pretty good idea, too. 82 00:04:13,850 --> 00:04:15,842 I'll try to remember to do that. 83 00:04:15,842 --> 00:04:17,050 So the muddy cards are great. 84 00:04:17,050 --> 00:04:21,890 Please today we'll do the same thing. 85 00:04:21,890 --> 00:04:27,690 So let's start with this topic. 86 00:04:31,750 --> 00:04:34,570 It's a subject which we constantly 87 00:04:34,570 --> 00:04:39,160 use throughout the course doing dynamics. 88 00:04:39,160 --> 00:04:40,880 You have to be able to figure out 89 00:04:40,880 --> 00:04:43,710 coordinate systems, degrees of freedom, 90 00:04:43,710 --> 00:04:45,070 drawing free body diagrams. 91 00:04:45,070 --> 00:04:52,930 So I'm going to do a few quick examples-- coordinates, fbd's. 92 00:04:56,140 --> 00:04:58,960 I picked some examples here just to emphasize 93 00:04:58,960 --> 00:05:02,010 a few different points. 94 00:05:02,010 --> 00:05:04,010 Here we have a slope. 95 00:05:04,010 --> 00:05:07,130 I've got a wheel. 96 00:05:07,130 --> 00:05:07,970 It's a rigid body. 97 00:05:12,230 --> 00:05:14,170 I pick a preliminary coordinate system. 98 00:05:14,170 --> 00:05:16,420 Sometimes you do that just to help you think about it. 99 00:05:19,050 --> 00:05:20,900 And now let's talk about degrees of freedom. 100 00:05:20,900 --> 00:05:22,730 What do we mean by degrees of freedom? 101 00:05:27,340 --> 00:05:39,640 I'm going to define it as the number of independent 102 00:05:39,640 --> 00:06:00,585 coordinates necessary to describe the motion. 103 00:06:05,640 --> 00:06:08,380 So it's the number of independent coordinates 104 00:06:08,380 --> 00:06:10,690 that you need. 105 00:06:10,690 --> 00:06:19,740 Now, with few exceptions, I can compute that 106 00:06:19,740 --> 00:06:28,270 by multiplying 6 times the number of rigid bodies 107 00:06:28,270 --> 00:06:33,750 in the problem plus 3 times the number of particles 108 00:06:33,750 --> 00:06:37,140 minus the constraints, the number of constraints. 109 00:06:37,140 --> 00:06:40,445 So this is the number of rigid bodies. 110 00:06:44,840 --> 00:06:47,885 This is the number of particles. 111 00:06:51,280 --> 00:06:53,205 And C, this is the constraints. 112 00:07:01,650 --> 00:07:03,650 So take a look at this problem. 113 00:07:03,650 --> 00:07:06,750 A wheel's a rigid body. 114 00:07:06,750 --> 00:07:08,970 The difference between a rigid body and a particle 115 00:07:08,970 --> 00:07:10,440 is a rigid body is big enough. 116 00:07:10,440 --> 00:07:12,600 It has mass at some extent. 117 00:07:12,600 --> 00:07:14,910 Its rotational inertia will matter. 118 00:07:14,910 --> 00:07:16,790 So here we've got a wheel. 119 00:07:16,790 --> 00:07:19,300 So we've got certainly one rigid body but no particles. 120 00:07:19,300 --> 00:07:21,676 So in this case, n is 1, m is 0. 121 00:07:21,676 --> 00:07:23,300 And so the number of degrees of freedom 122 00:07:23,300 --> 00:07:25,216 that we should come up with in this problem is 123 00:07:25,216 --> 00:07:28,620 going to look like 6 minus C. 124 00:07:28,620 --> 00:07:30,060 And so then the problem becomes-- 125 00:07:30,060 --> 00:07:32,470 let's identify what the constraints are in the problem. 126 00:07:36,070 --> 00:07:38,010 So I drew my initial little coordinate system 127 00:07:38,010 --> 00:07:42,150 just so that I can use language like x, y, and z direction. 128 00:07:42,150 --> 00:07:43,910 So z's coming out of the board. 129 00:07:43,910 --> 00:07:46,000 So in this problem, let's figure out 130 00:07:46,000 --> 00:07:47,330 how many constraints there are. 131 00:08:00,400 --> 00:08:02,555 How about the y direction? 132 00:08:05,709 --> 00:08:07,500 The simplest kind of constraints are things 133 00:08:07,500 --> 00:08:08,800 that just allow no motion. 134 00:08:08,800 --> 00:08:12,230 Can it move in the y direction? 135 00:08:12,230 --> 00:08:19,019 No, OK, so y, y dot, y double dot, are 0. 136 00:08:19,019 --> 00:08:20,185 So that's a hard constraint. 137 00:08:20,185 --> 00:08:22,555 You can't move through the wall in the y direction. 138 00:08:25,920 --> 00:08:30,971 You have to make assumptions when you're doing problems. 139 00:08:30,971 --> 00:08:33,179 You have to try to simplify things as much as you can 140 00:08:33,179 --> 00:08:34,303 so that you make them easy. 141 00:08:34,303 --> 00:08:37,500 So really I'm going to assume in this problem-- I 142 00:08:37,500 --> 00:08:41,210 haven't shown any constraints or wheels or guides or rollers. 143 00:08:41,210 --> 00:08:46,312 But I'm going to assume that it won't move in the z direction. 144 00:08:46,312 --> 00:08:47,520 So that's another constraint. 145 00:08:47,520 --> 00:08:49,740 So this implies 1. 146 00:08:49,740 --> 00:08:51,300 This implies another constraint. 147 00:08:56,090 --> 00:08:59,110 If I don't do this, if I don't assume that, 148 00:08:59,110 --> 00:09:02,370 then I just end up with another equation of motion. 149 00:09:02,370 --> 00:09:04,830 For every degree of freedom, you end up with a problem. 150 00:09:04,830 --> 00:09:07,180 You're going to need an equation of motion. 151 00:09:07,180 --> 00:09:09,020 So if I did not make this assumption, 152 00:09:09,020 --> 00:09:13,300 I'd say that the summation of the forces in the z direction 153 00:09:13,300 --> 00:09:18,000 is equal to 0. 154 00:09:18,000 --> 00:09:21,980 And that's equal to the mass times the acceleration 155 00:09:21,980 --> 00:09:25,670 of the body in the z acceleration. 156 00:09:25,670 --> 00:09:28,327 It's a vector, vector component. 157 00:09:28,327 --> 00:09:29,910 And then you just say, oh well, that's 158 00:09:29,910 --> 00:09:31,470 a trivial equation of motion. 159 00:09:31,470 --> 00:09:33,095 And so you could deal with it that way. 160 00:09:33,095 --> 00:09:36,110 But we'll just assume that we have no motion in the z. 161 00:09:36,110 --> 00:09:39,710 And that gives us another constraint. 162 00:09:39,710 --> 00:09:42,840 Now we can assume again that there's 163 00:09:42,840 --> 00:09:45,580 constraints in the problem, or it's well behaved, 164 00:09:45,580 --> 00:09:48,000 in that the thing won't fall over. 165 00:09:48,000 --> 00:09:49,910 It won't roll over. 166 00:09:49,910 --> 00:09:53,400 And it won't change direction running down the hill. 167 00:09:53,400 --> 00:10:06,410 So we'll assume no rotation about the x or y axes. 168 00:10:09,450 --> 00:10:11,665 So that implies two more constraints. 169 00:10:14,500 --> 00:10:19,560 And finally, the constraints can come in many flavors. 170 00:10:19,560 --> 00:10:22,800 Finally we know that in this problem 171 00:10:22,800 --> 00:10:36,570 that I need to think in terms of a rotation. 172 00:10:36,570 --> 00:10:40,120 So there's a positive rotation in that coordinate system. 173 00:10:40,120 --> 00:10:42,270 So now I have a rotational coordinate 174 00:10:42,270 --> 00:10:46,910 that I can think about. 175 00:10:46,910 --> 00:10:52,710 But this now says that if there's no slip, 176 00:10:52,710 --> 00:10:57,030 I can say that the distance it rolls down the hill 177 00:10:57,030 --> 00:11:00,490 is minus r theta. 178 00:11:00,490 --> 00:11:03,540 That's a constraint. 179 00:11:03,540 --> 00:11:07,640 x and theta are not independent of one another. 180 00:11:07,640 --> 00:11:11,840 We're looking for the number of independent coordinates 181 00:11:11,840 --> 00:11:15,040 required to completely describe the motion. 182 00:11:15,040 --> 00:11:17,840 x and r are not independent because of this no slip 183 00:11:17,840 --> 00:11:19,090 condition. 184 00:11:19,090 --> 00:11:22,500 And so that implies yet another. 185 00:11:22,500 --> 00:11:30,620 OK, so we've got one, two, three, four, five constraints. 186 00:11:30,620 --> 00:11:32,890 And we said that the number of degrees 187 00:11:32,890 --> 00:11:38,670 of freedom in this problem is equal to 6 minus 5, which is 1. 188 00:11:38,670 --> 00:11:40,120 So you take the single coordinate. 189 00:11:40,120 --> 00:11:42,305 You could have told me that long ago that that's 190 00:11:42,305 --> 00:11:43,570 what it's going to take. 191 00:11:43,570 --> 00:11:45,180 But this is the sort of thinking you 192 00:11:45,180 --> 00:11:48,195 have to go through to come up with all these constraints. 193 00:11:48,195 --> 00:11:50,070 So this is going to take a single coordinate. 194 00:11:50,070 --> 00:11:51,520 It could be x. 195 00:11:51,520 --> 00:11:54,520 It could be theta. 196 00:11:54,520 --> 00:11:56,220 But you don't actually need them both. 197 00:11:56,220 --> 00:11:58,946 You use them both for a while, because it's convenient. 198 00:11:58,946 --> 00:12:00,320 But in the final analysis, you'll 199 00:12:00,320 --> 00:12:01,903 be able to write an equation of motion 200 00:12:01,903 --> 00:12:05,320 just in terms of x or just in terms of theta. 201 00:12:05,320 --> 00:12:20,150 OK, free body diagram of our wheel-- we said no slip. 202 00:12:20,150 --> 00:12:22,660 So here's your slope. 203 00:12:22,660 --> 00:12:24,140 You know you've got mg. 204 00:12:26,970 --> 00:12:30,140 You know there's going to be a normal force from the slope. 205 00:12:30,140 --> 00:12:32,890 There may also be some tangential force 206 00:12:32,890 --> 00:12:38,660 that makes it impossible for it not to slip, some f. 207 00:12:38,660 --> 00:12:42,150 So there's the free body diagram, in this case. 208 00:12:42,150 --> 00:12:51,960 What if-- I'll do e here, or do a case ii here, slip allowed. 209 00:12:57,130 --> 00:13:05,160 Then how many constraints do we have? 210 00:13:05,160 --> 00:13:05,920 Just four. 211 00:13:05,920 --> 00:13:08,392 Because now you can no longer say that this is true. 212 00:13:08,392 --> 00:13:09,850 They're independent of one another. 