1 00:00:00,080 --> 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 --> 00:00:12,680 To make a donation or to view additional materials 6 00:00:12,680 --> 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 00:00:21,210 --> 00:00:25,760 PROFESSOR: OK, let's get on with today's lecture. 9 00:00:25,760 --> 00:00:31,850 And I want to look at a variety of different problems, 10 00:00:31,850 --> 00:00:36,156 different classes of problems. 11 00:00:36,156 --> 00:00:38,530 We're going to look at four different classes of problems 12 00:00:38,530 --> 00:00:43,490 and talk about the way you'd go about approaching them. 13 00:00:43,490 --> 00:00:52,320 We need very few fundamental laws. 14 00:00:52,320 --> 00:00:55,370 We use the same fundamental laws again and again. 15 00:01:00,432 --> 00:01:02,890 And the issue is really, how do you go about applying them? 16 00:01:02,890 --> 00:01:05,660 One is the sum-- this fundamental law's 17 00:01:05,660 --> 00:01:08,145 for rigid bodies. 18 00:01:13,990 --> 00:01:17,470 So the fundamental laws, they're always true. 19 00:01:17,470 --> 00:01:20,190 The sum of the external forces, vectors. 20 00:01:26,290 --> 00:01:28,260 Time-rated change of the momentum 21 00:01:28,260 --> 00:01:31,500 of the system with respect to an inertial frame. 22 00:01:31,500 --> 00:01:36,090 And we recognize this mass times acceleration. 23 00:01:36,090 --> 00:01:37,740 Newton's second law. 24 00:01:37,740 --> 00:01:43,875 The other one is the summation of external torques. 25 00:01:47,340 --> 00:01:51,040 And that one is the time derivative 26 00:01:51,040 --> 00:01:55,790 of angular momentum with respect to a point which you choose-- 27 00:01:55,790 --> 00:02:06,960 this is an A-- plus the velocity of A with respect 28 00:02:06,960 --> 00:02:12,410 to the inertial frame crossed with the momentum with respect 29 00:02:12,410 --> 00:02:13,730 to the inertial frame. 30 00:02:13,730 --> 00:02:15,780 So that's the torque equation. 31 00:02:15,780 --> 00:02:19,650 Now we have two special cases of this where 32 00:02:19,650 --> 00:02:22,060 the second term goes away. 33 00:02:22,060 --> 00:02:26,860 One is when you are at the center of gravity, 34 00:02:26,860 --> 00:02:30,040 in which case by definition the velocity of that point 35 00:02:30,040 --> 00:02:32,020 and the momentum are in the same direction, 36 00:02:32,020 --> 00:02:33,910 the cross product disappears. 37 00:02:33,910 --> 00:02:37,120 So there's two special cases we use all the time 38 00:02:37,120 --> 00:02:39,980 when it simplifies to that. 39 00:02:39,980 --> 00:02:42,120 One is when you're at the center of gravity. 40 00:02:42,120 --> 00:02:44,130 The other is when, for whatever reason, 41 00:02:44,130 --> 00:02:46,190 this velocity is parallel to that, 42 00:02:46,190 --> 00:02:47,960 and the cross product goes to zero. 43 00:02:47,960 --> 00:02:50,510 And sometimes the velocity is just plain zero. 44 00:02:50,510 --> 00:02:52,410 So there's some certain cases that we 45 00:02:52,410 --> 00:02:54,359 use lots of times when that term goes away. 46 00:02:54,359 --> 00:02:55,400 But sometimes it doesn't. 47 00:02:55,400 --> 00:02:57,090 You just have to put up with it. 48 00:03:01,080 --> 00:03:02,690 So those are our two fundamental laws. 49 00:03:02,690 --> 00:03:05,840 And the question really is how to go about applying them. 50 00:03:08,890 --> 00:03:11,806 So I'm going to look at four classes of problems. 51 00:03:17,320 --> 00:03:19,090 Let's name them right now. 52 00:03:19,090 --> 00:03:28,780 Pure-- and we'll do simplest to hardest-- pure rotation 53 00:03:28,780 --> 00:03:42,150 about a fixed axis through G, through the center of mass. 54 00:03:42,150 --> 00:03:42,800 Pretty trivial. 55 00:03:42,800 --> 00:03:45,050 We can all do those kind of problems. 56 00:03:45,050 --> 00:04:05,630 A second class, pure rotation about a fixed axis at A, 57 00:04:05,630 --> 00:04:09,680 which is not equal to G not at the center of mass. 58 00:04:09,680 --> 00:04:11,675 Again, pretty simple problems. 59 00:04:15,880 --> 00:04:34,420 Third class, no external-- I'll call it 60 00:04:34,420 --> 00:04:35,516 no external constraints. 61 00:04:43,660 --> 00:04:46,410 We'll have to give an example to see what I really mean. 62 00:04:46,410 --> 00:04:51,790 Well, I'll give you an example right now, which we'll do. 63 00:04:51,790 --> 00:04:57,800 You have this hockey puck with a string attached to it force. 64 00:04:57,800 --> 00:05:01,780 And this whole thing is on a frictionless surface. 65 00:05:01,780 --> 00:05:05,760 So it's constrained so it can't go through the surface. 66 00:05:05,760 --> 00:05:08,670 So no external constraints in the direction 67 00:05:08,670 --> 00:05:09,840 that the motion can happen. 68 00:05:09,840 --> 00:05:12,340 So this is a 2D problem. 69 00:05:12,340 --> 00:05:15,050 Can move in the x, y, and rotation. 70 00:05:15,050 --> 00:05:16,050 But it's not touching. 71 00:05:16,050 --> 00:05:17,920 There's no things constraining it 72 00:05:17,920 --> 00:05:20,146 in the directions of movement, which 73 00:05:20,146 --> 00:05:21,270 are allowed in the problem. 74 00:05:21,270 --> 00:05:23,230 That's just too much words to write up here. 75 00:05:23,230 --> 00:05:24,690 So this kind of problem. 76 00:05:24,690 --> 00:05:51,935 And the fourth are problems with moving points of constraint. 77 00:06:07,690 --> 00:06:12,610 You won't see a textbook with sections broken out like this. 78 00:06:12,610 --> 00:06:14,070 I'm using my own terminology. 79 00:06:14,070 --> 00:06:15,460 I just made it up last night as I 80 00:06:15,460 --> 00:06:17,780 was finishing off the lecture. 81 00:06:17,780 --> 00:06:20,680 But I'm trying to give you some insight, 82 00:06:20,680 --> 00:06:24,040 and this is the way I think about these things. 83 00:06:24,040 --> 00:06:27,060 So let's quickly go through examples 84 00:06:27,060 --> 00:06:29,720 from the first couple of types because they're 85 00:06:29,720 --> 00:06:30,810 especially easy. 86 00:06:33,840 --> 00:06:35,110 So what are we saying? 87 00:06:35,110 --> 00:06:37,110 They're pure rotation about a fixed 88 00:06:37,110 --> 00:06:46,400 axis through G. Center of mass is somewhere along this axle. 89 00:06:46,400 --> 00:06:52,420 The axle is fixed so object can spin around it. 90 00:06:52,420 --> 00:06:57,210 And these kind of problems are particularly simple, 91 00:06:57,210 --> 00:06:58,850 so I'm not going to dwell on them. 92 00:06:58,850 --> 00:07:05,080 But here's class one, class one problems. 93 00:07:05,080 --> 00:07:11,500 And it's basically-- these are rotors, typically rotors, 94 00:07:11,500 --> 00:07:12,595 of almost all kinds. 95 00:07:17,120 --> 00:07:25,600 So for these problems, you use momentum, angular momentum, 96 00:07:25,600 --> 00:07:27,175 with respect to the center of mass. 97 00:07:29,910 --> 00:07:43,290 You can express it as IG times omega x, omega y, omega z. 98 00:07:43,290 --> 00:07:45,140 All these problems can be expressed 99 00:07:45,140 --> 00:07:49,270 with a moment of inertia matrix times the rotation vector. 100 00:07:49,270 --> 00:07:52,740 And it will give you HX, HY, HZ. 101 00:07:52,740 --> 00:07:56,080 And you can use that second formula up there. 102 00:07:56,080 --> 00:07:58,640 The torques are dH dt. 103 00:07:58,640 --> 00:08:01,890 So it's fixed through G. So the second term doesn't appear. 104 00:08:04,460 --> 00:08:07,540 So to do these problems, the sum of the-- see, 105 00:08:07,540 --> 00:08:09,520 I'll just give an example here. 106 00:08:09,520 --> 00:08:16,415 For 2D planar motion, they get especially simple. 107 00:08:21,880 --> 00:08:31,170 Then, for 2D planar motion, that means omega here is 0, 108 00:08:31,170 --> 00:08:37,590 0, and we'll let z be the rotation axis. 109 00:08:37,590 --> 00:08:49,050 And this H then with respect to G just becomes Izz omega z. 110 00:08:49,050 --> 00:08:54,090 All those 2D planar motion problems boil down to this. 111 00:08:54,090 --> 00:09:09,130 And dHG dt then is Izz with respect to G omega z do, 112 00:09:09,130 --> 00:09:12,705 or more familiar notation. 113 00:09:15,370 --> 00:09:19,240 So all those planar motion problems, axis 114 00:09:19,240 --> 00:09:20,820 passing through G like this. 115 00:09:23,590 --> 00:09:31,300 For those problems, does this matrix 116 00:09:31,300 --> 00:09:37,400 have to be with respect to principal axes? 117 00:09:37,400 --> 00:09:41,560 Do you have your XYZ coordinate system? 118 00:09:41,560 --> 00:09:44,940 To do this style of problem, does 119 00:09:44,940 --> 00:09:48,869 that actually have to be expressed in principle 120 00:09:48,869 --> 00:09:50,285 coordinates so that it's diagonal? 121 00:09:55,510 --> 00:09:57,510 It doesn't have to be. 122 00:09:57,510 --> 00:10:00,330 You're worried about rotation about z. 123 00:10:00,330 --> 00:10:02,950 You'll find that that statement's still always true. 124 00:10:02,950 --> 00:10:05,180 Now, there may be-- the thing could definitely 125 00:10:05,180 --> 00:10:08,560 have imbalances and have unusual other torques. 