1 00:00:00,080 --> 00:00:02,430 The following content is provided under a Creative 2 00:00:02,430 --> 00:00:03,810 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 --> 00:00:17,305 at ocw.mit.edu. 8 00:00:21,800 --> 00:00:26,650 J. KIM VANDIVER: If I wanted you to remember three equations, 9 00:00:26,650 --> 00:00:29,530 really commit them to memory, that'd allow you to essentially 10 00:00:29,530 --> 00:00:35,012 do everything that we've done, they'd be the following three. 11 00:00:35,012 --> 00:00:41,450 The first would essentially be Newton's second law. 12 00:00:49,640 --> 00:00:53,500 We use that all the time, and in vector form, 13 00:00:53,500 --> 00:00:58,850 summation of the external forces on a rigid body 14 00:00:58,850 --> 00:01:02,710 must be equal to the time rate of change 15 00:01:02,710 --> 00:01:07,630 of the linear momentum of the rigid body. 16 00:01:07,630 --> 00:01:13,100 And that we know is equal to its mass times the acceleration 17 00:01:13,100 --> 00:01:19,060 of its center of mass with respect to an inertial frame, 18 00:01:19,060 --> 00:01:22,770 and this slash o means with respect to an inertial frame. 19 00:01:22,770 --> 00:01:25,070 That's Newton's second law, and of course we 20 00:01:25,070 --> 00:01:27,160 use it all the time. 21 00:01:27,160 --> 00:01:35,712 Now Euler added to what Newton worked out for us. 22 00:01:35,712 --> 00:01:38,020 And let me actually just point out over here. 23 00:01:38,020 --> 00:01:44,780 So P with respect to o then is M velocity 24 00:01:44,780 --> 00:01:49,955 of G with respect to the inertial frame-- vector, 25 00:01:49,955 --> 00:01:52,260 vector, vector. 26 00:01:52,260 --> 00:01:57,840 OK, what Euler taught us was that the summation 27 00:01:57,840 --> 00:02:07,530 of the torques about a point A is the time rate of change 28 00:02:07,530 --> 00:02:16,050 of the angular momentum with respect to that point 29 00:02:16,050 --> 00:02:24,900 plus the velocity of the point cross P with respect to o. 30 00:02:24,900 --> 00:02:32,150 So that's our general equation for angular momentum 31 00:02:32,150 --> 00:02:33,460 and [? R ?] for torques. 32 00:02:33,460 --> 00:02:46,330 And remember, H with respect to A is the cross product of R-- 33 00:02:46,330 --> 00:02:47,741 how do I want to write this guy? 34 00:02:57,595 --> 00:02:58,678 I don't want to save that. 35 00:03:15,130 --> 00:03:17,030 So we define angular momentum. 36 00:03:17,030 --> 00:03:22,090 It's just a cross product of the distance from the point 37 00:03:22,090 --> 00:03:29,810 that we're at to the-- I didn't write this right. 38 00:03:29,810 --> 00:03:40,590 This is A with respect to G cross P o, 39 00:03:40,590 --> 00:03:44,010 so that's our general definition of angular momentum. 40 00:03:44,010 --> 00:03:46,130 And finally, the third equation is the one 41 00:03:46,130 --> 00:03:49,760 that I showed you last time, which does appear in the book 42 00:03:49,760 --> 00:03:52,190 but not until you get into chapter 21, 43 00:03:52,190 --> 00:03:55,950 and that is that you can write H with respect 44 00:03:55,950 --> 00:04:11,365 to A as H with respect to G plus [? RG/A ?] cross P with respect 45 00:04:11,365 --> 00:04:12,082 to o. 46 00:04:15,910 --> 00:04:18,420 There's something wrong here. 47 00:04:18,420 --> 00:04:19,839 Matt, what have I done wrong? 48 00:04:19,839 --> 00:04:22,840 MATT: Do you want a sum that will give you 49 00:04:22,840 --> 00:04:24,302 the angle [INAUDIBLE]. 50 00:04:24,302 --> 00:04:26,260 J. KIM VANDIVER: Yeah, it just slipped my brain 51 00:04:26,260 --> 00:04:27,176 because I don't do it. 52 00:04:27,176 --> 00:04:28,670 I don't think about this. 53 00:04:28,670 --> 00:04:31,390 This is the-- yeah, it's a summation. 54 00:04:31,390 --> 00:04:34,230 There we go. 55 00:04:34,230 --> 00:04:37,780 In general, this is a summation, and it 56 00:04:37,780 --> 00:04:45,040 is the location of all the little particles i with respect 57 00:04:45,040 --> 00:04:55,220 to A cross the linear momentum of each particle [? mi ?] Vi 58 00:04:55,220 --> 00:04:57,490 with respect to o. 59 00:04:57,490 --> 00:04:59,890 And we add all that up, that's the definition 60 00:04:59,890 --> 00:05:01,300 of angular momentum. 61 00:05:01,300 --> 00:05:04,330 And we generally for rigid bodies then 62 00:05:04,330 --> 00:05:10,140 reduce that to saying that it's a mass moment of inertia 63 00:05:10,140 --> 00:05:17,510 matrix computed with respect to A times omega x, omega y, 64 00:05:17,510 --> 00:05:19,710 omega z. 65 00:05:19,710 --> 00:05:23,470 That's how we generally express this. 66 00:05:23,470 --> 00:05:28,400 But the third equation I want to come to is this one. 67 00:05:28,400 --> 00:05:30,266 Yeah? 68 00:05:30,266 --> 00:05:34,129 AUDIENCE: That's not [INAUDIBLE], 69 00:05:34,129 --> 00:05:36,170 but it's not really with respect to A in general. 70 00:05:39,606 --> 00:05:40,106 [INAUDIBLE] 71 00:05:45,050 --> 00:05:48,160 J. KIM VANDIVER: This is i with respect to A. Let's just 72 00:05:48,160 --> 00:05:52,590 do this because I've just slipped a bit this morning. 73 00:05:52,590 --> 00:05:56,230 OK, I'm thinking too much about where I'm going here. 74 00:05:56,230 --> 00:05:58,670 This equation I introduced last time. 75 00:05:58,670 --> 00:06:01,060 The book doesn't make a big deal about this at all, 76 00:06:01,060 --> 00:06:03,930 but it's really a fundamental equation 77 00:06:03,930 --> 00:06:08,870 that we proved from scratch very simply last time. 78 00:06:08,870 --> 00:06:13,120 And this equation allows you to solve lots of problems 79 00:06:13,120 --> 00:06:17,050 without even, for example, knowing the parallel axis 80 00:06:17,050 --> 00:06:17,780 theorem. 81 00:06:17,780 --> 00:06:20,850 It'll just get you through it even without knowing things 82 00:06:20,850 --> 00:06:21,670 like that. 83 00:06:21,670 --> 00:06:24,260 So with these three equations, you can do all the problems 84 00:06:24,260 --> 00:06:27,680 that we've done, and I'm going to do two or three examples 85 00:06:27,680 --> 00:06:30,160 this morning, and including coming back 86 00:06:30,160 --> 00:06:33,050 to a discussion of the parallel axis theorem. 87 00:06:33,050 --> 00:06:39,849 OK, so these three are the ones I recommend that you remember 88 00:06:39,849 --> 00:06:41,890 because everything else will flow from-- we know, 89 00:06:41,890 --> 00:06:43,790 like, we have special cases. 90 00:06:43,790 --> 00:06:58,380 So when A is at G, then this just 91 00:06:58,380 --> 00:07:01,490 reduces to H with respect to G. When 92 00:07:01,490 --> 00:07:04,840 there's no vel-- when A doesn't move, this term goes away. 93 00:07:04,840 --> 00:07:06,944 Those are all special cases, which you can just 94 00:07:06,944 --> 00:07:09,360 substitute in the numbers and discover for yourself if you 95 00:07:09,360 --> 00:07:10,796 don't remember those. 96 00:07:10,796 --> 00:07:12,170 But these are the three equations 97 00:07:12,170 --> 00:07:14,600 you want to have in your kit to use. 98 00:07:36,130 --> 00:07:38,940 So last time, I kind of classified problems 99 00:07:38,940 --> 00:07:40,520 in four different classes depending 100 00:07:40,520 --> 00:07:42,060 on the simplifications. 101 00:07:42,060 --> 00:07:45,720 And the class 4 problem was the more general case, 102 00:07:45,720 --> 00:07:51,630 and that's when the point A is allowed to move in general, 103 00:07:51,630 --> 00:07:56,160 and that last equation we had can 104 00:07:56,160 --> 00:07:58,060 be very handy in those cases. 105 00:07:58,060 --> 00:08:02,080 And I mentioned a problem last time, but I didn't work it out. 106 00:08:02,080 --> 00:08:06,450 And that's simply this problem, where 107 00:08:06,450 --> 00:08:11,330 you've got a block, maybe a box on a cart, 108 00:08:11,330 --> 00:08:15,300 and you're accelerating the cart. 109 00:08:15,300 --> 00:08:17,960 At what point will the block tip over? 110 00:08:17,960 --> 00:08:21,690 So this one I've put a nail right here, 111 00:08:21,690 --> 00:08:27,560 so the block can't slide, but it will tip about this point. 112 00:08:27,560 --> 00:08:30,730 So if I accelerate it slowly, nothing happens. 113 00:08:30,730 --> 00:08:35,179 If I accelerate it fast enough, it tips around that point. 114 00:08:35,179 --> 00:08:36,299 So that's the key. 115 00:08:36,299 --> 00:08:39,860 Identifying the point's have the work, so that's the key bit. 116 00:08:39,860 --> 00:08:42,929 What's the greatest acceleration that I can have 117 00:08:42,929 --> 00:08:47,460 so that this block will just not quite tip? 118 00:08:47,460 --> 00:08:50,134 That's the problem. 119 00:08:50,134 --> 00:08:51,550 If you didn't have the nail there, 120 00:08:51,550 --> 00:08:53,174 there's also the possibility that it'll 121 00:08:53,174 --> 00:08:54,562 slip without tipping over. 122 00:08:54,562 --> 00:08:56,380 OK, so that's another problem that we'll 123 00:08:56,380 --> 00:08:59,740 address at another moment. 