213 00:13:09,850 --> 00:13:13,800 They take on values not controlled by this formulation. 214 00:13:13,800 --> 00:13:16,255 So 6 minus 4 gives you 2. 215 00:13:16,255 --> 00:13:17,630 And you're going to end up having 216 00:13:17,630 --> 00:13:20,300 to have both x and theta probably as your chosen 217 00:13:20,300 --> 00:13:22,030 coordinates to do the problem. 218 00:13:22,030 --> 00:13:24,312 And you'll end up with two equations of motion. 219 00:13:41,700 --> 00:13:43,320 So I've got a hockey puck here. 220 00:13:46,140 --> 00:13:49,270 I've drawn kind of a 3D perspective of this. 221 00:13:49,270 --> 00:13:51,660 So it has a coordinate system out here. 222 00:13:51,660 --> 00:13:53,500 The z-axis is going like that. 223 00:13:53,500 --> 00:13:55,220 So here's my z. 224 00:13:55,220 --> 00:13:56,230 Here's my x. 225 00:13:56,230 --> 00:14:04,930 Here's my y in the plane of the ice 226 00:14:04,930 --> 00:14:07,896 that this thing is sliding on. 227 00:14:07,896 --> 00:14:11,490 So this is my z-coordinate. 228 00:14:11,490 --> 00:14:13,560 And I have string wrapped around it. 229 00:14:13,560 --> 00:14:17,224 And I've got a piece of string coming off like that. 230 00:14:17,224 --> 00:14:18,890 And I'm pulling on it with some tension. 231 00:14:21,950 --> 00:14:26,360 Because otherwise the only constraints 232 00:14:26,360 --> 00:14:29,540 are it's sitting on this icy surface, which 233 00:14:29,540 --> 00:14:31,490 I'm going to assume is-- well, I don't even 234 00:14:31,490 --> 00:14:33,130 have to assume it's frictionless. 235 00:14:33,130 --> 00:14:33,930 I could. 236 00:14:33,930 --> 00:14:36,170 So let's figure out how many equations of motion 237 00:14:36,170 --> 00:14:37,600 we're going to need here. 238 00:14:37,600 --> 00:14:42,230 So the number of degrees of freedom, 6 times-- this 239 00:14:42,230 --> 00:14:43,502 is a rigid body. 240 00:14:43,502 --> 00:14:44,335 It's not a particle. 241 00:14:44,335 --> 00:14:46,150 And there's only one of them. 242 00:14:46,150 --> 00:14:51,540 So it's 6 times 1 minus C and the number of constraints. 243 00:14:51,540 --> 00:14:55,400 So can it move into the table, into the surface? 244 00:14:55,400 --> 00:14:58,680 No, so that's a constraint in z. 245 00:14:58,680 --> 00:15:02,530 Are there any constraints in the x or y? 246 00:15:02,530 --> 00:15:05,360 It can rotate about z. 247 00:15:05,360 --> 00:15:08,680 But it can't rotate about the x or y-axes. 248 00:15:08,680 --> 00:15:11,340 All right, so constraint into z is one. 249 00:15:11,340 --> 00:15:14,385 Can't rotate about x or y-- two, three. 250 00:15:14,385 --> 00:15:15,580 Are there any others? 251 00:15:23,110 --> 00:15:25,800 Who thinks I may have missed one? 252 00:15:25,800 --> 00:15:29,920 OK, I'd say we've got 6 minus 3. 253 00:15:29,920 --> 00:15:32,665 We're going to need three equations of motion 254 00:15:32,665 --> 00:15:35,040 to be able to actually describe the motion of this thing. 255 00:15:37,560 --> 00:15:42,220 And we'd probably use-- this is a pretty good 256 00:15:42,220 --> 00:15:43,050 coordinate system. 257 00:15:43,050 --> 00:15:50,973 We'd probably use an x, a y, and some theta with respect 258 00:15:50,973 --> 00:15:51,473 to z-axis. 259 00:15:54,640 --> 00:16:10,400 All right, another quick example-- 260 00:16:10,400 --> 00:16:12,985 so we've got a rod leaning against a wall. 261 00:16:12,985 --> 00:16:20,192 It's length L. Actually, I don't want to do that. 262 00:16:23,030 --> 00:16:23,880 It's L long. 263 00:16:23,880 --> 00:16:28,690 It's got a center of mass here at the middle, uniform rod. 264 00:16:31,840 --> 00:16:36,230 Let this be x, y, z coming out of the board, 265 00:16:36,230 --> 00:16:40,940 maybe give it an angle here just to help us describe motions. 266 00:16:40,940 --> 00:16:43,410 You might start off with a preliminary little coordinate 267 00:16:43,410 --> 00:16:45,880 system just so you could think that, OK, no motion 268 00:16:45,880 --> 00:16:47,660 in the x, no rotation in this. 269 00:16:47,660 --> 00:16:49,160 But once you get the problem, you're 270 00:16:49,160 --> 00:16:50,900 ready to set up the equations of motion, 271 00:16:50,900 --> 00:16:53,930 you might decide, OK, I know I need two coordinates. 272 00:16:53,930 --> 00:16:56,860 And the ones I've preliminarily chosen aren't too good. 273 00:16:56,860 --> 00:16:59,060 Then you change, and you pick the really good ones. 274 00:16:59,060 --> 00:17:01,018 But I'm just making the preliminary assessments 275 00:17:01,018 --> 00:17:04,680 so I can assess the problem here. 276 00:17:04,680 --> 00:17:11,280 Again, the degrees of freedom-- 6 times 1 minus C. 277 00:17:11,280 --> 00:17:14,089 Because I only have one rigid body. 278 00:17:14,089 --> 00:17:19,975 Now, the constraints here are a little more subtle. 279 00:17:26,740 --> 00:17:30,530 Let me just discuss these two points. 280 00:17:30,530 --> 00:17:33,840 What can you say about the motion at point A? 281 00:17:33,840 --> 00:17:36,921 This is right here where it touches the wall. 282 00:17:36,921 --> 00:17:38,420 This thing is sliding down the wall. 283 00:17:38,420 --> 00:17:40,580 It might be frictionless, might not. 284 00:17:40,580 --> 00:17:42,820 Whether or not friction acts doesn't really 285 00:17:42,820 --> 00:17:45,000 change the number of degrees of freedom 286 00:17:45,000 --> 00:17:49,050 unless you invoke things like no slip. 287 00:17:49,050 --> 00:17:52,180 What can you say kinematically about motion, 288 00:17:52,180 --> 00:17:53,610 about the motion at A? 289 00:17:53,610 --> 00:17:55,660 Does the wall restrict the motion? 290 00:17:55,660 --> 00:17:56,890 AUDIENCE: Yes. 291 00:17:56,890 --> 00:17:58,987 PROFESSOR: In what direction? 292 00:17:58,987 --> 00:18:00,671 AUDIENCE: x. 293 00:18:00,671 --> 00:18:01,921 PROFESSOR: In which direction? 294 00:18:01,921 --> 00:18:02,410 AUDIENCE: x. 295 00:18:02,410 --> 00:18:03,785 PROFESSOR: In x direction, right. 296 00:18:03,785 --> 00:18:07,159 So is this body constrained in the x direction? 297 00:18:07,159 --> 00:18:09,380 AUDIENCE: [INAUDIBLE]. 298 00:18:09,380 --> 00:18:11,270 PROFESSOR: Pardon? 299 00:18:11,270 --> 00:18:12,260 I hear a no. 300 00:18:12,260 --> 00:18:14,890 I hear some yeses. 301 00:18:14,890 --> 00:18:18,466 What about at B? 302 00:18:18,466 --> 00:18:20,490 Is it constrained at B? 303 00:18:20,490 --> 00:18:21,281 In the-- 304 00:18:21,281 --> 00:18:21,780 AUDIENCE: y. 305 00:18:21,780 --> 00:18:24,720 PROFESSOR: y direction, OK. 306 00:18:24,720 --> 00:18:29,970 I didn't bring my big foam disk. 307 00:18:29,970 --> 00:18:33,960 But earlier in the term, I said that the definition 308 00:18:33,960 --> 00:18:40,820 of translation is that all points on a rigid body 309 00:18:40,820 --> 00:18:46,980 do what if you're rectilinear or curvilinear translation as 310 00:18:46,980 --> 00:18:48,713 opposed to rotation? 311 00:18:48,713 --> 00:18:50,040 AUDIENCE: Move in parallel. 312 00:18:50,040 --> 00:18:55,380 PROFESSOR: All points move in parallel, exactly right. 313 00:18:55,380 --> 00:18:59,560 That means that if we use real strict definitions 314 00:18:59,560 --> 00:19:03,520 of translation and rotation, that if I constrain 315 00:19:03,520 --> 00:19:07,680 the motion of any point on that object, 316 00:19:07,680 --> 00:19:12,260 that object is now not allowed to translate in that direction 317 00:19:12,260 --> 00:19:14,710 by the definition of translation. 318 00:19:14,710 --> 00:19:18,340 So this point constrains it in the x direction. 319 00:19:18,340 --> 00:19:22,170 This constrains it in the y direction. 320 00:19:22,170 --> 00:19:32,786 And we're going to-- so it's constrained in translation. 321 00:19:41,210 --> 00:19:44,410 And that implies 2. 322 00:19:44,410 --> 00:19:58,416 And b, we'll assume that z motion is 0. 323 00:19:58,416 --> 00:19:59,790 We'll just assume there's nothing 324 00:19:59,790 --> 00:20:02,290 going on out of the plane. 325 00:20:02,290 --> 00:20:04,040 So that gives me another one. 326 00:20:07,470 --> 00:20:13,440 And I'll assume no rotation. 327 00:20:13,440 --> 00:20:17,170 I'm not going to allow any rotation in this problem. 328 00:20:17,170 --> 00:20:34,350 I'm not interested in rotation about the x or about the y, 329 00:20:34,350 --> 00:20:35,750 about these axes. 330 00:20:35,750 --> 00:20:37,800 And that implies two more. 331 00:20:37,800 --> 00:20:41,705 So we have two, three, four, five. 332 00:20:43,842 --> 00:20:46,050 And I better not have any more than that or the thing 333 00:20:46,050 --> 00:20:47,970 can't move. 334 00:20:47,970 --> 00:20:51,930 So in this case, this is 6 minus 5 equals 1. 335 00:20:51,930 --> 00:20:54,250 I need a single coordinate to describe the motion. 336 00:20:54,250 --> 00:20:55,350 And if you look at it, you say, well, 337 00:20:55,350 --> 00:20:56,891 that's kind of intuitive and obvious. 338 00:20:56,891 --> 00:20:59,731 If I specify the x position here, 339 00:20:59,731 --> 00:21:00,980 I could figure everything out. 340 00:21:00,980 --> 00:21:03,605 If I know the length and the x, I could figure out where it is. 341 00:21:03,605 --> 00:21:05,615 If I know the y position and the length, 342 00:21:05,615 --> 00:21:07,120 I could figure out where it is. 343 00:21:07,120 --> 00:21:08,850 If I know theta, I could figure it out. 344 00:21:08,850 --> 00:21:10,467 I only need one. 345 00:21:10,467 --> 00:21:12,175 But that strict definition of translation 346 00:21:12,175 --> 00:21:13,280 is really helpful here. 347 00:21:13,280 --> 00:21:18,700 This thing, this object, is in pure rotation. 