126 00:10:08,560 --> 00:10:11,180 That will fall out in the problem. 127 00:10:11,180 --> 00:10:16,710 But for the motion around the axis of spin, this [INAUDIBLE]. 128 00:10:20,650 --> 00:10:24,310 Where you only have a 1 omega z component, just 129 00:10:24,310 --> 00:10:28,210 the one component, you will get-- it'll work out just fine. 130 00:10:28,210 --> 00:10:28,892 Yeah. 131 00:10:28,892 --> 00:10:32,748 AUDIENCE: If you do have some kind of Ixz and Iyz terms, 132 00:10:32,748 --> 00:10:38,020 you would end up with more-- 133 00:10:38,020 --> 00:10:39,920 PROFESSOR: You will end up with-- first 134 00:10:39,920 --> 00:10:43,345 of all, if you have the off-diagonal z terms, xz, 135 00:10:43,345 --> 00:10:48,630 yz terms, when you multiply that out, you will find components 136 00:10:48,630 --> 00:10:54,160 of H that are in the z direction as well as perhaps in the x 137 00:10:54,160 --> 00:10:55,070 and y. 138 00:10:55,070 --> 00:10:58,790 And those other two components tell you 139 00:10:58,790 --> 00:11:02,610 that H is not pointing in the same direction as omega, right? 140 00:11:02,610 --> 00:11:06,900 And that it tells you instantly that the thing is dynamically 141 00:11:06,900 --> 00:11:11,530 unbalanced and will have other torques that 142 00:11:11,530 --> 00:11:15,060 are trying to bend that axle. 143 00:11:15,060 --> 00:11:16,750 So they'll always appear. 144 00:11:20,872 --> 00:11:21,790 All right. 145 00:11:21,790 --> 00:11:22,450 Let's move on. 146 00:11:22,450 --> 00:11:25,650 I want to make sure we get through this today. 147 00:11:25,650 --> 00:11:26,595 Class two problems. 148 00:11:32,060 --> 00:11:34,370 These are basically-- these are the pure rotation 149 00:11:34,370 --> 00:11:37,130 around some point that's not through G. 150 00:11:37,130 --> 00:11:49,030 And again now, this is a fixed-- key here is fixed axis at A. 151 00:11:49,030 --> 00:11:50,290 It's not moving. 152 00:11:50,290 --> 00:11:52,510 This is what makes these problems simpler. 153 00:11:52,510 --> 00:11:55,200 For these kinds of problems, you do 154 00:11:55,200 --> 00:11:57,350 the sum of the torques with respect 155 00:11:57,350 --> 00:11:58,935 to A, the external torques. 156 00:12:13,060 --> 00:12:16,510 And because that point's fixed, the second terms don't appear, 157 00:12:16,510 --> 00:12:19,740 the v cross p terms. 158 00:12:19,740 --> 00:12:27,320 And you can write these as I, a moment of inertia matrix, 159 00:12:27,320 --> 00:12:37,020 times whatever the rotations are. 160 00:12:40,410 --> 00:12:43,750 In order to define the mass moment of inertia matrix, 161 00:12:43,750 --> 00:12:48,930 you must have chosen a set of coordinates 162 00:12:48,930 --> 00:12:52,350 attached to the body. 163 00:12:52,350 --> 00:12:54,370 And then with those coordinates, you've 164 00:12:54,370 --> 00:12:57,490 computed the moments of inertia for the body. 165 00:12:57,490 --> 00:13:01,170 And if you chose wisely, you get principal coordinates 166 00:13:01,170 --> 00:13:03,720 and you only get diagonal entries on the matrix. 167 00:13:03,720 --> 00:13:07,960 If you chose unwisely, you will get other stuff. 168 00:13:07,960 --> 00:13:12,350 But it's still a valid mass moment of inertia matrix. 169 00:13:12,350 --> 00:13:14,120 It just gives rise-- you have to deal 170 00:13:14,120 --> 00:13:15,950 with a bunch of other terms. 171 00:13:15,950 --> 00:13:22,490 So this will still yield the same answer. 172 00:13:22,490 --> 00:13:25,170 How do you get I with respect to A? 173 00:13:29,410 --> 00:13:33,700 These are opportunities when you can use parallel axis. 174 00:13:33,700 --> 00:13:34,765 Yeah. 175 00:13:34,765 --> 00:13:40,545 AUDIENCE: Isn't I omega just H dot dH dt? 176 00:13:40,545 --> 00:13:41,420 PROFESSOR: Excuse me. 177 00:13:45,280 --> 00:13:46,730 You need to finish that out. 178 00:13:46,730 --> 00:13:50,250 This is H. So the time derivative of H 179 00:13:50,250 --> 00:13:52,220 is the time derivative of this expression. 180 00:13:52,220 --> 00:13:54,450 You have to figure out the moments of inertia 181 00:13:54,450 --> 00:13:55,970 and the rotation rates. 182 00:13:55,970 --> 00:14:01,230 And you may get multiple terms, only one of which-- 183 00:14:01,230 --> 00:14:08,920 let's say that this will work out. 184 00:14:08,920 --> 00:14:10,230 You will get multiple terms. 185 00:14:10,230 --> 00:14:15,130 You will get torques that are not in the direction of spin. 186 00:14:15,130 --> 00:14:18,280 Again, these might be unbalanced. 187 00:14:18,280 --> 00:14:23,130 On the other hand, it may be, for 2D problems, which 188 00:14:23,130 --> 00:14:31,960 is common-- so for the 2D planar motion, which 189 00:14:31,960 --> 00:14:37,430 most of the problems we do are, then what you would do 190 00:14:37,430 --> 00:14:43,070 is you're saying omega is, say, 0, 0, omega z, 191 00:14:43,070 --> 00:14:44,640 which simplifies that. 192 00:14:44,640 --> 00:14:47,800 We're multiplying this thing out quite a lot. 193 00:14:47,800 --> 00:15:03,990 And if I with respect to G is diagonal, 194 00:15:03,990 --> 00:15:07,535 then that means you your G you chose principal axes. 195 00:15:12,740 --> 00:15:16,640 But then how do you get to I with respect to A? 196 00:15:16,640 --> 00:15:21,690 For 2D problems, really simple ones, 197 00:15:21,690 --> 00:15:23,791 how do you get to I with respect to A? 198 00:15:23,791 --> 00:15:24,790 AUDIENCE: Parallel axis. 199 00:15:24,790 --> 00:15:27,690 PROFESSOR: That's the classic case 200 00:15:27,690 --> 00:15:31,670 for using the parallel axis theorem. 201 00:15:31,670 --> 00:15:36,320 So for these 2D planar motion problems-- and planar motion 202 00:15:36,320 --> 00:15:47,120 problems, then you can use parallel axis. 203 00:15:47,120 --> 00:15:48,040 I'll do an example. 204 00:15:48,040 --> 00:15:51,070 So this is kind of the set up for this. 205 00:15:51,070 --> 00:15:53,330 An example of this on the homework. 206 00:15:53,330 --> 00:15:57,845 What problem on the homework is just perfect for this? 207 00:15:57,845 --> 00:16:02,054 It's 2D, planar motion, about a fixed point. 208 00:16:02,054 --> 00:16:03,720 AUDIENCE: Circle with the square cutout. 209 00:16:03,720 --> 00:16:07,380 PROFESSOR: Yeah, the cylinder, this [INAUDIBLE] disk 210 00:16:07,380 --> 00:16:11,520 with the square cutout with the pin at the top turning it 211 00:16:11,520 --> 00:16:13,440 into a pendulum. 212 00:16:13,440 --> 00:16:14,470 That's the fixed point. 213 00:16:17,440 --> 00:16:18,770 You can figure this out. 214 00:16:18,770 --> 00:16:20,870 You can use parallel axis. 215 00:16:20,870 --> 00:16:23,600 And we're going to do an example right now that's almost 216 00:16:23,600 --> 00:16:26,120 identical to that problem. 217 00:16:26,120 --> 00:16:27,560 And we started it last time. 218 00:16:27,560 --> 00:16:32,190 So the example I want to work is very similar. 219 00:16:32,190 --> 00:16:35,190 It's basically this problem is that pendulum. 220 00:16:38,130 --> 00:16:41,170 So let's just work it through quickly. 221 00:16:41,170 --> 00:16:44,830 We had done the setup last time. 222 00:16:58,080 --> 00:17:01,580 So last time I basically derived an example 223 00:17:01,580 --> 00:17:05,589 of the parallel axis theorem for my little stick here. 224 00:17:05,589 --> 00:17:12,575 And I'll give you some geometry, some values. 225 00:17:17,609 --> 00:17:26,690 So here's G. There's the point I wanted to rotate about, A. 226 00:17:26,690 --> 00:17:31,650 The distance between these two points is d. 227 00:17:31,650 --> 00:17:33,920 That's going to pop up in my parallel axis theorem. 228 00:17:36,924 --> 00:17:39,920 We've got a set of coordinates here. 229 00:17:39,920 --> 00:17:42,140 My G is, of course, in the center of this thing, 230 00:17:42,140 --> 00:17:43,240 the geometric center. 231 00:17:43,240 --> 00:17:47,870 So I have a body fixed set of axes, 232 00:17:47,870 --> 00:17:51,000 which I'm going to call z. 233 00:17:51,000 --> 00:17:55,520 And my x is in this direction. 234 00:17:55,520 --> 00:18:00,860 So that would make my y going into the board. 235 00:18:00,860 --> 00:18:02,720 So the y is kind of like that. 236 00:18:06,520 --> 00:18:09,400 And this has some properties. 237 00:18:09,400 --> 00:18:12,870 The dimension in this direction means 238 00:18:12,870 --> 00:18:17,530 A. The direction in this dimension is B. 239 00:18:17,530 --> 00:18:30,630 So it's a width of A, a thickness B, and a length L. 240 00:18:30,630 --> 00:18:34,580 So when you compute, these are-- ah, symmetry now. 241 00:18:37,830 --> 00:18:40,120 So the axes that I've chosen for this problem 242 00:18:40,120 --> 00:18:41,175 are they principal axes. 243 00:18:47,070 --> 00:18:50,160 There's three planes of symmetry in this problem, 244 00:18:50,160 --> 00:18:52,615 and I've got one principal axis perpendicular 245 00:18:52,615 --> 00:18:53,840 to every one of them. 246 00:18:53,840 --> 00:18:56,680 And all three pass through the center of mass, 247 00:18:56,680 --> 00:18:59,300 and they're orthogonal to one another. 