1 00:00:00,070 --> 00:00:02,430 The following content is provided under a Creative 2 00:00:02,430 --> 00:00:03,810 Commons license. 3 00:00:03,810 --> 00:00:06,050 Your support will help MIT OpenCourseWare 4 00:00:06,050 --> 00:00:10,140 continue to offer high quality educational resources for free. 5 00:00:10,140 --> 00:00:12,690 To make a donation or to view additional materials 6 00:00:12,690 --> 00:00:16,590 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,590 --> 00:00:17,260 at ocw.mit.edu. 8 00:00:21,050 --> 00:00:22,930 PROFESSOR: OK. 9 00:00:22,930 --> 00:00:26,820 I've been giving out the money cards 10 00:00:26,820 --> 00:00:30,610 for a few of the lectures, and two or three questions 11 00:00:30,610 --> 00:00:32,980 came up in those that I haven't addressed so far. 12 00:00:32,980 --> 00:00:35,179 I'm calling them loose ends. 13 00:00:35,179 --> 00:00:37,220 And I'm going to pick up a couple of those today, 14 00:00:37,220 --> 00:00:40,080 I think they'll help you consolidate 15 00:00:40,080 --> 00:00:43,010 the knowledge around the quiz. 16 00:00:43,010 --> 00:00:45,180 So I'm going to tie up a little loose ends there. 17 00:00:45,180 --> 00:00:48,710 And then the lecture topic I started last time, which 18 00:00:48,710 --> 00:00:51,650 is making this transition from thinking about angular 19 00:00:51,650 --> 00:00:56,150 momentum of particles to using the full angular momentum 20 00:00:56,150 --> 00:00:58,080 equations for rigid bodies, where 21 00:00:58,080 --> 00:01:00,565 we talk about mass moments of inertia 22 00:01:00,565 --> 00:01:01,740 and products of inertia. 23 00:01:01,740 --> 00:01:04,190 And that's where we'll pick up, there again today. 24 00:01:04,190 --> 00:01:07,215 Because that's where we're going for the next few lectures. 25 00:01:07,215 --> 00:01:07,715 OK. 26 00:01:18,270 --> 00:01:22,190 Let's pick up with a first example here. 27 00:01:22,190 --> 00:01:29,290 This is on the topic of basically finding 28 00:01:29,290 --> 00:01:31,310 equations of motion. 29 00:01:31,310 --> 00:01:32,970 And there's been a little confusion 30 00:01:32,970 --> 00:01:38,120 with people, who have asked me what do you mean find? 31 00:01:38,120 --> 00:01:39,437 Do you mean solve, et cetera. 32 00:01:39,437 --> 00:01:42,020 And so I'm going to go through, and just a real quick example, 33 00:01:42,020 --> 00:01:47,030 skipping some of the steps because my purpose 34 00:01:47,030 --> 00:01:51,810 is emphasizing the steps, not working out all the details. 35 00:01:51,810 --> 00:01:55,950 So finding equations of motion. 36 00:01:55,950 --> 00:01:58,240 Where does it begin? 37 00:01:58,240 --> 00:01:59,940 One of the really important steps 38 00:01:59,940 --> 00:02:07,660 is this, determine the number of independent 39 00:02:07,660 --> 00:02:08,770 coordinates you need. 40 00:02:18,330 --> 00:02:20,980 Because when you've done that, that tells you, 41 00:02:20,980 --> 00:02:27,890 basically-- it really starts finding the number 42 00:02:27,890 --> 00:02:28,980 of degrees of freedom. 43 00:02:28,980 --> 00:02:30,730 Should have put this in a different order. 44 00:02:30,730 --> 00:02:33,020 Degrees of freedom tells you the number 45 00:02:33,020 --> 00:02:35,110 of independent coordinates you need. 46 00:02:35,110 --> 00:02:39,590 This is 1, 2, and then 3, that leads you 47 00:02:39,590 --> 00:02:43,130 to the number of equations of motion that you need. 48 00:02:43,130 --> 00:02:46,520 So this is really an important step. 49 00:02:46,520 --> 00:02:52,640 Secondly, draw a free body diagram. 50 00:02:55,150 --> 00:03:01,350 And third, apply summation of forces. 51 00:03:05,110 --> 00:03:10,320 External vector equations gives you mass times acceleration, 52 00:03:10,320 --> 00:03:27,730 and summation of torques gives you DHDT plus this V cross 53 00:03:27,730 --> 00:03:30,470 P term. 54 00:03:30,470 --> 00:03:32,980 So this is just kind of the step by step. 55 00:03:32,980 --> 00:03:36,670 So let's apply it briefly. 56 00:03:46,480 --> 00:03:49,870 We've talked a lot about things on hills, so here's a cart. 57 00:03:49,870 --> 00:03:53,680 It's got wheels attached by a cord 58 00:03:53,680 --> 00:03:58,630 to a second mass that's sliding. 59 00:03:58,630 --> 00:04:02,685 m1, m2, doesn't stretch the cord in between. 60 00:04:11,240 --> 00:04:14,320 Let's think of these things as rigid bodies. 61 00:04:14,320 --> 00:04:15,670 So how many degrees of freedom? 62 00:04:15,670 --> 00:04:18,890 How many possible degrees of freedom? 63 00:04:18,890 --> 00:04:21,529 For the maximum possible, you have how many rigid bodies? 64 00:04:21,529 --> 00:04:22,990 2. 65 00:04:22,990 --> 00:04:26,030 How many degrees of freedom for rigid bodies possible? 66 00:04:26,030 --> 00:04:27,860 6 each. 67 00:04:27,860 --> 00:04:31,750 So we're at 6 times m plus 3 times 68 00:04:31,750 --> 00:04:33,750 n minus the number of constraints 69 00:04:33,750 --> 00:04:36,300 is the number of independent degrees of freedom. 70 00:04:36,300 --> 00:04:39,390 This is the number of rigid bodies, number of particles, 71 00:04:39,390 --> 00:04:44,010 so we have 6 times 2, 3 times 0 minus constraints, 72 00:04:44,010 --> 00:04:47,505 so this one comes out 12 minus the constraints. 73 00:04:50,032 --> 00:04:51,990 You have to figure out the constraints quickly. 74 00:04:51,990 --> 00:04:55,830 We're not going to allow rotation in any of the three 75 00:04:55,830 --> 00:04:56,740 directions on either. 76 00:04:56,740 --> 00:04:58,960 They're on carts, they're big, they're sliding, 77 00:04:58,960 --> 00:05:00,680 they're not rolling or any of that. 78 00:05:00,680 --> 00:05:04,540 So no rotation for 3, no rotation for 3 more. 79 00:05:04,540 --> 00:05:06,390 That's minus 6. 80 00:05:06,390 --> 00:05:14,000 So c equals minus 6, or c equals 6 for rotation. 81 00:05:14,000 --> 00:05:15,650 And then what else can we say? 82 00:05:19,330 --> 00:05:24,280 There's no-- I'll designate this the y direction so we 83 00:05:24,280 --> 00:05:25,670 can talk about directions here. 84 00:05:25,670 --> 00:05:30,020 And I'll designate this x in general. 85 00:05:30,020 --> 00:05:34,620 No acceleration at all in the y direction, right. 86 00:05:34,620 --> 00:05:36,010 Can't move in the y. 87 00:05:36,010 --> 00:05:39,260 So that gives us 1, 2 for each mass, 88 00:05:39,260 --> 00:05:42,050 plus 2 more, that's 8 constraints 89 00:05:42,050 --> 00:05:43,630 that we've come up with. 90 00:05:43,630 --> 00:05:51,340 Now a 9th constraint is the fact that these two 91 00:05:51,340 --> 00:05:52,530 are tied together. 92 00:05:52,530 --> 00:05:54,730 And so if you had just temporarily 93 00:05:54,730 --> 00:06:00,510 assigned a coordinate here, x1, and another one here, x2, 94 00:06:00,510 --> 00:06:04,284 we know for a fact that x1 has got to be equal to x2, 95 00:06:04,284 --> 00:06:06,075 and that gives you yet one more constraint. 96 00:06:10,800 --> 00:06:13,900 So that's 9. 97 00:06:13,900 --> 00:06:15,250 So we might just stop there. 98 00:06:15,250 --> 00:06:18,010 We say OK, that's a total of 9, so the number 99 00:06:18,010 --> 00:06:24,980 of degrees of freedom, 12 minus 9 is 3, 100 00:06:24,980 --> 00:06:29,100 and that implies that you need three equations of motion. 101 00:06:29,100 --> 00:06:33,810 Some confusion comes, you know, if something's not-- 102 00:06:33,810 --> 00:06:37,430 let me rephrase that. 103 00:06:37,430 --> 00:06:40,810 We haven't talked anything about the z direction. 104 00:06:40,810 --> 00:06:44,960 I haven't described any constraints in the z direction. 105 00:06:44,960 --> 00:06:50,040 If this is me in a car and I'm dragging a sled down a hill, 106 00:06:50,040 --> 00:06:53,460 or I'm in a vehicle and I'm dragging a sled down a hill. 107 00:06:53,460 --> 00:06:57,990 I don't know if you've ever been in a vehicle with a trailer 108 00:06:57,990 --> 00:07:00,690 on an icy road in the winter time, that's 109 00:07:00,690 --> 00:07:05,640 a dicey maneuver, going down a hill trying to put brakes on. 110 00:07:05,640 --> 00:07:09,280 So this thing could conceivably move in this direction. 111 00:07:09,280 --> 00:07:12,020 And you can either constrain it to be that way 112 00:07:12,020 --> 00:07:15,910 to make the problem simple, or you can just say it's possible, 113 00:07:15,910 --> 00:07:18,740 so we have three equations of motion, 114 00:07:18,740 --> 00:07:23,460 of which two, for now, the summation of the forces, 115 00:07:23,460 --> 00:07:27,980 external in the-- since I've got x and y, 116 00:07:27,980 --> 00:07:32,550 z must be this way-- in the y direction. 117 00:07:32,550 --> 00:07:35,300 And we'll make it in z direction, 118 00:07:35,300 --> 00:07:39,240 this is coordinate system 1, so this would be z1 or z1. 119 00:07:39,240 --> 00:07:53,610 The summation of forces, z1, is m1 z1 double dot. 120 00:07:53,610 --> 00:07:55,261 But I'm going to set that equal to 0, 121 00:07:55,261 --> 00:07:56,510 I just know there's no forces. 