348 00:21:18,700 --> 00:21:23,260 And if it's in pure rotation, it must rotate about some point. 349 00:21:23,260 --> 00:21:24,172 Where? 350 00:21:24,172 --> 00:21:25,255 You know how to find that? 351 00:21:28,500 --> 00:21:31,440 What's the velocity here? 352 00:21:31,440 --> 00:21:34,460 It's got gravity acting on it, so it's probably down. 353 00:21:34,460 --> 00:21:35,860 But it's parallel to the wall. 354 00:21:35,860 --> 00:21:36,990 It has to be. 355 00:21:36,990 --> 00:21:38,840 What's the velocity here? 356 00:21:38,840 --> 00:21:42,460 It's got to be parallel to the wall. 357 00:21:42,460 --> 00:21:45,370 If I draw perpendicular to that-- if I'm saying, 358 00:21:45,370 --> 00:21:47,600 this is rotation. 359 00:21:47,600 --> 00:21:51,036 All points in the body rotate at the same rate. 360 00:21:51,036 --> 00:21:52,910 But their speed is determined by the distance 361 00:21:52,910 --> 00:21:56,110 away from the center of rotation. 362 00:21:56,110 --> 00:21:58,520 But if it's pure rotation, there must be 363 00:21:58,520 --> 00:22:00,270 a center of rotation somewhere. 364 00:22:00,270 --> 00:22:03,610 And it must be perpendicular to any velocity vector. 365 00:22:03,610 --> 00:22:04,950 So you draw the perpendicular. 366 00:22:04,950 --> 00:22:07,680 You draw the perpendicular. 367 00:22:07,680 --> 00:22:12,159 And here is the instantaneous center of rotation, the ICR. 368 00:22:12,159 --> 00:22:14,284 There's a little short section in the book on that. 369 00:22:17,250 --> 00:22:21,840 When this thing drops down to here, same kind of arguments 370 00:22:21,840 --> 00:22:24,320 hold. 371 00:22:24,320 --> 00:22:30,050 But the center of rotation has changed locations. 372 00:22:30,050 --> 00:22:31,021 Yeah. 373 00:22:31,021 --> 00:22:36,320 AUDIENCE: So is this not translating x and y? 374 00:22:36,320 --> 00:22:41,910 PROFESSOR: So now let's talk about the center of mass. 375 00:22:41,910 --> 00:22:44,900 So she asked-- excuse me, I should repeat the question. 376 00:22:44,900 --> 00:22:48,500 You guys aren't holding me to that very well. 377 00:22:48,500 --> 00:22:52,540 Raise your hands if I don't repeat an important question. 378 00:22:52,540 --> 00:22:53,920 She says, is it not translating? 379 00:22:56,720 --> 00:23:01,750 We've determined that-- let me ask you, 380 00:23:01,750 --> 00:23:04,170 does the center of mass move? 381 00:23:07,060 --> 00:23:10,984 Does Newton's second law apply to the motion 382 00:23:10,984 --> 00:23:11,900 of the center of mass? 383 00:23:15,246 --> 00:23:16,680 AUDIENCE: [INAUDIBLE]? 384 00:23:16,680 --> 00:23:18,114 PROFESSOR: Yeah, it's got to. 385 00:23:18,114 --> 00:23:20,050 So the center of mass translates-- 386 00:23:20,050 --> 00:23:21,240 no doubt about that. 387 00:23:21,240 --> 00:23:24,530 Newton's second law applies to it. 388 00:23:24,530 --> 00:23:27,430 So we're not saying that there isn't motion 389 00:23:27,430 --> 00:23:29,236 of the system in the x and y. 390 00:23:29,236 --> 00:23:31,610 We're just saying that that motion is caused by rotation. 391 00:23:34,800 --> 00:23:38,717 It's not caused by what is strictly 392 00:23:38,717 --> 00:23:39,675 defined as translation. 393 00:23:44,540 --> 00:23:55,872 Free body diagram, the reason-- let's 394 00:23:55,872 --> 00:23:57,580 think about a free body diagram for this. 395 00:23:57,580 --> 00:24:00,210 Here's our rod. 396 00:24:00,210 --> 00:24:03,930 There must be-- I'll let it be frictionless to keep 397 00:24:03,930 --> 00:24:05,310 the problem simple for a moment. 398 00:24:05,310 --> 00:24:07,610 That means there must be just a normal force 399 00:24:07,610 --> 00:24:09,020 in the x direction here. 400 00:24:09,020 --> 00:24:10,710 I'll call it Nx. 401 00:24:10,710 --> 00:24:15,360 And there must be a normal force in the y, call it Ny. 402 00:24:15,360 --> 00:24:18,035 The center of mass, there must be an Mg. 403 00:24:22,550 --> 00:24:25,300 So now that I set the problem up this way, 404 00:24:25,300 --> 00:24:27,530 how many unknowns are there in the problem? 405 00:24:27,530 --> 00:24:31,514 If I want to calculate literally the motion of this thing, 406 00:24:31,514 --> 00:24:33,180 find an equation of motion and solve it, 407 00:24:33,180 --> 00:24:34,500 how many unknowns do I have? 408 00:24:42,190 --> 00:24:43,443 How many do you think? 409 00:24:43,443 --> 00:24:44,044 AUDIENCE: One. 410 00:24:44,044 --> 00:24:44,960 PROFESSOR: I hear one. 411 00:24:44,960 --> 00:24:46,105 I see two on the board. 412 00:24:48,900 --> 00:24:51,409 But is two the right answer? 413 00:24:51,409 --> 00:24:51,950 I hear three. 414 00:24:51,950 --> 00:24:53,280 Who said three? 415 00:24:53,280 --> 00:24:55,010 All right, what's the third one? 416 00:24:55,010 --> 00:24:56,850 AUDIENCE: Acceleration? 417 00:24:56,850 --> 00:24:58,820 PROFESSOR: How do you describe acceleration 418 00:24:58,820 --> 00:25:04,518 with a coordinate of some kind? 419 00:25:04,518 --> 00:25:07,520 So there's yet another unknown. 420 00:25:07,520 --> 00:25:09,890 It's probably the thing that you're trying to solve for. 421 00:25:09,890 --> 00:25:14,000 It's actually the motion itself described by theta for x or y, 422 00:25:14,000 --> 00:25:15,190 whatever you do. 423 00:25:15,190 --> 00:25:18,940 There's at least three unknowns in this problem the way 424 00:25:18,940 --> 00:25:22,580 you see it in this free body diagram. 425 00:25:22,580 --> 00:25:25,100 So you've got to figure out ways around that. 426 00:25:25,100 --> 00:25:26,780 This instantaneous center of rotation 427 00:25:26,780 --> 00:25:30,060 gives you one possible way around that. 428 00:25:30,060 --> 00:25:33,780 Because about an instantaneous center, it's not moving. 429 00:25:33,780 --> 00:25:34,620 It's an axis. 430 00:25:34,620 --> 00:25:36,650 It's not moving. 431 00:25:36,650 --> 00:25:38,240 We have a little formula that says 432 00:25:38,240 --> 00:25:42,434 the time rate of change of angular momentum-- torque 433 00:25:42,434 --> 00:25:44,850 is related to the time rate of change of angular momentum. 434 00:25:44,850 --> 00:25:46,975 And you can have a messy formula or a not so messy. 435 00:25:46,975 --> 00:25:52,250 And it's not so messy when the axis of rotation is stationary. 436 00:25:52,250 --> 00:25:54,870 So at this instant in time, the axis is stationary. 437 00:25:54,870 --> 00:26:00,530 You can say that the torques about this point, 438 00:26:00,530 --> 00:26:04,410 the ICR, summation of the torques with respect 439 00:26:04,410 --> 00:26:13,150 to the ICR, is equal to d, is now a rigid body with respect 440 00:26:13,150 --> 00:26:19,820 to the ICR, dt. 441 00:26:19,820 --> 00:26:22,860 OK, in this problem, what are the torques? 442 00:26:22,860 --> 00:26:24,730 Where do the torques come from? 443 00:26:24,730 --> 00:26:28,629 Is there any torque caused by Nx? 444 00:26:28,629 --> 00:26:30,420 No, because there's no moment on it, right? 445 00:26:30,420 --> 00:26:31,900 It's pointing right at the center. 446 00:26:31,900 --> 00:26:34,320 Same thing-- no torque caused by this. 447 00:26:34,320 --> 00:26:36,640 You want to find equations that get rid of unknowns. 448 00:26:36,640 --> 00:26:38,787 So neither unknown appear in this equation. 449 00:26:38,787 --> 00:26:40,120 Where does the torque come from? 450 00:26:44,620 --> 00:26:45,770 Gravity, right? 451 00:26:45,770 --> 00:26:50,700 And it's going to be some Mg times a moment arm. 452 00:26:50,700 --> 00:26:53,740 And the moment arm is going to be like that. 453 00:26:53,740 --> 00:26:56,026 So it's an L over 2 sine theta. 454 00:27:00,140 --> 00:27:02,030 And the sine you'll have to figure out 455 00:27:02,030 --> 00:27:06,420 from an r cross and f. 456 00:27:06,420 --> 00:27:12,300 So you have an i cross j, gives you a k. 457 00:27:12,300 --> 00:27:13,635 But it's in the minus direction. 458 00:27:13,635 --> 00:27:17,470 So I think it'll come out minus. 459 00:27:17,470 --> 00:27:18,750 But I could be wrong. 460 00:27:18,750 --> 00:27:20,610 I did that on the fly. 461 00:27:23,870 --> 00:27:25,690 So we're not going to go further with this. 462 00:27:25,690 --> 00:27:30,440 But the instantaneous centers of rotation could be really handy. 463 00:27:30,440 --> 00:27:36,634 OK, a final example in this stuff-- 464 00:27:36,634 --> 00:27:37,675 how are we doing on time? 465 00:27:52,170 --> 00:27:56,670 Couple of carts, so the floor constrains the motion. 466 00:27:56,670 --> 00:28:03,305 And I've got a spring and a dashpot and an M1 and an M2. 467 00:28:06,230 --> 00:28:08,173 And I want to figure out how many-- yeah? 468 00:28:08,173 --> 00:28:09,960 AUDIENCE: I was just curious. 469 00:28:09,960 --> 00:28:14,109 Why is torque negative in your earlier solution? 470 00:28:14,109 --> 00:28:15,900 PROFESSOR: Let's just figure out r cross f. 471 00:28:15,900 --> 00:28:21,610 So my r-- you're saying, why is the torque negative? 472 00:28:21,610 --> 00:28:31,140 The r is L/2 in the i hat. 473 00:28:31,140 --> 00:28:37,790 And the gravity crossed with the force, which is Mg in the-- 474 00:28:37,790 --> 00:28:43,180 but it's minus Mg in the j hat. 475 00:28:43,180 --> 00:28:45,130 So i cross j is positive k. 476 00:28:45,130 --> 00:28:47,863 But the minus sign comes from there. 477 00:28:47,863 --> 00:28:51,560 AUDIENCE: What is that over to the left [INAUDIBLE]? 478 00:28:51,560 --> 00:28:54,190 PROFESSOR: Well, I'm working on the center of mass here. 479 00:28:54,190 --> 00:28:55,220 That's my equation. 