248 00:18:59,300 --> 00:19:02,160 So those conditions are all satisfied 249 00:19:02,160 --> 00:19:06,890 for these to be principal axes for this rectangular body 250 00:19:06,890 --> 00:19:07,973 and its uniform density. 251 00:19:12,820 --> 00:19:16,580 So I'll give you the-- for bodies like this, 252 00:19:16,580 --> 00:19:19,060 in your book or any book on dynamics, 253 00:19:19,060 --> 00:19:26,322 you can look up then the properties, the Izz. 254 00:19:26,322 --> 00:19:27,780 And we're going to spin this thing. 255 00:19:27,780 --> 00:19:32,610 This thing, we're going to have it rotating about-- which 256 00:19:32,610 --> 00:19:34,240 one am I going to use? 257 00:19:34,240 --> 00:19:36,180 Yeah, around the z-axis. 258 00:19:36,180 --> 00:19:37,320 That's how I'll set it up. 259 00:19:37,320 --> 00:19:39,190 That's the way I drilled my hole, 260 00:19:39,190 --> 00:19:41,760 so it's going back and forth. 261 00:19:41,760 --> 00:19:44,100 So that the wide part of it's this way. 262 00:19:47,500 --> 00:19:59,070 So Izz with respect to G M L squared plus a squared over 12. 263 00:20:03,740 --> 00:20:04,240 Iyy. 264 00:20:29,090 --> 00:20:35,190 OK, now just to make a point, the dimensions 265 00:20:35,190 --> 00:20:50,560 L 32.1 centimeters, a 4.71, b, 1.25. 266 00:20:50,560 --> 00:20:54,990 And eventually my d, this offset would 267 00:20:54,990 --> 00:21:02,470 be, for example, where I've drilled that hole, is at 10.2. 268 00:21:02,470 --> 00:21:04,140 The point I want to make here, lots 269 00:21:04,140 --> 00:21:06,660 of times it would be nice if I could just 270 00:21:06,660 --> 00:21:10,440 make the simplification to call this a slender rod 271 00:21:10,440 --> 00:21:14,410 and be able to ignore these a and b dimensions in this, 272 00:21:14,410 --> 00:21:15,960 just to get quick answers. 273 00:21:15,960 --> 00:21:19,600 Do you think that's slender enough? 274 00:21:19,600 --> 00:21:22,780 It's not even 10 times-- the length of the width 275 00:21:22,780 --> 00:21:23,630 here isn't even 10. 276 00:21:23,630 --> 00:21:27,330 It's probably six or seven. 277 00:21:27,330 --> 00:21:29,810 So the key issue then, if you look at this, 278 00:21:29,810 --> 00:21:33,870 is really what's the ratio of a squared to L squared? 279 00:21:33,870 --> 00:21:36,650 That would tell you something about the relative importance 280 00:21:36,650 --> 00:21:39,540 of the a squared term to the L squared. 281 00:21:39,540 --> 00:21:41,170 So let me tell you about that. 282 00:21:41,170 --> 00:21:51,350 So a squared over L squared is 0.022. 283 00:21:51,350 --> 00:21:59,500 b squared over L squared is 0.002. 284 00:21:59,500 --> 00:22:02,600 So even with this kind of fat stick, 285 00:22:02,600 --> 00:22:05,460 the approximation of ML squared over 12 is pretty good. 286 00:22:05,460 --> 00:22:07,500 It's only 2% off. 287 00:22:07,500 --> 00:22:09,920 And this approximation, L squared over 12, 288 00:22:09,920 --> 00:22:14,440 is less than 2/10 of a percent off. 289 00:22:14,440 --> 00:22:17,600 So for roughly slender things, we oftentimes 290 00:22:17,600 --> 00:22:20,880 just say ML squared over 12 for spin about their center. 291 00:22:26,410 --> 00:22:29,370 We now need to apply parallel axis. 292 00:22:29,370 --> 00:22:32,410 I want to spin this around, let this rotate around, 293 00:22:32,410 --> 00:22:36,000 not around G, but now around a point off to the side. 294 00:22:36,000 --> 00:22:43,790 So we worked out last time that Izz with respect to A 295 00:22:43,790 --> 00:22:50,055 is IzzG plus Md squared. 296 00:23:10,520 --> 00:23:12,730 So in this particular problem, this Izz about G 297 00:23:12,730 --> 00:23:16,100 is approximately ML squared over 12. 298 00:23:16,100 --> 00:23:20,500 So just by way of example, to see what we might find out 299 00:23:20,500 --> 00:23:25,780 here, is let's let d equal L/2. 300 00:23:25,780 --> 00:23:28,560 That would be if I move this-- if I 301 00:23:28,560 --> 00:23:30,300 put my hole right at the very top, 302 00:23:30,300 --> 00:23:31,466 how would this thing behave? 303 00:23:31,466 --> 00:23:34,810 I don't have a hole right at the top, but I have one close. 304 00:23:34,810 --> 00:23:38,310 So this is just because the numbers are easy to work. 305 00:23:38,310 --> 00:23:41,850 What happens if you put in L/2 into this formula? 306 00:23:41,850 --> 00:23:53,650 Well then IzzA is ML squared over 12 plus M. L over 2 307 00:23:53,650 --> 00:23:55,150 squared is L squared over 4. 308 00:23:59,600 --> 00:24:04,510 And that's ML squared over 3, which 309 00:24:04,510 --> 00:24:06,960 is a number you'll run into again and again and again 310 00:24:06,960 --> 00:24:09,680 in mechanical engineering because examples like this 311 00:24:09,680 --> 00:24:10,350 are used a lot. 312 00:24:10,350 --> 00:24:13,630 The mass moment of inertia about a slender rod 313 00:24:13,630 --> 00:24:15,980 pinned at its end, ML squared over 3. 314 00:24:34,170 --> 00:24:38,380 So let's take this problem a little more towards completion. 315 00:24:38,380 --> 00:24:47,275 The sum of the external moments with respect to A dHA dt. 316 00:24:52,580 --> 00:24:56,580 And that's going to be d by dt. 317 00:24:56,580 --> 00:25:02,420 In this problem, the only rotation is omega z. 318 00:25:02,420 --> 00:25:10,380 So this is going to be Izz with respect to A omega z. 319 00:25:13,090 --> 00:25:22,380 And that just gives us IzzA omega z dot, or more familiar, 320 00:25:22,380 --> 00:25:28,460 IzzA theta double bond, if you want. 321 00:25:28,460 --> 00:25:29,730 Here's our problem. 322 00:25:29,730 --> 00:25:33,770 It simplifies to this slender rod. 323 00:25:33,770 --> 00:25:38,600 And let me do the more general case. 324 00:25:38,600 --> 00:25:39,970 Here's my rod. 325 00:25:39,970 --> 00:25:44,300 The pivot point that it's going around is here. 326 00:25:44,300 --> 00:25:50,310 This is d still, and this is G. And it's swinging with respect 327 00:25:50,310 --> 00:25:51,100 to this point. 328 00:25:51,100 --> 00:25:56,230 So here's the angle theta. 329 00:25:56,230 --> 00:25:59,160 So it swings about this point that you've fixed, 330 00:25:59,160 --> 00:26:06,490 and that point is d above the center of gravity, 331 00:26:06,490 --> 00:26:07,370 center of mass. 332 00:26:12,710 --> 00:26:18,670 So what are the external moments about this point? 333 00:26:18,670 --> 00:26:20,960 There's no torque right at the point, 334 00:26:20,960 --> 00:26:26,180 but our free body diagram of drawing this as a-- 335 00:26:26,180 --> 00:26:28,840 here's our point of rotation. 336 00:26:28,840 --> 00:26:33,590 Here it is displaced through an angle theta. 337 00:26:33,590 --> 00:26:36,835 The weight of the object can all be concentrated, 338 00:26:36,835 --> 00:26:39,210 thought of, for the purposes of the free body 339 00:26:39,210 --> 00:26:42,950 diagram as acting through G. 340 00:26:42,950 --> 00:26:49,970 So here's the mass at G, gravity acting down on it. 341 00:26:49,970 --> 00:26:56,980 And the length of this arm here about which it's swinging is D. 342 00:26:56,980 --> 00:27:04,160 So the torque about this is-- and it's pulling it back-- 343 00:27:04,160 --> 00:27:12,297 minus Mgd sine theta. 344 00:27:12,297 --> 00:27:14,130 Probably you've seen that many times before, 345 00:27:14,130 --> 00:27:17,960 including the recent homework. 346 00:27:17,960 --> 00:27:25,650 And that must be equal to Izz about A theta double dot. 347 00:27:25,650 --> 00:27:29,190 So we have an equation of motion. 348 00:27:29,190 --> 00:27:30,695 Just collecting the terms together. 349 00:27:40,900 --> 00:27:43,900 So this is a oscillator that, for this problem, 350 00:27:43,900 --> 00:27:47,260 has no external excitation. 351 00:27:47,260 --> 00:27:50,570 This is its equation of motion. 352 00:27:50,570 --> 00:27:51,850 And I need theta double dot. 353 00:27:54,670 --> 00:27:55,610 But is it linear? 354 00:27:59,714 --> 00:28:00,500 Is it linear? 355 00:28:00,500 --> 00:28:03,230 No, it's not a linear equation of motion. 356 00:28:03,230 --> 00:28:05,260 But for sure, it is an oscillator. 357 00:28:08,180 --> 00:28:12,190 And for anything that vibrates, you 358 00:28:12,190 --> 00:28:14,630 can have lots of nonlinear problems 359 00:28:14,630 --> 00:28:16,940 that exhibit vibration. 360 00:28:16,940 --> 00:28:20,280 You can think of them-- you can pose problems 361 00:28:20,280 --> 00:28:25,280 where you say, OK, what's their static equilibrium position? 362 00:28:25,280 --> 00:28:28,440 And think of a very small motion about that static equilibrium 363 00:28:28,440 --> 00:28:29,270 position. 364 00:28:29,270 --> 00:28:33,570 You can always linearize about the static equilibrium position 365 00:28:33,570 --> 00:28:36,270 and be able to come up with a linearized equation of motion 366 00:28:36,270 --> 00:28:40,090 that at least from that you can calculate the natural frequency 367 00:28:40,090 --> 00:28:45,710 of the system for small motions around its static equilibrium 368 00:28:45,710 --> 00:28:46,210 position. 