124 00:08:59,740 --> 00:09:01,620 OK, so let's do this problem quickly. 125 00:09:05,920 --> 00:09:13,490 So this is our problem, and I've got a handle here, 126 00:09:13,490 --> 00:09:16,426 and I'm pushing on it with a force. 127 00:09:16,426 --> 00:09:18,480 I've got a couple of wheels. 128 00:09:18,480 --> 00:09:20,424 This is M1. 129 00:09:20,424 --> 00:09:28,770 This is M2, and I'll have an inertial frame here X, Y. 130 00:09:28,770 --> 00:09:32,640 So the question is what's the maximum force 131 00:09:32,640 --> 00:09:36,220 that I can apply such that the block won't tip over? 132 00:09:36,220 --> 00:09:41,910 And I've got a little nub there, so this thing can't slide. 133 00:09:41,910 --> 00:09:45,880 OK, and this is my point A because that's the point 134 00:09:45,880 --> 00:09:48,720 I expect it to rotate around. 135 00:09:48,720 --> 00:09:52,560 The X, Y, and Z is coming out of the board. 136 00:09:52,560 --> 00:09:58,920 OK, so Newton's second, so we apply it first. 137 00:09:58,920 --> 00:10:00,630 From Newton's second law, we say the sum 138 00:10:00,630 --> 00:10:05,582 of the forces, in this case, on the system. 139 00:10:05,582 --> 00:10:10,270 By the system I mean M1 and M2. 140 00:10:10,270 --> 00:10:12,900 We want to remember that you can collect things together 141 00:10:12,900 --> 00:10:13,550 at times. 142 00:10:13,550 --> 00:10:15,930 So the sum of forces on the system, in this case, 143 00:10:15,930 --> 00:10:26,190 is then M1 plus M2, and it's the acceleration of that point, 144 00:10:26,190 --> 00:10:33,590 and in this case we are just going to do in the X-direction 145 00:10:33,590 --> 00:10:37,720 because we know that's the only one where there's any action. 146 00:10:37,720 --> 00:10:40,190 So the sum of the forces in the X-direction 147 00:10:40,190 --> 00:10:43,055 is the total mass of the system times the acceleration 148 00:10:43,055 --> 00:10:44,930 in the X-direction. 149 00:10:44,930 --> 00:10:47,390 So the X double dot we're looking 150 00:10:47,390 --> 00:10:53,370 for is F over M1 plus M2. 151 00:10:53,370 --> 00:10:56,120 So this we're trying to find out what the maximum value for this 152 00:10:56,120 --> 00:10:56,620 is. 153 00:11:05,200 --> 00:11:07,400 Now next we can apply the torque equation 154 00:11:07,400 --> 00:11:09,010 because that's the key one. 155 00:11:09,010 --> 00:11:11,550 And that's the sum of the torques-- now it's 156 00:11:11,550 --> 00:11:18,930 just on the box-- at with respect to point A. 157 00:11:18,930 --> 00:11:20,900 And we can say, well, looking at that, 158 00:11:20,900 --> 00:11:23,010 we need a free body diagram now of our box. 159 00:11:26,780 --> 00:11:29,790 You've got an M1g down. 160 00:11:36,760 --> 00:11:41,350 And you might have a normal force upwards, generally, 161 00:11:41,350 --> 00:11:44,810 but now we want to think about where 162 00:11:44,810 --> 00:11:50,000 are the forces going to be acting on the box 163 00:11:50,000 --> 00:11:53,770 when it's just at this point that it's just barely about-- 164 00:11:53,770 --> 00:11:55,270 just barely starts to tip. 165 00:11:58,640 --> 00:12:03,110 So we're hypothesizing that here at A, there'll 166 00:12:03,110 --> 00:12:07,300 be some upward force, right? 167 00:12:07,300 --> 00:12:10,330 And it's going to take care of the static equilibrium, 168 00:12:10,330 --> 00:12:15,690 but it also could enter into our moment calculation 169 00:12:15,690 --> 00:12:17,490 if we weren't clever with respect to where 170 00:12:17,490 --> 00:12:18,720 we calculate this point. 171 00:12:22,960 --> 00:12:25,780 So the sum of the torques about this point, 172 00:12:25,780 --> 00:12:29,370 we don't have to bring this one into it. 173 00:12:29,370 --> 00:12:31,320 We do have to consider this one. 174 00:12:31,320 --> 00:12:39,760 So the [? R ?] here cross that force, so we get a moment, Mg, 175 00:12:39,760 --> 00:12:42,836 and I need some dimensions on my box. 176 00:12:42,836 --> 00:12:53,260 We'll call it b and a height h. 177 00:12:53,260 --> 00:12:57,080 So the moment arm, this axon is b/2, 178 00:12:57,080 --> 00:12:59,780 and it's in the minus [? z ?] direction, 179 00:12:59,780 --> 00:13:08,680 so I get a minus M1g b/2 k hat. 180 00:13:08,680 --> 00:13:10,980 Those are my external torques on the box 181 00:13:10,980 --> 00:13:14,650 because the normal force supporting it on the floor 182 00:13:14,650 --> 00:13:15,710 is acting at A now. 183 00:13:19,060 --> 00:13:29,950 And this must be from our second formula, 184 00:13:29,950 --> 00:13:34,050 dH with respect to A/dt plus the velocity of A 185 00:13:34,050 --> 00:13:39,000 with respect to o, cross P with respect to o. 186 00:13:39,000 --> 00:13:49,990 And P with respect to o was [? M1VG ?] with respect 187 00:13:49,990 --> 00:13:55,710 to o, all right? 188 00:13:55,710 --> 00:13:59,060 So we need to know what is the velocity of A 189 00:13:59,060 --> 00:14:00,620 with respect to o. 190 00:14:03,650 --> 00:14:08,220 But we're only allowing motions in this direction. 191 00:14:08,220 --> 00:14:10,950 This has to be evaluated in the inertial frame, 192 00:14:10,950 --> 00:14:15,200 so we have a coordinate X here. 193 00:14:15,200 --> 00:14:17,960 So it's going to be some X dot, and I'll 194 00:14:17,960 --> 00:14:20,180 give you a capital I hat just remind you 195 00:14:20,180 --> 00:14:23,230 that's in the inertial frame. 196 00:14:23,230 --> 00:14:26,760 What is the velocity of G with respect to o? 197 00:14:34,253 --> 00:14:35,544 AUDIENCE: Would it be the same? 198 00:14:35,544 --> 00:14:37,008 J. KIM VANDIVER: Yeah. 199 00:14:37,008 --> 00:14:38,740 You know, in this condition we're 200 00:14:38,740 --> 00:14:41,130 saying we want the block not to quite tip over, 201 00:14:41,130 --> 00:14:43,810 and that means it's moving with the cart at exactly 202 00:14:43,810 --> 00:14:45,800 the same velocity, so it's the same. 203 00:14:45,800 --> 00:14:49,260 It's also equal to A with respect to o, 204 00:14:49,260 --> 00:14:51,260 but if that's the case, what's the cross product 205 00:14:51,260 --> 00:14:52,884 between velocity of A with respect to o 206 00:14:52,884 --> 00:14:54,811 and G with respect to o? 207 00:14:54,811 --> 00:14:56,330 Ah, so that one goes to zero. 208 00:14:59,680 --> 00:15:00,770 That's because of this. 209 00:15:09,170 --> 00:15:15,630 OK, so that leaves us with-- we don't 210 00:15:15,630 --> 00:15:17,780 have to deal with these terms, but we 211 00:15:17,780 --> 00:15:25,210 do need to compute H with respect to A. 212 00:15:25,210 --> 00:15:28,350 And now I'm going to use this formulation because it makes 213 00:15:28,350 --> 00:15:30,630 this problem easy to do and not have 214 00:15:30,630 --> 00:15:33,760 to deal with parallel axes or any of that. 215 00:15:33,760 --> 00:15:36,020 So I'm going to use my third equation, 216 00:15:36,020 --> 00:15:38,860 and it says that this is H with respect 217 00:15:38,860 --> 00:15:47,380 to G plus RG with respect to A cross P with respect to o. 218 00:15:52,280 --> 00:15:56,590 So you notice I chose this problem kind of on purpose 219 00:15:56,590 --> 00:15:59,250 today because I started saying, here's three 220 00:15:59,250 --> 00:16:00,990 really important equations. 221 00:16:00,990 --> 00:16:04,850 We use all three to do this problem. 222 00:16:04,850 --> 00:16:06,440 OK, so now we're exercising the third. 223 00:16:19,940 --> 00:16:26,720 So my block here, since we're dealing 224 00:16:26,720 --> 00:16:28,510 with angular momentum of rigid bodies, 225 00:16:28,510 --> 00:16:32,130 I now have to think in terms of a coordinate system attached 226 00:16:32,130 --> 00:16:39,980 to the block, and I'll formulate it 227 00:16:39,980 --> 00:16:42,840 so it's lined up the same way. 228 00:16:42,840 --> 00:16:50,480 So this is a x1 I'll call it, y1, and z1. 229 00:16:50,480 --> 00:17:03,550 But x1 and big X, these are in the same direction, 230 00:17:03,550 --> 00:17:06,599 and so is z and z1 and capital [? E. ?] 231 00:17:06,599 --> 00:17:12,020 And that means that my I hats are the same as the unit 232 00:17:12,020 --> 00:17:15,890 vectors associated with my coordinate system attached 233 00:17:15,890 --> 00:17:16,690 to the body. 234 00:17:16,690 --> 00:17:19,425 Remember, this is this body-fixed coordinate system 235 00:17:19,425 --> 00:17:22,260 that allows us to compute moments of inertia, 236 00:17:22,260 --> 00:17:25,470 and I'm going to have this located at G, OK? 237 00:17:49,660 --> 00:17:53,300 So the only rotation that we're allowing in this problem. 238 00:17:53,300 --> 00:17:57,550 This is basically a 2D planar motion problem. 