480 00:28:55,220 --> 00:28:56,550 And this thing is L long. 481 00:28:56,550 --> 00:28:59,860 So half of the length must be L/2. 482 00:28:59,860 --> 00:29:05,517 And I'm interested in this side of the right triangle. 483 00:29:05,517 --> 00:29:07,892 AUDIENCE: Why are you interested in that side rather than 484 00:29:07,892 --> 00:29:09,530 the side that connects it to the-- 485 00:29:09,530 --> 00:29:13,670 PROFESSOR: Because this side crossed with that gives me 0. 486 00:29:13,670 --> 00:29:14,420 There's no moment. 487 00:29:14,420 --> 00:29:16,140 So this is a moment equation. 488 00:29:16,140 --> 00:29:18,930 I'm trying to compute moments. 489 00:29:18,930 --> 00:29:19,660 Yeah? 490 00:29:19,660 --> 00:29:22,366 AUDIENCE: Should the moment come from the ICR load? 491 00:29:22,366 --> 00:29:23,532 PROFESSOR: Oh, you're right. 492 00:29:23,532 --> 00:29:24,823 I don't know what I'm thinking. 493 00:29:24,823 --> 00:29:26,440 I could have messed this up. 494 00:29:26,440 --> 00:29:27,700 It's got to be about the ICR. 495 00:29:27,700 --> 00:29:29,910 So the force is down. 496 00:29:29,910 --> 00:29:33,050 Ooh, it's got to be this one. 497 00:29:33,050 --> 00:29:36,450 I messed up-- good catch. 498 00:29:36,450 --> 00:29:41,910 So that's a cosine, still L/2. 499 00:29:41,910 --> 00:29:42,890 You've got a theta. 500 00:29:42,890 --> 00:29:48,790 You have a-- this is also theta. 501 00:29:48,790 --> 00:29:50,920 And we're looking now for this side. 502 00:29:50,920 --> 00:29:55,130 Eh, it's still sine theta, right? 503 00:29:55,130 --> 00:29:57,609 Does the sign still work out the right way? 504 00:29:57,609 --> 00:29:58,525 AUDIENCE: [INAUDIBLE]. 505 00:30:02,890 --> 00:30:05,840 PROFESSOR: Good, OK, I've got to keep rolling. 506 00:30:05,840 --> 00:30:09,720 I've got something else really fun I want to talk about. 507 00:30:09,720 --> 00:30:11,426 So let's do this example quickly. 508 00:30:11,426 --> 00:30:13,925 This is mostly to get you to think about free body diagrams. 509 00:30:17,210 --> 00:30:21,080 This-- two rigid bodies. 510 00:30:21,080 --> 00:30:23,950 The degrees of freedom quickly here-- 6 times 2. 511 00:30:23,950 --> 00:30:28,909 There's no particles-- minus c, so 12 minus c. 512 00:30:28,909 --> 00:30:29,450 Now how many? 513 00:30:29,450 --> 00:30:30,866 What does your intuition tell you? 514 00:30:30,866 --> 00:30:32,830 How many independent coordinates is it 515 00:30:32,830 --> 00:30:36,330 going to take to solve this problem? 516 00:30:36,330 --> 00:30:38,340 Hold up your fingers. 517 00:30:38,340 --> 00:30:44,137 I see two, one, two, one. 518 00:30:44,137 --> 00:30:45,300 OK, one or two. 519 00:30:45,300 --> 00:30:51,240 Well, I think you can find 10 constraints in this problem. 520 00:30:51,240 --> 00:30:52,715 If you assume it doesn't roll over 521 00:30:52,715 --> 00:30:55,100 and you assume it doesn't move, you can find 10. 522 00:30:55,100 --> 00:30:57,880 You're going to need two coordinates. 523 00:30:57,880 --> 00:31:01,290 Because just because you've got a spring and a dashpot, 524 00:31:01,290 --> 00:31:02,950 they don't fix. 525 00:31:02,950 --> 00:31:05,050 They don't say there's any particular relation 526 00:31:05,050 --> 00:31:07,644 between the motion of this and the motion of that. 527 00:31:07,644 --> 00:31:09,560 You're going to need an independent coordinate 528 00:31:09,560 --> 00:31:16,220 to describe the-- whoops, 2 times 6. 529 00:31:16,220 --> 00:31:18,240 This is 12 minus 10. 530 00:31:18,240 --> 00:31:22,250 You don't need an independent coordinate 531 00:31:22,250 --> 00:31:24,730 to describe the motion of each of these masses. 532 00:31:24,730 --> 00:31:27,720 And I'd probably choose a coordinate that, let's say, 533 00:31:27,720 --> 00:31:29,560 goes from the center of mass of this one. 534 00:31:29,560 --> 00:31:34,430 I'll call it x1, center of mass of this one call it x2. 535 00:31:34,430 --> 00:31:36,930 And because I've been doing problems, vibration problems 536 00:31:36,930 --> 00:31:39,520 and stuff like this, for a long time, 537 00:31:39,520 --> 00:31:44,180 I'll tell you it's smart to start your coordinates at 0 538 00:31:44,180 --> 00:31:47,030 when you have zero spring forces. 539 00:31:47,030 --> 00:31:50,319 Or from the static equilibrium position-- 540 00:31:50,319 --> 00:31:51,610 that's the good place to start. 541 00:31:51,610 --> 00:31:55,100 So then your answer, if you're at the static equilibrium 542 00:31:55,100 --> 00:32:00,220 position, then any non-zero answers that come out of it 543 00:32:00,220 --> 00:32:02,990 are movement around that point. 544 00:32:02,990 --> 00:32:06,730 That's what mother nature does. 545 00:32:06,730 --> 00:32:09,480 A spring hanging on a mass hanging on a spring, 546 00:32:09,480 --> 00:32:11,350 it has a static equilibrium position. 547 00:32:11,350 --> 00:32:13,554 The vibration occurs around it. 548 00:32:13,554 --> 00:32:14,970 Your car is sitting on the ground. 549 00:32:14,970 --> 00:32:18,840 The vibration is around its static equilibrium position. 550 00:32:18,840 --> 00:32:27,770 So you usually choose these at the static equilibrium. 551 00:32:27,770 --> 00:32:40,540 And you'll find that that makes for the simplest 552 00:32:40,540 --> 00:32:42,080 equations of motion. 553 00:32:42,080 --> 00:32:46,200 They tend to not have constants on the right hand 554 00:32:46,200 --> 00:32:51,730 side that's caused by gravity or offsets or things like that. 555 00:32:51,730 --> 00:32:54,990 So this thing, actually a problem very similar to this 556 00:32:54,990 --> 00:32:56,680 is on your homework. 557 00:32:56,680 --> 00:32:58,996 The homework says, go find what these 10 are. 558 00:32:58,996 --> 00:33:00,370 So you can name them pretty fast. 559 00:33:00,370 --> 00:33:02,970 That's why I'm not doing it for you. 560 00:33:02,970 --> 00:33:08,150 What I do want to do quickly here 561 00:33:08,150 --> 00:33:21,000 is just talk about how you assign free body 562 00:33:21,000 --> 00:33:22,500 diagrams for this problem. 563 00:33:32,230 --> 00:33:34,620 Because people, one of the most easiest thing 564 00:33:34,620 --> 00:33:37,650 to get confused about is figuring out 565 00:33:37,650 --> 00:33:40,010 the directions of the forces that 566 00:33:40,010 --> 00:33:43,890 come from the springs in the dashpots, which direction 567 00:33:43,890 --> 00:33:46,720 acts on each. 568 00:33:46,720 --> 00:33:51,620 So I've made my x's so that x1 is 569 00:33:51,620 --> 00:33:55,090 0 when this thing is at its static equilibrium position. 570 00:33:55,090 --> 00:34:08,210 And I'm going to start by assuming that x1 and x1 dot are 571 00:34:08,210 --> 00:34:09,730 positive. 572 00:34:09,730 --> 00:34:13,040 You just establish positive motions. 573 00:34:13,040 --> 00:34:17,234 And you deduce the direction of the spring and dashpot forces. 574 00:34:30,679 --> 00:34:34,030 OK, and you do them one at a time. 575 00:34:34,030 --> 00:34:36,560 The problems we're doing here are linear. 576 00:34:36,560 --> 00:34:39,480 Spring force is equal to kx. 577 00:34:39,480 --> 00:34:42,066 Dashpot forces are Bx dot. 578 00:34:42,066 --> 00:34:43,190 So they're linear problems. 579 00:34:43,190 --> 00:34:45,870 So the superposition holds. 580 00:34:45,870 --> 00:34:48,380 You can just do these conceptually one at a time 581 00:34:48,380 --> 00:34:49,374 and figure them out. 582 00:34:49,374 --> 00:34:50,790 And then you add them all together 583 00:34:50,790 --> 00:34:51,914 to get the complete answer. 584 00:34:51,914 --> 00:34:59,240 So if x1 moves in the positive direction, 585 00:34:59,240 --> 00:35:03,380 what is the direction that that spring 586 00:35:03,380 --> 00:35:06,400 puts on this mass as a result? 587 00:35:06,400 --> 00:35:07,750 Which direction is it? 588 00:35:07,750 --> 00:35:11,420 And my coordinate system is positive. 589 00:35:11,420 --> 00:35:13,230 Here's x1. 590 00:35:13,230 --> 00:35:14,680 We'll do the same thing here. 591 00:35:14,680 --> 00:35:15,510 Here's x2. 592 00:35:15,510 --> 00:35:20,080 So if it moves in the positive direction out to here, 593 00:35:20,080 --> 00:35:22,520 what is the spring going to do? 594 00:35:22,520 --> 00:35:25,315 No other motion allowed in the system, just that one. 595 00:35:25,315 --> 00:35:27,080 It moves a little bit. 596 00:35:27,080 --> 00:35:28,960 What does the spring do on that mass? 597 00:35:28,960 --> 00:35:29,922 AUDIENCE: Push back. 598 00:35:29,922 --> 00:35:34,760 PROFESSOR: Push back kx. 599 00:35:34,760 --> 00:35:37,890 And I let the sign be indicated by the direction of the arrow. 600 00:35:37,890 --> 00:35:40,730 And I'll use that when I write out my equation of motion. 601 00:35:40,730 --> 00:35:44,360 OK, now I'm going to assume a positive bx1, so 602 00:35:44,360 --> 00:35:45,810 its velocity in that direction. 603 00:35:45,810 --> 00:35:47,408 What does the dashpot do? 604 00:35:47,408 --> 00:35:49,360 AUDIENCE: [INAUDIBLE]. 605 00:35:49,360 --> 00:35:51,040 PROFESSOR: Resist or [INAUDIBLE]? 606 00:35:51,040 --> 00:35:51,804 AUDIENCE: Resist. 607 00:35:51,804 --> 00:35:53,053 PROFESSOR: Pushes back, right? 608 00:35:53,053 --> 00:35:56,514 OK, so you have another force here, bx1 dot. 609 00:35:59,300 --> 00:36:02,270 Now, this is if you have two bodies. 610 00:36:02,270 --> 00:36:06,000 You have to have two free body diagrams, one for each. 611 00:36:06,000 --> 00:36:11,100 But I need to know what's the effect of x2 and x2 dot 612 00:36:11,100 --> 00:36:13,440 on this body. 613 00:36:13,440 --> 00:36:15,990 Well, let's just go do the same thing. 