369 00:28:46,210 --> 00:28:48,420 So in this case, it's pretty easy to do. 370 00:28:48,420 --> 00:28:55,215 And you've seen it before for small theta. 371 00:28:58,450 --> 00:29:01,070 Sine theta is approximately equal to theta. 372 00:29:01,070 --> 00:29:04,630 So we are going to linearize the equation. 373 00:29:04,630 --> 00:29:09,140 This theta equals 0 is a static equilibrium position. 374 00:29:09,140 --> 00:29:11,710 So we're linearizing around 0. 375 00:29:11,710 --> 00:29:15,280 And around 0, that's the approximation. 376 00:29:15,280 --> 00:29:21,230 So you just substitute that in, IzzA theta double 377 00:29:21,230 --> 00:29:26,960 dot plus Mgd theta equals 0. 378 00:29:26,960 --> 00:29:29,160 There's your linearized equation of motion. 379 00:29:29,160 --> 00:29:31,980 I want an estimate of the natural frequency. 380 00:29:31,980 --> 00:29:36,300 So find omega n. 381 00:29:41,010 --> 00:29:42,870 So this is basically entering into solving 382 00:29:42,870 --> 00:29:44,220 differential equations. 383 00:29:44,220 --> 00:29:47,800 But I let mother nature tell me what the answer is. 384 00:29:47,800 --> 00:29:50,230 I do the example, and I say it oscillates. 385 00:29:50,230 --> 00:29:54,297 Looks a lot like sine omega t to me. 386 00:29:54,297 --> 00:29:55,630 Plug in [INAUDIBLE] to make a t. 387 00:29:55,630 --> 00:29:56,390 Let's find out. 388 00:30:00,450 --> 00:30:06,080 Some theta amplitude sine omega t. 389 00:30:06,080 --> 00:30:08,110 Plug it in. 390 00:30:08,110 --> 00:30:13,720 So you plug that in [INAUDIBLE] and you get minus omega squared 391 00:30:13,720 --> 00:30:21,860 IzzA plus Mgd. 392 00:30:21,860 --> 00:30:24,700 And all of this, you can factor out the theta 393 00:30:24,700 --> 00:30:30,400 0 sine omega t equals 0. 394 00:30:30,400 --> 00:30:32,150 That's what you get. 395 00:30:32,150 --> 00:30:34,240 And in general, theta, that's not 0, or else 396 00:30:34,240 --> 00:30:37,590 you'd have a trivial problem, not moving at all. 397 00:30:37,590 --> 00:30:41,840 But in order for this equation-- and this can be anything. 398 00:30:41,840 --> 00:30:45,360 We're doing the 0 minus 1 and plus 1 depending on the time. 399 00:30:45,360 --> 00:30:47,650 So in order to satisfy this equation, 400 00:30:47,650 --> 00:30:54,150 this part inside of the parentheses has to be 0. 401 00:30:54,150 --> 00:30:58,120 And when you just solve that, you 402 00:30:58,120 --> 00:31:04,365 find that omega squared equals Mgd/IzzA. 403 00:31:12,600 --> 00:31:17,490 And the reason I've gone to the bother of working this out 404 00:31:17,490 --> 00:31:22,900 in detail right to the end is that every one degree 405 00:31:22,900 --> 00:31:28,400 of freedom rotational oscillator that you will ever encounter-- 406 00:31:28,400 --> 00:31:40,080 sticks, wheels with static imbalances. 407 00:31:40,080 --> 00:31:41,148 Let me show you this one. 408 00:31:45,730 --> 00:31:47,650 It's an oscillator too. 409 00:31:47,650 --> 00:31:51,300 Any one degree of freedom oscillator, 410 00:31:51,300 --> 00:31:53,840 rotational oscillator, pendulum-- basically 411 00:31:53,840 --> 00:31:56,412 all pendula-- this is the formula 412 00:31:56,412 --> 00:31:57,495 for the natural frequency. 413 00:32:02,610 --> 00:32:06,781 So it's going to be of that form for any pendulum. 414 00:32:06,781 --> 00:32:11,465 So any 1dof pendulum. 415 00:32:14,740 --> 00:32:17,000 This is the generic answer. 416 00:32:19,520 --> 00:32:23,920 So that cutout problem for today has to come down to this. 417 00:32:23,920 --> 00:32:27,370 Or this is the distance between the mass center 418 00:32:27,370 --> 00:32:30,320 and the point of rotation. 419 00:32:30,320 --> 00:32:31,960 And that's your mass moment of inertia 420 00:32:31,960 --> 00:32:34,096 about the point of rotation. 421 00:32:34,096 --> 00:32:38,110 So that's worth knowing that one. 422 00:32:38,110 --> 00:32:39,770 OK, keep moving. 423 00:32:49,913 --> 00:32:53,120 Now things will begin to get interesting. 424 00:32:53,120 --> 00:32:59,340 These latter two classes are harder conceptually, 425 00:32:59,340 --> 00:33:01,590 but once you have a solution method for them, 426 00:33:01,590 --> 00:33:03,690 they're not all that hard. 427 00:33:03,690 --> 00:33:08,080 This one is the problem I described at the beginning. 428 00:33:08,080 --> 00:33:10,030 We've got this hockey puck like thing. 429 00:33:13,270 --> 00:33:16,640 And the string wrapped around it pulling on it 430 00:33:16,640 --> 00:33:17,840 with a known force. 431 00:33:17,840 --> 00:33:21,680 In this problem, they call it 150 newtons. 432 00:33:21,680 --> 00:33:27,086 The mass of this thing is 75 kilograms. 433 00:33:30,870 --> 00:33:33,135 It's on a frictionless surface. 434 00:33:36,550 --> 00:33:46,350 And we want to find its acceleration 435 00:33:46,350 --> 00:33:51,800 of the center of mass and the rotation 436 00:33:51,800 --> 00:33:54,380 around the center of mass. 437 00:33:54,380 --> 00:33:58,280 So find theta double dot and find the linear acceleration. 438 00:33:58,280 --> 00:34:00,960 That's basically the name of the problem. 439 00:34:00,960 --> 00:34:06,630 And they give you that it's 75 kilograms, 150 newtons, 440 00:34:06,630 --> 00:34:15,449 and kappa, the radius of gyration, is 0.15 meters. 441 00:34:15,449 --> 00:34:21,404 This is defined as the radius of gyration. 442 00:34:24,870 --> 00:34:28,320 So what's that? 443 00:34:28,320 --> 00:34:29,420 It's a radius of gyration. 444 00:34:29,420 --> 00:34:34,320 It's really appropriate, really only useful, 445 00:34:34,320 --> 00:34:39,949 for mostly 2D rotational problems 446 00:34:39,949 --> 00:34:41,960 around an axis of rotation. 447 00:34:41,960 --> 00:34:51,480 And what it means is this is the distance away from the center 448 00:34:51,480 --> 00:34:56,139 of rotation at which you could concentrate all of the mass 449 00:34:56,139 --> 00:35:01,580 and have the same mass moment of inertia. 450 00:35:01,580 --> 00:35:06,260 So this has-- here's G here at the center, uniform disk. 451 00:35:09,040 --> 00:35:17,330 We know that there is an Izz about G for this problem. 452 00:35:17,330 --> 00:35:19,260 And I need to pick a coordinate system so we 453 00:35:19,260 --> 00:35:20,870 can talk about things here. 454 00:35:20,870 --> 00:35:23,570 Let me get M out of the center. 455 00:35:23,570 --> 00:35:27,510 So I'm going to let z be upwards. 456 00:35:27,510 --> 00:35:30,690 And because I'm looking ahead and want 457 00:35:30,690 --> 00:35:36,140 to keep the equation simple, I'm going to make my x-axis 458 00:35:36,140 --> 00:35:41,590 here parallel to f so I only have to deal with one vector 459 00:35:41,590 --> 00:35:43,100 component equation. 460 00:35:43,100 --> 00:35:45,830 And then that makes the y-axis this way. 461 00:35:51,550 --> 00:35:56,280 So I'm interested in Izz because I'm spinning around the z-axis. 462 00:35:56,280 --> 00:35:59,100 I know that for a uniform disk, that's a principal axis, 463 00:35:59,100 --> 00:36:01,870 is the vertical one. 464 00:36:01,870 --> 00:36:05,830 And what I'm saying is that you can then set find. 465 00:36:05,830 --> 00:36:12,990 There's an IzzG that can be expressed as M kappa squared. 466 00:36:12,990 --> 00:36:21,700 So kappa, in effect, is just IzzG over M square root. 467 00:36:21,700 --> 00:36:23,450 Now, why do we use that kind of thing? 468 00:36:23,450 --> 00:36:28,009 Well, the way this problem was set up-- I 469 00:36:28,009 --> 00:36:29,300 actually took it out of a book. 470 00:36:29,300 --> 00:36:32,170 It wasn't a uniform disk. 471 00:36:32,170 --> 00:36:34,230 It's a pulley wheel or something. 472 00:36:34,230 --> 00:36:40,760 And it's got spokes in here and a rim. 473 00:36:40,760 --> 00:36:47,577 It's still axially-- it still has some axial symmetries. 474 00:36:47,577 --> 00:36:48,910 But it's getting a little messy. 475 00:36:48,910 --> 00:36:50,326 It's hard for you-- you can't just 476 00:36:50,326 --> 00:36:52,620 say that's MR squared over 2. 477 00:36:52,620 --> 00:36:56,250 It's got some other mass moment of inertia about the center. 478 00:36:56,250 --> 00:36:59,670 But it's got holes and stuff in it because of the spokes. 479 00:36:59,670 --> 00:37:06,010 So oftentimes, you'll be given the radius of gyration 480 00:37:06,010 --> 00:37:08,650 because it's a little difficult to give you 481 00:37:08,650 --> 00:37:13,316 a mathematical description of what the actual Izz is. 482 00:37:13,316 --> 00:37:14,440 That's often why you do it. 483 00:37:14,440 --> 00:37:17,346 And the thing is, you can measure. 484 00:37:17,346 --> 00:37:18,720 Rather than try to calculate, you 485 00:37:18,720 --> 00:37:21,320 can actually just measure the mass moment 486 00:37:21,320 --> 00:37:22,960 of inertia of something. 