122 00:07:56,510 --> 00:07:59,740 So this becomes a trivial equation of motion. 123 00:07:59,740 --> 00:08:03,620 And I add another one, summation of forces on the second mass. 124 00:08:03,620 --> 00:08:11,880 This is on m2 in the z2 direction, is m2 z2 double dot, 125 00:08:11,880 --> 00:08:14,020 and we set that equal to 0 also. 126 00:08:14,020 --> 00:08:16,410 So what it boils down to, I had 3 degrees 127 00:08:16,410 --> 00:08:19,860 of freedom, 2 trivial equations of motion, 128 00:08:19,860 --> 00:08:22,350 leaving me with just 1 equation of motion 129 00:08:22,350 --> 00:08:24,560 that's going to be meaningful. 130 00:08:24,560 --> 00:08:25,060 Yeah. 131 00:08:25,060 --> 00:08:31,431 AUDIENCE: If x1 is equal to x2, would that 132 00:08:31,431 --> 00:08:36,124 mean that the rod has to be entirely along the x-axis? 133 00:08:36,124 --> 00:08:37,606 So that would mean that-- 134 00:08:37,606 --> 00:08:39,971 PROFESSOR: So he asked if x1 equals x2, 135 00:08:39,971 --> 00:08:42,179 does that mean they both have to be along the x-axis? 136 00:08:42,179 --> 00:08:43,350 I'm assuming that. 137 00:08:43,350 --> 00:08:46,110 So I really am assuming this thing's going down the hill. 138 00:08:46,110 --> 00:08:49,250 I'm making a point about this z direction thing 139 00:08:49,250 --> 00:08:51,170 because it's just a subtlety that you 140 00:08:51,170 --> 00:08:56,070 have to decide on when you're figuring out how to actually 141 00:08:56,070 --> 00:08:57,710 analyze the situation. 142 00:08:57,710 --> 00:09:00,040 If you really were thinking about what 143 00:09:00,040 --> 00:09:04,460 happens when a vehicle is going down a steep, icy hill towing 144 00:09:04,460 --> 00:09:08,777 a trailer, maybe put the brakes on, maybe not. 145 00:09:08,777 --> 00:09:10,610 You could probably say well, unless I really 146 00:09:10,610 --> 00:09:15,690 have a disaster, we're not going to get rollovers and things 147 00:09:15,690 --> 00:09:16,200 like that. 148 00:09:16,200 --> 00:09:18,600 But you could imagine that it can get out 149 00:09:18,600 --> 00:09:19,840 of the x direction, right. 150 00:09:19,840 --> 00:09:22,167 It could start sliding into z. 151 00:09:22,167 --> 00:09:23,750 It's probably not going to go anywhere 152 00:09:23,750 --> 00:09:25,500 in the y, that's still a good assumption. 153 00:09:25,500 --> 00:09:28,860 So this is about modeling, and how complicated 154 00:09:28,860 --> 00:09:30,790 do you make the modeling. 155 00:09:30,790 --> 00:09:33,610 You actually have to make quite a few modeling decisions when 156 00:09:33,610 --> 00:09:37,510 you go to do this, and we tend, in class and examples, 157 00:09:37,510 --> 00:09:39,390 tend to really oversimplify problems 158 00:09:39,390 --> 00:09:42,680 so that we can do them. 159 00:09:42,680 --> 00:09:44,170 So I've boiled this down to where 160 00:09:44,170 --> 00:09:46,720 I'm going to end up with one significant equation of motion. 161 00:10:04,840 --> 00:10:11,100 And that one's going to say that the summation of the forces 162 00:10:11,100 --> 00:10:14,520 in the x direction-- and I'm just 163 00:10:14,520 --> 00:10:17,050 going to write mx double dot here, 164 00:10:17,050 --> 00:10:19,780 not putting down m1s or m2s because we're 165 00:10:19,780 --> 00:10:21,750 going to have to go to free body diagrams 166 00:10:21,750 --> 00:10:25,310 to figure out how to apply this. 167 00:10:28,544 --> 00:10:29,335 Free body diagrams. 168 00:10:32,800 --> 00:10:33,400 First mass. 169 00:10:39,920 --> 00:10:41,470 So this is m1 here. 170 00:10:48,420 --> 00:10:53,100 m1g, a normal force, no doubt a friction force, 171 00:10:53,100 --> 00:10:58,310 I'll call that f1, and a tension in the cord. 172 00:11:01,330 --> 00:11:03,870 And I'm assuming that tension's going to always be there. 173 00:11:03,870 --> 00:11:05,530 So if, again, in a situation where 174 00:11:05,530 --> 00:11:07,350 the trailer starts overtaking the car 175 00:11:07,350 --> 00:11:09,220 and the rope goes slack, we're not 176 00:11:09,220 --> 00:11:11,490 going to consider that one today. 177 00:11:11,490 --> 00:11:14,020 So there's your free body diagram for the first one 178 00:11:14,020 --> 00:11:19,790 and we're going to need to break mg into a couple of components. 179 00:11:19,790 --> 00:11:24,690 We're going to need the slope here, 180 00:11:24,690 --> 00:11:28,490 and that translates into an angle here. 181 00:11:28,490 --> 00:11:31,260 So there's our first free body diagram. 182 00:11:31,260 --> 00:11:40,010 And our second free body diagram, second mass, tension, 183 00:11:40,010 --> 00:11:46,838 normal force, another m2g. 184 00:11:49,782 --> 00:11:51,240 And this one's on wheels, and we're 185 00:11:51,240 --> 00:11:54,030 going to consider this one frictionless. 186 00:11:54,030 --> 00:11:56,280 So we don't have any friction force holding it back. 187 00:12:08,950 --> 00:12:10,454 So the reason I've kind of-- this 188 00:12:10,454 --> 00:12:12,120 seemed like a ridiculous simple problem, 189 00:12:12,120 --> 00:12:13,880 but the point I want to make is not 190 00:12:13,880 --> 00:12:17,240 the solution, not the particular problem, 191 00:12:17,240 --> 00:12:19,390 but an issue that crops up. 192 00:12:19,390 --> 00:12:21,280 How many unknowns are there in this problem. 193 00:12:24,064 --> 00:12:25,456 AUDIENCE: 2. 194 00:12:25,456 --> 00:12:29,560 PROFESSOR: Well, you know n1 immediately, 195 00:12:29,560 --> 00:12:36,060 or n2, or the friction force, or the tension, 196 00:12:36,060 --> 00:12:38,616 or, for that matter, the x double dot we're looking for. 197 00:12:38,616 --> 00:12:40,490 There's actually-- you start off this problem 198 00:12:40,490 --> 00:12:43,440 with five unknowns. 199 00:12:43,440 --> 00:12:48,070 But you're only looking to derive one equation of motion. 200 00:12:48,070 --> 00:12:50,914 So the fact that we're looking for one equation of motion 201 00:12:50,914 --> 00:12:52,330 doesn't say you don't have to deal 202 00:12:52,330 --> 00:12:54,610 with several intermediate equations to get there. 203 00:12:54,610 --> 00:12:57,860 That's just part of the work. 204 00:12:57,860 --> 00:13:11,700 So this is five unknowns to start with, n1, n2, f1, t, 205 00:13:11,700 --> 00:13:12,670 and x double dot. 206 00:13:17,340 --> 00:13:21,850 So if you'll find out that the summation of the forces on y 207 00:13:21,850 --> 00:13:33,100 and the y1 on the first mass, this one, this gives you n1. 208 00:13:33,100 --> 00:13:36,970 Summation of the forces in the y direction on m2, 209 00:13:36,970 --> 00:13:41,270 this one gives you m2 directly. 210 00:13:41,270 --> 00:13:45,400 From this flows directly, you know that the friction force 211 00:13:45,400 --> 00:13:47,710 is mu n1. 212 00:13:47,710 --> 00:13:49,504 See, that's a third equation. 213 00:13:49,504 --> 00:13:51,420 So this gives you one equation, this gives you 214 00:13:51,420 --> 00:13:53,503 another equation, this gives you a third equation. 215 00:13:53,503 --> 00:13:55,630 You have 5 unknowns, you need 5 equations. 216 00:13:55,630 --> 00:13:58,760 So there's 3 of them right off the top. 217 00:13:58,760 --> 00:14:07,790 So this leaves-- you solve for the those, this leaves t 218 00:14:07,790 --> 00:14:10,530 and x double dot to solve for. 219 00:14:18,990 --> 00:14:26,190 So now the sum of the forces on mass 1 in the x direction 220 00:14:26,190 --> 00:14:29,280 says m1 x double dot. 221 00:14:33,040 --> 00:14:37,390 And the sum of the forces on the second mass in the x direction 222 00:14:37,390 --> 00:14:42,080 gives you m2 x double dot. 223 00:14:42,080 --> 00:14:48,060 And I explicitly haven't said this yet, what I've done 224 00:14:48,060 --> 00:14:52,780 is, one of my requirements is x1 equals x2, 225 00:14:52,780 --> 00:14:56,070 and I'm just going to call them x. 226 00:14:56,070 --> 00:14:57,720 Both of these are exactly the same, 227 00:14:57,720 --> 00:14:59,870 both masses have to move with the same motion, 228 00:14:59,870 --> 00:15:01,030 is the assumption. 229 00:15:01,030 --> 00:15:05,935 So that means that x1 double dot equals x2 double dot equals 230 00:15:05,935 --> 00:15:07,310 x double dot, and that's what I'm 231 00:15:07,310 --> 00:15:10,590 assuming when I'm writing down these two equations. 232 00:15:10,590 --> 00:15:14,490 I can write those two equations, one from each free body 233 00:15:14,490 --> 00:15:15,140 diagram. 234 00:15:15,140 --> 00:15:18,130 I've already eliminated three of the unknowns. 235 00:15:18,130 --> 00:15:23,460 And now because I have two equations, each have t in them. 236 00:15:23,460 --> 00:15:35,150 Essentially, you eliminate t and solve for x double dot. 237 00:15:35,150 --> 00:15:39,360 And if you do that in this problem, 238 00:15:39,360 --> 00:15:40,815 you get your equation of motion. 