614 00:36:15,990 --> 00:36:17,730 Let's let x2 be positive. 615 00:36:17,730 --> 00:36:20,275 Now it's the only motion. 616 00:36:20,275 --> 00:36:26,690 I have a positive movement of x2 that stretches the spring. 617 00:36:26,690 --> 00:36:30,370 What is that spring force applied to this? 618 00:36:30,370 --> 00:36:31,530 What direction is it in? 619 00:36:36,620 --> 00:36:38,481 This body is moving in that way. 620 00:36:38,481 --> 00:36:39,773 AUDIENCE: Positive x direction. 621 00:36:39,773 --> 00:36:41,189 PROFESSOR: Which way is the spring 622 00:36:41,189 --> 00:36:42,387 going to pull on this thing? 623 00:36:42,387 --> 00:36:44,520 It's going to pull on it, right? 624 00:36:44,520 --> 00:36:50,950 So k, by amount kx2, and of positive velocity, 625 00:36:50,950 --> 00:36:53,850 to the right. 626 00:36:53,850 --> 00:36:56,040 It makes the dashpot open up. 627 00:36:56,040 --> 00:36:59,871 What direction is the force that dashpot puts on this? 628 00:36:59,871 --> 00:37:03,240 AUDIENCE: Positive x. 629 00:37:03,240 --> 00:37:06,160 PROFESSOR: bx2 dot. 630 00:37:06,160 --> 00:37:08,660 And now in this problem, except for gravity, 631 00:37:08,660 --> 00:37:12,040 I've got a normal force f and an Mg down. 632 00:37:12,040 --> 00:37:14,059 But all the action, all the motion, 633 00:37:14,059 --> 00:37:15,350 is in the horizontal direction. 634 00:37:15,350 --> 00:37:20,481 I can write out an equation of motion for the first rigid body 635 00:37:20,481 --> 00:37:20,980 here. 636 00:37:20,980 --> 00:37:26,020 And that's a sum of the forces in the x direction on body 637 00:37:26,020 --> 00:37:28,670 one-- I'll give this a 1 here-- in the x 638 00:37:28,670 --> 00:37:34,070 is equal to M1 x1 double dot. 639 00:37:34,070 --> 00:37:36,610 And now I can just write it out. 640 00:37:36,610 --> 00:37:45,730 It is k x2, because that one's positive, 641 00:37:45,730 --> 00:37:53,710 minus x1 plus b x2 dot minus x1 dot. 642 00:37:56,530 --> 00:38:00,200 And that has the signs right. 643 00:38:00,200 --> 00:38:04,250 And the whole key is just one at a time assume positive motions 644 00:38:04,250 --> 00:38:06,220 and deduce what happens, and then use 645 00:38:06,220 --> 00:38:10,330 the arrows, the direction of the arrows, to set the signs. 646 00:38:10,330 --> 00:38:13,520 And now there's your equation of motion. 647 00:38:13,520 --> 00:38:14,900 Now we can do the same thing, sum 648 00:38:14,900 --> 00:38:29,800 of the forces on 2 in the x 2 direction, M2 x2 double dot. 649 00:38:29,800 --> 00:38:32,440 And now we would do exactly the same thing. 650 00:38:32,440 --> 00:38:40,930 So positive motion of x1, what does it do over here? 651 00:38:43,899 --> 00:38:46,315 It gives me a force through the spring in which direction, 652 00:38:46,315 --> 00:38:47,476 positive or negative? 653 00:38:47,476 --> 00:38:48,316 AUDIENCE: Positive. 654 00:38:48,316 --> 00:38:49,690 PROFESSOR: Right, so now you just 655 00:38:49,690 --> 00:38:56,980 end up with a kx1, bx1 dot. 656 00:38:56,980 --> 00:39:03,710 And you'll find that kx2 bx2 dot. 657 00:39:03,710 --> 00:39:05,130 And you sum it up. 658 00:39:05,130 --> 00:39:12,080 And you'll end up-- this should switch around, k x1 minus x2 659 00:39:12,080 --> 00:39:16,320 plus b x1 dot minus x2 dot. 660 00:39:23,920 --> 00:39:25,720 I've got two equations of motion. 661 00:39:25,720 --> 00:39:26,530 And they're mixed. 662 00:39:26,530 --> 00:39:29,810 So each one has both coordinates in it. 663 00:39:29,810 --> 00:39:33,090 So this problem has two questions of motion, 664 00:39:33,090 --> 00:39:34,435 and they're coupled. 665 00:39:34,435 --> 00:39:35,435 They're not independent. 666 00:39:35,435 --> 00:39:37,870 You have to solve them together. 667 00:39:37,870 --> 00:39:41,400 OK, now we're going to move on to a subject which 668 00:39:41,400 --> 00:39:43,070 has come up in conversation. 669 00:39:43,070 --> 00:39:44,820 People have asked about this lots of time. 670 00:39:44,820 --> 00:39:50,470 And they say, what about the centrifugal force? 671 00:39:50,470 --> 00:39:53,040 And you sometimes use the term "fictitious force." 672 00:39:53,040 --> 00:39:59,150 How many of you use or heard the word used "fictitious force"? 673 00:39:59,150 --> 00:40:07,580 And how many of you heard us say that centripetal acceleration 674 00:40:07,580 --> 00:40:12,620 is not a force, it's an acceleration? 675 00:40:12,620 --> 00:40:14,645 And yet we love to talk about this concept 676 00:40:14,645 --> 00:40:19,840 of centrifugal force, which doesn't exist. 677 00:40:19,840 --> 00:40:21,020 But it trips us all up. 678 00:40:21,020 --> 00:40:22,680 Because it's handy to think about it. 679 00:40:22,680 --> 00:40:26,210 And so we're going to talk about fictitious forces now. 680 00:40:38,730 --> 00:40:40,160 They're handy. 681 00:40:40,160 --> 00:40:42,370 But they are dangerous. 682 00:40:42,370 --> 00:40:44,680 You really have to understand your fundamentals 683 00:40:44,680 --> 00:40:47,260 if you're going to use the concept of fictitious forces 684 00:40:47,260 --> 00:40:49,640 without getting yourself in trouble. 685 00:40:49,640 --> 00:40:52,220 OK, what is a fictitious force? 686 00:40:52,220 --> 00:40:57,800 Well, Newton's law, let's start there, Newton's second. 687 00:41:03,480 --> 00:41:11,030 Sum of the forces external on the body 688 00:41:11,030 --> 00:41:13,780 equals a mass times acceleration. 689 00:41:13,780 --> 00:41:15,070 It's a vector equation. 690 00:41:15,070 --> 00:41:19,360 So we can break it down into its components. 691 00:41:19,360 --> 00:41:57,935 So a fictitious force, you take the true acceleration times 692 00:41:57,935 --> 00:41:58,435 the mass. 693 00:41:58,435 --> 00:42:00,060 So you have a fictitious force. 694 00:42:00,060 --> 00:42:08,660 It's going to multiply the true acceleration times the mass 695 00:42:08,660 --> 00:42:25,540 and put it on the summation of forces side of the equation. 696 00:42:29,320 --> 00:42:30,320 That's really all it is. 697 00:42:30,320 --> 00:42:33,160 And you move it over to this side of the equation. 698 00:42:33,160 --> 00:42:35,620 And you think of it as a force. 699 00:42:35,620 --> 00:42:37,490 So what you've done is you've said 700 00:42:37,490 --> 00:42:44,220 that the summation of the forces minus Ma-- because to move it 701 00:42:44,220 --> 00:42:46,510 over here, you've got to subtract it from this side-- 702 00:42:46,510 --> 00:42:48,785 equals 0. 703 00:42:48,785 --> 00:42:50,160 And you're saying, I'm just going 704 00:42:50,160 --> 00:42:51,830 to think of this as a force. 705 00:42:51,830 --> 00:42:57,010 And it makes this whole equation be conveniently equal to 0. 706 00:42:57,010 --> 00:43:00,820 But now, that's kind of abstract. 707 00:43:00,820 --> 00:43:04,506 Let's see if we can figure out an example or two 708 00:43:04,506 --> 00:43:05,780 to illustrate this. 709 00:43:20,020 --> 00:43:22,030 I'm sure you've done this problem in physics. 710 00:43:22,030 --> 00:43:23,905 I'm going to pick a really elementary problem 711 00:43:23,905 --> 00:43:27,770 so you don't get hung up on the physics to start with. 712 00:43:27,770 --> 00:43:29,315 This is the elevator problem. 713 00:43:29,315 --> 00:43:31,220 You've got cables pulling it up. 714 00:43:37,730 --> 00:43:40,450 You've got some scales in here, and you're standing on it. 715 00:43:44,246 --> 00:43:45,370 Here's your center of mass. 716 00:43:45,370 --> 00:43:51,870 We'll call that A. And I need a coordinate system, 717 00:43:51,870 --> 00:43:54,130 my inertial system. 718 00:43:54,130 --> 00:43:57,690 Newton's law only applies in inertial systems. 719 00:43:57,690 --> 00:43:59,040 So here's my x. 720 00:43:59,040 --> 00:43:59,920 I called this z here. 721 00:43:59,920 --> 00:44:00,785 I'll call this y. 722 00:44:04,520 --> 00:44:06,710 So I need to write an equation of motion 723 00:44:06,710 --> 00:44:10,260 about this person riding up in the elevator. 724 00:44:12,976 --> 00:44:13,850 We take a look at it. 725 00:44:13,850 --> 00:44:15,016 How many degrees of freedom? 726 00:44:18,110 --> 00:44:20,824 Intuitively obvious-- it's probably how many? 727 00:44:20,824 --> 00:44:21,590 AUDIENCE: One. 728 00:44:21,590 --> 00:44:23,870 PROFESSOR: One-- all sorts of constraints. 729 00:44:23,870 --> 00:44:25,400 It's only going up. 730 00:44:25,400 --> 00:44:27,720 So we're going to say we need one degree of freedom. 731 00:44:31,640 --> 00:44:36,650 Free body diagrams-- well, this object 732 00:44:36,650 --> 00:44:40,430 has some force pushing up on it. 733 00:44:40,430 --> 00:44:42,040 This is the person. 734 00:44:42,040 --> 00:44:45,620 This is your mass. 735 00:44:45,620 --> 00:44:55,280 It has a force pulling down on it, mg. 736 00:44:55,280 --> 00:44:58,089 And that's it. 737 00:44:58,089 --> 00:44:59,630 Now this N is going to work out to be 738 00:44:59,630 --> 00:45:01,566 what the scales read, right? 739 00:45:01,566 --> 00:45:03,190 Because it's coming through the scales. 740 00:45:03,190 --> 00:45:04,917 So that's really what we're looking for. 741 00:45:11,130 --> 00:45:14,490 And the problem here is, define the weight on the scales. 742 00:45:14,490 --> 00:45:22,290 So we say that the summation of the forces in the y direction-- 743 00:45:22,290 --> 00:45:25,390 and in this problem, we would say 744 00:45:25,390 --> 00:45:28,960 it's the mass times the acceleration of point 745 00:45:28,960 --> 00:45:37,390 A with respect to O. And the sum of those 746 00:45:37,390 --> 00:45:45,530 forces-- you've got an N minus mg. 747 00:45:45,530 --> 00:45:47,125 And that's a pretty simple equation. 