487 00:37:22,960 --> 00:37:26,060 So how would I measure-- let's say I didn't know any formulas, 488 00:37:26,060 --> 00:37:28,510 but how would I measure the mass moment of inertia 489 00:37:28,510 --> 00:37:30,052 of this in the z direction? 490 00:37:36,450 --> 00:37:38,055 What experiment would you do? 491 00:37:38,055 --> 00:37:41,310 AUDIENCE: [INAUDIBLE] angular acceleration. 492 00:37:41,310 --> 00:37:42,740 PROFESSOR: She says apply torque. 493 00:37:42,740 --> 00:37:44,830 Measure the angular acceleration. 494 00:37:44,830 --> 00:37:46,620 Hang a weight off of it, known weight. 495 00:37:46,620 --> 00:37:48,590 Wrap a string around it. 496 00:37:48,590 --> 00:37:49,400 Known mass. 497 00:37:49,400 --> 00:37:52,120 Known G. Known torque around the center. 498 00:37:52,120 --> 00:37:54,620 Measure the angular acceleration. 499 00:37:54,620 --> 00:37:56,955 I theta double dot equals the torque. 500 00:37:56,955 --> 00:37:59,505 You know the torque, you know the measure 501 00:37:59,505 --> 00:38:00,810 of the theta double dot. 502 00:38:00,810 --> 00:38:03,910 Calculate I. I equals M kappa squared, and you could just 503 00:38:03,910 --> 00:38:06,910 say, well, the kappa for this system is. 504 00:38:06,910 --> 00:38:08,900 That's how you use it. 505 00:38:08,900 --> 00:38:11,380 I'll give you a very common example, really hard 506 00:38:11,380 --> 00:38:13,180 to calculate mass moment of inertia. 507 00:38:13,180 --> 00:38:15,679 A marine propeller. 508 00:38:15,679 --> 00:38:17,970 You actually do want to know the mass moment of inertia 509 00:38:17,970 --> 00:38:21,400 about its center for purposes of torsional oscillations 510 00:38:21,400 --> 00:38:23,290 on the shaft, et cetera. 511 00:38:23,290 --> 00:38:24,620 Hard to calculate. 512 00:38:24,620 --> 00:38:25,590 Pretty easy to measure. 513 00:38:28,140 --> 00:38:32,700 So how many degrees of freedom does this problem has? 514 00:38:32,700 --> 00:38:36,980 Well, when we say it's 2D, it's a rigid body. 515 00:38:36,980 --> 00:38:40,870 But it's 2D, which means it's lying on a plane. 516 00:38:40,870 --> 00:38:42,290 It's a planar motion problem. 517 00:38:42,290 --> 00:38:44,390 Only allowed to rotate in z. 518 00:38:44,390 --> 00:38:48,280 Not allowed to rotate in around the x-axis or the y-axis. 519 00:38:48,280 --> 00:38:52,570 So for the rigid body, six degrees of freedom possible. 520 00:38:52,570 --> 00:38:55,340 There's two immediately that you said it can't rotate. 521 00:38:55,340 --> 00:38:57,770 So we're down to four. 522 00:38:57,770 --> 00:39:06,070 It cannot translate in the z direction. 523 00:39:06,070 --> 00:39:09,750 We're down to three. 524 00:39:09,750 --> 00:39:11,300 So that leaves us what? 525 00:39:14,120 --> 00:39:18,340 So the degrees of freedom for this problem is 526 00:39:18,340 --> 00:39:20,822 6 minus 3 constraints is 3. 527 00:39:20,822 --> 00:39:23,030 That means we have to have three equations of motion. 528 00:39:23,030 --> 00:39:25,520 And they would account for what are the possible-- 529 00:39:25,520 --> 00:39:26,895 in other words, saying this, what 530 00:39:26,895 --> 00:39:29,170 are the possible motions now of this problem? 531 00:39:29,170 --> 00:39:30,520 AUDIENCE: [INAUDIBLE]. 532 00:39:30,520 --> 00:39:31,935 PROFESSOR: x, y, and z. 533 00:39:31,935 --> 00:39:33,030 Or rotation in z. 534 00:39:52,260 --> 00:39:54,140 So notice we've set the problem up. 535 00:39:54,140 --> 00:40:01,160 Now to go about solving it, we need a free body diagram. 536 00:40:01,160 --> 00:40:03,530 So here's my disk. 537 00:40:03,530 --> 00:40:04,650 Here's the force. 538 00:40:08,660 --> 00:40:13,420 Any other-- and there certainly has weight in the z direction, 539 00:40:13,420 --> 00:40:17,320 but there's no z acceleration. 540 00:40:17,320 --> 00:40:20,240 So in the plane of the board, that's 541 00:40:20,240 --> 00:40:21,970 the only external forces acting on this. 542 00:40:25,540 --> 00:40:28,690 And there's our G. 543 00:40:28,690 --> 00:40:31,510 Now this problem-- and I'll say generalize 544 00:40:31,510 --> 00:40:32,580 on this in a few minutes. 545 00:40:32,580 --> 00:40:40,030 This problem can always be restated as-- recast, 546 00:40:40,030 --> 00:40:42,710 let me put it that way-- as, here's your point 547 00:40:42,710 --> 00:40:49,060 G with a force acting on this center of mass. 548 00:40:49,060 --> 00:40:52,209 See, this force doesn't go through the center of mass. 549 00:40:52,209 --> 00:40:54,000 This force goes through the center of mass. 550 00:40:54,000 --> 00:40:58,070 I'm going to replace that problem with this problem. 551 00:40:58,070 --> 00:41:02,760 A pure moment acting about the center of mass. 552 00:41:02,760 --> 00:41:04,600 You can always make this transition. 553 00:41:04,600 --> 00:41:07,490 And I'll do the general case for you in just a minute. 554 00:41:07,490 --> 00:41:08,690 But you can always do this. 555 00:41:11,230 --> 00:41:14,330 So that's kind of my second point here. 556 00:41:14,330 --> 00:41:19,810 Third point, we need then-- this is our free body diagram. 557 00:41:19,810 --> 00:41:25,211 We need apply our laws of motion. 558 00:41:25,211 --> 00:41:31,420 So sum of the forces in the y are-- now remember, 559 00:41:31,420 --> 00:41:33,880 this is z coming out of the board. 560 00:41:33,880 --> 00:41:36,930 y this way, x this way. 561 00:41:36,930 --> 00:41:40,210 Sum of the forces in the y? 562 00:41:40,210 --> 00:41:45,570 Zero, M acceleration of G in the y direction, zero. 563 00:41:45,570 --> 00:41:48,020 So you know there's no acceleration in the y. 564 00:41:48,020 --> 00:41:49,910 So it has three degrees of freedom. 565 00:41:49,910 --> 00:41:52,340 That's the first equation of motion. 566 00:41:52,340 --> 00:41:57,020 It gives you a trivial result. So y double dot is zero. 567 00:41:57,020 --> 00:41:59,647 That's your first equation of motion. 568 00:41:59,647 --> 00:42:01,480 Then you have the second equation of motion, 569 00:42:01,480 --> 00:42:04,620 sum of the forces in the x direction 570 00:42:04,620 --> 00:42:09,090 equals just our F i hat. 571 00:42:09,090 --> 00:42:11,480 It's positive x direction because we 572 00:42:11,480 --> 00:42:15,310 were clever in how we set up the coordinate system. 573 00:42:15,310 --> 00:42:21,450 And that's got to be Mx double dot, i hat direction. 574 00:42:21,450 --> 00:42:26,020 So we know right away that x double dot is the force 575 00:42:26,020 --> 00:42:33,410 F divided by M. And that's 150 newtons over 75 kilograms, 576 00:42:33,410 --> 00:42:37,740 or 2 meters per second squared. 577 00:42:37,740 --> 00:42:38,920 All right. 578 00:42:38,920 --> 00:42:48,600 The third one, then, is the moment 579 00:42:48,600 --> 00:42:52,580 of inertia with respect to G in this problem-- 580 00:42:52,580 --> 00:43:02,254 excuse me, the angular momentum in some IzzG times omega z. 581 00:43:02,254 --> 00:43:03,670 And that's what we're looking for. 582 00:43:03,670 --> 00:43:10,920 So this is another way of-- IzzG theta dot k hat direction. 583 00:43:16,800 --> 00:43:20,680 And we're going to apply that the external torques, some 584 00:43:20,680 --> 00:43:28,000 of the torques, is dH respect to G dt. 585 00:43:28,000 --> 00:43:33,620 And that's going to give us IzzG theta double dot. 586 00:43:38,720 --> 00:43:40,690 So this third class of problems you 587 00:43:40,690 --> 00:43:45,110 are best just working with respect to the center of mass. 588 00:43:45,110 --> 00:43:46,360 That's kind of the point here. 589 00:43:46,360 --> 00:43:51,550 There's no points of contact. 590 00:43:51,550 --> 00:43:54,110 There's just known external forces. 591 00:43:54,110 --> 00:43:55,650 You have to deal with them. 592 00:43:55,650 --> 00:43:59,360 Do your work with respect to the center of mass. 593 00:43:59,360 --> 00:44:02,280 So we have force equations. 594 00:44:02,280 --> 00:44:04,720 We have moment equations. 595 00:44:04,720 --> 00:44:10,350 And basically you know Izz for this problem 596 00:44:10,350 --> 00:44:12,460 is kappa squared M. And you're given M, 597 00:44:12,460 --> 00:44:14,200 and you're given kappa. 598 00:44:14,200 --> 00:44:18,790 So you can now-- and what we have to-- actually, 599 00:44:18,790 --> 00:44:21,090 the last thing left here is to figure out the torque. 600 00:44:21,090 --> 00:44:24,400 What's the torque? 601 00:44:24,400 --> 00:44:29,410 Well, it's R in this direction crossed 602 00:44:29,410 --> 00:44:30,990 with F in that direction. 603 00:44:30,990 --> 00:44:41,670 So it's Rj cross with Fi, j cross i, minus RF, 604 00:44:41,670 --> 00:44:48,320 minus RF, k hat. 605 00:44:48,320 --> 00:44:57,160 So you can now solve for theta double dot as RF over Izz, 606 00:44:57,160 --> 00:44:59,480 or RF over M kappa squared. 607 00:45:22,620 --> 00:45:26,890 And that says-- minus says it's rotating this way, which 608 00:45:26,890 --> 00:45:28,430 is what you'd expect. 609 00:45:28,430 --> 00:45:30,570 Right? 610 00:45:30,570 --> 00:45:32,990 Now I meant to ask you a question before we started. 