239 00:16:14,550 --> 00:16:17,992 You eliminate t, solve for x1 double dot. 240 00:16:17,992 --> 00:16:19,700 Look at this, and you just look at things 241 00:16:19,700 --> 00:16:21,060 that doesn't make sense. 242 00:16:21,060 --> 00:16:25,120 This says the total mass times the acceleration is a system, 243 00:16:25,120 --> 00:16:27,640 it's one system. 244 00:16:27,640 --> 00:16:30,040 Mass times the acceleration of the center of gravity 245 00:16:30,040 --> 00:16:32,090 of the system, if you will, has got 246 00:16:32,090 --> 00:16:33,930 to be equal to the sums of the forces on it. 247 00:16:33,930 --> 00:16:39,870 Well, it's got m1 plus m2g sine theta pulling it down the hill, 248 00:16:39,870 --> 00:16:47,160 and it has minus m mu m1g cosine theta 249 00:16:47,160 --> 00:16:48,780 dragging it back up the hill. 250 00:16:48,780 --> 00:16:52,180 And that's the entire equation of motion, it makes sense. 251 00:16:52,180 --> 00:16:58,650 But the equation of motion, the thing you're looking for, 252 00:16:58,650 --> 00:17:01,410 is the one that ends up with this acceleration term in it. 253 00:17:01,410 --> 00:17:03,340 If you have multiple degrees of freedom, 254 00:17:03,340 --> 00:17:05,240 multiple coordinates-- if you have, 255 00:17:05,240 --> 00:17:08,470 let's say, three significant equations of motion 256 00:17:08,470 --> 00:17:11,000 that result, there won't necessarily 257 00:17:11,000 --> 00:17:13,180 be one in terms of each coordinate. 258 00:17:13,180 --> 00:17:16,036 They'll have the coordinates mixed in them. 259 00:17:16,036 --> 00:17:18,119 Like we did that that two mass system with springs 260 00:17:18,119 --> 00:17:22,200 the other day, each equation of motion had x1 and x2. 261 00:17:22,200 --> 00:17:24,200 They don't necessarily separate. 262 00:17:24,200 --> 00:17:26,730 They're coupled through their coordinates. 263 00:17:26,730 --> 00:17:29,670 This one, it's one equation of motion for the system, 264 00:17:29,670 --> 00:17:31,640 therefore you have only one coordinate. 265 00:17:31,640 --> 00:17:34,314 But that's not generally true of multiple degree 266 00:17:34,314 --> 00:17:35,105 of freedom systems. 267 00:17:38,100 --> 00:17:43,020 OK, that's your method, though, for a simple problem. 268 00:17:43,020 --> 00:17:47,900 I want to do a little more difficult problem that 269 00:17:47,900 --> 00:17:50,450 involves rotation. 270 00:18:00,964 --> 00:18:02,630 And this is the problem, I'm sure you've 271 00:18:02,630 --> 00:18:04,706 done this problem in physics. 272 00:18:04,706 --> 00:18:07,150 It's a classic problem that people do. 273 00:18:09,922 --> 00:18:26,300 A disk, a pulley, really, supporting two masses 274 00:18:26,300 --> 00:18:31,390 rotates about this point, which I'll call a here. 275 00:18:38,150 --> 00:18:43,760 So at some theta there's no slip, so theta is going 276 00:18:43,760 --> 00:18:47,120 to be related to movement, x. 277 00:18:47,120 --> 00:18:52,820 And I'm going to assign this one a coordinate, x1 going down. 278 00:18:52,820 --> 00:18:56,100 This one a coordinate, x2 going up. 279 00:18:56,100 --> 00:18:58,674 A little foreknowledge here, because you've 280 00:18:58,674 --> 00:18:59,465 worked the problem. 281 00:19:02,240 --> 00:19:05,020 So I want to solve for the motion of this system. 282 00:19:08,120 --> 00:19:13,700 Now again, we need to know the number of degrees of freedom. 283 00:19:13,700 --> 00:19:18,890 So it's the maximum possible, which is our 6m plus 284 00:19:18,890 --> 00:19:22,040 3n minus the constraints. 285 00:19:22,040 --> 00:19:27,092 And let's think about how we want to model this again. 286 00:19:27,092 --> 00:19:29,300 This time I'm just going to model these as particles, 287 00:19:29,300 --> 00:19:31,420 doesn't matter how big they are. 288 00:19:31,420 --> 00:19:33,780 My problem, really, they only go up and down. 289 00:19:33,780 --> 00:19:35,660 So I'm going to model them as particles. 290 00:19:35,660 --> 00:19:41,000 This is 6 times 0 plus 3 times 2 minus constraints. 291 00:19:41,000 --> 00:19:42,620 So 6 minus the constraints. 292 00:19:42,620 --> 00:19:44,980 So the issue is really how many constraints. 293 00:19:47,570 --> 00:19:55,230 Well, we're going to require x1 equal x2. 294 00:19:55,230 --> 00:19:57,756 Cord's taught, doesn't stretch. 295 00:19:57,756 --> 00:19:59,130 If this thing goes down, that has 296 00:19:59,130 --> 00:20:01,787 to go up exactly an equal amount, 297 00:20:01,787 --> 00:20:02,870 and that's one constraint. 298 00:20:11,710 --> 00:20:14,690 In that, leaving us 6 minus 1 is 5, leaving us 299 00:20:14,690 --> 00:20:17,654 with a lot of degrees of freedom here. 300 00:20:17,654 --> 00:20:19,570 Kind of back to the issue I was making before, 301 00:20:19,570 --> 00:20:24,190 are there any constraints in the-- I'll call it 302 00:20:24,190 --> 00:20:29,740 x, y, z directions here. 303 00:20:29,740 --> 00:20:33,030 Are there any constraints in the y or z directions 304 00:20:33,030 --> 00:20:36,280 on either of those masses? 305 00:20:36,280 --> 00:20:39,520 No, I haven't shown any, no tracks, no guides, no anything. 306 00:20:39,520 --> 00:20:42,500 So technically, there are no additional constraints 307 00:20:42,500 --> 00:20:43,710 in this problem. 308 00:20:43,710 --> 00:20:49,560 But if there's no forces in the y, x, I guess y this way, 309 00:20:49,560 --> 00:20:51,820 and no forces in the z, I'm going 310 00:20:51,820 --> 00:20:54,290 to end up with two trivial equations of motion 311 00:20:54,290 --> 00:20:56,980 for this one, 1 in the y, 1 in the z, 312 00:20:56,980 --> 00:20:59,460 for this one, 1 in the y, 1 in the z. 313 00:20:59,460 --> 00:21:01,880 So back to this issue of, there's 314 00:21:01,880 --> 00:21:05,480 a difference between constraints and trivial equations 315 00:21:05,480 --> 00:21:05,980 of motion. 316 00:21:05,980 --> 00:21:08,614 We're going have 4 to reveal equations of motion. 317 00:21:08,614 --> 00:21:10,280 So really, again, I'm going to come down 318 00:21:10,280 --> 00:21:13,860 to one significant equation of motion. 319 00:21:13,860 --> 00:21:23,110 So I have 1 constraint, 4 trivial EOMs, 320 00:21:23,110 --> 00:21:32,390 and 1 significant equation of motion. 321 00:21:32,390 --> 00:21:32,890 OK. 322 00:21:35,440 --> 00:21:39,700 Now, because I want to talk about rotation, 323 00:21:39,700 --> 00:21:41,500 we need to pick a coordinate. 324 00:21:41,500 --> 00:21:49,020 Now I can pick-- I can either let x1 equal x2 325 00:21:49,020 --> 00:21:52,000 and just let it be the sum x, a single coordinate, 326 00:21:52,000 --> 00:21:55,880 or theta, the rotation. 327 00:21:55,880 --> 00:21:57,960 I'm actually going to use x for a second. 328 00:22:05,250 --> 00:22:09,990 But there's 2 obvious ways to approach this problem. 329 00:22:09,990 --> 00:22:15,140 One is to draw a free body diagrams 330 00:22:15,140 --> 00:22:22,160 of each of these masses, sum the forces on each, and how many-- 331 00:22:22,160 --> 00:22:25,110 if I do that, how many unknowns do I end up with? 332 00:22:29,120 --> 00:22:30,990 We can draw the free-- here's the free body 333 00:22:30,990 --> 00:22:32,280 diagram for mass 1. 334 00:22:32,280 --> 00:22:33,185 What's it got on it? 335 00:22:33,185 --> 00:22:34,950 Well, m1g. 336 00:22:34,950 --> 00:22:37,064 What else is acting on it? 337 00:22:37,064 --> 00:22:38,972 AUDIENCE: [? Tension. ?] 338 00:22:38,972 --> 00:22:44,590 PROFESSOR: And here's the second one, m2g, 339 00:22:44,590 --> 00:22:47,100 and tension acting on it. 340 00:22:47,100 --> 00:22:50,090 So now we can sum, sum of the forces 341 00:22:50,090 --> 00:22:52,520 equals mass times the acceleration of each one, 342 00:22:52,520 --> 00:22:54,840 and the external forces are going to involve t. 343 00:22:54,840 --> 00:22:57,810 So you're going to end up with how many unknowns? 344 00:23:02,960 --> 00:23:04,290 How many unknowns? 345 00:23:04,290 --> 00:23:07,380 I can write two equations for the sum of the forces 346 00:23:07,380 --> 00:23:08,660 in the x direction. 347 00:23:08,660 --> 00:23:11,520 x double dot is certainly an unknown. 348 00:23:11,520 --> 00:23:12,020 What else? 349 00:23:12,020 --> 00:23:12,500 AUDIENCE: t. 350 00:23:12,500 --> 00:23:13,041 PROFESSOR: t. 351 00:23:13,041 --> 00:23:15,269 So I end up with this other [? end. ?] 352 00:23:15,269 --> 00:23:17,560 So that means I'm going to have to write two equations, 353 00:23:17,560 --> 00:23:18,900 I'm going to have to eliminate t, 354 00:23:18,900 --> 00:23:20,608 going to go through the same thing there. 355 00:23:20,608 --> 00:23:22,670 So I don't want to bother with that. 356 00:23:22,670 --> 00:23:25,485 Is there another way to do this problem? 