239 00:17:57,550 --> 00:18:00,510 It could conceivably have x and y translations 240 00:18:00,510 --> 00:18:05,600 and a z rotation constrained in the y, 241 00:18:05,600 --> 00:18:09,650 so it has possible rotation and translation. 242 00:18:09,650 --> 00:18:16,680 So the only omega we're considering is this omega z, 243 00:18:16,680 --> 00:18:22,415 and so we'll proceed to use it to compute H. 244 00:18:22,415 --> 00:18:23,880 So we use this formula. 245 00:18:29,510 --> 00:18:33,507 So H with respect to G is-- just I'm 246 00:18:33,507 --> 00:18:35,715 doing a couple of these things sort of carefully just 247 00:18:35,715 --> 00:18:38,970 to remind you of where these things all come from. 248 00:18:38,970 --> 00:18:46,580 This is I with respect to G, and in this case, 0 0 omega z. 249 00:18:46,580 --> 00:18:51,810 And we come out of this then the only-- 250 00:18:51,810 --> 00:18:55,540 are these principal axes? 251 00:18:55,540 --> 00:19:00,410 So if they are, this is diagonal, right, 252 00:19:00,410 --> 00:19:04,460 so I've located these through G. 253 00:19:04,460 --> 00:19:06,490 Taken into consideration the symmetries, 254 00:19:06,490 --> 00:19:09,460 I know because the symmetry of these are principal axes. 255 00:19:09,460 --> 00:19:14,140 Therefore, this is a diagonal mass moment of inertia matrix, 256 00:19:14,140 --> 00:19:17,140 so when I calculate to do out the multiplication, 257 00:19:17,140 --> 00:19:27,420 I get Izz with respect to G omega z k-hat. 258 00:19:27,420 --> 00:19:31,420 There's only one vector component of H 259 00:19:31,420 --> 00:19:35,250 that comes out of that, and that's Hz 260 00:19:35,250 --> 00:19:41,680 is that Izz G omega z k-hat. 261 00:19:41,680 --> 00:19:46,880 And there are two other possible components of H, 262 00:19:46,880 --> 00:19:48,390 but they're both 0. 263 00:19:48,390 --> 00:19:52,720 There's no other rotate-- there is no angular momentum 264 00:19:52,720 --> 00:19:53,950 in the x- or y-directions. 265 00:19:56,390 --> 00:19:56,890 OK. 266 00:20:08,970 --> 00:20:14,050 [? RG/A ?] in this problem is from here-- this 267 00:20:14,050 --> 00:20:29,960 is point A-- two there, so RG respect to A is b/2 i plus 268 00:20:29,960 --> 00:20:33,190 h/2 k. 269 00:20:33,190 --> 00:20:37,490 It's the location of just the position of G with respect 270 00:20:37,490 --> 00:20:44,000 to A. [? b/2 ?] over [? h/2 up. ?] So in order 271 00:20:44,000 --> 00:20:46,755 to complete my calculation of H of A, 272 00:20:46,755 --> 00:20:50,085 I need to get the second piece of the [? RG/A ?] cross P/o. 273 00:21:07,730 --> 00:21:20,170 So [? RG/A ?] is my b/2 i plus h/2 k cross. 274 00:21:20,170 --> 00:21:30,560 And then my linear momentum is M1 x1 dot 275 00:21:30,560 --> 00:21:33,064 in the i hat direction. 276 00:21:33,064 --> 00:21:35,230 And it'll either be a capitalize I or a lowercase i. 277 00:21:35,230 --> 00:21:37,110 They're the same, all right? 278 00:21:37,110 --> 00:21:39,970 So this is my second term the RG cross 279 00:21:39,970 --> 00:21:46,340 P. i cross i, these two-- this pair gives you nothing, 280 00:21:46,340 --> 00:21:52,260 so it's k cross i is j, and so you get Izz with respect 281 00:21:52,260 --> 00:21:55,590 to G omega z. 282 00:21:55,590 --> 00:22:11,790 And then a k term from here M1 h/2 283 00:22:11,790 --> 00:22:15,390 x dot, and it's also in the k. 284 00:22:19,190 --> 00:22:25,248 And that's my total angular momentum now with respect to A. 285 00:22:25,248 --> 00:22:26,660 AUDIENCE: [INAUDIBLE]? 286 00:22:26,660 --> 00:22:34,700 J. KIM VANDIVER: And I think-- just a sec-- k-- you're right. 287 00:22:34,700 --> 00:22:35,820 Wait a second. 288 00:22:35,820 --> 00:22:38,856 k cross i should give me a j, right? 289 00:22:38,856 --> 00:22:39,772 AUDIENCE: [INAUDIBLE]. 290 00:22:46,506 --> 00:22:48,430 Professor? 291 00:22:48,430 --> 00:22:52,890 J. KIM VANDIVER: Yeah, k cross i gives me a positive j, right? 292 00:22:52,890 --> 00:22:53,870 [INTERPOSING VOICES] 293 00:22:53,870 --> 00:22:57,496 [? AUDIENCE: RGA ?] should be a positive j based off 294 00:22:57,496 --> 00:22:58,620 of your [INAUDIBLE] system. 295 00:22:58,620 --> 00:23:01,340 J. KIM VANDIVER: Ah, yeah, yeah. 296 00:23:01,340 --> 00:23:06,510 This is a good catch. 297 00:23:06,510 --> 00:23:08,120 All right, so what happens now? 298 00:23:08,120 --> 00:23:13,155 Now we get j cross i, gives you a minus k. 299 00:23:19,670 --> 00:23:23,290 Now that should be OK, and let's see if that 300 00:23:23,290 --> 00:23:24,601 agrees with what I wrote here. 301 00:23:24,601 --> 00:23:26,100 Yes, they did it right in the paper. 302 00:23:26,100 --> 00:23:27,766 Just couldn't put it right on the board. 303 00:23:50,060 --> 00:23:54,370 So now the sum of the torques with respect to A, 304 00:23:54,370 --> 00:24:01,030 we've already figured out that it's minus M1 g b/2 k 305 00:24:01,030 --> 00:24:09,255 hat must be equal to the time rate of change 306 00:24:09,255 --> 00:24:11,810 from our second equation dH dt. 307 00:24:11,810 --> 00:24:14,910 And we've already figured out that the second term goes away. 308 00:24:14,910 --> 00:24:17,775 So it's just then the time rate of change of that expression. 309 00:24:28,157 --> 00:24:30,830 Oh, I think we could probably do it straight away, 310 00:24:30,830 --> 00:24:33,810 so the time rate of change, the derivative of this, 311 00:24:33,810 --> 00:24:35,570 you get omega z dot. 312 00:24:35,570 --> 00:24:37,080 You get an x double dot. 313 00:24:37,080 --> 00:24:40,590 The k doesn't rotate, so there's nothing that comes from that. 314 00:24:40,590 --> 00:24:41,990 If you take that time derivative, 315 00:24:41,990 --> 00:24:56,260 you get Izz with respect to G omega z dot minus M1 h/2 316 00:24:56,260 --> 00:25:02,090 x double dot, and these are still all in the k direction. 317 00:25:02,090 --> 00:25:05,300 And that's an equation that only has 318 00:25:05,300 --> 00:25:09,020 k hat terms on the left hand side, k on the right, 319 00:25:09,020 --> 00:25:13,500 so this is just one potential component of the torque. 320 00:25:13,500 --> 00:25:16,860 And so we can drop the k's at this point. 321 00:25:16,860 --> 00:25:20,470 And we now have an expression for the external torques 322 00:25:20,470 --> 00:25:24,850 in terms of the accelerations of the system. 323 00:25:24,850 --> 00:25:27,480 This is essentially theta double dot [? if it ?] tips, 324 00:25:27,480 --> 00:25:30,120 and that's the linear acceleration. 325 00:25:30,120 --> 00:25:33,150 But now then the key to the problem 326 00:25:33,150 --> 00:25:36,590 is we're trying to find out when just at the point 327 00:25:36,590 --> 00:25:42,242 it might tip but doesn't, so what's omega dot-- omega z dot? 328 00:25:42,242 --> 00:25:46,640 OK, so what we're looking for we let omega z 329 00:25:46,640 --> 00:25:48,985 dot or require that to be 0. 330 00:25:48,985 --> 00:25:51,150 I mean this makes this even simpler, 331 00:25:51,150 --> 00:25:57,260 and you find out then that M1 g [? b/2 ?] 332 00:25:57,260 --> 00:26:01,707 equals M1 h/2 x double dot. 333 00:26:01,707 --> 00:26:03,540 And I have to add minus signs on both sides, 334 00:26:03,540 --> 00:26:04,820 so I got rid of them. 335 00:26:04,820 --> 00:26:10,415 And the M1s go away, and I can solve for h for x double dot, 336 00:26:10,415 --> 00:26:13,870 and that's going to be g b/h. 337 00:26:19,140 --> 00:26:23,290 OK, and we started off saying what's the maximum force? 338 00:26:23,290 --> 00:26:28,700 So this is now x max, the acceleration maximum 339 00:26:28,700 --> 00:26:38,480 and the force maximum is just equal to M1 plus M2 340 00:26:38,480 --> 00:26:51,350 x dot max for-- and let's see if that makes sense. 341 00:26:51,350 --> 00:26:59,950 If b and h were equal-- that's sort of a common sense 342 00:26:59,950 --> 00:27:05,270 argument here-- b and h were equal, 343 00:27:05,270 --> 00:27:10,290 that tells you that you could accelerate 1g at 1g-- just 1g 344 00:27:10,290 --> 00:27:11,320 and that makes sense. 345 00:27:11,320 --> 00:27:16,182 The Mg here is [? putting a ?] restoring moment down of the MG 346 00:27:16,182 --> 00:27:21,760 [? b/2. ?] And the acceleration of this is putting 347 00:27:21,760 --> 00:27:25,870 an overturning moment on it in the other direction that would 348 00:27:25,870 --> 00:27:30,750 be Mg [? h/2, ?] and if h and b are the same, 349 00:27:30,750 --> 00:27:33,250 then it would be exactly 1g. 350 00:27:33,250 --> 00:27:36,240 Now so that's the answer the problem. 351 00:27:36,240 --> 00:27:41,630 If you want to do live dangerously from the beginning, 352 00:27:41,630 --> 00:27:45,410 you could have done this with a fictitious force. 353 00:27:45,410 --> 00:27:47,730 I just mention in passing I'm not recommending 354 00:27:47,730 --> 00:27:48,910 that you generally go there. 