748 00:45:51,400 --> 00:45:53,630 Now I'm going to specify. 749 00:45:53,630 --> 00:46:02,000 It's given that acceleration of A with respect to O 750 00:46:02,000 --> 00:46:03,125 is 1/4 of g. 751 00:46:06,590 --> 00:46:08,790 So that's what the cables in the elevator are doing. 752 00:46:08,790 --> 00:46:10,510 It's making this thing move, and it's 753 00:46:10,510 --> 00:46:12,630 going up at 1/4 of g, the N acceleration. 754 00:46:12,630 --> 00:46:14,290 So it's getting faster and faster. 755 00:46:16,830 --> 00:46:30,400 OK, so if I solve this for N, I will get-- 756 00:46:30,400 --> 00:46:32,276 let me stop there for a second. 757 00:46:35,872 --> 00:46:37,330 It's normally what I would just do. 758 00:46:37,330 --> 00:46:39,730 But since we're talking about fictitious forces, 759 00:46:39,730 --> 00:46:42,830 I need to go through that step for a second. 760 00:46:42,830 --> 00:46:45,035 So now I say that, well, the summation 761 00:46:45,035 --> 00:46:54,300 of the forces in the y direction here minus M acceleration of A 762 00:46:54,300 --> 00:47:00,200 with respect to O, that total is equal to 0. 763 00:47:00,200 --> 00:47:13,090 And that then is N minus Mg minus M times g over 4. 764 00:47:18,740 --> 00:47:24,960 And now I have taken this upward acceleration. 765 00:47:24,960 --> 00:47:26,510 And I've treated it like a force. 766 00:47:26,510 --> 00:47:28,850 I've just moved it to this other side, 767 00:47:28,850 --> 00:47:30,890 set the whole thing equal to 0. 768 00:47:30,890 --> 00:47:32,655 This is the fictitious force. 769 00:47:39,224 --> 00:47:41,640 I'm going to say, OK, that's all the forces in the system, 770 00:47:41,640 --> 00:47:50,210 solve for N. And of course you get N is M times g plus g/4. 771 00:47:50,210 --> 00:47:54,290 And that's 5/4 mg. 772 00:47:54,290 --> 00:47:57,280 And so you read 25% heavier. 773 00:47:57,280 --> 00:47:58,530 It's a really trivial example. 774 00:47:58,530 --> 00:48:01,850 But the notion is that you think of this 775 00:48:01,850 --> 00:48:03,730 as a force that's been applied. 776 00:48:03,730 --> 00:48:05,950 It's the mass times acceleration with a minus 777 00:48:05,950 --> 00:48:06,920 sign in front of it. 778 00:48:06,920 --> 00:48:09,420 It'll always turn out like that. 779 00:48:09,420 --> 00:48:11,390 It's minus the mass times the acceleration. 780 00:48:11,390 --> 00:48:16,190 Now, the acceleration can come from Coriolis acceleration. 781 00:48:16,190 --> 00:48:19,740 It can come from centripetal acceleration. 782 00:48:19,740 --> 00:48:21,650 So if we do this in a rotating thing, 783 00:48:21,650 --> 00:48:25,850 this fictitious force might be the centrifugal force, 784 00:48:25,850 --> 00:48:29,590 which is minus M times the centripetal acceleration. 785 00:48:29,590 --> 00:48:31,748 And we'll do an example like that. 786 00:48:43,973 --> 00:48:48,190 All right, stop-- there we go. 787 00:48:48,190 --> 00:48:49,920 OK, trivial example-- let's see if we can 788 00:48:49,920 --> 00:48:50,870 find something a little harder. 789 00:48:50,870 --> 00:48:51,950 How are we doing on time? 790 00:48:51,950 --> 00:48:53,470 We're doing pretty good. 791 00:48:53,470 --> 00:49:01,270 So let's do an example where the notion is really 792 00:49:01,270 --> 00:49:02,800 quite powerful. 793 00:49:02,800 --> 00:49:05,990 Now, I showed you this last time. 794 00:49:05,990 --> 00:49:09,180 And we talked a lot about-- we did the time derivative is 795 00:49:09,180 --> 00:49:10,080 the angular momentum. 796 00:49:10,080 --> 00:49:13,190 And we computed the torques that this thing 797 00:49:13,190 --> 00:49:17,280 exerts around different axes with respect 798 00:49:17,280 --> 00:49:18,890 to the point of attachment to this. 799 00:49:18,890 --> 00:49:20,700 Well, this is an example which, if you're 800 00:49:20,700 --> 00:49:22,400 comfortable with fictitious forces, 801 00:49:22,400 --> 00:49:26,840 you can figure out those torques really rather quickly. 802 00:49:26,840 --> 00:49:30,520 So I'm going to-- these two problems are identical. 803 00:49:30,520 --> 00:49:33,880 This problem, or with a shaft running like that, 804 00:49:33,880 --> 00:49:35,980 are really exactly identical. 805 00:49:35,980 --> 00:49:39,180 But I'm going to pretend that my thing is made this way. 806 00:49:39,180 --> 00:49:41,440 Because it makes it easier to see where these torques 807 00:49:41,440 --> 00:49:43,840 are coming from. 808 00:49:43,840 --> 00:49:45,660 So here's my z. 809 00:49:45,660 --> 00:49:48,760 And I've got my rotation about the z-axis. 810 00:49:48,760 --> 00:49:52,942 Here's this mass, my coordinates. 811 00:49:52,942 --> 00:49:56,740 Here's my r hat. 812 00:49:56,740 --> 00:50:01,550 Here's my z k hat, is this distance here. 813 00:50:01,550 --> 00:50:10,050 And I'm going to set conditions in the problem, my rotation 814 00:50:10,050 --> 00:50:10,990 rate. 815 00:50:10,990 --> 00:50:17,470 It's also theta dot about the k z-axis. 816 00:50:17,470 --> 00:50:24,410 And omega dot, theta double dot-- also in the k. 817 00:50:24,410 --> 00:50:25,270 It's not restricted. 818 00:50:25,270 --> 00:50:28,220 So I'm allowing this thing to accelerate. 819 00:50:28,220 --> 00:50:32,810 But r dot equals r double dot is 0. 820 00:50:32,810 --> 00:50:34,430 So it's not changing a position. 821 00:50:34,430 --> 00:50:36,180 It's just fixed. 822 00:50:36,180 --> 00:50:39,880 And I need to figure out the forces on this thing 823 00:50:39,880 --> 00:50:43,150 and talk about how we might consider some of them 824 00:50:43,150 --> 00:50:45,870 as fictitious forces. 825 00:50:45,870 --> 00:50:48,070 So let's think about in the r direction-- 826 00:50:48,070 --> 00:50:54,660 summation of the forces here in this r hat direction. 827 00:50:54,660 --> 00:50:56,040 So it's one vector component. 828 00:50:56,040 --> 00:51:01,900 I don't have to carry along all of the other baggage. 829 00:51:01,900 --> 00:51:08,946 It's equal to-- and I'll call this B. 830 00:51:08,946 --> 00:51:10,900 This is going to be the point about which I 831 00:51:10,900 --> 00:51:11,950 care about things. 832 00:51:11,950 --> 00:51:19,380 There's a fixed coordinate system here, Oxyz. 833 00:51:19,380 --> 00:51:22,837 But then a rotating coordinate system-- we'll call this A. 834 00:51:22,837 --> 00:51:24,420 And it's going to have my coordinates. 835 00:51:24,420 --> 00:51:27,820 I'll use polar coordinates, r theta z. 836 00:51:27,820 --> 00:51:28,930 But this one rotates. 837 00:51:28,930 --> 00:51:36,040 But it also has its origin right there coincident with O. 838 00:51:36,040 --> 00:51:38,610 So the sum in the r direction, then, we 839 00:51:38,610 --> 00:51:45,310 have the mass times the acceleration of B 840 00:51:45,310 --> 00:51:50,670 with respect to O. And the acceleration of B with respect 841 00:51:50,670 --> 00:51:56,440 to O is the acceleration of-- we'll 842 00:51:56,440 --> 00:52:01,860 just write out the whole formula using cylindrical coordinates. 843 00:52:07,570 --> 00:52:14,650 r double dot minus r theta dot squared, this is in the r hat, 844 00:52:14,650 --> 00:52:24,170 plus r theta double dot plus 2r dot theta 845 00:52:24,170 --> 00:52:26,570 dot in the theta hat direction. 846 00:52:26,570 --> 00:52:31,100 That's my full expression for-- whoops, not quite full. 847 00:52:31,100 --> 00:52:34,430 Forgot a term, right? 848 00:52:34,430 --> 00:52:36,600 There we go. 849 00:52:36,600 --> 00:52:40,850 That's my full expression for acceleration 850 00:52:40,850 --> 00:52:45,970 using cylindrical coordinates, but with a moving, 851 00:52:45,970 --> 00:52:48,890 possibly translating and rotating reference frame. 852 00:52:48,890 --> 00:52:50,998 This piece here counts for what? 853 00:52:54,910 --> 00:52:57,355 AUDIENCE: [INAUDIBLE]. 854 00:52:57,355 --> 00:52:58,822 PROFESSOR: Acceleration of what? 855 00:52:58,822 --> 00:53:01,756 In general, why do we have that term? 856 00:53:01,756 --> 00:53:04,690 AUDIENCE: [INAUDIBLE]. 857 00:53:04,690 --> 00:53:07,310 PROFESSOR: It's the translational acceleration 858 00:53:07,310 --> 00:53:08,629 of the A frame. 859 00:53:08,629 --> 00:53:09,670 So we can account for it. 860 00:53:09,670 --> 00:53:11,770 In this case, what is that? 861 00:53:11,770 --> 00:53:14,460 OK, so this term is 0. 862 00:53:14,460 --> 00:53:18,660 But this formula you really need to know. 863 00:53:18,660 --> 00:53:21,190 Because it accounts for all of the accelerations 864 00:53:21,190 --> 00:53:23,000 in the problem. 865 00:53:23,000 --> 00:53:24,770 You know that, you just go in and start 866 00:53:24,770 --> 00:53:26,260 assigning, picking out things. 867 00:53:26,260 --> 00:53:27,144 What's this? 868 00:53:29,800 --> 00:53:30,720 0. 869 00:53:30,720 --> 00:53:32,440 Is this 0? 870 00:53:32,440 --> 00:53:34,960 No, how about this one? 871 00:53:34,960 --> 00:53:37,550 Not 0-- I'm allowing it to accelerate. 872 00:53:37,550 --> 00:53:39,860 This one-- definitely 0. 873 00:53:39,860 --> 00:53:41,090 And this one? 874 00:53:41,090 --> 00:53:41,590 0. 875 00:53:41,590 --> 00:53:43,990 So I don't have many terms left in this. 876 00:53:46,680 --> 00:53:59,721 So this then is the mass times minus r theta dot squared. 877 00:53:59,721 --> 00:54:12,450 And this is r hat plus r theta double dot 878 00:54:12,450 --> 00:54:15,420 in the theta hat direction. 879 00:54:15,420 --> 00:54:17,606 Now I said I was summing the forces in the r. 880 00:54:24,360 --> 00:54:26,370 I put all of them in at the moment. 