357 00:23:32,270 --> 00:23:35,980 This is a problem where you can use angular momentum 358 00:23:35,980 --> 00:23:38,440 and not have to deal with t at all. 359 00:23:38,440 --> 00:23:40,029 So let's set that problem up. 360 00:23:44,620 --> 00:23:49,120 You know that the sum of the torques about that point 361 00:23:49,120 --> 00:23:55,380 a, with respect to point a, it's going to be derivative, 362 00:23:55,380 --> 00:23:57,910 and since we're dealing with particles here, 363 00:23:57,910 --> 00:24:01,422 of the angular momentum with respect to-- I'll 364 00:24:01,422 --> 00:24:04,540 just call it lowercase h for particles. 365 00:24:04,540 --> 00:24:09,480 Plus this velocity of a with respect 366 00:24:09,480 --> 00:24:16,920 to an inertial frame across a linear momentum with respect 367 00:24:16,920 --> 00:24:18,440 to an inertial frame. 368 00:24:18,440 --> 00:24:21,530 That's the full equation for sum of torques. 369 00:24:21,530 --> 00:24:23,830 What's velocity of a with respect to o in this problem? 370 00:24:26,930 --> 00:24:27,670 0. 371 00:24:27,670 --> 00:24:29,420 Fortunately, this is one of those problems 372 00:24:29,420 --> 00:24:32,770 where you can get rid of this difficult second term. 373 00:24:32,770 --> 00:24:35,990 So it's just torques as the time derivative of the angular 374 00:24:35,990 --> 00:24:37,100 momentum. 375 00:24:37,100 --> 00:24:44,080 So we need an expression, then, for both the sum of the torques 376 00:24:44,080 --> 00:24:45,970 with respect to a. 377 00:24:45,970 --> 00:24:47,390 Let's see, what would that be? 378 00:24:50,950 --> 00:24:57,290 So now the external torques with respect to-- I'll 379 00:24:57,290 --> 00:25:03,220 finish my-- oops, come here you-- free body diagram. 380 00:25:03,220 --> 00:25:06,870 So now what I really want is a free body diagram 381 00:25:06,870 --> 00:25:07,795 of the whole system. 382 00:25:10,990 --> 00:25:13,790 So here's the whole system created as one thing. 383 00:25:13,790 --> 00:25:20,850 You have a force down, m1g, another force down, m2g. 384 00:25:20,850 --> 00:25:25,630 Up here you have some normal force up, 385 00:25:25,630 --> 00:25:28,350 that's the support of the pin. 386 00:25:28,350 --> 00:25:38,820 You have tensions in these, but now this equation 387 00:25:38,820 --> 00:25:40,550 applies to the system. 388 00:25:40,550 --> 00:25:43,910 The ts are internal to the system, they are irrelevant. 389 00:25:46,640 --> 00:25:49,360 So I'm talking about this whole thing treated as a system, 390 00:25:49,360 --> 00:25:50,860 and I'm going to compute the moments 391 00:25:50,860 --> 00:25:53,620 about point a, which is right there, where that axle is. 392 00:25:53,620 --> 00:25:56,830 Does n create a moment at the axle? 393 00:25:56,830 --> 00:26:03,260 Nope, but the m1 and m2 times g create moments. 394 00:26:03,260 --> 00:26:04,160 Sure, OK. 395 00:26:07,467 --> 00:26:09,300 I'm going to have positive out of the board, 396 00:26:09,300 --> 00:26:13,370 the positive moment, positive angular direction. 397 00:26:13,370 --> 00:26:16,090 So the torques applied to this system 398 00:26:16,090 --> 00:26:22,630 are r cross t, so you're going to end up with-- I 399 00:26:22,630 --> 00:26:26,350 want to summarize these. 400 00:26:26,350 --> 00:26:33,734 An m1g, and I didn't write the radius on this problem, 401 00:26:33,734 --> 00:26:39,345 but at some radius, capital R. So the torques that are m1g 402 00:26:39,345 --> 00:26:49,190 are positive minus m2gR, k hat direction, 403 00:26:49,190 --> 00:26:53,460 and that must be equal to the time derivative of the angular 404 00:26:53,460 --> 00:26:57,570 momentum about a. 405 00:26:57,570 --> 00:27:04,470 Now we need an expression for the angular 406 00:27:04,470 --> 00:27:06,221 momentum with respect to a. 407 00:27:09,590 --> 00:27:16,260 Angular momentum is, in general, this 408 00:27:16,260 --> 00:27:19,940 is a r cross linear momentum, right. 409 00:27:19,940 --> 00:27:28,940 So R for mass 1 with respect to a cross 410 00:27:28,940 --> 00:27:34,650 the momentum of that second mass with respect 411 00:27:34,650 --> 00:27:35,880 to an inertial frame. 412 00:27:35,880 --> 00:27:38,340 And a and the inertial frame are the same thing, 413 00:27:38,340 --> 00:27:40,340 a sticks in the inertial frame. 414 00:27:40,340 --> 00:27:42,970 But angular momentum is always with respect 415 00:27:42,970 --> 00:27:44,690 to the inertial frame. 416 00:27:44,690 --> 00:27:50,070 Plus the second piece, which is R of m2 with respect to a 417 00:27:50,070 --> 00:27:54,760 crossed with p for mass 2 with respect to some inertial frame. 418 00:28:01,230 --> 00:28:03,230 I'm just going to give you the results for this. 419 00:28:07,020 --> 00:28:18,516 m1 plus m2 R x dot k. 420 00:28:18,516 --> 00:28:25,680 So x dot is this velocity, R cross and mass times 421 00:28:25,680 --> 00:28:30,360 velocity is momentum, so the perpendicular radius 422 00:28:30,360 --> 00:28:33,460 to that is the radius, R. So it's 423 00:28:33,460 --> 00:28:35,830 Rx dot times m, shouldn't surprise you, 424 00:28:35,830 --> 00:28:36,850 in the k hat direction. 425 00:28:36,850 --> 00:28:38,940 That's the total angular momentum 426 00:28:38,940 --> 00:28:43,970 that comes from these two particles with respect to a. 427 00:28:43,970 --> 00:28:45,785 And taking their time derivative. 428 00:28:50,000 --> 00:28:53,230 These are constant, that's a constant, this is not, 429 00:28:53,230 --> 00:28:55,660 this is a constant, but it doesn't change direction, 430 00:28:55,660 --> 00:28:57,420 so this one is pretty simple. 431 00:29:04,970 --> 00:29:12,050 Now I can set equal the sum of the external torques, that, 432 00:29:12,050 --> 00:29:14,990 to the time derivative of the angular momentum, 433 00:29:14,990 --> 00:29:16,845 just to fulfill this expression. 434 00:29:19,450 --> 00:29:26,170 And in so doing, I end up with a solution for x double dot. 435 00:29:26,170 --> 00:29:36,520 m1 minus m2, m1 plus m2, times g. 436 00:29:36,520 --> 00:29:37,935 Turns out the R goes away. 437 00:29:41,950 --> 00:29:46,130 So one equation, never had to mess with tension. 438 00:29:46,130 --> 00:29:52,070 This is a pretty nice, direct way of solving this problem. 439 00:29:52,070 --> 00:29:56,674 If you solve for g here, and you measure x double dot, 440 00:29:56,674 --> 00:29:59,380 this actually gives you an experimental way 441 00:29:59,380 --> 00:30:02,140 of determining acceleration of gravity. 442 00:30:02,140 --> 00:30:04,790 It's actually what this thing was used for a long time 443 00:30:04,790 --> 00:30:07,930 ago, before they had a lot of the measurement techniques 444 00:30:07,930 --> 00:30:09,170 and things that we do today. 445 00:30:09,170 --> 00:30:12,630 This is a way of determining the acceleration of gravity. 446 00:30:12,630 --> 00:30:14,580 So these two masses are quite close together. 447 00:30:14,580 --> 00:30:18,514 This number is pretty small, you can, however 448 00:30:18,514 --> 00:30:19,805 accurate your timing device is. 449 00:30:23,970 --> 00:30:27,520 Now, just to mention it, I neglected something in this. 450 00:30:27,520 --> 00:30:29,440 I assumed something and I didn't even say it. 451 00:30:29,440 --> 00:30:29,970 What was it? 452 00:30:29,970 --> 00:30:31,511 What would screw up this measurement? 453 00:30:31,511 --> 00:30:33,760 I'm trying to measure the acceleration of gravity, 454 00:30:33,760 --> 00:30:38,780 if I built this apparatus, would I get a very good measurement? 455 00:30:38,780 --> 00:30:41,160 AUDIENCE: The pulley would have to be massless. 456 00:30:41,160 --> 00:30:43,510 PROFESSOR: Yeah, the pulley would have to be massless. 457 00:30:43,510 --> 00:30:45,260 I've made an assumption about that, right. 458 00:30:45,260 --> 00:30:47,040 So how would you fix up this equation 459 00:30:47,040 --> 00:30:48,642 to account for the pulley? 460 00:30:51,946 --> 00:30:54,306 AUDIENCE: You'd have to take into account 461 00:30:54,306 --> 00:30:55,260 its moment of inertia. 462 00:30:55,260 --> 00:30:56,926 PROFESSOR: Yeah, you'd put in something. 463 00:30:56,926 --> 00:30:59,750 And where would that go into the problem? 464 00:30:59,750 --> 00:31:02,765 How would you account its inertia, moment of inertia 465 00:31:02,765 --> 00:31:03,410 in the problem? 466 00:31:03,410 --> 00:31:04,150 AUDIENCE: ha. 467 00:31:04,150 --> 00:31:06,250 PROFESSOR: Yeah, you'd just put it into ha. 468 00:31:06,250 --> 00:31:11,080 So this expression for h would end up with one more term, 469 00:31:11,080 --> 00:31:14,834 it's going to look like-- well, when you take the time 470 00:31:14,834 --> 00:31:16,250 derivative, you're going to end up 471 00:31:16,250 --> 00:31:17,620 with another piece over here. 472 00:31:17,620 --> 00:31:21,750 Some i about a theta double dot. 473 00:31:21,750 --> 00:31:24,260 You're going to have to relate theta double dot to x 474 00:31:24,260 --> 00:31:28,930 double dot, which you can, because x equals R theta. 475 00:31:28,930 --> 00:31:30,660 x double dot is R theta double dot. 476 00:31:30,660 --> 00:31:33,890 You could fix that and you'd have an equation of motion. 