355 00:27:48,910 --> 00:27:51,260 This was a very straightforward way of doing it. 356 00:27:51,260 --> 00:27:54,060 We just started with these three laws, 357 00:27:54,060 --> 00:27:58,890 and we just worked our way through the problem and all 358 00:27:58,890 --> 00:28:00,980 the way to the end, and it fell out. 359 00:28:00,980 --> 00:28:03,260 If you didn't want to live dangerously though, 360 00:28:03,260 --> 00:28:15,930 you could say that when you accelerate this, 361 00:28:15,930 --> 00:28:19,450 you can think of there being a fictitious force that 362 00:28:19,450 --> 00:28:25,340 is equal to minus the mass of this object 363 00:28:25,340 --> 00:28:28,460 times the acceleration on the center of gravity of force 364 00:28:28,460 --> 00:28:32,400 that is pulling it in the other direction. 365 00:28:32,400 --> 00:28:40,584 So you could think of there being a force M1X double dot 366 00:28:40,584 --> 00:28:42,500 opposite to the direction of the acceleration. 367 00:28:42,500 --> 00:28:45,640 This is minus the mass times the acceleration. 368 00:28:45,640 --> 00:28:47,710 It's the fictitious force. 369 00:28:47,710 --> 00:28:54,860 Here's your point A. Here's gravity M1 g. 370 00:28:54,860 --> 00:28:57,040 This is the distance b/2. 371 00:28:59,570 --> 00:29:03,960 This is the height h/2, and you could say just 372 00:29:03,960 --> 00:29:09,140 at that moment of balance, the sum of those two moment 373 00:29:09,140 --> 00:29:11,445 should be 0-- the sub of the external torques. 374 00:29:15,030 --> 00:29:21,950 And you would end up with a M1 x double dot 375 00:29:21,950 --> 00:29:29,100 h/2 minus M1 g [? b/2, ?] and you come up 376 00:29:29,100 --> 00:29:30,700 with the same answer. 377 00:29:30,700 --> 00:29:35,790 But for most of this that takes a real leap of faith 378 00:29:35,790 --> 00:29:39,340 to do that to convince yourself that you're right, right? 379 00:29:39,340 --> 00:29:40,735 You got to know a lot of dynamics 380 00:29:40,735 --> 00:29:43,170 and remember all the parts and pieces 381 00:29:43,170 --> 00:29:47,270 to be able to just go there. 382 00:29:47,270 --> 00:29:50,746 But if you do it just carefully, those three equations, 383 00:29:50,746 --> 00:29:51,620 it'll all worked out. 384 00:29:51,620 --> 00:29:52,280 Yeah? 385 00:29:52,280 --> 00:29:54,840 AUDIENCE: Why is omega z [INAUDIBLE]? 386 00:29:54,840 --> 00:29:56,440 J. KIM VANDIVER: Because as long as 387 00:29:56,440 --> 00:30:00,040 the condition satisfied that it doesn't tip over, 388 00:30:00,040 --> 00:30:03,336 what's the rotation rate of this block? 389 00:30:03,336 --> 00:30:04,260 AUDIENCE: 0. 390 00:30:04,260 --> 00:30:05,830 J. KIM VANDIVER: 0. 391 00:30:05,830 --> 00:30:09,820 So we're not solving for dynamics in this problem 392 00:30:09,820 --> 00:30:11,040 of the thing tipping. 393 00:30:11,040 --> 00:30:12,570 It isn't rolling on us. 394 00:30:12,570 --> 00:30:15,330 We're coming just up to the point 395 00:30:15,330 --> 00:30:17,970 that it tips, and say we're not going any farther. 396 00:30:17,970 --> 00:30:22,820 And so we just require the two moments-- the one essentially 397 00:30:22,820 --> 00:30:26,990 caused by the acceleration, this fictitious force, that's 398 00:30:26,990 --> 00:30:30,400 got to be just in equilibrium with the restoring moment 399 00:30:30,400 --> 00:30:32,230 provided by Mg. 400 00:30:32,230 --> 00:30:34,400 That's essentially the problem we've worked, 401 00:30:34,400 --> 00:30:36,880 but we've done it really carefully just using 402 00:30:36,880 --> 00:30:40,090 these three laws. 403 00:30:40,090 --> 00:30:43,700 And the real problem though if you didn't have the nail here 404 00:30:43,700 --> 00:30:48,210 is if now if you need to test your solution, 405 00:30:48,210 --> 00:30:53,330 could you ever actually reach that rate of acceleration 406 00:30:53,330 --> 00:30:55,870 before the thing starts sliding on you? 407 00:30:55,870 --> 00:30:57,070 And, you know, maybe not. 408 00:30:57,070 --> 00:31:01,305 This thing slides before it tips, OK? 409 00:31:01,305 --> 00:31:06,247 AUDIENCE: Would the fictitious force take into account M2? 410 00:31:06,247 --> 00:31:07,830 J. KIM VANDIVER: Take into account M2? 411 00:31:07,830 --> 00:31:09,934 AUDIENCE: Yes. 412 00:31:09,934 --> 00:31:11,600 J. KIM VANDIVER: So the angular momentum 413 00:31:11,600 --> 00:31:15,970 was all calculated with respect to the rigid body M1, right? 414 00:31:15,970 --> 00:31:19,050 The only reason M2 entered into it 415 00:31:19,050 --> 00:31:20,959 this is the way things happen in real life. 416 00:31:20,959 --> 00:31:22,000 It's just a real problem. 417 00:31:22,000 --> 00:31:24,519 It's a box on a cart, and is it going to tip over, 418 00:31:24,519 --> 00:31:25,560 and you're pushing on it. 419 00:31:25,560 --> 00:31:27,770 How hard can I push? 420 00:31:27,770 --> 00:31:29,790 To do that problem you just have to consider 421 00:31:29,790 --> 00:31:32,270 these to begin with so that you realize that, 422 00:31:32,270 --> 00:31:36,970 oh, the acceleration involves both of these masses, 423 00:31:36,970 --> 00:31:39,230 but the angular momentum part of the problem 424 00:31:39,230 --> 00:31:43,670 involves only the boxes tipping. 425 00:31:43,670 --> 00:31:47,440 So problems even as simple as this one looks has 426 00:31:47,440 --> 00:31:49,370 their little complications, and you've 427 00:31:49,370 --> 00:31:52,020 got to work through them. 428 00:31:52,020 --> 00:31:54,610 So just so we had to treat the body as a two-body system 429 00:31:54,610 --> 00:31:58,460 to start with, and then look at angular momentum for just 430 00:31:58,460 --> 00:31:59,050 the box. 431 00:31:59,050 --> 00:32:02,598 AUDIENCE: Does the pin or screw exert 432 00:32:02,598 --> 00:32:03,990 a force on the [INAUDIBLE]? 433 00:32:07,880 --> 00:32:10,060 J. KIM VANDIVER: Yes, absolutely, mm-hmm. 434 00:32:10,060 --> 00:32:12,430 Does it create a moment? 435 00:32:12,430 --> 00:32:16,380 No, and that's why we don't have to consider it. 436 00:32:16,380 --> 00:32:19,305 I could have put it in there and probably should have. 437 00:32:19,305 --> 00:32:27,260 If I had done a free body diagram here, 438 00:32:27,260 --> 00:32:31,490 you could still say that there's a reaction force upwards here. 439 00:32:31,490 --> 00:32:35,610 And I'll call it Ry and another one this direction 440 00:32:35,610 --> 00:32:37,570 I'll call Rx. 441 00:32:37,570 --> 00:32:41,600 And when computing moments about this point, neither of them 442 00:32:41,600 --> 00:32:42,580 enter into the problem. 443 00:32:42,580 --> 00:32:46,880 And that's why we like to do our calculations for angular 444 00:32:46,880 --> 00:32:53,010 momentum with respect to points around which the body rotates 445 00:32:53,010 --> 00:32:54,590 because that allows us to not have 446 00:32:54,590 --> 00:32:59,040 to solve for those unknown forces that pop up there, 447 00:32:59,040 --> 00:33:01,530 so yeah, indeed there are forces there. 448 00:33:01,530 --> 00:33:06,520 OK, got it figured out. 449 00:33:06,520 --> 00:33:09,660 So we've talked a little tiny bit about parallel axis 450 00:33:09,660 --> 00:33:13,900 theorem, and so I'm going to consider this a slender rod. 451 00:33:17,640 --> 00:33:22,700 I can spin it about one of its printable axes and calculate 452 00:33:22,700 --> 00:33:24,790 its angular momentum of whatever I need, 453 00:33:24,790 --> 00:33:29,190 and if I move it over to an axis that's-- I just move it over 454 00:33:29,190 --> 00:33:30,230 by a little bit. 455 00:33:30,230 --> 00:33:37,350 Let's say this is the x-axis here in my body, 456 00:33:37,350 --> 00:33:40,990 and this is the z, so it's at omega z. 457 00:33:40,990 --> 00:33:45,835 And I move my point and around which I am rotating over 458 00:33:45,835 --> 00:33:48,460 by a small amount A, and that's this distance between these two 459 00:33:48,460 --> 00:33:49,850 holes. 460 00:33:49,850 --> 00:33:52,200 And by parallel axis theorem, we could say, well, 461 00:33:52,200 --> 00:33:55,520 now there's a mass moment of inertia with respect to A, 462 00:33:55,520 --> 00:33:58,700 which is the mass moment of inertia with respect to g, 463 00:33:58,700 --> 00:34:03,530 which is [? Ml ?] squared over 12, if this is the length, 464 00:34:03,530 --> 00:34:06,040 plus the mass times this distance squared. 465 00:34:06,040 --> 00:34:08,760 We know how to do that. 466 00:34:08,760 --> 00:34:17,080 But now what happens if I take this stick, my slender rod, 467 00:34:17,080 --> 00:34:24,920 and here's my original x, y, z. 468 00:34:24,920 --> 00:34:29,270 And I want to move it, my point A, my stick now 469 00:34:29,270 --> 00:34:38,520 I'm going to move it over by an amount a and up by an amount c 470 00:34:38,520 --> 00:34:43,480 so that it's now up here with respect to this point. 