611 00:45:32,990 --> 00:45:34,222 But think about this. 612 00:45:34,222 --> 00:45:36,180 If I had, right at the beginning, had said, OK, 613 00:45:36,180 --> 00:45:37,020 this is a problem. 614 00:45:39,850 --> 00:45:43,220 If I grab this string and pull in this direction, 615 00:45:43,220 --> 00:45:45,717 will there be any motion in the y direction? 616 00:45:45,717 --> 00:45:46,800 I meant to start that way. 617 00:45:46,800 --> 00:45:48,930 I'm really kicking myself for not doing that. 618 00:45:48,930 --> 00:45:50,680 Because a lot of people think that there's possible, 619 00:45:50,680 --> 00:45:52,420 that it could move off in the y direction 620 00:45:52,420 --> 00:45:56,480 because it's not being loaded symmetrically. 621 00:45:56,480 --> 00:45:58,100 You're pulling on a side. 622 00:45:58,100 --> 00:46:01,127 Some people think it'll kind of try to move away like that. 623 00:46:01,127 --> 00:46:01,960 It doesn't, does it? 624 00:46:21,200 --> 00:46:24,026 So an important generalization. 625 00:46:24,026 --> 00:46:25,315 We've got a rigid body. 626 00:46:28,780 --> 00:46:30,310 You have a force acting on it. 627 00:46:33,390 --> 00:46:34,910 Has a mass center here. 628 00:46:37,460 --> 00:46:43,150 So perpendicular to that force is some distance. 629 00:46:43,150 --> 00:46:43,980 We'll call it d. 630 00:46:46,880 --> 00:46:53,450 You can always equate this problem to-- and set it up 631 00:46:53,450 --> 00:46:59,306 as-- a force. 632 00:46:59,306 --> 00:47:00,680 Conceptually, you can think of it 633 00:47:00,680 --> 00:47:02,544 as equal and opposite forces. 634 00:47:02,544 --> 00:47:03,710 But it's cancel one another. 635 00:47:03,710 --> 00:47:06,580 It's just like I've done nothing to this problem. 636 00:47:06,580 --> 00:47:12,880 And then a force acting at this distance F. 637 00:47:12,880 --> 00:47:14,060 This is our distance d. 638 00:47:14,060 --> 00:47:16,520 So this problem is identical to that problem. 639 00:47:16,520 --> 00:47:19,095 I've just added and taken away two more forces. 640 00:47:23,510 --> 00:47:25,450 So the total forces on the system 641 00:47:25,450 --> 00:47:31,030 are still F. And there's still an F operating at a lever arm 642 00:47:31,030 --> 00:47:32,310 d. 643 00:47:32,310 --> 00:47:36,980 But now, if I had put these two together, 644 00:47:36,980 --> 00:47:38,600 they are equal and opposite. 645 00:47:38,600 --> 00:47:43,630 And they form a couple, a moment, acting like that. 646 00:47:43,630 --> 00:47:53,170 So this is equivalent to-- there's G with an F on it 647 00:47:53,170 --> 00:47:59,480 and a moment M0. 648 00:47:59,480 --> 00:48:04,750 And this M0 is my D cross F. 649 00:48:04,750 --> 00:48:08,240 So that's the generalization for what I did up here. 650 00:48:08,240 --> 00:48:10,730 I went from that to that. 651 00:48:10,730 --> 00:48:12,260 And this is why you can do that. 652 00:48:15,400 --> 00:48:22,180 And so now if you have an object and lots of different forces 653 00:48:22,180 --> 00:48:26,570 acting on it, and this is Fi down here and here's G, 654 00:48:26,570 --> 00:48:32,770 you can draw a radius from G to this point. 655 00:48:32,770 --> 00:48:40,480 So I'll call that Ri with respect to G. Then 656 00:48:40,480 --> 00:48:48,840 the way to generalize this is that this is equal to some F 657 00:48:48,840 --> 00:49:00,050 total and a moment acting at G. 658 00:49:00,050 --> 00:49:05,170 And all that you have to do there is F total 659 00:49:05,170 --> 00:49:09,430 is the summation of the Fi's, vector sum. 660 00:49:09,430 --> 00:49:13,400 And the MG, the M with respect to G here, 661 00:49:13,400 --> 00:49:21,010 is the summation of the Ri with respect to G cross Fi. 662 00:49:21,010 --> 00:49:26,070 So that's the generalization for multiple forces on a body. 663 00:49:26,070 --> 00:49:30,940 So you are making an equivalent force acting at G and a moment 664 00:49:30,940 --> 00:49:33,300 acting at G. I shouldn't call this little g. 665 00:49:33,300 --> 00:49:34,300 That's really confusing. 666 00:49:36,980 --> 00:49:38,660 So that's the generalization when you 667 00:49:38,660 --> 00:49:43,080 need to do problems like this. 668 00:50:03,060 --> 00:50:08,950 Catch your breath while I scrub a board here. 669 00:50:08,950 --> 00:50:12,365 Now we've got to move on to these class four problems. 670 00:50:16,500 --> 00:50:17,155 Moving points. 671 00:50:30,070 --> 00:50:32,624 An example is the truck problem that you had. 672 00:50:38,700 --> 00:50:39,535 Known acceleration. 673 00:50:45,990 --> 00:50:50,670 So there's two common ways of doing this problem. 674 00:50:50,670 --> 00:50:53,470 You can do this problem by summing forces 675 00:50:53,470 --> 00:50:57,800 at the center of mass of that pipe 676 00:50:57,800 --> 00:51:01,650 and summing moments around it. 677 00:51:01,650 --> 00:51:04,900 But the moment around this comes from a friction force 678 00:51:04,900 --> 00:51:07,490 here, which you don't know. 679 00:51:07,490 --> 00:51:09,490 So that introduces an unknown that you 680 00:51:09,490 --> 00:51:10,830 have to then solve for. 681 00:51:10,830 --> 00:51:15,830 So if you work around G for this pipe, you can do it. 682 00:51:15,830 --> 00:51:18,760 You can work around G. You could say sum of the moments 683 00:51:18,760 --> 00:51:20,700 around G, sum of the force with respect to G. 684 00:51:20,700 --> 00:51:23,245 But you have to deal with unknown forces. 685 00:51:29,060 --> 00:51:41,260 So you're working with respect to G implies unknown forces, e, 686 00:51:41,260 --> 00:51:42,460 for example, friction. 687 00:51:45,760 --> 00:51:49,760 So you'd really maybe rather around the point of contact, 688 00:51:49,760 --> 00:51:54,370 A. Because if you sum your moments about A, 689 00:51:54,370 --> 00:51:55,980 the friction force has no moment arm, 690 00:51:55,980 --> 00:51:57,570 and it doesn't appear in the answer. 691 00:52:02,250 --> 00:52:04,940 But this gets trickier. 692 00:52:04,940 --> 00:52:10,670 This is a little more sophisticated, I'll just 693 00:52:10,670 --> 00:52:12,740 call it, step. 694 00:52:12,740 --> 00:52:18,680 And you need-- to do this, you need a little theorem. 695 00:52:18,680 --> 00:52:27,260 So to work with respect to A, you 696 00:52:27,260 --> 00:52:31,770 need to be able to say that the angular momentum with respect 697 00:52:31,770 --> 00:52:37,110 to A-- you now are working around a moving point, maybe 698 00:52:37,110 --> 00:52:38,880 accelerating. 699 00:52:38,880 --> 00:52:40,750 It's very handy to be able to say 700 00:52:40,750 --> 00:52:44,210 it's the angular momentum around G, which is easier 701 00:52:44,210 --> 00:52:52,980 to calculate, plus RGA, the distance 702 00:52:52,980 --> 00:52:56,490 from the center of mass to the point you're working on, 703 00:52:56,490 --> 00:53:00,280 cross the linear momentum of the system with respect 704 00:53:00,280 --> 00:53:02,720 to the inertial frame. 705 00:53:02,720 --> 00:53:03,630 We need this. 706 00:53:03,630 --> 00:53:05,670 This is a formula we need. 707 00:53:05,670 --> 00:53:17,690 And see why this is true? 708 00:53:17,690 --> 00:53:20,910 This is sort of thing-- this is a formula that's 709 00:53:20,910 --> 00:53:22,750 come out of the blue here. 710 00:53:22,750 --> 00:53:24,430 And why is it true? 711 00:53:24,430 --> 00:53:27,760 So I don't usually like to do proofs, but the proofs of this 712 00:53:27,760 --> 00:53:29,090 [INAUDIBLE] on the board. 713 00:53:29,090 --> 00:53:31,555 But the proof of this is really quite simple. 714 00:53:37,320 --> 00:53:40,430 Here's a little mass point Mi. 715 00:53:40,430 --> 00:53:45,940 And this radius is R of i with respect to A. 716 00:53:45,940 --> 00:53:53,160 This is R of G with respect to A. 717 00:53:53,160 --> 00:54:00,290 And therefore, this is R of i with respect to G is this one. 718 00:54:00,290 --> 00:54:04,960 And we know that this plus this gives us that. 719 00:54:04,960 --> 00:54:16,120 We can say Ri particle i with respect to A is RG with respect 720 00:54:16,120 --> 00:54:20,775 to A plus Ri with respect to G, all vectors. 721 00:54:45,170 --> 00:54:52,730 So the angular momentum of that body from the basic definition 722 00:54:52,730 --> 00:54:54,530 of angular momentum is the summation 723 00:54:54,530 --> 00:54:58,137 of all the little mass bits of the Ri with respect to A 724 00:54:58,137 --> 00:55:01,920 cross the linear momentum of each little mass bit. 725 00:55:04,560 --> 00:55:07,880 But we can expand that with that sum. 726 00:55:07,880 --> 00:55:13,370 So this is the summation of my RG with respect 727 00:55:13,370 --> 00:55:22,440 to A plus Ri with respect to G cross Pio, summation 728 00:55:22,440 --> 00:55:25,190 over all the mass bits. 729 00:55:25,190 --> 00:55:28,570 I'm going to expand this. 730 00:55:28,570 --> 00:55:33,010 And I can expand this into this times that, summations 731 00:55:33,010 --> 00:55:35,230 of this times that, and this times that. 732 00:55:35,230 --> 00:55:50,610 So that becomes a RGA cross summation of-- what's Pio? 733 00:55:50,610 --> 00:55:55,780 This is a little Mi's, Vi with respect to o. 734 00:55:55,780 --> 00:55:56,830 Each one has velocity. 