477 00:31:33,890 --> 00:31:37,340 But that means we need to know about i about a, 478 00:31:37,340 --> 00:31:42,720 that's where we're going to at the end of this lecture 479 00:31:42,720 --> 00:31:45,470 and for the next several lectures. 480 00:31:45,470 --> 00:31:46,900 OK. 481 00:31:46,900 --> 00:31:52,070 That's that example, and I've got two more brief ones 482 00:31:52,070 --> 00:31:54,050 that I wanted to talk about. 483 00:31:54,050 --> 00:31:55,841 Any last questions? 484 00:31:55,841 --> 00:31:56,340 Yeah. 485 00:31:56,340 --> 00:31:58,200 AUDIENCE: Can you explain again why 486 00:31:58,200 --> 00:31:59,991 you didn't take the tensions into account 487 00:31:59,991 --> 00:32:00,990 for your sum of torques? 488 00:32:00,990 --> 00:32:05,680 PROFESSOR: OK, so why did I not take the tensions into account? 489 00:32:12,520 --> 00:32:21,830 So I can write the equation of motion 490 00:32:21,830 --> 00:32:26,470 for this thing as a complete system. 491 00:32:26,470 --> 00:32:29,170 One, the masses and the pulley are all 492 00:32:29,170 --> 00:32:35,300 the same thing, the summation of the external torques on that, 493 00:32:35,300 --> 00:32:39,100 they're going to amount up to taking into account the time 494 00:32:39,100 --> 00:32:41,870 rate of change of the angular momentum of the system. 495 00:32:41,870 --> 00:32:43,550 Now, if I didn't understand that, 496 00:32:43,550 --> 00:32:46,614 I could have blindly gone ahead and put the ts in there, right, 497 00:32:46,614 --> 00:32:49,030 they would've been exactly equal and opposite with respect 498 00:32:49,030 --> 00:32:50,860 to a and it would have cancelled out. 499 00:32:50,860 --> 00:32:55,092 So either way, if you're not sure about that assumption, 500 00:32:55,092 --> 00:32:56,550 you could just put them in and they 501 00:32:56,550 --> 00:33:00,570 would appear in the torque equation, but as a minus 502 00:33:00,570 --> 00:33:02,800 tR and a plus tR and they'd cancel. 503 00:33:05,420 --> 00:33:11,590 I want to move on to a third example, 504 00:33:11,590 --> 00:33:17,850 and this is the third item that I want to clear up, 505 00:33:17,850 --> 00:33:19,190 loose ends I'm calling them. 506 00:33:19,190 --> 00:33:21,370 The muddy cards are really useful. 507 00:33:21,370 --> 00:33:25,100 I get questions in those that spark something. 508 00:33:25,100 --> 00:33:27,670 And this is a question that came up two or three times 509 00:33:27,670 --> 00:33:30,030 in the muddy cards and I haven't addressed it, 510 00:33:30,030 --> 00:33:34,540 and that is, we were working with rotor problems. 511 00:33:34,540 --> 00:33:37,580 And remember this problem. 512 00:33:37,580 --> 00:33:39,860 You have the rotor, it had an arm, 513 00:33:39,860 --> 00:33:42,770 I did it this way to make some things obvious. 514 00:33:42,770 --> 00:33:49,780 But this is the z direction, it's rotating about that axis. 515 00:33:49,780 --> 00:33:53,680 I've got a point mass up here. 516 00:33:53,680 --> 00:33:57,485 r hat, so this is R-- actually, I'm 517 00:33:57,485 --> 00:33:59,580 going to make it a capital R so it's easier 518 00:33:59,580 --> 00:34:01,200 to distinguish from the r hat. 519 00:34:01,200 --> 00:34:02,210 And this is z. 520 00:34:08,620 --> 00:34:18,929 This thing's rotating, it's got bearings here to keep it going. 521 00:34:18,929 --> 00:34:24,449 And we talked about torques, so this 522 00:34:24,449 --> 00:34:38,030 is my point A. I want to write the sum of the torques 523 00:34:38,030 --> 00:34:44,810 about A, time derivative of the angular momentum. 524 00:34:44,810 --> 00:34:47,060 We've done this problem before, so I'm just putting up 525 00:34:47,060 --> 00:34:51,860 a couple of points for review to clear up 526 00:34:51,860 --> 00:34:54,066 some possible misconceptions. 527 00:34:56,989 --> 00:35:01,140 That's this term, so what about point A now in this problem? 528 00:35:01,140 --> 00:35:03,101 What's the velocity at point A? 529 00:35:03,101 --> 00:35:06,600 0, so again we can get rid of this guy. 530 00:35:06,600 --> 00:35:08,600 I'm going to come back to this and do an example 531 00:35:08,600 --> 00:35:10,220 one of these days where this isn't 0, 532 00:35:10,220 --> 00:35:14,920 where it's really handy to be able to do a problem 533 00:35:14,920 --> 00:35:17,710 where that's not 0. 534 00:35:17,710 --> 00:35:18,210 OK. 535 00:35:21,100 --> 00:35:22,580 This is true. 536 00:35:22,580 --> 00:35:25,870 I need a free body diagram of our little mass, 537 00:35:25,870 --> 00:35:30,240 so here's my free body diagram. 538 00:35:30,240 --> 00:35:35,220 And it has possibly a force in the z direction. 539 00:35:35,220 --> 00:35:37,410 That comes from the rod, there's rods that's 540 00:35:37,410 --> 00:35:39,090 supporting this thing, right. 541 00:35:39,090 --> 00:35:41,670 There's possibly a force in the z direction. 542 00:35:41,670 --> 00:35:45,750 There's a force in the r hat direction, in the R direction. 543 00:35:45,750 --> 00:35:54,950 There's a force in theta direction going into the board. 544 00:35:54,950 --> 00:35:55,975 And there's mg. 545 00:36:01,110 --> 00:36:04,660 All sorts of forces on this thing. 546 00:36:04,660 --> 00:36:09,420 And the question was asked, when we did this problem before 547 00:36:09,420 --> 00:36:13,850 and did the time derivatives of the angular momentum, 548 00:36:13,850 --> 00:36:16,660 we found that we got-- there's three terms 549 00:36:16,660 --> 00:36:19,540 and I'll write them down here for you. 550 00:36:19,540 --> 00:36:22,750 I'm just saying in advance what we're going to do. 551 00:36:22,750 --> 00:36:24,570 When you solve this problem, you find out 552 00:36:24,570 --> 00:36:27,520 that it takes to torque to accelerate this shaft and spin. 553 00:36:27,520 --> 00:36:29,840 That the driving one, that's what makes it happen, 554 00:36:29,840 --> 00:36:30,910 makes it accelerate. 555 00:36:30,910 --> 00:36:34,810 We had two more terms that were torques at this point, that 556 00:36:34,810 --> 00:36:36,590 is what it takes to support this system. 557 00:36:36,590 --> 00:36:39,800 It's trying to bend out, it's trying to bend back, 558 00:36:39,800 --> 00:36:41,300 those are torques that show up here. 559 00:36:41,300 --> 00:36:43,910 And we actually get them when we work through this. 560 00:36:43,910 --> 00:36:48,180 But we don't get something that tells us 561 00:36:48,180 --> 00:36:51,062 about the moment the torque created 562 00:36:51,062 --> 00:36:52,270 this point caused by gravity. 563 00:36:55,890 --> 00:36:58,870 The question was, why don't we get the torque about this point 564 00:36:58,870 --> 00:36:59,680 caused by gravity. 565 00:36:59,680 --> 00:37:02,920 There's clearly mg down, there's clearly a moment arm. 566 00:37:02,920 --> 00:37:05,490 So mgR is the torque about this point. 567 00:37:05,490 --> 00:37:08,770 And if you were doing the statics problem in 2.001, 568 00:37:08,770 --> 00:37:10,890 there'd be a torque around this point caused 569 00:37:10,890 --> 00:37:13,015 by the weight of this thing just sitting there, not 570 00:37:13,015 --> 00:37:14,170 even spinning. 571 00:37:14,170 --> 00:37:19,430 And what we're doing here gives you no help with that. 572 00:37:19,430 --> 00:37:24,730 But just for the quick review of this problem, 573 00:37:24,730 --> 00:37:29,030 more in the line of helping you think about the quiz. 574 00:37:29,030 --> 00:37:33,660 This then is R, we'll call this point B 575 00:37:33,660 --> 00:37:37,270 and this is point A, remember this is RB with respect 576 00:37:37,270 --> 00:37:43,850 to A cross p with respect to o. 577 00:37:43,850 --> 00:37:46,610 And that's where our angular momentum comes from. 578 00:37:46,610 --> 00:37:57,460 In this problem that is r hat plus z k hat cross m times 579 00:37:57,460 --> 00:38:03,290 the velocity, which is R omega z. 580 00:38:03,290 --> 00:38:08,030 And that must be in the theta hat direction. 581 00:38:08,030 --> 00:38:19,070 When you multiply these out, omega z is theta dot. 582 00:38:19,070 --> 00:38:22,850 They're kind of interchangeable in this problem. 583 00:38:22,850 --> 00:38:26,010 So when you multiply this out, you get two terms. 584 00:38:26,010 --> 00:38:41,230 mR squared theta dot k hat minus mRz theta dot r hat. 585 00:38:41,230 --> 00:38:42,560 Two terms from this. 586 00:38:45,360 --> 00:38:53,330 And when you do the time derivative of the dhdt, 587 00:38:53,330 --> 00:38:54,200 you get three terms. 588 00:38:56,710 --> 00:39:02,017 mR squared theta double dot and the k. 589 00:39:02,017 --> 00:39:03,350 Now, why do you get three terms? 590 00:39:03,350 --> 00:39:06,970 Because this term has two variables in it that 591 00:39:06,970 --> 00:39:10,490 are functions of time, theta dot has a derivative, 592 00:39:10,490 --> 00:39:14,550 and r hat has a derivative, because it rotates. 