471 00:34:43,480 --> 00:34:45,889 So this stick has been like this, 472 00:34:45,889 --> 00:34:47,960 and now I'm going to then rotate it about that. 473 00:34:47,960 --> 00:34:49,350 That's the simple case. 474 00:34:49,350 --> 00:34:51,250 Now I want to rotate it about this axis 475 00:34:51,250 --> 00:34:54,409 when it's moved over to like that. 476 00:34:57,780 --> 00:35:04,280 So not only have I moved it-- so here's the original problem-- 477 00:35:04,280 --> 00:35:15,675 like this, but now I move it up and over-- maybe 478 00:35:15,675 --> 00:35:17,990 I'll do it there-- and now I spin it. 479 00:35:17,990 --> 00:35:21,270 And I'm interested in angular momentum about this point, 480 00:35:21,270 --> 00:35:24,990 so now it's dynamically unbalanced for sure. 481 00:35:24,990 --> 00:35:27,900 This thing is trying to wobble, and it's 482 00:35:27,900 --> 00:35:31,550 been pushed up and over, so is there 483 00:35:31,550 --> 00:35:34,160 a way to get-- how do we solve that problem? 484 00:35:34,160 --> 00:35:38,140 How do we compute the torques for this problem? 485 00:35:38,140 --> 00:35:41,010 Well, we could try to go at it with parallel access 486 00:35:41,010 --> 00:35:44,110 to start with, but you don't know how to do that in general 487 00:35:44,110 --> 00:35:45,130 for this problem. 488 00:35:45,130 --> 00:35:48,140 So what do we go to? 489 00:35:48,140 --> 00:35:49,970 We go back to those three equations, 490 00:35:49,970 --> 00:35:52,140 and just start there, and it'll all fall out. 491 00:35:52,140 --> 00:35:56,800 I'm going to do this kind of briefly for this problem, 492 00:35:56,800 --> 00:36:04,420 so I want to compute H with respect to A H with respect 493 00:36:04,420 --> 00:36:12,520 to G plus RG with respect to A cross P. 494 00:36:12,520 --> 00:36:19,150 And now this is [? RG/A ?] is this vector here, 495 00:36:19,150 --> 00:36:21,232 that position vector, OK? 496 00:36:24,430 --> 00:36:28,600 And the only rotation is omega z. 497 00:36:28,600 --> 00:36:30,710 This is our z-axis. 498 00:36:30,710 --> 00:36:39,030 So I can write this as Izz with respect to G, 0, 0, 499 00:36:39,030 --> 00:36:48,800 omega z plus my [? RG/A ?] cross P/o, and that in this problem-- 500 00:36:48,800 --> 00:36:50,990 let's get the vector right this time-- it's 501 00:36:50,990 --> 00:36:59,780 a in the i direction plus c in the k direction cross. 502 00:36:59,780 --> 00:37:06,360 Now we need to know what's P with respect to o, 503 00:37:06,360 --> 00:37:12,910 but that's just the velocity of this so the center of mass, 504 00:37:12,910 --> 00:37:16,270 which is now here. 505 00:37:16,270 --> 00:37:20,520 The velocity is omega cross r, so when 506 00:37:20,520 --> 00:37:22,810 you do that-- you've done that problem lots of times-- 507 00:37:22,810 --> 00:37:28,140 it's this distance, so perpendicular distance 508 00:37:28,140 --> 00:37:30,110 [? across ?] k. 509 00:37:30,110 --> 00:37:31,740 It's got to be going into the board 510 00:37:31,740 --> 00:37:32,864 if it's spinning like this. 511 00:37:32,864 --> 00:37:35,090 We know the answer's got to come out plus j, 512 00:37:35,090 --> 00:37:47,810 and it's r-- it's A omega z in the j-direction cross Ma omega 513 00:37:47,810 --> 00:37:50,610 z, but it is in the j hat direction. 514 00:37:50,610 --> 00:37:54,240 That's your linear momentum, and this is [? RG/A ?] 515 00:37:54,240 --> 00:37:55,535 cross with a linear momentum. 516 00:37:59,440 --> 00:38:04,150 This here is M velocity of G here with respect to o. 517 00:38:04,150 --> 00:38:11,890 That's this term, so we carry out this calculation 518 00:38:11,890 --> 00:38:19,000 and we get-- and I need a j, OK. 519 00:38:19,000 --> 00:38:25,280 So i cross j gives me a k, so I'll get 2, 520 00:38:25,280 --> 00:38:30,030 and this is a diagonal matrix because we 521 00:38:30,030 --> 00:38:34,710 began with a set of principal axes for our stick, 522 00:38:34,710 --> 00:38:40,180 so this term is Izz with respect to G. 523 00:38:40,180 --> 00:38:43,390 I shouldn't have written zz here, just i with respect to z. 524 00:38:43,390 --> 00:38:44,560 We multiply this out. 525 00:38:44,560 --> 00:38:54,190 We pick up just this term omega z k hat plus i cross j is k, 526 00:38:54,190 --> 00:39:03,370 so that's Ma squared omega [? zk, ?] and k cross j 527 00:39:03,370 --> 00:39:12,645 is minus i minus M ac omega z in the i hat direction. 528 00:39:15,330 --> 00:39:20,140 So this is my angular momentum, and reminding, it is a vector. 529 00:39:20,140 --> 00:39:22,715 This has got components in the k and the i directions. 530 00:39:52,570 --> 00:39:54,090 I've rewritten it here. 531 00:39:54,090 --> 00:39:58,970 This also could be written-- this H with respect to A 532 00:39:58,970 --> 00:40:11,192 is I with respect to A times the rotation vector. 533 00:40:11,192 --> 00:40:12,900 If we had just set out with this problem, 534 00:40:12,900 --> 00:40:15,420 and said we're going-- if we know the mass moment of inertia 535 00:40:15,420 --> 00:40:18,790 matrix computed with respect to this point 536 00:40:18,790 --> 00:40:23,130 and multiplied it by our rotation rate vector, 537 00:40:23,130 --> 00:40:26,570 we should have gotten the same answer. 538 00:40:26,570 --> 00:40:27,780 So we didn't do that. 539 00:40:27,780 --> 00:40:30,820 We just worked it out using just this formula, 540 00:40:30,820 --> 00:40:33,430 but we've essentially discovered the answer. 541 00:40:33,430 --> 00:40:35,250 This is the rotation around k. 542 00:40:37,760 --> 00:40:42,470 This is the Izz term with respect to A, 543 00:40:42,470 --> 00:40:45,850 and it really looks like the familiar parallel axis piece. 544 00:40:45,850 --> 00:40:49,640 It's the amount that you moved this axis over, 545 00:40:49,640 --> 00:40:53,509 and the axis you're spinning around, you moved it over A. 546 00:40:53,509 --> 00:40:55,300 And you know the parallel axis theorem just 547 00:40:55,300 --> 00:40:57,980 says add Ma squared, and you've got the answer. 548 00:40:57,980 --> 00:41:01,350 But we also moved it up, and it gave us another term. 549 00:41:01,350 --> 00:41:04,250 What do you suppose in that term is? 550 00:41:04,250 --> 00:41:11,207 So this is Izz with respect to A, this term. 551 00:41:11,207 --> 00:41:13,290 By parallel axis theorem, you're familiar with it. 552 00:41:13,290 --> 00:41:22,410 This one is I, and this is xz with respect 553 00:41:22,410 --> 00:41:26,040 in the A coordinate system. 554 00:41:26,040 --> 00:41:32,240 OK, we've created-- we've made this thing unbalanced. 555 00:41:32,240 --> 00:41:34,010 The angular momentum vector no longer 556 00:41:34,010 --> 00:41:36,640 points in the same direction as the rotation. 557 00:41:36,640 --> 00:41:38,755 It has a k component and an I component, 558 00:41:38,755 --> 00:41:40,880 and sure enough, when I spend that thing like that, 559 00:41:40,880 --> 00:41:47,610 I feel that additional torque out around my hand 560 00:41:47,610 --> 00:41:49,230 down here where I'm holding it. 561 00:41:49,230 --> 00:41:52,620 This thing is going-- trying to pull it back and forth. 562 00:41:52,620 --> 00:41:55,090 So we have essentially just shown-- 563 00:41:55,090 --> 00:41:58,790 we have just arrived at a more complicated parallel axis 564 00:41:58,790 --> 00:42:02,860 theorem by just using this basic formula. 565 00:42:02,860 --> 00:42:07,830 So Williams in the dynamics-- that second handout 566 00:42:07,830 --> 00:42:12,660 from Williams has actually given us a general formulation 567 00:42:12,660 --> 00:42:14,810 for parallel axis theorem. 568 00:42:18,720 --> 00:42:20,070 So this is just handy to know. 569 00:42:20,070 --> 00:42:28,200 I'll just give it to you, and if you have a problem some day, 570 00:42:28,200 --> 00:42:30,770 where you just like to go directly to this statement, 571 00:42:30,770 --> 00:42:32,650 so now I'm just going to say, in general, we 572 00:42:32,650 --> 00:42:37,410 could have moved over by A, up by c and actually 573 00:42:37,410 --> 00:42:42,980 into the board by b, so x, I'm moving A. y, I'm moving b. 574 00:42:42,980 --> 00:42:44,240 z, I'm moving c. 575 00:42:44,240 --> 00:42:48,310 OK, so you could have all those possible moves of the new point 576 00:42:48,310 --> 00:42:49,540 about which you're rotating. 577 00:42:49,540 --> 00:42:50,948 Yes? 578 00:42:50,948 --> 00:42:55,214 AUDIENCE: For the [INAUDIBLE] matrix i with respect to A, 579 00:42:55,214 --> 00:42:57,435 does that mean that you're rotating about point A? 580 00:42:57,435 --> 00:42:58,060 Or what is it-- 581 00:42:58,060 --> 00:43:00,796 J. KIM VANDIVER: You are rotating about point A. 582 00:43:00,796 --> 00:43:02,245 AUDIENCE: OK. 583 00:43:02,245 --> 00:43:04,660 But isn't point A just the [? immersion? ?] 584 00:43:11,520 --> 00:43:12,805 J. KIM VANDIVER: Let me think. 