735 00:55:56,830 --> 00:55:58,570 Each one has a mass. 736 00:55:58,570 --> 00:56:08,340 So this is Mi Vi with respect to o plus the summation 737 00:56:08,340 --> 00:56:13,982 RiG's cross Mi Vio's. 738 00:56:16,890 --> 00:56:19,580 Now, notice I pulled this one outside the summation. 739 00:56:19,580 --> 00:56:21,052 That's because this is a single-- 740 00:56:21,052 --> 00:56:22,010 this is a fixed number. 741 00:56:22,010 --> 00:56:23,468 It doesn't change in the summation. 742 00:56:23,468 --> 00:56:27,500 It's just the distance from my starting point to G. 743 00:56:27,500 --> 00:56:29,280 So I can pull it out and do the summation 744 00:56:29,280 --> 00:56:30,530 and then do the cross product. 745 00:56:33,680 --> 00:56:39,155 What is the summation of all the MVi's for the body? 746 00:56:44,590 --> 00:56:46,490 That's the momentum of each little particle. 747 00:56:46,490 --> 00:56:49,400 Add them all up, what do you get? 748 00:56:49,400 --> 00:56:54,520 This is RGA cross P with respect to o. 749 00:56:54,520 --> 00:56:56,620 That's this term. 750 00:56:56,620 --> 00:57:00,730 And this is all the little distances 751 00:57:00,730 --> 00:57:08,120 from the center of mass to the cross with the momentum 752 00:57:08,120 --> 00:57:09,100 of each little one. 753 00:57:13,460 --> 00:57:15,370 What's that? 754 00:57:15,370 --> 00:57:20,190 Well, this looks like a definition of angular momentum. 755 00:57:20,190 --> 00:57:24,580 This is the angular momentum of every little mass particle 756 00:57:24,580 --> 00:57:28,530 with respect to G added up. 757 00:57:28,530 --> 00:57:33,900 This is H with respect to G, which 758 00:57:33,900 --> 00:57:36,780 is what we set out to prove. 759 00:57:36,780 --> 00:57:43,920 So the H with respect to A is H with respect to G plus RGA 760 00:57:43,920 --> 00:57:46,570 cross the linear momentum of the body. 761 00:57:46,570 --> 00:57:47,928 Yeah. 762 00:57:47,928 --> 00:57:52,374 AUDIENCE: Can you also do this by writing Pi with respect 763 00:57:52,374 --> 00:57:54,350 to o, I mean the velocity part of that, 764 00:57:54,350 --> 00:57:57,808 as the sum of the velocities? 765 00:57:57,808 --> 00:58:00,344 PROFESSOR: You got to keep the M's in there. 766 00:58:00,344 --> 00:58:01,760 AUDIENCE: Well, yeah. [INAUDIBLE]. 767 00:58:01,760 --> 00:58:03,736 But instead of writing out R as a sum, 768 00:58:03,736 --> 00:58:08,676 you can write out the velocity as the sum of the velocities 769 00:58:08,676 --> 00:58:11,991 with respect to the origin. 770 00:58:11,991 --> 00:58:13,990 PROFESSOR: Are you talking about this term here? 771 00:58:13,990 --> 00:58:19,310 AUDIENCE: No, the definition of angular momentum. 772 00:58:19,310 --> 00:58:19,810 Yeah. 773 00:58:24,175 --> 00:58:25,130 The velocity-- 774 00:58:25,130 --> 00:58:26,020 PROFESSOR: If you could figure out-- 775 00:58:26,020 --> 00:58:27,880 if you had the angular momentum [INAUDIBLE] 776 00:58:27,880 --> 00:58:30,520 and multiplied by Ri, that is the angular momentum 777 00:58:30,520 --> 00:58:32,940 with respect to A. But I broke it 778 00:58:32,940 --> 00:58:37,110 apart so that I could show you that this formula I want to use 779 00:58:37,110 --> 00:58:37,875 has two pieces. 780 00:58:40,490 --> 00:58:42,430 I want to use that. 781 00:58:42,430 --> 00:58:49,020 So that if I can use that-- it's easy to get H with respect to G 782 00:58:49,020 --> 00:58:49,650 sometimes. 783 00:58:49,650 --> 00:58:52,160 It's really hard to know what to do with things that 784 00:58:52,160 --> 00:58:53,980 happen around this point A. 785 00:58:53,980 --> 00:58:55,832 So now let's go back. 786 00:58:55,832 --> 00:58:57,790 I think, to understand this, we need to go back 787 00:58:57,790 --> 00:59:00,190 to our truck problem. 788 00:59:00,190 --> 00:59:03,200 We now have a formula that you know where it comes from. 789 00:59:03,200 --> 00:59:08,890 We have our truck that is accelerating at x1 double dot. 790 00:59:08,890 --> 00:59:16,820 And we want to find out what's theta double dot for that pipe. 791 00:59:16,820 --> 00:59:18,272 What's it doing? 792 00:59:18,272 --> 00:59:19,730 So first we needed some kinematics. 793 00:59:25,780 --> 00:59:34,710 And in particular, what is the x2-- did I label this 794 00:59:34,710 --> 00:59:35,210 very well? 795 00:59:35,210 --> 00:59:36,390 I didn't. 796 00:59:36,390 --> 00:59:38,770 So here's my pipe. 797 00:59:38,770 --> 00:59:42,310 Here's my truck bed that it's in contact with. 798 00:59:42,310 --> 00:59:46,450 Truck's moving at x1 double dot, we know. 799 00:59:46,450 --> 00:59:49,310 In an inertial frame, the movement 800 00:59:49,310 --> 00:59:54,210 of the center of mass of my pipe is x2. 801 00:59:54,210 --> 01:00:02,630 And it has some angular rotation I'll call theta. 802 01:00:02,630 --> 01:00:06,750 So the movement of this guy I need 803 01:00:06,750 --> 01:00:10,260 to be able to express in terms of x1 and theta. 804 01:00:10,260 --> 01:00:14,590 Well, if this is fixed to the truck and the truck move 805 01:00:14,590 --> 01:00:18,540 forward, then x2 would be equal to x1. 806 01:00:18,540 --> 01:00:22,360 But in fact, it rolls back a little bit. 807 01:00:22,360 --> 01:00:23,820 And the distance this point moves 808 01:00:23,820 --> 01:00:25,790 if it rolls through an angle theta 809 01:00:25,790 --> 01:00:28,310 is it rolls backwards an amount R theta. 810 01:00:32,140 --> 01:00:35,180 And we're going to need to take two derivatives of that. 811 01:00:35,180 --> 01:00:42,200 x2 double dot equals x1 double dot minus r theta double dot. 812 01:00:42,200 --> 01:00:43,950 So that's a little kinematic relationship 813 01:00:43,950 --> 01:00:45,075 we need to do this problem. 814 01:00:55,440 --> 01:00:58,830 Next we need to apply a physical law, which 815 01:00:58,830 --> 01:01:03,720 is the one I've derived. 816 01:01:03,720 --> 01:01:06,760 So this is our physics here now, our physical law. 817 01:01:11,020 --> 01:01:16,680 And that's the external torques, dH with respect 818 01:01:16,680 --> 01:01:24,310 to A dt, plus VAo cross Po. 819 01:01:27,360 --> 01:01:34,656 And in this case, is VAo 0? 820 01:01:34,656 --> 01:01:38,140 No, in fact, it's x1 dot, right? 821 01:01:38,140 --> 01:01:40,480 It's in the i direction. 822 01:01:40,480 --> 01:01:44,164 How about what direction is P, the momentum of the pipe? 823 01:01:49,250 --> 01:01:52,355 Does it move in the y direction? 824 01:01:52,355 --> 01:01:52,855 Up, down? 825 01:01:52,855 --> 01:01:53,355 No. 826 01:01:53,355 --> 01:01:54,490 It only moves in the x. 827 01:01:54,490 --> 01:01:56,920 So this velocity is only in the x. 828 01:01:56,920 --> 01:01:59,740 This is in the x, or the i hat direction. 829 01:01:59,740 --> 01:02:01,910 This cross product is zero. 830 01:02:01,910 --> 01:02:03,730 So it just happens that they're parallel. 831 01:02:03,730 --> 01:02:04,890 So this thing goes to zero. 832 01:02:04,890 --> 01:02:07,025 We don't have to deal with it. 833 01:02:07,025 --> 01:02:08,650 That's because these guys are parallel. 834 01:02:13,350 --> 01:02:15,380 Parallel motion. 835 01:02:15,380 --> 01:02:17,080 But you did have to consider it. 836 01:02:17,080 --> 01:02:18,371 You did have to think about it. 837 01:02:18,371 --> 01:02:20,120 It's not just trivially 0. 838 01:02:20,120 --> 01:02:24,790 OK, so that means that the torques about A is just dHA dt. 839 01:02:24,790 --> 01:02:33,070 And HA for this problem is i-- well, 840 01:02:33,070 --> 01:02:41,030 it's HG, which is IzzG theta double dot. 841 01:02:41,030 --> 01:02:44,480 But I have to put in this-- theta dot, excuse me. 842 01:02:44,480 --> 01:02:46,550 This is just the H. I have to put 843 01:02:46,550 --> 01:02:51,690 in the second term, RG with respect 844 01:02:51,690 --> 01:02:55,747 to A cross P with respect to o. 845 01:02:59,330 --> 01:03:01,280 So here's my HA. 846 01:03:04,234 --> 01:03:05,400 I'll write it again up here. 847 01:03:05,400 --> 01:03:13,150 This is IzzG theta dot in the k. 848 01:03:13,150 --> 01:03:15,020 And now this second term. 849 01:03:15,020 --> 01:03:19,990 R is Rj. 850 01:03:19,990 --> 01:03:21,920 My y-axis is up. 851 01:03:21,920 --> 01:03:26,990 The radius of this thing is R. Here's the radius. 852 01:03:26,990 --> 01:03:37,050 So the moment arm is Rj crossed with the linear momentum 853 01:03:37,050 --> 01:03:42,070 of that piece of pipe, which is the mass of the pipe times 854 01:03:42,070 --> 01:03:48,140 x2 dot in the i direction. 855 01:03:48,140 --> 01:03:50,350 j cross i is minus k. 856 01:03:50,350 --> 01:03:56,330 So Izz with respect to G theta double dot-- theta dot. 857 01:03:56,330 --> 01:03:59,580 I keep taking the derivative a little too soon. 858 01:03:59,580 --> 01:04:06,050 k minus RM x2 dot k. 859 01:04:06,050 --> 01:04:12,300 And now some of the torques, d by dt of H with respect to A. 860 01:04:12,300 --> 01:04:14,380 Take the derivatives. 