593 00:39:14,550 --> 00:39:16,320 So one of the key bits of mathematics 594 00:39:16,320 --> 00:39:18,529 you have to learn in this course, 595 00:39:18,529 --> 00:39:20,570 I'm kind of giving you a little quiz review here, 596 00:39:20,570 --> 00:39:23,030 you need to know how to take the derivative of a rotating 597 00:39:23,030 --> 00:39:24,210 vector. 598 00:39:24,210 --> 00:39:26,250 And that's what we do here, gives us 599 00:39:26,250 --> 00:39:34,970 two terms minus mR z theta double dot r hat 600 00:39:34,970 --> 00:39:43,500 minus mRz theta dot squared theta hat. 601 00:39:43,500 --> 00:39:47,260 So three terms in this time derivative of the angular 602 00:39:47,260 --> 00:39:52,400 momentum, and they have to be equal to the external torques. 603 00:39:52,400 --> 00:39:56,470 This is equal to the summation of the torques about A, 604 00:39:56,470 --> 00:39:58,610 the external torques. 605 00:39:58,610 --> 00:40:01,980 Well, you'll need a torque in the k direction. 606 00:40:01,980 --> 00:40:04,320 That's what it takes to accelerate the thing, 607 00:40:04,320 --> 00:40:05,110 make it go faster. 608 00:40:09,590 --> 00:40:14,890 This mass has a force on it to make 609 00:40:14,890 --> 00:40:18,500 it go faster, that's this f in the theta hat direction. 610 00:40:18,500 --> 00:40:21,650 And that rods have a push on that mass, 611 00:40:21,650 --> 00:40:23,825 the mass pushes back on the rod. 612 00:40:23,825 --> 00:40:25,200 So if in the theta direction it's 613 00:40:25,200 --> 00:40:28,290 like that, the mass pushes back on the rod, it twists the rod, 614 00:40:28,290 --> 00:40:29,060 or tries to. 615 00:40:29,060 --> 00:40:33,936 That's a torque about this in the r hat direction. 616 00:40:33,936 --> 00:40:35,435 So there's centripetal acceleration, 617 00:40:35,435 --> 00:40:38,590 it takes force to cause centripetal acceleration. 618 00:40:38,590 --> 00:40:41,270 It's that force is inward. 619 00:40:41,270 --> 00:40:44,020 It's about a moment arm z, and so this gives you 620 00:40:44,020 --> 00:40:47,000 a torque about the point A in the theta hat direction. 621 00:40:47,000 --> 00:40:50,500 So these are three different terms, each one has a purpose. 622 00:40:50,500 --> 00:40:52,450 No work is done here, no work is done here, 623 00:40:52,450 --> 00:40:53,616 because there's no movement. 624 00:40:56,990 --> 00:40:59,830 Now, but gravity, we started this question 625 00:40:59,830 --> 00:41:01,620 as why doesn't gravity pop out of this. 626 00:41:04,220 --> 00:41:07,220 Because this only tells you about the time rate 627 00:41:07,220 --> 00:41:09,820 of change of angular momentum. 628 00:41:09,820 --> 00:41:13,810 Gravity has nothing to do with angular momentum. 629 00:41:13,810 --> 00:41:17,870 r cross p is all that angular momentum is. 630 00:41:17,870 --> 00:41:19,630 The linear momentum of little 631 00:41:19,630 --> 00:41:24,060 clumps of mass times the radius from the point you're 632 00:41:24,060 --> 00:41:26,000 computing the angular momentum. 633 00:41:26,000 --> 00:41:28,060 Has nothing to do with g, never will. 634 00:41:30,760 --> 00:41:35,970 You'll never get the g related static moment 635 00:41:35,970 --> 00:41:38,200 out of this equation. 636 00:41:38,200 --> 00:41:41,292 It's there, though, and if you were designing the system, 637 00:41:41,292 --> 00:41:42,750 you'd have to take it into account. 638 00:42:08,530 --> 00:42:12,270 So remember, I didn't bring it today, but I have my shaker. 639 00:42:12,270 --> 00:42:13,635 I've bolted it to the floor. 640 00:42:18,120 --> 00:42:22,720 Inside of that shaker is a little rotating mass. 641 00:42:22,720 --> 00:42:25,570 It has a little arm and eccentricity, it 642 00:42:25,570 --> 00:42:29,916 has some mass that I'm going to make m, 643 00:42:29,916 --> 00:42:35,250 it's rotating theta direction. 644 00:42:35,250 --> 00:42:39,000 And it rotates a constant speed. 645 00:42:39,000 --> 00:42:44,940 So it's some constant omega, theta double dot equals 0. 646 00:42:44,940 --> 00:42:47,040 So I just got my shaker bolted to the floors, 647 00:42:47,040 --> 00:42:49,650 it's putting a lot of vibration into the floor. 648 00:42:49,650 --> 00:42:51,920 And the question that someone came up 649 00:42:51,920 --> 00:42:55,660 with on a muddy card that was a really inside insightful 650 00:42:55,660 --> 00:42:59,960 question, why-- or they didn't say why-- 651 00:42:59,960 --> 00:43:03,810 they said, shouldn't the torque required 652 00:43:03,810 --> 00:43:07,960 to drive this thing somehow be affected by gravity? 653 00:43:11,710 --> 00:43:14,320 So does the torque that it takes to run 654 00:43:14,320 --> 00:43:17,710 this around and around depend on gravity, 655 00:43:17,710 --> 00:43:19,980 was the question that was asked. 656 00:43:19,980 --> 00:43:21,910 Let's take a quick look at that. 657 00:43:21,910 --> 00:43:25,000 We just discovered that dhdt doesn't tell you anything 658 00:43:25,000 --> 00:43:28,030 about torque from gravity, right? 659 00:43:28,030 --> 00:43:30,170 Well, let's see what happens then. 660 00:43:30,170 --> 00:43:35,000 So the summation of the external torques-- I'll call this now 661 00:43:35,000 --> 00:43:37,080 point A, where it's rotating about. 662 00:43:37,080 --> 00:43:39,720 This point now doesn't move in this problem. 663 00:43:39,720 --> 00:43:41,430 It's an inertial point. 664 00:43:41,430 --> 00:43:44,460 Summation of the torques with respect to A 665 00:43:44,460 --> 00:43:49,200 is dh with respect to A, dt, and there's no additional terms 666 00:43:49,200 --> 00:43:53,370 because that velocity point is 0. 667 00:43:53,370 --> 00:44:01,100 And that's d by dt, the torque is just r cross p, 668 00:44:01,100 --> 00:44:05,200 so that is me theta dot. 669 00:44:05,200 --> 00:44:07,435 That's the velocity, that's the momentum. 670 00:44:21,100 --> 00:44:22,805 I've left out something. 671 00:44:31,650 --> 00:44:36,670 So r cross p, I need an e squared in here. 672 00:44:36,670 --> 00:44:44,730 me squared theta dot in the k hat direction. 673 00:44:44,730 --> 00:44:46,590 I need to take the time derivative of that. 674 00:44:46,590 --> 00:44:48,923 That's a constant, that's a constant, that's a constant, 675 00:44:48,923 --> 00:44:50,940 it only comes from this term. 676 00:44:53,470 --> 00:44:58,980 And that gives me me squared theta double dot k hat 677 00:44:58,980 --> 00:45:01,960 direction, and that's got to be equal to the sum of the torques 678 00:45:01,960 --> 00:45:04,640 in the system, the external torques. 679 00:45:04,640 --> 00:45:05,390 And what are they? 680 00:45:09,790 --> 00:45:11,400 So torques about this point. 681 00:45:11,400 --> 00:45:16,250 So axial forces in this thing contribute no torques, 682 00:45:16,250 --> 00:45:19,290 transverse forces, external forces only 683 00:45:19,290 --> 00:45:25,100 come from the mg on this thing. 684 00:45:25,100 --> 00:45:28,510 So my torques on this system, there 685 00:45:28,510 --> 00:45:30,550 is some mechanical torque being applied. 686 00:45:30,550 --> 00:45:33,830 That's what I'm looking for. 687 00:45:33,830 --> 00:45:35,460 I've got a motor driving this thing, 688 00:45:35,460 --> 00:45:47,290 so there's some t of t in there, some torque, minus mge cosine 689 00:45:47,290 --> 00:45:49,270 theta is the moment arm. 690 00:45:49,270 --> 00:45:51,700 So there's this force, there's this moment arm 691 00:45:51,700 --> 00:45:54,570 is e cosine theta. 692 00:45:54,570 --> 00:45:57,260 So this is the external torque caused by gravity, 693 00:45:57,260 --> 00:46:01,766 but all of this equals what? 694 00:46:01,766 --> 00:46:03,665 What's theta double dot? 695 00:46:03,665 --> 00:46:04,165 0. 696 00:46:08,713 --> 00:46:17,340 The external torque is mge cosine omega t, 697 00:46:17,340 --> 00:46:19,950 theta is omega t. 698 00:46:19,950 --> 00:46:23,560 And so indeed, as this thing goes around, 699 00:46:23,560 --> 00:46:27,170 when it's coming up, you've got to apply enough torque 700 00:46:27,170 --> 00:46:29,020 to lift it against gravity. 701 00:46:29,020 --> 00:46:31,269 When it clears the top, gravity is helping it, 702 00:46:31,269 --> 00:46:32,560 it's going down the other side. 703 00:46:32,560 --> 00:46:35,640 So in fact, if you plotted the torque as a function of time 704 00:46:35,640 --> 00:46:40,040 for this system, it's like this. 705 00:46:40,040 --> 00:46:42,220 It's just lifting that mass up and down. 706 00:46:42,220 --> 00:46:43,970 Then, of course, if there's any friction 707 00:46:43,970 --> 00:46:45,320 in this thing, et cetera, it's going 708 00:46:45,320 --> 00:46:47,570 to have to apply a little bit of torque for that, too. 709 00:46:47,570 --> 00:46:50,640 But indeed, this is an insightful question 710 00:46:50,640 --> 00:46:55,850 that someone asked, is that the gravity does 711 00:46:55,850 --> 00:46:57,510 have to enter into this thing. 712 00:46:57,510 --> 00:47:00,220 So there will be a torque that the motor has to supply 713 00:47:00,220 --> 00:47:02,420 to drive this thing in gravity. 