585 00:43:12,805 --> 00:43:14,180 I'm trying to think of a good way 586 00:43:14,180 --> 00:43:18,570 to-- when you started this problem, 587 00:43:18,570 --> 00:43:22,680 we're spending about the middle-- about G, OK? 588 00:43:22,680 --> 00:43:24,840 And we can calculate angular momentum 589 00:43:24,840 --> 00:43:29,210 of this object with respect to G and learn some things, right? 590 00:43:29,210 --> 00:43:30,950 But now we've built a different device. 591 00:43:30,950 --> 00:43:36,520 This is an object now that is, in fact, spinning. 592 00:43:42,380 --> 00:43:44,140 Let's imagine it's a fan blade, and you've 593 00:43:44,140 --> 00:43:46,440 broken one blade off. 594 00:43:46,440 --> 00:43:49,230 The motor is here. 595 00:43:49,230 --> 00:43:52,590 A proper fan blade would have blades on both sides, 596 00:43:52,590 --> 00:43:53,760 so it's nice and balanced. 597 00:43:53,760 --> 00:43:56,460 The shaft has some length and it sticks out, 598 00:43:56,460 --> 00:43:58,430 and now you break off one of the blades. 599 00:43:58,430 --> 00:44:01,800 And so you have a system that's doing this, 600 00:44:01,800 --> 00:44:04,730 but this is going to put a lot of unusual loads on this point 601 00:44:04,730 --> 00:44:05,710 down here. 602 00:44:05,710 --> 00:44:11,790 So it is rotating at the same rotation rate about this point, 603 00:44:11,790 --> 00:44:15,820 but it is now a system whose center of mass is over here, 604 00:44:15,820 --> 00:44:16,690 and it's up. 605 00:44:16,690 --> 00:44:20,140 It's above this point, and it's out 606 00:44:20,140 --> 00:44:21,940 from this point, the center of mass. 607 00:44:21,940 --> 00:44:24,720 And now you compute the angular momentum with this point, 608 00:44:24,720 --> 00:44:26,460 and take its time derivative, you 609 00:44:26,460 --> 00:44:31,390 will find all the torques required to make this happen. 610 00:44:31,390 --> 00:44:32,306 AUDIENCE: [INAUDIBLE]? 611 00:44:34,710 --> 00:44:36,085 J. KIM VANDIVER: A little louder. 612 00:44:36,085 --> 00:44:39,550 AUDIENCE: The movement A, [? i-hat ?] Mc [? k-hat ?] 613 00:44:39,550 --> 00:44:40,050 [INAUDIBLE] 614 00:44:40,050 --> 00:44:42,770 J. KIM VANDIVER: Movement that you've moved the center of mass 615 00:44:42,770 --> 00:44:46,560 away from this axis of rotation. 616 00:44:46,560 --> 00:44:51,660 And you've moved it up and over. 617 00:44:51,660 --> 00:44:54,342 So actually, let's think about that for a minute. 618 00:44:54,342 --> 00:44:55,800 This is our beginning point, right? 619 00:44:55,800 --> 00:45:03,090 We're at G. If I just move it up, does it cause any problems? 620 00:45:03,090 --> 00:45:05,100 It's still balanced, and you'll find out 621 00:45:05,100 --> 00:45:07,760 that nothing's happened in this problem. 622 00:45:07,760 --> 00:45:10,700 H with respect to A is the same as it was before. 623 00:45:14,190 --> 00:45:15,875 But now you move it up and over, it 624 00:45:15,875 --> 00:45:17,230 starts causing complications. 625 00:45:17,230 --> 00:45:17,750 Yeah? 626 00:45:17,750 --> 00:45:24,185 AUDIENCE: [INAUDIBLE] H of A equals [? A ?] times omega z. 627 00:45:24,185 --> 00:45:28,145 So i doesn't have like direction, 628 00:45:28,145 --> 00:45:31,610 but you have an [? ixz ?] terms and [? xc ?] term, 629 00:45:31,610 --> 00:45:34,580 and you're multiplying it by a vector with an omega z term, 630 00:45:34,580 --> 00:45:37,550 so how are you getting an i hat from that? 631 00:45:37,550 --> 00:45:41,890 J. KIM VANDIVER: OK, so I haven't gone into this, 632 00:45:41,890 --> 00:45:44,090 and I'm not going to. 633 00:45:44,090 --> 00:45:48,530 The [? i ?] matrix is a thing called a tensor. 634 00:45:48,530 --> 00:45:53,590 And so this is a convention that you 635 00:45:53,590 --> 00:45:57,710 can make it a very mathematical and assigned unit vectors 636 00:45:57,710 --> 00:46:01,150 to each of these terms inside, but I haven't done that. 637 00:46:01,150 --> 00:46:07,810 But the convention here is that this, the H, angular momentum 638 00:46:07,810 --> 00:46:09,650 is a vector. 639 00:46:09,650 --> 00:46:12,526 It consists of three components. 640 00:46:17,930 --> 00:46:23,590 It has three potential vector components, Hx in the I, 641 00:46:23,590 --> 00:46:30,950 Hy in the j, and Hz in the k directions. 642 00:46:30,950 --> 00:46:34,630 And we, in a compact form, say we can express that 643 00:46:34,630 --> 00:46:37,490 by multiplying its mass moment of inertia matrix 644 00:46:37,490 --> 00:46:43,570 with respect to A times the vector that is its rotation. 645 00:46:43,570 --> 00:46:45,850 And I only have one component of this omega, 646 00:46:45,850 --> 00:46:49,140 but it is in the k direction. 647 00:46:49,140 --> 00:46:55,050 And when you take this vector multiplied by the top row, that 648 00:46:55,050 --> 00:46:58,900 gives you everything that's [? I. ?] You multiply 649 00:46:58,900 --> 00:47:00,710 this vector times the second row, 650 00:47:00,710 --> 00:47:02,950 it gives you everything is in the j direction, 651 00:47:02,950 --> 00:47:06,160 and the result is part of Hy in the j direction. 652 00:47:06,160 --> 00:47:09,400 You take this vector, and you multiply it by the third row, 653 00:47:09,400 --> 00:47:13,500 it gives you the z-directed components, OK? 654 00:47:13,500 --> 00:47:22,110 So this one, because when we multiply this out 655 00:47:22,110 --> 00:47:24,790 what's in this matrix-- is this matrix diagonal? 656 00:47:27,680 --> 00:47:28,870 No way. 657 00:47:28,870 --> 00:47:32,201 And we move it, and we've gone and done something strange 658 00:47:32,201 --> 00:47:32,700 here. 659 00:47:32,700 --> 00:47:35,110 It's not diagonal, and when you multiply 660 00:47:35,110 --> 00:47:38,730 this times this top row, there's a term here, 661 00:47:38,730 --> 00:47:41,010 which is not 0-- shouldn't write 0. 662 00:47:41,010 --> 00:47:42,060 Make an x here. 663 00:47:42,060 --> 00:47:46,715 This times this gives you an omega z times that in the i hat 664 00:47:46,715 --> 00:47:47,215 direction. 665 00:47:50,430 --> 00:47:55,020 OK this term comes from there being something there. 666 00:47:55,020 --> 00:47:57,970 OK, and I'm just going to give you a generalization that you 667 00:47:57,970 --> 00:48:01,120 can go back and read the Williams handout, which 668 00:48:01,120 --> 00:48:10,620 he says, if you want to rotate about some fixed point A. 669 00:48:10,620 --> 00:48:15,120 Then you can say that this matrix 670 00:48:15,120 --> 00:48:19,170 is equal to the one with respect to G, which in our case 671 00:48:19,170 --> 00:48:21,910 is diagonal. 672 00:48:21,910 --> 00:48:24,060 The original one that we had doesn't have to be, 673 00:48:24,060 --> 00:48:27,590 but if you pick principal axes, this will be diagonal. 674 00:48:27,590 --> 00:48:40,580 Plus M b squared plus c squared minus ab minus ac. 675 00:48:40,580 --> 00:48:41,900 It's symmetric. 676 00:48:41,900 --> 00:48:45,555 This is a squared plus c squared. 677 00:49:02,070 --> 00:49:04,220 So if we're rotating a system if you 678 00:49:04,220 --> 00:49:06,280 know its mass moment of inertia matrix 679 00:49:06,280 --> 00:49:10,270 with respect to G [INAUDIBLE] perhaps principal axes, 680 00:49:10,270 --> 00:49:13,650 and you want to rotate it about any other point 681 00:49:13,650 --> 00:49:17,510 where you've moved it by an a, by a b, by a c, 682 00:49:17,510 --> 00:49:20,560 this is essentially the general parallel axis theorem. 683 00:49:20,560 --> 00:49:23,450 And you can just plug it in and use that. 684 00:49:23,450 --> 00:49:26,600 And you can see what would happen in this case 685 00:49:26,600 --> 00:49:29,120 if you multiplied-- if you add these two together, 686 00:49:29,120 --> 00:49:30,330 they add term by term. 687 00:49:30,330 --> 00:49:33,050 This goes into this point on. 688 00:49:33,050 --> 00:49:39,820 This one adds to this, so in this problem, 689 00:49:39,820 --> 00:49:46,150 in our particular problem, this becomes 690 00:49:46,150 --> 00:49:55,490 you have an Ixx, Iyy, Izz with respect to G terms. 691 00:49:55,490 --> 00:50:01,945 It's diagonal to start with, and then in our problem 692 00:50:01,945 --> 00:50:05,220 that we did there, we didn't have a b. 693 00:50:05,220 --> 00:50:08,270 No b terms, so the b terms are all 0, 694 00:50:08,270 --> 00:50:12,570 but you do end up with a minus ac term over here. 695 00:50:18,160 --> 00:50:24,280 And if you multiply this by 0, 0, omega z, see what happens. 696 00:50:24,280 --> 00:50:29,590 You get minus ac in the I direction, 697 00:50:29,590 --> 00:50:32,695 and you need an M here. 698 00:50:38,110 --> 00:50:41,270 And you come down here, you get the Izz G omega 699 00:50:41,270 --> 00:50:43,870 z in the k direction, and that's those two terms. 