881 00:54:29,430 --> 00:54:31,465 It may be smarter to do that. 882 00:54:31,465 --> 00:54:33,330 Let me reverse course here. 883 00:54:43,030 --> 00:54:47,710 This is the vector summation of all the forces for a moment. 884 00:54:47,710 --> 00:54:53,140 And then we'll take the r component next. 885 00:54:53,140 --> 00:54:55,670 So just the r component, the summation 886 00:54:55,670 --> 00:55:01,000 of the forces in the r, we look at this, 887 00:55:01,000 --> 00:55:07,830 and we say, oh, it's just that minus Mr theta 888 00:55:07,830 --> 00:55:11,390 dot squared r hat. 889 00:55:11,390 --> 00:55:13,160 So that's the r direction. 890 00:55:13,160 --> 00:55:14,805 But now we need a free body diagram. 891 00:55:18,670 --> 00:55:19,270 Yeah? 892 00:55:19,270 --> 00:55:23,570 AUDIENCE: Why did you take the 2r theta [INAUDIBLE]? 893 00:55:23,570 --> 00:55:29,660 PROFESSOR: Oh, because that's r dot 0. 894 00:55:33,360 --> 00:55:36,580 So in my r direction, this is my acceleration. 895 00:55:36,580 --> 00:55:38,970 That's the only acceleration in the r direction. 896 00:55:38,970 --> 00:55:41,790 It's what we know to be centripetal. 897 00:55:41,790 --> 00:55:43,570 And the minus tells us it's inward. 898 00:55:47,230 --> 00:55:49,560 Free body diagram-- need that. 899 00:55:52,810 --> 00:55:59,108 So looking-- a side view. 900 00:56:03,500 --> 00:56:04,430 Here's your bead. 901 00:56:07,580 --> 00:56:10,070 I'm going to draw it as an unknown here. 902 00:56:10,070 --> 00:56:13,800 There is an unknown force in the r direction 903 00:56:13,800 --> 00:56:18,480 that comes from this bar holding it. 904 00:56:18,480 --> 00:56:20,940 It's applying-- there's got to be a force that 905 00:56:20,940 --> 00:56:22,510 makes this go in a circle. 906 00:56:22,510 --> 00:56:24,070 And that bar is what provides it. 907 00:56:24,070 --> 00:56:25,403 It's the only thing [INAUDIBLE]. 908 00:56:25,403 --> 00:56:27,157 I'll just call this unknown and r. 909 00:56:27,157 --> 00:56:29,240 And I'm just drawing it in the positive direction. 910 00:56:29,240 --> 00:56:32,620 The way you can do this, if you're not sure the direction, 911 00:56:32,620 --> 00:56:33,500 draw it positive. 912 00:56:33,500 --> 00:56:36,600 And the sign that falls out tells you 913 00:56:36,600 --> 00:56:38,680 what the right answer is. 914 00:56:38,680 --> 00:56:42,740 There'll be some force in the z provided by the rod. 915 00:56:42,740 --> 00:56:43,760 There'll be Mg. 916 00:56:46,540 --> 00:56:50,260 And that should be it. 917 00:56:50,260 --> 00:56:57,550 If we did a top view, then there'd 918 00:56:57,550 --> 00:57:02,675 be an unknown in the theta direction. 919 00:57:07,810 --> 00:57:11,590 You'd also see the N in the r direction. 920 00:57:11,590 --> 00:57:12,480 And anything else? 921 00:57:12,480 --> 00:57:16,770 No, the Mg is down where you can't see it. 922 00:57:16,770 --> 00:57:18,680 So these are your two free body diagrams. 923 00:57:24,300 --> 00:57:32,680 So now, in this direction, this is the acceleration. 924 00:57:32,680 --> 00:57:34,935 And the external forces are just that. 925 00:57:37,790 --> 00:57:40,905 So the sum of the forces is Nr. 926 00:57:43,870 --> 00:57:48,180 Now I want to treat it, bring in this concept 927 00:57:48,180 --> 00:57:49,410 of a fictitious force. 928 00:57:54,780 --> 00:57:56,992 I'm going to move the acceleration 929 00:57:56,992 --> 00:57:57,950 term to the other side. 930 00:58:22,170 --> 00:58:24,640 Nr-- unknown positive. 931 00:58:24,640 --> 00:58:28,530 Now I'm going to move that acceleration 932 00:58:28,530 --> 00:58:32,560 term over here minus-- ah, but now it's 933 00:58:32,560 --> 00:58:33,680 minus the acceleration. 934 00:58:36,490 --> 00:58:38,800 So this term looks like that, has a minus sign. 935 00:58:38,800 --> 00:58:48,580 If you move it over here, that actually becomes plus, 936 00:58:48,580 --> 00:58:49,295 equals 0. 937 00:58:51,930 --> 00:58:55,100 Now this is your fictitious force. 938 00:59:03,719 --> 00:59:05,510 Sometimes people call them inertial forces. 939 00:59:08,340 --> 00:59:13,500 And it acts like it's pulling out on the object. 940 00:59:13,500 --> 00:59:20,490 So the force that the mass appears to apply to the rod 941 00:59:20,490 --> 00:59:24,170 is this centrifugal force pulling out. 942 00:59:24,170 --> 00:59:26,090 And of course now we can solve for Nr. 943 00:59:26,090 --> 00:59:28,900 And we find it's minus Mr theta dot squared, which 944 00:59:28,900 --> 00:59:29,870 we knew all along. 945 00:59:29,870 --> 00:59:33,750 It's the force applied to the mass by the arm 946 00:59:33,750 --> 00:59:34,640 as it spins around. 947 00:59:34,640 --> 00:59:37,650 It has to pull in on it to make it go in that circle. 948 00:59:40,170 --> 00:59:44,230 Now, we could do the same thing in the theta direction. 949 00:59:44,230 --> 00:59:48,020 The theta direction will have an Mr theta double dot. 950 00:59:48,020 --> 00:59:52,350 And when we look at the theta free body diagram, 951 00:59:52,350 --> 00:59:53,920 it's plus N theta. 952 00:59:53,920 --> 00:59:55,650 You could solve that, and you immediately 953 00:59:55,650 --> 01:00:02,152 come up with a solution for the force in the theta direction. 954 01:00:02,152 --> 01:00:03,735 I'm just going to write that one down. 955 01:00:11,460 --> 01:00:13,970 When you solve for this one, you get a minus sign. 956 01:00:13,970 --> 01:00:19,430 This one ends up plus, Mr theta double dot. 957 01:00:19,430 --> 01:00:22,660 I'm not going to go through the gyrations of getting to this. 958 01:00:22,660 --> 01:00:25,170 You can do that one. 959 01:00:25,170 --> 01:00:30,646 Now, I want to do one quick thing with this. 960 01:00:34,540 --> 01:00:40,400 Once you develop confidence in knowing 961 01:00:40,400 --> 01:00:45,752 when you can use a fictitious force and not get in trouble, 962 01:00:45,752 --> 01:00:47,377 this is the sort of thing you might do. 963 01:00:53,720 --> 01:00:55,220 Here's my system. 964 01:00:55,220 --> 01:00:56,960 It's rotating. 965 01:00:56,960 --> 01:01:02,370 And I want a quick estimate of, what's the torque? 966 01:01:02,370 --> 01:01:08,340 What's the bending moment about this point caused by the fact 967 01:01:08,340 --> 01:01:11,660 that it is the centripetal acceleration? 968 01:01:11,660 --> 01:01:14,180 Well, centripetal acceleration is 969 01:01:14,180 --> 01:01:17,400 equivalent to having this fictitious force outward 970 01:01:17,400 --> 01:01:25,820 on this of an amount Mr theta dot squared. 971 01:01:25,820 --> 01:01:31,790 And this is the moment arm z. 972 01:01:31,790 --> 01:01:35,330 what's the torque that that causes about this point? 973 01:01:42,680 --> 01:01:45,046 This is just levers now, forces and lever arms. 974 01:01:45,046 --> 01:01:46,002 AUDIENCE: 0. 975 01:01:46,002 --> 01:01:48,680 PROFESSOR: No, not 0. 976 01:01:48,680 --> 01:01:50,910 About this point here-- this is O. 977 01:01:50,910 --> 01:01:53,260 And I've got that centrifugal force pulling out 978 01:01:53,260 --> 01:01:54,680 on this fictitious force. 979 01:01:54,680 --> 01:01:55,632 Yeah? 980 01:01:55,632 --> 01:01:57,445 AUDIENCE: zMR of theta dot squared? 981 01:01:57,445 --> 01:02:01,993 PROFESSOR: Yeah, in the-- this direction, 982 01:02:01,993 --> 01:02:03,870 which is theta, right? 983 01:02:03,870 --> 01:02:08,710 So that's the moment the torque about this point caused 984 01:02:08,710 --> 01:02:16,860 by that-- torque at O is minus Mr theta dot squared 985 01:02:16,860 --> 01:02:20,430 times z, the moment arm. 986 01:02:20,430 --> 01:02:23,192 And it's not minus. 987 01:02:23,192 --> 01:02:24,150 It's in that direction. 988 01:02:24,150 --> 01:02:25,676 So it's plus. 989 01:02:25,676 --> 01:02:29,190 It's in the theta hat direction, the torque, 990 01:02:29,190 --> 01:02:31,940 that that force is applying about this point. 991 01:02:35,580 --> 01:02:37,830 We could have solved this problem the way 992 01:02:37,830 --> 01:02:41,300 we did in the last lecture, very carefully 993 01:02:41,300 --> 01:02:45,660 going through the dh dt's and following it all out. 994 01:02:45,660 --> 01:02:49,140 And we would have gotten that answer for the torque 995 01:02:49,140 --> 01:02:50,827 except for a minus sign. 996 01:02:54,490 --> 01:02:57,640 Now, why the difference? 997 01:02:57,640 --> 01:03:00,900 This is the torque that the centripetal force 998 01:03:00,900 --> 01:03:03,170 causes down here. 999 01:03:03,170 --> 01:03:08,050 When you do dh dt, you get the torque 1000 01:03:08,050 --> 01:03:12,516 required to make what's happening happen. 1001 01:03:15,780 --> 01:03:18,730 It's just the opposite. 1002 01:03:18,730 --> 01:03:19,640 This is the torque. 1003 01:03:19,640 --> 01:03:21,130 This is putting about that. 1004 01:03:21,130 --> 01:03:23,730 There must be an equal and opposite torque 1005 01:03:23,730 --> 01:03:27,820 that this system puts on this arm out here 1006 01:03:27,820 --> 01:03:30,110 to keep it from flopping out. 1007 01:03:30,110 --> 01:03:32,030 So this is applying a torque here. 1008 01:03:32,030 --> 01:03:33,960 It must resist with a torque. 1009 01:03:33,960 --> 01:03:43,200 And so when you do dh dt, you're going 1010 01:03:43,200 --> 01:03:50,840 to end up with a minus Mr theta dot squared theta hat. 1011 01:03:53,600 --> 01:03:58,210 So you've got to be careful what you mean. 1012 01:03:58,210 --> 01:04:00,720 But from an engineering point of view, 1013 01:04:00,720 --> 01:04:02,930 if I were just trying to calculate the bending moment 1014 01:04:02,930 --> 01:04:06,030 down here and deciding whether or not this shaft sticking out 1015 01:04:06,030 --> 01:04:07,950 here is going to break off or not, 1016 01:04:07,950 --> 01:04:11,480 I could make a very quick estimate of the bending moment 1017 01:04:11,480 --> 01:04:14,785 by knowing this centrifugal force times the moment arm. 1018 01:04:14,785 --> 01:04:15,284 Yeah? 1019 01:04:15,284 --> 01:04:21,190 AUDIENCE: Is that supposed to be a z in the dh dt expression? 