714 00:47:02,420 --> 00:47:02,920 Yeah. 715 00:47:02,920 --> 00:47:06,798 AUDIENCE: Should that expression also have-- 716 00:47:06,798 --> 00:47:10,150 the expression for torque also me squared theta double dot-- 717 00:47:10,150 --> 00:47:12,980 PROFESSOR: Ah, now theta double dot is? 718 00:47:12,980 --> 00:47:14,830 Yeah, see, it would. 719 00:47:14,830 --> 00:47:17,450 If this thing was spinning up and I 720 00:47:17,450 --> 00:47:21,180 was trying to account for the torque required to spin it up, 721 00:47:21,180 --> 00:47:23,070 then here is. 722 00:47:23,070 --> 00:47:26,440 Then I would include that, this would be an equation of motion 723 00:47:26,440 --> 00:47:28,630 that says all these things are true, 724 00:47:28,630 --> 00:47:31,200 and I can solve for torque again. 725 00:47:31,200 --> 00:47:34,480 And it will allow me to decide how fast I could spin it up. 726 00:47:34,480 --> 00:47:36,230 If I have a dinky little motor, it 727 00:47:36,230 --> 00:47:37,770 doesn't spin up very fast, if I had 728 00:47:37,770 --> 00:47:40,061 a really powerful motor that could really put it to it, 729 00:47:40,061 --> 00:47:41,170 spin it up quickly. 730 00:47:41,170 --> 00:47:41,670 OK. 731 00:47:57,410 --> 00:48:00,490 So now I want to move on to the third topic, which 732 00:48:00,490 --> 00:48:03,640 is to kind of go back to where I left off last time, talking 733 00:48:03,640 --> 00:48:07,270 about we need to move on from particles to rigid bodies 734 00:48:07,270 --> 00:48:09,640 so we can do more interesting problems. 735 00:48:09,640 --> 00:48:12,530 So I want to pick up with the subject of angular 736 00:48:12,530 --> 00:48:13,865 momentum for rigid bodies. 737 00:48:32,580 --> 00:48:35,990 Now last time I just barely scratched the surface of this. 738 00:48:35,990 --> 00:48:38,060 And lots of muddy cards said I don't get it. 739 00:48:38,060 --> 00:48:41,010 I didn't expect you to get it with it being half baked 740 00:48:41,010 --> 00:48:42,980 and the first time you've seen it. 741 00:48:42,980 --> 00:48:46,840 So we're going to continue and we won't finish today. 742 00:48:46,840 --> 00:48:53,620 So let's think about a general rigid body. 743 00:48:53,620 --> 00:49:05,020 Here's my inertial system, got a body out here 744 00:49:05,020 --> 00:49:06,570 that's rotating about some point, 745 00:49:06,570 --> 00:49:08,880 A. A could even be outside the body 746 00:49:08,880 --> 00:49:10,135 and have it rotate about it. 747 00:49:13,520 --> 00:49:15,620 And attached to A is a reference frame. 748 00:49:21,070 --> 00:49:23,280 Little x, little y, little z. 749 00:49:23,280 --> 00:49:25,490 So my Axy frame. 750 00:49:25,490 --> 00:49:32,650 Now, I put up last time, there's two pages out of Williams 751 00:49:32,650 --> 00:49:37,190 which gives the equations for the moment 752 00:49:37,190 --> 00:49:43,920 and products of inertia in terms of summations of masses times 753 00:49:43,920 --> 00:49:45,750 particle locations. 754 00:49:45,750 --> 00:49:48,090 And in order to do that, Williams defines 755 00:49:48,090 --> 00:49:50,250 a coordinate system on this body, 756 00:49:50,250 --> 00:49:53,310 and that coordinate system is fixed to the body, 757 00:49:53,310 --> 00:49:55,470 rotates to the body, and Williams calls that 758 00:49:55,470 --> 00:49:57,050 coordinate system little oxyz. 759 00:50:00,160 --> 00:50:04,660 In his book, he calls the inertial frame big Oxyz. 760 00:50:04,660 --> 00:50:06,670 It's really hard to do that on a blackboard 761 00:50:06,670 --> 00:50:09,170 and how you'd be able to tell it apart, OK. 762 00:50:09,170 --> 00:50:15,630 So I'm going to depart, and my frame in here is an Axyz frame. 763 00:50:15,630 --> 00:50:17,860 But A and o, if you're reading that handout, 764 00:50:17,860 --> 00:50:19,220 are the same thing. 765 00:50:19,220 --> 00:50:20,370 A and little o. 766 00:50:20,370 --> 00:50:22,960 It's a frame fixed to the body that's rotating with it. 767 00:50:34,260 --> 00:50:39,680 We can write angular momentum for rigid bodies 768 00:50:39,680 --> 00:50:43,820 as a vector hx having a component in the i direction, j 769 00:50:43,820 --> 00:50:54,140 direction, and these coordinates as the product 770 00:50:54,140 --> 00:50:58,070 of a matrix of constants. 771 00:50:58,070 --> 00:51:02,930 And these constants are these moments of inertia and products 772 00:51:02,930 --> 00:51:03,765 of inertia terms. 773 00:51:18,410 --> 00:51:19,580 And so forth. 774 00:51:19,580 --> 00:51:21,410 I'll write out a couple more of these, iy. 775 00:51:31,060 --> 00:51:34,900 It's a symmetric matrix, and you multiply it 776 00:51:34,900 --> 00:51:44,000 by the components of the rotation 777 00:51:44,000 --> 00:51:48,990 that you are rotating this object, 778 00:51:48,990 --> 00:51:52,710 so here's a vector omega. 779 00:51:52,710 --> 00:51:57,220 This object is rotating about A, the direction, the axis 780 00:51:57,220 --> 00:51:59,270 of rotation is like that. 781 00:51:59,270 --> 00:52:02,060 And you can break this rotation rate 782 00:52:02,060 --> 00:52:04,960 into components in the xyz system. 783 00:52:04,960 --> 00:52:07,620 And that's what these are, these are the components of it. 784 00:52:07,620 --> 00:52:11,290 So you multiply out this matrix in a vector, 785 00:52:11,290 --> 00:52:14,030 you will get individual equations 786 00:52:14,030 --> 00:52:17,670 for the hx, hy, and hz components of the angular 787 00:52:17,670 --> 00:52:19,415 momentum of that object. 788 00:53:02,550 --> 00:53:04,230 Now let's consider, let's just do 789 00:53:04,230 --> 00:53:17,100 a case where the spin is only about the z-axis. 790 00:53:17,100 --> 00:53:19,315 We do lots of these problems, the book 791 00:53:19,315 --> 00:53:20,856 has a whole chapter on it and they're 792 00:53:20,856 --> 00:53:22,150 called planar motion problems. 793 00:53:22,150 --> 00:53:26,970 We just typically pick the spin around the z as a convention. 794 00:53:26,970 --> 00:53:35,350 And if you have a case like that, then h here is i times 795 00:53:35,350 --> 00:53:39,170 0, 0, omega z. 796 00:53:39,170 --> 00:53:53,830 Then you multiply that out, you get ixz omega z, iyz omega z, 797 00:53:53,830 --> 00:53:59,090 and izz omega z. 798 00:53:59,090 --> 00:54:03,560 Vector times the square matrix gives you back a vector. 799 00:54:03,560 --> 00:54:06,700 That's what you get back. 800 00:54:06,700 --> 00:54:13,450 And if you want to write h as a vector, which we frequently 801 00:54:13,450 --> 00:54:19,370 do, h, now this is with respect to A, 802 00:54:19,370 --> 00:54:23,460 and we'll find that i here as also with respect to A. 803 00:54:23,460 --> 00:54:29,140 Have to be very careful in your construction of this matrix. 804 00:54:29,140 --> 00:54:33,770 It has to do with the point about which you are computing 805 00:54:33,770 --> 00:54:35,410 your angular momentum. 806 00:54:35,410 --> 00:54:37,590 OK, if you want to write this as a vector, 807 00:54:37,590 --> 00:54:47,539 then this becomes hx i hat plus hy j hat plus hz k hat. 808 00:54:47,539 --> 00:54:49,330 That's just where the unit vectors come in. 809 00:54:49,330 --> 00:54:51,700 When you want to express this as a vector, 810 00:54:51,700 --> 00:54:57,740 you take these three components, and these are hx, hy, hz. 811 00:55:07,340 --> 00:55:09,650 This little double subscript, the first one 812 00:55:09,650 --> 00:55:13,770 tells you the component of h, this is hx, hy, hz. 813 00:55:13,770 --> 00:55:17,210 The second one tells you the axis of rotation 814 00:55:17,210 --> 00:55:19,720 about which the object is spinning to give you 815 00:55:19,720 --> 00:55:22,800 this piece of angular momentum. 816 00:55:22,800 --> 00:55:30,720 So ixz is hx spinning at rate omega z. 817 00:55:33,730 --> 00:55:39,045 Now, the direction of spin was? 818 00:55:39,045 --> 00:55:40,670 What's the unit vector in the direction 819 00:55:40,670 --> 00:55:42,706 of rotation for this problem? 820 00:55:48,020 --> 00:55:50,520 What's omega? 821 00:55:50,520 --> 00:55:53,950 We said we're going to start off with just-- direction, 822 00:55:53,950 --> 00:55:56,700 it's only spinning in z direction. 823 00:55:56,700 --> 00:55:59,420 So it's just spinning in z direction. 824 00:55:59,420 --> 00:56:04,000 But I multiply this thing out, I get three terms. 825 00:56:04,000 --> 00:56:06,660 And I get a term in the i, a j, and a k. 826 00:56:09,230 --> 00:56:19,920 Now these two terms, so this is i, x is z, 827 00:56:19,920 --> 00:56:34,210 omega is zi plus iyz omega zj plus izz omega zk. 828 00:56:34,210 --> 00:56:38,200 That's these three terms. 829 00:56:38,200 --> 00:56:44,980 These two terms exist because I've 830 00:56:44,980 --> 00:56:50,720 assumed that these off diagonal terms are not 0. 