700 00:50:43,870 --> 00:50:44,510 Here they are. 701 00:50:49,120 --> 00:50:51,280 Yeah, oops, I need the plus [? A ?] squared. 702 00:50:55,950 --> 00:50:56,930 Yeah? 703 00:50:56,930 --> 00:51:00,360 AUDIENCE: When you move it over to the end of the stick, 704 00:51:00,360 --> 00:51:05,260 normally like you're just shifting it parallely, 705 00:51:05,260 --> 00:51:06,730 but now there's gravity. 706 00:51:06,730 --> 00:51:08,531 Is that why you have those little things? 707 00:51:08,531 --> 00:51:09,655 J. KIM VANDIVER: Say again? 708 00:51:09,655 --> 00:51:11,613 AUDIENCE: So normally, if you just take a stick 709 00:51:11,613 --> 00:51:15,328 and you have a principal axis and you 710 00:51:15,328 --> 00:51:17,496 can't move that parallel to the end, 711 00:51:17,496 --> 00:51:18,966 you wouldn't get all of this but-- 712 00:51:18,966 --> 00:51:21,340 J. KIM VANDIVER: Yeah, and if you just move it-- so yeah, 713 00:51:21,340 --> 00:51:22,714 I'm glad you asked that question. 714 00:51:22,714 --> 00:51:28,870 So normally the types of problems in which we usually 715 00:51:28,870 --> 00:51:31,470 use the parallel axis theorem is you've 716 00:51:31,470 --> 00:51:34,150 been spinning it around one axis. 717 00:51:34,150 --> 00:51:36,600 It's a principal axis, everything's fine, 718 00:51:36,600 --> 00:51:40,060 and you want to just move it over, 719 00:51:40,060 --> 00:51:43,570 so we do that when we do pendulum problems. 720 00:51:43,570 --> 00:51:45,220 Here's the thing around the center, 721 00:51:45,220 --> 00:51:48,170 and we just want to know, how does this thing behave? 722 00:51:48,170 --> 00:51:50,870 When we move it up here, it becomes a pendulum, 723 00:51:50,870 --> 00:51:53,270 and we would like to compute the angular 724 00:51:53,270 --> 00:51:54,780 momentum around this point. 725 00:51:54,780 --> 00:51:57,260 We say, oh, we've moved it a distance d. 726 00:51:57,260 --> 00:52:02,870 The new mass moment of inertia in the z direction 727 00:52:02,870 --> 00:52:07,820 is the original plus Md squared, right? 728 00:52:07,820 --> 00:52:11,520 So when you only make one move, you only 729 00:52:11,520 --> 00:52:14,190 have an a, a b, or a c. 730 00:52:14,190 --> 00:52:16,290 You never get any of the off-diagonal terms 731 00:52:16,290 --> 00:52:18,690 because their products, see? 732 00:52:18,690 --> 00:52:21,950 But if you start doing more than one, 733 00:52:21,950 --> 00:52:24,130 you've introduced complications, and that's 734 00:52:24,130 --> 00:52:26,550 part of the reason I put this up is 735 00:52:26,550 --> 00:52:32,570 there's limits to the simple parallel axis theorem, 736 00:52:32,570 --> 00:52:34,800 but there is a general way of doing it. 737 00:52:34,800 --> 00:52:35,300 Yes? 738 00:52:35,300 --> 00:52:39,478 AUDIENCE: I believe her actual question involved 739 00:52:39,478 --> 00:52:40,891 what causes the torques? 740 00:52:40,891 --> 00:52:41,763 Is it gravity? 741 00:52:41,763 --> 00:52:42,304 For example-- 742 00:52:42,304 --> 00:52:43,720 AUDIENCE: Yeah, I was asking why-- 743 00:52:43,720 --> 00:52:44,910 AUDIENCE: [INAUDIBLE]. 744 00:52:44,910 --> 00:52:47,900 J. KIM VANDIVER: Ah, so what causes these torques when 745 00:52:47,900 --> 00:52:49,067 you have two? 746 00:52:49,067 --> 00:52:49,650 AUDIENCE: Yes. 747 00:52:49,650 --> 00:52:52,920 J. KIM VANDIVER: OK, we've talked about this a little bit 748 00:52:52,920 --> 00:52:53,420 before. 749 00:52:56,340 --> 00:52:59,370 When we calculate the angular momentum, 750 00:52:59,370 --> 00:53:01,540 it never involves gravity. 751 00:53:04,470 --> 00:53:08,410 Never gets in there, not in the angular momentum expression. 752 00:53:08,410 --> 00:53:11,810 So when you go to compute d, the time derivative of the angular 753 00:53:11,810 --> 00:53:13,643 momentum, you will not find torques 754 00:53:13,643 --> 00:53:15,532 that are caused by gravity. 755 00:53:15,532 --> 00:53:16,490 You just can't do them. 756 00:53:16,490 --> 00:53:19,830 It's not part-- but are the torques in the System I mean, 757 00:53:19,830 --> 00:53:20,830 why just hold this here. 758 00:53:20,830 --> 00:53:22,246 There's [? Mg ?] down, and there's 759 00:53:22,246 --> 00:53:25,870 a static moment caused by this thing trying 760 00:53:25,870 --> 00:53:27,420 to twist this down. 761 00:53:27,420 --> 00:53:29,300 When you sum the torques-- when you 762 00:53:29,300 --> 00:53:32,060 do the sum of the torques in the equation, 763 00:53:32,060 --> 00:53:36,390 you do your free body diagram, that term will appear. 764 00:53:36,390 --> 00:53:40,450 But it is balanced by some physical torque over here 765 00:53:40,450 --> 00:53:41,920 that balances it. 766 00:53:41,920 --> 00:53:46,480 You just won't find it from doing [? dh ?] dt. 767 00:53:46,480 --> 00:53:49,501 It's just part of the static equilibrium of the system. 768 00:53:49,501 --> 00:53:52,868 AUDIENCE: But the reason when you move the pen over, 769 00:53:52,868 --> 00:53:56,235 the reason you even have the shift is because of 770 00:53:56,235 --> 00:53:57,170 [? gravity? ?] 771 00:53:57,170 --> 00:53:58,170 J. KIM VANDIVER: No, no. 772 00:53:58,170 --> 00:54:04,670 So she's asking if is gravity the reason you-- 773 00:54:04,670 --> 00:54:07,670 I'm not even quite sure. 774 00:54:07,670 --> 00:54:12,725 AUDIENCE: How did you shift the z component of your [? plan? ?] 775 00:54:12,725 --> 00:54:16,200 J. KIM VANDIVER: OK, so let's just make 776 00:54:16,200 --> 00:54:18,100 this a real physical problem. 777 00:54:18,100 --> 00:54:21,470 I'm making an airplane engine, and I'm making an airplane 778 00:54:21,470 --> 00:54:23,750 with a propeller on it. 779 00:54:23,750 --> 00:54:28,320 And the bearing that supports the propeller shaft, 780 00:54:28,320 --> 00:54:31,200 if I put it really close to the propeller, 781 00:54:31,200 --> 00:54:35,290 OK, then it's doing this, and everything's fine. 782 00:54:35,290 --> 00:54:39,080 What if I extended the propeller shaft? 783 00:54:39,080 --> 00:54:42,800 OK, so now the bearing is back here, 784 00:54:42,800 --> 00:54:45,130 and now the propeller spins. 785 00:54:45,130 --> 00:54:50,260 Does that cause any torques, any loads back here on the bearing 786 00:54:50,260 --> 00:54:52,350 because I've extended it? 787 00:54:52,350 --> 00:54:55,170 No, and you could prove that just by going through this. 788 00:54:55,170 --> 00:54:57,740 You do [? dh ?] dt, and you find out the only torques 789 00:54:57,740 --> 00:55:00,330 when it's nice and balanced are those required 790 00:55:00,330 --> 00:55:04,700 to drive this propeller in the direction 791 00:55:04,700 --> 00:55:07,460 of the axis of rotation. 792 00:55:07,460 --> 00:55:14,020 But as soon as you do something to that propeller, like you 793 00:55:14,020 --> 00:55:18,510 mount it off-center, which rarely happens, 794 00:55:18,510 --> 00:55:21,350 but if you dinged-- if you broke off 795 00:55:21,350 --> 00:55:23,970 the tip of one of the blades of this propeller, 796 00:55:23,970 --> 00:55:27,650 you'd now have a propeller that looks like that, right? 797 00:55:27,650 --> 00:55:29,800 And now, even in the original system 798 00:55:29,800 --> 00:55:34,530 if your bearing is right close, this is spinning around, 799 00:55:34,530 --> 00:55:35,930 but is it putting a load? 800 00:55:35,930 --> 00:55:38,240 Sure there's a centrifugal force that 801 00:55:38,240 --> 00:55:39,760 is going around and around. 802 00:55:39,760 --> 00:55:41,960 It has nothing to do with gravity, 803 00:55:41,960 --> 00:55:45,264 nothing at all to do with gravity. 804 00:55:45,264 --> 00:55:48,960 OK, just because you now have the mass centers out here, 805 00:55:48,960 --> 00:55:51,310 it has momentum, the time rate of change 806 00:55:51,310 --> 00:55:56,502 of that linear momentum is a force making 807 00:55:56,502 --> 00:55:57,710 this thing going in a circle. 808 00:55:57,710 --> 00:56:01,770 OK, now if I extend the propeller shaft 809 00:56:01,770 --> 00:56:05,920 so it's like that, and I foolishly designed my airplane 810 00:56:05,920 --> 00:56:08,700 so it had a long propeller shaft sticking out there, 811 00:56:08,700 --> 00:56:11,860 if there's a little bit of unbalanced in this blade 812 00:56:11,860 --> 00:56:14,970 so that you're not spinning about the mass center. 813 00:56:14,970 --> 00:56:17,650 You're spinning about some other point. 814 00:56:17,650 --> 00:56:20,620 Now that centrifugal force going around is pulling 815 00:56:20,620 --> 00:56:24,360 is trying to bend this back and forth around this bearing. 816 00:56:24,360 --> 00:56:28,930 And because of the way in which we formulate angular momentum, 817 00:56:28,930 --> 00:56:31,340 if you formulate it about that point 818 00:56:31,340 --> 00:56:36,330 and take its time derivative, it will reveal those moments. 