1020 01:04:21,190 --> 01:04:24,190 Does there have to be a z in that expression? 1021 01:04:24,190 --> 01:04:25,508 PROFESSOR: A v, a velocity? 1022 01:04:25,508 --> 01:04:26,464 AUDIENCE: A z. 1023 01:04:26,464 --> 01:04:27,898 PROFESSOR: A z, yeah, right. 1024 01:04:27,898 --> 01:04:32,708 I'm just getting a little speedy in writing the equation. 1025 01:04:32,708 --> 01:04:34,640 Good catch. 1026 01:04:34,640 --> 01:04:41,110 All right, so now, we also have a theta direction thing here, 1027 01:04:41,110 --> 01:04:42,840 right? 1028 01:04:42,840 --> 01:04:46,820 We can think of a fictitious force in the theta direction. 1029 01:04:46,820 --> 01:04:49,650 And that was the Mr theta double dot term. 1030 01:04:49,650 --> 01:04:55,980 Does it generate torques that you could quickly compute? 1031 01:04:55,980 --> 01:04:58,570 So as this thing is trying to accelerate 1032 01:04:58,570 --> 01:05:02,380 in the positive acceleration, omega dot, theta double dot, 1033 01:05:02,380 --> 01:05:04,730 it's trying to accelerate into the board. 1034 01:05:04,730 --> 01:05:07,490 The bar is having to push that thing into the board. 1035 01:05:07,490 --> 01:05:10,580 There's a force in the positive theta hat direction. 1036 01:05:10,580 --> 01:05:15,630 But the fictitious inertial force 1037 01:05:15,630 --> 01:05:22,190 is the mass pushing back on the rod, pushing this way, 1038 01:05:22,190 --> 01:05:25,740 Mr theta double dot pushing this way. 1039 01:05:25,740 --> 01:05:28,170 What moments does that cause about this point? 1040 01:05:33,742 --> 01:05:34,658 AUDIENCE: [INAUDIBLE]. 1041 01:05:40,125 --> 01:05:42,380 PROFESSOR: So there's actually two moment arms. 1042 01:05:42,380 --> 01:05:46,730 If you have a force this way now, the force in, 1043 01:05:46,730 --> 01:05:48,550 I'll call it, the theta direction 1044 01:05:48,550 --> 01:05:53,240 is Mr theta double dot. 1045 01:05:53,240 --> 01:05:56,436 Now, this is my fictitious force. 1046 01:05:56,436 --> 01:05:58,010 This is this fictitious force. 1047 01:05:58,010 --> 01:06:00,900 It's in the minus theta hat direction. 1048 01:06:00,900 --> 01:06:04,330 It's pushing back. 1049 01:06:04,330 --> 01:06:08,090 It's in the minus theta hat direction, Mr theta double dot. 1050 01:06:08,090 --> 01:06:15,920 And a torque is some R cross with the force. 1051 01:06:15,920 --> 01:06:23,922 And this R is that, RBA. 1052 01:06:23,922 --> 01:06:31,930 And it's composed of r R hat and z k hat here. 1053 01:06:31,930 --> 01:06:37,300 So we have to do a little r r hat 1054 01:06:37,300 --> 01:06:51,950 plus z k hat crossed with minus Mr theta double dot theta hat. 1055 01:06:51,950 --> 01:06:55,460 So you get an r hat cross theta gives you a k. 1056 01:06:55,460 --> 01:06:58,085 A k cross theta hat gives you an r. 1057 01:06:58,085 --> 01:07:01,650 You're going to get two terms out of this. 1058 01:07:01,650 --> 01:07:07,130 One is going to look like Mr squared theta double dot. 1059 01:07:07,130 --> 01:07:09,400 And that's a torque. 1060 01:07:09,400 --> 01:07:13,150 And that is the one that it takes to-- that's 1061 01:07:13,150 --> 01:07:16,680 the torque about this axis. 1062 01:07:16,680 --> 01:07:19,780 There's a force times this moment arm. 1063 01:07:19,780 --> 01:07:21,260 That's a torque about this axis. 1064 01:07:21,260 --> 01:07:23,730 That's the actual torque it takes to speed this thing up. 1065 01:07:26,640 --> 01:07:30,860 And then there's a torque about this axis, which is 1066 01:07:30,860 --> 01:07:34,320 this length times this force. 1067 01:07:34,320 --> 01:07:35,830 And that's the other term. 1068 01:07:35,830 --> 01:07:40,800 That'll give you the term in the R direction. 1069 01:07:40,800 --> 01:07:41,750 It'll be a twist. 1070 01:07:41,750 --> 01:07:44,380 It'll be trying to twist this bar like that. 1071 01:07:44,380 --> 01:07:46,570 Because there's a force in this direction. 1072 01:07:46,570 --> 01:07:47,780 Yeah. 1073 01:07:47,780 --> 01:07:49,700 AUDIENCE: It seems like these forces 1074 01:07:49,700 --> 01:07:52,580 are coming from the systems. 1075 01:07:52,580 --> 01:07:54,980 They aren't even fictitious. 1076 01:07:54,980 --> 01:08:02,080 PROFESSOR: So she's asking, it's like the forces are real. 1077 01:08:02,080 --> 01:08:06,510 But the force-- I mean, this is why 1078 01:08:06,510 --> 01:08:11,030 fictitious forces, this concept of them, is so dangerous. 1079 01:08:11,030 --> 01:08:14,870 It's because it's really tempting 1080 01:08:14,870 --> 01:08:17,620 to start thinking of them like real forces. 1081 01:08:17,620 --> 01:08:22,500 So any time you get stuck, you go back to the basics. 1082 01:08:22,500 --> 01:08:26,830 And you say, Newton's law, F equals ma, 1083 01:08:26,830 --> 01:08:30,605 you see torques is dh dt plus that v cross p term 1084 01:08:30,605 --> 01:08:31,630 if you need it. 1085 01:08:31,630 --> 01:08:37,100 And you work it out carefully using all the vector math. 1086 01:08:37,100 --> 01:08:40,529 And then you are dealing-- and then 1087 01:08:40,529 --> 01:08:47,779 Coriolis, centripetal, Euler terms are only accelerations. 1088 01:08:47,779 --> 01:08:49,779 You treat them as pure accelerations. 1089 01:08:49,779 --> 01:08:50,850 That's all they are. 1090 01:08:53,580 --> 01:09:01,790 Now, to cause accelerations, you need to apply forces. 1091 01:09:01,790 --> 01:09:05,160 And we give these forces names. 1092 01:09:05,160 --> 01:09:07,350 Because it's helpful conceptually 1093 01:09:07,350 --> 01:09:11,080 to think about these things as forces sometimes. 1094 01:09:11,080 --> 01:09:14,445 But they are not real forces. 1095 01:09:17,490 --> 01:09:19,795 But it's quick and easy. 1096 01:09:19,795 --> 01:09:21,819 If you get too comfortable with them, 1097 01:09:21,819 --> 01:09:24,760 they are great assets to your intuition. 1098 01:09:24,760 --> 01:09:26,870 So I know immediately just looking 1099 01:09:26,870 --> 01:09:30,420 at this thing, as soon as I see a machine that 1100 01:09:30,420 --> 01:09:35,470 has a part that rotates like this, I say, rotational motion. 1101 01:09:35,470 --> 01:09:41,920 That's central, circular motion, must be the equivalent 1102 01:09:41,920 --> 01:09:45,290 of a centrifugal force. 1103 01:09:45,290 --> 01:09:48,000 This mass is going to pull in that direction. 1104 01:09:48,000 --> 01:09:49,770 Because it's going around and around. 1105 01:09:49,770 --> 01:09:51,960 And it's going to cause a moment about this thing. 1106 01:09:51,960 --> 01:09:54,326 I know for sure that's got to be there. 1107 01:09:54,326 --> 01:09:56,080 And if I'm going to do it rigorously, 1108 01:09:56,080 --> 01:09:57,960 I call it centripetal acceleration. 1109 01:09:57,960 --> 01:09:59,040 And I compute. 1110 01:09:59,040 --> 01:10:03,390 It's really the force that this bar puts on that mass 1111 01:10:03,390 --> 01:10:07,010 to make it go in a circle. 1112 01:10:07,010 --> 01:10:09,200 But it's handy to think of it as a force 1113 01:10:09,200 --> 01:10:12,760 if what I'm interested in is the force. 1114 01:10:12,760 --> 01:10:17,210 The other even better-- let's go back to the old really simple 1115 01:10:17,210 --> 01:10:18,670 demonstration. 1116 01:10:18,670 --> 01:10:21,920 This thing going around constant speed, 1117 01:10:21,920 --> 01:10:23,930 there's definitely a tension in the string. 1118 01:10:23,930 --> 01:10:25,100 And I am pulling. 1119 01:10:25,100 --> 01:10:27,690 The tension is inward, right? 1120 01:10:27,690 --> 01:10:31,230 That tension causes an acceleration 1121 01:10:31,230 --> 01:10:33,780 of r theta dot squared. 1122 01:10:33,780 --> 01:10:37,440 And that acceleration is inward. 1123 01:10:37,440 --> 01:10:39,220 But it's easy. 1124 01:10:39,220 --> 01:10:42,860 If you ask me, come on, Kim, quick, 1125 01:10:42,860 --> 01:10:45,760 tell me, what's the tension in the string, 1126 01:10:45,760 --> 01:10:49,000 I just say, that's easy-- Mr theta dot squared, 1127 01:10:49,000 --> 01:10:50,050 centrifugal force. 1128 01:10:59,230 --> 01:11:00,010 They're handy. 1129 01:11:00,010 --> 01:11:04,550 But any time you get stuck, go back and be really strict 1130 01:11:04,550 --> 01:11:06,810 and say, what's acceleration? 1131 01:11:06,810 --> 01:11:09,650 And what's real external forces? 1132 01:11:09,650 --> 01:11:13,330 Gravity is an external force. 1133 01:11:13,330 --> 01:11:16,580 The force that the bar puts on the mass, that's a real force. 1134 01:11:16,580 --> 01:11:20,040 But these other things are just accelerations. 1135 01:11:20,040 --> 01:11:23,030 All right, OK, hey, perfect. 1136 01:11:23,030 --> 01:11:25,400 I didn't get to the thing I really wanted to do. 1137 01:11:25,400 --> 01:11:26,400 So I'll do it next time. 1138 01:11:26,400 --> 01:11:29,027 And I'm going to show you this thing. 1139 01:11:29,027 --> 01:11:30,860 And I want you to fill out your muddy cards. 1140 01:11:30,860 --> 01:11:32,960 I'm giving you a couple minutes. 1141 01:11:32,960 --> 01:11:35,755 So this is a mechanical shaker. 1142 01:11:35,755 --> 01:11:37,300 You see it moving already. 1143 01:11:37,300 --> 01:11:42,720 And we're going to talk about that next time.