861 01:04:14,380 --> 01:04:27,456 IzzG theta double dot k minus RM x2 double dot k. 862 01:04:27,456 --> 01:04:29,830 And fortunately, none of these unit vectors are rotating. 863 01:04:29,830 --> 01:04:32,070 So we don't have to deal with any of that. 864 01:04:32,070 --> 01:04:34,710 And I'm almost to the end, but now I 865 01:04:34,710 --> 01:04:39,410 have to go back to that original kinematic relationship, which 866 01:04:39,410 --> 01:04:46,380 allows me to express x2 in terms of x1 and R theta. 867 01:04:46,380 --> 01:04:51,100 And if I substitute that in here and solve for it, 868 01:04:51,100 --> 01:05:10,620 I get theta double dot equals MR over IzzG plus MR squared. 869 01:05:13,870 --> 01:05:20,310 x double dot whole thing, x1 double dot. 870 01:05:20,310 --> 01:05:22,860 So you've accelerated x1 double dot. 871 01:05:22,860 --> 01:05:25,740 The thing starts rolling. 872 01:05:25,740 --> 01:05:28,440 It's actually rolling backwards, which is a plus theta 873 01:05:28,440 --> 01:05:31,820 direction, at this rate. 874 01:05:31,820 --> 01:05:37,010 We did this using this formula for HA. 875 01:05:37,010 --> 01:05:39,700 Now, the beauty of this formula is 876 01:05:39,700 --> 01:05:43,170 that it works for any points that 877 01:05:43,170 --> 01:05:46,870 are moving, even can be accelerating, 878 01:05:46,870 --> 01:05:48,925 all sorts of nasty conditions, it's true. 879 01:05:48,925 --> 01:05:54,380 It's based on the fundamental definition of angular momentum. 880 01:05:54,380 --> 01:05:55,965 So the book-- yeah. 881 01:05:55,965 --> 01:05:58,000 AUDIENCE: [INAUDIBLE] final point here. 882 01:05:58,000 --> 01:06:00,160 I just want to make sure i understood. 883 01:06:00,160 --> 01:06:04,620 In this first line here, [INAUDIBLE]. 884 01:06:04,620 --> 01:06:07,390 That stroke doesn't mean that V sub A was 0. 885 01:06:07,390 --> 01:06:10,940 It meant that that term is 0. 886 01:06:10,940 --> 01:06:13,370 PROFESSOR: This term's zero because these 887 01:06:13,370 --> 01:06:16,772 happen to be parallel, but the product is zero in this case. 888 01:06:16,772 --> 01:06:20,390 If it had not turned out to be-- they were different directions. 889 01:06:20,390 --> 01:06:21,650 It had to be a non-zero term. 890 01:06:21,650 --> 01:06:25,715 And you would have to bring it along and take its derivative 891 01:06:25,715 --> 01:06:26,840 along with the other stuff. 892 01:06:29,870 --> 01:06:32,430 But this is a really powerful method. 893 01:06:32,430 --> 01:06:36,750 Now, the book is reasonably good in lots of points. 894 01:06:36,750 --> 01:06:39,750 But in chapter 17, when it does problems 895 01:06:39,750 --> 01:06:43,910 that are kind of like this, it introduces something 896 01:06:43,910 --> 01:06:46,910 that I find it hard to digest what they're doing. 897 01:06:46,910 --> 01:06:48,525 There's particularly-- where's Matt? 898 01:06:48,525 --> 01:06:53,830 There's equation 1715. 899 01:06:53,830 --> 01:06:56,340 When you get to that bit of the book, just ignore it. 900 01:06:56,340 --> 01:06:59,510 Use this method. 901 01:06:59,510 --> 01:07:02,360 The author tries to give you a little trick that you can use. 902 01:07:02,360 --> 01:07:05,420 But the problem with tricks is you have to memorize them. 903 01:07:05,420 --> 01:07:09,800 So what I've shown you today is based on basic definition 904 01:07:09,800 --> 01:07:10,740 of angular momentum. 905 01:07:10,740 --> 01:07:15,950 That expression at the top is always usable, not just 906 01:07:15,950 --> 01:07:20,250 special little conditions, which is what the formula in the book 907 01:07:20,250 --> 01:07:21,500 is generated for. 908 01:07:25,710 --> 01:07:27,010 We've got a couple of minutes. 909 01:07:29,540 --> 01:07:31,780 Let you ask questions, and then I'll 910 01:07:31,780 --> 01:07:34,100 just pose a conceptual problem or two 911 01:07:34,100 --> 01:07:35,650 and ask you what method you'd use. 912 01:07:35,650 --> 01:07:37,420 Yeah. 913 01:07:37,420 --> 01:07:40,892 AUDIENCE: So [INAUDIBLE] when you [INAUDIBLE] dHA dt, 914 01:07:40,892 --> 01:07:44,690 is that equal to 0 because there is no torque in the system? 915 01:07:44,690 --> 01:07:49,605 PROFESSOR: Oh, you know, boy, I'm glad you caught that. 916 01:07:49,605 --> 01:07:52,080 Yeah, in order to be able to do this, 917 01:07:52,080 --> 01:07:56,550 we've got to know something here, right? 918 01:07:56,550 --> 01:07:58,020 Important catch. 919 01:07:58,020 --> 01:07:59,280 Thank you. 920 01:07:59,280 --> 01:08:02,580 Why is this true? 921 01:08:02,580 --> 01:08:05,490 AUDIENCE: [INAUDIBLE]. 922 01:08:05,490 --> 01:08:08,890 PROFESSOR: So we chose-- that's why 923 01:08:08,890 --> 01:08:16,590 we chose to work around point A. With respect to that point, 924 01:08:16,590 --> 01:08:19,350 there are no external forces that 925 01:08:19,350 --> 01:08:23,460 create moments on that pipe with respect to that point. 926 01:08:23,460 --> 01:08:26,250 And that's why you can say that the time 927 01:08:26,250 --> 01:08:28,310 derivative of this angular momentum 928 01:08:28,310 --> 01:08:32,060 is equal to zero because there are no external torques. 929 01:08:32,060 --> 01:08:35,660 If you had picked G to do this problem, 930 01:08:35,660 --> 01:08:39,437 would the sum of the torques about G be zero? 931 01:08:39,437 --> 01:08:41,520 No, you'd have to put that friction force in there 932 01:08:41,520 --> 01:08:44,600 and have RF and figure out F is. 933 01:08:44,600 --> 01:08:47,640 We completely avoided having to calculate the friction force. 934 01:08:47,640 --> 01:08:50,880 That's the point of being able to use techniques like this 935 01:08:50,880 --> 01:08:54,990 and make your computations around points of contact. 936 01:09:05,890 --> 01:09:10,390 So textbooks have lots of problems like this. 937 01:09:10,390 --> 01:09:14,774 You've got a box on a cart. 938 01:09:14,774 --> 01:09:17,149 And your kid's pushing it, and he gets a little exuberant 939 01:09:17,149 --> 01:09:19,689 and pushes a little too hard, accelerates the cart 940 01:09:19,689 --> 01:09:22,710 a little too fast, and the box falls over and breaks 941 01:09:22,710 --> 01:09:25,300 the lamp or whatever's in it. 942 01:09:25,300 --> 01:09:28,580 And if it's this way, and I accelerate it, 943 01:09:28,580 --> 01:09:30,620 it's much more tolerant. 944 01:09:30,620 --> 01:09:32,859 Falls over easier this way. 945 01:09:32,859 --> 01:09:35,640 But if I asked you, gave you a problem 946 01:09:35,640 --> 01:09:40,899 and said, calculate the maximum acceleration 947 01:09:40,899 --> 01:09:44,950 that I can put on this object such that it 948 01:09:44,950 --> 01:09:49,149 just barely-- just right at the edge of tipping over, 949 01:09:49,149 --> 01:09:50,149 but doesn't tip it over. 950 01:09:50,149 --> 01:09:52,648 What's that maximum acceleration that you don't tip it over? 951 01:09:52,648 --> 01:09:57,050 What method would you use? 952 01:09:57,050 --> 01:10:01,100 I gave you four classes of approaches to problems. 953 01:10:07,412 --> 01:10:09,740 AUDIENCE: [INAUDIBLE]. 954 01:10:09,740 --> 01:10:11,230 PROFESSOR: I hear a four. 955 01:10:11,230 --> 01:10:12,890 Anybody else want to bid here? 956 01:10:16,090 --> 01:10:17,050 More fours. 957 01:10:17,050 --> 01:10:20,168 Would three work? 958 01:10:20,168 --> 01:10:22,580 AUDIENCE: [INAUDIBLE]. 959 01:10:22,580 --> 01:10:23,270 PROFESSOR: Why? 960 01:10:23,270 --> 01:10:26,060 He says three would be complicated. 961 01:10:26,060 --> 01:10:28,400 Three means taking moments and forces with respect 962 01:10:28,400 --> 01:10:31,400 to the center of mass. 963 01:10:31,400 --> 01:10:34,378 If you do that, what do you have to deal with in this problem? 964 01:10:34,378 --> 01:10:35,294 AUDIENCE: [INAUDIBLE]. 965 01:10:36,857 --> 01:10:39,190 PROFESSOR: You have to, then-- around the center of mass 966 01:10:39,190 --> 01:10:40,910 there's a friction force. 967 01:10:40,910 --> 01:10:44,210 There's a normal force pushing up here. 968 01:10:44,210 --> 01:10:47,240 The way you do this problem is where it's just barely 969 01:10:47,240 --> 01:10:51,940 about to go, all of the force is pushing on this corner. 970 01:10:51,940 --> 01:10:52,440 Think of it. 971 01:10:52,440 --> 01:10:55,540 It's just lifting up a fraction. 972 01:10:55,540 --> 01:10:57,710 All of that contact point moves to right here. 973 01:10:57,710 --> 01:11:00,230 You have an upward force and a friction force, 974 01:11:00,230 --> 01:11:02,500 and they create a moment about G. 975 01:11:02,500 --> 01:11:04,120 But if you do the forces around G, 976 01:11:04,120 --> 01:11:06,970 you have to solve for those two. 977 01:11:06,970 --> 01:11:11,680 You do the forces around A. The trick here is figuring out 978 01:11:11,680 --> 01:11:15,540 where's A. But if you realize A is right here, 979 01:11:15,540 --> 01:11:19,350 and you do what we just did, this problem's 980 01:11:19,350 --> 01:11:21,500 just a piece of cake.