819 00:56:36,330 --> 00:56:38,880 It's really amazing that it can do that for you, 820 00:56:38,880 --> 00:56:42,433 but it has [? zero, ?] nothing to do with gravity, OK? 821 00:56:45,340 --> 00:56:51,400 So handy, hard to remember this from just a blackboard 822 00:56:51,400 --> 00:56:55,090 presentation, but it's in that second reading by Williams. 823 00:56:55,090 --> 00:56:57,645 He does it in a very-- he proves it 824 00:56:57,645 --> 00:57:01,849 in a very simple way using the summations of the MI [? RI's ?] 825 00:57:01,849 --> 00:57:02,390 and so forth. 826 00:57:02,390 --> 00:57:04,970 Proves it in a very simple way, but a very general 827 00:57:04,970 --> 00:57:08,880 handy formula that you can use. 828 00:57:08,880 --> 00:57:14,600 OK, so let's have another topic, which I'm not going to do. 829 00:57:14,600 --> 00:57:15,790 We've got a few minutes. 830 00:57:15,790 --> 00:57:17,059 You have [? money ?] cards. 831 00:57:17,059 --> 00:57:18,350 We've been thinking about that. 832 00:57:18,350 --> 00:57:19,741 And let's ask questions, yeah? 833 00:57:19,741 --> 00:57:20,616 AUDIENCE: [INAUDIBLE] 834 00:57:23,637 --> 00:57:25,053 J. KIM VANDIVER: Where did I the-- 835 00:57:25,053 --> 00:57:27,886 AUDIENCE: The plus [? MA ?] squared and the plus A squared. 836 00:57:27,886 --> 00:57:30,010 J. KIM VANDIVER: Ah, I was doing that kind of fast, 837 00:57:30,010 --> 00:57:31,410 and I probably even messed it up, 838 00:57:31,410 --> 00:57:44,550 so let me check the-- so I was doing for the example 839 00:57:44,550 --> 00:57:55,100 that [? RG/A ?] equals-- we moved it over by ai and up 840 00:57:55,100 --> 00:58:00,840 by ck, so and plus 0j. 841 00:58:00,840 --> 00:58:03,830 We didn't move it in the j direction, OK? 842 00:58:03,830 --> 00:58:06,450 So this is my amount that I've moved it. 843 00:58:06,450 --> 00:58:09,520 I've moved it an amount a, and an amount c. 844 00:58:09,520 --> 00:58:11,770 So the Williams formula would say, ah, 845 00:58:11,770 --> 00:58:15,760 that new mass moment of inertia matrix with respect to a 846 00:58:15,760 --> 00:58:20,950 becomes the original plus-- and now every place there's 847 00:58:20,950 --> 00:58:23,010 a b here, it goes to 0. 848 00:58:31,810 --> 00:58:41,660 And the remaining bits I add with those, 849 00:58:41,660 --> 00:58:51,940 so I should get a M-- whoops, this is a c squared, 850 00:58:51,940 --> 00:58:53,420 no second term. 851 00:58:53,420 --> 00:58:57,190 The third term is minus Mac. 852 00:58:57,190 --> 00:59:00,640 Over here, that one is 0. 853 00:59:00,640 --> 00:59:03,520 This one becomes a squared plus c squared-- 854 00:59:03,520 --> 00:59:05,670 AUDIENCE: --minus [INAUDIBLE]? 855 00:59:05,670 --> 00:59:08,120 J. KIM VANDIVER: Minus, yep, M. 856 00:59:08,120 --> 00:59:13,960 And this one is minus Mac 0 Izz and a squared plus b squared, 857 00:59:13,960 --> 00:59:15,090 so it's just a squared. 858 00:59:15,090 --> 00:59:17,658 So this is now correct. 859 00:59:17,658 --> 00:59:18,610 AUDIENCE: [INAUDIBLE]? 860 00:59:18,610 --> 00:59:19,610 J. KIM VANDIVER: Pardon? 861 00:59:19,610 --> 00:59:20,840 AUDIENCE: Times M in the bottom of it? 862 00:59:20,840 --> 00:59:22,298 J. KIM VANDIVER: Yeah, all of these 863 00:59:22,298 --> 00:59:29,750 needs M. Ma squared's M, M, M, M, 864 00:59:29,750 --> 00:59:33,800 so that's our problem when we did two shifts. 865 00:59:33,800 --> 00:59:37,350 But a second we shifted it in two directions, 866 00:59:37,350 --> 00:59:41,310 we get these off-diagonal terms, which 867 00:59:41,310 --> 00:59:44,940 means that if you are actually making this device rotate 868 00:59:44,940 --> 00:59:47,954 about point A, you will get these unbalanced moments-- 869 00:59:47,954 --> 00:59:49,120 this dynamically unbalanced. 870 00:59:51,970 --> 00:59:53,414 Yeah? 871 00:59:53,414 --> 00:59:56,039 AUDIENCE: Will we be trying to find our [? mass ?] moments 872 00:59:56,039 --> 01:00:01,776 of inertia directly, or should we just do [INAUDIBLE] equation 873 01:00:01,776 --> 01:00:05,185 and have them fall out like the integral with x times y 874 01:00:05,185 --> 01:00:07,679 [? vM? ?] [INAUDIBLE] 875 01:00:07,679 --> 01:00:10,220 J. KIM VANDIVER: He's kind of asking advice in whether or not 876 01:00:10,220 --> 01:00:15,590 we ought to be trying to find the inertia 877 01:00:15,590 --> 01:00:18,280 matrix about other points, right, 878 01:00:18,280 --> 01:00:21,290 like [? the I ?] with respect to A. 879 01:00:21,290 --> 01:00:24,850 Where I started today, I said you can basically 880 01:00:24,850 --> 01:00:27,360 do all the problems with these three equations, 881 01:00:27,360 --> 01:00:32,080 and this doesn't mention parallel axis theorem. 882 01:00:32,080 --> 01:00:36,937 We use this to find, in fact, to solve this problem. 883 01:00:36,937 --> 01:00:38,770 We didn't talk parallel axis theorem at all, 884 01:00:38,770 --> 01:00:42,220 it just-- the answer dropped out. 885 01:00:42,220 --> 01:00:43,800 And if we looked into that careful, 886 01:00:43,800 --> 01:00:45,030 we could generalize that. 887 01:00:45,030 --> 01:00:47,270 We could work now with that a bit and say, ah, 888 01:00:47,270 --> 01:00:49,030 there's a pattern to this. 889 01:00:49,030 --> 01:00:55,650 I'll bet this can be recast like this, and it can. 890 01:00:55,650 --> 01:00:59,200 So if you know this exists, and you 891 01:00:59,200 --> 01:01:02,500 don't want to have to grind through finding 892 01:01:02,500 --> 01:01:08,730 these terms that come from here, you 893 01:01:08,730 --> 01:01:14,850 can do the problems by finding the mass moment of inertia 894 01:01:14,850 --> 01:01:17,470 matrix with respect to A. 895 01:01:17,470 --> 01:01:26,460 You can use this form only when you have fixed axis rotation. 896 01:01:26,460 --> 01:01:30,220 The thing is the point A is [? about ?] which 897 01:01:30,220 --> 01:01:35,060 this rotation is occurring, then you can write down the formula 898 01:01:35,060 --> 01:01:37,700 this directly like that. 899 01:01:37,700 --> 01:01:41,780 So you that got to be careful when you apply 900 01:01:41,780 --> 01:01:43,750 the parallel axis theorem. 901 01:01:46,990 --> 01:01:51,530 The nice thing about these three forms 902 01:01:51,530 --> 01:01:54,120 is they're generally true. 903 01:01:54,120 --> 01:01:55,670 Point A can move. 904 01:01:55,670 --> 01:01:57,940 Point A can be accelerating. 905 01:01:57,940 --> 01:02:02,820 The problem that we did here, point A is accelerating. 906 01:02:02,820 --> 01:02:06,890 Not only moving, it's not even an inertial frame. 907 01:02:06,890 --> 01:02:11,031 We solved this problem-- this is a non-inertial frame problem. 908 01:02:11,031 --> 01:02:15,520 We went right at it and solved it directly, OK? 909 01:02:15,520 --> 01:02:18,600 So I would say, if you have any doubt to answer your question, 910 01:02:18,600 --> 01:02:21,760 when you're in doubt just use the formula. 911 01:02:21,760 --> 01:02:25,020 And then you will need to find moments of inertia 912 01:02:25,020 --> 01:02:26,940 with respect to the center of mass, 913 01:02:26,940 --> 01:02:29,830 and it's in your interest to use principal axes because they're 914 01:02:29,830 --> 01:02:30,330 easier. 915 01:02:30,330 --> 01:02:30,830 Yeah? 916 01:02:30,830 --> 01:02:32,780 AUDIENCE: If you were to say [INAUDIBLE]? 917 01:02:38,662 --> 01:02:40,120 J. KIM VANDIVER: Ah, so if you took 918 01:02:40,120 --> 01:02:43,930 the derivative of this, [? dh ?] dt, 919 01:02:43,930 --> 01:02:48,050 this one just gives you omega z dot. 920 01:02:48,050 --> 01:02:50,900 That's you're spin in the direction of rotation. 921 01:02:50,900 --> 01:02:54,530 This term, that unit vector rotates. 922 01:02:54,530 --> 01:03:01,720 This gives you two terms-- gives you an i term, omega dot i, 923 01:03:01,720 --> 01:03:07,600 and it gives you an omega j term, OK? 924 01:03:07,600 --> 01:03:10,730 And what those are, those are two torques. 925 01:03:10,730 --> 01:03:14,040 This problem, there's a torque caused 926 01:03:14,040 --> 01:03:16,190 by the centrifugal force trying to bend 927 01:03:16,190 --> 01:03:20,410 this in the j direction. 928 01:03:20,410 --> 01:03:24,650 And if this is accelerating, you know, a theta double dot term, 929 01:03:24,650 --> 01:03:28,415 there is a torque trying to bend this thing backwards. 930 01:03:28,415 --> 01:03:30,410 OK, get them both. 931 01:03:30,410 --> 01:03:36,610 All right [? money ?] cards and see you on Thursday. 932 01:03:36,610 --> 01:03:42,230 Thursday we're going to do a nasty problem, 933 01:03:42,230 --> 01:03:44,530 and it is it a lead in to a problem. 934 01:03:44,530 --> 01:03:47,130 You can do the same problem extraordinarily easily 935 01:03:47,130 --> 01:03:47,940 using energy. 936 01:03:47,940 --> 01:03:49,731 And that's kind of the purpose of doing it, 937 01:03:49,731 --> 01:03:52,060 so you see both ways of doing it.