831 00:56:50,720 --> 00:56:53,650 The problem we started with, we started with an example 832 00:56:53,650 --> 00:56:54,150 last time. 833 00:56:54,150 --> 00:56:56,850 Our bicycle wheel thing with the unbalanced masses on it, 834 00:56:56,850 --> 00:57:02,440 we use the Williams formulas to compute these different terms. 835 00:57:02,440 --> 00:57:09,840 If the off diagonal terms here are not 0, 836 00:57:09,840 --> 00:57:13,130 then when you write the angular momentum expression, 837 00:57:13,130 --> 00:57:15,660 you get parts of the angular momentum 838 00:57:15,660 --> 00:57:19,290 that are not in the direction of spin. 839 00:57:19,290 --> 00:57:22,000 That's a really important conclusion. 840 00:57:22,000 --> 00:57:26,290 So the off diagonal terms lead to angular momentum 841 00:57:26,290 --> 00:57:27,997 not in the direction of spin. 842 00:57:27,997 --> 00:57:29,580 And when you take the time derivative, 843 00:57:29,580 --> 00:57:32,280 you end up with torques, and they're right back 844 00:57:32,280 --> 00:57:33,520 to this problem up here. 845 00:57:36,650 --> 00:57:41,280 If you have off diagonal terms in this matrix, 846 00:57:41,280 --> 00:57:44,310 when you spin it around one of its axes, 847 00:57:44,310 --> 00:57:46,225 it is dynamically unbalanced. 848 00:57:52,370 --> 00:57:54,890 If these are not 0, you spin it around one 849 00:57:54,890 --> 00:57:58,230 of the axes of the system for which these are defining, 850 00:57:58,230 --> 00:58:00,810 in which these are defined, you find out 851 00:58:00,810 --> 00:58:04,800 that you get unbalanced torques in the system. 852 00:58:04,800 --> 00:58:07,290 So those two go together. 853 00:58:07,290 --> 00:58:30,610 Now, another way of saying that is any time 854 00:58:30,610 --> 00:58:34,220 you end up with the angular momentum vector not pointing 855 00:58:34,220 --> 00:58:37,790 in the same direction as the rotation, 856 00:58:37,790 --> 00:58:40,558 then the system is going to be dynamically unbalanced. 857 00:58:57,700 --> 00:59:01,500 Actually, I kind of want to keep Atwood's machine here. 858 01:00:15,130 --> 01:00:19,740 So this was our unbalanced bicycle wheel problem 859 01:00:19,740 --> 01:00:22,200 we had talked about last time. 860 01:00:22,200 --> 01:00:25,794 I can simulate that with this. 861 01:00:31,710 --> 01:00:35,190 I basically have drawn it like this. 862 01:00:35,190 --> 01:00:36,230 So this is the problem. 863 01:00:38,950 --> 01:00:42,140 This thing is definitely unbalanced, 864 01:00:42,140 --> 01:00:46,380 it's trying to do this as it goes around. 865 01:00:46,380 --> 01:00:48,890 And last time we actually worked up, 866 01:00:48,890 --> 01:00:54,830 from the William formulas, what the moment of inertia matrix 867 01:00:54,830 --> 01:00:55,350 looked like. 868 01:00:55,350 --> 01:01:04,190 So now this xyz system are attached and rotating 869 01:01:04,190 --> 01:01:06,500 with that frame. 870 01:01:06,500 --> 01:01:22,400 So my axis of spin is-- this one's a little exaggerated. 871 01:01:22,400 --> 01:01:25,260 That drawing is like this. 872 01:01:25,260 --> 01:01:30,580 The x is like that. 873 01:01:30,580 --> 01:01:36,240 So x is like this, z is like that, minus x minus z. 874 01:01:36,240 --> 01:01:39,100 So when this thing spins, that's the problem that's drawn there. 875 01:01:41,810 --> 01:01:46,600 And if I compute with those with Williams formulas, 876 01:01:46,600 --> 01:01:52,830 the various quantities-- so i with respect 877 01:01:52,830 --> 01:01:55,220 to A for this system. 878 01:02:01,070 --> 01:02:08,280 The first term, the ixx term, is summation miyi 879 01:02:08,280 --> 01:02:15,260 squared plus zi squared, and so forth. 880 01:02:15,260 --> 01:02:19,990 You get a bunch of terms, and I will write out one other one 881 01:02:19,990 --> 01:02:20,490 here. 882 01:02:20,490 --> 01:02:30,010 This term over in the corner is ixz, 883 01:02:30,010 --> 01:02:40,270 and that's minus summation of the mixizi and so forth. 884 01:02:40,270 --> 01:02:44,430 And if we went through and worked up each of these things, 885 01:02:44,430 --> 01:02:49,290 i with respect to A for this problem, 886 01:02:49,290 --> 01:03:04,406 comes out mz1 squared 0 minus mx1z1 0 minus mx1z1. 887 01:03:07,970 --> 01:03:18,844 The middle term mx1 squared plus z1 squared, 0, 0, 888 01:03:18,844 --> 01:03:24,160 and mx1 squared. 889 01:03:24,160 --> 01:03:27,020 So that's what this mass moment of inertia matrix 890 01:03:27,020 --> 01:03:29,090 looks like for these two particles. 891 01:04:06,190 --> 01:04:08,230 So now if I want to write the angular 892 01:04:08,230 --> 01:04:14,170 momentum of this system using this new notation, 893 01:04:14,170 --> 01:04:18,010 I would say that it's i computed with respect 894 01:04:18,010 --> 01:04:24,970 to A times my omega, and our case is 0, 0, omega z. 895 01:04:28,670 --> 01:04:33,940 And if we write that out, we do that, multiply that out, we 896 01:04:33,940 --> 01:04:48,403 end up with a minus mx1z1 omega z, 0, and mx1 squared omega z. 897 01:04:51,790 --> 01:04:57,850 These are our three components, hx, hy, hz. 898 01:05:08,690 --> 01:05:13,300 And if you wanted to write it as a vector, 899 01:05:13,300 --> 01:05:15,420 then you'd add the unit vectors. 900 01:05:15,420 --> 01:05:22,260 So the hx and the i plus 0 for the hy plus hz in the k. 901 01:05:30,970 --> 01:05:35,010 So now if you went and took the time derivative of those terms, 902 01:05:35,010 --> 01:05:38,010 what do you get? 903 01:05:38,010 --> 01:05:38,780 AUDIENCE: Torques. 904 01:05:38,780 --> 01:05:40,560 PROFESSOR: Torques. 905 01:05:40,560 --> 01:05:43,070 And you'll get 3 terms. 906 01:05:43,070 --> 01:05:46,000 When we did the example a minute ago, what we're doing here 907 01:05:46,000 --> 01:05:48,530 is not very different from that. 908 01:05:48,530 --> 01:05:51,590 You're going to get the torque that it takes to accelerate it 909 01:05:51,590 --> 01:05:53,830 around the spin axis, but you're also 910 01:05:53,830 --> 01:05:57,020 going to get the torque two derivatives of this one, which 911 01:05:57,020 --> 01:05:58,330 gives you two terms. 912 01:05:58,330 --> 01:06:00,000 And these are the moments of torques 913 01:06:00,000 --> 01:06:03,030 about that center of the axle, in this case, 914 01:06:03,030 --> 01:06:05,050 trying to twist the system around. 915 01:06:07,970 --> 01:06:12,080 Now, reach some closure here. 916 01:06:12,080 --> 01:06:13,570 We've got a good stopping point. 917 01:06:29,090 --> 01:06:34,073 Here's our system one last time. 918 01:06:34,073 --> 01:06:34,823 Here's the z-axis. 919 01:06:37,650 --> 01:06:40,550 The angular momentum that comes out of this, 920 01:06:40,550 --> 01:06:46,120 you have a component hz in the z direction, 921 01:06:46,120 --> 01:06:47,600 and you end up with a component-- 922 01:06:47,600 --> 01:06:57,150 it's got a minus in it-- in the x direction like this, 923 01:06:57,150 --> 01:07:01,910 so that the total h vector with respect to A looks like that. 924 01:07:01,910 --> 01:07:04,075 And it's not in the direction of spin, 925 01:07:04,075 --> 01:07:10,030 it's actually perpendicular to our bar here. 926 01:07:10,030 --> 01:07:13,050 And it's dynamically unbalanced. 927 01:07:13,050 --> 01:07:19,940 So just to-- how do we make the transition 928 01:07:19,940 --> 01:07:27,060 from that to rigid bodies? 929 01:07:27,060 --> 01:07:31,520 The Williams formulas, that are these, 930 01:07:31,520 --> 01:07:34,780 say that if you want the mass moment of inertia of a body, 931 01:07:34,780 --> 01:07:36,950 all you have to do is sum up all the little mass 932 01:07:36,950 --> 01:07:41,470 bits at the correct distances off of axes, 933 01:07:41,470 --> 01:07:42,360 and you will get it. 934 01:07:42,360 --> 01:07:46,410 So when you have particles, you can just add them up. 935 01:07:46,410 --> 01:07:49,130 When you have a rigid body, those summations 936 01:07:49,130 --> 01:07:50,980 become integrals. 937 01:07:50,980 --> 01:08:02,546 And, for example, izz is the integral 938 01:08:02,546 --> 01:08:03,670 of-- how should I say this. 939 01:08:08,340 --> 01:08:13,595 x squared plus y squared dm, every little mass bit. 940 01:08:20,250 --> 01:08:22,696 It looks like-- is there an exam some distance away? 941 01:08:22,696 --> 01:08:24,029 I see a lot of people vanishing. 942 01:08:24,029 --> 01:08:26,120 OK, so let me-- I'll tell you what. 943 01:08:26,120 --> 01:08:29,250 I'll just make it easy for it and let you go. 944 01:08:29,250 --> 01:08:33,399 Let me just say one thing to where we're going. 945 01:08:33,399 --> 01:08:40,399 For every rigid body, there is a different set 946 01:08:40,399 --> 01:08:48,810 of axes for which, when you go to make up this matrix, 947 01:08:48,810 --> 01:08:51,790 you can make a diagonal. 948 01:08:51,790 --> 01:08:53,910 And those are called the principal axes, 949 01:08:53,910 --> 01:08:55,520 and that's where we're going next, 950 01:08:55,520 --> 01:08:59,189 those play a really important role in what we want to do. 951 01:08:59,189 --> 01:09:00,920 OK.